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 1

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 intersect 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, $AB = 2\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$ . 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 $B = (a,\frac{1}{a})$ and $C=(\frac{1}{a},a)$ , 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 the centroid of $\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$ . It is known that the centroid is equidistant from the three vertices of $\triangle ABC$ . Because we have the coordinates of both $A$ and $G$ , we know that the distance from $G$ to any vertice of $\triangle ABC$ is $\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}$ . Therefore, $AG=BG=CG=2\sqrt{2}$ . It follows that from $\triangle ABG$ , where $AG=BG=2\sqrt{2}$ and $\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}$ , $[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ using the formula for the area of a triangle with sine $\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)$ . Because $\triangle ACG$ and $\triangle BCG$ are congruent to $\triangle ABG$ , they also have an area of $2\sqrt{3}$ . Therefore, $[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}$ . Squaring that gives us the answer of $\boxed{\textbf{(C) }108}$ .

Solution 4 (5-second sol)

WLOG let the centroid of the triangle be $(1, 1)$ . By symmetry, one of the vertices is $(-1, -1)$ . The distance between these two points is $2\sqrt2$ , so the height of the triangle is $3\sqrt 2$ , the side length is $2\sqrt6$ , and the area is $6\sqrt3$ , yielding an answer of $\boxed{\textbf{(C) }108}$ . -Stormersyle

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

Page

Toolbox

Search