I heard $\textrm{Alireza Dadgarnia}$ is the proposer. Can you add? and can someone confirm? This is a monster funny problem. I couldn't solve it during contest

Did IGO take place ? @below Like when. I cannot find it in the official website.

Yes of course

I definitely can't do this without GeoGebra Let $D$ be the foot of altitute from $A$ to $BC$, $AD \cap \Gamma = D' \ne A$, $AA \cap BC=A''$ and $A'$ be a point on $\Gamma$ such that $AA' \parallel BC$. By Butterfly Theorem with chord $AA,XX$, $MA''=MN$. Hence, $A'A' \cap BC=N$. Claim $\angle HJD = 90^{\circ}$ Let $\omega$ be the nine-point circle of $\triangle ABC$. $A'H \cap \omega=J'$ and $A'H \cap \Gamma = J'' \ne A'$($J',B$ lie on same side of $AH$.). (Note that $\angle HJ'D = 90^{\circ}$). By power of point, $AJ',D,A'$ are concyclic and $\angle D'DN = \angle D'A'N = 90^{\circ}$, $D'DA'N$ are concyclic. \begin{align*} \angle AJ'P &= \angle JPH - \angle JAH \\&= \angle HA'D - \angle HAJ'' \\&= \angle HA'D - \angle HA'D'\\&=\angle DA'D'=\angle D'ND \\&= \angle HNM \end{align*}. So, $J'=J$. Let $AM \cap \omega = K' \ne M$ Claim $K'=K.$ Let $JH \cap \omega = J''' \ne J$. Then, $J'''M \parallel AH$. By Reim's Theorem, $A,J,H,K'$ are concyclic. But $J'''P = DM$, $\angle PJH = \angle DK'M$. By angle chasing, $K',H,D,N$ are concyclic and $\angle JAK' + \angle NNK' = \angle HK'M$. So, $K'=K$. (I forgot to show that there are 2 possible $K$ and one of them is $D$. We can show as @below Lemma) Let $O_1,O_2$ be center of $\odot(KHJ),\odot(KMN)$ respectively. $Z = O_1O_2 \cap NH$. Since, $\angle NKH = 90^{\circ}$. We can easily get that $HO_1 \parallel NO_2$. By homothety, $P$ is exsimillcenter of $\odot(KHJ),\odot(KMN)$.

