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Difference between revisions of "1987 AIME Problems/Problem 9"

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== Problem ==
 
== Problem ==
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Triangle <math>\displaystyle ABC</math> has right angle at <math>\displaystyle B</math>, and contains a point <math>\displaystyle P</math> for which <math>\displaystyle PA = 10</math>, <math>\displaystyle PB = 6</math>, and <math>\displaystyle \angle APB = \angle BPC = \angle CPA</math>.  Find <math>\displaystyle PC</math>.
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[[Triangle]] <math>\displaystyle ABC</math> has [[right angle]] at <math>\displaystyle B</math>, and contains a [[point]] <math>\displaystyle P</math> for which <math>\displaystyle PA = 10</math>, <math>\displaystyle PB = 6</math>, and <math>\displaystyle \angle APB = \angle BPC = \angle CPA</math>.  Find <math>\displaystyle PC</math>.
  
 
[[Image:AIME_1987_Problem_9.png]]
 
[[Image:AIME_1987_Problem_9.png]]
 
== Solution ==
 
== Solution ==
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{{solution}}
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Let <math>PC = x</math>.
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Since the three [[angle]]s <math>\angle APB</math>, <math>\angle BPC</math> and <math>\angle CPA</math> are equal, each of them is equal to <math>120^\circ</math>.  By the [[Law of Cosines]] applied to triangles <math>\triangle APB</math>, <math>\triangle BPC</math> and <math>\triangle CPA</math> at their respective angles <math>P</math>, remembering that <math>\cos 120^\circ = -\frac12</math>, we have
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<math>AB^2 = 36 + 100 + 60 = 196</math>, <math>BC^2 = 36 + x^2 + 6x</math> and <math>CA^2 = 100 + x^2 + 10x</math>. 
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Then by the [[Pythagorean Theorem]], <math>AB^2 + BC^2 = CA^2</math> so that
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<math>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</math>
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and
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<math>4x = 132</math> so <math>x = 033</math>.
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== See also ==
 
== See also ==
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* [[1987 AIME Problems]]
 
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{{AIME box|year=1987|num-b=8|num-a=10}}
 
{{AIME box|year=1987|num-b=8|num-a=10}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 19:00, 15 February 2007

Problem

Triangle $\displaystyle ABC$ has right angle at $\displaystyle B$, and contains a point $\displaystyle P$ for which $\displaystyle PA = 10$, $\displaystyle PB = 6$, and $\displaystyle \angle APB = \angle BPC = \angle CPA$. Find $\displaystyle PC$.

AIME 1987 Problem 9.png

Solution

Let $PC = x$.

Since the three angles $\angle APB$, $\angle BPC$ and $\angle CPA$ are equal, each of them is equal to $120^\circ$. By the Law of Cosines applied to triangles $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$, remembering that $\cos 120^\circ = -\frac12$, we have

$AB^2 = 36 + 100 + 60 = 196$, $BC^2 = 36 + x^2 + 6x$ and $CA^2 = 100 + x^2 + 10x$.

Then by the Pythagorean Theorem, $AB^2 + BC^2 = CA^2$ so that

$x^2 + 10x + 100 = x^2 + 6x + 36 + 196$

and

$4x = 132$ so $x = 033$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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