If I'm not mistaken, it's easy to see that $\angle LFJ=\alpha/2$ and so $AFJL$ is cyclic. But $\angle JLA=90$ and so $\angle AFJ=90$. Thus, $AB=BS$ and hence $MS=AK$. Similarly, $MT=AL$, but $AK=AL$, so we are done.

Otherwise: $m(\angle{JFL})=m(\angle{JAL})=\frac{A}{2}\Rightarrow A,F,J,L$ are concyclic $\Rightarrow S,J,M,F$ are concyclic $\Rightarrow m(\angle{JST})=m(\angle{JTS})=\frac{A}{2}\Rightarrow JS=JT\Rightarrow SM=MT$.

Slightly different start to FelixD's approach: Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $AJ$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic.

Yes, we only need that <MFB= <A/2 and then AFJL cyclic then <AFJ=<ALJ=90 then <SFJ=<SMJ=90 then SFMJ cyclic with <MSJ=<A/2. Similar to <JTM=<A/2

By a simple angle chase, $\angle FBM = \angle ABJ = 90 + \angle B/2$ and $\angle FMB = \angle AJB = \angle C / 2$, so $\triangle BFM \sim \triangle BAJ$. Thus, $\triangle BFA \sim \triangle BMJ$ (since $BF/BA = BM/BJ$ and $\angle FBA = \angle MBJ = 90 - \angle B / 2$.) Since $BM \perp MJ$, we have $BF \perp AF$. Since $\angle ABF = \angle FBS = 90 - \angle B/2$, we have $BS = BA$. Similarly, $CA = CT$, so $SM = (s-c) + c = s = (s-b) + b = MT$.

The solutions are smaller than the problem statement.

Here is the attached figure.

It is well-known that $\angle BJC=90^{\circ}-\frac {1}{2}\angle A$ now $\angle LJC=\angle MJC=90^{\circ}-\angle MCJ=90^{\circ}-(90^{\circ}- \frac {1}{2}\angle C)=\frac {1}{2} \angle C$ so $\angle FJL=\angle BJC+\angle LJC=90^{\circ} -\frac {1}{2} \angle A+\frac {1}{2} \angle C$....(1) now $\angle MLJ=90^{\circ}- \angle LJC=90^{\circ}-\frac {1}{2} \angle C$....(2) so from (1),(2) we get in $\triangle LJF,\angle LFJ=\frac {1}{2} \angle A$ which inplies $AFJL$ is cyclic. since $\angle JLA=90^{\circ}$ we get $AFJ=90^{\circ}$ now $\angle FAB=\angle FAL- \angle A=\angle \frac {1}{2} \angle B$ and $\angle FSB=90^{\circ}-\angle FBS=90^{\circ}-\angle CBJ=\angle \frac {1}{2} \angle B$ hence $\angle FAB=\angle FSB$ so $\triangle FBS \cong \triangle FBA$[ASA] implying $AB=BS$ now $MS=MB+BS=BK+AB=AK$ simimlarly $MT=AL$ but $AK=AL$[tangent to the excircle from $A$] hence $MS=MT$ which means $M$ is the midpoint of $ST$ P.S.:Edited only to correct a typo.

Its quite easy to see that $A,K,J,G$ is cyclic. (1) $A,F,J,L$ is cyclic. (2) Since $\angle AKJ = 90^{\circ}$, this implies that $\angle AGJ = 90^{\circ}$ from (1) Similarly $\angle AFJ = 90^{\circ}$. Hence we get $A,F,K,J$ is cyclic (3) $A,G,L,J$ is cyclic (4) Thus from (1),(2),(3),(4) we get that all six points $A,G,F,J,K,L$ to be cyclic. Now, note that from triangle ABS, $AB = BS = c$. Similarly, $AC = CT = b$. Thus $SM = SB + BM = c + (s-c) = s$ Similarly $TM = TC + CM = b + (s-b) = s$. Hence proved.

It was really easy. Mine is almost similar as everyone. I proved that $AFJL$ and $AGJK$ are cyclic, which concludes that $AB=BS$ and $AC=CT$. Then $BM=s-c$ and $CM=s-b$. Now, we add these and $MS=MT$. This was quite easy for IMO 1 I guess.

