Solving Random ISLs And Sharygin Solutions! And INMO happened!!

Some of the ISLs I did before INMO :P 

[2005 G3]: Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$

Solution: Note that $$\Delta LDK \sim \Delta XBK$$ and $$\Delta ADY\sim \Delta XCY.$$

So we have $$\frac{BK}{DY}=\frac{XK}{LY}$$ and $$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$$

Hence $$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY} $$

$$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$$

$$\implies \Delta CBK\sim \Delta LDK$$

And we are done. We get that $$\angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2$$

Motivation: I took a hint on this. I had other angles but I didn't realise that $\Delta CBK\sim \Delta LDK.$
Probably I should have ratio chased a bit better. Will try that next time!

[2018 G4]: A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$.

Solution: Oh god this was so much a ggb problem. Not really but fineeeee...

Rename the point $T$ as $D.$

Claim: The three circles $(AC_1B),(AB_1C),(A_1BC)$ concur

Proof: We can simply ignore $A_2,B_2,C_2.$ Also, there might be configuration issues.

Define $H=(AC_1B)\cap (AB_1C).$

Note that $$\angle BHC=\angle AHB-\angle AHC$$

$$=\angle AC_1B-180+\angle AB_1C$$

$$=\angle ADB-180+\angle ADC$$

Now, we have to show that $$\angle BHC=180-\angle BA_1C=180-\angle BDC$$

But $$\angle BHC=\angle ADB+\angle ADC-180$$

So enough to show $$\angle ADB+\angle ADC-180=190-\angle BDC$$

Or show $$\angle ADB+\angle ADC+\angle BDC=360.$$

Claim: $H\in(A_1B_1C_1)$

Proof: Note that $$\angle A_1HC_1=\angle A_1HB+\angle BHC_1=\angle A_1CB+180-\angle C_1AB=\angle A_1B_1D+180-\angle C_1B_1D.$$

Where we used the fact that $A$ is centre of $B_1C_1D$ and $C$ is centre of $DB_1A_1.$

Now, we claim that $H$ is the concurrency point. So if we show $C_2,C,H$ collinear, then we are done as other collinearities follows similarly.

Enough to show that $$A_1HC=\angle A_1HC_2$$

or show that $$\angle A_1BC=\angle A_1HC_2$$

or show that $$\angle A_1C_1D=\angle A_1HC_2$$

But $$\angle A_1C_1H=\angle A_1HC_2$$

And done!

[2016 G4]: Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Here's the diagram!

Solution: Let the reflected point be $I'.$ Let $\angle ACB=C.$ We express all the angles in the form of $\angle C.$ 

By simple angle chase, we get $$ADI=C+C/2.$$

And so $$\angle I'DE=90+C+C/2.$$

Moreover, we have $$\angle I'CM=C+C/2.$$

Define $X=EM\cap I'C,$ we get that $$\angle 90-C-C/2\implies I'DEX\text{ is cyclic }.$$

Also, note that  as $II'||ED,$ $$\angle BDE=\angle EDI=\angle DII'$$

So $$\angle BDE=90-C-C/2=\angle BXI\text { as } BX=XC\implies BEDX \text{ is cyclic }. $$

Hence $BDEI'$ is cyclic.

Remark: It was pretty easy for G4. I think constructing $X$ wasn't that easy, but once you notice that angles are $90-C-C/2$ then everything breaks down..

[2001 N1]: Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]

Solution: Note that the condition $$a\equiv b\pmod n\implies \forall \text{ relatively prime} a, n|a^2-1.$$

Here is the construction. Assume $n>3.$ There are two cases.

Case 1: $m$ is odd, then take $a=2$

Case 2: $m$ is even and has all primes dividing $>3$

Then let $$m=2^{v_2(m)}\cdot p_1^{\alpha_1}\dots p_k^{\alpha_k}$$

Consider $$a\equiv 2\pmod p_i\implies a^2\equiv 4\pmod p_i$$

$$a\equiv 1 \pmod 2\implies a^2\equiv 1\pmod 2$$

Then note that $a^2$ is not $1\mod n.$

So $n$ has prime divisors only $2,3.$

Note that $v_3(n)\le 1$ else let $$a\equiv 2\pmod 3^{v_3(n)}.$$ 

Also $v_2(n)< 4$ else let $$a\equiv 3\pmod 2^{v_2(n)}$$

So the possible values of $n$ are $\boxed{2,3,4,6,12,24}$ for which we can verify.

