Let $I$ and $I_a,I_c$ be the incenter and the excenters of $\triangle ABC$ againts $A,C.$ Since $\omega$ is 9-point circle of $\triangle I_aI_bI_c,$ then $\omega$ cuts $II_a,I_aI_b$ at their midpoints $M,N$ $\Longrightarrow$ $MN \parallel BI_b$ $\Longrightarrow$ $\angle BI_bT=\angle TMN=\angle TCI_b.$ Since $\angle BTI_b=\angle CTI_b$ ($TM$ bisects $\angle BTC$), then $\triangle TBI_b \sim \triangle TI_bC$ $\Longrightarrow$ $\tfrac{TB}{TI_b}=\tfrac{TI_b}{TC}$ $\Longrightarrow$ $TB \cdot TC={TI_b}^2.$

nice and easy! my solution = as $ABMT,ATCM$ are concylic hence $\angle MTB=\angle MTC=\frac{\angle A}{2}$ which is due to $AM$ bisecting $\angle BAC$ and thus $\angle CTI_b=\angle I_bTC=180-\frac{\angle A}{2}$ also now let $\angle TCI_b=x$ than $\angle TCB=90+\frac{\angle C}{2}-x$ and thus $\angle CBT=180-\angle TCB=90-\frac{\angle C}{2}+x$ and so $\angle TBI_b=\frac{\angle B}{2}-\angle CBT=\frac{\angle A}{2}-x=\angle CI_bT$ and thus $\triangle TBI_b\sim \triangle TI_bC \Longrightarrow TB.TC={TI_b}^2$ so we are done

My solution: Let $D$ be the midpoint of arc $BC$ containing $A$ $\Rightarrow$ $D$ is the circumcenter of $\triangle{I_bBC}$ and $MB, MC$ are tangent to $(I_bBC)$ $\Rightarrow$ $I_bM$ is the symmedian of $\triangle{I_bBC}$ Moreover, $T$ lies on the symmedian such that $TI_b$ bisects $\angle{BTC}$ $\Rightarrow$ $\triangle{TBI_b}$$\sim$$\triangle{TI_bC}$ and the conclusion follows

Here's a solution: Observe that $\angle BTC = \angle A$ and since $MI_B$ is an angle bisector of $\angle BTC$, $\angle BTI_B = \angle CTI_B = 180 - \frac{1}{2}\angle A$. Now, let $D$ be the second intersection of the circumcircle $\tau$ of $CTI_B$ and $BC$, so we have $\angle CDI_B = \frac{1}{2} \angle A$, which means that $\triangle BCI_B \sim \triangle BI_BD$ and $BI_B$ is tangent to $\tau$, so $\angle TCI_B = \angle TI_BB$, so by AA symmetry, $\triangle BTI_B \sim \triangle I_BTC$, and the result follows. Here's a slightly more difficult problem based on the above: Let $I$ be the incenter of $\triangle ABC$. Show that it is possible to construct a right triangle with side lengths $IM, MT, TI_B$.

Extravert (mostly for convenience) to get Quote: $ABC$ is a triangle with circumcircle $w$. Let $I$ be the incenter, $M$ be the midpoint of arc $BC$ not including $A$, and let $N$ be the midpoint of arc $BAC$. If $NI\cap w=T$, show that $TI^2=TB\cdot TC.$ The problem now falls easily after drawing $(BIC)$: Let $IT\cap (BIC)=X, CT\cap (BIC)=Y$. It's trivial by angle chasing that $BT=TY$ and that $IT=TX$, now PoP kills it.

Let $N$ be the midpoint of arc $\widehat{BAC}$ of $\omega$ and let $I_c$ be the $C$-excenter. It is well-known that $B, C, I_b, I_c$ are inscribed in the circle $\Gamma$ of diameter $\overline{I_bI_c}$ with center $N.$ Moreover, note that $M$ is the midpoint of arc $\widehat{BC}$, implying that $TM$ and $TN$ are the internal and external bisectors of $\angle BTC$, respectively. Therefore, the inversion with power $TB \cdot TC$ combined with a reflection in $TM$ swaps $B$ and $C$ fixes lines $TM$ and $TN.$ It follows that the center of $\Gamma'$ (the image of $\Gamma$) is a point on line $TN.$ However, the center of $\Gamma'$ must also lie on the perpendicular bisector of $\overline{BC}$, implying that the center is $N$ itself. Therefore, $\Gamma$ is fixed under this inversion, and hence $I_b$ is fixed as well. Thus, $TI_b^2 = TB \cdot TC$ as desired. $\square$

the interestion between IbC and w is thw midpoint of arc ABC

Let $D$ be the midpoint of arc $BAC$. And $c$ is a circle with centre $D$ and radius $DB$. $c \cap I_bT=E$. As $ATM=90$ and $AI_b=AE$ so $TE=TI_b$ and it's enough to show that $TE^2=TB.TC$ We know that $MC$ and $MB$ are tangent to $c$. So we can simply show that $\triangle TCE \sim \triangle TEB$ and $TE^2=TB.TC$.

