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Difference between revisions of "1988 AHSME Problems/Problem 27"
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label("$C$",C,dir(C));
 
label("$C$",C,dir(C));
 
label("$D$",D,dir(D));
 
label("$D$",D,dir(D));
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label("$O$",O,NE);
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label("$O$",O,dir(45));
 
</asy>
 
</asy>
  
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==Solution==
 
==Solution==
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Using power of a point, we easily notice that the product of AB and CD has to be a perfect square. <math>\textbf{(D)}\ AB=9, CD=4</math>
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Let <math>E</math> and <math>F</math> be the intersections of lines <math>AB</math> and <math>BC</math> with the circle. One can prove that <math>BCDE</math> is a rectangle, so <math>BE=CD</math>.
 +
 
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In order for the area of trapezoid <math>ABCD</math> to be an integer, the expression <math>\frac{(AB+CD)BC}2=(AB+CD)BF</math> must be an integer, so <math>BF</math> must be rational.
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 +
By Power of a Point, <math>AB\cdot BE=BF^2\implies AB\cdot CD=BF^2</math>, so <math>AB\cdot CD</math> must be a perfect square. Among the choices, the only one where <math>AB\cdot CD</math> is a perfect square is <math>\textbf{(D)}\ AB=9, CD=4</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 21:22, 13 August 2023

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 $\textbf{(D)}\ AB=9, CD=4$

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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