Notice that triangles $OMC$ and $AOT$ are congruent. Hence we have triangle $OMT$ isosceles with $OT = OM$. Now we have that the medium point of $MT$ lies on the circles $\circ{OCM}$ and $\circ{AOT}$. Therefore this point ($L$) lies on these circles and we have: $\angle{OAL} =\frac{A}{2} = \angle{OTM}$ and hence we are done.

Here is my overcomplicated solution Let $P$ be the intersection point of the diagonals of the rectangle. Note that: 1. Triangles $ANC$ and $AMC$ are congruent (trivial) 2. Triangles $TAO$ and $OCM$ are congruent ($<OCM=90=<TAO(angle chase), OC=AO, CM=TA$) Thus $TO=MO$. 3. Triangles $AOM$ and $TOB$ are congruent ($TO=MO, AO=BO, AM=TB$) Thus $<BTO=<AMO$. Note that $<BTN=<BAC/2$ Thus we have to prove that $<AMT=<OTP$. We prove that triangles $TPO$ and $TMA$ are similar. Note that $BNPO$ is cyclic, thus $<NPO=<OBN$. Thus $<TPO=<TAM$ (AT is tangent and using the above fact+angle chasing). Note that now we are done by trig bash since we have to prove $TP/PO=AM/TA$ (use sines in triangles $BPO$ and $ANB$, and law of sines in triangle $ABC$)

My solution.

Attachments:

Since $ATBN$ is a rectangle. and $ANCM$ is cyclic So, $\angle{TNM}$ = $\angle {TNA}$ + $\angle {ANM}$ $=$ $\frac{\angle{A}}{2}$ + $\angle {C}$=$90$ $=$ $\angle {OCM}$ Also, triangles $AOC$ & $NCM$ are similar and $ACNT$ is a parallelogram. We get, $\frac{MC}{OC}$ $=$ $\frac {MN}{AC}$ $=$ $\frac {MN}{TN}$ Hence, triangles $TNM$ & $ OCM$ are simillar. Then $\angle {TMN}$ $=$ $\angle{OMC}$ $\Rightarrow $ $\angle{OMT}$ $=$ $\angle {NMC}$ $=$ $\frac {\angle{A}}{2}$ And we are done!

Here is the solution I came up with at the test !! Let $K$ be the middle of $AC$ , we know that the angle $AMC$ is equal to 90 degrees therefore $AK=KC=KM$ , therefore the angle $AMK$ is equal to $A/2$ so it suffices to prove that the angles $AMK$ and $TMO$ are equal which is equivalent to proving that the angles $AMT$ and $KMO$ are equal , we know that the angles $TAM$ and $OKM$ are both equal to $90+A$ , therefore we just need to prove that the triangles $TAM$ and $OKM$ are similar which is equivalent to proving that $TA/OK = AM/KM$ , we know that the angles $OAK$ and $CAM$ are both equal to $A/2$ and angles $AKO$ and $AMC$ are both equal to 90 degrees therefore the triangles $OAK$ and $CAM$ are similar and $CM/OK = AM/AK$ and we also know that $TA=BN=NC=CM$ and $KM = AK$ So if we change $TA$ and $CM$ and change $KM$ and $AK$ we’ll be done

spiral similarity kill this problem

Without loss of generality we can assume that $\triangle ABC$ is positively oriented. Let $L$ be the midpoint of $AB$. As $\angle (OA;AL)=\frac{1}{2} \angle (CA;AB)$ and $\angle ALO=\frac{\pi }{2}$, we have $R=R(L,\pi)\circ R(A,-\angle BAC)=R(O,\pi -\angle BAC)(*)$. But obviously $R:M\mapsto T$, so by $(*)$ we are done.