Motivation: Everything was simply motivated with the fact that $3|2^2-1$ where as other primes are just too big. CRT motivation was also the same, like I really wanted to construct $a.$
Case 2 still works because of the uniqueness of CRT..

[2011 C1]: Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0, 2^1, \cdots, 2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.

Determine the number of ways in which this can be done.

Solution: We will use induction. Denote $f(n)$ the number of distinct process.

We claim that the answer is $f(n)=(2n-1)!!.$ And by induction we will prove that $$f(n)=(2n-1)f(n-1).$$ Base case is true clearly.

Let the blocks be $$2^0,2^1,2^2,\dots 2^{n-2},2^{n-1}$$

Claim: At any step, the weight can never be equal.

Proof: Say $$\text{ left pan:} 2^{x_1}+\dots+2^{x_k}, \text{ Right pan:}2^{y_1}+\dots+2^{y_l}$$

And $x_1<\dots<x_k$ and $y_1<\dots<y_l.$

We compare both the sides with $v_2.$ Note that $v_2(\text{ left pan})=x_1$ and $v_2(\text{ right pan})=y_1.$

So sum of weight in left pan is never equal to sum of weight in right pan, as $x_1\ne y_1.$

Hence we can place the block $2^0$ in any of the pans and it will satisfy the condition. Since adding $1,$ we will have still have at least equality of both the pans. 

Now consider the blocks $$2^1,2^2,\dots 2^{n-2},2^{n-1}$$

Then the no of distinct process to arrange these blocks in $(2n-3)!!.$ Note that everything is same but multiplied by $2.$

Now consider any process, we have $2\times n-1$ ways to include the $2^0$ block( As if we are including $2^0$ in the first step then it must be on the left side).

Motivation: The answer was easy to guess. I think $2^0$ was very special and once you notice that you can place $2^0$ in both of the pans, induction then follows.

[2000 N4]: Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.

Motivation: The main motivation was that $$a+1|a^m+1|(a+1)^n.$$

But then by zsigmondy theorem, which states,

There’s a prime dividing $a^n + b^n$ and not $a^k + b^k$ for k < n with one
exception $(a, b, n) = (2, 1, 3), a>b>0$

Solution: By the second form of Zsigmondy theorem, we get that there exists a "new" prime factor of $a^m+1$, $p$ such that $(p,a+1)=1,$ but then $$a^m+1| (a+1)^n\implies p|(a+1)^n.$$

Now, let's see the exception case. Note that there is three exception case.

Case1: $a=1\implies (1,m,n)\text{ works}$

Case2: $m=1\implies (a,1,n)\text{ works}$

Case3: $3,2$ case. We get that $$(2,3,n)\implies 2+1|2^3+1=9|(2+1)^n.$$

Our solutions are $\boxed{(a,1,n),(1,m,n),(2,3,n);n>1}.$

[2018 C1]:Let $n\geqslant 3$ be an integer. Prove that there exists a set $S$ of $2n$ positive integers satisfying the following property: For every $m=2,3,...,n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$.

Solution:Let $F_1+F_2+\dots+F_n=S_n.$ Then consider set $\{F_1,F_2,\dots,F_{2n-2},F_{2n-1},S_{2n-3}+F_{2n-1}-F_{2n-2}=S_{2n-3}+F_{2n-3}\}.$

Then consider $$A_2=(S_{2n-3}+F_{2n-1}-F_{2n-2})+(F_{2n-2})=S_{2n-3}+F_{2n-1}.$$

And $$B_{2n-2}=F_1+F_2+\dots+F_{2n-3}+F_{2n-1}.$$

Then $$A_3=(S_{2n-3}+F_{2n-1}-F_{2n-2})+(F_{2n-3})+(F_{2n-4})=S_{2n-3}+F_{2n-1}$$

And $$B_{2n-1}=F_1+F_2+\dots+F_{2n-5}+F_{2n-2}+F_{2n-1}$$

Similarly, to get $A_4$ we replace $F_{2n-4}$ with $F_{2n-5}+F_{2n-6}.$

And so on.