An easy barycentric problem. We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula.

Murad.Aghazade wrote: An easy barycentric problem. We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula. Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof...

artsolver wrote: Murad.Aghazade wrote: An easy barycentric problem. We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula. Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof... Tenplusten gave the sketch of bary solution,and it is good idea to allow others to finish by themselves.For that reason it is not correct to deny others for skipping the long calculations.

In math publications leaving long calculations is acceptable, but you have to give final result or conclusion following from it.

It is just sum of some lemmas:).First note that $MC,MB$ are both tangent to circumcircle of $I_bBC$ by angle chasing.Also $BMCT$ is cyclic so $T$ is the midpoint of $I_b$ symmedian in triangle $BCI_B$ and so two triangles $TI_bC$ is similar to $TBI_b$ from which the relation follows.

Here is complex bash solution. $\omega$ is unit circle, $a=x^2$ $b=y^2$ $ c=z^2$. So $m=-yz$ and $Ib=xy+yz-xz$. We now want $t$. But its easy just solve $$\frac{m-t}{\bar m -\bar t }=\frac{m-Ib}{\bar m -\bar Ib }$$.We get $$t=\frac{x(xy+2yz-xz)}{z+2x-y}$$And now we need to prove $$|(t-y^2)(t-z^2)|=|Ib-t|^2$$. After substituting $t$ and $Ib$ we see that both sides are equal to $$\frac{(z+x)^2(x-y)^2(x-z)^2}{(z+2x-y)^2}$$and we are done. Calculations are not bad can be done under 20 minutes.

Here is a different approach. Denote by $N$ the midpoint of $\widehat{BCA}$. With elementary angle chasing we may show that $MN \parallel BI_B$. This implies $\angle TCI_B =\angle TMN =\angle TI_BB$. This combined with $\angle MTC =\angle MTB$ gives $\triangle I_BTC \sim \triangle I_BBT$, and the desired equality follows $\square$

observe that $TM$ bisects $\angle CTB$ so $\angle CTI_B=\angle BTI_B$ let $(BCI_B)$ intersect $MT$ at $U$ so $\angle CUB=180-\frac{\angle A}{2}$ combining with $\angle CI_B U=\angle CBU$ we are done because we have $\triangle I_B TC \sim \triangle TBI_B$

We will show that $T$ is the $I_b-$dumpty point of $\triangle BI_bC$ which implies the result since the dumpty point is the center of spiral similarity sending $\overline{CI_b}$ to $\overline{I_bC}$. Notice that the dumpty point is the intersection of the $I_b-$symmedian and $(BOC)$, where $O$ is the circumcenter of $\triangle BI_bC$. Now by Ceva's theorem $$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}\cdot\frac{\sin\angle I_bCM}{\sin\angle MCB}\cdot\frac{\sin\angle MBC}{\sin\angle MBI_b}=1$$Since $\angle MBC=\angle MCB$, $$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}=\frac{\sin\angle MBI_b}{\sin\angle I_bCM}=\frac{\cos\frac{C}{2}}{\sin\frac{B}{2}}=\frac{\sin\angle I_bCB}{\sin\angle I_bBC}$$which implies that $MT$ is a symmedian. Moreover, $$\angle BTC=\angle A=2\angle IAC=2\angle II_bC=\angle BOC$$Hence $BI_BOC$ is cyclic which completes the proof.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 35-38. Sincerely Jean-Louis

Solved with nukelauncher. First observe \(\measuredangle BTI_B=\measuredangle I_BTC\) since \(\overline{TM}\) bisects \(\angle BTC\). [asy][asy] size(5cm); defaultpen(fontsize(10pt)); pair B,A,C,I,M,NN,IB,T,X; B=dir(110); A=dir(210); C=dir(330); I=incenter(A,B,C); M=extension( (B+C)/2,origin,A,I); NN=extension( (C+A)/2,origin,B,I); IB=2NN-I; T=2*foot(origin,M,IB)-M; X=2*foot(origin,IB,C)-C; draw(circumcircle(A,C,IB),gray); draw(M--A,gray); draw(B--IB,dashed); draw(C--IB,dashed); draw(unitcircle); draw(A--B--C--cycle); draw(M--IB); dot("\(A\)",A,A); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(M\)",M,M); dot("\(N\)",NN,SW); dot("\(I_B\)",IB,S); dot("\(T\)",T,SE); dot("\(I\)",I,NW); [/asy][/asy] In addition, \[\measuredangle I_BTC=\measuredangle MTC=\measuredangle IAC=\measuredangle BI_BC=\measuredangle BI_BT+\measuredangle TI_BC,\]thus \(\measuredangle BI_BT=\measuredangle I_BTC+\measuredangle CI_BT=\measuredangle I_BCT\), so \[\triangle TBI_B\sim\triangle TI_BC.\]It follows that \(TB\cdot TC=TI_B^2\), as needed