Define $M'$ to be the reflection of $N$ across $AB$, note that this means that $M,M'$ are symmetric about $AN$. Next, let $E,F$ be the midpoints of $AB,AC$. Then, let the circle passing through $T,A,N,B$ centered at $E$ be walled $\omega_E$ and the circle passing through $A,N,C,M$ centered at $F$ is called $\omega_F$. Let $S=AC\cap w_E$. $\textbf{Claim 1: }$ $S\in MM'$. $\textbf{Proof: }$ Note that $MM'\parallel BC$, thus it clearly suffices to show that $SM'\parallel BC$. \[\angle ASM'=\angle ABM' = \angle ABN = \angle ACB\]thus $SM'\parallel BC$, so $S,M,M'$ must be collinear and we are done $\square$. $\textbf{Claim 3: }$ $O$ is the circumcenter of $\triangle TMM'$ $\textbf{Proof: }$ It suffices to show $OT=OM'$ since $OM'=OM$. Note that $AT=BN=BM'$, thus $\angle M'BA=\angle TAB$. Therefore, \[\angle M'BO = \angle M'BA +\angle OBA = \angle TAB +\angle OAB = \angle TAO\]Thus, by LoC \[OM'^2 = OB^2+BM'^2-2\cdot OB\cdot BM' \cdot \cos(\angle OBM')= OA^2+AT^2 - 2\cdot OA\cdot AT \cdot \cos(\angle OAT) = OT^2\]thus, $OM'=OT$ and we are done. (Alternatively we could've just said that $T,M'$ are symmetric about the perp. bis. of $AB$, and $O$ lies on the perp. bis, voila)$\square$ $\textbf{Claim 2: }$ $\angle TM'M=\angle ACB$ $\textbf{Proof: }$ Angle chase. \[\angle TM'M = \angle TM'S = \angle TBS = \angle TBA+\angle ABS=\angle BAN + (90-\angle BAS)=90-\angle BAN=90-\angle NAC = \angle ACB\]$\square$ Thus, we may extract that \[\angle TMO = 90-\frac12 \angle TOM = 90-\angle TM'M = 90-\angle ACB = \angle NAC = \frac12 \angle BAC\]and we're done !$\blacksquare$.

just complete the diagram let $X' ,T'$ be the reflection of $X,T$ over $AM$ notice that $XT'TX'$ is cyclic (nine point circle) and so $OT=OT'=OX=OX'$ $\angle MT'X=\angle BAC$ Ans since $\angle MT'C=\frac{1}{2}\angle BAC$ we have $\angle OXT=\angle CT'X=\frac{1}{2}\angle BAC $

Let D be where MN meets AC. ∠BAC/2 = ∠NAC = ∠CNM = ∠CMN so let's prove MCN and MOT are similar. note that proving MCO and MNT are similar will prove the last one. CO/NT = AO/AB = MC/MN ---> MCO and MNT are similar.

May have undershot a bit in search of suitable geometry problems We have $\angle MCA=\angle NCA=\angle ABC$, so $\overline{CM}$ is tangent to to $(ABC)$. $\overline{AT}$ is also trivially tangent. Further we have $CM=CN=BN=AT$, and obviously $OA=OR$, so by LL congruence $\triangle OCM \cong \triangle OAT \implies \angle TOA=\angle MOC$ and $OT=OM$. As such, $$\angle TOM=\angle AOC=2\angle B \implies \angle OMT=90^\circ-\frac{\angle TOM}{2}=90^\circ-\angle B=\frac{\angle BAC}{2}$$as desired. $\blacksquare$

Notice that $\angle OCM=\angle OCA+\angle ACM=\angle NAC+\angle ACN=90^\circ$. Therefore we have $\triangle OAT\cong\triangle OCM$ because of right angle and $OC=OA, AT=CM$. Let $K=AC\cap TM$. The spiral similarity at $O$ sends $A$ to $C$ and $T$ to $M$. Therefore $K$ lies on $(OTA)$ and $(OCM)$. So $\angle OMT=\angle OMK=\angle OCK=\angle OAC=1/2\angle BAC$.

