Irregularly Inscribing a Circle ... of What Radius?Date: 12/24/2011 at 09:42:14 From: huey Subject: find the radius of a circle given a inscribed hexagon A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle. Find the radius of the circle. No angles are given, and it is a irregular hexagon, so I don't know how to start. Date: 12/26/2011 at 19:38:57 From: Doctor Greenie Subject: Re: find the radius of a circle given a inscribed hexagon Hi, Huey -- I haven't found an analytical solution yet, but a numerical one using a graphing calculator was not too difficult. If we draw the radii to each vertex of the hexagon and the perpendiculars from the center of the circle to the midpoint of each side of the hexagon, then we end up with 12 right triangles with a common vertex at the center of the circle, and all with hypotenuse r, where r is the radius of the circle. Four of these triangles have opposite side length 2/2 = 1, four have opposite side length 7/2 = 3.5, and four have opposite side length 11/2 = 5.5. Then, since the central angles of all these 12 triangles add up to 360 degrees, we have (in degree measure) arcsin(1/r) + arcsin(3.5/r) + arcsin(5.5/r) = 90 A graphing calculator easily determines that the solution to this equation is r = 7, so the radius of the circle is 7. If you need an analytical solution, write back. I'm still playing with possible approaches to an analytical solution, and I will ask other volunteers here if they have any ideas. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ Date: 12/27/2011 at 09:55:53 From: huey Subject: find the radius of a circle given a inscribed hexagon Hi, I do really need analytical solutions. Thank you. Huey Date: 04/11/2012 at 14:55:17 From: Doctor Floor Subject: Re: find the radius of a circle given a inscribed hexagon Hi, Huey, I've had your question on my watchlist. This is probably of no help any more, but still I was eager to find a solution -- and just succeeded. So here we go: We had an inscribed hexagon with side lengths 2, 2, 7, 7, 11 and 11. Note that we may change the order of the sides without changing the radius of the circumcircle. Now if we change into the order 2, 7, 11, 11, 7, 2 we get a line symmetric hexagon. The line of symmetry of course is a diameter. So we may cut the hexagon in half and find two cyclic quadrilaterals with sides 2, 7, 11, and x, where x denotes the diameter of the circle. More precisely, we may consider one of the quadrilaterals. Name the vertices ABCD, such that AB = x, BC = 11, CD = 7 and DA = 2. Of course, triangles ABC and ABD are right, having AB as hypotenuse. Hence we have: AC = sqrt(x^2 - 121) BD = sqrt(x^2 - 4) Now we may apply Ptolemy's theorem: AB*DC + BC*AD = AC*BD From the Dr. Math Archives, see http://mathforum.org/library/drmath/view/54956.html This gives: 7x + 22 = sqrt(x^2 - 121)*sqrt(x^2 - 4) Squaring both sides: 49x^2 + 308x + 484 = x^4 - 125x^2 + 484 0 = x^4 - 174x^2 - 308x 0 = x(x^3 - 174x - 308) We may check for rational solutions of x^3 - 174x - 308 = 0 by the rational root theorem: http://mathforum.org/library/drmath/view/56425.html As 308 = 2*2*7*11, the number of possibilities is limited. We succeed with x = 14. This means that 14 is a possible diameter length for our circle. Now factor: x^3 - 174x - 308 = (x - 14)(x^2 + 14x + 22) Note that the second factor has two negative roots. So x = 14 is THE solution we are looking for. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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