Let $ABC$ be an acute-angled triangle with $AB<AC$, incentre $I$, and let $M$ be the midpoint of major arc $BAC$. Suppose the perpendicular line from $A$ to segment $BC$ meets lines $BI$, $CI$, and $MI$ at points $P$, $Q$, and $K$ respectively. Prove that the $A-$median line in $\triangle AIK$ passes through the circumcentre of $\triangle PIQ$. Proposed by Pranjal Srivastava and Rohan Goyal
Problem
Source: India IMOTC 2024 Day 3 Problem 1
Tags: geometry, incenter, excenter
31.05.2024 08:37
31.05.2024 09:03
31.05.2024 11:05
This problem reminds me of ISL 2017 G3 which I solved it by trig bash. Unsurprisingly, this problem can also be solved by trig bash too.
Attachments:

31.05.2024 15:40
Define $I_a$ as excenter and $N=AI\cap(ABC)$. See that $\triangle IPQ \sim \triangle I_aBC$. Doing complex bash taking unit circle as $(ABC)$ is easier than #2 complex solution.
31.05.2024 18:39
By trivial angles we see that $AI$ is tangent to $(PQI)$, now let $E$ the antipode of $I$ in $(IPQ)$ and let $F \ne I$ a point on $(IPQ)$ such that $AI=AF$ (so that $AF$ is tangent to $(IPQ)$), now let $I'$ the reflection of $I$ over $A$. Let $I_B, I_C$ the $B,C$ excenters of $\triangle ABC$ respectively. Note that $-1=(I, F; P, Q) \overset{K}{=} (MI \cap (IPQ), KF \cap (IPQ); Q, P) \overset{I}{=} (M, (KF \cap (IPQ))I \cap AM: I_C, I_B)$ But $M$ is known to be the midpoint of $I_CI_B$ and $EI \parallel I_BI_C$ is also true, therefore $KF \cap (IPQ)=E$, now if we extend this line $KF$ to $(A,AI)$, since we have $\angle IFE=90$, we must in fact have $I', F, E, K$ colinear, now take homothety centered at $I$ with scale $0.5$ and notice its equivalent to the problem, thus we are done .
02.06.2024 10:30
I like this problem because it is one of the few problems i introduces a lot of points ! Solution:Let $I_A,I_B,I_C$ the excenters;$O_1$ the circumcenter of $(IPQ)$; $T$ the midpoint of $KI$,$T'$ the ,midpoint of $MI$;$S$ the midpoint of $PQ$,$N$ is the midpoint of the little arc. By angle chasing we can see that : 1)$AI$ is tangent to $(IPQ)$; 2)$IQP \sim I_AI_CI_B$ From these 2 observations we can see that $O_1SIA$ is cyclic,so $\angle O_1IS=\angle O_1AS$.It is enough to prove that $\angle O_1IS=\angle TAK$.From similarity we have that $$\angle O_1IS=\angle O_1IQ-\angle QIS=90-\angle QPI-\angle QIS=\frac{B}{2}-\angle I_CI_AM$$. Observe that $IKA \sim IMN$ so $T \sim T'$ so $\angle TAK=\angle T'NM$. So all we have to prove is that $\angle I_CI_AM+\angle T'NM=\frac{B}{2}$. But $T'N\parallel I_AM$ (midlines) so we have that : $$\angle I_CI_AM+\angle T'NM=\angle I_CI_AA+\angle AI_AM+\angle T'NM=\angle I_CI_AA+\angle INT'+\angle T'NM=\angle I_CI_AA+\angle INM=\frac{C}{2}+\angle ABM=\frac{B}{2}$$so the problem is solved $\blacksquare$
01.08.2024 05:46
Let $I_a$, $I_b$, and $I_c$ be the respective excenters, let $AI$, $BI$, $CI$, and $BC$ meet the perpendicular bisector of $BC$ at $N$, $P'$, $Q'$, and $S$, and let $H$ be the $I_a$-Humpty point of $I_aI_bI_c$. Claim: $H$ lies on $(IP'Q')$ Since we have that $I_aCHB$ is a harmonic quadrilateral we get that $SHP'B$ is concyclic as $\angle SHB=\angle CBI_a=\angle SP'Q$. Thus $H$ is the Miquel point of $CSBIP'Q'$, finishing the claim. Let $I_aM$ meet $(IP'Q')$ again at $R$ which is the $I$-antipode as $IH\perp I_aM$. To finish the problem notice that $IMP'Q'N$ is similar to $IKPQA$ and homothety with ratio $\tfrac{1}{2}$ at $I$ gives the desired result.
