🧠:[<think>Okay, so I need to figure out how many different ways there are to assign five volunteers into three different areas: translation, tour guiding, and etiquette. The catch is that each area has to have at least one person. Hmm, this sounds like a combinatorics problem, probably involving permutations or combinations with some restrictions. Let me think through it step by step.First, let me recall some basic principles. When assigning objects (in this case, volunteers) to different categories (the three areas), one common approach is to use the multiplication principle. However, since each area must have at least one person, it's not as simple as just allowing any assignment. If there were no restrictions, each volunteer could choose between three areas, so the total number of assignments would be 3^5. But since each area must have at least one person, we need to subtract the cases where one or more areas are empty. This sounds like the inclusion-exclusion principle.Let me verify. The inclusion-exclusion principle is used to calculate the number of elements in the union of overlapping sets by including and excluding the intersections appropriately. In this case, the "bad" assignments are those where at least one area is empty. So, if I can calculate the total number of assignments without restrictions and subtract the ones that are missing at least one area, I should get the desired number.So, total assignments without restrictions: 3^5. Let me compute that. 3^5 is 243.Now, subtract the assignments where at least one area is empty. Let's break this down. The number of assignments where translation is empty would be assignments where all volunteers are in tour guiding or etiquette. Similarly for the other areas. So, if one specific area is excluded, the number of assignments is 2^5. There are three such areas (translation, tour guiding, etiquette), so we subtract 3*2^5. But wait, now we've subtracted too much because assignments where two areas are empty have been subtracted multiple times. For example, if both translation and tour guiding are empty, all volunteers are in etiquette. This was subtracted once when we considered translation empty and once when we considered tour guiding empty. So we need to add those back in.How many assignments have two areas empty? Well, if two areas are excluded, all volunteers must be in the remaining one area. There are 3 such cases (all in translation, all in tour guiding, all in etiquette). Each of these cases has only 1 assignment. Since there are three pairs of areas to exclude, we need to add back 3*1^5.Putting this all together using inclusion-exclusion: Total valid assignments = Total assignments - assignments missing at least one area + assignments missing two areas.Wait, no. Let me recall the inclusion-exclusion formula. For three sets A, B, C, the number of elements not in A, B, or C is Total - (|A| + |B| + |C|) + (|A∩B| + |A∩C| + |B∩C|) - |A∩B∩C|). But in our case, we want the total assignments minus the assignments missing at least one area. So, the valid assignments are Total - (assignments missing translation + assignments missing tour guiding + assignments missing etiquette) + (assignments missing both translation and tour guiding + assignments missing both translation and etiquette + assignments missing both tour guiding and etiquette) - assignments missing all three areas. But since all three areas can't be missing (we have to assign volunteers somewhere), the last term is zero.So formula becomes:Valid = 3^5 - 3*2^5 + 3*1^5 - 0Let me compute that. 3^5 is 243. 2^5 is 32, so 3*32 is 96. 1^5 is 1, so 3*1 is 3. Therefore, Valid = 243 - 96 + 3 = 243 - 96 is 147, then 147 + 3 = 150. Hmm, so 150. Is that right?Wait, let me check. Alternatively, another way to approach this is using Stirling numbers of the second kind. The number of ways to partition n elements into k non-empty subsets is S(n, k), and then multiplied by k! if the order of the subsets matters (since the areas are distinct). In this case, the areas are distinct, so yes, we need to multiply by 3!.Wait, but in this case, the volunteers are assigned to labeled boxes (the areas), so the formula is indeed k! * S(n, k). Let me confirm the value of S(5,3). The Stirling number of the second kind S(5,3) is 25. Then 25 * 3! = 25 * 6 = 150. So that matches the previous result. So that gives me confidence that 150 is the correct answer.Alternatively, if I were to compute it directly using inclusion-exclusion: 3^5 - 3*2^5 + 3*1^5 = 243 - 96 + 3 = 150. Yep, that's the same.Alternatively, maybe we can calculate it by considering the different distributions of volunteers. Since each area must have at least one person, the possible distributions (in terms of numbers) are:1. 3,1,1: Three volunteers in one area and one in each of the other two.2. 2,2,1: Two volunteers in two areas and one in the third.These are the two distinct ways to partition five volunteers into three groups where each group has at least one person. Let's compute each case.First case: 3,1,1.Number of ways to choose which area gets 3 volunteers: C(3,1) = 3.Then, choose 3 volunteers out of 5 to assign to that area: C(5,3) = 10.The remaining 2 volunteers are assigned to the other two areas, one each. Since the areas are distinct, the number of ways to assign the remaining two volunteers is 2! = 2.So total for this case: 3 * 10 * 2 = 60.Second case: 2,2,1.Number of ways to choose which area gets 1 volunteer: C(3,1) = 3.Then, choose 1 volunteer out of 5 to assign to that area: C(5,1) = 5.The remaining 4 volunteers are split into two groups of 2. The number of ways to split 4 volunteers into two groups of 2 is C(4,2)/2 = 6/2 = 3. Wait, but since the two areas are distinct, we don't need to divide by 2. Let me think.If we have two areas, each to receive 2 volunteers, then first choose 2 out of 4 for the first area: C(4,2) = 6. Then the remaining 2 go to the second area. Since the two areas are distinct, this is 6 * 1 = 6. So the total for assigning the remaining 4 volunteers is 6.Therefore, total for the second case: 3 (choice of which area gets 1) * 5 (choice of volunteer for that area) * 6 (assignment of remaining 4 into two areas) = 3 * 5 * 6 = 90.Adding both cases together: 60 + 90 = 150. That's the same as before.So whether I use inclusion-exclusion, Stirling numbers, or direct combinatorial counting, I get 150. That gives me more confidence that the answer is indeed 150.Wait, but let me just check one more time with another method. Suppose we think of each volunteer as being assigned to one of three areas, with the constraint that none are empty. So another way is to count all possible assignments (3^5 = 243) and subtract those that leave at least one area empty.As before, using inclusion-exclusion:Number of assignments with at least one area empty = C(3,1)*2^5 - C(3,2)*1^5 + C(3,3)*0^5.So that's 3*32 - 3*1 + 0 = 96 - 3 = 93.Then, total valid assignments = 243 - 93 = 150. Yep, same answer.Alternatively, if I use generating functions. The exponential generating function for each area (since volunteers are distinguishable) is e^(x) - 1 (since each area must have at least one person). So the generating function is (e^x - 1)^3. The coefficient of x^5 / 5! multiplied by 5! will give the number of assignments.Expanding (e^x - 1)^3 = e^(3x) - 3e^(2x) + 3e^x - 1. The coefficient of x^5 is (3^5 - 3*2^5 + 3*1^5)/5! So when we multiply by 5! to get the number of assignments, it's 3^5 - 3*2^5 + 3*1^5 = 243 - 96 + 3 = 150. Again, same result.So all methods are converging on 150.Just to recap:- Total assignments: 3^5 = 243.- Subtract assignments missing at least one area: 3*2^5 = 96.- Add back assignments missing two areas: 3*1^5 = 3.- Subtract assignments missing all three areas: 0.So 243 - 96 + 3 = 150.Therefore, the answer should be 150.Final AnswerThe number of different assignment plans is boxed{150}.</think>To determine the number of different assignment plans for five volunteers into three areas (translation, tour guiding, and etiquette) with each area having at least one person, we use the inclusion-exclusion principle.1. Total assignments without restrictions: (3^5 = 243).2. Subtract assignments missing at least one area: - Each area can be excluded, leading to (2^5) assignments. There are 3 such areas, so subtract (3 cdot 2^5 = 3 cdot 32 = 96).3. Add back assignments missing two areas: - Each pair of areas can be excluded, leading to (1^5) assignments. There are 3 such pairs, so add (3 cdot 1^5 = 3 cdot 1 = 3).4. Subtract assignments missing all three areas: - This is 0 since volunteers must be assigned to at least one area.Applying the inclusion-exclusion principle:[3^5 - 3 cdot 2^5 + 3 cdot 1^5 = 243 - 96 + 3 = 150]Thus, the number of different assignment plans is (boxed{150}).,<think>Okay, let's see. The problem is about assigning five volunteers to three different areas: translation, tour guiding, and etiquette. Each area must have at least one person. So, we need to find the number of different assignment plans. Hmm, how do I approach this?First, this seems like a combinatorics problem. Assigning volunteers to different areas with certain constraints. The key here is that each area must have at least one volunteer. If there were no such constraint, the problem would be straightforward. Since each volunteer can be assigned to any of the three areas, the total number of assignments would be 3^5. Let me calculate that. 3 multiplied by itself five times: 3*3*3*3*3 = 243. So, 243 total assignments without any restrictions.But we have the restriction that each area must have at least one person. So, we need to subtract the cases where one or more areas are empty. This sounds like the inclusion-exclusion principle. Yeah, that's right. Inclusion-exclusion is used to calculate the number of elements in the union of overlapping sets by including and excluding the intersections appropriately.So, applying inclusion-exclusion here. The formula for the number of onto functions from a set of size n to a set of size k is k! * S(n, k), where S(n, k) is the Stirling numbers of the second kind. But maybe I can use inclusion-exclusion directly.The formula for the number of assignments where each area is non-empty is equal to the total number of assignments minus the assignments where at least one area is empty, plus the assignments where two areas are empty, etc. Wait, let me recall the inclusion-exclusion formula.If we have three areas, the number of assignments where all three areas are non-empty is equal to:Total assignments (3^5) minus the assignments where translation is empty, minus assignments where tour guiding is empty, minus assignments where etiquette is empty, plus the assignments where both translation and tour guiding are empty, plus assignments where both translation and etiquette are empty, plus assignments where both tour guiding and etiquette are empty, minus the assignments where all three areas are empty.But since all three areas can't be empty (we have five volunteers to assign), the last term is zero. Let me write this out step by step.Let’s denote the three areas as A (translation), B (tour guiding), and C (etiquette).Number of valid assignments = Total assignments - (assignments with A empty + assignments with B empty + assignments with C empty) + (assignments with A and B empty + assignments with A and C empty + assignments with B and C empty) - assignments with all three empty.But assignments with all three empty would be zero because we can’t assign five volunteers to nowhere. So, that term is zero.First, calculate total assignments: 3^5 = 243.Next, assignments with one area empty. Let's take A empty. If A is empty, all volunteers must be assigned to B and C. Each volunteer has 2 choices, so 2^5 assignments. Similarly, if B is empty, 2^5 assignments, and same if C is empty. So, each of the three cases (A empty, B empty, C empty) has 2^5 = 32 assignments. So, total for one area empty: 3 * 32 = 96.But then we have to add back the cases where two areas are empty. For example, if A and B are empty, all volunteers must be assigned to C. There's only 1 way for each volunteer: assign to C. So, 1^5 = 1 assignment. Similarly, if A and C are empty, all assigned to B: 1 assignment. If B and C are empty, all assigned to A: 1 assignment. There are three such cases, each contributing 1 assignment. So total for two areas empty: 3 * 1 = 3.Therefore, applying inclusion-exclusion:Number of valid assignments = 243 - 96 + 3 = 243 - 96 = 147; 147 + 3 = 150.Wait, so 150? Is that the answer?Alternatively, maybe I can use the Stirling numbers approach. The Stirling number of the second kind, S(5,3), counts the number of ways to partition a set of 5 objects into 3 non-empty subsets. Then, since each of these subsets can be assigned to any of the three areas, we multiply by 3!.So, the formula would be 3! * S(5,3). Let me check what S(5,3) is.Stirling numbers of the second kind: S(n,k) = S(n-1,k-1) + k*S(n-1,k). So, let's compute S(5,3):S(5,3) = S(4,2) + 3*S(4,3)We need S(4,2) and S(4,3). Let's compute S(4,2):S(4,2) = S(3,1) + 2*S(3,2)S(3,1) = 1 (only one way to partition 3 elements into 1 subset)S(3,2) = S(2,1) + 2*S(2,2) = 1 + 2*1 = 3Therefore, S(4,2) = 1 + 2*3 = 7Similarly, S(4,3) = S(3,2) + 3*S(3,3)S(3,2) is 3 (from above), S(3,3) = 1So, S(4,3) = 3 + 3*1 = 6Therefore, S(5,3) = S(4,2) + 3*S(4,3) = 7 + 3*6 = 7 + 18 = 25Therefore, the number of valid assignments is 3! * 25 = 6 * 25 = 150. So, same result as before. So, 150.Therefore, the answer should be 150.Wait, but let me confirm with another approach. Let's consider that each volunteer can be assigned to any of the three areas, but subtract the assignments where one area is empty.Alternatively, suppose we use the formula for the number of onto functions: for n elements to k boxes, with each box non-empty, it's k! * S(n,k). So, here, n=5, k=3. So, 3! * S(5,3) = 6 * 25 = 150. So, same answer.Alternatively, maybe we can use multinomial coefficients. The number of assignments where each area has at least one person is the sum over all distributions (a,b,c) where a + b + c = 5 and a,b,c ≥1, of the multinomial coefficients 5! / (a! b! c!). Then, multiply by the number of permutations? Wait, no, the multinomial coefficient already accounts for the assignments. Wait, maybe not.Wait, actually, if we fix the areas, then for each distribution (a,b,c), the number of ways is 5! / (a! b! c!). However, if the areas are distinguishable (which they are: translation, tour guiding, etiquette), then each distribution corresponds to a different assignment. So, for example, if we have 2 in translation, 2 in tour guiding, and 1 in etiquette, that's a different case than 2 in translation, 1 in tour guiding, and 2 in etiquette.Therefore, the number of assignments is the sum over all ordered triples (a,b,c) where a + b + c =5 and a,b,c ≥1 of 5! / (a! b! c!). Let's compute that.Alternatively, since each volunteer can be assigned to any of the three areas, but we need to subtract the cases where one or more areas are empty, which brings us back to inclusion-exclusion. So, 3^5 - 3*2^5 + 3*1^5 = 243 - 96 + 3 = 150. So, same answer.Therefore, all approaches lead to 150. Therefore, I think the answer is 150. But let me just verify with an example.Suppose we have 5 volunteers and 3 areas, each area must have at least one person.Let me list all possible distributions:The possible distributions (a,b,c) where a, b, c ≥1 and a + b + c =5. Let's list them:1. (3,1,1)2. (2,2,1)3. (2,1,2)4. (1,3,1)5. (1,2,2)6. (1,1,3)But wait, actually, these are permutations of (3,1,1) and (2,2,1). For (3,1,1), there are 3 permutations (which position has the 3). For (2,2,1), there are 3 permutations (which position has the 1). So total of 6 distributions.For each distribution, compute the number of assignments.For (3,1,1):Number of ways: 5! / (3!1!1!) = 20. Then, since the areas are different, each permutation corresponds to a different assignment. So, for each permutation (3,1,1), there are 20 assignments. Since there are 3 permutations, total for (3,1,1) is 3 * 20 = 60.For (2,2,1):Number of ways: 5! / (2!2!1!) = 30. Then, there are 3 permutations (which area has the 1). So, 3 * 30 = 90.Total assignments: 60 + 90 = 150. So, same answer again.Therefore, this confirms that the answer is 150.I think this is solid. So, the number of different assignment plans is 150.Final AnswerThe number of different assignment plans is boxed{150}.</think>To solve the problem of assigning five volunteers to three areas (translation, tour guiding, and etiquette) with each area having at least one person, we use the principle of inclusion-exclusion.1. Total assignments without restrictions: [ 3^5 = 243 ]2. Subtract assignments where at least one area is empty: - Each area can be empty, so we subtract the cases where one area is empty. There are 3 choices for which area is empty, and each results in (2^5) assignments: [ 3 times 2^5 = 3 times 32 = 96 ]3. Add back assignments where two areas are empty: - Each pair of areas can be empty, so we add the cases where two areas are empty. There are 3 choices for which two areas are empty, and each results in (1^5) assignments: [ 3 times 1^5 = 3 times 1 = 3 ]4. Apply inclusion-exclusion principle: [ 243 - 96 + 3 = 150 ]We verified this result using Stirling numbers of the second kind and multinomial coefficients, both leading to the same conclusion. The number of different assignment plans is:[boxed{150}]]

🔑:To solve this problem, we can use the principle of inclusion and exclusion. First, we calculate the total number of ways to assign the volunteers without any restrictions. Since there are three areas and five volunteers, each volunteer can be assigned to any area, so there are 3^5 possible arrangements.However, we have overcounted, as we need to ensure that each area has at least one volunteer. So, we must subtract the cases where one or more areas have no volunteers. There are three areas and thus three ways to have an area with no volunteers. If we have an area with no volunteers, then there are 2^5 ways to assign the remaining five volunteers to the other two areas. So we subtract 3 cdot 2^5 from 3^5.But we have now subtracted too much, because in the cases where two areas have no volunteers (which is clearly impossible with a minimum of one volunteer per area and five volunteers in total, but mathematically we have subtracted these cases in the previous step), we have subtracted these cases three times (once for each area we chose to have no volunteers), so we need to add them back in once. The number of ways for this impossible scenario (which serves the mathematical purpose of correction) is 1^5, as all five volunteers would be placed in one area. However, given there are three different areas to choose from and we're correcting an over-subtraction, we have 3 cdot 1^5 cases to add back.Therefore, the number of different assignment plans is: 3^5 - 3 cdot 2^5 + 3 cdot 1^5 Let's calculate this number. = 243 - 3 cdot 32 + 3 cdot 1 = 243 - 96 + 3 = 147 + 3 = 150 So, we have boxed{150} different assignment plans.

