Difference between revisions of "1986 AHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
| + | Since we are given that <math>\triangle{PAB}\sim\triangle{PCA}</math>, we have the following | ||
| + | <math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</math>. | ||
| + | |||
| + | Solving for <math>PA</math> in <math>\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}</math> gives us <math>PA=\frac{4PC}{3}</math>. | ||
| + | |||
| + | We also have <math>\frac{PA}{PC+7}=\frac{3}{4}</math>. Substituting <math>PA</math> in for our expression yields | ||
| + | |||
| + | <math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math> | ||
| + | |||
| + | <math>\frac{16PC}{3}=3PC+21</math> | ||
| + | |||
| + | <math>\frac{7PC}{3}=21</math> | ||
| + | |||
| + | <math>PC=9\implies\boxed{\textbf{(C)}}</math> | ||
== See also == | == See also == | ||
Revision as of 21:42, 10 February 2018
Problem
In 
and side 
is extended, as shown in the figure, to a point 
so that 
is similar to 
. The length of 
is
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]](http://latex.artofproblemsolving.com/6/1/3/613a6e628f3f9a1136da0c89d1d1c6c7d50a09db.png)
Solution
Since we are given that 
, we have the following
.
Solving for 
in 
gives us 
.
We also have 
. Substituting in for our expression yields
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
