Difference between revisions of "1988 AHSME Problems/Problem 27"
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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)); | 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); | draw(unitcircle); | ||
Line 14: | Line 13: | ||
label("$C$",C,dir(C)); | label("$C$",C,dir(C)); | ||
label("$D$",D,dir(D)); | label("$D$",D,dir(D)); | ||
− | label("$O$",O, | + | label("$O$",O,dir(45)); |
</asy> | </asy> | ||
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==Solution== | ==Solution== | ||
− | + | 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>. | |
+ | |||
+ | 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. | ||
+ | |||
+ | 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, , and is tangent to the circle with center and diameter . In which one of the following cases is the area of an integer?
Solution
Let and be the intersections of lines and with the circle. One can prove that is a rectangle, so .
In order for the area of trapezoid to be an integer, the expression must be an integer, so must be rational.
By Power of a Point, , so must be a perfect square. Among the choices, the only one where is a perfect square is
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.