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1961 AMC 12 Problem 38

Attempt Problem 38 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

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38.

Triangle ABCABC is inscribed in a semicircle of radius rr so that its base ABAB coincides with diameter AB.AB. Point CC does not coincide with either AA or B.B. Let s=AC+BC.s=AC+BC. Then, for all permissible positions of C:C:

s2≤8r2s^2\le8r^2

s2=8r2s^2=8r^2

s2≥8r2s^2\ge8r^2

s2≤4r2s^2\le4r^2

s2=4r2s^2=4r^2

Answer: A
Concepts:inscribed angleright triangleCauchy-Schwarz Inequality
Difficulty rating: 1300
Small Hint:

The angle subtended by the diameter is a right angle

Big Hint:

If the legs are pp and q,q, compare (p+q)2(p+q)^2 with 2(p2+q2)2(p^2+q^2)

Solution:

By Thales’ theorem, ABCABC is right at C.C. Put p=ACp=AC and q=BC.q=BC. Then p2+q2=AB2=4r2. p^2+q^2=AB^2=4r^2. Since (p−q)2≥0,(p-q)^2\ge0, we have 2pq≤p2+q2.2pq\le p^2+q^2. Therefore s2=(p+q)2≤2(p2+q2)=8r2. s^2=(p+q)^2\le2(p^2+q^2)=8r^2.

Thus, the correct answer is A.

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Problem 38 in Other Years

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