Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 11"

m (A =/= B)
(→‎Solution)
 
(3 intermediate revisions by the same user not shown)
Line 4: Line 4:
 
==Solution==
 
==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
 +
\newline
 +
\newline
 +
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
 +
\newline
 +
\newline
 +
\newline
 +
\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}
 +
\newline
 +
\newline
 +
\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>.
  
  

Latest revision as of 18:33, 9 February 2017

Problem

Let $\triangle ABC$ be an equilateral triangle. Two points $D$ and $E$ are chosen on $\overline{AB}$ and $\overline{AC}$, respectively, such that $AD = CE$. Let $F$ be the intersection of $\overline{BE}$ and $\overline{CD}$. The area of $\triangle ABC$ is 13 and the area of $\triangle ACF$ is 3. If $\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, compute $p+q+r$.

Solution

Let $AD = CE = a$, and $EA = DB = b$. Note that we want to compute the ratio $\frac{a}{b}$.

Assign a mass of $a$ to point $A$. This gives point $C$ a mass of $b$ and point $D$ a mass of $a + \frac{a^2}{b}$. Thus, $\frac{CF}{FD} = \frac{a+\frac{a^2}{b}}{b}$.

Since the ratio of areas of triangles that share an altitude is simply the ratio of their bases, we have that:

$13 \cdot \frac{a}{a+b} \cdot \frac{a+\frac{a^2}{b}}{a+\frac{a^2}{b} + b} = 3 \newline \newline \newline 13 \cdot \frac{a}{a+b} \cdot \frac{ab + a^2}{ab + a^2 + b^2} = 3 \newline \newline \newline 13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 \newline \newline \newline \frac{a^2}{ab+a^2+b^2} = \frac{3}{13} \newline \newline \newline \frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} \newline \newline \newline (\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}$.

By the quadratic formula, we find that $\frac{b}{a} = \frac{\sqrt{129}-3}{6}$, so $\frac{a}{b} = \frac{6}{\sqrt{129}-3} = \frac{3 + \sqrt{129}}{20}$.

Thus, our final answer is $3 + 129 + 20 = \boxed{152}$.


Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_4_2006-2007_Problems/Problem_11&oldid=83381"