1988 AMC 12 Problem 27

Attempt Problem 27 of the 1988 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

In the figure, AB⊥BC,AB\perp BC, BC⊥CD,BC\perp CD, and BCBC is tangent to the circle with center OO and diameter AD.AD. In which one of the following cases is the area of ABCDABCD an integer?
AB=3,AB=3, CD=1CD=1
AB=5,AB=5, CD=2CD=2
AB=7,AB=7, CD=3CD=3
AB=9,AB=9, CD=4CD=4
AB=11,AB=11, CD=5CD=5
Answer: D
Concepts:circle tangencypower of a pointtrapezoid area
Small hint:
Let the tangent point be MM and use the diameter to form a rectangle inside the trapezoid
Big hint:
Power of point BB relates half of BCBC to ABAB and CDCD
Solution:
The tangent point is the midpoint of BC,BC, and the tangent-secant relation gives (BC2)2=AB⋅CD.(\frac{BC}{2})^2=AB\cdot CD. Thus BC=2AB⋅CD,BC=2\sqrt{AB\cdot CD}, and the trapezoid area is Area=AB+CD2BC=(AB+CD)AB⋅CD. \begin{aligned} \text{Area} &=\frac{AB+CD}{2}BC\\ &=(AB+CD)\sqrt{AB\cdot CD}. \end{aligned} Only AB=9, CD=4AB=9,\ CD=4 makes the product under the radical a square; the area is 13⋅6=78.13\cdot6=78.
Thus the correct answer is D.
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