The lines joining the incenter of a triangle to the vertices divide the triangle into three triangles. If one of these triangles is similar to the initial one,determine the angles of the triangle.
Problem
Source: Croatian NMC 2005, 1st Grade
Tags: geometry, incenter
nayel
08.05.2007 14:18
let $I$ be the incenter. without loss of generality, $\triangle IBC\sim\triangle ABC$ with $\angle IBC=\frac{B}{2}=C, \angle ICB=\frac{C}{2}=A, \angle BIC=90+\frac{A}{2}$ so after solving, the answer we get $A=\frac{2\cdot 90}{7}, B=\frac{5\cdot 90}{4},C=\frac{4\cdot 90}{7}$ and permutations.
MellowMelon
09.05.2007 00:14
Those values don't add up to 180 degrees.
I got the three angles being $\frac{180}{7}, \frac{2 \cdot 180}{7}, \frac{4 \cdot 180}{7}$ degrees. Then the triangle containing the second and third angles bisected has angles equal to the first and second angles of the original triangle, and we have AA similarity.
frill
31.05.2012 22:25
Let $ I $ be the incenter with $ \angle IBC=\frac{B}{2},\angle ICB=\frac{C}{2},\angle BIC=A+\frac{B}{2}+\frac{C}{2}>A$ so WLOG $ \angle BIC = B$ so $ 2A + C = B$ So $\angle IBC=\frac{B}{2} = A + \frac{C}{2}>A$ thus $\angle IBC=C$ so $B=2C$ and so $2A+C=2C$ so the angles are $A$, $2A$ and $4A$ i.e. $\frac{180}{7}$, $\frac{360}{7}$ and $\frac{720}{7}$