Very direct by the lemma that if $BI \cap EF = U$, then $\angle BUC = 90^{\circ}$ (this is since $I$, $F$, $U$, $C$ are concyclic by angle chase). So $X$ must be one of the points $BI \cap EF$ and $CI \cap EF$ since $\angle BXC = 90^{\circ}$ and there are at most two intersection points between a circle (here - with diameter $BC$) and a line (here $EF$). Without loss of generality $X = BI \cap EF$. Then $\angle XBC = 45^{\circ}$, so $\angle ABC = 90^{\circ}$, hence $BDIE$ is a square, so $DE$ is the perpendicular bisector of $BI$. As $MB = MI$ by angle chasing (known as the Trillium theorem), we have that $M$ lies on this perpendicular bisector, done!

As above we say $<ABC=90$. Let now $S$ be the A-Sharky_Devil point then it is well know that $S,D,M$ are collinear then: $<FSA=180-<AIF=90+<A/2$ but also $<MSA=<MSC+<CSA=<A/2+90$ so we have that $S,F,D,M$ are collinear the same if $<ACB=90$

My problem. Let $K$ be the foot of the perpendicular from $D$ to $EF$ and let $I$ be the incentre of $\triangle ABC$. We begin by proving that $BKXC$ is cyclic. Let $N$ be the midpoint of $DE$. From $\triangle BDF$ and $\triangle AFE$ isosceles we get: $$\angle DFE=180^{\circ}-\angle BFD-\angle EFA=180^{\circ}-\left(90^{\circ}-\frac{\angle B}{2}\right)-\left(90^{\circ}-\frac{\angle A}{2}\right)=90^{\circ}-\frac{\angle C}{2}$$Thus, using $\angle FKD=90^{\circ}$, we get $\angle KDF=\frac{\angle C}{2}$. We also have $\angle ECN=\angle ACI=\frac{\angle C}{2}$ so triangles $ECN$ and $KDF$ are similar which gives: $$\frac{FK}{FD}=\frac{EN}{EC}=\frac{ED}{2 \cdot EC}$$Similarly we get: $$\frac{EK}{ED}=\frac{FD}{2 \cdot FB}$$Multiplying the two expressions gives: $$\frac{FK}{FB}=\frac{EK}{EC}$$Combining this with $\angle BFK=\angle KEC=90^{\circ}-\frac{\angle A}{2}$ we get triangle $BFK$ and $CEK$ are similar so $\angle FKB=\angle CKE$ which shows line $EF$ is the external-angle bisector of $\angle BKC$. Therefore, $X$ lies on both the perpendicular bisector of $BC$ and the external angle bisector of $\angle BKC$ (and these lines are distinct) thus is the midpoint of arc $BKC$ and in particular lies on $\odot BKC$. Now we show that we must have either $\angle B=90^{\circ}$ or $\angle C=90^{\circ}$. Assume to the contrary that $\angle B,\angle C \neq 90^{\circ}$. By the above cyclicity and the condition in the problem statement $\angle BKC=\angle BXC=90^{\circ}$ . Using that $BD$ and $EF$ are the internal and external bisectors of $\angle BKC$ we get: $$\angle FKB=\angle BKD=\angle DKC=\angle CKE=45^{\circ}$$Hence $BK$ bisects $\angle FKD$, but $B$ also lies on the perpendicular bisector of $DF$ and if these lines are distinct this forces $B$ be the midpoint of arc $DF$ not containing $K$ in $\odot FKD$. If this were the case, we'd have $\angle B=180^{\circ}-\angle FKD=90^{\circ}$ which contradicts our initial assumption. Thus in fact these two lines must coincide giving $\triangle KFD$ isosceles with $KF=KD$. Similarly we get $KE=KD$ so $K$ is the circumcentre of $\triangle DEF$ but this means $K$ is in fact the incentre of $\triangle ABC$ which is absurd as $K$ lies on $EF$. WLOG we consider the case where $\angle B=90^{\circ}$. Observing that $A,I,M$ are collinear we get: $$\angle CDI=90^{\circ}=\angle CBA=\angle CMA=\angle CMI$$Hence $MDIC$ is cyclic so: $$\angle MDC=\angle MIC=180^{\circ}-\angle CIA=180^{\circ}-\left(90^{\circ}+\frac{\angle B}{2}\right)=45^{\circ}$$We also have $\angle FDB=90^{\circ}-\frac{\angle B}{2}=45^{\circ}$ so $\angle FDB=\angle MDC$ and thus $M,D,F$ are collinear as required.

