Question

Let $ABC$ be an equilateral triangle inscribed in circle $O.$ $M$ is a point on arc $BC.$ Lines $AM,$ $BM,$ and $CM$ are drawn. Then $AM$ is:

equal to $BM+CM$

less than $BM+CM$

greater than $BM+CM$

equal to, less than, or greater than $BM+CM,$ depending upon the position of $M$

none of these

Answer: A

Concepts:cyclic quadrilateral Ptolemy’s Theorem equilateral triangle

Difficulty rating: 1990

Small Hint:

Apply Ptolemy’s theorem to cyclic quadrilateral $ABMC$

Big Hint:

Use $AB=BC=CA$ to cancel the common side length

Solution:

In cyclic quadrilateral $ABMC,$ Ptolemy’s theorem gives $\begin{aligned} AM\cdot BC &=AB\cdot CM\\ &\quad+AC\cdot BM. \end{aligned}$ Since $ABC$ is equilateral, $AB=BC=AC.$ Dividing by this common length yields $AM=CM+BM.$

Therefore, the correct answer is A.

1957 AMC 12 Problem 48

Answer: A

Difficulty rating: 1990

Small Hint:

Big Hint:

Solution:

Problem 48 in Other Years