1989 AHSME Problems/Problem 15
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)
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