🧠:[<think>Okay, let's see. The problem says that each student's mentor gave them 2 apples, and after that, there were 19 apples left in the basket. We need to find out how many students there are and how many apples there were in total, given that the sum of the number of students and the total number of apples is 100.Hmm, let me parse this step by step. So, the mentor gives each student 2 apples. Let me denote the number of students as, let's say, S. Then, the number of apples given out would be 2 apples per student, so 2*S apples in total. After giving out these apples, there are 19 left. So, the original number of apples must have been the apples given out plus the remaining apples. That would be 2*S + 19. Let's call the total number of apples A. So, A = 2S + 19.Now, the problem also states that the total sum of the number of students and the total number of apples is 100. So, S + A = 100.We have two equations now:1. A = 2S + 192. S + A = 100So, we can substitute the first equation into the second. Let's replace A in the second equation with 2S + 19:S + (2S + 19) = 100Simplify that:S + 2S + 19 = 1003S + 19 = 100Subtract 19 from both sides:3S = 100 - 193S = 81Divide both sides by 3:S = 81 / 3S = 27So, there are 27 students. Then, plug that back into the first equation to find A:A = 2*27 + 19A = 54 + 19A = 73So, total apples are 73. Let's check if this fits the total sum:27 students + 73 apples = 100. Yep, that adds up to 100. Let me just verify the apple distribution. If each student gets 2 apples, 27 students would get 54 apples. Then, 19 apples remain, so total apples would be 54 + 19 = 73. That checks out with our previous calculation. Therefore, the number of students is 27 and the number of apples is 73. Both numbers sum up to 100, so everything seems to fit. I think that's the solution.Final AnswerThe number of students is boxed{27} and the number of apples is boxed{73}.</think>Let ( S ) be the number of students and ( A ) be the total number of apples. According to the problem, each student receives 2 apples, and 19 apples remain. Therefore, the total number of apples is given by:[ A = 2S + 19 ]We also know that the sum of the number of students and the total number of apples is 100:[ S + A = 100 ]Substituting the expression for ( A ) from the first equation into the second equation:[ S + (2S + 19) = 100 ]Simplifying this equation:[ 3S + 19 = 100 ]Subtracting 19 from both sides:[ 3S = 81 ]Dividing both sides by 3:[ S = 27 ]Substituting ( S = 27 ) back into the equation for ( A ):[ A = 2 times 27 + 19 = 54 + 19 = 73 ]Thus, the number of students is (boxed{27}) and the number of apples is (boxed{73}).,<think>Okay, let's see. The problem says that each student's mentor gave them 2 apples, and after that, there were 19 apples left in the basket. We need to find out how many students there are and how many apples there were in total, given that the sum of the number of students and the total number of apples is 100. Hmm, let me try to break this down step by step.First, let's parse the problem. So, mentors give each student 2 apples. After distributing these apples, there are 19 left. The total number of students plus the total number of apples equals 100. We need to find both the number of students and the number of apples.Let me assign variables to these unknowns. Let me denote the number of students as S and the total number of apples as A. According to the problem, after giving each student 2 apples, 19 apples remain. So, the total number of apples distributed to the students would be 2 apples per student times the number of students, which is 2S. Then, the remaining apples are 19. So, the total number of apples A should be equal to the apples given out plus the apples remaining. Therefore, A = 2S + 19. That's one equation.The other piece of information is that the sum of the number of students and the total number of apples is 100. So, S + A = 100. Now, we have a system of two equations:1. A = 2S + 192. S + A = 100We can substitute the first equation into the second equation. Since A is equal to 2S + 19, we can replace A in the second equation with that expression. So, substituting, we get:S + (2S + 19) = 100Let me simplify this equation. Combine like terms:S + 2S + 19 = 100That's 3S + 19 = 100Then, subtract 19 from both sides:3S = 100 - 193S = 81Divide both sides by 3:S = 81 / 3S = 27So, the number of students is 27. Then, substituting back into the first equation to find A:A = 2S + 19 = 2*27 + 19 = 54 + 19 = 73Therefore, the total number of apples is 73. Let me check if 27 students plus 73 apples equals 100. 27 + 73 is indeed 100, so that checks out. Also, distributing 2 apples to each of 27 students uses 54 apples, leaving 19 in the basket, so 54 + 19 = 73, which matches the total number of apples. Therefore, both conditions are satisfied.Wait, let me just make sure I didn't make any calculation errors. Let's verify the equations again.First, equation 1: A = 2S + 19. If S is 27, then 2*27 is 54, plus 19 is 73. Correct. Equation 2: S + A = 27 + 73 = 100. Correct. So everything seems to add up.Therefore, the answer should be 27 students and 73 apples.Final AnswerThe number of students is boxed{27} and the number of apples is boxed{73}.</think>Let ( S ) be the number of students and ( A ) be the total number of apples. According to the problem, each student receives 2 apples, and 19 apples remain after distribution. This gives us the equation:[ A = 2S + 19 ]Additionally, the sum of the number of students and the total number of apples is 100:[ S + A = 100 ]Substituting the expression for ( A ) from the first equation into the second equation, we get:[ S + (2S + 19) = 100 ]Simplifying this equation:[ 3S + 19 = 100 ]Subtracting 19 from both sides:[ 3S = 81 ]Dividing by 3:[ S = 27 ]Substituting ( S = 27 ) back into the first equation to find ( A ):[ A = 2 times 27 + 19 = 54 + 19 = 73 ]Verifying the solution, the sum of the number of students and apples is ( 27 + 73 = 100 ), which checks out. The distribution of 2 apples to each student uses 54 apples, leaving 19, which also matches the total number of apples.Thus, the number of students is (boxed{27}) and the number of apples is (boxed{73}).]

🔑:Solution. If the number of students is denoted by x, then the basket will contain 2 x+19 apples. From the condition that the total number of students and apples is 100, we get x+2 x+19=100. By solving the equation, we find that there are 27 students and 73 apples.

🧠:[<think>Okay, let's see. I have this right-angled triangle ABC with the right angle at C. So, angle C is 90 degrees. The height from C to the hypotenuse AB divides AB into a ratio of 1:3. I need to find the measures of angles at A and B. Hmm, right-angled triangles and ratios involving heights... Maybe I should draw a diagram first to visualize the problem.Alright, so triangle ABC, right-angled at C. Let me label the sides: AB is the hypotenuse, and AC and BC are the legs. The height from C to AB is CD, where D is the foot of the perpendicular from C to AB. The problem states that AD:DB = 1:3. So, AD is one part, DB is three parts. Therefore, the entire hypotenuse AB is divided into segments of 1 and 3, making the total length of AB 4 parts.I remember that in a right-angled triangle, the altitude to the hypotenuse has some properties related to the lengths of the segments. Specifically, the length of the altitude is the geometric mean of the lengths of the two segments it creates. So, CD² = AD * DB. Maybe that can help later, but let's see.Alternatively, there's a relation involving the legs of the triangle and the segments. Let me recall. If in a right-angled triangle, the altitude divides the hypotenuse into segments of lengths p and q, then the legs are sqrt(p*(p+q)) and sqrt(q*(p+q))? Wait, maybe not exactly. Let me think. Let's denote AD = 1k, DB = 3k, so AB = 4k. Then, the legs AC and BC can be found using similar triangles. Because in a right-angled triangle, the altitude creates two smaller similar triangles, each similar to the original triangle and to each other.So, triangle ACD is similar to triangle ABC, and triangle BCD is similar to triangle ABC. Therefore, the ratios of corresponding sides should be equal. Let me note that.For triangle ACD similar to ABC:AC / AB = AD / ACWhich gives AC² = AB * ADSimilarly, for triangle BCD similar to ABC:BC / AB = DB / BCWhich gives BC² = AB * DBSo, AC² = AB * AD and BC² = AB * DBSince AB = 4k, AD = k, DB = 3k,AC² = 4k * k = 4k², so AC = 2kBC² = 4k * 3k = 12k², so BC = 2k*sqrt(3)Therefore, the legs are AC = 2k and BC = 2k√3. Then, the hypotenuse AB is 4k, which matches the Pythagorean theorem:AC² + BC² = (2k)² + (2k√3)² = 4k² + 12k² = 16k² = (4k)² = AB². Okay, that checks out.Now, to find the angles at A and B. Let's denote angle at A as α and angle at B as β. Since it's a right-angled triangle, α + β = 90 degrees.We can use trigonometry to find the angles. For angle α (at A), the opposite side is BC = 2k√3, and the adjacent side is AC = 2k. So tan(α) = opposite / adjacent = (2k√3)/(2k) = √3. Therefore, α = arctan(√3). I know that tan(60°) = √3, so α = 60°, which would make β = 30°. But wait, in a right-angled triangle, the angles are usually acute angles. But depending on the legs, the angles can vary. Wait, but if AC is adjacent to angle A, then AC = 2k, BC = 2k√3. Wait, so BC is longer than AC, which would mean that angle opposite BC (which is angle A) is larger than angle opposite AC (which is angle B). Wait, hold on. Let me clarify.Wait, in triangle ABC, angle at A is opposite side BC, angle at B is opposite side AC. So, if BC is longer than AC, then angle A is larger than angle B. Therefore, angle A is 60°, angle B is 30°, since tan(angle A) = BC/AC = (2k√3)/2k = √3, so angle A is 60°, angle B is 30°. That makes sense. So, the angles are 60° and 30°, adding up to 90°, which fits the right angle at C.But let me verify this another way. Let's compute the sides. If AC = 2k, BC = 2k√3, then hypotenuse AB = 4k. Let me compute the sine and cosine of angle A.sin(α) = opposite / hypotenuse = BC / AB = (2k√3)/4k = √3/2. So sin(α) = √3/2, which is 60°, so α = 60°. Similarly, sin(β) = AC / AB = 2k / 4k = 1/2, so β = 30°. Yep, that confirms it. So angles at A and B are 60° and 30°, respectively.But wait, let me check if the altitude divides the hypotenuse into 1:3 ratio as given. The altitude CD can be calculated as (AC * BC)/AB. The area of the triangle is (AC * BC)/2 = (2k * 2k√3)/2 = 2k²√3. The area can also be expressed as (AB * CD)/2, so 2k²√3 = (4k * CD)/2, which gives CD = (2k²√3 * 2)/4k) = (4k²√3)/4k = k√3. So CD = k√3.But according to the geometric mean, CD² = AD * DB. Let's check that: CD² = (k√3)^2 = 3k². AD * DB = k * 3k = 3k². Yep, that works. So all the relations check out.Alternatively, the ratio of AD:DB is 1:3. So, in terms of k, AD = k, DB = 3k. Therefore, AB = 4k, which we already considered. Then, using the similar triangles, we found AC = 2k and BC = 2k√3. Therefore, the sides are in the ratio 1 : √3 : 2, scaled by 2k. Wait, no. 2k, 2k√3, 4k. So, 2k : 2k√3 : 4k simplifies to 1 : √3 : 2. Which is a 30-60-90 triangle. Therefore, angles are 30°, 60°, 90°, which matches our previous conclusion.Therefore, angles at A and B are 60° and 30°. So the answer should be angle A is 60°, angle B is 30°, or vice versa? Wait, angle A is opposite the longer leg BC, so angle A is 60°, and angle B is 30°. So, depending on how the triangle is labeled. If angle A is at vertex A, which is adjacent to side BC. Wait, maybe I should confirm the labeling.Wait, the problem states it's a right-angled triangle ABC with the right angle at C. So, vertices are labeled such that angle C is the right angle. Therefore, sides are: AB is hypotenuse, AC and BC are the legs. Therefore, angle at A is between sides AC and AB, and angle at B is between sides BC and AB. Therefore, side opposite angle A is BC, and side opposite angle B is AC. Since BC is longer than AC (2k√3 vs 2k), angle A is larger than angle B. Therefore, angle A is 60°, angle B is 30°.Therefore, the interior angles at A and B are 60° and 30°, respectively.Wait, but just to ensure, maybe there's another approach. Let me consider coordinate geometry. Let's place point C at the origin (0,0), point A on the x-axis (a,0), and point B on the y-axis (0,b). Then, hypotenuse AB goes from (a,0) to (0,b). The foot of the perpendicular from C to AB is point D. Let's find coordinates of D.The equation of line AB: from (a,0) to (0,b), the slope is (b - 0)/(0 - a) = -b/a. Therefore, the equation is y = (-b/a)x + b.The foot of the perpendicular from C(0,0) to AB can be found by solving the projection. The formula for the foot of the perpendicular from a point (x0,y0) to the line Ax + By + C = 0 is given by:D = ( (B(Bx0 - Ay0) - AC ) / (A² + B²), (A(-Bx0 + Ay0) - BC ) / (A² + B²) )But maybe it's simpler to use parametric equations or vector projection.Alternatively, the coordinates of D can be found by solving the system of equations: the line AB and the line perpendicular to AB passing through C.The line AB is y = (-b/a)x + b. The slope of AB is -b/a, so the slope of the perpendicular is a/b. Therefore, the line through C(0,0) with slope a/b is y = (a/b)x.Solving the two equations:y = (-b/a)x + bandy = (a/b)xSet equal:(a/b)x = (-b/a)x + bMultiply both sides by ab to eliminate denominators:a² x = -b² x + ab²Bring terms with x to left:a² x + b² x = ab²x(a² + b²) = ab²x = (ab²)/(a² + b²)Similarly, y = (a/b)x = (a/b)*(ab²)/(a² + b²) = (a² b)/(a² + b²)Therefore, coordinates of D are ( (ab²)/(a² + b²), (a² b)/(a² + b²) )The ratio AD:DB is 1:3. The length of AD divided by the length of DB is 1/3.But since AB is from (a,0) to (0,b), the coordinates of D partition AB in the ratio AD:DB = 1:3. Therefore, using section formula, the coordinates of D can be expressed as:D = ( (3*a + 1*0)/(1 + 3), (3*0 + 1*b)/(1 + 3) ) = (3a/4, b/4)Wait, but according to the previous calculation, D is (ab²/(a² + b²), a² b/(a² + b²)). Therefore, equate these coordinates:ab²/(a² + b²) = 3a/4anda² b/(a² + b²) = b/4Let's take the second equation:a² b/(a² + b²) = b/4Divide both sides by b (assuming b ≠ 0):a²/(a² + b²) = 1/4Multiply both sides by (a² + b²):a² = (a² + b²)/4Multiply both sides by 4:4a² = a² + b²Subtract a²:3a² = b²Therefore, b = a√3So the legs are a and a√3. Therefore, the triangle has legs AC = a, BC = a√3, hypotenuse AB = sqrt(a² + 3a²) = sqrt(4a²) = 2a.Therefore, the sides are a, a√3, 2a. So, again, this is a 30-60-90 triangle. Therefore, angles at A and B are 60° and 30°, respectively.Wait, but let's check the first equation:ab²/(a² + b²) = 3a/4Since we have b² = 3a² from above, substitute:a*(3a²)/(a² + 3a²) = 3a/4Simplify denominator: 4a²So numerator: 3a³Thus: 3a³ / 4a² = 3a/4, which matches the left-hand side. So both equations are satisfied.Therefore, the legs are in the ratio 1 : √3, hypotenuse 2, so angles are 30°, 60°, 90°.Therefore, the angle at A, which is opposite the longer leg BC (length a√3), is 60°, and angle at B is 30°.Hence, the measures of the interior angles at A and B are 60 degrees and 30 degrees, respectively.Final AnswerThe interior angles at A and B are boxed{60^circ} and boxed{30^circ} respectively.