🧠:Okay, let's try to figure out how to find the 100th smallest integer in the set S, which consists of numbers of the form 2^x + 2^y + 2^z where x, y, z are distinct non-negative integers. Hmm, so each element of S is the sum of three distinct powers of 2. Since exponents are distinct, each term is different. First, I need to understand how these numbers are ordered. Since powers of 2 grow exponentially, the smallest numbers in S should be the ones where the exponents x, y, z are as small as possible. Let's start by listing some small elements to see the pattern.The smallest exponent is 0, so 2^0 = 1. Then 2^1 = 2, 2^2 = 4, 2^3 = 8, etc. So the smallest element in S would be 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7. The next ones would involve the next smallest exponents. Let me list them in order:1. 1 + 2 + 4 = 7 (exponents 0,1,2)2. 1 + 2 + 8 = 11 (0,1,3)3. 1 + 4 + 8 = 13 (0,2,3)4. 2 + 4 + 8 = 14 (1,2,3)5. 1 + 2 + 16 = 19 (0,1,4)6. 1 + 4 + 16 = 21 (0,2,4)7. 2 + 4 + 16 = 22 (1,2,4)8. 1 + 8 + 16 = 25 (0,3,4)9. 2 + 8 + 16 = 26 (1,3,4)10. 4 + 8 + 16 = 28 (2,3,4)11. 1 + 2 + 32 = 35 (0,1,5)12. 1 + 4 + 32 = 37 (0,2,5)13. 2 + 4 + 32 = 38 (1,2,5)14. 1 + 8 + 32 = 41 (0,3,5)15. 2 + 8 + 32 = 42 (1,3,5)16. 4 + 8 + 32 = 44 (2,3,5)17. 1 + 16 + 32 = 49 (0,4,5)18. 2 + 16 + 32 = 50 (1,4,5)19. 4 + 16 + 32 = 52 (2,4,5)20. 8 + 16 + 32 = 56 (3,4,5)... and so on.Wait a second, each time we increase the highest exponent, we generate a block of numbers. For example, when the highest exponent is 2 (i.e., 4), we have only one number: 7. When the highest exponent is 3 (8), we have three numbers: 11,13,14. Wait, no, earlier when exponents go up to 2, it's 0,1,2. Then when exponents go up to 3, we have combinations where the highest is 3, so exponents can be 0,1,3; 0,2,3; 1,2,3. That gives three numbers. Similarly, when the highest exponent is 4 (16), there are C(4,2) combinations? Wait, no. For each new highest exponent, say n, we need to choose two exponents from the previous numbers 0 to n-1. So for highest exponent n, the number of elements is C(n, 2). Because we need to pick two distinct exponents less than n. Wait, no, since we have three exponents: x, y, z, all distinct, with the highest being n. So if the highest is n, then the other two exponents can be any two distinct numbers from 0 to n-1. So for each n >= 2, the number of elements with highest exponent n is C(n, 2). Because we need to choose two exponents from 0 to n-1. For example, when n=3, C(3,2)=3, which matches the three numbers: 11,13,14. For n=4, C(4,2)=6, but in my earlier list, when highest exponent is 4 (16), there are 6 numbers: 19,21,22,25,26,28. Wait, 19 is 0,1,4; 21 is 0,2,4; 22 is 1,2,4; 25 is 0,3,4; 26 is 1,3,4; 28 is 2,3,4. So yes, 6 numbers. Similarly, for n=5, C(5,2)=10, so there should be 10 numbers with highest exponent 5. Let me check: 35,37,38,41,42,44,49,50,52,56. Yes, 10 numbers. So the pattern is, for each n >= 2, the number of elements in S with highest exponent n is C(n,2). Therefore, the total number of elements in S up to highest exponent m is the sum from k=2 to m of C(k,2). But wait, we need to confirm that. Let's compute the cumulative count:- n=2: C(2,2)=1 element (7)- n=3: C(3,2)=3 elements (11,13,14). Cumulative: 1+3=4- n=4: C(4,2)=6 elements. Cumulative: 4+6=10- n=5: C(5,2)=10 elements. Cumulative: 10+10=20- n=6: C(6,2)=15 elements. Cumulative: 20+15=35- n=7: C(7,2)=21 elements. Cumulative: 35+21=56- n=8: C(8,2)=28 elements. Cumulative: 56+28=84- n=9: C(9,2)=36 elements. Cumulative: 84+36=120Wait, so if we go up to n=9, cumulative count is 120. So the 100th element would be in the block where the highest exponent is 9, since up to n=8 we have 84 elements, and the next block (n=9) adds 36 elements, so elements 85 to 120 are in this block. Therefore, the 100th element is the (100 - 84) = 16th element in the n=9 block. But let's confirm the cumulative counts step by step:- For n=2: highest exponent 2, sum C(2,2)=1, total 1- n=3: sum 3, total 1+3=4- n=4: sum 6, total 4+6=10- n=5: sum 10, total 10+10=20- n=6: sum 15, total 20+15=35- n=7: sum 21, total 35+21=56- n=8: sum 28, total 56+28=84- n=9: sum 36, total 84+36=120Yes, that's correct. So the 100th element is the 16th element in the n=9 block. Now, we need to figure out what the elements in the n=9 block are. The n=9 block consists of all numbers of the form 2^a + 2^b + 2^9 where a and b are distinct integers less than 9. To find the 16th element in this block, we need to order all combinations of a and b (with a < b < 9) in ascending order. The numbers in the n=9 block are 2^a + 2^b + 512. Since a and b are less than 9, their exponents go up to 8. To order these numbers, we need to consider the sum 2^a + 2^b. The smaller this sum is, the smaller the total number. So we need to list all pairs (a,b) where 0 <= a < b <=8, ordered by the sum 2^a + 2^b. Then the 16th element in this ordered list will be the 16th smallest number in the n=9 block.So first, let's list all pairs (a,b) with 0 <= a < b <=8. There are C(9,2) = 36 pairs. We need to sort these 36 pairs by the value of 2^a + 2^b. Let's see how to do this efficiently.Note that 2^a + 2^b is equivalent to a binary number with 1s at positions a and b. So ordering these numbers is equivalent to ordering binary numbers with exactly two 1s. The order would be from the smallest such number to the largest.Alternatively, we can think of 2^a + 2^b where a < b as numbers formed by two bits in a 9-bit number (since the highest exponent is 8). The order of these numbers is the same as the order of their binary representations, which corresponds to their integer values. Therefore, the numbers 2^a + 2^b are ordered by their size as a increases and b increases.But to list them in order, perhaps the easiest way is to generate all possible pairs (a,b) with a < b, compute 2^a + 2^b, and then sort them. However, manually doing this for 36 pairs might be tedious, but maybe we can find a pattern.Alternatively, note that the sum 2^a + 2^b is determined by the positions of a and b. The smallest sum is when a and b are the smallest possible. So the first few pairs are:1. (0,1): 1 + 2 = 32. (0,2): 1 + 4 = 53. (0,3): 1 + 8 = 94. (0,4): 1 + 16 = 175. (0,5): 1 + 32 = 336. (0,6): 1 + 64 = 657. (0,7): 1 + 128 = 1298. (0,8): 1 + 256 = 2579. (1,2): 2 + 4 = 610. (1,3): 2 + 8 = 1011. (1,4): 2 + 16 = 1812. (1,5): 2 + 32 = 3413. (1,6): 2 + 64 = 6614. (1,7): 2 + 128 = 13015. (1,8): 2 + 256 = 25816. (2,3): 4 + 8 = 1217. (2,4): 4 + 16 = 2018. (2,5): 4 + 32 = 3619. (2,6): 4 + 64 = 6820. (2,7): 4 + 128 = 13221. (2,8): 4 + 256 = 26022. (3,4): 8 + 16 = 2423. (3,5): 8 + 32 = 4024. (3,6): 8 + 64 = 7225. (3,7): 8 + 128 = 13626. (3,8): 8 + 256 = 26427. (4,5): 16 + 32 = 4828. (4,6): 16 + 64 = 8029. (4,7): 16 + 128 = 14430. (4,8): 16 + 256 = 27231. (5,6): 32 + 64 = 9632. (5,7): 32 + 128 = 16033. (5,8): 32 + 256 = 28834. (6,7): 64 + 128 = 19235. (6,8): 64 + 256 = 32036. (7,8): 128 + 256 = 384Now, let's sort these sums in ascending order. Let's list them:1. 3 (0,1)2. 5 (0,2)3. 6 (1,2)4. 9 (0,3)5. 10 (1,3)6. 12 (2,3)7. 17 (0,4)8. 18 (1,4)9. 20 (2,4)10. 24 (3,4)11. 33 (0,5)12. 34 (1,5)13. 36 (2,5)14. 40 (3,5)15. 48 (4,5)16. 65 (0,6)17. 66 (1,6)18. 68 (2,6)19. 72 (3,6)20. 80 (4,6)21. 96 (5,6)22. 129 (0,7)23. 130 (1,7)24. 132 (2,7)25. 136 (3,7)26. 144 (4,7)27. 160 (5,7)28. 192 (6,7)29. 257 (0,8)30. 258 (1,8)31. 260 (2,8)32. 264 (3,8)33. 272 (4,8)34. 288 (5,8)35. 320 (6,8)36. 384 (7,8)Wait, hold on, when I sorted the sums, let's see:Original list with their sums:1. 32. 53. 64. 95. 106. 127. 178. 189. 2010. 2411. 3312. 3413. 3614. 4015. 4816. 6517. 6618. 6819. 7220. 8021. 9622. 12923. 13024. 13225. 13626. 14427. 16028. 19229. 25730. 25831. 26032. 26433. 27234. 28835. 32036. 384Wait, but when you sort them numerically, the order should be:3,5,6,9,10,12,17,18,20,24,33,34,36,40,48,65,66,68,72,80,96,129,130,132,136,144,160,192,257,258,260,264,272,288,320,384.Yes, that seems correct. So the 16th element in this sorted list is 65, which corresponds to (0,6). Then the 16th element in the n=9 block is 65 + 512 = 577. Therefore, the 100th smallest element in S is 577?Wait, but hold on. Let's check the cumulative counts again. The n=9 block has 36 elements, and the 100th element is the 16th element in this block. But according to the sorted sums, the 16th sum is 65 (from pair (0,6)), so adding 512 gives 65 + 512 = 577. Is that correct?Wait, but let me double-check the sorted list. Let's count:1. 32. 53. 64. 95. 106. 127. 178. 189. 2010. 2411. 3312. 3413. 3614. 4015. 4816. 6517. 6618. 6819. 7220. 8021. 9622. 12923. 13024. 13225. 13626. 14427. 16028. 19229. 25730. 25831. 26032. 26433. 27234. 28835. 32036. 384Yes, the 16th element is 65. Therefore, the 16th element in the n=9 block is 65 + 512 = 577. Therefore, the 100th smallest element of S is 577. But let me confirm once more. Let's recount the cumulative counts:n=2: 1 element (7)n=3: 4 elements (7,11,13,14)n=4: 10 elements (up to 28)n=5: 20 elements (up to 56)n=6: 35 elements (up to 35th element)Wait, hold on, wait. Wait, when n=2, highest exponent is 2, sum is 7. Then n=3 adds 3 elements (total 4), n=4 adds 6 (total 10), n=5 adds 10 (total 20), n=6 adds 15 (total 35), n=7 adds 21 (total 56), n=8 adds 28 (total 84), n=9 adds 36 (total 120). Therefore, elements 85 to 120 are in the n=9 block. Therefore, the 100th element is the 16th element in n=9 block (100 - 84 = 16). Then, as per the sorted list, the 16th sum is 65, so the number is 65 + 512 = 577. But wait, let's verify if the sorted sums in the n=9 block are correctly ordered. Let's check the 16th element. Let's list the sums again:1. 32. 53. 64. 95. 106. 127. 178. 189. 2010. 2411. 3312. 3413. 3614. 4015. 4816. 6517. 6618. 6819. 7220. 8021. 9622. 12923. 13024. 13225. 13626. 14427. 16028. 19229. 25730. 25831. 26032. 26433. 27234. 28835. 32036. 384Yes, the 16th is 65. So 65 + 512 = 577.But to ensure there are no mistakes in the sorted list, let's verify a few positions. For example, the first few elements:1. (0,1): 1 + 2 = 32. (0,2): 1 + 4 = 53. (1,2): 2 + 4 = 64. (0,3): 1 + 8 = 95. (1,3): 2 + 8 = 106. (2,3): 4 + 8 = 127. (0,4): 1 + 16 = 178. (1,4): 2 + 16 = 189. (2,4): 4 + 16 = 2010. (3,4): 8 + 16 = 2411. (0,5): 1 + 32 = 3312. (1,5): 2 + 32 = 3413. (2,5): 4 + 32 = 3614. (3,5): 8 + 32 = 4015. (4,5): 16 + 32 = 4816. (0,6): 1 + 64 = 65... yes, that's correct. So up to the 15th element, we have all combinations with exponents up to 5. Then (0,6) is next. So 16th is 65. Then the rest follow. Therefore, the 16th element in the n=9 block is indeed 577. Therefore, the 100th smallest element of S is 577. But wait a second, just to be thorough, let's confirm with an example. Let's take n=6. The cumulative count up to n=6 is 35. Let's list elements up to n=6. Wait, elements with highest exponent 6 would be combinations of two exponents from 0 to 5, so C(6,2)=15 elements. So the elements would be:For n=6:1. 0,1,6: 1 + 2 + 64 = 672. 0,2,6: 1 + 4 + 64 = 693. 0,3,6: 1 + 8 + 64 = 734. 0,4,6: 1 + 16 + 64 = 815. 0,5,6: 1 + 32 + 64 = 976. 1,2,6: 2 + 4 + 64 = 707. 1,3,6: 2 + 8 + 64 = 748. 1,4,6: 2 + 16 + 64 = 829. 1,5,6: 2 + 32 + 64 = 9810. 2,3,6: 4 + 8 + 64 = 7611. 2,4,6: 4 + 16 + 64 = 8412. 2,5,6: 4 + 32 + 64 = 10013. 3,4,6: 8 + 16 + 64 = 8814. 3,5,6: 8 + 32 + 64 = 10415. 4,5,6: 16 + 32 + 64 = 112But wait, these numbers need to be sorted. So if we generate all these numbers and sort them, they should be in order. Let me sort them:67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112.So ordered, that's:67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112.So the numbers in the n=6 block are ordered as such. Therefore, the 35th element (since cumulative up to n=6 is 35) is 112. Then moving on to n=7:For n=7, highest exponent is 7 (128). The elements are combinations of two exponents from 0-6. There are C(7,2)=21 elements. Let's list some:0,1,7: 1 + 2 + 128 = 1310,2,7: 1 + 4 + 128 = 1330,3,7: 1 + 8 + 128 = 1370,4,7: 1 + 16 + 128 = 1450,5,7: 1 + 32 + 128 = 1610,6,7: 1 + 64 + 128 = 1931,2,7: 2 + 4 + 128 = 1341,3,7: 2 + 8 + 128 = 1381,4,7: 2 + 16 + 128 = 1461,5,7: 2 + 32 + 128 = 1621,6,7: 2 + 64 + 128 = 1942,3,7: 4 + 8 + 128 = 1402,4,7: 4 + 16 + 128 = 1482,5,7: 4 + 32 + 128 = 1642,6,7: 4 + 64 + 128 = 1963,4,7: 8 + 16 + 128 = 1523,5,7: 8 + 32 + 128 = 1683,6,7: 8 + 64 + 128 = 2004,5,7: 16 + 32 + 128 = 1764,6,7: 16 + 64 + 128 = 2085,6,7: 32 + 64 + 128 = 224Now, sort these numbers:131, 133, 134, 137, 138, 140, 145, 146, 148, 152, 161, 162, 164, 168, 176, 193, 194, 196, 200, 208, 224.Therefore, the n=7 block adds 21 elements starting from 131 up to 224. So cumulative count up to n=7 is 35 + 21 = 56. Then n=8 block:Highest exponent 8 (256). Combinations of two exponents from 0-7. C(8,2)=28 elements. Let's list a few:0,1,8: 1 + 2 + 256 = 2590,2,8: 1 + 4 + 256 = 2610,3,8: 1 + 8 + 256 = 2650,4,8: 1 + 16 + 256 = 2730,5,8: 1 + 32 + 256 = 2890,6,8: 1 + 64 + 256 = 3210,7,8: 1 + 128 + 256 = 3851,2,8: 2 + 4 + 256 = 2621,3,8: 2 + 8 + 256 = 2661,4,8: 2 + 16 + 256 = 2741,5,8: 2 + 32 + 256 = 2901,6,8: 2 + 64 + 256 = 3221,7,8: 2 + 128 + 256 = 3862,3,8: 4 + 8 + 256 = 2682,4,8: 4 + 16 + 256 = 2762,5,8: 4 + 32 + 256 = 2922,6,8: 4 + 64 + 256 = 3242,7,8: 4 + 128 + 256 = 3883,4,8: 8 + 16 + 256 = 2803,5,8: 8 + 32 + 256 = 2963,6,8: 8 + 64 + 256 = 3283,7,8: 8 + 128 + 256 = 3924,5,8: 16 + 32 + 256 = 3044,6,8: 16 + 64 + 256 = 3364,7,8: 16 + 128 + 256 = 4005,6,8: 32 + 64 + 256 = 3525,7,8: 32 + 128 + 256 = 4166,7,8: 64 + 128 + 256 = 448Sorting these numbers:259, 261, 262, 265, 266, 268, 273, 274, 276, 280, 289, 290, 292, 296, 304, 321, 322, 324, 328, 336, 352, 385, 386, 388, 392, 400, 416, 448.So the n=8 block adds 28 elements from 259 to 448. Therefore, cumulative count up to n=8 is 56 + 28 = 84. Then the n=9 block has 36 elements, which are numbers of the form 2^a + 2^b + 512. We need to find the 16th element here.Earlier, we determined that the 16th smallest sum of 2^a + 2^b (with a < b <=8) is 65, which comes from a=0, b=6. Therefore, the total number is 65 + 512 = 577. But let's confirm this with actual generation. Let's list all elements in the n=9 block:For n=9 (highest exponent 9, which is 512). The other two exponents are a and b, where 0 <= a < b <=8. The numbers are 2^a + 2^b + 512. The sums 2^a + 2^b are as previously listed:Sorted sums:3,5,6,9,10,12,17,18,20,24,33,34,36,40,48,65,66,68,72,80,96,129,130,132,136,144,160,192,257,258,260,264,272,288,320,384.Adding 512 to each:3 + 512 = 5155 + 512 = 5176 + 512 = 5189 + 512 = 52110 + 512 = 52212 + 512 = 52417 + 512 = 52918 + 512 = 53020 + 512 = 53224 + 512 = 53633 + 512 = 54534 + 512 = 54636 + 512 = 54840 + 512 = 55248 + 512 = 56065 + 512 = 57766 + 512 = 57868 + 512 = 58072 + 512 = 58480 + 512 = 59296 + 512 = 608129 + 512 = 641130 + 512 = 642132 + 512 = 644136 + 512 = 648144 + 512 = 656160 + 512 = 672192 + 512 = 704257 + 512 = 769258 + 512 = 770260 + 512 = 772264 + 512 = 776272 + 512 = 784288 + 512 = 800320 + 512 = 832384 + 512 = 896Now, listing these numbers in order:515, 517, 518, 521, 522, 524, 529, 530, 532, 536, 545, 546, 548, 552, 560, 577, 578, 580, 584, 592, 608, 641, 642, 644, 648, 656, 672, 704, 769, 770, 772, 776, 784, 800, 832, 896.Therefore, the 16th element in this list is 577. Hence, the 100th smallest element of S is 577.But just to ensure that we haven't missed any numbers in between, let's cross-verify. For example, up to n=8, cumulative count is 84. The next block (n=9) starts at 85. So the 85th element is 515, 86th is 517, 87th is 518, 88th is 521, 89th is 522, 90th is 524, 91st is 529, 92nd is 530, 93rd is 532, 94th is 536, 95th is 545, 96th is 546, 97th is 548, 98th is 552, 99th is 560, 100th is 577. Yes, exactly. So the 100th element is 577. Therefore, the answer should be 577.Final AnswerThe 100th smallest element of ( S ) is (boxed{577}).

✅:To determine the 100th smallest element of the set ( S ), which consists of integers of the form ( 2^x + 2^y + 2^z ) where ( x, y, z ) are pairwise distinct non-negative integers, we need to analyze the structure and ordering of these numbers.1. Understanding the Structure: - Each element of ( S ) is the sum of three distinct powers of 2. - The numbers in ( S ) are ordered by the sum of their binary representations, which grow exponentially.2. Cumulative Count Analysis: - For each highest exponent ( n ), the number of elements in ( S ) is given by ( binom{n}{2} ). - Cumulative counts up to certain exponents: - ( n = 2 ): 1 element - ( n = 3 ): 4 elements - ( n = 4 ): 10 elements - ( n = 5 ): 20 elements - ( n = 6 ): 35 elements - ( n = 7 ): 56 elements - ( n = 8 ): 84 elements - ( n = 9 ): 120 elements3. Identifying the Block: - The 100th element falls in the block with highest exponent ( n = 9 ) (since cumulative count up to ( n = 8 ) is 84). - This block has 36 elements, and the 100th element is the 16th element in this block.4. Listing and Sorting Elements in the Block: - The elements in the block with highest exponent 9 are of the form ( 2^a + 2^b + 512 ) where ( 0 leq a < b leq 8 ). - Sorting the sums ( 2^a + 2^b ) for ( 0 leq a < b leq 8 ): - The 16th smallest sum is 65 (from pair ( (0, 6) )). - Adding 512 to this sum gives ( 65 + 512 = 577 ).5. Verification: - The sorted list of elements in the ( n = 9 ) block confirms the 16th element is indeed 577.Thus, the 100th smallest element of ( S ) is (boxed{577}).

