[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = 826.0603120827556, xmax = 1079.7812335474573, ymin = 942.4726413207709, ymax = 1121.4713573431247; /* image dimensions */ pen fueaev = rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745); pen zzttqq = rgb(0.6,0.2,0.); pen ffqqtt = rgb(1.,0.,0.2); pen ttqqqq = rgb(0.2,0.,0.); pen qqzzff = rgb(0.,0.6,1.); pen wwccqq = rgb(0.4,0.8,0.); filldraw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229)--cycle, fueaev, zzttqq); /* draw figures */ draw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791), zzttqq); draw((917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229), zzttqq); draw((1016.4379799979403,1022.3895892670229)--(947.0656746274594,1105.0213195778317), zzttqq); draw(circle((967.0388536639019,1051.3533846178484), 57.26408231772152), ffqqtt); draw(circle((956.8566499375142,1049.562472969398), 28.632041158861085), ttqqqq); draw(circle((919.8235705142946,1054.6141521154225), 27.709034166988225), qqzzff); draw((845.91100825702,1023.5549216479395)--(896.2927272340726,1035.6632414844441)); draw((870.808680629589,1030.8290089453244)--(871.3950548615035,1028.3891541870591)); draw((896.2927272340726,1035.6632414844441)--(946.6744462111252,1047.7715613209486)); draw((921.1903996066416,1042.9373287818291)--(921.7767738385561,1040.4974740235637)); draw((947.0656746274594,1105.0213195778317)--(966.8432394557349,1022.7285054894069)); draw(circle((926.6235921378332,1049.7690770090317), 33.450500170506814), wwccqq); draw((947.0656746274594,1105.0213195778317)--(946.5042589782556,1022.8674962443777)); draw((975.0964736974341,1071.632874192071)--(917.2484989135296,1023.067421711791)); draw((928.625707321889,1054.3382095571908)--(1016.4379799979403,1022.3895892670229)); draw((947.0656746274594,1105.0213195778317)--(845.91100825702,1023.5549216479395)); draw((845.91100825702,1023.5549216479395)--(917.2484989135296,1023.067421711791)); /* dots and labels */ dot((947.0656746274594,1105.0213195778317),dotstyle); label("$A$", (948.1809314251065,1107.809461571948), NE * labelscalefactor); dot((917.2484989135296,1023.067421711791),dotstyle); label("$B$", (918.3478120880482,1025.838086944889), NE * labelscalefactor); dot((1016.4379799979403,1022.3895892670229),dotstyle); label("$C$", (1017.6056670786347,1025.2804585460653), NE * labelscalefactor); dot((946.6744462111252,1047.7715613209486),dotstyle); label("$H$", (947.9021172256947,1050.6525506925361), NE * labelscalefactor); dot((946.5042589782556,1022.8674962443777),dotstyle); label("$D$", (947.623303026283,1025.559272745477), NE * labelscalefactor); dot((975.0964736974341,1071.632874192071),dotstyle); label("$E$", (976.3411655656943,1074.3517576425363), NE * labelscalefactor); dot((928.625707321889,1054.3382095571908),dotstyle); label("$F$", (929.7791942639303,1057.0652772790068), NE * labelscalefactor); dot((845.91100825702,1023.5549216479395),dotstyle); label("$S$", (846.9713770386376,1026.3957153437125), NE * labelscalefactor); dot((918.861249095803,1082.3064707607587),dotstyle); label("$G$", (920.0206972845187,1085.225511419595), NE * labelscalefactor); dot((966.8432394557349,1022.7285054894069),dotstyle); label("$M$", (967.9767395833414,1025.559272745477), NE * labelscalefactor); dot((946.8700604192923,1076.3964404493902),dotstyle); label("$N$", (947.9021172256947,1079.0915990325364), NE * labelscalefactor); dot((946.7654547120856,1061.08913862816),dotstyle); label("$K$", (947.9021172256947,1063.756818064889), NE * labelscalefactor); dot((940.4989904395748,1073.0618507626966),dotstyle); label("$T$", (941.4893906392242,1075.745828639595), NE * labelscalefactor); dot((896.2927272340726,1035.6632414844441),dotstyle); label("$P$", (897.4367471321661,1038.3847259184183), NE * labelscalefactor); /* end of picture */ [/asy][/asy] This problem really should be thought of in terms of the excentral triangle, so that's what this diagram is; the original triangle is here $DEF$, and we also have $D \to K, K \to T, I \to H, M \to N, N \to P$ because I like these labels . Note that by inversion in $(AEHF)$, $(S,K;E,F)=-1$, so $S$ is the foot of the external angle bisector, which is line $BC$. Then we know that $$(D,T;E,F)_{\omega}\stackrel{N}{=}(K,S;E,F)=-1.$$ Now, we want to show that $(ABC), (TKH), (NDP)$ are concurrent. This would imply the result, since a homothety at $H$ with power $1/2$ takes $(ABC)$ to the 9 point circle $\omega$. Let $G=AS \cap (ABC)$, and I claim $G$ lies on both $(TKH)$ and $(NDP)$. First, note that by power of a point, $$SA*SG=SB*SC=SE*SF,$$so $G \in (AEFH).$ Furthermore, $G,H,M$ and the $A-$antipode on $(ABC)$ are collinear, since $\angle AGH=90$. The simplest way to show that $G,T,K,H$ are cyclic is by inverting in the circle of diameter $BC$; the circle $(AEFH)$ is preserved since $ME$ and $MF$ are tangent to it, so $G \to H$. Furthermore, $$(D,T;E,F)=-1 \implies (S,T',E,F)=-1 \implies T'=K.$$Therefore $GHTK$ are concyclic with $MH*MG=MK*MT.$ Now, note that $H$ is the orthocenter of triangle $ASM$. This is because $\angle MGA=90$ and $\angle ADM=90$. Then, the 9-point circle of triangle $ASM$ contains both $D$ and $G$ as feet of altitudes, and $N$ and $P$ as midpoints of $AH,SH$. Thus, $DGNP$ is cyclic, so $G=L_1$ is the desired point.

MAN, IDK

whatshisbucket wrote: Me trying to do a geometry problem is like trying to eat soup with a toothpick. Some things never change.

