💡Okay, so I have this math problem here: If the product of two numbers, a and b, is a square, and a and b are coprime, then both a and b must themselves be squares. Hmm, interesting. I need to figure out why that's true.First, let me recall what it means for two numbers to be coprime. Coprime means that the greatest common divisor (GCD) of a and b is 1. In other words, a and b share no common prime factors. So, if I factor a and b into their prime components, there won't be any overlapping primes between them.Now, the product ab is a square. A square number has even exponents in its prime factorization. For example, 36 is a square because it's 2^2 * 3^2. Each prime factor has an even exponent. So, if ab is a square, then when I factor ab into primes, all the exponents should be even.But since a and b are coprime, their prime factors are entirely separate. That means the prime factors of a don't interfere with the prime factors of b. So, if I write a as a product of primes, say a = p1^k1 * p2^k2 * ... * pn^kn, and b as q1^m1 * q2^m2 * ... * qm^mm, then all the pi's and qj's are distinct primes.When I multiply a and b together, ab = p1^k1 * p2^k2 * ... * pn^kn * q1^m1 * q2^m2 * ... * qm^mm. Since ab is a square, each exponent in this product must be even. That means each ki and each mj must be even. Because if any of them were odd, then ab wouldn't be a square.Wait, let me think about that again. If a and b are coprime, then the exponents in a and b are independent of each other. So, if ab is a square, it's because both a and b individually have exponents that are even. Because if, say, a had an odd exponent for one of its primes, then ab would also have an odd exponent for that prime, right? But ab is a square, so all exponents must be even. Therefore, a must have even exponents in its prime factorization, making a a square. Similarly, b must also have even exponents, making b a square.Let me try to put this into a more formal argument. Suppose a and b are coprime integers, and ab is a perfect square. Then, in the prime factorization of ab, every prime has an even exponent. Since a and b are coprime, their prime factors are distinct. Therefore, the exponents in a and b must each individually be even. Hence, a and b must themselves be perfect squares.Is there another way to think about this? Maybe using the fundamental theorem of arithmetic, which states that every integer greater than 1 can be represented uniquely as a product of prime factors, up to ordering. Since a and b are coprime, their prime factors are unique and don't overlap. So, when you multiply them, the exponents for each prime in a and b just add up. Since the sum is even, and they don't share any primes, each exponent must be even on its own.Let me test this with an example. Suppose a = 4 and b = 9. They are coprime because their GCD is 1. Their product is 36, which is 6^2, a perfect square. Both a and b are squares: 4 is 2^2 and 9 is 3^2. That works.Another example: a = 2 and b = 8. Wait, but 2 and 8 are not coprime because their GCD is 2. So, that doesn't fit the condition. Let me pick a = 2 and b = 3. They are coprime. Their product is 6, which is not a square. So, this doesn't satisfy the condition that ab is a square.Wait, let's find another example where ab is a square and a and b are coprime. Let's take a = 4 and b = 9 again. Their product is 36, which is a square. Both a and b are squares. Another example: a = 25 and b = 16. Their product is 400, which is 20^2. Both 25 and 16 are squares.What if a and b are not squares? Let's say a = 2 and b = 8. But as I saw earlier, they are not coprime. So, that doesn't fit. How about a = 2 and b = 32? Again, not coprime. Hmm, seems like if a and b are coprime and their product is a square, then they must individually be squares.Wait, let me try a = 1 and b = 1. Their product is 1, which is a square. Both a and b are squares. Another example: a = 1 and b = 4. Their product is 4, a square. Both are squares. Okay, seems consistent.Is there a case where a and b are coprime, ab is a square, but a or b is not a square? Let me think. Suppose a = 2 and b = 2. But then they are not coprime. If a = 2 and b = 8, same issue. If a = 3 and b = 12, not coprime. Hmm, seems like whenever ab is a square and a and b are coprime, a and b have to be squares.Maybe I can think about it in terms of exponents. If a and b are coprime, then in the prime factorization of a, all exponents are independent of those in b. So, if ab is a square, then each exponent in a must be even, and each exponent in b must be even. Therefore, a and b are squares.Another angle: suppose a is not a square. Then, in its prime factorization, there is at least one prime with an odd exponent. Since a and b are coprime, b doesn't have that prime. Therefore, in ab, that prime would still have an odd exponent, making ab not a square. Contradiction. Therefore, a must be a square. Similarly, b must be a square.Yeah, that makes sense. So, if ab is a square and a and b are coprime, then both a and b must be squares.

💡First, I need to determine how many letters Sam wrote on Tuesday. I know that Sam wrote some letters on Tuesday and 3 letters on Wednesday. The average number of letters he wrote per day over these two days is 5.To find the total number of letters written over the two days, I can multiply the average by the number of days:Total letters = 5 letters/day * 2 days = 10 letters.Since Sam wrote 3 letters on Wednesday, I can subtract that from the total to find out how many letters he wrote on Tuesday:Letters on Tuesday = Total letters - Letters on Wednesday = 10 letters - 3 letters = 7 letters.Therefore, Sam wrote 7 letters on Tuesday.

