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 = | + | 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 = BC \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> |
+ | |||
+ | == See also == | ||
+ | |||
+ | {{AHSME box|year=1957|num-b=47|num-a=49}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 13:49, 27 July 2024
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.