1967 AMC 12 Problem 29

Attempt Problem 29 of the 1967 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 1967 AMC 12 solutions, or check the answer key.

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

29.

ABAB is a diameter of a circle. Tangents ADAD and BCBC are drawn so that ACAC and BDBD intersect in a point on the circle. If AD=aAD=a and BC=b,BC=b, a≠b,a\ne b, the diameter of the circle is:
∣a−b∣\lvert a-b\rvert
12(a+b)\dfrac12(a+b)
ab\sqrt{ab}
aba+b\dfrac{ab}{a+b}
12⋅aba+b\dfrac12\cdot\dfrac{ab}{a+b}
Answer: C
Concepts:circletangent linesimilarityinscribed angle
Small hint:
Let the intersection point on the circle be PP; then ∠APB=90∘\angle APB=90^\circ
Big hint:
Use the parallel tangents to compare right triangles ADBADB and BCABCA
Solution:
Let d=ABd=AB be the diameter. Since the intersection of ACAC and BDBD lies on the circle, those two lines are perpendicular by Thales’ theorem. Also the tangents ADAD and BCBC are parallel. The resulting right triangles ADBADB and BCABCA are similar, so da=bd. \frac da=\frac bd. Hence d2=abd^2=ab and d=ab.d=\sqrt{ab}.
Therefore, the correct answer is C.
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