If the area of the maximal regular heptagon which can be inscribed in the unit square is , where and are positive integers, submit .
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Suppose that p = sin 1 4 1 π = cos 7 3 π . If we put z = e 7 3 π i , then z + z − 1 = 2 p and 0 = z 7 + 1 = ( z + 1 ) ( z 6 − z 5 + z 4 − z 3 + z 2 − z + 1 ) = ( z + 1 ) z 3 [ z 3 + z − 3 − z 2 − z − 2 + z + z − 1 − 1 ] = ( z + 1 ) z 3 [ ( z + z − 1 ) 3 − ( z + z − 1 ) 2 − 2 ( z + z − 1 ) − 1 ] = ( z + 1 ) z 3 ( 8 p 3 − 4 p 2 − 4 p + 1 ) so that 8 p 3 − 4 p 2 − 4 p + 1 = 0 . If we now put t = 2 8 1 π , so that p = sin 2 t , then 4 sin t ( cos 7 t + cos t + sin 5 t ) = 4 sin t cos 7 t + 4 sin t sin 5 t + sin 2 t = 2 ( sin 8 t − sin 6 t ) + 2 ( cos 4 t − cos 6 t ) + 2 sin 2 t = 2 ( − sin 6 t + cos 4 t + sin 2 t ) = 2 ( sin 1 0 t − sin 6 t + sin 2 t ) = 2 [ ( 1 6 p 5 − 2 0 p 3 + 5 p ) − ( 3 p − 4 p 3 ) + p ] = 3 2 p 5 − 3 2 p 3 + 6 p = ( 8 p 3 − 4 p 2 − 4 p + 1 ) ( 4 p 2 + 2 p − 1 ) + 1 = 1 using the fact that 1 4 t = 2 1 π , so that sin 8 t = cos 6 t and cos 4 t = sin 1 0 t , and also using the identities sin 3 u ≡ 3 sin u − 4 sin 3 u and sin 5 u ≡ 1 6 sin 5 u − 2 0 sin 3 u + 5 sin u .