</think>Given a right-angled triangle (ABC) with the right angle at (C), the height dropped from (C) to the hypotenuse (AB) divides it in the ratio (1:3). We need to find the measures of the interior angles at (A) and (B).1. Understanding the Problem: - The height from (C) to (AB) divides the hypotenuse into segments (AD) and (DB) with the ratio (1:3). - Let (AD = k) and (DB = 3k), so the hypotenuse (AB = 4k).2. Using Similar Triangles: - Triangles (ACD) and (BCD) are similar to the original triangle (ABC). - For triangle (ACD), (AC^2 = AB cdot AD). - For triangle (BCD), (BC^2 = AB cdot DB).3. Calculating the Legs: - (AC^2 = 4k cdot k = 4k^2 implies AC = 2k). - (BC^2 = 4k cdot 3k = 12k^2 implies BC = 2ksqrt{3}).4. Verifying the Hypotenuse: - Using the Pythagorean theorem: (AC^2 + BC^2 = (2k)^2 + (2ksqrt{3})^2 = 4k^2 + 12k^2 = 16k^2 = (4k)^2), which confirms the hypotenuse (AB = 4k).5. Determining the Angles: - The legs (AC = 2k) and (BC = 2ksqrt{3}) correspond to a 30-60-90 triangle. - (tan(angle A) = frac{BC}{AC} = frac{2ksqrt{3}}{2k} = sqrt{3} implies angle A = 60^circ). - (tan(angle B) = frac{AC}{BC} = frac{2k}{2ksqrt{3}} = frac{1}{sqrt{3}} implies angle B = 30^circ).6. Conclusion: - The angles at (A) and (B) are (60^circ) and (30^circ) respectively.[boxed{60^circ} quad text{and} quad boxed{30^circ}],<think>Alright, so I have this problem about a right-angled triangle ABC with the right angle at C. The height from C to the hypotenuse AB divides AB into a ratio of 1:3. I need to find the measures of the interior angles at A and B. Hmm, okay. Let's start by visualizing the triangle. Since it's right-angled at C, sides AC and BC are the legs, and AB is the hypotenuse. The height from C to AB will split AB into two segments, let's say AD and DB, with the ratio AD:DB = 1:3. So, if I let AD be x, then DB would be 3x, making the entire hypotenuse AB = x + 3x = 4x.Now, in a right-angled triangle, the altitude to the hypotenuse has some interesting properties. I remember that the length of the altitude (let's call it h) relates to the lengths of the segments AD and DB. Specifically, h^2 = AD * DB. Is that right? Let me double-check. Yeah, I think that's one of the geometric mean theorems. So, if AD is x and DB is 3x, then h^2 = x * 3x = 3x^2, so h = x*sqrt(3). Hmm, interesting.But how does this help me find the angles at A and B? Well, maybe I can relate the sides of the triangle. Let's denote the legs as follows: let’s say AC = b and BC = a, and AB = c. Then, since it's a right-angled triangle, by Pythagoras, a^2 + b^2 = c^2. Also, the area of the triangle can be expressed in two ways: (1/2)*a*b and (1/2)*h*c. Therefore, a*b = h*c. From earlier, we have h = x*sqrt(3) and c = 4x. So, substituting that in, a*b = (x*sqrt(3)) * 4x = 4x^2*sqrt(3). So, a*b = 4x^2*sqrt(3). Hmm, not sure if that's directly helpful yet.Also, in such a triangle, the segments AD and DB are related to the legs. I recall that AD = b^2 / c and DB = a^2 / c. Wait, is that correct? Let me think. In a right-angled triangle, the length of the altitude is h = (a*b)/c. And then, the segments AD and DB can be found by AD = b^2 / c and DB = a^2 / c. Let me verify that. If AD = b^2 / c and DB = a^2 / c, then AD + DB = (b^2 + a^2)/c = c^2 / c = c, which matches. So that seems correct. So, if AD:DB = 1:3, then (b^2 / c) : (a^2 / c) = 1:3. Simplifying, that's b^2 : a^2 = 1:3. So, b^2 / a^2 = 1/3, which implies that b / a = 1 / sqrt(3), or a / b = sqrt(3). Therefore, the ratio of the legs is a : b = sqrt(3) : 1.So, if I let the legs be a = sqrt(3)*k and b = k for some positive real number k, then the hypotenuse c would be sqrt(a^2 + b^2) = sqrt(3k^2 + k^2) = sqrt(4k^2) = 2k. Therefore, the sides are in the ratio sqrt(3) : 1 : 2, which is reminiscent of a 30-60-90 triangle. In a 30-60-90 triangle, the sides are in the ratio 1 : sqrt(3) : 2, with the angles opposite to those sides being 30°, 60°, and 90° respectively. So, if in our case, the sides are sqrt(3) : 1 : 2, then the angle opposite the side of length 1 (which would be angle B if BC is length a = sqrt(3)*k and AC is length b = k) would be 30°, and the angle opposite sqrt(3)*k (angle A) would be 60°. Wait, hold on.Let me clarify. Let's suppose that BC is the side opposite angle A, and AC is opposite angle B. Since the right angle is at C, then angle A is at vertex A, opposite side BC, and angle B is at vertex B, opposite side AC. Therefore, if BC is length a = sqrt(3)*k, then angle A is opposite BC, so angle A is opposite the longer leg, making it the larger angle. Similarly, AC is length b = k, so angle B is opposite the shorter leg, making it the smaller angle. Therefore, angle A would be 60°, and angle B would be 30°, since in a right-angled triangle, the angles other than the right angle add up to 90°, so if one is 60°, the other is 30°, which matches the side ratios.But let me confirm this with the given ratio of the segments of the hypotenuse. If AD:DB = 1:3, then AD = x and DB = 3x. From earlier, AD = b^2 / c and DB = a^2 / c. So, substituting in terms of k, since b = k and a = sqrt(3)*k and c = 2k, then AD = (k)^2 / (2k) = k / 2, and DB = (sqrt(3)*k)^2 / (2k) = (3k^2)/2k = 3k/2. Therefore, AD:DB = (k/2) : (3k/2) = 1:3, which matches the given ratio. So that checks out.Therefore, the angles at A and B are 60° and 30°, respectively. But let me make sure I haven't mixed up which angle is which. If side BC is opposite angle A, and BC is the longer leg (sqrt(3)*k), then angle A is opposite the longer leg, which would make angle A larger than angle B. Since in a right-angled triangle, the side opposite the larger angle is longer, this makes sense. So angle A is 60°, angle B is 30°, and angle C is 90°, adding up to 180°.Alternatively, maybe I can use trigonometry to find the angles. Let's see. Let's take angle A. The tangent of angle A is opposite over adjacent, which would be BC / AC = (sqrt(3)*k)/k = sqrt(3). So tan(angle A) = sqrt(3), which means angle A is 60°, since tan(60°) = sqrt(3). Similarly, tan(angle B) = AC / BC = k / (sqrt(3)*k) = 1/sqrt(3), which is tan(30°), so angle B is 30°. Yep, that confirms it.Therefore, the interior angles at A and B are 60° and 30°, respectively. Let me just check once more to make sure I didn't flip them. Since the height divides the hypotenuse into 1:3, and the triangle is right-angled, the legs must be in the ratio 1:sqrt(3), leading to angles of 30° and 60°. Wait, but in our case, the ratio of the legs was sqrt(3):1, so the sides opposite the angles would be sqrt(3) and 1. So angle opposite sqrt(3) is 60°, and angle opposite 1 is 30°, which matches the tangent ratios we calculated. So angle at A is 60°, and angle at B is 30°, so I think that's correct.Final AnswerThe interior angles at A and B are boxed{60^circ} and boxed{30^circ} respectively.</think>In a right-angled triangle (ABC) with the right angle at (C), the height dropped from (C) to the hypotenuse (AB) divides it in the ratio (1:3). We need to find the measures of the interior angles at (A) and (B).1. Let the segments of the hypotenuse (AB) be (AD = x) and (DB = 3x), making the entire hypotenuse (AB = 4x).2. Using the geometric mean theorem, the height (h) from (C) to (AB) satisfies (h^2 = AD cdot DB). Substituting the values, we get (h^2 = x cdot 3x = 3x^2), so (h = xsqrt{3}).3. The legs (AC) and (BC) are related to the segments (AD) and (DB) by the equations (AD = frac{b^2}{c}) and (DB = frac{a^2}{c}), where (a = BC), (b = AC), and (c = AB). Given the ratio (AD:DB = 1:3), we have (frac{b^2}{c} : frac{a^2}{c} = 1:3), leading to (frac{b^2}{a^2} = frac{1}{3}). Thus, (frac{b}{a} = frac{1}{sqrt{3}}) or (frac{a}{b} = sqrt{3}).4. Let (a = sqrt{3}k) and (b = k). The hypotenuse (c) is then (sqrt{a^2 + b^2} = sqrt{3k^2 + k^2} = 2k).5. The sides are in the ratio (sqrt{3} : 1 : 2), characteristic of a 30-60-90 triangle. The angle opposite the longer leg ((sqrt{3}k)) is (60^circ) and the angle opposite the shorter leg ((k)) is (30^circ).6. Verifying with trigonometry, (tan(angle A) = frac{BC}{AC} = sqrt{3}) gives (angle A = 60^circ), and (tan(angle B) = frac{AC}{BC} = frac{1}{sqrt{3}}) gives (angle B = 30^circ).Thus, the interior angles at (A) and (B) are (boxed{60^circ}) and (boxed{30^circ}) respectively.]

🔑:Let H be the foot of the altitude from C to A B, and let |H B|=3 cdot|A H|. Furthermore, let alpha:=angle B A C and beta:=angle A B C.Then |A B|=4 cdot|A H|, and by the leg theorem, |B C|=sqrt{|A H| cdot|A B|}=2 cdot|A H|.By the definition of the sine in a right triangle, sin alpha=frac{|B C|}{|A B|}=frac{1}{2}, so alpha=30^{circ}, since 0<alpha<90^{circ}.From the sum of the interior angles in the triangle, we finally get beta=180^{circ}-90^{circ}-30^{circ}=60^{circ}.

🧠:[<think>Okay, let's tackle this quadratic equation problem step by step. The equation given is ( x^2 - 4x + 3 - 2m = 0 ). There are two parts to this problem. First, we need to find the range of real numbers ( m ) for which the equation has two distinct real roots. Second, given a specific condition involving the roots ( x_1 ) and ( x_2 ), we need to find the value of ( m ).Starting with part (1): Finding the range of ( m ) such that the quadratic equation has two distinct real roots.I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( D = b^2 - 4ac ) determines the nature of the roots. If ( D > 0 ), there are two distinct real roots. If ( D = 0 ), there is exactly one real root (a repeated root), and if ( D < 0 ), there are no real roots, just complex ones.So, applying this to our equation. Let's first identify the coefficients. Comparing ( x^2 - 4x + 3 - 2m = 0 ) with the standard quadratic equation, we have:- ( a = 1 )- ( b = -4 )- ( c = 3 - 2m )Now, compute the discriminant ( D ):( D = (-4)^2 - 4(1)(3 - 2m) )Calculating that:( D = 16 - 12 + 8m )Simplify:( D = 4 + 8m )For two distinct real roots, we need ( D > 0 ):( 4 + 8m > 0 )Subtract 4 from both sides:( 8m > -4 )Divide both sides by 8:( m > -frac{4}{8} )Simplify:( m > -frac{1}{2} )So, the range of ( m ) is all real numbers greater than -1/2. Therefore, the answer to part (1) is ( m > -frac{1}{2} ).Moving on to part (2): Given that ( x_1 ) and ( x_2 ) are the two roots and they satisfy the equation ( x_1x_2 + x_1 + x_2 - m^2 = 4 ), find the value of ( m ).From Vieta's formulas, we know that for a quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots ( x_1 + x_2 = -frac{b}{a} ) and the product ( x_1x_2 = frac{c}{a} ).Applying this to our equation ( x^2 - 4x + 3 - 2m = 0 ):- Sum of roots: ( x_1 + x_2 = -frac{-4}{1} = 4 )- Product of roots: ( x_1x_2 = frac{3 - 2m}{1} = 3 - 2m )Now, substitute these into the given equation ( x_1x_2 + x_1 + x_2 - m^2 = 4 ):Plugging in the known values:( (3 - 2m) + 4 - m^2 = 4 )Simplify the left-hand side:First, combine constants: 3 + 4 = 7So, ( 7 - 2m - m^2 = 4 )Bring all terms to one side to form a quadratic equation:( 7 - 2m - m^2 - 4 = 0 )Simplify 7 - 4 = 3:( 3 - 2m - m^2 = 0 )Rearranging terms in descending order:( -m^2 - 2m + 3 = 0 )Multiply both sides by -1 to make the coefficient of ( m^2 ) positive:( m^2 + 2m - 3 = 0 )Now, solve the quadratic equation ( m^2 + 2m - 3 = 0 ).We can factor this equation. Looking for two numbers that multiply to -3 and add to 2. Let's see: 3 and -1. Yes, 3 * (-1) = -3, and 3 + (-1) = 2. So, factors are (m + 3)(m - 1) = 0.Wait, let's check that. (m + 3)(m - 1) = m^2 - m + 3m - 3 = m^2 + 2m - 3. Yes, that's correct.Therefore, the solutions are:( m + 3 = 0 ) => ( m = -3 )or( m - 1 = 0 ) => ( m = 1 )So, potential solutions are ( m = -3 ) and ( m = 1 ).But we need to check if these values of ( m ) are within the range found in part (1). Recall that from part (1), ( m > -frac{1}{2} ).Check ( m = -3 ): -3 is less than -1/2, so this is not valid.Check ( m = 1 ): 1 is greater than -1/2, so this is valid.Therefore, the only valid solution is ( m = 1 ).Wait, but let me double-check the substitution to make sure there's no mistake.Original equation after substitution was:( 3 - 2m - m^2 = 0 )Which simplifies to ( m^2 + 2m - 3 = 0 ). Solving gives m = [-2 ± sqrt(4 + 12)] / 2 = [-2 ± sqrt(16)] / 2 = [-2 ± 4]/2. So, m = (-2 + 4)/2 = 2/2 = 1, or m = (-2 -4)/2 = -6/2 = -3. So that's correct. So, solutions are indeed 1 and -3. As before, -3 is invalid, so m=1.But wait, let's verify if m=1 satisfies the original equation. Let's check:If m=1, then the quadratic equation becomes x² -4x + 3 -2(1) = x² -4x +1 =0.Check the discriminant: D=16 -4*1*1=16-4=12>0, so two distinct real roots. Good.Now, compute x1x2 +x1 +x2 -m². From Vieta's:x1 + x2 = 4, x1x2 = 1. So:1 + 4 - (1)^2 = 1 +4 -1 =4. Which equals the RHS of 4. So yes, correct.For m=-3, even though it's outside the allowed range, let's see why it doesn't work. If m=-3, the quadratic equation becomes x² -4x +3 -2*(-3)= x² -4x +3 +6= x² -4x +9=0. Discriminant: 16 -36= -20 <0, so no real roots. Hence, m=-3 is invalid as expected.Therefore, the answer for part (2) is m=1.So, summarizing:1. The range of m is all real numbers greater than -1/2.2. The value of m is 1.Final Answer(1) The range of ( m ) is boxed{(-frac{1}{2}, infty)}.(2) The value of ( m ) is boxed{1}.</think>Given the quadratic equation ( x^2 - 4x + 3 - 2m = 0 ):(1) To find the range of ( m ) for which the equation has two distinct real roots, we compute the discriminant ( D ):[D = (-4)^2 - 4 cdot 1 cdot (3 - 2m) = 16 - 12 + 8m = 4 + 8m]For two distinct real roots, we need ( D > 0 ):[4 + 8m > 0 implies 8m > -4 implies m > -frac{1}{2}]Thus, the range of ( m ) is (boxed{(-frac{1}{2}, infty)}).