💡Alright, so I have this problem about a school with n classes and k students. The conditions are that every two students have attended exactly one common class, and each class has fewer than k students. Also, it's given that k-1 is not a perfect square, and I need to prove that there exists a student who has attended at least sqrt(k) classes.Hmm, okay. Let me try to unpack this step by step.First, the setup: There are n classes and k students. Each pair of students shares exactly one common class. That sounds a bit like a combinatorial design, maybe something similar to a projective plane or a block design. In combinatorics, a projective plane has the property that every two points lie on exactly one line, which is similar to every two students sharing exactly one class here.Also, each class has fewer than k students. So, no class is attended by all students, which makes sense because if a class had k students, then every pair of students would share that class, but they might share others as well, which contradicts the condition that they share exactly one common class.Now, the goal is to show that there exists a student who has attended at least sqrt(k) classes. So, I need to find a lower bound on the number of classes a student must attend, given these conditions.Since k-1 is not a perfect square, that might be a clue. Maybe in the case where k-1 is a perfect square, there's a more symmetric structure, but since it's not, we can use that to show that some student must have a higher number of classes.Let me think about how to model this. Maybe I can represent the students as points and the classes as lines, where each line (class) connects some points (students). The condition that every two students share exactly one class translates to every two points lying on exactly one line, which is the definition of a projective plane.In a projective plane of order n, each line contains n+1 points, each point is on n+1 lines, and there are n^2 + n + 1 points and lines. The number of points is n^2 + n + 1, so if k were equal to n^2 + n + 1, then each student would be in n+1 classes. But in our problem, k is arbitrary, and k-1 is not a perfect square.Wait, so maybe the structure isn't a projective plane, but it's something similar. Let me recall that in a projective plane, the number of points is a square plus a linear term, but here we just have k students, so perhaps it's a more general design.Let me think about the parameters. Let's denote by r the number of classes a student attends. Since every two students share exactly one class, the number of pairs of students is C(k, 2) = k(k-1)/2. Each class, which has size less than k, say size s_i for class i, contributes C(s_i, 2) pairs of students who share that class.Since every pair of students shares exactly one class, the sum over all classes of C(s_i, 2) must equal C(k, 2). So, we have:Sum_{i=1 to n} [s_i choose 2] = [k choose 2]Which is:Sum_{i=1 to n} (s_i(s_i - 1))/2 = k(k - 1)/2Multiplying both sides by 2:Sum_{i=1 to n} s_i(s_i - 1) = k(k - 1)Also, since each class has fewer than k students, s_i < k for all i.Now, let's think about the average number of students per class. The total number of student-class memberships is Sum_{i=1 to n} s_i. Let's denote this as T. Then, T = Sum_{i=1 to n} s_i.We also know that each student attends r classes, so T = k * r.So, Sum_{i=1 to n} s_i = k * rWe have two equations:1. Sum_{i=1 to n} s_i = k * r2. Sum_{i=1 to n} s_i(s_i - 1) = k(k - 1)Let me try to relate these two. Let's denote S = Sum_{i=1 to n} s_i^2. Then, from equation 2:Sum_{i=1 to n} s_i^2 - Sum_{i=1 to n} s_i = k(k - 1)So,S - T = k(k - 1)But T = k * r, so:S - k * r = k(k - 1)Thus,S = k(k - 1) + k * r = k(k - 1 + r)Now, we can use the Cauchy-Schwarz inequality on the sum of squares. Recall that (Sum s_i)^2 <= n * Sum s_i^2. So,(T)^2 <= n * SBut T = k * r, and S = k(k - 1 + r). So,(k * r)^2 <= n * k(k - 1 + r)Divide both sides by k:k * r^2 <= n(k - 1 + r)But I don't know what n is, so maybe I can find another relation.Alternatively, let's think about the variance. The sum of squares S is related to the variance