Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 11"
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Let <math>AD = CE = a</math>, and <math>EA = DB = b</math>. Note that we want to compute the ratio <math>\frac{a}{b}</math>. | Let <math>AD = CE = a</math>, and <math>EA = DB = b</math>. Note that we want to compute the ratio <math>\frac{a}{b}</math>. | ||
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<math>13 \cdot \frac{a}{a+b} \cdot \frac{a+\frac{a^2}{b}}{a+\frac{a^2}{b} + b} = 3 | <math>13 \cdot \frac{a}{a+b} \cdot \frac{a+\frac{a^2}{b}}{a+\frac{a^2}{b} + b} = 3 | ||
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13 \cdot \frac{a}{a+b} \cdot \frac{ab + a^2}{ab + a^2 + b^2} = 3 | 13 \cdot \frac{a}{a+b} \cdot \frac{ab + a^2}{ab + a^2 + b^2} = 3 | ||
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13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 | 13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 | ||
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\frac{a^2}{ab+a^2+b^2} = \frac{3}{13} | \frac{a^2}{ab+a^2+b^2} = \frac{3}{13} | ||
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\frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} | \frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} | ||
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(\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}</math>. | (\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}</math>. |
Latest revision as of 19:33, 9 February 2017
Problem
Let be an equilateral triangle. Two points
and
are chosen on
and
, respectively, such that
. Let
be the intersection of
and
. The area of
is 13 and the area of
is 3. If
, where
,
, and
are relatively prime positive integers, compute
.
Solution
Let , and
. Note that we want to compute the ratio
.
Assign a mass of to point
. This gives point
a mass of
and point
a mass of
. Thus,
.
Since the ratio of areas of triangles that share an altitude is simply the ratio of their bases, we have that:
.
By the quadratic formula, we find that , so
.
Thus, our final answer is .