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>. | ||
− | [asy] | + | [[asy]] |
unitsize(0.2 cm); | unitsize(0.2 cm); | ||
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label("<math>C</math>", C, SE); | label("<math>C</math>", C, SE); | ||
label("<math>P</math>", P, NE); | label("<math>P</math>", P, NE); | ||
− | [/asy] | + | [[/asy]] |
== Solution == | == Solution == |
Revision as of 12:38, 9 December 2021
Contents
[hide]Problem
Triangle has right angle at , and contains a point for which , , and . Find .
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, NW); label("", B, SW); label("", C, SE); label("", P, NE); /asy
Solution
Let . Since , each of them is equal to . By the Law of Cosines applied to triangles , and at their respective angles , remembering that , we have
Then by the Pythagorean Theorem, , so
and
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.