Difference between revisions of "2020 CIME II Problems/Problem 1"
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x+2=3y | x+2=3y | ||
\end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}</cmath> | \end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}</cmath> | ||
− | Therefore, the perimeter is <math>1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}</math> | + | Therefore, the perimeter is <math>1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}</math>\\ |
+ | ~bhargavakanakapura | ||
==See also== | ==See also== |
Latest revision as of 02:08, 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
using
as the bisected angle. Using
as the bisected angle, we have that
These two equations form a system of equations:
Therefore, the perimeter is
\\
~bhargavakanakapura
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.