My problem. Let $W$ be the midpoint of $\overline{BC}$, let $X$ be the point on $\Omega$ opposite $M$. Observe that $\overline{KD}$ passes through $X$, and thus lines $BC$, $MK$, $XA$ concur at the orthocenter of $\triangle DMX$, which we call $S$. Denote by $I_A$ the $A$-excenter of $ABC$. Next, let $E$ be the foot of the altitude from $I$ to $\overline{XI_A}$; observe that $E$ lies on the circle centered at $M$ through $I$, $B$, $C$, $I_A$. Then, $S$ is the radical center of $\omega$, $\Omega$, and the circle with diameter $\overline{IX}$; hence line $SI$ passes through $E$; accordingly $I$ is the orthocenter of $\triangle XSI_A$; denote by $L$ the foot from $X$ to $\overline{SI_A}$. [asy][asy] size(10cm); pair A = dir(160); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = extension(A, I, B, C); pair M = circumcenter(B, I, C); pair I_A = 2*M-I; pair X = -M; pair K = foot(M, D, X); pair S = extension(M, K, B, C); pair N = midpoint(I--S); pair W = midpoint(B--C); pair E = foot(I, X, I_A); draw(A--B--C--cycle, orange); draw(unitcircle, orange); draw(CP(M, I), olive); draw(A--I_A, orange); draw(M--X, orange); draw(S--E, olive); draw(X--I_A, olive); draw(B--S, orange); pair L = foot(X, I_A, S); draw(M--S--X--cycle, heavyred+1); draw(X--K, heavyred+1); draw(A--M, heavyred+1); draw(S--W, heavyred+1); draw(S--I_A, orange); draw(circumcircle(K, I, D), palered+1); draw(circumcircle(M, A, N), palered+1); draw(X--L, orange+1); pair T = midpoint(I--L); draw(T--M, orange); dot("$A$", A, dir(150)); dot("$B$", B, dir(B)); dot("$C$", C, dir(-10)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$I_A$", I_A, dir(I_A)); dot("$X$", X, dir(X)); dot("$K$", K, dir(K)); dot("$S$", S, dir(S)); dot("$N$", N, dir(135)); dot("$W$", W, dir(-45)); dot("$E$", E, dir(45)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T)); /* Source generated by TSQ */ [/asy][/asy] We claim that this $L$ lies on both the circumcircle of $\triangle KID$ and $\triangle MAN$. It lies on the circumcircle of $\triangle MAN$ since this circle is the nine-point circle of $\triangle XSI_A$. For the other, note that $\triangle MWI \sim \triangle MIX$, according to which $\angle IWM = \angle MIX = 180^{\circ} - \angle LIM = 180^{\circ} - \angle MLI$, enough to imply that quadrilateral $MWIL$ is cyclic. But lines $IL$, $DK$, and $WM$ meet at $X$, implying the conclusion. All that remains to show is that the midpoint $T$ of $\overline{IL}$ lies on $\Omega$. But this follows from the fact that $\overline{TM} \parallel \overline{LI_A} \implies \angle MTX = 90^{\circ}$, thus the problem is solved. Remark: Some additional facts about this picture: the point $T$ is the contact point of the $A$-mixtilinear incircle, while the point $K$ is such that $\overline{AK}$ is an $A$-symmedian.

Since $AK$ is a symmedian in $\triangle ABC$, we have $-1=(AK;BC)=(SD;BC)$, so $S$ is the foot of the external angle bisector in $\angle ABC$. Denote $I_A, I_B, I_C$ as the excenters; invert about $I$ swapping $A$ and $I_A$. Since $\triangle ABC$ is the orthic triangle of $\triangle I_AI_BI_C$, we have the following mappings: $K$ maps to the harmonic conjugate of $I_A$ in $\odot(I_AI_BI_C)$. $M$ maps to the point on $\odot(I_AI_BI_C)$ with $I_AM'\perp \overline{I_BI_C}$. $D$ maps to the reflection of $I_A$ over $I_BI_C$. $N$ maps to the reflection of $I$ over $I_AY$, where $Y$ is the midpoint of $\overline{I_BI_C}$. It is well-known that $D'K'$ passes through $Y$ by reflection about $I_BI_C$, and we claim that $Y\in\odot(I_AN'M')$; this is just because $\angle I_AN'Y=\angle I_AIY=\pi-\angle M'IY=\pi-\angle YM'I_A$. Since $Y$ lies on $\odot(ABC)$, which is the nine-point circle of $\odot(I_AI_BI_C)$, inverting back yields the desired conclusion.

EulerMacaroni wrote: Sigh I solved it except I thought I didn't and then I submitted something wrong Can sort of relate to you, except for #2.

I coord-bashed the whole thing... oops...

v_Enhance wrote: My problem. Let $W$ be the midpoint of $\overline{BC}$, let $X$ be the point on $\Omega$ opposite $M$. Observe that $\overline{KD}$ passes through $X$, and thus lines $BC$, $MK$, $XA$ concur at the orthocenter of $\triangle DMX$, which we call $S$. Denote by $I_A$ the $A$-excenter of $ABC$. Next, let $E$ be the foot of the altitude from $I$ to $\overline{XI_A}$; observe that $E$ lies on the circle centered at $M$ through $I$, $B$, $C$, $I_A$. Then, $S$ is the radical center of $\omega$, $\Omega$, and the circle with diameter $\overline{IX}$; hence line $SI$ passes through $E$; accordingly $I$ is the orthocenter of $\triangle XSI_A$; denote by $L$ the foot from $X$ to $\overline{SI_A}$. [asy][asy] size(10cm); pair A = dir(160); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = extension(A, I, B, C); pair M = circumcenter(B, I, C); pair I_A = 2*M-I; pair X = -M; pair K = foot(M, D, X); pair S = extension(M, K, B, C); pair N = midpoint(I--S); pair W = midpoint(B--C); pair E = foot(I, X, I_A); draw(A--B--C--cycle, orange); draw(unitcircle, orange); draw(CP(M, I), olive); draw(A--I_A, orange); draw(M--X, orange); draw(S--E, olive); draw(X--I_A, olive); draw(B--S, orange); pair L = foot(X, I_A, S); draw(M--S--X--cycle, heavyred+1); draw(X--K, heavyred+1); draw(A--M, heavyred+1); draw(S--W, heavyred+1); draw(S--I_A, orange); draw(circumcircle(K, I, D), palered+1); draw(circumcircle(M, A, N), palered+1); draw(X--L, orange+1); pair T = midpoint(I--L); draw(T--M, orange); dot("$A$", A, dir(150)); dot("$B$", B, dir(B)); dot("$C$", C, dir(-10)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$I_A$", I_A, dir(I_A)); dot("$X$", X, dir(X)); dot("$K$", K, dir(K)); dot("$S$", S, dir(S)); dot("$N$", N, dir(135)); dot("$W$", W, dir(-45)); dot("$E$", E, dir(45)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T)); /* Source generated by TSQ */ [/asy][/asy] We claim that this $L$ lies on both the circumcircle of $\triangle KID$ and $\triangle MAN$. It lies on the circumcircle of $\triangle MAN$ since this circle is the nine-point circle of $\triangle XSI_A$. For the other, note that $\triangle MWI \sim \triangle WIX$, according to which $\angle IWM = \angle MIX = 180^{\circ} - \angle LIM = 180^{\circ} - \angle MLI$, enough to imply that quadrilateral $MWIL$ is cyclic. But lines $IL$, $DK$, and $WM$ meet at $X$, implying the conclusion. All that remains to show is that the midpoint $T$ of $\overline{IL}$ lies on $\Omega$. But this follows from the fact that $\overline{TM} \parallel \overline{LI_A} \implies \angle MTX = 90^{\circ}$, thus the problem is solved. Remark: Some additional facts about this picture: the point $T$ is the contact point of the $A$-mixtilinear incircle, while the point $K$ is such that $\overline{AK}$ is an $A$-symmedian. Can I ask how you are supposed to produce such a diagram on the actual contest?

