# 1963 AHSME Problems/Problem 32

## Problem

The dimensions of a rectangle $R$ are $a$ and $b$, $a < b$. It is required to obtain a rectangle with dimensions $x$ and $y$, $x < a, y < a$, so that its perimeter is one-third that of $R$, and its area is one-third that of $R$. The number of such (different) rectangles is: $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$

## Solution

Using the perimeter and area formulas, $$2(x+y) = \frac{2}{3}(a+b)$$ $$x+y = \frac{a+b}{3}$$ $$xy = \frac{ab}{3}$$ Dividing the second equation by the last equation results in $$\frac1y + \frac1x = \frac1b + \frac1a$$ Since $x,y < a$, $\tfrac1a < \tfrac1x, \tfrac1y$. Since $a < b$, $\tfrac1b < \tfrac1a$. That means $$\tfrac1x + \tfrac1y > \tfrac1a + \tfrac1a > \tfrac1a + \tfrac1b$$ This is a contradiction, so there are $\boxed{\textbf{(A)}\ 0}$ rectangles that satisfy the conditions.

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