1957 AHSME Problems/Problem 48

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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 \[BM + CM = AM \implies \boxed{\textbf{(A)}}.\]

See also

1957 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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