Let ABC be an acute triangle. The interior angle bisectors of ∠ABC and ∠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 ∠BAC=60∘.
Problem
Source: Romanian IMO Team Selection Test TST 1999, problem 2
Tags: geometry, incenter, geometry proposed
25.09.2005 01:40
(∃)K∈(BC) such that the triangle MLK is equilateral⟹m(^BKM)=m(^MKL)=m(^CKL)=60∘ a.s.o. I used a remarkable geometrical locus: Let a fixed line d, a fixed point A∉d and a mobile point M∈d. Then the geometrical locus of the point L for which the triangle AML is equilateral is the reunion of the two lines l1, l2 which are symmetrically w.r.t. the line d′, A∈d′, d′⊥d and m(^d,d′)=60∘.
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)
09.10.2014 15:34
Let I be the incenter of ABC. If part:Assume that ∠A=60o, then we have a2=b2−bc+c2⟹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 ∠ILM=∠IML=30o and ∠BIM=∠CIL=60o.Thus NLM is equilateral.
21.03.2020 07:03
Could anyone solve another part of the problem? i.e. "if the triangle KLM is equilateral ,then ∠BAC=60∘.