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

Revision as of 12:46, 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.

@@ Line 17: / Line 17: @@
 draw(C--P);
-label("<math>A</math>", A, NW);
+label("$A$", A, NW);
-label("<math>B</math>", B, SW);
+label("$B$", B, SW);
-label("<math>C</math>", C, SE);
+label("$C$", C, SE);
-label("<math>P</math>", P, NE);
+label("$P$", P, NE);
-<asy>
+</asy>
 == Solution ==

1987 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1987 AIME Problems/Problem 9"

Revision as of 12:46, 9 December 2021

Contents

Problem

Solution

Note

See also