R-Cyclocta!

Geometry Level pending

The cyclic octagon A B C D E F G H A B C D E F G H has side lengths, 6 6 , 6 6 , 6 6 , 6 6 , 7 7 , 7 7 , 7 7 , and 7 7 in that order. Find the radius r r of the circle that circumscribes A B C D E F G H A B C D E F G H .


The answer is 8.496.

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2 solutions

Chew-Seong Cheong
May 27, 2020

Arrange the sides of the octagon to be alternating lengths of 6 6 and 7 7 . This won't affect the radius r r of the circle. Then due to symmetry the central angle extended by the 6 7 r r 6-7-r-r quadrilateral is 9 0 90^\circ . Let the central angle extended by side of length 7 7 be θ \theta . By cosine rule , we have:

{ 2 r 2 ( 1 cos θ ) = 7 2 cos θ = 1 49 2 r 2 . . . ( 1 ) 2 r 2 ( 1 cos ( 9 0 θ ) ) = 2 r 2 ( 1 sin θ ) = 6 2 sin θ = 1 36 2 r 2 . . . ( 2 ) \begin{cases} 2r^2 (1-\cos \theta) = 7^2 & \implies \cos \theta = 1 - \dfrac {49}{2r^2} & ...(1) \\ 2r^2 (1-\cos (90^\circ - \theta)) = 2r^2 (1-\sin \theta) = 6^2 & \implies \sin \theta = 1 - \dfrac {36}{2r^2} & ...(2) \end{cases}

( 1 ) 2 + ( 2 ) 2 : ( 1 49 2 r 2 ) 2 + ( 1 36 2 r 2 ) 2 = cos 2 θ + sin 2 θ 4 9 2 + 3 6 2 4 r 4 49 + 36 r 2 + 1 = 0 3697 4 r 4 85 r 2 + 1 = 0 4 r 4 340 r 2 + 3697 = 0 \begin{aligned} (1)^2 + (2)^2: \quad \left(1 - \frac {49}{2r^2}\right)^2 + \left(1 - \frac {36}{2r^2}\right)^2 & = \cos^2 \theta + \sin^2 \theta \\ \frac {49^2 + 36^2}{4r^4} - \frac {49+36}{r^2} + 1 & = 0 \\ \frac {3697}{4r^4} - \frac {85}{r^2} + 1 & = 0 \\ 4r^4 - 340 r^2 + 3697 & = 0 \end{aligned}

Solving for r 2 r^2 , we have r 2 { 72.19848481 r 8.497 12.80151519 r 3.578 (too small, rejected) r^2 \approx \begin{cases} 72.19848481 & \implies r \approx \boxed{8.497} \\ 12.80151519 & \implies r \approx \red{3.578 \small \text{ (too small, rejected)}} \end{cases}

Let the angle opposite to each 6 6 unit chord be α α , and the angle opposite to each 7 7 unit chord be β β . Then 4 ( α + β ) = 360 ° α + β = 90 ° 4(α+β)=360\degree\implies α+β=90\degree .

So, 2 r sin α 2 = 6 , 2 r sin β 2 = 2 r sin ( 45 ° α 2 ) = 7 2r\sin \frac{α}{2}=6, 2r\sin \frac{β}{2}=2r\sin (45\degree-\frac{α}{2})=7

2 r cos α 2 = 6 + 7 2 \implies 2r\cos \frac{α}{2}=6+7\sqrt 2 .

Therefore, 4 r 2 = 6 2 + ( 6 + 7 2 ) 2 = 170 + 84 2 4r^2=6^2+(6+7\sqrt 2)^2=170+84\sqrt 2

r = 1 2 170 + 84 2 8.496969 \implies r=\frac{1}{2}\sqrt {170+84\sqrt 2}\approx \boxed {8.496969}

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