Let $S = BI \cap EF$, $T = CI \cap EF$. Let $(BC)$ denote the circle diameter $\overline{BC}$. Claim 1: $S, T \in (BC)$. Proof: $\measuredangle TFB = \measuredangle AFE = \measuredangle CIB = \measuredangle TIB$ since $\angle AFE = (180 - \hat{A})/2 = \hat{B}/2 + \hat{C}/2 = 180 - \angle CIB$. Thus, $TFBI$ cyclic and $\measuredangle CTB = \measuredangle ITB = \measuredangle IFB = 90$. $T \in (BC)$ follows similarly. $\square$ Case 1: $X \neq S, T$. Then $X, T, S$ are the intersections of $(BC), EF$ hence two must be equal. Thus $S = T$. But now $I = CI \cap BI \Rightarrow I \in EF$, yet $AB \perp IE, AC \perp IF \Rightarrow AB, AC \perp FIE \Rightarrow AB \parallel AC$, contradiction. Case 2: $X = S$. Then $\angle IBC = \angle XBC = 45$ so $\angle ABC = 90$, ie $\overline{AC}$ is a diameter. Let $N$ denote the midpoint of arc $AB$ not containing $C$. Since $\measuredangle ANI = \measuredangle ANC = 90$, $N$ is the $A$ Sharky-Devil point, with $N - D - M$. Furthermore, $BI$ is the perpendicular bisector of $\overline{DF}$ hence $DF \perp BI$. By Incentre-Excentre, we also have $MB = MI, CB = CI$ hence $MN$ is the perpendicular bisector of $\overline{BI}$, so $MN \perp BI$. Thus, $DF \parallel MN$. Since $N - D - M$ we get $N - F - D - M$. Case 3: $X = T$. This follows identically as above to get $E - D - M$. $\blacksquare$

I had a relatively short complex bash on contest, I'll post when I come home.

Step 1: See who precisely is X First We have that X is on the circle of diameter (BC) and there are only 2 points that satisfy this But from Iranian lemma we have that if we intersect BI with EF in B' then $\angle{BB'C}=90^{\circ}\ $ So B' is on the circle with diameter BC and similary C' is on (BC) So X is one of B' or C'. WLOG suppose that X=B' then 45=$\angle{B'BC} $=$\angle{IBC}= $=B/2 so angle ABC is 90 degrees Step 2: Finish From $\angle{ABC}=90^{\circ}\ $ we have AC is diameter so $\angle{IMC}=90^{\circ}\ $,but $\angle{IDC}=90^{\circ}\ $ so we get that IMDC is cyclic so$\angle{MDC}\ $=$\angle{MIC}\ $=180-$\angle{AIC}\ $=45=$\angle{BDF}\ $(since triangle is isosceles with 90 in B)so we get that M,D,F are collinear and we finish

İt is clear that when triangle $ABC$ be isosceles it is impossible.Let $|AB|>|AC|$ then we prove that when exits a point $X$ then $\angle$$ACB$$=90^{\circ}$ Let $\angle$$A=2$$\alpha$,$\angle$$B=2$$\beta$,$\angle$$C=2$$\omega$ We know $|CE|+|FB|=|CB|$ We use sinus law in triangles $EXC$ and $FBX$ then we get $|CE|=$$\frac{CX\sin(135-\alpha-2×\omega)}{cos(\alpha)}$ and $|FB|=$$\frac{BXsin(135-\alpha-2×\beta)}{cos(\alpha)}$ $|CX|=|BX|$ and $|BC|=$$\sqrt{2}$×$|CX|$ then we get $\frac{sin(135-\alpha-2×\omega)+sin(135-\alpha-2×\beta)}{cos(\alpha)}$ $=\sqrt{2}$ $\rightarrow$ $\frac{sin(45)×cos(\omega-\beta)}{cos(\alpha)}=$$\sqrt{2}$ from there we get $cos(\omega-\beta)=cos(\alpha)$ $\rightarrow$ $\omega-\beta=\alpha$ $\angle$$C$$=90^{\circ}$ now let's prove that $E,M,D$ collinear. From triangle $CEF$ we get $\angle$$CDE=45^{\circ}$ then we show that $\angle$$MDB=45^{\circ}$ we are done. İn quaraderal $IDMB$ is circlic,because $\angle$$IDB=90^{\circ}$=$\angle$$ACB$$=$$\angle$$AMB$ on the other hand $\angle$$MIB$$=180-$$\angle$$AIB$$=$$45$ and that is why $IDMB$is circlic $\angle$$MIB$$=$$\angle$$MDB$$=45^{\circ}$ we are done

