Difference between revisions of "1989 AHSME Problems/Problem 15"
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<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> | ||
+ | |||
+ | |||
+ | == 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 \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 20:37, 24 March 2023
Contents
Problem
In ,
,
,
, and
is on
with
. Find the ratio of
.
Solution
Drop the altitude
from
through
, and let
be
. Then by Pythagoras
and after subtracting the first equation from the second,
. Therefore the desired ratio is
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 | |
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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