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?

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

Diagram

[asy] size(15cm); Label f;  f.p=fontsize(6);  xaxis(-8,8,Ticks(f, 2.0));  yaxis(-8,8,Ticks(f, 2.0));  real f(real x)  {  return 1/x;  }  draw(graph(f,-8,-0.125)); draw(graph(f,0.125,8)); [/asy]

Solution 1

Without loss of generality, 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

Without loss of generality, let the centroid of $\triangle ABC$ be $G = (-1,-1)$. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$. Then, point $B$ must be the reflection of $C$ across the line $y=x$, so let $B = \left(a,\frac{1}{a}\right)$ and $C=\left(\frac{1}{a},a\right)$, 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

Without loss of generality, 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 $\boxed{\textbf{(C) }108}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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

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