Our general $A_m$ construction is,

$$A_m=(S_{2n-3}+F_{2n-1}-F_{2n-2})+F_{2n-3}+F_{2n-5}+\dots+F_{2n-2m+1}+F_{2n-2m}$$

Motivation: It's kind of like "process thing." The main motivation was like $F_{n+1}=F_n+F_{n-1}.$

So, I can change $F_{k+1}\rightarrow F_k+F_{k-1}$ increasing $m$ but the sum is constant.

[2003 G3]:Let $ABC$ be a triangle and let $P$ be a point in its interior. Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively. Suppose that\[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\]Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$. Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$.

Solution: We begin with a lemma.

Lemma: Let $I,J,K$ be the $A,B,C$ excentre. And $D,E,F$ be the feet from $I,J$ on $BC,AC.$ Let $P=ID\cap JE.$ Then $P$ is the circuemcentre of $IJK.$

Proof: Note that $\angle CID=C/2=\angle CJE\implies P\text { lies on the perpendicular bisector of } IJ$ Then $\angle JPI=180-C.$

But note that $$\angle BIC=B/2+C/2, \angle AJC=C/2+A/2\implies JKI=180-(B/2+C/2+C/2+A/2)=B/2+A/2=90-C/2=\frac{1}{2}\angle JPI.$$

Hence $P$ is the circumcentre.

Now, note that $$AP^2+PD^2=AF^2+PF^2+BP^2-BD^2\implies AF^2+PF^2-BD^2=PE^2\implies $$

$$AP^2-BD^2=PE^2\implies AP^2-BD^2=AP^2-AE^2\implies BD=AE.$$

Similarly, we get $CD=EF, BF=CE.$

Hence $D,F,E$ are the excircle touching points. In other words, $P-D- \text{ excentres }$ are collinear.

Using the lemma, we are done!

Motivation: Seeing the condition, I instantly felt Apollonius theorem vibes. Although it didn't work, I went in the same direction. Getting $CD=EF$ finishes the problem as the lemma is well known.. :P

And then one can first find out properties about $P$ and then can try to prove $CD=EF, BF=CE.$

Now, let's get into my Sharygin 2022 Solutions!! I did 12 complete :P.

Sharygin P13: Eight points in a general position are given in the plane. The areas of all 56 triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols "+" and "−" between them in such a way

that the obtained sum is equal to zero.

Proof: We begin with a lemma.

Lemma: Given $ABCD$ a quadrilateral, we consider areas of the triangle. Then it is possible to insert the symbols $+$ and $-$ between them in such a way that the obtained sum is equal to zero.

Proof: Note that $[ABD]+[ABC]=[ACD]+[ABC].$ So assign $[ABD],[ABC]$ as positive and $[ACD],[ABC]$ negative.

Claim: Given $8$ points $ABCDEFGH,$ we can choose $17$ tuples of four points, such no two tuples share three points.

Proof: Just consider the following contruction.

$$ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$$$$BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH $$

Let the eight points be $A,B,C,D,E,F,G,H.$ Then consider the following quadrilaterals$$ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$$$$BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH. $$Then note that by Lemma we can assign sign to the areas of each of the triangles of any of quadrilateral such that sum is $0.$ Now note since no to quadrilateral share three vertices, we get that$$[ABCD]+[ABEF]+[ABGH]+[ACEG]+[ACEF]+[ACFH]+[ADEH]+[ADFG]+$$$$[BCEH]+[BCFG]+[BDEG]+[BDHF]$$$$+[CDEF]+[CDGH]+[EFGH]=\text{ Sum of all } 56 \text{ triangles.}$$Let the $56$ triangles arranged in a row be $T_1,T_2,\dots T_{56}.$ So the sign of $T_1$ is positive, then we can proceed by noticing where $T_1$ belongs to from the $8$ quadrilateral i.e $ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG$

$BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH. $

And then proceed with assigning the signs ( using the two above lemmas).

