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Difference between revisions of "1989 AHSME Problems/Problem 15"
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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>
 
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>
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== Solution 2 (Trig) ==
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 +
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 =\frac{19}{30}.</math> Let <math>AD=x.</math> Using laws of cosines on <math>\bigtriangleup ABD</math> yields <math>25+x^2-2 \cdot 5 \cdot x \cdot \cos A = 25.</math> Fortunately, we know that <math>\cos A =\frac{19}{30}.</math> Plugging this information back into our equation yields <math>x=\frac{19}{3}.</math> Then, we know that <math>DC=9-\frac{19}{3}=\frac{8}{3} \implies\frac{AD}{DC}=\boxed{\frac{19}{8}}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 19:37, 24 March 2023

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 $49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A \implies \cos A =\frac{19}{30}.$ Let $AD=x.$ Using laws of cosines on $\bigtriangleup ABD$ yields $25+x^2-2 \cdot 5 \cdot x \cdot \cos A = 25.$ Fortunately, we know that $\cos A =\frac{19}{30}.$ Plugging this information back into our equation yields $x=\frac{19}{3}.$ Then, we know that $DC=9-\frac{19}{3}=\frac{8}{3} \implies\frac{AD}{DC}=\boxed{\frac{19}{8}}.$

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

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