Also first irl contest solve but fuzzed up ioqm

I showed that $A$ lies on $(EFII_1)$, too

Clearly $I_1 = BE_1 \cap BF_1$ because $BE_1$ is the angle bisector of $\angle F_1E_1D_1$ as $$\angle FE_1B = \angle F_1CB = \angle FCD = \angle FAD = \angle BAD_1 = \angle BE_1D_1$$and similarly $CF_1$ is the angle bisector of $\angle D_1F_1E_1$. $\textbf{Lemma 1}$ In any $\triangle XYZ$ with incenter $J$, $\angle YJZ = 90^{\circ}+\frac{\angle YXZ}{2}$. $\textbf{Proof}$ Omitted, just angle chase. $\blacksquare$ Now, notice that $$\angle EDF = \angle EDA + \angle ADF = \angle FCA + \angle EBA = \angle F_1CA + \angle ABE_1 = \angle E_1D_1F_1$$using the two given cyclites, then using that $I_1 = FF_1 \cap EE_1$ and $\textbf{Lemma 1}$, $$\angle EI_1F = \angle E_1I_1F_1 = 90^{\circ} + \frac{\angle E_1D_1F_1}{2} = 90^{\circ} + \frac{\angle EDF}{2} = \angle EIF$$giving the desired cyclicity. $\blacksquare$ Edit (regarding configuration issues pointed out below): You can observe that both $I_1, I$ lie inside $\triangle DEF$ as $D$ lies in the interior of $BC$, I did not include it previously because it's an AoPS post.

bora_olmez wrote: Clearly $I_1 = BE_1 \cap BF_1$ because $BE_1$ is the angle bisector of $\angle F_1E_1D_1$ as $$\angle FE_1B = \angle F_1CB = \angle FCD = \angle FAD = \angle BAD_1 = \angle BE_1D_1$$and similarly $CF_1$ is the angle bisector of $\angle D_1F_1E_1$. $\textbf{Lemma 1}$ In any $\triangle XYZ$ with incenter $J$, $\angle YJZ = 90^{\circ}+\frac{\angle YXZ}{2}$. $\textbf{Proof}$ Omitted, just angle chase. $\blacksquare$ Now, notice that $$\angle EDF = \angle EDA + \angle ADF = \angle FCA + \angle EBA = \angle F_1CA + \angle ABE_1 = \angle E_1D_1F_1$$using the two given cyclites, then using that $I_1 = FF_1 \cap EE_1$ and $\textbf{Lemma 1}$, $$\angle EI_1F = \angle E_1I_1F_1 = 90^{\circ} + \frac{\angle E_1D_1F_1}{2} = 90^{\circ} + \frac{\angle EDF}{2} = \angle EIF$$giving the desired cyclicity. $\blacksquare$ ig there might pop up some configuration issue, so its better to show that $A$ lies on $\odot EFII_1$, which is just another bit of trivial angle chase. my proof was exactly identical to bora's proof, but i just added the last fact, anyways my solution is not even going to be checked and will be thrown like any other random piece of paper, ig ill stop doin MO and concentrate IOQM level bashy sums first

Anyone proved the similarity part using homothety?

Walkthrough. $DE || D_1 E_1$ and $DF || D_1 F_1$ $E_1E$ and $F_1F$ are bisectors of $\triangle D_1E_1F_1$ Complete using $\angle{QIR} = \frac{\pi}{2} + \frac{\angle{QPR}}{2}$ when $I$ is incenter of $\triangle PQR$

NTistrulove wrote: Anyone proved the similarity part using homothety? DEF is just made larger to match up with D1E1F1. Use angle chasing and parallel lines to prove it.