of the class sizes.But maybe another approach is better. Let's consider the number of classes a student attends, r. We want to find a lower bound on r.Suppose, for contradiction, that every student attends at most m classes, where m < sqrt(k). Then, r <= m for all students.But wait, actually, the problem says to prove that there exists a student who has attended at least sqrt(k) classes, so assuming the contrary would be that all students attend at most m classes where m < sqrt(k). But since we need to prove at least sqrt(k), maybe m <= sqrt(k) - 1.But let's see.If every student attends at most m classes, then T = k * r <= k * m.But from equation 1, T = Sum s_i = k * r.From equation 2, S = k(k - 1 + r).We also have from Cauchy-Schwarz:(T)^2 <= n * SSo,(k * r)^2 <= n * k(k - 1 + r)Simplify:k^2 * r^2 <= n * k(k - 1 + r)Divide both sides by k:k * r^2 <= n(k - 1 + r)But we don't know n, so maybe we can find another relation.Alternatively, let's use the fact that each class has size less than k, so s_i <= k - 1 for all i.But we might need another approach.Let me think about the number of pairs. Each class contributes C(s_i, 2) pairs, and the total is C(k, 2).So, Sum C(s_i, 2) = C(k, 2)Which is Sum s_i(s_i - 1)/2 = k(k - 1)/2So, Sum s_i(s_i - 1) = k(k - 1)Let me denote s_i = a_i, so Sum a_i(a_i - 1) = k(k - 1)Also, Sum a_i = k * rWe can write Sum a_i^2 - Sum a_i = k(k - 1)So, Sum a_i^2 = k(k - 1) + Sum a_i = k(k - 1) + k * r = k(k - 1 + r)Now, using Cauchy-Schwarz:(Sum a_i)^2 <= n * Sum a_i^2Which is (k * r)^2 <= n * k(k - 1 + r)So,k^2 * r^2 <= n * k(k - 1 + r)Divide both sides by k:k * r^2 <= n(k - 1 + r)But we need to relate n and r.Alternatively, let's think about the number of classes n.Each class has size s_i < k, so the maximum number of classes is at least C(k, 2)/C(s_i, 2). But since s_i < k, C(s_i, 2) <= C(k - 1, 2).So, n >= C(k, 2)/C(k - 1, 2) = [k(k - 1)/2] / [(k - 1)(k - 2)/2] = k / (k - 2)But that's just a rough lower bound.Alternatively, maybe we can use the fact that in such a design, the number of classes n is equal to the number of students k, but that's not necessarily true.Wait, in a projective plane, the number of lines equals the number of points, but here it's not necessarily a projective plane.Alternatively, let's think about the dual problem. If we consider classes as points and students as lines, then each line (student) contains r points (classes), and each pair of lines intersects in exactly one point (since every two students share exactly one class). So, this is similar to a projective plane, but again, the parameters might not fit.In the dual, we have that the number of points n is equal to the number of lines k, and each line contains r points, each point is in s lines, where s is the number of classes a student attends. Wait, no, in the dual, the number of lines would be n, and the number of points would be k.Wait, maybe I'm complicating things.Let me try another approach. Suppose that every student attends at most m classes, where m < sqrt(k). Then, the total number of student-class memberships T = k * m.But from equation 1, T = Sum s_i = k * r, so r <= m.But we also have Sum s_i^2 = k(k - 1 + r)Using Cauchy-Schwarz:(T)^2 <= n * Sum s_i^2So,(k * r)^2 <= n * k(k - 1 + r)Simplify:k^2 * r^2 <= n * k(k - 1 + r)Divide by k:k * r^2 <= n(k - 1 + r)But we need to find a relation between n and r.Alternatively, let's think about the number of classes n. Each class has size s_i < k, so the total number of classes n must satisfy Sum s_i = k * r, and since each s_i < k, n > k * r / (k - 1). Because if each class has at most k - 1 students, then the number of classes needed to cover k * r student-class memberships is at least k * r / (k - 1).So, n >= k * r / (k - 1)But from the earlier inequality:k * r^2 <= n(k - 1 + r)Substitute n >= k * r / (k - 1):k * r^2 <= (k * r / (k - 1))(k - 1 + r)Simplify the right-hand side:(k * r / (k - 1))(k - 1 + r) = k * r + (k * r * r)/(k - 1) = k * r + (k * r^2)/(k - 1)So,k * r^2 <= k * r + (k * r^2)/(k - 1)Subtract k * r from both sides:k * r^2 - k * r <= (k * r^2)/(k - 1)Factor left side:k * r(r - 1) <= (k * r^2)/(k - 1)Divide both sides by k * r (assuming r > 0):r - 1 <= r / (k - 1)Multiply both sides by (k - 1):(r - 1)(k - 1) <= rExpand left side:r(k - 1) - (k - 1) <= rBring all terms to left:r(k - 1) - (k - 1) - r <= 0Factor r:r(k - 1 - 1) - (k - 1) <= 0Wait, that's:r(k - 2) - (k - 1) <= 0So,r(k - 2) <= k - 1Thus,r <= (k - 1)/(k - 2)But (k - 1)/(k - 2) = 1 + 1/(k - 2)Since k >= 2, this is greater than 1 but less than 2 for k > 3.Wait, this seems problematic because r is supposed to be an integer greater than or equal to 1.But if k > 3, then (k - 1)/(k - 2) < 2, so r <= 1 + something less than 1, so r <= 1.But that can't be because if r = 1, then each student is in only one class, but then two students would share that class, but there are k students, so we would need at least C(k, 2) classes, each containing exactly two students, but that would mean each class has size 2, and the total number of classes would be C(k, 2), which is k(k - 1)/2. But in that case, each student is in (k - 1) classes, which contradicts r = 1.So, this suggests that our assumption that every student attends at most m classes where m < sqrt(k) leads to a contradiction because it forces r to be less than 2, which is impossible unless k is very small.Wait, but k could be small. Let's test with small k.If k = 2, then k - 1 = 1, which is a perfect square, but the problem states that k - 1 is not a perfect square, so k = 2 is excluded.If k = 3, then k - 1 = 2, not a perfect square. Let's see:n classes, 3 students, each pair shares exactly one class, each class has less than 3 students, so each class has 2 students.How many classes? Each pair must share exactly one class, so there are C(3, 2) = 3 classes, each with 2 students. So, each student is in 2 classes. So, r = 2.sqrt(k) = sqrt(3) ≈ 1.732, so r = 2 >= sqrt(3), which satisfies the conclusion.Similarly, for k = 4, k - 1 = 3, not a perfect square.Wait, k = 4, k - 1 = 3, not a perfect square.In this case, how many classes? Each pair of students shares exactly one class, and each class has less than 4 students, so at most 3.But in reality, for k = 4, the structure is similar to a projective plane of order 1, which is the Fano plane, but that has 7 points and 7 lines. Wait, no, for k = 4, it's different.Wait, actually, for k = 4, we can have each student in 3 classes, each class containing 3 students, but that would mean each pair shares exactly one class. Let me check:Number of pairs: C(4, 2) = 6.Each class of size 3 contributes C(3, 2) = 3 pairs.So, to cover 6 pairs, we need 2 classes, but each class has 3 students, so each student is in 2 classes, but then each pair would share only one class if the classes are arranged properly.Wait, no, if we have two classes, each with 3 students, then some pairs would share both classes, which contradicts the condition.So, actually, for k = 4, we need more classes.Wait, perhaps each student is in 3 classes, and each class has 3 students, but arranged so that each pair shares exactly one class.This is similar to the Fano plane, but the Fano plane has 7 points and 7 lines, each line has 3 points, each point is in 3 lines, and each pair of points lies on exactly one line.But for k = 4, it's not possible to have such a structure because the Fano plane requires 7 points.So, maybe for k = 4, it's not possible to have such a system where every two students share exactly one class with each class having less than k students. Or perhaps it's possible with more classes.Alternatively, maybe each class has 2 students, so we have C(4, 2) = 6 classes, each with 2 students, and each student is in 3 classes. Then, each pair shares exactly one class, which is satisfied because each pair is in exactly one class. So, in this case, r = 3, which is greater than sqrt(4) = 2. So, the conclusion holds.Wait, but in this case, each class has size 2, which is less than k = 4, so it satisfies the condition.So, for k = 4, the conclusion holds because each student is in 3 classes, which is greater than sqrt(4) = 2.Similarly, for k = 5, k - 1 = 4, which is a perfect square, so the problem's condition excludes k = 5 because k - 