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

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}$.

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1991 AHSME Problems/Problem 19

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