Challenging myself? [Jan 15-Jan 27]

Ehh INMO was trash. I think I will get 17/0/0/0-1/3-5/10-14, which is def not good enough for qualifying from 12th grade. Well, I really feel sad but let's not talk about it and focus on EGMO rather. 

INMO 2023 P1

Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square.

I will use Atul's sol, cause it's the exact same as mine. 

Proof: Consider the graph $G$ induced by the elements of $S$ and edges being if the products are perfect squares. Note that if $xy = a^2$ and $xz = b^2$, then $yz = \left( \frac{ab}{x} \right)^2$, since its an integer and square of a rational number its a perfect square and so $yz$ is an edge too. So the graph is a bunch of disjoint cliques, say with sizes $c_1, c_2, \cdots, c_k$. Then $\sum_{i=1}^k c_i^2 = 2023$, which implies $k \geqslant 4$ since $2023 \equiv 7 \pmod 8$. Now pick one element from each of these cliques and it works.

I am starting with Harmonic DGX: 

A few stuffs to note for me.

2016 G2

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.

Proof: Define $T$ as the mixtillinear in touch point. Define $S$, $P$ as the midpoint of major arc, minor arc $BC$. 

Claim: $BFIT, CEIT$ harmonic

Proof: Introduce the midpoint of arc $AB, BC$ ( say $M_c,M_b$) and angle chase. We have $M_cB,M_cI$ tangent to $BFIT$. And we have $T-F-M_c$.

Define $H=ST\cap BC$.

Claim: $AH,CF,BE$ concur

Proof: By ceva, we just have to show $$\frac{BF}{CE}=\frac{BH}{CH}.$$

For which, we use the fact that $BFIT, CEIT$ are harmonic.

Claim: $SX,EF,BC,TP$ concur

Proof: Note that $(Z,H;B,C)=-1$ where $Z=EF\cap BC$. Moreover, we know that $$\angle BTH=HTC$$ and $\angle HTP=90\implies Z-T-P$.

As $SH\perp ZP, ZH\perp PS\implies X-H-P$.

Since $(S,P;B,C)=-1$, we get $(X,T;B,C)=-1$. 

Now define $A'$ as the intersection of the parallel line through $A$ to $BC$ and $(ABC)$. 

Claim: $T-D-A'$

Proof: We have it because $$\angle BTA=CTD$$ and $BA=CA'$. 

Note that $(AM\cap (ABC),A';B,C)=-1$ ( just reflect over perpendicular bisector of $BC$) . But we have $(X,T;B,C)=(XD\cap (ABC),A';B,C)=-1$. So $ XD\cap (ABC)=AM\cap (ABC)$.

This was a pretty nice problem (sort of config based).. but I would recommend to go through it in future, cause pretty nice config geo!

USA EGMO TST 2023 P1

Let $ABC$ be a triangle with $AB+AC=3BC$. The $B$-excircle touches side $AC$ and line $BC$ at $E$ and $D$, respectively. The $C$-excircle touches side $AB$ at $F$. Let lines $CF$ and $DE$ meet at $P$. Prove that $\angle PBC = 90^{\circ}$. 

Proof: If $BC=a$. Note that $CE=CD=\frac{AB+BC+CA}{2}-C=a$. So $C$ is the centre of $(BDE)$ and hence $\angle BDE=90$. So $C$ is midpoint of $BE$. 

Define $I$ as the incentre of $ABC$. Let the incircle touch $CA$ at $M$, $BC$ at $L$. Let $M$ antipode be $N$ and $L$ antipode be $J$. 

Claim:$B-N-J-D$ collinear ( By symmetry we also get $C-J-F$)

Proof: By homothety, $B-N-D$. Note that if $\angle BCA=2\theta$, we get $\angle BED=\theta\implies CI||EA$.

We have $\angle EBD=90-\theta$ and $\angle ILM =\theta\implies \Delta LMJ \text{ is similar to } \Delta EDB$. 

Note that $\angle MJN=90$. Since $JL\perp BC\implies \angle LJB=90-\angle LBJ=\theta$. And $\angle LJN=\angle JLM=\theta\implies J-N-B$.

Define $P'$ as the intersection of the perpendicular at $B$ to $BC$ and $CE$.  Define $C'=CI\cap BC$. Since $LJ||BP'$ and $I$ is midpoint of $NJ\implies C'$ is midpoint of $BP'$. 

