Equality with Fermat Point

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Equality with Fermat Point

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Source: 2012 Baltic Way, Problem 11

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nsato

#1 h Nov 22, 2012, 4:10 PM • 4 Y

Y by soheil74, dmusurmonov, Adventure10, Mango247

Let $ABC$ be a triangle with $\angle A = 60^\circ$ . The point $T$ lies inside the triangle in such a way that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$ . Let $M$ be the midpoint of $BC$ . Prove that $TA + TB + TC = 2AM$ .

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#2 h Nov 22, 2012, 4:47 PM • 2 Y

Y by Adventure10, Mango247

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

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#3 h Nov 24, 2012, 10:22 PM • 2 Y

Y by Adventure10, Mango247

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

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vslmat

#4 h Nov 25, 2012, 10:26 AM • 3 Y

Y by soheil74, Adventure10, Mango247

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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vlwk

#5 h Jul 28, 2016, 10:03 AM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 3 times. Last edited by vlwk, Jul 28, 2016, 10:15 AM

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KRIS17

#6 h Aug 28, 2019, 2:31 PM • 1 Y

Y by Adventure10

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?

This post has been edited 1 time. Last edited by KRIS17, Aug 28, 2019, 2:32 PM

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#7 h Aug 28, 2019, 2:57 PM • 1 Y

Y by Adventure10

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

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KRIS17

#8 h Aug 28, 2019, 3:58 PM • 1 Y

Y by Adventure10

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?

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LKira

#9 h Aug 28, 2019, 4:27 PM • 2 Y

Y by Adventure10, Mango247

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

This post has been edited 1 time. Last edited by LKira, Aug 28, 2019, 4:27 PM

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KRIS17

#10 h Aug 28, 2019, 4:40 PM • 2 Y

Y by Adventure10, Mango247

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 $A$ .

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LKira

#11 h Aug 28, 2019, 5:56 PM • 1 Y

Y by Adventure10

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

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KRIS17

#12 h Aug 28, 2019, 9:21 PM • 1 Y

Y by Adventure10

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).

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1079 posts

#13 h Aug 10, 2021, 10:55 PM

Y by

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 $AX=AA'$ .

$[asy]import olympiad; size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,M,a,I,x,X,T,N,K,O; A=dir(120);B=dir(210);C=dir(330);M=midpoint(B--C);a=2M-A;path w=circumcircle(a,B,C);I=incenter(A,B,C);x=foot(a,A,I);X=2x-a;T=intersectionpoints(A--X,w)[0];N=2M-X;K=extension(X,N,A,I);O=(0,0); draw(A--B--C--cycle,deep);draw(w,deep);draw(A--X,med);draw(A--a,med);draw(B--X--C,deep);draw(B--T,med);draw(C--T,med);draw(circumcircle(A,B,C),deep);draw(A--N,deep); dot("$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$",O,N); [/asy]$

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#14 h Apr 6, 2025, 8:23 AM

Y by

I will provide nice lemma which technically solves the problem.

Lemma: Let $X$ be the point in circumcircle of equilateral triangle $ABC$ . Let's assume $X$ lies on minor arc $BC$ , Then we have $AX=BX+CX$ .

Let's construct equilateral triangles $AZB$ and $AYC$ outside of the $ABC$ . Let $X=ZB \cap YC$ . Let $T'=ZC \cap BY$ . It's easy to conclude that $T'=T$ . Since $AZXC$ is a isosceles trapezoid, we have that $AT=ZC$ . But $ZC=ZT+TC=TB+TA+TC$ , by our lemma and we are done.

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