A not very nice solution uses coordinates: taking origin at M(0,0), C(2a,0),B(-2a,0), A(0,2h) then let X(p,q) with p>0 and q>0 because the minor arc MA is in the first quadrant. If T(u,v) then up+vq=0 because MX is perpendicular to MT. After a lot of computations one get tan(MTB-CTM) = a/h, which in fact does not depends on X(p,q).

Denote E be the midpoint of BT, then the quadrilateral XMET inscribed the circle whose diameter is XT. Since ME || CT, $ \angle MTB - \angle MTC = \angle MTE - \angle EMT = \angle MXE - \angle EXT = \angle MXE - \angle EXB = \angle MXB = \angle MAB$

Let the perpendiculers from $ B$ and $ C$ to $ XM$ and $ TM$ respectively intersect at $ P$. $ \angle BPC = 90^0\implies BM = CM = PM$. So $ P$ is the reflection of $ C$ at $ TM$ and of $ B$ at $ XM$. $ XT = XB = XP\implies$ $ X$ is the circumcenter of $ TBP$. $ \angle MTB - \angle CTM = \angle MTB - \angle PTM = \angle BTP = \frac {1}{2}\angle BXP =$ $ = \angle BXM = \angle BAM\ \ \text{(Const.)}$

Νotice here that the condition AB=AC is absolutely redundant.... Hence the problem can be generalized to any triangle ABC,using the same method of proof as mr.danh's idea,which is also the same as mine. Cheers, Nick

Sorry for the revival; why is XMET cyclic in mr.danh's solution? mr.danh wrote: then the quadrilateral XMET inscribed the circle whose diameter is XT

AwesomeToad wrote: Sorry for the revival; why is XMET cyclic in mr.danh's solution? $TX=BX$ so, $\triangle TXB$ is isosceles since $E$ is the midpoint $XE$ is the median as well as altitude of the triangle So, $\angle TEX=90^\circ$ it is given $\angle TMX=90^\circ$ So, $TEMX$ is cyclic with $TX$ being the diameter

Let $C' $ be the symetrical of $C$ wrt $TM$ and $L=AM\cap TC'$ and $Y$ the reflexion of $X$ wrt the midpoint of $AB$. and $E$ the midpoint of $TB$ then $ \angle{MTB}-\angle{CTM} = \angle LTB$ it is enough to show that $ATLB$ is cyclic to conclude that $\angle{MTB}-\angle{CTM} = \angle MAB$ we have: $\angle ALB=\frac{\pi}{2}-\angle TMB - \angle MTC = \frac{\pi }{2}- \angle AYX -\angle TME = \frac{\pi}{2}-\angle AYX -\frac{\angle TXB}{2}=\angle XBT -\angle ABX =\angle ABT $ Hence $ATBL$ is cyclic which leads to \[ \angle{MTB}-\angle{CTM}= \angle LTB = \angle MAB \]