🔑:Step 1: Identify the problem and its key components.We need to determine the 100th smallest element of the set ( S ), where ( S ) is defined as the set of integers of the form ( 2^x + 2^y + 2^z ) where ( x, y, z ) are pairwise distinct non-negative integers.Step 2: Understand the nature of the set ( S ).The key observation here is that every number in ( S ) has exactly three ones in its binary representation. For example, the integer ( 2^3 + 2^2 + 2^0 = 1000_2 + 0100_2 + 0001_2 = 1101_2 = 13_{10} ).Step 3: Counting integers with specific bit length.Each integer with exactly three 1's in its binary representation can be represented as choosing 3 positions to place the ones in otherwise zero-bit binary strings of varying lengths. Specifically:[binom{d}{3}]counts the number of such integers with exactly three 1's and total of ( d ) bits.Step 4: Find the range of valid values for the 100th element.Calculate:[binom{9}{3} = frac{9 times 8 times 7}{3 times 2 times 1} = 84][binom{10}{3} = frac{10 times 9 times 8}{3 times 2 times 1} = 120]Since ( binom{9}{3} = 84 ) and ( binom{10}{3} = 120 ), the 100th element falls within integers that have ten bits, as those with nine bits are not sufficient.Therefore, it falls within the integers of the form ( 2^9 + 2^x + 2^y ) for ( y < x < 9 ).Step 5: Determine specific positioning within the set.Out of 120 such numbers, the first 84 are from nine-bit integers. Thus, we require:[100 - 84 = 16]so we seek the 16th smallest integer of form ( 2^9 + 2^x + 2^y ).Step 6: Find the 16th integer from those with the added bit (2^9).We can treat this as finding the 16th smallest number of the form ( 2^x + 2^y ) with ( y < x < 9 ). Calculating:[binom{6}{2} = frac{6 times 5}{2 times 1} = 15]Since ( binom{6}{2} = 15 ), we are left with the smallest number of the form ( 2^9 + 2^6 + 2^0 ).Conclusion: The 100th smallest element of ( S ) is:[2^9 + 2^6 + 2^0 = 512 + 64 + 1 = 577][boxed{577}]

🧠:Okay, so I need to prove that in triangle ABC, the angle bisector of angle B meets the circumcircle at D, and then show that BD squared is greater than BA times BC. Hmm, let's start by drawing a diagram to visualize the problem. Triangle ABC with angle bisector from B meeting the circumcircle again at D. So BD is the angle bisector extended beyond the circumcircle? Wait, no. The angle bisector of angle B would start at B and split angle B into two equal parts. But since D is on the circumcircle, it's not the point where the bisector meets AC, but rather the other intersection point with the circumcircle. So BD is actually the angle bisector that starts at B and goes through the circumcircle again at D. So D is diametrically opposite some point? Not necessarily. Let me recall some properties of angle bisectors and circumcircles.First, maybe using the Angle Bisector Theorem. The theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So in triangle ABC, if the angle bisector of angle B meets AC at point E, then AB/BC = AE/EC. But here, the angle bisector meets the circumcircle at D, so maybe extending BE beyond E to meet the circumcircle again at D. So BD is longer than BE. Maybe there are some properties related to cyclic quadrilaterals here.Since D is on the circumcircle of triangle ABC, quadrilateral ABDC is cyclic. Wait, no, because ABC is already a triangle, so the circumcircle passes through A, B, C. Adding D as another point on the circle, so ABCD is cyclic. But BD is the angle bisector. Maybe using power of a point? The Power of a Point theorem states that for a point outside a circle, the product of the lengths of the segments from the point to the circle is equal. But here, B is on the circle, so power of point B with respect to the circumcircle is zero. Hmm, maybe that's not directly helpful.Alternatively, maybe using the theorem related to the length of a chord in terms of angles. The length of a chord is 2R sin θ, where R is the radius and θ is the angle subtended at the center. But BD is a chord subtended by angle... Let's see. If BD is the angle bisector, then the angles at B are split into two equal parts. Let angle ABC be 2β, so each part is β. Then, the angle subtended by BD at the center would relate to the arcs it cuts. Hmm, perhaps using inscribed angles.Alternatively, maybe Stewart's Theorem. Stewart's Theorem relates the length of a cevian to the sides of the triangle. If BD is a cevian, but BD is not just a cevian; it's extended to the circumcircle. So if E is the point where the angle bisector meets AC, then BD is longer than BE. But how does that help?Wait, let me recall that if a cevian is extended to meet the circumcircle again, then there are some relations. For example, in some cases, the product of the lengths of the cevian and its extension can be related to the sides. Alternatively, maybe use the formula for the length of BD in terms of the sides of the triangle and the angle.Alternatively, maybe use coordinates or trigonometry. Let me try trigonometric approach. Let's denote angle ABC as 2β, so angle ABD = angle CBD = β. Let’s set the circumradius as R. Then, in the circumcircle, BD is a chord subtended by angle... Let's see. The arc BD would correspond to the angle at A and C. Wait, perhaps not directly. Let me recall that in a circle, the length of a chord is 2R sin θ, where θ is half the angle subtended at the center. Wait, more precisely, the length of chord BD is 2R sin(θ), where θ is half the measure of the arc BD. Hmm.Alternatively, use the Law of Cosines on triangle ABD and CBD. Since BD is the angle bisector, angles at B are equal. Let me denote BA = c, BC = a, and BD = d. Then in triangles ABD and CBD, angle at B is β. Applying the Law of Cosines to both triangles:In triangle ABD:AD² = AB² + BD² - 2 AB * BD * cos βIn triangle CBD:CD² = BC² + BD² - 2 BC * BD * cos βBut since D is on the circumcircle of ABC, points A, B, C, D are concyclic. Therefore, angles subtended by the same chord are equal. For instance, angle ABD equals angle ACD because they subtend arc AD. Wait, maybe that's not directly applicable. Alternatively, angle ADB equals angle ACB because they subtend the same arc AB.Alternatively, use Power of a Point for point D with respect to the circumcircle. Wait, D is on the circumcircle, so its power is zero. Hmm.Wait, maybe considering triangle ABD and triangle CBD. Since D is on the circumcircle, angle BAD is equal to angle BCD? Wait, maybe. Let me think. Since ABCD is cyclic, angle BAD and angle BCD are supplementary? Wait, no. In cyclic quadrilaterals, opposite angles are supplementary. So angle BAD + angle BCD = 180 degrees. Hmm, not sure if that helps.Alternatively, use the formula for the length of an angle bisector. The length of the angle bisector from B to AC is given by a formula, but here BD is extended to the circumcircle, so perhaps we can relate BD to the angle bisector length.The formula for the angle bisector length is: If BE is the angle bisector, then BE = (2ac/(a + c)) cos β, where a and c are the adjacent sides. Wait, in triangle ABC, if sides AB = c, BC = a, then the angle bisector BE is given by BE = (2ac/(a + c)) cos β. But BD is longer than BE. So BD = BE + ED. But how to find ED?Alternatively, since D is on the circumcircle, BD * BE = BA * BC? Wait, maybe applying the Power of a Point theorem from point E. If E is on BD, then the power of E with respect to the circumcircle would be EA * EC = ED * EB. Wait, but E is the intersection point of the angle bisector and AC. So EA/EC = AB/BC = c/a, from the Angle Bisector Theorem. So EA = (c/(a + c)) * AC, and EC = (a/(a + c)) * AC. Then, EA * EC = (ac/(a + c)^2) * AC². On the other hand, Power of a Point gives EA * EC = ED * EB. But EB is the length from E to B, which is BE, and ED is the length from E to D. So ED * BE = (ac/(a + c)^2) * AC². Hmm, but I don't know AC. Maybe express AC in terms of a, c, and angles? Using the Law of Cosines: AC² = AB² + BC² - 2 AB * BC cos(2β) = c² + a² - 2ac cos(2β). Therefore, ED * BE = (ac/(a + c)^2)(c² + a² - 2ac cos 2β). But BE is the angle bisector, which we know is (2ac/(a + c)) cos β. Therefore, ED * (2ac/(a + c)) cos β = (ac/(a + c)^2)(c² + a² - 2ac cos 2β). Then, solving for ED:ED = [ (ac/(a + c)^2)(c² + a² - 2ac cos 2β) ] / [ (2ac/(a + c)) cos β ) ]Simplify:ED = [ (c² + a² - 2ac cos 2β) / (2(a + c)) cos β ) ]Simplify numerator:c² + a² - 2ac cos 2β = c² + a² - 2ac(2 cos² β - 1) = c² + a² - 4ac cos² β + 2ac= (c² + a² + 2ac) - 4ac cos² β = (a + c)^2 - 4ac cos² βTherefore, ED = [ (a + c)^2 - 4ac cos² β ) ] / [ 2(a + c) cos β ) ] = [ (a + c) - (4ac cos² β)/(a + c) ) ] / (2 cos β )Hmm, this seems complicated. Maybe there's a better approach.Alternatively, use inversion. Maybe invert with respect to point B. But that might be overcomplicating.Alternatively, use coordinates. Let's place point B at the origin, point A along the x-axis, and point C somewhere in the plane. Let me try coordinate geometry.Let’s set up coordinate system: Let’s place point B at (0,0). Let’s let BA = c, so point A is at (c, 0). Let’s let angle at B be 2β, so angle between BA and BC is 2β. Then point C can be placed at (a cos 2β, a sin 2β), where BC = a. Then, the angle bisector of angle B will go through a point D on the circumcircle. The angle bisector from B will have direction given by the angle β from the x-axis. So parametric equations of the angle bisector is x = t cos β, y = t sin β, where t > 0.We need to find the point D where this line intersects the circumcircle again. The circumcircle of triangle ABC can be found using coordinates. Let's compute the circumcircle equation.Points A (c, 0), B (0,0), C (a cos 2β, a sin 2β). The circumcircle can be determined by finding the perpendicular bisectors of AB and BC.Midpoint of AB is (c/2, 0), and the perpendicular bisector is the line perpendicular to AB (which is horizontal), so vertical line x = c/2.Midpoint of BC is ( (a cos 2β)/2, (a sin 2β)/2 ). The slope of BC is (a sin 2β - 0)/(a cos 2β - 0) = tan 2β. Therefore, the perpendicular bisector of BC has slope -cot 2β. So the equation of the perpendicular bisector of BC is:y - (a sin 2β)/2 = -cot 2β (x - (a cos 2β)/2 )Now, the circumcircle center is at the intersection of x = c/2 and the above line. Substitute x = c/2 into the equation:y - (a sin 2β)/2 = -cot 2β (c/2 - (a cos 2β)/2 )Simplify:y = (a sin 2β)/2 - cot 2β * ( (c - a cos 2β)/2 )= (a sin 2β)/2 - (cos 2β / sin 2β) * ( (c - a cos 2β)/2 )= [ a sin 2β - (cos 2β)(c - a cos 2β)/ sin 2β ] / 2So y-coordinate of center is:[ a sin² 2β - c cos 2β + a cos² 2β ] / (2 sin 2β )= [ a (sin² 2β + cos² 2β) - c cos 2β ] / (2 sin 2β )= (a - c cos 2β) / (2 sin 2β )Therefore, the center of the circumcircle is at (c/2, (a - c cos 2β)/(2 sin 2β )).Now, the radius R can be computed as the distance from the center to point B (0,0):R² = (c/2)^2 + [ (a - c cos 2β)/(2 sin 2β ) ]²But maybe instead of computing R, let's find the point D on the angle bisector (parametric line x = t cos β, y = t sin β) that lies on the circumcircle.The equation of the circumcircle is:(x - c/2)^2 + (y - (a - c cos 2β)/(2 sin 2β ))^2 = R²But since we know points A, B, C lie on the circle, maybe substitute the parametric equations into the circle equation and solve for t.Let’s substitute x = t cos β, y = t sin β into the circle equation:( t cos β - c/2 )² + ( t sin β - (a - c cos 2β)/(2 sin 2β ) )² = R²But R² is the distance squared from the center to B:R² = (c/2)^2 + [ (a - c cos 2β)/(2 sin 2β ) ]²Therefore, the equation becomes:( t cos β - c/2 )² + ( t sin β - (a - c cos 2β)/(2 sin 2β ) )² = (c/2)^2 + [ (a - c cos 2β)/(2 sin 2β ) ]²Subtracting the right-hand side from both sides:( t cos β - c/2 )² - (c/2)^2 + ( t sin β - (a - c cos 2β)/(2 sin 2β ) )² - [ (a - c cos 2β)/(2 sin 2β ) ]² = 0Simplify each term:First term: ( t cos β - c/2 )² - (c/2)^2 = t² cos² β - t c cos β + c²/4 - c²/4 = t² cos² β - t c cos βSecond term: [ t sin β - k ]² - k² where k = (a - c cos 2β)/(2 sin 2β )Expand: t² sin² β - 2 t sin β k + k² - k² = t² sin² β - 2 t sin β kTherefore, overall equation:t² cos² β - t c cos β + t² sin² β - 2 t sin β k = 0Combine like terms:t² (cos² β + sin² β) - t (c cos β + 2 sin β k ) = 0Since cos² β + sin² β = 1:t² - t (c cos β + 2 sin β k ) = 0Factor t:t [ t - (c cos β + 2 sin β k ) ] = 0Solutions t = 0 (which is point B) and t = c cos β + 2 sin β kSo the parameter t for point D is t = c cos β + 2 sin β kRecall that k = (a - c cos 2β)/(2 sin 2β )Substitute k into t:t = c cos β + 2 sin β * (a - c cos 2β)/(2 sin 2β )Simplify:t = c cos β + [ sin β (a - c cos 2β) ] / sin 2βNote that sin 2β = 2 sin β cos β, so:t = c cos β + [ sin β (a - c cos 2β) ] / (2 sin β cos β )Cancel sin β:t = c cos β + (a - c cos 2β)/(2 cos β )So:t = c cos β + a/(2 cos β ) - c cos 2β/(2 cos β )Simplify term by term:First term: c cos βThird term: - c cos 2β/(2 cos β ) = - c (2 cos² β - 1)/(2 cos β ) = - c (2 cos² β)/(2 cos β ) + c/(2 cos β ) = - c cos β + c/(2 cos β )Therefore, substituting back:t = c cos β + a/(2 cos β ) - c cos β + c/(2 cos β )Simplify:c cos β - c cos β cancels. So:t = a/(2 cos β ) + c/(2 cos β ) = (a + c)/(2 cos β )Therefore, the parameter t for point D is (a + c)/(2 cos β )Therefore, BD = t = (a + c)/(2 cos β )So BD squared is [(a + c)/(2 cos β )]^2We need to prove BD² > BA * BC, which is a * c. So:[(a + c)^2 / (4 cos² β )] > a cMultiply both sides by 4 cos² β:(a + c)^2 > 4 a c cos² βSo need to show that (a + c)^2 > 4 a c cos² βWhich simplifies to (a + c)^2 / (4 a c ) > cos² βBut since angle at B is 2β, and in triangle ABC, using the Law of Cosines:AC² = AB² + BC² - 2 AB * BC cos(2β ) = c² + a² - 2 a c cos 2βAlso, by the Law of Cosines, cos 2β = (a² + c² - AC² )/(2 a c )But maybe there's another approach. Let's recall that in any triangle, the angles relate to the sides via the Law of Cosines. Also, note that 0 < 2β < π, so 0 < β < π/2, hence cos β is positive.Alternatively, express cos² β in terms of other variables. Since 2β is the angle at B, which is fixed in triangle ABC. Maybe relate cos² β to the sides via some identity.Alternatively, note that (a + c)^2 - 4 a c cos² β = a² + 2 a c + c² - 4 a c cos² β= a² + c² + 2 a c (1 - 2 cos² β )But 1 - 2 cos² β = - cos 2β, so:= a² + c² - 2 a c cos 2βBut from the Law of Cosines, a² + c² - 2 a c cos 2β = AC²Therefore, (a + c)^2 - 4 a c cos² β = AC²Therefore, (a + c)^2 = AC² + 4 a c cos² βSo the inequality (a + c)^2 > 4 a c cos² β is equivalent to AC² + 4 a c cos² β > 4 a c cos² βWhich simplifies to AC² > 0, which is always true unless AC = 0, which isn't possible in a triangle. Therefore, the inequality (a + c)^2 > 4 a c cos² β holds, hence BD² = [(a + c)/(2 cos β )]^2 > a cTherefore, BD² > BA * BC. Hence proved.Wait, so the key step was to express BD in terms of a, c, and β, then use the identity derived from the Law of Cosines to show that (a + c)^2 - 4 a c cos² β equals AC², which is positive, hence the inequality holds. That's elegant. So the critical part was computing BD using coordinates, then manipulating the inequality to relate it to AC².Alternatively, maybe there's a more straightforward way using power of a point or some geometric properties, but the coordinate approach worked here. Let me verify again.BD squared is [(a + c)/(2 cos β )]^2. Then, BA * BC = a c. So [(a + c)^2 / (4 cos² β )] > a c. Multiply both sides by 4 cos² β:(a + c)^2 > 4 a c cos² βWhich, as shown, reduces to AC² > 0. Which is true.So the proof is complete. Therefore, BD² > BA * BC.Final Answerboxed{BD^2 > BA cdot BC}

✅:To prove that ( BD^2 > BA cdot BC ) in triangle ( ABC ) where the angle bisector of (angle B) meets the circumcircle at ( D ), we start by considering the coordinates and trigonometric relationships.1. Setup and Coordinates: - Place ( B ) at the origin ((0, 0)). - Let ( BA = c ) and ( BC = a ). - The angle (angle ABC = 2beta), so the angle bisector splits it into two angles of (beta) each. - The coordinates of ( A ) are ((c, 0)) and ( C ) are ((a cos 2beta, a sin 2beta)).2. Equation of the Circumcircle: - The circumcircle of ( triangle ABC ) passes through ( A ), ( B ), and ( C ). - The center of the circumcircle is found by the perpendicular bisectors of ( AB ) and ( BC ).3. Parametric Form of the Angle Bisector: - The angle bisector from ( B ) has the parametric equations ( x = t cos beta ) and ( y = t sin beta ).4. Finding Point ( D ): - Substitute the parametric equations into the circumcircle equation and solve for ( t ). - The parameter ( t ) for point ( D ) is found to be ( frac{a + c}{2 cos beta} ).5. Length of ( BD ): - The length ( BD ) is ( frac{a + c}{2 cos beta} ). - Therefore, ( BD^2 = left( frac{a + c}{2 cos beta} right)^2 ).6. Inequality Verification: - We need to show ( left( frac{a + c}{2 cos beta} right)^2 > ac ). - Simplifying, this reduces to ( (a + c)^2 > 4ac cos^2 beta ). - Using the Law of Cosines in ( triangle ABC ), we find ( (a + c)^2 - 4ac cos^2 beta = AC^2 ), which is always positive.Thus, the inequality ( BD^2 > BA cdot BC ) holds true.[boxed{BD^2 > BA cdot BC}]

🔑:1. Identify key angle relationships in the triangles: - The angle bisector of angle B meets the circumcircle of triangle ABC at D. Hence, angle ABD = angle DBC because BD is the angle bisector of angle ABC. - angle BAC = angle BDC because both angles subtend arc BC.2. Establish triangle similarity: - Since angle BAD = angle DBC and angle BAC = angle BDC, we have triangle BAD sim triangle BDC by AA (Angle-Angle) similarity criterion.3. Write the ratios of corresponding sides from the similar triangles triangle BAD and triangle BDC: [ frac{BA}{BD} = frac{BD}{BC} ]4. Cross-multiply to find an equation involving BD^2: [ BA cdot BC = BD^2 ]5. Introduce a point E such that BE < BD to argue that BD is greater than some other segment in the triangle: We construct a point E on BD such that BE = BD. [ BE < BD ] Here, by construction, BE indeed represents a segment smaller than BD.6. Calculate the product involving such configurations: Since BE < BD, based on the bisector relationship and similarity-derived equation, it directly follows: [ BD cdot BE < BD cdot BD ] [ BE cdot BD = BA cdot BC ] Plugging this into the inequality: [ BD^2 > BA cdot BC ] Conclusion:[boxed{BD^2 > BA cdot BC}]

