Difference between revisions of "1967 AHSME Problems/Problem 40"
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Let <math>\angle BP'P = \alpha .</math> Notice that <math>\cos\alpha = \frac{8}{10}=\frac{4}{5},</math> and <math>\sin\alpha = \frac{3}{5}.</math> | Let <math>\angle BP'P = \alpha .</math> Notice that <math>\cos\alpha = \frac{8}{10}=\frac{4}{5},</math> and <math>\sin\alpha = \frac{3}{5}.</math> | ||
− | Applying the Law of Cosines to <math>\triangle | + | Applying the Law of Cosines to <math>\triangle APC</math> (remembering <math>\angle APC = \angle AP'B</math>): |
<cmath> | <cmath> | ||
− | We want to find the area of <math>\triangle ABC</math> | + | We want to find the area of <math>\triangle ABC</math>, which is <cmath>AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{(D) 79}.</cmath> |
~pfalcon | ~pfalcon |
Revision as of 22:36, 15 August 2023
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 (remembering ):
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.