Let $BK$ and $BI$ intersect the circumcirle again at $T$ and $D$. Claim: $I$ is the orthocenter of $BKM$ This is equivalent to showing that the orthocenter of $BIM$ lies on $AC$. The most straightforward way to prove this is to let $A=(0,0)$, $B=(1,0)$, $C=(2,0)$, $M=(1,1/2)$, and $I=(1/\phi^2,1/\phi^2)$ and then use standard bashing methods to compute the orthocenter of $BIM$. The orthocenter comes out to $(\sqrt{5}/(5+3\sqrt{5}),0)$ as wanted. Claim: $AFDB$ is a trapezoid By reflection it is sufficient to show that $\angle IKH=\angle HKC$. But, $\angle HKC=\angle BMK-\angle BCA=\frac{1}{2}\angle ABC=\angle IKH$ Claim: $FKHD$ is concyclic $$\angle HDF=\angle HBA=\angle HBM=180^{\circ}-\angle FKH$$ Claim: $M$, $S$, and $T$ are collinear $$\angle SFT=\angle HFT=\angle HFD+\angle DFT=\angle BKH+\angle KBH=90^{\circ}$$ Claim: $TM$ is tangent to the incircle It is sufficient to show that $\angle BMI=\angle IMT$ but since $MI\perp BT$ and $MB=MT$ the result follows.

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Solution by erkosfobiladol. I realized that it's similar to above. Anyway Let $D$ be the tangency point of $(I)$ with $BC$. $SM\cap (ABC)=\{S,R\}, \ BI\cap AC=L$. $\textbf{Claim 1:} \ K,I,D$ are collinear. Let $ID\cap AC=K'$. $AB=1, AC=2, BC=\sqrt{5}$. Hence $CD=\frac{\sqrt{5}+1}{2}$ $CK'=\frac{CD}{sin B}=\frac{\frac{\sqrt{5}+1}{2}}{\frac{2}{\sqrt{5}}}=\frac{\sqrt{5}+5}{4}$ $BKLM$ is cyclic since $\angle BMK=90-\frac{\angle B}{2}=\angle BLK$. $CK=\frac{CB.CM}{CL}=\frac{\frac{5}{2}}{\frac{2\sqrt{5}}{\sqrt{5}+1}}=\frac{5\sqrt{5}+5}{4\sqrt{5}}=\frac{\sqrt{5}+5}{4}=CK'$ as desired. $\textbf{Claim 2:} \ AF\parallel BI$. By angle chasing, we get that $KL=KI$. Let $F'$ be the point which satisfies $ILF'A$ is an isosceles trapezoid. $F'\in KI$ and $\overline{MHK}$ is the perpendicular bisector of $AF'$. Thus $MF'=MA\implies F'\in (ABC)\implies F'=F$ $\textbf{Claim 3:} \ B,K,R$ are collinear. $H,K,A,B$ are on circle with diameter $BK$. Hence $\angle KBS=\angle CBS+\angle KBC=\angle CBS+\angle ABC-\angle ABK=\angle CFS+\angle ABC-\angle AHK=\angle CFA-\angle SFA+\angle ABC-\angle KHF$ $=180-90=90$ Thus $BK$ passes through the antipode of $S$ which is $R$. $\textbf{Claim 4:} \ MR$ is tangent to $(I)$. $MI\perp BK$ since $I$ is the orthocenter of $MKB$. And $MR=MB$ which gives that $IR=IB$. Hence $MB$ is the reflection of $\overline{MDB}$ with respect to $MI$. Thus $MR$ is also tangent to $(I)$ as desired.$\blacksquare$