We will prove the result for an arbitrary $\triangle ABC.$ Define $\Gamma \equiv \odot(X, XB)$ and $\omega \equiv \odot(ABM).$ Let $S$ be the reflection of $T$ in $M$ and let $MT$ meet $\omega$ for a second time at $Y.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 937.3756443093395, xmax = 1122.3822298074242, ymin = 958.746170568615, ymax = 1051.0863181805378; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((981.6398411686986,1019.9779690940252),4.894354113705946,-41.70511190447346,-22.045779161991184)--(981.6398411686986,1019.9779690940252)--cycle, qqwuqq); draw(arc((1036.6505840120194,970.956380225791),4.894354113705946,138.29488809552655,157.95422083800884)--(1036.6505840120194,970.956380225791)--cycle, qqwuqq); draw(arc((976.8060711531035,995.1907717416409),4.894354113705946,78.96519747630964,98.62453021878137)--(976.8060711531035,995.1907717416409)--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((976.8060711531035,995.1907717416409)--(1041.4843540276145,995.7435775781753)); draw(circle((992.8090020683957,1014.8258301910363), 25.330403040433268), linetype("2 2") + blue); draw(circle((1013.9145017344058,1000.8191459995926), 37.532841845441226), linetype("2 2") + blue); draw((1036.6505840120194,970.956380225791)--(971.7035024023855,1028.83251438248)); draw((976.8060711531035,995.1907717416409)--(981.6398411686986,1019.9779690940252), red); draw((1041.4843540276145,995.7435775781753)--(981.6398411686986,1019.9779690940252), red); draw((976.8060711531035,995.1907717416409)--(1036.6505840120194,970.956380225791), red); draw((1036.6505840120194,970.956380225791)--(1041.4843540276145,995.7435775781753), red); draw((971.7035024023855,1028.83251438248)--(976.8060711531035,995.1907717416409)); /* dots and labels */ dot((990.,1040.),linewidth(3.pt) + dotstyle); label("$A$", (990.7241041487343,1041.9713196788784), NW * labelscalefactor); dot((976.8060711531035,995.1907717416409),linewidth(3.pt) + dotstyle); label("$B$", (974.5727355735047,993.1697945016816), SW * labelscalefactor); dot((1041.4843540276145,995.7435775781753),linewidth(3.pt) + dotstyle); label("$C$", (1042.4411126168939,993.1697945016816), SE * labelscalefactor); dot((1009.145212590359,995.4671746599081),linewidth(3.pt) + dotstyle); label("$M$", (1008.5069240951992,992.1909236789404), S * labelscalefactor); label("$\omega$", (960,1010), NE * labelscalefactor,blue); dot((1013.9145017344058,1000.8191459995926),linewidth(3.pt) + dotstyle); label("$X$", (1015.5221649915111,1000.1850353979937), E * labelscalefactor); dot((971.7035024023855,1028.83251438248),linewidth(3.pt) + dotstyle); label("$Y$", (967.3520911855517,1029.8774503544778), NE * labelscalefactor); label("$\Gamma$", (1042,1032), NE * labelscalefactor,blue); dot((981.6398411686986,1019.9779690940252),linewidth(3.pt) + dotstyle); label("$T$", (981.6513640015929,1022.2730164414478), N * labelscalefactor); dot((1036.6505840120194,970.956380225791),linewidth(3.pt) + dotstyle); label("$S$", (1037.7099036403113,967.3928628361625), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $BTCS$ is a parallelogram because its diagonals bisect one another. Hence, $\measuredangle CTM = \measuredangle BST.$ Meanwhile, $S \in \Gamma$ because $XM$ is the perpendicular bisector of $\overline{ST}.$ Then since $\measuredangle YBX = \measuredangle YMX = 90^{\circ}$, it follows that $YB$ is tangent to $\Gamma.$ Thus, the Alternate Segment Theorem yields $\measuredangle BST = \measuredangle YBT.$ Hence, \begin{align*} \measuredangle MTB - \measuredangle CTM = \measuredangle MTB - \measuredangle YBT = \measuredangle BYT. \end{align*}But clearly $\measuredangle BYT = \measuredangle BYM = \measuredangle BAM$ is fixed, as desired. $\square$

Here is an analytic solution: We use complex numbers and set $(ABM)$ to be the unit circle. Furthermore, let $a=-1,b=1$. Then it is clear that $c=2m-1$. Now since $TM \perp XM$, we have \[\frac{t-m}{x-m}=-\frac{\bar t - \frac{1}{m}}{\frac{1}{x}-\frac{1}{m}} \implies t-m=mx \bar t - x \implies \bar t = \frac{t-m+x}{mx}\]Now since $XT=XB$, we have \[(t-x)(\bar t - \tfrac{1}{x})=(1-x)(1-\tfrac{1}{x}) \implies t \bar t - \frac{t}{x}- \bar t x + x +\frac{1}{x} -1=0\]Plugging in our first equation into our second, we get \[t\left(\frac{t-m+x}{mx}\right)-\frac{t}{x}-\left(\frac{t-m+x}{m}\right) + x + \frac{1}{x} -1=0 \implies t^2 - 2mt + (m-1)x^2+m=0\]Now let $r$ be the root that lies within angle $AMB$ and call the other root $s$. It is clear that these roots are reflections about $M$ (and thus $r,m,s$ are collinear), so now consider the following complex number which has argument $\angle MTB - \angle CTM$ \[\frac{\frac{s-r}{1-r}}{\frac{2m-1-r}{s-r}}=-\frac{(s-r)^2}{(1-r)(1-s)}=\frac{4m^2-4((m-1)x^2+m)} {(m-1)(1-x^2)}=\frac{4(m-x^2)}{1-x^2}\]Now notice that since $|x|=1$, $x^2$ lies on the unit circle, and so we see that the argument of this complex number is $\angle MXB=\angle MAB$ which is independent of $X$, as desired. $\Box$

We do not use complex numbers

When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial.

My solution:(similar MStang and mr.danh) Let $Q$ be the midpoint of $BT,$ then $\angle XQT =90,$ also $\angle XMT=90\to XMQT$ is cyclic. Also $MQ$ is the midline of $\triangle CBT\to MQ\parallel CT.$ Then $\angle MTB-\angle CTM=\angle MXQ-\angle QMT=\angle MXQ-\angle QXT=\angle MXQ-\angle BXQ=\angle MXB=\angle MAB.$ As desired.

I generally don't like geometry, but the locus of $T$ as $X$ varies on $\omega_{ABM}$ is (showed in black) a cute shape, which surprisingly is independent of $\angle BAM$: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(27.1cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 22.8, ymin = -5.76, ymax = 6.3; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((6.38,1.), 4.011234224026315), linewidth(2.) + blue); draw((9.02,4.02)--(3.74,-2.02), linewidth(2.) + green); draw((3.74,-2.02)--(14.408602941635356,-1.9233120679801736), linewidth(2.) + green); draw((9.02,4.02)--(9.074301470817678,-1.9716560339900866), linewidth(2.) + green); draw((xmin, -0.5299145299145296*xmin + 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Let $N$ be the midpoint of $BT$. Now since $M$ is the midpoint of $BC$, we hace $MN||CT$ so $\angle CTM = \angle TMN$. Also $\angle TNX = 90 = \angle XMT$, hence $\omega_{TNMX}$ exists. So coupled with this, we get $\angle CTM = \angle TMN = \angle TXN$ (call this $\spadesuit$) Now, $\angle BTM $ $$= \angle BTX - \angle XTM = \angle BTX + \angle TXM - 90 = \angle TXM + (\angle BTX - 90)$$$$= \angle TXM - \angle TXN = \angle NXM = \angle NXB + \angle BXM = \angle TXN + \angle BXM \overset{\spadesuit}{=} \angle CTM + \angle BAM$$, hence $\angle BTM - \angle CTM = \angle BAM$, independent of $X$, as desired.

Let $N$ be not the midpoint of $BT$.

