🧠:Okay, so I need to figure out the largest possible value of the greatest number in a circle of twelve distinct natural numbers where one of them is 1, and any two neighboring numbers differ by either 10 or 7. Hmm, let's start by breaking down the problem step by step.First, the numbers are arranged in a circle, which means that the sequence is cyclic. Each number has two neighbors, and each pair of neighbors must differ by either 10 or 7. All numbers must be distinct natural numbers. The challenge is to arrange these numbers such that the largest number in the circle is as big as possible. Since one of the numbers is 1, I need to see how we can build up from 1 using differences of 10 or 7, and then continue around the circle, ensuring all numbers are distinct and the differences alternate appropriately.Let me think. If we start at 1, the next number can be either 1 + 10 = 11 or 1 - 10. But since we're dealing with natural numbers, 1 - 10 would be negative, which isn't allowed. So the next number after 1 must be 11. Then from 11, we can either add 10 or 7, or subtract 10 or 7. But again, we need to stay within natural numbers. So from 11, adding 10 gives 21, adding 7 gives 18; subtracting 10 gives 1, but 1 is already used, and subtracting 7 gives 4. So possible next numbers after 11 are 21, 18, or 4. But we need to choose numbers that will help maximize the largest number. So maybe we want to add as much as possible early on? Let's try that.Starting with 1, then 11. Then from 11, add 10 to get 21. Then from 21, add 10 to get 31, but wait, but each step must differ by either 10 or 7. However, if we keep adding 10, we can get large numbers, but we have to close the circle eventually. Since it's a circle of 12 numbers, we need to make sure that after 12 steps, we come back to 1. So the total sum of differences around the circle must be zero, right? Because each step is either +10, -10, +7, or -7, but when we go all the way around, the total difference should be zero. So the sum of all the differences around the circle must be a multiple of 12? Wait, no. Wait, actually, the total sum of the differences will be zero because we end up back at the starting number. So, if we denote the differences as d1, d2, ..., d12, then d1 + d2 + ... + d12 = 0. Each di is either ±10 or ±7. So the sum of these twelve differences must equal zero.Therefore, the problem reduces to finding a sequence of twelve differences, each either +10, -10, +7, or -7, such that their sum is zero, and when we apply these differences starting from 1, we get twelve distinct natural numbers, with the largest one being as big as possible.But we also need to ensure that when applying these differences, we don't repeat any numbers and that all numbers are natural. So, starting at 1, then adding or subtracting 10 or 7 each time. Hmm. Let me think about how to model this.Maybe I can model this as a graph where each node is a number, and edges represent the possible differences. Since we have a circle, it's a cyclic graph. So each node is connected to two others, one via +10 or -10, and the other via +7 or -7. But given that the numbers must be distinct, we can't revisit a number once it's been used.Alternatively, since the numbers must form a cycle, the sequence must return to 1 after 12 steps. So starting at 1, applying twelve differences (each ±10 or ±7) and ending up back at 1. Therefore, the total sum of differences is zero.So first, we need to find a combination of twelve numbers, each being ±10 or ±7, such that their sum is zero. Then, we need to arrange these differences in such a way that starting from 1, applying each difference in order, we never repeat a number, and all numbers are natural.Moreover, to maximize the largest number, we probably want to have as many positive differences (i.e., +10 and +7) as possible early in the sequence, and then negative differences (i.e., -10 and -7) later to bring the sum back to 1. However, since the sum of all differences must be zero, the number of positive and negative differences must balance out. Let's formalize this.Let’s denote:Let x be the number of +10s,y be the number of +7s,z be the number of -10s,w be the number of -7s.Then, we have:x + y + z + w = 12 (since there are 12 differences),and10x + 7y -10z -7w = 0 (since the total sum is zero).We can rewrite the second equation as:10(x - z) + 7(y - w) = 0.We need to find non-negative integers x, y, z, w satisfying these equations.Let me rearrange the second equation:10(x - z) = -7(y - w)Which implies that 10 divides the right-hand side, so 7(y - w) must be divisible by 10. Since 7 and 10 are coprime, (y - w) must be divisible by 10. Let’s denote y - w = 10k, where k is an integer. Then, substituting back:10(x - z) = -7 * 10k => x - z = -7kSo we have:x = z -7kandy = w +10kSince x, y, z, w are non-negative integers, we must have:z >=7k,w >= -10k, but since w is non-negative, this implies 10k <= w, but k can be negative. Hmm, this might get complicated. Let me think of possible values of k.Since y - w =10k, and y and w are non-negative integers, then k must be such that 10k is an integer, which it is since k is integer. Also, y = w +10k >=0, so w >= -10k. But since w >=0, this implies that -10k <=0, so k >=0. Wait, because if k were negative, then y = w +10k would require w >= -10k, but w >=0, so -10k <=0 => k >=0. Therefore, k must be non-negative.Similarly, x = z -7k. Since x >=0, z >=7k.Therefore, k must be a non-negative integer such that:z >=7k,w >=0,and x = z -7k,y = w +10k,with x + y + z + w =12.Substituting x and y:(z -7k) + (w +10k) + z + w =12Simplify:2z + 2w +3k =12So 2(z + w) +3k =12.Therefore, 3k =12 -2(z +w)Thus, 12 -2(z +w) must be divisible by 3 and non-negative.So 12 -2(z +w) >=0 => z +w <=6.Also, 12 -2(z +w) must be divisible by 3. Let’s denote m = z +w. Then 12 -2m ≡0 mod3 => 12 ≡2m mod3. Since 12 ≡0 mod3, so 0≡2m mod3 => 2m≡0 mod3 => m≡0 mod3, since 2 and 3 are coprime. Therefore, m must be a multiple of 3, and m <=6. So possible m are 0, 3, 6.Thus:Case 1: m=0. Then 3k=12-0=12 =>k=4. But then z +w=0 => z=w=0. Then x= z -7k=0 -28= -28, which is impossible, since x must be non-negative. So this case is invalid.Case 2: m=3. Then 3k=12 -6=6 =>k=2. Then z +w=3. And x= z -14, y= w +20. Since x and y must be non-negative:x= z -14 >=0 => z >=14. But z +w=3, so z <=3. Contradiction. So invalid.Case3: m=6. Then 3k=12 -12=0 =>k=0. Then z +w=6. And x=z -0=z, y=w +0=w. So the equations become:x + y + z +w =12 => z + w + z +w=12 => 2(z +w)=12 => z +w=6. Which matches.Therefore, when k=0, the equations reduce to:x = z,y =w,and x + y + z +w =12,10x +7y -10z -7w=0.But substituting x=z and y=w, the second equation becomes 10z +7w -10z -7w=0, which is 0=0. So all solutions with x=z, y=w, and x + y + z +w=12, which is x + y + x + y=12 => 2x +2y=12 =>x + y=6. Therefore, x and y can be any non-negative integers such that x + y=6.Therefore, the possible solutions are when x + y=6, z=x, w=y.So for example, if x=6, y=0, then z=6, w=0. That would mean 6 +10s, 0 +7s, 6 -10s, 0 -7s. Wait, but in terms of differences:Number of +10s =x=6,Number of +7s=y=0,Number of -10s=z=6,Number of -7s=w=0.So total differences: 6(+10) and 6(-10). But this would mean the differences alternate +10 and -10, but in some order. However, if we have six +10s and six -10s, then the total sum is zero. Similarly, if we have x=5, y=1, z=5, w=1. So five +10s, one +7, five -10s, one -7. Then total sum is 5*10 +1*7 -5*10 -1*7=0.So all these solutions are valid. Therefore, the possible combinations are any x and y where x + y=6, with z=x, w=y. So for x from 0 to 6, y=6-x.But now, the challenge is to arrange these differences in a sequence such that starting from 1, applying each difference in turn, we never repeat a number, and all numbers are natural. Also, since we want the largest possible number, we need to maximize the peak value in the sequence.To maximize the maximum number, we need to have as many large increases as possible before decreasing. However, we need to make sure that we can return to 1 after 12 steps without repeating any numbers. So the key is to cluster the positive differences (i.e., +10 and +7) early in the sequence and the negative differences (i.e., -10 and -7) later. However, since the sequence is a circle, the starting point is arbitrary. But since 1 is fixed, we have to start from there.Wait, but the circle can be traversed in either direction, but the problem states that the numbers are written in a circle, so direction is fixed. But since it's a circle, the actual sequence can be rotated, but since 1 is specified, we have to start at 1. So maybe the sequence must start with 1 and then proceed with the differences.But perhaps not. The numbers are in a circle, so the neighbors of 1 could be either 1+10=11 or 1+7=8, but 1-10 is negative, and 1-7 is negative, so the neighbors of 1 must be 11 and 8. Wait, no. The problem says "any two neighboring numbers differ either by 10 or by 7". So each pair of neighbors must differ by 10 or 7 in either direction. Therefore, from 1, the next number can be 11 or 8 (since 1+10=11, 1+7=8). Similarly, the previous number (since it's a circle) can also be 11 or 8. So 1 has two neighbors: one is 11 or 8, and the other is also 11 or 8? Wait, no. Wait, in a circle, each number has two neighbors. For number 1, one neighbor is previous and one is next. Both must differ by 10 or 7. So the previous number could be 1+10=11 or 1+7=8, and the next number could also be 11 or 8. But since all numbers must be distinct, the previous and next numbers can't both be 11 or both be 8. So 1 must be between two numbers: one is 11 and the other is 8. So for example, the sequence could be ..., 8, 1, 11, ... or ..., 11, 1, 8, ... But both 8 and 11 must be adjacent to 1.But 8 and 11 are distinct, so they can be placed next to 1. Then from 8, the next number can be 8+10=18 or 8+7=15 or 8-10= -2 (invalid) or 8-7=1 (already used). So the next number after 8 can only be 18 or 15. Similarly, from 11, the next number can be 11+10=21, 11+7=18, 11-10=1 (used), or 11-7=4. So next after 11 can be 21, 18, or 4.So depending on the direction we take from 1, we can go to 11 or 8. Let's try to build two possible paths: one going through 11 first and one going through 8 first. Let's see which one allows for a larger maximum number.First path: 1 -> 11. From 11, we want to go to the largest possible number to maximize the maximum. So 11 +10=21. Then from 21, add 10 to get 31, and so on. But we need to make sure that we can come back to 1 after 12 steps. Let's try this path:1, 11, 21, 31, 41, 51, 61, ... Wait, but we have to use 12 numbers. If we keep adding 10 each time, after 12 steps, we would have 1 + 10*11=111, which is way too large, and we can't get back to 1. So this approach won't work. We need to have a balance of positive and negative differences.Alternatively, maybe alternate adding and subtracting? But we need to arrange the differences such that we go up as much as possible and then come back down. However, since the total sum must be zero, the number of +10s and +7s must equal the number of -10s and -7s. Wait, but earlier we saw that the total number of positive differences (x + y) equals 6, and the total number of negative differences (z + w)=6. So we need to have six positive steps and six negative steps.Therefore, in the sequence of differences, half must be positive and half negative. To maximize the largest number, we need to cluster the positive differences early and the negative differences late. However, we must ensure that after the positive differences, the negative differences bring us back down to 1 without repeating any numbers.Let me try constructing such a sequence. Starting at 1, let's use as many +10s and +7s as possible to climb up, then use -10s and -7s to come back down.Suppose we start with 1, then go +10 to 11, then +10 to 21, +10 to 31, +10 to 41, +10 to 51, +10 to 61. That's six +10s. Then we need six -10s to come back: 61 -10=51, but 51 is already used. So that's a problem. Repeating numbers. Therefore, this approach doesn't work.Alternatively, instead of using all +10s, maybe use a mix of +10s and +7s to climb higher, and then use a mix of -10s and -7s to come back down without overlapping.But how?Let me think. Suppose we use as many large increments as possible first. Let's try:Start at 1.1 +10=1111 +10=2121 +10=3131 +10=4141 +10=5151 +10=61 (six +10s)Now we need to come back down with six steps. Let's try:61 -10=51 (already used)Nope. So can't do that. Maybe alternate -10 and -7?61 -7=54 (new)54 -10=44 (new)44 -7=37 (new)37 -10=27 (new)27 -7=20 (new)20 -10=10 (new)10 -7=3 (new)Wait, but we need six steps to come back. Wait, starting from 61, if we do six steps:But after 61, we need six steps to get back to 1. Let's try:61 -7=5454 -10=4444 -7=3737 -10=2727 -7=2020 -10=1010 -7=3But that's seven steps, which is too many. Wait, we already used six steps to go up, so total steps would be 6 up +6 down=12. But starting at 1, going up six steps, then down six steps. But when we go down, the sixth step should land back at 1. Let me check:Start at 1.1 +10=11 (1)11 +10=21 (2)21 +10=31 (3)31 +10=41 (4)41 +10=51 (5)51 +10=61 (6)Now, need to go back down:61 -7=54 (7)54 -10=44 (8)44 -7=37 (9)37 -10=27 (10)27 -7=20 (11)20 -10=10 (12)10 -7=3 (13)Wait, but we only have 12 numbers. So from step 1 to step 12, we need to end at 1. But in the above, after 6 steps up and 6 steps down, we end at 10 -7=3, which is not 1. So that doesn't work. Moreover, we have a conflict because the total sum of differences must be zero. If we have six +10s and six steps that include some combination of -10s and -7s, the total sum would be 6*10 + (sum of six negative differences) =0. So sum of negative differences must be -60. Let's see:If we have six -10s, sum is -60. So total sum would be 60 -60=0. But if we use some -7s, then sum of negative differences would be less than -60, so it can't balance the +60. Therefore, to have the total sum zero, if we use six +10s, we must use six -10s. Similarly, if we use six +7s, we need six -7s. But in the problem, x + y=6 and z +w=6. So the number of positive and negative differences are each 6.But earlier, when x + y=6 and z +w=6, but x=z and y=w. Wait, no, in the case when k=0, x=z and y=w. So the number of +10s equals the number of -10s, and the number of +7s equals the number of -7s. Therefore, to have the total sum zero, we need equal numbers of +10 and -10, and equal numbers of +7 and -7.Therefore, for example, if we have three +10s and three -10s, and three +7s and three -7s, total differences would be 3*10 +3*7 -3*10 -3*7=0. But x + y=6 (3+3=6) and z +w=6 (3+3=6). So this is valid. Alternatively, other distributions like 4 +10s, 2 +7s, 4 -10s, 2 -7s. Then total sum is 4*10 +2*7 -4*10 -2*7=0. Also valid.Therefore, there are multiple ways to balance the differences. But how does this affect the maximum number?To maximize the maximum number, we need to maximize the sum of the positive differences early in the sequence. Since +10 is larger than +7, we should use as many +10s as possible in the first half, and then use -10s and -7s in the second half.But we need to ensure that the path doesn't cross itself (i.e., repeat numbers) and that it closes back to 1.Alternatively, think of this as a graph traversal where each step is +10, +7, -10, or -7, and we need to traverse 12 distinct nodes starting and ending at 1.But this seems complex. Maybe another approach is to model the problem as a system of equations or inequalities.Alternatively, consider the problem as a circular sequence where each number is previous ±10 or ±7. Let's denote the numbers as a1, a2, ..., a12, with a1=1. Each ai+1 = ai ±10 or ±7, and a12 + difference =1.We need all ai to be distinct natural numbers.To maximize the maximum ai, we need to have a run of as many +10s and +7s as possible before needing to decrease.But again, the challenge is closing the loop. Let's attempt to construct such a sequence manually.Start with 1. Let's choose the next number to be 11 (difference +10). Then from 11, to maximize, go to 21 (+10). From 21, go to 31 (+10). Then 41, 51, 61. Now we've used six steps, reaching 61. Now we need to come back down in six steps. But if we use six -10s, we would go 61-10=51, which is already used. So we can't do that. Therefore, we need to intersperse some -7s to avoid repetition.Alternatively, after climbing up with +10s, use some -7s to step down to new numbers.For example:1, 11, 21, 31, 41, 51, 61, 54 (-7), 47 (-7), 37 (-10), 30 (-7), 23 (-7), 16 (-7). Wait, but 16 -7=9, which is not 1. Hmm, that's only 12 numbers if we count correctly:1 (1), 11 (2), 21 (3), 31 (4), 41 (5), 51 (6), 61 (7), 54 (8), 47 (9), 37 (10), 30 (11), 23 (12). Then the next number should be 16, but we need to end at 1. This doesn't work.Alternatively, after 61, maybe go -10 to 51 (used), so invalid. Go -7 to 54 (new). Then from 54, go -10 to 44 (new). Then from 44, -10 to 34 (new). Then from 34, -10 to 24 (new). Then 24 -10=14, then 14 -10=4. Then 4 -10= -6 invalid. This path isn't working.Alternatively, from 61, go -7 to 54. Then -10 to 44. -7 to 37. -10 to 27. -7 to 20. -10 to 10. -7 to 3. But again, this is seven steps down, but we only have six steps left. And we end at 3, not 1.This approach isn't working. Maybe starting with a different combination of +10 and +7.What if instead of using all +10s, we use some +7s to make the descent possible?For example:1, 11 (+10), 21 (+10), 31 (+10), 38 (+7), 48 (+10), 55 (+7), 62 (+7), 52 (-10), 42 (-10), 35 (-7), 28 (-7), 21... Wait, 21 is already used. So that's a problem.Alternatively:1, 11, 21, 31, 24 (-7), 17 (-7), 27 (+10), 37 (+10), 47 (+10), 40 (-7), 33 (-7), 26 (-7), 19 (-7). Hmm, ends at 19, not 1.This is tricky. Maybe a different strategy. Let's consider that the sequence must consist of six +10/+7 and six -10/-7. To maximize the peak, we want to have as many high additions as possible before subtracting.But how to arrange them so that we don't repeat numbers and return to 1.Alternatively, perhaps use a pattern that goes up, peaks, then goes down. For example, increase to a peak, then decrease back. But in a circle, the peak would be somewhere, and then the numbers go down and back up. But since it's a circle, the sequence must be such that the end connects back to the beginning.Alternatively, think of the numbers as being arranged in a circle where the differences alternate between increases and decreases. But not sure.Another idea: since we have to have six positive and six negative differences, maybe interleave them. For example, +10, -10, +10, -10,... But this would just oscillate between numbers. For example:1, 11, 1 (repeat), which is invalid.Similarly, +10, -7, +10, -7,... Let's try:1 +10=11, 11 -7=4, 4 +10=14, 14 -7=7, 7 +10=17, 17 -7=10, 10 +10=20, 20 -7=13, 13 +10=23, 23 -7=16, 16 +10=26, 26 -7=19. Then the next step would be 19 +10=29, but we need to end at 1. This isn't working.Alternatively, maybe use a combination of +10s and +7s on the way up and a different combination on the way down.Wait, but since the total number of +10s equals the number of -10s, and similarly for +7s and -7s, we have to have symmetry in the differences. For example, if we use three +10s and three +7s, we need three -10s and three -7s.Therefore, perhaps the path goes up three times with +10s and three times with +7s, then down three times with -10s and three times with -7s. Let's try:Start at 1.+10: 11+10:21+10:31+7:38+7:45+7:52Now we've used six steps up: three +10s and three +7s. Total increase: 3*10 +3*7=30 +21=51. So from 1 to 52. Now we need to decrease by 51 to get back to 1. Since we need to use six steps, each being -10 or -7. The total decrease needed is 51. Let's see:Let’s use three -10s and three -7s. Total decrease:3*10 +3*7=30 +21=51. Perfect.So the down path would be:52 -10=4242 -10=3232 -10=2222 -7=1515 -7=88 -7=1So the full sequence would be:1, 11, 21, 31, 38, 45, 52, 42, 32, 22, 15, 8, 1But wait, that's 13 numbers. Wait, no, starting at 1, we have 12 numbers: 1 (1), 11 (2), 21 (3), 31 (4), 38 (5), 45 (6), 52 (7), 42 (8), 32 (9), 22 (10), 15 (11), 8 (12). Then the 13th number would be 1, but since it's a circle of 12 numbers, the 12th number should connect back to 1. So the sequence should be:1, 11, 21, 31, 38, 45, 52, 42, 32, 22, 15, 8And then 8 connects back to 1 via -7 (since 8 +7=15, but 15 is already in the sequence. Wait, no. Wait, the differences are between neighbors. So between 8 and 1, the difference must be either 10 or 7. 8 and 1 differ by 7, which is allowed. Yes. So this sequence works.Now let's check if all numbers are distinct:1, 11, 21, 31, 38, 45, 52, 42, 32, 22, 15, 8. Yes, all are distinct natural numbers. The largest number is 52. Can we get higher?Let's see if we can rearrange the differences to get a higher peak. For example, if we use more +10s earlier. Suppose we use four +10s and two +7s on the way up, then four -10s and two -7s on the way down.Total increase:4*10 +2*7=40 +14=54Total decrease needed:54, which would require 4*10 +2*7=54. So yes.Let's try:Up:1 +10=1111 +10=2121 +10=3131 +10=4141 +7=4848 +7=55Now six steps up: four +10s, two +7s. Total increase 54. Current number:55.Down:55 -10=4545 -10=3535 -10=2525 -10=1515 -7=88 -7=1Sequence:1,11,21,31,41,48,55,45,35,25,15,8Check numbers: all distinct. Largest number 55. That's better than 52.But wait, let's verify the connections:From 55, subtract 10 to get45 (new)45-10=35 (new)35-10=25 (new)25-10=15 (new)15-7=8 (new)8-7=1 (back to start). All numbers are distinct. Yes. So this sequence has a maximum of 55.Can we go higher? Let's try five +10s and one +7 on the way up.Total increase:5*10 +1*7=50 +7=57Then decrease with five -10s and one -7:5*10 +1*7=57.Sequence:Up:1 +10=1111 +10=2121 +10=3131 +10=4141 +10=5151 +7=58Down:58 -10=4848 -10=3838 -10=2828 -10=1818 -10=88 -7=1Sequence:1,11,21,31,41,51,58,48,38,28,18,8Check numbers: all distinct. Largest number 58. Great.Can we push further? Six +10s on the way up.Total increase:6*10=60Then decrease with six -10s. But as before, this would go:1,11,21,31,41,51,61,51,41,31,21,11,1. But duplicates occur immediately at step 8 (51 again). So invalid.But if we use six +10s and six -10s, but interleave them differently. However, it's difficult because stepping back would repeat numbers. So 61-10=51 which is already there. So not possible.Therefore, the previous attempt with five +10s and one +7 gives a maximum of 58. Can we do better?What if we use five +10s and one +7, but arrange the +7 earlier to allow for a larger peak?For example:Up:1 +10=1111 +10=2121 +10=3131 +7=3838 +10=4848 +10=58Total increase:3*10 +1*7 +2*10=30 +7 +20=57. Wait, that's the same as before. The total increase is still 57, leading to 58. Then the down steps would still need to total -57.Alternatively, if we arrange the +7 earlier:1 +10=1111 +7=1818 +10=2828 +10=3838 +10=4848 +10=58Total increase:1*7 +4*10=7 +40=47. Wait, no. Wait, the sequence would be:1→11(+10), 11→18(+7), 18→28(+10), 28→38(+10), 38→48(+10), 48→58(+10). So total increase:10 +7 +10 +10 +10 +10=57. Same as before. Then down steps:58 -10=48 (used). So invalid. So regardless of where the +7 is placed in the up sequence, the problem of repeating 48 occurs.Alternatively, use a different down sequence. Instead of all -10s, use some -7s.Wait, if total decrease needed is 57, and we have six steps to do it. Let’s denote the number of -10s as a and -7s as b, so a + b=6 and 10a +7b=57.Solve for a and b:From a + b=6 => a=6 -b.Substitute into 10a +7b=57:10(6 -b) +7b=5760 -10b +7b=5760 -3b=57-3b= -3 => b=1, a=5.Therefore, we need five -10s and one -7.So starting from 58:58 -10=48 (used in step 7, but step 7 is 58, step 8 would be 48, which was already at step 6 (48). So conflict. Therefore, can't do this.Alternatively, arrange the down steps to avoid repetition:From 58, maybe:58 -7=51 (new)51 -10=41 (new)41 -10=31 (new)31 -10=21 (new)21 -10=11 (new)11 -10=1.But this uses five -10s and one -7. Let's check:Start at 58:58 -7=51 (step7)51 -10=41 (8)41 -10=31 (9)31 -10=21 (10)21 -10=11 (11)11 -10=1 (12)So the full sequence:1,11,21,31,41,51,58,51,41,31,21,11. Wait, but 51 is repeated at step7 and step8. So invalid.Alternatively, from 58, do:58 -10=48 (new)48 -7=41 (new)41 -10=31 (new)31 -7=24 (new)24 -10=14 (new)14 -7=7 (new)7 -10= -3 invalid. Not working.Alternatively, 58 -10=48 (step7)48 -10=38 (step8)38 -10=28 (step9)28 -10=18 (step10)18 -10=8 (step11)8 -7=1 (step12)Sequence:1,11,21,31,41,51,58,48,38,28,18,8. All distinct? Check:1,11,21,31,41,51,58,48,38,28,18,8. Yes, all distinct. The differences are:From 1 to11: +1011 to21:+1021 to31:+1031 to41:+1041 to51:+1051 to58:+758 to48:-1048 to38:-1038 to28:-1028 to18:-1018 to8:-108 to1:-7 (since 8 -7=1)Wait, but the last difference is from 8 to1, which is -7. But according to the problem, any two neighboring numbers differ by either 10 or7. So 8 and1 differ by7, which is allowed. So this sequence works. The differences used are five +10s, one +7, five -10s, one -7. The total sum is 5*10 +1*7 -5*10 -1*7=0. Perfect. All numbers are distinct natural numbers, and the largest number is58. But wait, in this sequence, from 51 we go to58 (+7), then from58 to48 (-10), etc. So the numbers are indeed distinct. Let's verify each step:1. 12. 1+10=113.11+10=214.21+10=315.31+10=416.41+10=517.51+7=588.58-10=489.48-10=3810.38-10=2811.28-10=1812.18-10=813.8-7=1 (close the circle)Yes, 12 distinct numbers with largest 58. Can we do better?Let's try with more +7s in the up sequence. For example, four +10s and two +7s.Total increase:4*10 +2*7=54Total decrease needed:54, which requires 4*10 +2*7=54 (i.e., four -10s and two -7s).Let's construct the sequence:Up:1 +10=1111 +10=2121 +10=3131 +10=4141 +7=4848 +7=55Down:55 -10=4545 -10=3535 -10=2525 -10=1515 -7=88 -7=1Sequence:1,11,21,31,41,48,55,45,35,25,15,8. All distinct. Largest number55. Which is less than58. So not better.What if we use three +10s and three +7s. Total increase:3*10 +3*7=51Total decrease:3*10 +3*7=51. Let's try:Up:1 +10=1111 +10=2121 +10=3131 +7=3838 +7=4545 +7=52Down:52 -10=4242 -10=3232 -10=2222 -7=1515 -7=88 -7=1Sequence:1,11,21,31,38,45,52,42,32,22,15,8. Largest number52. Still less than58.Alternatively, can we interleave +10s and +7s in a way that allows for a higher peak?For example:1 +10=1111 +7=1818 +10=2828 +10=3838 +10=4848 +10=58Then down steps:58 -10=48 (repeat). Not good.Alternatively:1 +10=1111 +7=1818 +10=2828 +7=3535 +10=4545 +10=55Total increase:10+7+10+7+10+10=54. Then down steps need to total -54. Let's use four -10s and two -7s:55 -10=45 (repeat)Not good.Alternatively:55 -7=48 (new)48 -10=38 (new)38 -7=31 (new)31 -10=21 (new)21 -7=14 (new)14 -10=4 (new)But then need to get back to1. 4 to1 is difference of3, which is invalid. So this doesn't work.Alternatively, different down steps:55 -10=45 (new)45 -10=35 (new)35 -10=25 (new)25 -10=15 (new)15 -7=8 (new)8 -7=1Sequence:1,11,18,28,35,45,55,45,... Wait, 45 repeats. So invalid.This is challenging. Let's go back to the sequence that gave us58. Is there a way to tweak that to get a higher number?In the previous successful sequence, we had:1,11,21,31,41,51,58,48,38,28,18,8.The up steps were five +10s and one +7, and the down steps were five -10s and one -7. Total increase57, total decrease57.Is there a way to have a higher total increase?Suppose we use five +10s and two +7s, but wait, that would require seven steps up. No, we only have six steps up. Wait, no, x + y=6. So total of six positive differences.Wait, if we use six positive differences, say six +10s, but that requires six -10s, which causes repeats. Alternatively, five +10s and one +7, totaling six positive differences, then six negative differences of five -10s and one -7.But that's what we already did.Alternatively, can we use four +10s and two +7s, totaling six positive differences (total increase4*10 +2*7=54), then six negative differences of four -10s and two -7s (total decrease54). Wait, but this would require:From the peak number, which would be1 +54=55, then decreasing by54 using four -10s and two -7s. But we tried this and got a maximum of55, which is less than58.So the sequence with peak58 seems better. Is there a way to push this further?Wait, let's try another arrangement. Suppose we use five +10s and one +7 in the up phase, but place the +7 earlier to reach higher numbers.For example:1 +10=1111 +10=2121 +10=3131 +10=4141 +7=4848 +10=58Same as before. Peak58.Alternatively:1 +7=88 +10=1818 +10=2828 +10=3838 +10=4848 +10=58Total increase:7 +10*5=57. Then down steps:58 -10=48 (used)Same problem.Alternatively, interleave +7s and +10s differently:1 +10=1111 +7=1818 +10=2828 +10=3838 +10=4848 +10=58Same peak. Then down steps must be five -10s and one -7. But again, step down from58 would be to48, which is already present.Wait, but in our previous successful sequence, the +7 was used to jump from51 to58, avoiding the immediate repetition. Then the down steps started from58, going to48 (new), and so on. So that worked because48 was not used before step7.So in that sequence:1,11,21,31,41,51,58,48,38,28,18,8The number48 is first introduced at step8, after58. So no repetition.So the key was to use the +7 at the last up step to jump to a new number58, then step down via -10 to48 which hadn't been used yet.Therefore, maybe we can use this strategy to push further. Suppose we use a +7 earlier to create a larger jump.Wait, but how? Let's think.Suppose we have:1 +10=1111 +10=2121 +10=3131 +10=4141 +10=5151 +7=58Then down steps as before:58→48→38→28→18→8→1.Alternatively, if we can find a way to insert a +7 earlier to make room for a larger number. But it seems difficult.Alternatively, use two +7s in the up phase. For example:1 +10=1111 +7=1818 +10=2828 +10=3838 +10=4848 +7=55Then down steps need to total decrease of55 -1=54. So five -10s and one -7:55 -10=4545 -10=3535 -10=2525 -10=1515 -10=55 -7= -2 (invalid)No good. Alternatively, four -10s and two -7s:55 -10=4545 -10=3535 -10=2525 -10=1515 -7=88 -7=1Sequence:1,11,18,28,38,48,55,45,35,25,15,8. All distinct. Largest number55. Less than58.Alternatively, use three +7s and three +10s:1 +10=1111 +7=1818 +7=2525 +10=3535 +10=4545 +10=55Then down steps:55 -10=45 (used). Not good.Alternatively, different down steps:55 -7=4848 -10=3838 -7=3131 -10=2121 -7=1414 -10=4But then need to get back to1. 4 to1 needs -3, invalid.Another idea: can we have a larger peak by using a +7 earlier to allow more +10s later?For example:1 +7=88 +10=1818 +10=2828 +10=3838 +10=4848 +10=58Total increase:7 +10*5=57. Then down steps:58 -10=48 (used). Doesn't work.Alternatively, from58, do:58 -7=51 (new)51 -10=41 (new)41 -10=31 (new)31 -10=21 (new)21 -10=11 (new)11 -10=1Sequence:1,8,18,28,38,48,58,51,41,31,21,11. But 11 is already used as the second number. Wait, the sequence is:1 (1)8 (2)18 (3)28 (4)38 (5)48 (6)58 (7)51 (8)41 (9)31 (10)21 (11)11 (12)But 11 was already at step2. Wait, no, step2 is8, step12 is11. Wait, original sequence starts at1, next is8, then18,..., then11 at step12. Then 11 neighbors1 and8. But 11 is distinct from1 and8. Wait, check all numbers:1,8,18,28,38,48,58,51,41,31,21,11. All distinct. Yes. Then the difference between11 and1 is10, which is allowed. So this sequence works. Largest number58. Same as before.But wait, this sequence has a different path but same maximum. So no improvement.Another approach: Maybe use a mix of +10s and +7s in both up and down phases to weave through the numbers without repeating.For example:1 +10=1111 +7=1818 +10=2828 +7=3535 +10=4545 +7=52Then down:52 -10=4242 -7=35 (repeat)Not good.Alternatively:52 -7=45 (repeat)Also bad.Alternatively, different down steps:52 -10=4242 -10=3232 -7=2525 -10=1515 -7=88 -10= -2 invalid.No good.Alternatively:52 -10=4242 -7=3535 -10=2525 -7=1818 -10=88 -7=1Sequence:1,11,18,28,35,45,52,42,35,... Repeats35. Invalid.Hmm. This is tough. Let's consider that the maximum number achieved so far is58. Is there a way to get higher?Let's try using six positive differences as five +10s and one +7, leading to a total increase of57, then decreasing with five -10s and one -7. This gives us the sequence with58. If we try to use a different combination, say four +10s and two +7s, leading to54, then decrease with four -10s and two -7s. Which gives peak55.Alternatively, let's consider other distributions of +10s and +7s. For example, if x=6, y=0, z=6, w=0. This would mean six +10s and six -10s. But as we saw earlier, this causes immediate repeats. For example, 1,11,21,31,41,51,61,51,41,... which repeats numbers.Alternatively, x=5, y=1, z=5, w=1. This is what we did to get58. Similarly, x=4, y=2, z=4, w=2, which gives peak55. So the maximum seems to be58.Wait, let's verify if there's another combination where the total increase is higher. Suppose we have x=5, y=1, total increase5*10 +1*7=57. To get a higher increase, we would need more than57. For example, if we could use x=5, y=2, but x+y=7, which exceeds6. Not allowed. Similarly, x=6 would need y=0, but that causes repeats.Therefore, the maximum possible increase with six positive differences is57, achieved by five +10s and one +7. This leads to a peak of1 +57=58. Then decreasing by57 via five -10s and one -7.Thus, the largest possible number is58.But wait, let's confirm this with another example. Suppose we start with1 and take a different path that still uses five +10s and one +7, but arranged differently:1 +10=1111 +7=1818 +10=2828 +10=3838 +10=4848 +10=58Then down steps:58 -10=48 (used). Not good.But if we arrange the +7 earlier in the up phase:1 +10=1111 +10=2121 +7=2828 +10=3838 +10=4848 +10=58Then down steps:58 -10=48 (used). Same issue.But in the previous successful sequence, the +7 was used at the last up step:1→11→21→31→41→51→58 (with +7 from51 to58). Then down steps avoid repetition by going to48, which was not yet used. Because in this sequence,48 is introduced only after58. So that's the key. By placing the +7 at the last up step, we jump to58, then step down to48, which is new.If we use the +7 earlier, then the step down would revisit a number already used.Therefore, the critical factor is to use the +7 at the last possible moment to jump to a new high, allowing the subsequent steps to decrease without repetition.Therefore, the sequence:1,11,21,31,41,51,58,48,38,28,18,8is valid and reaches58.Is there a way to use two +7s in the up phase to get a higher number? Let's try:1 +10=1111 +10=2121 +10=3131 +7=3838 +10=4848 +7=55Total increase:3*10 +2*7=44. Then need to decrease by44 with six steps. Let's use four -10s and two -7s:4*10 +2*7=54. Wait, no, total decrease needed is44. So:Let’s solve for a and b:10a +7b=44, a +b=6.a=6 -b.10(6 -b) +7b=4460 -10b +7b=4460 -3b=44-3b= -16b=16/3≈5.33. Not integer. Therefore, impossible. So no solution. Therefore, can't decrease by44 in six steps with integer combinations of -10 and -7.Similarly, other combinations might not work. Therefore, the maximum achievable peak with six positive differences is58, as in the previous example.Therefore, after thorough exploration, the largest possible value of the greatest number written is58.

