Let $ABC$ be an acute angled triangle with $AC>AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA,$ and $AB$ at $D, E,$ and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying \[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]Prove that the circumcircles of $\triangle AXY, \triangle AEF$ and $\triangle ABC$ meet at a point $Z\ne A$. Proposed by Atul Shatavart Nadig and Shantanu Nene
Problem
Source: India IMOTC 2024 Day 2 Problem 2
Tags: geometry
31.05.2024 08:23
31.05.2024 09:22
31.05.2024 13:44
This problem is so cute Congrats to the proposers!
Attachments:

01.06.2024 18:52
Solved with MrOreoJuice and old friend Executioner230607. Solution: Let $T \coloneqq EF \cap BC$ and denote the inverse of any point $\bullet$ as $\bullet '$ across $\omega$ Claim: $X,I,C$ are collinear, similarly $Y, I ,B$ are collinear. Proof: Note that \[\measuredangle XDE = \measuredangle XDI + \measuredangle IDE = \measuredangle IEX + \measuredangle DEI = \measuredangle DEX \]hence $\overline{XD} = \overline{XE}$ therefore $X$ lies on the perpendicular bisector of $\overline{DE}$ which is also $CI$. $\square$ Claim: $XQED$ is cyclic, similarly $YQFD$ is cyclic. Proof: By Sharky-Devil and by using the fact that $Q$ is the Miquel Point of $BCFE$ we have, \begin{align*} \angle DQE &= \angle DQC + \angle CQE \\ &= \angle A/2 + \angle BQF \\ &= \angle A/2 + \angle BTF \\ &= \angle A/2 + (\angle B - \angle BFT) \\ &= \angle A/2 + (\angle B - \angle AFE) \\ &=90^\circ - \angle C \end{align*}and \begin{align*} \angle DXE &= 2\angle DXI \\ &=2(\angle DIC - 45^\circ) \\ &=2(90^\circ - \angle C/2 - 45^\circ) \\ &= 90^\circ - \angle C \end{align*}therefore $\angle DQE = \angle DXE$. $\square$ Consider an inversion around $\omega$. $Q'$ is foot of $D$ to $\overline{EF}$. $X'$ is intersection of $CI$ and $(Q'ED)$ which is intersection of perpendicular bisector of $DE$ with semicircle with diameter $DE$ (and from our original diagram this intersection lies inside $\triangle DEF$). $A'$ is midpoint of $\overline{EF}$. $Y'$ has a similar definition as $X'$. Therefore in this inverted figure, we need to prove $A'X'Y'Q'$ is cyclic. We will now complex bash the inverted problem with a little bit of synthetic magic! Observe that \[\measuredangle X'QY' = \measuredangle X'QA' + \measuredangle A'QY' = \measuredangle X'DE + \measuredangle Y'DF = -45^\circ + (-45)^\circ = 90^\circ\] We now need to show that $\measuredangle XA'Y' = 90^\circ$. Start by computing $X'$ and $Y'$. Note that $X'$ is precisely the clock-wise rotation of $D$ across midpoint of $\overline{DE}$. Similarly, $Y'$ is counter-clockwise rotation of $D$ across midpoint of $\overline{DF}$. This yields \[x' = \left( \frac{e-d}{2}\right)i + \frac{e+d}{2} \qquad \qquad y' = \left( \frac{d-f}{2}\right)i + \frac{d+f}{2}\]To show that $A'X' \perp A'Y'$ it is enough to show that $\frac{x'-a'}{y'-a'}$ is purely imaginary. Using $a' = \frac{e+f}{2}$ This is true since \[\frac{x'-a'}{y'-a'}= \frac{\left(\frac{e-d}{2}\right)i + \frac{d-f}{2}}{\left(\frac{d-f}{2}\right)i + \frac{d-e}{2}} = i\]and thus we're done. $\blacksquare$
06.06.2024 20:28
Just spam inversion lol Let's do inversion around $\omega$.Let $Z=(AEF)\cap (ABC)$ Then $A\to A'$=midpoint of $EF$,$Z\to Z'$=the foot from $D$ to $EF$ and since $\angle IDX=\angle IEX=45$ we have that $\angle IX'D'=\angle IX'E'=45$ so $D'X'E'=90$ so $X'$ is on the circle with diameter $DE$.Since $X'I'$ is the bisector of $D'X'E'$ we have that $X'I'$ intersects the circle with diameter $DE$ in the middle of the arc $DE$.Since $I$ and the middle of the arc are on the perpendicular bisector of $DE$ then $X'$ is the middle of the bigger arc $\widehat{DE}$ and we want to prove that $X'Y'Z'A'$ is cyclic Let's restate the problem as follows: We have a triangle $ABC$ with $D$ the foot from $A,M$ the midpoint of $BC$,$X$ the midpoint of arc $\widehat{ADC}$,$Y$ the midpoint of arc $\widehat{ADB}$ and we want to prove $XDMY$ cyclic.For this we will do inversion at $D$ with power $1$ The problem becomes: Triangle $A'B'C',D$ the foot from $A'$,since $\angle ADX=45$ we have $\angle A'DX'=45$ so $DX'$ is the exterior bisector of $\angle ADC$ with $X'\in A'C'$,similary $Y'$ and we want $X',Y',M'$ collinear For this we do menelaus: We want $\frac{A'Y'}{Y'B'}\cdot \frac{B'M'}{M'C'}\cdot \frac{C'X'}{X'A'}=1$. By exterior angle theorem $\frac{A'X'}{X'C'}=\frac{A'D}{DC'},\frac{A'Y'}{Y'B'}=\frac{A'D}{DB'}$ and so we want to prove that $\frac{M'B'}{M'C'}=\frac{DB'}{DC'}$,but $M'B'=DB'+DM'=\frac{1}{DB}+\frac{1}{DM},M'C'=\frac{1}{DM}-\frac{1}{DC}$,$\frac{DB'}{DC'}=\frac{DC}{DB}$ and plugging in gives the result.