delegat wrote: Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. Author: Farzan Barekat, Canada Let $N$ be the midpoint of $AB.$ Define $\omega,\Omega$ to be the circles $(AMB), \mathcal{C}(X,XB).$ Let $Y$ be the antipode of $X$ in $\omega.$ Clearly $T \in \overline{MY}, T \in \Omega$ by the conditions. Also, $\measuredangle XBY=\pi/2$ implies that $YB$ is tangent to $\Omega.$ Claim: $\measuredangle YBT=\measuredangle CTM.$ Proof: Let $L$ be the reflection of $T$ over $M.$ Then $TBLC$ is a parallelogram. Further, since $TM=ML$ and $\measuredangle XMT=\pi/2,$ hence $TX=XL.$ So, $L \in \Omega.$ Since $YB$ is tangent to this circle, hence $\measuredangle YBT=\measuredangle BLT=\measuredangle CTL.$ $\square$ Thus we have $$\measuredangle MTB-\measuredangle CTM=\measuredangle MTB-\measuredangle YBT=\measuredangle TYB=\measuredangle MAB$$which is independent of the position of $X.$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1086.5975564873966, xmax = 1190.1243804115663, ymin = -879.117740078539, ymax = 608.3405920287947; /* image dimensions */ draw(arc((-586.1713229414271,-304.6162131768956),108.41533032852203,92.83654263451801,108.26645574818497)--(-586.1713229414271,-304.6162131768956)--cycle, linewidth(0.2) + red); draw(arc((-603.0349055618302,35.73566225078713),108.41533032852203,-35.66215225285588,-20.232239139188955)--(-603.0349055618302,35.73566225078713)--cycle, linewidth(0.2) + red); /* draw figures */ draw(circle((-353.0915854617429,10.809060298607557), 392.19799485807476), linewidth(0.4)); draw((-586.1713229414271,-304.6162131768956)--(-120.01184798205874,326.2343337741107), linewidth(0.4)); draw((-120.01184798205874,326.2343337741107)--(337.5007033357593,-310.9150475089905), linewidth(0.2)); draw((337.5007033357593,-310.9150475089905)--(-586.1713229414271,-304.6162131768956), linewidth(0.4)); draw((-120.01184798205874,326.2343337741107)--(-124.3353098028339,-307.76563034294304), linewidth(0.4)); draw((-728.1424189633996,125.50927590610821)--(21.959248039913632,-103.89115530889305), linewidth(0.2)); draw((-586.1713229414271,-304.6162131768956)--(-728.1424189633996,125.50927590610821), linewidth(0.4)); draw((-728.1424189633996,125.50927590610821)--(-124.3353098028339,-307.76563034294304), linewidth(0.4)); draw(circle((21.959248039913632,-103.89115530889305), 640.4009214688908), linewidth(0.2) + linetype("2 2")); draw((-603.0349055618302,35.73566225078713)--(354.3642859561624,-651.2669229366732), linewidth(0.4)); draw((337.5007033357593,-310.9150475089905)--(354.3642859561624,-651.2669229366732), linewidth(0.4)); draw((-603.0349055618302,35.73566225078713)--(337.5007033357593,-310.9150475089905), linewidth(0.4)); draw((-586.1713229414271,-304.6162131768956)--(354.3642859561624,-651.2669229366732), linewidth(0.4)); draw((-603.0349055618302,35.73566225078713)--(-586.1713229414271,-304.6162131768956), linewidth(0.4)); draw((-124.3353098028339,-307.76563034294304)--(337.5007033357593,-310.9150475089905), linewidth(0.2)); draw((-603.0349055618302,35.73566225078713)--(-728.1424189633996,125.50927590610821), linewidth(0.2)); /* dots and labels */ dot((-353.0915854617429,10.809060298607557),dotstyle); label("$N$", (-377.5612961388625,42.4125677139054), NE * labelscalefactor); dot((-120.01184798205874,326.2343337741107),dotstyle); label("$A$", (-110.85958353069825,348.1437992403399), NE * labelscalefactor); dot((-120.0118479820587,326.2343337741107),linewidth(4pt) + dotstyle); dot((-586.1713229414271,-304.6162131768956),linewidth(4pt) + dotstyle); label("$B$", (-622.5799426813222,-360.89246110819965), NE * labelscalefactor); dot((-124.3353098028339,-307.76563034294304),dotstyle); label("$M$", (-141.2158760226844,-365.2290743213406), NE * labelscalefactor); dot((337.5007033357593,-310.9150475089905),linewidth(4pt) + dotstyle); label("$C$", (346.6531104556648,-293.6749563045155), NE * labelscalefactor); dot((21.959248039913632,-103.89115530889305),dotstyle); label("$X$", (56.1000251752257,-98.52736171317432), NE * labelscalefactor); dot((21.95924803991367,-103.89115530889306),linewidth(4pt) + dotstyle); dot((-728.1424189633996,125.50927590610821),linewidth(4pt) + dotstyle); label("$Y$", (-783.0346315675349,129.14483197672368), NE * labelscalefactor); dot((-603.0349055618302,35.73566225078713),linewidth(4pt) + dotstyle); label("$T$", (-652.9362351733084,12.056275221919), NE * labelscalefactor); dot((354.3642859561624,-651.2669229366732),linewidth(4pt) + dotstyle); label("$L$", (368.33617652136917,-696.9799851266205), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

