solved with anurag27826 Let $D,E,N$ be the midpoint of arc $AB,AC,BC$. $R,S$ be the feet of $N$ on $AB,AC$. $R-M-H-S$ are collinear and lie on the simson line of $N$.Note that $DE \perp AI$ and $MH \perp AI \implies DE \parallel R-H-M-S$ . By converse of reims we have $R-B'-D$ are collinear and $C'-S-E$ are also collinear. By reims $B'RC'S$ is cyclic. By radical axis theorem on $(ABC),(ARNS),(B'RC'S)$ we get $B'-H-C'$ collinear.

Let $N_A$ be the midpoint of arc $\widehat{BC}$ not containing $A$ and define $N_B,N_C$ similarly. Let $P_B$ be the foot from $N_A$ to $\overline{AB}$ and let $P_C$ be the foot from $N_A$ to $\overline{AC}$. Then the Simson line from $N_A$ to $\triangle ABC$ is $\overline{P_BMP_C}$, and $\overline{P_BP_C}$ is perpendicular to $\overline{AI}$ because $AP_BN_AP_C$ is a kite, so $P_B=\overline{MH}\cap\overline{AB},P_C=\overline{MH}\cap\overline{AC}$. Let $X=\overline{BI}\cap\overline{MH}$. Then $\measuredangle P_BB'B=\measuredangle P_BXB=90^{\circ}-\measuredangle BIH=\measuredangle ICB=\measuredangle N_BCB=\measuredangle N_BB'B$. Thus $B'=\overline{P_BN_B}\cap (ABC)$ and $C'$ is similarly $\overline{P_CN_C}\cap (ABC)$. Then by DIT on $N_BB'C'N_C$ and $\overline{MH}$, we want to show $(\overline{N_CN_B}\cap\overline{MH},H),(P_B,P_C),$ and the two intersections of $(ABC)$ and $\overline{MH}$ are all pairs under an involution. Consider the inversion around $H$ with radius $\sqrt{-HA\cdot HN_A}$. By Power of a point from $H$ to $(ABC)$ and $(AP_BN_AP_C)$(which is a circle because $\measuredangle AP_BN_A=\measuredangle AP_CN_A=90^{\circ}$), the last two pairs are pairs under this involution. $\overline{N_BN_C}\perp \overline{AI}\perp\overline{MH}$ so $\overline{N_CN_B}\cap\overline{MH}$ is the point at infinity of $\overline{MH}$, so $H$ and this point swap under the inversion, as desired.

Assume that $T,T'$ are midpoints of arc $BC$ Where $A,T'$ are on same arc.Let $MH \cap AB,AC = X,Y$ respectively.And let $MH \cap IB,IC = P,Q$ respectively. Note that :$\angle CC'Q =\angle QYA = 90 - \angle A/2 = \angle T'AC$.So $C',Q,T'$ are collibear. Similar for $P,B',T'$.Now remove $A$ and move $I$ on $C(T,TB)$ which means $H$ moves on $(TM)$ and $HM$ has degree 1. $BI$ has degree 1 too so $P$ has defree at most 2 and $B'=T'P \cap (TBC)$ has degree at most 2.Similar for $C'$ so the collinearity of $B',C',H$ has degree at most 3 since the line $B'C'$ have degree at most 2. So we need to check 4 points(other than the ones which gives us the degree bounds of course) where the midpoints of arc $BC$ in $C(T,TB)$ and the intersections of a line through $T$ tangent to $(TBC)$ are ok and we're done.

I'm afraid it is unsuitably easy for an ARMO G11 P4. Simple angle chasing and monge theorum kills it instantly.

Call the vertices of T(b) and T(c) E,X and F,Y respectively such that A,B,E and A,C,F are collinear. Let N and S be the north and south poles of circle (ABC). In triangle AEF, AH is the altitude and the angle bisector, so <AEF=<AFE=90-α/2. We have <FCY=γ/2 and <EBX=β/2 so <BXE=γ/2 and <CYF=β/2. <CYX=180-<CYF=180-β/2=180-IBC=CBX. So C,B,X,Y are concyclic. <CC'N=<CBN=90-α/2=<AFE=CC'Y. So C',Y,N and similarly B',X,N are colinear. B,X,C,Y are concyclic so MX•MY=MB•MC=MN•MS. So N,S,X,Y are concyclic. Note that <SB'N=<SC'N. Therefore, In triangle NXY, the feet of perpendicular lines from S to the sides of the triangle are B',H,C' And from Simpson's theorems we get that points B,H',C are collinear and we are done.

