$\left( \exists\right) K\in (BC)$ 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 $m\left( \widehat {d,d'}\right)=60^{\circ}.$

If the angles BKM, MKL and LKC are equal, then the triangle would be isosceles (ML is parallel to BC)

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.

Could anyone solve another part of the problem? i.e. "if the triangle $KLM$ is equilateral ,then $\angle BAC = 60^\circ$.

ΒΛΕΠΕ ΣΧΗΜΑ:

Attachments:

ΑΝΤΊΣΤΡΟΦΟ:

Attachments: