1991 AHSME Problems/Problem 19

Revision as of 13:29, 13 December 2016 by E power pi times i (talk | contribs) (Solution)

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$\boxed{\textbf{(B) } 128}$.

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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