Difference between revisions of "1989 AHSME Problems/Problem 15"
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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 01:59, 4 February 2016
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
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.