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

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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>.
 
[[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]]
+
[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("<math>A</math>", A, NW);
 +
label("<math>B</math>", B, SW);
 +
label("<math>C</math>", C, SE);
 +
label("<math>P</math>", P, NE);
 +
[/asy]
 +
 
 
== Solution ==
 
== Solution ==
 
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
 
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

Revision as of 12:38, 9 December 2021

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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