Wait how do you bash this? It doesn't seem very feasible...

It's not... took 8 pages and cancelling was such a huge pain.

I figured out that $AK$ is the $A$-symmedian and $AS$ is the $A$-external bisector. Is that 1 point?

Consider an inversion about $I$ mapping each point $X$ to $X'$. As $\angle BAI = \angle IBC$, we have $\angle A'B'I = \angle IC'B'$, so $I$ is the orthocenter of $\triangle A'B'C'$. $A'I$ meets $(IB'C')$ at $D'$, which must be the reflection of $A'$ over $B'C'$, and the circle of diameter $D'M'$ cuts $(A'B'C')$ at some point $K'$. Note that the reflection of $K'$ across $B'C'$ lies on $(IM'K')$ and $(IB'C')$, so $S'$ is the reflection of $K'$ across $B'C'$. Thus $S'$ lies on the circle of diameter $A'I$. Let $R'$ be the midpoint of $B'C'$. It is well known that $S'$ is then the intersection of $A'R'$, the circle through $A',B'$ tangent to $B'C'$, the circle through $A',C'$ tangent to $B'C'$, circle $(IB'C')$, and the circle of diameter $AI'$. It is also the projection of $I$ onto $A'R'$. (Thus $K'$ is the corresponding point for symmetric triangle $D'B'C'$.) Now $N'$ is the reflection of $I$ across $S'$, which is the reflection of $I$ across $A'R'$. We claim that $R$ is the desired point, so that $R$ lies on circles $(KID)$, $(MAN)$, and such that the midpoint of $IR$ lies on $\Omega$. Equivalently, we claim $R'$ lies on $K'D'$, $R'$ lies on $(M'A'N')$, and the reflection of $I$ over $R'$ lies on $\Omega'$. The first claim is true, as was discussed earlier. The third claim is also true, since $I$ is the orthocenter of $A'B'C'$ and $R'$ is the midpoint of $B'C'$. To show that the second is true, note that $R'M'=R'I$, by symmetry across $B'C'$, and $R'I=R'N'$, by symmetry across $A'R'$. So $R'$ is the center of a circle through $M',I',N'$, and \[ \angle M'R'N' = 2(180-\angle M'IN') = 2(\angle A'IN') = 180 - \angle M'A'N' \]so $R'$ lies on $(M'A'N')$, as desired.

MSTang wrote: I figured out that $AK$ is the $A$-symmedian and $AS$ is the $A$-external bisector. Is that 1 point? a bunch of people told me no; i did the same

strategos21 wrote: It's not... took 8 pages and cancelling was such a huge pain. Think ur confusing this with jmo

Oh right. sorry.

joshualee2000 wrote: strategos21 wrote: It's not... took 8 pages and cancelling was such a huge pain. Think ur confusing this with jmo strategos21 wrote: Oh right. sorry. Phew. You scared me for a moment there.

Oops. I thought I proved that $I_A$, $L$, $S$ could not be collinear. Still solved it, but my proof ended up being 4 squished pages long.

Can someone answer my question in post #7?

jasonhu4 wrote: Can I ask how you are supposed to produce such a diagram on the actual contest? With ruler and compass

Let $\omega_1$ and $\omega_2$ be the circles centered at $M$ and $N$ which have radiis $MI$ and $NS=NA$, resp. $\omega_1 \cap \omega_2= \{I,P\}$. Let $I_A$ be A-excenter and $M'$ be $M$ antipode with respect to the circle $\Omega$. Let $F=PI\cap \Omega$. It's well known that $B,I,C$ and $I_A$ are concyclic, where the circle $(BICI_A)$ is centered at $M$. CLAIM 1: The line $AS$ is the radical axis of $\omega_1$ and $(KMD)$

Proof

CLAIM 2: Lines $SA$ and $KD$ intersects at $M'$

Proof

CLAIM 3: $P\in (MAN)$

Proof

CLAIM 4: $KMI_AP$ is cyclic

Proof

CLAIM 5: $P\in (KID)$:

Proof

So we proved that $P \in \{L_1,L_2\}$. For finishing the problem, we use radical axes theorem on circles $(KID)$, $\omega_1$ and $(KDM)$ to get $F-I-M'$, so $\angle MFM'= \angle MFI=90^{\circ}$. Since $MI=MP$ we get $IF=FP$, As desired. $\blacksquare$

Attachments:

Let $I_A$ be the $A$-excenter of $ABC,$ $P$ the midpoint of $BC,$ $S' = MP \cap \Omega,$ and let $\Psi$ denote the map from $\sqrt{bc}$ inversion. Under $\Psi,$ recall $(I, I_A)$ swap, and since line $BC$ and $\Omega$ swap, $(K,P),(M,D)$ also swap. So, $\Psi(S) = \Psi(\overleftrightarrow{MK}) \cap \Psi(\overleftrightarrow{BC}) = (ADP) \cap \Omega.$ But then since $\angle{S'PD} = 90 = \angle{S'AM} = \angle{S'AD},$ $S'$ lies on $(ADP),$ hence $\Psi{(S)} = S'.$ It follows that $\angle{SAM} = \angle{MAS'} = 90,$ and so $S,A,S'$ are collinear. Also from this, $ASKD$ is cyclic, and $K,D,S'$ are collinear. Now, let $X$ be the foot of the perpendicular from $I$ to $SI_A,$ and $X' = S'I \cap SI_A.$ Since $N$ is the midpoint of right triangle $\triangle{SAI},$ $N$ is the circumcenter, so $NA = NI.$ Also, from Fact 5, $M$ is the midpoint of $II_A,$ so $MN$ is the $I$-midline in $\triangle{SII_A}.$ Then, $X$ is the reflection of $I$ across $MN,$ so $$\angle{NXM} = \angle{NIM} = 180 - \angle{AIN} = \angle{NAM},$$implying $MXNA$ is cyclic. Now from properties of inversion, $\angle{AIS'} = \angle{A\Psi(S')\Psi(I)} = \angle{ASI_A},$ so it follows that $AIX'S$ is cyclic. But this means $\angle{IX'S} = 90,$ hence $X = X'.$ Now, since $ASKD$ is cyclic by angles, and $SA, XI, DK$ concur at $S',$ the converse of radical axis theorem implies $DKXI$ is cyclic. Hence $X$ lies on both $(KID)$ and $(MAN).$ Finally, denote $J$ as the midpoint of $XI,$ which satisfies $90 = \angle{IJM} = \angle{S'JM},$ implying that $J$ lies on $\Omega.$ The conclusion follows.

