Let $BI$, $CI$ intersect $EF$ at $S$, $R$. By the Iran lemma $BCRS$ is cyclic with center $M$. Obviously $BN$, $CN$ are tangents to $(BIC)$ so $IN$ is the A-symmedian of $\triangle BIC$. Now because $BC$ and $RS$ are antiparallel $K$ is the midpoint of $RS$ as desired.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11cm); 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 = 1.3684958677685986, xmax = 22.158495867768593, ymin = -12.348071074380178, ymax = 3.447928925619838; /* image dimensions */ /* draw figures */ draw((9.800495867768594,0.7765289256198357)--(8.08,-8.16), blue); draw((8.08,-8.16)--(20.84,-8.2), blue); draw((20.84,-8.2)--(9.800495867768594,0.7765289256198357), blue); draw(circle((14.470732018210908,-4.756486190720132), 7.2405360283990605)); draw(circle((11.906019829124801,-5.010392945637503), 3.1615853126020785)); draw((12.588024351080941,-3.0349734212138957)--(14.46,-8.18), red); draw((9.800495867768594,0.7765289256198357)--(11.906019829124801,-5.010392945637503)); draw((9.800495867768594,0.7765289256198357)--(14.493429511792344,2.4840142617575824)); draw((14.493429511792344,2.4840142617575824)--(14.46,-8.18)); draw((8.08,-8.16)--(9.509889986408975,-4.154927352103933)); draw((9.509889986408975,-4.154927352103933)--(20.84,-8.2)); draw((8.08,-8.16)--(15.666158861336946,-1.9150194373543004)); draw((15.666158861336946,-1.9150194373543004)--(20.84,-8.2)); draw((8.801447232659761,-4.412688434913765)--(15.666158861336946,-1.9150194373543004), red); draw((8.801447232659761,-4.412688434913765)--(11.906019829124801,-5.010392945637503)); draw((11.906019829124801,-5.010392945637503)--(13.900619849576985,-2.5573963196159077)); draw((11.906019829124801,-5.010392945637503)--(14.46,-8.18)); draw((3.2907540958251396,-8.144986689955566)--(11.906019829124801,-5.010392945637503)); draw((10.815671706586325,-8.168575773374878)--(14.493429511792344,2.4840142617575824)); draw((11.906019829124801,-5.010392945637503)--(11.896108952076682,-8.1719627239877)); draw((3.2907540958251396,-8.144986689955566)--(8.08,-8.16)); /* dots and labels */ dot((9.800495867768594,0.7765289256198357),linewidth(3.pt) + dotstyle); label("$A$", (9.068495867768597,1.0499289256198359), NE * labelscalefactor); dot((8.08,-8.16),linewidth(3.pt) + dotstyle); label("$B$", (7.7484958677685976,-8.872071074380173), NE * labelscalefactor); dot((20.84,-8.2),linewidth(3.pt) + dotstyle); label("$C$", (20.992495867768596,-8.982071074380174), NE * labelscalefactor); dot((11.906019829124801,-5.010392945637503),linewidth(3.pt) + dotstyle); label("$I$", (12.126495867768597,-5.902071074380171), NE * labelscalefactor); dot((13.900619849576985,-2.5573963196159077),linewidth(3.pt) + dotstyle); label("$E$", (13.930495867768597,-2.1620710743801674), NE * labelscalefactor); dot((8.801447232659761,-4.412688434913765),linewidth(3.pt) + dotstyle); label("$F$", (8.166495867768598,-4.076071074380169), NE * labelscalefactor); dot((14.46,-8.18),linewidth(3.pt) + dotstyle); label("$M$", (14.854495867768597,-8.894071074380173), NE * labelscalefactor); dot((14.493429511792344,2.4840142617575824),linewidth(3.pt) + dotstyle); label("$N$", (14.722495867768597,2.8319289256198377), NE * labelscalefactor); dot((12.588024351080941,-3.0349734212138957),linewidth(3.pt) + dotstyle); label("$K$", (12.060495867768596,-2.9320710743801683), NE * labelscalefactor); dot((9.509889986408975,-4.154927352103933),linewidth(3.pt) + dotstyle); label("$D$", (9.552495867768597,-3.856071074380169), NE * labelscalefactor); dot((15.666158861336946,-1.9150194373543004),linewidth(3.pt) + dotstyle); label("$G$", (15.778495867768596,-1.5020710743801668), NE * labelscalefactor); dot((3.2907540958251396,-8.144986689955566),linewidth(3.pt) + dotstyle); label("$J$", (3.3704958677685983,-8.982071074380174), NE * labelscalefactor); dot((10.815671706586325,-8.168575773374878),linewidth(3.pt) + dotstyle); label("$L$", (10.894495867768597,-8.872071074380173), NE * labelscalefactor); dot((11.896108952076682,-8.1719627239877),linewidth(3.pt) + dotstyle); label("$O$", (12.346495867768597,-8.806071074380174), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $\textit{Proof}$ Let $(I)$ tagents $\overline{BC}$ at $O$, $\overline{BI}\cap\overline{FE}=G$,$\overline{CI}\cap\overline{EF}=D$. The line passes through $I$ and parallel to $\overline{EF}$ meets $\overline{CB}$ at $J$, $\overline{NI}\cap\overline{BC}=L$. $\boxed{\textbf{Claim:}\ BDGC\quad\text{cyclic the cirlcle with radius}\quad \frac{BC}{2}}$ $\textit{Proof}$ We have $\angle EGI=90^o-\angle AIG=90^o-(180^o-\angle AIB)=90^o+\frac{\angle ACB}{2}-90^o=\angle ECI$ And hence $EGCI$ cyclic so $\angle IGC=\angle IEC=90^o\implies \overline{BG}\bot\overline{GC}$ Similarly we have $\overline{CD}\bot\overline{DB}$ and the complete the proof. $\quad\blacksquare$ And so $MD=MG$ (1). We have $\angle JIB=\angle DGB=\angle ICB\implies\quad JI^2=JB\cdot JC$ Easy to see that $\angle OIM=\angle IMN=180^o-\angle IMX=180^o-\angle XIN=\angle AIN$ with $X=\overline{NM}\cap\overline{AI}$ And so $\bigtriangleup IOM\sim\bigtriangleup IAN\implies \angle JMI=\angle IAN=\angle JIL\implies JI^2=JL\cdot JM$ So we have $JB\cdot JC=JL\cdot JM$ and since $M$ is the midpoint of $\overline{BC}\overset{\text{Maclaurin's theorem}}{\implies} (JL,BC)=-1$ Now we have $$-1=I(JL,BC)\overset{EF}{=}I(\infty_{EF}K,GD)\implies KG=KD\quad (2)$$By $(1),(2)$ we get that $\overline{MK}\bot\overline{DG}\equiv\overline{EF}$ and so the end the proof . \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]

