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"

(→‎Solution)
(→‎Solution 2 (Trig))
Line 31: Line 31:
 
== Solution 2 (Trig) ==
 
== Solution 2 (Trig) ==
  
−
Using laws of cosines on <math>\bigtriangleup ABC</math> yields <math>49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A</math>
+
Using laws of cosines on <math>\bigtriangleup ABC</math> yields <math>49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A \implies \cos A = \displaystyle\frac{19}{30}.</math> Now we use law of cosines on <math>\bigtriangleup ABD</math> getting <math>25=25+AD^2-2 \cdot 5 \cdot AD \cos A.</math> Fortunately, we know that <math>\cos A=\displaystyle\frac{19}{3)}</math> so plugging that in our first equation yields <math>AD=\displaystyle\frac{19}{3}.</math> Then, we know <math>DC=\displaystyle\frac{8}{3} \implies \displaystyle\frac{AD}{DC}=\displaystyle\frac{19}{8}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:57, 21 June 2020

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 \[\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}\]


Solution 2 (Trig)

Using laws of cosines on $\bigtriangleup ABC$ yields $49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A \implies \cos A = \displaystyle\frac{19}{30}.$ Now we use law of cosines on $\bigtriangleup ABD$ getting $25=25+AD^2-2 \cdot 5 \cdot AD \cos A.$ Fortunately, we know that $\cos A=\displaystyle\frac{19}{3)}$ so plugging that in our first equation yields $AD=\displaystyle\frac{19}{3}.$ Then, we know $DC=\displaystyle\frac{8}{3} \implies \displaystyle\frac{AD}{DC}=\displaystyle\frac{19}{8}.$

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=126148"