Easy problem it can be solving with only using angel change and prove$A,E,I_1,I,F$

This was just simple angle chase

NTistrulove wrote: Anyone proved the similarity part using homothety? The triangles are not similar. Two sides are parallel but the third side may or may not be.

bora_olmez wrote: Edit (regarding configuration issues pointed out below): You can observe that both $I_1, I$ lie inside $\triangle DEF$ as $D$ lies in the interior of $BC$, I did not include it previously because it's an AoPS post. Are you sure? Doesn't seem to be true here

2nd IRL Contest solve after STEMS P5 :") Here's a proof without parallelism. Claim :$BE \cap CF \equiv I_1$ Proof Note that, $$\angle BE_1F_1=\angle BCF_1=\angle DCF=\angle DAF=\angle D_1AB=\angle BE_1D_1 \implies \text{BE bisects } \angle D_1E_1F_1$$Similarly $CF$ is another angle bisector ,so the claim is proved $\blacksquare$ We observe that $\angle I_1EA+\angle I_1FA=\angle BEA+\angle CFA=\angle BDA+\angle CDA=180 \implies I_1 \in \odot(\triangle AEF)$[This part was motivated by ELMOSL 2013 G3] Now,observe \begin{align*}\angle EDF &=\angle EDA+\angle FDA \\ &=\angle EBA+\angle FCA \\ &=180-(\angle A+\angle AEB)+180-(\angle A +\angle CFA)\\ &=360-(2\angle A+\angle CFA+\angle BEA)\\ &=360-(180+2\angle A)\\ &=180-2\angle A \end{align*}Finally $\angle EIF=90+\frac{\angle EDF}{2}=90+\frac{180-2\angle A}{2}=180- \angle A \implies I \in \odot(\triangle AEF)$ Thus as $I,I_1 \in \odot(\triangle AEF)$ we are done $\blacksquare$ [This was my solution in the contest where I did find out the parallel lines but ended up not using them.]

nice problem it was main thing to see was to see that ff1 is the bisector which can be proved easily by angle chase and e is the miquel point.

Honestly learning so much geo didn’t pay off, this was trivial, probably the easiest inmo geo in the recent years anyways my sol is the same as all above and i left the fact the A also lies on this circle as a remark which might help also directed angles which might help with configuration issues @below No because this might still be a nice problem but it isn't hard but ok if u say so ALSO WHAT MAKAR IS VIEWING THIS THREAD ORZ ORZ ORZ SIR

Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem.

Master_of_Aops wrote: NTistrulove wrote: Anyone proved the similarity part using homothety? The triangles are not similar. Two sides are parallel but the third side may or may not be. Oh... I fudged up

Is there another contest or does it just have 3 pbs this year?

Flash_Sloth wrote: Is there another contest or does it just have 3 pbs this year? This year INMO had 3 problems and we had 2.5 hours to attempt them, mainly because of covid

I did the parallel proof and showed angle D1=D. However, I did not have time to prove angle F1I1E1= angle FI1E1. I assumed that. How much may I lose for not showing this proof?

DebayuRMO wrote: Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem. Even I also spent 1.5 hours to get everything in place and write it completely and I must say it was a very nice problem (at least for me). If a problem has a short solution or solution using simple stuffs doesn't mean it is easy... observing stuffs which makes a problem trivial are sometimes tough. Btw congratulations for solving it!!

Pratik12 wrote: DebayuRMO wrote: Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem. Even I also spent 1.5 hours to get everything in place and write it completely and I must say it was a very nice problem (at least for me). If a problem has a short solution or solution using simple stuffs doesn't mean it is easy... observing stuffs which makes a problem trivial are sometimes tough. Btw congratulations for solving it!! Absolutely !

All angles are directed modulo $180^\circ$ when we use $\measuredangle$. We directly give a super awesome claim which is just enough to annihilate the problem. Claim: $DF \parallel D_1F_1$ and $D_1E_1 \parallel DE$ and $I_1$ is the intersection of $CF_1$ and $BE_1$. Proof: This will be angle chase \[\measuredangle F_1D_1A = \measuredangle F_1CA = \measuredangle FCA = \measuredangle FDA \implies DF \parallel D_1F_1\]One can analogously chase to prove that other pair of lines are parallel as well. Now, we show that $F_1C$ bisects angle $\angle D_1F_1E_1$. Just see that \[\measuredangle CF_1E_1 = \measuredangle CBE_1 = \measuredangle DBE =\measuredangle DAE = \measuredangle D_1AC = \measuredangle D_1F_1C\]and the other one follows symmetrically. $\square$ With the above claim, we can deduce that $\angle F_1D_1E = \angle FDE$. By a well-known lemma, we know that \[\angle F_1I_1E_1 = \angle FI_1E = \angle 90^\circ + \frac{\angle F_1D_1E_1}{2} = \angle FIE \implies \angle EI_1F = \angle EIF\]and thus we are done. $\blacksquare$

