IMO 1992 SL 8

Show that in the plane there exists a convex polygon of 1992

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IMO 1992 SL 8

Origin: IND

Problem

Show that in the plane there exists a convex polygon of 1992 sides satisfying the following conditions: (i) its side lengths are 1, 2, 3, . . ., 1992 in some order; (ii) the polygon is circumscribable about a circle. Alternative formulation. Does there exist a 1992-gon with side lengths 1, 2, 3, . . ., 1992 circumscribed about a circle? Answer the same question for a 1990-gon.

Solution

For simplicity, we shall write n instead of 1992. Lemma. There exists a tangent n-gon A1A2 . . . An with sides A1A2 = a1, A2A3 = a2, . . . , AnA1 = an if and only if the system x1 + x2 = a1, x2 + x3 = a2, , . . . , xn + x1 = an (1) has a solution (x1, . . . , xn) in positive reals. Proof. Suppose that such an n-gon A1A2 . . . An exists. Let the side AiAi+1 touch the inscribed circle at point Pi (where An+1 = A1). Then x1 = A1Pn = A1P1, x2 = A2P1 = A2P2, . . . , xn = AnPn−1 = AnPn is clearly a positive solution of (1). Now suppose that the system (1) has a positive real solution (x1, . . . , xn). Let us draw a polygonal line A1A2 . . . An+1 touching a circle of radius r at points P1, P2, . . . , Pn respectively such that A1P1 = An+1Pn = x1 and AiPi = AiPi−1 = xi for i = 2, . . . , n. Observe that OA1 = OAn+1 =

x2 1 + r2 and the function f(r) = \angleA1OA2 + \angleA2OA3 + \cdot \cdot \cdot + \angleAnOAn+1 = 2(arctan x1 r + \cdot \cdot \cdot + arctan xn r ) is continuous. Thus A1A2 . . . An+1 is a closed simple polygonal line if and only if f(r) = 360◦. But such an r exists, since f(r) \to0 O P1 P2 Pn−1 Pn A1 A2 A3 An−1 An An+1 when r \to\inftyand f(r) \to\inftywhen r \to0. This proves the second direction of the lemma. For n = 4k, the system (1) is solvable in positive reals if ai = i for i \equiv1, 2 (mod 4), ai = i + 1 for i \equiv3 and ai = i −1 for i \equiv0 (mod 4). Indeed, one solution is given by xi = 1/2 for i \equiv1, xi = 3/2 for i \equiv3 and xi = i−3/2 for i \equiv0, 2 (mod 4). Remark. For n = 4k + 2 there is no such n-gon. In fact, solvability of the system (1) implies a1 + a3 + \cdot \cdot \cdot = a2 + a4 + \cdot \cdot \cdot , while in the case n = 4k + 2 the sum a1 + a2 + \cdot \cdot \cdot + an is odd.