1 is a perfect square.Wait, no, the problem says that k - 1 is not a perfect square, so k = 5 is excluded because 5 - 1 = 4, which is a perfect square.So, for k = 6, k - 1 = 5, not a perfect square.In this case, let's see:Number of pairs: C(6, 2) = 15.Each class has size less than 6, so at most 5.If we have classes of size 5, then each class contributes C(5, 2) = 10 pairs. To cover 15 pairs, we need at least 2 classes, but then some pairs would be covered twice, which is not allowed.So, classes must be smaller.If we have classes of size 3, each contributes 3 pairs. To cover 15 pairs, we need 5 classes. Each student would be in r classes, and since each class has 3 students, the total number of student-class memberships is 5 * 3 = 15, so r = 15 / 6 = 2.5, which is not an integer. So, that's not possible.Alternatively, if we have classes of size 4, each contributes 6 pairs. To cover 15 pairs, we need at least 3 classes (3 * 6 = 18, which is more than 15), but then some pairs would be covered multiple times.Alternatively, mix of class sizes.But this is getting complicated. Maybe it's better to think in terms of the initial approach.So, going back, we have:Sum s_i = k * rSum s_i^2 = k(k - 1 + r)Using Cauchy-Schwarz:(k * r)^2 <= n * k(k - 1 + r)So,k^2 * r^2 <= n * k(k - 1 + r)Divide by k:k * r^2 <= n(k - 1 + r)But we need to find a lower bound on r.Assume for contradiction that r <= m < sqrt(k). Then, we have:k * m^2 <= n(k - 1 + m)But we also have that n >= k * r / (k - 1), as earlier.So,k * m^2 <= (k * m / (k - 1))(k - 1 + m)Simplify:k * m^2 <= (k * m / (k - 1))(k - 1 + m) = k * m + (k * m^2)/(k - 1)Subtract k * m from both sides:k * m^2 - k * m <= (k * m^2)/(k - 1)Factor left side:k * m(m - 1) <= (k * m^2)/(k - 1)Divide both sides by k * m (assuming m > 0):m - 1 <= m / (k - 1)Multiply both sides by (k - 1):(m - 1)(k - 1) <= mExpand:m(k - 1) - (k - 1) <= mBring all terms to left:m(k - 1) - m - (k - 1) <= 0Factor m:m(k - 1 - 1) - (k - 1) <= 0Which is:m(k - 2) - (k - 1) <= 0So,m(k - 2) <= k - 1Thus,m <= (k - 1)/(k - 2)But (k - 1)/(k - 2) = 1 + 1/(k - 2)Since k >= 3 (because k - 1 is not a perfect square, and k = 2 would make k - 1 = 1, which is a perfect square, but k = 2 is excluded as we saw earlier), so for k >= 3, (k - 1)/(k - 2) < 2.So, m <= 1 + 1/(k - 2) < 2But m is an integer, so m <= 1But if m <= 1, then each student is in at most 1 class, which is impossible because then the number of pairs covered would be C(s_i, 2) for each class, but if each student is in only one class, then the number of pairs is Sum C(s_i, 2). But since each pair must be covered exactly once, and each class contributes C(s_i, 2) pairs, the total must be C(k, 2). However, if each student is in only one class, then the number of classes is k (since each class can have multiple students), but then the number of pairs covered would be Sum C(s_i, 2), which would have to equal C(k, 2). But if each class has s_i students, and Sum s_i = k (since each student is in one class), then Sum C(s_i, 2) = Sum (s_i^2 - s_i)/2. To maximize this sum, we would have one class with all k students, but that's not allowed since s_i < k. So, the maximum Sum C(s_i, 2) would be when one class has k - 1 students and the rest have 1, but then Sum C(s_i, 2) = C(k - 1, 2) + (k - (k - 1)) * C(1, 2) = C(k - 1, 2) + 0 = (k - 1)(k - 2)/2, which is less than C(k, 2) = k(k - 1)/2. So, it's impossible to cover all pairs if each student is in only one class.Therefore, our assumption that m <= 1 leads to a contradiction, meaning that there must be at least one student who attends at least 2 classes.But wait, we wanted to show that there exists a student who attends at least sqrt(k) classes. So, maybe we need a different approach.Alternatively, let's consider the number of classes a student attends, r. We have:Sum s_i = k * rSum s_i^2 = k(k - 1 + r)Using the Cauchy-Schwarz inequality:(Sum s_i)^2 <= n * Sum s_i^2So,(k * r)^2 <= n * k(k - 1 + r)Thus,k^2 * r^2 <= n * k(k - 1 + r)Divide both sides by k:k * r^2 <= n(k - 1 + r)But we also have that n >= k * r / (k - 1), as earlier.So,k * r^2 <= (k * r / (k - 1))(k - 1 + r)Simplify:k * r^2 <= k * r + (k * r^2)/(k - 1)Subtract k * r from both sides:k * r^2 - k * r <= (k * r^2)/(k - 1)Factor left side:k * r(r - 1) <= (k * r^2)/(k - 1)Divide both sides by k * r (assuming r > 0):r - 1 <= r / (k - 1)Multiply both sides by (k - 1):(r - 1)(k - 1) <= rExpand:r(k - 1) - (k - 1) <= rBring all terms to left:r(k - 1) - r - (k - 1) <= 0Factor r:r(k - 2) - (k - 1) <= 0So,r(k - 2) <= k - 1Thus,r <= (k - 1)/(k - 2)But as before, (k - 1)/(k - 2) = 1 + 1/(k - 2) < 2 for k > 3.So, r <= 1 + 1/(k - 2) < 2, meaning r <= 1, which is impossible as we saw earlier.Therefore, our initial assumption that every student attends at most m classes where m < sqrt(k) must be false. Hence, there must exist at least one student who attends at least sqrt(k) classes.Wait, but in the case where k - 1 is a perfect square, say k - 1 = t^2, then perhaps the structure is more symmetric, and every student attends exactly t + 1 classes, which would be less than sqrt(k) if t + 1 < sqrt(k). But since k - 1 = t^2, then k = t^2 + 1, so sqrt(k) = sqrt(t^2 + 1) ≈ t + 1/(2t). So, t + 1 > sqrt(k), which contradicts. Wait, no, actually, t + 1 > sqrt(t^2 + 1) because (t + 1)^2 = t^2 + 2t + 1 > t^2 + 1 for t > 0.So, in the case where k - 1 is a perfect square, the number of classes per student would be t + 1, which is greater than sqrt(k). But the problem states that k - 1 is not a perfect square, so we can't have that symmetric case, which allows us to conclude that some student must have more classes.Wait, but in the case where k - 1 is a perfect square, say k - 1 = t^2, then k = t^2 + 1, and sqrt(k) = sqrt(t^2 + 1) < t + 1. So, in that case, if every student attended exactly t + 1 classes, that would be more than sqrt(k). But the problem says that k - 1 is not a perfect square, so we can't have that exact structure, which might allow for a lower bound.But in our earlier approach, we showed that assuming all students attend less than sqrt(k) classes leads to a contradiction, implying that at least one student must attend at least sqrt(k) classes.Therefore, the conclusion is that there exists a student who has attended at least sqrt(k) classes.

💡Okay, so I've got this math problem here with four options, A through D, and I need to figure out which one is correct. Let me take it step by step because I want to make sure I understand each part properly.Starting with **Option A**: It says √2 + √8 = √10. Hmm, I remember that when adding square roots, they need to have the same radicand, which is the number under the square root symbol. So, √2 is straightforward, but √8 can probably be simplified. Let me think: 8 is 4 times 2, and since 4 is a perfect square, √8 is the same as √(4*2) which is √4 times √2, and √4 is 2. So, √8 simplifies to 2√2. Now, going back to the original expression: √2 + √8 becomes √2 + 2√2. Since both terms have √2, I can add their coefficients: 1√2 + 2√2 equals 3√2. So, √2 + √8 equals 3√2, not √10. That means **Option A** is incorrect.Moving on to **Option B**: It states that a³ · a⁴ = a¹². I recall the rule for multiplying exponents with the same base: you add the exponents. So, a³ times a⁴ should be a^(3+4) which is a⁷. But here, it's saying a¹², which is way too high. That doesn't make sense because 3 + 4 is 7, not 12. Therefore, **Option B** is also incorrect.Next is **Option C**: It claims that (a - b)² = a² - b². I remember the formula for squaring a binomial: (a - b)² equals a² - 2ab + b². So, it's not just a² - b²; there's an extra middle term, -2ab. Without that, the equation isn't correct. Hence, **Option C** is wrong.Finally, **Option D**: It says (-2ab²)³ = -8a³b⁶. Let me break this down. When raising a product to a power, each factor inside the parentheses is raised to that power. So, (-2)³ is -8 because a negative number raised to an odd power remains negative. Then, a³ is straightforward, and (b²)³ is b^(2*3) which is b⁶. Putting it all together, (-2ab²)³ equals -8a³b⁶. That matches exactly what's given in **Option D**, so this one is correct.So, after going through each option carefully, I can confidently say that **Option D** is the correct one.