Sharygin P11: Let $ABC$ be a triangle with $\angle A=60^o$ and $T$ be a point such that $\angle ATB=\angle BTC=\angle ATC$. A circle passing through $B,C$ and $T$ meets $AB$ and $AC$ for the second time at points $K$ and $L$.Prove that the distances from $K$ and $L$ to $AT$ are equal.

Proof: Note that $\angle ATB=\angle CTA=120\implies \angle CAT=60-\angle TAB=\angle TBA\implies \Delta ATB\sim\Delta CTA.$

Hence$$\frac{AT}{TB}=\frac{CT}{AT}.$$Now, since $BCTLK$ is cyclic. So $\angle BLC=180-\angle BTC=60\implies ALB\text{ is equilateral. }\implies AL=AB.$

Similarly $AK=AC.$

Define $G,N,M,I$ as the feet of altitude to $AT$ from $K,C,B,L$ respectively.

Note that$$\frac{KG}{BM}=\frac{CN}{LI}=\frac{AC}{AB}$$Note that$$\frac{KG}{BM}\times\frac{CN}{LI}=\left(\frac{AC}{AB}\right)^2$$Lemma:$$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2$$

Proof: Note that $\Delta CTN\sim \Delta BTM $ as $\angle CTN= \angle BTM, \angle CNT=\angle BMT=90\implies$$$\frac{CN}{BM}=\frac{BT}{CT}$$Note that$$\frac{AB}{AC}=\frac{BT}{AT}=\frac{CT}{AT}\implies \frac{BT}{CT}=\left(\frac{AC}{AB}\right)^2$$But $\frac{CN}{BM}=\frac{BT}{CT}.$

Hence$$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2$$

Now, since$$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2,\frac{KG}{BM}\times\frac{CN}{LI}=\left(\frac{AC}{AB}\right)^2\implies \frac{KG}{LI}=1.$$

Sharygin P15: A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$. Let $I$ be the incenter of $ABC$. Prove that $AI^2 = AD \cdot AE$.

Proof: Define $M_A:= AI\cap (ABC).$

We begin with the following claim.

Claim: $AI$ bisects $\angle DAE$

Proof: We know that $AM_A$ bisects $\angle BAC.$ Hence $M_AB=M_AC.$ Since$$DE||BC \implies M_AD=M_AE \implies AM_a \text{ bisects }\angle DAE $$

Define $B':= AC\cap DE, C':= AB\cap DE.$ Define $G:= (I)\cap DE, J:= (I)\cap BC, H:= (I)\cap AB.$ We state the following lemma, which is true for any tangential quad satisfying the fact that $I\in GH.$

Lemma: $\Delta AIB \sim \Delta AB'I$

Proof: Since$$C'B'||BC \implies \angle BC'I+\angle C'BI=90^{\circ}\implies \angle C'IB=90^{\circ}$$$$\implies  \angle C'BI=\angle C'IH=\angle C'IG.$$Note that $I$ is the $A-$excentre in $\Delta AC'B'.$ Letting $\angle AB'C'=C, \angle AC'B'=B,$ we get$$\angle C'IA=180-\angle AC'I-\angle C'AI= 180-(\angle AC'B'+90-\angle AC'B'/2 )- A/2$$$$=180-(90+B/2)-A/2=C/2= 90-(90-C/2)=\angle GIB'$$$$\implies \angle AIB'=\angle CIG'=\angle ABI $$And note that $AI$ bisects $\angle BAC.$

Hence$$\Delta AIB \sim \Delta AB'I$$

Hence $AI^2=AB'\cdot AB$ and similarly $AI^2=AC'\cdot AC.$

Now we are ready to invert!

Invert wrt $(A,AI)$ followed by reflection wrt $AI.$ Then it's enough to show that $D\leftrightarrow E.$

Now, note that $B\leftrightarrow B', C\leftrightarrow C' \implies (ABC)\leftrightarrow B'C'\implies DE\leftrightarrow (ABC).$

Now let's try to find out where $D$ will go. Note that $D^{*}$ ( the inverted point $D$) would lie on $(ABC).$ Moreover, $AI$ would also bisect $\angle DAD^{*}.$ But $E\in (ABC)$ and $AI \text { bisects } \angle DAD^{*}. $

Hence $D^{*}=E.$ Hence $AI^2=AD\cdot AE.$

Sharygin P16: Let $ABCD$ be a cyclic quadrilateral, $E= AC \cap BD , F = AD \cap BC .$ The bisectors of angles $AFB$ and $AEB$ meet $CD$ at points $X, Y.$ Prove that $A, B, X, Y$ are concyclic.