k12byda5h wrote: I definitely can't do this without GeoGebra There are some ways to do the problem without GeoGebra like the first official solution: Let $D$ be the intersection of $AH$ and $BC$. Denote $\Omega$ by the circle with diameter $PM$. It's obvious that $D$ lies on $\Omega$. Also since $\triangle ABC$ is acute, $H$ lies on the segment $PD$ and so inside of $\Omega$. $N$ lies on the extension of $DM$ and so outside of $\Omega$. We claim that there are at most two possible cases for $K$. The following lemma proves our claim. Lemma. Given a circle $\omega$ and four points $A$, $B$, $C$, and $D$, such that $A$ and $B$ lie on the circle, $C$ inside and $D$ outside of the circle. There are exactly two points like $K$ on $\omega$ so that the circumcircles of triangles $\triangle ACK$ and $\triangle BDK$ are tangent to each other. Proof. Invert the whole diagram at center $A$ with an arbitrary radius, the images of points and circles are denoted by primes. Since $A$ lies on $\omega$, $\omega'$ is a line, passes through $B'$ and $K'$. Notice that $C'$ and $D'$ lie on the different sides of $\omega'$. Since the circumcircles of triangles $\triangle ACK$ and $\triangle BDK$ are tangent to each other, we have $C'K'$ is tangent to the circumcircle of triangle $\triangle B'D'K'$. It means $\angle C'K'B'=\angle B'D'K'$. Let $X$ and $Y$ be two arbitrary points, lie on $\omega'$, and the different sides of $B'$. First assume that $K'\equiv B'$ so $\angle C'B'Y=\angle C'K'B'> 0=\angle K'D'B'$ and when $K'$ moves along the ray $\overrightarrow{B'X}$, $\angle C'K'B'$ decreases and $\angle K'D'B'$ increases. It yields there is exactly one point $K'$ on the ray $\overrightarrow{B'X}$ so that $\angle C'K'B'=\angle B'D'K'$. In the same way, we get there is only one possible case for $K'$ on the ray $\overrightarrow{B'Y}$ and the result follows. $\square$ Denote $\omega_1$ and $\omega_2$ by the circumcircles of triangles $\triangle AJP$ and $\triangle HND$. Let $\mathcal{H}$ be the indirect homothety that sends $\omega_1$ to $\omega_2$. Notice that $J$ and $N$ lie on the different sides of $AH$. Now since the arc $AP$ of $\omega_1$ is equal to the arc $HD$ of $\omega_2$ and $AP\parallel HD$, $\mathcal{H}$ sends $A$ to $D$ and $P$ to $H$ therefore $(A,H)$ and $(P,D)$ are anti-homologous pairs. Let $L$ be the anti-homologous point of $J$ under $\mathcal{H}$. It's well-known that the pairs of anti-homologous points lie on a circle so $ALHJ$ and $LPJD$ are cyclic quadrilaterals. Let $E$ be the reflection of $A$ over the point $M$. We claim that $HDEN$ is cyclic. $A'$ lies on $\Gamma$ so that $AA'\parallel BC$. We know that $(A'X,BC)=-1$ hence $NA'$ is tangent to $\Gamma$. Also by symmetry $NE$ is tangent to the circumcircle of triangle $\triangle CEB$. Now since $HE$ is the diameter of this circle, we have $\angle NEH=90^\circ=\angle NDH$ and our claim is proved. The line $AM$ meets the circumcircle of triangle $\triangle PDM$ again at $L'$. We have $$AL'\cdot AM=AP\cdot AD\Longrightarrow AL'\cdot AE=AH\cdot AD$$it follows that $L'HDEN$ is cyclic so $L'\equiv L$. We have \begin{align*} \angle PJH=\angle AJH-\angle AJP&=\angle HLM-\angle HND\\ &=\angle HLM-\angle HLD=\angle DLM=\angle DJM \end{align*}therefore $\angle HJD=90^\circ$. From this, we can conclude that the cirucmcircles of triangles $\triangle DHJ$ and $\triangle DMN$ are tangent to each other and the common external tangents of them are concurrent at $H$ since the tangent line to the circumcircle of triangle $\triangle DHJ$ through $H$ is parallel to $DMN$. So the problem is proved for $K\equiv D$, now suppose that $K\ne D$. Since $\angle AHL=\angle LNM$ the circumcircles of triangles $\triangle LHJ$ and $\triangle LMN$ are tangent to each other. So $L\equiv K$. Denote $O_1$ and $O_2$ by the circumcenters of triangles $\triangle LHJ$ and $\triangle LMN$. It's obvious that $O_1$, $L$, and $O_2$ are collinear so $\angle O_1LH+\angle O_2LN=90^\circ$. It yields $$\angle HO_1L=180^\circ-2\angle O_1LH=2\angle O_2LN=180^\circ -\angle LO_2N \Longrightarrow O_1H\parallel O_2N$$therefore the direct homothety that sends $\left(O_1\right)$ to $\left(O_2\right)$, sends $H$ to $N$ and the conclusion follows.

k12byda5h wrote: I definitely can't do this without GeoGebra . well , the problem isnt that hard if you can make good observations... . . definetly $p_4$ was harder. and last years $p_5,p_4$ was also harder. since i didnt solve any (or saying it better , i couldn't even draw the diagram) . . . first , we do some observations. let $AM$ meet the "nine_point circle" at $G$. and let $AH$ meet $BC$ at $D$ . we prove $k \in \{G,D\}$. . first we tend to find $J,N$ as they are really bad points. here is how

now , let $HN,MJ$ meet at $R$ , we will prove $N',R,A$ are collinear. its clear that $RJHA$ is cyclic . first note that $APJ,RMN$ are similar triangles (trivial by some angle chasing) so we conclude $HJA,RNR'$ are similar as well. now as $N'RNR'$ is a parallelogram clearly $$RNR'=HJA=HRA$$so $AR \parallel NR' \parallel RN'$ so the claim is proven. . . now lets go dig in $J$