First, notice that $\triangle OCM \cong \triangle OAT$ because $\angle OAT = \angle OCM =90^\circ$, $OA=OC$, and $AT=CN=CM$. Now let $D$ be the midpoint of $\overline{MT}$. Because $ODAT$ and $ODMC$ are both cyclic, $\measuredangle ODC = \measuredangle ODA$, so $D$ lies on $\overline{AC}$. Now $\angle OMT = \angle OCA = \frac 12 \angle ABC,$ as needed.

turko.arias wrote: Let $ABC$ be an isosceles triangle ($AB = AC$) with its circumcenter $O$. Point $N$ is the midpoint of the segment $BC$ and point $M$ is the reflection of the point $N$ with respect to the side $AC$. Suppose that $T$ is a point so that $ANBT$ is a rectangle. Prove that $\angle OMT = \frac{1}{2} \angle BAC$. Proposed by Ali Zamani Complex numbers make it trivial $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Working in complex numbers, where $\odot (ABC)$ is the C.U. and $a=1, bc=1$ and $o=0$ $N$ is the midpoint of $BC:$ $$\Rightarrow n=\frac{b+c}{2}$$$M$ is the reflection of $N$ in $AC:$ $$\Rightarrow m=c+1-c\overline{n}$$$$\Rightarrow m=\frac{2bc+b-c}{2b}$$$TANB$ is a parallelogram (in that order)$:$ $$\Rightarrow t=b+1-\frac{b+c}{2}$$$$\Rightarrow t=\frac{b-c+2}{2}$$$\angle OMT=\frac{1}{2}\angle BAC \Leftrightarrow \angle OMT=\angle BAO:$ $$\Leftrightarrow \frac{(a-b)(m-0)}{(a-0)(m-t)}\in \mathbb{R}$$$$\Leftrightarrow \frac{(1-b)(2bc+b-c)}{(2b)(\frac{2bc+b-c}{2b}-\frac{b-c+2}{2})}\in \mathbb{R}$$$$\Leftrightarrow \frac{(1-b)(2bc+b-c)}{(2bc+b-c)-(b^2-bc+2b)}\in \mathbb{R}$$$$\Leftrightarrow \frac{(1-b)(2bc+b-c)}{3bc-b-b^2-c}\in \mathbb{R}$$Since $bc=1:$ $$\Leftrightarrow \frac{(1-b)(2+b-c)}{3-b-b^2-c}\in \mathbb{R}$$$$\Leftrightarrow 1\in \mathbb{R}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Note that $$\measuredangle ACM = \measuredangle NCA = \measuredangle ABC $$Which means $CM$ is tangent to $(ABC)$ $$\measuredangle NCM = 2(\measuredangle NCA) = 2(90^{\circ}- \measuredangle NAC)= -2\measuredangle NAC= 2 \measuredangle CAN= \measuredangle NAM$$$$\measuredangle NCM+ \measuredangle MAN= \measuredangle NAM+ \measuredangle MAN= 0^{\circ}$$Which means $NAMC$ is cyclic. Now note: $OA=OC$ $\angle OAT= \angle OCM= 90^{\circ}$ $AT=BN=NC=CM$ Hence $\triangle OCM \cong \triangle OAT$ Finally $$\measuredangle TOM =\measuredangle TON+ \measuredangle NOM= \measuredangle TOA+ \measuredangle NOC +\measuredangle MOC= \measuredangle NOC= 2 \measuredangle NAC$$$$\measuredangle OTM= \measuredangle TMO= \measuredangle NAC= \tfrac{1}{2}\measuredangle BAC$$

This was a bit annoying back when I solved it IDK why. Consider homothety at $N$ with factor $\dfrac{1}{2}$. \begin{align*} T\mapsto T' (\text{midpt. of }AB)\\ M\mapsto M' (\text{foot}(N,AC))\\ O\mapsto O' (\text{midpt. of }NO) .\end{align*} $\angle OMT = \angle O'M'T'\implies$ proving $AT'O'M'$ finishes. But $O'A$ is the angle bisector of $\angle T'AM'$, so proving $O'T'=O'M'$ finishes by Fact $5$. Now $P' = \text{foot}(N,AB)$. $O'M'=O'P'$ due to symmetry. $P$ be the reflection of $N$ over $P'$. So $PT \parallel AB$ and as $N$ is the orthocenter of $\triangle ANB\implies T\in \odot(ANB)$ and $P\in \odot(ANB)\implies ATPB$ isosceles trapezium $\implies OT = OP \implies O'T' = O'P' = O'M'$.

