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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).
AsFK=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💜

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