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).

The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is

the line segment from $A$ to $B$

the line passing through $A$ and $B$

the perpendicular bisector of the line segment from $A$ to $B$

an ellipse having positive area

a parabola

Answer: A

Concepts:triangle inequality

Difficulty rating: 1140

Small Hint:

Apply the triangle inequality to $AP,$ $PB,$ and $AB$

Big Hint:

Recall when equality holds in the triangle inequality

Solution:

The triangle inequality gives $AP+PB\ge AB.$ Equality holds exactly when $A,$ $P,$ $B$ are collinear with $P$ between $A$ and $B,$ including the endpoints. Thus the locus is the segment from $A$ to $B.$

Therefore, the correct answer is A.

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Full Exam

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