WakeUp wrote: Slightly different start to FelixD's approach: Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $A$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic. Cool solution, bro. But ... too many letters as for me. Ты ещё не задолбался точечки отмечать? ( Use russian translator, bro)

Notice that $MK$ and $ML$ are parallel to the internal bisectors of $B$ and $C$ respectively. Then $MK \perp BJ$, $ML \perp CJ$, so $M$ is the orthocenter of $JFG$. Since $JM \perp BC$ from tangency, and $JM \perp FG$ from the orthocenter, $BC \parallel FG$. Now we can prove that $MF \parallel AG$ and $MG \parallel AF$ with any of the methods above, or by considering the excenters $I_b, I_c$, the fact tht $I_bI_c$ is antiparallel to $BC$ and a short angle chase. With this, $FMG$ is the medial triangle of $AST$ and we are done.

So this takes about five minutes in barycentric coordinates.

Hmm I should learn how to do geometry.

You can actually prove that AMLT and AMKS are isoceles trapeziods, so SM=AK, MT=AL, AK=AL so done. Use simple trigonometry to prove it.

COMPLEX NUMBERS ROCK!!!!!!!!!!!!!!!!! Let the excircle be the unit circle and $m=1$. Then $s=\frac{2k}{k+l}$ and $t=\frac{2l}{k+l}$

Menelaus gives $\frac{AG}{GT} \cdot \frac{TM}{MB} \cdot \frac{BK}{KA}=1$. Similarly, $\frac{AF}{FS}\cdot \frac{SM}{MC} \cdot \frac{CL}{LA}=1$. Since $AK=AL$, $BM=BK$, $CM=CL$, it suffices to show $\frac{AG}{GT}=\frac{AF}{FS} \Longleftrightarrow FG \parallel BC$. Let $U,V$ denote the midpoints of $MK,ML$. Then $GFUV$ is cyclic, so $FG$ is antiparallel to $UV$ which is parallel to $KL$ which is antiparallel to $BC$ with respect to lines $JF,JG$, thus $FG \parallel BC$.

it is Simple problem and i saw it A few minutes ago $<JFL=90-[(B+C)/2]=A/2$ so $ALJF$ is cyclic and so $<AFJ=90$ and we see $FMJS,MGTJ$ are cyclic[lemma(1)]. and we get $Q$ be the intersection of the lines $KM,FJ$ and $P$ be the intersection of the lines $LM,JG$ . we see $FQPG$ is cyclic because $<FQG=<GPF=<90$ so $<MGJ=<JFM$ and we use to lemma(1) and easy to see that $<MGJ=<MTJ=<LFJ=<JSM$ so $JS=JT$ we are done.

let $\angle MJL = 2\alpha, \angle KJM = 2\beta$.J is center of circle and C is intersection of tangents So JC is bisector of $\angle MJL$.Analogously we can say BJ is bisector of $\angle MJK$. J is center of circle and C is intersection of tangents So we get $\angle LJP$=$\angle MJP$=$\angle CLP$=$\angle CMP$=$\angle FMB$=$\alpha.$ Analogously we get $\angle MJQ$=$\angle KJQ$=$\angle BMQ$=$\angle BKQ$=$\angle GMC$=$\beta$. $\triangle MKJ$ is isosceles triangle and BJ is bisector so BQ=median and altitude.So FQ is median and altitude so $\triangle KFM$ is isosceles.And we get $\angle FKM= \angle FMK = \alpha+\beta$.And $\angle KFQ=90-\alpha-\beta$.Analogously $\angle PGL=90-\alpha-\beta$.And we know GL is bisector so $\angle MGP=90-\alpha-\beta$.So we get $\angle KFQ=\angle MGP=90- \alpha - \beta$.So $KFGJ$ is cyclic.Analogously $JFGL$ is cyclic FGJ is common so we can say $KLFGJ$ is cyclic.$\angle AKJ+\angle ALJ=90+90=180$ so $AKJL$ is cyclic.We know $KLFGJ$ is cyclic $KJL$ is common so $AKLFGJ $ is cyclic ==> $AGLJ$ is cyclic .$\angle ALJ=90$ so $\angle AGJ=90$. We know $CP$ is bisector of $\angle MCL$ so $CG$ is bisector of $\angle ACT$. $CG$ is bisector and altitude of $\triangle ACT$ so $AC=CT$. Analogously $SB=ST$ and $BM=BK(tangents)$ so $SM=AK$. Analogously $AL=MT$. So $MT=AL=tangent=AK=SM ==>> MT=MS$.