✅:To determine the largest possible value of the greatest number in a circle of twelve distinct natural numbers where one number is 1 and neighboring numbers differ by either 10 or 7, we follow this reasoning:1. Constraints: The sequence must form a closed cycle with twelve distinct natural numbers. The sum of differences (each ±10 or ±7) around the circle must be zero. 2. Balancing Differences: Let ( x ) be the number of +10s, ( y ) the number of +7s, ( z ) the number of -10s, and ( w ) the number of -7s. The equations are: - ( x + y + z + w = 12 ) - ( 10(x - z) + 7(y - w) = 0 ) Solving these leads to ( x = z ) and ( y = w ), with ( x + y = 6 ). Thus, six positive and six negative differences are needed.3. Maximizing the Peak: To maximize the largest number, prioritize using +10s early. However, the sequence must close back to 1 without repeating numbers. 4. Optimal Sequence: - Ascending Phase: Use five +10s and one +7 to reach the peak: [ 1 xrightarrow{+10} 11 xrightarrow{+10} 21 xrightarrow{+10} 31 xrightarrow{+10} 41 xrightarrow{+10} 51 xrightarrow{+7} 58 ] - Descending Phase: Use five -10s and one -7 to return to 1 without repeats: [ 58 xrightarrow{-10} 48 xrightarrow{-10} 38 xrightarrow{-10} 28 xrightarrow{-10} 18 xrightarrow{-10} 8 xrightarrow{-7} 1 ]5. Verification: The sequence ( 1, 11, 21, 31, 41, 51, 58, 48, 38, 28, 18, 8 ) contains distinct natural numbers, closes correctly, and achieves the maximum peak of 58.Answer: The largest possible value of the greatest number written is (boxed{58}).

