A Hexagon Inscribed within a Circle
Date: 2/29/96 at 18:3:21 From: Heather Huang Subject: Hexagon in a circle Hello - My name is Heather Huang. I was wondering if you could help me solve a problem. Here goes: A hexagon is inscribed within a circle. Three consecutive sides have a length of 3 and the other three consecutive sides have a length of 5. A chord is drawn within the circle which divides the hexagon into two trapezoids. One of the trapezoids has three sides that measure 3 and the other trapezoid has three sides that measure 5. The length of the chord is m/n. What is the value of m + n? If you could help me solve this problem I would be eternally grateful! Heather
Date: 5/29/96 at 2:50:10 From: Doctor Pete Subject: Hexagon in a circle Hello, Heather - Let 2p be the central angle subtended by a side of length 3, and let 2q be the central angle subtended by a side of length 5. Then we have the following relations: sin(p) = 3/(2R),  sin(3p)=x/(2R),  sin(q) = 5/(2R),  p+q = Pi/3,  where R is the radius of the circumscribed circle, and x is the length we wish to find. Then  gives sin(3p) = sin(p)cos(2p)+sin(2p)cos(p) = sin(p)(cos^2(p)-sin^2(p))+2*sin(p)cos^2(p) = 3*sin(p)cos^2(p)-sin^3(p) = 3*sin(p)(1-sin^2(p))-sin^3(p) = 3*sin(p)-4*sin^3(p) = x/2R, and substituting , we get 9/(2R)-4*27/(2R)^3 = x/2R ==> x = 9-27/R^2. To find R, we substitute  into : sin(q) = sin(Pi/3-p) = sin(Pi/3)cos(p)-sin(p)cos(Pi/3) = (sqrt(3*(1-sin^2(p)))-sin(p))/2 = 5/(2R), and we now substitute  into this to get 5/2R = (sqrt(3*(1-(3/2R)^2))-3/2R)/2 ==> 5 = (sqrt(3*(4R^2-9))-3)/2 ==> 13 = sqrt(3*(4R^2-9)) ==> 169 = 3*(4R^2-9) ==> R^2 = (169/3+9)/4 = 196/12 = 49/3. Therefore x = 9-27/(49/3) = 360/49, and the sum of the numerator and denominator is 409. (Hope that's right!) -Doctor Pete, The Math Forum
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