The structure of the most un-elegant solution in this topic. I will compute these step by step: 1) Let $a,b,c$ be the side lengths of $\Delta ABC$.We can compute $AM$ by Pythagoras. 2) Let $AX=x$, We can compute $BX$ by Pythagoras. 3) We can find $MX$ by Ptolemy over $ABMX$. 4)We can compute $MT$ by Pythagoras over $\Delta MTX$($TX=BX$ will be useful here.) 5) Note that $\angle XMA =\angle TMB$. The side lengths of $\Delta MXA$ is known, hence both the angles is known. 6) By cosine rule we can find BT and TC and hence $\angle BTM \& \angle CTM$. 7) If we subtract these angles, we will get a relation in which there is not $x$ ! That means it is independent of $X$. I know it is near impossible to do but It is the first thing that came to my mind (Geo is a lie without bash)

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); pair A = dir(55), B = dir(235), M = dir(305), C = 2M-B, X = dir(355), D = dir(175), T = intersectionpoint(D--M,arc(X,abs(B-X),90,225)), E = 2M-T; dot("$A$", A, dir(55)); dot("$B$", B, dir(235)); dot("$M$", M, dir(270)); dot("$C$", C, dir(20)); dot("$X$", X, dir(5)); dot("$D$", D, dir(175)); dot("$T$", T, dir(210)); dot("$E$", E, dir(280)); draw(A--B--C--A--M--X^^unitcircle, purple); draw(T--X^^E--X^^B--X, blue); draw(T--B--E--C--T, orange); draw(X--A--D--2B-D^^D--E, heavygreen); draw(arc(circumcenter(T,B,E),abs(circumcenter(T,B,E)-T),165,315), magenta); [/asy][/asy] Let $\overline{MT}$ meet $(ABM)$ again at $D$, and $E$ be the reflection of $T$ over $M$. Claim: $X$ is the circumcenter of $(TBE)$. Proof. Note that $\triangle XMT \cong \triangle XME$ by SAS, so $TX = EX$. Also we are given that $TX = BX$ so we are done. $\square$ In addition, $\angle DBX = \angle DMX = 90^\circ$, so $\overline{DB}$ is tangent to $(TBE)$. It follows that \begin{align*}\angle MTB - \angle CTM &= \angle MTB - \angle BEM = \angle MTB - \angle DBT \\ &= \angle MDB = \angle BAM,\end{align*}which does not depend on $X$ as desired. $\blacksquare$

Solution from Twitch Solves ISL: Construct parallelogram $CTBS$ whose diagonals meet at $M$. Also, let $N$ be the midpoint of $\overline{BT}$. [asy][asy] pair A = dir(180); pair B = dir(0); pair M = dir(35); pair C = 2*M-B; pair X = dir(100); filldraw(A--B--C--cycle, invisible, red); draw(A--M, red); filldraw(unitcircle, invisible, blue); pair Y = -X; draw(CP(X, B), mediumgreen); pair T = IP(M--Y, CP(X, B)); pair S = 2*M-T; draw(S--T, mediumgreen); draw(B--X--M, mediumgreen); pair N = midpoint(B--T); draw(B--T, deepgreen); draw(T--X--S, mediumgreen); draw(X--N, deepgreen); filldraw(circumcircle(X, M, T), invisible, dotted+deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$M$", M, dir(M)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$T$", T, dir(T)); dot("$S$", S, dir(S)); dot("$N$", N, dir(N)); /* TSQ Source: A = dir 180 B = dir 0 M = dir 35 C = 2*M-B X = dir 100 A--B--C--cycle 0.1 lightred / red A--M red unitcircle 0.1 cyan / blue Y := -X CP X B mediumgreen T = IP M--Y CP X B S = 2*M-T S--T mediumgreen B--X--M mediumgreen N = midpoint B--T B--T deepgreen T--X--S mediumgreen X--N deepgreen circumcircle X M T 0.1 yellow / dotted deepgreen */ [/asy][/asy] We first eliminate $C$ from the diagram by noting that \[ \angle CTM = \angle MSB = \angle TSB = \frac{1}{2} \angle TXB = \angle NXB. \]Also, noting that $XMTN$ are cyclic (as $\angle XMT = \angle XNT = 90^{\circ}$), we have \[ \angle MTB = \angle MTN = \angle NXM. \]Thus $\angle MTB - \angle CTM = \angle NXM - \angle NXB = \angle MXB$ which is fixed.