🔑:1. Let's label the numbers around the circle for convenience, starting with 1 and proceeding clockwise. We denote these numbers as a_1=1, a_2, ldots, a_{12}.2. Notice that for any 1 leq i leq 6, the following inequalities hold: [ a_{i+1} - a_{1} = (a_{i+1} - a_{i}) + (a_{i} - a_{i-1}) + cdots + (a_{2} - a_{1}) leq 10i ] This is because each pairwise difference a_{j+1} - a_{j} (for 1 leq j leq i) can be at most 10.3. Similarly, for any 1 leq i leq 6, we also have: [ a_{13-i} - a_1 = (a_{13-i} - a_{12-i}) + cdots + (a_{12} - a_1) leq 10i ] Thus, the maximum value of a_7 (which is halfway around the circle from a_1) can be calculated by: [ a_{7} leq a_{1} + 6 cdot 10 = 1 + 60 = 61 ]4. If a_7 = 61, then each difference a_{i+1} - a_i = 10. This includes a_{2} - a_{1}, a_{3} - a_{2}, ldots, a_{7} - a_{6} and a_7 - a_8, a_8 - a_9, ldots, a_{12} - a_1. If all these differences are 10, we would have: [ a_2 = a_{12} = a_1 + 10 = 11 ] This leads to a contradiction since the sequence cannot cycle back to the starting value correctly if all changes are 10.5. Therefore, a_{7} < 61. The logical next highest upper bound by allowing for some differences of 7 instead is: [ a_{7} leq 1 + 5 cdot 10 + 7 = 1 + 50 + 7 = 58 ] Hence, the maximum possible value of a_{7} is 58.6. Finally, validate that it is possible to position 58 correctly on the circle. Consider the numbers arranged as follows, maintaining the differences of 10 and 7: [ 1, 11, 21, 31, 41, 51, 58, 48, 38, 28, 18, 8 ] This configuration satisfies all conditions of the problem: the differences between adjacent numbers are either 10 or 7, and among twelve distinct natural numbers, one of which is 1. Therefore, a maximum value of 58 can indeed be achieved. Conclusion:[boxed{58}]

🧠:Okay, let's see. The problem is: If you roll four fair 6-sided dice, what is the probability that at least three of them will show the same value? Hmm. Alright, so I need to calculate the probability that when rolling four dice, at least three come up with the same number. That could be three of a kind or four of a kind. So the problem is combining those two scenarios. Let me think about how to approach this.First, I know that when calculating probabilities involving dice, it's helpful to figure out the total number of possible outcomes and then the number of favorable outcomes. The total number of outcomes when rolling four dice is 6^4, since each die has 6 possibilities and they are independent. Let me confirm that: 6*6*6*6 = 1296. Yeah, that's right.Now, the favorable outcomes are those where at least three dice show the same value. So this includes two cases: exactly three dice show the same number and the fourth is different, or all four dice show the same number. I need to calculate the number of outcomes for each case and then add them together before dividing by the total number of outcomes to get the probability.Let me start with the easier case: all four dice showing the same number. How many such outcomes are there? Well, there are 6 possible numbers (1 through 6), and for each number, there's exactly one outcome where all four dice show that number. So that's 6 outcomes.Next, the case where exactly three dice show the same number. Here, three dice are the same, and the fourth is different. Let's break this down. First, choose the value that appears three times. There are 6 choices for this. Then, choose the value that is different. Since it has to be different from the first one, there are 5 choices here. Now, we also need to consider which die is the one showing the different number. Since there are four dice, the different number could be in any one of the four positions. So we need to multiply by 4 to account for the different positions the single different die can occupy.So putting that together: 6 (choices for the triple) * 5 (choices for the single different number) * 4 (positions for the different number). Let me calculate that: 6*5*4 = 120. So there are 120 outcomes where exactly three dice show the same number.Therefore, the total number of favorable outcomes is the sum of the two cases: 6 (four of a kind) + 120 (three of a kind) = 126. Then, the probability would be 126 / 1296. Let me simplify that fraction. Both numerator and denominator are divisible by 6: 126 ÷6=21, 1296 ÷6=216. So 21/216. Then, 21 and 216 are both divisible by 3: 21 ÷3=7, 216 ÷3=72. So the simplified probability is 7/72. Wait, is that correct? Let me check again.Hold on, maybe I made a mistake in the calculation. Let me go through each step again.Total outcomes: 6^4 = 1296. Correct.Four of a kind: 6 outcomes. Correct, one for each number.Three of a kind: Let's see. Choose which number is tripled: 6 options. Then, choose a different number for the fourth die: 5 options. Then, choose which of the four dice is the single one: 4 options. So 6*5*4=120. Yes, that seems right. So 120. Then total favorable is 6+120=126. 126/1296=7/72≈0.097222... So approximately 9.72%.But wait, let me think again if there's a possibility of overcounting or undercounting. For example, when we count three of a kind, are there any overlapping cases where another three of a kind could be counted again? But since we have exactly three the same and the fourth different, each such outcome is unique. So when we count all triples with a single different number, there's no overlap. Similarly, four of a kind is entirely separate. So I think that's correct.Alternatively, maybe another way to compute this. The probability of at least three is the sum of the probabilities of exactly three and exactly four. Let me confirm the numbers again.For exactly four: 6/1296=1/216≈0.00463.For exactly three: 120/1296=5/54≈0.09259.Adding those: 1/216 + 5/54. Let's convert to common denominators. 1/216 + (5/54)*(4/4)=1/216 + 20/216=21/216=7/72≈0.09722. Same as before. So that seems consistent.Wait, but I want to make sure that in the three of a kind calculation, there isn't a miscalculation. Let's take a small example. Suppose we have four dice. Let's fix a value, say, 1. How many ways can exactly three 1s appear? The three 1s can be in any three positions, and the fourth die must be a non-1. So the number of ways is C(4,3)*5. C(4,3)=4, so 4*5=20. Then, since there are 6 possible numbers for the triple, total is 6*20=120. That's the same as before. So that's correct.Alternatively, if I think of it as: first select which three dice are the same. The number of ways to choose three dice out of four is C(4,3)=4. Then, assign a number to those three: 6 choices. Then, assign a different number to the remaining die: 5 choices. So total is 4*6*5=120. That's the same result.Therefore, the calculation seems correct.So the total probability is (120 +6)/1296=126/1296=7/72≈0.0972, which is approximately 9.72%.Therefore, the probability is 7/72.But wait, let me check once more with a different approach to ensure there's no mistake.Another way to compute this is to compute the number of favorable outcomes for at least three of a kind. So, for each number from 1 to 6, count the number of ways that number appears at least three times, then subtract the overlap where two different numbers each appear three or more times. However, in four dice, can two different numbers each appear at least three times? That would require that one number appears three times and another appears once, but since there are four dice, if one number is three times, the other is once. So there's no overlap where two numbers each appear three or more times. Because 3+3=6, which is more than four. So, in other words, in four dice, the maximum number of overlaps would be if two numbers each appear two times, but not three. Therefore, when counting for each number the number of times it appears at least three times, we don't have overlapping cases where two numbers are each counted. So, we can safely use inclusion-exclusion here.So, for each number, the number of dice rolls where that number appears at least three times is C(4,3)*1^3*5^1 + C(4,4)*1^4*5^0. That is, for exactly three: 4*5=20, and exactly four: 1. So total for each number is 21. Since there are 6 numbers, 6*21=126. But wait, but this counts all the cases where any number is at least three times. But since, as we saw, there's no overlap, this is correct. So total is 126, which matches our previous calculation. Therefore, 126/1296=7/72. So that's the same answer.Therefore, this confirms that the probability is 7/72.So, after verifying through multiple methods, I think the answer is 7/72. Let me just make sure that I didn't miss any other cases. For instance, is there a case where three are of one number and three are of another? No, because with four dice, you can't have two sets of three. So the only possibilities are exactly three of one number and one of another, or four of one number. Hence, the cases are correctly accounted for. Therefore, I think the answer is correct.Final AnswerThe probability is boxed{dfrac{7}{72}}.

✅:To determine the probability that at least three out of four fair 6-sided dice show the same value, we consider two cases: exactly three dice showing the same value and all four dice showing the same value.1. Total number of possible outcomes: [ 6^4 = 1296 ]2. Case 1: Four of a kind: - There are 6 possible values (one for each number from 1 to 6). - Number of outcomes: [ 6 ]3. Case 2: Exactly three of a kind: - Choose the value for the three dice: 6 choices. - Choose a different value for the fourth die: 5 choices. - Choose which die shows the different value: 4 choices. - Number of outcomes: [ 6 times 5 times 4 = 120 ]4. Total favorable outcomes: - Sum of outcomes for both cases: [ 6 + 120 = 126 ]5. Probability calculation: - Probability is the ratio of favorable outcomes to total outcomes: [ frac{126}{1296} = frac{7}{72} ]Thus, the probability that at least three of the four dice will show the same value is boxed{dfrac{7}{72}}.

🔑:To solve this problem, we need to find the probability that at least three of the four dice show the same value. This can happen in either of two ways:1. Exactly three dice show the same value, and one die shows a different value.2. All four dice show the same value.We will calculate the number of favorable outcomes for each case and then find the probability.# 1. Exactly three dice show the same value, and one die shows a different value:- Choosing the value for the three dice: There are 6 possible values (since the dice are 6-sided).- Choosing which three dice will show this value: This can be done in ( binom{4}{3} ) ways.- Choosing the value for the remaining die: The value must be different from the value shown by the three dice. So, there are 5 choices.Thus, the number of favorable outcomes for this case is:[ 6 times binom{4}{3} times 5 = 6 times 4 times 5 = 120 ]# 2. All four dice show the same value:- Choosing the value for the four dice: There are 6 possible values.Thus, the number of favorable outcomes for this case is:[ 6 ]# Total number of favorable outcomes:Combining both cases, we get:[ 120 + 6 = 126 ]# Total number of possible outcomes:Since each die is 6-sided and the dice are rolled independently:[ 6^4 = 1296 ]# Probability:The probability of at least three dice showing the same value is the ratio of the favorable outcomes to the total outcomes:[ frac{126}{1296} ]Simplifying the fraction:[ frac{126}{1296} = frac{7}{72} ]Thus, the probability is:[ boxed{frac{7}{72}} ]

🧠:Okay, let's see. I need to find the range of the function ( y = -x^2 - 2x + 3 ) on the interval ( -5 leqslant x leqslant 0 ). The options are given, but I should work through it step by step. First, the function is a quadratic, right? Since the coefficient of ( x^2 ) is negative (-1), the parabola opens downward. That means the function has a maximum value at its vertex and the range will go down from there to negative infinity if the domain is all real numbers. But here, the domain is restricted to ( -5 leq x leq 0 ), so the range will be between the minimum and maximum values on that interval.To find the range, I need to determine the maximum and minimum values of the function on the given interval. For a quadratic function, the vertex is either the maximum or minimum point. Since the parabola opens downward, the vertex is the maximum point. Then, the minimum will occur at one of the endpoints of the interval, either at x = -5 or x = 0. Let me check that.First, let's find the vertex. The formula for the x-coordinate of the vertex of a parabola ( y = ax^2 + bx + c ) is ( x = -frac{b}{2a} ). In this case, a = -1, b = -2. Plugging in, we get:( x = -frac{-2}{2*(-1)} = frac{2}{-2} = -1 ).So the vertex is at x = -1. Since this is within the interval [-5, 0], this will give the maximum value of the function on that interval. Let me compute y at x = -1:( y = -(-1)^2 -2*(-1) + 3 = -1 + 2 + 3 = 4 ).So the maximum value is 4.Now, to find the minimum, I need to evaluate the function at the endpoints x = -5 and x = 0.First, at x = -5:( y = -(-5)^2 -2*(-5) + 3 = -25 + 10 + 3 = -12 ).Then, at x = 0:( y = -(0)^2 -2*(0) + 3 = 0 - 0 + 3 = 3 .So, the function takes the values -12 at x = -5 and 3 at x = 0. Since the parabola is decreasing from the vertex (x = -1) to the right, but since the interval goes up to x = 0, which is to the right of the vertex, the value at x = 0 is 3. However, on the left side, from x = -5 to x = -1, the function is increasing because the parabola opens downward, so it increases from x = -5 up to x = -1 (the vertex), then decreases after that.Wait, let me confirm that. Since the parabola opens downward, the function increases to the left of the vertex and decreases to the right. So on the interval from x = -5 (left end) to x = -1 (vertex), the function is increasing, and from x = -1 to x = 0 (right end), it's decreasing.Therefore, the minimum value on the interval would be at x = -5, which is -12, and the maximum at x = -1 is 4. However, at x = 0, the value is 3. But since 3 is higher than -12 but lower than 4, which is the maximum. So the range should be from -12 to 4. That is, the minimum is -12 and the maximum is 4.Looking at the options:(A) ( (-infty, 4] ) – Hmm, this is considering all values less than or equal to 4, but since the domain is restricted, the actual minimum is -12, so this is incorrect.(B) ( [3, 12] ) – 12 isn't even a value here. At x = -5, we have -12, so this is wrong.(C) ( [-12, 4] ) – This includes the minimum at -12 and the maximum at 4, which matches our calculations.(D) ( [4, 12] ) – Again, 12 isn't a value here, and 4 is the maximum, but the minimum is -12, so this is wrong.Therefore, the correct answer should be option C.But wait, let me double-check my calculations to ensure I didn't make any mistakes.Calculating y at x = -5:( y = -(-5)^2 -2*(-5) + 3 = -25 +10 +3 = (-25 +10) +3 = (-15) +3 = -12. Correct.At x = 0:( y = 0 -0 +3 = 3. Correct.At x = -1:( y = -1 -2*(-1) +3 = -1 +2 +3 = 4. Correct.So the maximum is indeed 4, and the minimum is -12.Therefore, the range is from -12 to 4, inclusive. So option C is the correct answer.But just to make sure, maybe I should sketch a quick graph of the function on the interval to visualize.The function is a downward-opening parabola with vertex at (-1, 4). The interval from x = -5 to x = 0. At x = -5, which is far left, the function is at -12. Moving towards the vertex at x = -1, it increases up to 4. Then from x = -1 to x = 0, it decreases again down to 3. So the lowest point is at x = -5, the highest at x = -1. So the range is all y-values between -12 and 4. Hence, [-12, 4], which is option C.Yes, that makes sense. The answer should be C.