An old problem (lemma) comes to rescue... see: https://artofproblemsolving.com/community/c6h23677p150733

Let $I$ be the incenter. By Iran Lemma, $\angle BXC=90^{\circ}$ implies $XI$ passes through $B$ or $C$. WLOG $XI$ passes through $B$. Then, $\angle ABC=90^{\circ}$, so $DF$ is the perpendicular bisector of $BI$, which passes through $M$ by Incenter-Excenter.

Claim 1: By lemma 255 (255th problem in Sharygin olympiad) ABC is rectangular triangle. Claim 2: I (incenter) is an orthocenter in triangle FXM.

A proof for $\angle B=90^\circ$ or $\angle C=90^\circ$ without using the lemma. Let $EF\cap BC=T$. Since $A, I, M$ are collinear, $\angle FTB=\angle DIM$. Since $(T,D;B,C)=-1$ and $\angle BXC=90^\circ$, $\angle TXB=\angle DXB=\angle XBC-\angle FTB=45^\circ-\angle FTB$. Therefore, $\angle DXM=90^\circ-\angle FTB-2\angle TXB=\angle FTB$. Hence, $XIDM$ is a cyclic quadrilateral and since $DI\parallel XM$, it is a isosceles trapezoid. Let $K$ be the midpoint of $BC$ and let $A$-excircle touch $BC$ at $N$. Let $J$ be $A$-excenter. Let $r=DI$, $r_a=JN$. Since $IM=JM$, $ID+MK=JN-MK$, so $MK=\dfrac{r_a-r}{2}$. Then, $\dfrac{a}{2}=XK=MK+ID=\dfrac{r_a+r}{2}\implies r_a+r=a$. It is easy to see that in any triangle $r_a\cdot r=(u-b)\cdot (u-c)$, so we have either $(r,r_a)=(u-b,u-c)$ or $(r,r_a)=(u-c,u-b)$, which respectively implies that $\angle B=90^\circ$ or $\angle C=90^\circ$. [asy][asy] size(16cm); pair A = dir(120), B = dir(210), C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C), E = foot(I, A, C), F = foot(I, A, B); pair T = extension(B, C, E, F); pair K = (B + C) / 2; pair M = extension(A, I, circumcenter(A, B, C), K); pair X = extension(E, F, circumcenter(A, B, C), K); pair N = 2 * K - D; pair J = 2 * M - I; draw(F -- D -- E -- T -- C -- A -- B); draw(X -- I -- D -- M -- X, linewidth(2)); draw(A -- J, dashed); draw(J -- N); dot("$A$", A, dir(120)); dot("$B$", B, dir(230)); dot("$C$", C, dir(330)); dot("$D$", D, dir(220)); dot("$E$", E, dir(60)); dot("$F$", F, dir(150)); dot("$I$", I, dir(150)); dot("$T$", T, dir(180)); dot("$M$", M, dir(220)); dot("$X$", X, dir(120)); dot("$K$", K, dir(140)); dot("$N$", N, dir(300)); dot("$J$", J, dir(300)); [/asy][/asy]

hukilau17 wrote:

Sniped My solution on contest was essentially identical, however I calculated $x$ first and got $(d+ie)(d-if)=0$ or analogously $(d-ie)(d+if)=0$. Checking $M-D-E$ or $M-D-F$ collinear is now trivial. Motivation is that when $BC$ is fixed, the choice for point $A$ lies on a conic, the result that $M$ is on $DE$ $\textbf{or}$ $DF$ is a dead giveaway that the conic is degenerate, ie. two lines. It is now obvious that the condition for $X-E-F$ collinear must factor, and the finish is trivial.