. . building out of user Dadgarnia (which i am not going to write again) we have there are two cases for $K$ , one of em clearly is $K=D$ , the other one we provei s $G=K$. by some easy angle chasing $A,R,G,H,J$ are cyclic , so we prove $$GNM+GJH=HGM$$. take note that : $$GNM+GJH=GNM+GAH=HGM+(GHA-GNM)$$so we only need to prove $GHA,GNN'$ are similar. we prove $GPA,GMN'$ are similar. we prove $G,P,N'$ are collinear which solves the problem. clearly $PGA=90$ so we need $N'GA=90$ so we need $P$ be the orthocenter of $AMN'$ so we only need to prove $MP$ is the altitude of $AN'$ which is trivial bc $AN'$ is tangent to the circle. . now we reach the problem it self. which is easy . . we only prove the tangent from $H$ to $O_{AHJ}$ is parallel to tangent of $N$ to $O_{GMN}$ since $H,G,N$ are not collinear its enough. . let the tangent from $H$ be called $x$ and the tangent from $N$ be called $y$ . the angle between $x,BC$ clearly is $90-ARH$ and the angle between $y,BC$ is $MGN$. clearly $ARH=HGM$ so we only need to prove $MGN=PGH$ which we allready know is true , since $PGH,GMN$ are similar. . . verry nice angle chasing exercise maybe the hardest i have seen

It seems that this problem has many redundant factors. Let me rewrite this problem as follows Let $D$ be any point on circle $\omega$ with diameter $PM$. $H$ is any point on $PD$. Circle diameter $DH$ meets $\omega$ again at $J$. A circle goes through $K$, $M$ and is tangent to $(KHJ)$ meets $DM$ again at $N$. Prove that exsimilicenter of $(KJH)$ and $(KMN)$ lies on line $HN$.

Mr.C wrote: building out of user Dadgarnia (which i am not going to write again)we have there are two cases for $K$ , one of em clearly is $K=D$ , the other one we provei s $G=K$. Finding $J$ by good observation is understandable. But $K$ is ... (It's the point of the tangent of 2 circles on the circle and point $K$ isn't appear in the problem. It's somehow really hard to claim).

k12byda5h wrote: Mr.C wrote: building out of user Dadgarnia (which i am not going to write again)we have there are two cases for $K$ , one of em clearly is $K=D$ , the other one we provei s $G=K$. Finding $J$ by good observation is understandable. But $K$ is ... (It's the point of the tangent of 2 circles on the circle and point $K$ isn't appear in the problem. It's somehow really hard to claim). i dont know ... my problem was $j$ tbh , my first guess for $K$ was $G$. and well its correct. but it took a little time for me to guess $J$ . . . if you dont start with re_defining $k,j$ i think this problem can be a huge pain. but really try to guess $K$ . (assuming you drawed the median) the best place which is logical for $K$ to be and it looks like it to me seems to be $G$ , i also add that i have absolutly no reasoanble comment on how i made that observation. i just did since i felt it looked like it

Redefine $K$ as the projection of $P$ in $AM$, and $L=(AKH)\cap (MP)$. Let $D=AH\cap BC$. Step 1: Prove $J$ is the point given in the statement Clearly it's enough to prove that $\angle AJP=\angle HNM$. Step 1.1: Prove $\angle AJP=\angle HKD$

Step 1.2: Prove $HKND$ is cyclic

So $J$ is the point given in the statement. Step 2: Prove that $\omega _1$ and $\omega _2$ are tangent

