Olympiad Maths Prep

Track / Stage 6 / 37 of 400 #1037 of 2000

Problem 1037

National olympiad, first round
Geometry Difficulty 6.0 Prove it

313. Consider a circle in which a regular (2n+1)(2 n+1)-gon A1A2A2n+1A_{1} A_{2} \ldots A_{2 n+1} is inscribed. Let AA be an arbitrary point on the arc A1A2n+1A_{1} A_{2 n+1}.
a) Prove that the sum of the distances from AA to the vertices with even indices is equal to the sum of the distances from AA to the vertices with odd indices.

b) Construct equal circles that are tangent to the given circle at points A1,A2,,A2n+1A_{1}, A_{2}, \ldots, A_{2 n+1}. Prove that the sum of the tangents drawn from AA to the circles tangent to the given circle at the vertices with even indices is equal to the sum of the tangents drawn to the circles tangent to the given circle at the vertices with odd indices.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

313. a) Let AA be an arbitrary point on the circle ( AA is on the arc A1A2n+1A_{1} A_{2 n+1} ). Denote the side of the polygon by aa, and the length of the diagonal connecting vertices through one, by bb. By Ptolemy's theorem (problem II.237) for the quadrilateral AAkAk+1Ak+2A A_{k} A_{k+1} A_{k+2} we have: AAka+AAk+2a=AAk+1b(k=1,2,,2n1)\left|A A_{k}\right| a + \left|A A_{k+2}\right| a = \left|A A_{k+1}\right| b (k=1,2, \ldots, 2 n-1). Similar relations can be written for the quadrilaterals A2nA2n+1AA1A_{2 n} A_{2 n+1} A A_{1} and A2n+1AA1A2:A_{2 n+1} A A_{1} A_{2}:

AA1a+AA2n+1b=AA2naAA2n+1a+AA1b=AA2a \begin{gathered} \left|A A_{1}\right| a + \left|A A_{2 n+1}\right| b = \left|A A_{2 n}\right| a \\ \left|A A_{2 n+1}\right| a + \left|A A_{1}\right| b = \left|A A_{2}\right| a \end{gathered}

By adding all these equations, leaving the vertices with even numbers on the right and the odd ones on the left, we obtain the required statement.
b) Our statement follows from part a) and the result of problem I. 206. (A similar formula can be obtained in the case of internal tangency.)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.