Difference between revisions of "2020 CIME II Problems/Problem 1"
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==Solution== | ==Solution== | ||
| + | For simplicity, let <math>AE=x</math> and <math>AF=y</math>. By the angle bisector theorem, we have that <cmath>\frac{AB}{AE}=\frac{BC}{BE}\Longrightarrow\frac{y+1}{x}=\frac{3}{2}</cmath> using <math>\angle B</math> as the bisected angle. Using <math>\angle C</math> as the bisected angle, we have that <cmath>\frac{AC}{AF}=\frac{BC}{BF}\Longrightarrow\frac{x+2}{y}=3</cmath>These two equations form a system of equations: <cmath>\left\{\begin{matrix} | ||
| + | 2y+2=3x | ||
| + | \\ | ||
| + | x+2=3y | ||
| + | \end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}</cmath> | ||
==See also== | ==See also== | ||
Revision as of 01:04, 30 November 2021
Problem
Let 
be a triangle. The bisector of 
intersects 
at 
, and the bisector of 
intersects 
at 
. If 
, 
, and 
, then the perimeter of 
can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
For simplicity, let 
and 
. By the angle bisector theorem, we have that ![\[\frac{AB}{AE}=\frac{BC}{BE}\Longrightarrow\frac{y+1}{x}=\frac{3}{2}\]](http://latex.artofproblemsolving.com/7/4/8/748cfa6dd52c445bf5084933c08bffa670028c78.png)
using 
as the bisected angle. Using 
as the bisected angle, we have that ![\[\frac{AC}{AF}=\frac{BC}{BF}\Longrightarrow\frac{x+2}{y}=3\]](http://latex.artofproblemsolving.com/b/2/6/b261c2136a9267ba7d9cfd6f4d617652e226a64e.png)
These two equations form a system of equations: ![\[\left\{\begin{matrix} 2y+2=3x \\ x+2=3y \end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}\]](http://latex.artofproblemsolving.com/d/4/8/d4842ef9ca4d22fd26eeb026a15aef8bdce0f26f.png)
See also
| 2020 CIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
