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.

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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1963 AHSME Problems/Problem 32

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