Now define $Y$ as midpoint of $NJ$. Note that $C-I-Y-C'$ ( as $CI$ is perpendicular bisector of $LM$ and $LMNJ$ rectangle).

Note that $\Delta BCY$ is similar to $\Delta BED$. So $E-D-P'$. So $P'=P$. And we are done! 

USAJMO 2011 P5

Points $A,B,C,D,E$ lie on a circle $\omega$ and point $P$ lies outside the circle.

The given points are such that

(i) lines $PB$ and $PD$ are tangent to $\omega$,

(ii) $P, A, C$ are collinear,

and (iii) ${DE} \parallel {AC}$.

Prove that ${BE}$ bisects ${AC}$.

Proof: Note that $$(A,C;B,D)=-1\text { projecting through } E \text { on } AC\implies (A,C; BE\cap AC, P_{\infty})=-1.$$ Hence  $BE\cap AC$ is the midpoint of $AC$. Hence $BE$ bisects $AC$.

APMO 2013

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of ${AC}$ such that ${PB}$ and ${PD}$ are tangent to $\omega$. The tangent at $C$ intersects ${PD}$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between ${AQ}$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.

Proof: Note that $$-1=(A,E;C,D)=(A,RE\cap \omega; C,D)$$ But we know $$(A,C;B,D)=-1\implies RE\cap \omega=B.$$

China TST 2002

Let $ABCD$ be a quadrilateral. Point $E$ is the intersection of lines $AB$ and $CD$

while point $F$ is the intersection of lines $BC$ and $DA$. The diagonals of the quadrilateral meet at $P$, and point $O$ is the foot from $P$ to $\overline{EF}$.

Prove that $\angle BOC = \angle AOD$.

Proof: Let $DB\cap FE=X, FP\cap DC=Y$. Note that $$-1=(D,C;Y,E)=(D,B;P,Y)\implies PO\text{ bisects } \angle DOB.$$ Similarly we get $PO \text{ bisects } \angle AOC.$ Hence  $\angle BOC = \angle AOD$.

ELMO Shortlist 2019 G1

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$.

Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$.

Prove that $BQ$ bisects segment $\overline{EF}$.

Proof: Let the parallel line through $B$ to $EF$ intersect $(ABC)$ at $X.$ Let $EF\cap BC=T, TA\cap (ABC)=R.$ Then note that $(AREF), R-E-X.$ 

So $$-1=(B,C;T,D)\text{ projecting through } A \text { on } (ABC)$$ $$\implies -1= (BC; RP)\text{ projecting through } A \text { on } (ABC)\text{ projecting through } E \text { on } (ABC)$$ $$\implies (BE\cap (ABC)A;XQ)=-1\text{ projecting through } E \text { on } (ABC)\implies  (EF;\infty_{EF}BQ\cap EF)=-1.$$

We have $$FM\cdot FX=MA\cdot MD=MY\cdot MF\implies EYXF\text{ is cyclic}.$$

The following problem was done with Atul and Archit :D

SAGF part 1 P2

In an acute-angled triangle, $ABC$ $w$ is a circumscribed circle, and $O$ is inside the triangle and $OB = OC$. The point $D$ is selected on the side of $AB$,so that $OD \parallel BC$. The straight line $AO$ repeatedly intersects the circles $w$ and $(COD)$ at the points $A',P$, respectively. $M$ is the middle of the side of $BC$. Circle $(MAA')$ intersects the line $BC$ at the point $L$. The straight line $AL$ repeatedly intersects the circle $w$ at the point $K$. Prove that $\angle APK = \angle AA'M$

Proof: We start with the following lemma.

Lemma:Let $ABC$ be a triangle, $O$ be a point on the perpendicular bisector of $BC$. Let $D,E$ denote the intersection of line parallel $AB,AC$. Define $P=(DOC)\cap (EOP).$ Then points $A,O$ and $P$ are collinear.

Proof: Invert about $O$. The inverted problem is: Give $ABC$ triangle with $O$ be a point on the perpendicular bisector of $BC$. Let $D,E$ denote the intersection of line parallel $(AOB),(AOC)$. We have to show that $AO, CD,BE$ concur.