We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We get that $J = (-a:b:c)$, $M = (0:s-b:s-c)$, $L = (b-s:0:s)$. We first compute $S$ and then find $MS$. We simply need the length $MS$ to be symmetric around $b$ and $c$. We can see that $KM$ is the line \[\begin{vmatrix}0&s-b&s-c \\ b-s&0&s \\ x&y&z \end{vmatrix}= 0 \Rightarrow (s-b)(b-s)z = (s-b)sx + (b-s)(s-c)y\]Since $BJ$ is parametrized by $(-a:y:c)$, we can see that \[F = \left(-a:\frac{(s-b)c-as}{s-c}:c \right)\]This means that \[S = \left(0:\frac{(s-b)c-as}{s-c}:c \right) = (0:(s-b)c-as):(s-c)c) = (0:a+c:-c)\]Notice now that $\overrightarrow{SM} = \left(0, \frac{s}{a}, \frac{-s}{a} \right)$. Thus $MS = s$ as desired. $\blacksquare$

Solved with fuzimiao, Jndd, bronzetruck2016, a_smart_alecks. $AKJL$ is cyclic because $\angle AKJ = \angle ALJ = 90^{\circ}$. By the converse of Pascal, $A$, $F$, $K$, $J$, $L$, $G$ lie on a conic because $AK \cap FJ = B$, $FL \cap GK = M$, and $GJ \cap AL = C$. Since $AKJL$ is cyclic, we conclude that $AFKJLG$ is also cyclic with diameter $AJ$. Next, $ACT$ is isosceles because $CJ$ bisects $\angle ACT$ and $\angle JGA = 90^{\circ}$. Therefore, $C$ and $G$ is on the perpendicular bisector of $AT$. Similarly, $B$ and $F$ are on the perpendicular bisector of $AS$. Both of these perpendicular bisectors go through $J$. From this, $SJ = AJ = TJ$. However, this means that the foot of $J$ to $ST$ is the midpoint of $ST$, so $M$ is the midpoint of $ST$ and we are done.

Claim: $AFKJLG$ is cyclic. Firstly, note that since $\overline{FJ}$ and $\overline{JG}$ are the perpendicular bisectors of $\overline{KM}$ and $\overline{ML}$, it follows that $\triangle KFM$ and $\triangle MGL$ are isosceles. Angle chasing gives us $\angle KFL = \angle A = \angle AGL$, so it follows that $KFAGL$ is cyclic. Furthermore, $AKJL$ is cyclic with diameter $AJ$, so the claim follows. So, $\angle JFA = \angle JGA = 90^{\circ}$. Because $\overline{BJ}$ and $\overline{CJ}$ are angle bisectors, it follows that $\triangle ABS$ and $\triangle ACT$ are isosceles, so \[ SB + BM = AB + BM = \text{semiperimeter} = AC + CM = TC + CM.\]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ Solved with Blueberryfaygo Claim $1$: $LA = KA$, $BK = MB$, $CL = MC$ Proof: Trivial because they're tangents to the A-excircle. $\blacksquare$ Claim $2$: $M$ is the orthocenter of $\Delta FJG$ Proof: First, notice that because $\Delta MCL$ is isosceles and $JC$ is the angle bisector of $\angle MCL$, we know that $JC \perp ML$. This is equivalent to saying $JG \perp FM$, since $JC$ and $ML$ are the same lines as $JG$ and $FM$, respectively. Now, since $JG \perp FM$, we know $FM$ is an altitude of $\Delta FJG$. Similarly, $GM$ is also an altitude of $\Delta FJG$. Since $FM$ and $GM$ intersect at point $M$, that means $M$ is the orthocenter of $\Delta FJG$ and we are done. $\blacksquare$ Claim $3$: $\frac{AF}{FS} = \frac{AG}{GT}$ Proof: Because $M$ is the orthocenter of $\Delta FJG$, then $JM$ is an altitude, meaning that $JM \perp FG$. But we also know that $JM \perp BC$ because the A-excircle has center at $J$ and is tangent to $BC$ at $M$. So, $FG \parallel BC$, which means $\Delta AFG \sim \Delta AST$. The desired ratios equality follows. $\blacksquare$ Main Proof: Use Menelaus' Theorem on $\Delta ACS$ and $\Delta ABT$, and divide the equations so we get $$ \frac{\frac{AF}{FS} \cdot \frac{SM}{MC} \cdot \frac{CL}{LA}}{\frac{AG}{GT} \cdot \frac{TM}{MB} \cdot \frac{BK}{KA}} = 1. $$Now, applying Claims $1$ and $3$ to simplify it, we get $$ \frac{SM}{TM} = 1, $$so we are done. $\blacksquare$

