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

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

Revision as of 12:30, 13 December 2016

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 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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