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

Revision as of 16:31, 21 June 2020

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 $\angle DBA=\arcsin\left(\frac{12}{13}\right).$ We want $\sin(\angle DBE)=\sin(180-\angle DBC)=\sin(180- \arcsin\left(\frac{3}{5}\right)-\arcsin\left(\frac{12}{13}\right))$

@@ Line 32: / Line 32: @@
 == 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>
+We have <math>\angle ABC = \arcsin\left(\frac{3}{5}\right)</math> and <math>\angle DBA=\arcsin\left(\frac{12}{13}\right).</math> We want <math>\sin(\angle DBE)=\sin(180-\angle DBC)=\sin(180- \arcsin\left(\frac{3}{5}\right)-\arcsin\left(\frac{12}{13}\right))</math>
 == 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

Page

Toolbox

Search