Claim 1: $FG \parallel BC$ Proof: Clearly $BJ \perp MK$ and $CJ \perp ML$, meaning $M$ is the orthocentre of $\triangle FGJ$. So $MJ \perp BC$ and $MJ \perp FG$ implies that $FG \parallel BC$. Claim 2: $A$, $G$, $L$, $J$, $K$, $F$ are concylic Proof: $\angle GFJ = \angle MBJ = \angle MKJ = \angle GKJ \implies JKFG$ is cyclic. Similarly, $JLGF$ is cyclic. Thus with $AKJL$ cyclic as well, $A$, $G$, $L$, $J$, $K$, $F$ are concylic. Claim 3: $ AT \parallel FM$ Proof: $ \angle GFM = \angle CML = \angle CLM = \angle ALF = \angle AGF$. Claim 4: $F$ is the midpoint of $\overline{AS} $ Proof: $ \angle AKJ = \angle AFJ = \angle AFB = 90^{\circ} = \angle SFB$ and $ \angle FBS = \angle MBJ = \angle KBJ = \angle FBA $. So, $\triangle FBS \cong \triangle FBA $, which implies $ SF = FA$. Hence, claim 3 and 4 follows that $M$ is the midpoint of $ST$. $\square$

By right angles, notice that $AKJL$ is a cyclic quad. Let the circumcircle of this quadrilateral be $\omega.$ Then we can angle chase to get $\angle JFM= \frac{A}{2}$ and $\angle LKJ=\frac{A}{2}.$ This implies that $F$ lies on $\omega.$ By symmetry $G$ lies on $\omega$ too. Then by cyclic quads, $\frac{B}{2}=\angle BJK=\angle BAF.$ By definition then vertical angles, $\angle KBJ=\angle FBA=90-\frac{B}{2}.$ Therefore $ABF$ is a right triangle and therefore $FJ \perp AS.$ By definition then vertical angles again, $\angle SBF=90-\frac{B}{2}.$ Therefore by ASA, triangles $SBF$ and $ABF$ are congruent. By symmetry, triangles $ACG$ and $TCG$ are congruent too. By tangents excircle properties, $SM=SB+SM=AB+BK=s$ and $AL=AC+CL=CM+CT=MT=s$ where $s$ is the semiperimeter of ABC. Therefore $SM=MT$ and we are done.

Let's solve this easy angle chase problem. First we notice that $\angle SMJ = \angle SFJ$. So, quadrilateral $FSJM$ is cyclic. Now, Since $\angle CML = 180^{\circ} - \frac{180^{\circ}-\angle C}{2}$ and $$\angle FSJ = 90^{\circ}- \angle SJF = 90^{\circ}-\angle SMF = 90^{\circ}-\angle CML = 90^{\circ} - \frac{\angle C}{2}$$ Also, $$\angle ACJ = \angle C + \frac{180^{\circ}-\angle C}{2}= 90^{\circ} +\frac{\angle C}{2}$$ Therefore, quadrilateral $ACJS$ is cyclic. Since $A$, $I$ and $J$ are collinear, we get, $$\angle JSM = \angle JSC = \angle JAC = \angle IAC = \frac{\angle C}{2}$$ Similarly, $\angle JTM = \frac{\angle C}{2}$. Hence by congruency, we get that $M$ is the midpoint of $ST$.

bad formatting because i copied my solution from math dash let JC meet ML at X, then JXM is right, so JFL = 90 - BJX = 90 - BJC = a/2. since JAL is also a/2, we see that (AFJL) is cyclic, and since (AKJL) is also cyclic so is (AFJKL). now AFJ is right, then so is SFJ, so (SFJM) is cyclic. since JFM = JFL = JAL = a/2, we have JSM = JFM = a/2. repeating this argument gives JTM = a/2. since JMS = JMT = 90, by AAS congruency we have JSM and JTM are congruent, so SM = TM.

We present three claims, all of which may be proven by angle chasing with respect to the angles of the triangle $ABC$. Claim 1: $A, F, G, J, K, L$ are concylic. Claim 2: $AFMG$ is a parallelogram. Claim 3: $FG \parallel ST$. Note that from Claim 2, $AM$ bisects $FG$. But now taking the homothety at $A$ sending $FG$ to $ST$ (which exists by Claim 3), $AM$ bisects $ST$, so $M$ is the midpoint of $ST$. $\square$ Remark(due to Rijul Saini): The above claim 2 and 3 may essentially be summarised as Iran Lemma on the $A-$ excircle.

