1988 AHSME Problems/Problem 27

Problem

In the figure, $AB \perp BC, BC \perp CD$ , and $BC$ is tangent to the circle with center $O$ and diameter $AD$ . In which one of the following cases is the area of $ABCD$ an integer?

$[asy] pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O); draw(A--B--C--D--A); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$O$",O,dir(45)); [/asy]$

$\textbf{(A)}\ AB=3, CD=1\qquad \textbf{(B)}\ AB=5, CD=2\qquad \textbf{(C)}\ AB=7, CD=3\qquad\\ \textbf{(D)}\ AB=9, CD=4\qquad \textbf{(E)}\ AB=11, CD=5$

Solution

Let $E$ and $F$ be the intersections of lines $AB$ and $BC$ with the circle. One can prove that $BCDE$ is a rectangle, so $BE=CD$ .

In order for the area of trapezoid $ABCD$ to be an integer, the expression $\frac{(AB+CD)BC}2=(AB+CD)BF$ must be an integer, so $BF$ must be rational.

By Power of a Point, $AB\cdot BE=BF^2\implies AB\cdot CD=BF^2$ , so $AB\cdot CD$ must be a perfect square. Among the choices, the only one where $AB\cdot CD$ is a perfect square is $\textbf{(D)}\ AB=9, CD=4$

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1988 AHSME Problems/Problem 27

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