Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 4"
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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>. | 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>. | ||
| − | Now, using the [[Shoelace Theorem]] (or your chosen alternative) to calculate the [[area]] of the triangle we get 665 as our answer. | + | 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/Problem 3 | Previous Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]] |
| − | *[[Mock AIME 1 2006-2007/Problem 5 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Next Problem]] |
*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] | ||
Latest revision as of 14:21, 5 November 2012
has all of its vertices on the parabola 
. The slopes of 
and 
are 
and 
, respectively. If the 
-coordinate of the triangle's centroid is 
, find the area of .
Solution
If a triangle in the Cartesian plane has vertices 
and 
then its centroid has coordinates 
. Let our triangle have vertices 
and 
. Then we have by the centroid condition that 
. From the first slope condition we have 
and from the second slope condition that 
. Then 
, 
and 
, so our three vertices are 
and 
.
Now, using the Shoelace Theorem (or your chosen alternative) to calculate the area of the triangle we get 
as our answer.