15.06.2024 13:44
Rijul saini wrote: Let $ABC$ be an acute angled triangle with $AC>AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA,$ and $AB$ at $D, E,$ and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying \[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]Prove that the circumcircles of $\triangle AXY, \triangle AEF$ and $\triangle ABC$ meet at a point $Z\ne A$. Proposed by Atul Shatavart Nadig and Shantanu Nene $\textbf{Claim 1:}$ Let $L$ be foot of altitude from $D$ to $EF$; $Z$ be the second intersection of circles $(AEF)$, $(ABC)$ $\hspace{1.5cm}$Then $Z$, $L$, $I$ are colinear Proof : $\, \,$Cuz $\angle FLD = 90^\circ$ and $L(FD, BC) = -1$ so $LD$ is the internal bisector of $\angle BLC$ (Cuz $D$ lie on segment $BC$) $\, \,$Which means $\angle FLB = \angle ELC$ $\, \,$Combine with $\angle AFE = \angle AEF$, we see that $\triangle LFB \sim \triangle LEC$ $\, \,$In the order hand, cuz $Z = (AEF) \cap (ABC)$ ($Z$ \neq $A$) so $Z$ is center of spiral similarity carries $(AEF)$ to $(ABC)$, $\, \,$which means $\triangle ZFB \sim^+ \triangle ZEC$ $\, \,$Thus, $\frac{LF}{LE} = \frac{FB}{EC} = \frac{ZF}{ZE}$. $\, \,$So we have that $ZL$ is the internal bisector of $\angle FZE$ and $Z$, $L$, $I$ are colinear (Cuz $L$ always lie on segment $EF$) $\textbf{Claim 2:}$ $X$, $Z$, $E$, $D$ are concyclic; $Y$, $Z$, $F$, $D$ are concyclic Proof : $\, \,$Cuz $Z$ is center of spiral similarity carries $EF$ to $BC$ and $\frac{LF}{LE} = \frac{FB}{EC} = \frac{DB}{DC}$ $\, \,$So $\triangle ZLD \sim^+ \triangle ZFB$. Lead to $\angle DZL = \angle BZF$ $\, \,$Therefore, $\angle FZD = \angle FZI - \angle DZL = \frac12 \angle BAC - \angle BZF = \frac12 \angle BAC + \angle FZA - \angle BZA $ $\hspace{3cm}= \frac12 \angle BAC + \angle ACB - \angle AEF = \angle BAC + \angle ACB - 90^\circ = 90^\circ - \angle ABC$ $\, \,$In the order hand, we see that $\angle FDY = \angle FDI + \angle IDY = \frac12 \angle ABC + 45^\circ, \angle DFR = \frac12 \angle ABC + 45^\circ$ $\, \,$so $\angle DYF = 90^\circ - \angle ABC$ $\, \,$Thus, $\angle DZF = 90^\circ - \angle ABC = \angle DYF$. Lead to $Y$, $Z$, $F$, $D$ are concyclic. Similar, $X$, $Z$, $E$, $D$ are concyclic $\textbf{Claim 3:}$ Let $X'$, $Y'$ be points inside $\angle FDE$ satisfying $\triangle DX'E$, $\triangle DY'F$ are right isosceles triangles at $X$, $Y$ $\hspace{1.5cm}$Then $A'$, $X'$, $Y'$, $L$ are concyclic, with $A'$ is the midpoint of segment $EF$ Proof : $\, \,$Let $U$, $V$ be the reflections of $E$, $F$ across $X'$, $Y'$, respectively $\, \,$We easy to see that $\triangle EDU$, $\triangle FDV$ are right isosceles triangles at $D$ $\, \,$Lead to $D$ is a center of rotation with $90^\circ$ carries $UF$ to $EV$ $\, \,$Therefore $EV \perp FU$ and $EV = FU$. Which means $\triangle X'A'Y'$ is a right isosceles triangle (Cuz $\overrightarrow{A'X'} = \frac12 \overrightarrow{EV}; \overrightarrow{A'Y'} = \frac12 \overrightarrow{FU}$) $\, \,$But we ez to see that $\angle X'LY' = \angle X'LD + \angle Y'LD = \angle X'ED + \angle Y'FD = 45^\circ + 45^\circ = 90^\circ$ $\, \,$Thus, $X'$, $Y'$, $A'$, $L$ are concyclic on a circle with diameter $X'Y'$ $\textbf{From these claim we see that :}$ $\, \,$Let $f$ be the inversion about the incircle of $\triangle ABC$. $\, \,$Cuz $\angle AZI = \angle IA'L = 90^\circ$ so $ID^2 = \overline{IA'}.