✅:The function ( y = -x^2 - 2x + 3 ) is a downward-opening parabola. To determine its range on ( -5 leqslant x leqslant 0 ):1. Vertex Calculation: The vertex of ( y = ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Here, ( a = -1 ), ( b = -2 ), so: [ x = -frac{-2}{2(-1)} = -1 ] The vertex is at ( x = -1 ), which lies within the interval ([-5, 0]). Evaluating ( y ) at ( x = -1 ): [ y = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4 ] This is the maximum value (4) on the interval.2. Endpoint Evaluation: - At ( x = -5 ): [ y = -(-5)^2 - 2(-5) + 3 = -25 + 10 + 3 = -12 ] This is the minimum value (-12). - At ( x = 0 ): [ y = -(0)^2 - 2(0) + 3 = 3 ] This value (3) is less than the maximum (4) and greater than the minimum (-12).3. Range Determination: Since the function increases from ( x = -5 ) to the vertex ( x = -1 ), reaching 4, and then decreases to 3 at ( x = 0 ), the range spans from the minimum (-12) to the maximum (4). Final Answer: (boxed{C}) ([-12, 4])

🔑:1. Rewrite the given function in vertex form: The given function is [ y = -x^2 - 2x + 3 ] To complete the square, we first rewrite it in the form: [ y = -(x^2 + 2x) + 3 ] 2. Complete the square: Inside the parentheses, complete the square for (x^2 + 2x): [ x^2 + 2x = (x+1)^2 - 1 ] Therefore, [ y = -((x+1)^2 - 1) + 3 = -(x+1)^2 + 1 + 3 = -(x+1)^2 + 4 ] So, the function in vertex form is: [ y = -(x+1)^2 + 4 ]3. Determine the range of (y): Consider the interval given: (-5 leq x leq 0). Let's find the values of (y) within this interval: - When ( x = -5 ): [ y = -((-5)+1)^2 + 4 = -(-4)^2 + 4 = -16 + 4 = -12 ] - When ( x = 0 ): [ y = -(0+1)^2 + 4 = -1 + 4 = 4 ] 4. Analyze the vertex form: The vertex form (y = -(x+1)^2 + 4) shows that the parabola opens downwards (since the coefficient of ((x+1)^2) is negative), and the vertex (( -1, 4 )) is the maximum point. 5. Find the minimum value of (y): Since ( (x+1)^2 ) is always non-negative and the maximum value of ( -(x+1)^2) is 0, the minimum value will occur when ( (x+1)^2 ) is maximized: - When ( x = -5 ), ( (x+1)^2 = 16 ) and ( y = -16 + 4 = -12 ).6. Conclusion: The range of (y) for (-5 leq x leq 0) is therefore: [ [-12, 4] ] Hence, the correct answer is:[boxed{mathrm{C}}]

🧠:Okay, let's tackle this geometry problem. So, we have triangle MNP, and MM₁ and PP₁ are the angle bisectors. The problem states that the lengths of the perpendiculars dropped from vertex N to the lines MM₁ and PP₁ are equal, and we need to prove that triangle MNP is isosceles. Hmm, interesting. Let me try to visualize this first.First, let me recall what an angle bisector does. An angle bisector of a triangle is a line that splits an angle into two equal parts. So, MM₁ bisects angle M, and PP₁ bisects angle P. The perpendiculars from N to these bisectors are equal in length. Our goal is to show that this condition implies that triangle MNP must have two equal sides, making it isosceles.Since the problem involves angle bisectors and perpendiculars, maybe using some properties related to angle bisectors, distances from a point to a line, or perhaps trigonometry. Let me jot down what I know:1. The distance from a point to a line can be calculated using the formula: if the line is ax + by + c = 0 and the point is (x₀, y₀), the distance is |ax₀ + by₀ + c| / sqrt(a² + b²). But maybe there's a more geometric approach here.2. In a triangle, the angle bisector theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. For example, in triangle MNP, angle bisector MM₁ would divide side NP into segments NM₁ and M₁P such that NM₁ / M₁P = NM / MP.3. If two distances from a point to two different lines are equal, then the point lies on the angle bisector of the angle formed by those two lines. Wait, but here the distances are from the same point N to two different angle bisectors. So maybe not directly applicable, but perhaps similar logic.Let me try to draw triangle MNP. Let's denote the vertices as M, N, P. Let's assume it's not isosceles initially. Then, angles at M and P are different. The angle bisectors MM₁ and PP₁ would then split angles M and P into two equal parts each. The perpendiculars from N to MM₁ and PP₁ are equal in length. We need to show that this equality implies that two sides must be equal, so the triangle is isosceles.Hmm. Let me consider coordinates. Maybe placing the triangle in a coordinate system would help. Let me assign coordinates to the points. Let me place point N at the origin (0, 0) for simplicity. Let me set point M along the x-axis at (m, 0), and point P somewhere in the plane at (p, q). Then, angle bisectors MM₁ and PP₁ can be found using angle bisector formulas. The perpendicular distances from N (which is at (0,0)) to these bisectors are equal. Maybe this approach could work.Alternatively, using vectors or trigonometry. Let me think. Let's denote angles at M as α, at N as β, and at P as γ. Since it's a triangle, α + β + γ = 180°. The angle bisectors from M and P would split angles α and γ into α/2 and γ/2, respectively. The lengths of the perpendiculars from N to these bisectors are equal. Maybe using trigonometric expressions for these distances.Let me recall that the distance from a point to a line defined by an angle. Suppose we have a line making an angle θ with the x-axis, then the distance from a point (x, y) to this line can be expressed in terms of sine and cosine. Alternatively, in a triangle, if we consider the distance from a vertex to an angle bisector, maybe we can express it using the sides and angles.Alternatively, using area. The area of a triangle can be expressed as 1/2 * base * height. Maybe if we can relate the areas formed by these perpendiculars.Wait, the perpendiculars from N to MM₁ and PP₁ are the heights in the triangles formed by N and the bisectors. So, maybe the areas of triangles NMM₁ and NPP₁ can be related through these heights. But I need to check.Wait, but MM₁ is the angle bisector from M to side NP, so M₁ is a point on NP. Similarly, PP₁ is the angle bisector from P to side MN, so P₁ is a point on MN. So, MM₁ connects M to M₁ on NP, and PP₁ connects P to P₁ on MN.Therefore, the perpendiculars from N to MM₁ and PP₁ are the heights of triangles NMM₁ and NPP₁ with respect to bases MM₁ and PP₁. However, since these bases are different, the areas would be (1/2)*MM₁*height_from_N_to_MM₁ and (1/2)*PP₁*height_from_N_to_PP₁. Since the heights are given to be equal, but unless we know something about the lengths of MM₁ and PP₁, we can't directly relate the areas.Alternatively, maybe using trigonometric relationships. Let's consider triangle N and the two angle bisectors. Let me denote the angle at M as 2α (so the bisector splits it into α and α), and the angle at P as 2γ (split into γ and γ). Then, angles at N is β = 180° - 2α - 2γ.Now, the angle bisectors MM₁ and PP₁. Let me try to find the equations for these bisectors and then compute the distances from N to each bisector.Alternatively, maybe using the formula for the distance from a point to a line in terms of angles. Suppose we have a line making an angle θ with respect to a reference line, and a point at some distance and angle from that line. The perpendicular distance can be expressed using sine of the angle between the line and the line connecting the point to the line.Wait, if we consider the angle between the bisector and the sides of the triangle. For example, the angle bisector from M to NP makes an angle α with side MN and α with side MP. Similarly for the bisector from P.But perhaps this is getting too vague. Let me try to be more precise.Let me consider triangle MNP with vertices M, N, P. Let's denote the angle at M as ∠M = 2α, so the bisector MM₁ divides it into two angles of α each. Similarly, let ∠P = 2γ, so the bisector PP₁ divides it into two angles of γ each. Then, angle at N is ∠N = 180° - 2α - 2γ.Now, we need to find the lengths of the perpendiculars from N to MM₁ and PP₁. Let's denote these perpendiculars as h₁ and h₂ respectively. The problem states that h₁ = h₂, and we need to show that triangle MNP is isosceles, i.e., either 2α = 2γ (so α = γ) making sides opposite to these angles equal (i.e., NP = MN), or some other pair of sides equal.Wait, actually in triangle MNP, sides opposite angles M, N, P are NP, MP, and MN respectively. So, if angles at M and P are equal (2α = 2γ), then sides opposite them, which are NP and MN, would be equal. So NP = MN, making the triangle isosceles with apex at N? Wait, no. Wait, if angles at M and P are equal, then sides opposite them, which are NP and MN, would be equal. So NP = MN, so sides MN and NP are equal, making triangle MNP isosceles with equal sides MN and NP, and base MP. Alternatively, if angles at N is equal to another angle, but the problem states that the perpendiculars from N to the bisectors are equal.Alternatively, maybe using the formula for the distance from a point to a line in terms of sine of angles. Let me recall that in a triangle, the distance from a vertex to a line can be calculated using the formula involving the length of another side and the sine of the included angle.Wait, suppose we have a line (angle bisector) MM₁. The distance from N to MM₁ can be expressed as NM * sin(angle between NM and MM₁). Similarly, the distance from N to PP₁ can be expressed as NP * sin(angle between NP and PP₁). If these distances are equal, then NM * sin(θ₁) = NP * sin(θ₂), where θ₁ is the angle between NM and MM₁, and θ₂ is the angle between NP and PP₁.Let me try to find θ₁ and θ₂. Since MM₁ is the angle bisector of angle M (which is 2α), then the angle between MM₁ and side MN is α. So, in triangle MM₁N, angle at M is α, angle at N is β, and angle at M₁ is 180° - α - β. Wait, maybe not. Let me think.Wait, in triangle MNP, angle at M is 2α. The angle bisector MM₁ splits it into two angles of α each. So, in triangle MM₁N, angle at M is α, angle at N is β, and angle at M₁ is 180° - α - β. Similarly, in triangle PP₁N, angle at P is γ, angle at N is β, and angle at P₁ is 180° - γ - β.But perhaps this is getting too convoluted. Let me see. The angle between NM and MM₁ is α, since MM₁ bisects angle M into two α angles. Therefore, the angle between side NM (which is adjacent to angle M) and the bisector MM₁ is α.Similarly, angle at P is 2γ, so the angle bisector PP₁ splits it into two angles of γ each. Therefore, the angle between side NP (which is adjacent to angle P) and the bisector PP₁ is γ.Therefore, the distance from N to MM₁ is NM * sin(α), and the distance from N to PP₁ is NP * sin(γ). The problem states that these distances are equal:NM * sin(α) = NP * sin(γ)But in triangle MNP, by the Law of Sines, we have:NM / sin(2γ) = NP / sin(2α) = MP / sin(β)Because in triangle MNP, the Law of Sines gives:NM / sin(∠P) = NP / sin(∠M) = MP / sin(∠N)Given that ∠M = 2α and ∠P = 2γ, so:NM / sin(2γ) = NP / sin(2α)Therefore, NM / NP = sin(2γ) / sin(2α)But from the given condition, NM * sin(α) = NP * sin(γ)Let me write that as:(NM / NP) = (sin(γ) / sin(α))From the Law of Sines ratio, we have NM / NP = sin(2γ) / sin(2α)Therefore, combining these two equations:sin(2γ) / sin(2α) = sin(γ) / sin(α)Simplify the left side using double-angle identity:sin(2γ) = 2 sinγ cosγsin(2α) = 2 sinα cosαTherefore:[2 sinγ cosγ] / [2 sinα cosα] = sinγ / sinαSimplify:( sinγ cosγ ) / ( sinα cosα ) = sinγ / sinαCancel sinγ / sinα from both sides:cosγ / cosα = 1Therefore:cosγ = cosαWhich implies that γ = α, since angles α and γ are both between 0 and 90 degrees (since they are halves of angles in a triangle, which must be less than 90 degrees if the original angles are less than 180). Therefore, α = γ.If α = γ, then the original angles at M and P were 2α and 2γ, which are equal. Therefore, angles at M and P are equal, which by the Law of Sines implies that sides opposite them are equal. The side opposite angle M is NP, and the side opposite angle P is MN. Therefore, NP = MN, making triangle MNP isosceles with NP = MN.Hence, proved.Wait, let me check my steps again to make sure I didn't make a mistake.1. Assigned angles at M and P as 2α and 2γ, with bisectors splitting them into α and γ.2. Expressed distances from N to MM₁ and PP₁ as NM sinα and NP sinγ, respectively.3. Equated them: NM sinα = NP sinγ.4. Used Law of Sines on triangle MNP: NM / sin2γ = NP / sin2α → NM / NP = sin2γ / sin2α.5. Substituted into the previous equation: (sin2γ / sin2α) * sinα = sinγ.6. Simplified using double-angle identities:sin2γ / sin2α = (2 sinγ cosγ) / (2 sinα cosα) = (sinγ cosγ) / (sinα cosα)So, substituting back:(NM / NP) * sinα = (sin2γ / sin2α) * sinα = (sinγ cosγ / sinα cosα) * sinα = (sinγ cosγ) / cosα = sinγTherefore:(sinγ cosγ) / cosα = sinγAssuming sinγ ≠ 0 (which it can't be, since γ is an angle in a triangle, so 0 < γ < 90°), we can divide both sides by sinγ:cosγ / cosα = 1Therefore, cosγ = cosα → γ = α (since both angles are between 0 and 90°, so arccos is injective there).Therefore, γ = α, so angles at M and P are equal (2α = 2γ), hence triangle is isosceles with sides opposite to these angles equal (NP = MN).Yes, this seems correct. So, the key steps were expressing the distances using trigonometry, applying the Law of Sines, and then simplifying using double-angle identities to find that the angles must be equal, leading to the triangle being isosceles.I think that's a solid proof. Let me see if there's another way to approach it, maybe using coordinate geometry or properties of angle bisectors, but this trigonometric method seems straightforward and logical.Alternatively, using coordinates: Let's place point N at the origin (0,0), point M at (c,0), and point P at (d,e). Then, find equations of angle bisectors MM₁ and PP₁, compute the perpendicular distances from N to these lines, set them equal, and derive conditions on the coordinates that force the triangle to be isosceles. However, this might involve more computation, but let's sketch it.Let me denote:- N(0,0), M(c,0), P(d,e)Angle bisector from M to NP: The angle bisector theorem tells us that MM₁ divides NP into segments proportional to adjacent sides. So, NM / MP = length of NM / length of MP. Wait, NM is the length from M to N, which is sqrt((c - 0)^2 + 0^2) = c. MP is the length from M to P, which is sqrt((d - c)^2 + e^2). Therefore, the point M₁ divides NP in the ratio NM : MP = c : sqrt((d - c)^2 + e^2). Hmm, this might get complicated.Alternatively, the angle bisector from M can be found using the formula for the angle bisector in coordinates. The angle bisector from M to NP can be represented parametrically or with a slope. Similarly for the bisector from P.Then, finding the equations of these bisectors, computing the distance from N(0,0) to each, setting them equal, and solving for the coordinates to see that the triangle must be isosceles. But this would involve a lot of algebra. The trigonometric approach is more straightforward.Alternatively, using vector projections. The distance from a point to a line can be expressed as the magnitude of the cross product divided by the length of the direction vector of the line. But this might also get involved.Given that the trigonometric method worked, and the steps are logical, I think that's sufficient. The key insight was recognizing that the distances can be expressed in terms of sides and sines of the bisected angles, then relating them via the Law of Sines and simplifying using double-angle identities. This leads directly to the conclusion that the angles at M and P must be equal, hence the triangle is isosceles.So, I think this is a valid proof. Let me check once more for any possible errors.Wait, in step where I equated NM sinα = NP sinγ, is this accurate? Let's see. The distance from N to MM₁ is indeed NM multiplied by the sine of the angle between NM and MM₁. Since MM₁ is the angle bisector, the angle between NM and MM₁ is α (half of angle M). Similarly, the angle between NP and PP₁ is γ. Therefore, the distances are NM sinα and NP sinγ. Yes, that seems correct.And then applying the Law of Sines to relate NM and NP. Since in triangle MNP, Law of Sines gives NM / sin(angle P) = NP / sin(angle M). Angle P is 2γ, angle M is 2α. So NM / sin2γ = NP / sin2α. Therefore, NM / NP = sin2γ / sin2α. Substituting into NM sinα = NP sinγ gives (sin2γ / sin2α) sinα = sinγ. Then, replacing sin2γ and sin2α with 2 sinγ cosγ and 2 sinα cosα, we get (2 sinγ cosγ / 2 sinα cosα) * sinα = sinγ. Simplify numerator and denominator: (sinγ cosγ / sinα cosα) * sinα = sinγ. The sinα cancels, leaving (sinγ cosγ / cosα) = sinγ. Divide both sides by sinγ (non-zero), get cosγ / cosα = 1 → cosγ = cosα → γ = α. All steps seem valid.Therefore, the proof holds, and triangle MNP must be isosceles with NM = NP.Final Answerboxed{text{Triangle } MNP text{ is isosceles}}

✅:To prove that triangle (MNP) is isosceles given that the lengths of the perpendiculars dropped from vertex (N) to the angle bisectors (MM_1) and (PP_1) are equal, we can follow these steps:1. Identify the angles and bisectors: - Let the angles at vertices (M) and (P) be (2alpha) and (2gamma) respectively, which are bisected by (MM_1) and (PP_1) into angles (alpha) and (gamma). - The angle at vertex (N) is (180^circ - 2alpha - 2gamma).2. Express the perpendicular distances: - The distance from (N) to (MM_1) is (NM sin(alpha)). - The distance from (N) to (PP_1) is (NP sin(gamma)). - Given these distances are equal, we have (NM sin(alpha) = NP sin(gamma)).3. Apply the Law of Sines: - In triangle (MNP), the Law of Sines gives (frac{NM}{sin(2gamma)} = frac{NP}{sin(2alpha)}). - This implies (frac{NM}{NP} = frac{sin(2gamma)}{sin(2alpha)}).4. Relate the ratios: - From the given condition, (frac{NM}{NP} = frac{sin(gamma)}{sin(alpha)}). - Equating the two ratios: (frac{sin(2gamma)}{sin(2alpha)} = frac{sin(gamma)}{sin(alpha)}).5. Simplify using double-angle identities: - Using (sin(2theta) = 2 sin(theta) cos(theta)), we get (frac{2 sin(gamma) cos(gamma)}{2 sin(alpha) cos(alpha)} = frac{sin(gamma)}{sin(alpha)}). - Simplifying, (frac{cos(gamma)}{cos(alpha)} = 1), leading to (cos(gamma) = cos(alpha)).6. Conclude equal angles: - Since (cos(gamma) = cos(alpha)), it follows that (gamma = alpha). - Therefore, the original angles at (M) and (P) are equal ((2alpha = 2gamma)), making triangle (MNP) isosceles with (NM = NP).Thus, we conclude that triangle (MNP) is isosceles.[boxed{text{Triangle } MNP text{ is isosceles}}]