💡Alright, so I've got this geometry problem here, and it's pretty involved. Let me try to unpack it step by step. First, I need to visualize the problem. There's a triangle ABC inscribed in a circle ω. Angle A is 60 degrees, and angle B is 75 degrees. Okay, so angle C must be 45 degrees because the sum of angles in a triangle is 180 degrees. That gives me a rough idea of the triangle's shape.Next, the angle bisector of angle A meets BC at E and the circumcircle ω at D. So, E is the point where the bisector intersects BC, and D is the other intersection point of the bisector with the circumcircle. I remember that angle bisectors have some proportional properties, so maybe that will come into play.Then, we're told about reflections. The reflections of A across D and C are D' and C', respectively. Reflections... okay, so D' is the mirror image of A over D, and C' is the mirror image over C. That means AD = DD' and AC = CC'. I need to keep that in mind because reflections often imply some symmetry or congruent triangles.The tangent to ω at A meets BC at P. Tangents have properties related to angles in circles, like the tangent-chord angle being equal to the inscribed angle. So, maybe I can use that to find some angle relationships involving P.Next, the circumcircle of APD' meets AC at F ≠ A. So, we're creating another circle that goes through A, P, and D', and it intersects AC again at F. This F is a new point we need to consider.Finally, we need to prove that the circumcircle of C'FE is tangent to BC at E. Tangent implies that the circle touches BC at exactly one point, E, and that the radius at E is perpendicular to BC. So, to prove tangency, I might need to show that the angle between BC and the radius at E is 90 degrees or use some power of a point theorem.Let me think about how to approach this. Maybe starting with some angle chasing. Since we have several points defined by reflections and intersections, perhaps there are cyclic quadrilaterals or similar triangles involved.First, let's note down all the given angles:- Angle A = 60°- Angle B = 75°- Therefore, angle C = 45°Since AD is the angle bisector of angle A, it splits angle A into two 30° angles. So, angle BAD = angle CAD = 30°.Now, since ABC is inscribed in ω, points A, B, C, D are all on ω. So, perhaps some properties of cyclic quadrilaterals can be used here.Looking at the reflection points, D' is the reflection of A over D, meaning that D is the midpoint of AD'. Similarly, C' is the reflection of A over C, so C is the midpoint of AC'. This might imply that AD' = 2AD and AC' = 2AC.The tangent at A meets BC at P. Tangent properties tell us that angle PAB is equal to angle ACB because the tangent makes an angle equal to the inscribed angle over the same arc. Since angle ACB is 45°, angle PAB should also be 45°. Wait, but angle PAB is part of triangle PAB, which might help us find some lengths or other angles.Now, the circumcircle of APD' intersects AC at F. So, points A, P, D', F are concyclic. This means that angles subtended by the same chord are equal. Maybe angle AFP equals angle AD'P or something like that.I think I need to find some relationships between these points. Maybe using power of a point from P with respect to ω or the circumcircle of C'FE.Alternatively, since we need to prove tangency, perhaps showing that the power of E with respect to the circumcircle of C'FE is zero, which would imply that E lies on the circle and the tangent condition is satisfied.Wait, but E is already on BC and on the angle bisector. Maybe I can find some similar triangles involving E, F, and C'.Another thought: since C' is the reflection