1987 AIME Problems/Problem 9: Difference between revisions

Latest revision as of 12:37, 11 December 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.

Video Solution by Pi Academy

https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh

~ smartschoolboy9

@@ Line 1: / Line 1: @@
 == Problem ==
+[[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);
+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 <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
+<cmath>AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x</cmath>
+Then by the [[Pythagorean Theorem]], <math>AB^2 + BC^2 = CA^2</math>, so
+<cmath>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</cmath>
+and
+<cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath>
+=== Note ===
+This is the [[Fermat point]] of the triangle.
+== Video Solution by Pi Academy ==
+https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh
+~ smartschoolboy9
 == See also ==
-* [[1987 AIME Problems]]
+{{AIME box|year=1987|num-b=8|num-a=10}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

1987 AIME (Problems • Answer Key • Resources)
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