🧠:Okay, let's try to solve this trigonometric equation. Hmm, the equation is:(cos 4x cos (pi + 2x) - sin 2x cos left(frac{pi}{2} - 4xright) = frac{sqrt{2}}{2} sin 4x).First, I need to simplify both sides of the equation using trigonometric identities. Let me start by recalling some identities that might help here.Starting with the first term: (cos 4x cos (pi + 2x)). The argument (pi + 2x) in the cosine function can be simplified using the identity (cos(pi + theta) = -cos theta). So, (cos(pi + 2x) = -cos 2x). Therefore, the first term becomes (cos 4x times (-cos 2x) = -cos 4x cos 2x).Now, moving to the second term: (-sin 2x cos left(frac{pi}{2} - 4xright)). The cosine of (frac{pi}{2} - 4x) can be simplified using the identity (cosleft(frac{pi}{2} - thetaright) = sin theta). So, (cosleft(frac{pi}{2} - 4xright) = sin 4x). Therefore, the second term becomes (-sin 2x times sin 4x).Putting these together, the left-hand side (LHS) simplifies to:(-cos 4x cos 2x - sin 2x sin 4x).Wait, this looks like the formula for (-cos(4x + 2x)) because (cos A cos B + sin A sin B = cos(A - B)), but here we have negative signs. Let me check that identity again.The identity is (cos(A + B) = cos A cos B - sin A sin B). Wait, so if I have (-cos 4x cos 2x - sin 2x sin 4x), factor out a negative sign: (-(cos 4x cos 2x + sin 4x sin 2x)). The expression inside the parentheses is (cos(4x - 2x)) because (cos(A - B) = cos A cos B + sin A sin B). So, (cos(4x - 2x) = cos 2x). Therefore, the entire LHS becomes (-cos 2x).Wait, so the left-hand side simplifies to (-cos 2x)? Let me verify step by step.Original LHS:(cos 4x cos (pi + 2x) - sin 2x cos left(frac{pi}{2} - 4xright))After applying identities:(-cos 4x cos 2x - sin 2x sin 4x)Yes, that's correct. Then, combining those terms: (-(cos 4x cos 2x + sin 4x sin 2x)) which is (-cos(4x - 2x)) because of the cosine of difference identity. Wait, no, the cosine of difference is (cos(A - B) = cos A cos B + sin A sin B). So, if it's (cos(4x - 2x)), that would be (cos 2x). But here, we have (cos 4x cos 2x + sin 4x sin 2x = cos(4x - 2x) = cos 2x). So, the expression becomes (-cos 2x).Therefore, the left-hand side simplifies to (-cos 2x). So the equation now is:(-cos 2x = frac{sqrt{2}}{2} sin 4x).Wait, that seems simpler. Now, let's write the equation as:(-cos 2x = frac{sqrt{2}}{2} sin 4x)But perhaps we can express both sides in terms of sine or cosine of the same angle. Let's recall that (sin 4x = 2 sin 2x cos 2x). So, substituting that in, the right-hand side (RHS) becomes:(frac{sqrt{2}}{2} times 2 sin 2x cos 2x = sqrt{2} sin 2x cos 2x).So, the equation is now:(-cos 2x = sqrt{2} sin 2x cos 2x).Hmm, let's rearrange this. Let's bring all terms to one side:(-cos 2x - sqrt{2} sin 2x cos 2x = 0)Factor out (cos 2x):(cos 2x (-1 - sqrt{2} sin 2x) = 0)So, this product equals zero when either factor is zero. Therefore, we have two cases:1. (cos 2x = 0)2. (-1 - sqrt{2} sin 2x = 0 Rightarrow sin 2x = -frac{1}{sqrt{2}})Let's solve each case separately.Case 1: (cos 2x = 0)The general solution for (cos theta = 0) is (theta = frac{pi}{2} + kpi), where (k) is an integer. Here, (theta = 2x), so:(2x = frac{pi}{2} + kpi)Divide both sides by 2:(x = frac{pi}{4} + frac{kpi}{2})Case 2: (sin 2x = -frac{1}{sqrt{2}})First, simplify (-frac{1}{sqrt{2}}) as (-frac{sqrt{2}}{2}). The general solution for (sin theta = -frac{sqrt{2}}{2}) is:(theta = frac{5pi}{4} + 2kpi) or (theta = frac{7pi}{4} + 2kpi), where (k) is an integer.Here, (theta = 2x), so:1. (2x = frac{5pi}{4} + 2kpi)2. (2x = frac{7pi}{4} + 2kpi)Divide both sides by 2:1. (x = frac{5pi}{8} + kpi)2. (x = frac{7pi}{8} + kpi)So, combining all solutions, we have:From Case 1: (x = frac{pi}{4} + frac{kpi}{2})From Case 2: (x = frac{5pi}{8} + kpi) and (x = frac{7pi}{8} + kpi)But wait, let's check if these solutions are overlapping or if there are any restrictions. Also, we need to ensure that we didn't divide by zero or make any incorrect steps.Looking back at the step where we factored out (cos 2x), we assumed that either (cos 2x = 0) or the other term is zero. That seems valid.Now, check if the solutions from Case 1 are already included in Case 2 or vice versa. Let's test with specific k values.Take x = π/4 (k=0 in Case 1). Then 2x = π/2, so cos2x=0. Let's check in the original equation.Original equation:Left-hand side: cos4x cos(π + 2x) - sin2x cos(π/2 -4x)If x = π/4:4x = π, 2x = π/2.So:cos(π) * cos(π + π/2) - sin(π/2) * cos(π/2 - π)cos(π) = -1cos(π + π/2) = cos(3π/2) = 0sin(π/2) = 1cos(π/2 - π) = cos(-π/2) = cos(π/2) = 0Therefore, LHS = (-1)(0) - (1)(0) = 0 - 0 = 0RHS: (√2)/2 sin4x = (√2)/2 sin(π) = 0So, 0=0, which holds. So x=π/4 is a valid solution.Similarly, check x=5π/8 (k=0 in Case 2):Compute 2x = 5π/4sin2x = sin5π/4 = -√2/2, which satisfies Case 2.Check in original equation:4x = 5π/2, 2x=5π/4LHS: cos(5π/2) * cos(π +5π/4) - sin(5π/4) * cos(π/2 -5π/2)Simplify:cos(5π/2) = cos(2π + π/2) = cos(π/2) = 0cos(π +5π/4) = cos(9π/4) = cos(π/4) = √2/2sin(5π/4) = -√2/2cos(π/2 -5π/2) = cos(-2π) = cos2π=1Therefore, LHS = 0*(√2/2) - (-√2/2)*1 = 0 + √2/2 = √2/2RHS: (√2)/2 sin4x = (√2)/2 sin(5π/2) = (√2)/2 *1 = √2/2So, √2/2 = √2/2, which holds. Therefore, x=5π/8 is a valid solution.Similarly, check x=7π/8:4x=7π/2, 2x=7π/4cos(7π/2)=cos(3π + π/2)=cos(π/2)=0cos(π +7π/4)=cos(11π/4)=cos(3π/4)= -√2/2sin(7π/4)= -√2/2cos(π/2 -7π/2)=cos(-3π)=cos3π= -1Thus, LHS=0*(-√2/2) - (-√2/2)*(-1) = 0 - (√2/2) = -√2/2Wait, but the RHS is (√2)/2 sin4x = (√2)/2 sin(7π/2)= (√2)/2*(-1)= -√2/2So, LHS= -√2/2 and RHS= -√2/2, which holds. So x=7π/8 is also a valid solution.Therefore, all solutions found are valid. However, we need to check if there are any overlaps between the solutions from Case 1 and Case 2.Case 1 solutions are x=π/4 + kπ/2. Let's write some of them:k=0: π/4k=1: π/4 + π/2 = 3π/4k=2: π/4 + π = 5π/4k=3: π/4 + 3π/2 = 7π/4But these are spaced by π/2 each time.Case 2 solutions are x=5π/8 +kπ and x=7π/8 +kπ.For example, k=0: 5π/8,7π/8k=1: 5π/8 +π=13π/8,7π/8 +π=15π/8k=2: 5π/8 +2π=21π/8, etc.Now, check if any of these solutions coincide with Case 1 solutions.For example, π/4 = 2π/8, 3π/4=6π/8, 5π/4=10π/8,7π/4=14π/8Case 2 solutions at 5π/8,7π/8,13π/8,15π/8,... none of these coincide with the Case 1 solutions, which are at 2π/8,6π/8,10π/8,14π/8, etc. So no overlap.Therefore, all solutions are distinct, and the general solution is the union of both cases.Therefore, the solutions are:x = π/4 + kπ/2, where k is any integer,andx =5π/8 +kπ,x=7π/8 +kπ, where k is any integer.But let's express all solutions in a compact form if possible.Alternatively, note that x=5π/8 +kπ and x=7π/8 +kπ can be combined as x=5π/8 +kπ/4, but we need to check.Wait, actually, when k varies over integers, x=5π/8 +kπ and x=7π/8 +kπ can be written as x=5π/8 +kπ and x=7π/8 +kπ, which are two distinct sequences separated by π. However, perhaps we can write them as x= (5π/8 + kπ) and x=(7π/8 +kπ). Alternatively, since 5π/8 and 7π/8 differ by π/4, but adding π each time, maybe not. Let me see.Alternatively, we can express all solutions together. The first set is x=π/4 +kπ/2, which is every π/2 starting at π/4. The second set is every π starting at 5π/8 and 7π/8. Let's see:The first set: x=π/4, 3π/4,5π/4,7π/4,...Second set: x=5π/8,7π/8,13π/8,15π/8,21π/8,23π/8,...These don't overlap with the first set. For example, π/4=2π/8, which is different from 5π/8 or 7π/8. Similarly, 3π/4=6π/8, which is different. 5π/4=10π/8, which isn't in the second set. Therefore, all solutions are distinct.Therefore, the general solution is:x = π/4 + (kπ)/2,x =5π/8 +kπ,x=7π/8 +kπ,where k is any integer.But let me check if x=5π/8 +kπ and x=7π/8 +kπ can be written as x=5π/8 + (π/2) *2k, but that might not simplify. Alternatively, combining these two into one expression.Note that 5π/8 and 7π/8 are separated by π/4, but when we add multiples of π, each subsequent solution will be π apart. Therefore, they cannot be combined into a simpler expression. So, the solution set must be written as three separate series:1. x=π/4 +kπ/2,2. x=5π/8 +kπ,3. x=7π/8 +kπ.However, sometimes, trigonometric equations can be expressed in a more compact form. Let's see if these can be combined.Wait, let's check the period of the original equation. The original equation involves functions like sin4x and cos4x, so the period is π/2. Therefore, the solutions should repeat every π/2. But our current solutions are:First series: spaced by π/2,Second and third series: spaced by π.But π is longer than π/2, so perhaps these solutions can be incorporated into the first series? Wait, perhaps not. Let's check specific solutions.First series: x=π/4 +kπ/2. For k=0: π/4, k=1: 3π/4, k=2:5π/4, k=3:7π/4, etc.Second series: x=5π/8 +kπ. For k=0:5π/8, k=1:13π/8, k=2:21π/8,...Third series: x=7π/8 +kπ. For k=0:7π/8, k=1:15π/8, k=2:23π/8,...If we convert these to eighths:First series: 2π/8, 6π/8,10π/8,14π/8,18π/8,22π/8,...Second series:5π/8,13π/8,21π/8,29π/8,...Third series:7π/8,15π/8,23π/8,31π/8,...Looking at these, they are distinct and do not overlap. So, each series adds unique solutions. Therefore, the general solution must include all three series.Alternatively, is there a way to express all solutions using a single expression? Let's think.The solutions from Case 1 are x=π/4 +kπ/2.The solutions from Case 2 can be written as x=5π/8 +kπ and x=7π/8 +kπ. Notice that 5π/8 = π/4 + 3π/8 and 7π/8 = π/4 + 5π/8. Hmm, not sure if that helps.Alternatively, note that 5π/8 = π/2 + π/8 and 7π/8 = π/2 + 3π/8. Still not helpful.Alternatively, think about all solutions as x= π/8 (1 + 2k) where k is integer. Wait, no. Let me see:Looking at the solutions:First series: π/4,3π/4,5π/4,7π/4,... which is π/4 + kπ/2. In terms of π/8, this is 2π/8,6π/8,10π/8,14π/8,...Second series:5π/8,13π/8,21π/8,...Third series:7π/8,15π/8,23π/8,...So, all solutions in terms of π/8:2π/8, 5π/8,6π/8,7π/8,10π/8,13π/8,14π/8,15π/8,...So, these are every π/8 except multiples of π/8 that are 0, π/8, 4π/8, 8π/8, etc. Hmm, but this might not lead to a simple expression.Alternatively, since the equation reduces to (-cos 2x = frac{sqrt{2}}{2} sin4x), which can be written as:(sin4x + sqrt{2}cos2x = 0)Wait, no, if we move all terms to one side:(-cos2x - sqrt{2}sin2xcos2x = 0), but this might not help.Alternatively, maybe use substitution. Let me set y = 2x. Then the equation becomes:(-cos y = frac{sqrt{2}}{2} sin2y)Since (sin2y = 2sin y cos y), substitute:(-cos y = frac{sqrt{2}}{2} times 2 sin y cos y)Simplify:(-cos y = sqrt{2} sin y cos y)Bring all terms to one side:(-cos y - sqrt{2} sin y cos y = 0)Factor out (cos y):(cos y (-1 - sqrt{2} sin y) = 0)Which gives the same cases as before:1. (cos y = 0 Rightarrow y = frac{pi}{2} + kpi Rightarrow x = frac{pi}{4} + frac{kpi}{2})2. (-1 - sqrt{2} sin y =0 Rightarrow sin y = -frac{1}{sqrt{2}} Rightarrow y = frac{5pi}{4} + 2kpi) or (y = frac{7pi}{4} + 2kpi)Which translates to:For y = 5π/4 +2kπ ⇒ x =5π/8 +kπFor y =7π/4 +2kπ ⇒x=7π/8 +kπTherefore, same solutions as before. Hence, the general solution is correctly given by the three series.Therefore, combining all solutions, the answer is:x = π/4 + (kπ)/2,x =5π/8 +kπ,x=7π/8 +kπ,for any integer k.Alternatively, if we want to present them as combined:x = π/4 + kπ/2,andx =5π/8 +kπ,x =7π/8 +kπ,where k ∈ ℤ.But since the problem might expect the solutions in a certain form, perhaps expressed together. However, these are separate solutions and can't be combined further without redundancy. So, the final answer should list all three families of solutions.Just to ensure there's no mistake, let's verify another solution from Case 1. Take x=3π/4 (k=1 in Case 1):Compute 2x=3π/2, 4x=3π.LHS: cos4x cos(π +2x) - sin2x cos(π/2 -4x)cos4x=cos3π=-1cos(π +2x)=cos(π +3π/2)=cos5π/2=0sin2x=sin3π/2=-1cos(π/2 -4x)=cos(π/2 -3π)=cos(-5π/2)=cos5π/2=0Thus, LHS=(-1)(0) - (-1)(0)=0 -0=0RHS= (√2)/2 sin4x= (√2)/2 sin3π=0So, 0=0, which holds. Therefore, x=3π/4 is valid.Another check for Case 2: x=13π/8 (k=1 in second series):2x=13π/4, which is equivalent to 13π/4 - 2π=13π/4 -8π/4=5π/4, so sin2x=sin5π/4=-√2/2, which satisfies Case 2.Compute LHS:4x=13π/2, which is 6π +π/2, so cos4x=cosπ/2=0cos(π +2x)=cos(π +13π/4)=cos(17π/4)=cos(π/4)=√2/2sin2x=sin13π/4=sin(3π +π/4)= -√2/2cos(π/2 -4x)=cos(π/2 -13π/2)=cos(-6π)=1Thus, LHS=0*(√2/2) - (-√2/2)*1=0 +√2/2=√2/2RHS= (√2)/2 sin4x= (√2)/2 sin(13π/2)= (√2)/2*(-1)= -√2/2Wait, hold on, there's a mistake here. Wait, sin4x=sin(13π/2). Let's compute 13π/2 divided by 2π: 13π/2 =6π +π/2. So, sin(13π/2)=sin(π/2)=1. Wait, no, sin(6π +π/2)=sin(π/2)=1. Therefore, RHS=(√2)/2 *1=√2/2But LHS=√2/2, so √2/2=√2/2, which holds. Wait, but earlier when I computed x=7π/8, I got LHS=-√2/2 and RHS=-√2/2. Wait, but here, x=13π/8:4x=13π/2, which is equivalent to π/2 (since 13π/2 -6π=π/2). Therefore, sin4x=sin(π/2)=1.Therefore, RHS=(√2)/2*1=√2/2, and LHS=√2/2. So, equality holds. Wait, but earlier I thought that when x=7π/8, which is 7π/8=14π/16, but perhaps miscalculation. Let me recheck x=7π/8:x=7π/8:2x=7π/4, which is sin2x=sin7π/4=-√2/24x=7π/2, which is sin4x=sin7π/2=sin(3π +π/2)=-1Wait, sin7π/2=sin(π/2)=1 but with multiple rotations. 7π/2=3π +π/2. sin(3π +π/2)=sin(π/2) with a negative sign because sin(3π +θ)= -sinθ. Therefore, sin7π/2= -sinπ/2= -1Thus, RHS= (√2)/2*(-1)= -√2/2Now, LHS:cos4x=cos7π/2=0cos(π +2x)=cos(π +7π/4)=cos(11π/4)=cos(3π/4)= -√2/2sin2x=sin7π/4= -√2/2cos(π/2 -4x)=cos(π/2 -7π/2)=cos(-3π)=cos3π= -1Thus, LHS=0*(-√2/2) - (-√2/2)*(-1)=0 - (√2/2)= -√2/2, which equals RHS. So, correct.Therefore, all solutions are valid.In conclusion, the solutions are:x = π/4 + (kπ)/2,x =5π/8 +kπ,x=7π/8 +kπ,for any integer k.Alternatively, since 5π/8 = π/4 +3π/8 and 7π/8= π/4 +5π/8, but it's better to present them as separate series.Therefore, the final answer is:(x = frac{pi}{4} + frac{kpi}{2}), (x = frac{5pi}{8} + kpi), (x = frac{7pi}{8} + kpi) for any integer (k).But to check if there's a more concise way, sometimes solutions can be combined using different expressions. For example, x=5π/8 +kπ and x=7π/8 +kπ can be written as x=5π/8 +kπ/4 with certain k, but I don't think so. Let me see:If we set k=2m and k=2m+1 for the second and third series:x=5π/8 +2mπ and x=7π/8 +2mπ and x=5π/8 + (2m+1)π and x=7π/8 + (2m+1)π. But this complicates more. Alternatively, perhaps note that 5π/8 and7π/8 can be represented as π/8(5 +8k) and π/8(7 +8k), but again, not simpler.Therefore, the most straightforward way is to list all three families of solutions.

✅:The given equation is:[cos 4x cos (pi + 2x) - sin 2x cos left(frac{pi}{2} - 4xright) = frac{sqrt{2}}{2} sin 4x]Step-by-Step Solution:1. Simplify Trigonometric Expressions: - Use (cos(pi + theta) = -cos theta) to rewrite (cos(pi + 2x)) as (-cos 2x). - Use (cosleft(frac{pi}{2} - thetaright) = sin theta) to rewrite (cosleft(frac{pi}{2} - 4xright)) as (sin 4x).2. Substitute Simplified Terms: [ cos 4x (-cos 2x) - sin 2x sin 4x = frac{sqrt{2}}{2} sin 4x ] Simplify the left-hand side (LHS): [ -cos 4x cos 2x - sin 2x sin 4x ]3. Combine Terms Using Trigonometric Identity: Recognize the identity (cos(A + B) = cos A cos B - sin A sin B): [ -(cos 4x cos 2x + sin 4x sin 2x) = -cos(4x - 2x) = -cos 2x ] Thus, the equation simplifies to: [ -cos 2x = frac{sqrt{2}}{2} sin 4x ]4. Express in Terms of Sine and Factor: Substitute (sin 4x = 2 sin 2x cos 2x): [ -cos 2x = sqrt{2} sin 2x cos 2x ] Rearrange and factor: [ cos 2x (1 + sqrt{2} sin 2x) = 0 ]5. Solve for Each Factor: - Case 1: (cos 2x = 0) [ 2x = frac{pi}{2} + kpi implies x = frac{pi}{4} + frac{kpi}{2}, quad k in mathbb{Z} ] - Case 2: (1 + sqrt{2} sin 2x = 0 implies sin 2x = -frac{1}{sqrt{2}}) [ 2x = frac{5pi}{4} + 2kpi quad text{or} quad 2x = frac{7pi}{4} + 2kpi ] Divide by 2: [ x = frac{5pi}{8} + kpi quad text{or} quad x = frac{7pi}{8} + kpi, quad k in mathbb{Z} ]Final Solutions:[x = frac{pi}{4} + frac{kpi}{2}, quad x = frac{5pi}{8} + kpi, quad x = frac{7pi}{8} + kpi quad text{for any integer } k]

🔑:1. Expanding the left-hand side of the equation using trigonometric properties: [ cos 4x cos (pi + 2x) - sin 2x cos left( frac{pi}{2} - 4x right) ]2. Use trigonometric identities to simplify each term: - For ( cos(pi + 2x) ): [ cos(pi + 2x) = -cos 2x ] - For ( cos left( frac{pi}{2} - 4x right) ): [ cos left( frac{pi}{2} - 4x right) = sin 4x ]3. Substitute these identities back into the equation: [ cos 4x (-cos 2x) - sin 2x (sin 4x) = frac{sqrt{2}}{2} sin 4x ]4. Simplify the expression: [ - cos 4x cos 2x - sin 2x sin 4x = frac{sqrt{2}}{2} sin 4x ]5. Apply the product-to-sum identity for the terms on the left: [ cos A cos B = frac{1}{2} [cos(A+B) + cos(A-B)] ] [ sin A sin B = frac{1}{2} [cos(A-B) - cos(A+B)] ]6. Notice that ( cos 4x cos 2x ) and ( sin 2x sin 4x ) can be written as: [ cos 4x cos 2x = frac{1}{2} (cos(6x) + cos(2x)) ] [ sin 2x sin 4x = frac{1}{2} (cos(2x) - cos(6x)) ]7. Substitute these identities back into the equation: [ - left( frac{1}{2} (cos(6x) + cos(2x)) right) - left( frac{1}{2} (cos(2x) - cos(6x)) right) = frac{sqrt{2}}{2} sin 4x ]8. Simplify to obtain: [ -frac{1}{2}(cos 6x + cos 2x) - frac{1}{2}(cos 2x - cos 6x) = frac{sqrt{2}}{2} sin 4x ] [ -frac{1}{2} (cos 6x + cos 2x) - frac{1}{2} (cos 2x - cos 6x) = frac{sqrt{2}}{2} sin 4x ] [ -frac{1}{2} cos 6x - frac{1}{2} cos 2x - frac{1}{2} cos 2x + frac{1}{2} cos 6x = frac{sqrt{2}}{2} sin 4x ] [ - cos 2x = frac{sqrt{2}}{2} sin 4x ]9. Observe that: [ cos 2x + sqrt{2} sin 2x cos 2x = 0 ]10. Factor: [ cos 2x (1 + sqrt{2} sin 2x) = 0 ]11. Set each factor to zero and solve: - For ( cos 2x = 0 ): [ cos 2x = 0 implies 2x = frac{pi}{2} + pi n implies x = frac{pi}{4} + frac{pi n}{2} = frac{pi}{4} (2n + 1),; n in mathbb{Z} ] - For ( 1 + sqrt{2} sin 2x = 0 ): [ sin 2x = -frac{sqrt{2}}{2} ] which gives, [ 2x = (-1)^{k+1} frac{pi}{4} + pi k implies x = (-1)^{k+1} frac{pi}{8} + frac{pi k}{2},; k in mathbb{Z} ] Conclusion:(boxed{x_{1} = frac{pi}{4}(2 n + 1),; x_{2} = (-1)^{k+1} frac{pi}{8} + frac{pi k}{2},; n, k in mathbb{Z}})

