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

Revision as of 09:37, 16 February 2017

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 the triangle be $(-1,-1)$ , and let it be $I$ . By symmetry, $A = (1,1)$ , 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 $\boxed{\textbf{(C) } 108}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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 ==Solution==
+WLOG, let the centroid of the triangle be <math>(-1,-1)</math>, and let it be <math>I</math>. By symmetry, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

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

Revision as of 09:37, 16 February 2017

Problem 24

Solution

See Also