Step 3: Prove $K$ is the point given in the statement

Step 4: Prove the statement

buratinogigle wrote: It seems that this problem has many redundant factors. Let me rewrite this problem as follows Let $D$ be any point on circle $\omega$ with diameter $PM$. $H$ is any point on $PD$. Circle diameter $DH$ meets $\omega$ again at $J$. A circle goes through $K$, $M$ and is tangent to $(KHJ)$ meets $DM$ again at $N$. Prove that exsimilicenter of $(KJH)$ and $(KMN)$ lies on line $HN$. Let $S$ $\equiv$ $(KDH)$ $\cap$ $(KMN)$ $(S \not \equiv K)$. We have $$(SN, SH) \equiv (SN, SK) + (SK, SH) \equiv (MN, MK) + (PK, PD) \equiv 0 \pmod \pi$$hence $H, S, N$ are collinear. From this, we have $$(SM, SP) \equiv (SM, SK) + (SK, SP) \equiv (NM, NK) + (HK, HP) \equiv (NM, NK) + (PK, PD) + (KH, KP)$$$$\equiv (NM, NK) + (PK, PD) + (KH, KP) \equiv (NM, NK) + (SK, SH) + (SH, SP)$$$$\equiv (NM, NK) + (SK, SP) \equiv 0 \pmod \pi$$hence $P, S, M$ are collinear. Let $D'$ be reflection of $D$ in midpoint of $PM;$ $O_1, O_2$ be center of $(KHJ), (KMN)$. Then $$(O_1H, O_1K) + (O_2K, O_2N) \equiv 2 (JH, JK) + 2 (MK, MD) \equiv 2 (MD', MK) + 2 (MK, MD)$$$$\equiv 2 (MD', MD) \equiv 0 \pmod \pi$$or $O_1H$ $\parallel$ $O_2N,$ which means the external homothetic center of $(KHJ)$ and $(KMN)$ lies on $HN$

Solved with p_square, Aryan23 Let the circle with diameter $MP$ (The nine point circle) be $\omega$. Let $AM \cap \omega = K'$ and $(AK'H) \cap \omega = J'$, we will show that $K = K'$ and $J = J'$. Let $A'$ be the point on $(ABC)$ with $AA' || BC$ and let $A''$ be the reflection of $A'$ across $BC$. Since $AX$ is median in $\triangle ABC$, $A'X$ is symmedian in $\triangle A'BC$ and so the tangents to $(ABC)$ at $X,A'$ meet at $N$ and so $NA'$ is tangent to $(ABC)$ Claim: $HKND$ is cyclic $AP.AD = AK.AM \implies (2AP)(AD) = (AK)(2AM) \implies AH.AD = AK.AA''$ and so $KHDA''$ is cyclic. So, it suffices to show that $HDA''N$ is cyclic. Let $H'$ be the reflection of $H$ across $BC$, which is well known to lie on $(ABC)$ So, $\angle DNA'' = \angle DNA' = |\angle ABC - \angle ACB|$ and $\angle DHA'' = \angle DH'A' = \angle AH'A' = \angle ACA' = |\angle ABC - \angle ACB|$ This means $\angle DNA'' = \angle DHA''$ and so $HDA''N$ is indeed cyclic, proving the claim. $\blacksquare$ Now, $\angle HNM = \angle HND = \angle HKD = \angle AHK - \angle ADK = \angle AJK - \angle PDK = \angle AJK - \angle PJK = \angle AJP$ and so $J' = J$ Let $T$ be a point on the tangent to $(AKHJ)$ at $K$. Note that $\angle HKN = 180 - \angle HDN = 90^\circ$ Then, $\angle TKN = 90 - \angle TKH = 90 - \angle KAH = 90 - \angle DAM = \angle DMA = \angle DMK$ and so $T$ lies on the tangent to $(KMN)$ at $K$ as well. Therefore, the two circles $(KHJ), (KMN)$ are tangent to each other. So, $K' = K$ as well. Now, let $O_1, O_2$ be circumcenters of $\omega_1, \omega_2$ respectively. Obviously, it suffices to show that $O_1H || O_2N$ by homothety and so this means its enough to show $\angle O_1HA + \angle O_2NM = 90^\circ$. This is true since $\angle O_1HA + \angle O_2NM = (90 - \frac{\angle AO_1H}{2}) + (90 - \frac{\angle MO_2N}{2}) = (90 - \angle AJH) + (90 - \angle MKN) = 180 - (\angle AJH + \angle MKN) = 180 - (\angle HKM + \angle MKN) = 180 - \angle HKN = 180 - 90 = 90^\circ$, as desired. $\blacksquare$