🧠:Okay, let's try to solve this problem step by step. The function f(x) has several properties given, and we need to find f(1/2011). Let's list out the given conditions first to make sure I don't miss anything.1. f(0) = 0.2. For all real numbers x, f(x) + f(1 - x) = 1.3. For all real numbers x, f(x/3) = (1/2)f(x).4. The function f is non-decreasing on the interval [0, 1]. So, if 0 ≤ x1 < x2 ≤ 1, then f(x1) ≤ f(x2).Our target is f(1/2011). Let's think about how to use these properties to find the value.First, the third condition seems important because 1/2011 is a fraction that can be related to powers of 3. Since 3^7 is 2187, which is just over 2000, maybe 3^7 is 2187, so 1/3^7 is 1/2187. But 1/2011 is a bit larger than 1/2187. Hmm. Maybe we can use some recursive relation here.Also, the second condition tells us that f(1 - x) = 1 - f(x). That's a symmetry condition. For example, if we know f at some point x, we know f at 1 - x. Since we need f(1/2011), perhaps 1 - 1/2011 = 2010/2011, so f(2010/2011) = 1 - f(1/2011). But I don't know if that helps yet.The third condition, f(x/3) = (1/2)f(x), can be rewritten as f(x) = 2f(x/3). So, if we can express x in terms of x/3, maybe we can iterate this. For example, if we need f(1/2011), maybe we can write 1/2011 as (1/3^k) * something, but I need to check if that's possible.Wait, 2011 is a prime number, right? So, 2011 doesn't divide any power of 3, so 1/2011 can't be expressed exactly as a multiple of 1/3^k. Hmm, so maybe we need another approach.Since the function is defined on all real numbers, but the non-decreasing condition is only given on [0,1]. So, the function is non-decreasing between 0 and 1. Maybe the function has a specific structure on [0,1], like a Cantor function or something similar. Let me recall that the Cantor function is constant on the middle-third intervals, and it's defined using base 3 expansions. Also, the Cantor function satisfies f(x/3) = (1/2)f(x), which is similar to the third condition here. Moreover, the Cantor function is non-decreasing. So perhaps this function f is the Cantor function?Wait, let's check the other conditions. The Cantor function satisfies f(0) = 0, f(1) = 1, and f(1 - x) = 1 - f(x). Let me verify that. If we take the Cantor function, which is symmetric around x = 1/2? Let's see, if you take x in the Cantor set, then 1 - x is also in the Cantor set, and since the Cantor function is based on ternary expansions, flipping the digits would correspond to 1 - x. So, yes, f(1 - x) = 1 - f(x) for the Cantor function. So maybe this is the Cantor function.If that's the case, then f(1/3) = 1/2 f(1) = 1/2*1 = 1/2. Wait, but according to the third condition, f(x/3) = (1/2)f(x). Let's check if that holds for the Cantor function. The Cantor function at x = 1/3 is 1/2, and at x = 1, f(1) = 1. Then f(1/3) = 1/2, which is (1/2)f(1). Similarly, f(1/9) = 1/4, which is (1/2)f(1/3). So, yes, this property holds. So the given function f is indeed the Cantor function.Therefore, f(1/2011) would be the value of the Cantor function at 1/2011. But how do we compute that? Since 1/2011 is not a triadic rational (i.e., a number of the form k/3^n), the Cantor function isn't straightforward to compute there. Wait, but perhaps we can use the properties given here to express 1/2011 in terms of base 3?Alternatively, since the function is non-decreasing and satisfies f(x/3) = (1/2)f(x), maybe we can use the ternary expansion of 1/2011 to find its value.Wait, let me recall that the Cantor function can be defined as follows: express x in base 3, and if there is a 1 in the expansion, replace all digits after the first 1 with 0. Then replace all 2s with 1s and interpret the result as a binary number. That gives the Cantor function value at x. So, for example, 1/3 in base 3 is 0.1, which becomes 0.1 in binary, so 1/2. 2/3 is 0.2 in base 3, which becomes 0.1 in binary, so also 1/2. Then 1/9 is 0.01 in base 3, becomes 0.01 in binary, which is 1/4. Similarly, 2/9 is 0.02, becomes 0.01 in binary, 1/4. Then 1/27 is 0.001 in base 3, becomes 0.001 in binary, 1/8, etc.So, for numbers that have only 0s and 2s in their ternary expansion, the Cantor function is determined by replacing 2s with 1s and reading as binary. For numbers that have a 1 in their ternary expansion, the Cantor function value is the same as the first occurrence where a 1 appears, replacing the rest with 0s.But 1/2011 is a number between 0 and 1, and we need to find its ternary expansion. However, 2011 is not a power of 3, so this might get complicated. Alternatively, maybe we can find an n such that 3^n * (1/2011) is in [0,1], but that might not help.Wait, the functional equation f(x/3) = (1/2)f(x). Let's try to use that. Let me set x = 1/2011. Then f(1/(3*2011)) = (1/2)f(1/2011). Similarly, f(1/(3^2*2011)) = (1/2)f(1/(3*2011)) = (1/2)^2 f(1/2011), and so on. So, recursively, we can write f(1/(3^n *2011)) = (1/2)^n f(1/2011).But since the function is non-decreasing on [0,1], we can consider the values approaching 0. As n approaches infinity, 1/(3^n *2011) approaches 0. Since f is non-decreasing and f(0) = 0, the limit as x approaches 0 from the right of f(x) must be 0. Therefore, lim_{n→∞} f(1/(3^n *2011)) = 0. But according to the functional equation, this limit is also lim_{n→∞} (1/2)^n f(1/2011) = 0, which is always true regardless of f(1/2011). So that doesn't give us new information.Alternatively, maybe we need to use the fact that f is non-decreasing to bound f(1/2011). Let's see. Let's consider the interval [0, 1/3]. Since f(x/3) = (1/2) f(x), then for x in [0, 1/3], we can write x = y/3 where y is in [0,1]. Then f(y/3) = (1/2)f(y). So, for x in [0,1/3], f(x) = (1/2)f(3x). Similarly, for x in [1/3, 2/3], we can use the symmetry condition. Wait, but we need to see how the function behaves on different intervals.Alternatively, let's consider partitioning the interval [0,1] into thirds. The function is non-decreasing, so f(0) = 0, f(1) = 1. At x = 1/3, f(1/3) = 1/2 f(1) = 1/2. Similarly, f(2/3) = 1 - f(1 - 2/3) = 1 - f(1/3) = 1 - 1/2 = 1/2. So on [0,1/3], the function increases from 0 to 1/2. Wait, but f(1/3) is 1/2. Wait, hold on. If f(1/3) = 1/2, then since the function is non-decreasing, on [0,1/3], it goes from 0 to 1/2. Then on [1/3, 2/3], it must stay constant at 1/2? But f(2/3) is also 1/2. Then on [2/3,1], it goes from 1/2 to 1. Hmm, this is similar to the Cantor function.Wait, but the standard Cantor function is constant on the middle third [1/3, 2/3], but here, according to the given function's properties, maybe it's not constant? Wait, let's check.Given that f is non-decreasing on [0,1], and f(1/3) = 1/2, and f(2/3) = 1/2. So between 1/3 and 2/3, the function is constant at 1/2. Then from 2/3 to 1, it increases from 1/2 to 1. Similarly, from 0 to 1/3, it increases from 0 to 1/2. Then we can further divide each interval into thirds.For example, on [0,1/3], applying the functional equation f(x) = (1/2)f(3x). So for x in [0,1/9], f(x) = (1/2)f(3x) = (1/2)*(1/2)f(9x) = (1/4)f(9x). Wait, but if x is in [0,1/9], then 9x is in [0,1/9*9) = [0,1). But we need to know how f behaves there. However, since f is non-decreasing, and at x=1/9, f(1/9) = (1/2)f(1/3) = (1/2)*(1/2) = 1/4. Similarly, f(2/9) = (1/2)f(2/3) = (1/2)*(1/2) = 1/4. So on [1/9, 2/9], the function is constant at 1/4? Wait, but f(1/9) = 1/4 and f(2/9) = 1/4. Then between 1/9 and 2/9, the function is constant. Similarly, between 2/9 and 1/3, it increases from 1/4 to 1/2. Wait, no, wait.Wait, actually, if we take x in [0,1/3], f(x) = (1/2)f(3x). So, for x in [0,1/3], 3x is in [0,1]. So, the function on [0,1/3] is a scaled version of the function on [0,1], scaled by 1/2. Similarly, the function on [2/3,1] can be determined by the symmetry condition f(x) = 1 - f(1 - x). So, the function on [2/3,1] is a mirror image of the function on [0,1/3], scaled by 1/2 and shifted.Therefore, the function f is constructed similarly to the Cantor function, which is also known for its self-similar structure. Each third interval is a scaled version of the entire function. So, perhaps the function f is indeed the Cantor function, and the value f(1/2011) can be determined by finding the ternary expansion of 1/2011 and applying the Cantor function rules.But calculating 1/2011 in base 3 is going to be tedious. Alternatively, maybe there's a pattern or a recursive way to calculate it.Alternatively, let's note that 1/2011 is less than 1/3, so it's in the first interval [0,1/3]. Therefore, f(1/2011) = (1/2)f(3*(1/2011)) = (1/2)f(3/2011). Now, 3/2011 is still less than 1/3 (since 3/2011 ≈ 0.00149, and 1/3 ≈ 0.3333), so 3/2011 is in [0,1/3]. Therefore, f(3/2011) = (1/2)f(9/2011). Similarly, 9/2011 ≈ 0.00447, still less than 1/3. So, f(9/2011) = (1/2)f(27/2011). Continuing this way, each time multiplying the numerator by 3.We can keep doing this until we get a value in [1/3, 2/3], where the function might take a different form. Wait, but when does 3^k / 2011 exceed 1/3? Let's solve for k in 3^k / 2011 ≥ 1/3. That is, 3^k ≥ 2011/3 ≈ 670.333. Let's compute 3^6 = 729, which is greater than 670.333. Therefore, when k=6, 3^6 = 729, so 729/2011 ≈ 0.3625, which is in [1/3, 2/3] since 1/3 ≈ 0.3333.Therefore, after 6 iterations, we reach x = 729/2011 ≈ 0.3625. Then, f(729/2011) = 1/2 f(3*(729/2011)) = 1/2 f(2187/2011). But wait, 2187/2011 ≈ 1.0875, which is greater than 1. However, the function is defined on all real numbers, but our non-decreasing condition is only on [0,1]. Let me check the original problem statement. It says f is defined on R, but the non-decreasing condition is for 0 ≤ x1 < x2 ≤ 1.So, for x > 1, we can use the given functional equations to express f(x) in terms of values within [0,1]. Let's see. For example, if x > 1, then 1 - x < 0, so f(1 - x) = 1 - f(x). But f(1 - x) is defined since the function is on all R. However, f(0) = 0, and maybe we can use the other conditions to extrapolate.Wait, but 2187/2011 is greater than 1. Let's compute f(2187/2011). Let's use the relation f(x) + f(1 - x) = 1. So, f(2187/2011) = 1 - f(1 - 2187/2011). Compute 1 - 2187/2011 = (2011 - 2187)/2011 = (-176)/2011. So f(-176/2011) is involved here. But we know f(0) = 0, and f is non-decreasing on [0,1], but for negative arguments, how does f behave? The problem statement doesn't specify the behavior for x < 0, except through the functional equations.Wait, let's check the third condition: f(x/3) = (1/2)f(x). If x is negative, then x/3 is more negative. But unless there's some constraint on negative x, we might need to relate it to positive x via the other equations. For example, using the second condition f(x) + f(1 - x) = 1. Let's take x = -176/2011. Then 1 - x = 1 + 176/2011 = (2011 + 176)/2011 = 2187/2011. So f(-176/2011) + f(2187/2011) = 1. But f(-176/2011) is unknown. However, using the third condition, for x = -176/2011, f(-176/(2011*3)) = (1/2)f(-176/2011). But this might not help unless we can find a connection to the [0,1] interval.Alternatively, maybe we can avoid dealing with negative x by using another approach. Let's see. Starting from f(729/2011):f(729/2011) = 1/2 f(2187/2011). Then, since 2187/2011 > 1, we can write f(2187/2011) = 1 - f(1 - 2187/2011) = 1 - f(-176/2011). But unless we know something about f(-176/2011), this seems stuck.Alternatively, maybe we can use the scaling property again. Since f(x) = 2 f(x/3), even for x >1. Let's check: f(2187/2011) = 2 f(2187/(2011*3)) = 2 f(729/2011). Wait, but then f(729/2011) = (1/2) f(2187/2011) implies that f(2187/2011) = 2 f(729/2011). So substituting back, we get f(729/2011) = (1/2)(2 f(729/2011)) = f(729/2011). Which is a tautology. Not helpful.Alternatively, maybe there is a periodicity or another relation. Wait, perhaps combining multiple functional equations. Let's see. For example, f(x) + f(1 - x) = 1, so f(2187/2011) = 1 - f(-176/2011). If we can express f(-176/2011) in terms of other values.But f(-176/2011) is a negative argument. Let's use the third equation again: f(-176/2011) = 2 f(-176/(2011*3)) = 2 f(-176/6033). Continuing this way, we keep getting more negative arguments, which might not be helpful unless we can relate it back to positive arguments.Alternatively, perhaps the function is defined such that for x outside [0,1], it's extended using the functional equations. However, without more information, it's difficult to proceed this way.Wait, maybe there's another approach. Since the function is non-decreasing on [0,1], and we know its values at certain points, maybe we can bound f(1/2011) between two values that we know.For example, let's note that 1/2011 is a very small positive number. Since the function is non-decreasing, f(1/2011) ≥ f(0) = 0 and f(1/2011) ≤ f(1/3) = 1/2. But that's a very loose bound.Wait, but using the functional equation f(x/3) = (1/2) f(x), we can write f(1/2011) = 2 f(1/(3*2011)) = 2 f(1/6033). Similarly, f(1/6033) = 2 f(1/18099), and so on. Each time, the argument gets smaller, approaching 0. Since f is non-decreasing and f(0) = 0, the value of f at these small points can't be negative, but they can be very small. However, this seems like an infinite loop.Alternatively, let's try to find an n such that 3^n /2011 is approximately 1, but 3^n ≈2011. Let's compute 3^6=729, 3^7=2187. So 3^7=2187 is close to 2011. So if we take n=7, then 3^7=2187, so 3^7 /2011≈1.0875. Then, using the functional equation:f(1/2011) = (1/2)^7 f(3^7 /2011) = (1/128) f(2187/2011).But 2187/2011 ≈1.0875 >1. Then, using the symmetry condition:f(2187/2011) = 1 - f(1 - 2187/2011) = 1 - f(-176/2011).But again, we end up with f(-176/2011), which is problematic.Alternatively, maybe we can relate this to the interval [0,1] by considering modulo 1 or something. Wait, 2187/2011 =1 + 176/2011. So, 2187/2011 =1 + 176/2011. Maybe we can write f(1 + 176/2011) in terms of f(176/2011)?But the problem statement doesn't specify how the function behaves for x >1. However, if we use the relation f(x) + f(1 - x) =1, even for x >1:Take x = 2187/2011, then 1 - x = -176/2011, so f(2187/2011) + f(-176/2011) =1.But we still don't know f(-176/2011). Alternatively, perhaps we can use the third condition on x = -176/2011:f(-176/2011) = 2 f(-176/(3*2011)) =2 f(-176/6033).This seems to lead us into an infinite regression.Alternatively, perhaps there's a pattern where after a certain number of iterations, the argument wraps around into [0,1], but since 2011 and 3 are coprime, this might not happen.Alternatively, maybe we can use the fact that the function is non-decreasing on [0,1], so for x in [0,1], f(x) is between 0 and 1. Then, using the functional equations, maybe we can establish that f(x) is uniquely determined as the Cantor function, hence f(1/2011) is the Cantor function evaluated at 1/2011.But how do we compute that? The Cantor function assigns to each x in [0,1] a value based on its ternary expansion. Specifically, if we write x in base 3 as 0.a_1a_2a_3..., then the Cantor function value is the number obtained by replacing every '2' with '1' and interpreting the result as a binary number. However, if there is a '1' in the ternary expansion, the Cantor function stops at the first '1' and replaces the rest with '0's.But since 1/2011 is a rational number, its ternary expansion is either terminating or repeating. Let's try to find the ternary expansion of 1/2011. This might be tedious, but perhaps there's a pattern.Alternatively, since we know that 3^7 =2187, which is just a bit larger than 2011, perhaps 1/2011 in base 3 is 0.0000001... (base 3), but this is a rough guess.Alternatively, let's consider that 1/2011 ≈0.000496 approximately. Let's convert 0.000496 to base 3.Multiply by 3: 0.000496 *3=0.001488, integer part 0.Multiply fractional part by3: 0.001488*3=0.004464, integer part 0.Continue: 0.004464*3=0.013392, integer part 0.0.013392*3=0.040176, integer part 0.0.040176*3=0.120528, integer part 0.0.120528*3=0.361584, integer part 0.0.361584*3=1.084752, integer part 1.Now subtract 1: 0.084752*3=0.254256, integer part 0.0.254256*3=0.762768, integer part 0.0.762768*3=2.288304, integer part 2.Subtract 2: 0.288304*3=0.864912, integer part 0.0.864912*3=2.594736, integer part 2.Subtract 2: 0.594736*3=1.784208, integer part 1.Subtract 1: 0.784208*3=2.352624, integer part 2.And so on.So, the ternary expansion of 1/2011 is 0.00000010020212... (base 3). Notice that the first '1' appears at the 7th digit after the decimal. So, in base 3, the expansion starts with six 0s, then a 1, followed by other digits.According to the definition of the Cantor function, when there is a '1' in the ternary expansion, the function value is determined by replacing the first '1' with a '1' in binary, and ignoring the rest. So, the first seven digits after the decimal are 0000001..., so replacing that with binary would give 0.0000001 (binary), which is 1/2^7 = 1/128.But wait, let's check: The ternary expansion starts with six 0s, then a 1. So, the position of the first '1' is at the 7th place. In the Cantor function, this corresponds to a binary fraction with a 1 at the 7th place, which is 1/2^7 = 1/128.However, the standard Cantor function replaces all 2s with 1s and disregards any digits after the first 1. So, even if after the first 1 there are more digits, they don't affect the value. Therefore, f(1/2011) = 1/128.But wait, let's confirm this with the functional equations. If f(1/2011) = 1/128, then according to the scaling relation, f(1/(3^k *2011)) = (1/2)^k *1/128.But when k=7, 3^7=2187, so 1/(3^7 *2011)=1/(2187*2011) which is very small. But the function at this point would be (1/2)^7 *1/128=1/(128*128)=1/16384. However, as x approaches 0, f(x) approaches 0, which is consistent.But how does this relate to the ternary expansion? The ternary expansion of 1/2011 has the first '1' at the 7th position, so according to the Cantor function definition, the value is the binary fraction with a 1 at the 7th position, which is 1/2^7=1/128. Therefore, this seems correct.Hence, the answer should be 1/128, which is 1/2^7. Therefore, f(1/2011)=1/128.But let me check this with another approach to make sure.Using the functional equations:Starting with x=1/2011.Since 1/2011 is in [0,1/3], we have f(1/2011)=1/2 f(3/2011).3/2011 is still in [0,1/3], so f(3/2011)=1/2 f(9/2011).9/2011 is still in [0,1/3], so f(9/2011)=1/2 f(27/2011).Continuing this:27/2011 ≈0.0134, still in [0,1/3].f(27/2011)=1/2 f(81/2011).81/2011 ≈0.0403, still in [0,1/3].f(81/2011)=1/2 f(243/2011).243/2011≈0.1208, still in [0,1/3].f(243/2011)=1/2 f(729/2011).729/2011≈0.3625, which is in [1/3,2/3].Now, for x=729/2011 in [1/3,2/3], we can use the relation f(x) + f(1 -x)=1. Therefore, f(729/2011)=1 - f(1 -729/2011)=1 - f(1282/2011).Now, 1282/2011≈0.6375, which is in [1/3,2/3] as well. Wait, but how does this help?Alternatively, since x=729/2011 is in [1/3,2/3], and the function is non-decreasing, f(729/2011) must be between f(1/3)=1/2 and f(2/3)=1/2. Wait, but f(1/3)=1/2 and f(2/3)=1/2, so since the function is non-decreasing on [0,1], and constant on [1/3,2/3], then f(x)=1/2 for all x in [1/3,2/3]. Therefore, f(729/2011)=1/2.Therefore, backtracking:f(729/2011)=1/2.Thus, f(243/2011)=1/2 *1/2=1/4.f(81/2011)=1/2 *1/4=1/8.f(27/2011)=1/2 *1/8=1/16.f(9/2011)=1/2 *1/16=1/32.f(3/2011)=1/2 *1/32=1/64.f(1/2011)=1/2 *1/64=1/128.Yes! This matches the previous result. Therefore, f(1/2011)=1/128.So, the key steps are:1. Recognize the function resembles the Cantor function, which is non-decreasing and satisfies the given functional equations.2. Use the functional equation f(x/3) = (1/2)f(x) repeatedly until the argument falls into the interval [1/3, 2/3], where the function is constant at 1/2.3. Backtrack the computations to find the original value.This confirms that f(1/2011) = 1/128.

✅:To solve for ( fleft(frac{1}{2011}right) ), we utilize the given properties of the function ( f(x) ):1. Functional Equation: ( fleft(frac{x}{3}right) = frac{1}{2}f(x) )2. Non-Decreasing: ( f ) is non-decreasing on ([0, 1])3. Symmetry: ( f(x) + f(1 - x) = 1 )4. Boundary Condition: ( f(0) = 0 )Step-by-Step Process:1. Repeated Application of Scaling: - Since ( frac{1}{2011} ) is in ([0, frac{1}{3}]), apply ( fleft(frac{x}{3}right) = frac{1}{2}f(x) ): [ fleft(frac{1}{2011}right) = frac{1}{2}fleft(frac{3}{2011}right) ] - Continue applying the scaling property until the argument exceeds (frac{1}{3}): [ begin{aligned} fleft(frac{3}{2011}right) &= frac{1}{2}fleft(frac{9}{2011}right) fleft(frac{9}{2011}right) &= frac{1}{2}fleft(frac{27}{2011}right) fleft(frac{27}{2011}right) &= frac{1}{2}fleft(frac{81}{2011}right) fleft(frac{81}{2011}right) &= frac{1}{2}fleft(frac{243}{2011}right) fleft(frac{243}{2011}right) &= frac{1}{2}fleft(frac{729}{2011}right) end{aligned} ] - At ( frac{729}{2011} approx 0.3625 ), the argument enters ([frac{1}{3}, frac{2}{3}]).2. Interval Transition: - For ( x in [frac{1}{3}, frac{2}{3}] ), the function is constant at ( frac{1}{2} ) due to non-decreasing nature and symmetry: [ fleft(frac{729}{2011}right) = frac{1}{2} ]3. Backtracking Values: - Substitute back the computed values: [ begin{aligned} fleft(frac{243}{2011}right) &= frac{1}{2} times frac{1}{2} = frac{1}{4} fleft(frac{81}{2011}right) &= frac{1}{2} times frac{1}{4} = frac{1}{8} fleft(frac{27}{2011}right) &= frac{1}{2} times frac{1}{8} = frac{1}{16} fleft(frac{9}{2011}right) &= frac{1}{2} times frac{1}{16} = frac{1}{32} fleft(frac{3}{2011}right) &= frac{1}{2} times frac{1}{32} = frac{1}{64} fleft(frac{1}{2011}right) &= frac{1}{2} times frac{1}{64} = frac{1}{128} end{aligned} ]Conclusion:By systematically applying the functional equations and leveraging the non-decreasing property, we determine that ( fleft(frac{1}{2011}right) = frac{1}{128} ).[boxed{dfrac{1}{128}}]

🔑:Given: - f(0)=0 - f(x) + f(1-x) = 1- fleft(frac{x}{3} right) = frac{1}{2} f(x)- f(x) is non-decreasing on [0, 1]We need to find fleft(frac{1}{2011}right).1. Evaluate f(1): From f(x) + f(1-x) = 1, set x = 1, we get: [ f(1) + f(0) = 1 ] Since f(0) = 0: [ f(1) = 1 ]2. Evaluate fleft(frac{1}{3}right) and fleft(frac{1}{2}right): From fleft(frac{x}{3}right) = frac{1}{2} f(x), set x = 1, we get: [ fleft(frac{1}{3}right) = frac{1}{2} f(1) = frac{1}{2} times 1 = frac{1}{2} ] Consider x = frac{1}{2}: [ fleft(frac{1}{2} right) + fleft(1 - frac{1}{2}right) = 1 ] Since 1 - frac{1}{2} = frac{1}{2}: [ fleft(frac{1}{2} right) + fleft(frac{1}{2}right) = 1 implies 2fleft(frac{1}{2}right) = 1 implies fleft(frac{1}{2} right) = frac{1}{2} ]3. Determine the behavior on x in left[frac{1}{3}, frac{1}{2}right]: Given f(x) is non-decreasing and left[frac{1}{3}, frac{1}{2}right]: [ fleft(frac{1}{3}right) = frac{1}{2} quad text{and} quad fleft(frac{1}{2}right) = frac{1}{2} ] It means in the interval xin left[frac{1}{3}, frac{1}{2}right], f(x) = frac{1}{2}.4. Evaluate fleft(frac{1}{2011}right): Use the functional equation repeatedly. Firstly, we have: [ fleft(frac{1}{2011}right) = fleft(frac{1}{3} cdot frac{3}{2011}right) = frac{1}{2} fleft(frac{3}{2011}right) ] Next, break frac{3}{2011}: [ fleft(frac{3}{2011}right) = fleft(frac{1}{3} cdot frac{9}{2011}right) = frac{1}{2} fleft(frac{9}{2011} right) ] Following the same step to break: [ fleft(frac{9}{2011}right) = fleft(frac{1}{3} cdot frac{27}{2011}right) = frac{1}{2} fleft(frac{27}{2011} right) ] Continue with the pattern, we generalize: [ fleft(frac{3^k}{2011} right) = left(frac{1}{2}right)^k fleft(frac{3^k}{2011} right) ] Choose k=6 sufficiently large, noting frac{3^6}{2011} approx frac{729}{2011}, which lies between frac{1}{3} and frac{1}{2}. Since f(x) = frac{1}{2} in this range: [ fleft(frac{3^6}{2011} right) = frac{1}{2} ] Therefore, [ fleft(frac{1}{2011}right) = left(frac{1}{2}right)^6 fleft( frac{729}{2011}right) = left(frac{1}{2}right)^6 cdot frac{1}{2} = frac{1}{128} ] Conclusion:[boxed{frac{1}{128}}]

🧠:Okay, so I have this problem here: A triangle is divided into 1000 smaller triangles. What is the minimum number of distinct points that can be the vertices of these triangles? Hmm, interesting. Let me try to figure this out step by step.First, I need to recall some basics about triangulation and dividing triangles into smaller ones. I remember that when you divide a triangle into smaller triangles, you have to add points inside the original triangle and connect them with edges so that all the regions formed are triangles. The question is asking for the minimum number of distinct vertices needed to create 1000 small triangles. So, we need to minimize the number of points used while maximizing the number of triangles created. Let me start with a simpler case. If a triangle is divided into smaller triangles without adding any points, it's just one triangle. If I add a single point inside the triangle and connect it to the three vertices, then the original triangle is divided into three smaller triangles. So, with one interior point, we get three triangles. That seems like a formula might be involved here. Maybe the number of triangles relates to the number of points added?Wait, there's a formula for the number of triangles formed when you add points inside a triangle. I think Euler's formula might come into play here. Euler's formula states that for a planar graph, the number of vertices minus the number of edges plus the number of faces equals 2. In this case, the original triangle is part of the graph, so maybe we can apply Euler's formula to find a relationship between the number of vertices and the number of triangles.Let me try to recall Euler's formula: V - E + F = 2. Here, F includes all the faces, including the outer face. But in our case, we're dealing with a triangulated triangle, so all inner faces are triangles, and the outer face is the original triangle. Wait, but the problem states that the triangle is divided into 1000 smaller triangles. Does that include the original triangle or not? Hmm, the wording says "divided into 1000 smaller triangles," so probably the original triangle is divided into 1000 smaller ones, meaning the original triangle is not counted as one of the 1000. Therefore, the 1000 triangles are all the inner faces, and the outer face is the original triangle. But maybe I need to clarify that.Alternatively, maybe the formula for the number of triangles in a triangulation. Let me think. If you have a triangulated convex polygon with n vertices, the number of triangles formed is n - 2. But that's for a convex polygon divided into triangles by non-intersecting diagonals. However, in our case, the original triangle is being divided into smaller triangles with additional vertices inside. So, maybe a different formula.Wait, when you triangulate a polygon with V vertices (including the original ones) and E edges, then the number of triangles T is given by T = 2V - E - 2 or something like that? Maybe not. Let's try to apply Euler's formula properly.Let's suppose we have the original triangle, which has 3 vertices and 3 edges. Then we add some number of points inside and on the edges. Let's denote the total number of vertices as V, edges as E, and faces as F. In the triangulation, each face (except the outer face) is a triangle. The original triangle is the outer face. So, F = 1000 + 1 = 1001 faces. Euler's formula says V - E + F = 2, so V - E + 1001 = 2 => V - E = -999 => E = V + 999.But we also know that in a triangulation, each edge is shared by two faces, except for the edges on the boundary, which are part of only one face (the outer face). Let's count the edges. Each triangle has three edges, but each interior edge is shared by two triangles. Let me denote the number of edges on the boundary as 3 (the original triangle's edges). The total number of edges can also be calculated as (3T + B)/2, where T is the number of triangles and B is the number of boundary edges. Wait, let me check.If there are T triangles, each contributing 3 edges, that's 3T edges. But each interior edge is shared by two triangles, so they are counted twice, and the boundary edges are counted once. Therefore, the total number of edges E is (3T + B)/2, where B is the number of boundary edges. In our case, the original triangle has 3 edges, so B = 3. So, E = (3*1000 + 3)/2 = (3003)/2 = 1501.5. Wait, that can't be, since E must be an integer. So, there's an inconsistency here. Hmm, maybe my approach is wrong.Wait, maybe the formula is E = (3T + B)/2. Let's check with a simple example. Suppose we have the original triangle divided into three smaller triangles by adding an interior point connected to all three vertices. Then T = 3, B = 3. So E = (9 + 3)/2 = 6. The original triangle has 3 edges, and the three new edges from the interior point to the vertices, totaling 6 edges. That's correct. So, yes, E = (3T + B)/2. But in this case, T = 3, so 3T = 9, plus B = 3, divided by 2 is 6. Correct. So for T = 1000, E = (3*1000 + 3)/2 = 3003/2 = 1501.5. But that's not an integer. Therefore, that's a problem. So perhaps there's a mistake in my reasoning here.Wait, maybe B is not 3? Wait, the original triangle's edges might be subdivided. If we add points on the edges, then the boundary edges are subdivided into multiple edges. So, if points are added on the edges, then B would be the number of edges on the boundary after subdivision. So perhaps B is not necessarily 3? Hmm, that complicates things. So maybe my initial assumption that B = 3 is incorrect if we allow adding points on the edges.Therefore, if we allow adding points on the edges, the boundary edges can be split into multiple edges. Therefore, B would be the total number of edges on the boundary after subdivision. So, in that case, the formula E = (3T + B)/2 still holds, but B is variable. Then, we can use Euler's formula: V - E + F = 2. Since F = T + 1 (1000 triangles plus the outer face), so F = 1001. Therefore:V - E + 1001 = 2 => V - E = -999 => E = V + 999.But also, E = (3*1000 + B)/2 = (3000 + B)/2.Therefore, setting these equal:V + 999 = (3000 + B)/2Multiply both sides by 2:2V + 1998 = 3000 + BTherefore, B = 2V + 1998 - 3000 = 2V - 1002.But B is the number of boundary edges, which must be at least 3 (if no points are added on the original edges) and can be more if we add points on the original edges.But in a triangulation, the boundary edges form a polygon. Since we started with a triangle, even after adding points on the edges, the boundary is still a triangle, but each edge can be divided into multiple edges. However, in a triangulation, all the boundary edges must be part of the original triangle's edges. Wait, actually, if you add a point on an edge, it splits the edge into two edges. So the original three edges can be split into multiple edges, but the entire boundary is still a triangle, but with subdivided edges.Wait, but actually, the boundary of the triangulation is a polygon. If you start with a triangle and add points on its edges, the boundary becomes a polygon with more edges. For example, if you add one point on each edge of the original triangle, the boundary would be a hexagon. Wait, no. Let me see. If you split each edge of the original triangle into two edges by adding a point in the middle, then each original edge becomes two edges, so the boundary would have 3*2=6 edges, forming a hexagon? But actually, no, the three original edges are each split into two, but the figure is still topologically a triangle, but with subdivided edges. However, the outer face is a polygon with six edges? Wait, that can't be. Let me think.Suppose you have the original triangle ABC. If you add a point D on edge AB, then edge AB is split into AD and DB. Similarly, add points E on BC and F on AC. Then, the boundary edges would be AD, DB, BE, EC, CF, FA. So, the boundary is a hexagon? Wait, but that's not possible because the original triangle's edges are each split into two, but the figure is still a triangle. Wait, but in terms of the planar graph, the outer face is bounded by those six edges. But that would make the outer face a hexagon. But that contradicts the fact that we started with a triangle and only subdivided its edges. So, maybe my understanding is off here.Wait, no. If you add a point on each edge of the original triangle and connect them appropriately, the outer face can still be a triangle. Wait, for example, if you add a point on each edge and connect them in a way that forms a smaller triangle inside, then the outer boundary is still the original triangle. Wait, no. If you connect the three points on the edges, forming an inner triangle, then the original edges are still there, but the outer face is still bounded by the original three edges. The inner edges are part of the inner triangles. So, in that case, the outer face is the original triangle with its original edges, even if they are subdivided. Wait, maybe the subdivision of edges doesn't affect the outer face's polygon structure? Hmm, I need to clarify this.In planar graph terms, the outer face is bounded by the edges that are not shared by two triangles. So, if you have a triangulation where the original triangle's edges are subdivided, then the outer face would have a boundary composed of those subdivided edges. For example, if you split each original edge into two, then the outer face's boundary would have six edges. Therefore, the outer face is a hexagon. But that contradicts the idea that we are only subdividing the original triangle into smaller triangles. Wait, maybe not. If you split each edge into two and connect the midpoints, you create four smaller triangles. The original triangle is divided into four smaller triangles: three at the corners and one in the center. Wait, in that case, the outer faces would be the three corner triangles, each adjacent to the original edges. Wait, but the original edges are split into two, so each corner triangle has two edges that are part of the original edges. Hmm, this is getting confusing. Let me visualize it.Original triangle ABC. Add a midpoint D on AB, E on BC, F on AC. Connect D, E, F. Then, you have four smaller triangles: ADF, DBE, EFC, and DEF. The outer faces are ADF, DBE, EFC, each adjacent to the original edges. Each original edge is split into two, so the boundary edges are AD, DB, BE, EC, CF, FA. So, the outer face is actually a hexagon? But how? Because each original edge is split into two, and the outer face is the combination of the three original edges split into two each. But in reality, the outer face would consist of the three original edges, each split into two, making six edges. However, each of these edges is part of the boundary of the outer face, which is the original triangle. Wait, but the outer face is still a triangle, but each edge is divided into two. So, in the planar embedding, the outer face is a triangle, but each edge is represented by two edges in the graph. Hmm, maybe in planar graph terms, the outer face can have multiple edges but still form a triangle. That doesn't quite make sense. Maybe my confusion arises from the difference between geometric edges and graph edges.Perhaps I need to step back. Let me consider the formula again. If the original triangle is divided into smaller triangles with vertices added on the edges and inside, then the total number of triangles is 1000, and we need to find the minimum number of vertices. The minimal number of vertices would correspond to the most efficient way of adding points to create as many triangles as possible with as few points as possible.I remember that in a triangulation, each time you add a point inside a triangle, you can connect it to the three vertices, splitting the original triangle into three. But if you add a point on an edge, you split the edge into two and can connect it to form two triangles. But maybe there's a more efficient way. Alternatively, using a structure where each new point is part of multiple triangles.Wait, another thought: the problem is similar to finding the minimum number of vertices in a planar triangulation with 1000 faces. But planar triangulations require that every face (including the outer face) is a triangle. However, in our case, the outer face is the original triangle, so it's not necessarily a triangulation of the entire plane, but a subdivision of the original triangle into smaller triangles.I recall that in a triangulated planar graph without multiple edges or loops, the number of edges can be related to the number of faces. For a maximal planar graph (every face is a triangle), the formula is E = 3V - 6. But in our case, the outer face is a triangle, and all inner faces are triangles. So, perhaps similar formulas apply here.Let me see. If we have a planar graph where all inner faces are triangles and the outer face is also a triangle. Then, using Euler's formula V - E + F = 2. Here, F is the number of faces, which is 1000 inner triangles plus the outer face, so F = 1001. Therefore:V - E + 1001 = 2 => V - E = -999 => E = V + 999.In a maximal planar graph (triangulation), we have E = 3V - 6. But in our case, the outer face is a triangle, so maybe similar? Let's check.In a maximal planar graph with V vertices, E = 3V - 6. But in our case, since the outer face is fixed as a triangle, maybe the formula is slightly different. Let me think.If the graph is triangulated, then each face (including the outer face) is a triangle. Then, each face is bounded by three edges, and each edge is shared by two faces. So, 3F = 2E. But in our case, the outer face is a triangle, so 3(F - 1) + 3 = 2E, since the outer face is also a triangle. Wait, so:Each inner face (there are F - 1 = 1000 of them) is a triangle, contributing 3 edges each, and the outer face is also a triangle, contributing 3 edges. Each edge is shared by two faces, except the edges on the outer face, which are shared by only one face. Wait, no. If the graph is embedded in the plane with all faces (including outer) being triangles, then every edge is shared by two faces. But in our problem, the original triangle is divided into smaller triangles, so the outer face is the original triangle. If we are not allowing multiple edges or vertices on the outer face, then maybe all edges are shared by two triangles except the three original edges. Wait, this is confusing.Alternatively, let me use the standard formula. If all inner faces are triangles and the outer face is a triangle, then the total number of edges can be calculated as follows. Each triangle has three edges, and each inner edge is shared by two triangles. The outer edges are part of only one triangle each (the original triangle's edges). Therefore:Total edges from triangles: 3 * (number of inner triangles + 1) = 3*(1000 + 1) = 3003. But each inner edge is counted twice, and the outer edges are counted once. Therefore:Total edges E = (3*1001 - 3)/2 + 3 = (3003 - 3)/2 + 3 = 3000/2 + 3 = 1500 + 3 = 1503. Wait, let me verify this formula.If we have T inner triangles and the outer face, which is also a triangle. Each inner edge is shared by two triangles, and the outer edges are only part of the outer triangle. So the total number of edges is (3*(T + 1) - B)/2 + B, where B is the number of boundary edges. Here, B = 3 (original edges). So:E = (3*(1000 + 1) - 3)/2 + 3 = (3003 - 3)/2 + 3 = 3000/2 + 3 = 1500 + 3 = 1503.So E = 1503.But from Euler's formula, we had E = V + 999. Therefore:1503 = V + 999 => V = 1503 - 999 = 504.Therefore, the number of vertices is 504. But wait, this includes the original three vertices of the triangle. So, the minimum number of distinct points is 504.But this seems too straightforward. Let me check with a smaller example. Suppose we have T = 1 (the original triangle). Then using the formula:E = (3*(1 + 1) - 3)/2 + 3 = (6 - 3)/2 + 3 = 3/2 + 3 = 4.5. Which is not possible. Hmm, so maybe the formula is not applicable for T = 1.Wait, actually, T = 1 corresponds to the original triangle, so number of edges is 3, vertices 3. Let's plug into Euler's formula:V - E + F = 3 - 3 + 2 = 2, which works (F = 2: the outer face and the inner face? Wait, no. If there's only the original triangle, then F = 1 (the outer face). Wait, maybe my earlier assumption about F was wrong.Wait, confusion arises here. When we talk about planar graphs, the "faces" include the outer face. So, in the case of the original triangle, it's just one face (the outer face), right? But if we triangulate the interior, then each subdivision adds inner faces.Wait, let's re-examine Euler's formula. For a connected planar graph, V - E + F = 2. If we have a single triangle, then V = 3, E = 3, F = 1 (the outer face). Then 3 - 3 + 1 = 1 ≠ 2. That's a problem. Wait, no, actually, in planar graph theory, the outer face is also counted. Wait, no. Wait, if you have a triangle drawn in the plane, it divides the plane into two regions: the inside and the outside. But in planar graph theory, usually, the "infinite" face is considered as one face, and the inner face is another. So, for a single triangle, F = 2. Then, 3 - 3 + 2 = 2, which satisfies Euler's formula. Ah, right. So even a single triangle has two faces: the inner and the outer. Wait, but in our problem statement, when they say the triangle is divided into 1000 smaller triangles, does that include the original triangle or not? If the original triangle is divided into 1000 smaller ones, then the original triangle is replaced by 1000 smaller triangles, so the total number of inner faces is 1000, and the outer face is still the "infinite" face. Wait, no. Wait, if you subdivide a triangle into smaller triangles, all the faces except the outer face are the smaller triangles. So, if the original triangle is subdivided into 1000 smaller triangles, then F = 1000 (inner faces) + 1 (outer face) = 1001. Therefore, in that case, using Euler's formula:V - E + F = 2 => V - E + 1001 = 2 => V - E = -999 => E = V + 999.In the case of the original triangle (before subdivision), V = 3, E = 3, F = 2 (the inner and outer face). Then, 3 - 3 + 2 = 2, which works. If we divide the original triangle into three smaller triangles by adding an interior point connected to the three vertices, then V = 4, E = 6, F = 4 (three inner triangles and the outer face). Check Euler's formula: 4 - 6 + 4 = 2, which holds.So, in this case, T = 3 (the three inner triangles). Then, according to the formula, E = V + (F - 2). Wait, no. Wait, when we have T = 3 inner triangles, F = 4. Then, according to the previous equation, E = V + (F - 2) = 4 + (4 - 2) = 6. Which matches.But in our earlier calculation for T = 1000, we found that E = 1503 and V = 504. But let's verify with the smaller case. For T = 3:Using the formula for edges:E = (3*(T + 1) - B)/2 + B.Assuming B = 3 (original edges, not subdivided). Then:E = (3*4 - 3)/2 + 3 = (12 - 3)/2 + 3 = 9/2 + 3 = 4.5 + 3 = 7.5. Which is not an integer, which is wrong. But in reality, when we add one interior point connected to three vertices, E = 6. So, this formula isn't working here. Therefore, my formula is incorrect.So, perhaps my approach is flawed. Let me think differently.Maybe instead of trying to use Euler's formula directly, I can use the relationship between the number of triangles and the number of vertices in a triangulation.In a triangulation, each new vertex added inside can increase the number of triangles. For example, adding a vertex inside a triangle and connecting it to the three vertices creates three triangles. So, each internal vertex adds two triangles. Wait, no. If you start with one triangle and add an internal vertex connected to all three, you get three triangles. So, adding one vertex gives three triangles. Then, if you add another vertex inside one of those triangles and connect it to the three vertices of that triangle, you split that triangle into three, so total triangles become 3 + 2 = 5. Wait, each time you add a vertex inside a triangle, you replace one triangle with three, so the total increases by two. So, the number of triangles increases by two for each internal vertex after the first one. Wait, let's check.Original triangle: 1 triangle.Add first internal vertex: split into 3 triangles. Total triangles = 3. So, two more triangles.Add second internal vertex: choose one of the three triangles and split it into three. Total triangles = 3 + 2 = 5.Third internal vertex: split another triangle into three. Total = 5 + 2 = 7.So, each internal vertex after the first adds two triangles.Therefore, the formula would be: T = 1 + 2 + 2 + ... + 2 (n times), where n is the number of internal vertices. Wait, no. The first internal vertex adds 2 triangles (from 1 to 3), then each subsequent internal vertex adds 2 triangles. So, T = 1 + 2k, where k is the number of internal vertices. Wait, if k = 1, T = 3. If k = 2, T = 5. k=3, T=7. So, T = 1 + 2k. Therefore, to get T = 1000, we need k = (1000 -1)/2 = 499.5. Which is not possible, since k must be an integer. Therefore, this suggests that this model isn't scalable, or maybe it's a different way of adding vertices.But this seems to contradict the earlier example where adding a vertex connected to three edges can create three triangles. Maybe there's a different way to add vertices that create more triangles. For example, if we add a vertex on an edge, splitting the edge into two, and then connecting it appropriately.Alternatively, consider that when you add a vertex on an edge, it splits the edge into two and allows for more triangles. Maybe adding points on edges can help create more triangles with fewer vertices.Alternatively, there is a formula for the number of triangles in terms of the number of vertices. Let me recall that in a triangulation, the number of triangles T is related to the number of vertices V by T = 2V - 4 - B, where B is the number of vertices on the boundary. Wait, not sure. Let me check some references in my mind.Wait, in a convex polygon triangulated into triangles, the number of triangles is n - 2, where n is the number of vertices. But that's for a convex polygon divided into triangles without adding new vertices. In our case, we have a triangle subdivided into smaller triangles with added vertices. So, maybe the formula is similar but different.Alternatively, in a triangulation of a polygon with V vertices (including the original), the number of triangles is 2V - 2 - B, where B is the number of boundary vertices. Wait, not sure. Let me try to derive it.Suppose we have a triangle subdivided into smaller triangles. Let V be the total number of vertices, E the edges, and T the number of triangles. The original triangle has 3 vertices on the boundary. Let B be the number of boundary vertices (on the original triangle's edges). Let I be the number of interior vertices (inside the original triangle). So, V = B + I.In such a triangulation, each interior vertex has degree at least 6? Wait, no. In a planar triangulation, each interior vertex must have degree at least 3. But in our case, since all faces are triangles, each vertex must have degree at least 3.Wait, let's use Euler's formula again. V - E + F = 2. F = T + 1, as before. Each face (except the outer face) is a triangle, and the outer face is a polygon with B edges. Wait, but the outer face is the original triangle, so if we added points on the edges, then the outer face is a polygon with more edges. Wait, confusion again.Alternatively, let's consider that the boundary is a polygon with B edges, which are the subdivided edges of the original triangle. Then, using the formula for the number of edges in a triangulation:Each triangle has three edges, each interior edge is shared by two triangles, and each boundary edge is part of one triangle. So, total edges E = (3T + B)/2.From Euler's formula: V - E + (T + 1) = 2 => V - E + T + 1 = 2 => V - E + T = 1.But E = (3T + B)/2, so:V - (3T + B)/2 + T = 1 => V - (3T + B)/2 + T = 1 => V - (3T + B - 2T)/2 = 1 => V - (T + B)/2 = 1 => 2V - T - B = 2.So, 2V - T - B = 2.But also, we have V = B + I, where I is the number of interior vertices.So substituting V = B + I into the equation:2(B + I) - T - B = 2 => 2B + 2I - T - B = 2 => B + 2I - T = 2 => T = B + 2I - 2.Therefore, T = B + 2I - 2.But we need T = 1000. So:1000 = B + 2I - 2 => B + 2I = 1002.We want to minimize the total number of vertices V = B + I. So, since B + 2I = 1002, we can express V as:V = (1002 - 2I) + I = 1002 - I.To minimize V, we need to maximize I. Since B must be at least 3 (the original three vertices), and B = 1002 - 2I, we have 1002 - 2I ≥ 3 => 2I ≤ 999 => I ≤ 499.5. Since I must be an integer, I ≤ 499. Therefore, the maximum I is 499, which gives B = 1002 - 2*499 = 1002 - 998 = 4. Wait, but B must be at least 3. Wait, if I = 499, then B = 4. But B is the number of boundary vertices. Wait, but if we started with three vertices on the boundary, how can B be 4?Wait, perhaps B is the number of boundary edges? No, earlier we considered B as the number of boundary vertices, but actually, in the formula above, T = B + 2I - 2, where B is the number of boundary edges. Wait, no, wait. Let me re-examine.Wait, the formula we derived was T = B + 2I - 2, where B is the number of boundary vertices. Wait, in the derivation, B was introduced as the number of boundary vertices. Wait, let me check the steps again.We started with V = B + I, where B is the number of boundary vertices and I the number of interior vertices.From the equation 2V - T - B = 2, substituting V = B + I gives:2(B + I) - T - B = 2 => 2B + 2I - T - B = 2 => B + 2I - T = 2 => T = B + 2I - 2.Therefore, T = B + 2I - 2.So, yes, B is the number of boundary vertices.But originally, the triangle has three boundary vertices. If we add vertices on the edges, then the boundary vertices increase. Each time we add a vertex on an edge, it splits the edge into two, and the new vertex becomes a boundary vertex. So, B is the total number of vertices on the boundary (original edges). Therefore, B can be any integer greater than or equal to 3. So, in order to maximize I (and minimize V), we need to set B as small as possible. But according to the equation B + 2I = 1002, to maximize I, we need to minimize B.But B must be at least 3, right? So, the minimum B is 3. Let's check.If B = 3, then 3 + 2I = 1002 => 2I = 999 => I = 499.5. But I must be an integer. Therefore, impossible. Therefore, the next possible B is 4.If B = 4, then 4 + 2I = 1002 => 2I = 998 => I = 499. Therefore, V = B + I = 4 + 499 = 503.But is B = 4 possible? B is the number of boundary vertices. How can a triangle with subdivided edges have 4 boundary vertices?Wait, the boundary vertices are the ones lying on the original triangle's edges. If you add a vertex on one edge, then B becomes 4. For example, original triangle has vertices A, B, C. Add a vertex D on edge AB. Then, the boundary vertices are A, D, B, C. Wait, but edge AB is split into AD and DB, so vertices on the boundary are A, D, B, C (assuming we don't add vertices on the other edges). So, B = 4. Then, the original triangle's edges have 4 boundary vertices. But in this case, the other edges (BC and AC) still have their original vertices. So, adding a vertex on one edge increases B by 1. Similarly, adding a vertex on another edge increases B by another 1.But in our case, to get B = 4, we need to add one vertex on one edge, keeping the other two edges with their original vertices. So, yes, B = 4 is possible. Similarly, B = 5 would require adding one vertex on two edges, and so on.Therefore, if B = 4, then I = 499, and V = 4 + 499 = 503. However, the formula gives T = B + 2I - 2 = 4 + 2*499 - 2 = 4 + 998 - 2 = 1000, which matches. So, this seems correct.But wait, if B = 4, that means there are four boundary vertices. Then, the original triangle's edges have been subdivided such that one edge has an extra vertex, making two boundary vertices on that edge, and the other two edges remain with one vertex each. So, the total boundary vertices are 2 (from the subdivided edge) + 1 + 1 = 4.But in such a case, the original triangle's subdivided edge has two segments, and the other edges are not subdivided. Then, when triangulating, we need to ensure that all faces are triangles. However, if one edge is split into two, creating a four-sided polygon on the boundary, how do we triangulate that? Wait, no. The boundary is still a triangle but with subdivided edges. Wait, if you have four boundary vertices, how is the outer face still a triangle?Wait, maybe I'm mixing up the concepts here. If B is the number of boundary vertices, then the outer face (the original triangle) is a polygon with B edges. Wait, but a polygon with B vertices has B edges. So, if B = 4, then the outer face is a quadrilateral, which contradicts the requirement that all inner faces are triangles and the outer face is also a triangle. Therefore, there's a mistake in my reasoning here.Hold on, in the problem statement, it's specified that the triangle is divided into smaller triangles. Therefore, the outer face must remain a triangle. Therefore, we cannot have the outer face being a quadrilateral or any other polygon. Therefore, the original edges cannot be subdivided in such a way that the outer face becomes a different polygon. Therefore, all subdivisions must be such that the original edges are preserved as edges of the outer face, even if they are split into smaller edges. Wait, this is confusing.Alternatively, if we add a vertex on an edge of the original triangle, we split that edge into two, but both of the new edges are still part of the boundary of the outer face. Therefore, the outer face is still a triangle, but each original edge is represented by a path of edges in the graph. For example, if we add a vertex on edge AB, creating edges AD and DB, then the outer face is still triangle ABC, but edge AB is represented by AD and DB. So, in terms of the planar graph, the outer face is still a triangle, but the edge AB is subdivided into two edges. Therefore, the number of boundary edges increases, but the outer face remains a triangle.But in planar graph terms, the outer face is a triangle, so it must be bounded by three edges. But if we subdivide an edge, then the outer face would be bounded by more edges. Wait, no. The outer face is a triangle, which is a 3-sided polygon. If we subdivide an edge of the original triangle, the outer face would still be a triangle, but the subdivided edge is part of the boundary.Wait, I'm getting tangled up here. Let me try to think differently.If the outer face must remain a triangle, then the original three edges must each be single edges. Therefore, we cannot add any vertices on the edges of the original triangle. Therefore, all added vertices must be strictly inside the original triangle. Therefore, B = 3 (the original three vertices), and V = 3 + I.From the equation T = B + 2I - 2, with B = 3:T = 3 + 2I - 2 = 2I + 1.We need T = 1000, so 2I + 1 = 1000 => 2I = 999 => I = 499.5. But I must be an integer. Therefore, impossible. Hence, this suggests that if we cannot add any vertices on the edges, it's impossible to have 1000 triangles with integer vertices. Therefore, we must add some vertices on the edges to make B larger, allowing I to be integer.Therefore, to make T = B + 2I - 2 = 1000, with B ≥ 3 and I integer, we need B + 2I = 1002. So, B must be even because 1002 - B must be even for I to be integer. Wait, 1002 - B must be divisible by 2. Therefore, B must be even. Since B ≥ 3, the smallest even B is 4. Then, B = 4, I = (1002 - 4)/2 = 499. Therefore, V = B + I = 4 + 499 = 503.Similarly, if B = 6, then I = (1002 - 6)/2 = 498, so V = 6 + 498 = 504, which is worse (higher). So, the minimal V occurs when B is minimal even number greater than or equal to 4. Since B must be at least 3, but B needs to be even to make I integer. Therefore, the minimal B is 4, leading to V = 503.But how can B = 4? If we add one vertex on one edge of the original triangle, then the number of boundary vertices becomes 4. The original triangle has vertices A, B, C. Add a vertex D on edge AB. Then, the boundary vertices are A, D, B, C. Wait, but the outer face is still a triangle with vertices A, B, C, but the edge AB is split into AD and DB. So, in terms of the graph, the outer face is still bounded by the edges AD, DB, BC, CA. Wait, but that's four edges, which would make the outer face a quadrilateral, which contradicts the requirement that the outer face is a triangle.Ah, here's the crux. If we subdivide an edge of the original triangle, the outer face becomes a quadrilateral, which is not allowed because we must keep the original triangle as the outer face. Therefore, perhaps my initial assumption that B is the number of boundary vertices is incorrect. Alternatively, in the problem statement, maybe the original triangle's edges can be subdivided, but the outer face is still considered a triangle. However, in planar graph terms, subdividing an edge would change the outer face's structure.This is a critical point. If the problem allows subdivision of the original triangle's edges but still considers the outer face as a triangle (i.e., the subdivided edges are part of the original triangle's boundary), then the outer face would have more edges, but visually, it's still considered a triangle. However, mathematically, in planar graph terms, the outer face would be a polygon with more edges. Therefore, this seems contradictory.Therefore, perhaps the problem requires that the original triangle is divided into smaller triangles without adding any vertices on the original edges. In other words, all added vertices are strictly inside the original triangle. In that case, B = 3 (original vertices), and the equation T = 2I + 1 = 1000 leads to I = 499.5, which is impossible. Therefore, the problem must allow adding vertices on the original edges.But if we allow adding vertices on the original edges, then the outer face becomes a polygon with more edges. However, the problem states that the original triangle is divided into smaller triangles. So, perhaps the outer face is still considered the original triangle, and the subdivisions are internal. Wait, but subdividing the edges would technically change the outer face.This is a bit ambiguous. Depending on the interpretation, the answer could differ. If we are not allowed to add vertices on the edges, then it's impossible to reach 1000 triangles. Therefore, the problem must allow adding vertices on the edges.Assuming that vertices can be added on the edges, and the outer face is still considered a triangle with the original three edges (even if they are subdivided into smaller edges), then the formula T = B + 2I - 2 still holds, with B being the number of boundary vertices (on the original edges). Therefore, to minimize V = B + I, we need to minimize B. Since B must be even (to make 1002 - B even for I to be integer), the minimal B is 4, leading to V = 503.But earlier, we found that with V = 504, but that was under a different assumption. So, which is correct?Wait, going back to the initial equations:We have T = B + 2I - 2 = 1000.Therefore, B + 2I = 1002.To minimize V = B + I, express V = 1002 - I.To minimize V, maximize I. Since B = 1002 - 2I, and B must be at least 3, but when B is the number of boundary vertices, which is the number of vertices on the original triangle's edges. If we add one vertex on one edge, then B = 4 (original three vertices plus one new vertex on an edge). Wait, no. If we add a vertex on an edge, say between A and B, then we have two vertices on that edge: A and D (the new vertex). But D is a new boundary vertex. So, the total boundary vertices are A, D, B, C. Therefore, B = 4.So, B is the number of vertices on the boundary (original edges), including the original vertices and the added ones. So, each time we add a vertex on an edge, B increases by 1. So, starting with B = 3, adding k vertices on the edges gives B = 3 + k.Therefore, in the equation B + 2I = 1002, B = 3 + k, so 3 + k + 2I = 1002 => k + 2I = 999.We need to minimize V = B + I = 3 + k + I.Since k + 2I = 999, we can express k = 999 - 2I.Therefore, V = 3 + (999 - 2I) + I = 1002 - I.To minimize V, maximize I. The maximum I is when k is minimized. Since k ≥ 0, the minimal k is 0, leading to I = 999/2 = 499.5, which is not possible. Therefore, k must be at least 1, making I = (999 - 1)/2 = 499. Then, k = 999 - 2*499 = 999 - 998 = 1.Therefore, B = 3 + k = 4, I = 499, V = 4 + 499 = 503.Therefore, the minimal number of vertices is 503.But let's verify this with Euler's formula.Given V = 503, B = 4, I = 499.Then, T = 1000.From the equation E = (3T + B)/2.E = (3*1000 + 4)/2 = 3004/2 = 1502.From Euler's formula: V - E + F = 2.F = T + 1 = 1001.So, 503 - 1502 + 1001 = 2.503 - 1502 = -999; -999 + 1001 = 2. Correct.Therefore, this checks out.But how does this triangulation look? We have 4 boundary vertices (three original and one added on an edge), and 499 interior vertices. Each interior vertex contributes to two additional triangles, and the boundary vertex allows for the extra triangle.But how is the single boundary vertex added? Let's say we add one vertex on one edge, splitting it into two. Then, the original triangle is now a quadrilateral on the boundary? No, wait, the boundary is still a triangle but with a subdivided edge.Wait, perhaps not. If we have a vertex added on an edge, the outer face must still be a triangle. But if we split an edge into two, the outer face would require that both subdivided edges are part of the boundary.But in planar terms, the outer face is a triangle with vertices A, B, C, but edge AB is split into two edges, AD and DB. Therefore, the outer face is bounded by AD, DB, BC, and CA. Wait, that's four edges, making the outer face a quadrilateral, which is not a triangle. This is contradictory.Therefore, my earlier assumption that B is the number of boundary vertices might be incorrect. Perhaps B is the number of boundary edges instead.Let me redefine.Let B be the number of boundary edges (edges on the original triangle), and V_b be the number of boundary vertices. In a polygon, the number of edges equals the number of vertices. So, if we have a polygon with V_b vertices, it has V_b edges.In that case, the formula T = B + 2I - 2 might need to be adjusted.Wait, let's rederive the equations with B as the number of boundary edges.Using Euler's formula: V - E + F = 2.F = T + 1.E = (3T + B)/2.Therefore, V - (3T + B)/2 + T + 1 = 2.Multiply through by 2: 2V - 3T - B + 2T + 2 = 4 => 2V - T - B = 2 => T = 2V - B - 2.But we also know that the number of boundary edges B is related to the number of boundary vertices V_b. Since each boundary edge is part of the boundary polygon, which has V_b vertices, then B = V_b.Additionally, the total number of vertices V is equal to the number of interior vertices I plus the number of boundary vertices V_b: V = I + V_b.Therefore, substituting V = I + V_b into T = 2V - B - 2:T = 2(I + V_b) - V_b - 2 = 2I + 2V_b - V_b - 2 = 2I + V_b - 2.Given T = 1000:1000 = 2I + V_b - 2 => 2I + V_b = 1002.We need to minimize V = I + V_b.Express V = I + V_b = I + (1002 - 2I) = 1002 - I.To minimize V, we need to maximize I. The maximum I occurs when V_b is minimized.V_b, the number of boundary edges (which equals the number of boundary vertices), must be at least 3 (a triangle). So, V_b ≥ 3.Therefore, 2I = 1002 - V_b.To maximize I, minimize V_b.Minimum V_b = 3:2I = 1002 - 3 = 999 => I = 499.5, which is not an integer.Next, V_b = 4:2I = 1002 - 4 = 998 => I = 499.Then, V = 499 + 4 = 503.Similarly, V_b = 5:2I = 1002 - 5 = 997 => I = 498.5, not integer.V_b = 6:2I = 1002 - 6 = 996 => I = 498. V = 6 + 498 = 504.Therefore, the minimal V occurs at V_b = 4, I = 499, V = 503.So, this is consistent with the previous result. Therefore, the minimal number of vertices is 503.But now, considering that V_b is the number of boundary edges (which equals the number of boundary vertices), when V_b = 4, the original triangle has been subdivided such that one edge is split into two edges, making the boundary a quadrilateral with four edges and four vertices. However, this contradicts the requirement that the outer face is a triangle. Therefore, there seems to be a disconnect between the mathematical model and the problem's geometric interpretation.The problem states that a triangle is divided into 1000 smaller triangles. If we allow subdivision of the original edges, the outer face becomes a polygon with more edges, but the problem still refers to it as a triangle. This suggests that perhaps the original triangle's edges are not subdivided, and all subdivisions are internal. However, as previously noted, this leads to a non-integer number of interior vertices.Therefore, there must be a different approach. Let's consider that the original edges can be subdivided, but the outer face is still considered a triangle with the original three edges, and the subdivisions are considered internal to the original triangle. In this case, the formula T = 2I + 1 would apply, but since we need T = 1000, I = 999/2 = 499.5, which is impossible. Therefore, the problem must allow adding vertices on the edges, changing the number of boundary edges, and hence the outer face's structure.Given that the problem doesn't specify that the outer face must remain a triangle, but only that the original triangle is divided into smaller triangles, it's permissible to have the original edges subdivided, with the outer face being a polygon. However, the problem statement might be implying that the division is entirely within the original triangle, and the outer boundary is still the original triangle. This is a common phrasing in such problems, where the original edges remain intact, and all subdivisions are inside. But in that case, adding vertices on the edges would subdivide them, but the outer boundary would still be considered the original triangle.This is a bit of a wording ambiguity. Assuming that vertices can be added on the edges, but the original triangle's edges are preserved as the boundary, then each time we add a vertex on an edge, we're splitting it into two edges, but both are still part of the boundary. Therefore, the outer face remains a triangle, but with subdivided edges.In planar graph terms, the outer face is a triangle with three edges, but each edge can be divided into multiple edges. However, in this case, the number of boundary edges B is 3, but each original edge is split into k parts. So, B = 3k, but this complicates the formula.Alternatively, perhaps the formula T = 2V - B - 2 applies, where B is the number of vertices on the boundary.But given the time I've spent on this and the consistent result of V = 503 when allowing subdivision of edges, I think the answer is 504 or 503. Wait, but in the first calculation using Euler's formula directly, we got V = 504. But in the later derivation, we got V = 503. There's a discrepancy here.Let me cross-verify.If V = 503, B = 4, I = 499.Then, from the formula E = (3T + B)/2 = (3000 + 4)/2 = 3004/2 = 1502.From Euler's formula: V - E + F = 503 - 1502 + 1001 = 2, which works.But in this case, B is the number of boundary edges, which is 4. But a triangle with 4 boundary edges is a quadrilateral, which is a contradiction. Therefore, this suggests that B in this case is not the number of boundary edges but the number of boundary vertices.If B is the number of boundary vertices (V_b), then the number of boundary edges is also V_b. So, with V_b = 4, we have a quadrilateral as the outer face, which contradicts the original triangle.Therefore, this indicates that my earlier assumption is wrong, and we cannot have B = 4 if the outer face must remain a triangle.Therefore, the problem must require that the outer face remains a triangle, meaning B = 3 (the original three vertices), and any vertices added on the edges are considered internal vertices. But this leads to the impossible I = 499.5.Therefore, the problem must allow the outer face to be a polygon with more edges, accepting that the original edges are subdivided. In that case, the minimal V is 503, even though the outer face is a quadrilateral. However, this contradicts the problem statement's wording of dividing a triangle into smaller triangles, which usually implies that the outer boundary remains a triangle.Given this confusion, perhaps the answer expected is based on a formula where the number of vertices V satisfies T = 2V - 4, leading to V = (T + 4)/2. For T = 1000, V = (1000 + 4)/2 = 502. But this formula is for a maximal planar graph where all faces are triangles, including the outer face. In our case, the outer face is a triangle, so maybe the formula is V = (T + 2)/2. For example, for T = 3, V = (3 + 2)/2 = 2.5, which is incorrect.Alternatively, in a triangulation with V vertices, the number of triangles is 2V - 4. So, 2V - 4 = 1000 => V = 502.But this formula assumes that the graph is triangulated, including the outer face. However, in our problem, the outer face is a triangle, so it's already part of the triangulation. Therefore, the number of inner triangles would be 2V - 4 - 1 = 2V - 5. Wait, not sure.Alternatively, in a triangulation of a polygon with V vertices, the number of triangles is V - 2. But that's for a convex polygon. For example, a triangle (3 vertices) can be triangulated into 1 triangle, which is V - 2 = 1. Wait, no, that's not correct. A triangle is already a triangle. If you have a convex polygon with n sides, the number of triangles in a triangulation is n - 2. So, a triangle (n=3) would give 1 triangle, which is correct. A quadrilateral (n=4) would give 2 triangles, etc.But in our case, the original triangle is divided into smaller triangles with added vertices inside and on edges. This is a different scenario. The formula for the number of triangles in terms of the number of vertices added is not straightforward.Let me look for a pattern.Case 1: Original triangle. V = 3, T = 1.Case 2: Add one interior point connected to all three vertices. V = 4, T = 3.Case 3: Add another interior point in one of the small triangles, connected to three vertices. V = 5, T = 5.Case 4: Continue this process. Each new interior point adds two triangles. So, the number of triangles follows T = 1 + 2(n-3), where n is the number of vertices. Wait, for n=3, T=1; n=4, T=3=1+2(1); n=5, T=5=1+2(2); n=6, T=7=1+2(3); so T=1+2(V-3)=2V-5.But for V=4, T=3=2*4-5=3, which works. For V=5, T=5=2*5-5=5, works. For V=6, T=7=2*6-5=7, works. So, general formula T=2V-5.But wait, according to this, T=2V-5. Therefore, solving for V when T=1000:2V -5 = 1000 => 2V=1005 => V=502.5. Not possible. Therefore, this suggests that the formula is not applicable here or requires adjustment.But this formula assumes that each new vertex is added inside a triangle, splitting it into three, which adds two new triangles each time. However, this only holds if we don't add vertices on the edges. If we add vertices on the edges, we can create more triangles with fewer vertices.For example, adding a vertex on an edge allows two triangles to share that edge, each using the new vertex. This can create more triangles without adding as many vertices.Therefore, to minimize the number of vertices, we should add as many vertices on edges as possible, since each edge vertex can contribute to more triangles.This is similar to creating a grid-like structure inside the triangle. For example, a triangular grid where each edge is divided into n segments, leading to (n+1) vertices per edge. The number of small triangles in such a grid is n². But for n=31, 31²=961, which is close to 1000. For n=32, 32²=1024. So, 32 segments per edge would give 1024 triangles, requiring 32+1=33 vertices per edge. The total number of vertices would be the number of vertices on the edges plus the interior vertices.But this approach might not give the minimal number of vertices, as it's a structured grid. Maybe a more efficient configuration exists.In a triangular grid with m points per edge, the total number of vertices is m(m+1)/2. Wait, no. For a triangular grid with k subdivisions per edge, the number of vertices is (k+1)(k+2)/2. For example, k=1: 6 vertices; k=2: 10 vertices. The number of small triangles would be k². So, for k=31, 31²=961 triangles, vertices= (32)(33)/2=528. For k=32, 32²=1024 triangles, vertices= (33)(34)/2=561. But we need 1000 triangles. Since 31²=961 and 32²=1024, we can't get exactly 1000 with this method. Also, the number of vertices is higher than the 503 we found earlier. Therefore, this method is not efficient.Therefore, a better approach is needed. Returning to the earlier formula where V=503, this seems more efficient. However, the confusion about the outer face remains.Given that the problem doesn't specify that the outer face must remain a triangle, and only requires that the original triangle is divided into 1000 smaller triangles, it's permissible to add vertices on the edges and have the outer face as a polygon. In that case, using the derived formula V=503 is correct.But to ensure correctness, let's cross-validate with another method.Suppose we have a triangulation with V vertices, E edges, T triangles.Euler's formula: V - E + T + 1 = 2 => V - E + T = 1.In a triangulation, each internal edge is shared by two triangles. The boundary edges are part of only one triangle. Each triangle has three edges, so total edge count is (3T + B)/2, where B is the number of boundary edges.Therefore, E = (3T + B)/2.Substitute into Euler's formula:V - (3T + B)/2 + T = 1 => 2V - 3T - B + 2T = 2 => 2V - T - B = 2 => T = 2V - B - 2.We need to minimize V, given T=1000.Thus, 1000 = 2V - B - 2 => 2V - B = 1002 => B = 2V - 1002.But B ≥ 3, so 2V - 1002 ≥ 3 => 2V ≥ 1005 => V ≥ 502.5. Since V must be integer, V ≥ 503.If V=503, then B=2*503 -1002=1006 -1002=4.So, with V=503, B=4. This matches the previous result.Therefore, the minimal number of vertices is 503.Therefore, the answer is boxed{503}.

✅:To determine the minimum number of distinct vertices required to divide a triangle into 1000 smaller triangles, we use Euler's formula and planar graph theory.1. Euler's Formula: For a planar graph, ( V - E + F = 2 ), where ( V ) is the number of vertices, ( E ) is the number of edges, and ( F ) is the number of faces (including the outer face).2. Faces: The original triangle is divided into 1000 smaller triangles, so there are ( F = 1000 + 1 = 1001 ) faces (including the outer face).3. Edges: Each triangle contributes 3 edges, but each interior edge is shared by two triangles. The boundary edges (original edges of the triangle) are counted once. The formula for edges is ( E = frac{3T + B}{2} ), where ( T ) is the number of triangles and ( B ) is the number of boundary edges. Here, ( T = 1000 ).4. Euler's Formula Application: Using Euler's formula ( V - E + F = 2 ): [ V - frac{3 cdot 1000 + B}{2} + 1001 = 2 ] Simplifying, we get: [ V - frac{3000 + B}{2} = -999 implies 2V - 3000 - B = -1998 implies 2V - B = 1002 ]5. Boundary Edges: The number of boundary edges ( B ) must be an integer and at least 3. To minimize ( V ), we need to find the smallest ( B ) that allows ( V ) to be an integer.6. Solving for ( V ): - Rearranging ( 2V - B = 1002 ), we get ( B = 2V - 1002 ). - The smallest ( B ) that makes ( V ) an integer is 4 (since ( B ) must be even to keep ( V ) integer). - Substituting ( B = 4 ), we get ( 2V - 4 = 1002 implies 2V = 1006 implies V = 503 ).7. Verification: - With ( V = 503 ), ( B = 4 ), and using the edge formula ( E = frac{3 cdot 1000 + 4}{2} = 1502 ). - Check Euler's formula: ( 503 - 1502 + 1001 = 2 ), which holds true.Thus, the minimum number of distinct vertices required is (boxed{503}).

🔑:1. Sum of Angles Consideration: - We start with the fact that the sum of the internal angles of a triangle is 180^circ. - Since the given triangle is divided into 1000 smaller triangles, the total sum of their internal angles is: 1000 times 180^circ = 180000^circ 2. Calculation of Excess Angle: - We need to find out how much of this total angle sum comes from intersections at internal points versus boundaries. - Since a triangle only has three vertices contributing 180^circ, we need to determine the extra angles in excess of these vertices.3. Internal Vertices Contribution: - Each internal vertex, where only angles of the triangles meet, contributes 360^circ to the total angle sum. - First, calculate the excess angle: 180000^circ - 180^circ = 179820^circ - Now, calculate how many such internal vertices are needed to account for this excess: frac{179820^circ}{360^circ} = 499.5 - Since vertices must be whole points, we round down to 499 complete internal vertices.4. Boundary Vertices Contribution: - Each boundary vertex in the subdivision will contribute 180^circ (just as in the original triangle). - For the minimal configuration, we also need to include boundary vertices. Considering the existing vertices of the initial triangle, we add these three: 499 text{ (internal points)} + 3 text{ (original vertices)} + 1 text{ (extra)} = 503 5. Conclusion: - By combining both groups of vertices, we conclude that: - 499 internal vertices - 3 original vertices (of the initial triangle) - 1 additional point contributing to boundary.Thus, the minimal number of distinct points where the vertices of these triangles can be located is: boxed{503}

🧠:Okay, so I need to find all functions ( f: mathbb{R}_{+}^{*} rightarrow mathbb{R}_{+}^{*} ) such that for every positive real number ( x ), there's exactly one positive real number ( y ) satisfying the inequality ( x f(y) + y f(x) leq 2 ). Hmm, that's interesting. Let me start by understanding the problem step by step.First, the domain and codomain are both positive real numbers. So, ( f ) takes positive real numbers and maps them to positive real numbers. The condition is that for each ( x ), there is exactly one ( y ) such that ( x f(y) + y f(x) leq 2 ). That "exactly one" part is crucial here. It means that for each ( x ), the equation ( x f(y) + y f(x) = 2 ) must have exactly one solution ( y ), right? Because if there were more than one ( y ) satisfying the inequality, but the problem states there's exactly one. Wait, actually, the problem says "there exists exactly one ( y ) such that ( x f(y) + y f(x) leq 2 )." So, does that mean that for each ( x ), the set of ( y ) satisfying the inequality is a singleton? That is, only one ( y ) makes the inequality true? But since it's an inequality, maybe there could be a range of ( y ) values that satisfy it, but the problem says only one does. That suggests that the inequality must actually be an equality for exactly one ( y ). Because if the inequality were strict for some ( y ), maybe there's a region around that ( y ) where it's also true? Hmm, but ( x f(y) + y f(x) ) is a function of ( y ), so maybe depending on ( f ), the set of ( y ) satisfying the inequality could be an interval or something. However, the problem states that for each ( x ), there is exactly one ( y ) that satisfies it. Therefore, likely, the inequality must hold as equality for exactly one ( y ), and for all other ( y ), the expression is greater than 2. That seems to make sense. So, the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ) for each ( x ). So, maybe the inequality is tight only at that point. Therefore, the problem reduces to finding functions ( f ) such that for each ( x ), the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ).So, perhaps this is a functional equation where the equation ( x f(y) + y f(x) = 2 ) has a unique solution ( y ) for each ( x ). Let me note that the equation can be rewritten as ( x f(y) = 2 - y f(x) ). Hmm, but both ( f(x) ) and ( f(y) ) are positive, so the right-hand side must also be positive. Therefore, ( 2 - y f(x) > 0 implies y f(x) < 2 implies y < frac{2}{f(x)} ). Similarly, ( x f(y) = 2 - y f(x) > 0 implies x f(y) > 0 ), which is always true since ( x ) and ( f(y) ) are positive. So, the constraint here is that ( y < frac{2}{f(x)} ).But how does this help? Maybe I need to consider the equation ( x f(y) + y f(x) = 2 ) and think about how to solve for ( y ) in terms of ( x ). If for each ( x ), there's exactly one ( y ), maybe this equation defines ( y ) as a function of ( x ). Let's denote this unique ( y ) by ( y(x) ). So, for each ( x ), we have ( x f(y(x)) + y(x) f(x) = 2 ). Since this ( y(x) ) is unique, maybe there's a functional relationship here that can help us determine ( f ).Alternatively, perhaps the function ( f ) has some symmetry or specific form. Let me consider possible simple functions and see if they satisfy the condition. For example, suppose ( f(x) = frac{c}{x} ) for some constant ( c > 0 ). Let's check if this works.Let ( f(x) = frac{c}{x} ). Then, substituting into the equation:( x f(y) + y f(x) = x cdot frac{c}{y} + y cdot frac{c}{x} = frac{c x}{y} + frac{c y}{x} ).Set this equal to 2: ( frac{c x}{y} + frac{c y}{x} = 2 ).Multiply both sides by ( xy ): ( c x^2 + c y^2 = 2 xy ).Which simplifies to ( c x^2 - 2 xy + c y^2 = 0 ).This is a quadratic equation in terms of ( y ): ( c y^2 - 2 x y + c x^2 = 0 ).The discriminant of this quadratic is ( ( -2x )^2 - 4 cdot c cdot c x^2 = 4 x^2 - 4 c^2 x^2 = 4 x^2 (1 - c^2) ).For the quadratic to have exactly one real solution, the discriminant must be zero. So,( 4 x^2 (1 - c^2) = 0 implies 1 - c^2 = 0 implies c^2 = 1 implies c = 1 ).Since ( c > 0 ), we have ( c = 1 ). Therefore, if ( f(x) = frac{1}{x} ), then the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ). Let's check that solution. With ( c = 1 ), the quadratic equation becomes ( y^2 - 2 x y + x^2 = 0 ), which factors as ( (y - x)^2 = 0 ), so the unique solution is ( y = x ). Therefore, for each ( x ), the unique ( y ) is ( x ) itself. So, substituting back, this works. Therefore, ( f(x) = frac{1}{x} ) is a solution.But the question is to find all such functions. Is this the only solution? Let's check another possible function. Suppose ( f(x) = k cdot frac{1}{x} ), where ( k ) is a positive constant. Then, similar to above, substituting into the equation:( x cdot frac{k}{y} + y cdot frac{k}{x} = frac{k x}{y} + frac{k y}{x} ).Setting equal to 2: ( k left( frac{x}{y} + frac{y}{x} right ) = 2 ).Multiply both sides by ( xy ): ( k (x^2 + y^2) = 2 xy ).Then, ( k x^2 - 2 xy + k y^2 = 0 ).Again, discriminant is ( 4 x^2 - 4 k^2 x^2 = 4 x^2 (1 - k^2) ). For discriminant to be zero, ( 1 - k^2 = 0 implies k = 1 ). So, only ( k = 1 ) works, leading to ( f(x) = 1/x ). Therefore, scaling by another constant doesn't work unless ( k = 1 ). So that suggests that ( f(x) = 1/x ) is the only function of the form ( k/x ) that works.But could there be other functions besides ( f(x) = 1/x ) that satisfy the condition? Let's consider if there are other possibilities.Suppose that ( f ) is a function such that for each ( x ), the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y = g(x) ). Then, for each ( x ), ( y = g(x) ) must satisfy ( x f(g(x)) + g(x) f(x) = 2 ).Additionally, since the problem states that for each ( x ), there is exactly one ( y ), and since the equation is symmetric in ( x ) and ( y ) (if we swap ( x ) and ( y ), the equation remains the same), perhaps the function ( g ) is the inverse function of itself or has some symmetry. For instance, if ( g(x) = x ), then substituting back, we get ( x f(x) + x f(x) = 2 implies 2 x f(x) = 2 implies x f(x) = 1 implies f(x) = 1/x ). So this gives the same solution as before. Therefore, if ( y = x ) is the unique solution for each ( x ), then ( f(x) = 1/x ).But maybe there exists a different function ( g ) such that ( y = g(x) ) is not equal to ( x ), but still satisfies the condition. Let's suppose that such a ( g ) exists. Then, we have the equation ( x f(g(x)) + g(x) f(x) = 2 ). If ( g ) is invertible, then perhaps we can relate ( f(g(x)) ) and ( f(x) ). However, this seems a bit abstract. Maybe we can consider the system of equations. Let me think.Alternatively, let's suppose that ( y ) is a function of ( x ), say ( y = g(x) ). Then, for each ( x ), ( x f(g(x)) + g(x) f(x) = 2 ). If ( g ) is invertible, then we can write ( x = g^{-1}(y) ). So, substituting back, we get ( g^{-1}(y) f(y) + y f(g^{-1}(y)) = 2 ). But this seems a bit convoluted.Alternatively, maybe if we consider that the equation ( x f(y) + y f(x) = 2 ) is symmetric in ( x ) and ( y ), so if ( y ) is the unique solution for a given ( x ), then ( x ) should be the unique solution for ( y ). Therefore, ( g ) should be an involution, i.e., ( g(g(x)) = x ). If that's the case, then we can write the equation for ( y = g(x) ), and then for ( x = g(y) ), we have ( y f(x) + x f(y) = 2 ), which is the same as the original equation. So, that's consistent. Therefore, ( g ) is an involution.If ( g ) is an involution, then ( g(g(x)) = x ). If we further suppose that ( g(x) = x ), then that's the trivial involution, leading to ( f(x) = 1/x ). But are there other involutions ( g ) that could work?Suppose we let ( g(x) = k/x ) for some constant ( k ). Then, ( g(g(x)) = g(k/x) = k / (k/x) = x ), so this is an involution. Let's see if such a ( g ) can satisfy the functional equation.Let ( y = g(x) = k/x ). Then, substituting into the equation:( x f(k/x) + (k/x) f(x) = 2 ).So, we have ( x f(k/x) + (k/x) f(x) = 2 ).If we can find a function ( f ) that satisfies this equation for all ( x ), then that might be a solution. Let's try to assume a form for ( f ). For example, suppose ( f(x) = c / x ), as before. Let's see if this works with ( g(x) = k/x ).Substituting ( f(x) = c / x ) into the equation:( x cdot frac{c}{k/x} + frac{k}{x} cdot frac{c}{x} = x cdot frac{c x}{k} + frac{k c}{x^2} = frac{c x^2}{k} + frac{k c}{x^2} ).Set this equal to 2:( frac{c x^2}{k} + frac{k c}{x^2} = 2 ).Multiply both sides by ( k x^2 ):( c x^4 + k^2 c = 2 k x^2 ).Rearranged:( c x^4 - 2 k x^2 + k^2 c = 0 ).This is a quadratic in ( x^2 ):( c (x^2)^2 - 2 k x^2 + k^2 c = 0 ).The discriminant is ( ( -2 k )^2 - 4 cdot c cdot k^2 c = 4 k^2 - 4 c^2 k^2 = 4 k^2 (1 - c^2) ).For this quadratic to have exactly one solution for ( x^2 ), the discriminant must be zero, so:( 4 k^2 (1 - c^2) = 0 implies 1 - c^2 = 0 implies c = 1 ).Then, substituting ( c = 1 ), the equation becomes:( x^4 - 2 k x^2 + k^2 = 0 ).Which factors as:( (x^2 - k)^2 = 0 implies x^2 = k implies x = sqrt{k} ).But this must hold for all ( x in mathbb{R}_{+}^{*} ), which is only possible if ( k ) is such that all ( x ) satisfy ( x = sqrt{k} ), which is impossible unless ( k ) is variable, which it's not. Therefore, this approach doesn't work unless ( k ) is fixed, but we need the equation to hold for all ( x ). Hence, assuming ( f(x) = c / x ) and ( g(x) = k / x ) doesn't lead us to a valid solution unless ( k = x^2 ), which is not a constant. Therefore, this suggests that the only possible solution with ( f(x) = c / x ) is when ( g(x) = x ), leading to ( c = 1 ), as before.Therefore, maybe the only solution is ( f(x) = 1/x ). But we need to confirm whether there are other functions beyond reciprocal functions that satisfy the condition.Let me consider another approach. Let's assume that the function ( f ) is continuous. Maybe that can help. Although the problem doesn't specify continuity, sometimes assuming continuity can help find possible solutions, and then we can check if they work without the assumption.Alternatively, let's analyze the equation ( x f(y) + y f(x) = 2 ). For each ( x ), this is an equation in ( y ). Let's consider solving for ( y ):( x f(y) + y f(x) = 2 ).Let me treat this as an equation in ( y ). Let's rearrange it:( x f(y) = 2 - y f(x) implies f(y) = frac{2 - y f(x)}{x} ).But ( f(y) ) must be positive, so ( 2 - y f(x) > 0 implies y < frac{2}{f(x)} ). Therefore, for each ( x ), the possible ( y ) must be less than ( 2 / f(x) ).However, since ( f(y) ) is positive, the right-hand side ( frac{2 - y f(x)}{x} ) must also be positive. Which gives ( 2 - y f(x) > 0 implies y < frac{2}{f(x)} ), as above.But how does this help? Maybe I can think of this as a functional equation where ( f(y) ) is expressed in terms of ( y ), ( x ), and ( f(x) ). But this seems complicated because ( x ) and ( y ) are related through the equation. Let's consider specific cases. For example, let’s suppose that for each ( x ), the unique ( y ) is related to ( x ) in some way. Maybe ( y ) is proportional to ( x ), such as ( y = k x ). Let's try that.Suppose that for each ( x ), the unique ( y ) is ( y = k x ), where ( k ) is a positive constant. Then, substituting into the equation:( x f(k x) + k x f(x) = 2 ).Divide both sides by ( x ):( f(k x) + k f(x) = frac{2}{x} ).This is a functional equation for ( f ). Let's see if we can solve this. Let's denote ( x ) by ( t ), so:( f(k t) + k f(t) = frac{2}{t} ).Suppose we let ( k = 1 ). Then the equation becomes ( f(t) + f(t) = 2 / t implies 2 f(t) = 2 / t implies f(t) = 1 / t ). Which is the solution we found earlier. So that works. But if ( k neq 1 ), can we find another solution?Let’s suppose ( k neq 1 ). Let’s make a substitution. Let ( g(t) = f(t) ). Then the equation is ( g(k t) + k g(t) = frac{2}{t} ).This looks like a functional equation that might be solvable by iterative substitution or assuming a particular form. Let's assume ( g(t) = C / t ), similar to before. Let’s test this:( g(k t) = C / (k t) ), so substituting into the equation:( C / (k t) + k (C / t) = frac{2}{t} ).Multiply both sides by ( t ):( C / k + k C = 2 ).Factor out ( C ):( C (1/k + k) = 2 implies C = 2 / (k + 1/k) = 2k / (k^2 + 1) ).Therefore, if we set ( C = 2k / (k^2 + 1) ), then ( g(t) = C / t ) satisfies the equation. But in this case, we have ( f(t) = C / t ), where ( C = 2k / (k^2 + 1) ).However, recall that we assumed that the unique solution ( y ) is ( k x ). But for the original problem, for each ( x ), there exists exactly one ( y ), which might depend on ( x ). But if we fix ( k ), then ( y = k x ) for all ( x ). However, this might not satisfy the uniqueness condition unless the equation ( x f(y) + y f(x) = 2 ) indeed has only ( y = k x ) as a solution.Wait, but if we set ( f(t) = C / t ), then the equation ( x f(y) + y f(x) = 2 ) becomes:( x cdot (C / y) + y cdot (C / x) = 2 implies (C x)/y + (C y)/x = 2 ).Which is similar to the previous case. If we set ( y = k x ), then:( (C x)/(k x) + (C k x)/x = 2 implies C / k + C k = 2 ).Which is the same equation as before, giving ( C = 2k / (k^2 + 1) ). But then, for the equation ( (C x)/y + (C y)/x = 2 ), the solutions for ( y ) would generally be multiple unless the discriminant is zero. Wait, let's check that.Let me write the equation ( (C x)/y + (C y)/x = 2 ) as:Multiply both sides by ( xy ):( C x^2 + C y^2 = 2 xy ).Rearranged:( C x^2 - 2 xy + C y^2 = 0 ).Which is a quadratic in ( y ):( C y^2 - 2 x y + C x^2 = 0 ).The discriminant is ( ( -2x )^2 - 4 cdot C cdot C x^2 = 4 x^2 - 4 C^2 x^2 = 4 x^2 (1 - C^2) ).For there to be exactly one solution ( y ), the discriminant must be zero:( 4 x^2 (1 - C^2) = 0 implies 1 - C^2 = 0 implies C = 1 ).But if ( C = 1 ), then regardless of ( k ), we have ( C = 1 ). Wait, but earlier, if ( C = 2k / (k^2 + 1) ), setting ( C = 1 ), we get:( 2k / (k^2 + 1) = 1 implies 2k = k^2 + 1 implies k^2 - 2k + 1 = 0 implies (k - 1)^2 = 0 implies k = 1 ).Therefore, only when ( k = 1 ), ( C = 1 ), which gives us the previous solution ( f(x) = 1 / x ). So even if we tried to assume ( y = k x ), the only possibility that works is ( k = 1 ), leading to ( f(x) = 1 / x ).Therefore, this suggests that even if we try to impose a linear relationship between ( y ) and ( x ), the only solution that gives a unique ( y ) is when ( y = x ) and ( f(x) = 1/x ).Another approach: suppose that for each ( x ), the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ). Let's consider the function ( h_x(y) = x f(y) + y f(x) ). The equation ( h_x(y) = 2 ) must have exactly one positive real solution. Let's analyze the behavior of ( h_x(y) ).Since ( f(y) ) is positive for all ( y > 0 ), ( h_x(y) ) is the sum of two positive terms. Let's consider the behavior as ( y to 0^+ ) and ( y to +infty ).As ( y to 0^+ ), ( x f(y) ) tends to ( x cdot f(0^+) ), but ( y ) is approaching 0, so ( f(y) ) might approach some limit. However, ( f: mathbb{R}_{+}^{*} to mathbb{R}_{+}^{*} ), so ( f(y) ) is defined for ( y > 0 ), but we don't know the behavior as ( y to 0^+ ). Similarly, as ( y to +infty ), ( y f(x) ) tends to infinity since ( f(x) > 0 ), and ( x f(y) ) depends on the behavior of ( f(y) ). If ( f(y) ) is bounded below by some positive constant, then ( x f(y) to +infty ) as ( y to +infty ). However, if ( f(y) ) approaches zero as ( y to +infty ), then ( x f(y) ) might approach zero, and ( y f(x) ) would still approach infinity. Therefore, depending on ( f ), ( h_x(y) ) might tend to infinity as ( y to +infty ).But for the equation ( h_x(y) = 2 ) to have exactly one solution, the function ( h_x(y) ) must cross the value 2 exactly once. So, the function ( h_x(y) ) must be decreasing from some upper value to 2 and then increasing back to infinity, or increasing from some lower value to 2 and then decreasing. Wait, but given the behavior as ( y to 0^+ ) and ( y to +infty ), let's think.Suppose that as ( y to 0^+ ), ( h_x(y) to x f(0^+) + 0 cdot f(x) = x f(0^+) ). If ( f(0^+) ) is finite, then ( h_x(y) ) approaches ( x f(0^+) ). If ( f(0^+) = +infty ), then ( h_x(y) to +infty ). Similarly, as ( y to +infty ), ( h_x(y) to x f(+infty) + infty cdot f(x) ). If ( f(+infty) ) is finite, then ( h_x(y) to +infty ). If ( f(+infty) = 0 ), then ( h_x(y) to 0 + infty cdot f(x) = +infty ). So regardless of the behavior of ( f(y) ), unless ( f(y) ) tends to negative infinity (which it can't because it's positive), ( h_x(y) ) tends to ( +infty ) as ( y to +infty ).Therefore, ( h_x(y) to +infty ) as ( y to +infty ). As ( y to 0^+ ), depending on ( f ), it can approach a finite limit or infinity. For the equation ( h_x(y) = 2 ) to have exactly one solution, the function ( h_x(y) ) must attain the value 2 exactly once. If ( h_x(y) ) is decreasing from ( +infty ) to some minimum and then increasing back to ( +infty ), then it might cross the value 2 twice. But we need exactly one crossing, so the function ( h_x(y) ) must be strictly monotonic or have a minimum or maximum exactly at 2.Wait, for ( h_x(y) ) to cross 2 exactly once, it should be either strictly decreasing from ( +infty ) to some value below 2 and then increasing back to ( +infty ), but that would cross 2 twice. Alternatively, if ( h_x(y) ) is strictly decreasing from ( +infty ) to a minimum value at some ( y ), and then increasing back to ( +infty ). If the minimum value is exactly 2, then it touches 2 at one point. But the problem states that for every ( x ), there exists exactly one ( y ). So, for each ( x ), ( h_x(y) ) must have a unique minimum at some ( y ), and that minimum is exactly 2. Therefore, the function ( h_x(y) ) must have its minimum value 2, achieved at exactly one ( y ).Therefore, the problem reduces to finding functions ( f ) such that for each ( x ), the function ( h_x(y) = x f(y) + y f(x) ) has its minimum value 2, achieved at exactly one ( y ).So, if we can show that ( h_x(y) ) has a unique minimum at ( y = g(x) ), and that this minimum is 2, then this would satisfy the condition. Let's try this approach.To find the minimum of ( h_x(y) ), take the derivative with respect to ( y ):( h_x'(y) = x f'(y) + f(x) ).Set this equal to zero for critical points:( x f'(y) + f(x) = 0 implies x f'(y) = -f(x) implies f'(y) = - frac{f(x)}{x} ).But this seems problematic because the left-hand side ( f'(y) ) is a function of ( y ), while the right-hand side is a function of ( x ). For this equation to hold for all ( x ) and ( y ), the only possibility is that both sides are constants. Wait, but if ( f'(y) = - frac{f(x)}{x} ), unless there's a relationship between ( x ) and ( y ).Wait, actually, when we take the derivative, we're treating ( x ) as a constant and ( y ) as a variable. So, for each fixed ( x ), the critical point ( y ) must satisfy ( f'(y) = - frac{f(x)}{x} ).But this creates a relationship between ( x ) and ( y ). Let's denote ( y = g(x) ), the critical point. Then, we have:( f'(g(x)) = - frac{f(x)}{x} ).Additionally, at this critical point ( y = g(x) ), the function ( h_x(y) ) attains its minimum value 2. Therefore:( h_x(g(x)) = x f(g(x)) + g(x) f(x) = 2 ).So, we have two equations:1. ( f'(g(x)) = - frac{f(x)}{x} )2. ( x f(g(x)) + g(x) f(x) = 2 )This system of equations must hold for all ( x > 0 ).This seems complicated, but maybe we can assume that ( g(x) = x ), which worked earlier. Let's check:If ( g(x) = x ), then equation 1 becomes:( f'(x) = - frac{f(x)}{x} ).This is a differential equation: ( f'(x) = - frac{f(x)}{x} ).Solving this:( frac{df}{dx} = - frac{f}{x} implies frac{df}{f} = - frac{dx}{x} implies ln |f| = - ln |x| + C implies f = frac{K}{x} ).Where ( K = e^C ) is a positive constant. Since ( f: mathbb{R}_{+}^{*} to mathbb{R}_{+}^{*} ), ( K > 0 ).Then, equation 2 becomes:( x f(x) + x f(x) = 2 x f(x) = 2 implies x f(x) = 1 implies f(x) = frac{1}{x} ).Which gives ( K = 1 ). Therefore, this recovers the solution ( f(x) = 1/x ).Therefore, if we assume that the critical point ( y = g(x) = x ), then the only solution is ( f(x) = 1/x ). But what if ( g(x) neq x )? Is there another function ( g(x) ) such that the system of equations 1 and 2 is satisfied?To explore this, let's suppose that there exists a function ( g ) such that ( y = g(x) ) is not equal to ( x ). Then, we have the two equations:1. ( f'(g(x)) = - frac{f(x)}{x} )2. ( x f(g(x)) + g(x) f(x) = 2 )This is a system of functional equations. Let's see if we can relate these equations. Perhaps differentiate equation 2 with respect to ( x ):Differentiating equation 2:( frac{d}{dx} [x f(g(x)) + g(x) f(x)] = frac{d}{dx} 2 implies 0 ).Compute the derivative:Left-hand side:( f(g(x)) + x f'(g(x)) cdot g'(x) + g'(x) f(x) + g(x) f'(x) ).Set equal to zero:( f(g(x)) + x f'(g(x)) g'(x) + g'(x) f(x) + g(x) f'(x) = 0 ).But from equation 1, ( f'(g(x)) = - f(x)/x ). Substitute this into the equation:( f(g(x)) + x (- f(x)/x ) g'(x) + g'(x) f(x) + g(x) f'(x) = 0 ).Simplify:( f(g(x)) - f(x) g'(x) + g'(x) f(x) + g(x) f'(x) = 0 ).Notice that the terms ( - f(x) g'(x) ) and ( + g'(x) f(x) ) cancel each other. Therefore, we are left with:( f(g(x)) + g(x) f'(x) = 0 implies f(g(x)) = - g(x) f'(x) ).But ( f(g(x)) > 0 ) and ( g(x) > 0 ), so the right-hand side must also be positive. Therefore, ( - f'(x) > 0 implies f'(x) < 0 ). So, the function ( f ) is strictly decreasing.So, we now have:( f(g(x)) = - g(x) f'(x) ).But from equation 1:( f'(g(x)) = - frac{f(x)}{x} ).Therefore, we have two equations:1. ( f(g(x)) = - g(x) f'(x) )2. ( f'(g(x)) = - frac{f(x)}{x} )These look like a system of differential-difference equations. They relate the function ( f ) at ( x ) and ( g(x) ), and its derivatives. Solving such equations can be quite challenging, but perhaps assuming a particular form for ( g(x) ) could help.Alternatively, let's suppose that ( g(x) = k x ), as before. Let’s see if this leads us somewhere.Assume ( g(x) = k x ). Then, equation 1:( f(k x) = - k x f'(x) ).Equation 2:( f'(k x) = - frac{f(x)}{x} ).So, we have:1. ( f(k x) = - k x f'(x) )2. ( f'(k x) = - frac{f(x)}{x} )Let’s try to express these equations in terms of a single variable. Let’s denote ( t = k x ). Then, ( x = t / k ).Equation 1 becomes:( f(t) = - k (t / k) f'(t / k) implies f(t) = - t f'(t / k) ).Equation 2 becomes:( f'(t) = - frac{f(t / k)}{ t / k } implies f'(t) = - frac{k}{t} f(t / k) ).So, we have:1. ( f(t) = - t f'(t / k) )2. ( f'(t) = - frac{k}{t} f(t / k) )This seems a bit symmetric. Let’s substitute equation 1 into equation 2.From equation 1: ( f(t / k) = - (t / k) f'(t / k^2) ).Substitute into equation 2:( f'(t) = - frac{k}{t} [ - (t / k) f'(t / k^2) ] = frac{k}{t} cdot frac{t}{k} f'(t / k^2) = f'(t / k^2) ).Therefore, ( f'(t) = f'(t / k^2) ).This suggests that ( f' ) is a periodic function with period ( ln(k^2) ) in the logarithmic scale. However, given that ( f ) is defined for all positive real numbers and must be strictly decreasing (since ( f'(x) < 0 )), the only periodic function that is strictly decreasing is a constant function. But a constant function can't be periodic and strictly decreasing unless it's a contradiction. Therefore, the only possibility is that ( f' ) is constant, which would imply ( f ) is linear, but ( f ) maps ( mathbb{R}_{+}^{*} ) to ( mathbb{R}_{+}^{*} ), so a linear function would eventually become negative, which is not allowed. Hence, the only way this holds is if ( k^2 = 1 implies k = 1 ), since ( k > 0 ).If ( k = 1 ), then ( g(x) = x ), which brings us back to the previous case where ( f(x) = 1/x ).Therefore, this suggests that the only solution when assuming ( g(x) = k x ) is ( k = 1 ), leading to ( f(x) = 1/x ).Alternatively, maybe there is a non-linear function ( g(x) ) that satisfies the system. Let's explore another approach.Suppose that the function ( f ) is reciprocal, i.e., ( f(x) = c / x ). Wait, we already did this earlier and found that ( c = 1 ). So, no new solutions there.Alternatively, suppose ( f(x) = c x^n ), a power function. Let's see if this could work.Let ( f(x) = c x^n ), where ( c > 0 ) and ( n ) is a real number. Substitute into the equation ( x f(y) + y f(x) = 2 ):( x (c y^n) + y (c x^n) = 2 implies c x y^n + c y x^n = 2 ).Factor out ( c x y^n ):( c x y^n (1 + (y / x)^{n - 1}) = 2 ).This seems messy. Let's see if we can find ( n ) such that this equation has a unique solution ( y ) for each ( x ).Alternatively, suppose ( n = -1 ). Then ( f(x) = c / x ), which we already considered, leading to ( c = 1 ).Suppose ( n = 1 ). Then ( f(x) = c x ). Substitute into the equation:( x (c y) + y (c x) = 2 implies 2 c x y = 2 implies c x y = 1 implies y = 1 / (c x) ).So, for each ( x ), ( y = 1 / (c x) ). But this must hold for all ( x ). However, in this case, the equation ( x f(y) + y f(x) = 2 ) reduces to ( 2 c x y = 2 implies x y = 1/c ). So, for each ( x ), there's exactly one ( y = 1/(c x) ). Therefore, this satisfies the condition. Wait, but this seems to work for any ( c > 0 ). Wait, is this correct?Wait, if ( f(x) = c x ), then for each ( x ), the equation becomes ( 2 c x y = 2 implies x y = 1/c implies y = 1/(c x) ). So, for each ( x ), there's exactly one ( y = 1/(c x) ). Therefore, this function ( f(x) = c x ) satisfies the condition. But wait, earlier when we considered ( f(x) = c x ), we found that the equation ( x f(y) + y f(x) = 2 ) becomes linear in ( y ), and thus has exactly one solution. But does this contradict our previous analysis where assuming a quadratic equation? Wait, no. In this case, if ( f(x) = c x ), then substituting into the equation gives a linear equation in ( y ), so it has exactly one solution, which is ( y = 1/(c x) ). Therefore, this would seem to be another solution.But hold on, the problem states that ( f: mathbb{R}_{+}^{*} to mathbb{R}_{+}^{*} ). If ( f(x) = c x ), then ( f(x) ) is positive for ( x > 0 ), as required. And for each ( x ), ( y = 1/(c x) ) is positive. So this seems to satisfy the conditions. However, earlier when I considered ( f(x) = c / x ), I found that only ( c = 1 ) works, but that was under the assumption that the equation was quadratic. But if ( f(x) = c x ), the equation becomes linear, so the uniqueness is automatic.Wait, but here's a problem. If ( f(x) = c x ), then substituting into the original inequality ( x f(y) + y f(x) leq 2 ), we get ( x (c y) + y (c x) = 2 c x y leq 2 implies c x y leq 1 ). Therefore, the inequality holds for all ( y leq 1/(c x) ). But the problem states that there exists exactly one ( y ) such that the inequality holds. However, in this case, all ( y leq 1/(c x) ) satisfy the inequality. Since ( y ) must be positive, the set of solutions is the interval ( (0, 1/(c x)] ). But the problem requires that there is exactly one ( y ) in ( mathbb{R}_{+}^{*} ) satisfying the inequality. Therefore, this is a contradiction unless the interval ( (0, 1/(c x)] ) is a singleton, which would require ( 1/(c x) = 0 ), but that's impossible. Therefore, this suggests that our previous conclusion was wrong.Wait, this is a critical mistake. Earlier, I thought that if ( f(x) = c x ), then the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ). However, the problem states that there must be exactly one ( y ) such that the inequality ( x f(y) + y f(x) leq 2 ) holds. If ( f(x) = c x ), then ( x f(y) + y f(x) = 2 c x y leq 2 implies x y leq 1/c ). So, all ( y ) such that ( y leq 1/(c x) ) satisfy the inequality. Therefore, there are infinitely many ( y ) satisfying the inequality, which contradicts the problem's requirement of exactly one ( y ).Therefore, my previous reasoning was incorrect. I confused the equation ( x f(y) + y f(x) = 2 ) with the inequality. The problem requires that there's exactly one ( y ) making the inequality true. If the inequality is ( x f(y) + y f(x) leq 2 ), then even if the equation has one solution, the inequality might have a range of solutions. However, the problem states that for each ( x ), there exists exactly one ( y ) such that the inequality holds. Therefore, the inequality must hold only at that specific ( y ), meaning that ( x f(y) + y f(x) leq 2 ) only when ( y ) is that unique value. Therefore, actually, the inequality must be tight only at that point, and for all other ( y ), ( x f(y) + y f(x) > 2 ). Therefore, the equation ( x f(y) + y f(x) = 2 ) must have exactly one solution, and for all other ( y ), the expression is greater than 2. Therefore, the function ( x f(y) + y f(x) ) must attain its minimum value of 2 at exactly one point ( y ) for each ( x ), and be greater than 2 elsewhere.Therefore, when I considered ( f(x) = c x ), the expression ( x f(y) + y f(x) = 2 c x y ), which is minimized when ( y ) is as small as possible, but since ( y ) is in ( mathbb{R}_{+}^{*} ), the infimum is 0, but the expression can take any positive value. However, the equation ( 2 c x y = 2 implies y = 1/(c x) ). However, the inequality ( 2 c x y leq 2 implies y leq 1/(c x) ), so there are infinitely many solutions, which violates the problem's condition. Therefore, ( f(x) = c x ) is not a valid solution.This suggests that my earlier mistake was not considering the difference between the equation and the inequality. So, returning, the problem requires that the inequality ( x f(y) + y f(x) leq 2 ) is satisfied by exactly one ( y ) for each ( x ). Therefore, this is equivalent to saying that the equation ( x f(y) + y f(x) = 2 ) has exactly one solution ( y ), and for all other ( y ), ( x f(y) + y f(x) > 2 ). Therefore, the expression ( x f(y) + y f(x) ) must achieve its minimum value of 2 at exactly one point ( y ), and be greater than 2 otherwise. Therefore, the function ( h_x(y) = x f(y) + y f(x) ) must have a unique minimum at some ( y ), and that minimum must be exactly 2.Thus, we can approach this problem by considering that ( h_x(y) ) must have its minimum value 2, attained at exactly one ( y ). As before, taking the derivative and setting it to zero gives the critical point. This leads us back to the earlier differential equations. Since this path led us to ( f(x) = 1/x ) when assuming ( y = x ), and other attempts to find different solutions led to contradictions or infinite solutions, it's plausible that ( f(x) = 1/x ) is the only solution.To confirm this, let's verify that ( f(x) = 1/x ) indeed satisfies the original condition. For any ( x > 0 ), substitute ( f(x) = 1/x ):The expression becomes ( x cdot f(y) + y cdot f(x) = x cdot (1/y) + y cdot (1/x) = x/y + y/x ).We need to check that there's exactly one ( y > 0 ) such that ( x/y + y/x leq 2 ).But ( x/y + y/x geq 2 ) by AM ≥ GM, since ( x/y ) and ( y/x ) are positive. Equality holds if and only if ( x/y = y/x implies x^2 = y^2 implies y = x ), since ( x, y > 0 ).Therefore, ( x/y + y/x leq 2 ) if and only if ( y = x ), and in that case, equality holds. Therefore, for each ( x ), there is exactly one ( y = x ) satisfying the inequality (and it's actually an equality). Thus, ( f(x) = 1/x ) satisfies the condition.Now, to check if there are other solutions, suppose there exists another function ( f neq 1/x ) satisfying the condition. For such a function, there must exist some ( x ) where ( f(x) neq 1/x ). However, given the analysis above, assuming different forms for ( f ) either leads to contradictions or doesn't satisfy the uniqueness condition. The key step was recognizing that for the inequality to have exactly one solution, the expression must achieve its minimum value of 2 at that ( y ), and the function ( f(x) = 1/x ) is the only one that makes this minimum achievable uniquely at ( y = x ) due to the AM-GM inequality.Moreover, if we consider the earlier differential equation approach, the only solution that emerges is ( f(x) = 1/x ). Therefore, combining these observations, it's highly likely that ( f(x) = 1/x ) is the unique solution.Final AnswerThe only such function is (boxed{dfrac{1}{x}}).

✅:To find all functions ( f: mathbb{R}_{+}^{*} longrightarrow mathbb{R}_{+}^{*} ) such that for every ( x in mathbb{R}_{+}^{*} ), there exists exactly one ( y in mathbb{R}_{+}^{*} ) satisfying ( x f(y) + y f(x) leq 2 ), we analyze the problem step by step.1. Understanding the Condition: - The inequality ( x f(y) + y f(x) leq 2 ) must have exactly one solution ( y ) for each ( x ). - This implies the equation ( x f(y) + y f(x) = 2 ) must have exactly one solution ( y ) for each ( x ), and the inequality must be strict for all other ( y ).2. Testing Possible Functions: - Consider ( f(x) = frac{c}{x} ). Substituting into the equation gives ( frac{c x}{y} + frac{c y}{x} = 2 ). This simplifies to a quadratic equation in ( y ), which must have a unique solution. Solving for ( c ) shows ( c = 1 ), leading to ( f(x) = frac{1}{x} ).3. Verification with AM-GM Inequality: - For ( f(x) = frac{1}{x} ), the equation becomes ( frac{x}{y} + frac{y}{x} = 2 ). By the AM-GM inequality, ( frac{x}{y} + frac{y}{x} geq 2 ), with equality if and only if ( x = y ). Thus, ( y = x ) is the unique solution.4. Uniqueness and Differential Equations: - Considering the function ( h_x(y) = x f(y) + y f(x) ), we find that it must have a unique minimum at 2. Solving the resulting differential equations and analyzing critical points confirms that ( f(x) = frac{1}{x} ) is the only solution.5. Contradiction with Other Forms: - Testing other forms like ( f(x) = cx ) or different power functions leads to contradictions or infinite solutions, violating the uniqueness condition.Thus, the only function satisfying the given condition is ( f(x) = frac{1}{x} ).[boxed{dfrac{1}{x}}]

🔑:```markdownStep 1: Initial function analysis and constraintsGiven the function ( f: mathbb{R}_{+}^{*} rightarrow mathbb{R}_{+}^{*} ) with the property that for all ( x in mathbb{R}_{+}^{*} ), there exists exactly one ( y in mathbb{R}_{+}^{*} ) such that:[x f(y) + y f(x) leq 2]To find such ( f ), consider the form of the inequality. Let us investigate the constraints imposed by the inequality.Step 2: Assume a suitable form for ( y ) in terms of ( x )Notice that ( f(x) in mathbb{R}_{+}^{*} ), meaning it is a positive function. We examine simple forms to see what constraints they impose. Since the inequality must hold for a unique ( y ), suppose ( y = f^{-1}(k/x) ) to simplify the inequality.Step 3: Substituting and simplifyingLet's substitute ( y = 1 ):[x f(1) + f(x) leq 2]We see this must hold for all ( x in mathbb{R}_{+}^{*} ).Step 4: Exploring property of the solutionAssume ( f(y) ) takes specific simple forms like constant functions or linear multiples inversely related to ( x ). Consider an interesting edge case ( f(x) = frac{c}{x} ) for a constant ( c ):If ( f(x) = frac{k}{x} ), substitute back into the main inequality,[x f(y) + y f(x) leq 2 implies x frac{k}{y} + y frac{k}{x} leq 2 implies kleft(frac{x}{y} + frac{y}{x}right) leq 2]Use the AM-GM (Arithmetic Mean - Geometric Mean) inequality:[frac{x}{y} + frac{y}{x} geq 2,]which holds with equality if and only if ( x = y ). Now we set [k(2) leq 2 implies k = 1]Thus, a candidate solution is ( f(x) = frac{1}{x} ).Step 5: Verifying the solutionChecking if ( f(x) = frac{1}{x} ) satisfies the original property for all ( x, y in mathbb{R}_{+}^{*} ):Given ( f(x) = frac{1}{x} ),[x f(y) + y f(x) = x frac{1}{y} + y frac{1}{x} = frac{x}{y} + frac{y}{x}]Again by AM-GM inequality:[frac{x}{y} + frac{y}{x} geq 2 quad text{(with equality if ( x = y ))}]Since the equality in AM-GM holds exactly when ( x = y ):Our verification shows the function ( f(x) = frac{1}{x} ) fits entirely the given condition over ( mathbb{R}_{+}^{*} ). Therefore, this is indeed our solution.Conclusion:The only function ( f: mathbb{R}_{+}^{*} to mathbb{R}_{+}^{*} ) that satisfies the given condition is:[boxed{f(x) = frac{1}{x}}]```