Let $ N',M'$ be the antipode of $N$ and the midpoint of the $EF$ Since $N'I \perp EF$ it suffices to show that $\frac{N'M}{N'N}=\frac{IK}{IN}$. We have $\angle IAN =90$° then $ \frac{IK}{IN} =\frac{IM'}{IA} $ more $NBN'C\sim AFIE$ so $ \frac{IM'}{IA} = \frac{N'M}{N'N} $ therefore the result follows. RH HAS

DNCT1 wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11cm); 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 = 1.3684958677685986, xmax = 22.158495867768593, ymin = -12.348071074380178, ymax = 3.447928925619838; /* image dimensions */ /* draw figures */ draw((9.800495867768594,0.7765289256198357)--(8.08,-8.16), blue); draw((8.08,-8.16)--(20.84,-8.2), blue); draw((20.84,-8.2)--(9.800495867768594,0.7765289256198357), blue); draw(circle((14.470732018210908,-4.756486190720132), 7.2405360283990605)); draw(circle((11.906019829124801,-5.010392945637503), 3.1615853126020785)); draw((12.588024351080941,-3.0349734212138957)--(14.46,-8.18), red); draw((9.800495867768594,0.7765289256198357)--(11.906019829124801,-5.010392945637503)); draw((9.800495867768594,0.7765289256198357)--(14.493429511792344,2.4840142617575824)); draw((14.493429511792344,2.4840142617575824)--(14.46,-8.18)); draw((8.08,-8.16)--(9.509889986408975,-4.154927352103933)); draw((9.509889986408975,-4.154927352103933)--(20.84,-8.2)); draw((8.08,-8.16)--(15.666158861336946,-1.9150194373543004)); draw((15.666158861336946,-1.9150194373543004)--(20.84,-8.2)); draw((8.801447232659761,-4.412688434913765)--(15.666158861336946,-1.9150194373543004), red); draw((8.801447232659761,-4.412688434913765)--(11.906019829124801,-5.010392945637503)); draw((11.906019829124801,-5.010392945637503)--(13.900619849576985,-2.5573963196159077)); draw((11.906019829124801,-5.010392945637503)--(14.46,-8.18)); draw((3.2907540958251396,-8.144986689955566)--(11.906019829124801,-5.010392945637503)); draw((10.815671706586325,-8.168575773374878)--(14.493429511792344,2.4840142617575824)); draw((11.906019829124801,-5.010392945637503)--(11.896108952076682,-8.1719627239877)); draw((3.2907540958251396,-8.144986689955566)--(8.08,-8.16)); /* dots and labels */ dot((9.800495867768594,0.7765289256198357),linewidth(3.pt) + dotstyle); label("$A$", (9.068495867768597,1.0499289256198359), NE * labelscalefactor); dot((8.08,-8.16),linewidth(3.pt) + dotstyle); label("$B$", (7.7484958677685976,-8.872071074380173), NE * labelscalefactor); dot((20.84,-8.2),linewidth(3.pt) + dotstyle); label("$C$", (20.992495867768596,-8.982071074380174), NE * labelscalefactor); dot((11.906019829124801,-5.010392945637503),linewidth(3.pt) + dotstyle); label("$I$", (12.126495867768597,-5.902071074380171), NE * labelscalefactor); dot((13.900619849576985,-2.5573963196159077),linewidth(3.pt) + dotstyle); label("$E$", (13.930495867768597,-2.1620710743801674), NE * labelscalefactor); dot((8.801447232659761,-4.412688434913765),linewidth(3.pt) + dotstyle); label("$F$", (8.166495867768598,-4.076071074380169), NE * labelscalefactor); dot((14.46,-8.18),linewidth(3.pt) + dotstyle); label("$M$", (14.854495867768597,-8.894071074380173), NE * labelscalefactor); dot((14.493429511792344,2.4840142617575824),linewidth(3.pt) + dotstyle); label("$N$", (14.722495867768597,2.8319289256198377), NE * labelscalefactor); dot((12.588024351080941,-3.0349734212138957),linewidth(3.pt) + dotstyle); label("$K$", (12.060495867768596,-2.9320710743801683), NE * labelscalefactor); dot((9.509889986408975,-4.154927352103933),linewidth(3.pt) + dotstyle); label("$D$", (9.552495867768597,-3.856071074380169), NE * labelscalefactor); dot((15.666158861336946,-1.9150194373543004),linewidth(3.pt) + dotstyle); label("$G$", (15.778495867768596,-1.5020710743801668), NE * labelscalefactor); dot((3.2907540958251396,-8.144986689955566),linewidth(3.pt) + dotstyle); label("$J$", (3.3704958677685983,-8.982071074380174), NE * labelscalefactor); dot((10.815671706586325,-8.168575773374878),linewidth(3.pt) + dotstyle); label("$L$", (10.894495867768597,-8.872071074380173), NE * labelscalefactor); dot((11.896108952076682,-8.1719627239877),linewidth(3.pt) + dotstyle); label("$O$", (12.346495867768597,-8.806071074380174), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $\textit{Proof}$ Let $(I)$ tagents $\overline{BC}$ at $O$, $\overline{BI}\cap\overline{FE}=G$,$\overline{CI}\cap\overline{EF}=D$. The line passes through $I$ and parallel to $\overline{EF}$ meets $\overline{CB}$ at $J$, $\overline{NI}\cap\overline{BC}=L$. $\boxed{\textbf{Claim:}\ BDGC\quad\text{cyclic the cirlcle with radius}\quad \frac{BC}{2}}$ $\textit{Proof}$ We have $\angle EGI=90^o-\angle AIG=90^o-(180^o-\angle AIB)=90^o+\frac{\angle ACB}{2}-90^o=\angle ECI$ And hence $EGCI$ cyclic so $\angle IGC=\angle IEC=90^o\implies \overline{BG}\bot\overline{GC}$ Similarly we have $\overline{CD}\bot\overline{DB}$ and the complete the proof. $\quad\blacksquare$ And so $MD=MG$ (1). We have $\angle JIB=\angle DGB=\angle ICB\implies\quad JI^2=JB\cdot JC$ Easy to see that $\angle OIM=\angle IMN=180^o-\angle IMX=180^o-\angle XIN=\angle AIN$ with $X=\overline{NM}\cap\overline{AI}$ And so $\bigtriangleup IOM\sim\bigtriangleup IAN\implies \angle JMI=\angle IAN=\angle JIL\implies JI^2=JL\cdot JM$ So we have $JB\cdot JC=JL\cdot JM$ and since $M$ is the midpoint of $\overline{BC}\overset{\text{Maclaurin's theorem}}{\implies} (JL,BC)=-1$ Now we have $$-1=I(JL,BC)\overset{EF}{=}I(\infty_{EF}K,GD)\implies KG=KD\quad (2)$$By $(1),(2)$ we get that $\overline{MK}\bot\overline{DG}\equiv\overline{EF}$ and so the end the proof . \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \] Nice solution!

