Difference between revisions of "1967 AHSME Problems/Problem 40"
m (→Solution) |
m (Fixed a couple typos) |
||
Line 39: | Line 39: | ||
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>\begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}</cmath> | <cmath>\begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}</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 23:36, 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.