Competition problem — Combinatorics

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Combinatorics Difficulty 5.5 AIME, harder Find the answer

A merchant uses a faulty scale to measure goods, that is, a scale whose arms are not of equal length. Two customers request goods of equal weight. The merchant places the goods for the first customer in the pan labeled $A$ , and the weights in the pan labeled $B$ . For the second customer, the merchant places the goods in the pan labeled $B$ , and the weights in the other pan. Question: Under this procedure, does the merchant gain, or lose?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

If the arms of the balance are $l_{1}$ and $l_{2}$ , the weights of the goods are $p_{1}$ and $p_{2}$ , and the weight used for measuring the goods is $q$ , then

$p_{1} l_{1}=q l_{2} \text { and } p_{2} l_{2}=q l_{1} \text {, }$

from which

$p_{1}+p_{2}=q\left(\frac{l_{2}}{l_{1}}+\frac{l_{1}}{l_{2}}\right)$

But

$\frac{l_{2}}{l_{1}}+\frac{l_{1}}{l_{2}}=\frac{l_{1}^{2}+l_{2}^{2}}{l_{1} l_{2}}=\frac{\left(l_{1}-l_{2}\right)^{2}+2 l_{1} l_{2}}{l_{1} l_{2}}=2+\frac{\left(l_{1}-l_{2}\right)^{2}}{l_{1} l_{2}}>2$

Thus

$p_{1}+p_{2}>2 q$

so the trader loses. $\Phi$

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