🔑:Let's prove that the triangle (MNP) is isosceles, given that segment lengths from vertex (N) perpendicular to the bisectors (MM_1) and (PP_1) are equal.# 1. First Method1. Introduce Perpendiculars and Extend Them: Let (ND) and (NE) be the perpendiculars dropped from vertex (N) to the lines (MM_1) and (PP_1) respectively. 2. Construct Extensions to Intersect (MP): Extend the perpendiculars (NE) and (ND) until they intersect line (MP) at points (T) and (S) respectively. 3. Identify Key Triangles: Notice that triangles (NPT) and (NMS) are isosceles. This is because (P E) and (M D) are both heights bisecting the angles at (P) and (M) respectively. - Therefore, we have: [ NP = PT quad text{and} quad NM = MS ] 4. Utilize the Given Equal Perpendicular Lengths: Given that (NE = ND), we know: [ NT = 2 cdot NE = 2 cdot ND = NS ] 5. Compare Angles: Due to the above equality: [ angle NTS = angle NST ] 6. Conclude Triangle Equality: Hence, triangles (NPT) and (NMS) are congruent. Consequently, we have: [ MN = NP ]# Conclusion:Therefore, triangle (MNP) is isosceles with (MN = NP).[boxed{text{MN = NP}}]# 2. Second Method1. Introduce Intersection Point: Let (O) be the intersection point of the angle bisectors of triangle (MNP).2. Utilize Right Triangles: Consider the right triangles (ONE) and (OND) with the common hypotenuse (ON). 3. Equal Angles from Isosceles Properties: From the equality of these right triangles' legs, we know that: [ angle NOP_1 = angle NOM_1 ] 4. Equalize Bisector and Opposite Angles: Using this information in combination with: [ angle P_1OM = angle M_1OP quad text{and} quad angle MNO = angle PNO ] It follows that: [ angle NMO = angle NPO ]5. Final Angle Equality: Hence, we deduce that: [ angle MNP = angle NPM ]# Conclusion:Thus, confirming that triangle (MNP) is isosceles with (MN = NP).[boxed{text{MN = NP}}]

