Let $ABC$ be an acute triangle. The interior angle bisectors of $\angle ABC$ and $\angle ACB$ meet the opposite sides in $L$ and $M$ respectively. Prove that there is a point $K$ in the interior of the side $BC$ such that the triangle $KLM$ is equilateral if and only if $\angle BAC = 60^\circ$.
Problem
Source: Romanian IMO Team Selection Test TST 1999, problem 2
Tags: geometry, incenter, geometry proposed
Virgil Nicula
25.09.2005 01:40
$\left( \exists\right) K\in (BC)$ such that the triangle $MLK$ is equilateral$\Longrightarrow m(\widehat {BKM})=m(\widehat {MKL})=m(\widehat {CKL})=60^{\circ}\ a.s.o.$ I used a remarkable geometrical locus: Let a fixed line $d$, a fixed point $A\not\in d$ and a mobile point $M\in d$. Then the geometrical locus of the point $L$ for which the triangle $AML$ is equilateral is the reunion of the two lines $l_1,\ l_2$ which are symmetrically w.r.t. the line $d',\ A\in d',\ d'\perp d$ and $m\left( \widehat {d,d'}\right)=60^{\circ}.$
lazybitch
24.03.2008 00:25
If the angles BKM, MKL and LKC are equal, then the triangle would be isosceles (ML is parallel to BC)
Sardor
09.10.2014 15:34
Let $ I $ be the incenter of $ ABC $. If part:Assume that $ \angle A=60^o $, then we have $ a^2=b^2-bc+c^2 \implies BM+CL=BC $.If the lines perpendicular to $ BI,CI $ through $ M,L $, respectively, intersects $ N $ , then $ N $ must lies on $ BC $.On the other hand $ AMIL $ is cyclic and $ \angle ILM= \angle IML= 30^o $ and $ \angle BIM= \angle CIL=60^o $.Thus $ NLM $ is equilateral.
Alpha314159
21.03.2020 07:03
Could anyone solve another part of the problem? i.e. "if the triangle $KLM$ is equilateral ,then $\angle BAC = 60^\circ$.
