Maths Olympiad Prep

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Problem 1742

IMO Shortlist mid-range; USAMO P2/P5
Combinatorics Difficulty 8.1 Prove it Team Selection Test for IMO 2011 · Turkey · 2011

Let AA and BB be sets with 201122011^2 and 20102010 elements, respectively. Show that there is a function f:A×ABf: A \times A \to B satisfying the condition f(x,y)=f(y,x)f(x,y) = f(y,x) for all (x,y)A×A(x,y) \in A \times A such that for every function g:ABg: A \to B there exists (a1,a2)A×A(a_1, a_2) \in A \times A with g(a1)=f(a1,a2)=g(a2)g(a_1) = f(a_1, a_2) = g(a_2) and a1a2a_1 \neq a_2.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let p=2011p = 2011, A={(i,j):0i,j<p}A = \{(i,j) : 0 \le i, j < p\} and B={k:0k<p1}B = \{k : 0 \le k < p-1\}. We define ff as follows:
f((i1,j1),(i2,j2))={j1j2i1i2(modp)if i1i2 and j1j2i1i21(modp)0otherwise f((i_1, j_1), (i_2, j_2)) = \begin{cases} \dfrac{j_1 - j_2}{i_1 - i_2} \pmod{p} & \text{if } i_1 \neq i_2 \text{ and } \dfrac{j_1 - j_2}{i_1 - i_2} \neq -1 \pmod{p} \\ 0 & \text{otherwise} \end{cases}
In other words, we send a pair of points to the slope of the line passing through them unless they lie on a line with slope \infty or 1-1, in which case we send it to 00.

Let g:ABg: A \to B be a function. Since AB=p2p1>p\dfrac{|A|}{|B|} = \dfrac{p^2}{p-1} > p, there is a value in BB that is taken at least p+1p+1 times by gg. Since there are exactly pp lines with slope equal to this value, there is at least a pair of points lying on the same one of these lines.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.