of A over C, then AC = CC', and triangle ACC' is isosceles. Similarly, AD' is twice AD, so triangle ADD' is also isosceles.Perhaps looking into spiral similarities or inversion properties might help, but that might be too advanced for this problem.Let me try to find coordinates for these points to compute distances and angles. Maybe placing the triangle in a coordinate system would help. Let's assume point A is at (0,0), and side AC is along the x-axis. Then, point C is at some coordinate, say (c,0), and point B is somewhere in the plane.Given the angles at A and B, maybe I can compute coordinates using trigonometry. But this might get messy, and I'm not sure if it's the most efficient way.Alternatively, using complex numbers could be an option, but again, it might complicate things.Wait, maybe using the Law of Sines and Cosines in various triangles could help. For example, in triangle ABC, I can find the ratios of the sides. Since angle A is 60°, angle B is 75°, and angle C is 45°, the sides are proportional to the sines of these angles.Let me denote the sides as follows:- a = BC- b = AC- c = ABBy the Law of Sines:a / sin(60°) = b / sin(75°) = c / sin(45°)This gives me ratios between the sides. Maybe assigning a specific length to one side would make calculations easier. Let's say AC = 1 for simplicity. Then, using the Law of Sines, I can find the lengths of AB and BC.Calculating:sin(60°) = √3 / 2 ≈ 0.866sin(75°) = sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 ≈ 0.966sin(45°) = √2 / 2 ≈ 0.707So, if AC = 1 (which is side b), then:a / (√3 / 2) = 1 / (sin75°) = 1 / (√6/4 + √2/4) ≈ 1 / 0.966 ≈ 1.035Therefore, a ≈ (√3 / 2) * 1.035 ≈ 0.866 * 1.035 ≈ 0.896Similarly, c / (√2 / 2) = 1 / sin75°, so c ≈ (√2 / 2) * 1.035 ≈ 0.707 * 1.035 ≈ 0.732So, sides are approximately:- BC ≈ 0.896- AC = 1- AB ≈ 0.732Not sure if these approximate values will help, but maybe exact expressions would be better.Alternatively, keeping AC = 1, then:a = BC = (sin60° / sin75°) * AC = (√3 / 2) / (sin75°)But sin75° = sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4So, a = (√3 / 2) / (√6/4 + √2/4) = (√3 / 2) * (4 / (√6 + √2)) = (2√3) / (√6 + √2)Rationalizing the denominator:Multiply numerator and denominator by (√6 - √2):(2√3)(√6 - √2) / [(√6 + √2)(√6 - √2)] = (2√3√6 - 2√3√2) / (6 - 2) = (2√18 - 2√6) / 4 = (2*3√2 - 2√6)/4 = (6√2 - 2√6)/4 = (3√2 - √6)/2So, BC = (3√2 - √6)/2Similarly, AB = c = (sin45° / sin75°) * AC = (√2/2) / (sin75°) = (√2/2) / (√6/4 + √2/4) = (√2/2) * (4 / (√6 + √2)) = (2√2) / (√6 + √2)Again, rationalizing:(2√2)(√6 - √2) / [(√6 + √2)(√6 - √2)] = (2√12 - 2√4) / (6 - 2) = (2*2√3 - 2*2)/4 = (4√3 - 4)/4 = (√3 - 1)So, AB = √3 - 1Okay, so now I have exact lengths:- AC = 1- BC = (3√2 - √6)/2- AB = √3 - 1Good, now maybe I can find coordinates for points A, B, and C.Let me place point A at (0,0), and point C at (1,0). Then, point B is somewhere in the plane. Let me find its coordinates.Given that AB = √3 - 1 and AC = 1, and angle at A is 60°, I can use the Law of Cosines to find coordinates of B.Wait, actually, since angle at A is 60°, and sides AB and AC are known, maybe it's easier to place point B in polar coordinates.From point A at (0,0), point B is at a distance AB = √3 - 1 at an angle of 60° from AC.So, coordinates of B would be:x = AB * cos(60°) = (√3 - 1) * 0.5y = AB * sin(60°) = (√3 - 1) * (√3/2)Calculating:x = (√3 - 1)/2 ≈ (1.732 - 1)/2 ≈ 0.366y = (√3 - 1)(√3)/2 = (3 - √3)/2 ≈ (3 - 1.732)/2 ≈ 0.634So, point B is approximately at (0.366, 0.634)Point C is at (1,0), point A is at (0,0).Now, I need to find point E, which is where the angle bisector of angle A meets BC.The angle bisector