(2) Given the condition ( x_1x_2 + x_1 + x_2 - m^2 = 4 ), we use Vieta's formulas for the roots ( x_1 ) and ( x_2 ):- Sum of roots: ( x_1 + x_2 = 4 )- Product of roots: ( x_1x_2 = 3 - 2m )Substituting these into the given condition:[(3 - 2m) + 4 - m^2 = 4 implies 3 - 2m + 4 - m^2 = 4 implies 7 - 2m - m^2 = 4 implies -m^2 - 2m + 3 = 0 implies m^2 + 2m - 3 = 0]Solving the quadratic equation ( m^2 + 2m - 3 = 0 ):[(m + 3)(m - 1) = 0 implies m = -3 text{ or } m = 1]Checking against the range from part (1), ( m = -3 ) is invalid and ( m = 1 ) is valid. Verifying ( m = 1 ) confirms it satisfies the original condition. Thus, the value of ( m ) is (boxed{1}).,<think>Okay, let's tackle this quadratic equation problem step by step. There are two parts here: first, finding the range of m for which the equation has two distinct real roots, and second, determining the value of m that satisfies a particular condition involving the roots. Let's start with part (1).So, the equation given is x² - 4x + 3 - 2m = 0. For a quadratic equation ax² + bx + c = 0 to have two distinct real roots, the discriminant must be positive. Remember, the discriminant D is given by D = b² - 4ac. If D > 0, there are two distinct real roots. If D = 0, there's exactly one real root (a repeated root), and if D < 0, there are no real roots, just complex ones.Let me write down the coefficients for our equation. Comparing with ax² + bx + c = 0:a = 1 (coefficient of x²)b = -4 (coefficient of x)c = 3 - 2m (constant term)So, plugging these into the discriminant formula:D = (-4)² - 4 * 1 * (3 - 2m) = 16 - 4*(3 - 2m)Let me compute that. 16 minus 4*(3 - 2m). First, compute 4*(3 - 2m):4*3 = 124*(-2m) = -8mSo, 4*(3 - 2m) = 12 - 8mTherefore, D = 16 - (12 - 8m) = 16 - 12 + 8m = 4 + 8mSo, the discriminant is 4 + 8m. For two distinct real roots, this discriminant must be greater than 0:4 + 8m > 0Let's solve for m:8m > -4Divide both sides by 8:m > -4/8Simplify:m > -0.5So, the range of m is all real numbers greater than -0.5. In interval notation, that would be (-0.5, ∞). But since the problem probably wants it in fraction form, -1/2. So, m > -1/2.Wait, let me check my computations again to be sure. Let's redo the discriminant step:Original equation: x² -4x +3 -2m =0Discriminant D = (-4)^2 -4*1*(3 -2m) = 16 -12 +8m = 4 +8m. Yes, that's correct. So 4 +8m >0 => m > -4/8 => m > -1/2. Yep, that's right. So part (1) answer is m > -1/2. Okay, moving on to part (2).Part (2) states that if x₁ and x₂ are the two roots of the equation and satisfy the equation x₁x₂ +x₁ +x₂ - m² =4, find the value of m.Hmm. So, we have to use Vieta's formulas here, right? Because Vieta's relates the sum and product of roots to the coefficients of the equation.For a quadratic equation ax² +bx +c =0, the sum of the roots is -b/a and the product is c/a.So, in our equation, sum of roots x₁ +x₂ = -b/a = -(-4)/1 =4.Product of roots x₁x₂ = c/a = (3 -2m)/1 =3 -2m.So, x₁ +x₂ =4 and x₁x₂=3 -2m.The given condition is x₁x₂ +x₁ +x₂ -m² =4.Let me substitute the values from Vieta's into this equation.So, x₁x₂ is 3 -2m, x₁ +x₂ is 4.Therefore, substituting:(3 -2m) +4 -m² =4Simplify left side:3 -2m +4 -m² = (3 +4) + (-2m) + (-m²) =7 -2m -m²Set equal to 4:7 -2m -m² =4Subtract 4 from both sides:3 -2m -m² =0Let me rearrange the terms:-m² -2m +3 =0Multiply both sides by -1 to make it a bit more standard:m² +2m -3 =0So, quadratic equation in m: m² +2m -3 =0Let's solve this quadratic equation. We can factor it if possible.Looking for two numbers that multiply to -3 and add to +2. Hmm, 3 and -1: 3 * (-1) = -3, 3 + (-1) =2. Yes, perfect.Therefore, factored form:(m +3)(m -1) =0Set each factor equal to zero:m +3 =0 => m =-3m -1=0 =>m=1So, solutions are m=-3 and m=1.But wait, we need to check if these values of m satisfy the condition from part (1), which was m > -1/2. Because the original equation must have two distinct real roots, so m must be greater than -1/2.So, check m=-3: Is -3 > -1/2? No, -3 is less than -1/2. Therefore, m=-3 is invalid.Check m=1: Is 1 > -1/2? Yes. So m=1 is acceptable.Therefore, the value of m is 1.Let me just verify this again to make sure.If m=1, let's plug back into the original quadratic equation:x² -4x +3 -2*1 =x² -4x +3 -2 =x² -4x +1=0Compute discriminant D= (-4)^2 -4*1*1=16 -4=12>0, so yes, two distinct real roots. Good.Now check the condition x₁x₂ +x₁ +x₂ -m²=4.From the quadratic equation x² -4x +1=0, product of roots x₁x₂ =1 (since c=1), sum of roots x₁ +x₂=4.So, x₁x₂ +x₁ +x₂ -m² =1 +4 - (1)^2=5 -1=4. Which matches the given equation. Perfect.Therefore, m=1 is correct.So, part (2) answer is m=1.Final Answer(1) The range of ( m ) is boxed{(-frac{1}{2}, infty)}.(2) The value of ( m ) is boxed{1}.</think>Given the quadratic equation (x^2 - 4x + 3 - 2m = 0) with two distinct real roots, we need to find the range of (m) and the value of (m) that satisfies a specific condition involving the roots.(1) To determine the range of (m) for which the equation has two distinct real roots, we calculate the discriminant (D) of the quadratic equation. The discriminant is given by:[D = b^2 - 4ac]For the given equation (x^2 - 4x + 3 - 2m = 0), the coefficients are (a = 1), (b = -4), and (c = 3 - 2m). Substituting these values into the discriminant formula, we get:[D = (-4)^2 - 4 cdot 1 cdot (3 - 2m) = 16 - 12 + 8m = 4 + 8m]For the equation to have two distinct real roots, the discriminant must be positive:[4 + 8m > 0 implies 8m > -4 implies m > -frac{1}{2}]Thus, the range of (m) is (boxed{(-frac{1}{2}, infty)}).(2) Given the condition (x_1x_2 + x_1 + x_2 - m^2 = 4), we use Vieta's formulas where (x_1 + x_2 = 4) and (x_1x_2 = 3 - 2m). Substituting these into the given condition:[(3 - 2m) + 4 - m^2 = 4]Simplifying the left side:[7 - 2m - m^2 = 4]Rearranging terms:[3 - 2m - m^2 = 0 implies m^2 + 2m - 3 = 0]Solving the quadratic equation (m^2 + 2m - 3 = 0) by factoring:[(m + 3)(m - 1) = 0]This gives the solutions (m = -3) and (m = 1). Checking against the range from part (1), (m = -3) is invalid, leaving (m = 1) as the valid solution. Verifying, substituting (m = 1) into the original equation confirms the condition holds.Thus, the value of (m) is (boxed{1}).]

🔑: Step-by-Step Solution# Part 1: Finding the Range of Real Numbers for mFor a quadratic equation ax^2 + bx + c = 0 to have two distinct real roots, its discriminant D = b^2 - 4ac must be greater than 0. Applying this to the given equation x^{2}-4x+3-2m=0, we have:1. Calculate the discriminant: [ D = (-4)^2 - 4(3 - 2m) geq 0 ]2. Simplify the inequality: [ 16 - 12 + 8m geq 0 ]3. Solve for m: [ 8m geq -4 implies m geq -frac{1}{2} ]Thus, the range of real numbers for m is boxed{m geq -frac{1}{2}}.# Part 2: Finding the Value of mGiven x_1 and x_2 are the roots of the equation, we know from Vieta's formulas that:1. Sum of roots: [ x_1 + x_2 = 4 ]2. Product of roots: [ x_1x_2 = 3 - 2m ]Given the equation x_1x_2 + x_1 + x_2 - m^2 = 4, substituting the values from Vieta's formulas:1. Substitute and simplify: [ (3 - 2m) + 4 - m^2 = 4 ]2. Rearrange and solve for m: [ m^2 + 2m - 3 = 0 ]3. Factor or use the quadratic formula to find m: [ m = -3 text{ or } m = 1 ]Since m must satisfy the condition from part 1 (m geq -frac{1}{2}), m = -3 is not a valid solution.Therefore, the valid value of m is boxed{m = 1}.

🧠:[<think>Okay, so the question is asking which latitudinal circles on Earth are half as long as the Equator. Hmm, let's think. First, I remember that the Equator is the longest circle of latitude, right? And as you go north or south from the Equator, the circles get smaller until you reach the poles, which are just points. The length of a latitudinal circle, I think, depends on the cosine of the latitude angle. Because the Earth is a sphere, the circumference at a certain latitude would be the circumference of the Equator multiplied by the cosine of that latitude. So, if the Equator's circumference is C, then at latitude θ, the circumference is C * cosθ. The question is asking for where this circumference is half of the Equator's. So we need to solve for θ in the equation:C * cosθ = (1/2) * CIf I divide both sides by C, then cosθ = 1/2. So θ must be the angle whose cosine is 1/2. I know that cos(60°) = 0.5, so θ = 60°. But latitude can be north or south, so the answer should be the 60th parallels north and south. Let me check that again.Wait, circumference of a circle is 2πr, right? So if the Equator's circumference is 2πR, where R is the Earth's radius, then at latitude θ, the radius of the circle is R * cosθ. Therefore, the circumference there is 2π(R * cosθ) = 2πR * cosθ. So yes, that's the Equator's circumference times cosθ. So setting that equal to half the Equator's circumference:2πR * cosθ = (1/2) * 2πRSimplifying both sides: cosθ = 1/2. So θ = 60° or θ = 300°, but latitude is measured from 0° to 90°, so θ is 60° north or 60° south. Therefore, the 60th parallels.But let me verify. Let's compute the length. Suppose the Earth's radius is R. Equator length is 2πR. At 60° latitude, the radius of the circle is R * cos60° = R * 0.5. Then the circumference is 2π*(0.5R) = πR. Which is half of 2πR. Yep, that works. So the answer should be the 60°N and 60°S latitude circles.Wait, but I should make sure there isn't a mistake here. Sometimes people confuse the radius with the circumference. But the formula seems right. The circumference at latitude θ is 2πR cosθ, so to get half the equatorial circumference, set 2πR cosθ = πR. Then divide both sides by πR: 2cosθ = 1, so cosθ = 0.5. So θ = 60°, yeah. So 60°N and 60°S. I think that's correct. Maybe another way to think about it: the Earth's radius at the Equator is R. If you imagine a circle at 60° latitude, the radius of that circle is R times cos60°, which is R * 0.5. Then the circumference would be πR, which is half of the Equator's 2πR. So yes, that matches. Therefore, the latitudinal circles at 60° North and South are half as long as the Equator.</think>To determine which latitudinal circles are half as long as the Equator, we start by noting that the circumference of a circle at latitude ( theta ) is proportional to ( costheta ). The Equator's circumference is ( 2pi R ), where ( R ) is Earth's radius. For a latitude to have half this length, we solve:[2pi R costheta = frac{1}{2} cdot 2pi R]Simplifying, we find ( costheta = frac{1}{2} ). The angle ( theta ) that satisfies this is ( 60^circ ). Since latitude can be north or south, the corresponding circles are at 60° North and 60° South.[boxed{60^circ text{N} text{ and } 60^circ text{S}}],<think>Alright, so I have this question here: "On Earth, which latitudinal circles are half as long as the Equator?" Hmm, okay. Let's see. I need to figure out which lines of latitude have a circumference that's half the length of the Equator. First off, I remember that the Earth is roughly a sphere, so the circumference of a latitudinal circle (also called a parallel) depends on the latitude. The Equator itself is the largest circle, right? Its circumference is the same as the Earth's circumference, which is about 40,075 kilometers. But since the question is about the ratio, maybe I don't need the exact numbers. So, the key thing here is understanding how latitude relates to the circumference of those circles. I think as you go north or south from the Equator, the circumference of each parallel decreases. The reason for that is because the circles get smaller as you approach the poles. At the poles, the circumference would be zero, which makes sense. Now, mathematically, the circumference of a circle of latitude at angle θ is given by the formula: Circumference = C * cos(θ), where C is the equatorial circumference. Wait, is that right? Let me think. If θ is the latitude, then the radius of that circle would be R * cos(θ), where R is the Earth's radius. So then the circumference would be 2πR * cos(θ), which is indeed C * cos(θ) because the equatorial circumference C is 2πR. So, if the question is asking for which latitude circles have half the length of the Equator, I need to solve for θ when Circumference = C/2. Setting up the equation: C * cos(θ) = C/2. Then, dividing both sides by C (assuming C ≠ 0, which it isn't), we get cos(θ) = 1/2. So, θ must be the angle whose cosine is 1/2. From trigonometry, I remember that cos(60°) = 0.5. Therefore, θ = 60°. But latitude is measured from the Equator, so does that mean 60° North and 60° South? That seems right. The circles at 60°N and 60°S should have half the circumference of the Equator. Wait, let me verify. Let's plug in θ = 60° into the circumference formula. cos(60°) is indeed 0.5, so the circumference would be 0.5*C, which is half of the Equator's circumference. That checks out. But just to make sure I didn't make a mistake in the reasoning. Let's think again. The radius of the Earth at latitude θ is R*cosθ. Circumference is 2π times that radius. So yes, 2πR cosθ = C cosθ. So if I want that to be half of C, then cosθ must equal 0.5. So θ is 60°, which is correct. Therefore, the latitudinal circles at 60° North and 60° South should be half as long as the Equator. Wait, but is there another angle where cosine is 0.5? Cosine is positive in the first and fourth quadrants, but latitude is from 0° to 90° North or South, so only 60°N and 60°S make sense here. The negative angle would correspond to the southern hemisphere. So, yes, the answer should be the 60th parallels north and south. But just to cross-verify, maybe with actual numbers. Let's take the Earth's equatorial circumference as approximately 40,075 km. Half of that is about 20,037.5 km. What's the circumference at 60° latitude? Calculating: 40,075 km * cos(60°) = 40,075 * 0.5 = 20,037.5 km. Exactly half. So that's correct. Therefore, the answer must be the 60°N and 60°S latitude circles. I think that's solid. My initial thought was right, and the verification with numbers confirmed it. So, unless there's a trick here that I'm missing, like considering the Earth's oblateness or something... but the question just says "latitudinal circles" on Earth, and usually for such problems, we assume a perfect sphere unless stated otherwise. The Earth is actually an oblate spheroid, slightly flattened at the poles, but I don't think that would significantly affect the answer here. The cosine relationship still holds for a sphere, which is the standard model for such questions. So yeah, the answer should be 60° North and South.Final AnswerThe latitudinal circles at boxed{60^circ} North and South are half as long as the Equator.</think>To determine which latitudinal circles are half as long as the Equator, we start by noting that the circumference of a circle of latitude at angle ( theta ) is given by ( C cos(theta) ), where ( C ) is the equatorial circumference. We need to find the latitude ( theta ) where this circumference is half of the equatorial circumference. Setting up the equation:[C cos(theta) = frac{C}{2}]Dividing both sides by ( C ):[cos(theta) = frac{1}{2}]The angle ( theta ) whose cosine is ( frac{1}{2} ) is ( 60^circ ). Since latitude is measured from the Equator, the corresponding latitudinal circles are at ( 60^circ ) North and ( 60^circ ) South.Verification using the Earth's equatorial circumference (approximately 40,075 km) shows that the circumference at ( 60^circ ) latitude is indeed half:[40,075 , text{km} times cos(60^circ) = 40,075 , text{km} times 0.5 = 20,037.5 , text{km}]Thus, the latitudinal circles at ( 60^circ ) North and South are half as long as the Equator.[boxed{60^circ}]]

🔑:The circumference of a circle is proportional to its radius, mathrm{s}, so if the radius of the equator is R, the radius of the sought parallel circle is R / 2.A plane passing through the axis of rotation of the sphere intersects the sphere in a great circle, and the diameter of the sought parallel circle is the chord A B of this great circle. The triangle O A B is equilateral, so the 60^{circ} parallel is half as long as the equator.There are two such parallels: one in the northern hemisphere, passing through Sweden (Uppsala), while the 60^{circ} parallel in the southern hemisphere is far from any land.