Attachments:

05.08.2024 17:20
Trig bash for storage.
10.09.2024 15:18
I finally have time to submit by contest solution for this problem. This was only geometry problem I fully solved in IMOTC, I have very fun solving. I found little bit different solution with introducing new harmonic cross ratios. At first look this looks like 17G3, but indeed this is very different and amazing problme!! Let $A_0 = AI \cap (ABC)$ Claim: : $AI$ is tangent to $(IPQ)$ Proof. Note that $\measuredangle BPK = 90 - \frac{\measuredangle B}{2}$, $\measuredangle PQC = 90 - \frac{\measuredangle C}{2}$ and $\measuredangle QIP = 90 - \frac{\measuredangle A}{2}$. Also $$\measuredangle AIQ = \measuredangle A_0IC = 90 - \frac{ \measuredangle B}{2} = \measuredangle IPQ$$which give us $AI$ tangent to $(IPQ)$. Define $X= AM \cap BC$ and $G$ is $I-$dumpty point of $\triangle IPQ$. Claim: : $\overline{X-G-I}$ Proof. Let $AN$ be tangent to $(IPQ)$ other then $AI$. It's well known that $\overline{N-G-I}$. If $A'= AI \cap BC$ then following claims are follow from well known results $$(IB,IC;IA',IX) =(IP,IQ;IA,IG)=-1$$Hence $\overline{X-G-I}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.cm); 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 = -6.068409913689637, xmax = 27.799478144164162, ymin = -13.781839061307451, ymax = 4.066378941012234; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((5.52,-0.82)--(5.,-6.68)--(18.06,-6.74)--cycle, linewidth(0.8) + ffxfqq); /* draw figures */ draw((5.52,-0.82)--(5.,-6.68), linewidth(0.8) + ffxfqq); draw((5.,-6.68)--(18.06,-6.74), linewidth(0.8) + ffxfqq); draw((18.06,-6.74)--(5.52,-0.82), linewidth(0.8) + ffxfqq); draw(circle((11.541038154299478,-4.307361747480499), 6.958059529302715), linewidth(0.8)); draw((5.52,-0.82)--(5.487250638057658,-7.948444449449819), linewidth(0.8)); draw((11.573004399783407,2.6506243528555724)--(-4.623131731048777,-6.635789593884921), linewidth(0.8)); draw((5.,-6.68)--(-4.623131731048777,-6.635789593884921), linewidth(0.8)); draw(circle((6.314486759525287,-5.0658494289780505), 1.4226958630714122), linewidth(0.8) + red); draw((5.487250638057658,-7.948444449449819)--(11.573004399783407,2.6506243528555724), linewidth(0.8)); draw((5.52,-0.82)--(11.509071908815542,-11.265347847816571), linewidth(0.8)); draw((7.548698020741028,-4.358187022537834)--(-4.623131731048777,-6.635789593884921), linewidth(0.8) + qqwuqq); draw((7.548698020741028,-4.358187022537834)--(5.,-6.68), linewidth(0.8)); draw((5.505871553368096,-3.8952918835443042)--(18.06,-6.74), linewidth(0.8)); /* dots and labels */ dot((5.52,-0.82),dotstyle); label("$A$", (5.161259542112825,-0.5646174894337418), NE * labelscalefactor); dot((5.,-6.68),dotstyle); label("$B$", (4.684247034255729,-7.342170205236651), NE * labelscalefactor); dot((18.06,-6.74),dotstyle); label("$C$", (17.841842042647286,-6.408020710683171), NE * labelscalefactor); dot((7.548698020741028,-4.358187022537834),linewidth(4.pt) + dotstyle); label("$I$", (7.526446560237591,-4.02295817139769), NE * labelscalefactor); dot((11.509071908815542,-11.265347847816571),linewidth(4.pt) + dotstyle); label("$A_0$", (11.561177355862194,-10.840261929522022), NE * labelscalefactor); dot((11.573004399783407,2.6506243528555724),linewidth(4.pt) + dotstyle); label("$M$", (11.720181525147893,2.853972150208781), NE * labelscalefactor); dot((5.487250638057658,-7.948444449449819),linewidth(4.pt) + dotstyle); label("$K$", (5.936404867380606,-8.01793792470087), NE * labelscalefactor); dot((-4.623131731048777,-6.635789593884921),linewidth(4.pt) + dotstyle); label("$X$", (-4.577745826636213,-6.14963893559391), NE * labelscalefactor); dot((5.495150407944361,-6.228927870777472),linewidth(4.pt) + dotstyle); label("$P$", (5.578645486487784,-6.0701368509510605), NE * labelscalefactor); dot((5.505871553368096,-3.8952918835443042),linewidth(4.pt) + dotstyle); label("$Q$", (5.578645486487784,-3.5061946212191684), NE * labelscalefactor); dot((4.907904056900141,-4.852334512692914),linewidth(4.pt) + dotstyle); label("$N$", (4.7836246400592906,-4.321090988808375), NE * labelscalefactor); dot((6.228301038820584,-4.605260767615373),linewidth(4.pt) + dotstyle); label("$G$", (6.31403976943414,-4.440344115772649), NE * labelscalefactor); dot((6.314486759525286,-5.0658494289780505),linewidth(4.pt) + dotstyle); label("$O$", (6.393541854076989,-4.897481102469032), NE * labelscalefactor); dot((6.517974329399343,-6.153315735993827),linewidth(4.pt) + dotstyle); label("$L$", (6.174911121309154,-6.010510287468923), NE * labelscalefactor); dot((8.89020849021879,-6.6978723207820146),linewidth(4.pt) + dotstyle); label("$A'$", (9.116488253094577,-6.4875227953260195), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $L$ be midpoint of $IK$. Claim: : $$(M,L;I,K)=\frac{A_0I}{A_0A}$$Proof. $(M,L;I,K) = \frac{MI}{MK} / \frac{LI}{LK}$ but as $L$ is midpoint we get $\frac{MI}{MK}$. From $AK \parallel MA_0$ we get $\triangle AIK \sim \triangle A_0IM$ $$\frac{MI}{IK} = \frac{A_0I}{AI} \Rightarrow \frac{IK}{MI} = \frac{AI}{A_0I} \Rightarrow \frac{MI}{MK} = \frac{AA_0}{A_0I}$$$$ \frac{MI}{MK} = \frac{A_0I}{A_0A}$$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.50872409290906cm); 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 = -56.802987153065544, xmax = 105.70573693984352, ymin = -84.39286145450127, ymax = 1.2484731907055522; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((6.806886433199685,-8.072026173044682)--(3.186294770417865,-19.130225512849663)--(37.383746248259556,-20.595392615088233)--cycle, linewidth(0.8) + ffxfqq); draw((29.40745007646128,-56.985950327590295)--(21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356)--cycle, linewidth(0.4) + qqwuqq); /* draw figures */ draw((6.806886433199685,-8.072026173044682)--(3.186294770417865,-19.130225512849663), linewidth(0.8) + ffxfqq); draw((3.186294770417865,-19.130225512849663)--(37.383746248259556,-20.595392615088233), linewidth(0.8) + ffxfqq); draw((37.383746248259556,-20.595392615088233)--(6.806886433199685,-8.072026173044682), linewidth(0.8) + ffxfqq); draw(circle((20.338091171105066,-18.62412347112488), 17.159261611364997), linewidth(0.8)); draw((21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356), linewidth(0.4)); draw((3.186294770417865,-19.130225512849663)--(-15.402460368442137,-18.3338026414356), linewidth(0.8)); draw((6.806886433199685,-8.072026173044682)--(29.40745007646128,-56.985950327590295), linewidth(0.8)); draw((21.07259450977063,-1.4805892995267258)--(29.40745007646128,-56.985950327590295), linewidth(0.4) + yqqqyq); draw((22.74318071774964,-12.605735963489321)--(-15.402460368442137,-18.3338026414356), linewidth(0.8) + blue); draw((6.806886433199685,-8.072026173044682)--(19.6035878324395,-35.76765764272303), linewidth(0.8)