# 1995 AJHSME Problems/Problem 9

## Problem

Three congruent circles with centers $P$, $Q$, and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$. The area of the rectangle is $[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("P",P,W); label("Q",Q,W); label("R",R,W); [/asy]$ $\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

## Solution 1

If circle $Q$ has diameter $4$, then so do congruent circles $P$ and $R$. Draw a diameter through $P$ parallel to $AD$. The diameter will be congruent to $AD$, and thus $AD = 4$, which is the height of the rectangle.

Draw a horizontal line $PQR$ that extends to the sides of the rectangle. This line is $2$ diameters long, so it has length $4\cdot 2 = 8$. It is parallel and congruent to $AB$, so the width of the rectangle is $8$.

Thus, the area of the rectangle is $lw = 4\cdot 8 = 32$, and the answer is $\boxed{C}$.

## Solution 2

The area of circle $P$ and circle $R$ almost fill the rectangle. Circle $P$ has radius $\frac{4}{2} = 2$, and so does circle $R$. Thus, the sum of their areas is $\pi \cdot 2^2 + \pi \cdot 2^2 = 8\pi$. Since $\pi \approx 3.14$, the area of the circles is just over $24$. The area of the rectangle is greater than the area of the circles. $64$ would be too high, as the two circles appear to take up much more than half the area of the rectangle. But $24$ is too low, as it is less than the area of the two circles. Thus, the only reasonable answer is $\boxed{C}$, which is $32$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 