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Problem 1268

AIME late

Combinatorics Difficulty 5.4 Prove it Junior Balkan Mathematical Olympiad · JBMO

Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\frac{a b}{a+b}$ . Continue this procedure until only one segment remains on the board. Prove:

a) the length of the last remaining segment does not depend on the order of the deletions.

b) for every positive integer $n$ , the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$ .

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

Solution:

a) Observe that $\frac{1}{\frac{a b}{a+b}} = \frac{1}{a} + \frac{1}{b}$ . Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \ldots, a_{n}$ , and $c$ is the length of the last remaining segment, then $\frac{1}{c} = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \ldots + \frac{1}{a_{n}}$ , proving a).

b) From a) and the equation $\frac{1}{n} = \frac{1}{2n} + \frac{1}{3n} + \frac{1}{6n}$ it follows that if the lengths of the starting segments are $2n, 3n$ and $6n$ , then the length of the last remaining segment is $n$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.