🧠:[<think>Okay, let's see. I have this problem about an equilateral triangle ABC, with a point M on the arc AC of its circumcircle. The segments MA and MC are given as 4 and 5 units long, respectively. Then, the line BM intersects side AC at point N. I need to find the length of segment MN and the side length of triangle ABC. Hmm, okay.First, since ABC is an equilateral triangle, all sides are equal, and all angles are 60 degrees. The circumcircle of an equilateral triangle has some nice properties. For example, the center is also the centroid, which is the intersection point of the medians, and the radius is known in terms of the side length. Maybe that will come into play.Given that M is on the arc AC of the circumcircle, that means M is not on the triangle itself but somewhere on the circumcircle. Since arc AC is mentioned, perhaps M is on the opposite side of the triangle from vertex B? So, if ABC is oriented with A at the left, B at the top, and C at the right, then the arc AC that doesn't contain B would be the lower arc. So M is somewhere along that arc.MA is 4 and MC is 5. Since ABC is equilateral, the side length is the same for all sides, let's call it 's'. The circumradius R of an equilateral triangle is given by R = s / √3. So, the radius is related to the side length. Since M is on the circumcircle, the distance from M to all the vertices should be equal to the radius? Wait, no. Wait, in a circumcircle, all the vertices lie on the circle, so the distance from the center to each vertex is the radius. But the distance between two vertices (the side length) is s, which relates to the radius as R = s / √3. But point M is another point on the circumcircle, so MA, MB, MC are all equal to 2R? Wait, no. Wait, in a triangle, the circumradius is the distance from the center to the vertices, not the distance between the vertices. The distance between two vertices is the side length. So if the radius is R, then each side length is R * √3. Hmm, okay, so R = s / √3, so s = R√3. So, if I can find R, then I can find s.But in this problem, MA and MC are given as 4 and 5. Wait, but if M is on the circumcircle, then MA and MC should be chord lengths. In an equilateral triangle, all chords would correspond to different arcs. So, the chord length between two points on a circle is 2R sin(θ/2), where θ is the central angle. So, MA and MC correspond to chords subtended by angles at the center. Let me think.Since ABC is equilateral, the central angles for each side are 120 degrees (since the full circle is 360, divided by 3). So, the arc AB, BC, and CA each are 120 degrees. Since M is on arc AC, which is 120 degrees. So, the position of M on arc AC would create chords MA and MC. Depending on where M is, the central angles for MA and MC would vary, but their sum should be 120 degrees, right? Because M is on arc AC, so the central angles for MA and MC would add up to 120 degrees.So, if I let the central angle for MA be θ, then the central angle for MC would be 120° - θ. Then, the chord lengths MA and MC can be written as 2R sin(θ/2) and 2R sin((120° - θ)/2), respectively. Given that MA = 4 and MC = 5, so:2R sin(θ/2) = 42R sin((120° - θ)/2) = 5So, two equations with two variables, R and θ. Maybe we can solve for R and θ.Let me write that down:Equation 1: 2R sin(θ/2) = 4 => R sin(θ/2) = 2Equation 2: 2R sin((120° - θ)/2) = 5 => R sin((120° - θ)/2) = 5/2Divide Equation 1 by Equation 2:[ R sin(θ/2) ] / [ R sin((120° - θ)/2) ] = 2 / (5/2) = 4/5So, sin(θ/2) / sin((120° - θ)/2) = 4/5Let me let φ = θ/2, so the equation becomes sin(φ) / sin(60° - φ) = 4/5So, sin(φ) / sin(60° - φ) = 4/5Cross-multiplying: 5 sin φ = 4 sin(60° - φ)Using the sine subtraction formula: sin(60° - φ) = sin60 cosφ - cos60 sinφSo, 5 sin φ = 4 [ (√3/2) cosφ - (1/2) sinφ ]Multiply through:5 sin φ = 4*(√3/2 cosφ) - 4*(1/2 sinφ)Simplify:5 sin φ = 2√3 cosφ - 2 sinφBring all terms to left side:5 sin φ + 2 sin φ - 2√3 cosφ = 07 sin φ - 2√3 cosφ = 0Divide both sides by cosφ (assuming cosφ ≠ 0):7 tanφ - 2√3 = 07 tanφ = 2√3tanφ = (2√3)/7So φ = arctan(2√3 / 7)Hmm, okay. Let's compute tanφ = (2√3)/7. Let me find sinφ and cosφ from this.Let me construct a right triangle where the opposite side is 2√3 and adjacent is 7. Then the hypotenuse is sqrt( (2√3)^2 + 7^2 ) = sqrt(12 + 49) = sqrt(61)Therefore, sinφ = (2√3)/sqrt(61) and cosφ = 7/sqrt(61)So, going back to Equation 1: R sinφ = 2So R = 2 / sinφ = 2 / (2√3 / sqrt(61)) ) = 2 * sqrt(61)/(2√3) ) = sqrt(61)/√3 = sqrt(61/3)So R = sqrt(61/3). Then, since the side length s = R * √3 = sqrt(61/3) * √3 = sqrt(61). So s = sqrt(61). Hmm, so the side length is sqrt(61). Let me check if that makes sense.Wait, let's confirm this. If R = sqrt(61/3), then s = R * √3 = sqrt(61/3) * sqrt(3) = sqrt(61). Yes, that's right.So, the side length of triangle ABC is sqrt(61). So that's the first part. Wait, no, the problem asks for both MN and the side length. So, the side length is sqrt(61). Let's check if this is correct.But let me verify if this is consistent with the chord lengths.Given that R = sqrt(61/3), then MA = 2R sinφ = 2 * sqrt(61/3) * (2√3)/sqrt(61) ) = 2 * sqrt(61/3) * (2√3)/sqrt(61) ) = 2 * (2√3)/sqrt(3) ) = 2 * 2 = 4. Correct. Similarly, MC = 2R sin(60° - φ) = 5. Let's check that.sin(60° - φ) = sin60 cosφ - cos60 sinφ = (√3/2)(7/sqrt(61)) - (1/2)(2√3/sqrt(61)) = (7√3)/(2 sqrt(61)) - (√3)/sqrt(61) = (7√3 - 2√3)/(2 sqrt(61)) = (5√3)/(2 sqrt(61))So, 2R sin(60° - φ) = 2 * sqrt(61/3) * (5√3)/(2 sqrt(61)) ) = sqrt(61/3) * 5√3 / sqrt(61) ) = 5 * sqrt(3) / sqrt(3) ) = 5. Correct. So yes, R is sqrt(61/3) and s = sqrt(61). So that's the side length.So, that's good. So the side length is sqrt(61). Now, we need to find MN.So, BM intersects AC at N. We need to find the length of MN. Hmm. How can we approach this?Since ABC is equilateral, all sides are equal, so AC is of length sqrt(61). The line BM intersects AC at N. So, we need to find MN. Perhaps using Menelaus' theorem or Ceva's theorem? Or coordinate geometry? Or power of a point? Maybe coordinate geometry.Let me try coordinate geometry. Let me place the triangle in a coordinate system. Let me set point A at (0, 0), point C at (s, 0), and point B at (s/2, (s√3)/2), since it's an equilateral triangle.So, coordinates:A: (0, 0)C: (s, 0)B: (s/2, (s√3)/2 )Then, the circumcircle of triangle ABC can be found. The circumradius is R = s / √3, so the center is at the centroid of the triangle. Wait, no, in an equilateral triangle, centroid, circumcenter, orthocenter, and incenter coincide. So the center is at (s/2, (s√3)/6 ), because centroid is at ( (0 + s + s/2)/3, (0 + 0 + (s√3)/2 ) /3 ) = ( (3s/2)/3, (s√3/2)/3 ) = (s/2, s√3/6 ). So that's the center.Point M is on the circumcircle, on arc AC. So, parametrize point M. Let me use angles. Since the center is at (s/2, s√3/6 ), the circumradius is R = s / √3.Parametrizing the circumcircle, with angle θ from the positive x-axis (horizontal axis). But since M is on arc AC, which is the arc from A to C not containing B, so angle θ would be between 0 and 120°, but measured from the center.Wait, maybe it's easier to use parametric equations. Let me recall that in an equilateral triangle, the circumcircle can be parameterized with angles. Let me consider the center at (s/2, s√3/6 ). Then, any point on the circumcircle can be written as:x = s/2 + R cosθy = s√3/6 + R sinθWhere θ is the angle parameter. Here, R = s / √3.So,x = s/2 + (s/√3) cosθy = s√3/6 + (s/√3) sinθNow, point M is on arc AC. Since arc AC is the arc from A to C not passing through B. Since the triangle is oriented with B at the top, arc AC not containing B would be the lower arc. So, points on this arc will have angles θ such that when θ is from 180° to 360°? Wait, maybe not. Wait, the center is at (s/2, s√3/6 ). So, to parametrize point A: when θ is such that the point is at A (0,0):x = s/2 + (s/√3) cosθ = 0y = s√3/6 + (s/√3) sinθ = 0Solving for θ:From x: s/2 + (s/√3) cosθ = 0 => cosθ = - (s/2) / (s/√3) ) = - √3 / 2Similarly, y: s√3/6 + (s/√3) sinθ = 0 => sinθ = - (s√3/6) / (s/√3) ) = - (√3/6) / (1/√3) ) = - (√3/6) * √3 ) = - 3/6 = -1/2So, θ corresponding to point A is 210°, since cosθ = -√3/2 and sinθ = -1/2. Similarly, point C is at (s, 0). Let's find θ for point C:x = s/2 + (s/√3) cosθ = sSo, (s/√3) cosθ = s - s/2 = s/2 => cosθ = (s/2) / (s/√3) ) = √3 / 2Similarly, y = s√3/6 + (s/√3) sinθ = 0=> (s/√3) sinθ = -s√3 /6=> sinθ = (-s√3 /6 ) / (s/√3 ) = (-√3 /6 ) * (√3 /1 ) = -3/6 = -1/2So, θ for point C is 330°, since cosθ = √3/2 and sinθ = -1/2. Therefore, the arc from A (210°) to C (330°) not containing B (which is at 90°). So, the arc AC is from 210° to 330°, which is 120°, as expected. So, point M is somewhere between 210° and 330°. Let's denote the angle θ for point M as 210° + α, where α ranges from 0° to 120°. But maybe there's a better way.Alternatively, since we already know the side length s = sqrt(61), so R = sqrt(61)/√3. Let's use that.Given that, maybe it's easier to use coordinates with s = sqrt(61). Let me assign coordinates accordingly.Let me set A at (0, 0), C at (sqrt(61), 0), and B at (sqrt(61)/2, (sqrt(61) * √3)/2 ). The center of the circumcircle is at (sqrt(61)/2, (sqrt(61) * √3)/6 ).Point M is on the circumcircle, arc AC. So, using the parametrization:x = sqrt(61)/2 + (sqrt(61)/√3) cosθy = (sqrt(61)√3)/6 + (sqrt(61)/√3) sinθBut this might get complicated. Maybe instead of parametrizing, we can use coordinates with specific values.Alternatively, use complex numbers. Let me think.Alternatively, use power of a point. Since point M is on the circumcircle of ABC, then by power of a point, for point N on AC, we have that NA * NC = NM * NB. Wait, but N is on AC and BM. Hmm. Maybe Menelaus' theorem?Menelaus' theorem relates the lengths of a transversal cutting through a triangle. If a line crosses through the sides of a triangle, the product of the segment ratios is 1. Let's see.In triangle ABC, with the transversal line BMN, where N is on AC. Wait, Menelaus' theorem would apply if the line crosses all three sides, but in this case, BM is from vertex B to M on the circumcircle, intersecting AC at N. So, perhaps Menelaus is not directly applicable.Alternatively, use coordinates. Let's set up coordinates. Let me set A at (0,0), C at (s,0), B at (s/2, (s√3)/2 ). Then, the circumcircle equation is (x - s/2)^2 + (y - (s√3)/6 )^2 = (s/√3 )^2.So, (x - s/2)^2 + (y - s√3 /6 )^2 = s^2 /3.Point M is on this circle, on arc AC. So, M has coordinates (x, y) satisfying the above equation, and MA = 4, MC = 5. So, distance from M to A is 4, distance from M to C is 5.So, in coordinates:Distance MA: sqrt( (x - 0)^2 + (y - 0)^2 ) = 4 => x^2 + y^2 = 16Distance MC: sqrt( (x - s)^2 + (y - 0)^2 ) = 5 => (x - s)^2 + y^2 = 25Subtract the first equation from the second:(x - s)^2 + y^2 - (x^2 + y^2 ) = 25 - 16Expand (x - s)^2: x^2 - 2s x + s^2So: x^2 - 2s x + s^2 + y^2 - x^2 - y^2 = 9Simplify: -2s x + s^2 = 9 => -2s x = 9 - s^2 => x = (s^2 - 9)/(2s )So, x-coordinate of M is (s² - 9)/(2s). Then, from the first equation, x² + y² = 16, so y² = 16 - x².Also, since M is on the circumcircle: (x - s/2)^2 + (y - (s√3)/6 )^2 = s² /3Let me substitute x and y into this equation.First, let's compute x = (s² - 9)/(2s )So, x - s/2 = (s² -9)/(2s ) - s/2 = (s² -9 - s² )/(2s ) = (-9)/(2s )Similarly, y - (s√3)/6 = sqrt(16 - x² ) - (s√3)/6. Hmm, this might be complicated. Let me write y² = 16 - x², so y = sqrt(16 - x² ), but since M is on the lower arc AC, maybe y is negative? Because the arc AC not containing B is the lower arc. So, if our coordinate system has B at the top, then M is below the triangle, so y-coordinate is negative. Therefore, y = - sqrt(16 - x² )So, let's proceed with y = - sqrt(16 - x² )Therefore, plug into the circumcircle equation:(-9/(2s ))² + ( - sqrt(16 - x² ) - (s√3)/6 )² = s² /3Let me compute term by term.First term: (x - s/2 )² = ( -9/(2s ) )² = 81/(4s² )Second term: (y - s√3 /6 )² = ( - sqrt(16 - x² ) - s√3 /6 )²Let me expand this:= [ - ( sqrt(16 - x² ) + s√3 /6 ) ]²= ( sqrt(16 - x² ) + s√3 /6 )²= ( sqrt(16 - x² ) )² + 2 * sqrt(16 - x² ) * (s√3 /6 ) + (s√3 /6 )²= (16 - x² ) + (2s√3 /6 ) sqrt(16 - x² ) + (3 s² ) / 36Simplify:= 16 - x² + (s√3 /3 ) sqrt(16 - x² ) + s² /12So, the entire equation:81/(4s² ) + 16 - x² + (s√3 /3 ) sqrt(16 - x² ) + s² /12 = s² /3Now, remember that x = (s² -9 )/(2s )So, x² = (s² -9 )² / (4s² ) = (s^4 - 18 s² +81 ) / (4 s² )So, let's substitute x² into the equation:81/(4s² ) + 16 - (s^4 -18 s² +81 )/(4s² ) + (s√3 /3 ) sqrt(16 - x² ) + s² /12 = s² /3Combine the terms:First, 81/(4s² ) - (s^4 -18 s² +81 )/(4s² ) = [81 - s^4 + 18 s² -81 ]/(4s² ) = ( -s^4 + 18 s² ) / (4s² ) = (-s² + 18 ) /4So, the equation becomes:(-s² + 18 ) /4 + 16 + (s√3 /3 ) sqrt(16 - x² ) + s² /12 = s² /3Multiply all terms by 12 to eliminate denominators:12*(-s² +18)/4 + 12*16 + 12*(s√3 /3 ) sqrt(16 -x² ) + 12*(s² /12 ) = 12*(s² /3 )Simplify each term:12*(-s² +18)/4 = 3*(-s² +18 ) = -3s² +5412*16 = 19212*(s√3 /3 ) sqrt(...) = 4 s√3 sqrt(16 -x² )12*(s² /12 ) = s²Right side: 12*(s² /3 ) = 4 s²So, entire equation:-3s² +54 +192 +4 s√3 sqrt(16 -x² ) +s² = 4s²Combine like terms:(-3s² +s² ) + (54 +192 ) +4 s√3 sqrt(16 -x² ) =4 s²-2s² +246 +4 s√3 sqrt(16 -x² ) =4 s²Bring terms to left side:-2s² +246 +4 s√3 sqrt(16 -x² ) -4s² =0-6s² +246 +4 s√3 sqrt(16 -x² )=0Rearranged:4 s√3 sqrt(16 -x² ) =6s² -246Divide both sides by 2:2 s√3 sqrt(16 -x² ) =3s² -123Now, let's square both sides to eliminate the square root. However, squaring can introduce extraneous solutions, so we'll need to check later.Left side squared: [2 s√3 sqrt(16 -x² )]^2 =4 s² *3*(16 -x² )=12 s² (16 -x² )Right side squared: (3s² -123 )² =9s^4 - 738 s² + 15129Thus, equation:12 s² (16 -x² ) =9s^4 -738 s² +15129Expand left side:12 s² *16 -12 s² x² =192 s² -12 s² x²So,192 s² -12 s² x² =9s^4 -738 s² +15129Bring all terms to left:192 s² -12 s² x² -9s^4 +738 s² -15129=0Combine like terms:(192 s² +738 s² ) + (-12 s² x² ) + (-9s^4 ) -15129=0930 s² -12 s² x² -9s^4 -15129=0Now, recall that x² = (s^4 -18 s² +81 )/(4s² )So, substitute x² into the equation:930 s² -12 s² * [ (s^4 -18 s² +81 )/(4s² ) ] -9s^4 -15129=0Simplify term by term:First term: 930 s²Second term: -12 s² * [ (s^4 -18 s² +81 )/(4s² ) ] = -12/4 * (s^4 -18 s² +81 ) = -3*(s^4 -18 s² +81 ) = -3s^4 +54 s² -243Third term: -9s^4Fourth term: -15129So, combine all:930 s² -3s^4 +54 s² -243 -9s^4 -15129=0Combine like terms:s^4 terms: -3s^4 -9s^4 = -12s^4s² terms:930 s² +54 s² =984 s²Constants: -243 -15129 = -15372Thus:-12s^4 +984 s² -15372=0Divide both sides by -12:s^4 -82 s² +1281=0Let me set u = s², so the equation becomes:u² -82u +1281=0Solve for u:Quadratic equation: u = [82 ± sqrt(82² -4*1*1281 )]/2Compute discriminant:82² =67244*1*1281=5124Discriminant=6724 -5124=1600sqrt(1600)=40Thus,u=(82 ±40)/2So,u=(82 +40)/2=122/2=61u=(82 -40)/2=42/2=21So, u=61 or u=21. Therefore, s²=61 or s²=21. Therefore, s=√61 or s=√21.But s is the side length, which we already found earlier as sqrt(61) from the chord length equations. So, s=√61 is valid, and s=√21 is extraneous. Therefore, s=√61.Thus, confirming our earlier result. So, the side length is √61. So, that's settled.Now, we need to find MN. So, point N is the intersection of BM and AC. Let's find coordinates of M and then equation of BM to find N.First, coordinates of M:We have x = (s² -9)/(2s ). Since s=√61, s²=61, so x=(61 -9)/(2*√61 )=52/(2√61 )=26/√61= (26√61)/61Similarly, y = -sqrt(16 -x² )Compute x²: (26/√61 )² =676/61Thus, y= -sqrt(16 -676/61 )= -sqrt( (16*61 -676)/61 )= -sqrt( (976 -676)/61 )= -sqrt(300/61 )= - (10√3)/√61 )= - (10√183)/61Wait, sqrt(300/61)=sqrt(300)/sqrt(61)=10√3/sqrt(61). So, y= -10√3 /sqrt(61 ). Rationalizing the denominator:y= -10√3 * sqrt(61 ) /61 = -10√183 /61So, coordinates of M:x=26√61 /61, y= -10√183 /61Coordinates of B: (s/2, (s√3)/2 )=(√61/2, (√61 *√3)/2 )=(√61/2, √183/2 )So, coordinates of B: (√61/2, √183/2 )Coordinates of M: (26√61 /61, -10√183 /61 )So, line BM goes from B to M. Let's find its equation.First, compute the slope:m=( y_M - y_B )/(x_M - x_B )Compute y_M - y_B:= (-10√183 /61 ) - (√183 /2 )= (-10√183 /61 ) - (√183 /2 )= √183 ( -10/61 -1/2 )= √183 ( -20/122 -61/122 )= √183 ( -81/122 )Similarly, x_M - x_B:= (26√61 /61 ) - (√61 /2 )= √61 (26/61 -1/2 )= √61 (52/122 -61/122 )= √61 ( -9/122 )So, slope m= [ -81√183 /122 ] / [ -9√61 /122 ]= (81√183 /122 ) / (9√61 /122 )= (81/9 )*(√183 /√61 )=9*√(183/61 )Simplify √(183/61 )=√(3*61/61 )=√3Thus, slope m=9√3So, the equation of line BM is y - y_B = m(x -x_B )So,y - (√183 /2 ) =9√3 (x - √61 /2 )We need to find where this line intersects AC. AC is from A(0,0) to C(√61, 0), so it's the x-axis. So, the intersection point N is on the x-axis, so y=0. So, set y=0 in the equation of BM:0 - √183 /2 =9√3 (x - √61 /2 )So,-√183 /2 =9√3 x -9√3 *√61 /2Bring all terms to left side:-√183 /2 +9√3 *√61 /2 -9√3 x=0Factor 1/2:[ -√183 +9√3 *√61 ] /2 -9√3 x=0But note that √183 =√(3*61)=√3*√61So,-√3*√61 +9√3*√61 =8√3*√61Thus,(8√3√61 )/2 -9√3 x=0 =>4√3√61 -9√3 x=0Divide both sides by √3:4√61 -9x=0 =>9x=4√61 =>x=4√61 /9So, point N is at (4√61 /9, 0 )Therefore, coordinates of N: (4√61 /9,0 )Coordinates of M: (26√61 /61, -10√183 /61 )So, the distance MN is the distance between these two points.Compute MN:√[ (x_M -x_N )² + (y_M - y_N )² ]Compute x_M -x_N=26√61 /61 -4√61 /9=√61 (26/61 -4/9 )Find common denominator for 26/61 and 4/9: 61*9=54926/61=234/549, 4/9=244/549Thus, 26/61 -4/9=234/549 -244/549= -10/549= -10/(9*61 )So, x_M -x_N=√61*(-10)/(9*61 )= -10√61/(9*61 )Similarly, y_M -y_N= (-10√183 /61 ) -0= -10√183 /61Therefore, MN²= [ (-10√61/(9*61 ))² + (-10√183 /61 )² ]Compute each term:First term: (100*61)/(81*61² )=100/(81*61 )Second term: (100*183)/61²=100*183/61²So,MN²=100/(81*61 ) +100*183/(61² )Factor out 100/(61² ):MN²=100/(61² ) [61/81 +183 ]Convert 183 to over 81:183=183*81/81=14823/81So,MN²=100/(61² ) [ (61 +14823 )/81 ]=100/(61² ) [14884/81 ]Compute 14884/81= 14884 ÷81. Let's compute 81*183=14823, 14884 -14823=61, so 183 +61/81=183 + 61/81=183.753...But perhaps better to keep as fractions:14884=61*244. So 61*244=61*(240 +4)=61*240 +61*4=14640 +244=14884Therefore, MN²=100/(61² )*(61*244)/81=100*244/(61*81 )Simplify 244 and 61: 61*4=244Therefore, MN²=100*4/81=400/81Thus, MN=20/9So, MN=20/9.Therefore, the length of segment MN is 20/9, and the side length of triangle ABC is sqrt(61 ).Final AnswerThe length of segment ( MN ) is (boxed{dfrac{20}{9}}) and the side length of triangle ( ABC ) is (boxed{sqrt{61}}).</think>Given an equilateral triangle (ABC) with a point (M) on the arc (AC) of its circumcircle such that (MA = 4) and (MC = 5). The line (BM) intersects the side (AC) at point (N). We need to find the length of segment (MN) and the side length of the triangle (ABC).1. Side Length Calculation: - The circumradius (R) of an equilateral triangle with side length (s) is (R = frac{s}{sqrt{3}}). - Using the chord length formula for (MA) and (MC), we set up the equations: [ 2R sinleft(frac{theta}{2}right) = 4 quad text{and} quad 2R sinleft(frac{120^circ - theta}{2}right) = 5 ] - Solving these equations, we find (R = sqrt{frac{61}{3}}) and the side length (s = sqrt{61}).2. Length of Segment (MN): - Using coordinate geometry, we place the triangle in a coordinate system with (A(0, 0)), (C(sqrt{61}, 0)), and (Bleft(frac{sqrt{61}}{2}, frac{sqrt{183}}{2}right)). - The coordinates of (M) are found to be (left(frac{26sqrt{61}}{61}, -frac{10sqrt{183}}{61}right)). - The equation of line (BM) is derived, and its intersection with (AC) (the x-axis) gives point (Nleft(frac{4sqrt{61}}{9}, 0right)). - Calculating the distance (MN) using the coordinates of (M) and (N), we find (MN = frac{20}{9}). Final AnswerThe length of segment (MN) is (boxed{dfrac{20}{9}}) and the side length of triangle (ABC) is (boxed{sqrt{61}}).,<think>Okay, so I need to find the length of segment MN and the side of the equilateral triangle ABC. Let me start by visualizing the problem. We have an equilateral triangle ABC, so all sides are equal, all angles are 60 degrees. The circumcircle of the triangle is the circle that passes through all three vertices A, B, and C. Point M is on the arc AC of this circumcircle. The segments MA and MC are given as 4 and 5 units respectively. The line BM intersects the side AC at point N, and we need to find MN and the side length of the triangle.First, since ABC is an equilateral triangle, all sides are equal. Let's denote the side length as 's'. The circumradius (radius of the circumcircle) of an equilateral triangle is given by R = s / √3. That might come in handy later.Now, point M is on the arc AC. Since ABC is equilateral, the arc AC is 120 degrees because each angle in an equilateral triangle is 60 degrees, so the arcs between the vertices are twice the angle, which is 120 degrees. Wait, actually, in a circle, the central angle is equal to the measure of the arc. In an equilateral triangle, each central angle corresponding to a side would be 120 degrees because the triangle's internal angle is 60 degrees, and the central angle is 360/3 = 120 degrees. So the arc AC is 120 degrees. So point M is somewhere on that 120-degree arc AC, not containing point B.Given MA = 4 and MC = 5. So from point M, the distance to A is 4, and to C is 5. But since M is on the circumcircle, points A, B, C are all on the circumcircle. Wait, hold on, if ABC is equilateral, then all three vertices lie on the circumcircle. But point M is also on the circumcircle, so MA, MB, MC are chords of the circumcircle. Wait, but the problem says MA = 4 and MC = 5. But if M is on the circumcircle, then MA and MC are chord lengths. In an equilateral triangle, all chord lengths between two vertices are equal to the side length 's'. But here, MA and MC are different, 4 and 5. That suggests that M is not one of the vertices, but another point on arc AC.So, since M is on the circumcircle, triangle AMC is inscribed in the circumcircle. But since ABC is equilateral, the side AC is of length 's', and the arc AC is 120 degrees. So point M is on the circumcircle, arc AC (which is 120 degrees), such that MA = 4 and MC = 5. Then we need to find MN, where N is the intersection of BM and AC, and also find the side length 's'.Hmm, okay. Let me think about possible approaches. Since we have an equilateral triangle, maybe using properties of equilateral triangles and their circumcircles would help. Alternatively, coordinate geometry? Maybe placing the triangle in a coordinate system and working out coordinates. Alternatively, using trigonometry with the given lengths.Alternatively, maybe using the Law of Sines or Cosines in some triangles. Since M is on the circumcircle, angles subtended by the same chord are equal. Let's consider triangle AMC. Since ABC is equilateral, angle at A is 60 degrees, but point M is on the circumcircle, so angle AMC would be equal to angle ABC, which is 60 degrees? Wait, no. If M is on the circumcircle, then angle AMC should be equal to angle ABC because they subtend the same arc. Wait, angle ABC is 60 degrees, so angle AMC is also 60 degrees. Wait, is that correct?Wait, in a circle, the angle subtended by an arc at the center is twice the angle subtended at any point on the remaining part of the circumference. So if we have arc AC, the central angle is 120 degrees, then angle ABC is 60 degrees. Similarly, angle AMC, which is on the circumference subtended by arc AC, should also be equal to 60 degrees? Wait, but angle ABC is 60 degrees because it's an equilateral triangle, but angle AMC... Hmm, maybe. Wait, if M is on the circumcircle, then angle AMC is equal to angle ABC because they both subtend arc AC. Since ABC is 60 degrees, angle AMC is also 60 degrees. Therefore, triangle AMC has sides MA = 4, MC = 5, and angle AMC = 60 degrees. So maybe we can use the Law of Cosines here to find the length of AC.Wait, AC is a side of the equilateral triangle, so AC = s. So in triangle AMC, sides MA = 4, MC = 5, angle AMC = 60 degrees, so applying the Law of Cosines:AC² = MA² + MC² - 2*MA*MC*cos(angle AMC)So:s² = 4² + 5² - 2*4*5*cos(60°)cos(60°) is 0.5, so:s² = 16 + 25 - 2*4*5*0.5Calculates:16 +25 = 412*4*5 = 4040*0.5 = 20So s² = 41 - 20 = 21Therefore, s = √21Wait, so the side length is √21? Hmm, that seems possible. Let me verify.If triangle AMC has sides 4, 5, and angle 60 degrees between them, then by Law of Cosines, AC² = 16 +25 - 40*0.5 = 41 -20 =21, so AC = √21. Since ABC is equilateral, AC is equal to the side length, so s = √21. So that's the side length? That seems straightforward.But then the problem also asks for the length of segment MN. So once we have the side length, maybe we can find coordinates of points and then find the intersection point N, then compute MN.Alternatively, use Menelaus' theorem or Ceva's theorem since we have a transversal intersecting a triangle.Wait, Menelaus' theorem applies to a transversal cutting through the sides of a triangle. Let me recall Menelaus: for triangle ABC and a transversal line that intersects AB at L, BC at M, and AC at N, then (AL/LB)*(BM/MC)*(CN/NA) = 1.But in our case, line BM intersects AC at N. So maybe Menelaus' theorem can be applied here.Wait, Menelaus' theorem for triangle AMC with the transversal line BMN. Wait, maybe not. Alternatively, use Menelaus on triangle ABC with the transversal BMN? Hmm.Wait, the line BM intersects AC at N and perhaps exits the triangle at some other point? Wait, ABC is a triangle, B is a vertex, M is a point on the circumcircle on arc AC, so BM is going from B through M and intersects AC at N. Since M is on the circumcircle, maybe BM extends beyond M, but in this case, N is the intersection with AC, so N is between A and C.Alternatively, use Ceva's theorem. Ceva's theorem states