We claim that $\triangle BTI_B\sim\triangle I_bTC$. Obviously, we have that$$\angle BTM=\angle BAM=\angle MAC=\angle MTC\implies \angle I_bTB=\angle CTI_b.$$We obtain that $\angle I_bCT=\angle I_bCA-\angle TCA$ and $$ \angle BI_bT=\angle AII_b-\angle AMI_b=180^\circ-\angle AIB -\angle TCA=90^\circ-0.5\angle ACB-\angle TCA=\angle I_bCT-\angle TCA,$$hence we conclude that $\angle I_bCT=\angle BI_bT$. $TB\cdot TC=TI_b^2$ follows directly from $\triangle BTI_B\sim\triangle I_bTC$.

We claim triangles $CTI_B, I_BTB$ are similar, which clearly implies the result. Relabel $M \rightarrow M_A$ and let $M_B$ be the arc midpoint of minor arc $BC$. First remark $\angle CTM_A = \angle BTM_A = A/2$ so $\angle CTI_B = \angle BTI_B$. Next let $\angle M_B BT = x$. Straightforward angle chasing yields $\angle ACI_B = 90$, so $\angle TCI_B = 90-\frac{C}{2}-\angle ACT = 90-\frac{C}{2}-\frac{A}{2}-x = \frac{B}{2}-x$. But $\angle TI_B B = \frac{2(B/2)-2x}{2}=\frac{B}{2}-x$ by standard angle facts. So $\angle TCI_B = \angle TI_B B$, so we're done.

It is easy to see that $\angle BTI_B = \angle CTI_B$. Let $N$ be the midpoint of the arc $\widehat{AC}$ not containing $B$. Becuase $NIMC$ is a kite, we get that $NM\perp IC$ and we know that $I_BC\perp IC \implies NM\parallel I_BC$. So we get $\angle NBT = \angle NMT = \angle TI_BC \implies \triangle BTI_B \sim \triangle I_BTC \implies TB\cdot TC = TI_B^2$.

It's enough to show, that: $$\triangle BTI_B\sim\triangle CTI_B$$Claim. $MT$ bisects $\angle CTB$.

Claim. $\angle TCI_B = \angle TI_BB$.

Therefore $\triangle BTI_B\sim\triangle CTI_B$ by $ASA$. $\blacksquare$

Nice and cute, notice first that, as $M$ is the midpoint of the arc, then $\angle BTI_b=\angle CTI_b$ but notice we want $TI_b/TC=TB/TI_b$ and thus we just need one more angle to prove $BTI_b \sim I_bTC$. We'll show $\angle I_bBT=\angle TI_bC$, but $\angle I_bBT=B/2-\angle TBC$ and as $\angle BTI_b=90+B/2+C/2=\angle TBC+\angle TI_bC+\angle BCI_b=\angle TBC+\angle TI_bC+90+C/2$ and thus $\angle TBC \angle TI_bC=B/2=\angle TBI_b+\angle TBC$, proving our claim.

Note that $\angle CTI_b=$ $\angle I_bTB=$ $\pi$ $-\frac{\angle BAC}{2}=$ $\pi$ $-\angle BI_bC$ $\implies$ $\angle TBI_b=$ $\angle TI_bC.$ Hence $\triangle BTI_b\sim \triangle I_bTC\implies |TB|\cdot |TC|=|TI_b|^2$ $\blacksquare$

We will prove BTIb and IbTC are similar and it will follow TB . TC = TIb^2. ∠BTIb = 180 - ∠MTB = 180 - ∠CTM = ∠IbTC. Let ∠TBIb = x. ∠ACT = ∠B/2 + x and ∠ACIb = 90 - ∠C/2 so ∠TCIb = 90 - ∠C/2 - ∠B/2 - x. ∠CIbT = ∠A/2 - ∠TCIb = ∠A/2 - 90 + ∠C/2 + ∠B/2 + x = x = ∠TBIb so BTIb and IbTC are similar. we're Done.

Too easy for Iran TST 1. think about using simillar triangle theory 2. just use angle moving for the evidemce of proof.

Easy problem Claim 1: $MB$ and $MC$ are tangent to $(BI_BC)$

Now this combined with $\measuredangle BTI_B = \measuredangle CTI_B$ gives that $T$ is $I_B-$ dumpty point so this implies the conclusion as $\triangle BTI_B \sim \triangle I_BTC$

We uploaded our solution https://calimath.org/pdf/IranTST2015-2.pdf on youtube https://youtu.be/GClFR42aUoU.