Here is another way to finish once we get that WLOG $\angle ABC=90^{\circ}$. We will show that $\angle MDC=\angle BDF=45^{\circ}$. Let $O$ be the circumcenter of $(ABC)$. Since $\angle ABC=90^{\circ}$, $O$ is also the midpoint of $AC$. Let $N$ be the midpoint of $BC$. Then $O, M, N$ are collinear and $ON$ is midline in $\triangle ABC$, so $ON=\frac{c}{2}$. Thus, $MN=OM-ON=OC-ON=\frac{b}{2}-\frac{c}{2}$. In any triangle $ABC$, $BD=s-b$ (where $s$ is its semiperimeter). So, $ND=NB-BD=\frac{a}{2}-(s-b)=\frac{a}{2}-\frac{a-b+c}{2}=\frac{b-c}{2}$. Finally, $\angle MND=90^\circ$ and $MN=ND$, so $\angle MDN=45^\circ$.

Ab_Rin wrote: Claim 1: By lemma 255 (255th problem in Sharygin olympiad) ABC is rectangular triangle. Claim 2: I (incenter) is an orthocenter in triangle FXM. Which lemma do u mean? could u please give the link for that lemma?

contesthunter wrote: Ab_Rin wrote: Claim 1: By lemma 255 (255th problem in Sharygin olympiad) ABC is rectangular triangle. Claim 2: I (incenter) is an orthocenter in triangle FXM. Which lemma do u mean? could u please give the link for that lemma? Yep, but it is in russian, you may use translator: https://m.vk.com/@botayemnoteveryday-o-svyazi-dvuh-zamechatelnyh-zadach

What is the placement of this problem on the shortlist?

Using the well known Iranian Configuration we get that the two points $X$ on $EF$ such that $\angle BXC = 90^{\circ}$ are the intersections of the $A$-intouch line, $B$-bisector, and $C$-midline or $A$-intouch line, $C$-bisector, and $B$-midline. Then we have one of $B$ or $C$ is a right angle and we can finish with one more application of the Iranian Lemma.

Non iranian solution WLOG Let $90\geq\angle{ABC}>\angle {ACB}$. In this configuration, we have to prove that $\overline{F,D,M}$. Suppose that its correct. By angle chasing we get $\angle{AMC}=90^\circ$, $\angle{ABC}=90^\circ$. So, we have to prove that, if we can take a point on $EF$,$X$ which satisfies the condition in question, $ABC$ is a right triangle. First, we are gonna prove that in a right triangle condition is correct. Let $\angle{ABC}$ be $90^\circ$. Let $T$ be the point that satisfies $TBC$ is a right isosceles triangle and $\angle T=90^\circ$. Its clear that $T=X$. Let midpoint of $BC$ be $R$, Let $RX\cap AC$ be $Y$. We know that $AEM$ is isosceles so $\overline{X,E,F}$ $\Longleftrightarrow |EY|=|XY|$. Lets use the same distances $$XY=XR-YR=\frac{BC-AB}{2}=\frac{CE-AE}{2}=EY\blacksquare$$Lets prove that theres no a point on $AC$ such that $\overline{X,E',F'}$ without $A$. Let $A'$ satisfies the conditions. Let incenter of $A'BC$ and incenter of $ABC$ be $I',I$ respectively. Its clear that $\overline{C,I,I'}$. So if $A'\in [AC]$ when $A'$ goes to $C$, $\angle{AE'X}$ decreases because of $\angle{EXE'}$ increasing. Similarly $\angle{BA'E'}$ increasing so in isosceles triangle $A'E'F',\angle{A'F'E'}=\angle{A'E'F'}$ values dicreases. Because of that $\angle{E'F'X}=\angle{AE'X}+\angle{AE'F'}$ decreases. Thus, $\angle{E'F'X}<180^\circ$. Similarly we can prove the other side so for a segment $BC$, the suitable $A$ points satisfies that $ABC$ is a right triangle. We're done.

