Move if needed. I thought I had this done, but I realized I didn't. Please Help Let ABC be a trinagle and let I be the incenter. Let N, M be the midpoints of the sides AB and CA respectively. The lines BI and CI meet MN at K and L respectively. Prove that AI+BI+CI>BC+KL. Tahnks
Problem
Source: JBMO 1997
Tags: geometry, incenter, inequalities, inradius, triangle inequality, angle bisector
mathgeniuse^ln(x)
20.02.2007 23:26
I tried using some sort of form of Triangle Inequality but I couldn't figure anything out.
scorpius119
21.02.2007 04:48
The triangle inequality gives $BI+CI>BC$, so the result will follow from the stronger inequality $AI>KL$. A bit of angle chasing reveals $MK=MB$ and $NL=NC$. So we can express $KL$ in terms of these lengths with
\[KL=MK+NL-MN=MB+NC-MC=\frac{AB+AC-BC}{2}\]
which happens to be the length of the tangent line from $A$ to the incircle. Let $r$ be the inradius, then $AI=\sqrt{KL^{2}+r^{2}}>KL$, as desired.
quantumbyte
23.04.2011 05:29
Can someone explain to me how a bit of angle chasing tells us that MK=MB?
jwang780
23.04.2011 06:44
The fact that BI is an angle bisector and that alt. int. angles of parallel lines are congruent show that MKB is an iscoceles triangle.
cheeseyicecream
13.06.2011 03:28
How do we know K and L are located between M and N?