🧠:Okay, so I need to figure out how many three-digit numbers n = abc exist such that the digits a, b, c can form the sides of an isosceles triangle, which includes equilateral triangles. The options are 45, 81, 165, 216. Hmm, let's break this down step by step.First, let's recall that a three-digit number has digits a, b, c where a is from 1 to 9 (since it can't be zero), and b and c are from 0 to 9. But wait, the problem states that the digits a, b, c must form the sides of an isosceles triangle. However, in a triangle, each side must be a positive integer, so actually, all three digits a, b, c must be at least 1. Because if any digit is zero, then one side of the triangle would be zero, which isn't allowed. Wait, hold on, the digits can be zero? But in a three-digit number, the first digit can't be zero, but the other two can. But if the digits form the sides of a triangle, then all three sides must be positive integers. So that means even if the three-digit number has a zero in the tens or units place, those digits would be zero, which can't be the length of a side. Therefore, actually, all digits a, b, c must be at least 1. Therefore, the three-digit number abc has a from 1 to 9, and b and c from 1 to 9 as well. So, all three digits must be between 1 and 9 inclusive. That's an important point. So the three-digit numbers we're considering are from 111 to 999, but only those where all digits are at least 1.Now, the problem is to count how many such numbers have digits that can form an isosceles triangle. An isosceles triangle has at least two sides equal. So, the three digits must satisfy the triangle inequality (sum of any two sides must be greater than the third) and also have at least two sides equal.Wait, but when they say "can form the sides of an isosceles triangle," does that mean that the three digits can be arranged in some order to form such a triangle? For example, if the digits are 2, 2, 3, that's a valid isosceles triangle. Similarly, 3, 3, 3 is equilateral and also counts. But even if the digits are in a different order, like 2, 3, 2, that's still the same triangle. So, the key is that regardless of the order of the digits, as long as they can form an isosceles triangle, the number counts. So the number abc is considered valid if any permutation of its digits forms a valid isosceles triangle. So the digits don't have to be in any specific order; the important thing is the multiset {a, b, c} can form an isosceles triangle.Therefore, first, we need to consider all three-digit numbers with digits a, b, c from 1 to 9, and count how many such numbers have digits that can form an isosceles triangle when arranged appropriately. So we can approach this problem by first considering all possible triplets (a, b, c) where a, b, c ∈ {1, 2, ..., 9}, and then determining how many such triplets satisfy the triangle inequality and have at least two equal numbers. Then, since each three-digit number corresponds to a unique ordered triplet (a, b, c) with a ≠ 0, but here we already have a, b, c ≥ 1, so each number from 111 to 999 with digits 1-9. However, some triplets correspond to the same multiset. But since the problem is asking for the number of three-digit numbers, not the number of distinct triangles, we need to count all ordered triplets (a, b, c) where a, b, c are digits from 1 to 9, forming a number abc, such that the multiset {a, b, c} can form an isosceles triangle.So, first, let's formalize the conditions:1. The three digits a, b, c must satisfy the triangle inequality. That is, the sum of any two digits must be greater than the third. So, for any ordering of the digits as x, y, z where x ≤ y ≤ z, we must have x + y > z.2. At least two of the digits must be equal (since it's isosceles, including equilateral). So, the multiset {a, b, c} must have at least two equal elements.Therefore, the problem reduces to counting all ordered triplets (a, b, c) with a, b, c ∈ {1, 2, ..., 9}, such that:- The multiset {a, b, c} has at least two equal elements.- The multiset {a, b, c} satisfies the triangle inequality.Then, since each such triplet corresponds to a unique three-digit number (as a is non-zero), the total count is the number of such triplets.Alternatively, since the digits can be arranged in any order to form the triangle, but the number itself is a specific ordering, we need to consider all permutations. However, perhaps it's easier to first compute the number of unordered multisets {a, b, c} that satisfy the conditions, and then for each such multiset, count how many ordered triplets (i.e., three-digit numbers) correspond to it. Then, sum over all such multisets.But perhaps that's complicated. Let's think step by step.First, let's consider that the triangle must be isosceles, so two sides are equal, or all three are equal. Let's handle the equilateral case first.Equilateral triangles: All three sides equal. So digits a = b = c. Each digit can be from 1 to 9, so there are 9 such numbers: 111, 222, ..., 999.Now, for the isosceles (but not equilateral) case. Here, exactly two sides are equal. So, the multiset {a, a, b} where a ≠ b. However, we need to ensure that these satisfy the triangle inequality. For a triangle with two equal sides a and a third side b, the triangle inequality requires that:a + a > b (since the two equal sides are the legs, and the third side is the base). Also, the third side must satisfy |a - a| < b < a + a, but since |a - a| = 0, so effectively, b < 2a. But since we need a, b ≥ 1, and since the triangle inequality requires that the sum of any two sides must exceed the third. So, in this case, with sides a, a, b, the inequalities are:1. a + a > b (which simplifies to b < 2a)2. a + b > a (which simplifies to b > 0, which is already satisfied since b ≥ 1)3. a + b > a (same as above)So the main constraint is b < 2a. But since we also need that the three sides can form a triangle, so even if two sides are equal, but if the third is too long, it's invalid. So in the case of two equal sides, the third side must be less than twice the length of the equal sides.But also, note that in the case where the two equal sides are not the longest sides. Wait, if we have sides a, a, b, then depending on whether b is the longest side or not. If b is longer than a, then the triangle would have sides a, a, b with b > a. Then, the triangle inequality requires that a + a > b. So in this case, b must be less than 2a. If b is less than or equal to a, then the triangle has sides a, a, b with b ≤ a, and the triangle inequality requires that a + b > a, which is b > 0, which is always true. Therefore, if we have two equal sides and a third side, the third side must be less than twice the length of the equal sides.Therefore, for each pair (a, b), where a is the length of the two equal sides, and b is the third side (either equal or different), but in the non-equilateral case, b ≠ a. So, for each a from 1 to 9, and b from 1 to 9, b ≠ a, and such that b < 2a. Then, we can compute how many such ordered triplets there are.Wait, but in the case where the two equal sides are not the longest sides. For example, if a = 3 and b = 4, then the sides would be 3, 3, 4. Then, the triangle inequality holds because 3 + 3 > 4, 3 + 4 > 3, etc. But if a = 3 and b = 7, then 3 + 3 = 6, which is not greater than 7, so that's invalid.Alternatively, if the two equal sides are the longer sides. For example, suppose we have sides a, a, b where a > b. Then, the triangle inequality is automatically satisfied because a + b > a (since b > 0) and a + a > b (since a > b). But in this case, if a > b, then the third side is smaller, so the main inequality to check is a + b > a, which is b > 0, which is always true. Therefore, in this case, when the two equal sides are the longer sides, the triangle inequality is automatically satisfied as long as b ≥ 1. However, if the two equal sides are not the longer sides, i.e., if the third side is longer, then we need to check that the sum of the two equal sides is greater than the third side.Therefore, to summarize:For a triangle with two equal sides (a, a, b):- If b > a: need b < 2a.- If b < a: automatically satisfies triangle inequality.- If b = a: equilateral, which we already considered.Therefore, for each a from 1 to 9, the possible b's (for the non-equilateral isosceles case) are:- When b ≠ a: - If b < a: all b from 1 to a - 1. - If b > a: b from a + 1 to 2a - 1 (since b must be less than 2a).But note that b must be between 1 and 9. So for each a, the valid b's are:- For b < a: 1 ≤ b ≤ a - 1.- For b > a: a + 1 ≤ b ≤ min(2a - 1, 9).Therefore, the number of possible b's for each a is:Number of b < a: (a - 1) numbers.Number of b > a: (min(2a - 1, 9) - a) numbers. Let's compute that:For a given a, the upper limit for b > a is 2a - 1, but cannot exceed 9. So:If 2a - 1 ≤ 9, then upper limit is 2a - 1.Otherwise, upper limit is 9.So when is 2a - 1 ≤ 9?Solving 2a - 1 ≤ 9:2a ≤ 10a ≤ 5.Therefore, for a from 1 to 5, the upper limit for b is 2a - 1. For a from 6 to 9, the upper limit is 9.Therefore, for each a:If 1 ≤ a ≤ 5: - b > a: from a + 1 to 2a - 1. Number of such b's is (2a - 1) - a = a - 1.If 6 ≤ a ≤ 9: - b > a: from a + 1 to 9. Number of such b's is 9 - a.Therefore, for each a, the number of possible b's where b ≠ a and the triangle is isosceles is:For a from 1 to 5: (a - 1) [b < a] + (a - 1) [b > a] = 2(a - 1).For a from 6 to 9: (a - 1) [b < a] + (9 - a) [b > a].Therefore, total number of possible b's for each a is:For a = 1: 0 [since b < 1 is impossible] + 0 [since 2*1 -1 =1, which is equal to a, but b must be greater than a, which is 1, so from 2 to 1, which is impossible]. Wait, maybe I need to check.Wait, let's plug in a =1:Case a =1.For b < a: 1 ≤ b <1, which is impossible. So 0.For b > a: a +1 = 2 to 2a -1 =1. But since 2a -1 =1, but a +1 =2. So this interval is from 2 to1, which is invalid. Therefore, 0.So for a =1, there are 0 possible b's where b ≠1 and the triangle is isosceles. Makes sense because if a=1, then the two equal sides are 1,1, and the third side must be less than 2*1=2. But the third side must be at least 1. So b can only be 1, which is equilateral. But since we are considering non-equilateral, b≠1. Therefore, no solutions here.Similarly, a=2:For a=2:b < a: b=1.b > a: from 3 to 3 (since 2a -1=3). So b=3.So total b's:1 +1=2.So for a=2, we can have b=1 and b=3.Similarly, let's test this.For a=2, the two equal sides are 2,2. Then, the third side can be 1 or 3. Wait, but if the third side is 1, then the sides are 1,2,2, which satisfies the triangle inequality (1 + 2 > 2, 2 + 2 >1). If the third side is 3, then 2 + 2 >3 (4>3), which is true. So both 1 and 3 are valid. So yes, two possibilities.Similarly, a=3:For a=3:b < a: b=1,2.b > a: from 4 to 5 (since 2a -1=5). So b=4,5.Total b's:2 +2=4.So a=3, b=1,2,4,5.Similarly, sides (3,3,1), (3,3,2), (3,3,4), (3,3,5). Let's check triangle inequality:For b=1: 1 +3 >3 (4>3, okay), 3+3>1.For b=4:3+3>4, okay.For b=5:3+3>5 (6>5, okay). So all valid.Continuing this way, for a=4:b <4:1,2,3.b >4: from5 to7 (since 2*4-1=7). So 5,6,7.Total b's:3 +3=6.Similarly, a=5:b <5:1,2,3,4.b >5: from6 to9 (since 2*5-1=9). So 6,7,8,9.Wait, 2a -1=9, which is equal to 9. So from6 to9, which is 4 numbers.So total b's:4 +4=8.For a=6:Now, for a=6:b <6:1,2,3,4,5.b >6: from7 to9. So 7,8,9. That's 3 numbers.Total b's:5 +3=8.Similarly, a=7:b <7:1-6.b >7:8,9.Total b's:6 +2=8.a=8:b <8:1-7.b >8:9.Total b's:7 +1=8.a=9:b <9:1-8.b >9: impossible.Total b's:8 +0=8.Wait, so from a=6 to a=9, each has 8 possible b's? Wait, let's check:For a=6:b <6: 5 numbers (1-5).b >6: 3 numbers (7-9). Total 5+3=8.a=7:b <7:6 numbers (1-6).b >7:2 numbers (8-9). Total 6+2=8.a=8:b <8:7 numbers (1-7).b >8:1 number (9). Total 7+1=8.a=9:b <9:8 numbers (1-8).b >9:0. Total 8.So from a=6 to a=9, each a contributes 8 b's. However, wait for a=5:a=5:b <5:4 numbers (1-4).b >5:4 numbers (6-9). Total 8.Similarly, a=5 also has 8.So the number of possible b's for each a is:a=1:0a=2:2a=3:4a=4:6a=5:8a=6:8a=7:8a=8:8a=9:8So now, for each a, the number of possible b's where b≠a, and the multiset {a,a,b} forms an isosceles triangle.But remember, each such multiset {a,a,b} can be arranged in different ways as three-digit numbers. For example, {2,2,3} can be arranged as 223, 232, 322, 233, 323, 332. Wait, no. Wait, {a,a,b} has two a's and one b. So the number of distinct permutations is 3: aab, aba, baa. So for each multiset {a,a,b}, there are 3 distinct three-digit numbers. However, we need to be careful about leading digits. Wait, in the three-digit number, the first digit (a in the number abc) can be any of the digits, so if the multiset is {a,a,b}, then the possible three-digit numbers are all permutations where the first digit is a or b. However, since a and b are both between 1 and 9, there's no issue with leading zeros. Therefore, for each multiset {a,a,b}, there are 3 different three-digit numbers. However, if a = b, which is the equilateral case, then all permutations are the same number, so each equilateral number is counted once (like 111, 222, etc.).But in our current case, we are considering the non-equilateral isosceles triangles, i.e., exactly two sides equal. So for each a and b (a ≠ b), the multiset {a,a,b} can be arranged into 3 different three-digit numbers.However, wait, the problem asks for the count of three-digit numbers, not the count of distinct triangles. So even if different numbers correspond to the same triangle (same multiset), they are considered different numbers. Therefore, we need to count all ordered triplets (a, b, c) where the multiset {a, b, c} can form an isosceles triangle. So for example, the multiset {2,2,3} corresponds to 3! / 2! = 3 distinct numbers: 223, 232, 322.Therefore, for each multiset {a,a,b} with a ≠ b, there are 3 three-digit numbers. Similarly, for the equilateral case, each multiset {a,a,a} corresponds to 1 three-digit number.Therefore, to compute the total number of three-digit numbers:- First, the equilateral case: 9 numbers.- Second, the isosceles (non-equilateral) case: for each a from 1 to 9, and each valid b (as computed above), multiply by 3 (the number of permutations). However, wait, but we have to be careful not to double-count. Because when we fix a and count the number of valid b's, then for each such b, we can have three numbers: aab, aba, baa. However, when a and b are different, these are distinct numbers.However, the problem is that when we iterate a from 1 to 9 and for each a, count the number of b's, then multiply by 3, we might be overcounting. Wait, no, because for each a and b, with a ≠ b, the numbers aab, aba, baa are distinct and valid three-digit numbers. Therefore, for each a and each valid b (where b can be less than or greater than a), there are 3 numbers. However, we need to ensure that when we count for different a's and b's, we don't have overlapping multisets. For example, if a=2 and b=3, we get the multiset {2,2,3}, and if a=3 and b=2, we get the multiset {3,3,2}. These are different multisets, so they correspond to different sets of numbers. Therefore, the total count for non-equilateral isosceles triangles would be the sum over all a from 1 to 9 of (number of valid b's for each a) multiplied by 3.But wait, let's think:Suppose for a=2, we have b=1 and b=3. Each of these b's gives 3 numbers. For b=1: numbers 221, 212, 122. For b=3: numbers 223, 232, 322. These are all distinct numbers. Similarly, for a=3 and b=1, we get numbers 331, 313, 133. So each combination of a and b (a ≠ b) gives unique numbers. Therefore, the total number of such numbers is 3 times the total number of valid (a, b) pairs.But we have to compute the total number of valid (a, b) pairs where a is from 1 to 9, b is from 1 to 9, b ≠ a, and either b < a or b > a with b < 2a (for a ≤5) or b ≤9 (for a ≥6).Earlier, we computed for each a, the number of valid b's:a=1:0a=2:2a=3:4a=4:6a=5:8a=6:8a=7:8a=8:8a=9:8So summing these up:For a=1:0a=2:2a=3:4 → total so far 6a=4:6 → total 12a=5:8 → total 20a=6:8 → 28a=7:8 →36a=8:8→44a=9:8→52Total valid (a, b) pairs:52.Therefore, total numbers for non-equilateral isosceles:52 *3=156.Adding the equilateral case:9.Total numbers:156 +9=165.But wait, the answer options are 45, 81, 165, 216. So 165 is option C. So the answer should be C.But let's double-check to make sure we didn't make a mistake.First, let's verify the count of valid (a, b) pairs:For each a:a=1:0a=2: b=1,3 →2a=3: b=1,2,4,5 →4a=4: b=1,2,3,5,6,7 → Wait, wait, earlier we said for a=4:b <4:1,2,3 (3 numbers)b >4:5,6,7 (3 numbers)Total 6. Yes.Wait, but earlier when we broke it down for a=5:b <5:1,2,3,4 (4 numbers)b >5:6,7,8,9 (4 numbers)Total 8.Similarly, for a=6:b <6:1,2,3,4,5 (5 numbers)b >6:7,8,9 (3 numbers)Total 8.Same for a=7-9: each has 8.So total pairs:a=1:0a=2:2a=3:4a=4:6a=5:8a=6:8a=7:8a=8:8a=9:8Adding these:0 +2=2+4=6+6=12+8=20+8=28+8=36+8=44+8=52Yes, total 52 pairs.Multiply by 3 gives 156.Add 9 equilateral: total 165.Therefore, the answer is 165, which is option C.But let's check for possible mistakes.One possible mistake is in the calculation of the number of valid b's for each a.For example, when a=6, the valid b's are 1-5 and 7-9. So 5 +3=8. Correct.For a=7, valid b's are 1-6 and 8-9:6 +2=8. Correct.For a=8:1-7 and 9:7 +1=8. Correct.For a=9:1-8:8. Correct.Another possible mistake is thinking that all ordered triplets (a, b, c) with two equal digits and satisfying triangle inequality are valid. But in our approach, we considered that for each a and b (where a is the repeated digit), we count the number of b's such that the triangle inequality holds, and then multiply by 3 for the permutations.But is there a case where the same three-digit number is counted multiple times? For example, consider the multiset {2,2,3}. When a=2 and b=3, we count it as 223, 232, 322. When a=3 and b=2, we would have the multiset {3,3,2}, which is a different multiset, leading to different numbers: 332, 323, 233. So no overlap. Therefore, each multiset is unique based on the choice of a and b, and thus there is no overcounting.Another check: Let's take a small a and see.For a=2, b=1: numbers 221, 212, 122. These are valid as they form the sides 2,2,1 which satisfy 2+2>1, 2+1>2, etc.For a=2, b=3: numbers 223, 232, 322. Sides 2,2,3. 2+2>3 (4>3), so valid.Similarly, for a=3, b=5: numbers 335, 353, 533. Sides 3,3,5: 3+3>5 (6>5), valid.But wait, if a=5 and b=9: numbers 559, 595, 955. Sides 5,5,9. Check triangle inequality:5+5>9? 10>9, yes. Valid.If a=5 and b=10: but b can't be 10. So all b's are up to 9.But wait, a=5, b=9 is allowed because 9 < 2*5=10. So yes.But what about a=6 and b=9: {6,6,9}. Check triangle inequality:6+6>9 (12>9), yes. So valid.Similarly, a=9 and b=8: {9,9,8}. 9+8>9 (17>9), yes. Also 9+9>8 (18>8). Valid.Therefore, all the cases we considered satisfy the triangle inequality.Therefore, our calculation seems correct.Another way to approach this problem: first count all possible three-digit numbers with digits from 1-9, then subtract those that don't form isosceles triangles. But that might be more complicated.Alternatively, consider all possible triplets (a, b, c) where at least two are equal, and satisfy the triangle inequality. For each such triplet, count all permutations. But as we did, it's equivalent.Another check: total number of three-digit numbers with all digits 1-9 is 9*9*9=729.But we are only counting those with at least two equal digits and satisfy triangle inequality. The answer is 165, which is much less than 729. The options given include 165, which seems plausible.Another sanity check: the number of three-digit numbers with exactly two digits equal (excluding equilateral) is C(3,2)*9*8=3*9*8=216. But this is without considering the triangle inequality. Since 216 is one of the options, but our answer is 165, which is less, it makes sense because we're imposing the triangle inequality condition.Therefore, after applying the triangle inequality constraints, the count reduces from 216 to 165. But wait, wait: actually, the total number of three-digit numbers with exactly two digits equal is 3*9*8=216, and the equilateral numbers are 9. So total with at least two equal digits is 216 +9=225. Then, out of these 225, how many satisfy the triangle inequality?But according to our previous calculation, it's 165. So the difference is 225 -165=60 numbers that have at least two equal digits but do not satisfy the triangle inequality.For example, take a=1 and b=1 (equilateral:111). That's valid. But any other number with two 1's and a third digit greater than or equal to 2 would not satisfy the triangle inequality. For example, 112: sides 1,1,2. Check 1+1>2? 2>2? No, it's equal, so not valid. Therefore, numbers like 112,113,...,119 are invalid. Similarly, 121, 211, etc. So for a=1 and b≥2, the numbers would be invalid. In our calculation, for a=1, we had 0 valid b's (excluding equilateral), which matches, because any b≠1 would result in 1,1,b with b≥2, which doesn't satisfy 1+1>b.Similarly, for a=2 and b=4: 2,2,4. Check 2+2>4? 4>4? No. Therefore, invalid. But wait, in our calculation for a=2, b can be 1 or 3. So b=4 is not allowed. Wait, for a=2, the valid b's when b>2 are from 3 to 2a-1=3. So only b=3. Therefore, 2,2,4 is not allowed, which is correct because 2+2=4 is not greater than 4. So our calculation correctly excludes such cases.Therefore, the count of 165 seems accurate.Hence, the answer is C) 165.

