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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] = 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>.
 
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>.
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== Solution 2 ==
 
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Solution by Arjun Vikram
 
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Extend lines <math>AD</math> and <math>CE</math> to meet at a new point <math>F</math>. Now, we see that <math>FAC\sim FDE \sim ACB</math>. Using this relationship, we can see that <math>AF=\frac{15}4</math>, (so <math>FD=\frac{63}4</math>), and the ratio of similarity between <math>FDE</math> and <math>FAC</math> is <math>\frac{63}{15}</math>. This ratio gives us that <math>\frac{63}5</math>. By the Pythagorean Theorem, <math>DB=13</math>. Thus, <math>\frac{DE}{DB}=\frac{63}{65}</math>, and the answer is <math>63+65=\boxed{\textbf{(B) } 128}</math>.
 
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== Solution 3 (Trig)==
 
== Solution 3 (Trig)==
  
 
We have <math>\angle ABC = \arcsin\left(\frac{3}{5}\right)</math> and <math>\angle DBA=\arcsin\left(\frac{12}{13}\right).</math> Now we are trying to find <math>\sin(\angle DBE)=\sin\left(180^{\circ}-\angle DBC=\sin(180^{\circ}-\arcsin\left(\frac{3}{5}\right)\right)-\arcsin\left(\frac{12}{13}\right)=\sin(\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{12}{13}\right)).</math> Now we use the <math>\sin</math> angle sum identity, which states <math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).</math> Using this identity yields <math>\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\left(\arcsin\left(\frac{3}{5}\right)\right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).</math>
 
We have <math>\angle ABC = \arcsin\left(\frac{3}{5}\right)</math> and <math>\angle DBA=\arcsin\left(\frac{12}{13}\right).</math> Now we are trying to find <math>\sin(\angle DBE)=\sin\left(180^{\circ}-\angle DBC=\sin(180^{\circ}-\arcsin\left(\frac{3}{5}\right)\right)-\arcsin\left(\frac{12}{13}\right)=\sin(\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{12}{13}\right)).</math> Now we use the <math>\sin</math> angle sum identity, which states <math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).</math> Using this identity yields <math>\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\left(\arcsin\left(\frac{3}{5}\right)\right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).</math>
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== Solution 4 (Really simple similar triangles)==
 
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Triangle <math>ABC</math> has a right angle at <math>C</math>, <math>AC = 3</math> and <math>BC = 4</math>. Triangle <math>ABD</math> has a right angle at <math>A</math> and <math>AD = 12</math>. Points <math>C</math> and <math>D</math> are on opposite sides of <math>\overline{AB}</math>. The line through <math>D</math> parallel to <math>\overline{AC}</math> meets <math>\overline{CB}</math> extended at <math>E</math>. If <math>DE/DB = m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, then <math>m + n =</math>
 
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~ math31415926535
 
  
 
== See also ==
 
== See also ==

Latest revision as of 16:59, 25 June 2025

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 3 (Trig)

We have $\angle ABC = \arcsin\left(\frac{3}{5}\right)$ and $\angle DBA=\arcsin\left(\frac{12}{13}\right).$ Now we are trying to find $\sin(\angle DBE)=\sin\left(180^{\circ}-\angle DBC=\sin(180^{\circ}-\arcsin\left(\frac{3}{5}\right)\right)-\arcsin\left(\frac{12}{13}\right)=\sin(\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{12}{13}\right)).$ Now we use the $\sin$ angle sum identity, which states $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).$ Using this identity yields $\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\left(\arcsin\left(\frac{3}{5}\right)\right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).$

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AHSME Problems and Solutions

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