My original solution: Let $P$ and $Q$ be the midpoint of minor arc $BC$ and midpoint of $EF$ respectively. It suffices to prove that $KM\parallel IP$, that is $$\frac{NK}{KI}=\frac{NM}{MP}$$But since $EF$ is parallel to $AN$, and the quadrilaterals $AEIF, NBPC$ are similar with points $Q$ corresponding to $M$ in the similarity, then $$\frac{NK}{KI}=\frac{AQ}{QI}=\frac{NM}{MP}$$as we needed. Nice solutions up there though!

thinker123 wrote: OP wrote: Own. IMO 2021 Malaysian Training Camp 2 Contradict?? how is that a contradiction?

megarnie wrote: thinker123 wrote: OP wrote: Own. IMO 2021 Malaysian Training Camp 2 Contradict?? how is that a contradiction? like how can the problem be their "own" as well as from IMO 2021 Malaysian training camp 2?

Because they proposed the problem for the training camp.

Simply check their status and they qualed for imo in Malaysia from 2015-2018. It makes sense that they would propose problems for the training camp.

Hi, this is my proposal for the training camps indeed, and I am from Malaysia

navi_09220114 wrote: Hi, this is my proposal for the training camps indeed, and I am from Malaysia orz

Let $AI\cap EF=R$ and $D$ be the antipode of $N$, then we know that $N$,$M$,$D$ and $A$,$I$,$D$ are collinear. Since $AD \perp EF$ we can finish with proving that $ID//KM$. Then we will prove that $NM/MD=NK/KI$. Since $AI\perp FE$ and $\angle DAN=90$ we have $EF//AN$, then $NK/KI=AR/RI$. It is easy to see that $AFI$ and $NCD$ triangles are similar and since $FR\perp AI$ and $CM\perp ND$ we have $AR/RI=NM/MD$, which means that $NK/KI=AR/RI=NM/MD$, so we are done:).

Let the incircle be tangent to $BC$ at $D$, $S$ be the midpoint of the arc $BC$ not containing $A$ and $L$ be the foot of the altitude from $M$ to $EF$. $EF\cap BC=X$. e will show that $I,L,N$ are collinear. \[cos^2\frac{A}{2}=\frac{cos\frac{A}{2}.NC}{\frac{1}{cos A}.NC}=\frac{NM}{NS}\overset{?}{=} \frac{ML}{SI}=\frac{ML}{SB}=ML.cos\frac{A}{2}.\frac{1}{MB}\]\[ML\overset{?}{=}MB.cos\frac{A}{2}\]\[MB\overset{?}{=}\frac{ML}{cos\frac{A}{2}}=\frac{sin\frac{B-C}{2}.XM}{sin90-\frac{A}{2}}=\frac{BF}{XB}.XM=\frac{BD}{XB}.XM\]Since $(X,D;B,C)=-1,$ we have \[\frac{XB}{BD}=\frac{XM}{MB}\]Which gives the desired result.$\blacksquare$