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Difference between revisions of "2020 CIME II Problems/Problem 1"
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==Solution==
 
==Solution==
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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}{CE}\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}
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2y+2=3x
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\\
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x+2=3y
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\end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}</cmath>
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Therefore, the perimeter is <math>1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}</math>\\
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~bhargavakanakapura
  
 
==See also==
 
==See also==

Latest revision as of 02:08, 30 November 2021

Problem

Let $ABC$ be a triangle. The bisector of $\angle ABC$ intersects $\overline{AC}$ at $E$, and the bisector of $\angle ACB$ intersects $\overline{AB}$ at $F$. If $BF=1$, $CE=2$, and $BC=3$, then the perimeter of $\triangle ABC$ can be expressed in the form $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

For simplicity, let $AE=x$ and $AF=y$. By the angle bisector theorem, we have that \[\frac{AB}{AE}=\frac{BC}{CE}\Longrightarrow\frac{y+1}{x}=\frac{3}{2}\] using $\angle B$ as the bisected angle. Using $\angle C$ as the bisected angle, we have that \[\frac{AC}{AF}=\frac{BC}{BF}\Longrightarrow\frac{x+2}{y}=3\]These two equations form a system of equations: \[\left\{\begin{matrix} 2y+2=3x \\ x+2=3y \end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}\] Therefore, the perimeter is $1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}$\\ ~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

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