Difference between revisions of "1986 AHSME Problems/Problem 16"
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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>. | 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>. | ||
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We also have <math>\frac{PA}{PC+7}=\frac{3}{4}</math>. Substituting <math>PA</math> in for our expression yields | We also have <math>\frac{PA}{PC+7}=\frac{3}{4}</math>. Substituting <math>PA</math> in for our expression yields |
Revision as of 22:50, 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
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.