Problem
ALG-B1-M06-P022 An independent sum
Seven numbers from the interval \((0,1)\) are given. For any chosen four of them, their squares are taken, and for the other three the values \(2x-x^2\) are taken. The sum of the resulting seven numbers does not depend on the choice of the four numbers. Prove that among the seven given numbers there are four equal ones.
Compare two sums that differ only by swapping the applied functions for two numbers \(x\) and \(y\).
Take any two given numbers \(x\) and \(y\). Consider two choices of four numbers that differ only in this: in one choice \(x\) is in the square group and \(y\) is in the other group, while in the second choice they are swapped. The other five contributions are the same.
By the condition, the sums are equal, so \(x^2+(2y-y^2)=y^2+(2x-x^2)\). Moving everything to one side gives \(2x^2-2y^2-2x+2y=0\), that is \((x-y)(x+y-1)=0\).
Thus any two numbers in the set are either equal or have sum \(1\). Fix one number \(x\). Then every other number is either \(x\) or \(1-x\). Hence there are at most two distinct values among the seven numbers. By the pigeonhole principle, one of them occurs at least four times.
Source analysis: the main objects are several quantities and two functions applied to each; the first approach is to try to compute the sum itself; the hidden observation is that it is enough to swap two arguments; the key ideas are comparing two choices and the pigeonhole principle. The geometric context is replaced by pure algebra.