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 | + | Since we are given that <math>\triangle{PAB}\sim\triangle{PCA}</math>, we have <math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</math>. |
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− | <math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</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 <math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math> |
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− | <math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math> | ||
− | <math>\frac{16PC}{3}=3PC+21</math> | + | Which we can further simplify to <math>\frac{16PC}{3}=3PC+21</math> |
Latest revision as of 23:51, 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
.
Solving for in
gives us
.
We also have . Substituting
in for our expression yields
Which we can further simplify to
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 |
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