\overline{IA} = \overline{IL}.\overline{IZ}$ $\, \,$Therefore, $f$ : $A \leftrightarrow A'$, $Z \leftrightarrow L$, $(ZFD) \leftrightarrow (DF)$, $(ZED) \leftrightarrow (DE)$. $\, \,$Combine with $X$ and $X'$ lie on perpendicular bisector of segment $DF$, $Y$ and $Y'$ lie on perpendicular bisector of segment $DE$ $\, \,$we see that $f$ : $X \leftrightarrow X'$, $Y \leftrightarrow Y'$. $\, \,$But in Claim 3 we proved that $X'$, $Y'$, $L$, $A'$ are concyclic $\, \,$Thus, $A$, $Z$, $X$, $Y$ are concyclic
Attachments:
india-imotc-2024.pdf (72kb)
22.07.2024 13:17
complex is cool, but i am too lazy to type
27.07.2024 17:09
Same configuration and series of lemmas used in USA TSTST 2020 P2 Edit: Well it wasn't nearly as simple as I thought it would be after inversion coz u need to angle chase a bunch afterwards.
29.07.2024 02:50
L13832 wrote: Same configuration and series of lemmas used in USA TSTST 2020 P2 Also similar to Kazakhstan National Olympiad 2024 (9 grade), P5
28.08.2024 12:56
Invert with respect to the incircle and restate in terms of $\Delta DEF$. Inverted Problem wrote: In $\Delta ABC$, $X$ and $Y$ are midpoints of arcs $AB$ and $AC$ of the circles $(AB)$ and $(AC)$ lying on the same side of the side as the opposite vertex. Let $D$ be the foot of altitude from $A$ onto $BC$ and $M$ be the midpoint of $BC$. Prove that $X,Y,M,D$ are concyclic. Note that by simple angle chasing, $\angle XDY=90$° so it suffices to show that $\angle XMY=90°$°. Note that upon fixing $AB,AC,BC$; $AY,CY,BX,AX$ are known in terms of $a,b,c$. Since angle $BAY$ can be worked out, by cosine formula we can calculate $BY$ and similarly $CX$. By apollonius theorem in $\Delta BYC$, we can calculate $YM$ and similarly $XM$. Lastly, one can also find out $XY$ using cosine fomula in $\Delta AXY$ so we are done by converse of pythagoras in $\Delta XMY$.
11.12.2024 19:43
Invert around the incenter. Note that $(AEF)\cap (ABC)$ is $A-$sharky devil point which swaps with the foot of the altitude from $D$ to $EF$ under this inversion. New Problem Statement: $ABC$ is a triangle with altitude $AD$ where $D$ lies on $BC$. Let $X,Y$ be outside $\triangle ABC$ such that $\measuredangle AXB=90=\measuredangle CYA$ and $AX=AB,YA=YC$. If $M$ is the midpoint of $BC$, then prove that $D,M,X,Y$ are concyclic. Invert at $D$ with radius $\sqrt{-DB.DC}$. New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ and $D,E,F$ lie on $BC,CA,AB$ respectively. $X,Y$ lie on $CF,BE$ such that $DX,DY$ bisect $\measuredangle ADC,\measuredangle BDA$ respectively. Prove that $BC,EF,XY$ are concurrent. Let $H$ be the orthocenter of $\triangle ABC$. Set $EF\cap BC=Q$. \[\frac{QB}{QC}.\frac{XC}{XH}.\frac{YH}{YB}=\frac{QB}{QC}.\frac{DC}{DH}.\frac{DH}{DB}=\frac{QB}{QC}.\frac{DC}{DB}=1\]As desired.$\blacksquare$