At first we define points $E$ the intersection of line $BI$ and circumcircle of $\triangle ABC$ , $D$ such that $BC$ is tangent to incircle of $\triangle ABC$ at $D$ and $P$ such that $MP$ is tangent to incircle of $\triangle ABC$(Different from $MD$). If $AB=c$ , it's obvious that $AC=2c$ and $BC=\sqrt{5}c$ so we have $DB=\frac{\sqrt5 - 1}{2}c$ and $DC=\frac{\sqrt5 + 1}{2}c$! We want to show that $P,M,S$ are collinear. $Claim 1$ : $F,K,I,D$ are collinear! $Proof$ : $ABHK$ is cyclic because $\angle KAB=\angle KHB=90^{\circ}$. We need to show $ABDHK$ is cyclic pentagon and for this we will prove $AHDB$ is cyclic! We want $\angle AHB=\angle ADB$ and we already have $\angle AEH=\angle ACD \implies$ So it suffices to show $\triangle AHE\sim\triangle ADC$ that It's equivalent to show $\frac{AE}{HE}=\frac{AC}{DC}$! $\frac{AE}{HE}=\frac{EC}{HB}=\frac{2HM}{HB}$ ($E$ is midpoint of arc $AC$ and $H$ is midpoint of $BE$)$=2tan\frac{B}{2}=2(\frac{1-cosB}{sinB})=2(\frac{1-\frac{1}{\sqrt5}}{\frac{2}{\sqrt5}})=\sqrt5 - 1=\frac{4}{\sqrt5 +1}=\frac{AC}{DC}$ $\implies$ So $F,K,I,D$ are collinear! $Claim 2$ : $EF || BC$! $Proof$ : We need to show $\angle ECB=\angle FBC$ and because $\angle ECB=\angle BID$($IECD$ is cyclic) $\implies$ So it suffices to show $\triangle BDI\sim\triangle FDB$ that it's equivalent to show $DB^2=DI.DF$! For $DI \rightarrow$ It's easy to see that $DI=\frac{AB-BC+CA}{2}=\frac{3-\sqrt5}{2}c$ ($\angle BAC=90^{\circ}$) For $DF \rightarrow$ $\triangle BFC$ is right-angled and $FD$ is altitude from $F$ to $BC \implies$ So $DF^2=DB.DC=(\frac{\sqrt5 - 1}{2}c).(\frac{\sqrt5 + 1}{2}c)=c^2$ which means $DF$=$c$ $\implies DI.DF=(\frac{3-\sqrt5}{2}c).c=(\frac{\sqrt5 - 1}{2c})^2=DB^2$ $\implies$ So $EF || BC$! $Claim 3$ : $AF||BE$! $Proof$ : $BF=EC=EA$($EFCB$ is isosceles trapezoid and $E$ is midpoint of arc $AC$) $\implies$ $AFEB$ is isosceles trapezoid $\implies$ So $AF||BE$! $Claim 4(Last Claim)$ : $P,M,S$ are collinear! $Proof$ : We need to show $\angle PMD+\angle DMS=180^{\circ}$. Because of the $Claim 1$, Obviously $I$ is the orthocenter of $\triangle MBK$ and also $M$ is the circumcenter of $\triangle ABC$ $\implies$ So it suffices to show $\angle BAD+\angle BFS=90^{\circ}$! Now because of $Claim3$, $\angle BFH=\angle EAH=\angle CAD$(by $\triangle AHE\sim\triangle ADC$) and $\angle BAD+ \angle CAD=90^{\circ}$! Finally we are done with this nice problem (:

a geometric solution for $FI\perp BC$ firstly we define points $X$ , $Y$ ,$N$ intersection of $BI$ with $AC$ and circumcircle of $\triangle ABC$ and the midpoint of $AC$ respectively since $AN=AB$ and $\angle YAN=\angle XBA$ and $\angle YNA=\angle A=90$ so $$\triangle AYN\sim\triangle ABX$$and so $BX=AY$ $I$ is the incenter of $\triangle ABC$ so $YI=AY=BX$ then $IH=HX$ $FI\cap BC=D$ $\angle BID=\angle KIH=\angle KXH=C+B/2\implies \angle BID=90-\angle IBD\implies FI\perp BC $ and the rest can be solved similar to other solutions posted

