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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>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>.
 +
 
 +
<asy>
 +
unitsize(0.2 cm);
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 +
pair A, B, C, P;
 +
 
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A = (0,14);
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B = (0,0);
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C = (21*sqrt(3),0);
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P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180));
 +
 
 +
draw(A--B--C--cycle);
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draw(A--P);
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draw(B--P);
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draw(C--P);
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label("$A$", A, NW);
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label("$B$", B, SW);
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label("$C$", C, SE);
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label("$P$", P, NE);
 +
</asy>
  
−
[[Image:AIME_1987_Problem_9.png]]
 
 
== Solution ==
 
== Solution ==
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{{solution}}
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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
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<cmath>AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x</cmath>
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 +
Then by the [[Pythagorean Theorem]], <math>AB^2 + BC^2 = CA^2</math>, so
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<cmath>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</cmath>
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 +
and
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<cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath>
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=== Note ===
 +
This is the [[Fermat point]] of the triangle.
 +
 
 
== See also ==
 
== See also ==
−
* [[1987 AIME Problems]]
 
−
 
 
{{AIME box|year=1987|num-b=8|num-a=10}}
 
{{AIME box|year=1987|num-b=8|num-a=10}}
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:44, 5 February 2024

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 $PC$.

[asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180)); draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, NE); [/asy]

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

\[4x = 132 \Longrightarrow x = \boxed{033}.\]

Note

This is the Fermat point of the triangle.

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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