The cyclic octagon A B C D E F G H has side lengths, 6 , 6 , 6 , 6 , 7 , 7 , 7 , and 7 in that order. Find the radius r of the circle that circumscribes A B C D E F G H .
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Let the angle opposite to each 6 unit chord be α , and the angle opposite to each 7 unit chord be β . Then 4 ( α + β ) = 3 6 0 ° ⟹ α + β = 9 0 ° .
So, 2 r sin 2 α = 6 , 2 r sin 2 β = 2 r sin ( 4 5 ° − 2 α ) = 7
⟹ 2 r cos 2 α = 6 + 7 2 .
Therefore, 4 r 2 = 6 2 + ( 6 + 7 2 ) 2 = 1 7 0 + 8 4 2
⟹ r = 2 1 1 7 0 + 8 4 2 ≈ 8 . 4 9 6 9 6 9
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Arrange the sides of the octagon to be alternating lengths of 6 and 7 . This won't affect the radius r of the circle. Then due to symmetry the central angle extended by the 6 − 7 − r − r quadrilateral is 9 0 ∘ . Let the central angle extended by side of length 7 be θ . By cosine rule , we have:
⎩ ⎪ ⎨ ⎪ ⎧ 2 r 2 ( 1 − cos θ ) = 7 2 2 r 2 ( 1 − cos ( 9 0 ∘ − θ ) ) = 2 r 2 ( 1 − sin θ ) = 6 2 ⟹ cos θ = 1 − 2 r 2 4 9 ⟹ sin θ = 1 − 2 r 2 3 6 . . . ( 1 ) . . . ( 2 )
( 1 ) 2 + ( 2 ) 2 : ( 1 − 2 r 2 4 9 ) 2 + ( 1 − 2 r 2 3 6 ) 2 4 r 4 4 9 2 + 3 6 2 − r 2 4 9 + 3 6 + 1 4 r 4 3 6 9 7 − r 2 8 5 + 1 4 r 4 − 3 4 0 r 2 + 3 6 9 7 = cos 2 θ + sin 2 θ = 0 = 0 = 0
Solving for r 2 , we have r 2 ≈ { 7 2 . 1 9 8 4 8 4 8 1 1 2 . 8 0 1 5 1 5 1 9 ⟹ r ≈ 8 . 4 9 7 ⟹ r ≈ 3 . 5 7 8 (too small, rejected)