Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 11"
m (A =/= B) |
(→Solution) |
||
Line 5: | Line 5: | ||
{{solution}} | {{solution}} | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Assign a mass of <math>a</math> to point <math>A</math>. This gives point <math>C</math> a mass of <math>b</math> and point <math>D</math> a mass of <math>a + \frac{a^2}{b}</math>. Thus, <math>\frac{CF}{FD} = \frac{a+\frac{a^2}{b}}{b}</math>. | ||
+ | |||
+ | Since the ratio of areas of triangles that share an altitude is simply the ratio of their bases, we have that: | ||
+ | |||
+ | <math>13 \cdot \frac{a}{a+b} \cdot \frac{a+\frac{a^2}{b}}{a+\frac{a^2}{b} + b} = 3 | ||
+ | \newline | ||
+ | 13 \cdot \frac{a}{a+b} \cdot \frac{ab + a^2}{ab + a^2 + b^2} = 3 | ||
+ | \newline | ||
+ | 13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 | ||
+ | \newline | ||
+ | \frac{a^2}{ab+a^2+b^2} = \frac{3}{13} | ||
+ | \newline | ||
+ | \frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} | ||
+ | \newline | ||
+ | (\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}</math>. | ||
+ | |||
+ | By the quadratic formula, we find that <math>\frac{b}{a} = \frac{\sqrt{129}-3}{6}</math>, so <math>\frac{a}{b} = \frac{6}{\sqrt{129}-3} = \frac{3 + \sqrt{129}}{20}</math>. | ||
+ | |||
+ | Thus, our final answer is <math>3 + 129 + 20 = \boxed{152}</math>. | ||
Revision as of 18:29, 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 .