🧠:Okay, so I need to prove that angle AOC₁ is equal to angle MPB₁ in this triangle setup. Let me start by understanding the given information and the figure.First, triangle ABC has points B₁ on AC and C₁ on AB. Then, BB₁ and CC₁ intersect at point P. There's also an incircle of triangle AB₁C₁ with center O, and M is the point where this incircle touches B₁C₁. The key given condition is that OP is perpendicular to BB₁. I need to show that angle AOC₁ equals angle MPB₁.Let me sketch the figure mentally. Triangle ABC, with B₁ somewhere on AC and C₁ on AB. The lines BB₁ and CC₁ meet at P, which is likely inside the triangle. The incenter O of AB₁C₁ is the intersection of the angle bisectors of that triangle. M is the touch point of the incircle on B₁C₁. Since OP is perpendicular to BB₁, that gives a right angle at the intersection of OP and BB₁.Hmm. So perhaps I need to relate the angles at O and M with point P. Let me recall some properties of incenters and tangency points. The incenter is equidistant from all sides, and the tangency points divide the sides into segments related to the triangle's semiperimeter. Also, angles involving the incenter often relate to half the angles of the triangle.But here, we have triangle AB₁C₁, so O is its incenter. Then, OM is perpendicular to B₁C₁ because the radius at the point of tangency is perpendicular to the tangent line. So OM ⊥ B₁C₁. Also, OP is perpendicular to BB₁ by given.So maybe there's some relationship between OP and OM? Or maybe similar triangles involving O, P, M, etc.Let me consider coordinates. Sometimes coordinate geometry can help, but maybe synthetic methods are better here. Let me see.Alternatively, since we have perpendicular lines, maybe using slopes or dot products in coordinate geometry. But setting up coordinates might be messy. Let's see if synthetic geometry can work.First, note that O is the incenter of AB₁C₁. So AO is the angle bisector of angle A in triangle AB₁C₁. Similarly, the incenter lies at the intersection of the angle bisectors. So AO, B₁O, and C₁O are the angle bisectors.But how does point P come into play here? P is the intersection of BB₁ and CC₁. Also, OP is perpendicular to BB₁.Perhaps there's a way to relate OP to other lines in the figure. Since OP ⊥ BB₁, and OM ⊥ B₁C₁, maybe triangles OMP and something else are similar?Wait, point M is on B₁C₁. So OM is perpendicular to B₁C₁. So if I can connect OP and OM in some way.Alternatively, maybe we can use properties of cyclic quadrilaterals if certain angles are right angles. For instance, since OP ⊥ BB₁ and OM ⊥ B₁C₁, perhaps points O, M, P, and some other point form a cyclic quadrilateral? Not sure yet.Wait, let's think about the angles we need to relate: angle AOC₁ and angle MPB₁.Angle AOC₁ is at point O, between points A, O, C₁. Since O is the incenter of AB₁C₁, angle OAC₁ is half of angle BAC₁ (since AO is the angle bisector). Similarly, angle OC₁A is half of angle AC₁B₁? Wait, maybe not exactly. Let me recall: the incenter's angles are formed by the angle bisectors.Wait, in triangle AB₁C₁, the incenter O lies at the intersection of the bisectors of angles at A, B₁, and C₁. So AO bisects angle A (of triangle AB₁C₁), which is the same as angle BAC of the original triangle, since C₁ is on AB and B₁ is on AC. Wait, no. Triangle AB₁C₁ has vertex at A, with sides AB₁ and AC₁. Wait, no: if C₁ is on AB, then AC₁ is a segment from A to C₁ on AB, and B₁ is on AC, so AB₁ is a segment from A to B₁ on AC. Therefore, triangle AB₁C₁ is a smaller triangle inside ABC.So angle at A in triangle AB₁C₁ is the same as angle BAC in triangle ABC, right? Wait, no. Because in triangle ABC, angle at A is between sides AB and AC. But in triangle AB₁C₁, the sides are AB₁ and AC₁. So angle at A is still the same angle, between AB and AC. Wait, but AB₁ is part of AC, and AC₁ is part of AB. So actually, the angle at A in triangle AB₁C₁ is the same as angle BAC in triangle ABC. So AO bisects angle BAC.But then, since O is the incenter, AO is the angle bisector of angle A in triangle AB₁C₁, which is the same as angle A in triangle ABC. So AO bisects angle BAC into two equal parts.Similarly, the incenter O has other angle bisectors: from B₁ and C₁. But maybe focusing on AO first.Now, angle AOC₁. Let's see. In triangle AB₁C₁, O is the incenter. So angle AOC₁ would be the angle between the bisector of angle A and the bisector of angle C₁. Hmm. Wait, no. The incenter is where all three angle bisectors meet, so angle AOC₁ is formed by the bisectors of angles at A and C₁. Wait, but angle at C₁ in triangle AB₁C₁ is angle AC₁B₁. Hmm. So angle bisector at C₁ would split angle AC₁B₁ into two equal parts.Similarly, angle AOC₁ would be the angle between the bisector of angle A and the bisector of angle C₁.Alternatively, maybe using trigonometric identities or the inradius.Alternatively, maybe considering triangle AOC₁. If I can find some properties about this triangle or relate it to another triangle.Now, angle MPB₁ is at point P, between points M, P, B₁. So this is an angle formed by the lines PM and PB₁. Since M is the tangency point on B₁C₁, and P is the intersection of BB₁ and CC₁.Given that OP is perpendicular to BB₁, maybe triangle OPB₁ is a right triangle with right angle at P. So OP is the altitude from O to BB₁.Hmm. Since OP is perpendicular to BB₁, and O is the incenter of AB₁C₁. Maybe O lies on some altitude or median related to BB₁?Alternatively, perhaps we can use harmonic division or projective geometry concepts, but that might be more advanced.Wait, perhaps Ceva's theorem? Since BB₁ and CC₁ intersect at P. Ceva's theorem states that for concurrent lines, the product of certain ratios equals 1. But here, BB₁ and CC₁ are two cevians intersecting at P, but we need a third to apply Ceva. Maybe not directly applicable here.Alternatively, Menelaus' theorem? If there's a transversal cutting across the triangle.Alternatively, since M is the point of tangency, we know that B₁M = (AB₁ + B₁C₁ - AC₁)/2, but maybe that's too formulaic. Wait, in a triangle, the lengths from the vertices to the points of tangency are equal to (perimeter/2 - opposite side). So in triangle AB₁C₁, the tangency point M on B₁C₁ would give B₁M = (AB₁ + B₁C₁ - AC₁)/2. Hmm, but not sure how that helps here.Alternatively, since M is the tangency point, OM is perpendicular to B₁C₁. So OM is the radius perpendicular to the side. So OM is the inradius.Given that OP is perpendicular to BB₁, and OM is perpendicular to B₁C₁, perhaps quadrilateral OMPB₁ has some properties. If two of its sides are perpendicular to BB₁ and B₁C₁, but unless those lines are related, maybe not.Alternatively, perhaps reflecting points. Sometimes, reflecting the incenter over a side gives a point on the circumcircle or something, but not sure.Wait, let's think about the angles we need to prove equal: angle AOC₁ and angle MPB₁.If I can express both angles in terms of other angles in the figure and show they are equal.First, angle AOC₁. In triangle AB₁C₁, O is the incenter. So AO and C₁O are angle bisectors. So angle AOC₁ is equal to 180 degrees minus half of angle A minus half of angle C₁. Wait, in a triangle, the angle at the incenter between two angle bisectors is equal to 90 degrees plus half the original angle. Wait, formulaically, in any triangle, the angle at the incenter between two angle bisectors is 90° + half the angle opposite. Wait, maybe.Wait, let me recall: In triangle XYZ, with incenter I, angle XIY = 90° + (angle XZY)/2. Is that right?Wait, yes, in triangle XYZ, the angle at the incenter I between the angle bisectors of X and Y is equal to 90° + (angle Z)/2. So similarly, in triangle AB₁C₁, angle AOC₁ would be 90° + (angle B₁)/2. Wait, is that correct?Wait, let me verify. In triangle ABC, incenter I. The angle at I between angle bisectors from A and B is angle AIB = 90° + (angle C)/2. Yes, that's a standard result.So in triangle AB₁C₁, angle AOC₁ (where O is the incenter) would be 90° + (angle B₁)/2. Wait, angle B₁ in triangle AB₁C₁ is angle AB₁C₁. Which is the same as angle AB₁C in the original triangle, since B₁ is on AC. Hmm. So angle AB₁C is part of the original triangle ABC.Alternatively, angle at B₁ in triangle AB₁C₁ is angle AB₁C₁, which is the same as angle AB₁C in triangle ABC because C₁ is on AB. Wait, maybe not exactly. Let's think.In triangle AB₁C₁, angle at B₁ is between sides AB₁ and B₁C₁. Since AB₁ is part of AC in triangle ABC, and B₁C₁ is a segment from B₁ to C₁, which is on AB. So angle at B₁ in triangle AB₁C₁ is angle AB₁C₁. In triangle ABC, angle at B₁ would be along AC. So angle AB₁C₁ is different from angle ABC.Hmm, perhaps this approach is getting too convoluted. Let me try a different angle.Since OP is perpendicular to BB₁, and O is the incenter, maybe there's a reflection or rotation that maps one angle to the other.Alternatively, consider cyclic quadrilaterals. If two angles are equal, maybe their corresponding points lie on a circle.Alternatively, since both angles AOC₁ and MPB₁ are angles that we need to relate, perhaps showing that their sides are respectively perpendicular or something.Wait, angle AOC₁: sides OA and OC₁. Angle MPB₁: sides PM and PB₁. If I can relate OA to PM and OC₁ to PB₁ through some isometry or similarity.Alternatively, consider triangles AOC₁ and MPB₁. If they are similar, then their angles would be equal. But to show similarity, we need corresponding angles equal or sides in proportion with included angles.Alternatively, maybe triangle AOC₁ is similar to triangle MPB₁.Wait, let me think about the points involved. O is the incenter, M is the tangency point, P is the intersection of cevians. Maybe there's a homothety or inversion that maps one to the other.Alternatively, using power of a point. Since M is the tangency point, power of point P with respect to the incircle. The power would be PM², since PM is tangent to the incircle. So power of P is PM² = PB₁ * something? Not sure.Alternatively, since OP is perpendicular to BB₁, which is a cevian, perhaps O lies on the circumcircle of some triangle related to BB₁ and its perpendiculars.Alternatively, coordinate geometry. Let's try setting coordinates.Let me place point A at the origin (0,0), point B at (c,0), and point C at (d,e). Then, points B₁ on AC and C₁ on AB can be parameterized. Let me define coordinates:Let’s set coordinate system with A at (0,0), B at (1,0), and C at (0,1) for simplicity. Wait, but then AC is from (0,0) to (0,1), and AB is from (0,0) to (1,0). Then points B₁ is on AC, so let’s say B₁ is at (0, b) for some 0 < b < 1. Similarly, point C₁ is on AB, so let's say C₁ is at (c, 0) for some 0 < c < 1.Then lines BB₁ and CC₁ intersect at P.First, let's compute coordinates of P. BB₁ is the line from B(1,0) to B₁(0, b). The equation of BB₁ can be found. Similarly, CC₁ is the line from C(0,1) to C₁(c, 0). Let's compute their intersection P.Equation of BB₁: passing through (1,0) and (0, b). The slope is (b - 0)/(0 - 1) = -b. So equation is y = -b(x - 1) = -bx + b.Equation of CC₁: passing through (0,1) and (c, 0). Slope is (0 - 1)/(c - 0) = -1/c. Equation is y = (-1/c)x + 1.To find P, set the two equations equal:-bx + b = (-1/c)x + 1Multiply both sides by c to eliminate denominator:-bc x + bc = -x + cBring all terms to left:(-bc x + x) + (bc - c) = 0x(-bc + 1) + c(b - 1) = 0x(1 - bc) = c(1 - b)Thus, x = [c(1 - b)] / (1 - bc)Then y = -b x + b = -b [c(1 - b)/(1 - bc)] + b = [-bc(1 - b) + b(1 - bc)] / (1 - bc)Simplify numerator:- bc + b²c + b - b²c = -bc + bThus, y = (b - bc) / (1 - bc) = b(1 - c)/(1 - bc)Therefore, coordinates of P are:x = c(1 - b)/(1 - bc)y = b(1 - c)/(1 - bc)Now, let's find the incenter O of triangle AB₁C₁.Triangle AB₁C₁ has vertices at A(0,0), B₁(0,b), and C₁(c,0). The incenter O can be found using the formula that weights the sides.In a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), the incenter coordinates are ( (a x₁ + b x₂ + c x₃)/(a + b + c), (a y₁ + b y₂ + c y₃)/(a + b + c) ), where a, b, c are the lengths of the sides opposite the respective vertices.In triangle AB₁C₁:- Side opposite A is B₁C₁. Let's compute its length.Coordinates of B₁: (0, b), C₁: (c,0). So length B₁C₁ = sqrt((c - 0)^2 + (0 - b)^2) = sqrt(c² + b²). Let's call this length a.- Side opposite B₁ is AC₁. Coordinates of A(0,0) and C₁(c,0). Length AC₁ = c. Let's call this length b.- Side opposite C₁ is AB₁. Coordinates of A(0,0) and B₁(0,b). Length AB₁ = b. Let's call this length c.Wait, in the formula, a, b, c are the lengths opposite to vertices A, B, C. But in triangle AB₁C₁, the vertices are A, B₁, C₁. So the side opposite A is B₁C₁, length sqrt(b² + c²). The side opposite B₁ is AC₁, length c. The side opposite C₁ is AB₁, length b.Therefore, the incenter O has coordinates:( (sqrt(b² + c²)*0 + c*0 + b*c) / (sqrt(b² + c²) + c + b), (sqrt(b² + c²)*0 + c*b + b*0) / (sqrt(b² + c²) + c + b) )Wait, no. Wait, in the formula, the coordinates are weighted by the lengths opposite each vertex. For vertex A, the weight is the length opposite A, which is B₁C₁ = sqrt(b² + c²). For vertex B₁, the weight is AC₁ = c. For vertex C₁, the weight is AB₁ = b.Therefore, coordinates of O:x = (sqrt(b² + c²)*0 + c*0 + b*c) / (sqrt(b² + c²) + c + b)Wait, no. Wait, for vertex A (0,0), the weight is the side opposite A, which is B₁C₁ = sqrt(b² + c²). So the x-coordinate is (sqrt(b² + c²)*0 + c*0 + b*c) / (sqrt(b² + c²) + c + b). Wait, actually, the formula is ( (a x_A + b x_B + c x_C ) / (a + b + c) ), where a, b, c are the lengths opposite to A, B, C.But in triangle AB₁C₁, vertices are A(0,0), B₁(0,b), C₁(c,0). Let’s denote them as A, B₁, C₁. Then the side opposite A is B₁C₁, length a = sqrt(b² + c²). The side opposite B₁ is AC₁, length c. The side opposite C₁ is AB₁, length b.Therefore, the incenter coordinates O would be:x = (a * x_A + b * x_{B₁} + c * x_{C₁}) / (a + b + c)Similarly for y.So plugging in:x = (a * 0 + b * 0 + c * c) / (a + b + c) = c² / (sqrt(b² + c²) + b + c)y = (a * 0 + b * b + c * 0) / (a + b + c) = b² / (sqrt(b² + c²) + b + c)So coordinates of O are ( c² / D, b² / D ) where D = sqrt(b² + c²) + b + c.Now, the tangency point M on B₁C₁. In triangle AB₁C₁, the incircle touches B₁C₁ at M. The coordinates of M can be found using the formula that the tangency point divides the side into segments proportional to the adjacent sides.In any triangle, the tangency point on a side is located at a distance from the vertex equal to (perimeter/2 - opposite side). So in triangle AB₁C₁, the tangency point M on B₁C₁ is located at distance from B₁ equal to (AB₁ + B₁C₁ - AC₁)/2.Wait, AB₁ is length b, B₁C₁ is sqrt(b² + c²), and AC₁ is c. So the length from B₁ to M is (b + sqrt(b² + c²) - c)/2. Similarly, from C₁ to M is (c + sqrt(b² + c²) - b)/2.Alternatively, since B₁ is (0, b) and C₁ is (c, 0), the coordinates of M can be found by moving from B₁ towards C₁ by the length (b + sqrt(b² + c²) - c)/2.But maybe parameterizing the line B₁C₁.Parametric equations for B₁C₁: from (0, b) to (c, 0). The vector from B₁ to C₁ is (c, -b). The length of B₁C₁ is sqrt(c² + b²). So the unit vector in that direction is (c / sqrt(c² + b²), -b / sqrt(c² + b²)).Therefore, the distance from B₁ to M is (b + sqrt(b² + c²) - c)/2. Therefore, the coordinates of M are:B₁ + distance * unit vectorSo:x = 0 + [(b + sqrt(b² + c²) - c)/2] * (c / sqrt(c² + b²))y = b + [(b + sqrt(b² + c²) - c)/2] * (-b / sqrt(c² + b²))Simplify:x = [c(b + sqrt(b² + c²) - c)] / [2 sqrt(c² + b²)]y = b - [b(b + sqrt(b² + c²) - c)] / [2 sqrt(c² + b²)]This seems complicated. Maybe keeping it symbolic for now.Now, we need to find angle AOC₁ and angle MPB₁ and show they are equal.First, angle AOC₁. Points A(0,0), O(c²/D, b²/D), C₁(c,0). So vectors OA and OC₁ can be considered.Vector OA is from A to O: (c²/D, b²/D)Vector OC₁ is from O to C₁: (c - c²/D, 0 - b²/D) = (c(1 - c/D), -b²/D)The angle between vectors OA and OC₁ can be found using the dot product.Similarly, angle MPB₁: points M, P, B₁. Need coordinates of M, P, B₁.Coordinates of P were found earlier: x = c(1 - b)/(1 - bc), y = b(1 - c)/(1 - bc)Coordinates of M as above. It's going to be messy, but perhaps with the coordinates, we can compute the angles and show equality.Alternatively, maybe there's a symmetry or ratio that makes this easier.Alternatively, note that OP is perpendicular to BB₁. Let's compute the slope of OP and BB₁ and verify they are negative reciprocals.Coordinates of O: (c²/D, b²/D). Coordinates of P: (c(1 - b)/(1 - bc), b(1 - c)/(1 - bc)).Slope of OP: [y_P - y_O]/[x_P - x_O] = [b(1 - c)/(1 - bc) - b²/D] / [c(1 - b)/(1 - bc) - c²/D]Similarly, slope of BB₁: from B(1,0) to B₁(0,b). Slope is (b - 0)/(0 - 1) = -b.Given that OP is perpendicular to BB₁, the product of their slopes should be -1. So:Slope of OP * (-b) = -1 => Slope of OP = 1/b.So if we compute the slope of OP from coordinates, it should equal 1/b. This might give an equation relating b and c. Maybe this can help us find a relationship between b and c.But this seems complicated. Let me see.First, compute slope of OP:Numerator: y_P - y_O = [b(1 - c)/(1 - bc)] - [b² / D]Denominator: x_P - x_O = [c(1 - b)/(1 - bc)] - [c² / D]Where D = sqrt(b² + c²) + b + c.This looks very messy, but maybe substituting D.Alternatively, since this condition must hold for any triangle, maybe there is a specific ratio or relation between b and c that satisfies OP ⊥ BB₁. Perhaps assuming specific values for b and c to simplify.Wait, maybe take a special case where b = c. Let's assume b = c, so D = sqrt(b² + b²) + b + b = b√2 + 2b.Then O's coordinates become (b² / (b√2 + 2b), b² / (b√2 + 2b)) = (b / (√2 + 2), b / (√2 + 2)).Coordinates of P: If b = c, then x = b(1 - b)/(1 - b²) = b(1 - b)/[(1 - b)(1 + b)] = b/(1 + b)Similarly, y = b(1 - b)/(1 - b²) = same as x, so P is (b/(1 + b), b/(1 + b))Coordinates of M: In this case, since b = c, length from B₁ to M is (b + sqrt(2b²) - b)/2 = (b + b√2 - b)/2 = (b√2)/2. So M is located at distance b√2 / 2 from B₁ along B₁C₁.Since B₁ is (0,b) and C₁ is (b,0), the direction from B₁ to C₁ is (b, -b). Unit vector is (1/√2, -1/√2). So moving a distance of b√2 / 2 from B₁:M = (0, b) + (b√2 / 2)(1/√2, -1/√2) = (0, b) + (b/2, -b/2) = (b/2, b/2)So M is at (b/2, b/2)Now, check OP perpendicular to BB₁.Slope of BB₁ is -b (since from (1,0) to (0,b)). Slope of OP: from O(b/(√2 + 2), b/(√2 + 2)) to P(b/(1 + b), b/(1 + b))Compute slope:[y_P - y_O]/[x_P - x_O] = [b/(1 + b) - b/(√2 + 2)] / [b/(1 + b) - b/(√2 + 2)] = same numerator and denominator, so slope is 1. But slope of BB₁ is -b. For them to be perpendicular, 1 * (-b) = -1 => b = 1.So if b = c = 1, then the condition holds. Let's set b = c = 1.Then coordinates:A(0,0), B(1,0), C(0,1). B₁ is on AC: (0,1). Wait, but b = 1, so B₁ is (0,1), which is point C. Similarly, C₁ is on AB at (1,0), which is point B. But then triangle AB₁C₁ becomes triangle ACB, which is the original triangle. So P would be intersection of BB₁ (which is BC) and CC₁ (which is CB). So P would be point B or C? Wait, this is degenerate.Hmm, setting b = c = 1 leads to degenerate triangle. So perhaps this approach is flawed. Maybe b = c but less than 1.Wait, let's try b = c = t, where t is between 0 and 1. Then coordinates:O(t² / (t√2 + 2t), t² / (t√2 + 2t)) = (t / (√2 + 2), t / (√2 + 2))P(t/(1 + t), t/(1 + t))M(t/2, t/2)Slope of OP: [t/(1 + t) - t/(√2 + 2)] / [t/(1 + t) - t/(√2 + 2)] = same numerator and denominator, so slope is 1.Slope of BB₁ is -t. For OP to be perpendicular to BB₁, slope product must be -1:1 * (-t) = -1 => t = 1.But t=1 is degenerate. Therefore, in this specific case, the only solution is degenerate, which suggests that assuming b = c isn't helpful unless we allow t=1, but that's the triangle itself. So perhaps this special case doesn't help.Alternatively, maybe take b = c but less than 1 and see if the angles still hold. Wait, but if slope of OP is 1 and slope of BB₁ is -t, then unless t =1, they are not perpendicular. Hence, only when t=1, which is degenerate. Therefore, this suggests that the case b = c is not helpful unless degenerate.Alternatively, take another specific case. Let me try b = 1/2, c =1/2.So b = c =1/2. Then D = sqrt( (1/2)^2 + (1/2)^2 ) +1/2 +1/2 = sqrt(1/2) +1 ≈ 0.707 +1 =1.707Coordinates of O: ( (1/2)^2 /1.707 , (1/2)^2 /1.707 ) ≈ (0.25/1.707, 0.25/1.707) ≈ (0.146, 0.146)Coordinates of P: x = (1/2)(1 -1/2)/(1 - (1/2)(1/2)) = (1/2)(1/2)/(1 -1/4) = (1/4)/(3/4)=1/3. Similarly, y=1/3.Coordinates of M: For b = c =1/2, in triangle AB₁C₁, the tangency point M on B₁C₁.B₁ is (0,1/2), C₁ is (1/2,0). The side B₁C₁ is from (0,1/2) to (1/2,0). The length from B₁ to M is (AB₁ + B₁C₁ - AC₁)/2. AB₁ =1/2, B₁C₁ = sqrt( (1/2)^2 + (1/2)^2 )=sqrt(1/2)=√2/2, AC₁=1/2.So length B₁M = (1/2 + √2/2 -1/2)/2= (√2/2)/2=√2/4≈0.3536.Coordinates of M: Starting at B₁(0,1/2), moving along B₁C₁ towards C₁(1/2,0) by √2/4. The direction vector is (1/2, -1/2), which has length sqrt( (1/2)^2 + ( -1/2)^2 )=sqrt(1/2)=√2/2. So unit vector is (1/2)/(√2/2), (-1/2)/(√2/2))=(1/√2, -1/√2)Thus, moving √2/4 along this direction:M = B₁ + √2/4*(1/√2, -1/√2)= (0,1/2) + (1/4, -1/4)= (1/4, 1/4)So M is at (1/4,1/4)Now, check if OP is perpendicular to BB₁.Coordinates of O: (0.146,0.146) ≈ (1/6.85,1/6.85)Coordinates of P: (1/3,1/3)Slope of OP: (1/3 -0.146)/(1/3 -0.146) ≈ (0.333 -0.146)/(0.333 -0.146)=0.187/0.187=1. So slope is 1.Slope of BB₁: from B(1,0) to B₁(0,1/2). Slope is (1/2 -0)/(0 -1)= -1/2.Product of slopes:1*(-1/2)= -1/2 ≠ -1. So not perpendicular. Contradicts the given condition. Therefore, in this case, OP is not perpendicular to BB₁, which violates the problem's condition. So this case is invalid.Therefore, our assumption of b = c =1/2 doesn't satisfy the given condition. Thus, such a case isn't applicable. Hence, coordinate approach with specific values may not be the best route unless we can derive general relations.Alternatively, since this seems too calculation-heavy, perhaps there's a synthetic approach using known theorems or properties.Let me recall that in a triangle, if the incenter's perpendicular to a cevian, there might be properties related to equal angles or symmetries.Alternatively, since OP is perpendicular to BB₁, and M is the tangency point, perhaps PM is the tangent from P to the incircle. Then, PM = PB₁*sin(angle) or something.Wait, PM is tangent to the incircle, so power of point P with respect to the incircle is PM² = (distance from P to O)^2 - r², where r is the inradius. But not sure.Alternatively, since OP is perpendicular to BB₁, and we need to relate angle AOC₁ to angle MPB₁, maybe there's an orthocenter or reflection involved.Alternatively, consider triangle OPB₁. It's right-angled at P. So OP is perpendicular to PB₁ (but PB₁ is BB₁, which is the same line). So triangle OPB₁ is right-angled at P.Similarly, in triangle AB₁C₁, O is the incenter, M is the tangency point. So OM is perpendicular to B₁C₁.Therefore, OM and OP are both perpendicular to lines involving B₁: OM ⊥ B₁C₁ and OP ⊥ BB₁. Maybe quadrilateral OMPB₁ has some orthocentric properties.Alternatively, consider that both O and M lie on the incircle related to B₁C₁. Wait, O is the center, so it's not on the incircle. M is on the incircle.Alternatively, since PM is tangent to the incircle, and OM is the radius, then PM is perpendicular to OM. Wait, yes! Because the radius at the point of tangency is perpendicular to the tangent line. So PM ⊥ OM.So angle OMP is 90 degrees. Similarly, we already have OP ⊥ PB₁ (which is BB₁), so angle OPB₁ is 90 degrees.Therefore, quadrilateral OMPB₁ has two right angles at M and P. If two opposite angles are 90 degrees, is it cyclic? No, because in a cyclic quadrilateral, the sum of opposite angles is 180°, not necessarily each being 90°. However, if two adjacent angles are 90°, it's not necessarily cyclic either.But maybe considering triangles OMP and OPB₁.Wait, since PM ⊥ OM and OP ⊥ PB₁, maybe there's some similar triangles here.Alternatively, consider that triangles OMP and PBO share some properties. Wait, not sure.Alternatively, since PM is tangent to the incircle, and OP is perpendicular to BB₁, perhaps there is a reflection that maps one line to another.Wait, another approach: angle chasing.Let me try to express both angles AOC₁ and MPB₁ in terms of other angles and see if they can be related.First, angle AOC₁. In triangle AB₁C₁, O is the incenter. So angle AOC₁ is equal to angle A + angle C₁ of triangle AB₁C₁ divided by 2, plus 90°, or some combination. Wait, need to recall the formula for the angle at the incenter.In triangle ABC, angle BIC = 90° + (angle A)/2, where I is the incenter. Similarly, in triangle AB₁C₁, angle AOC₁ would be 90° + (angle B₁)/2.Wait, yes, in any triangle, the angle at the incenter between two angle bisectors is 90° plus half the third angle.So in triangle AB₁C₁, angle at O between AO and C₁O is angle AOC₁ = 90° + (angle B₁)/2.Similarly, angle MPB₁ is an angle in triangle MPB₁. Let's see if it can be expressed as 90° plus half of some angle.But angle MPB₁ is at point P, between M and B₁. If we can relate this angle to angle B₁ of triangle AB₁C₁, then maybe equality can be shown.Alternatively, since PM is tangent to the incircle, and OM is the radius, angle OMP is 90°. Also, OP is perpendicular to PB₁, angle OPB₁ is 90°. So we have two right angles at M and P.Perhaps quadrilateral OMPB₁ is cyclic? If two opposite angles are 90°, but in cyclic quadrilaterals, opposite angles sum to 180°, so two right angles would satisfy that. Wait, if quadrilateral has four vertices with two right angles at M and P, then it could be cyclic. If OMPB₁ is cyclic, then angle MPB₁ would be equal to angle MOB₁, because they subtend the same arc.But wait, in a cyclic quadrilateral, angles subtended by the same chord are equal. So if OMPB₁ is cyclic, angle MPB₁ would equal angle MOB₁.But angle MOB₁ is the angle at O between M and B₁. Since OM is perpendicular to B₁C₁ and O is the incenter, perhaps angle MOB₁ relates to the incenter's angle.Alternatively, since O is the incenter, angle MOB₁ is equal to 90° + (angle A)/2 or something similar. Wait, need to think.But maybe this is leading somewhere. If OMPB₁ is cyclic, then angle MPB₁ = angle MOB₁. But angle MOB₁ is the angle at O between M and B₁.But M is the tangency point, so OM is the inradius and perpendicular to B₁C₁. So angle MOB₁ is the angle between OM (which is perpendicular to B₁C₁) and OB₁.But OB₁ is the angle bisector of angle B₁ in triangle AB₁C₁. So angle MOB₁ is equal to angle between the angle bisector of B₁ and the perpendicular to B₁C₁.Alternatively, in triangle AB₁C₁, the incenter O, so angle OB₁M is equal to half of angle B₁.Wait, angle at B₁ is angle AB₁C₁. The angle bisector at B₁ divides it into two equal parts. So angle OB₁C₁ = (angle AB₁C₁)/2.Since OM is perpendicular to B₁C₁, angle MOB₁ = 90° - angle OB₁M = 90° - (angle AB₁C₁)/2.But angle AOC₁ was earlier thought to be 90° + (angle AB₁C₁)/2. If angle MOB₁ = 90° - (angle AB₁C₁)/2, then angle AOC₁ + angle MOB₁ = 180°.But if angle MPB₁ = angle MOB₁ (if quadrilateral is cyclic), then angle AOC₁ = 180° - angle MPB₁. But we need angle AOC₁ = angle MPB₁, which would imply 180° = 2 angle MPB₁, which would mean angle MPB₁ = 90°, which isn't necessarily the case.Hmm, perhaps this approach isn't correct.Wait, but earlier we derived that angle AOC₁ = 90° + (angle B₁)/2. If angle MPB₁ can be shown to be equal to 90° + (angle B₁)/2, then they would be equal.Alternatively, maybe angle MPB₁ is related to angle B₁ through some other geometric relation.Given that PM is tangent to the incircle, which has center O. Then, PM is perpendicular to OM. So triangle PMO is right-angled at M.Similarly, OP is perpendicular to PB₁, so triangle OPPB₁ is right-angled at P.So we have two right triangles: PMO and OPB₁.Maybe these two triangles are similar?In triangle PMO: right angle at M.In triangle OPB₁: right angle at P.If they are similar, then the corresponding angles would be equal. For similarity, we need angles to match. For example, angle at O in both triangles.Angle at O in triangle PMO is angle POM, and in triangle OPB₁ is angle OPB₁.If angle POM = angle OPB₁, then similarity could hold.Alternatively, since both triangles share angle at O (if we consider angle POM and angle POB₁), but not sure.Alternatively, ratios of sides. In triangle PMO: PM/MO = tan(angle POM). In triangle OPB₁: OP/PB₁ = tan(angle OPB₁).If angle POM = angle OPB₁, then tan(angle POM) = tan(angle OPB₁), so PM/MO = OP/PB₁. If this ratio holds, then triangles are similar.Alternatively, if we can show PM/MO = OP/PB₁, then triangles PMO and OPB₁ are similar.But how to show this ratio?Given that O is the incenter of AB₁C₁, MO is the inradius r. Also, PM is the length of the tangent from P to the incircle, which is sqrt(PO² - r²) by the power of point P.But OP is given, and PB₁ is a segment from P to B₁.But this seems again too algebraic. Let me see:Power of point P with respect to the incircle is PM² = PO² - r².So PM = sqrt(PO² - r²).But in triangle OPB₁, which is right-angled at P, we have OP² + PB₁² = OB₁².Wait, no, triangle OPB₁ is right-angled at P, so OP² + PB₁² = OB₁².But OB₁ is the distance from O to B₁. Since O is the incenter of AB₁C₁, OB₁ is the distance from O to vertex B₁, which can be computed, but I'm not sure.Alternatively, maybe using coordinates again, but this is time-consuming. Alternatively, recall that in triangle AB₁C₁, the inradius is r = area / semiperimeter.But this might not help directly.Alternatively, since PM is tangent to the incircle, and PB₁ is a secant line, by power of a point, PM² = PB₁ * PC₁, but PC₁ is not necessarily related here. Wait, PB₁ is from P to B₁, but PC₁ is from P to C₁. Not sure.Wait, power of point P with respect to the incircle is PM² = PO² - r².But how to relate PO and PB₁?Alternatively, in triangle PMO and triangle OPB₁:If we can show that PM / OP = MO / PB₁, then by similarity, triangles would be similar.PM / OP = MO / PB₁Cross-multiplying: PM * PB₁ = OP * MOBut PM = sqrt(PO² - r²), MO = r. So:sqrt(PO² - r²) * PB₁ = OP * rSquare both sides:(PO² - r²) * PB₁² = OP² * r²PO² * PB₁² - r² * PB₁² = OP² * r²Bring all terms to left:PO² * PB₁² - OP² * r² - r² * PB₁² = 0Factor:PO² * PB₁² - r² (OP² + PB₁²) = 0But OP² + PB₁² = OB₁² from triangle OPB₁ (right-angled at P).Thus:PO² * PB₁² - r² * OB₁² = 0But OB₁ is the distance from O to B₁, which can be expressed.In triangle AB₁C₁, O is the incenter. The distance from O to B₁ is equal to the length of the angle bisector of B₁, but not sure.Alternatively, in triangle AB₁C₁, using the formula for the distance from incenter to a vertex:The distance from O to B₁ is sqrt(r² + (s - AC₁)^2 ), where s is the semiperimeter.In triangle AB₁C₁, semiperimeter s = (AB₁ + B₁C₁ + AC₁)/2.Then, the distance from O to B₁ is sqrt(r² + (s - AC₁)^2 ).But AC₁ is a side of triangle AB₁C₁. So s - AC₁ = (AB₁ + B₁C₁ + AC₁)/2 - AC₁ = (AB₁ + B₁C₁ - AC₁)/2 = B₁M.Therefore, the distance from O to B₁ is sqrt(r² + (B₁M)^2 ). But B₁M is the length from B₁ to M, which is the tangent segment, which is equal to s - AC₁.But B₁M = PM is not necessarily, because PM is the tangent from P to the incircle, which is different.Wait, no. B₁M is the tangent from B₁ to the incircle, which is equal to s - AC₁ in triangle AB₁C₁. But PM is the tangent from P to the incircle.This seems complicated. Perhaps another approach.Recall that in triangle AB₁C₁, the incenter O, and M is the touch point on B₁C₁. Then, line OM is perpendicular to B₁C₁. Line OP is perpendicular to BB₁. We need to relate angles AOC₁ and MPB₁.Maybe using the fact that both angles are equal to 90° plus half of angle B₁.But earlier, angle AOC₁ = 90° + (angle B₁)/2. If angle MPB₁ can also be shown to be 90° + (angle B₁)/2, then they are equal.Let me see angle MPB₁. It is the angle at P between points M and B₁.In triangle MPB₁, angle at P. If we can express this angle in terms of angle B₁.Since PM is tangent to the incircle, and OM is perpendicular to B₁C₁, then PM is the tangent, and angle between tangent PM and PB₁ is equal to the angle in the alternate segment. Wait, yes! There's a theorem that the angle between a tangent and a chord is equal to the angle in the alternate segment.So angle between tangent PM and chord PB₁ is equal to the angle that PB₁ makes with the chord in the alternate segment. The alternate segment here would be the segment of the circle opposite to the angle. Since the incircle is tangent to B₁C₁ at M, the alternate segment would be the arc MB₁ on the incircle.But the angle in the alternate segment would be the angle between the chord MB₁ and the tangent at M, which is equal to the angle that chord MB₁ makes with the other tangent at M. Wait, perhaps not directly applicable.Wait, the tangent at M is PM. The angle between PM and PB₁ is equal to the angle that PB₁ makes with the other tangent from M, which is the angle in the alternate segment. But since the incircle is tangent to B₁C₁ at M, the angle between PM and MB₁ is equal to the angle in the alternate segment, which would be angle MOB₁.Wait, maybe more precisely, the angle between tangent PM and chord MB₁ is equal to the angle that MB₁ makes with the other tangent at M, which is the angle in the alternate segment. But since there's only one tangent at M, this might not hold.Alternatively, by the alternate segment theorem, the angle between tangent PM and chord MB₁ is equal to the angle that MB₁ makes with the opposite side of the circle. But since the incircle is tangent to B₁C₁ at M, the alternate segment would be the arc MB₁ on the incircle, but the incircle is tangent to other sides as well.Wait, maybe the alternate segment theorem is not straightforward here because the incircle is tangent to all three sides, not just one.Alternatively, focusing on triangle PMB₁. Angle MPB₁ is the angle we need. If we can relate it to angle AOC₁.Given that angle AOC₁ = 90° + (angle B₁)/2, and angle MPB₁ needs to be equal to that.Since OP is perpendicular to BB₁, and O is the incenter, perhaps there's an isosceles triangle or reflection that links these angles.Alternatively, since OP is perpendicular to BB₁, which passes through P, and M is the tangency point, perhaps there is a rotation or reflection that swaps O and M or something.Alternatively, construct a line from O perpendicular to BB₁, which is OP, and since M is the tangency point, which is related to B₁C₁.Another thought: since OP is perpendicular to BB₁ and OM is perpendicular to B₁C₁, then OP and OM are each perpendicular to different lines. If we can relate the directions of BB₁ and B₁C₁, perhaps through some angle relation.If I can show that angle between OP and OM is equal to the angle between BB₁ and B₁C₁, then maybe some parallelogram or rotational symmetry exists.Alternatively, since OP ⊥ BB₁ and OM ⊥ B₁C₁, the angle between OP and OM is equal to the angle between BB₁ and B₁C₁. Because if two lines are perpendicular to two other lines, the angle between the first two is equal to the angle between the latter two.So angle between OP and OM is equal to angle between BB₁ and B₁C₁.But angle between BB₁ and B₁C₁ is angle at B₁ between BB₁ and B₁C₁, which is angle CB₁C₁ in triangle ABC. Not sure.Alternatively, perhaps using trigonometric identities in triangle OPB₁ and OMB₁.Given the complexity, maybe it's better to return to the coordinate approach but proceed symbolically.Recall that:Coordinates of O: (c²/D, b²/D), where D = sqrt(b² + c²) + b + c.Coordinates of P: (c(1 - b)/(1 - bc), b(1 - c)/(1 - bc))Coordinates of M: ( [c(b + sqrt(b² + c²) - c)] / [2 sqrt(b² + c²)] , [b(b + sqrt(b² + c²) - c) ] / [2 sqrt(b² + c²)] )Wait, this is very complicated, but perhaps we can compute vectors for the angles.First, angle AOC₁. This is the angle at O between points A, O, C₁.Vector OA: from O to A: (0 - c²/D, 0 - b²/D) = (-c²/D, -b²/D)Vector OC₁: from O to C₁: (c - c²/D, 0 - b²/D) = (c(1 - c/D), -b²/D)The angle between these vectors can be found using the dot product:cos(angle AOC₁) = (OA . OC₁) / (|OA| |OC₁|)Compute OA . OC₁:(-c²/D)(c(1 - c/D)) + (-b²/D)(-b²/D)= -c³(1 - c/D)/D + b⁴/D²|OA| = sqrt( (c²/D)^2 + (b²/D)^2 ) = (1/D) sqrt(c⁴ + b⁴)|OC₁| = sqrt( [c(1 - c/D)]^2 + [ -b²/D ]^2 )= sqrt( c²(1 - c/D)^2 + b⁴/D² )This is getting very messy. Maybe there's a simplification I'm missing.Similarly, angle MPB₁ is the angle at P between M, P, B₁.Coordinates of PM: from P to M: (x_M - x_P, y_M - y_P)Coordinates of PB₁: from P to B₁: (0 - x_P, b - y_P)The angle between vectors PM and PB₁ can be found using the dot product:cos(angle MPB₁) = (PM . PB₁) / (|PM| |PB₁|)This would involve similar complex computations.Given the complexity of the coordinate approach, it's clear that this path is too calculation-heavy and may not lead to an elegant proof. Therefore, a synthetic geometric approach is necessary.Let me think differently. Since OP is perpendicular to BB₁, and we need to relate angle AOC₁ to angle MPB₁, perhaps there's a way to show that both angles are equal by relating them to a common third angle.Given that O is the incenter, and M is the point of tangency, perhaps considering the excenter or other triangle centers.Alternatively, since OP is perpendicular to BB₁, maybe O lies on the circumcircle of triangle BPB₁ or some related circle.Alternatively, consider homothety. The incenter O is located at a certain position, and under some homothety or inversion, it could map to M or P, preserving angles.Alternatively, use the fact that both angles AOC₁ and MPB₁ are related to the angle bisectors or perpendiculars.Wait, another idea: Since OP is perpendicular to BB₁, and OM is perpendicular to B₁C₁, then lines OP and OM are altitudes of some triangle. Maybe triangle OB₁P or similar.Alternatively, since PM is tangent to the incircle, then PM is equal to the length of the tangent from P to the circle. Similarly, other tangent lengths could be equal.But perhaps the key lies in noting that both angles AOC₁ and MPB₁ are right angles plus some other angle, and due to the perpendicularity condition, these angles are equal.Recall that angle AOC₁ = 90° + (angle B₁)/2 as per the incenter angle property. If angle MPB₁ can be shown to also be 90° + (angle B₁)/2, then they are equal.To find angle MPB₁, consider triangle MPB₁. Since PM is tangent to the incircle at M, and OM is the radius, then angle OMP = 90°. Also, OP is perpendicular to BB₁ (angle OPB₁ = 90°). So we have two right angles involving O, M, P, B₁.Perhaps quadrilateral OMPB₁ has some cyclic properties or the angles relate through some other geometric relations.Since angle OMP = 90° and angle OPB₁ = 90°, points O, M, P, B₁ lie on a circle with diameter OP. Wait, no. In a circle, if two angles subtended by the same chord are right angles, then the chord must be the diameter. But here, angles at M and P are both 90°, so the circle would have OM and OP as diameters, but that's only possible if M and P are endpoints of the diameter, which they aren't.Alternatively, if quadrilateral OMPB₁ is cyclic, then the sum of opposite angles must be 180°. Angle at M is 90°, angle at P is 90°, so sum is 180°, which satisfies cyclic quadrilateral condition. Therefore, OMPB₁ is cyclic.Therefore, points O, M, P, B₁ lie on a circle. Therefore, angle MPB₁ = angle MOB₁ because they subtend the same arc MB₁.Therefore, angle MPB₁ = angle MOB₁.Now, angle MOB₁ is the angle at O between M and B₁. Since OM is perpendicular to B₁C₁ and OB₁ is the angle bisector of angle AB₁C₁, angle MOB₁ can be related to the incenter's angles.In triangle AB₁C₁, the incenter O, so angle OB₁C₁ = (angle AB₁C₁)/2. Since OM is perpendicular to B₁C₁, angle MOB₁ = 90° - angle OB₁M.But angle OB₁M is equal to angle OB₁C₁, which is (angle AB₁C₁)/2.Therefore, angle MOB₁ = 90° - (angle AB₁C₁)/2.But earlier, angle AOC₁ was supposed to be 90° + (angle AB₁C₁)/2. Therefore, angle MPB₁ = angle MOB₁ = 90° - (angle AB₁C₁)/2, which is not equal to angle AOC₁. Contradicts what we need to prove.Wait, this suggests a mistake in the reasoning. Earlier, we thought angle AOC₁ = 90° + (angle AB₁C₁)/2, but if angle MPB₁ = 90° - (angle AB₁C₁)/2, then they are supplementary, not equal. Hence, this approach must be wrong.But according to the standard incenter angle formula, in triangle AB₁C₁, angle AOC₁ = 90° + (angle B₁)/2, where angle B₁ is angle AB₁C₁. If angle MPB₁ is equal to angle MOB₁ = 90° - (angle B₁)/2, then angle AOC₁ + angle MPB₁ = 180°, which would imply that they are supplementary, not equal. But the problem states that they are equal, so this suggests a flaw in the reasoning.Therefore, perhaps the earlier assumption that angle AOC₁ = 90° + (angle B₁)/2 is incorrect. Let me verify this formula.In a triangle XYZ, the angle at the incenter between the angle bisectors of X and Y is equal to 90° + (Z)/2. So in triangle AB₁C₁, angle at O between AO and C₁O is angle AOC₁ = 90° + (angle B₁)/2. So this part seems correct.But then angle MPB₁ = angle MOB₁ = 90° - (angle B₁)/2. Hence, angle AOC₁ = 180° - angle MPB₁.But the problem states that angle AOC₁ = angle MPB₁, which would require 180° - angle MPB₁ = angle MPB₁, hence angle MPB₁ = 90°, which isn't necessarily the case.This suggests that there is a mistake in the assumption that angle MPB₁ = angle MOB₁. But if quadrilateral OMPB₁ is cyclic, then angles subtended by the same arc should be equal. So angle MPB₁ and angle MOB₁ should subtend the same arc, which would be arc MMB₁ or something. Wait, in cyclic quadrilateral OMPB₁, angle MPB₁ and angle MOB₁ both subtend arc MOB₁. But wait, angle MPB₁ is at point P, subtended by arc MOB₁, and angle MOB₁ is at point O, subtended by arc MPB₁. Wait, no. In a cyclic quadrilateral, angles subtended by the same chord are either equal or supplementary, depending on their position.Specifically, in cyclic quadrilateral OMPB₁, angle MPB₁ and angle MOB₁ are subtended by chord MB₁. Since P and O are on opposite sides of chord MB₁, these angles are supplementary. Therefore, angle MPB₁ + angle MOB₁ = 180°. But we previously had angle MOB₁ = 90° - (angle B₁)/2, so angle MPB₁ = 180° - (90° - (angle B₁)/2) = 90° + (angle B₁)/2.Ah, this resolves the contradiction! So angle MPB₁ = 90° + (angle B₁)/2, which is equal to angle AOC₁. Hence, angle AOC₁ = angle MPB₁, as required.Let me verify this step again:1. Since quadrilateral OMPB₁ is cyclic (because angles at M and P are right angles), then angles subtended by the same chord from opposite sides are supplementary.2. Angle MPB₁ and angle MOB₁ are subtended by chord MB₁ from points P and O, which are on opposite sides of the chord. Hence, angle MPB₁ + angle MOB₁ = 180°.3. Earlier, we found angle MOB₁ = 90° - (angle B₁)/2.4. Therefore, angle MPB₁ = 180° - (90° - (angle B₁)/2) = 90° + (angle B₁)/2.5. Since angle AOC₁ = 90° + (angle B₁)/2, we have angle AOC₁ = angle MPB₁.Therefore, the proof is complete.Final Answerboxed{angle AOC_1 = angle MPB_1}

✅:To prove that (angle AOC_1 = angle MPB_1) given the conditions of the problem, we can follow this reasoning:1. Understanding the Given Conditions: - Points (B_1) and (C_1) are on sides (AC) and (AB) of triangle (ABC). - Lines (BB_1) and (CC_1) intersect at point (P). - (O) is the incenter of triangle (AB_1C_1), and (M) is the point of tangency of the incircle with (B_1C_1). - (OP) is perpendicular to (BB_1).2. Key Properties and Relationships: - The incenter (O) of triangle (AB_1C_1) implies (AO), (B_1O), and (C_1O) are angle bisectors. - (OM) is perpendicular to (B_1C_1) because (M) is the tangency point. - (OP) is perpendicular to (BB_1) by given condition.3. Cyclic Quadrilateral: - Since (OM perp B_1C_1) and (OP perp BB_1), quadrilateral (OMPB_1) has two right angles at (M) and (P). - This implies that quadrilateral (OMPB_1) is cyclic (opposite angles sum to (180^circ)).4. Angle Relationships in Cyclic Quadrilateral: - In cyclic quadrilateral (OMPB_1), angles subtended by the same chord are supplementary. - Angle (angle MPB_1) and (angle MOB_1) are subtended by chord (MB_1) from points (P) and (O), respectively.5. Calculating Angles: - (angle MOB_1 = 90^circ - frac{1}{2} angle AB_1C_1) because (OM perp B_1C_1) and (OB_1) is the angle bisector. - Since (angle MPB_1 + angle MOB_1 = 180^circ) (cyclic quadrilateral), we find (angle MPB_1 = 90^circ + frac{1}{2} angle AB_1C_1).6. Incenter Angle Property: - In triangle (AB_1C_1), the angle at the incenter (O) between (AO) and (C_1O) is (angle AOC_1 = 90^circ + frac{1}{2} angle AB_1C_1).7. Conclusion: - Both (angle AOC_1) and (angle MPB_1) are equal to (90^circ + frac{1}{2} angle AB_1C_1).Thus, we have shown that (angle AOC_1 = angle MPB_1).[boxed{angle AOC_1 = angle MPB_1}]

🔑:Let's proceed step-by-step to prove that angle AOC_1 = angle MPB_1.1. Setting Up the Problem: Consider triangle ABC with points B_1 on AC and C_1 on AB. The segments BB_1 and CC_1 intersect at point P. Let O be the center of the incircle of triangle AB_1C_1 and M be the point where this incircle touches segment B_1C_1. It is given that lines OP and BB_1 are perpendicular.2. Identify Angles Involved: We denote the angle angle AB_1C_1 = beta. To find the required angles: - angle AOC_1 can be identified using properties of triangle AB_1C_1. - angle MPB_1 involves the geometry of the points P and M and their angles with O and B_1.3. Calculating angle AOC_1: Since O is the incenter of triangle AB_1C_1, the angle angle AOC_1 can be determined using the incenter properties: [ angle AOC_1 = 90^circ + frac{beta}{2} ]4. Determining angle MB_1O: Because M is the touchpoint of the incircle with B_1C_1, the angle angle MB_1O made by radius OM and side B_1C_1 in triangle AB_1C_1 is: [ angle MB_1O = frac{beta}{2} ]5. Using the Perpendicularity: Since OP is perpendicular to BB_1 and point P lies on BB_1, there is a right-angle property at point P: [ angle OPB_1 = 90^circ ]6. Angles from P and M: Considering points P and M related to circle geometry, the inscribed angles subtended by diameter OB_1. The angles angle MPO and angle MB_1O subtended from similar arcs: [ angle MPO = angle MB_1O = frac{beta}{2} ]7. Final Angle Calculation: The angle angle MPB_1 is the sum of angles B_1: [ angle MPB_1 = angle OPB_1 + angle MPO ] Combining all angles: [ angle MPB_1 = 90^circ + frac{beta}{2} ]8. Conclusion: Since both angles are formed with the same components 90^circ + frac{beta}{2}: [ angle AOC_1 = angle MPB_1 = 90^circ + frac{beta}{2} ]Hence, we have proved:[boxed{angle AOC_1 = angle MPB_1}]

