Difference between revisions of "1967 AHSME Problems/Problem 40"

Latest revision as of 14:48, 11 November 2023

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 $50.$

@@ Line 153: / Line 153: @@
 ~proloto
+==Solution 4(Answer Choices, Approximation)==
+Let <math>s</math> be the side length of <math>ABC.</math> Notice that <math>s\le 14</math> by the triangle inequality. This means that
+<cmath>[ABC]\le\dfrac{14^2\sqrt{3}}{2}\approx 84.87.</cmath> This automatically rules out choices <math>A, B,</math> and <math>C.</math> Now, we will look at if the area is <math>50</math>. By the equilateral triangle area formula, <math>s</math> would equal <math>10\sqrt{\dfrac{2}{\sqrt{3}}}\approx 10.75.</math> This is very close to <math>10.</math> If <math>s=10,</math> <math>\angle APB=90</math> and <math>\angle APC, \angle BPC<90</math> by the Pythagorean theorem and Pythagorean inequalities. Thus, <cmath>\angle APB+\angle APC+\angle BPC<270.</cmath>
+<math>\angle APB+\angle APC+\angle BPC</math> needs to be <math>360,</math> and it probably cannot increase by more than <math>90</math> by just adding <math>0.75</math> to <math>s</math> (more rigorous proof below) Thus, the only viable answer choice is <math>\boxed{79}.</math>
+*In fact, for <math>s<\sqrt{136}\approx 11.66</math> (<math>[ABC]<34\sqrt{3}\approx58.88</math>) <math>\angle APC, \angle BPC<90</math> still holds. As <math>\angle APB <180,</math> <cmath>\angle APB+\angle APC+\angle BPC<360.</cmath> Thus, we know for sure that the nearest integer to the area cannot be <math>50.</math>
 == See also ==

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Last 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

Page

Toolbox

Search