✅:To determine how many three-digit numbers ( n = overline{abc} ) exist where digits ( a, b, c ) can form the sides of an isosceles (including equilateral) triangle, we analyze the following:1. Equilateral Triangles: All three digits are equal. There are 9 such numbers (111, 222, ..., 999).2. Isosceles (Non-Equilateral) Triangles: Exactly two digits are equal. For each pair of equal digits ( a ), the third digit ( b ) must satisfy the triangle inequality: - If ( b < a ), it automatically satisfies the inequality. - If ( b > a ), ( b ) must be less than ( 2a ).Calculation for Valid ( b ) Values:- For ( a = 1 ): No valid ( b ) (since ( b neq 1 ) would violate the triangle inequality).- For ( a = 2 ): ( b = 1, 3 ) (2 values).- For ( a = 3 ): ( b = 1, 2, 4, 5 ) (4 values).- For ( a = 4 ): ( b = 1, 2, 3, 5, 6, 7 ) (6 values).- For ( a = 5 ): ( b = 1, 2, 3, 4, 6, 7, 8, 9 ) (8 values).- For ( a = 6 ) to ( a = 9 ): ( b ) ranges from ( 1 ) to ( a-1 ) and ( a+1 ) to ( 9 ), yielding 8 valid values each.Total Valid ( (a, b) ) Pairs:- Summing valid pairs: ( 0 + 2 + 4 + 6 + 8 + 8 + 8 + 8 + 8 = 52 ).Counting Permutations:- Each valid pair ( (a, b) ) corresponds to 3 permutations (e.g., ( aab, aba, baa )).- Total non-equilateral isosceles numbers: ( 52 times 3 = 156 ).Final Count:- Adding equilateral cases: ( 156 + 9 = 165 ).Thus, the number of such three-digit numbers is (boxed{C}).

🔑: Problem:设三位数 n=overline{abc}，若以 a、b、c 为三条边的长可以构成一个等腰(含等边)三角形，则这样的三位数 n 有 ( ).(A) 45 个 (B) 81 个 (C) 165 个 (D) 216 个1. Constraints on values of a, b, c: To form a triangle, the side lengths a, b, c must be positive. Thus, a, b, c in {1, 2, cdots, 9}.2. Case of Equilateral Triangle: - If a = b = c, each digit must be the same for it to be an equilateral triangle. - The number of such three-digit numbers is equal to the number of choices for one digit (since all three digits are identical): [ n_1 = binom{9}{1} = 9 ]3. Case of Isosceles Triangle (non-equilateral): - In an isosceles triangle (i.e., exactly two sides are equal), only two different digits can be used among a, b, c. Assume a = b neq c (similar arguments apply when b = c neq a and c = a neq b). - We consider 9 - 1 = 8 possibilities for c, multiplied by 3 ways to assign (a, b, c). However, not all combinations work.4. Counting Valid Combinations: - Consider distinct combinations (a, b, c): [ text{Total combinations} = binom{9}{2} = 36 ] - But we need to subtract invalid combinations that cannot form a triangle due to the triangle inequality c < 2a or c < 2b when a neq b = c. These invalid conditions give us 20 cases: [ begin{aligned} &a=9: b in {1, 2, 3, 4} implies 4 text{ cases}, &a=8: b in {1, 2, 3, 4} implies 4 text{ cases}, &a=7: b in {1, 2, 3} implies 3 text{ cases}, &a=6: b in {1, 2, 3} implies 3 text{ cases}, &a=5: b in {1, 2} implies 2 text{ cases}, &a=4: b in {1, 2} implies 2 text{ cases}, &a=3: b in {1} implies 1 text{ case}, &a=2: b in {1} implies 1 text{ case}. end{aligned} ] - Hence, the number of valid combinations is: [ text{Valid isosceles combinations} = 2 binom{9}{2} - 20 ] - We multiply by the number of ways to place two equal sides and one different side (which is 3): [ n_2 = 3 left( binom{9}{2} - 10 right) = 3 left( 36 - 10 right) = 156 ]5. Total number of three-digit numbers: [ n = n_1 + n_2 = 9 + 156 = 165 ]Conclusion:[boxed{text{C}}]

