Difference between revisions of "1957 AHSME Problems/Problem 48"
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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}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 19:18, 16 October 2021
Problem
Let be an equilateral triangle inscribed in circle . is a point on arc . Lines , , and are drawn. Then is:
Solution
Since quadrilateral is inscribed in circle , thus it is a cyclic quadrilateral. By Ptolemy's Theorem, Because is equilateral, we cancel out , , and to get that
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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