Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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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 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== | ||
+ | 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 16:03, 31 January 2021
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?
Diagram
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 .
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
Without loss of generality, let the centroid of be . Assuming we don't know one vertex is we let the vertices be
Since the centroid coordinates are the average of the vertex coordinates, we have that and
We also know that the centroid is the orthocenter in an equilateral triangle, so Examining slopes, we simplify the equation to . From the equation we get that . These equations are starting to resemble Vieta's:
are the roots of the equation . This factors as for the points . The side length is clearly , so the square of the area is
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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