Solved with paper and pencil geogebra. Nevertheless a nice problem! We rephrase the problem a bit: Rephrased IGO Advanced 2020 P5 wrote: Consider an acute-angled triangle $\triangle ABC$ ($AC>AB$) with its orthocenter $H$ and circumcircle $\Gamma$. Points $M$,$P$ are midpoints of $BC$ and $AH$ respectively. The line $\overline{AM}$ meets $\Gamma$ again at $X$ and point $N$ lies on the line $\overline{BC}$ so that $\overline{NX}$ is tangent to $\Gamma$. Let $K$ be the intersection of $AM$ and $(MP)$ and let $J$ be the intersection of $(AKH)$ and $(MP)$. Prove that $\angle AJP=\angle HNM$, $(KHJ)$ and $(KMN)$ are externally tangent to each other, the common external tangents of $(KHJ)$ and $(KMN)$ meet on the line $\overline{NH}$. Firstly we do some simple observations: $(MP)$ is actually the nine-point circle of $\triangle ABC$, $OMXN$ is cyclic as $\measuredangle NMO=\measuredangle NXO=90^\circ$. We denote foot from $A$ to $BC$ as $D$, let $O$ be the circumcentre of $(ABC)$, let $N_9$ be the centre of the nine-point circle of $\triangle ABC$, let $E$ be the intersection of $(MP)$ and $(OMXN)$. Claim. $E$ lies on $KN$. Proof. We angle chase; \begin{align*} \measuredangle MEK&= \measuredangle MDK\\&=\measuredangle MKD+\measuredangle KMD\\&=\measuredangle MPD+\measuredangle CAM+\measuredangle BCA\\&=OAD+\measuredangle CAM+\measuredangle BCA\\&=\measuredangle CAD+\measuredangle OAX+\measuredangle BCA\\&=90^\circ+\measuredangle OAX\\&=\measuredangle ACX\\&=\measuredangle MXN\\&=\measuredangle MEN. \end{align*} Now note that as $OE\perp KN$ and $N_9$ lies on the perpendicular bisector of $KE$, we get that $\measuredangle HKN=90^\circ$, therefore $KHDN$ is cyclic. For the part $(a)$; \begin{align*} \measuredangle PJA&= \measuredangle KJA+\measuredangle PJK\\&=\measuredangle KHA+\measuredangle KDH\\&=\measuredangle HKD\\&=\measuredangle HNM. \end{align*}For the part $(b)$; Let $\ell$ be the tangent from $K$ to $(KMN)$, we show that $\ell$ is also tangent to $(KHJ)$; \begin{align*} \measuredangle (\ell,KA)&=\measuredangle (\ell,KM)\\&= \measuredangle KNM\\&=\measuredangle KHD\\&=\measuredangle KHA. \end{align*}For the part $(c)$; By homothety, it is sufficient to show that $O_1H\parallel O_2N$, where $O_1,O_2$ denote the centres of $(KHJ),(KMN)$, respectively. \begin{align*} \measuredangle HO_1K&=2\measuredangle HAK\\&= 2\measuredangle DMA\\&=2\measuredangle NMK\\&=\measuredangle NO_2K. \end{align*}

