Indija TS 2008

Yes. Draw $AK$ parallel to $BC$ and $AK=BD$. Join $K$ to $E$ and $F$. Choose $L$ on $BC$ with $L$ between $B$ and $C$, such that $BD=LC$. Note that $KB$ is parallel to $AD$ and $KB=AD$. Hence $\angle KBE=\angle APE=60^{\circ }$. On the other hand, $KB=AD=BE$. Thus $KBE$ is an equilateral triangle. This gives $\angle BEK=60^{\circ }=\angle BQR$. Thus $CF$ is parallel to $EK$ but $CF=BE=EK$. We conclude that $FCEK$ is a parallelogram. Since $CL$ is parallel to $AK$ and $CF$ is parallel to $KE$ we have $\angle LCF=\angle AKE$.

Consider the triangles $LCF$ and $AKE$. We have $LC=AK$, $CF=KE$ and the included angles are also the same. Hence $LCF$ is congruent to $AKE$. Moreover $LF$ is parallel to $CA$. Thus $$\frac{BF}{FA}=\frac{BL}{LC}=\frac{CD}{BD}=\frac{1}{\lambda },\text{ say.}$$ Consider the transversal $FQC$ of the triangle $ABD$. By Menelaus' theorem $$\frac{BF}{FA}\cdot \frac{AP}{PD}\cdot \frac{CD}{CB}=1.$$ Thus $$\frac{AP}{PD}=\frac{FA}{BF}\cdot \frac{CB}{CD}=\lambda \left(\frac{CD+DB}{CD}\right)=\lambda (1+\lambda ).$$ Similarly, we get $$\frac{BQ}{QE}=\frac{CR}{RF}=\lambda (1+\lambda ).$$ We thus have $$\frac{AP}{PD}=\frac{BQ}{QE}=\frac{CR}{RF}.$$ This equality implies $$\frac{AD}{PD}=\frac{BE}{QE}=\frac{CF}{RF}.$$ Using $AD=BE=CF$, we get $PD=QE=RF$. But then $AP=BQ=CR$ and hence $AR=BP=CQ$. Consider the triangles $ABP$ and $BCQ$. Observe that $\angle BPA=120^{\circ }=\angle CQB$, $BP=CQ$ and $AP=BQ$. Thus $ABP$ and $BCQ$ are congruent triangles. This gives $AB=BC$. Similarly, we obtain $BC=CA$.

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Alternative solution.

Let us write $PQ=QR=RP=x$, $BE=CF=AD=y$, $AR=u$, $BP=v$ and $CQ=w$. Using the collinearity of the points $A$, $F$, $B$ with respect to the triangle $PQR$, Menelaus' theorem gives $$\frac{RA}{AP}\cdot \frac{PB}{BQ}\cdot \frac{QF}{FR}=1.$$ (we are using only the lengths). This may be written in the form $$\frac{u}{u+x}\cdot \frac{v}{v+x}\cdot \frac{y-w}{y-(w+x)}=1.$$ Simplification gives $$u+v=\frac{(x+u)(x+v)(x+w)-uvw}{yx}-x.$$ Note that the expression on the right side is symmetric in $u,v,w$. It follows that $$u+v=v+w=w+u.$$ Thus $u=v=w$ and hence $AP=BQ=CR$. As in the earlier solution, it follows that the triangles $ABP,BCQ,CAR$ are congruent, and hence $AB=BC=CA$.