IMO 1970 SL 1

Origin: BEL

Problem

Consider a regular 2n-gon and the n diagonals of it that pass through its center. Let P be a point of the inscribed circle and let a1, a2, . . . , an be the angles in which the diagonals mentioned are visible from the point P. Prove that n  i=1 tan2 ai = 2ncos2 \pi 2n sin4 \pi 2n .

Solution

Denote respectively by R and r the radii of the circumcircle and incircle, by A1, . . . , An, B1, . . . , Bn ,the vertices of the 2n-gon and by O its center. Let P ′ be the point symmetric to P with respect to O. Then AiP ′BiP is a parallelogram, and applying cosine theorem on triangles AiBiP and PP ′Bi yields 4R2 = PA2 i + PB2 i −2PAi \cdot PBi cos ai 4r2 = PB2 i + P ′B2 i −2PBi \cdot P ′Bi cos \anglePBiP ′. Since AiP ′BiP is a parallelogram, we have that P ′Bi = PAi and \anglePBiP ′ = \pi −ai. Subtracting the expression for 4r2 from the one for 4R2 yields 4(R2 −r2) = −4PAi \cdot PBi cos ai = −8S\triangleAiBiP cot ai, hence we conclude that tan2 ai = 4S2 \triangleAiBiP (R2 −r2)2 . (1) Denote by Mi the foot of the perpendicular from P to AiBi and let mi = PMi. Then S\triangleAiBiP = Rmi. Substituting this into (1) and adding up these relations for i = 1, 2, . . ., n, we obtain n  i=1 tan2 ai = 4R2 (R2 −r2)2

n  i=1 m2 i

. Note that all the points Mi lie on a circle with diameter OP and form a regular n-gon. Denote its center by F. We have that m2 i = \parallel−−\to PMi\parallel2 = \parallel−−\to FMi −−−\to FP\parallel2 = \parallel−−\to FMi2\parallel+ \parallel−−\to FP 2\parallel−2⟨−−\to FMi, −−\to FP⟩= r2/2 −2⟨−−\to FMi, −−\to FP⟩. From this it follows that n i=1 m2 i = 2n(r/2)2 −2 n i=1⟨−−\to FMi, −−\to FP⟩= 2n(r/2)2 −2⟨n i=1 −−\to FMi, −−\to FP⟩= 2n(r/2)2, because n i=1 −−\to FMi = −\to0 . Thus n  i=1 tan2 ai = 4R2 (R2 −r2)2 2n r = 2n (r/R)2 (1 −(r/R)2)2 = 2ncos2 \pi 2n sin4 \pi 2n . Remark. For n = 1 there is no regular 2-gon. However, if we think of a 2-gon as a line segment, the statement will remain true.