Proof: Define $K:=EX\cap AB, J:=FY\cap AB.$ We begin with the following claim about $K-E-X$ and $F-J-Y.$

Claim: $K-E-X||F-J-Y.$

Proof: This is simply angle chase. Let $\angle BDC=\theta, \angle ADB=\alpha, \angle ACB=\alpha, \angle ACD=\beta.$

Note that $\angle EDX=90-(\theta+\beta)/2, \angle DFY=90-\alpha-(\theta+\beta)/2.$ And we get that$$\angle EXD=90+\beta/2-\theta/2=\angle FYD\implies FY||EX$$

Define $G:= AB\cap CD.$ We begin with the following lemma.

Lemma: $GX=GK$

Proof: Note that$$\Delta AKE\sim \Delta DXE \implies \angle GXK=\angle GKX$$

Now to prove that $AXYB$ is cyclic, by Power of Point, it's enough to show that$$GA\cdot GB=GD\cdot GC=GX\cdot GY.$$This motivates use to invert wrt $(G, GA\cdot GB).$

Let $K'$ and $X'$ be the inverted image of $K,X$ respectively. Note that after inversion, we will have$$\angle GXK=\angle GKX=\angle GX'K'\implies K'X'||KX.$$So it's enough to show that $K'-X'-F$ collinear and we will be done as by using our \claim, we get $K'=J,X'=Y.$

Define $Q$ as the Miquel point of $ABCD.$ Hence we have $QFBA$ cyclic and we also have $Q\in FG\implies GQ\cdot GF=GA\cdot GB\implies Q\text{ is the inverted image of }F.$

Now, to show that $F-K'-X',$ it is enough to show that $AGKX$ is cyclic.

But first we proove the following lemma.

Lemma: $AK/KB=DX/XC$

Proof: Note that$$\frac{AK}{KB}\cdot \frac{XC}{DX}=\frac{AE}{EB}\cdot \frac{EC}{ED}=\frac{AE\cdot EC}{EB\cdot ED}=1$$

Using the well know fact that there is a spiral similarity centred at $Q$ which take $DC\leftrightarrow AB\implies \Delta AQB\sim \Delta DQC.$

Using the lemma and the similarity, we get that $\angle QKG=\angle QXG.$

This implies that$$ QKXG\text { is cyclic}.$$We are done!

Sharygin P18: The products of the opposite sidelengths of a cyclic quadrilateral $ABCD$ are

equal. Let $B'$ be the reflection of $B$ about $AC$. Prove that the circle passing through $A,B', D$ touches $AC$

Proof: Since$$ AB\times CD=BC\times AD\implies \frac{AB}{BC}\times \frac{CD}{AD}=1\implies (A,C;B,D)=-1.$$So $ABCD$ is a harmonic quadrilateral.

Lemma: Let $ABDC$ be an harmonic quadrilateral, let $H_A$ denote the $A-$ humpty point. Then $H_A$ is the reflection point of $D$ wrt $BC.$

Proof: Let $D'$ be the reflection of $H_A$ wrt $BC.$ Note that$$180-A=\angle BH_AC=\angle BD'C\implies D'\in (ABC)$$$$\angle H_ABC=\angle BCD'=\angle BAD'\implies AD, AH_A \text { isogonal }\implies D'\in \text { symmedian } $$But since $ABDC$ is harmonic, we get that $D$ lies on the symmedian. Hence $D'=D.$

Hence by our Lemma , we get $B'$ as the $A$ humpty point of the triangle. And by our definition of Humpty point, we get that the circle passing through $A,B', D$ touches $AC .$

Sharygin P19: Let $I$ be the incenter of triangle $ABC$, and $K$ be the common point of $BC$ with the external bisector of angle $A$. The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$ . Prove that $\angle BAX = \angle CAY$