Let $N$ be the midpoint of $BT$. Observe that since $\angle XMT = \angle XNT = 90$, the quadrilateral $XMNT$ is cyclic. Furthermore, $NM \parallel CT$. On the one hand, note that $$\angle MTB = \angle MTN = MXN.$$ On the other hand, note that $$\angle CTM = \angle NMT = \angle NXT = \angle NXB.$$Now, note that $$\angle MTB - \angle CTM = \angle MXN - \angle NXB = \angle BXM = \angle BAM$$which does not depend on $X$, as desired.

[asy][asy] unitsize(3cm); pair A, B, C, M, T, T1, X; A = dir(90); B = dir(210); C = dir(330); M = (B + C) / 2; X = rotate(20, (A + B) / 2) * M; T = IP(CP(X, B), Line(M, rotate(90, M) * X, 10), 0); T1 = 2 * M - T; draw(A--B--C--A^^unitcircle, heavyred); draw(circumcircle(A, B, M), heavygreen); draw(CP(X, T), blue); draw(X--T^^X--B^^X--M^^X--T1^^T--T1, orange); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$M$", M, S); dot("$X$", X, E); dot("$T$", T, NW); dot("$T'$", T1, SE); [/asy][/asy] Let $T'$ be the reflection of $T$ over $M$. Note that $TBT'C$ is a parallelogram. Additionally, as $\angle TMX = 90^\circ$, $\triangle XTT'$ is isosceles, so we have $XT = XB = XT'$. Hence, $X$ is the circumcenter of $\triangle BTT'$. Now, \begin{align*} \angle BTM - \angle CTM &= \angle BTT' - \angle CTT' \\ &= \angle BTT' - \angle BT'T \\ &= \frac{1}{2}(\angle BXT' - \angle BXT) \\ &= \frac{1}{2}((\angle BXM + \angle MXT) - (\angle MXT - \angle BXM)) \\ &= \frac{1}{2}(2\angle BXM) \\ &= \angle BAM, \end{align*}which is fixed, as desired. $\Box$

AMN300 wrote: We do not use complex numbers

yes I think the locus of T is a circle passing through $A,M$ and midpoint of $AB$ When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial.

Dunno if this has been posted before. Complex bash with $a=-1, b=1$ so that $|m|=|x|=1$ and $c=2m-1$. Since the foot from $T$ to $\overline{MX}$ is $M$, $$m=\frac{1}{2}(t+m+x-mx\overline{t}) \implies \overline{t}=\frac{t-m+x}{mx}.$$Since $\overline{TX}=\overline{BX}$ we should have $\frac{t-x}{b-x}$ on the unit circle, and upon substituting our expression for $\overline{t}$ $$\frac{t-x}{b-x}=\frac{\frac{1}{b}-\frac{1}{x}}{\overline{t}-\frac{1}{x}} \implies t^2-2mt+x^2m+m-x^2.$$Let $r$ be the unwanted root. By Vieta's, it follows that the argument of $\angle{MTB}-\angle{CTM}$ is (for $k \in \mathbb{R}$) $$\frac{\frac{m-t}{1-t}}{\frac{2m-1-t}{m-t}}=k_1 \cdot \frac{\frac{r-t}{1-t}}{\frac{r-1}{r-t}}=k_1 \cdot \frac{(r-t)^2}{-(rt-r-t+1)} = k_1 \cdot \frac{4m^2-4m-4x^2m+4x^2}{-(x^2m-x^2-m+1)} = k_2 \cdot \frac{m-x^2}{1-x^2}$$or $\frac{m-x^2}{1-x^2}$. Since $x^2 \in (ABM)$, this is equivalent to $\angle{MXB}=\angle{MAB}$ which does not depend on $X$, done.

