# 1997 JBMO Problems/Problem 3

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## Problem

Let $ABC$ be a triangle 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 $AI+BI+CI>BC+KL$.

## Solution $[asy] size(9.22 cm); pair B=(0,0), A=(50,120), C=(140,0), N=(25,60), M=(95,60), I=(60,40), K=(90,60), L=(20,60); draw(B--A--C--B); draw(circle(I,40)); dot(A); label("A",(50,125)); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(N); label("N",(24,65)); dot(M); label("M",M,NE); dot(I); label("I",I,S); dot(K); label("K",K,NW); dot(L); label("L",L,NW); draw(L--C,dotted); draw(K--B,dotted); draw(L--M); draw(anglemark(C,B,A,200)); draw(anglemark(N,K,I,200)); draw(anglemark(A,C,B,200)); draw(anglemark(A,C,B,150)); draw(anglemark(C,L,K,400)); draw(anglemark(C,L,K,450)); [/asy]$

First, by SAS Similarity, $\triangle ANM \sim \triangle ABC,$ so $NM \parallel BC$ and $MN = \tfrac12 BC.$ That means $\angle IBC = \angle IKN,$ and since $\angle IBN = \angle IBC,$ $\triangle NBK$ is an isosceles triangle. Similarly, $\angle MLC = \angle LCB = \angle LCM,$ making $\triangle MLC$ an isosceles as well. Thus, $ML = MC$ and $NB = NK.$

By the Triangle Inequality, $AI + IB > AB,$ and $AI + IC > AC$, and $BI + IC > BC.$ That means \begin{align*} 2(AI + IB + IC) &> AB + AC + BC \\ AI + IB + IC &> NK + ML + MN \\ &> NK + (MN + LN) + MN \\ &> KL + 2 MN \\ &> KL + BC. \end{align*}

## See Also

 1997 JBMO (Problems • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 All JBMO Problems and Solutions
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