trivial by the Iranian lemma, WLOG assume $\measuredangle B=90^{\circ}$ we have that $A-I-M$ are collinear $\angle IDC=\angle IMC=\angle ABC$ and we have $\angle ECI=\angle EMI $ and we have that $\angle ECI+\angle CAI=45^{\circ}\implies \angle CEM=\angle CDM=\angle FDM=45^{\circ}$ thus we are done [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 = -24.658927636937037, xmax = 28.130071976249802, ymin = -15.105304760536413, ymax = 10.091201643415133; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); draw((2.8201671052187764,-13.863668164495618)--(2.5652999246292856,-13.084510196333175)--(1.7861419564668428,-13.339377376922666)--(2.0410091370563337,-14.118535345085109)--cycle, linewidth(0.4)); draw((0.5960471031587434,-11.853790157486085)--(0.5960471031587437,-11.034007003784655)--(-0.22373605054268642,-11.034007003784655)--(-0.22373605054268664,-11.853790157486085)--cycle, linewidth(0.4)); draw((-4.062791415445341,-11.853790157486086)--(-4.062791415445341,-11.034007003784657)--(-4.8825745691467715,-11.034007003784657)--(-4.8825745691467715,-11.853790157486086)--cycle, linewidth(0.4)); /* draw figures */ draw((-4.8825745691467715,7.047647764291851)--(-4.8825745691467715,-11.853790157486086), linewidth(0.4)); draw((-4.8825745691467715,-11.853790157486086)--(8.964592843259439,-11.853790157486086), linewidth(0.4)); draw((-4.8825745691467715,7.047647764291851)--(8.964592843259439,-11.853790157486086), linewidth(0.4)); draw(circle((-0.22373605054268664,-7.194951638882), 4.658838518604084), linewidth(0.4)); draw(circle((2.041009137056334,-2.403071196597118), 11.715464148487989), linewidth(0.4)); draw((-4.8825745691467715,7.047647764291851)--(2.0410091370563337,-14.118535345085109), linewidth(0.4)); draw((-4.882574569146771,-7.194951638881999)--(2.0410091370563337,-14.118535345085109), linewidth(0.4)); draw((-4.8825745691467715,-11.853790157486086)--(2.0410091370563337,-14.118535345085109), linewidth(0.4)); draw((2.0410091370563337,-14.118535345085109)--(8.964592843259439,-11.853790157486086), linewidth(0.4)); draw((3.534490936065777,-4.441679719954091)--(2.0410091370563337,-14.118535345085109), linewidth(0.4)); draw((-0.22373605054268664,-11.853790157486085)--(3.534490936065777,-4.441679719954091), linewidth(0.4)); draw((-0.22373605054268664,-7.194951638882)--(-0.22373605054268664,-11.853790157486085), linewidth(0.4)); draw((-0.22373605054268664,-7.194951638882)--(8.964592843259439,-11.853790157486086), linewidth(0.4)); draw(circle((4.370428396358375,-9.524370898184042), 5.1509747670491475), linewidth(0.4) + ffqqtt); /* dots and labels */ dot((-4.8825745691467715,7.047647764291851),dotstyle); label("$A$", (-5.413743297722508,7.540635044242123), NE * labelscalefactor); dot((-4.8825745691467715,-11.853790157486086),dotstyle); label("$B$", (-5.645612988556418,-12.168288676640222), NE * labelscalefactor); dot((8.964592843259439,-11.853790157486086),dotstyle); label("$C$", (9.116757327869184,-11.472679604138492), NE * labelscalefactor); dot((-0.22373605054268664,-7.194951638882),linewidth(4pt) + dotstyle); label("$I$", (-0.0807404085425789,-6.8739307359326105), NE * labelscalefactor); dot((-0.22373605054268664,-11.853790157486085),linewidth(4pt) + dotstyle); label("$D$", (-0.7763494810443088,-12.477448264418767), NE * labelscalefactor); dot((3.534490936065777,-4.441679719954091),linewidth(4pt) + dotstyle); label("$E$", (3.706464541744617,-4.13013939439801), NE * labelscalefactor); dot((-4.882574569146771,-7.194951638881999),linewidth(4pt) + dotstyle); label("$F$", (-4.718134225220778,-6.8739307359326105), NE * labelscalefactor); dot((2.0410091370563337,-14.118535345085109),linewidth(4pt) + dotstyle); label("$M$", (1.3104777364608808,-14.796145172757866), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

We see from Iran Lemma that $X \in BI \cup CI$. WLOG $X \in BI \implies \angle ABC = \pi/2$. But now let $DF \cap AI = M'$ where $I$ is the incentre, then $\angle AM'C = \pi/2 = \angle ABC$, so $M'$ lies on the circumcentre of $\Delta ABC$ as well as the angle bisector of $\angle BAC$, which yields $M = M'$, as desired. $\square$

WLOG let $\angle B =\pi/2$. It can be observed that $\angle FDI = \pi/4$. We can also observe $\angle CDM = \pi/4= \angle MIC $, by using the fact that $IDMC$ are concylic. We are done

Iran's lemma and Iran's lemma agian!!