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 <math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</math>. | ||
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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>. | ||
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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 <math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math> | ||
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+ | Which we can further simplify to <math>\frac{16PC}{3}=3PC+21</math> | ||
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+ | <math>\frac{7PC}{3}=21</math> | ||
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+ | <math>PC=9\implies\boxed{\textbf{(C)}}</math> | ||
== See also == | == See also == | ||
{{AHSME box|year=1986|num-b=15|num-a=17}} | {{AHSME box|year=1986|num-b=15|num-a=17}} | ||
− | [[Category: | + | [[Category: Intermediate Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:51, 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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