Geos are my life support! ft life update

Just a compilation of $10$ very nice and hard Geo problems and solutions :P. Without diagrams ( cause I am a lazy person).

Problem 1[ China TST]: Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.

Proof: 

Define $S=AC\cap EF,~~T=FE\cap BD.$

Note that $$(E,R;D,C)=(S,P;A,C)=-1 .$$

Since $\angle POS=90,$ by Right Angles and Bisectors harmonic lemma, we get $OP$ bisecting $\angle COA.$ 

Similarly , we get $$((B,D;P,T)=-1. $$ Since $\angle POT=90,$ by Right Angles and Bisectors harmonic lemma, we get $OP$ bisecting $\angle BOD.$ 

Problem 2[ISL 2002]: The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.

Proof: 

• Note that $(B,C;K,Z)=-1.$
•   Note that $(A,D;M,A_{infty})=(J,K;N,K')=-1$
• but we have $(J,K;E,F)=-1.$ So find out tangeny, conclude.

Problem 3[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 $$(A,C;B,D)=-1\implies RE\cap \omega=B$$

Problem 4[ISL 2001]: Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?

Proof: Let $\angle ABQ=x$, so $\angle ABC = 2x$.Also let $QB=\sin 60.$ ( Scaling won't affect the  

Here we have, $AB+BP=AQ+QB.$ We will Note that, by angle bisector theorem we have $AQ=frac{AB}{AB+AC}\cdot BC.$

Now we find all the lenghts in terms of $\sin.$

Using $\sin$ law in $QBA,$ we get,

$$\frac{QB}{\sin 60}=\frac{AB}{\sin (120-x)}=\frac{AQ}{\sin x}.$$

Since we had $QB=\sin 60\implies AB=sin(120-x),~~AQ=sin(x).$.

Now, we find $AC,BC.$ Again using $\sin$ law on $\Delta ABC,$

we get $$\frac{BC}{\sin 60}=\frac{AB}{\sin(120-x)}=\frac{\sin(120-x)}{\sin(120-x)}\implies BC=\frac{\sin(120-x)\cdot \sin 60 }{\sin(120-2x)}$$

Similarly, we get $$ \frac{AC}{sin (2x)}=\frac{AB}{\sin(120-2x)}\implies AC=\frac{\sin(120-x)\cdot sin(2x)}{\sin(120-2x)}.$$

Now putting all the values we get

$$sin\left(60\right)+sin\left(x\right)=sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right) $$

Simplifying the LHS, we get

$$sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right)$$

$$=\sin(120-x)+\frac{\left(sin\left(120-x\right)\left(sin\:60\right)\right)}{\left(sin\left(120-2x\right)+sin\left(2x\right)\right)}$$

$$=sin\left(60+x\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right)$$

Note that $\sin(60+x)=2\sin(\frac{60+x}{2})\cos(\frac{60+x}{2}).$

 Also simplifying the RHS, we get $sin\left(60\right)+sin\left(x\right)=2\sin(\frac{x+60}{2})\cos(\frac{x-60}{2}).$

Since LHS=RHS we cancel out $\sin(\frac{x+60}{2})\ne 0.$

We get $$cos\left(\frac{x}{2}-30\right)=cos\:\left(\frac{x}{2\:}+30\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right) $$

So we have 

$$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$

Note that here RHS is

$$ 1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$

$$=1+\frac{sin\left(60\right)}{\left(2sin\left(2x+30\right)+cos\left(30\right)\right)}$$

$$1+\frac{1}{2sin\left(2x+30\right)} $$

So we have $$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=\frac{1+sin\left(2x+30\right)}{2sin\left(2x+30\right)}$$

Cross multiplying, we get 

$$cos\left(\frac{x}{2}-30\right)2sin\left(2x+30\right)$$
$$=cos\:\left(\frac{x}{2\:}+30\right)+2sin\left(2x+30\right)cos\:\left(\frac{x}{2\:}+30\right) $$

$$ =sin\left(\frac{5x}{2}\right)+sin\left(60+\frac{3x}{2}\right)=cos\:\left(\frac{x}{2\:}+30\right)+sin\left(\frac{5x}{2}+60\right)+sin\left(\frac{3x}{2}\right)$$

Now taking $x/2=y,$ we get 

$$\sin(5y+60)+\sin(60-y)+\sin(3y)$$ $$=\sin(60+3y)+\sin(5y)\implies \sin(5y+60)-\sin(5y)+\sin(60-y)=\sin(60+3y)-\sin(3y).$$

So, we have,

$$\cos (5y+30)+\sin(60-y)=\cos(3y+30)\implies \sin(60-5y)+\sin(60-y)=\cos(3y+30)$$ $$\implies 2\cdot \cos(3y+30)\cos(60-2y)=\cos (3y+30). $$

Here $\cos(6-2y)=1/2$ is eliminated, so $\cos(30+3y)=0\implies y=20=x/2\implies \angle ABC=2x=80\implies \angle ACB=40.$

Problem 5[ISL 2011]: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.