Here is an another approach : Lemma : Let $\triangle ABC$ with circumcircle $\omega$ be a right-angled triangle such that $AC=2AB$ and also let $I$ be its incenter and $D$ the touchpoint of incircle with the side $BC$. Now if $T'$ be the mid-point of the bigger arc $AC$ in $\omega$ , then points $I , D , T'$ are collinear and also if $IT'$ cuts $\omega$ for the second time at $F$ , then we have $\angle CBF=90-\frac{\angle B}{2}$. Proof : Let points $M , S$ be mid-points of sides $BC$ and $AC$ respectively and also suppose that lines $MS$ and $ID$ intersect each other at point $T'$ , then for showing that $T'$ lies on the circle $\omega$ , since quadrilateral $T'DSC$ is cyclic , its enough to show that $DS \parallel BT$ which $T$ is the mid-point of smaller arc $AC$ , so if $K$ be the intersection point of $AT$ and $BC$ , then we can get : $$\frac{AK}{CK}=\frac{AB}{BC} \implies CK=\frac{2AB.BC}{AB+BC}$$$$CD=\frac{CB+CA-AB}{2}=\frac{AB+BC}{2} \implies \frac{CK}{CB}=\frac{2AB}{AB+BC}=\frac{AB}{\frac{AB+BC}{2}}=\frac{CS}{CD}$$And by Thales theorem , we are done and also since $\angle BCT'=\angle BAT'=90-\angle T'AC=\frac{\angle B}{2}$ , we get $BI \parallel CT'$ and $\angle FBC=90-\frac{\angle B}{2}$. Now in our main problem , suppose that the line passing trough points $I , D , T'$ (Defined as like as the Lemma) , intersect with $\omega$ at point $F$ for the second time and also $FH$ cuts $\omega$ at $S$ ; then by Pascal theorem for $SFT'TBC$ , lines $HD , TT' , SC$ are concurrent at some point $R$. So if $Q$ be the intersection point of $BT$ and $AC$ ( $T$ is the mid-point of smaller arc $AC$ ) , then while $RM \perp AC$ , we have $\angle RMH=\angle BQC=90+\frac{\angle B}{2}$ and by lemma , $\angle HSC=\angle FSC=\angle FBC=90-\frac{\angle B}{2}$. Thus , quadrilateral $RSMH$ is cyclic and since $DIHM$ is cyclic too , we have : $$\angle RHM=\angle DHM=\angle DIM=\angle MSC , \angle BSC=90 \implies \angle SMB=2\angle DIM$$Now while if $P$ be the touchpoint of second tangent line passing trough $M$ with the incircle , then $IDMP$ is a kite and $\angle DMP=180-2\angle DIM$ , so points $S , M , P$ are collinear and $S$ , is the desired point of problem. So for completing our proof , it's enough to show that if $F$ be the intersection point of the semi-line $IK$ with $\omega$ , then it is collinear with $I , K , D , T'$ , and for that , while $KHNT$ is a cyclic quadrilateral ( $N$ is the mid-point of $AC$. ) , we have : $$MN.MT=MH.MK=MN.MT'$$and since $DT'CN$ is cyclic too , we can get : $$MN.MT'=MC.MD=MB.MD=MH.MK$$Which means $BDHK$ is cyclic and $\angle BHK=\angle BDK=90$ , so we're done.

