Difference between revisions of "1988 AHSME Problems/Problem 27"
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==Solution== | ==Solution== | ||
| − | Using power of a point, we easily notice that the product of AB and CD have to be a perfect square. \textbf{(D)}\ AB=9, CD=4 | + | Using power of a point, we easily notice that the product of AB and CD have to be a perfect square. \$textbf{(D)}\ AB=9, CD=4$ |
== See also == | == See also == | ||
Revision as of 01:46, 6 February 2016
Problem
In the figure, 
, and 
is tangent to the circle with center 
and diameter 
.
In which one of the following cases is the area of 
an integer?
![[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); 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]](http://latex.artofproblemsolving.com/d/8/d/d8d58321572892427739e31ba1afdfbaaf71f9de.png)
Solution
Using power of a point, we easily notice that the product of AB and CD have to be a perfect square. $textbf{(D)}\ AB=9, CD=4$
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