Proof: Redefine $F=(AEI)\cap AC,~~(BID)\cap BC=G.$ We will show $E-F-G-D.$

We will use angles to prove.

Note that $$A/2=\angle IAF=\angle IEF=\angle BED=\angle IED\implies F\in ED. $$

Also $$B/2=\angle IBC=\angle IDG\implies G\in ED.$$

Again note that $$ \angle EAF=\angle EIF\implies IF||BC.$$

Similarly we can show $IG||BC$

Also by angle chase note that $$\angle IFG=\angle IGF\implies IF=IG$$

And angle chase gives $CF=CG.$

\\

Since $CI$ is the angle bisector. Hence $\Delta IFG\cong \Delta CFG.$

\\ 

Now define $J=(AEIF)\cap (IBGD).$ Also consider $P.$

So $$\angle IFP=180-\angle AFI-\angle PFC=180-180-\angle AEI-\angle IAC  $$ $$=180-C-A/2=B+A/2.$$

Similarly we get $$\angle IGP=180-C-B/2=A+B/2$$

So $\angle P=360-(B+A/2)-(B/2+A)-C=180-A/2-B/2.$

So $$\angle FJG=360-(180-\angle FJI)-(180-\angle IBG)=\angle FJI+\angle IBG=A/2+B/2. $$

Hence $(JPFG).$

Then we will show $I-J-P$ collinear.

Note that $\angle BAD=\angle PFC=A/2.$

But due to congruency stuff, we have $$\angle GFC=B/2+A/2\implies \angle GFP=B/2\implies GJP=B/2$$ But $\angle IJG=180-B/2.$

Clearly by radical axis we have $AE,BD,I-J-P$ concurrent. Now, enough to show $I-J-K$ collinear.

We will show $AJBK$ cyclic.

For this note that $\angle AKB=180-2C.$\\

Now note that $$ \angle AJB=\angle AJI+\angle BJI=\angle AEI+\angle IDB=2C.$$ Hence $AJBK$ cyclic.

Also $\angle IJB=C=\angle KAB=\angle KIJ\implies K-I-J.$

And we are done!

Problem 6[USATST 2019]: Let $ABC$ be an acute triangle with circumcircle $\Omega$ and orthocenter $H$. Points $D$ and $E$ lie on segments $AB$ and $AC$ respectively, such that $AD = AE$. The lines through $B$ and $C$ parallel to $\overline{DE}$ intersect $\Omega$ again at $P$ and $Q$, respectively. Denote by $\omega$ the circumcircle of $\triangle ADE$.

a. Show that lines $PE$ and $QD$ meet on $\omega$.

b. Prove that if $\omega$ passes through $H$, then lines $PD$ and $QE$ meet on $\omega$ as well.

Proof: 

Part a. Let $EP\cap \Omega=Z,~~D'=ZQ\cap \omega.$

Note that $PC=QB.$

We get$$\angle PZQ=\frac{\widehat{PC}+\widehat{QC}}{2}=\frac{\widehat{QB}+\widehat{QC}}{2}=A$$Hence $Z-D'-Q\implies D'=D.$

Part b. Define $H_c,H_b$ as the reflection of orthocentre wrt $BA, AB.$

Claim: $H_b-D-P$ and $H_c-E-Q.$

Proof: Note that$$\angle AH_bD=\angle DHA=\angle DEA=\angle ADE=\angle ABP=\angle AH_bP\implies H_b-D-P.$$Similarly, we get $H_c-E-Q.$

Let $G=EQ\cap DP.$

Now, clearly $\angle H_cHH_b=180-\angle .$ So enough to show that $H_bHH_cG$ is cyclic.

So enough to show that$$\frac{\widehat{PQ}}{2}=\angle GH_bH=\angle HH_cE=\frac{\widehat{BQ}}{2},$$which is true.

Problem 7[ISL 2004]: Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$.

Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.

Proof: Let $K=EG\cap AB, I= BC\cap GE, H=AE\cap GB.$ By pascal on $EEBCAG\implies D-F-I.$ Then by brokards on $AEBG,$ we get that $AB\perp HF$ but $AB\perp DF\implies H\in d.$

Now, let $G'=CF\cap \Gamma.$ By angle chase, we get$$\angle CFH=\angle GBC=\angle GG'C\implies GG'||d$$and we are done.

Problem 8[EGMO 2012]: Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)

Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Proof: Simple angle chase!

Let$$\angle OCB = \theta \implies \angle COB=180-2\theta\implies \angle BAC= 90-\theta \implies FKE =180-2\theta .$$We also have$$\angle EDC=90-\theta , \angle FDB=90-\theta \implies \angle FDE=2\theta \implies (FKED).$$

As$$FK=KE\implies \angle FDK=\angle EDK \implies \angle KDE= \theta \implies KD\perp BC.$$

Problem 9[EGMO 2012]: Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$. Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$. Let $L$ be the reflection of $K$ in the line $AB$, and let $M$ be the reflection of $K$ in the line $BC$. Let $E$ be the second point of intersection of $\Gamma $ with the circumcircle of triangle $BLM$.

