Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AHSME Problems/Problem 15"

m
Line 5: Line 5:
 
<asy>
 
<asy>
 
draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0));
 
draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0));
−
dot((0,0));
+
dot((0,0));dot((9,0));dot((3,4));dot((6,0));
−
dot((9,0));
 
−
dot((3,4));
 
−
dot((6,0));
 
 
label("A", (0,0), W);
 
label("A", (0,0), W);
 
label("B", (3,4), N);
 
label("B", (3,4), N);
Line 17: Line 14:
  
 
== Solution ==
 
== 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>
 +
Drop the altitude <math>h</math> from <math>B</math> through <math>AD</math>, and let <math>AD</math> be <math>2x</math>. Then by Pythagoras <cmath>\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}</cmath> and after subtracting the first equation from the second, <math>x=3\tfrac16</math>. Therefore the desired ratio is <cmath>\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 00:59, 4 February 2016

Problem

In $\triangle ABC$, $AB=5$, $BC=7$, $AC=9$, and $D$ is on $\overline{AC}$ with $BD=5$. Find the ratio of $AD:DC$.

[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]

$\textrm{(A)}\ 4:3\qquad\textrm{(B)}\ 7:5\qquad\textrm{(C)}\ 11:6\qquad\textrm{(D)}\ 13:5\qquad\textrm{(E)}\ 19:8$

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] Drop the altitude $h$ from $B$ through $AD$, and let $AD$ be $2x$. Then by Pythagoras \[\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}\] and after subtracting the first equation from the second, $x=3\tfrac16$. Therefore the desired ratio is \[\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}\]

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_15&oldid=75292"