); draw((6.806886433199685,-8.072026173044682)--(17.686135369787962,-80.52158874888097), linewidth(0.8) + yqqqyq); draw((17.686135369787962,-80.52158874888097)--(21.07259450977063,-1.4805892995267258), linewidth(0.8)); draw((29.40745007646128,-56.985950327590295)--(21.07259450977063,-1.4805892995267258), linewidth(0.4) + qqwuqq); draw((21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356), linewidth(0.4) + qqwuqq); draw((-15.402460368442137,-18.3338026414356)--(29.40745007646128,-56.985950327590295), linewidth(0.4) + qqwuqq); /* dots and labels */ dot((6.806886433199685,-8.072026173044682),dotstyle); label("$A$", (5.091474077156273,-6.8578892200322885), NE * labelscalefactor); dot((3.186294770417865,-19.130225512849663),dotstyle); label("$B$", (1.6581911737849555,-22.307662285203232), NE * labelscalefactor); dot((37.383746248259556,-20.595392615088233),dotstyle); label("$C$", (36.3724960856505,-18.96974835137), NE * labelscalefactor); dot((9.799725588417735,-14.549364957855785),linewidth(4.pt) + dotstyle); label("$I$", (9.669184614984697,-12.96150327047019), NE * labelscalefactor); dot((19.6035878324395,-35.76765764272303),linewidth(4.pt) + dotstyle); label("$A_0$", (19.87366435556056,-33.751938629774294), NE * labelscalefactor); dot((21.07259450977063,-1.4805892995267258),linewidth(4.pt) + dotstyle); label("$M$", (21.781043746322403,-1.0403820782086615), NE * labelscalefactor); dot((-15.402460368442137,-18.3338026414356),linewidth(4.pt) + dotstyle); label("$X$", (-15.222116434457359,-16.01331029568914), NE * labelscalefactor); dot((29.40745007646128,-56.985950327590295),linewidth(4.pt) + dotstyle); label("$J$", (29.792037187522144,-56.25901544076407), NE * labelscalefactor); dot((22.74318071774964,-12.605735963489321),linewidth(4.pt) + dotstyle); label("$R$", (23.11620931985569,-11.817075636013085), NE * labelscalefactor); dot((7.82409198858344,-14.846032169857061),linewidth(4.pt) + dotstyle); label("$G$", (5.85442583346101,-13.438348118160652), NE * labelscalefactor); dot((8.07908311814851,-16.544126946930962),linewidth(4.pt) + dotstyle); label("$L$", (6.331270681151471,-17.825320716912895), NE * labelscalefactor); dot((12.092612003507117,-19.511810761109412),linewidth(4.pt) + dotstyle); label("$A'$", (13.102467518356015,-18.492903503679543), NE * labelscalefactor); dot((17.686135369787962,-80.52158874888097),linewidth(4.pt) + dotstyle); label("$Z$", (18.061653934336807,-79.71978194713475), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Key Claim: : $A,G,L$ are collinear. Proof. Let $Z = AG \cap MA_0$. Note that $$(M,Z;A_0,\infty{\overline{AK}}) \Rightarrow \frac{MA_0}{M\infty{AK}}/ \frac{ZA_0}{Z\infty{AK}} \Rightarrow \frac{MA_0}{ZA_0} $$ If they are collinear then $$\frac{A_0I}{A_0A} = (M,L;I,K)\stackrel{A}{=}(M,AG \cap MA_0;A_0,\infty{\overline{AK}}) = \frac{MA_0}{ZA_0}$$ Let $J$ be $A$-excenter of $\triangle ABC$. Claim: : $\triangle ZA_0A \sim \triangle MA_0J $ Proof. Observe that in $\sqrt{bc}$ inversion $I \leftrightarrow J$ and $M \leftrightarrow X$ , hence $AI\cdot AJ = AM \cdot AX$. by well known result , we have $I$ as orthocenter of $\triangle JMX$. Hence $MJ \perp IX$ but as $IX \perp AZ$ , we have $AZ \parallel MJ$ Therefore we get $$\frac{MA_0}{JA_0} = \frac{ZA_0}{AA_0} \Rightarrow \frac{MA_0}{IA_0} = \frac{ZA_0}{AA_0}$$
30.11.2024 12:00
Solved this a few days ago, really cool problem.