We first prove that $AEIF$ is concyclic. Notice that $\angle FIE =90°+\dfrac{1}{2}\angle FDE=90°+\dfrac{1}{2}(\angle BDE -\angle BDF)= 90°+\dfrac{1}{2}(180° - \angle BAC -\angle BAC)=180° -\angle BAC$ Now $\angle AF_1C =\angle AD_1C$ and $\angle F_1FA =180°-\angle CFA =180°-\angle CDA=\angle D_1DA$ implies that $\triangle AF_1F$ and $\triangle CD_1D$ are similar. So $$\angle F_1 AB= \angle BCD_1=\angle BAD_1$$So B is the midpoint of arc $F_1D_1$ that doesn't contain $E_1$. So $BE_1$ is the angle bisector of $\angle F_1E_1D_1$. Similarly $CF_1$ is the angle bisector of $\angle E_1F_1D_1$ .So, $B,I_1,E$ and $E_1$ lie in a line. It's easy to see that $\angle F_1AE_1=2\angle BAC$ due to the bisectors. So, $\angle FI_1E= 90°+\dfrac{1}{2}\angle F_1D_1E_1= 90°+\dfrac{1}{2}( 180°-\angle F_1AE_1)=90°+\dfrac{1}{2}(180°-2\angle BAC)=180°-\angle BAC$ Therfore $AFI_1E$ are concyclic and thus I lies in the circle as proved before and the claim is proved.$\blacksquare$

Check the video solution at https://www.youtube.com/watch?v=Bi7BVaIRlVQ

Guys I found a solution heavily using the properties of miquel point ! INMO 2022 P1 wrote: Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$. Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$. Let $AD$, $BE$, and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$, $E_1(\ne B)$ and $F_1(\ne C)$, respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$, respectively. Prove that $E,F, I, I_1$ are concyclic. First note that $I_1 \equiv BE \cap CF$ by a simple angle chase. Now Consider the complete quadrilateral,$BFAECI_1$ clearly $D(\odot(\triangle ABE)\cap \odot(\triangle ACF))$ is its miquel point. Since $D \in BC$ we get that quadrilateral $AEI_1F$ must be cyclic.Now its well known that if $O$ is the centre of $\odot(AEI_1F)$,$D \in \odot(\triangle OEF)$(Check E.G.M.O proposition $10.14$) and $OD$ bisects $\angle EDF$(E.G.M.O proposition $10.15$). So by the well known Fact 5/Incentre-Excentre Lemma we get that $I \equiv OD \cap \odot(AEI_1F)$.The End $\blacksquare$

Well for an hour I just thought $D$ is inside of $ABC$ $\angle D'F'C = \angle D'AC = \angle DAE= \angle DBE = \angle CDE' = \angle E'F'C \implies CF'$ is angle bisector of $\angle E'F'D'$. $\angle D'E'B = \angle D'AB = \angle DAF = \angle DCF = \angle BCF' = \angle F'E'B \implies BE'$ is angle bisector of $\angle F'E'D'$. $\angle FI'E = \angle F'I'E' = \angle 180 - \angle I'E'F' - \angle I'F'E' = \angle 180 - \angle A$. $\angle FIE = \angle 90 + \frac{\angle FDE}{2} = \angle 90 + \angle FCA + \angle EBA = \angle 90 + \frac{\angle 180 - \angle 2A}{2} = \angle 180 - \angle A$. Now we have $\angle FI'E = \angle FIE$ which implies $II'E'F'$ is cyclic.

anyone with inversion at A?

