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

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 $\boxed{108}.$

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

@@ Line 1: / Line 1: @@
-==Problem 24==
+==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?
 <math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math>
-==Solution==
+==Diagram==
-WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</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>.
+<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 <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==
-WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other vertex of the parabola. 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 = (a,1/a)</math> and <math>C=(1/a,a)</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>
+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==
+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==
 {{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

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