Difference between revisions of "1967 AHSME Problems/Problem 40"
(Created page with "== Problem == Located inside equilateral triangle <math>ABC</math> is a point <math>P</math> such that <math>PA=8</math>, <math>PB=6</math>, and <math>PC=10</math>. To the neare...") |
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== Solution == | == Solution == | ||
<math>\fbox{D}</math> | <math>\fbox{D}</math> | ||
+ | |||
+ | <asy> | ||
+ | draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); | ||
+ | label("$A$",(0,10),N); | ||
+ | label("$B$",(-9.5,-5.2),N); | ||
+ | label("$C$",(9.5,-5.2),N); | ||
+ | |||
+ | dot((-3,0)); | ||
+ | label("$P$",(-3,-2),N); | ||
+ | draw((-3,0)--(0,10)); | ||
+ | draw((-3,0)--(-8.66,-5)); | ||
+ | draw((-3,0)--(8.66,-5)); | ||
+ | |||
+ | dot((-9,7.5)); | ||
+ | label("$P'$",(-9.2,7.5),N); | ||
+ | draw((-9,7.5)--(0,10)); | ||
+ | draw((-9,7.5)--(-8.66,-5)); | ||
+ | draw((-9,7.5)--(-3,0)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle. | ||
+ | |||
+ | Rotate <math>\triangle ABC</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> | ||
+ | |||
+ | Therefore, <math>\triangle APP'</math> is equilateral, so <math>P'P=8</math>, which means <math>\angle P'PB = 90^{\circ}.</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 APB</math>, | ||
+ | <cmath> | ||
+ | |||
+ | 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 | ||
== See also == | == See also == |
Revision as of 15:43, 2 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.