Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 11"
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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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\newline | \newline | ||
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 | ||
| + | \newline | ||
\newline | \newline | ||
13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 | 13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 | ||
| + | \newline | ||
\newline | \newline | ||
\frac{a^2}{ab+a^2+b^2} = \frac{3}{13} | \frac{a^2}{ab+a^2+b^2} = \frac{3}{13} | ||
| + | \newline | ||
\newline | \newline | ||
\frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} | \frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} | ||
\newline | \newline | ||
| + | \newline | ||
(\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}</math>. | (\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}</math>. | ||
Revision as of 18:30, 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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 
.