theorem tells us that BE/EC = AB/AC = (√3 - 1)/1 = √3 - 1So, BE = (√3 - 1) * ECBut BE + EC = BC = (3√2 - √6)/2So, let me denote EC = x, then BE = (√3 - 1)xThus, (√3 - 1)x + x = (3√2 - √6)/2x(√3 - 1 + 1) = (3√2 - √6)/2x√3 = (3√2 - √6)/2Therefore, x = (3√2 - √6)/(2√3) = [3√2 - √6]/(2√3)Rationalizing:Multiply numerator and denominator by √3:[3√6 - √18]/(2*3) = [3√6 - 3√2]/6 = (√6 - √2)/2So, EC = (√6 - √2)/2Therefore, BE = (√3 - 1) * EC = (√3 - 1)(√6 - √2)/2Let me compute that:(√3 - 1)(√6 - √2) = √3*√6 - √3*√2 - 1*√6 + 1*√2 = √18 - √6 - √6 + √2 = 3√2 - 2√6 + √2 = (3√2 + √2) - 2√6 = 4√2 - 2√6Therefore, BE = (4√2 - 2√6)/2 = 2√2 - √6So, BE = 2√2 - √6 and EC = (√6 - √2)/2Now, coordinates of E can be found using the ratio BE/EC = √3 - 1Since E divides BC in the ratio BE:EC = √3 - 1 : 1Coordinates of B: ( (√3 - 1)/2 , (3 - √3)/2 )Coordinates of C: (1, 0)Using the section formula, coordinates of E:x = [ (√3 - 1)*1 + 1*( (√3 - 1)/2 ) ] / (√3 - 1 + 1) = [ (√3 - 1) + (√3 - 1)/2 ] / √3Wait, no, the section formula is:If a point divides the line segment joining (x1,y1) and (x2,y2) in the ratio m:n, then the coordinates are:( (m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n) )Here, E divides BC in the ratio BE:EC = √3 - 1 : 1So, m = √3 - 1, n = 1Coordinates of B: ( (√3 - 1)/2 , (3 - √3)/2 )Coordinates of C: (1, 0)Thus,x_E = [ (√3 - 1)*1 + 1*( (√3 - 1)/2 ) ] / (√3 - 1 + 1 ) = [ (√3 - 1) + (√3 - 1)/2 ] / √3Simplify numerator:(√3 - 1) + (√3 - 1)/2 = (2(√3 - 1) + (√3 - 1))/2 = (3√3 - 3)/2Thus,x_E = (3√3 - 3)/(2√3) = [3(√3 - 1)]/(2√3) = [3(√3 - 1)]/(2√3)Multiply numerator and denominator by √3:[3(√3 - 1)√3]/(2*3) = [3(3 - √3)]/6 = (9 - 3√3)/6 = (3 - √3)/2Similarly, y_E:y_E = [ (√3 - 1)*0 + 1*( (3 - √3)/2 ) ] / √3 = [ (3 - √3)/2 ] / √3 = (3 - √3)/(2√3)Rationalizing:(3 - √3)/(2√3) * √3/√3 = (3√3 - 3)/6 = (√3 - 1)/2So, coordinates of E are ( (3 - √3)/2 , (√3 - 1)/2 )Alright, now moving on to point D, which is where the angle bisector meets the circumcircle again.Since AD is the angle bisector, and D is on the circumcircle, perhaps we can find its coordinates.Alternatively, since reflections are involved, maybe it's better to find D' first.Reflection of A over D is D', so D is the midpoint of AD'.Similarly, reflection of A over C is C', so C is the midpoint of AC'.Given that, if I can find D, then D' is just the reflection, so coordinates of D' would be (2x_D - 0, 2y_D - 0) = (2x_D, 2y_D)But to find D, since it's on the circumcircle and on the angle bisector, perhaps parametrize the angle bisector and find its other intersection with ω.The angle bisector from A is the line from (0,0) through E, which we have coordinates for.So, parametric equation of AD: starting at A(0,0), going through E( (3 - √3)/2 , (√3 - 1)/2 )So, direction vector is ( (3 - √3)/2 , (√3 - 1)/2 )Parametrize as:x = t*(3 - √3)/2y = t*(√3 - 1)/2We need to find t such that (x,y) lies on the circumcircle ω.But what's the equation of ω? Since points A, B, C are on ω, we can find its equation.Given points A(0,0), B( (√3 - 1)/2 , (3 - √3)/2 ), and C(1,0), we can find the circumcircle.The general equation of a circle is x² + y² + Dx + Ey + F = 0Plugging in A(0,0):0 + 0 + 0 + 0 + F = 0 => F = 0So, equation becomes x² + y² + Dx + Ey = 0Plugging in C(1,0):1 + 0 + D*1 + E*0 = 0 => 1 + D = 0 => D = -1So, equation is x² + y² - x + Ey = 0Now, plug in point B( (√3 - 1)/2 , (3 - √3)/2 ):[ ( (√3 - 1)/2 )² + ( (3 - √3)/2 )² ] - ( (√3 - 1)/2 ) + E*( (3 - √3)/2 ) = 0Compute each term:First term: ( (√3 - 1)/2 )² = (3 - 2√3 + 1)/4 = (4 - 2√3)/4 = (2 - √3)/2Second term: ( (3 - √3)/2 )² = (9 - 6√3 + 3)/4 = (12 - 6√3)/4 = (6 - 3√3)/2So, sum of first two