Random high-powered config-throwing, I guess. Let $I_AI_BI_C$ be the excentral triangle. It suffices to show that one of $L_1, L_2$ lies on the circumcircle of $I_AI_BI_C$, $(ABC)$ is the nine-point circle of this triangle and $I$ is the orthocenter. Let $N_A$ be the midpoint of arc $BAC$, which is also the midpoint of $I_AI_C$. Let $M_A$ be the midpoint of $BC$. Since $\angle MKN_A=90^\circ$, $D$ lies on $KN_A$. By Ratio Lemma, we have that $\frac{BK}{KC} = \frac{BK}{KC} \cdot \frac{BN_A}{N_AC} = \frac{BD}{CD} = \frac{BA}{AC}$, so $K$ is the intersection of the $A$-symmedian with $(ABC)$. Applying Radical Axis on $(ABC)$, $(MM_ADK)$, and $(N_AADM_A)$. we have that $MK$, $M_AD$, and $AN_A$ concur, i.e. $S$ is the intersection of the $A$-external angle bisector with $BC$. It is well-known (say, by Brocard on $I_BCBI_C$) that $I$ is the orthocenter of $\triangle I_AN_AS$. Thus, we have that $(MAN)$ is the nine-point circle of $\triangle I_AN_AS$. Let $Q$ be the $I_A$-Queue point in $\triangle I_AI_BI_C$, defined as $I_AS\cap (I_AI_BI_C)\cap IN_A$ (which is well-known to exist). Using the nine-point circle $(MAN)$, we have that $Q$ lies on $(MAN)$. From here, we claim that $Q$ is the desired point. All we need to do now is show that $Q$ lies on $(KID)$. To this end, consider an inversion at $I$ with power $-II_A\cdot IA$. This inversion swaps $BC$ with $(II_CI_A'I_B)$, where $I_A'$ is the reflection of $I_A$ over $I_BI_C$ (we apply Reflecting the Orthocenter implicitly here); in particular, $D$ goes to $I_A'$. Since inversion preserves cross-ratio, $K$ is mapped to the intersection of $(I_AI_BI_C)$ with the $I_A$-symmedian in $\triangle I_AI_BI_C$, call it $K^*$. $Q$ goes to $N_A$. It suffices to show that $I_A'$, $K^*$, and $N_A$ are collinear. But this is clear, since $K^*$ is the $I_A'$-Humpty point in $\triangle I_A'I_BI_C$: it lies on the "(BHC)" circle of that triangle, and it satisfies $K^*I_B/K^*I_C=I_A'I_B/I_A'I_C$. Since a Humpty point lies on the corresponding median, we're done.

We split the solution into four parts. Part 1: Preliminary observations If $P$ is the midpoint of $BC$, then $\sqrt{bc}$ inversion and reflection over the $A-$ angle bisector swaps $P$ and $K$. Hence, $AK$ is the $A-$ symmedian. Then, we have $$-1 = (AK;BC) \buildrel M\over{=} (DS;BC)$$ so $AS$ is the $A-$ external angle bisector. Let $X$ be the midpoint of arc $\widehat{BAC}$ so that $X$ lies on $AS$. Then $X$ lies on $KD$, since $\angle MKD = 90^\circ$. Let $Y$ be the $A-$ mixtilinear intouch point. It is well known that $X$ lies on $YI$, so $\angle MYI = 90^\circ$. Part 2: $Y$ lies on $MN$ Let $MN$ intersect $AS$ at $F$ and $BC$ at $Q$. By Menelaus on $\triangle DIS$, $$\frac{MD}{MI}\cdot \frac{NI}{NS}\cdot \frac{QS}{QD} = 1.$$ But since $MI^2 = MB^2 =MD\cdot MI$, degenerate triangles $\triangle MDI$ and $\triangle MIA$ are homothetic at $M$. Hence, $\frac{AI}{IM} = \frac{ID}{DM}$, which rearranges to $\frac{AI}{ID} = \frac{MI}{MD}$. Then we have $$\frac{QS}{QD} = \frac{MI}{MD} = \frac{AI}{ID}$$ so $QI\parallel AS$. It follows that $QSFI$ is a parallelogram, so $FI\parallel BC$. But then $FI\perp XM$ and $MI\perp FX$, so $I$ is the orthocenter of $\triangle XFM$. But $Y$ is the foot of the altitude from $M$ to $XI$, so $Y$ must be the foot of the altitude from $X$ to $MF$, as desired. Now, let $Z$ be the reflection of $I$ over $Y$. Part 3: $Z$ lies on $(MAN)$ Since $\triangle IAS$ is right, $N$ is the circumcenter of $\triangle IAS$. Hence, we have $$\measuredangle MZN = \measuredangle NIM = \measuredangle NIA = \measuredangle IAN = \measuredangle MAN$$ as desired. Part 4: $Z$ lies on $(KID)$ Since $\angle MYI = 90^\circ$, $\triangle MIZ$ is isosceles. Hence, $Z$ lies on $(BIC)$. Then PoP at $X$ gives $$XI\cdot XZ = XB^2 = XD\cdot XK$$ as desired.