$\text{nice configuration, but it can be solved easily by angle chasing}$

oof what a great problem

I find this problem very interesting. I've discovered a generalization as follows: when we replace the incenter with an arbitrary point $P$ that satisfies the property of being the tangent point at $P$ of the circumcircle of triangle $PBC$, with $AM$ being perpendicular. Let $ABC$ be a triangle inscribed in the circle $\omega$. Take points $M$ and $N$ on $\omega$ such that $MN$ is parallel to $BC$. Let $K$ be the orthogonal projection of $N$ onto $BC$. Call $\ell$ the line passing through $K$, perpendicular to $AM$. Point $P$ is such that $AM$ is perpendicular to the tangent at $P$ of the circumcircle of triangle $PBC$. $\omega_b$ is the circle passing through the intersection of lines $\ell$, $AB$, and $PB$. $\omega_c$ is the circle passing through the intersection of lines $\ell$, $AC$, and $PC$. $\omega_b$ and $\omega_c$ intersect $\omega$ again at $S$ and $T$, respectively. Prove that the lines $ST$, $AN$, and $\ell$ are concurrent.

buratinogigle wrote: I find this problem very interesting. I've discovered a generalization as follows: when we replace the incenter with an arbitrary point $P$ that satisfies the property of being the tangent point at $P$ of the circumcircle of triangle $PBC$, with $AM$ being perpendicular. Let $ABC$ be a triangle inscribed in the circle $\omega$. Take points $M$ and $N$ on $\omega$ such that $MN$ is parallel to $BC$. Let $K$ be the orthogonal projection of $N$ onto $BC$. Call $\ell$ the line passing through $K$, perpendicular to $AM$. Point $P$ is such that $AM$ is perpendicular to the tangent at $P$ of the circumcircle of triangle $PBC$. $\omega_b$ is the circle passing through the intersection of lines $\ell$, $AB$, and $PB$. $\omega_c$ is the circle passing through the intersection of lines $\ell$, $AC$, and $PC$. $\omega_b$ and $\omega_c$ intersect $\omega$ again at $S$ and $T$, respectively. Prove that the lines $ST$, $AN$, and $\ell$ are concurrent. $\textbf{Solution:}$ Let $\ell$ cut $PB,PC$ at $X,Y$ and $AB,AC$ at $U,V$. It is easy to see that $\ell$ is the Simson line of $N$ wrt. $\triangle{ABC}$ so $NU,NV \perp AB,AC$. Since $AUNV,ABNC$ are cyclic so by radical axis theorem we are left to show $USVT$ is cyclic. Since the tangent from $P$ to $(PBC)$ is perpendicular to $AM$ so $\angle{BSU}=\angle{BXU}=\angle{PXK} = \angle{PCB} \implies$ $SU$ cut $CP$ at $Q$ lie on $(ABC)$. Since $\angle{QST}=\angle{QCT}=\angle{UVT} \implies$ $USVT$ is cyclic. $Q.E.D.$

Nice problem! Let $N$ be a midpoint of a minor arc of $BC$, $S$ be a midpoint of a major arc of $BAC$, line $MH$ intersects lines $BI, AB, AC, CI$ at points $P,Q,R,T$ (look at picture). Since $\angle NHM = \angle NAS = 90^{\circ}$ we have $AS \parallel MH$. Also, from circles $\angle SC'C = \angle SAC = \angle ART = \angle TC'C$, so $S, T, C'$ are collinear. By analogue, $S, B', P$ are collinear. Now, since $NS$ is a diameter of $(ABC)$, we have $\angle PB'N = 90^{\circ} = \angle PHN$, so points $P, B', H, N$ are concyclic. By analogue, $T, C', N, H$ are also concyclic. Let's note, that since $AS \parallel MH$ we have $\angle AQR = \frac{\angle B + \angle C}{2}$, so $\angle BPT = \angle BQR - \angle PBQ = \frac{\angle C}{2} = \angle BCT$, because $BI, CI$ are angles bisectors. So, $PBTC$ is inscribed. Now, by PoP $PM \cdot MT = BM \cdot MC = NM \cdot MS$m so $PSTN$ is also inscribed!. Now look at second picture. We have proved everything is inscribed in the picture. Then, in fact, there remains the statement opposite to the statement about the Miquel point, which is proved in the same way by angle chasing. P.S. 1) second picture should be first 2) I'm sorry for my english...