Using angle chase, notice that \[\angle LFJ = \angle XBJ - \angle CXL = 90 - \frac B2 - \frac C2 = \frac A2 = \angle LAJ,\] so $AFJL$ is cyclic. Notice that $\angle ALJ = \angle AKJ = 90$ implies $AFKJL$ is also cyclic. Instead exploiting symmetry like a normal person, we can instead notice that the converse of Pascal's on hexagon $AKGJFL$ implies these six points lie on a conic, which must be a circle as we know five of the points define a circle. Hence we have $\angle AFB = \angle AGC = 90$, giving us isosceles trapezoids $AXKS$ and $AXLT$, so \[XS = AK = AL = XT. \quad \blacksquare\] [asy][asy] size(330); pair A, B, C, J, K, L, F, G, S, T, X; A = dir(120); B = dir(210); C = dir(330); J = extension(A, incenter(A, B, C), B, rotate(90, B)*incenter(A, B, C)); K = foot(J, A, B); L = foot(J, A, C); X = foot(J, B, C); F = extension(B, J, X, L); G = extension(C, J, X, K); S = extension(A, F, B, C); T = extension(A, G, B, C); draw(A--K--J--L--A--J--X^^K--G--J--F--L^^A--S--T--cycle); draw(circumcircle(A, K, L), dashed); draw(circumcircle(X, K, L), dotted); draw(A--K^^S--X, red+linewidth(1.5)); draw(A--L^^T--X, blue+linewidth(1.5)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$K$", K, W); dot("$L$", L, E); dot("$J$", J, SE); dot("$X$", X, N); dot("$F$", F, NW); dot("$G$", G, NE); dot("$S$", S, W); dot("$T$", T, E); [/asy][/asy]

Claim 1: $AF\bot BF$ and $AG\bot CG$ Prove: Let $P=AC\cap KM$ and $Q=AB\cap LM$. Clearly, $BJ$ is a perpendicular bissector of $KM$ e $CJ$ of $LM$, so $\angle KFB=\angle BFQ$ and $\angle LGC=\angle CGP$. Also, by ceva and then menelaus with the points $K$, $M$ and $L$ and $Q$, $M$ and $L$ likewise $P$, $M$ and $P$, we get that $(A, B; K, Q)=(A, C; P, L)=-1$, so we´re done $\blacksquare$ Claim 2: $FG//BC$ It´s easy to see that $GK\bot FJ$ and $FL\bot GJ$, therefore $M$ is the orthocenter of $\triangle FJG$ so $JM\bot FG$, but $JM\bot BC$. $\blacksquare$. Hence, from the Claim 1, we get that $AGMF$ is a paralelogram. We want to prove that $(S,T;M, P_{\infty, BC})=-1$ but $(S, T; M, P_{\infty, BC})\stackrel{A}{=}(F, G; AM\cap FG, P_{\infty, FG})=-1$ so we´re done. $\blacksquare$

Note that $$\angle JFL = \angle JAL = \angle KAJ = \angle KGJ,$$so $AKJG$ and $AFJL$ are cyclic. Then $$\angle AGM = \angle AJK = \angle AJL = \angle AFL = 90^\circ - \frac{\angle A}{2}$$and $$\angle FAG = \angle FAJ + \angle GAJ = \angle FLJ + \angle GKJ = \angle FMG = 90^\circ + \frac{\angle{A}}{2},$$so $AFMG$ is a parallelogram. We also have $AS \perp FJ$ and $AT \perp JG$, so since $\angle ABF = \angle FBS$ and $\angle ACG = \angle GCT$, we get $SF = AF = MG$ and $FM = AG = GT$. Then $$\triangle SFM \cong \triangle MGT,$$so $SM = MT$, as desired.

This is really a problem about triangle $MKL$. We claim that $BA=BS$ and $CA=CT$, which will clearly finish. Note that if we show that $\triangle BKM\sim\triangle BSA$, we will win. Thus it suffices to show that $KM\parallel AF$, or $\angle AFJ=90^{\circ}$, or $F\in(AKJL)$. Now let's invert around the excircle. Then $F'\in (JLM)$, so \[\angle JF'L=\angle JML=90^{\circ}-\angle K,\]so since $F'$ lies on $BJ$, it is exactly $BJ\cap KL$, as desired. $\blacksquare$

Notice that $BJ$ is the perpendicular bisector of $MK$ so $FK=FM.$ Therefore $\angle MFK = 2 \angle JFM = 2(90^\circ - \angle BJC) = \angle A.$ Thus $KFAL$ is cyclic. Similarly $LGAK$ is cyclic and it follows that pentagon $FAGLK$ is cyclic. Thus,$$\angle FAG = 180^\circ - \angle FLG = 90^\circ +\frac12 \angle JGL = 90^\circ + \frac{\angle A}{2}.$$Also, $\angle AGK = \angle ALK = 90^\circ - \frac{\angle A}{2},$ so $\angle FAG + \angle AGK = 180^\circ \implies AS \parallel KG.$ But $BK = BM$ so by homothety $BS=BA.$ Therefore $MS = AK = s$ where $s$ denotes the semiperimeter. Similarly $MT=s$ too, and we are done. QED