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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 4"

 
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4. <math>\triangle ABC</math> has all of it's verticies on the parabola <math>y=x^{2}</math>. The slopes of <math>AB</math> and <math>BC</math> are <math>10</math> and <math>-9</math>, respectively. If the x-coordinate of the triangle's centroid is <math>1</math>, find the area of <math>\triangle ABC</math>.
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<math>\triangle ABC</math> has all of its [[vertex| vertices]] on the [[parabola]] <math>y=x^{2}</math>. The slopes of <math>AB</math> and <math>BC</math> are <math>10</math> and <math>-9</math>, respectively. If the <math>x</math>-coordinate of the triangle's centroid is <math>1</math>, find the area of <math>\triangle ABC</math>.
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==Solution==
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If a [[triangle]] in the [[Cartesian plane]] has vertices <math>(a_1, a_2), (b_1, b_2)</math> and <math>(c_1, c_2)</math> then its [[centroid]] has coordinates <math>\left(\frac{a_1 + b_1 + c_1}{3}, \frac{a_2 + b_2 + c_2}{3}\right)</math>.  Let our triangle have vertices <math>A(a, a^2), B(b, b^2)</math> and <math>C(c, c^2)</math>.  Then we have by the centroid condition that <math>a + b + c = 3</math>.  From the first [[slope]] condition we have <math>10 = \frac{b^2 - a^2}{b - a} = b + a</math> and from the second slope condition that <math>-9 = \frac{c^2 - b^2}{c - b} = c + b</math>.  Then <math>c = (a + b + c) - (a + b) = -7</math>, <math>b = (b + c) - c = -2</math> and <math>a = (a + b) - b = 12</math>, so our three vertices are <math>(-7, 49), (-2, 4)</math> and <math>(12, 144)</math>.
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Now, using the [[Shoelace Theorem]] (or your chosen alternative) to calculate the [[area]] of the triangle we get <math>\boxed{665}</math> as our answer.
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*[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 5 | Next Problem]]
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*[[Mock AIME 1 2006-2007]]

Latest revision as of 14:21, 5 November 2012

$\triangle ABC$ has all of its vertices on the parabola $y=x^{2}$. The slopes of $AB$ and $BC$ are $10$ and $-9$, respectively. If the $x$-coordinate of the triangle's centroid is $1$, find the area of $\triangle ABC$.


Solution

If a triangle in the Cartesian plane has vertices $(a_1, a_2), (b_1, b_2)$ and $(c_1, c_2)$ then its centroid has coordinates $\left(\frac{a_1 + b_1 + c_1}{3}, \frac{a_2 + b_2 + c_2}{3}\right)$. Let our triangle have vertices $A(a, a^2), B(b, b^2)$ and $C(c, c^2)$. Then we have by the centroid condition that $a + b + c = 3$. From the first slope condition we have $10 = \frac{b^2 - a^2}{b - a} = b + a$ and from the second slope condition that $-9 = \frac{c^2 - b^2}{c - b} = c + b$. Then $c = (a + b + c) - (a + b) = -7$, $b = (b + c) - c = -2$ and $a = (a + b) - b = 12$, so our three vertices are $(-7, 49), (-2, 4)$ and $(12, 144)$.

Now, using the Shoelace Theorem (or your chosen alternative) to calculate the area of the triangle we get $\boxed{665}$ as our answer.

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