Paramizo_Dicrominique wrote: buratinogigle wrote: I find this problem very interesting. I've discovered a generalization as follows: when we replace the incenter with an arbitrary point $P$ that satisfies the property of being the tangent point at $P$ of the circumcircle of triangle $PBC$, with $AM$ being perpendicular. Let $ABC$ be a triangle inscribed in the circle $\omega$. Take points $M$ and $N$ on $\omega$ such that $MN$ is parallel to $BC$. Let $K$ be the orthogonal projection of $N$ onto $BC$. Call $\ell$ the line passing through $K$, perpendicular to $AM$. Point $P$ is such that $AM$ is perpendicular to the tangent at $P$ of the circumcircle of triangle $PBC$. $\omega_b$ is the circle passing through the intersection of lines $\ell$, $AB$, and $PB$. $\omega_c$ is the circle passing through the intersection of lines $\ell$, $AC$, and $PC$. $\omega_b$ and $\omega_c$ intersect $\omega$ again at $S$ and $T$, respectively. Prove that the lines $ST$, $AN$, and $\ell$ are concurrent. $\textbf{Solution:}$ Let $\ell$ cut $PB,PC$ at $X,Y$ and $AB,AC$ at $U,V$. It is easy to see that $\ell$ is the Simson line of $N$ wrt. $\triangle{ABC}$ so $NU,NV \perp AB,AC$. Since $AUNV,ABNC$ are cyclic so by radical axis theorem we are left to show $USVT$ is cyclic. Since the tangent from $P$ to $(PBC)$ is perpendicular to $AM$ so $\angle{BSU}=\angle{BXU}=\angle{PXK} = \angle{PCB} \implies$ $SU$ cut $CP$ at $Q$ lie on $(ABC)$. Since $\angle{QST}=\angle{QCT}=\angle{UVT} \implies$ $USVT$ is cyclic. $Q.E.D.$ Why is $\ell$ the Simson line?

qwerty123456asdfgzxcvb wrote: Paramizo_Dicrominique wrote: buratinogigle wrote: I find this problem very interesting. I've discovered a generalization as follows: when we replace the incenter with an arbitrary point $P$ that satisfies the property of being the tangent point at $P$ of the circumcircle of triangle $PBC$, with $AM$ being perpendicular. Let $ABC$ be a triangle inscribed in the circle $\omega$. Take points $M$ and $N$ on $\omega$ such that $MN$ is parallel to $BC$. Let $K$ be the orthogonal projection of $N$ onto $BC$. Call $\ell$ the line passing through $K$, perpendicular to $AM$. Point $P$ is such that $AM$ is perpendicular to the tangent at $P$ of the circumcircle of triangle $PBC$. $\omega_b$ is the circle passing through the intersection of lines $\ell$, $AB$, and $PB$. $\omega_c$ is the circle passing through the intersection of lines $\ell$, $AC$, and $PC$. $\omega_b$ and $\omega_c$ intersect $\omega$ again at $S$ and $T$, respectively. Prove that the lines $ST$, $AN$, and $\ell$ are concurrent. $\textbf{Solution:}$ Let $\ell$ cut $PB,PC$ at $X,Y$ and $AB,AC$ at $U,V$. It is easy to see that $\ell$ is the Simson line of $N$ wrt. $\triangle{ABC}$ so $NU,NV \perp AB,AC$. Since $AUNV,ABNC$ are cyclic so by radical axis theorem we are left to show $USVT$ is cyclic. Since the tangent from $P$ to $(PBC)$ is perpendicular to $AM$ so $\angle{BSU}=\angle{BXU}=\angle{PXK} = \angle{PCB} \implies$ $SU$ cut $CP$ at $Q$ lie on $(ABC)$. Since $\angle{QST}=\angle{QCT}=\angle{UVT} \implies$ $USVT$ is cyclic. $Q.E.D.$ Why is $\ell$ the Simson line? Ok no problem if you don't see it. Here is a tutorial: 1. Let $NX \perp AB$ at $X$. What do you have? 2. Do you see the property of $\angle{KXN}$? What does it equal to? 3. Now what do you have to do to prove $KX \perp AM$? 4. Is the problem done yet?

