Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
− | + | 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>. | |
==Solution 2== | ==Solution 2== | ||
− | + | 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>. | |
~minor LaTeX edit by dolphin7 | ~minor LaTeX edit by dolphin7 | ||
==Solution 3== | ==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 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>. | |
==Solution 4 (5-second solution)== | ==Solution 4 (5-second solution)== | ||
− | + | Without loss of generality, let the centroid of the triangle be <math>(1, 1)</math>. By symmetry, the other vertex is <math>(-1, -1)</math>. The distance between these two points is <math>2\sqrt2</math>, so the height of the triangle is <math>3\sqrt 2</math>, the side length is <math>2\sqrt6</math>, and the area is <math>6\sqrt3</math>, yielding an answer of <math>\boxed{\textbf{(C) }108}</math>. | |
-Stormersyle | -Stormersyle | ||
Revision as of 14:20, 13 January 2020
Contents
[hide]Problem 24
The vertices of an equilateral triangle lie on the hyperbola , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
Solution 1
Without loss of generality, let the centroid of be . 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, , so , so since is isosceles and , then by Law of Cosines, . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to . Therefore, the area of the triangle is , so the square of the area of the triangle is .
Solution 2
Without loss of generality, let the centroid of be . Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let . Then, point must be the reflection of across the line , so let and , where . Because is the centroid, the average of the -coordinates of the vertices of the triangle is . So we know that . Multiplying by and solving gives us . So and . So , and finding the square of the area gives us . ~minor LaTeX edit by dolphin7
Solution 3
Without loss of generality, let the centroid of be and let point be . It is known that the centroid is equidistant from the three vertices of . Because we have the coordinates of both and , we know that the distance from to any vertice of is . Therefore, . It follows that from , where and , using the formula for the area of a triangle with sine . Because and are congruent to , they also have an area of . Therefore, . Squaring that gives us the answer of .
Solution 4 (5-second solution)
Without loss of generality, let the centroid of the triangle be . By symmetry, the other vertex is . The distance between these two points is , so the height of the triangle is , the side length is , and the area is , yielding an answer of . -Stormersyle
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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