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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...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
 
<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
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<asy>
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draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle);
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label("$A$",(0,10),N);
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label("$B$",(-9.5,-5.2),N);
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label("$C$",(9.5,-5.2),N);
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dot((-3,0));
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label("$P$",(-3,-2),N);
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draw((-3,0)--(0,10));
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draw((-3,0)--(-8.66,-5));
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draw((-3,0)--(8.66,-5));
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dot((-9,7.5));
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label("$P'$",(-9.2,7.5),N);
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draw((-9,7.5)--(0,10));
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draw((-9,7.5)--(-8.66,-5));
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draw((-9,7.5)--(-3,0));
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</asy>
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Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle.
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Rotate <math>\triangle ABC</math> <math>60^{\circ}</math> about A. Note that <math>\triangle PAC \cong \triangle P'AB</math>, so
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<cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath>
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Therefore, <math>\triangle APP'</math> is equilateral, so <math>P'P=8</math>, which means <math>\angle P'PB = 90^{\circ}.</math>
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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>
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Applying the Law of Cosines to <math>\triangle APB</math>,
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<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)\\&= 16-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}</cmath>
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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>
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 +
~pfalcon
  
 
== See also ==
 
== See also ==

Revision as of 15:43, 2 January 2021

Problem

Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$, $PB=6$, and $PC=10$. To the nearest integer the area of triangle $ABC$ is:

$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution

$\fbox{D}$

[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 $6^2+8^2=10^2.$ That makes us want to construct a right triangle.

Rotate $\triangle ABC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$, so \[\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.\]

Therefore, $\triangle APP'$ is equilateral, so $P'P=8$, which means $\angle P'PB = 90^{\circ}.$

Let $\angle BP'P = \alpha .$ Notice that $\cos\alpha = \frac{8}{10}=\frac{4}{5},$ and $\sin\alpha = \frac{3}{5}.$

Applying the Law of Cosines to $\triangle APB$, \begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 16-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}

We want to find the area of $\triangle ABC$', which is \[AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{(D)79}.\]

~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

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