Claim. $\triangle OAT \cong \triangle OCM$. Proof. Note that $OC = OA$, as both are the circumradii of $\triangle ABC$. In addition, $AT = BN = NC = CM$, due to reflections. Lastly, $\angle OAT = 90^{\circ}$, but $\angle OCM = \angle OCA + \angle ACM = \angle NAC + \angle ACN = 90^{\circ}$, as well. Thus, $\triangle OCM \cong \triangle OAT$, as desired. $\blacksquare$ Claim. $\triangle AOC \sim \triangle TOM$. Proof. Note that $\angle AOC = \angle AOC - \angle COM = \angle TOA$, $AO = OC$, and $TO = OM$, then $\triangle TOM \sim \triangle AOC$, as desired. $\blacksquare$ Therefore: \[ \tfrac{1}{2}\angle BAC = \angle NAC = \angle OCA = \angle OMT, \]as desired.

First, we have the following claim. Claim : Triangles $\triangle TAO$ and $\triangle OCM$ are similar. Proof : Simply notice that due to the reflection, $$\angle ACN = \angle ACM = 90-\alpha$$Then, $\angle OAC= \angle ACO$ (by the Isosceles triangle) which gives us that $$\angle OCM = \angle OTA = 90^\circ$$Next, $$OA=OC \text{ and } MC=CN=TA$$Thus, the required claim is in fact true. Now, notice that $$\angle TOM = \angle TOA + \angle AOM = \angle AOM + \angle MOC = \angle AOC$$Thus, $$\angle OMT = \angle OTM = \frac{180-(180-2\alpha)}{2}=\alpha=\frac{A}{2}=\frac{1}{2} \angle BAC$$as required.

I think I overcooked this Let $P$ be the midpoint of $AB$, $Q$ be the midpoint of $ON$, and $R$ be the foot from $N$ to $AC$. Claim: $A, P, Q, R$ are concyclic. Proof. We have: $\frac{NQ}{QA} = \frac{\frac{\cancel{R} \cos{\frac{A}{2}}}{2}}{\frac{\cancel{R} \cos{\frac{A}{2}}}{2} + \cancel{R}} = \frac{\cos{\frac{A}{2}}}{\cos{\frac{A}{2}} + 2}$ Hence, by m-n theorem in $\Delta{APN}$, we have: $(2 \cos{\frac{A}{2}} + 2) \cot{\angle{PQA}} = (\cancel{\cos{\frac{A}{2}}} + 2) \cot{\frac{A}{2}} - \cancel{(\cos{\frac{A}{2}}) \cot{\frac{A}{2}}}$ $\implies \cot{\angle{PQA}} = \frac{\cot{\frac{A}{2}}}{\cos{\frac{A}{2}} + 1}$ $\implies \boxed{\tan{\angle{PQA}} = \sin{\frac{A}{2}} - \tan{\frac{A}{2}}}$. Let $S$ be the foot of the perpendicular from $P$ to $AC$. We have: $\tan{\angle{PRA}} = \tan{\angle{PRS}} = \frac{PS}{SR} = \frac{\frac{\cancel{c}}{2} \sin{A}}{\cancel{c} \cos^2 \frac{A}{2} - \frac{\cancel{c}}{2} \cos{A}}$ $\implies \boxed{\tan{\angle{PRA}} = \frac{\sin{A}}{2 \cos^2{\frac{A}{2}} - \cos{A}}}$. Now it is easy to check that $\tan{\angle{PQA}} = \tan{\angle{PRA}}$. This proves the claim. The claim implies that $\angle{QRP} = \angle{QAP} = \angle{OAB} = \frac{1}{2} \angle{BAC}$. By a homothety centered at $N$ with ratio $2$, we then have $\angle{OMT} = \frac{1}{2} \angle{BAC}$ as desired.