💡Okay, so I have this problem where there's a pentagon ABCDE inscribed in a circle. The sides AB, BC, CD, and DE are all 4 units long, and the side AE is 1 unit. I need to compute the product (1 - cos angle B)(1 - cos angle ACE). Hmm, okay, let's try to figure this out step by step.First, since ABCDE is inscribed in a circle, it's a cyclic pentagon. That means all its vertices lie on a single circle. I remember that in cyclic polygons, opposite angles have certain properties, but I'm not sure exactly how that applies here. Maybe I can use the Law of Cosines or something related to circles?Looking at the given sides, four of them are equal (AB=BC=CD=DE=4) and one is different (AE=1). So, the pentagon isn't regular, but it has some symmetry. Maybe the chords corresponding to these sides have some relationships?Let me label the points: A, B, C, D, E in order around the circle. Since AB=BC=CD=DE=4, the arcs subtended by these chords should be equal, right? Because in a circle, equal chords subtend equal arcs. So, arcs AB, BC, CD, and DE are equal. That leaves arc EA, which is subtended by chord AE=1, so it must be a different arc.Let me denote the measure of each equal arc AB, BC, CD, DE as θ. Then, the measure of arc EA would be something else, let's say φ. Since the total circumference of the circle is 360 degrees, we have 4θ + φ = 360°. But I don't know the actual measures yet, so maybe I can relate them using the chord lengths.I recall that the length of a chord in a circle is related to the radius and the subtended angle by the formula: chord length = 2R sin(α/2), where α is the measure of the arc in radians and R is the radius. Wait, actually, the formula is chord length = 2R sin(α/2). So, for chord AB, which is 4 units, we have 4 = 2R sin(θ/2). Similarly, for chord AE, which is 1 unit, we have 1 = 2R sin(φ/2).So, from chord AB: 4 = 2R sin(θ/2) => sin(θ/2) = 2/R.From chord AE: 1 = 2R sin(φ/2) => sin(φ/2) = 1/(2R).Hmm, okay, so sin(θ/2) = 2/R and sin(φ/2) = 1/(2R). Maybe I can relate θ and φ somehow.Earlier, I noted that 4θ + φ = 360°, but that's in degrees. Maybe I should convert that to radians for consistency? Let me see, 360° is 2π radians. So, 4θ + φ = 2π.But I don't know θ or φ yet. Maybe I can express φ in terms of θ or vice versa.From 4θ + φ = 2π, we can write φ = 2π - 4θ.So, sin(φ/2) = sin((2π - 4θ)/2) = sin(π - 2θ) = sin(2θ), because sin(π - x) = sin x.So, sin(φ/2) = sin(2θ). But earlier, we have sin(φ/2) = 1/(2R). So, sin(2θ) = 1/(2R).But from chord AB, sin(θ/2) = 2/R. Let me write that down:sin(θ/2) = 2/R => R = 2 / sin(θ/2).Then, sin(2θ) = 1/(2R) = 1/(2*(2 / sin(θ/2))) = sin(θ/2)/4.So, sin(2θ) = sin(θ/2)/4.Hmm, okay, so sin(2θ) = (1/4) sin(θ/2). Let me write that equation:sin(2θ) = (1/4) sin(θ/2).I need to solve for θ. This seems a bit tricky. Maybe I can use double-angle identities or something.I know that sin(2θ) = 2 sinθ cosθ. So, substituting that in:2 sinθ cosθ = (1/4) sin(θ/2).Hmm, so 2 sinθ cosθ = (1/4) sin(θ/2).I can write sinθ in terms of sin(θ/2). Using the double-angle identity again, sinθ = 2 sin(θ/2) cos(θ/2). So, substituting that in:2*(2 sin(θ/2) cos(θ/2)) * cosθ = (1/4) sin(θ/2).Simplify:4 sin(θ/2) cos(θ/2) cosθ = (1/4) sin(θ/2).Assuming sin(θ/2) ≠ 0, which it can't be because θ is an arc between two points on the circle, so θ must be between 0 and 2π, so sin(θ/2) is positive.So, we can divide both sides by sin(θ/2):4 cos(θ/2) cosθ = 1/4.So, 4 cos(θ/2) cosθ = 1/4.Multiply both sides by 4:16 cos(θ/2) cosθ = 1.Hmm, okay, so 16 cos(θ/2) cosθ = 1.I need to solve for θ. Maybe I can express cosθ in terms of cos(θ/2). Let's