terms: (2 - √3)/2 + (6 - 3√3)/2 = (8 - 4√3)/2 = 4 - 2√3Third term: - ( (√3 - 1)/2 ) = (-√3 + 1)/2Fourth term: E*( (3 - √3)/2 )Putting it all together:4 - 2√3 + (-√3 + 1)/2 + E*(3 - √3)/2 = 0Multiply everything by 2 to eliminate denominators:8 - 4√3 + (-√3 + 1) + E*(3 - √3) = 0Simplify:8 - 4√3 - √3 + 1 + E*(3 - √3) = 0Combine like terms:9 - 5√3 + E*(3 - √3) = 0Solve for E:E*(3 - √3) = -9 + 5√3E = (-9 + 5√3)/(3 - √3)Rationalize denominator:Multiply numerator and denominator by (3 + √3):E = [ (-9 + 5√3)(3 + √3) ] / [ (3 - √3)(3 + √3) ] = [ (-27 - 9√3 + 15√3 + 5*3) ] / (9 - 3) = [ (-27 + 6√3 + 15) ] / 6 = [ (-12 + 6√3) ] / 6 = (-2 + √3)So, E = -2 + √3Therefore, the equation of ω is:x² + y² - x + (-2 + √3)y = 0Now, back to parametrizing AD:x = t*(3 - √3)/2y = t*(√3 - 1)/2Plug into circle equation:[ t²*(3 - √3)²/4 + t²*(√3 - 1)²/4 ] - t*(3 - √3)/2 + (-2 + √3)*t*(√3 - 1)/2 = 0Compute each term:First term: t² [ (3 - √3)² + (√3 - 1)² ] / 4Compute (3 - √3)² = 9 - 6√3 + 3 = 12 - 6√3Compute (√3 - 1)² = 3 - 2√3 + 1 = 4 - 2√3Sum: (12 - 6√3) + (4 - 2√3) = 16 - 8√3So, first term: t²*(16 - 8√3)/4 = t²*(4 - 2√3)Second term: - t*(3 - √3)/2Third term: (-2 + √3)*t*(√3 - 1)/2Compute (-2 + √3)(√3 - 1):= -2√3 + 2 + 3 - √3 = (-2√3 - √3) + (2 + 3) = -3√3 + 5So, third term: (-3√3 + 5)t / 2Putting it all together:t²*(4 - 2√3) - t*(3 - √3)/2 + t*(-3√3 + 5)/2 = 0Combine the linear terms:[ - (3 - √3)/2 + (-3√3 + 5)/2 ] t = [ (-3 + √3 - 3√3 + 5) / 2 ] t = [ (2 - 2√3) / 2 ] t = (1 - √3) tSo, equation becomes:t²*(4 - 2√3) + (1 - √3) t = 0Factor out t:t [ t*(4 - 2√3) + (1 - √3) ] = 0Solutions: t = 0 (which is point A) and t = -(1 - √3)/(4 - 2√3)Simplify t:Multiply numerator and denominator by (4 + 2√3):t = -(1 - √3)(4 + 2√3) / [ (4 - 2√3)(4 + 2√3) ] = -(4 + 2√3 - 4√3 - 6) / (16 - 12) = -( -2 - 2√3 ) / 4 = (2 + 2√3)/4 = (1 + √3)/2So, t = (1 + √3)/2Therefore, coordinates of D:x = t*(3 - √3)/2 = (1 + √3)/2 * (3 - √3)/2 = [ (1)(3) + (1)(-√3) + √3*3 + √3*(-√3) ] / 4 = [ 3 - √3 + 3√3 - 3 ] / 4 = (2√3)/4 = √3/2Similarly, y = t*(√3 - 1)/2 = (1 + √3)/2 * (√3 - 1)/2 = [ (1)(√3) + (1)(-1) + √3*√3 + √3*(-1) ] / 4 = [ √3 - 1 + 3 - √3 ] / 4 = (2)/4 = 1/2So, D is at (√3/2, 1/2)Therefore, reflection D' is the reflection of A over D, so D' = (2*√3/2, 2*1/2) = (√3, 1)Similarly, reflection C' is the reflection of A over C, which is (2*1, 2*0) = (2, 0)So, C' is at (2,0)Now, moving on to point P, which is where the tangent at A meets BC.The tangent at A to ω. The tangent at a point on a circle is perpendicular to the radius at that point. Since A is at (0,0), the radius OA is along the x-axis (since O is the center of ω). Wait, actually, we need to find the center O of ω to determine the tangent.Wait, earlier, we found the equation of ω: x² + y² - x + (-2 + √3)y = 0The general form is x² + y² + Dx + Ey + F = 0, so center is at (-D/2, -E/2) = (1/2, (2 - √3)/2 )So, center O is at (1/2, (2 - √3)/2 )Therefore, the radius OA is from O(1/2, (2 - √3)/2 ) to A(0,0). The slope of OA is [ ( (2 - √3)/2 - 0 ) / (1/2 - 0) ] = (2 - √3)/1 = 2 - √3Therefore, the tangent at A is perpendicular to OA, so its slope is -1/(2 - √3) = -(2 + √3)/ ( (2 - √3)(2 + √3) ) = -(2 + √3)/ (4 - 3) = -(2 + √3)/1 = -2 - √3So, the tangent at A has slope -2 - √3 and passes through A(0,0), so its equation is y = (-2 - √3)xNow, find point P where this tangent meets BC.First, find equation of BC.Points B( (√3 - 1)/2 , (3 - √3)/2 ) and C(1,0)Slope of BC:m = [ 0 - (3 - √3)/2 ] / [ 1 - (√3 - 1)/2 ] = [ (-3 + √3)/2 ] / [ (2 - √3 + 1)/2 ] = [ (-3 + √3)/2 ] / [ (3 - √3)/2 ] = (-3 + √3)/(3 - √3)Multiply numerator and denominator by (3 + √3):[ (-3 + √3)(3 + √3) ] / [ (3 - √3)(3 + √3) ] = [ -9 - 3√3 + 3√3 + 3 ] / (9 - 3) = [ -6 ] / 6 = -1So, slope of BC is -1Equation of BC: passing through C(1,0):y - 0 = -1(x - 1) => y = -x + 1So, equation of BC is y = -x + 1Now, tangent at A is y = (-2 - √3)xFind intersection P:Set (-2 - √3)x = -x + 1Bring all terms to left:(-2 - √3)x + x - 1 = 0 => (-1 - √3)x - 1 = 0 => x = -1 / (1 + √3 )Rationalize denominator:x = -1 / (1 + √3 ) * (1 - √3)/(1 - √3) = (-1 + √3)/(1 - 3) = (-1 + √3)/(-2) = (1 - √3)/2Thus, x = (1 - √3)/2Then, y = (-2 - √3)x = (-2 - √3)(1 - √3)/2Compute:(-2)(1) + (-2)(-√3) + (-√3)(1) + (-√3)(-√3) = -2 + 2√3 - √3 + 3 = ( -2 + 3 ) + (2√3 - √3 ) = 1 + √3Thus, y = (1 + √3)/2Therefore, point P is at ( (1 - √3)/2 , (1 + √3)/2 )Now, we need to find the circumcircle of APD'Points A(0,0), P( (1 - √3)/2 , (1 + √3)/2 ), D'(√3, 1)Find the equation of the circumcircle passing through these three points.Let me denote the general equation as x² + y² + Gx + Hy + K = 0Plugging in A(0,0):0 + 0 + 0 + 0 + K = 0 => K = 0So, equation becomes x² + y² + Gx + Hy = 0Plugging in P( (1 - √3)/2 , (1 + √3)/2 ):[ (1 - √3)² / 4 + (1 + √3)² / 4 ] + G*(1 - √3)/2 + H*(1 + √3)/2 = 0Compute each term:(1 - 2√3 + 3)/4 + (1 + 2√3 + 3)/4 = (4 - 2√3)/4 + (4 + 2√3)/4 = (8)/4 = 2So, first part is 2Second part: G*(1 - √3)/2 + H*(1 + √3)/2Thus, equation becomes:2 + G*(1 - √3)/2 + H*(1 + √3)/2 = 0Multiply everything by 2:4 + G*(1 - √3) + H*(1 + √3) = 0Equation (1): G*(1 - √3) + H*(1 + √3) = -4Now, plug in D'(√3, 1):(√3)² + 1² + G*√3 + H*1 = 0 => 3 + 1 + G√3 + H = 0 => 4 + G√3 + H = 0Equation (2): G√3 + H = -4Now, we have two equations:1) G*(1 - √3) + H*(1 + √3) = -42) G√3 + H = -4Let me solve equation (2) for H:H = -4 - G√3Plug into equation (1):G*(1 - √3) + (-4 - G√3)*(1 + √3) = -4Expand:G*(1 - √3) -4*(1 + √3) - G√3*(1 + √3) = -4Compute each term:G*(1 - √3 - √3 - 3) -4*(1 + √3) = -4Simplify inside G:1 - √3 - √3 - 3 = -2 - 2√3So,G*(-2 - 2√3) -4 -4√3 = -4Bring constants to right:G*(-2 - 2√3) = -4 + 4 + 4√3 = 4√3Thus,G = (4√3) / (-2 - 2√3) = (4√3) / [ -2(1 + √3) ] = (-2√3)/(1 + √3)Rationalize denominator:Multiply numerator and denominator by (1 - √3):G = (-2√3)(1 - √3) / (1 - 3) = (-2√3 + 6)/(-2) = (2√3 - 6)/2 = √3 - 3So, G = √3 - 3Then, from equation (2):H = -4 - G√3 = -4 - (√3 - 3)√3 = -4 - (3 - 3√3) = -4 -3 + 3√3 = -7 + 3√3Thus, equation of circumcircle of APD' is:x² + y² + (√3 - 3)x + (-7 + 3√3)y = 0Now, find point F where this circle intersects AC again.AC is the x-axis from (0,0) to (1,0). So, y = 0.Plug y = 0 into the circle equation:x² + 0 + (√3 - 3)x + 0 = 0 => x² + (√3 - 3)x = 0Factor:x(x + √3 - 3) = 0Solutions: x = 0 (point A) and x = 3 - √3Thus, point F is at (3 - √3, 0)So, F is at (3 - √3, 0)Now, we have points C'(2,0), F(3 - √3, 0), and E( (3 - √3)/2 , (√3 - 1)/2 )We need to prove that the circumcircle of C'FE is tangent to BC at E.First, let's find the equation of the circumcircle of C'FE.Points C'(2,0), F(3 - √3, 0), E( (3 - √3)/2 , (√3 - 1)/2 )Let me denote the general equation as x² + y² + Mx + Ny + P = 0Plugging in C'(2,0):4 + 0 + 2M + 0 + P = 0 => 2M + P = -4 -- Equation (3)Plugging in F(3 - √3, 0):(3 - √3)² + 0 + M*(3 - √3) + 0 + P = 0Compute (3 - √3)² = 9 - 6√3 + 3 = 12 - 6√3So, 12 - 6√3 + M*(3 - √3) + P = 0 -- Equation (4)Plugging in E( (3 - √3)/2 , (√3 - 1)/2 ):[ (3 - √3)/2 ]² + [ (√3 - 1)/2 ]² + M*(3 - √3)/2 + N*(√3 - 1)/2 + P = 0Compute each term:(3 - √3)² /4 = (9 - 6√3 + 3)/4 = (12 - 6√3)/4 = (6 - 3√3)/2(√3 - 1)² /4 = (3 - 2√3 + 1)/4 = (4 - 2√3)/4 = (2 - √3)/2So, sum of first two terms: (6 - 3√3)/2 + (2 - √3)/2 = (8 - 4√3)/2 = 4 - 2√3Then, M*(3 - √3)/2 + N*(√3 - 1)/2 + P = 0Thus, equation becomes:4 - 2√3 + M*(3 - √3)/2 + N*(√3 - 1)/2 + P = 0 -- Equation (5)Now, we have three equations:3) 2M + P = -44) 12 - 6√3 + M*(3 - √3) + P = 05) 4 - 2√3 + M*(3 - √3)/2 + N*(√3 - 1)/2 + P = 0Let me solve equations 3 and 4 first.From equation 3: P = -4 - 2MPlug into equation 4:12 - 6√3 + M*(3 - √3) + (-4 - 2M) = 0Simplify:12 - 6√3 + 3M - M√3 -4 -2M = 0Combine like terms:(12 - 4) + (-6√3) + (3M - 2M) - M√3 = 0 => 8 - 6√3 + M - M√3 = 0Factor M:M(1 - √3) + 8 - 6√3 = 0Solve for M:M(1 - √3) = -8 + 6√3M = (-8 + 6√3)/(1 - √3)Multiply numerator and denominator by (1 + √3):M = [ (-8 + 6√3)(1 + √3) ] / (1 - 3) = [ (-8 -8√3 +6√3 + 18) ] / (-2) = [ (10 - 2√3) ] / (-2) = -5 + √3So, M = -5 + √3Then, from equation 3: P = -4 - 2M = -4 - 2*(-5 + √3) = -4 +10 - 2√3 = 6 - 2√3Now, plug M and P into equation 5:4 - 2√3 + (-5 + √3)*(3 - √3)/2 + N*(√3 - 1)/2 + (6 - 2√3) = 0Simplify term by term:First term: 4 - 2√3Second term: (-5 + √3)(3 - √3)/2Compute (-5 + √3)(3 - √3):= -15 +5√3 +3√3 - (√3)^2 = -15 +8√3 -3 = -18 +8√3So, second term: (-18 +8√3)/2 = -9 +4√3Third term: N*(√3 -1)/2Fourth term: 6 - 2√3Putting it all together:(4 - 2√3) + (-9 +4√3) + N*(√3 -1)/2 + (6 - 2√3) = 0Combine like terms:4 -9 +6 + (-2√3 +4√3 -2√3) + N*(√3 -1)/2 = 0Simplify:1 + 0√3 + N*(√3 -1)/2 = 0Thus,N*(√3 -1)/2 = -1Multiply both sides by 2:N*(√3 -1) = -2Solve for N:N = -2 / (√3 -1 ) = -2(√3 +1 ) / ( (√3 -1)(√3 +1 ) ) = -2(√3 +1 ) / (3 -1 ) = -2(√3 +1 ) / 2 = -(√3 +1 )So, N = -√3 -1Thus, the equation of the circumcircle of C'FE is:x² + y² + (-5 + √3)x + (-√3 -1)y + (6 - 2√3) = 0Now, to prove that this circle is tangent to BC at E, we need to show that the circle touches BC at E and that the tangent at E is the same as BC.Alternatively, we can compute the power of E with respect to the circle and show that it's zero, but since E is on the circle, that's already satisfied. To show tangency, we need to show that the tangent at E to the circle is the same as BC.The tangent at E to the circle is perpendicular to the radius at E. The radius is the line from the center of the circle to E.First, find the center of the circumcircle of C'FE.The general equation is x² + y² + (-5 + √3)x + (-√3 -1)y + (6 - 2√3) = 0Center is at ( -M/2, -N/2 ) = ( (5 - √3)/2 , (√3 +1)/2 )So, center O' is at ( (5 - √3)/2 , (√3 +1)/2 )Now, compute the slope of the radius O'E.Coordinates of E: ( (3 - √3)/2 , (√3 -1)/2 )Coordinates of O': ( (5 - √3)/2 , (√3 +1)/2 )Slope of O'E:[ ( (√3 +1)/2 - (√3 -1)/2 ) / ( (5 - √3)/2 - (3 - √3)/2 ) ] = [ ( (√3 +1 - √3 +1 ) / 2 ) / ( (5 - √3 -3 + √3 ) / 2 ) ] = [ (2)/2 ) / (2/2 ) ] = (1)/(1) = 1So, slope of O'E is 1Therefore, the tangent at E is perpendicular to O'E, so its slope is -1But the equation of BC is y = -x +1, which has slope -1Thus, the tangent at E to the circumcircle of C'FE is the same as BC, meaning the circle is tangent to BC at E.Hence, proved.

💡Okay, so I need to figure out how many children were on the playground at recess. The problem says there were 27 boys and 35 girls. Hmm, okay, so I guess I need to add those two numbers together to get the total number of children. Let me think about how to do that step by step.First, I'll write down the number of boys: 27. Then, I'll write down the number of girls: 35. Now, I need to add these two numbers. I can do this by adding the tens and the ones separately. So, 20 plus 30 is 50, and 7 plus 5 is 12. Then, I add those two results together: 50 plus 12 equals 62. So, there are 62 children in total on the playground.Wait, let me double-check that to make sure I didn't make a mistake. If I add 27 and 35 directly, 27 plus 35 is indeed 62. Yeah, that seems right. So, the total number of children on the playground at recess is 62.

