olymmath

IMC 2011 Problem 10

    Let F=A0A1...AnF=A_0A_1...A_n be a convex polygon in the plane. Define for all 1≤k≤n−11 \leq k \leq n-1 the operation fkf_k which replaces FF with a new polygon fk(F)=A0A1..Ak−1Ak′Ak+1...Anf_k(F)=A_0A_1..A_{k-1}A_k^\prime A_{k+1}...A_n where Ak′A_k^\prime is the symmetric of AkA_k with respect to the perpendicular bisector of Ak−1Ak+1A_{k-1}A_{k+1} . Prove that (f1∘f2∘f3∘...∘fn−1)n(F)=F(f_1\circ f_2 \circ f_3 \circ...\circ f_{n-1})^n(F)=F .

    Dataset text AI-MO AoPS. Quoted with source.

    Official text: International Mathematics Competition for University Students 2011 official paper, page 1.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator imc/2011/p10 · Tag SNWZDY924C · Edit this page on GitHub · Keys: j next, k previous