Show that the lines $KH$, $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Proof: Note that $B$ is the circumcentre of $(KLM).$ So we have $\angle KEM=\angle BEM.$ Let $KL\cap AB=Z.$

Note that$$\angle BMK=\angle BKL=\angle BMK+90-\angle KBZ=90-\angle BMC+90-\angle KBZ=180-(\angle CBK+\angle KBZ)=\angle B. $$

So$$ \angle MBK=2\angle B\implies \angle BEM=90-\angle B.$$

Now let $H_A$ be the reflection of $H$ over $AB.$ Clearly $H_AM, HK, BC$ concur.

So enough to show that $H_A-M-E.$ Note that $\angle H_AEB=\angle H_ACB=\angle HCB=90-\angle B.$ Hence $H_A-M-E.$

Problem 10[Iran TST 2018]: In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively. The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$ lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$.

Proof: Note that $BPQC$ is cyclic. Now define$$ J=PC\cap BQ,~~J'=PB\cap CQ.$$

Claim: $PJ'QJ$ is $\omega.$

To show this it's enough to show that, $PM,~~PQ$ tangent to it at points $P,Q.$

Note that$$\angle JPM=\angle PCM=\angle PQB\implies PM\text{~~is tangent to~~} J'PJQ\text{~~at~~} P. $$Similarly we get $QM$ is tangent to $J'PJQ$ at $Q.$

Hence $PJ'QJ$ is $\omega.$

Now, consider the parallel line through $J$ to $BC.$ Let it intersect $MP,~~MQ$ at $X',Y'$ respectively.

Note that$$ \angle PJ'X'=\angle PCM=\angle MPC=\angle X'PJ\implies X'PJ\text{~~is isosceles~~}$$$$\implies XJ'\text{~~is tangent to}\omega.$$Similarly, we get $Y'JQ$ isosceles $\implies Y'J $ is tangent to $\omega.$ Angle chase gives us $X'-J-Y'.$

Hence if we show $B-X'-F$ and $C-Y'-E,$ we will be done.

Claim: $B-X'-F$ are collinear

Note that $B-K$ is the polar of $C$ by brokards. ( where $K=PQ\cap JJ'$)

Also, note that $C\in $ polar of $X'$ wrt $\omega.$ So by la-hire, we get $X'$ lying in the polar of $C.$ But we also know that $F$ is the tangency point of $C$ to $\omega.$ Hence $F$ lies in the polar of $C.$

Hence $B-X'-K-F.$ Similarly, we get $C-Y'-K-E.$ Hence $X'=X,~~Y'=Y.$

And we are done!

Yee that's it for today's blog! How were they? 

I think I should do more hard  ISL G's and EGMO geos and post some solutions here! Sorry for those who were expecting a GT blogpost, I kinda procrastinated with GT today :pleading:...

Also, lemme add my life update. I gave IGO and EGMO TSTs. IGO went okayish and EGMO TST was literally very bad and I genuinely feel I underperformed. We basically had it on Nov 21 and Nov 28. I was so upset about the TSTs that I didn't even do the math on my own for like 3 weeks regretting how bad my TSTs went. I think I would blame my thought process on that two tst days. My brain wasn't trying much and arghhh.. it's just bad.  What happened was, I was not brave about my claims. One has to be brave to make claims and believe that it would be right. Anyways, there are loopholes that I need to fix. I hope nobody goes through what I went through after TSTs.. But then it was a new experience. I did learn many new things and did a whole bunch of ISLs!!! 

More details: Day 1 was bad. Okay sorry. I should say horrible. I knew P1 had weighted am-gm but then I couldn't find the right way. I tried P1 for like 2 hours and then tried P2 for 2 hours which was very interesting ( was proposed by Rohan bhaiya and Pranjal Bhaiya) but then the progress wasnt nice. I at the end had 30 mins for P3 where I misread the problem. I did notice that it uses the EGMO lemma and stated it, but didn't help much. The solution to P3 was incredible. As Atul says, black magic. 

After the tests, I came to know that had I attached my rough files for P1, I would have got 5 marks or even more, I was just so close. :( ( I got 0 in P1).

Day 2 would have gone better had it been day 1. But I sort of had this mindset of " I am the dumbest creature" and it went bad. I tried the combo and did get progress on that. But again, that progress was considered useless. 
P2 which was pretty nice geo which I should say was solvable by me, had I been brave enough. I guess the concurrency point but then $AI\cap BC$ point was the game changer ( which I didn't add). I didn't try P3 much. 

That's my experience. Not a valuable one but yeah..At the end, I think having a calm mind really helps! 

Moral of the story: Try geos.

Though, we, EGMO TST girls are invited with the four EGMO team members to the EGMO training camp which is like 3 months long!!! I am so excited!!
Fun fact: I know all the EGMO 2022 team members personally :P and all of them are just so proooo and kind and sweet. 

Sunaina💜