1971 AMC 12 Problem 5

Attempt Problem 5 of the 1971 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 1971 AMC 12 solutions, or check the answer key.

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

5.

Points A,A, B,B, Q,Q, D,D, and CC lie on the circle shown, and the measures of arcs BQBQ and QDQD are 42∘42^\circ and 38∘,38^\circ, respectively. The sum of the measures of angles PP and QQ is:
80∘80^\circ
62∘62^\circ
40∘40^\circ
46∘46^\circ
None of these
Answer: C
Concepts:circleinscribed anglearcangle chasing
Small hint:
Use the external-secant formula for angle PP
Big hint:
Adding angle QQ cancels the unknown intercepted arc
Solution:
Let mAC⌢=u.m\overset{\frown}{AC}=u. The external-secant formula gives m∠P=12(mBD⌢−u), m\angle P=\frac12\left(m\overset{\frown}{BD}-u\right), while the inscribed-angle theorem gives m∠Q=u2.m\angle Q=\frac{u}{2}. Therefore m∠P+m∠Q=12mBD⌢=12(42∘+38∘)=40∘. \begin{aligned} m\angle P+m\angle Q &=\frac12m\overset{\frown}{BD}\\ &=\frac12(42^\circ+38^\circ)\\ &=40^\circ. \end{aligned}
Therefore, the correct answer is C.
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