Fairly easy for 11.4 ig We first define some points. Let $MH \cap AB,AC = D,E$ resp., $MH \cap BI,CI = F,G$, $J = (BDF) \cap (ABC)$ and $K = (CEG) \cap (ABC)$, let $L,P,Q$ be the arc midpoints opp. to $A,C,B$ respectively. Let $(BIC) \cap (BDF) = N$ and $(BIC) \cap (CEG) = O$. Claim : $I-D-N, I-E-O$

Clearly it suffices to prove $JDEK$ cyclic, but this follows from, $\measuredangle DJK =\frac{ \measuredangle C}{2} + \measuredangle BCK = \measuredangle GCK = \measuredangle GEK = \measuredangle DEK$ so done

Let $MH$ meet $AB,AC$ at $P,Q$. Let $D,E,F$ be midpoints of minor arcs $AB,AC,BC$. Let $MH$ meet $BI,CI$ at $P',Q'$. Claim $1: P,B',D$ and $Q,C',E$ are collinear. Proof $:$ Note that $\angle HP'I = 90 - \frac{A+B}{2} = \frac{C}{2} = \angle PP'B = \angle PB'B = \angle DAB$ so $P,B',D$ are collinear. we prove the other part with same approach. Claim $ 2: B'PC'Q$ is cyclic. Proof $:$ Note that $\angle B'PQ = \angle B'BE = \angle B'C'E = \angle B'C'Q$. Claim $3: APFQ$ is cyclic. Proof $:$ Note that $\angle BPM = \angle BP'P + \angle P'BP = 90 - \frac{A}{2} = \angle BFM$ so $BMFP$ is cyclic. similarly $CMFQ$ is cyclic so $\angle APF = 90 = \angle AQF$ so $APFQ$ is cyclic. Note that $B'C'$ is the Radical axis of $ABC$ and $B'PC'Q$ so we need to prove $P_{ABC}(H) = P_{B'PC'Q}(H)$. $P_{ABC}(H) = AH.HF = HP.HQ = P_{B'PC'Q}(H)$ so $H$ lies on Radical axis of $ABC$ and $B'PC'Q$ which is $B'C'$ so $B',H,C'$ are collinear as wanted.