Shayan-TayefehIR wrote: Here is an another approach : Lemma : Let $\triangle ABC$ with circumcircle $\omega$ be a right-angled triangle such that $AC=2AB$ and also let $I$ be its incenter and $D$ the touchpoint of incircle with the side $BC$. Now if $T'$ be the mid-point of the bigger arc $AC$ in $\omega$ , then points $I , D , T'$ are collinear and also if $IT'$ cuts $\omega$ for the second time at $F$ , then we have $\angle CBF=90-\frac{\angle B}{2}$. Proof : Let points $M , S$ be mid-points of sides $BC$ and $AC$ respectively and also suppose that lines $MS$ and $ID$ intersect each other at point $T'$ , then for showing that $T'$ lies on the circle $\omega$ , since quadrilateral $T'DSC$ is cyclic , its enough to show that $DS \parallel BT$ which $T$ is the mid-point of smaller arc $AC$ , so if $K$ be the intersection point of $AT$ and $BC$ , then we can get : $$\frac{AK}{CK}=\frac{AB}{BC} \implies CK=\frac{2AB.BC}{AB+BC}$$$$CD=\frac{CB+CA-AB}{2}=\frac{AB+BC}{2} \implies \frac{CK}{CB}=\frac{2AB}{AB+BC}=\frac{AB}{\frac{AB+BC}{2}}=\frac{CS}{CD}$$And by Thales theorem , we are done and also since $\angle BCT'=\angle BAT'=90-\angle T'AC=\frac{\angle B}{2}$ , we get $BI \parallel CT'$ and $\angle FBC=90-\frac{\angle B}{2}$. Now in our main problem , suppose that the line passing trough points $I , D , T'$ (Defined as like as the Lemma) , intersect with $\omega$ at point $F$ for the second time and also $FH$ cuts $\omega$ at $S$ ; then by Pascal theorem for $SFT'TBC$ , lines $HD , TT' , SC$ are concurrent at some point $R$. So if $Q$ be the intersection point of $BT$ and $AC$ ( $T$ is the mid-point of smaller arc $AC$ ) , then while $RM \perp AC$ , we have $\angle RMH=\angle BQC=90+\frac{\angle B}{2}$ and by lemma , $\angle HSC=\angle FSC=\angle FBC=90-\frac{\angle B}{2}$. Thus , quadrilateral $RSMH$ is cyclic and since $DIHM$ is cyclic too , we have : $$\angle RHM=\angle DHM=\angle DIM=\angle MSC , \angle BSC=90 \implies \angle SMB=2\angle DIM$$Now while if $P$ be the touchpoint of second tangent line passing trough $M$ with the incircle , then $IDMP$ is a kite and $\angle DMP=180-2\angle DIM$ , so points $S , M , P$ are collinear and $S$ , is the desired point of problem. So for completing our proof , it's enough to show that if $F$ be the intersection point of the semi-line $IK$ with $\omega$ , then it is collinear with $I , K , D , T'$ , and for that , while $KHNT$ is a cyclic quadrilateral ( $N$ is the mid-point of $AC$. ) , we have : $$MN.MT=MH.MK=MN.MT'$$and since $DT'CN$ is cyclic too , we can get : $$MN.MT'=MC.MD=MB.MD=MH.MK$$Which means $BDHK$ is cyclic and $\angle BHK=\angle BDK=90$ , so we're done. Impressive! This is a very unique and fun problem. I was trying for a while to come up with a synthetic solution but couldn't.

Here is a synthetic and short solution. Let $U = BI \cap (ABC)$, $V = BI \cap AC$. $D = (I) \cap BC$, $D'$ is the reflection of $D$ across $M$. $T$ is the reflection of $D$ across $I$. $AT \cap (I) = Z$. Note that $A,T,Z,D'$ are collinear and $M$ is the center of $DZD'$. Note that $AC=2AB \implies ABV \cong NCU \implies BV = UC = UI$. Additionally, $\angle HKV = \angle ABH = \angle IBD = \angle IKH \implies$ $ HI = HV$. Thus $H$ is the midpoint of $UB$. Because $IKV$ is an isosceles triangle, pairs $A,F$ and $B,U$ are symmetric wrt $K-H-M$. Now, $\angle SBD' = \angle SFU = \angle BAH = \angle BKH = \angle HIM \implies BS \parallel IM \parallel AD'$. We can finish because $MZ$ is tangent to $(I)$ and $\angle SMD' = 2 \angle SBD' = 2 \angle ZD'M = 2 \angle ZMD \implies S \in ZM$.

If $D$ is the $\text{A-intouch point}$ and note that $ID\cap AC=L$ .Note that $L \equiv K$ and $MK$ is angle bisector of $\angle DKC$.Now apply Cartesian co-ordinates on $\triangle ABC$ with $M$ as the origin.The bash isn't easy as it took me 3 hours when I was taking is at a practice test given by TC but eventually works out.

Let ray $BI$ intersect $CA$ and $(ABC)$ at $J$ and $B'$. Then $H$ is the midpoint of $BB'$, and by trig it is easy to show that $B'J = BI$ so $H$ is also the midpoint of $IJ$. Therefore, by symmetry we see that $A$ and $F$ are reflections over line $KM$. We also have $\triangle BAJ \sim \triangle BHM$, so by spiral similarity, $\triangle BHA \sim \triangle BMJ$. Hence, \[ \measuredangle JIM = \measuredangle MJB = \measuredangle HAB = \measuredangle B'FH = \measuredangle B'BS, \]and hence $IM \parallel BS$. Since $MBS$ is an isosceles triangle, it follows that $IM$ bisects the external angle $\angle SMB$. This shows that $SM$ is tangent to the incircle.