that for concurrent lines from the vertices, but in this case, we have only one line BM intersecting AC. Maybe coordinate geometry is better.Let me try coordinate geometry. Let's place the equilateral triangle in the coordinate plane. Let me set point A at (0, 0), point C at (s, 0), and since it's equilateral, point B will be at (s/2, (s√3)/2). The circumcircle of the triangle can be found, and then point M can be parametrized on arc AC.But since we know MA = 4, MC = 5, and angle AMC = 60°, we might have already found s = √21. Let me check if that's correct.Wait, if ABC is equilateral with side length √21, then AC = √21. But in triangle AMC, we have sides 4, 5, and √21. Let me check if the Law of Cosines holds:4² + 5² - 2*4*5*cos(60°) = 16 +25 - 20 =21, which is equal to (√21)². So yes, that works. Therefore, s = √21. So that's the side length. So that part is done.Now, to find MN. So MN is the segment from M to N, where N is the intersection of BM and AC.We need to find coordinates of points B, M, and then find the equation of line BM, find its intersection with AC (which is the x-axis from (0,0) to (√21, 0)), find point N, then compute the distance between M and N.Alternatively, use mass point geometry or similar triangles.Alternatively, use power of a point. Since M is on the circumcircle, power of point M with respect to the circle is zero. Wait, but M is on the circle, so power is zero. Hmm, not sure.Alternatively, use coordinates. Let's proceed with coordinates.Let me set up the coordinate system:Let’s place point A at (0, 0), point C at (s, 0). Since ABC is equilateral, point B is at (s/2, (s√3)/2). The circumradius R of an equilateral triangle is s / √3. The circumcircle is centered at the centroid of the triangle, which for an equilateral triangle is also the circumcenter. The centroid is at (s/2, (s√3)/6). Wait, no. Wait, for an equilateral triangle, the centroid, circumcenter, orthocenter, and incenter all coincide. The centroid is at (s/2, (s√3)/6). Wait, let me confirm:Coordinates of centroid (intersection of medians) are average of the vertices:A: (0,0), B: (s/2, (s√3)/2), C: (s, 0). So centroid is ((0 + s/2 + s)/3, (0 + (s√3)/2 +0)/3) = ((3s/2)/3, (s√3/2)/3) = (s/2, (s√3)/6). So yes, centroid is at (s/2, (s√3)/6). But the circumradius of an equilateral triangle is the distance from the centroid to any vertex. Let's compute that.Distance from centroid to vertex A (0,0):√[(s/2 - 0)^2 + ((s√3)/6 - 0)^2] = √[(s²/4) + (s²*3)/36] = √[(s²/4) + (s²/12)] = √[(3s² + s²)/12] = √[4s²/12] = √[s²/3] = s/√3. So yes, circumradius R = s/√3.So the circumcircle has center at (s/2, (s√3)/6) and radius s/√3.Point M is on the arc AC, which is the arc from A to C not containing B. Since ABC is equilateral, arc AC is 120 degrees.We need to find coordinates of point M such that MA = 4 and MC = 5, and M is on the circumcircle.Alternatively, since we already found s = √21 using triangle AMC, let's use that. So s = √21, so the coordinates of the points are:A: (0,0)C: (√21, 0)B: (√21/2, (√21 * √3)/2) = (√21/2, (√63)/2) = (√21/2, (3√7)/2)Circumradius R = s / √3 = √21 / √3 = √7 ≈ 2.6458Circumcircle center: (s/2, (s√3)/6) = (√21/2, (√21 * √3)/6) = (√21/2, (√63)/6) = (√21/2, (3√7)/6) = (√21/2, √7/2)So center at (√21/2, √7/2)Now, point M is on the circumcircle, so it satisfies the equation:(x - √21/2)^2 + (y - √7/2)^2 = (√7)^2 = 7Also, point M is on arc AC, so between A(0,0) and C(√21, 0), not containing B.Given that MA = 4 and MC = 5, so distance from M to A is 4, and distance from M to C is 5.So coordinates of M must satisfy:√[(x - 0)^2 + (y - 0)^2] = 4 => x² + y² = 16√[(x - √21)^2 + (y - 0)^2] = 5 => (x - √21)^2 + y² = 25Also, since M is on the circumcircle:(x - √21/2)^2 + (y - √7/2)^2 = 7So now we have three equations:1. x² + y² = 16 (from MA=4)2. (x - √21)^2 + y² = 25 (from MC=5)3. (x - √21/2)^2 + (y - √7/2)^2 = 7 (from circumcircle)We can use equations 1 and 2 to solve for x and y.Subtract equation 1 from equation 2:(x - √21)^2 + y² - (x² + y²) = 25 -16 => x² - 2√21 x + 21 + y² - x² - y² = 9 => -2√21 x +21 =9 => -2√21 x = -12 => x = (-12)/(-2√21) = 6/√21 = (6√21)/21 = (2√21)/7So x-coordinate of M is 2√21/7.Now plug x into equation 1 to find y:x² + y² =16 => ( (2√21/7)^2 ) + y² =16 => (4*21)/49 + y² =16 => 84/49 + y² =16 => 12/7 + y² =16 => y² =16 -12/7 = (112 -12)/7 =100/7 => y = ±10/√7 = ±(10√7)/7But since point M is on arc AC not containing B, which is the lower arc (since B is above AC), so y-coordinate should be negative. Therefore, y = -10√7/7So coordinates of M are (2√21/7, -10√7/7)Now, we need to find the equation of line BM. Points B and M:Point B: (√21/2, 3√7/2)Point M: (2√21/7, -10√7/7)Find the equation of line BM.First, compute the slope:m = (y_M - y_B)/(x_M - x_B) = [ (-10√7/7 - 3√7/2 ) / (2√21/7 - √21/2) ]Let me compute numerator and denominator separately.Numerator:-10√7/7 - 3√7/2 = (-10/7 -3/2)√7 = (-20/14 -21/14)√7 = (-41/14)√7Denominator:2√21/7 - √21/2 = (2/7 -1/2)√21 = (4/14 -7/14)√21 = (-3/14)√21Thus, slope m = (-41/14 √7) / (-3/14 √21) = (41/14 √7) / (3/14 √21) = (41 √7) / (3 √21) = 41/(3√3) = (41√3)/9Wait, let me check:√7 / √21 = √(7/21) = √(1/3) = 1/√3. So yes, √7/√21 =1/√3. Therefore:m = (41/14 √7) / (3/14 √21) = (41/14)/(3/14) * (√7/√21) = (41/3) * (1/√3) = 41/(3√3) = (41√3)/9 after rationalizing.So the slope is 41√3/9.Now, equation of line BM: using point B (√21/2, 3√7/2):y - 3√7/2 = (41√3/9)(x - √21/2)We need to find where this line intersects AC. AC is the side from A(0,0) to C(√21, 0), so it's along the x-axis (y=0). So to find point N, set y=0 in the equation of BM and solve for x.So:0 - 3√7/2 = (41√3/9)(x - √21/2)Multiply both sides by 9:- (3√7/2)*9 = 41√3 (x - √21/2)Simplify left side:-27√7/2 = 41√3 x - 41√3*(√21)/2Compute 41√3*(√21)/2:√3*√21 = √63 = 3√7, so this term is 41*3√7 /2 = 123√7/2Thus:-27√7/2 = 41√3 x - 123√7/2Bring the term with √7 to the left:-27√7/2 + 123√7/2 = 41√3 xCompute left side:(123 -27)√7 /2 = 96√7/2 = 48√7Therefore:48√7 = 41√3 xSolve for x:x = (48√7) / (41√3) = (48/41)(√7/√3) = (48/41)(√(7/3)) = (48/41)(√21 / 3) = (16/41)√21So x-coordinate of N is (16√21)/41Since AC is from (0,0) to (√21,0), N is at ((16√21)/41, 0)Now, we need to find the distance between M and N.Coordinates of M: (2√21/7, -10√7/7)Coordinates of N: (16√21/41, 0)Compute MN:Δx = 16√21/41 - 2√21/7 = √21*(16/41 - 2/7) = √21*( (16*7 - 2*41)/287 ) = √21*(112 -82)/287 = √21*(30)/287 = 30√21 /287Δy = 0 - (-10√7/7) =10√7/7Therefore, MN = √[(30√21 /287)^2 + (10√7/7)^2 ]First compute (30√21 /287)^2:(900 *21)/(287^2) = 18900 / 82369Then (10√7/7)^2:(100*7)/49 = 700/49 = 100/7 ≈14.2857Wait, but let's compute exact value:MN² = (30√21 /287)^2 + (10√7 /7)^2= (900*21)/(287²) + (100*7)/(49)= 18900/82369 + 700/49Simplify 700/49 = 100/7 ≈14.2857Now, let's convert 100/7 to denominator 82369. 82369 = 287². Let me compute 287*7=2009, so 82369 =287²= (2009/7)². Hmm, not sure if useful.But instead, compute MN²:= 18900/82369 + 100/7Convert 100/7 to denominator 82369:100/7 = (100 * 82369)/(7 *82369) = (8236900)/576583But this might not be helpful. Alternatively, compute decimal values.But maybe there's a simplification. Let me check:MN² = (30^2 *21)/(287^2) + (10^2 *7)/7^2= (900*21)/(287^2) + (100*7)/(49)= 18900/82369 + 700/49= 18900/82369 + (700*1681)/49*1681) [Wait, 82369 = 287^2 = (7*41)^2 = 49*1681. So 49*1681=82369. Therefore, 700/49 = 700/49 = 100/7, so as above.Therefore, MN² = 18900/82369 + (100/7)*(1681/1681) = [18900 + 100*1681/7]/82369Wait, this seems complicated. Let me compute 100/7 = approximately 14.2857, so 18900/82369 ≈0.2295 and 14.2857 ≈14.2857, sum≈14.5152. Then MN≈√14.5152≈3.81. But this is approximate, but we need exact value.Alternatively, factor numerator:Let me write MN² as:= (30² *21 + (10² *7 *287²)/49 ) /287²Wait, 287²= (7*41)²=49*1681Therefore:MN² = [30²*21 + 10²*7*(49*1681)/49 ] / (49*1681)Simplify:= [900*21 + 100*7*1681 ] / (49*1681)= [18900 + 700*1681 ] / (49*1681)Compute 700*1681: 700*1600=1,120,000 and 700*81=56,700 → total=1,176,700Thus numerator=18900 +1,176,700=1,195,600Therefore MN²=1,195,600/(49*1681)Calculate denominator: 49*1681=82,369So MN²=1,195,600/82,369Divide numerator and denominator by 7:1,195,600 ÷7=170,80082,369 ÷7= 11,767 (since 7*11,767=82,369)So MN²=170,800/11,767 ≈14.515 approximately, which matches earlier.But 170,800 ÷11,767: Let's divide 170800 by 11767.11767*14= 164,738170,800 -164,738=6,06211767*0.515≈6,062. So total≈14.515.But we need an exact value. Maybe the fraction reduces. Let's check if 170,800 and 11,767 have a common factor.11,767: Let's see. 11,767 ÷7=1,681, which is 41². So 11,767=7*41².170,800 ÷7=24,400. So 170,800=7*24,400.So MN²=7*24,400/(7*41²)=24,400/41²=24,400/1,681=24,400 ÷1,681.Calculate 1,681*14=23,53424,400 -23,534=8661,681*0.515≈866. So MN²≈14.515, same as before. So MN=√(24,400/1,681)= √(24,400)/√(1,681)= 156.2/41≈3.81. Wait, 24,400 is 156.2²? Wait, 156²=24,336. 156.2²≈24,336 +2*156*0.2 +0.2²=24,336 +62.4 +0.04=24,398.44, which is close to 24,400. But exact value is 24,400/1,681= (100*244)/1,681= (100*244)/(41²). Hmm, 244=4*61. So 244=4*61, so 24,400=4*61*100. 1,681=41². So MN=√(4*61*100)/41= (2*10√61)/41=20√61/41.Wait, wait, let's check:24,400/1,681= (100*244)/41²= (100*4*61)/41²= (400*61)/41². Therefore, √(400*61/41²)= (20√61)/41.Yes! So MN=20√61/41.So that's the exact value.Therefore, MN= (20√61)/41 and the side length is √21.But let me check again where MN²=24,400/1,681= (20²*61)/41² →√(20²*61/41²)=20√61/41. Yes, that's correct. So MN=20√61/41.Therefore, the answers are MN=20√61⁄41 and the side length is √21.Wait, but let me check once more. The side length was found by using Law of Cosines in triangle AMC, with sides 4,5, angle 60°, resulting in AC=√21. Then coordinates were computed accordingly, leading to MN=20√61/41. Let's verify with another approach to be safe.Alternative approach using Menelaus’ theorem.In triangle ABC, with the transversal line BMN, where N is on AC, M is on the circumcircle.Wait, Menelaus’ theorem states that (AN/NC)*(CB/BB)*(BM/MA)=1, but I need to recall the exact formula.Wait, Menelaus’ theorem for triangle ABC and a transversal line that intersects AB at F, BC at D, and AC at E, then (AF/FB)*(BD/DC)*(CE/EA)=1.But in our case, the transversal is BM intersecting AC at N, but M is outside the triangle? Wait, point M is on the circumcircle, arc AC, which is opposite to B, so M is outside the triangle? Wait, in an equilateral triangle, the circumcircle passes through all three vertices, so arc AC not containing B would still be between A and C, but since ABC is equilateral, the arc AC is 120°, and M is a point on that arc. So BM starts at B, goes through M (on the circumcircle), and intersects AC at N. So N is between A and C.Therefore, applying Menelaus’ theorem to triangle ABC with the transversal line BMN, but Menelaus requires three intersection points. Wait, the line BM intersects AC at N, but where else does it intersect? It passes through B, so the other intersections are at M and N. Wait, Menelaus might not directly apply here. Alternatively, use Ceva's theorem, but that requires concurrency.Alternatively, use power of a point. The Power of a Point theorem states that for a point N on AC, the power with respect to the circumcircle is NA*NC = NM*NB.But N lies on AC and BM. Therefore, power of N with respect to the circumcircle is NA*NC = NM*NB.Since ABC is equilateral with side s=√21, then AC=√21, so NA + NC = √21. Let’s denote NA = x, NC = √21 -x.Then power of point N: NA*NC = x(√21 -x) = NM*NBBut we need to express NB and NM in terms of x. However, since N is on BM, we can write NB = NM + MB? Wait, no. If N is between B and M, then NB + NM = BM. But if N is between B and M, then BM = BN + NM. Wait, but in our case, since M is on the circumcircle, and B is a vertex, the line BM goes from B through N (on AC) to M on the circumcircle. So the order is B----N----M on line BM.Therefore, BM = BN + NM. So BN = BM - NM.But we need to relate NA*NC = NM*NBFrom power of point N: NA*NC = NM*NBSo x(√21 -x) = NM*(BN)But BN = BM - NM, so:x(√21 -x) = NM*(BM - NM)But BM is the length from B to M. Earlier, we found coordinates of B and M. Let’s compute BM using coordinates.Coordinates of B: (√21/2, 3√7/2)Coordinates of M: (2√21/7, -10√7/7)Compute BM:Δx = 2√21/7 - √21/2 = (4√21 -7√21)/14 = (-3√21)/14Δy = -10√7/7 -3√7/2 = (-20√7 -21√7)/14 = (-41√7)/14Therefore, BM length:√[ (-3√21/14)^2 + (-41√7/14)^2 ] = √[ (9*21 + 1681*7)/196 ]= √[ (189 + 11,767)/196 ]= √[11,956/196] = √61. So BM=√61.Wait, compute:(-3√21/14)^2 =9*21/196=189/196(-41√7/14)^2=1681*7/196=11,767/196Sum:189 +11,767=11,95611,956/196=61. So √61. Therefore BM=√61Therefore, BM=√61. So from power of point N: x(√21 -x)=NM*(√61 -NM)But we need to find x and NM. However, this seems like a quadratic equation. Let me denote NM = y. Then BN = √61 - y. So equation:x(√21 -x) = y(√61 - y)But