recall that cosθ = 2 cos²(θ/2) - 1.So, substituting that in:16 cos(θ/2)*(2 cos²(θ/2) - 1) = 1.Let me let x = cos(θ/2) for simplicity.So, 16x*(2x² - 1) = 1.Expanding:32x³ - 16x = 1.Bring all terms to one side:32x³ - 16x - 1 = 0.So, we have a cubic equation: 32x³ -16x -1 = 0.Hmm, solving a cubic equation might be a bit involved. Maybe I can try rational roots. The possible rational roots are ±1, ±1/2, ±1/4, ±1/8, ±1/16, ±1/32.Let me test x=1/2:32*(1/2)^3 -16*(1/2) -1 = 32*(1/8) - 8 -1 = 4 -8 -1 = -5 ≠ 0.x=1/4:32*(1/4)^3 -16*(1/4) -1 = 32*(1/64) -4 -1 = 0.5 -4 -1 = -4.5 ≠0.x=1/8:32*(1/8)^3 -16*(1/8) -1 = 32*(1/512) -2 -1 = 1/16 -2 -1 ≈ -2.9375 ≠0.x= -1/2:32*(-1/2)^3 -16*(-1/2) -1 = 32*(-1/8) +8 -1 = -4 +8 -1=3≠0.x= -1/4:32*(-1/4)^3 -16*(-1/4) -1 = 32*(-1/64) +4 -1 = -0.5 +4 -1=2.5≠0.x=1/16:32*(1/16)^3 -16*(1/16) -1=32*(1/4096) -1 -1≈0.0078 -1 -1≈-1.9922≠0.x= -1/8:32*(-1/8)^3 -16*(-1/8) -1=32*(-1/512)+2 -1≈-0.0625 +2 -1≈0.9375≠0.x=1:32*1 -16*1 -1=32-16-1=15≠0.x=-1:-32 -(-16) -1=-32+16-1=-17≠0.So, none of the rational roots work. Maybe I need to use the method for solving cubics or approximate the solution.Alternatively, maybe there's a trigonometric identity I can use to solve 16 cos(θ/2) cosθ =1.Wait, let's think differently. Maybe I can use multiple-angle identities or something else.Alternatively, since this is getting complicated, maybe I can use the Law of Cosines on triangles ABC and ACE.Let me consider triangle ABC. It has sides AB=4, BC=4, and AC is a diagonal. Let's denote AC as x.Applying the Law of Cosines on triangle ABC:x² = AB² + BC² - 2*AB*BC*cos(angle ABC)So, x² = 4² + 4² - 2*4*4*cos(angle B)x² = 16 + 16 - 32 cos(angle B)x² = 32(1 - cos(angle B))Okay, so that's one equation involving x and angle B.Now, let's consider triangle ACE. It has sides AC=x, CE=x, and AE=1. So, triangle ACE is isoceles with sides x, x, and base 1.Applying the Law of Cosines on triangle ACE:AE² = AC² + CE² - 2*AC*CE*cos(angle ACE)1² = x² + x² - 2x² cos(angle ACE)1 = 2x²(1 - cos(angle ACE))So, 1 = 2x²(1 - cos(angle ACE))But from triangle ABC, we have x² = 32(1 - cos(angle B)). So, substituting that into the equation:1 = 2*(32(1 - cos(angle B)))*(1 - cos(angle ACE))Simplify:1 = 64(1 - cos(angle B))(1 - cos(angle ACE))So, (1 - cos(angle B))(1 - cos(angle ACE)) = 1/64Wait, that seems straightforward. So, the product is 1/64.But wait, did I skip something? Let me double-check.In triangle ABC, sides AB=BC=4, so it's isoceles, and angle ABC is angle B. So, applying Law of Cosines gives x² = 32(1 - cos(angle B)).In triangle ACE, sides AC=CE=x, and AE=1. So, applying Law of Cosines gives 1 = 2x²(1 - cos(angle ACE)).Substituting x² from the first equation into the second gives 1 = 2*32*(1 - cos(angle B))*(1 - cos(angle ACE)) => 1 = 64*(1 - cos(angle B))*(1 - cos(angle ACE)).Therefore, (1 - cos(angle B))(1 - cos(angle ACE)) = 1/64.So, that seems correct. I didn't need to go through the circle and chord lengths after all. Maybe I overcomplicated it initially by trying to relate the arcs and chord lengths. The Law of Cosines approach was more straightforward.So, the final answer is 1/64.**Final Answer**boxed{dfrac{1}{64}}

