Construct equilateral triangle $BCU$ outside triangle $ABC$, $AU$ intersects $(O)$ at $I$, we easily have $A,T,U$ are collinear and $B,C,U,T$ are concyclic, this leads to $TB+TC=TU \Rightarrow TA+TB+TC=AU$. Construct parallelogram $ABNC$, now we only have to prove that $AU=AN$. Notice that $UB, UC$ are tangents of $(O)$ at $B,C$, so we have $ABIC$ is a harmonic quadrilateral and $AI$ is the symmedian of triangle $ABC$. We have $\widehat{BAI}=\widehat{CAM}$ and after several angle calculations, we have $\widehat{AUN}=\widehat{ANU}$, hence proved

As before, construct the parallelogram $ABNC$; additionally, construct the equilateral triangle $\Delta ABP$, $C$ and $P$ on different sides of $AB$. We see that a $60^\circ$ rotation about $B$ will map $A$ to $P$ and $U$ to $C$, hence $AU=PC$ (Torricelli problem). On the other side we see that $\Delta PAC\equiv\Delta NCA$ (s.a.s.), and $AN=PC$, hence $AU=PC=AN$, done. Best regards, sunken rock

Another proof: Let $O$ be the circumcenter of $ABC$. Is is obvious that $T$ must lie on the circumcircle of $BOC, AT$ meets this circle again at $S$. Then $\Delta BSC$ is equilateral. If we choose a point $F$ on $AS$ so that $BF = BT$ then $\Delta BTF$ is also equilateral. But then it is easy to see that $\Delta BTC\cong\Delta BFS$, hence $TC = FS$. Thus $TA + TB + TC = AS$ and to complete the proof it remains to show that $AS = 2.AM$ Notice that $SB, SC$ are in fact tangents to the circumcircle of $ABC$ and $AS$ is the A-symmedian, then $\angle BAS = \angle MAC$. By sinus law in $AMC$: $AM/sinC = MC/sinMAC$ and in triangle $ABS$: $AS/sinC = BS/sinBAS = 2. MC/sin MAC$. Indeed, $AS = 2.AM$ q.e.d. Note; In general, the relationship between A-symmedian $AS$ and median $AM$ is $AM = AS. cosA$

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Let $AB=c$, $BC=a$, $AC=b$, $AM=d$. By Stewart's Theorem we have \[AM^2=(2d)^2=2b^2+2c^2-a^2=2b^2+2c^2-(b^2+c^2-2bc\cos 60^{\circ})=b^2+c^2+bc.\]Hence it suffices to show $b^2+c^2+bc=(TA+TB+TC)^2$. Now Cosine Rule on $\triangle{ATB}$, $\triangle{BTC}$, $\triangle{CTA}$ yields \begin{align} c^2&=TA^2+TB^2-2TA \cdot TB\cos 120^{\circ} \nonumber \\ &=TA^2+TB^2+TA \cdot TB \\ b^2&=TA^2+TC^2+TA \cdot TC \\ b^2+c^2-bc&=b^2+c^2-2bc\cos 60^{\circ} \nonumber \\ &=a^2 \nonumber \\ &=TB^2+TC^2+TB \cdot TC \end{align}Now $(1)+(2)-(3)$ gives $bc=2TA^2+TA \cdot TB + TA \cdot TC - TB \cdot TC$. Denote this as $(4)$, then $(1)+(2)+(4)$ gives $b^2+c^2+bc=4TA^2+TB^2+TC^2+2TA\cdot TB+2TA\cdot TC-TB\cdot TC$. It suffices to show this is equivalent to $(TA+TB+TC)^2=TA^2+TB^2+TC^2+2TA\cdot TB+2TA\cdot TC+2TB\cdot TC \iff TA^2=TB\cdot TC$. To prove this, extend $AT$ to $D$ such that $AT=TD$ and extend $BT$ to $E$ such that $TE=TC$. Then $\angle{CTA}=\angle{ATB}=\angle{DTE} \implies \triangle{DTE} \equiv \triangle{ATC}$. Therefore $\angle{ADE}=\angle{TDE}=\angle{TAC}=60^{\circ}-\angle{BAP}=\angle{ABP}=\angle{ABE} \implies ABDE$ is cyclic, so by Power of a Point $AP \cdot PD=BP \cdot PE \iff PA^2=PB\cdot PC$, as desired. Hence done.

The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle! How come no one observed this? Am I missing something?

KRIS17 wrote: The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle! How come no one observed this? Am I missing something? Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral

Pluto1708 wrote: KRIS17 wrote: The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle! How come no one observed this? Am I missing something? Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60) So why can't we use Central angle theorem?

KRIS17 wrote: Pluto1708 wrote: KRIS17 wrote: The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle! How come no one observed this? Am I missing something? Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60) So why can't we use Central angle theorem? $T$ lies on circumcircle of $BOC,$ not circumcenter Look at the figure at post 4

LKira wrote: KRIS17 wrote: Pluto1708 wrote: Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60) So why can't we use Central angle theorem? $T$ lies on circumcircle of $BOC,$ not circumcenter Look at the figure at post 4 True, but my point is that $O$ and $T$ happen to be one and the same as per the given inputs in the problem using central angle theorem on Point $T$ and vertex $A$.

It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60) So why can't we use Central angle theorem?[/quote] $T$ lies on circumcircle of $BOC,$ not circumcenter Look at the figure at post 4[/quote] True, but my point is that $O$ and $T$ happen to be one and the same as per the given inputs in the problem using central angle theorem on Point $T$ and vertex $A$.[/quote] Did you look at the figure at post 4 ? O and T coincide is just one small case, not the whole problem

Even though most people have given the solution in the general case, I still believe that the problem indirectly asks about the special case where $T$ coincides with circumcenter (due to the inputs given in the problem).

Let $X$ be the point on $AT$ such that $XBC$ is equilateral triangle. Let $A'$ be the reflection of $A$ over $M$. By Ptolemy, $TX=TB+TC$. Hence, we need $AX=AA'$. Reflect diagram over $BC$, note that $X$ goes to $N$, the midpoint of arc $BAC$ and $A'O$ goes to $AK$, where $K$ is the midpoint of arc $BC$ as $\angle BAC=60^\circ$. Thus, $A'X\parallel AN\perp AK$. Also $K$ lies on the perpendicular bisector of $A'X$ as it is center of $(BTC)$. We conclude that $AX=AA'$. [asy][asy]import olympiad; size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,M,a,I,x,X,T,N,K,O; A=dir(120);B=dir(210);C=dir(330);M=midpoint(B--C);a=2M-A;path w=circumcircle(a,B,C);I=incenter(A,B,C);x=foot(a,A,I);X=2x-a;T=intersectionpoints(A--X,w)[0];N=2M-X;K=extension(X,N,A,I);O=(0,0); draw(A--B--C--cycle,deep);draw(w,deep);draw(A--X,med);draw(A--a,med);draw(B--X--C,deep);draw(B--T,med);draw(C--T,med);draw(circumcircle(A,B,C),deep);draw(A--N,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$A'$",a,dir(a)); dot("$X$",X,dir(X)); dot("$T$",T,dir(T)); dot("$N$",N,dir(N));dot("$K$",K,dir(K));dot("$O$",O,N); [/asy][/asy]