🧠:Okay, so I need to prove these two set equalities involving unions and intersections over arbitrary index sets. Let me start by understanding what each side of the equations represents and how they might be equivalent. The first equation is:bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}where the right-hand side intersection is over all "paths" ( j_{infty} = (j_{i})_{i in mathfrak{I}} ), with each ( j_i in mathfrak{J}_i ).Similarly, the second equation is:bigcap_{i in mathfrak{I}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{I}} A_{i j_{i}}Alright, let me tackle the first one first. Maybe if I can get through that, the second will be similar but dual.First, let's parse the left-hand side (LHS) of the first equation. It's a union over all ( i in mathfrak{I} ) of the intersection over ( j in mathfrak{J}_i ) of ( A_{ij} ). So for each ( i ), we take the intersection of sets ( A_{ij} ) across ( j in mathfrak{J}_i ), and then take the union of all these intersections.The right-hand side (RHS) is an intersection over all paths ( j_{infty} ), where each path is a choice of ( j_i in mathfrak{J}_i ) for each ( i in mathfrak{I} ), of the union over ( i in mathfrak{I} ) of ( A_{i j_i} ). So for each path (which is a function assigning to each ( i ) some ( j_i )), we take the union of the ( A_{i j_i} ) across ( i ), and then intersect all those unions.I need to show that these two sets are equal. Let me try to see what it means for an element to be in each side.Suppose ( x ) is in the LHS. Then there exists some ( i_0 in mathfrak{I} ) such that ( x in bigcap_{j in mathfrak{J}_{i_0}} A_{i_0 j} ). That means for this particular ( i_0 ), ( x ) is in every ( A_{i_0 j} ) for all ( j in mathfrak{J}_{i_0} ).Now, we need to see if ( x ) is in the RHS. To be in the RHS, ( x ) must be in every union ( bigcup_{i in mathfrak{I}} A_{i j_i} ) for every possible path ( j_{infty} ). So for any path ( j_{infty} ), which includes a choice of ( j_i ) for each ( i ), ( x ) must be in at least one ( A_{i j_i} ) for some ( i ).Wait, but if ( x ) is in the intersection over all such unions, then for every path ( j_{infty} ), there exists some ( i ) (depending on the path) such that ( x in A_{i j_i} ).But on the LHS, we have that for ( i_0 ), ( x ) is in all ( A_{i_0 j} ), so in particular, for any path ( j_{infty} ), if we look at the component ( j_{i_0} ) in that path, then ( x in A_{i_0 j_{i_0}} ). Therefore, for every path, ( x ) is in ( A_{i_0 j_{i_0}} ), so the union over ( i ) would include ( x ), since ( x ) is in ( A_{i_0 j_{i_0}} ). Therefore, ( x ) is in the union for each path, hence in the intersection over all paths of such unions. Thus, LHS is a subset of RHS.Conversely, suppose ( x ) is in the RHS. Then for every path ( j_{infty} ), there exists some ( i ) (possibly depending on the path) such that ( x in A_{i j_i} ). We need to show that there exists some ( i_0 in mathfrak{I} ) such that ( x in bigcap_{j in mathfrak{J}_{i_0}} A_{i_0 j} ).Wait, this direction might be trickier. Suppose ( x ) is in the RHS. That means for every possible path ( j_{infty} ), ( x ) is in at least one ( A_{i j_i} ). Suppose, for contradiction, that for every ( i in mathfrak{I} ), there exists some ( j_i in mathfrak{J}_i ) such that ( x notin A_{i j_i} ). Then, if we construct the path ( j_{infty} ) by choosing these ( j_i ) for each ( i ), this would be a path where ( x ) is not in any ( A_{i j_i} ), contradicting the fact that ( x ) is in the RHS. Therefore, our assumption is wrong; there must exist some ( i_0 in mathfrak{I} ) such that for all ( j in mathfrak{J}_{i_0} ), ( x in A_{i_0 j} ). Therefore, ( x ) is in the intersection over ( j in mathfrak{J}_{i_0} ) of ( A_{i_0 j} ), hence in the union over ( i ) of such intersections. Therefore, RHS is a subset of LHS.Therefore, the two sets are equal. That seems to work. Let me check again.For LHS ⊆ RHS: Take ( x ) in LHS. Then there exists ( i_0 ) such that ( x ) is in every ( A_{i_0 j} ). For any path ( j_{infty} ), ( x ) is in ( A_{i_0 j_{i_0}}} ), so ( x ) is in the union over ( i ) of ( A_{i j_i} ). Hence, ( x ) is in the intersection over all such unions. Thus, LHS ⊆ RHS.For RHS ⊆ LHS: Suppose ( x ) is in RHS. Then for every path ( j_{infty} ), there exists some ( i ) where ( x in A_{i j_i} ). Suppose for contradiction that ( x ) is not in LHS. Then for every ( i ), there is some ( j_i in mathfrak{J}_i ) such that ( x notin A_{i j_i} ). Let’s construct a path ( j_{infty} ) by choosing these ( j_i ) for each ( i ). Then, for this path, there would be no ( i ) with ( x in A_{i j_i} ), which contradicts ( x ) being in RHS. Therefore, such an ( i_0 ) must exist where ( x in bigcap_{j in mathfrak{J}_{i_0}} A_{i_0 j} ), so ( x ) is in LHS. Hence, equality holds.Okay, that seems solid. Now moving on to the second equality:bigcap_{i in mathfrak{I}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{I}} A_{i j_{i}}.Here, the LHS is an intersection over all ( i in mathfrak{I} ) of the union over ( j in mathfrak{J}_i ) of ( A_{ij} ). So for each ( i ), we take the union of all ( A_{ij} ), and then intersect all these unions across ( i ).The RHS is a union over all paths ( j_{infty} ) (same as before, each path is a selection of ( j_i in mathfrak{J}_i ) for each ( i )) of the intersection over ( i in mathfrak{I} ) of ( A_{i j_i} ). So each path gives a set which is the intersection of ( A_{i j_i} ) across ( i ), and then we take the union of all these intersections.Let me again analyze both directions.Suppose ( x ) is in the LHS. Then, for every ( i in mathfrak{I} ), ( x ) is in ( bigcup_{j in mathfrak{J}_i} A_{ij} ). That means, for each ( i ), there exists at least one ( j_i in mathfrak{J}_i ) such that ( x in A_{i j_i} ). Now, if we can choose such ( j_i ) for each ( i ), then the path ( j_{infty} = (j_i)_{i in mathfrak{I}} ) would satisfy ( x in bigcap_{i in mathfrak{I}} A_{i j_i} ), hence ( x ) is in the union over all such paths. Therefore, LHS ⊆ RHS.Conversely, suppose ( x ) is in the RHS. Then there exists some path ( j_{infty} ) such that ( x in bigcap_{i in mathfrak{I}} A_{i j_i} ). This means, for this particular path, ( x ) is in every ( A_{i j_i} ) for all ( i ). Therefore, for each ( i ), since ( x in A_{i j_i} ), which is a subset of ( bigcup_{j in mathfrak{J}_i} A_{ij} ), it follows that ( x ) is in each ( bigcup_{j in mathfrak{J}_i} A_{ij} ), hence in their intersection. Therefore, RHS ⊆ LHS.Hence, equality holds.Wait, let me verify the first direction again. If ( x ) is in LHS, then for every ( i ), there exists some ( j_i ) such that ( x in A_{i j_i} ). So if we can choose such ( j_i ) for all ( i ), forming a path ( j_{infty} ), then ( x ) would be in the intersection over ( i ) of ( A_{i j_i} ), hence in the union over all such intersections. So that's correct.But here's a catch: the Axiom of Choice. Because for each ( i ), we need to choose a ( j_i ) such that ( x in A_{i j_i} ). If the index set ( mathfrak{I} ) is infinite, this requires choosing infinitely many ( j_i )'s simultaneously. So, this step implicitly uses the Axiom of Choice. Since the problem statement doesn't specify any restrictions on the index sets, we can assume that we're working in a framework where the Axiom of Choice is allowed. Therefore, the proof is valid.Similarly, in the first equality, when we constructed the path ( j_{infty} ) by choosing ( j_i ) for each ( i ) such that ( x notin A_{i j_i} ), if possible, that also requires the Axiom of Choice if ( mathfrak{I} ) is infinite. However, in that case, the contradiction arises from assuming that such a path exists, so perhaps that part doesn't require the Axiom of Choice, but rather uses the negation of the conclusion to derive a contradiction. Hmm, maybe that direction doesn't require AC, but the other direction does? Let me think.In the first equality's RHS to LHS direction: If ( x ) is in RHS, then for every path ( j_{infty} ), ( x ) is in some ( A_{i j_i} ). To show ( x ) is in LHS, we supposed for contradiction that for every ( i ), there's some ( j_i ) where ( x notin A_{i j_i} ). Then, constructing such a path ( j_{infty} ) would lead to a contradiction. Here, constructing the path requires choosing such ( j_i ) for each ( i ), which again might need AC if ( mathfrak{I} ) is infinite. Therefore, this direction also uses AC.But in standard set theory, when proving set equalities, we generally allow the use of the Axiom of Choice unless specified otherwise. So I think the proofs are valid.So both equalities hold, using these arguments with possible invocation of the Axiom of Choice for the infinite cases. But since the problem didn't specify finiteness, we can proceed.Let me recap:For the first equality:- LHS is the union over ( i ) of the intersections over ( j in mathfrak{J}_i ).- RHS is the intersection over all paths of the unions over ( i ) of ( A_{i j_i} ).- Showed mutual inclusion by considering elements and constructing appropriate paths or using contradiction.For the second equality:- LHS is the intersection over ( i ) of the unions over ( j in mathfrak{J}_i ).- RHS is the union over all paths of the intersections over ( i ) of ( A_{i j_i} ).- Similarly, showed mutual inclusion by using the existence of paths (with AC) or element membership.Therefore, both equalities are established.I should check if there are any edge cases or if my reasoning missed something. For example, what if some ( mathfrak{J}_i ) are empty? Wait, the problem states "arbitrary sets", so we need to ensure that even if some ( mathfrak{J}_i ) are empty, the equality holds. Let's consider that.First equation: If for some ( i ), ( mathfrak{J}_i ) is empty, then the intersection ( bigcap_{j in mathfrak{J}_i} A_{ij} ) would be the universal set (since the intersection over an empty family is everything). Therefore, the LHS would include that universal set, making the LHS itself the universal set. On the RHS, a "path" ( j_{infty} ) requires choosing ( j_i in mathfrak{J}_i ) for each ( i ). But if some ( mathfrak{J}_i ) is empty, then there are no such paths. Therefore, the intersection over all paths would be the universal set (since the intersection over an empty family is everything). Hence, both sides would be the universal set. So equality holds in that case.Similarly, if all ( mathfrak{J}_i ) are non-empty, the previous reasoning applies. If some ( mathfrak{J}_i ) is empty, then depending on the side, the result is universal set.For the second equation: If any ( mathfrak{J}_i ) is empty, then the union ( bigcup_{j in mathfrak{J}_i} A_{ij} ) is empty (since the union over an empty family is empty). Therefore, the LHS intersection would be empty. On the RHS, the union over all paths ( j_{infty} ): but if any ( mathfrak{J}_i ) is empty, there are no paths (since you can't choose ( j_i ) from an empty set). Therefore, the union over all paths is empty. Hence, both sides are empty, so equality holds.Another edge case: if all ( mathfrak{J}_i ) are singletons. Then paths are uniquely determined, so the intersection over paths would just be the single path, and similar for unions. Then both sides should collapse appropriately.Testing with a simple example might help. Let's take a finite case.First equality example: Let ( mathfrak{I} = {1, 2} ), ( mathfrak{J}_1 = {a, b} ), ( mathfrak{J}_2 = {c} ).Compute LHS: ( bigcup_{i in {1,2}} bigcap_{j in mathfrak{J}_i} A_{ij} ).For ( i=1 ), ( bigcap_{j in {a,b}} A_{1j} = A_{1a} cap A_{1b} ).For ( i=2 ), ( bigcap_{j in {c}} A_{2j} = A_{2c} ).Thus, LHS = ( (A_{1a} cap A_{1b}) cup A_{2c} ).RHS: ( bigcap_{j_{infty}} bigcup_{i in {1,2}} A_{i j_i} ).A path ( j_{infty} ) is a choice of ( j_1 in {a,b} ) and ( j_2 in {c} ). So there are two possible paths: ( (a, c) ) and ( (b, c) ).For each path, compute ( bigcup_{i} A_{i j_i} ):- For path (a,c): ( A_{1a} cup A_{2c} ).- For path (b,c): ( A_{1b} cup A_{2c} ).Intersection of these two sets: ( (A_{1a} cup A_{2c}) cap (A_{1b} cup A_{2c}) ).Distributive law: ( (A_{1a} cap A_{1b}) cup A_{2c} ), which matches LHS. So equality holds here.Second equality example: Same ( mathfrak{I} ), ( mathfrak{J}_i ).LHS: ( bigcap_{i in {1,2}} bigcup_{j in mathfrak{J}_i} A_{ij} = (A_{1a} cup A_{1b}) cap A_{2c} ).RHS: ( bigcup_{j_{infty}} bigcap_{i in {1,2}} A_{i j_i} ).Possible paths are (a,c) and (b,c).For each path, compute the intersection:- (a,c): ( A_{1a} cap A_{2c} ).- (b,c): ( A_{1b} cap A_{2c} ).Union of these: ( (A_{1a} cap A_{2c}) cup (A_{1b} cap A_{2c}) ).Distributive law: ( (A_{1a} cup A_{1b}) cap A_{2c} ), which matches LHS. So equality holds.Good, the examples work. So my reasoning seems correct.Therefore, the two set equalities hold in general, by mutual inclusion, using arguments that may involve the Axiom of Choice for infinite index sets, but valid in ZFC.

✅:To establish the given set equalities, we analyze each direction of inclusion using systematic reasoning.First Equality:bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}Proof:1. Left to Right Inclusion (⊆): - Let ( x in bigcup_{i} bigcap_{j} A_{ij} ). There exists ( i_0 in mathfrak{I} ) such that ( x in bigcap_{j in mathfrak{J}_{i_0}} A_{i_0 j} ). - For any path ( j_{infty} = (j_i)_{i in mathfrak{I}} ), ( x in A_{i_0 j_{i_0}} ) (since ( x ) is in all ( A_{i_0 j} )). - Thus, ( x in bigcup_{i} A_{i j_i} ) for every path ( j_{infty} ), so ( x in bigcap_{j_{infty}} bigcup_{i} A_{i j_i} ).2. Right to Left Inclusion (⊇): - Let ( x in bigcap_{j_{infty}} bigcup_{i} A_{i j_i} ). For every path ( j_{infty} ), ( x in A_{i j_i} ) for some ( i ). - Suppose ( x notin bigcup_{i} bigcap_{j} A_{ij} ). Then for each ( i ), there exists ( j_i in mathfrak{J}_i ) with ( x notin A_{i j_i} ). Constructing such a path ( j_{infty} ) leads to a contradiction, as ( x ) would not be in any ( A_{i j_i} ). - Hence, such an ( i_0 ) must exist where ( x in bigcap_{j} A_{i_0 j} ), proving ( x ) is in the left-hand side.Second Equality:bigcap_{i in mathfrak{I}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{I}} A_{i j_{i}}Proof:1. Left to Right Inclusion (⊆): - Let ( x in bigcap_{i} bigcup_{j} A_{ij} ). For each ( i ), ( x in A_{i j_i} ) for some ( j_i in mathfrak{J}_i ). - By the Axiom of Choice, select such ( j_i ) for all ( i ), forming a path ( j_{infty} ). Then ( x in bigcap_{i} A_{i j_i} ), so ( x in bigcup_{j_{infty}} bigcap_{i} A_{i j_i} ).2. Right to Left Inclusion (⊇): - Let ( x in bigcup_{j_{infty}} bigcap_{i} A_{i j_i} ). There exists a path ( j_{infty} ) such that ( x in bigcap_{i} A_{i j_i} ). - For each ( i ), ( x in A_{i j_i} subseteq bigcup_{j} A_{ij} ). Hence, ( x in bigcap_{i} bigcup_{j} A_{ij} ).Conclusion:Both equalities hold by mutual inclusion, with the second requiring the Axiom of Choice for general index sets.Final AnswerBoth equalities are established as follows:1. boxed{bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}}2. boxed{bigcap_{i in mathfrak{J}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{J}} A_{i j_{i}}}

🔑:1. Consider the relation to prove: bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}, where the intersection is taken over all "paths" j_{infty} = left(j_{i}, i in mathfrak{I}right) with j_{i} in mathfrak{J}_{i}. 2. To simplify the proof, we will translate the set notation into indicator functions. Let I_{i j} = I_{i j}(omega) be the indicator function of the set A_{i j}. The union and intersection relations in terms of indicator functions can be stated as: max_{i in mathfrak{I}} min_{j in mathfrak{J}_{i}} I_{i j}(omega) = min_{j_{infty}} max_{i in mathfrak{I}} I_{i j_{i}}(omega), quad omega in Omega. 3. We need to demonstrate this relation. For a fixed omega in Omega, consider: max_{i in mathfrak{I}} min_{j in mathfrak{J}_{i}} I_{i j} = 1. This is true if and only if: exists i in mathfrak{I} , forall j in mathfrak{J}_{i}: I_{i j}(omega) = 1. 4. Interpreting this condition, it means there is at least one index i in mathfrak{I} for which all A_{i j} (where j in mathfrak{J}_{i}) contains the element omega. So, omega is in: bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j}. 5. This condition also translates to: min_{j_{infty}} max_{i in mathfrak{I}} I_{i j_{i}}(omega) = 1. This condition is met if: forall j_{infty}, exists i in mathfrak{I}: I_{i j_{i}}(omega) = 1. The interpretation of this is that for any choice of j_{infty}, there exists an i such that omega in A_{i j_i}. Thus, omega is in: bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}. 6. Having shown the equivalence in indicator function terms for a fixed omega, we conclude: bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}}. 7. For the second relation to prove: bigcap_{i in mathfrak{J}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{J}} A_{i j_{i}}. This follows directly from the first relation by considering complements. Transpose the variables and sets accordingly to transform the first equation into the second one. Therefore, the second relation is also valid.Conclusion: boxed{bigcup_{i in mathfrak{I}} bigcap_{j in mathfrak{J}_{i}} A_{i j} = bigcap_{j_{infty}} bigcup_{i in mathfrak{I}} A_{i j_{i}} bigcap_{i in mathfrak{J}} bigcup_{j in mathfrak{J}_{i}} A_{i j} = bigcup_{j_{infty}} bigcap_{i in mathfrak{J}} A_{i j_{i}}}.