Track / Stage 6 / 127 of 400 #1127 of 1964

Problem 1127

National olympiad, first round

Number theory Difficulty 6.2 Prove it

A seller wants to cut a piece of cheese into parts that can be divided into two piles of equal weight. He can cut any piece of cheese in the same ratio $a:(1-a)$ by weight, where $0<a<1$ . Is it true that for any interval of length 0.001 within the interval $(0,1)$ , there exists a value of $a$ such that he can achieve the desired result with a finite number of cuts?

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's call a number $a$ suitable if $01+1024 \cdot 0.001>2$ , we have

$\begin{gathered} \frac{b_{n-1,10}}{b_{n, 10}}=\sqrt[1024]{\frac{b_{n-1}}{b_{n}}}\frac{1}{1.001}>0.999. \end{gathered}$

Since $b_{n-1} / b_{n}>1.5$ , there exists a natural number $N$ such that $b_{N}<0.001^{1024}$ and, consequently, $b_{N, 10}<0.001$ . In the set of suitable numbers $b_{N, 10}<b_{N-1,10}<\ldots<b_{0,10}$ , the first number is less than 0.001, the last is greater than 0.999, and the differences between adjacent numbers are less than 0.001. Therefore, this set has a non-empty intersection with each interval of length 0.001 within the interval $(0,1)$ .

## Answer

Correct.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.