Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Russia

A convex heptagon is given. One calculates the cosines of four arbitrary its angles, the sines of three remaining angles, and then finds the sum of these 7 numbers. It happens that this sum is independent of the choice of the four angles. Prove that this heptagon has four equal angles.

Solution

Consider one of the sums described in the problem. Now, swap the arguments of one sine and one cosine (call these arguments α\alpha and β\beta, respectively); the sum changes by

(sin⁡β+cos⁡α)−(sin⁡α+cos⁡β)=2(sin⁡(β−π/4)−sin⁡(α−π/4)).(\sin \beta + \cos \alpha) - (\sin \alpha + \cos \beta) = \sqrt{2}(\sin(\beta - \pi/4) - \sin(\alpha - \pi/4)).

Since, by the condition, the value of the sum does not change, we get that

sin⁡(α−π/4)=sin⁡(β−π/4).\sin(\alpha - \pi/4) = \sin(\beta - \pi/4).

Since α,β∈(0,π)\alpha, \beta \in (0, \pi), this can happen only if α−π/4=β−π/4\alpha - \pi/4 = \beta - \pi/4 or α−π/4=π−(β−π/4)\alpha - \pi/4 = \pi - (\beta - \pi/4), that is, if β=α\beta = \alpha or β=3π/2−α\beta = 3\pi/2 - \alpha.

Thus, if α\alpha is an arbitrary angle of the heptagon, then each of its other angles is either equal to α\alpha or 3π/2−α3\pi/2 - \alpha. Therefore, the angles of the heptagon take no more than two distinct values, so 4 of them must be equal.

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