🧠:Okay, let's try to figure out this problem step by step. So, the question is: If log base 4 of (x + 2y) plus log base 4 of (x - 2y) equals 1, then what's the minimum value of |x| - |y|. Hmm, alright. I need to find the minimum of |x| - |y| given that logarithmic equation. Let me start by recalling some logarithm properties.First, I remember that log_a b + log_a c = log_a (b*c) when the bases are the same. So maybe I can combine those two logarithms into a single logarithm. Let me write that down:log₄[(x + 2y)(x - 2y)] = 1Simplifying the argument inside the log, that's a difference of squares, right? So (x + 2y)(x - 2y) = x² - (2y)² = x² - 4y². Therefore, the equation becomes:log₄(x² - 4y²) = 1Now, converting the logarithmic equation to its exponential form, log base 4 of something equals 1 means that 4^1 = that something. So:x² - 4y² = 4Alright, so now we have a hyperbola equation here: x²/4 - y² = 1. But maybe it's better to keep it as x² - 4y² = 4 for now. The next step is to find the minimum value of |x| - |y|. Hmm, okay.So we need to minimize |x| - |y| subject to the constraint x² - 4y² = 4. Let's think about how to approach this. Since we have a constraint and an objective function, this seems like an optimization problem. Maybe I can use methods like substitution or Lagrange multipliers? Let me see.Since it's a quadratic constraint, substitution might work. Let's try expressing one variable in terms of the other. Let's solve the constraint equation for x²:x² = 4 + 4y²So x = ±√(4 + 4y²). Therefore, |x| is √(4 + 4y²) since the square root is always non-negative. So |x| = √(4(1 + y²)) = 2√(1 + y²). So the expression we need to minimize is:|x| - |y| = 2√(1 + y²) - |y|Hmm, okay. Now we can treat this as a function of y, say f(y) = 2√(1 + y²) - |y|, and find its minimum. Since |y| is involved, maybe we can consider y ≥ 0 and y ≤ 0 separately. But since the function is even in y (because replacing y with -y doesn't change the value), we can assume y ≥ 0 without loss of generality. So let's set y ≥ 0, so |y| = y. Then f(y) becomes 2√(1 + y²) - y.So, we need to find the minimum of f(y) = 2√(1 + y²) - y for y ≥ 0. To find the minimum, we can take the derivative of f(y) with respect to y, set it to zero, and solve for y.Let's compute f'(y):f'(y) = 2*(1/(2√(1 + y²)))*2y - 1 = (2y)/√(1 + y²) - 1Wait, hold on. Let me recast that. The derivative of 2√(1 + y²) is 2*(1/(2√(1 + y²)))*2y? Wait, no. Wait, the derivative of sqrt(u) is (1/(2√u)) * u'. So, if u = 1 + y², then derivative of 2*sqrt(u) is 2*(1/(2√u))*du/dy = (1/√u)*(2y) = (2y)/√(1 + y²). Then subtract derivative of y, which is 1. So f'(y) = (2y)/√(1 + y²) - 1.Set this equal to zero:(2y)/√(1 + y²) - 1 = 0So,(2y)/√(1 + y²) = 1Multiply both sides by √(1 + y²):2y = √(1 + y²)Now, square both sides to eliminate the square root:(2y)² = (√(1 + y²))²4y² = 1 + y²Subtract y² from both sides:3y² = 1So y² = 1/3Therefore, y = ±1/√3. But since we assumed y ≥ 0, we take y = 1/√3.Now, let's check if this critical point is a minimum. We can use the second derivative test or check the behavior around the point. Let's compute the second derivative.First, f'(y) = (2y)/√(1 + y²) - 1Compute f''(y):Differentiate the first term: d/dy [ (2y)/√(1 + y²) ]Let me write that as 2y*(1 + y²)^(-1/2). Using the product rule:Derivative = 2*(1 + y²)^(-1/2) + 2y*(-1/2)*(1 + y²)^(-3/2)*(2y)Simplify:= 2/(√(1 + y²)) + 2y*(-y)/( (1 + y²)^(3/2) )= 2/(√(1 + y²)) - 2y²/( (1 + y²)^(3/2) )Factor out 2/( (1 + y²)^(3/2) ):= [2(1 + y²) - 2y²] / (1 + y²)^(3/2)= [2 + 2y² - 2y²] / (1 + y²)^(3/2)= 2 / (1 + y²)^(3/2)So the second derivative f''(y) = 2 / (1 + y²)^(3/2). Since this is always positive for all y, the function is concave upward everywhere. Therefore, the critical point at y = 1/√3 is a minimum.Therefore, the minimum value of f(y) = 2√(1 + y²) - y occurs at y = 1/√3. Let's compute this value.First, compute √(1 + y²):y = 1/√3, so y² = 1/3, so 1 + y² = 4/3. Thus, √(4/3) = 2/√3. Therefore,2√(1 + y²) = 2*(2/√3) = 4/√3Then subtract y = 1/√3:4/√3 - 1/√3 = 3/√3 = √3.So the minimum value of |x| - |y| is √3.Wait, but hold on. Let me verify this because sometimes when dealing with absolute values and substitutions, you can have issues. Let's confirm.We assumed y ≥ 0, so y = 1/√3, which is positive. Then x² = 4 + 4y² = 4 + 4*(1/3) = 4 + 4/3 = 16/3. Therefore, x = ±4/√3. Therefore, |x| = 4/√3. Then |x| - |y| = 4/√3 - 1/√3 = 3/√3 = √3. So that checks out. So that's the minimum value. So √3 is the answer. Wait, but the problem says "the minimum value of |x| - |y|", so is this the minimum? Let me see.Wait, but maybe there's a lower value? Let me check. Let's see. If y is larger, then |y| increases, but |x| also increases because x² = 4 + 4y². So maybe as y increases, |x| - |y| could decrease or increase? Let's take the limit as y approaches infinity. Then |x| = 2√(1 + y²) ≈ 2|y|. So |x| - |y| ≈ 2|y| - |y| = |y|, which goes to infinity. So as y increases, |x| - |y| increases. On the other hand, as y approaches zero, |x| - |y| = 2√(1 + 0) - 0 = 2 - 0 = 2. Wait, but 2 is larger than √3 ≈ 1.732. So the function has a minimum at √3 and increases towards 2 as y approaches zero, and increases to infinity as y approaches infinity. So yes, the minimum is indeed √3.Therefore, the answer should be √3. Wait, but let me check if there's another approach. Maybe using Lagrange multipliers?Let's try that method. We need to minimize f(x, y) = |x| - |y| subject to the constraint g(x, y) = x² - 4y² - 4 = 0.But the absolute values complicate things. Let's consider different cases based on the signs of x and y. However, since we have |x| and |y|, maybe we can assume x ≥ 0 and y ≥ 0 because of the symmetry. Let's see.If x is positive, then |x| = x. If x is negative, |x| = -x, but since x² is the same for x and -x, maybe the minimum occurs in the positive x region. Similarly for y. So let's assume x ≥ 0 and y ≥ 0. Then f(x, y) = x - y.So we need to minimize x - y subject to x² - 4y² = 4 with x ≥ 0, y ≥ 0.Set up the Lagrangian: L = x - y + λ(x² - 4y² - 4)Take partial derivatives:∂L/∂x = 1 + 2λx = 0 → 1 + 2λx = 0∂L/∂y = -1 - 8λy = 0 → -1 - 8λy = 0∂L/∂λ = x² - 4y² - 4 = 0From the first equation: 1 + 2λx = 0 → λ = -1/(2x)From the second equation: -1 - 8λy = 0 → 8λy = -1 → λ = -1/(8y)Set the two expressions for λ equal: -1/(2x) = -1/(8y) → 1/(2x) = 1/(8y) → 8y = 2x → 4y = x → x = 4yNow substitute x = 4y into the constraint equation:x² - 4y² = 4 → (16y²) - 4y² = 4 → 12y² = 4 → y² = 4/12 = 1/3 → y = 1/√3 (since y ≥ 0)Then x = 4*(1/√3) = 4/√3Therefore, x - y = 4/√3 - 1/√3 = 3/√3 = √3. So same result as before. So the minimum is indeed √3.But wait, we assumed x ≥ 0 and y ≥ 0. What if x is negative? Let's see. Suppose x is negative. Then |x| = -x. If y is positive, then |y| = y. So |x| - |y| = -x - y. But if x is negative, then in the equation x² - 4y² =4, x² is positive, so x can be any real number as long as x² =4 +4y². But if x is negative, then |x| - |y| = -x - |y|. If we take x negative and y positive, then to minimize -x - y, we need to maximize x + y. But since x is negative, making x more negative (i.e., x decreasing) would make -x larger, but y is positive. However, x² =4 +4y², so if x becomes more negative, y would have to increase as well. Let's see.Suppose x is negative. Let x = -a where a >0. Then the equation becomes a² -4y²=4. Then |x| - |y| = a - |y|. So similar to the positive x case but with x replaced by a. So in this case, a = |x|. So even if x is negative, |x| - |y| is a - |y| where a² =4 +4y², same as the positive x case. Therefore, the minimum would be the same. Similarly, if y is negative, |y| = -y, but since in the expression |x| - |y|, it's |x| minus |y|. So if y is negative, |y| = -y, so |x| - |y| = |x| - (-y) = |x| + y. But since y is negative, that's |x| plus a negative number. To minimize |x| + y, you might want to minimize |x| and take y as negative as possible. But let's see.Suppose y is negative. Then |y| = -y. So |x| - |y| = |x| - (-y) = |x| + y. However, in the constraint x² -4y²=4. If y is negative, say y = -b where b >0, then the equation becomes x² -4b²=4. So |x| - |y| = |x| - b. Wait, no: original expression is |x| - |y|. If y is negative, |y| = | -b | = b, so |x| - |y| = |x| - b. So similar to the case when y is positive, except y is replaced by b. So the analysis is the same. Therefore, regardless of the sign of y, the expression |x| - |y| is equivalent to |x| - b where b = |y|. So in effect, we can assume y is non-negative, and x can be positive or negative, but |x| is positive. Therefore, the minimum occurs at |x| - |y| where |x| and |y| are related by |x|² =4 +4|y|², and the minimum is achieved when the derivative is zero, which we already found to be √3.Therefore, the minimum value is √3. So I think that's the answer.Wait, just to double-check, let's plug in some other points. For example, if y = 0, then x² =4 +0=4, so x=±2. Then |x| - |y| =2 -0=2, which is bigger than √3 ≈1.732. If y=1, then x²=4+4(1)=8, so x=±2√2≈2.828, then |x| - |y|≈2.828 -1≈1.828, which is still bigger than √3. If y=1/√3≈0.577, then x=4/√3≈2.309, so |x| - |y|≈2.309 -0.577≈1.732, which is √3. So yes, that's indeed the minimum.Another point, say y=0.5. Then x²=4 +4*(0.25)=4 +1=5, so x≈±2.236. Then |x| - |y|≈2.236 -0.5≈1.736, which is slightly larger than √3≈1.732. So that checks out. So the minimum is indeed achieved at y=1/√3, x=4/√3, giving |x| - |y|=√3.Therefore, the answer is √3, which in boxed form is boxed{sqrt{3}}.Final AnswerThe minimum value of ( |x| - |y| ) is boxed{sqrt{3}}.

✅:Given the equation (log _{4}(x+2 y) + log _{4}(x-2 y) = 1), we start by combining the logarithms using the property (log_a b + log_a c = log_a (bc)):[log_4[(x + 2y)(x - 2y)] = 1]This simplifies to:[log_4(x^2 - 4y^2) = 1]Converting the logarithmic equation to its exponential form, we get:[x^2 - 4y^2 = 4]We need to minimize (|x| - |y|) subject to the constraint (x^2 - 4y^2 = 4). Expressing (x) in terms of (y) from the constraint equation, we have:[x = pm sqrt{4 + 4y^2}]Thus, (|x| = 2sqrt{1 + y^2}). The objective function to minimize becomes:[|x| - |y| = 2sqrt{1 + y^2} - |y|]Considering (y geq 0) (due to symmetry), we define (f(y) = 2sqrt{1 + y^2} - y). Taking the derivative and setting it to zero:[f'(y) = frac{2y}{sqrt{1 + y^2}} - 1 = 0]Solving for (y):[frac{2y}{sqrt{1 + y^2}} = 1 implies 2y = sqrt{1 + y^2} implies 4y^2 = 1 + y^2 implies 3y^2 = 1 implies y = frac{1}{sqrt{3}}]Substituting (y = frac{1}{sqrt{3}}) back into the constraint equation to find (x):[x^2 = 4 + 4left(frac{1}{sqrt{3}}right)^2 = 4 + frac{4}{3} = frac{16}{3} implies x = pm frac{4}{sqrt{3}}]Thus, (|x| - |y| = frac{4}{sqrt{3}} - frac{1}{sqrt{3}} = frac{3}{sqrt{3}} = sqrt{3}).Verification using the second derivative test confirms this is a minimum. Therefore, the minimum value of (|x| - |y|) is (boxed{sqrt{3}}).

