# 1967 AHSME Problems/Problem 40

## 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 1

$[asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label("A",(0,10),N); label("B",(-9.5,-5.2),N); label("C",(9.5,-5.2),N); dot((-3,0)); label("P",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5)); dot((-9,7.5)); label("P'",(-9.2,7.5),N); draw((-9,7.5)--(0,10)); draw((-9,7.5)--(-8.66,-5)); draw((-9,7.5)--(-3,0)); [/asy]$

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

Rotate $\triangle APC$ $60^{\circ}$ 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 $\triangle APC$ (remembering $\angle APC = \angle AP'B$): \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*}

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

## Solution 2 (Magic Formula)

Fun formula: Given a point whose distances from the vertices of an equilateral triangle are $a$, $b$, and $c$, the side length of the triangle is:

$$s=\sqrt{\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)}$$

Given that the area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$, the answer is:

\begin{align*} [ABC] &= \frac{\sqrt{3}}{4}\cdot\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{6\cdot 16(3^2 4^2+4^2 5^2+5^2 3^2)-3\cdot16(3^4+4^4+5^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96(144+400+225)-48(81+256+625)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-48\cdot962}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-96\cdot481}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot288}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot96\cdot3}\right)\\ &= 25\sqrt{3}\pm36 \approx \{6.5, \text{or } 78.5\} \end{align*}

$6.5$ is not a choice, therefore the answer is $\boxed{\textbf{(D) }79}$.

(Note that the $6.5$ answer is actually the solution for when point $P$ is exterior to $\triangle ABC$.)

~proloto

## Solution 3

Rotate $P$ and $B$ $60^{\circ}$ CCW around $A$, becoming $X$ and $C$. Rotate $P$ and $C$ $60^{\circ}$ CCW around $B$, becoming $Y$ and $A$. Rotate $P$ and $A$ $60^{\circ}$ CCW around $C$, becoming $Z$ and $B$:

$[asy] import graph; import geometry; size(12cm); pair A, B, C, P, X, Y, Z; // Define the equilateral triangle ABC real a = sqrt(100+48*sqrt(3)); A = (0, 0); B = rotate(60)*A + (a, 0); C = rotate(120)*B + (a, 0); // Define point P using given distances pair[] P_candidates = intersectionpoints(Circle(A,8), Circle(B,6)); for (pair candidate : P_candidates) { if (length(candidate - C) < 10.1 && length(candidate - C) > 9.9) { P = candidate; break; } } // Rotate C and P about A through 60 degrees to get B and X X = rotate(60,A)*P; // Rotate A and P about B through 60 degrees to get C and Y Y = rotate(60,B)*P; // Rotate B and P about C through 60 degrees to get A and Z Z = rotate(60,C)*P; // Draw the triangle and the segments draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); // Connect X, Y, Z to P and to the vertices of the triangle draw(X--P, dashed); draw(Y--P, dashed); draw(Z--P, dashed); draw(X--A, dashed); draw(X--C, dashed); draw(Y--A, dashed); draw(Y--B, dashed); draw(Z--B, dashed); draw(Z--C, dashed); // Label the points label("A", A, SW); label("B", B, SE); label("C", C, N); label("P", P, NNE*2); label("X", X, NW); label("Y", Y, S); label("Z", Z, E); // Add the distances label("8", (A+P)/2, NW); label("6", (B+P)/2, NE); label("10", (C+P)/2, N); // Add right angle marks draw(rightanglemark(C,X,P,15)); draw(rightanglemark(P,B,Z,15)); draw(rightanglemark(A,P,Y,15)); [/asy]$

Notice that since $\triangle AXC\cong\triangle APB$, $\triangle BYA\cong\triangle BPC$, and $\triangle CZB\cong\triangle CPA$, then

$$[AYBZCX]=2\cdot[ABC]$$

Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles:

\begin{align*}[ABC] &= \frac{1}{2}[AYBZCX] = \frac{1}{2}\left(\underbrace{3\cdot\frac12\cdot6\cdot8}_{\text{3 right triangles}}+\underbrace{\frac{\sqrt{3}}{4}\left(6^2+8^2+10^2\right)}_{\text{3 equilateral triangles}}\right)\\ &= \frac{1}{2}\left(72+\frac{\sqrt{3}}{4}\cdot200)\right) = 36+25\sqrt{3}\\ &\approx \boxed{\textbf{(D) }79} \end{align*}

~proloto

## Solution 4(Answer Choices, Approximation)

Let $s$ be the side length of $ABC.$ Notice that $s\le 14$ by the triangle inequality. This means that $$[ABC]\le\dfrac{14^2\sqrt{3}}{2}\approx 84.87.$$ This automatically rules out choices $A, B,$ and $C.$ Now, we will look at if the area is $50$. By the equilateral triangle area formula, $s$ would equal $10\sqrt{\dfrac{2}{\sqrt{3}}}\approx 10.75.$ This is very close to $10.$ If $s=10,$ $\angle APB=90$ and $\angle APC, \angle BPC<90$ by the Pythagorean theorem and Pythagorean inequalities. Thus, $$\angle APB+\angle APC+\angle BPC<270.$$ $\angle APB+\angle APC+\angle BPC$ needs to be $360,$ and it probably cannot increase by more than $90$ by just adding $0.75$ to $s$ (more rigorous proof below) Thus, the only viable answer choice is $\boxed{79}.$

• In fact, for $s<\sqrt{136}\approx 11.66$ ($[ABC]<34\sqrt{3}\approx58.88$) $\angle APC, \angle BPC<90$ still holds. As $\angle APB <180,$ $$\angle APB+\angle APC+\angle BPC<360.$$ Thus, we know for sure that the nearest integer to the area cannot be $50.$

## See also

 1967 AHSC (Problems • Answer Key • Resources) Preceded byProblem 39 Followed byLast Problem 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.