💡Okay, so I need to figure out this problem about money being invested at a certain interest rate and how long it takes to double. Let me try to understand what's being asked here.First, the problem mentions that if money is invested at r percent interest, compounded annually, the amount will double in approximately some years. Then, it gives an example with Pat's parents investing 5,000 at 8 percent interest, compounded annually, and after 18 years, the investment grows to 20,000. The question is asking for the constant value in the formula used to calculate the number of years it takes for the investment to double.Hmm, okay. So, I think this is related to the Rule of 72, which I've heard about before. From what I remember, the Rule of 72 is a quick way to estimate how long it takes for an investment to double, given a fixed annual rate of interest. The formula is something like dividing 72 by the interest rate to get the approximate number of years.Let me write that down to make it clearer:Years to double ≈ 72 / rWhere r is the annual interest rate in percentage.In this case, the interest rate is 8 percent. So, plugging that into the formula:Years to double ≈ 72 / 8Years to double ≈ 9So, according to the Rule of 72, it should take approximately 9 years for the investment to double at 8 percent interest.Now, the problem mentions that after 18 years, the investment grew from 5,000 to 20,000. Let's see if that makes sense with our calculation. If it doubles every 9 years, then after 9 years, it would go from 5,000 to 10,000, and after another 9 years (so 18 years total), it would double again to 20,000. That matches perfectly with the given information.So, it seems like the Rule of 72 is the right approach here, and the constant value in the formula is 72.But just to be thorough, maybe I should verify this with the actual compound interest formula to ensure that the Rule of 72 isn't giving a misleading approximation in this case.The compound interest formula is:A = P(1 + r)^tWhere:A = the amount of money accumulated after t years, including interest.P = the principal amount (the initial amount of money).r = annual interest rate (in decimal form).t = time in years.In this problem:A = 20,000P = 5,000r = 8% = 0.08t = 18 yearsLet's plug these values into the formula to see if it holds:20,000 = 5,000(1 + 0.08)^18First, divide both sides by 5,000 to simplify:4 = (1.08)^18Now, let's calculate (1.08)^18 to see if it equals 4.I can use logarithms to solve for t in the equation:(1.08)^t = 4Taking the natural logarithm of both sides:ln((1.08)^t) = ln(4)Using the power rule of logarithms, this simplifies to:t * ln(1.08) = ln(4)Now, solve for t:t = ln(4) / ln(1.08)Calculating the natural logs:ln(4) ≈ 1.3863ln(1.08) ≈ 0.0770So,t ≈ 1.3863 / 0.0770 ≈ 18.05 yearsWow, that's really close to the 18 years given in the problem. So, the Rule of 72 gave us an approximation of 9 years per doubling, and over two doublings (18 years), it matches almost exactly with the precise calculation.This makes me more confident that the Rule of 72 is a reliable approximation method for estimating the doubling time of an investment at a given interest rate.Therefore, the constant value in the formula to calculate the number of years it takes for the investment to double is indeed 72.

💡First, substitute ( x = 6 ) into the inner fraction:[frac{3}{6} = frac{1}{2}]Next, substitute this back into the main fraction:[2 - frac{1}{2} = frac{4}{2} - frac{1}{2} = frac{3}{2}]Finally, simplify the division by a fraction by multiplying by its reciprocal:[frac{3}{frac{3}{2}} = 3 times frac{2}{3} = 2]The final answer is (boxed{text{(B)} 2}).

💡First, identify the total number of marbles Connie had initially, which is 143.Next, determine how many marbles Connie has left after giving some to Juan, which is 70.To find out how many marbles Connie gave to Juan, subtract the number of marbles she has left from the total number she started with.So, 143 (initial marbles) minus 70 (remaining marbles) equals 73 marbles given to Juan.Therefore, Connie gave Juan 73 marbles.