🔑:# Problem:若 log_{4}(x+2y) + log_{4}(x-2y) = 1, 则 |x| - |y| 的最小值是 qquad # Solution:1. Given the equation: [ log_{4}(x+2y) + log_{4}(x-2y) = 1 ] Using the logarithm property log_{a}(mn) = log_{a}(m) + log_{a}(n), we can rewrite the equation as: [ log_{4}left[(x+2y)(x-2y)right] = 1 ]2. Simplify inside the logarithm: [ log_{4}(x^2 - (2y)^2) = 1 ] [ log_{4}(x^2 - 4y^2) = 1 ]3. Convert the logarithmic equation to its exponential form: [ x^2 - 4y^2 = 4 ] This can be rewritten as: [ x^2 - 4y^2 = 4 ] which represents a hyperbola equation.4. To ensure the arguments of the original logs are positive, we need: [ x + 2y > 0 quad text{and} quad x - 2y > 0 ]5. This means: [ x > 2y quad text{and} quad x > -2y ] Combining these inequalities and considering the hyperbola x^2 - 4y^2 = 4, it simply implies: [ x > 2|y| ]6. Since x > 0 and we need to find the minimum value of x - y, transform the original equation by setting u = x - y and substitute x = u + y into the hyperbola equation: [ (u + y)^2 - 4y^2 = 4 ] Expanding and simplifying: [ u^2 + 2uy + y^2 - 4y^2 = 4 ] [ u^2 + 2uy - 3y^2 = 4 ]7. This forms a quadratic equation in y: [ 3y^2 - 2uy + (4 - u^2) = 0 ] For y to have real values, the discriminant Delta of this quadratic must be non-negative, where Delta is given by: [ Delta = (-2u)^2 - 4(3)(4 - u^2) = 4u^2 - 12(4 - u^2) ] Simplify the discriminant: [ Delta = 4u^2 - 48 + 12u^2 = 16u^2 - 48 ] 8. For the discriminant to be non-negative: [ 16u^2 - 48 geq 0 ] [ 16u^2 geq 48 ] [ u^2 geq 3 ] [ u geq sqrt{3} ]9. Therefore, the minimum value of u = x - y = sqrt{3}.So, the minimum value of |x| - |y| is:[boxed{sqrt{3}}]

🧠:Okay, so I need to figure out which of the given functions are infinitesimal as ( x rightarrow 0 ), and then determine their order relative to ( x ). Let me start by recalling what an infinitesimal is. An infinitesimal is a function that approaches zero as ( x ) approaches a certain point, in this case, zero. So all these functions should approach zero when ( x ) approaches zero. Then, their order relative to ( x ) tells us how fast they approach zero compared to ( x ) itself.First, let me check each function one by one to see if they are indeed infinitesimal as ( x rightarrow 0 ).1) (10x): As ( x ) approaches 0, (10x) will definitely approach 0. So this is an infinitesimal.2) (x^3): Similarly, (x^3) as ( x rightarrow 0 ) will go to 0. Since any positive power of ( x ) will go to 0, this is also an infinitesimal.3) (sqrt{3x}): Hmm, here we have a square root. Let's see. If ( x ) approaches 0 from the positive side (since the square root of a negative number isn't real), then ( sqrt{3x} ) will approach 0. But if ( x ) approaches 0 from the negative side, the square root would be imaginary. However, typically when considering limits like ( x rightarrow 0 ), unless specified otherwise, we consider the right-hand limit (i.e., ( x rightarrow 0^+ )). So assuming ( x ) is approaching 0 from the positive side, ( sqrt{3x} ) does approach 0, making it an infinitesimal.4) (tan frac{x}{5}): The tangent function is continuous at 0, since ( tan 0 = 0 ). As ( x rightarrow 0 ), ( frac{x}{5} rightarrow 0 ), so ( tan frac{x}{5} rightarrow tan 0 = 0 ). Therefore, this is also an infinitesimal.5) (log(1+x)): The natural logarithm of 1 is 0. As ( x rightarrow 0 ), ( 1 + x rightarrow 1 ), so ( log(1+x) rightarrow log 1 = 0 ). Hence, this is an infinitesimal as well.So all five functions are infinitesimal as ( x rightarrow 0 ). Now, the next part is to determine their order relative to ( x ). The order of an infinitesimal function ( f(x) ) relative to ( x ) is determined by the limit ( lim_{x rightarrow 0} frac{f(x)}{x^k} ), where ( k ) is the order. If ( k = 1 ), then we compare to ( x ). If the limit is a non-zero finite number, then the order is the same as ( x^k ). If the limit is zero, then the order is higher than ( x^k ), and if the limit is infinity, the order is lower than ( x^k ).But in this case, the question is to determine whether each function's order is higher than ( x ), lower than ( x ), or the same as ( x ). So for each function, we can compute the limit ( lim_{x rightarrow 0} frac{f(x)}{x} ). If the limit is a non-zero constant, then the order is the same as ( x ). If the limit is zero, the order is higher than ( x ), meaning it goes to zero faster than ( x ). If the limit is infinity, then the order is lower than ( x ), meaning it goes to zero slower than ( x ).Wait, but actually, order is typically considered in terms of the exponent. So if a function behaves like ( x^k ) as ( x rightarrow 0 ), then ( k ) is the order. So if ( k > 1 ), the order is higher than ( x ), and if ( k < 1 ), the order is lower than ( x ). However, when comparing to ( x ), the standard is to see if the limit ( lim_{x rightarrow 0} frac{f(x)}{x} ) exists. If it's a non-zero constant, then it's the same order. If it's zero, then ( f(x) ) is of higher order (i.e., smaller), and if it's infinity, lower order.Therefore, let's compute this limit for each function.1) (10x):Compute ( lim_{x rightarrow 0} frac{10x}{x} = lim_{x rightarrow 0} 10 = 10 ). So this is a non-zero constant, hence the same order as ( x ).2) (x^3):( lim_{x rightarrow 0} frac{x^3}{x} = lim_{x rightarrow 0} x^2 = 0 ). Therefore, this limit is zero, which means ( x^3 ) is of higher order than ( x ).3) (sqrt{3x}):First, note that ( sqrt{3x} = (3x)^{1/2} = 3^{1/2} x^{1/2} ). Then, compute ( lim_{x rightarrow 0} frac{3^{1/2} x^{1/2}}{x} = lim_{x rightarrow 0} 3^{1/2} x^{-1/2} = 3^{1/2} lim_{x rightarrow 0} frac{1}{x^{1/2}} ). As ( x rightarrow 0^+ ), ( x^{1/2} rightarrow 0 ), so ( frac{1}{x^{1/2}} rightarrow infty ). Therefore, the limit is infinity. Thus, ( sqrt{3x} ) is of lower order than ( x ), meaning it approaches zero slower than ( x ).4) (tan frac{x}{5}):We can use the Taylor series expansion for ( tan y ) around ( y = 0 ). Recall that ( tan y approx y + frac{y^3}{3} + cdots ). So ( tan frac{x}{5} approx frac{x}{5} + frac{1}{3} left( frac{x}{5} right)^3 + cdots ). Therefore, the leading term is ( frac{x}{5} ), so as ( x rightarrow 0 ), ( tan frac{x}{5} approx frac{x}{5} ). Therefore, ( lim_{x rightarrow 0} frac{tan frac{x}{5}}{x} = lim_{x rightarrow 0} frac{frac{x}{5} + cdots}{x} = frac{1}{5} ). Hence, the limit is ( frac{1}{5} ), which is a non-zero constant. Therefore, ( tan frac{x}{5} ) has the same order as ( x ).5) (log(1+x)):Again, using the Taylor series expansion for ( log(1 + x) ) around ( x = 0 ), we have ( log(1 + x) approx x - frac{x^2}{2} + frac{x^3}{3} - cdots ). The leading term is ( x ), so ( lim_{x rightarrow 0} frac{log(1 + x)}{x} = lim_{x rightarrow 0} frac{x - frac{x^2}{2} + cdots}{x} = lim_{x rightarrow 0} 1 - frac{x}{2} + cdots = 1 ). Therefore, this limit is 1, a non-zero constant. Hence, ( log(1 + x) ) has the same order as ( x ).So summarizing:1) 10x: Same order as x.2) x³: Higher order than x.3) √(3x): Lower order than x.4) tan(x/5): Same order as x.5) log(1+x): Same order as x.But let me double-check each of these to make sure I didn't make any mistakes.Starting with 1) 10x over x is 10, so same order. That seems straightforward.2) x³ over x is x², which tends to 0. Hence, higher order. Correct.3) √(3x) over x is proportional to x^{-1/2}, which tends to infinity. So lower order. Correct.4) For tan(x/5), using the limit, we can also compute directly using L’Hospital’s Rule if we don't recall the Taylor series. The limit is lim_{x→0} [tan(x/5)/x] = lim_{x→0} [ (1/5) sec²(x/5) / 1 ] by L’Hospital (since numerator and denominator both approach 0). sec²(0) = 1, so the limit is 1/5. Hence, same order. Correct.5) Similarly for log(1+x)/x, applying L’Hospital’s Rule gives lim_{x→0} [1/(1+x) /1] = 1. Hence, same order. Correct.Therefore, all steps check out.So the conclusion is:1) Same order as x.2) Higher order.3) Lower order.4) Same order.5) Same order.Therefore, the answer should list each of these in order.Final Answer1) Same order as ( x ): (boxed{10x}); 2) Higher order than ( x ): (boxed{x^{3}}); 3) Lower order than ( x ): (boxed{sqrt{3x}}); 4) Same order as ( x ): (boxed{tan frac{x}{5}}); 5) Same order as ( x ): (boxed{log (1+x)}).

✅:To determine which of the given functions are infinitesimal as ( x rightarrow 0 ) and their order relative to ( x ), we analyze each function:1. (10x): - As ( x rightarrow 0 ), (10x rightarrow 0). - The limit ( lim_{x rightarrow 0} frac{10x}{x} = 10 ), which is a non-zero constant. Hence, it has the same order as ( x ).2. (x^3): - As ( x rightarrow 0 ), (x^3 rightarrow 0). - The limit ( lim_{x rightarrow 0} frac{x^3}{x} = lim_{x rightarrow 0} x^2 = 0 ). Hence, it has a higher order than ( x ).3. (sqrt{3x}): - As ( x rightarrow 0^+ ), (sqrt{3x} rightarrow 0). - The limit ( lim_{x rightarrow 0} frac{sqrt{3x}}{x} = lim_{x rightarrow 0} frac{3^{1/2} x^{1/2}}{x} = 3^{1/2} lim_{x rightarrow 0} x^{-1/2} = infty ). Hence, it has a lower order than ( x ).4. (tan frac{x}{5}): - As ( x rightarrow 0 ), (tan frac{x}{5} rightarrow 0). - Using the Taylor series expansion, (tan frac{x}{5} approx frac{x}{5}). The limit ( lim_{x rightarrow 0} frac{tan frac{x}{5}}{x} = frac{1}{5} ), which is a non-zero constant. Hence, it has the same order as ( x ).5. (log(1+x)): - As ( x rightarrow 0 ), (log(1+x) rightarrow 0). - Using the Taylor series expansion, (log(1+x) approx x). The limit ( lim_{x rightarrow 0} frac{log(1+x)}{x} = 1 ), which is a non-zero constant. Hence, it has the same order as ( x ). Final Answer1) Same order as ( x ): (boxed{10x}); 2) Higher order than ( x ): (boxed{x^{3}}); 3) Lower order than ( x ): (boxed{sqrt{3x}}); 4) Same order as ( x ): (boxed{tan frac{x}{5}}); 5) Same order as ( x ): (boxed{log (1+x)}).

🔑:To determine the order of infinitesimals of the given expressions as ( x rightarrow 0 ), we compare each expression to ( x ) by computing the limit of their ratio to ( x ).1. Analyze ( 10x ): [ lim_{x rightarrow 0} frac{10x}{x} = lim_{x rightarrow 0} 10 = 10 ] Since the limit is a finite non-zero constant, ( 10x ) is an infinitesimal of the same order as ( x ).2. Analyze ( x^3 ): [ lim_{x rightarrow 0} frac{x^3}{x} = lim_{x rightarrow 0} x^2 = 0 ] Since the limit is zero, ( x^3 ) is an infinitesimal of a higher order than ( x ).3. Analyze ( sqrt{3x} ): [ lim_{x rightarrow 0} frac{sqrt{3x}}{x} = lim_{x rightarrow 0} sqrt{frac{3}{x}} = +infty ] Since the limit tends to infinity, ( sqrt{3x} ) is an infinitesimal of a lower order than ( x ).4. Analyze ( tanleft(frac{x}{5}right) ): [ lim_{x rightarrow 0} frac{tanleft(frac{x}{5}right)}{x} = lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x cosleft(frac{x}{5}right)} = left( lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x} right) cdot left( lim_{x rightarrow 0} frac{1}{cosleft(frac{x}{5}right)} right) ] Using the small angle approximation, ( sinleft(frac{x}{5}right) approx frac{x}{5} ): [ lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x} = frac{1}{5} quad text{and} quad lim_{x rightarrow 0} frac{1}{cosleft(frac{x}{5}right)} = 1 ] Therefore, [ lim_{x rightarrow 0} frac{tanleft(frac{x}{5}right)}{x} = frac{1}{5} cdot 1 = frac{1}{5} ] Since the limit is a finite non-zero constant, ( tanleft(frac{x}{5}right) ) is an infinitesimal of the same order as ( x ).5. Analyze ( lg(1+x) ): [ lim_{x rightarrow 0} frac{log(1+x)}{x} = lim_{x rightarrow 0} log(1+x)^{1/x} ] Using the approximation ( log(1+x) approx x ) for small ( x ): [ lim_{x rightarrow 0} frac{log(1+x)}{x} = 1 ] Therefore, ( log(1+x) ) is an infinitesimal of the same order as ( x ).Conclusion:- Same order as ( x ): ( 10x ), ( tanleft(frac{x}{5}right) ), ( log(1+x) )- Higher order than ( x ): ( x^3 )- Lower order than ( x ): ( sqrt{3x} )[boxed{text{The functions are classified as follows:}}]

🧠:Alright, so I need to show that the polynomial ( x^4 + 2x^2 + 2x + 2 ) can't be factored into the product of two quadratics with integer coefficients. Hmm, okay. Let me start by recalling how polynomial factorization works, especially for quartic polynomials into quadratics. First, if it can be factored into two quadratics with integer coefficients, then we should be able to write it as ( (x^2 + ax + b)(x^2 + cx + d) ), where a, b, c, d are integers. Then, multiplying these out should give us the original polynomial. So let me set that up.Multiplying the two quadratics:( (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd )Now, this should equal the given polynomial ( x^4 + 2x^2 + 2x + 2 ). Therefore, we can equate the coefficients of corresponding powers of x:1. Coefficient of ( x^4 ): 1 (matches on both sides)2. Coefficient of ( x^3 ): a + c = 0 (since there's no x^3 term in the original polynomial)3. Coefficient of ( x^2 ): ac + b + d = 24. Coefficient of x: ad + bc = 25. Constant term: bd = 2So, we have a system of equations here:1. ( a + c = 0 )2. ( ac + b + d = 2 )3. ( ad + bc = 2 )4. ( bd = 2 )Our goal is to find integers a, b, c, d satisfying all these equations. If no such integers exist, then the polynomial can't be factored as such.Starting with the first equation: ( a + c = 0 ), so ( c = -a ). That simplifies things because we can substitute c with -a in the other equations.Let's do that substitution:Second equation becomes: ( a(-a) + b + d = 2 ) → ( -a² + b + d = 2 )Third equation: ( a d + b (-a) = 2 ) → ( a d - a b = 2 ) → ( a(d - b) = 2 )Fourth equation: ( b d = 2 )So now, the variables we have to solve for are a, b, d, with c = -a.Let's note that the fourth equation is ( b d = 2 ). Since b and d are integers, the possible pairs (b, d) are the integer pairs that multiply to 2. The possible pairs are:(1, 2), (2, 1), (-1, -2), (-2, -1)So four possibilities for (b, d). Let's consider each case one by one and see if we can find integer a such that the other equations are satisfied.Case 1: b = 1, d = 2Then, from the third equation: ( a(d - b) = a(2 - 1) = a(1) = a = 2 ). So a = 2.Then, substitute into the second equation: ( -a² + b + d = -4 + 1 + 2 = -1 ≠ 2 ). Not valid. So this case doesn't work.Case 2: b = 2, d = 1Third equation: ( a(d - b) = a(1 - 2) = a(-1) = -a = 2 → a = -2 )Then, second equation: ( -a² + b + d = -4 + 2 + 1 = -1 ≠ 2 ). Again, invalid.Case 3: b = -1, d = -2Third equation: ( a(d - b) = a(-2 - (-1)) = a(-1) = -a = 2 → a = -2 )Check second equation: ( -a² + b + d = -4 + (-1) + (-2) = -7 ≠ 2 ). Nope.Case 4: b = -2, d = -1Third equation: ( a(d - b) = a(-1 - (-2)) = a(1) = a = 2 )Second equation: ( -a² + b + d = -4 + (-2) + (-1) = -7 ≠ 2 ). Also invalid.So none of the possible (b, d) pairs in the fourth equation lead to a valid solution in the second equation. Therefore, there are no integer solutions for a, b, c, d. Hence, the polynomial can't be factored into two quadratics with integer coefficients.Wait a second, but let me double-check my steps to make sure I didn't miss any possible pairs for b and d. The product bd = 2. So in integers, the factors are 1*2, 2*1, (-1)*(-2), (-2)*(-1). So that's all four cases. So I think I covered all possibilities.Alternatively, maybe there's another approach. Maybe trying to factor it modulo some prime. For example, if the polynomial factors over integers, then it must factor modulo any prime. Let's try modulo 2. Reducing the polynomial modulo 2: ( x^4 + 0x^2 + 0x + 0 = x^4 ). So modulo 2, the polynomial is x^4. If it factors into two quadratics modulo 2, then x^4 = (x^2 + a x + b)(x^2 + c x + d) mod 2. Since coefficients are 0 or 1.But in mod 2, possible quadratics are limited. Let's see:Possible quadratic factors mod 2:x^2, x^2 + 1, x^2 + x, x^2 + x + 1.But x^2 * x^2 = x^4. So x^4 can be factored as (x^2)(x^2) mod 2. Alternatively, perhaps other combinations. Let's check:(x^2 + a x + b)(x^2 + c x + d) mod 2.Multiply out:x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.Set equal to x^4 mod 2. So coefficients must be:- x^3 term: a + c ≡ 0 mod 2- x^2 term: ac + b + d ≡ 0 mod 2- x term: ad + bc ≡ 0 mod 2- constant term: bd ≡ 0 mod 2So, in mod 2, we have:From the constant term: bd ≡ 0 mod 2. So either b or d must be 0 mod 2. But in mod 2, 0 is 0 and 1 is 1. So possible b and d are 0 or 1. Wait, but in our original polynomial mod 2, the constant term is 0, but in the original problem, the constant term is 2, which is 0 mod 2. So yeah, if we factor x^4 mod 2, then the constant terms of the quadratics must multiply to 0 mod 2.Therefore, either b or d is 0 mod 2. Let's consider possible pairs (b, d):If b = 0, then d can be 0 or 1.But if b = 0, then the constant term is 0* d = 0. Similarly, if d = 0, then b*0 = 0.But let's see possible factorizations:If we take (x^2)(x^2), then constants are 0 and 0. Then, coefficients would be:x^4 + 0x^3 + 0x^2 + 0x + 0. Which matches x^4. So that works.Alternatively, suppose (x^2 + x)(x^2 + x). Then, multiplying them out:x^4 + 2x^3 + x^2. But mod 2, 2x^3 ≡ 0, so it's x^4 + x^2. Which is not equal to x^4, so that doesn't work.Alternatively, (x^2 +1)(x^2 +1) mod 2. Multiply out: x^4 + 2x^2 +1 ≡ x^4 + 1 mod 2. Not equal to x^4.Or (x^2 +x +1)(x^2 +x +1) mod 2. That would be x^4 + 2x^3 + 3x^2 + 2x +1 ≡ x^4 + x^2 +1 mod 2. Not x^4.Alternatively, (x^2)(x^2 + x +1). Then, product is x^4 + x^3 + x^2. Which is x^4 +x^3 +x^2 mod 2, not equal to x^4.So the only possible factorization mod 2 is (x^2)(x^2). So if the original polynomial factors into two quadratics over integers, then mod 2 it must factor as (x^2)(x^2). Therefore, each quadratic factor must reduce to x^2 mod 2. Which would mean that in the integer factorization, the coefficients of x in both quadratics must be even, and the constant terms must be even as well. But in our original polynomial, the constant term is 2, which is even. However, if both quadratics have even constant terms, then their product would have a constant term divisible by 4. But our original polynomial has constant term 2, which is not divisible by 4. Contradiction. Therefore, the polynomial cannot factor into two quadratics with integer coefficients.Wait, that's a different argument. Let me elaborate. If we suppose that the polynomial factors as (x^2 + a x + b)(x^2 + c x + d) with integer coefficients, then mod 2, each quadratic would have to be congruent to x^2 mod 2, since that's the only factorization mod 2. Therefore, the linear coefficients a and c must be even (so that a x ≡ 0 mod 2, same with c x), and the constants b and d must be even (so that b ≡ 0 mod 2 and d ≡ 0 mod 2). Therefore, b and d are even integers, so their product b d is divisible by 4. However, the constant term of the original polynomial is 2, which is not divisible by 4. Contradiction. Hence, no such factorization exists.This seems like a more straightforward argument. So even though the previous approach by equating coefficients didn't find any solutions, this modulo 2 argument shows that such a factorization is impossible because it would require the constant term to be a multiple of 4, but it's 2.Therefore, the polynomial cannot be factored into two quadratics with integer coefficients.But to make sure, let's cross-verify both approaches.In the first approach, by assuming the factorization, we derived that the constants b and d must multiply to 2, which gives possible pairs (1,2), (2,1), (-1,-2), (-2,-1). However, according to the modulo 2 argument, both b and d must be even, but in all those pairs, either both are odd or one is even and the other is odd. Wait, actually, in the pairs (1,2) and (2,1), one is even and one is odd. In the pairs (-1,-2) and (-2,-1), similarly. But the modulo 2 argument suggests that both b and d must be even, which would make their product divisible by 4. But since the constant term is 2, which is 2 mod 4, that's impossible. Therefore, the constant term 2 cannot be expressed as a product of two even integers because their product would be divisible by 4. Hence, the contradiction.So this aligns with the first method. In the first method, all possible pairs for b and d have one even and one odd, which would mean that their product is even (2), but modulo 4, if one is even (say 2) and the other is 1, then the product is 2 mod 4. However, if both were even, the product would be 0 mod 4, which doesn't match. But in our case, the original polynomial has constant term 2, which is 2 mod 4, but if the factorization required both b and d to be even (as per the modulo 2 reduction), then the product would be 0 mod 4, contradiction. Wait, this is a bit confusing.Wait, if we follow the modulo 2 argument, in order for the quadratics to reduce to x^2 mod 2, their constant terms must be 0 mod 2, i.e., even. Therefore, both b and d must be even. But then bd would be divisible by 4, but the original constant term is 2, which is not divisible by 4. Therefore, impossible.But in the first approach, we considered possible pairs where bd = 2, which are (1,2), (2,1), (-1,-2), (-2,-1). Each of these pairs has one even and one odd number. Therefore, in such cases, one of b or d is even, and the other is odd. However, according to the modulo 2 argument, both need to be even. Therefore, these cases cannot happen, hence there's a contradiction.Therefore, the combination of these two approaches confirms that there are no integer solutions. Thus, the polynomial cannot be factored as required.Another angle: suppose we use the rational root theorem. If the polynomial has a linear factor with integer coefficients, then it would have a root which is a divisor of 2. Testing possible roots: x = ±1, ±2.Let's evaluate the polynomial at x=1: 1 + 2 + 2 + 2 = 7 ≠0x=-1: 1 + 2 - 2 + 2 = 3 ≠0x=2: 16 + 8 + 4 + 2 = 30 ≠0x=-2: 16 + 8 -4 + 2 = 22 ≠0Therefore, there are no linear factors with integer coefficients. So if the polynomial factors, it must be as a product of two quadratics. But we already showed that's impossible via the above methods.Alternatively, suppose someone tries to factor it over the reals. But since we need integer coefficients, that's a different question. But just to check, maybe the polynomial is irreducible over integers but reducible over reals. But that's beyond our current concern.Alternatively, maybe considering the polynomial modulo another prime, like 3. Let's see. Let's compute the polynomial mod 3.Original polynomial: x^4 + 2x^2 + 2x + 2.Modulo 3, coefficients become:x^4 + 2x^2 + 2x + 2.If we can show that it's irreducible mod 3, then it would be irreducible over integers. Let's check if it factors into quadratics mod 3.Possible quadratic factors mod 3: There are several possibilities. Let's list all monic quadratics mod 3 and see if any multiply to give our polynomial.But maybe a better approach is to check for roots mod 3. If it has a root, then it would factor into a linear and a cubic. But since it has no integer roots, maybe it has roots mod 3.Testing x=0: 0 + 0 + 0 + 2 = 2 ≠0 mod3x=1: 1 + 2 + 2 + 2 = 7 ≡1 mod3≠0x=2: 16 + 8 +4 +2=30≡0 mod3. Wait, 16 mod3 is 1, 8 mod3 is 2, 4 mod3 is 1, 2 mod3 is 2. So 1 + 2 +1 +2=6≡0 mod3. So x=2 is a root mod3.Therefore, the polynomial factors as (x - 2)(something) mod3. But since we are working mod3, x - 2 ≡x +1 mod3. So divide the polynomial by x +1 mod3.Let's perform polynomial division mod3.Divide ( x^4 + 2x^2 + 2x + 2 ) by x +1 mod3.Using synthetic division:Coefficients: 1 (x^4), 0 (x^3), 2 (x^2), 2 (x), 2 (constant)Root x = -1 (which is 2 mod3). So we use synthetic division with root -1:Bring down 1Multiply by -1: -1 mod3 = 2Add to next coefficient: 0 + 2 = 2Multiply by -1: 2*(-1)= -2 mod3=1Add to next coefficient: 2 +1=3≡0 mod3Multiply by -1:0*(-1)=0Add to next coefficient: 2 +0=2Multiply by -1:2*(-1)= -2≡1 mod3Add to last coefficient: 2 +1=3≡0 mod3So the result is x^3 + 2x^2 +0x +2, with remainder 0. Therefore, the polynomial factors as (x +1)(x^3 + 2x^2 + 2) mod3.Now, check if the cubic factors further. Let's check if it has a root mod3. The cubic is x^3 + 2x^2 + 2.Test x=0:0 +0 +2=2≠0x=1:1 +2 +2=5≡2≠0x=2:8 + 8 +2=18≡0 mod3. So x=2 is a root.Thus, factor as (x -2)(quadratic) mod3. Again, x -2 ≡x +1 mod3. So divide the cubic by x +1.Using synthetic division again on x^3 + 2x^2 +0x +2:Coefficients:1,2,0,2Root x=-1:Bring down 1Multiply by -1: -1Add to next coefficient:2 + (-1)=1Multiply by -1: -1Add to next coefficient:0 + (-1)= -1≡2 mod3Multiply by -1: -2≡1 mod3Add to last coefficient:2 +1=3≡0 mod3So the cubic factors as (x +1)(x^2 +x +2) mod3.Therefore, the original polynomial factors mod3 as (x +1)^2 (x^2 +x +2) mod3. Since x^2 +x +2 is irreducible mod3 (check if it has roots: x=0 gives 2≠0, x=1 gives 1+1+2=4≡1≠0, x=2 gives4+2+2=8≡2≠0), so it's irreducible.Therefore, the polynomial factors into a product of a square of a linear factor and an irreducible quadratic mod3. But in the original problem, if the polynomial factored into two quadratics over integers, then mod3 it would have to factor into two quadratics or a linear times a cubic. But we found that mod3, it factors into (x +1)^2 (x^2 +x +2). So to factor into two quadratics mod3, we would need to pair the linear factors differently. However, since there's a repeated linear factor, but we only have two linear factors and one quadratic, we can't combine them into two quadratics. The only way to factor into two quadratics would be if the original polynomial factored as (x +1)^2 times (x^2 +x +2), which is a cubic times a linear, but not two quadratics. Therefore, the polynomial cannot be factored into two quadratics mod3, which implies that it cannot be factored into two quadratics over the integers.This gives another angle of contradiction. Since if it factored into two quadratics over integers, then mod3 it would have to factor into two quadratics or a product involving a linear factor. But in reality, it factors into a linear squared times an irreducible quadratic, which means that over integers, it cannot be written as a product of two quadratics. Therefore, another contradiction.So combining all these methods: coefficient comparison, modulo 2 argument, modulo 3 argument, and checking for linear factors, we can confidently conclude that the polynomial ( x^4 + 2x^2 + 2x + 2 ) cannot be factored into the product of two quadratics with integer coefficients.

