1972 AMC 12 Problem 26

Attempt Problem 26 of the 1972 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 1972 AMC 12 solutions, or check the answer key.

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

26.

In the circle shown, MM is the midpoint of arc CAB⌢,\overset{\frown}{CAB}, and segment MPMP is perpendicular to chord ABAB at P.P. If the measure of chord ACAC is xx and that of segment APAP is x+1,x+1, then segment PBPB has measure equal to:
3x+23x+2
3x+13x+1
2x+32x+3
2x+22x+2
2x+12x+1
Answer: E
Concepts:arcchordrectanglesymmetry
Small hint:
Copy arc CA⌢\overset{\frown}{CA} to an arc MN⌢\overset{\frown}{MN} ending at a new point NN between MM and BB
Big hint:
Drop NQ⊥ABNQ\perp AB; use equal arcs to identify a rectangle and equal horizontal end segments
Solution:
Choose NN on arc MB⌢\overset{\frown}{MB} so that arcs MN⌢\overset{\frown}{MN} and CA⌢\overset{\frown}{CA} are equal, and drop NQ⊥AB.NQ\perp AB. Then MN=AC=x.MN=AC=x. The remaining equal arcs AM⌢\overset{\frown}{AM} and NB⌢\overset{\frown}{NB} give equal corresponding chord projections, so QB=AP=x+1.QB=AP=x+1. Also MNPQMNPQ is a rectangle, hence PQ=MN=x.PQ=MN=x. Therefore PB=PQ+QB=x+(x+1)=2x+1. \begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1. \end{aligned}
Therefore, the correct answer is E.
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