# Difference between revisions of "1991 AHSME Problems/Problem 19"

## Problem

$[asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]$

Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$. Triangle $ABD$ has a right angle at $A$ and $AD=12$. Points $C$ and $D$ are on opposite sides of $\overline{AB}$. The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$. If $$\frac{DE}{DB}=\frac{m}{n},$$ where $m$ and $n$ are relatively prime positive integers, then $m+n=$

$\text{(A) } 25\quad \text{(B) } 128\quad \text{(C) } 153\quad \text{(D) } 243\quad \text{(E) } 256$

## Solution 1

Solution by e_power_pi_times_i

Let $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$, respectively, and let $DE = x$ and $BE^2 = 169-x^2$. Then, $[FDEC] = x(4+\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \dfrac{x\sqrt{169-x^2}}{2} + 30 + \dfrac{(x-3)(4+\sqrt{169-x^2})}{2}$. So, $4x+x\sqrt{169-x^2} = 60 + x\sqrt{169-x^2} - 3\sqrt{169-x^2}$. Simplifying $3\sqrt{169-x^2} = 60 - 4x$, and $1521 - 9x^2 = 16x^2 - 480x + 3600$. Therefore $25x^2 - 480x + 2079 = 0$, and $x = \dfrac{48\pm15}{5}$. Checking, $x = \dfrac{63}{5}$ is the answer, so $\dfrac{DE}{DB} = \dfrac{\dfrac{63}{5}}{13} = \dfrac{63}{65}$. The answer is $\boxed{\textbf{(B) } 128}$.

## Solution 2

Solution by Arjun Vikram

Extend lines $AD$ and $CE$ to meet at a new point $F$. Now, we see that $FAC\sim FDE \sim ACB$. Using this relationship, we can see that $AF=\frac{15}4$, (so $FD=\frac{63}4$), and the ratio of similarity between $FDE$ and $FAC$ is $\frac{63}{15}$. This ratio gives us that $\frac{63}5$. By the Pythagorean Theorem, $DB=13$. Thus, $\frac{DE}{DB}=\frac{63}{65}$, and the answer is $63+65=\boxed{\textbf{(B) } 128}$.

## Solution 3 (Trig)

We have $\angle ABC = \arcsin(\displaystyle\frac{3}{5})$ and $\angle DBA=\arcsin(\displaystyle\frac{12}{13}).$