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

(Created page with "== Problem== Let <math>ABC</math> be an equilateral triangle inscribed in circle <math>O</math>. <math>M</math> is a point on arc <math>BC</math>. Lines <math>\overline{AM}...")
 
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Since quadrilateral <math>ABMC</math> is inscribed in circle <math>O</math>, thus it is a cyclic quadrilateral. By Ptolemy's Theorem, <cmath>AC \cdot MB + MC \cdot AB = MC \cdot AM.</cmath> Because <math>\triangle ABC</math> is equilateral, we cancel out <math>AB</math>, <math>AC</math>, and <math>BC</math> to get that <cmath>BM + CM = AM \implies \boxed{\textbf{(A)}}.</cmath>
 
Since quadrilateral <math>ABMC</math> is inscribed in circle <math>O</math>, thus it is a cyclic quadrilateral. By Ptolemy's Theorem, <cmath>AC \cdot MB + MC \cdot AB = MC \cdot AM.</cmath> Because <math>\triangle ABC</math> is equilateral, we cancel out <math>AB</math>, <math>AC</math>, and <math>BC</math> to get that <cmath>BM + CM = AM \implies \boxed{\textbf{(A)}}.</cmath>
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== See also ==
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{{AHSME box|year=1957|num-b=47|num-a=49}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Revision as of 18:18, 16 October 2021

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 (Problems • Answer Key • Resources)
Preceded by
Problem 47 		Followed by
Problem 49
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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