Difference between revisions of "1988 AHSME Problems/Problem 27"

Revision as of 02:53, 7 October 2019

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$",O,N); [/asy]$

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

Solution

Using power of a point, we easily notice that the product of AB and CD has to be a perfect square. $\textbf{(D)}\ AB=9, CD=4$

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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Difference between revisions of "1988 AHSME Problems/Problem 27"

Revision as of 02:53, 7 October 2019

Problem

Solution

See also

@@ Line 5: / Line 5: @@
 <asy>
-defaultpen(fontsize(10pt)+linewidth(.8pt));
 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);