Let $P_b$ and $P_c$ denote the intersections of sides $AB$ and $AC$ with $\overline{HM}$ and let $Q_b$ and $Q_c$ denote the intersections of lines $\overline{BI}$ and $\overline{CI}$ with $\overline{HM}$. Also, let $N$ denote the $BC$-arc midpoint in $(ABC)$. [asy][asy] size(8 cm); import geometry; pair foot(pair P, pair A, pair B) { return foot(triangle(A,B,P).VC); } pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair M= midpoint(B--C); pair H = foot(M,A,I); pair Qb = intersectionpoint(line(B,I),line(H,M)); pair Qc = intersectionpoint(line(C,I),line(H,M)); pair Pb = intersectionpoint(line(A,B),line(H,M)); pair Pc = intersectionpoint(line(A,C),line(H,M)); pair B1 = intersectionpoints(circle(B,Pb,Qb),circle(A,B,C))[1]; pair C1 = intersectionpoints(circle(C,Pc,Qc),circle(A,B,C))[0]; pair N = intersectionpoints(line(A,I),circle(A,B,C))[0]; filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); draw(Qb--Pc,blue); draw(Qb--I,green); draw(C--I,green); draw(circumcircle(B,Pb,Qb),orange+dotted); draw(circumcircle(C,Pc,Qc),orange+dotted); draw(circumcircle(B1,C1,Pb),darkgreen+dashed); draw(B1--C1,blue+dashed); draw(A--N,blue); markrightangle(A,H,M,grey); filldraw(circumcircle(A,B,C), tfil, tri); filldraw(circumcircle(A,Pb,Pc), tfil ,sec+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(260)); dot("$C$", C, dir(C)); dot("$I$", I, dir(90)); dot("$H$", H, dir(H)); dot("$M$", M, dir(M)); dot("$P_b$", Pb, dir(270)); dot("$P_c$", Pc, dir(10)); dot("$Q_b$", Qb, dir(Qb)); dot("$Q_c$", Qc, dir(100)); dot("$B'$", B1, dir(B1)); dot("$C'$", C1, dir(C1)); dot("$N$", N, dir(N)); [/asy][/asy] Claim : Points $A$ , $P_b$ , $P_c$ and $N$ are concyclic, with $N$ the $P_bP_c$-arc midpoint of $(AP_bP_c)$. Proof : First note that, \[\measuredangle CP_cM = \measuredangle MP_cA = \measuredangle AP_bM = \measuredangle BP_bM\]Thus, \[P_cC = \frac{\sin \angle P_cMC MC}{\sin \angle CP_cM} = \frac{\sin \angle P_bMB MB}{\sin \angle BP_bM} = P_bB\]Further, $NB = NC$ and $\measuredangle NBP_b = \measuredangle NCP_c$. Thus, $\triangle NBP_b \cong \triangle NCP_c$. So, \[\measuredangle P_b NP_c = \measuredangle BNC = \measuredangle BAC = \measuredangle P_bAP_c\]and $NP_b = NP_c$ which proves the claim. Claim : Points $B'$ , $C'$ , $P_b$ and $P_c$ are concyclic. Proof : We note that, \[\measuredangle P_cQ_cC = \measuredangle HQ_cI = 90 + \measuredangle HIC = \measuredangle IBC\]Thus, we continue to angle chase, \begin{align*} \measuredangle B'CP_c &= \measuredangle B'C'C + \measuredangle CC'P_c\\ &= \measuredangle B'C'C + \measuredangle CQ_CP_c\\ &= \measuredangle B'BC + \measuredangle CBI\\ &= \measuredangle IBB'\\ &= \measuredangle B'BQ_b \\ &= \measuredangle B'P_bQ_b\\ &= \measuredangle B'P_bP_c \end{align*}from which the claim is clear. Now, simply note that by the Radical Center Theorem on circles $(ABNC)$ , $(AP_bNP_c)$ and $(B'P_bC'P_c)$, we have that lines $\overline{AN}$ , $\overline{P_bP_c}$ and $\overline{B'C'}$ must concur. It is clear that the former two intersect at $H$, which implies that $\overline{B'C'}$ passes through $H$ as well, as desired. Remark : This is an extremely rich configuration. In particular, circles $(Q_bBH)$ and $(Q_cCH)$ intersect again at the $A-$excenter $I_a$. Further, points $Q_b$ , $B'$ , $I$ , $P_c$ , $C$ and $Q_c$ , $C'$ , $I$ , $P_b$ and $B$ lie on the same circle.

Let $MH\cap AB=K,MH\cap AC=L,MH\cap IB=P,MH\cap CI=Q$. Let $N,S$ be the midpoints of arcs $BAC,BC$. \[\measuredangle NB'B=90+\frac{\measuredangle A}{2}=180-\measuredangle AKL=\measuredangle 180-\measuredangle BB'P\]Thus, $P,B',N$ are collinear. Similarly we can conlude that $C',Q,N$ are collinear. Note that $\measuredangle IPQ=\frac{\measuredangle C}{2}=\measuredangle BCI=\measuredangle BCQ$ hence $B,C,P,Q$ are concyclic. We have $MN.MS=MB.MC=MP.MQ$ so $N,S,P,Q$ are concyclic. $S$ is the miquel point of $B'HQN$ so $B'H\cap NQ$ lies on $(SHQ)$ and $(SB'N)$ which is $C'$ as desired.$\blacksquare$