zacchro wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = 826.0603120827556, xmax = 1079.7812335474573, ymin = 942.4726413207709, ymax = 1121.4713573431247; /* image dimensions */ pen fueaev = rgb(0.9568627450980393,0.9176470588235294,0.8980392156862745); pen zzttqq = rgb(0.6,0.2,0.); pen ffqqtt = rgb(1.,0.,0.2); pen ttqqqq = rgb(0.2,0.,0.); pen qqzzff = rgb(0.,0.6,1.); pen wwccqq = rgb(0.4,0.8,0.); filldraw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229)--cycle, fueaev, zzttqq); /* draw figures */ draw((947.0656746274594,1105.0213195778317)--(917.2484989135296,1023.067421711791), zzttqq); draw((917.2484989135296,1023.067421711791)--(1016.4379799979403,1022.3895892670229), zzttqq); draw((1016.4379799979403,1022.3895892670229)--(947.0656746274594,1105.0213195778317), zzttqq); draw(circle((967.0388536639019,1051.3533846178484), 57.26408231772152), ffqqtt); draw(circle((956.8566499375142,1049.562472969398), 28.632041158861085), ttqqqq); draw(circle((919.8235705142946,1054.6141521154225), 27.709034166988225), qqzzff); draw((845.91100825702,1023.5549216479395)--(896.2927272340726,1035.6632414844441)); draw((870.808680629589,1030.8290089453244)--(871.3950548615035,1028.3891541870591)); draw((896.2927272340726,1035.6632414844441)--(946.6744462111252,1047.7715613209486)); draw((921.1903996066416,1042.9373287818291)--(921.7767738385561,1040.4974740235637)); draw((947.0656746274594,1105.0213195778317)--(966.8432394557349,1022.7285054894069)); draw(circle((926.6235921378332,1049.7690770090317), 33.450500170506814), wwccqq); draw((947.0656746274594,1105.0213195778317)--(946.5042589782556,1022.8674962443777)); draw((975.0964736974341,1071.632874192071)--(917.2484989135296,1023.067421711791)); draw((928.625707321889,1054.3382095571908)--(1016.4379799979403,1022.3895892670229)); draw((947.0656746274594,1105.0213195778317)--(845.91100825702,1023.5549216479395)); draw((845.91100825702,1023.5549216479395)--(917.2484989135296,1023.067421711791)); /* dots and labels */ dot((947.0656746274594,1105.0213195778317),dotstyle); label("$A$", (948.1809314251065,1107.809461571948), NE * labelscalefactor); dot((917.2484989135296,1023.067421711791),dotstyle); label("$B$", (918.3478120880482,1025.838086944889), NE * labelscalefactor); dot((1016.4379799979403,1022.3895892670229),dotstyle); label("$C$", (1017.6056670786347,1025.2804585460653), NE * labelscalefactor); dot((946.6744462111252,1047.7715613209486),dotstyle); label("$H$", (947.9021172256947,1050.6525506925361), NE * labelscalefactor); dot((946.5042589782556,1022.8674962443777),dotstyle); label("$D$", (947.623303026283,1025.559272745477), NE * labelscalefactor); dot((975.0964736974341,1071.632874192071),dotstyle); label("$E$", (976.3411655656943,1074.3517576425363), NE * labelscalefactor); dot((928.625707321889,1054.3382095571908),dotstyle); label("$F$", (929.7791942639303,1057.0652772790068), NE * labelscalefactor); dot((845.91100825702,1023.5549216479395),dotstyle); label("$S$", (846.9713770386376,1026.3957153437125), NE * labelscalefactor); dot((918.861249095803,1082.3064707607587),dotstyle); label("$G$", (920.0206972845187,1085.225511419595), NE * labelscalefactor); dot((966.8432394557349,1022.7285054894069),dotstyle); label("$M$", (967.9767395833414,1025.559272745477), NE * labelscalefactor); dot((946.8700604192923,1076.3964404493902),dotstyle); label("$N$", (947.9021172256947,1079.0915990325364), NE * labelscalefactor); dot((946.7654547120856,1061.08913862816),dotstyle); label("$K$", (947.9021172256947,1063.756818064889), NE * labelscalefactor); dot((940.4989904395748,1073.0618507626966),dotstyle); label("$T$", (941.4893906392242,1075.745828639595), NE * labelscalefactor); dot((896.2927272340726,1035.6632414844441),dotstyle); label("$P$", (897.4367471321661,1038.3847259184183), NE * labelscalefactor); /* end of picture */ [/asy][/asy] This problem really should be thought of in terms of the excentral triangle, so that's what this diagram is; the original triangle is here $DEF$, and we also have $D \to K, K \to T, I \to H, M \to N, N \to P$ because I like these labels . Note that by inversion in $(AEHF)$, $(S,K;E,F)=-1$, so $S$ is the foot of the external angle bisector, which is line $BC$. Then we know that $$(D,T;E,F)_{\omega}\stackrel{N}{=}(K,S;E,F)=-1.$$ Now, we want to show that $(ABC), (TKH), (NDP)$ are concurrent. This would imply the result, since a homothety at $H$ with power $1/2$ takes $(ABC)$ to the 9 point circle $\omega$. Let $G=AS \cap (ABC)$, and I claim $G$ lies on both $(TKH)$ and $(NDP)$. First, note that by power of a point, $$SA*SG=SB*SC=SE*SF,$$so $G \in (AEFH).$ Furthermore, $G,H,M$ and the $A-$antipode on $(ABC)$ are collinear, since $\angle AGH=90$. The simplest way to show that $G,T,K,H$ are cyclic is by inverting in the circle of diameter $BC$; the circle $(AEFH)$ is preserved since $ME$ and $MF$ are tangent to it, so $G \to H$. Furthermore, $$(D,T;E,F)=-1 \implies (S,T',E,F)=-1 \implies T'=K.$$Therefore $GHTK$ are concyclic with $MH*MG=MK*MT.$ Now, note that $H$ is the orthocenter of triangle $ASM$. This is because $\angle MGA=90$ and $\angle ADM=90$. Then, the 9-point circle of triangle $ASM$ contains both $D$ and $G$ as feet of altitudes, and $N$ and $P$ as midpoints of $AH,SH$. Thus, $DGNP$ is cyclic, so $G=L_1$ is the desired point. OH NO

Let $I_A$ be the $A$-excenter and $M'$ be the midpoint of major arc $\widehat{BAC}$, and let $F$ be the midpoint of $\overline{BC}$. As $AM'FD, DFMK, AM'MK$ are all cyclic along diameter circles, it follows that $\overline{XA}$ passes through $S$. Claim. [Characterization of $L$] There exists $i \in \{1, 2\}$ such that $L_i = \overline{M'I} \cap \overline{I_AS}$ satisfies $\angle IL_iS = 90^\circ$. Proof. First, $D$ is the orthocenter of triangle $M'SM$ as $\overline{MD} \perp \overline{SM'}$ and $\overline{SD} \perp \overline{MM'}$. Now, I claim that $I$ is the orthocenter of triangle $M'SI_A$. As $\overline{I_AI} \perp \overline{M'S}$, it suffices to show that $AI \cdot AI_A = SA \cdot AM'$. This follows because $\triangle SAB \sim \triangle CAM'$ and hence $$AI \cdot AI_A = AB \cdot AC = AS \cdot AM'.$$Now define $L$ to be the foot of the altitude, which lies on $\overline{M'I}$; $L$ lies on the nine-point circle $(MAN)$. Furthermore, $ASKD$ is cyclic, implying that $M'I \cdot M'L_1 = M'A \cdot M'S = M'D \cdot M'K$, hence $L$ lies on $(KID)$ and $L \in \{L_1, L_2\}$, as needed. $\blacksquare$ Now just let $L$ be the point we identified. For $E = \overline{IL_1} \cap \Omega$, we have $\angle IEM = \angle IL_1I_A = 90^\circ$, so $E$ is the midpoint of $\overline{IL}$ by homothety.