This problem is so beautiful, finally solved it!!! Solution. Let $SM \cap {(ABC)}=R$, $D$ be the $A$-intouch point and $E$ be the $B$-intouch point. Observations: $M$ is the circumcenter of $\triangle ABC$ $(AKHDB)$ $I$ is the orthocenter of $\triangle MKB$ $BI\cap (ABC)=M_B$, here $M_B$ is midpoint of minor-arc $AC$ Claim I: $\overline{R-M-S}$ $$\measuredangle RFS=\measuredangle SFA=\measuredangle SFK+\measuredangle KFA=\measuredangle GBK+\measuredangle KBH=90^{\circ}$$this gives us that $RS$ is the diameter of $(ABC)$. Claim II: $\overline{F-K-I-D}$ Let $AB=1, BC=2, AC=\sqrt{5} \implies BD=\frac{\sqrt{5}-1}{2}$ and $\frac{\sqrt{5}+1}{2}$ Let $K'$ be the phantom point such that $ID \cap AC=K'$. $CK'=\frac{\sqrt{5}+5}{4}$ and $CK=\frac{CM \cdot MB}{CL}=\frac{\sqrt{5}+5}{4}=CK'$ So, $K'\equiv K$ Claim III: $\overline{B-K-R}$ $$\angle SBK=\angle SBC+\angle KBC=\angle SBC+\angle ABC-\angle ABK=\angle SFC+\angle ABC-\angle ABK=\angle AFC-\angle SFA+\angle ABC-\angle KHF=90^{\circ}$$The last angle chasing step came from the fact that $MK$ is the perpendicular bisector of $AF$. Note that $MI\perp BK$ because $M$ is the circumcenter of $\triangle ABC$ and because $MR=MB$ we have $\angle RMI=\angle BMI$ and the result follows.

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Here is a solution using poles and polars. However, in the final step, I used trig. If there is a way to prove it without using trig, please tell me. All pole and polars are respect to the incircle. Let \( X \) be the midpoint of \( AC \), and \( IP \perp MX \). It is well-known that \( MX \) is the polar of \( H_p \), the orthocenter of \( \triangle AIB \). Since \( X \) is the midpoint, \( \angle AXB = 45^\circ \). We have: $$ C_1I = AC_1 = AB_1 = B_1I = NX = XP = PH_p = H_pN $$At this step, we need \( IH = HN \). The reasoning use only geometry is revealed by the comment 5 ; however, I solved this using trig and obtained \( BI = NB' \), which is equivalent. As a result, \( H \) is the intersection of the diagonals of the rectangle \( AXPC_1 \). Consequently, \( B_1, H, H_p \) and \( C_1, H, X \) are collinear. Since \( X \) lies on \( MX \), which is the polar of \( H_p \), applying La Hire’s theorem gives \( B_1H_p \) as the polar of \( X \). Let \( Y \) be the point where \( XC_1 \) meets the incircle. Then \( (X, H; Y, C_1) \) forms a harmonic division. Also, let \( R = XM \cap B_1H_p \). Since \( XI \perp RH_p \) and \( IH_p \perp XR \), this implies that \( IR \perp BX \) and they meet at \( Q \). Since \( R \) lies on the polar of \( H_p \) and \( B \) lies on \( BQ \), the polar of \( P \), we apply La Hire’s theorem again to deduce that \( R \) lies on \( A_1C_1 \). Thus, \( A_1C_1, H_pH \), and \( XM \) are concurrent. Let \( X' = XM \cap A_1Y \). We have: $$ R(X, H; Y, C_1) = R(X', RH \cap A_1Y; Y, A_1) = -1 $$Since \( X' \) lies on the polar of \( H_p \), applying La Hire's theorem tells us that \( H_p \) lies on the polar of \( X' \), so \( RH \cap A_1Y \) is \( H_p \). Thus, \( A_1, H_p, Y \) are collinear and form the polar of \( M \). Therefore, \( MY \) is tangent to the incircle. The last thing to prove is that \( S, M, Y \) are collinear. Although I couldn’t find a geometric solution for this, the trigonometric approach is quite straightforward. We need \( \angle SMB + \angle BMY = 180^\circ \) iff \( \angle A_1IM = \angle BB'S = \angle NB'X + \angle XB'S = \angle NB'X + \angle XC_1P \) (just angle chasing), which is equivalent to showing their sines are equal, which can be done easily.

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