Note that as $OB=OC$ we have$$\angle DAB=\angle DOB=\angle EOC=\angle EAC.$$Define $\ell_1$ as the reflection of $DB$ over the angle bisector of $\angle B$ and $\ell_2$ as the reflection of $CE$ over the angle bisector of $\angle C$. We claim that $\ell_1 \parallel \ell_2 \parallel OP$

To see this,$$\angle AOD=\theta=\angle ABD\implies \angle AOE=180-\theta=\angle ACE.$$Let $X$ be on $\ell_1$ such that $D$ and $X$ are on opposite sides of $AB$. Similarly, define $Y$ be on $\ell_2$ such that $E$ and $Y$ are on opposite sides of $AC$. So note that$$\angle CBX=\theta,\angle BCY=180-\theta\implies \ell_1 \parallel \ell_2.$$

Note that $\angle ABX=\angle B+\theta$. And$$180-\angle OAB-180-\angle ODB=\angle DBC=\angle B+\theta\implies OA||l_1||l_2.$$So $BX\cap CY=P_{\infty}, A-O-P_{\infty}\text{ collinear}.$ By Jacobi theorem, we get $A-O-P_{\infty},CD,BE$ concur. Hence, we get $A-O-P$ collinear.

Redefine the problem in the following way: Define trinagle $ABC$, $O$ be a point on the perpendicular bisector of $BC$. Let $D,F$ denote the intersection of line parallel $AB,AC$. Define $P=(DOC)\cap (EOP).$ Define $M$ as the midpoint of $BC$. Define $I=AB\cap (DOCP), H=AC\cap (BOFP)$. Define $L'= BC\cap IH$. Define $E=AO\cap BC, A'=AO\cap (ABC)$. We will shwo that $L',A',M,A$ cyclic i.e $L'=L$. Note that $(L',E;B,C)=-1$. And hence we have$$ME\cdot EL'=EC\cdot EB=AE\cdot EA'\implies L'MAA'\text{ is cyclic}\implies L'=L.$$

Due to this, we get that $K$ is the miquel point of $BCHI$(as $K=(ABC)\cap AL$). Since $K$ is the miquel point, we have $(ABCK),(AIHK),(KCHL)$ cyclic. We actually don't need all the cyclicities, but we'll prove them because why not.

Note that $(B,C;E,L)=-1\implies LB\cdot LC=LM\cdot LE.$ But $LB\cdot LC=LA\cdot LK\implies AKEM$ is cyclic.

$P\in (AKHI)$ as$$\angle IPH=\angle IPO+\angle OPH=\angle ADO+\angle AFO=180-\angle A\implies P\in(AKI).$$

Define $G=DF\cap AL$, then $KFGH$ cyclic, this is because$$\angle KGF=\angle KLC=\angle KLB=\angle KIB=\angle KHC.$$The last equality follows as $K$ is the spiral center taking $BI$ to $CH$.

So by PoP, we get$$AK\cdot AG=AF\cdot AH=AO\cdot AP=AD\cdot AI$$which gives $KDIG,KGOP$ cyclic.

$EKLP$ cyclic as$$\angle KLE=\angle KLB=\angle KIB=\angle KIA=\angle KPA=\angle KPE.$$

$AKME$ cyclic as$$LM\cdot LE=LB\cdot LC=LA\cdot LK\implies AKME\text{ is cyclic}.$$

Now to finish proving $\angle APK=\angle AA'M$, note that$$\angle APK=\angle AIK=\angle BIK=\angle BLK=\angle MLA=\angle AA'M$$so we're done.

GOTEEM 2020

Let $ABC$ be a scalene triangle. The incircle of $\triangle ABC$ is tangent to sides ${BC}$, ${CA}$, ${AB}$ at $D$, $E$, $F$, respectively. Let $G$ be a point on the incircle of $\triangle ABC$ such that $\angle AGD = 90^\circ$. If lines $DG$ and $EF$ intersect at $P$,prove that $AP$ is parallel to $BC$.

Proof: Let $D'$ be the antipode of $D$ wrt incircle. Note that $$(D',G;E,F)=-1\text { projecting through } D \text { on } EF\implies (DD'\cap EF, P;E,F)=-1.$$ Let $M$ be the midpoint of $BC$. Define $X=EF\cap BC$. We have $EF,DD',AM$ concur at say $R$.  Projecting through $R$ on $BC$,  $$-1=(R,P;E,F)=(M,PA\cap BC;B,C)\implies PA||BC.$$

AHH I have more to post but I will post in different blogpost! Bie people!