Let $M_A$ be the midpoint of $\overline{BC}$, $M_1$ be the midpoint of arc $BAC$, and let $L$ be the reflection of $I$ over the second intersection of $\overline{M_1I}$ with $(ABC)$. I claim that $L$ is one of the intersection points. First, observe that $K,D,M_1$ are collinear, so the internal bisectors of $\angle BAC$ and $\angle BKC$ concur at $D$, and hence $ABKC$ is harmonic and $\overline{AK}$ is a symmedian. Now, $(B,C;D,S)\stackrel{M}{=}(B,C;A,K)=-1$ (projecting through $A$ also works), so by Ceva-Menelaus $S$ is the foot of the external $\angle A$-bisector, so $S,A,M_1$ collinear. Let the incircle touch $\overline{BC}$ at $T$. We now invert about the incircle and denote images with $\bullet'$, so $A,B,C$ are sent to the midpoints of the intouch triangle and $(ABC)$ is sent to the 9-point circle $\omega$ of the intouch triangle. $M_1$ is sent to the point on $\omega$ such that $\angle IM_1'A'=90^\circ$. Then $L$ is sent to the midpoint of $I$ and the second intersection $X'$ of $\overline{IM_1'}$ with $\omega$. $D$ and $M$ are sent to the second intersection of $\overline{IA'}$ with $(IB'C')$ and $\omega$ respectively. $S$ is sent to the point on $(IB'C')$ such that $\angle IA'S'=90^\circ$, and $N'$ is the reflection of $I$ over $S'$. Observe that $\angle A'M'X'=90^\circ$, so since $(IB'C')$ and $\omega$ are reflections of each other across $\overline{B'C'}$ it follows that $X'$ is the orthocenter of $\triangle TB'C'$, so it is well-known that $L'$ is actually the midpoint of $\overline{B'C'}$. Furthermore, $D'$ is evidently the point on $(TB'C')$ such that $\overline{TD'} \parallel \overline{BC}$. Finally, since $M_1,D,K$ are collinear, $K'$ is the second intersection of $(IM_1'D')$ with $\omega$. We may now restate the inverted problem with reference triangle $TB'C'$ as follows. Restated wrote: Let $ABC$ be a triangle with orthocenter $H$. Let the midpoint of $\overline{BC}$ be $M$, the reflection of $A$ over $M$ be $A' \in (ABC)$, and $D$ the point on $(ABC)$ such that $\overline{AD} \parallel \overline{BC}$. Let $B',C'$ be the reflections of $A$ over $B,C$, and let $M'$ be the midpoint of $\overline{B'C'}$, equivalently the reflection of $A$ over $M$. Note that $HBCM'$ is cyclic, and let $\overline{DA'M}$ intersect $(HBC)$ again at $X$. Let $\overline{HMA'}$ intersect $(HBC)$ again at $Y$, and $(A'DY)$ intersect $(HBC)$ again at $K$. Finally, let $R$ be the reflection of $A'$ over $\overline{AM}$. Prove that $M,K,D$ are collinear (equivalent to $KIDL$ cyclic), and $MM'XR$ cyclic (equivalent to $MANL$ cyclic). For the first part, instead define $K'$ as the intersection of segment $\overline{MD}$ with $(HBC)$, i.e. the intersection of line $\overline{MD}$ with $(HBC)$ that doesn't lie on the $A$-altitude (the other intersection does). Also let $H'$ be the reflection of $H$ over $\overline{BC}$. Then it suffices to show that $A'DYK'$ is cyclic. We have $$MK'\cdot MD=MH'\cdot MD=MB\cdot MC=MH'\cdot MY=MA'\cdot MY,$$as desired. For the second part, we reflect everything over $M$, defining $H'$ as before. We then want to show that the reflection $H''$ of $H$ over $\overline{AM}$ lies on $(AMH')$. Note that $$\measuredangle H''MA=\measuredangle AMH=\measuredangle H''A'H,$$and since $\angle HH''A'=\angle HH'A'=90^\circ$ we have $(HH'H''A)$ cyclic, so $\measuredangle H''A'H=\measuredangle H''H'H=\measuredangle H''H'A$, hence $AMH'H''$ is cyclic as desired. $\blacksquare$ Remark: In theory after correctly guessing point $L$ (this can perhaps be motivated by $\sqrt{bc}$ inversion and then extraversion, swapping the incircle and $A$-excircle? idk I used ggb) it is possible to straight up complex bash everything (all points are directly computable). I do not want to see if it's possible in practice.

Speedsolved this for a challenge took me just 18 minutes. Really easy problem since its a very familiar picture. We let $L$ denote the major arc midpoint of $BC$ is $\Gamma$, and $I' = \overline{LI} \cap (BIC) \neq I$. Then, it is well known and easy to see that points $A$ , $S$ and $L$ , points $L$ , $I$ and $I'$ and points $L$ , $D$ , $K$ are collinear. We now show that $I'$ lies on all the circles we want. First, it is well known that $I'$ lies on $(ASI)$. Then, we prove the following. Claim : $I'$ is one of $L_1$ and $L_2$ i.e $I'$ lies on both circles $(AMN)$ and $(IDK)$. Proof : To see why $I'$ lies on $(DIK)$, we considering the power of $L$. Noting that $ADKS$ is cyclic due to the right angles, \[LI\cdot LI' = LA \cdot LS = LD \cdot LK\]which indeed implies that $I'$ lies on $(DIK)$. Further, we know that since $M$ is well known to be the center of $(BIC)$, $MI=MI'$ and thus, \[\measuredangle ANI' = 2\measuredangle ASL' = 2\measuredangle IAI' = 2\measuredangle MII' = \measuredangle IMI' = \measuredangle AMI'\]from which it also follows that $I'$ lies on $(AMN)$. Thus, $I'$ is indeed one of $L_1$ and $L_2$ as claimed. Then, consider the midpoint $N_I$ of $II'$. Returning to our observation that $\triangle MII'$ is isosceles, we know that $MN_I \perp II'$. But this immediately implies that $MI_NAL$ is cyclic and thus, $N_I$ must lie on $\Gamma$. Thus, the midpoint of one of $IL_1$ and $IL_2$ lies on $\Gamma$ as we wanted.

