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

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(→‎Solution 3)
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We are given that the vertex of the hyperbola is the centroid of the triangle. Hence, WLOG, the centroid of the triangle is at <math>(1,1)</math>. Mark the centroid to be point <math>D</math>.
 
We are given that the vertex of the hyperbola is the centroid of the triangle. Hence, WLOG, the centroid of the triangle is at <math>(1,1)</math>. Mark the centroid to be point <math>D</math>.
  
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The length of <math>AD=(\sqrt{(1-(-1)}^2+(1-(-1)}^2)\implies \sqrt{8}\implies 2\sqrt{2}</math>.
+
The length of <math>AD=\sqrt{{1-(-1)}^2+{(1-(-1)}^2} \implies \sqrt{8} \implies 2\sqrt{2}</math>.
  
 
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>
 
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>
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Using the formula for the area of an equilateral triangle...
 
Using the formula for the area of an equilateral triangle...
  
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<math>\frac{2\sqrt{6}^2}{\sqrt{3}}{4}\implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}</math>
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<math>\frac{(2\sqrt{6})^2 \sqrt{3}}{4} \implies \frac{24\sqrt{3}}{4}\implies 6\sqrt{3}</math>
  
 
Hence, the area squared is <math>=({6\sqrt{3}})^2 \implies 108\implies \boxed{C}</math>.
 
Hence, the area squared is <math>=({6\sqrt{3}})^2 \implies 108\implies \boxed{C}</math>.

Revision as of 18:50, 14 June 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 $\triangle ABC$ 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 is $=({6\sqrt{3}})^2 \implies 108\implies \boxed{C}$.

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

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