💡Alright, so I have this problem here about vector spaces and bases. It's a multiple-choice question with options A, B, C, and D. The question states that {a, b, c} is a basis for the space, and I need to determine which of the options are correct. Let me try to understand each option one by one.Starting with **Option A**: It says that the vectors a + b, b + c, and c + a can always form a basis of the space. Hmm, okay. So, if {a, b, c} is a basis, that means they are linearly independent and span the space. Now, if I take combinations of these vectors like a + b, b + c, and c + a, are these new vectors also going to be linearly independent? I think I need to check if these new vectors are linearly independent. If they are, then they can form a basis.To check linear independence, I can set up an equation like x(a + b) + y(b + c) + z(c + a) = 0 and see if the only solution is x = y = z = 0. Let's expand this:x(a + b) + y(b + c) + z(c + a) = xa + xb + yb + yc + zc + za = (x + z)a + (x + y)b + (y + z)c = 0.Since {a, b, c} are linearly independent, the coefficients must all be zero:x + z = 0,x + y = 0,y + z = 0.This gives us a system of equations:1. x + z = 02. x + y = 03. y + z = 0Let me solve this system. From equation 1, z = -x. From equation 2, y = -x. Plugging y = -x and z = -x into equation 3: (-x) + (-x) = 0 => -2x = 0 => x = 0. Then y = -x = 0 and z = -x = 0. So the only solution is x = y = z = 0, which means the vectors a + b, b + c, and c + a are linearly independent. Therefore, they can form a basis. So, Option A seems correct.Moving on to **Option B**: It states that if a is perpendicular to b, and b is perpendicular to c, then a is perpendicular to c. Hmm, okay. So, in vector terms, if a · b = 0 and b · c = 0, does that imply a · c = 0? I think this is about orthogonality. Let me think of an example. Suppose in 3D space, let’s say a is along the x-axis, b is along the y-axis, and c is along the z-axis. Then, a is perpendicular to b, and b is perpendicular to c. But a is not necessarily perpendicular to c; in this case, a and c are also perpendicular because they are along different axes. Wait, in this case, a and c are perpendicular. But is this always true?Wait, maybe not. Let me think of another example. Suppose a is along the x-axis, b is along the y-axis, and c is in the x-y plane but not along the x or y-axis. So, a · b = 0, and b · c = 0 only if c is along the x-axis, but if c has a component in the y-direction, then b · c wouldn't be zero. Hmm, maybe my example isn't correct.Wait, let's think algebraically. If a · b = 0 and b · c = 0, does that mean a · c = 0? I don't think so. For example, in 3D, let’s take a = (1, 0, 0), b = (0, 1, 0), and c = (0, 0, 1). Then, a · b = 0, b · c = 0, and a · c = 0. So, in this case, it's true. But what if c is not along the z-axis? Suppose c = (1, 0, 1). Then, a · b = 0, b · c = 0 (since b is (0,1,0) and c is (1,0,1), their dot product is 0*1 + 1*0 + 0*1 = 0). But a · c = 1*1 + 0*0 + 0*1 = 1, which is not zero. So, in this case, a and c are not perpendicular. Therefore, Option B is not necessarily true. So, Option B is incorrect.Now, **Option C**: It says that for any vector p in space, there always exist ordered real numbers (x, y, z) such that p = x a + y b + z c. Well, this is essentially the definition of a basis. If {a, b, c} is a basis, then by definition, any vector in the space can be expressed uniquely as a linear combination of a, b, and c. So, this should always be true. Therefore, Option C is correct.Finally, **Option D**: It states that there exist ordered real numbers such that c = x a + y b. Hmm, so this is saying that c can be expressed as a linear combination of a and b. But wait, if {a, b, c} is a basis, then they are linearly independent. That means c cannot be expressed as a linear combination of a and b alone, otherwise, the set {a, b, c} would be linearly dependent, which contradicts the fact that they form a basis. Therefore, Option D is incorrect.So, summarizing:- Option A: Correct- Option B: Incorrect- Option C: Correct- Option D: IncorrectTherefore, the correct options are A and C.**Final Answer**The correct options are boxed{A} and boxed{C}.

💡Okay, so I have this problem here about a semi-circular iron sheet being used to form the lateral surface of a cone. The base radius of the cone is given as 4 cm, and I need to find the length of the generatrix, which I think is another word for the slant height of the cone. The options are A: 4 cm, B: 8 cm, C: 12 cm, and D: 16 cm.Alright, let me start by recalling some basic concepts about cones and their lateral surfaces. I remember that when you form a cone from a flat sheet, like a semi-circle in this case, the circumference of the base of the cone is related to the length of the arc of the semi-circle. So, the semi-circular iron sheet has a certain radius, let's call it 'r'. The length of the arc of a semi-circle is half the circumference of a full circle, which would be π times the diameter, or π times 2r, but since it's a semi-circle, it's just πr. Now, this arc length becomes the circumference of the base of the cone. The base of the cone is a circle with a radius of 4 cm, so its circumference should be 2π times the radius, which is 2π*4 cm, or 8π cm.Wait, hold on. If the arc length of the semi-circle is πr and that equals the circumference of the base of the cone, which is 8π cm, then I can set up the equation πr = 8π. If I divide both sides by π, that cancels out, and I get r = 8 cm. So, the radius of the semi-circular sheet is 8 cm. But the question is asking for the length of the generatrix of the cone. I think the generatrix is the slant height, which is the distance from the base to the tip of the cone along its surface. In a cone, the slant height is related to the radius of the base and the height of the cone by the Pythagorean theorem: slant height squared equals radius squared plus height squared.But wait, do I know the height of the cone? I don't think so. Maybe I don't need it. Let me think again.When you form the cone from the semi-circular sheet, the radius of the semi-circle becomes the slant height of the cone. Because when you roll up the semi-circle into a cone shape, the radius of the original semi-circle becomes the slant height. So, if the radius of the semi-circle is 8 cm, then the slant height of the cone is also 8 cm.Let me double-check that. The circumference of the base of the cone is 8π cm, and that's equal to the length of the arc of the semi-circle, which is πr. So, solving πr = 8π gives r = 8 cm. Since the radius of the semi-circle becomes the slant height of the cone, the generatrix is 8 cm.Looking back at the options, B is 8 cm, so that should be the answer.But just to make sure I didn't make any mistakes, let me visualize it. Imagine taking a semi-circle with radius 8 cm and rolling it up to form a cone. The arc of the semi-circle becomes the circumference of the base of the cone. Since the arc length is π*8 = 8π, and the base circumference is 2π*4 = 8π, that matches up perfectly. So, the slant height, which is the original radius of the semi-circle, is indeed 8 cm.I think that makes sense. So, the length of the generatrix is 8 cm, which is option B.