Here's much easier proof. We don't need to find the exact position of $K$. After we get $\angle HJD = 90^\circ$, simple angle chase completes the proof. Let $D = AH \cap BC$, $E = AH \cap \Gamma \ne A$. Claim 1: $\angle HJD = 90^\circ$. Proof: Recall $\angle AJP = \angle DNE$. Let $\Phi$ be inversion at $H$ swapping $A,D$. Then $\Phi$ swaps $D,E$. From $\angle AJP = \angle DNE$ we obtain $\Phi(J) \in \odot(DEN)$. But also $\Phi(J) \in \Gamma$. Let $A' \in \odot(ABC)$ such that $AA' \parallel BC$. Note $NA'$ is tangent to $\Gamma$. Thus $A' \in \odot(DEN)$. It follows $$\Phi(J) = A'$$Hence, $$ \angle HJD = \angle HAA' = 90^\circ $$proving our Claim. $\square$ Claim 2: $\angle HKN = 90^\circ$. Proof: We just do simple angle chase. Let $\ell$ be common tangent to $\omega_1,\omega_2$ at $K$. Then (undirected angles), $$ \angle HKN = \angle(KH, \ell) + \angle (KN,\ell) = \angle KJH + (180^\circ - \angle KMN) = (\angle KJD - 90^\circ) + \angle KMD = 90^\circ $$as desired! $\square$ Now the following easy Lemma finishes: Lemma: Let $\omega_1,\omega_2$ be two circles externally tangent at some point $K$. Let $H,N$ be points on $\omega_1,\omega_2$ respectively such that $\angle HKN = 90^\circ$. Let $\mathcal H$ be homothety through exsimilicenter of $\omega_1,\omega_2$ sending $\omega_1 \to \omega_2$. Then $\mathcal H$ sends $H$ to $N$. Proof: Let $N'$ be antipode of $N$ in $\omega_2$. From $\angle HKN = 90^\circ$ we have $H-K-N'$ collinear. So homothey at $K$ sending $\omega_1 \to \omega_2$ sends $H \to N'$. Thus tangent to $H,N'$ at $\omega_1,\omega_2$ respectively are parallel. Hence tangent at $N$ to $\omega_2$ is parallel to tangent at $H$ to $\omega_1$. Our Lemma follows. $\blacksquare$

Beautiful beautiful!! Let define we points first. $D$ be feet of altitude from $A$ to $BC$.Let $H_A$ be $A$-humpty point. $T$ be point on $BC$ such that $AT$ is tangent to $(ABC)$. Consider point $W$ on nine point circle of $\triangle ABC$ and also on $AM$. $A_0 = AM \cap (BHC)$ and $A'=(AH)\cap(BHC)$. Claim : $H,D,W,N,A_0$ cyclic If you reflect $X$ above perpendicular bisector of $BC$ as $X'$, then $X'T$ is tangent to $(ABC)$. Hence $MT=MN$. It's well know that $T$ is center of apollonius circle with $A$ and $H_A$ both lie on it. Therefore $TW \perp AM$. Now by using power of point $$MT\cdot MD = MW \cdot MA = MA_0 \cdot MN$$hence $W,D,N,A_0$ cyclic. And by $$ AH \cdot AD = AH_A \cdot AM = AW \cdot AX$$$H \in (WDA_0)$. Let $Q = DA_0 \cap (BHC)$ As $$\measuredangle HND = \measuredangle HA_0Q = \measuredangle HA'Q$$In $\sqrt{HA\cdot HD}$. $P \leftrightarrow A'$ and $(BHC) \leftrightarrow (Nine point circle)$. So if $Q'$ is image of $Q$. Then $\measuredangle PQ'A = \measuredangle HNM$ with $Q'$ lie on nine point circle. Therefore $Q' \equiv P$. Key observation is, in $\sqrt{HA\cdot HD}$ $$ H \leftrightarrow D , W \leftrightarrow A_0 , H_A \leftrightarrow M , N \leftrightarrow N'$$Hence $$\measuredangle WA_0D = \measuredangle WND = \measuredangle H_AN'A_0$$Which give us $DA_0$ is tangent to $(H_AA_0N')$ at $A_0$. Observe this is same as $(PHW)$ tangent to $(WMN)$ at $W$. Therefore $W \equiv K$. Now consider intersection of common tangent as $Y$. let $L = YK \cap (KMN)$. Now $N,H,L$ are collinear, iff $KH \parallel LN$. As $KL$ pass through center, it equivlent to $\measuredangle HKN = 90$ But as $W \in (HDN)$ we have $\measuredangle HKN = 90$.