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Difference between revisions of "1987 AIME Problems/Problem 9"
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== Problem ==
 
== Problem ==
−
[[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>.
+
[[Triangle]] <math>ABC</math> has [[right angle]] at <math>B</math>, and contains a [[point]] <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>.  Find <math>PC</math>.
  
 
[[Image:AIME_1987_Problem_9.png]]
 
[[Image:AIME_1987_Problem_9.png]]
 
== Solution ==
 
== Solution ==
−
Let <math>PC = x</math>.
+
Let <math>PC = x</math>. Since <math>\angle APB = \angle BPC = \angle CPA</math>, 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
  
−
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
+
<cmath>AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x</cmath>
  
−
<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>. 
+
Then by the [[Pythagorean Theorem]], <math>AB^2 + BC^2 = CA^2</math>, so
  
−
Then by the [[Pythagorean Theorem]], <math>AB^2 + BC^2 = CA^2</math> so that
+
<cmath>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</cmath>  
−
 
 
−
<math>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</math>
 
  
 
and
 
and
  
−
<math>4x = 132</math> so <math>x = 033</math>.
+
<cmath>4x = 132 \Longrightarrow x = 033</cmath>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1987|num-b=8|num-a=10}}
 
{{AIME box|year=1987|num-b=8|num-a=10}}
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 18:03, 3 October 2007

Problem

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

AIME 1987 Problem 9.png

Solution

Let $PC = x$. Since $\angle APB = \angle BPC = \angle CPA$, 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

\[AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x\]

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

\[x^2 + 10x + 100 = x^2 + 6x + 36 + 196\]

and

\[4x = 132 \Longrightarrow x = 033\]

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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