Construct $N$ the midpoint of $TB$. Since $\overline{MN} \parallel \overline{TC}$, we can calculate \begin{align*} \angle MTB - \angle MTC = \angle MTN - \angle TMN = \angle MXN - \angle NXB = \angle MXB = \angle MAB, \end{align*}which is constant done yay kill me now misread this problem for a good 50 minutes

Solved with MrOreoJuice and Fakesolver19. [asy][asy] //meh asy, as always... size(210); pair A=(0,6.5),B=(-3,0),C=(3,0),M=(0,0),T=(-3.246,3.0711); pair X=(1.828,1.9321), TT=(3.246,-3.0711); draw(A--B--C--cycle, orange); draw(circumcircle(A,B,M), orange); draw(T--B--TT--C--cycle, magenta); draw(X--T, red); draw(X--B, red); draw(X--TT, red); draw(X--M, orange); draw(circumcircle(B,T,TT), red); draw(T--TT, magenta); dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,E); dot("$M$",M,S); dot("$T'$",TT,S); dot("$X$",X,NE); dot("$T$",T,NW); [/asy][/asy] Let $T'$ be the reflection of $T$ over $M$. Note that $TBT'C$ is a parallelogram and $X$ is circumcenter of $(TBT')$. \begin{align*} \angle MTB - \angle CTM &= \angle T'TB - \angle TT'B \\ &= \dfrac{\angle T'XB - \angle TXB}{2} \\ &= \dfrac{\angle TXT' - 2\angle TXB}{2} \\ &= \dfrac{2\angle TXM - 2\angle TXB}{2}\\ &= \dfrac{2\angle BXM}{2}\\ &= \angle BAM \end{align*}which does not depend on $X$ as desired. $\blacksquare$

Let $N$ be the midpoint of $BT$. We have that $\angle XMT=\angle XNT=90$ so $XMNT$ is cyclic. Also $CT\parallel MN$ thus $\angle CTM=\angle TMN$. In turn $\angle TMN=\angle TXN=\angle BXN$. Also $\angle MTB=\angle MXN$, so we can get that $$\angle MTB-\angle CTM= \angle MXN-\angle BXN = \angle MXB=\angle MAB$$which doesn't depend on $X$.

Denote by $N$ midpoint of $BT.$ Clearly $CT\parallel MN, XN\perp BT,$ hence $$\angle BTM-\angle CTM=\angle NTM-\angle NMT=\angle NXM-\angle NXT=\angle BXM=\angle BAM.$$

Let $N$ be midpoint of $TB$. $BXT$ is isosceles so $\angle XNT = 90 = XMT$ so $XMNT$ is cyclic. $\angle MTB - \angle MTC = \angle MXE - \angle EXT = \angle MXE - \angle BXE = \angle MXB = \angle MAB$ and $\angle MAB$ is fixed and independent from $X$. we're Done.

Let $N$ be the midpoint of $\overline{BT}$ and note $MNTX$ is cyclic and $\overline{MN}\parallel\overline{CT}.$ Hence, $\angle BTM=\angle BXM$ and $$\angle MTC=\angle NMB=\angle TXN=\angle NXB$$so $$\angle MTB-\angle CTM=\angle BXM=\tfrac{1}{2}\angle A.$$$\square$

All angles are in degrees. Let $MT$ intersect the circle again at $U$. Note that $\angle XBU = \angle XMU = 90$, thus $AXBU$ is a rectangle. Since $\angle XBM + \angle TMB = \angle XBM + \angle UMB = \angle XBM + \angle ABX = \angle MBA$, we have $$\angle MTB = 180 - \angle TBX - \angle XBM - \angle TMB = 180 - \angle MBA - \angle TBX.$$ Let $N$ be the midpoint of $BT$. Notice that $\angle XMT = \angle XNT = 90$, so $XMNT$ is cyclic. Since $MN\parallel CT$, we have $$\angle MTC = \angle TMN = \angle TXN = 90 - \angle XBT.$$This implies that $\angle BTM - \angle MTC = 90 - \angle MBA$, a constant value.

Let $N$ denote the midpoint of $BT.$ Then, $\angle TNX=\angle TMX=90^\circ$ so $TNMX$ is cyclic. Then, $\angle MTB-\angle CTM=\angle NTM-\angle TMN$ because $NM || TC.$ Now, $\angle NTM-\angle TMN=\angle NXM-\angle TXN=\angle BXM=\angle BAM$ which is fixed so we're done.

