Difference between revisions of "1967 AHSME Problems/Problem 40"
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Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle. | Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle. | ||
− | Rotate <math>\triangle | + | Rotate <math>\triangle APC</math> <math>60^{\circ}</math> about A. Note that <math>\triangle PAC \cong \triangle P'AB</math>, so |
<cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath> | <cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath> | ||
Revision as of 11:15, 3 January 2021
Problem
Located inside equilateral triangle is a point such that , , and . To the nearest integer the area of triangle is:
Solution
Notice that That makes us want to construct a right triangle.
Rotate about A. Note that , so
Therefore, is equilateral, so , which means
Let Notice that and
Applying the Law of Cosines to ,
We want to find the area of ', which is
~pfalcon
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Problem 40 | |
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.