Difference between revisions of "1987 AIME Problems/Problem 9"
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== Problem == | == 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 == | == 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. | ||
== See also == | == See also == | ||
− | |||
− | |||
{{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}} |
Revision as of 17:44, 5 February 2024
Contents
[hide]Problem
Triangle has right angle at , and contains a point for which , , and . Find .
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.