Let $T_1$ be the reflection of $T$ in $M$, ray $MT$ meet $(ABM)$ again at $Y$, and the circle centered at $X$ with radius $XB = XT$ meet $(ABM)$ again at $Q$. Because $XM \perp TT_1$, we know $XT = XT_1$, so $BTQT_1$ is cyclic. Now, observe $$\angle YBX = \angle YMX = \angle TMX = 90^{\circ}$$which implies $YB$ is tangent to $(BTQT_1)$. Similarly, we deduce that $YQ$ is tangent to $(BTQT_1)$, so $\overline{TT_1Y}$ is the $T$-Symmedian of $BQT$. Now, since $M$ is the midpoint of $TT_1$, we know it's the $T$-Dumpty point of $BQT$, yielding $MBT \overset{+}{\sim} MTQ$. Thus, because $BTCT_1$ is clearly a parallelogram, $$\angle MTB - \angle CTM = \angle MQT - \angle BT_1T$$$$= \angle MQT - \angle BQT = \angle MQB = \angle MAB$$which finishes. $\blacksquare$ Config Stuff: Let ray $XT$ meet $(ABM)$ again at $P$. The Incenter-Excenter Lemma implies $T$ is the incenter of $BPQ$. Moreover, since $\angle XMT = 90^{\circ}$, we know the $P$-Mixtilinear Incircle of $BPQ$ touches $(BPQ)$ at $M$. In fact, $MBT \overset{+}{\sim} MTQ$ is actually a well-known mixtilinear lemma.

Let $D$ be the midpoint of $BT$. So $TDMX$ is cyclic and $MD\parallel CT$. So the desired angle quantity is $\angle BAM$ which is fixed.

Let $E = \overline{TM} \cap (X)$. Notice that $\angle MTB - \angle CTM = \angle MTB - \angle BET = \frac 12(\widehat{BE} - \widehat{TB})$ because $TCEB$ is a parallelogram. However, setting $D = \overline{XM} \cap (X)$ as the arc midpoint of $\widehat{TE}$, we have $$\frac 12(\widehat{BE} - \widehat{TB})= \widehat{BD} = \angle BXM = \angle BAM$$is fixed regardless of $X$, as needed.

time to go through my writeup list Let $N$ be the midpoint of $BT$, with $XMNT$ cyclic (since $\angle TNX=\angle TMX=90^\circ$) and $MN \parallel TC$ (since $MN$ is the $B$-midsegment in $\triangle BTC$). Now we angle chase: \[\angle MTC=\angle NMT=\angle TXN=\angle NXB\]and $\angle BTM=\angle NXM$, so we are done since $\angle BTM - \angle MTC = \angle BXM = 90 - \angle ACM$. $\square$

Quote: When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial. LOL i made the exact same mistake on ggb, even though i READ TX=BX I didn't construct it properly and constructed it as on the circle with radius BX but with center B, so i was like "why is this not working" My solution is the same as above, but without typos: @above I think it should be MTC=NMT=..., and it should also be BTM=NXM, hence BTM-MTC=BXM=BAM, which is independent.

Let $\overline{MT}$ intersect $(ABM)$ again at $X'$ (the $X$-antipode) and the circle $\omega$ centered at $X$ passing through $B$ and $T$ at $D$. Then since $\overline{XM} \perp \overline{TD}$, $M$ is the midpoint of chord $\overline{DT}$, hence $BDCT$ is a parallelogram. On the other hand, since $\angle X'BX=90^\circ$, $\overline{X'B}$ is tangent to $\omega$, hence $\angle X'BT=\angle BDT=\angle CTM$, hence $\angle MTB-\angle CTM=\angle BX'M=\angle BAM$ which is fixed. $\blacksquare$

Let $S$ be such that $SBTC$ is a parallelogram. Then \[2(\angle MTB-\angle CTM)=\angle BXS-\angle BTX=(\angle BXM+\angle MXS)-(\angle MXT-\angle BXM)=2\angle BXM=2\angle BAM\]done.

Let Omega be circle with center X and radius XB. Let TM intersect Omega at Y. Let P be point on Omega(Arc TBY) such that Angle PTY=CTY After that observe TY||BP (Angle chase for this) We are done