1966 AMC 12 Problem 31

Attempt Problem 31 of the 1966 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 1966 AMC 12 solutions, or check the answer key.

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

31.

Triangle ABCABC is inscribed in a circle with center O′.O'. A circle with center OO is inscribed in triangle ABC.ABC. AOAO is drawn, and extended to intersect the larger circle in D.D. Then we must have:
CD=BD=O′DCD=BD=O'D
AO=CO=ODAO=CO=OD
CD=CO=BDCD=CO=BD
CD=OD=BDCD=OD=BD
O′B=O′C=ODO'B=O'C=OD
Answer: D
Concepts:angle chasingcircumcircle, circumcenter, and circumradiusincircle, incenter, and inradius
Small hint:
Line ADAD bisects ∠A,\angle A, so DD is the midpoint of arc BCBC
Big hint:
Compare angles in triangle CODCOD to show CD=ODCD=OD
Solution:
Because AOAO is an angle bisector, the inscribed angles ∠BAD\angle BAD and ∠CAD\angle CAD are equal. Hence arcs BDBD and CDCD, and therefore chords BDBD and CD,CD, are equal.
Let ∠BAD=α\angle BAD=\alpha and ∠BCO=β.\angle BCO=\beta. Then ∠OCD=α+β.\angle OCD=\alpha+\beta. In triangle AOC,AOC, the exterior angle ∠COD\angle COD also equals α+β.\alpha+\beta. Thus CD=OD,CD=OD, so CD=OD=BD.CD=OD=BD.
Therefore, the correct answer is D.
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