MathDB

5

Part of 2011 IMC

Problems(2)

IMC 2011 Day 1 Problem 5

Source:

7/30/2011
Let nn be a positive integer and let VV be a (2n−1)(2n-1)-dimensional vector space over the two-element field. Prove that for arbitrary vectors v1,…,v4n−1∈V,v_1,\dots,v_{4n-1} \in V, there exists a sequence 1≤i1<⋯<i2n≤4n−11\leq i_1<\dots<i_{2n}\leq 4n-1 of indices such that vi1+⋯+vi2n=0.v_{i_1}+\dots+v_{i_{2n}}=0.
vectoralgebrapolynomialfunctionabstract algebraanalytic geometryinduction
IMC 2011 Day 2, Problem 5

Source:

8/1/2011
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.
inductiongeometryperpendicular bisectorIMCcollege contests