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Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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==Problem 24==
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==Problem==
 
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
 
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
  
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==Diagram==
 
==Diagram==
 
<asy>
 
<asy>
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size(5cm);
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size(15cm);
 
Label f;  
 
Label f;  
 
f.p=fontsize(6);  
 
f.p=fontsize(6);  
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</asy>
 
</asy>
  
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==Solution 1==
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==Solution 1 (Law of Cosines)==
  
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Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. 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, <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>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</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>.
+
WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. 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, <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 the [[Law of Cosines]], <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</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>.
 +
 
 +
 
 +
Note: We could’ve also noticed that the centroid divides the median into segments of ratio <math>2:1.</math> ~peelybonehead
  
 
==Solution 2==
 
==Solution 2==
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Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.
+
WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.
 +
<math>\newline</math>
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Alternatively, from <math>a + \frac{1}{a} = -4</math>, we obtain <math>a^2 - 2 + \frac{1}{a^2} = (-4)^2 - 4 \implies \left|a - \frac{1}{a} \right| = \sqrt{12}</math>. Since <math>BC</math> lies on the line <math>x + y = -4</math>, which forms an isosceles right triangle with the coordinate axes, <math>BC = \sqrt{12} \sqrt{2} = \sqrt{24}</math>. Hence the squared area is <math>\left(\frac{\sqrt{3}}{4} s^2 \right)^2 = \frac{3}{16} \cdot 24^2 = \boxed{108}</math>.
  
 
==Solution 3==
 
==Solution 3==
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Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertice of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.
+
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertex of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.
 +
 
 +
==Solution 4==
 +
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math>. Assuming we don't know one vertex is <math>(-1, -1)</math> we let the vertices be <math>A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).</math>
 +
 
 +
Since the centroid coordinates are the average of the vertex coordinates, we have that <math>\frac{x_1+x_2+x_3}{3}=1</math> and <math>\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.</math>
 +
 
 +
We also know that the centroid is the orthocenter in an equilateral triangle, so <math>CG \perp AB.</math> Examining slopes, we simplify the equation to <math>x_1x_2x_3 = -1</math>. From the equation <math>\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,</math> we get that <math>x_1x_2+x_1x_3+x_2x_3 = -3</math>. These equations are starting to resemble Vieta's:
 +
 
 +
<math>x_1+x_2+x_3=3</math>
 +
 
 +
<math>x_1x_2+x_1x_3+x_2x_3 = -3</math>
 +
 
 +
<math>x_1x_2x_3=-1</math>
 +
 
 +
<math>x_1,x_2,x_3</math> are the roots of the equation <math>x^3 - 3x^2 - 3x + 1 = 0</math>. This factors as <math>(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,</math> for the points <math>(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)</math>. The side length is clearly <math>\sqrt{24}</math>, so the square of the area is <math>\boxed{108}.</math>
  
 +
<math>\sim\textbf{Leonard\_my\_dude}\sim</math>
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 04:06, 15 October 2024

Problem

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 (Law of Cosines)

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 the 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 $\boxed{\textbf{(C) } 108}$.


Note: We could’ve also noticed that the centroid divides the median into segments of ratio $2:1.$ ~peelybonehead

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 $B = \left(a,\frac{1}{a}\right)$ 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 $\boxed{\textbf{(C) } 108}$. $\newline$ Alternatively, from $a + \frac{1}{a} = -4$, we obtain $a^2 - 2 + \frac{1}{a^2} = (-4)^2 - 4 \implies \left|a - \frac{1}{a} \right| = \sqrt{12}$. Since $BC$ lies on the line $x + y = -4$, which forms an isosceles right triangle with the coordinate axes, $BC = \sqrt{12} \sqrt{2} = \sqrt{24}$. Hence the squared area is $\left(\frac{\sqrt{3}}{4} s^2 \right)^2 = \frac{3}{16} \cdot 24^2 = \boxed{108}$.

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 vertex 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

Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$. Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).$

Since the centroid coordinates are the average of the vertex coordinates, we have that $\frac{x_1+x_2+x_3}{3}=1$ and $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.$

We also know that the centroid is the orthocenter in an equilateral triangle, so $CG \perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$. From the equation $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$. These equations are starting to resemble Vieta's:

$x_1+x_2+x_3=3$

$x_1x_2+x_1x_3+x_2x_3 = -3$

$x_1x_2x_3=-1$

$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$. This factors as $(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,$ for the points $(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)$. The side length is clearly $\sqrt{24}$, so the square of the area is $\boxed{108}.$

$\sim\textbf{Leonard\_my\_dude}\sim$

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 10 Problems and Solutions

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