Let $X$ be the midpoint of $\widehat{BAC}$. Claim: $X$, $D$, and $K$ are collinear. Proof. Obvious as, $$\angle DKM = 90 = \angle XKM$$so the claim follows. $\square$ Claim: $X$, $A$ and $S$ are collinear. Proof. Note that, \begin{align*} -1 = (AK,BC) \end{align*}from $\sqrt{bc}$ inversion. Then projecting at $M$ we find, \begin{align*} -1 = (DS,BC) \end{align*}and hence $S = \overline{XA} \cap \overline{BC}$. $\square$ Claim: $DKAS$ is cyclic. Proof. Note that, $$\angle MKD = 90 = \angle MAS$$so the claim follows. $\square$ Now redefine $L = \overline{XI} \cap (BIC)$. Claim: $L$ lies on $(KID)$. Proof. It is equivalent to show $L$ lies on $(AIS)$ by radical axis, but this is well known. $\square$ Claim: $L$ lies on $(MAN)$. Proof. Note that, \begin{align*} \angle LMA = 180 - 2\angle LIM = 180 - 2\angle ASL = 180 - \angle ANL \end{align*}so the claim follows. However then from properties of mixintillinear touch point $T$ namely $M$, $I$, $T$ collinear, the midpoint of $NL$ is $T$ so we're done.

The KID has grown into a MAN, thanks to the OTIS walkthrough. Let $I_A$ be the $A$-excenter of $\triangle ABC,$ and let $X$ be the antipode of $M$ on $\Omega.$ First note that $X,A,S$ are collinear by radical center on $(AXD), (MKD), \Omega.$ Now, we see that $I_A I \perp XS$ since $\angle XAM = \angle XAI_A = 90^\circ$. Claim 1: $SI \perp XI_A$ and thus $I$ is the orthocenter of $\triangle XSI_A.$ Proof: When we rephrase the problem in terms of the excentral triangle $I_A I_B I_C,$ this claim is just a well-known Humpty point property. This means that $(MAN)$ is the nine-point circle of $\triangle XSI_A$ (since $N$ is the midpoint of $IS,$ $M$ is the midpoint of $II_A,$ and $A$ is the foot from $I_A$ to $SX$), so $(MAN)$ passes through the foot of the altitude from $X$ to $SI_A$, call it $L.$ Claim 2: The midpoint of $IL$ lies on $\Omega.$ Proof: Note that $\Omega$ is the circle with diameter $XM.$ Thus the intersection of $\Omega$ with $\overline{XLI}$ is the foot of the altitude from $M$ to $IL.$ Since $M$ is the midpoint of $II_A,$ it follows that $MI = ML$ and that the midpoint of $IL$ lies on $\Omega,$ as claimed. Now we go for the kill; we just need to show that $L$ lies on $(KID),$ and we win. Invert with respect to $(BIC);$ the point $K$ gets sent to $S,$ the point $I$ to itself, and the point $D$ to point $A.$ Moreover, by the previous claim, since $MI = ML,$ we see that $L$ gets sent to itself. Thus it suffices to show that $ASLI$ is cyclic, which follows from $\angle SAI = \angle SLI = 90^\circ.$ Therefore, we have identified a point $L$ that lies on both $(MAN)$ and $(KID)$ and for which the midpoint of $IL$ lies on $\Omega,$ and we are done.

Really nice! Let $X$ $Y$ and $F$ be the midpoint of $BC$, midpoint of arc $BAC$ and $(YAI) \cap (BIC)$ respectively. First, we note that $XDKM$ and $YADX$ is cyclic. So, by radical axes $Y-A-S$ collinear. Furthermore $Y-D-K$ collinear as well because \[90 = \measuredangle DKM = \measuredangle YDM.\] Now, again radical axis on $(YAIF), (ABC), (BIFC)$ gives $S-N-I-F$ collinear. Now, let $AI \cap YF = I_A.$ We notice \[\measuredangle IFI_A = \measuredangle I_AAY = 90.\] So $II_A$ is the diameter of $(BIC)$ implying that $I_A$ is the $A-$excenter. Furthermore, this also implies that $I$ is the orthocenter of $\triangle SYI_A$ implying that $(FAN)$ is the nine point circle of $\triangle SYI_A.$ Let $L$ be the feet of $Y$ to $I_AS.$ Then, notice as $D$ is the orthocenter of $\triangle YSM$ \[YI \cdot YL = YA \cdot YS = YD \cdot YK\] So $L \in (KID)$ and also $L \in (FAN).$ Now it suffices to show that $M \in (FAN).$ But, $M$ is just the midpoint of $II_A$ which implies that $M$ lies on $(FAN)$ as needed. Finally to finish let $P$ be the midpoint of $IL.$ We note that $M-P-N$ collinear. Furthermore as \[\measuredangle YPM = \measuredangle YLI_A = 90 = \measuredangle YAM\] we get $P \in (ABC)$ as needed.

mathtastic wrote: MAN, IDK how does this have more upvotes that i will ever see in my life bro :skull:

