# Difference between revisions of "1957 AHSME Problems/Problem 48"

## Problem

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

$[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair B = (1,0); pair C = dir(120); pair A = dir(240); pair M = dir(90 - 18); draw(Circle(O,1)); draw(A--C--M--B--cycle); draw(B--C); draw(A--M); dot(O); label("A",A,SW); label("B",B,E); label("M",M,NE); label("C",C,NW); label("O",O,SE);[/asy]$

$\textbf{(A)}\ \text{equal to }{BM + CM}\qquad \textbf{(B)}\ \text{less than }{BM + CM}\qquad \\ \textbf{(C)}\ \text{greater than }{BM+CM}\qquad \\ \textbf{(D)}\ \text{equal, less than, or greater than }{BM + CM}\text{, depending upon the position of } {M}\qquad \\ \textbf{(E)}\ \text{none of these}$

## Solution

Since quadrilateral $ABMC$ is inscribed in circle $O$, thus it is a cyclic quadrilateral. By Ptolemy's Theorem, $$AC \cdot MB + MC \cdot AB = MC \cdot AM.$$ Because $\triangle ABC$ is equilateral, we cancel out $AB$, $AC$, and $BC$ to get that $$BM + CM = AM \implies \boxed{\textbf{(A)}}.$$