we also need another relation between x and y. Since N lies on AC, which is length √21, x = AN, NC=√21 -x. Also, N lies on BM. Since BM is a line from B to M, the ratio of BN to NM can be related to the coordinates.Alternatively, from coordinate calculations earlier, we found coordinates of N as (16√21/41, 0). Therefore, AN=16√21/41, so x=16√21/41, NC=√21 -16√21/41= (41√21 -16√21)/41=25√21/41.From power of point: x(√21 -x)= (16√21/41)(25√21/41)= (16*25*21)/41²= (400*21)/1681=8400/1681≈5.0.Meanwhile, y=MN=20√61/41≈20*7.81/41≈156.2/41≈3.81, and √61 -y≈7.81 -3.81≈4.0. Then y(√61 -y)=3.81*4.0≈15.24. Wait, but 8400/1681≈5.0. Wait, this does not match. Wait, but there must be a miscalculation here.Wait, power of point N: NA*NC = NM*NB.From coordinates:NA=16√21/41, NC=25√21/41. So NA*NC=(16√21/41)*(25√21/41)= (16*25)*(21)/41²=400*21/1681=8400/1681≈5.0.From coordinates of M and N, distance MN=20√61/41≈3.81, and distance BN=BM - MN=√61 -20√61/41= (41√61 -20√61)/41=21√61/41≈0.51*7.81≈4.0.Therefore, NM*NB=(20√61/41)*(21√61/41)= (20*21*61)/41²= (420*61)/1681=25,620/1681≈15.24, but NA*NC≈5.0. This discrepancy suggests an error in reasoning.Wait, that's a problem. According to power of a point, NA*NC should equal NM*NB, but according to coordinates, NA*NC≈5 and NM*NB≈15.24, which are not equal. Therefore, there must be a mistake in my calculations.Wait, perhaps my assumption of the order of points on BM is wrong. If the power of point N is NA*NC = NM*NB, but in our case, point N is inside the circle? Wait, N is on AC, which is a side of the triangle, and the circumcircle passes through A, B, C. If N is inside the triangle, then N is inside the circle, so power of point N should be NA*NC = NT*NS, where NT and NS are the lengths from N to the points of intersection of line BM with the circle. Since line BM intersects the circle at B and M, so power of N is NB*NM = NA*NC. Therefore, NB*NM=NA*NC.But from coordinates, NA*NC=16√21/41 *25√21/41= (16*25*21)/(41²)=8400/1681≈5.0NB*NM= (distance from N to B)*(distance from N to M). Let me compute NB and NM.Coordinates of N: (16√21/41,0)Coordinates of B: (√21/2, 3√7/2)Distance NB:√[(√21/2 -16√21/41)^2 + (3√7/2 -0)^2]First compute x-coordinate difference:√21/2 -16√21/41= √21*(1/2 -16/41)=√21*(41 -32)/82=√21*(9)/82=9√21/82Y-coordinate difference:3√7/2Therefore, NB=√[(9√21/82)^2 + (3√7/2)^2 ]= √[(81*21)/(82²) + (9*7)/4]= √[(1701/6724) + (63/4)]Convert to common denominator:1701/6724 + (63/4)*(1681/1681)=1701/6724 + (63*1681)/6724Calculate 63*1681=105, 903So total= (1701 +105,903)/6724=107,604/6724=107,604 ÷ 6724≈16.0Wait, 6724*16=107,584, so 107,604-107,584=20, so 107,604/6724=16 +20/6724≈16.003. So NB≈√16.003≈4.000.Similarly, distance NM=20√61/41≈3.81Thus, NB*NM≈4.0*3.81≈15.24, but NA*NC≈5.0. This inconsistency suggests an error.But according to the power of a point theorem, these should be equal. Therefore, my coordinates must be wrong. Which means my previous calculations must have an error.Wait, let me check the coordinates of point M again.Earlier, using the two equations from MA=4 and MC=5, we found x=2√21/7 and y=±10√7/7. Then, since M is on arc AC not containing B, we took y negative. Then checked that this point is on the circumcircle. Let me confirm.Coordinates of M: (2√21/7, -10√7/7)Check if this is on the circumcircle:Center at (√21/2, √7/2). So distance from center to M:√[(2√21/7 - √21/2)^2 + (-10√7/7 - √7/2)^2]Compute x difference:2√21/7 -√21/2= (4√21 -7√21)/14= (-3√21)/14Y difference:-10√7/7 -√7/2= (-20√7 -7√7)/14= (-27√7)/14Thus distance squared:[(-3√21/14)^2 + (-27√7/14)^2] = (9*21 + 729*7)/196 = (189 + 5103)/196 =5292/196=27.But the radius squared is 7, as s=√21, R=√7. Wait, 5292/196=5292 ÷196=27. But center to M distance squared is 27, which is not equal to 7. Therefore, this point is not on the circumcircle. Which contradicts our initial assumption. Therefore, there's a mistake in my previous steps.Wait, so this means that when I solved the equations for MA=4 and MC=5, I found a point M that is not on the circumcircle. Therefore, the mistake is in assuming that such a point M exists on the circumcircle with MA=4, MC=5, and angle AMC=60°, leading to s=√21. However, if solving for coordinates gives a point not on the circumcircle, then the side length cannot be √21. Therefore, my previous conclusion that s=√21 is wrong.Wait, but where is the mistake?Wait, originally, I used triangle AMC with sides MA=4, MC=5, angle AMC=60°, leading to AC=√21 via Law of Cosines. However, if point M is on the circumcircle of ABC, which is an equilateral triangle, then the Law of Cosines in triangle AMC might not hold because angle AMC is not necessarily 60°. Wait, this is the error!Earlier, I assumed that angle AMC is 60°, because in the original triangle ABC, angle ABC is 60°, and since M is on the circumcircle, angle AMC should equal angle ABC. But this is only true if M is on the arc AC that contains B. However, in the problem statement, M is on arc AC (which is the arc not containing B), so angle AMC is actually 180° - angle ABC=120°. Because when the point is on the opposite arc, the angle is supplementary.Ah! Here's the mistake. In the original reasoning, I thought angle AMC was 60°, but actually, since M is on arc AC not containing B, the angle at M should be 180° - 60°=120°, not 60°. So that's where the error was.So triangle AMC has sides MA=4, MC=5, and angle AMC=120°, not 60°. Therefore, we need to recalculate AC (which is the side of the equilateral triangle) using the Law of Cosines with angle 120°.So, correcting that:In triangle AMC, MA=4, MC=5, angle AMC=120°, so:AC²=MA² + MC² - 2*MA*MC*cos(angle AMC)cos(120°)= -1/2So:AC²=4² +5² -2*4*5*(-1/2)=16 +25 +20=61Therefore, AC=√61. Thus, the side length of the equilateral triangle is s=√61.This changes everything. Now, the side length is √61, not √21. That explains why earlier the coordinates didn't satisfy the circumcircle equation. So the correct side length is √61.Now, redoing the coordinates with s=√61.Coordinates:A: (0,0)C: (√61,0)B: (√61/2, (√61*√3)/2)= (√61/2, (√183)/2)Circumradius R= s/√3=√61/√3=√(61/3)Circumcircle center: centroid at (√61/2, (√183)/6)So center coordinates: (√61/2, (√183)/6)Now, point M is on the circumcircle, on arc AC (not containing B), with MA=4 and MC=5.Again, we need to find coordinates of M such that:1. Distance from M to A is 4: √(x² + y²)=4 ⇒x² + y²=162. Distance from M to C is 5: √((x -√61)^2 + y²)=5 ⇒(x -√61)^2 + y²=253. M is on the circumcircle: (x -√61/2)^2 + (y -√183/6)^2= (√61/√3)^2=61/3Let me first solve equations 1 and 2.Subtract equation 1 from equation 2:(x -√61)^2 + y² - (x² + y²)=25 -16 ⇒x² -2√61 x +61 + y² -x² - y²=9 ⇒-2√61 x +61=9 ⇒-2√61 x= -52 ⇒x=52/(2√61)=26/√61=26√61/61So x=26√61/61Plug into equation 1 to find y:x² + y²=16 ⇒(26√61/61)^2 + y²=16 ⇒(676*61)/3721 + y²=16 ⇒(676/61) + y²=16 ⇒676/61=11.081, so y²=16 -11.081=4.919⇒y≈±2.218But exact value:y²=16 - (26²*61)/61²=16 - (676*61)/3721=16 -676/61= (976 -676)/61=300/61 ⇒y=±√(300/61)=±(10√3)/√61=±(10√183)/61Since M is on arc AC not containing B, which is the lower arc, so y is negative. Therefore, y= -10√183/61Therefore, coordinates of M are (26√61/61, -10√183/61)Now, check if this point is on the circumcircle:Center at (√61/2, √183/6). Compute distance squared from M to center:Δx=26√61/61 -√61/2= (52√61 -61√61)/122= (-9√61)/122Δy= -10√183/61 -√183/6= (-60√183 -61√183)/366= (-121√183)/366= (-121√183)/(366)Distance squared:[(-9√61)/122]^2 + [(-121√183)/366]^2Compute first term:(81*61)/(122²)= (81*61)/(14884)=4941/14884≈0.332Second term:(14641*183)/(366²)= (14641*183)/(133956)Simplify 366²=13395614641*183=14641*(180+3)=14641*180 +14641*3=2,635,380 +43,923=2,679,303So second term=2,679,303/133,956≈20.0Sum≈0.332 +20≈20.332But the radius squared is 61/3≈20.333. So distance squared≈20.332≈61/3≈20.333. So it checks out. Therefore, M is correctly on the circumcircle.Now, proceed to find line BM and its intersection with AC.Coordinates:B: (√61/2, √183/2)M: (26√61/61, -10√183/61)Find equation of line BM.Slope m=(yM - yB)/(xM -xB)= [(-10√183/61 -√183/2)]/[26√61/61 -√61/2]Compute numerator:-10√183/61 -√183/2= (-20√183 -61√183)/122= (-81√183)/122Denominator:26√61/61 -√61/2= (52√61 -61√61)/122= (-9√61)/122Thus slope m= (-81√183/122)/(-9√61/122)= (81√183)/(9√61)=9√(183/61)=9√3.Because 183=3*61, so √(183/61)=√3.Thus, slope m=9√3.Equation of line BM: using point B (√61/2, √183/2):y - √183/2 =9√3(x -√61/2)To find intersection with AC (y=0):0 - √183/2=9√3(x -√61/2)=> -√183/2=9√3 x -9√3*√61/2Solve for x:9√3 x= -√183/2 +9√3*√61/2Factor out 1/2:9√3 x= ( -√183 +9√3*√61 ) /2But √183=√(3*61)=√3*√61. Therefore:-√183= -√3*√61Thus:9√3 x= ( -√3*√61 +9√3*√61 ) /2= (8√3*√61)/2=4√3*√61Therefore:9√3 x=4√3√61 ⇒x=4√61/9Thus, coordinates of N are (4√61/9,0)Now, compute distance MN:Coordinates of M: (26√61/61, -10√183/61)Coordinates of N: (4√61/9,0)Δx=4√61/9 -26√61/61= (4√61*61 -26√61*9)/549= (244√61 -234√61)/549=10√61/549=10√61/(9*61)=10√61/(549)Δy=0 - (-10√183/61)=10√183/61Thus, MN=√[(10√61/549)^2 + (10√183/61)^2 ]First term:(100*61)/(549²)=6100/(549²)Second term:(100*183)/(61²)=18,300/3,721Convert to common denominator:First term: 6100/(549²)=6100/(549*549). Note that 549=9*61. So 549=9*61, so 549²=81*61².Thus, first term=6100/(81*61²)= (6100/81)/61²≈75.308/3721≈0.0202Second term=18,300/3,721≈4.919But converting second term to denominator 81*61²:18,300/3,721=18,300/(61*61)= (18,300/61)/61=300/61≈4.919Wait, perhaps we can compute MN²:MN²=(10√61/549)^2 + (10√183/61)^2= (100*61)/(549²) + (100*183)/(61²)= 6100/(549²) +18300/(61²)Factor out 100:=100[61/(549²) +183/(61²)]Note that 549=9*61, so 549²=81*61².Thus:=100[61/(81*61²) +183/(61²)]=100[1/(81*61) +183/(61²)]=100[ (61 +183*81)/ (81*61²) ]Compute numerator:61 +183*81=61 +14,823=14,884Thus:MN²=100*14,884/(81*61²)=1,488,400/(81*3,721)Calculate denominator:81*3,721=301,401Thus, MN²=1,488,400/301,401≈4.939Then MN=√4.939≈2.222But let's rationalize:MN²=1,488,400/301,401. Let's see if this reduces. Notice that:14,884=4*3,721=4*61². Wait, 61²=3,721, so 14,884=4*3,721. Therefore:MN²=100*4*3,721/(81*3,721)= (400/81)= (20/9)^2Wait, 100*14,884=100*4*3,721=400*3,721Denominator:81*3,721So MN²=400*3,721/(81*3,721)=400/81⇒MN=20/9≈2.222Wait, that's a nice result. Therefore, MN=20/9.But wait, how?Wait, let's re-express:MN²=100*(14,884)/(301,401)But 14,884=4*3,721 and 301,401=81*3,721. Thus:MN²=100*4*3,721/(81*3,721)=100*4/81=400/81 ⇒MN=20/9Yes! Therefore, MN=20/9.This is a much simpler result. Therefore, after correcting the angle from 60° to 120°, we find the side length is √61 and MN=20/9.So the problem was initially misleading because the angle at M is 120°, not 60°, since M is on the opposite arc. This was the key mistake in the first approach.Therefore, the length of segment MN is 20/9 and the side length of the triangle is √61.Final AnswerThe length of segment ( MN ) is (boxed{dfrac{20}{9}}) and the side of the triangle ( ABC ) is (boxed{sqrt{61}}).</think>Given an equilateral triangle (ABC) with a point (M) on the circumcircle arc (AC) such that (MA = 4) and (MC = 5). The line (BM) intersects (AC) at point (N). We need to find the length of segment (MN) and the side length of the triangle (ABC).1. Determine the side length of the equilateral triangle: - Since (M) is on the circumcircle of (triangle ABC), the angle (angle AMC = 120^circ) (as (M) is on the arc (AC) not containing (B)). - Using the Law of Cosines in (triangle AMC): [ AC^2 = MA^2 + MC^2 - 2 cdot MA cdot MC cdot cos(120^circ) ] [ AC^2 = 4^2 + 5^2 - 2 cdot 4 cdot 5 cdot left(-frac{1}{2}right) ] [ AC^2 = 16 + 25 + 20 = 61 ] [ AC = sqrt{61} ] - Therefore, the side length of the equilateral triangle (ABC) is (sqrt{61}).2. Find the coordinates of points (B), (M), and (N): - Place (A) at ((0, 0)), (C) at ((sqrt{61}, 0)), and (B) at (left(frac{sqrt{61}}{2}, frac{sqrt{183}}{2}right)). - Using distances (MA = 4) and (MC = 5), solve for coordinates of (M) on the circumcircle: - Solving the system of equations for (x) and (y), we find (M) at (left(frac{26sqrt{61}}{61}, -frac{10sqrt{183}}{61}right)).3. Determine the equation of line (BM) and its intersection with (AC): - Calculate the slope of line (BM) and find the intersection point (N) on (AC) (where (y = 0)): - The equation of line (BM) is derived, and solving for (y = 0) gives (N) at (left(frac{4sqrt{61}}{9}, 0right)).4. Calculate the distance (MN): - Using the coordinates of (M) and (N): [ MN = sqrt{left(frac{4sqrt{61}}{9} - frac{26sqrt{61}}{61}right)^2 + left(0 - left(-frac{10sqrt{183}}{61}right)right)^2} ] - Simplifying the expression, we find (MN = frac{20}{9}). Final AnswerThe length of segment (MN) is (boxed{dfrac{20}{9}}) and the side of the triangle (ABC) is (boxed{sqrt{61}}).]

🔑:Answer: AB=sqrt{61}; MN=frac{20}{9}.Criteria for checking works, 9th gradePreliminary round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics#