Let $X$ be the second intersection of $BV$ with the circle. Since $AU,BX$ are symmetric chords in the circle wrt a line which passes through the center (the perpendicular bisector of $AB$), they have equal lengths, so in order to prove that $BT+TC=AU=BX=BT+TX$, it suffices to show that $TX=TC$. A simple angle chase will show that both $\angle TCX$ and $\angle TXC$ are equal to $\angle A$, so we're done. P.S. The condition on $\angle A$ ensured the fact that $BX=BT+TX$, and not $BX=TX-BT$, for example.

I note $x=m(\widehat {BAU}),y=m(\widehat {CAU})$. Thus, $m(\widehat {BTC})=2(x+y)=2A$ and $\frac{TB}{\sin (C-y)}=\frac{TC}{\sin (B-x)}=\frac{a}{\sin 2A}=\frac{R}{\cos A}\Longrightarrow$ $TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$ $=2R\cdot \cos \frac{B-C-x+y}{2}=2R\sin (C+x)=AU.$

We also discussed this in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=32558 (in this topic, the problem was given in the weaker form "prove that $TB+TC\leq 2R$, where R is the circumradius of triangle ABC", but most of the solutions actually solved it in the stronger form). darij

Let denote by $X$ a point of intersection of tangents in points $B,C$ to circumcircle of $\Delta ABC$ and by $Y$ an image of $C$ in symetry with respect to an angle bisector of $\angle BAU$ (obviosuly $Y$ lies on circumcircle of $\Delta ABC$). If $\alpha_1=\angle BAU, \alpha_2=\angle UAC$ then by simple angle chasing we get that $\angle WVT=2\alpha_1$ and $\angle VWT=2\alpha_2\Longrightarrow \angle BTC=2(\alpha_1+\alpha_2)=\angle BOC$. Now it is easy to see that $TBCX$ is cyclic so by Ptolemy's theorem we get: $BC\cdot TX = CT\cdot BX+BT\cdot CX=(BT+TC)BX$. But $\angle YAU=\angle BTX$, $\angle AUY=\angle BXT$ hence $\Delta AUY$ is similar to $\Delta TXB \Longrightarrow$ $\frac{TX}{BX}=\frac{AU}{BC}\iff \frac{BC\cdot TX}{BX}=AU \Longrightarrow AU=BT+TC$. qed

Too easy for an IMO, isn't it! Just trigonometry.

I found really good solution: Consider $X$ intersection $BT$ with cicrle around $ABC$. Now by easy angle chase we have $TX=TC$ . Now because $BVA$ is isosceles so is $VUX$ . Now we have $BX=AU$ and $BX=BT+TX=BT+TC$.

It's also necessary to argue that T is inside the circle. If the angle BAC isn't the smallest in the triangle, T could be outside the circle, and then BT + TC > AU. That's easy to see.

My solution ( essentially the same as Wolowizard but still posting as it is an IMO problem ). Proof. Firstly let us define a useful point. Let $K =^{Def} BT \cap (ABC)$. We proceed by the following claim. Claim: $TK = TC$. $\rightarrow$ $TK=TC \iff \angle TCK = \angle TKC$. Since $ K \in (ABC) \implies \angle TKC = \angle BKC = \angle BAC$. Hence it suffices to show that $\angle TCK = \angle A$. Since $ \triangle WAC$ is isosceles and so $\angle WCA= \angle CAW \implies \angle TCA = \angle CAW $. So it is now sufficient to prove that $\angle ACK = \angle BAV$. Which is again true as $\triangle VBA$ is isosceles. Now we just need to show that $AU = BK$. $\triangle VAB \sim \triangle VKU \implies \triangle VKU$ is isosceles. Therefore $\boxed { AU = AV + VU = VB + VK = BK = BT + TK = BT +TC} \blacksquare$ Note: For those who are thinking that where was the first condition used, as larkl correctly points out that if $\angle BAC$ is not the smallest then $T$ may lie outside $(ABC)$.

Let $CW$ meet $(ABC)$ again at $B'$. Simple angle chasing shows that $WB'=WU$. So, $AU=AW+WU=AW+WB'=CW+WB'=TC+TB'$. Extend $BT$ to meet $(ABC)$ again at $X$. Then, $$\angle B'BT=\angle B'BA+\angle ABT=\angle B'BA+\angle BAU=\angle B'CA+\angle BAU=\angle CAU+\angle BAU=\angle BAC=\angle BB'C=\angle BB'T$$Hence, triangle $BB'T$ is isosceles. Thus, $AU=TC+TB'=TC+TB$ as wanted.

Almost feel bad for reviving it, but nobody mentioned that this is just follows from symmetry, no angle chase necessary. Just can't pass by. Let $BV, CW$ meet $(ABC)$ at $X,Y$. Then $BX=CY$ like reflections of the same chord $AU$. So it's obvious that $BT+TC=BX=AU$.

Let $L,M,N$ be the midpoints of segments $\overline{AU},\overline{AB},\overline{AC}$, respectively. Claim: $O$ is the $T \text{-excenter}$ of $\triangle TVW$. proof: An easy angle chase gives that $\overline{OV},\overline{OW}$ are the external angle bisectors of $\angle OVT,\angle OWT$, respectively. This proves the claim $\square$ Now as $L$ is the foot of perpendicular from $O$ onto line $\overline{VW}$, so it follows that $TV + VL = TW + WL = \text{semiperimeter of } \triangle TVW$. So we obtain that \begin{align*} BT + CT &= (BV-TV) + (CW + TW) = BV + CW + (TW - TV) = AV + AW + (VL - WL) \\ & = (AV + VL) + (AW - WL) = AL + AL = 2AL = AU \end{align*}This completes the proof of the problem. $\blacksquare$ [asy][asy] size(300); pair C = dir(-30), B = dir(-140), A = dir(110), U = dir(-110) ; dot("$A$",A,dir(90)); dot("$B$",B,dir(B)); dot("$C$",C); draw(unitcircle); pair O = (0,0), M = 0.5*A + 0.5*B, N = 0.5*A + 0.5*C ; dot("$O$",O); dot("$M$",M,dir(180)); dot("$N$",N); draw(B--A--C); dot("$U$",U,dir(-90)); draw(A--U); pair V = extension(A,U,O,M), W = extension(A,U,O,N), T = extension(B,V,C,W); dot("$V$",V,dir(30)); dot("$W$",W,dir(-140)); dot("$T$",T,dir(100)); draw(B--V); draw(C--T); draw(O--M); draw(W--N); pair L = 0.5*A + 0.5*U ; dot("$L$",L,dir(-40)); draw(O--L); draw(B--C); [/asy][/asy]

avatarofakato wrote: Let denote by $X$ a point of intersection of tangents in points $B,C$ to circumcircle of $\Delta ABC$ and by $Y$ an image of $C$ in symetry with respect to an angle bisector of $\angle BAU$ (obviosuly $Y$ lies on circumcircle of $\Delta ABC$). If $\alpha_1=\angle BAU, \alpha_2=\angle UAC$ then by simple angle chasing we get that $\angle WVT=2\alpha_1$ and $\angle VWT=2\alpha_2\Longrightarrow \angle BTC=2(\alpha_1+\alpha_2)=\angle BOC$. Now it is easy to see that $TBCX$ is cyclic so by Ptolemy's theorem we get: $BC\cdot TX = CT\cdot BX+BT\cdot CX=(BT+TC)BX$. But $\angle YAU=\angle BTX$, $\angle AUY=\angle BXT$ hence $\Delta AUY$ is similar to $\Delta TXB \Longrightarrow$ $\frac{TX}{BX}=\frac{AU}{BC}\iff \frac{BC\cdot TX}{BX}=AU \Longrightarrow AU=BT+TC$. qed can someone please tell what was the motivation behind constructing $Y$ ?

Let $D=BT\cap (ABC)$ and $E=CT\cap (ABC)$. Note that $\angle TCD=\angle ECD=\angle ECA+\angle ACD=\angle ABV+\angle WCA=\angle CAW+\angle WAB=\angle CAB=\angle CDT$, thus $TC=TD$, similarly $TB=TE$. Also as $CAEU$ is a isosceles trapezoid, hence $$AU=UW+AW=CW+WE=TC+TW+WE=TC+TE=TC+TB.$$We are done.

Evan's video made me try this. I agree with him, this might be the easiest (kinda) recent IMO #2 Let $BT \cap \odot(ABC) = X$ Notice that $ABUX$ is an isosceles trapezoid because $$\angle BAU = \angle BAV = \angle ABV = \angle ABU$$Now $\angle BTC = 2 \cdot \angle BAC$ which follows trivially from angle chasing. This implies that $$\angle BTC = 2\cdot \angle TXC \implies TB=TC$$Now $$AU = BX = TB + TC \ \blacksquare$$

Let $D=\overline{BT}\cap (ABC).$ Since $$\angle BAU=\angle VBA=\angle DUA,$$$ABUD$ is a cyclic isosceles trapezoid. Thus, $$\angle TDC=\angle BAU+\angle UAC=\angle ACD+\angle ACW=\angle DCT$$and $TC=TD.$ Then, $$AU=AV+VU=TB+VT+VD=TB+VT+TC-VT=TB+TC.$$$\square$

Let $BT$ meet $(ABC)$ again at $B_1$. Observe that $$\angle BTC = 180 - \angle TBC - \angle TCB = \angle BAC + \angle ABT + \angle ACT$$$$= \angle BAC + \angle BAV + \angle CAW = 2 \angle BAC$$and $$\angle TB_1C = \angle BB_1C = \angle BAC$$so $$\angle TCB_1 = \angle BTC - \angle TB_1C = 2 \angle BAC - \angle BAC = \angle BAC = \angle TB_1C$$which yields $TB_1 = TC$. Thus, we know $$TB + TC = TB + TB_1 = BB_1$$so it suffices to show $ABUB_1$ is an isosceles trapezoid. Indeed, we have $$\angle ABB_1 = \angle ABV = \angle VAB = \angle UAB = \angle UB_1B$$so $AB \parallel UB_1$, which finishes since $ABUB_1$ is cyclic. $\blacksquare$ Remarks: This problem seems scarier than it actually is. Once you find $\angle BTC = 2 \angle A$, however, adding $B_1$ follows somewhat naturally, since we'd like to force $\angle BTC$ to be an exterior angle in order to form some isosceles triangle(s).

All you need to do is to construct a parallelogram isosceles trapezoid! Let $D = \overline{BT} \cap (ABC)$ and $E = \overline{CT} \cap (ABC)$. Observe that both $ABCD$ and $ABCE$ are isosceles trapezoids. Furthermore, by a simple arc chase, $DEBC$ is an isosceles trapezoid too. Ergo $$BT+CT=CE=AU.$$

Let $\angle BAU=\theta_b$ and $\angle CAU=\theta_c$. Of course, $$\angle ABT=\angle BCU=\theta_b$$and $$\angle ACT=\angle CBU=\theta_c.$$Note that $\theta_b+\theta_c=\alpha.$ We have $$\angle TBC=\beta-\theta_b$$and $$\angle TCB=\gamma-\theta_c,$$which means that $$\angle BTC=180-\beta-\gamma+\theta_b+\theta_c=2\alpha,$$so $BTOC$ is cyclic. WLOG the circumradius of $\triangle ABC$ is $\frac{1}{2}$. Then, $a=\sin\alpha$ etc. Thus, by law of sines, $$BT+CT$$$$=a(\frac{\sin\beta-\theta_b+\sin\gamma-\theta_c}{\sin2\alpha})$$and $$AU=\sin\angle ABU=\sin\beta+\theta_c.$$Since $$\sin2\alpha=2\sin\alpha\cos\alpha=2a\cos\alpha$$, it suffices to show that $$\frac{\sin\beta-\theta_b+\sin\gamma-\theta_c}{2\cos\alpha}=\sin\beta+\theta_c.$$By sum to product, we have $$\sin\beta-\theta_b+\sin\gamma-\theta_c$$$$2\sin(\frac{\beta+\gamma-\alpha}{2})\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2})$$$$=2\cos\alpha\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2}).$$Therefore, the desired equation simplifies to just $$\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2})=\sin\beta+\theta_c.$$However, we have $$(\beta+\theta_c)-(\frac{\beta-\theta_b-\gamma+\theta_c}{2})$$$$=\frac{\beta+\gamma+\theta_b+\theta_c}{2}=90,$$hence done.

Sketch: $\ell_1$ and $\ell_2$ denote the perpendicular bisectors of $AB$ and $AC$ respectively. Reflect $U$ over $\ell_1$ to get $U'$ and over $\ell_2$ to get $U''$. $UCAU''$ and $UBAU'$ are isosceles trapeziums. We also get that $\overline{T-C-U''}$ and $\overline{T-B-U'}$ are collinear from a litlil bit of angel chessing. Then chasing some more angels, we get $BCU'U''$ is also an isosceles trapezium $\implies AU = U''C = U''T + TC = TB + TC$.

Let $CT$ meet the circumcircle of $ABC$ at another point $C'$, we claim that $UC'AC$ is an isoceles trapezoid : Proof : This follows from $\angle{UCA} = \angle{UCC'} + \angle{C'CA} = \angle{UAC'} + \angle{WCA} = \angle{UAC'} + \angle{WAC} = \angle{UAC'} + \angle{UAC} = \angle{C'AC}$ and the fact that $U$ and $C'$ are on the same side of $AC$. Now an isoceles trapezoid has equal diagonals, thus $CC' = AU$ and thus $CT + TC' = AU$, hence we need to prove that $TC' = BT$ Proof : This follows from $\angle{TBC'} = \angle{TBA} + \angle{ABC'} = \angle{LBA} + \angle{ACC'} = \angle{LAB} +\angle{ACW} = \angle{UAB} + \angle{WAC} = \angle{UAB} + \angle{UAC} = \angle{BAC} = \angle{BC'C} = \angle{BC'T}$ Thus triangle $TBC'$ is isoceles at $T$, hence $TB = TC'$ Summarizing : $AU = CC' = CT + TC' = TC + TB$

Virgil Nicula wrote: I note $x=m(\widehat {BAU}),y=m(\widehat {CAU})$. Thus, $m(\widehat {BTC})=2(x+y)=2A$ and $\frac{TB}{\sin (C-y)}=\frac{TC}{\sin (B-x)}=\frac{a}{\sin 2A}=\frac{R}{\cos A}\Longrightarrow$ $TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$ $=2R\cdot \cos \frac{B-C-x+y}{2}=2R\sin (C+x)=AU.$ Why BTC are equal 2x+2y ?

guptaamitu1 wrote: Let $L,M,N$ be the midpoints of segments $\overline{AU},\overline{AB},\overline{AC}$, respectively. Claim: $O$ is the $T \text{-excenter}$ of $\triangle TVW$. proof: An easy angle chase gives that $\overline{OV},\overline{OW}$ are the external angle bisectors of $\angle OVT,\angle OWT$, respectively. This proves the claim $\square$ Now as $L$ is the foot of perpendicular from $O$ onto line $\overline{VW}$, so it follows that $TV + VL = TW + WL = \text{semiperimeter of } \triangle TVW$. So we obtain that \begin{align*} BT + CT &= (BV-TV) + (CW + TW) = BV + CW + (TW - TV) = AV + AW + (VL - WL) \\ & = (AV + VL) + (AW - WL) = AL + AL = 2AL = AU \end{align*}This completes the proof of the problem. $\blacksquare$ [asy][asy] size(300); pair C = dir(-30), B = dir(-140), A = dir(110), U = dir(-110) ; dot("$A$",A,dir(90)); dot("$B$",B,dir(B)); dot("$C$",C); draw(unitcircle); pair O = (0,0), M = 0.5*A + 0.5*B, N = 0.5*A + 0.5*C ; dot("$O$",O); dot("$M$",M,dir(180)); dot("$N$",N); draw(B--A--C); dot("$U$",U,dir(-90)); draw(A--U); pair V = extension(A,U,O,M), W = extension(A,U,O,N), T = extension(B,V,C,W); dot("$V$",V,dir(30)); dot("$W$",W,dir(-140)); dot("$T$",T,dir(100)); draw(B--V); draw(C--T); draw(O--M); draw(W--N); pair L = 0.5*A + 0.5*U ; dot("$L$",L,dir(-40)); draw(O--L); draw(B--C); [/asy][/asy] Wonderful solution, thank you

Let $BV$ meet the circumcircle again at $C'$. Claim: $TC = TC'$ Proof. Angle chasing. Then we have: $AU = AV + VU = BV + VC' = BC' = BT + TC' = TB + TC$ as desired.

WLOG assume that $W$ is between $T, C.$ Let $D=BV \cap (\triangle ABC), E=CW \cap (\triangle ABC).$ Then due to symmetry from the perpendicular bisectors we have that arcs $AE=CU,$ and arcs $AD=BU.$ Adding these yields arcs $ED=BC,$ therefore $TE=TB.$ But note that $$TB+TC=AU \iff TB+TC=CE \iff TB=TE,$$so we are done. QED