✅:To demonstrate that ( x^4 + 2x^2 + 2x + 2 ) cannot be factored into the product of two quadratics with integer coefficients, we analyze the polynomial through multiple systematic approaches: 1. Coefficient Comparison MethodAssume the polynomial factors as ( (x^2 + ax + b)(x^2 + cx + d) ) with integer coefficients. Expanding and equating coefficients yields:1. ( a + c = 0 ) (since the ( x^3 )-term is absent).2. ( ac + b + d = 2 ).3. ( ad + bc = 2 ).4. ( bd = 2 ).From ( a + c = 0 ), substitute ( c = -a ). The system simplifies to:- ( -a^2 + b + d = 2 ),- ( a(d - b) = 2 ),- ( bd = 2 ).Possible integer pairs for ( (b, d) ) are ( (1, 2), (2, 1), (-1, -2), (-2, -1) ). Testing each case:- Case 1: ( b = 1, d = 2 ): ( a = 2 ), but ( -a^2 + b + d = -1 neq 2 ).- Case 2: ( b = 2, d = 1 ): ( a = -2 ), but ( -a^2 + b + d = -1 neq 2 ).- Case 3: ( b = -1, d = -2 ): ( a = -2 ), but ( -a^2 + b + d = -7 neq 2 ).- Case 4: ( b = -2, d = -1 ): ( a = 2 ), but ( -a^2 + b + d = -7 neq 2 ).No valid solutions exist, indicating no such factorization. 2. Modular Arithmetic (Modulo 2)Reducing the polynomial modulo 2 gives ( x^4 ). If it factored into quadratics, each quadratic must reduce to ( x^2 mod 2 ), requiring even linear and constant terms. Thus, constants ( b ) and ( d ) must be even. However, ( bd = 2 ) implies ( b ) and ( d ) are ( pm1, pm2 ), but two even numbers multiply to ( 4k ), contradicting the constant term 2 (not divisible by 4). Contradiction. 3. Modular Arithmetic (Modulo 3)Testing roots modulo 3:- ( x = 2 ) is a root, leading to factorization ( (x + 1)^2(x^2 + x + 2) mod 3 ). The quadratic ( x^2 + x + 2 ) is irreducible mod 3. Hence, the polynomial cannot split into two quadratics modulo 3, contradicting a potential integer factorization. 4. Rational Root TheoremTesting possible integer roots ( pm1, pm2 ) confirms no linear factors. Thus, if reducible, it must factor into quadratics, which is disproven above. ConclusionThrough coefficient analysis, modular arithmetic, and root testing, we conclusively show that ( x^4 + 2x^2 + 2x + 2 ) cannot be factored into two quadratics with integer coefficients.Final Answerboxed{x^4 + 2x^2 + 2x + 2 text{ cannot be factored into the product of two quadratics with integer coefficients.}}

🔑:Let us assume, for the sake of contradiction, that the polynomial ( x^4 + 2x^2 + 2x + 2 ) can indeed be factored into the product of two quadratic polynomials with integer coefficients. Specifically, let us write:[ x^4 + 2x^2 + 2x + 2 = (x^2 + ax + b)(x^2 + cx + d) ]1. Coefficient of ( x^3 ): - The coefficient of ( x^3 ) in ( x^4 + 2x^2 + 2x + 2 ) is ( 0 ). Therefore, we must have: [ (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd ] Matching the coefficients of ( x^3 ): [ a + c = 0 quad Rightarrow quad c = -a ]2. Coefficient of ( x ): - The coefficient of ( x ) in ( x^4 + 2x^2 + 2x + 2 ) is ( 2 ). Therefore, we have: [ ad + bc = 2 ] Substituting ( c = -a ): [ ad + b(-a) = 2 quad Rightarrow quad ad - ab = 2 quad Rightarrow quad a(d - b) = 2 ]3. Coefficient of constant term: - The constant term of ( x^4 + 2x^2 + 2x + 2 ) is ( 2 ). Therefore, we need: [ bd = 2 ] We must find integer pairs ((b, d)) such that their product is 2. These pairs are: [ (b, d) = (1, 2), (-1, -2), (2, 1), (-2, -1) ]4. Coefficient of ( x^2 ): - The coefficient of ( x^2 ) in ( x^4 + 2x^2 + 2x + 2 ) is ( 2 ). Therefore, from: [ ac + b + d = 2, ] substituting ( c = -a ): [ (-a^2) + b + d = 2 quad Rightarrow quad -a^2 + b + d = 2 quad Rightarrow quad a^2 = b + d - 2 ]Now, we shall test the possible values of ( (b, d) ): Case 1: ((b, d) = (1, 2)) 1. ( a^2 = 1 + 2 - 2 = 1 ) - Therefore, ( a^2 = 1 ) implies ( a = 1 ) or ( a = -1 ).2. Let ( a = 1 ), then: [ a(d - b) = 2 Rightarrow 1(2 - 1) = 1 neq 2 ]3. Let ( a = -1 ), then: [ a(d - b) = 2 Rightarrow (-1)(2 - 1) = -1 neq 2 ] Case 2: ((b, d) = (-1, -2)) 1. ( a^2 = -1 - 2 - 2 = -5 ) (Impossible because ( a^2 ) needs to be non-negative) Case 3: ((b, d) = (2, 1)) 1. ( a^2 = 2 + 1 - 2 = 1 ) - Therefore, ( a^2 = 1 ) implies ( a = 1 ) or ( a = -1 ).2. Let ( a = 1 ), then: [ a(d - b) = 2 Rightarrow 1(1 - 2) = -1 neq 2 ]3. Let ( a = -1 ), then: [ a(d - b) = 2 Rightarrow (-1)(1 - 2) = 1 neq 2 ] Case 4: ((b, d) = (-2, -1)) 1. ( a^2 = -2 - 1 - 2 = -5 ) (Impossible because ( a^2 ) needs to be non-negative)Given that we have exhausted all possible pairs ((b, d)) and all lead to contradictions or impossible cases, we conclude that our assumption is wrong and the polynomial ( x^4 + 2x^2 + 2x + 2 ) cannot be factored into the product of two quadratics with integer coefficients.[boxed{text{No factorization into quadratics with integer coefficients}}]

🧠:Okay, so I need to find the total surface area of a regular triangular pyramid (which is a tetrahedron, right?) where the base is an equilateral triangle with side length 'a', and the dihedral angle at the base is 60 degrees. Hmm, let me think. Total surface area would include the base area plus the areas of the three triangular faces. Since it's a regular triangular pyramid, the three lateral faces should be congruent. But wait, is it a regular tetrahedron? Because in a regular tetrahedron, all faces are equilateral triangles, and all dihedral angles are the same. But here, the dihedral angle at the base is given as 60 degrees. If it's a regular tetrahedron, the dihedral angle is actually arccos(1/3) which is approximately 70.5 degrees, so this isn't a regular tetrahedron. Therefore, the lateral faces might not be equilateral triangles. So, the pyramid has a regular base (equilateral triangle) but the lateral edges might not be equal. Hmm, okay.First, let me recall that a dihedral angle is the angle between two planes. In this case, the dihedral angle at the base is the angle between the base and one of the lateral faces. So, each of the three dihedral angles between the base and the lateral faces is 60 degrees. The total surface area is the sum of the base area and the three lateral face areas.The base area is straightforward. Since the base is an equilateral triangle with side length 'a', its area is (√3/4)a². Now, the challenge is to find the area of each lateral face. For that, I need to find the slant height of the pyramid or the height of the triangular face. Wait, the lateral faces are triangles, but since the base is regular, they should be isosceles triangles. To find their area, I need the base of each triangle (which is 'a') and their height (the slant height). Alternatively, if I can find the length of the other two sides of the lateral triangle (the edges from the apex to the base vertices), maybe I can compute the area. But perhaps it's easier to find the slant height.Alternatively, maybe I can relate the dihedral angle to the height of the pyramid. Let me think. The dihedral angle between the base and a lateral face is 60 degrees. How is the dihedral angle related to other dimensions of the pyramid?I remember that the dihedral angle can be related to the angle between the normal vectors of the two planes. Alternatively, using some trigonometric relationships. Let me visualize the pyramid. The base is an equilateral triangle. Let's denote the apex of the pyramid as point S, and the base triangle as ABC. The dihedral angle along edge AB is 60 degrees. So, this angle is formed between the base ABC and the lateral face SAB.To find the dihedral angle, we can consider the angle between the two planes. One way to compute this is by finding the angle between the normals of the two planes. Alternatively, using the formula involving the height of the pyramid and the radius of the base's incircle or something like that.Wait, another approach: the dihedral angle can be related to the angle between the lateral face and the base. If I can find the height of the pyramid, then I can relate it to the dihedral angle. Let me recall that in a pyramid, the dihedral angle between the base and a lateral face can be found using the formula involving the height of the pyramid and the apothem of the base.The apothem of the base (which is an equilateral triangle) is the distance from the center of the base to the midpoint of a side. For an equilateral triangle with side length 'a', the apothem is (√3/6)a. Wait, the apothem is also equal to the radius of the inscribed circle. So, if I can find the relationship between the height of the pyramid (h) and the apothem (r = √3 a /6), then the dihedral angle θ between the base and the lateral face is given by tanθ = h / r. Is that correct?Wait, maybe. Let me visualize a cross-section through the pyramid along the plane that bisects one of the edges of the base and goes through the apex. In this cross-section, we have a triangle formed by the height of the pyramid (h), the apothem of the base (r = √3 a /6), and the slant height (the distance from the apex to the midpoint of a base edge). The angle between the base and the lateral face in this cross-section would be the dihedral angle. So, in this right triangle, the angle adjacent to the apothem is θ, so tanθ = h / r. Therefore, h = r tanθ.Given that θ is 60 degrees, so tanθ = tan60° = √3. Therefore, h = r * √3. Since r = (√3 a)/6, substituting:h = (√3 a /6) * √3 = (3a)/6 = a/2.So, the height of the pyramid is a/2. Okay, that seems manageable.Now, if the height of the pyramid is h = a/2, then we can find the slant height. Wait, the slant height (let's denote it as l) is the distance from the apex to the midpoint of a base edge. But in the cross-section, that's the hypotenuse of the right triangle with legs h and r. Wait, but in the cross-section, we already have h and r. Wait, the slant height is actually the hypotenuse of that triangle. Wait, in the cross-section, the slant height would be sqrt(h² + r²). Wait, no: hold on, the cross-section is a triangle with height h, base edge (from the center of the base to the midpoint of a side) which is r = √3 a /6, and the hypotenuse is the slant height l. Wait, but in reality, the cross-section is a triangle where h is perpendicular to the base, and r is the distance from the center to the midpoint of a side. Wait, actually, if we take the cross-section through the apex and the midpoint of a base edge, we have a triangle where the base is the apothem (r = √3 a /6), the height is h, and the hypotenuse is the slant height l. Therefore, l = sqrt(h² + r²). But we already have h = a/2 and r = √3 a /6. Let's compute l:l = sqrt( (a/2)² + ( (√3 a /6 )² ) )Compute each term:(a/2)² = a²/4(√3 a /6 )² = (3a²)/36 = a²/12So, l = sqrt( a²/4 + a²/12 ) = sqrt( (3a² + a²)/12 ) = sqrt(4a²/12) = sqrt(a²/3) = a/√3Hmm, so the slant height is a/√3. Wait, but the slant height is the height of the lateral face, which is an isosceles triangle with base 'a'. So, the area of each lateral face would be (base * slant height)/2 = (a * (a/√3))/2 = a²/(2√3). Since there are three lateral faces, total lateral surface area is 3 * a²/(2√3) = (3a²)/(2√3) = (a²√3)/2.Adding the base area, which is (√3/4)a², so total surface area is (√3/4)a² + (√3/2)a² = (√3/4 + 2√3/4)a² = (3√3/4)a².Wait, but let me double-check this. If the slant height is a/√3, then the lateral face is a triangle with base a and height a/√3, so area is (a * a/√3)/2 = a²/(2√3). Yes, that's correct. Then three of them would be 3*(a²/(2√3)) = (3a²)/(2√3). Rationalizing the denominator: (3a²)/(2√3) * (√3/√3) = (3a²√3)/(6) = (a²√3)/2. Then base area is (√3/4)a². Adding them together: (√3/4 + √3/2)a² = (√3/4 + 2√3/4)a² = (3√3/4)a². So total surface area is (3√3/4)a². Hmm, that seems reasonable.Wait, but let me confirm if the slant height is indeed a/√3. Let's re-derive it. We had h = a/2, and r = √3 a /6. Then slant height l is sqrt(h² + r²) = sqrt( (a²/4) + ( (3a²)/36 )) = sqrt( a²/4 + a²/12 ). Converting to twelfths: 3a²/12 + a²/12 = 4a²/12 = a²/3. Therefore, sqrt(a²/3) = a/√3. Yes, that's correct.Alternatively, maybe there's another way to compute the lateral face area. Since the dihedral angle is 60 degrees, perhaps using trigonometry related to the dihedral angle. Let me think. The dihedral angle is the angle between the two planes. So, between the base and the lateral face. How can we relate this angle to the slant height or the height of the pyramid?Alternatively, if we consider the dihedral angle, which is 60 degrees, and the angle between the lateral face and the base. If we take a cross-sectional view as before, the dihedral angle is the angle between the base and the lateral face. In that cross-section, the angle between the base and the lateral face is 60 degrees. If we have a right triangle in that cross-section with height h and base r, then the angle between the hypotenuse (slant height) and the base is 60 degrees. Therefore, tan(theta) = h / r, which is what we had earlier. So, tan(60°) = h / r => h = r tan(60°). Since tan(60°) is √3, h = r√3. Given that r is the apothem of the base, which is (√3 a)/6, then h = (√3 a /6) * √3 = (3a)/6 = a/2. So, that's consistent with our earlier result.Therefore, the slant height l is sqrt(h² + r²) = sqrt( (a/2)^2 + (√3 a /6)^2 ) = a/√3, as before.So, each lateral face is an isosceles triangle with base a and height a/√3. So, area of each lateral face is (a * a/√3)/2 = a²/(2√3). Three of them give (3a²)/(2√3) = (√3 a²)/2. Adding the base area (√3 a² /4) gives total surface area (3√3 a²)/4.But let me confirm if the lateral faces are indeed triangles with height a/√3. Alternatively, maybe the lateral edges are of some length, and we can compute the area using those. Let me consider the edge length from the apex to a base vertex. Let's denote the apex as S, and the base triangle as ABC. The edge SA, for example. Let's compute the length SA. To find SA, we can consider the right triangle from the apex S to the base vertex A, with height h = a/2. But the distance from the center of the base to a vertex is the circumradius of the base triangle. For an equilateral triangle, the circumradius R is (a)/√3. So, the distance from the center of the base to vertex A is R = a/√3. Then, the edge SA is the hypotenuse of a right triangle with legs h = a/2 and R = a/√3. So, SA = sqrt( (a/2)^2 + (a/√3)^2 ) = sqrt( a²/4 + a²/3 ) = sqrt( (3a² + 4a²)/12 ) = sqrt(7a²/12 ) = (a/2)sqrt(7/3). Wait, that complicates things. But if SA is the edge length, then the lateral face triangle SAB has sides SA, SB, and AB = a. So, if SA and SB are both (a/2)sqrt(7/3), then the triangle SAB is an isoceles triangle with sides (a/2)sqrt(7/3), (a/2)sqrt(7/3), and a. The area of such a triangle can be computed using Heron's formula. Let's try that.First, compute semi-perimeter: s = ( (a/2)sqrt(7/3) + (a/2)sqrt(7/3) + a ) / 2 = ( a sqrt(7/3) + a ) / 2 = a ( sqrt(7/3) + 1 ) / 2.Then, area = sqrt( s(s - a sqrt(7/3)/2 )(s - a sqrt(7/3)/2 )(s - a ) )This seems messy. Maybe there's a better way. Alternatively, since we know the height of the triangle is a/√3, as computed earlier, then area is (base * height)/2 = (a * (a/√3))/2 = a²/(2√3). Which is the same as before. Therefore, Heron's formula should give the same result. Let me check.Let me compute the terms step by step. Let’s denote SA = SB = l = (a/2)sqrt(7/3). Then AB = a.First, compute semi-perimeter: s = (l + l + a)/2 = (2l + a)/2 = l + a/2.Then area squared is s(s - l)(s - l)(s - a). So,Area² = s(s - l)^2(s - a)Substituting s = l + a/2,Area² = (l + a/2)( (l + a/2 - l )^2 )( l + a/2 - a )= (l + a/2)( (a/2)^2 )( l - a/2 )= (l + a/2)(l - a/2)(a²/4)= (l² - (a/2)^2)(a²/4 )Now, l = (a/2)sqrt(7/3), so l² = (a²/4)(7/3) = 7a²/12Therefore, l² - (a/2)^2 = 7a²/12 - a²/4 = 7a²/12 - 3a²/12 = 4a²/12 = a²/3Therefore, Area² = (a²/3)(a²/4 ) = a⁴/12Therefore, Area = sqrt(a⁴/12 ) = a²/(2√3 )Which matches our previous result. So, even though calculating via edge lengths seems complicated, it still gives the same area. Therefore, the area of each lateral face is indeed a²/(2√3 ), so three of them give (√3 a²)/2, and adding the base area gives (3√3 a²)/4.Therefore, the total surface area is (3√3 /4 )a².But let me just verify once more if my reasoning about the dihedral angle is correct. Dihedral angle is 60 degrees, which led me to compute the pyramid height as a/2. Then, from there, I calculated slant height and the areas. Alternatively, is there another way to compute the lateral face area directly from the dihedral angle?I remember that the dihedral angle can be related to the angle between the faces. So, if the dihedral angle is θ, then there's a relationship between θ, the angle of the lateral face at the apex, and other dimensions. Alternatively, using some trigonometric identities involving the dihedral angle.Alternatively, consider the dihedral angle formula. The dihedral angle between two faces can be calculated using the formula:cosθ = (cosα - cosβ cosγ ) / (sinβ sinγ )where α, β, γ are the face angles at the common edge. But maybe this is more complicated.Alternatively, for a pyramid, the dihedral angle between the base and a lateral face can be related to the angle between the slant edge and the base. Wait, maybe not. Let me think.Alternatively, consider the normal vectors to the base and the lateral face. The dihedral angle is the angle between these normals. The normal to the base is vertical (along the pyramid's height), and the normal to the lateral face can be found using the orientation of the lateral face.Alternatively, consider the following approach: Take the base triangle ABC and apex S. The dihedral angle along edge AB is 60 degrees. To find this dihedral angle, we can compute the angle between the two planes: the base ABC and the lateral face SAB.To compute this angle, we can find two vectors lying on each plane, perpendicular to the edge AB. For the base ABC, a vector perpendicular to AB can be the normal vector of the base, which is vertical (since the base is horizontal). For the lateral face SAB, a vector perpendicular to AB can be found by taking the cross product of vectors SA and SB, but perhaps it's easier to consider another approach.Alternatively, take a point on edge AB, say midpoint M of AB. Then, in the base ABC, the normal vector is vertical (upwards). In the lateral face SAB, the normal vector can be found by taking the cross product of vectors SM and SA (or SM and SB). Since SM is the median of the face SAB, but perhaps this is getting too involved.Alternatively, since we already found the height of the pyramid and the slant height, and verified the area through multiple methods, perhaps the answer is correct.But let me cross-verify the total surface area with another approach. Suppose we consider the lateral face area as 3*(1/2)*a*l, where l is the slant height. If l is a/√3, then total lateral area is 3*(1/2)*a*(a/√3 ) = (3a²)/(2√3 ) = (√3 a²)/2. Adding the base area (√3 a² /4 ), total is (3√3 a² )/4. So, this seems consistent.Alternatively, if I consider that in a regular tetrahedron, all dihedral angles are arccos(1/3) ≈ 70.5°, but here the dihedral angle is 60°, which is smaller, meaning the pyramid is "flatter". Therefore, the lateral faces are more inclined, which would result in a shorter slant height. Wait, but in our calculation, the slant height is a/√3 ≈ 0.577a, whereas in a regular tetrahedron with edge length a, the slant height (which is the same as the edge length) would be a, but in our case, the slant height is indeed shorter. Wait, but in a regular tetrahedron, all edges are equal, but here, the base edges are a, and the lateral edges are different. So, perhaps the slant height here is indeed shorter. Therefore, the calculation seems plausible.Therefore, after thorough analysis, I believe the total surface area is (3√3 /4 )a².Final AnswerThe total surface area is boxed{dfrac{3sqrt{3}}{4}a^2}.