Proof: To prove that $\angle BAX=\angle CAY,$ it is enough to show that $\angle XAI=\angle YAI$ as $\angle BAI=\angle CAI.$

Since we have $90$ degree and angle bisectors, by Right angles and bisectors lemma, it's enough to show that $(K,I;X,Y)=-1.$

Projecting from $I_A$ on $BC.$ We get $(K,I;X,Y)=(K,D;B,C).$

But note that $(K,D;B,C)=-1$ as $\angle KAD=90^{\circ},\angle BAD=\angle CAD.$ So by Right angles and bisectors lemma, we get $(K,D;B,C)=-1.$

Sharygin 22: Chords $A_1 A_2 , A_3 A_4 , A_5 A_6$ of a circle $\Omega$ concur at point $O.$ Let $B_i$ be the second

common point of $\Omega$ and the circle with diameter $OA_i$ . Prove that chords $B_1 B_2 , B_3 B_4 , B_5 B_6$ concur.

Proof: Given $A_1A_2,A_3A_4,A_5A_6$ concur at $O.$ Let $A$ denote the centre of $\Omega.$

We present a stronger claim.

Claim: We claim that $AO,B_1B_2,B_3B_4,B_5B_6$ concur.

Note that proving this claim, we will be done.

So, to prove the stronger claim, Enough to show $AO,B_1B_2,B_3B_4$ concur and by symmetry, we are done.

So let's remove $B_5,B_6,A_5,A_6.$

Claim: $A_1B_2,A_2B_1,A_3B_4,B_3A_4$ concur

Proof: Since all of the points lie on a circle, so it's enough to show that$$(A_1,A_2;A_3,A_4)=(B_2,B_1;B_4,B_3)$$But$$(B_2,B_1;B_4,B_3)=\frac{B_2B_4}{B_1B_4}\frac{B_3B_1}{B_3B_2}=(B_1,B_2;B_3,B_4).$$Now, let $L,M,N,P$ be the antipodes of $A_1,A_2,A_3,A_4$ wrt $\Omega.$

Then, note that$$(A_1,A_2;A_3,A_4)\stackrel{A}{=}(L,M;N,P)\stackrel{O}{=}(B_1,B_2;B_3,B_4)=(B_2,B_1;B_4,B_3)$$

Let the concurrency point of $A_1B_2,A_2B_1,A_3B_4,B_3A_4$ be $H.$ Let $\ell$ be the polar of $H$ wrt $\Omega.$

Define $G=A_1B_1\cap B_2A_2,H=B_1B_2\cap A_1A_2 B_1B_2\cap A_1A_2=J.$

Note that, by brokard's theorem, $GJ$ is the polar of $H\implies GJ\in \ell.$ Similarly, $I\in \ell.$

Note that $AH\cap \ell=F$ is the miquel point of the quadrilateral $A_1B_1B_2A_2,A_3B_4B_3A_4.$ ( As $\angle GFA=90$ and miquel point lies on polar. )

By the definition of miquel point, we get $F\in (GB_1B_2).$ But$$\angle OB_1G=\angle OB_2G=90\implies O\in(GB_1B_2)\implies \angle GFO=90\implies F-H-O-A. $$Similarly, we get $F,O\in (IB_3B_4).$

Claim: $B_1B_2,B_4B_3,OF$ concur

Proof: We use radical axis theorem on $(OB_1B_2),(OB_3B_4),\Omega.$

But $OF=OA.$

Hence $AO,B_1B_2,B_3B_4$ concur. And we are done!

I personally loveeddd all the problems I have posted the solutions to :P. My fav was def P22 :P.. JUST SO CLASSSICCCC.. And it's not randomly spammed geometry.. They were really very beautiful and were worth solving!

INMO: Oh is it me who took 1hr for the Geo? Anyways did p1 correctly, I did get a hunch on how to approach ( is it only me who got the hunch? Like taking D as the feet of altitude from A to BC) Anyways, then my p3, I literally fake solved it :(.. T_T.. My p2 mein i had progress but woops, I somehow didnt notice mod 3( cause i got N=5 doesnt work).. I approached with v2 method, but idk, I think I will get 17+ 1 or 2 max.. Congrats to everyone!!

Sunaina