Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2017 AMC 10B Problems/Problem 24

Problem 24

The vertices of an equilateral triangle lie on the hyperbola $xy=1$, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution

WLOG, let the centroid of $\triangle ABC$ be $I = (-1,-1)$. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must hit the graph exactly three times. Therefore, $A = (1,1)$, so $AI = BI = CI = 2\sqrt{2}$, so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$, then by Law of Cosines, $AB = 2\sqrt{6}$. Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$, so the square of the area of the triangle is $\boxed{\textbf{(C) } 108}$.

Solution 2

WLOG, let the centroid of $\triangle ABC$ be $G = (-1,-1)$. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let $A = (1,1)$. Then, point $B$ must be the reflection of $C$ across the line $y=x$, so let $B = (a,1/a)$ and $C=(1/a,a)$, where $a <-1$. Because $G$ is the centroid, the average of the $x$-coordinates of the vertices of the triangle is $-1$. So we know that $a + 1/a+ 1 = -3$. Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$. So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$. So $BC=2\sqrt{6}$, and finding the square of the area gives us $\boxed{\textbf{(C) } 108}$.

Solution 3

WLOG, let a vertex $A$ of equilateral triangle $ABC$ be at $(-1,-1)$ on hyperbola $xy=1$.

We are given that the vertex of the hyperbola is the centroid of the triangle. Hence, WLOG, the centroid of the triangle is at $(1,1)$. Mark the centroid to be point $D$.

The length of $AD=\sqrt{(1-(-1))^2+(1-(-1))^2\implies \sqrt{8}\implies 2\sqrt{2}$ (Error compiling LaTeX. Unknown error_msg).

Now, using the information that $AD$ is $\frac{2}{3}$ the height of equilateral triangle $ABC$(centroid), we find that the height of equilateral triangle $ABC$ is $3\sqrt{2}$

Hence, since the height of triangle $ABC=3\sqrt{2}$, its base is $=2sqrt{6}$

Using the formula for the area of an equilateral triangle...

$\frac{2\sqrt{6}}^2}{\sqrt{3}}{4}\implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg)

Hence, the area squares $={6\sqrt{3}}^2 \implies 108 \implies \boxed{\textbf{(C) } 108$ (Error compiling LaTeX. Unknown error_msg)

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&oldid=86058"