1977 AMC 12 Problem 5

Attempt Problem 5 of the 1977 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 1977 AMC 12 solutions, or check the answer key.

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

5.

The set of all points PP such that the sum of the (undirected) distances from PP to two fixed points AA and BB equals the distance between AA and BB is
the line segment from AA to BB
the line passing through AA and BB
the perpendicular bisector of the line segment from AA to BB
an ellipse having positive area
a parabola
Answer: A
Concepts:triangle inequality
Small hint:
Apply the triangle inequality to AP,AP, PB,PB, and ABAB
Big hint:
Recall when equality holds in the triangle inequality
Solution:
The triangle inequality gives AP+PB≥AB.AP+PB\ge AB. Equality holds exactly when A,A, P,P, BB are collinear with PP between AA and B,B, including the endpoints. Thus the locus is the segment from AA to B.B.
Therefore, the correct answer is A.
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