✅:To find the total surface area of a regular triangular pyramid with a base side length (a) and a dihedral angle at the base of (60^circ):1. Base Area Calculation: - The base is an equilateral triangle with side length (a). - Area of the base: (frac{sqrt{3}}{4}a^2).2. Height of the Pyramid: - The dihedral angle at the base is (60^circ). - The apothem of the base (distance from the center to the midpoint of a side) is (frac{sqrt{3}a}{6}). - Using the relationship (tan(60^circ) = frac{h}{text{apothem}}), we find the height (h) of the pyramid: [ h = frac{sqrt{3}a}{6} times sqrt{3} = frac{a}{2} ]3. Slant Height Calculation: - The slant height (l) is the distance from the apex to the midpoint of a base edge. - Using the Pythagorean theorem with the height (h) and apothem: [ l = sqrt{left(frac{a}{2}right)^2 + left(frac{sqrt{3}a}{6}right)^2} = sqrt{frac{a^2}{4} + frac{a^2}{12}} = sqrt{frac{a^2}{3}} = frac{a}{sqrt{3}} ]4. Lateral Face Area Calculation: - Each lateral face is an isosceles triangle with base (a) and height (frac{a}{sqrt{3}}). - Area of one lateral face: (frac{1}{2} times a times frac{a}{sqrt{3}} = frac{a^2}{2sqrt{3}}). - Total area of three lateral faces: (3 times frac{a^2}{2sqrt{3}} = frac{sqrt{3}a^2}{2}).5. Total Surface Area: - Sum of the base area and the lateral faces area: [ text{Total Surface Area} = frac{sqrt{3}}{4}a^2 + frac{sqrt{3}}{2}a^2 = frac{sqrt{3}}{4}a^2 + frac{2sqrt{3}}{4}a^2 = frac{3sqrt{3}}{4}a^2 ]The total surface area of the regular triangular pyramid is (boxed{dfrac{3sqrt{3}}{4}a^2}).

🔑:1. Identify given information and relationships: - The side length of the base of the regular triangular pyramid (tetrahedron) is ( a ). - The dihedral angle at the base is ( 60^circ ).2. Calculate the radius of the inscribed circle in the base triangle: - For an equilateral triangle ( ABC ) with side ( a ), the radius ( r ) of the inscribed circle can be obtained using the formula: [ r = frac{a sqrt{3}}{6} ]3. Draw diagrams and identify segments: - Consider a perpendicular dropped from the apex ( S ) to the center ( H ) of the base ( triangle ABC ). This line ( SH ) represents the height of the pyramid. - Apothem ( SK ) is drawn perpendicular from ( S ) to the midpoint ( K ) of one side of the base ( triangle ABC ).4. Determine the apothem ( (SK) ). - The apothem of the pyramid is calculated by recognizing the relationship with the radius ( r ) and angle ( 60^circ ): [ SK = frac{r}{cos 60^circ} = frac{a sqrt{3}}{6} cdot frac{1}{frac{1}{2}} = frac{a sqrt{3}}{3} ]5. Calculate the lateral surface area: - The lateral surface area of the pyramid is given by the perimeter of the base multiplied by the apothem, divided by 2 (since each triangular face of the pyramid is included): [ S_{text{lateral}} = frac{1}{2} cdot text{perimeter} cdot l ] - The perimeter ( p ) of the equilateral triangle base is: [ p = 3a ] - Thus, [ S_{text{lateral}} = frac{1}{2} cdot 3a cdot frac{a sqrt{3}}{3} = frac{a sqrt{3}}{3} cdot frac{3a}{2} = frac{a^2 sqrt{3}}{2} ]6. Calculate the base area: - The area ( S_{text{base}} ) of an equilateral triangle is given by: [ S_{text{base}} = frac{sqrt{3}}{4}a^2 ]7. Total surface area: - Combine the lateral surface area and the base area to get the total surface area: [ S_{text{total}} = S_{text{base}} + S_{text{lateral}} ] [ S_{text{total}} = frac{sqrt{3}}{4}a^2 + frac{a^2 sqrt{3}}{2} ]8. Simplify the expression: [ S_{text{total}} = frac{a^2 sqrt{3}}{4} + frac{2a^2 sqrt{3}}{4} = frac{3a^2 sqrt{3}}{4} ] Conclusion:[boxed{frac{3a^2 sqrt{3}}{4}}]

🧠:Okay, so I need to figure out how many white patches there are on a soccer ball that's made up of 32 patches total—white hexagons and black pentagons. The problem says each black patch borders only white ones, and each white patch borders three black and three white ones. Alright, let's break this down step by step.First, let's note that there are two types of patches: black pentagons and white hexagons. The total number is 32. So, if I let B be the number of black pentagons and W be the number of white hexagons, then B + W = 32. That's one equation. My goal is to find W.Now, the problem also gives information about how the patches border each other. Each black patch (which is a pentagon) borders only white patches. Since it's a pentagon, each black patch has 5 edges, right? Each edge is shared with a neighboring patch. So, each black pentagon borders 5 white hexagons. However, when we count the total number of edges adjacent to black patches, we have to consider that each edge is shared between two patches. But wait, maybe we need to count the number of adjacencies or something else.Alternatively, maybe we can use Euler's formula here. Euler's formula relates the number of vertices, edges, and faces in a convex polyhedron. The formula is V - E + F = 2, where V is vertices, E is edges, and F is faces. In this case, the soccer ball is a polyhedron with 32 faces (the patches). So, F = 32. If we can find the number of edges and vertices, we might be able to use this formula. But how?Wait, each face is a polygon. The black ones are pentagons and the white ones are hexagons. Each pentagon has 5 edges and each hexagon has 6 edges. But each edge is shared by two faces. So, the total number of edges E can be calculated as (5B + 6W)/2. Similarly, each vertex is where three patches meet—since in a typical soccer ball pattern, each vertex is the meeting point of two hexagons and one pentagon. Wait, is that always the case? Hmm.Wait, the problem says each white hexagon borders three black and three white patches. So, for a white hexagon, each of its six edges borders either a black or white patch. Specifically, three of its edges are adjacent to black pentagons, and three are adjacent to white hexagons. So, each white hexagon has three edges adjacent to black pentagons and three edges adjacent to white hexagons.Similarly, each black pentagon borders only white hexagons. So, each black pentagon's five edges are adjacent to white hexagons. Now, let's try to count the total number of edges adjacent to black pentagons. Each black pentagon has 5 edges, so total edges from black pentagons would be 5B. However, each of these edges is shared with a white hexagon. Since each white hexagon has 3 edges adjacent to black pentagons, the total number of edges adjacent to black pentagons is also 3W. Therefore, 5B = 3W. Is that correct?Yes, because each edge of a black pentagon is shared with exactly one white hexagon, and each white hexagon contributes 3 edges to black pentagons. So, 5B (total edges from black pentagons) must equal 3W (total edges from white hexagons adjacent to black pentagons). So, we have the equation 5B = 3W.We already have B + W = 32. So now, we can solve these two equations. Let me write them out:1. B + W = 322. 5B = 3WFrom the second equation, we can express B in terms of W: B = (3/5)W. Plugging that into the first equation:(3/5)W + W = 32Combine the terms:(3/5)W + (5/5)W = (8/5)W = 32Multiply both sides by 5/8:W = 32 * (5/8) = 20So, W = 20. Then B = 32 - 20 = 12.Wait, but let's confirm this with Euler's formula to check if this is consistent. Let's see. If there are 12 pentagons and 20 hexagons.Total number of edges E = (5*12 + 6*20)/2 = (60 + 120)/2 = 180/2 = 90 edges.Now, how many vertices V? Each pentagon has 5 vertices, each hexagon has 6. But each vertex is shared by multiple faces. In the soccer ball pattern, each vertex is where two hexagons and one pentagon meet. Wait, but is that the case here? Let's think.Each white hexagon borders three black pentagons and three white hexagons. So, for the vertices, each edge between a pentagon and a hexagon contributes two vertices (the endpoints of the edge). Wait, maybe not. Let's think differently.Alternatively, each edge is shared by two faces, and each vertex is shared by three faces. Let me use the formula:In Euler's formula, V - E + F = 2.We have F = 32, E = 90, so V = E - F + 2 = 90 - 32 + 2 = 60.But let's compute V another way. Each face contributes a certain number of vertices, but each vertex is shared by multiple faces. Each pentagon has 5 vertices, each hexagon has 6. So total vertices counted per face would be 12*5 + 20*6 = 60 + 120 = 180. But each vertex is shared by how many faces? In a typical soccer ball (truncated icosahedron), each vertex is shared by three faces: two hexagons and one pentagon. Is that the case here?Given that each white hexagon borders three pentagons and three hexagons, so each edge of a hexagon is either adjacent to a pentagon or another hexagon. So, the vertices where a hexagon meets a pentagon would be shared between the hexagon, pentagon, and another hexagon? Wait, maybe.Alternatively, let's compute the number of vertices. If each vertex is where three edges meet, then each vertex is shared by three faces. The total number of vertices can be calculated as (number of edges * 2) / 3, because each edge has two vertices, and each vertex is on three edges. Wait, not sure. Wait, each edge has two vertices, so total vertices if each vertex is degree three would be (2E)/3. Let's check:If E = 90, then (2*90)/3 = 60. Which matches the earlier result from Euler's formula. So, V = 60.So that checks out. So the numbers are consistent with Euler's formula. Therefore, the calculation seems correct.But let's just go through another angle to confirm. If there are 12 pentagons, each with 5 edges, and each edge is shared with a hexagon. So, 12*5 = 60 edges adjacent to pentagons. Each hexagon has 3 edges adjacent to pentagons, so total edges adjacent to pentagons from hexagons is 20*3 = 60. Which matches. Therefore, the equations are consistent.So, the number of white hexagons is 20. So, the answer should be 20.Wait, but just to make sure, let me think again. So, B = 12, W = 20.Each black pentagon borders 5 white hexagons. So, each pentagon is adjacent to 5 hexagons. So, total adjacencies from pentagons to hexagons is 12*5 = 60. But each hexagon is adjacent to 3 pentagons, so total adjacencies from hexagons to pentagons is 20*3 = 60. So, they match. Therefore, this is consistent.Also, each white hexagon borders three white hexagons. So, each white hexagon has three edges adjacent to other white hexagons. So, total adjacencies between white hexagons is 20*3 = 60. Since each adjacency is between two hexagons, the total number of edges between white hexagons is 60/2 = 30. Similarly, total edges from white hexagons are 20*6 = 120. But each hexagon's edges are split into 3 adjacent to pentagons and 3 adjacent to hexagons. So, 3*20 = 60 edges adjacent to pentagons and 3*20 = 60 edges adjacent to hexagons. But these 60 edges adjacent to hexagons are shared between two hexagons each, so total unique edges between hexagons are 30, which matches the 60/2 = 30.Therefore, all the counts check out. The total edges would be edges adjacent to pentagons (60) plus edges adjacent to hexagons (30), totaling 90, which matches our previous calculation. So, everything is consistent.Therefore, the number of white patches is 20. I think that's solid. The key was setting up the equation based on the edges each type contributes and then solving the system. Using Euler's formula to confirm adds another layer of verification. So, yeah, 20 white hexagons.

✅:The soccer ball is composed of 32 patches: black pentagons (B) and white hexagons (W). We derive the solution as follows:1. Total Patches: ( B + W = 32 ).2. Edge Relationships: - Each black pentagon borders 5 white hexagons, contributing ( 5B ) edges. - Each white hexagon borders 3 black pentagons, contributing ( 3W ) edges. Since these counts represent the same edges, we equate them: ( 5B = 3W ).3. Solving the System: From ( 5B = 3W ), express ( B = frac{3}{5}W ). Substitute into ( B + W = 32 ): [ frac{3}{5}W + W = 32 implies frac{8}{5}W = 32 implies W = 20. ]4. Verification via Euler’s Formula: - Total edges: ( frac{5B + 6W}{2} = frac{5(12) + 6(20)}{2} = 90 ). - Vertices: ( V = E - F + 2 = 90 - 32 + 2 = 60 ). - Consistency check: Each vertex is shared by 3 edges, so ( V = frac{2E}{3} = 60 ), confirming validity.Final Answer: There are (boxed{20}) white patches.

🔑:1. Let us denote the number of white patches by ( x ). 2. Then the number of black patches will be ( 32 - x ).3. Given that each white patch borders 3 black patches, the total number of boundaries between white and black patches can be expressed as ( 3 cdot x ).4. Similarly, each black patch borders 5 white patches, so the total number of boundaries between white and black patches from the perspective of the black patches is ( 5 cdot (32 - x) ).5. Since both expressions represent the same total number of boundaries, we set them equal to each other to form an equation: [ 3 cdot x = 5 cdot (32 - x) ]6. Expanding and simplifying the equation: [ 3x = 160 - 5x ]7. Adding ( 5x ) to both sides to combine like terms: [ 3x + 5x = 160 ]8. This simplifies to: [ 8x = 160 ]9. Solving for ( x ) by dividing both sides by 8: [ x = frac{160}{8} = 20 ] Conclusion:The number of white patches is:[boxed{20}]