Difference between revisions of "2017 AMC 10B Problems/Problem 24"

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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 $AD=\sqrt{(1-(-1)}^2+(1-(-1))^2\implies \sqrt{8}\implies 2\sqrt{2}$ .

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 $=2\sqrt{6}$

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

$\frac{2\sqrt{6}^2}{\sqrt{3}}{4}\implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}$

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

@@ Line 19: / Line 19: @@
 Now, using the information that <math>AD</math> is <math>\frac{2}{3}</math> the height of equilateral triangle <math>ABC</math>(centroid), we find that the height of equilateral triangle <math>ABC</math> is <math>3\sqrt{2}</math>
-Hence, since the height of triangle <math>ABC=3\sqrt{2}</math>, its base is <math>=2sqrt{6}</math>
+Hence, since the height of triangle <math>ABC=3\sqrt{2}</math>, its base is <math>=2\sqrt{6}</math>
 Using the formula for the area of an equilateral triangle...
-<math>\frac{2\sqrt{6}}^2}{\sqrt{3}}{4}\implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}</math>
+<math>\frac{2\sqrt{6}^2}{\sqrt{3}}{4}\implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}</math>
 Hence, the area squared <math>={6\sqrt{3}}^2 \implies 108\implies \boxed{\textbf{(C) } 108</math>

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
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