Seeing the involvement of $I$ and $I_a$, it is fairly natural to restate in terms of the excentral triangle. Also the required condition can be stated more naturally using homothety centered of the orthocenter of the excentral triangle with factor $2$ which I directly state. Restated Problem wrote: $\Delta ABC$ has orthic triangle $\Delta DEF$ and orthocenter $H$. $AH \cap EF=K$ and $AH \cap (DEF)=N$. $T$ is on the nine point circle such that $DFTE$ is harmonic. $X$ is the $A$ expoint and midpoint of $HX$ is $G$. Prove that $(ABC),(DNH),(HKT)$ are concurrent at a point. First notice that:- $$-1=(XD;BC) \stackrel{A}{=} (XK;FE)\stackrel{M}=(D,MK \cap (DEF);FE)$$implying $\overline{M-K-T}$ are collinear. Let $L$ be midpoint of $EF$ and $Q$ be the $A$ queue point. Note that $ML.MN=ME^2=MF^2$ and since it is well known that $ME,MF$ are tangent to $(AEF)$, this implies $ME^2=MH.MQ=ML.MN$ and since $TKLN$ is cyclic, $ML.MN=MT.MK=MH.MQ$ implying $HQTK$ is cyclic. Now it remains to show $DNQG$ is cyclic. We do $\sqrt{-HA.HD}$ inversion Then $D \longleftrightarrow A, N \longleftrightarrow H', Q \longleftrightarrow M $ Where $H'$ is reflection of $H$ over $BC$. Since $AH \perp XM$ and $XH_A \perp AM$ $\implies H$ is the orthocenter of $AXM$ so reflection of $H$ over $BC$ is also its reflection over $XM$ implying $H'$ lies on $(AXM)$ or that $X$ lies on $(AH'M)$. Inverting back, we get that the $A$ humpty point $H_A$ lies on $(DNQ)$. Now by power of point, since it is easy to see that $(AH_ADX)$ is cyclic with diameter $AX$:- $$HN.HD=\frac{HA.HD}{2}=\frac{HH_A.HX}{2}=HH_A.HG$$which implies that $G \in (DNH_A)$ and so $(DH_ANQG)$ is cyclic which is what we needed.

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Let $V$ be $A-$mixtilinear touch point and $I'$ be the reflection of $I$ with respect to $V$. Claim: $P,A,S$ are collinear. Proof: Since $P,S,D,M$ is an orthogonal system and $\measuredangle PAM=90$, conclusion follows.$\square$ Claim: $MV$ passes through the midpoint of $IS$. Proof: \[\frac{\sin KMV}{\sin VMD}\overset{?}{=} \frac{MI}{MS}=\frac{MB}{MS}=\frac{\sin DSM}{\sin \frac{A}{2}}\iff \frac{\sin KPV}{\sin VPA}\overset{?}{=} \frac{\sin MPK}{\sin \frac{A}{2}}\]Note that by $MI^2=MD.MA$, we see $MDI\sim MIA$. \[\frac{\sin DPI}{\sin IPA}.\frac{\sin \frac{A}{2}}{\sin MPD}=\frac{ID}{IA}.\frac{PA}{PD}.\frac{\frac{MI}{MP}}{\sin MPD}=\frac{MD}{MI}.\sin MDK.\frac{MI}{MP}.\frac{MP}{MK}=1\]Which completes the proof.$\square$ Claim: $I',M,A,N$ are concyclic. Proof: Since $NI=NS$ and $\measuredangle IAS=90$, we have $NA=NS=NI=NI'$. \[\measuredangle MAN=\measuredangle IAN=\measuredangle NIA=180-\measuredangle MIN=180-\measuredangle NI'M\]Thus, $I'\in (MAN)$.$\square$ Claim: $I',K,D,I$ are concyclic. Proof: Invert at $M$ with radius $MI$. $I',K,D,I$ swap with $I',S,A,I$ which lie on the circle with diameter $SI$ hence $I'\in (KDI)$ as desired.$\blacksquare$

This is a very wonderful problem by Evan Chen! Definitely make my day although I lost my MacBook (sad ) W.L.O.G. let the midpoint of $IL_2$; call it as $M_2$; lies on $(ABC)$. Define $G$ as the second intersection of $KD$ and $(ABC)$ and $I_A$ as the $A$-excenter of $\triangle{ABC}$. It's well known that $M$ is the center of $(IBI_AC)$. Let $J$ be the second intersection of $SI$ with $(IBI_AC)$. By the well-known $\sqrt{bc}$ inversion, it's easy to prove that $AK$ is the $A$-symmedian of $\triangle{ABC}$; just note that $M_{BC}$ swaps each other with $K$ since $K=(DM)\cap(ABC)$ goes to $K^*=(DM)\cap BC$ (by the fact that $M$ and $D$ swaps each other, so $(DM)$ is fixed under this inversion). Projecting the harmonic bundle $(A,K;B,C)$ by taking the perspectivity at $D$ to the line $BC$, we have $-1=(A,K;B,C)\overset{D}{=}(M,G;B,C)$, so $G$ is the midpoint of arc $BC$ containing $A$. Next, projecting the harmonic bundle $(M,G;B,C)$ by taking the perspectivity at $K$, we have $-1=(M,G;B,C)\overset{K}{=}(S,D;B,C)$. This means that $SA$ is the external angle bisector of $\angle{BAC}$, so $\angle{SAI}=90^\circ$ and $\overline{S,A,G}$. Now by Power of Point, we may have $SI\cdot SJ=SB\cdot SC=SA\cdot SG$, so $AIJG$ is cyclic, causing $\angle{IJG}=90^\circ$. But $\angle{IJI_A}=90^\circ$, so $\overline{G,J,I_A}$. Now see that $I_AI\perp SG$ and $SI\perp GI_A$, so $I$ is the orthocenter of $\triangle{GSI_A}$. Redefine $L_2$ to be the foot of altitude from $G$ to $SI_A$, so $L_2$ lies on $(IBI_AC)$. See that we can derive $SK\cdot SM=SB\cdot SC=SL_2\cdot SI_A$ by Power of Point, so this causes $MKL_2I_A$ to be cyclic. Now see that $\angle{L_2ID}=\angle{L_2II_A}=90^\circ-\angle{L_2I_AI}$ and $\angle{L_2KD}=360^\circ-90^\circ-\angle{L_2KM}=270^\circ-(180^\circ-\angle{L_2I_AM})=90^\circ+\angle{L_2I_AI}$, so it's easy to conclude that $L_2$ lies on $(KID)$ then. Next, $ASL_2I$ is trivially cyclic with center $N$ since that $\angle{SAI}=\angle{SL_2I}=90^\circ$. So we have $\angle{ANL_2}=2\angle{ASL_2}=2\angle{ASI_A}=2(90^\circ-\angle{SI_AA})=180^\circ-2\angle{L_2I_AM}=\angle{I_AML_2}$, so $L_2$ also lies on $(MAN)$. Returning back to the original definition, then $L_2$ as one of the intersection points between $(KID)$ and $(MAN)$ is also the intersection of $SI_A$ and $(IBI_AC)$ other than $I_A$. For the final act, notice that $NI=NL_2$ and $MI=ML_2$, so $MINL_2$ is a kite, meaning that $MN$ is perpendicular with $IL_2$ at its midpoint, which is $M_2$. But since we have $\angle{GM_2M}=\angle{IM_2M}=90^\circ$, therefore $M_2$ lies on $(ABC)$, as desired. $\blacksquare$ Note. The solution could be simpler if I just noticed that $(MANL_2)$ is the nine-point circle of $\triangle{GSI_A}$ (holy crap )

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