@above, could you share

Quite a nice problem.. It is very easy to get intimidated by the sheer amount of lines and absurd sounding points defined in the diagram..

Mahdi_Mashayekhi wrote: Well for an hour I just thought $D$ is inside of $ABC$ whoa same Observe that $\angle EDA=\angle EBA=\angle E_1BA=\angle E_1D_1A$, and likewise $\angle FDA=\angle F_1D_1A$, so $\angle EDF=\angle E_1D_1F_1 \implies \angle EIF=\angle E_1I_1F_1$. Thus it suffices to show that $E,E_1,I_1$ and $F,F_1,I_1$ are collinear. By symnmetry, it suffices to show that $\overline{E_1C}$ bisects $\angle F_1E_1D_1$. But this follows since $\angle F_1E_1C=\angle EBD=\angle EAD=\angle CE_1D_1$. $\blacksquare$ no i dont care about config issues

Just notice that $\angle F_1E_1B = \angle F_1CB = \angle BAD_1 = \angle BE_1D_1$, hence $I_1 = \overline{BE_1} \cap \overline{CF_1}$. Then $\angle EI_1F = \angle EIF$ as $\overline{FD} \parallel \overline{F_1D_1}$ by Reim's theorem and similar.

The key claim is that $AEFII_1$ is cyclic. To show $I \in (AEF)$, notice $\angle FDB = \angle EDC = \angle A$, so \[\angle EIF = 90 + \frac{\angle EDF}{2} = 90 + \frac{180-2\angle A}{2} = 180-\angle A.\] The angle bisector of $\angle D_1E_1F_1$ is $E_1B$, as \[\angle D_1E_1B = \angle DAF = \angle DCF = \angle BE_1F_1.\]Similarily, the angle bisector $\angle D_1F_1E_1$ is $F_1C$, making $I_1 = BE \cap CF$. From here, it's easy to find $AEI_1F$ cyclic, as desired. $\blacksquare$

was lying on my Geogebra so i'll upload that as well.

Let $I_{1}^{'} $ be the intersection of $BE_1$ and $CF_1$ By quick angle chasing we get that triangles $\triangle DBF, \triangle DEC$ are simmilar $\iff \triangle DBF \sim \triangle DEC \implies \frac{DB}{DE}=\frac{DF}{DC}$ combining with $\angle BDE=\angle CDE$ we get that triangles $\triangle DBE , \triangle DFC$ are also simmilar $\iff \triangle DBE \sim \triangle DFC$ So now we angle chase and we get that $E_1I_{1}^{'}$ is the angle bisector of $\angle F_1E_1D_1$ we also get that $F_1I_{1}^{'} $is the angle bisector of $\angle F_1E_1D_1$ so $I_{1}^{'}=I_1$ hence $BE_1 \cap CF_1 \cap AD_1 = \{I_1\}$ Also know by $\triangle DBE \sim \triangle DFC$ we get that quadrilaterals $BDF_1I_1$ and $CDEI_1$ are cyclic $\implies$ By Miquel Theorem that the quadrilateral $AFEI_1$ is also cyclic We also have that $\angle FIE=90 + \frac{\angle FDE}{2}=90+\frac{180-\angle FDB-\angle EDC}{2}=90+\frac{180-2 \cdot \angle A}{2}=180-\angle A \implies \angle FIE=180- \angle A$ which means that $AFIE$ is cyclic hence $I \in \odot (AFEI_1) \implies EFII_1$ is cyclic My 70th post

Can't believe in $2022$ you solve this problem and you are $INMO$ $AWARDEE$ $\angle FCA = \angle FDA$ but $\angle FCA = \angle F_1CA = \angle F_1DA$ implies $\angle FDA = \angle F_1D_1A$ Similarly $\angle EDA = \angle E_1D_1A$ Combining both, we get $\angle FDE = \angle F_1D_1E_1$ Now,$$\angle EI_1F = \angle E_1I_1F_1 = 90^{\circ} + \frac{\angle E_1D_1F_1}{2} = 90^{\circ} + \frac{\angle EDF}{2} = \angle EIF$$Hence proved!

reminded me to old times