1989 AHSME Problems/Problem 15
Contents
Problem
In 
, 
, 
, 
, and 
is on 
with 
. Find the ratio of 
.
![[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); dot((0,0));dot((9,0));dot((3,4));dot((6,0)); label("A", (0,0), W); label("B", (3,4), N); label("C", (9,0), E); label("D", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/8/5/e/85ef7a332d4265d8f052349a58f701f82e2d07aa.png)
Solution
![[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); draw((3,4)--(3,0),dotted); dot((0,0));dot((9,0));dot((3,4));dot((6,0));//dot((3,0)); label("A", (0,0), W); label("B", (3,4), N); label("C", (9,0), E); label("D", (6,0), S); label("$h$", (3,1.5),W); draw(rightanglemark((3,1),(3,0),(4,0),10)); label("$x$",(1.5,0),S);label("$x$",(4.5,0),S); [/asy]](http://latex.artofproblemsolving.com/e/5/d/e5d60ae241b0cfde5d142e61e289305386313b26.png)
Drop the altitude 
from 
through 
, and let be 
. Then by Pythagoras ![\[\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}\]](http://latex.artofproblemsolving.com/1/f/2/1f2b96de36cf9055c9f76b18f168b760805f193b.png)
and after subtracting the first equation from the second, 
. Therefore the desired ratio is ![\[\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}\]](http://latex.artofproblemsolving.com/0/1/4/0142543b27824e55ee97d53ec92a27a6f18bde17.png)
Solution 2 (Trig)
Using laws of cosines on 
yields 
Let 
Using laws of cosines on 
yields 
Fortunately, we know that 
Plugging this information back into our equation yields 
Then, we know that 
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
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