1964 AMC 12 Problem 33

Attempt Problem 33 of the 1964 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 1964 AMC 12 solutions, or check the answer key.

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

33.

PP is a point interior to rectangle ABCDABCD and such that PA=3PA=3 inches, PD=4PD=4 inches, and PC=5PC=5 inches. Then PB,PB, in inches, equals:
232\sqrt3
323\sqrt2
333\sqrt3
424\sqrt2
22
Answer: B
Concepts:rectangleBritish Flag Theoremdistance formula
Small hint:
Use the rectangle identity PA2+PC2=PB2+PD2PA^2+PC^2=PB^2+PD^2
Big hint:
Substitute the three known distances and solve for PB2PB^2
Solution:
For any point in a rectangle, the British flag theorem gives PA2+PC2=PB2+PD2.PA^2+PC^2=PB^2+PD^2. Thus 9+25=PB2+16,9+25=PB^2+16, so PB2=18PB^2=18 and PB=32.PB=3\sqrt2.
Therefore, the correct answer is B.
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