It is given $5$ numbers $1$, $3$, $5$, $7$, $9$. We get the new $5$ numbers such that we take arbitrary $4$ numbers(out of current $5$ numbers) $a$, $b$, $c$ and $d$ and replace them with $\frac{a+b+c-d}{2}$, $\frac{a+b-c+d}{2}$, $\frac{a-b+c+d}{2}$ and $\frac{-a+b+c+d}{2}$. Can we, with repeated iterations, get numbers: $a)$ $0$, $2$, $4$, $6$ and $8$ $b)$ $3$, $4$, $5$, $6$ and $7$
Problem
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2014
Tags: combinatorics, iterations, Invariants
Kaskade
16.09.2018 18:10
The answer to both parts is no
For part a, simply consider the sum of the five numbers.
We observe that $\frac{a+b+c-d}{2}+\frac{a+b-c+d}{2}+\frac{a-b+c+d}{2}+\frac{-a+b+c+d}{2}=a+b+c+d$ and then the problem is obvious
For part b, simply consider the sum of the squares of the five numbers.
Again, observe that $\left(\frac{a+b+c-d}{2}\right)^2+\left(\frac{a+b-c+d}{2}\right)^2+\left(\frac{a-b+c+d}{2}\right)^2+\left(\frac{-a+b+c+d}{2}\right)^2=a^2+b^2+c^2+d^2$ and again the problem is obvious.
hansu
16.09.2018 18:27
@above Nice solution! But what motivated you to consider the sum of the squares of the terms in the second part?
Kaskade
16.09.2018 18:30
hansu wrote: @above Nice solution! But what motivated you to consider the sum of the squares of the terms in the second part? I was thinking about possible invariants, then noticed that squaring each one will produce three positive cross terms and three negative, so they should all cancel by symmetry.
wizixez
27.12.2024 16:05
Invariant:$\sum x^2_i$