Difference between revisions of "1987 AIME Problems/Problem 9"
(→Solution) |
Megahertz13 (talk | contribs) (→Video Solution by MegaMath) |
||
| (8 intermediate revisions by 5 users not shown) | |||
| Line 2: | Line 2: | ||
[[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> | |
| + | 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 | 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 | ||
| Line 15: | Line 35: | ||
<cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | <cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | ||
| − | == Note == | + | === Note === |
| − | This is the Fermat point of the triangle | + | This is the [[Fermat point]] of the triangle. |
== See also == | == See also == | ||
Latest revision as of 17:44, 5 February 2024
Contents
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("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, NE); [/asy]](http://latex.artofproblemsolving.com/8/7/b/87b171f8fbb8055a3f8d854d493f599eec30c1fc.png)
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
![\[AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x\]](http://latex.artofproblemsolving.com/3/8/1/3811f0cadac72e5493473cc57a2249d160ce488c.png)
Then by the Pythagorean Theorem, 
, so
![\[x^2 + 10x + 100 = x^2 + 6x + 36 + 196\]](http://latex.artofproblemsolving.com/6/8/8/688be5159b368f96110b866f2f826348c7f9c0d2.png)
and
![\[4x = 132 \Longrightarrow x = \boxed{033}.\]](http://latex.artofproblemsolving.com/b/9/6/b96d8f130f6036d0500a38ca00ce1281338256c8.png)
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. 
