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 $\angle ABC = \arcsin\left(\frac{3}{5}\right)$ 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 $\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\left(\arcsin\left(\frac{3}{5}\right)\right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).$

Solution 4 (Really simple similar triangles)

By Pythagoras on right triangle $ABD$ , $BD = 13$ . Draw the line through $D$ parallel to $CE$ . Let this line intersect $AC$ at $F$ . [asy] unitsize(0.3 cm);

pair A, B, C, D, E, F;

A = (0,3); B = (4,0); C = (0,0); D = scale(12/5)*rotate(90)*(B - A) + A; E = (D + reflect(B,C)*(D))/2; F = extension(A,C,D, D + C - E);

draw(A--C--E--D); draw(A--B--D--cycle); draw(A--F--D);

label(" $A$ ", A, W); label(" $B$ ", B, S); label(" $C$ ", C, SW); label(" $D$ ", D, NE); label(" $E$ ", E, SE); label(" $F$ ", F, NW); [/asy] Since $\angle DAB = 90^\circ$ , $\angle FAD + \angle CAB = 90^\circ$ . Hence, right triangles $ABC$ and $DAF$ are similar. Then $\frac{AF}{4} = \frac{12}{5},$ so $AF = 48/5$ .

Since quadrilateral $CEDF$ is a rectangle, $DE = CF = AC + AF = 3 + 48/5 = 63/5$ . Therefore, $\frac{DE}{DB} = \frac{63/5}{13} = \frac{63}{65}.$ Then $m = 63$ , $n = 65$ , and $m + n = 63 + 65 = \boxed{128}$ . The answer is (B).

~ math31415926535

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