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Announcing the Unofficial TC 2022! ( Month 1 problem solving )

So it's been 1 month since the INMO results were out! So here's my past 1 month's journey into problem-solving. 

There hasn't been much in my life, although the Sharygin correspondence round results came out! And I qualified for the final round. I just made the cutoff though. I got 85 marks and the (unofficial) cutoff was 84 marks. I am one of the seven students selected for the final round and the only kid from India in my grade! They, however, are not inviting Indian students in the final round and have asked us to conduct the final round in India if an organization agrees.

Monthly reflection:

  • Till May 16, I tried the Awesome Math Application ( which I got accepted into woohoo! And I am taking courses in the second and third season hehe)
  • Then I was completing 108 algebra problems along with MBL problems till June 1
  • June 1- June 10: Since I was feeling very guilty for not doing enough problems from the Sophie psets, I did a few Sophie psets problems. BTW Thankyou so much to Ananda, Atul and Sanika di for working so hard in making and grading our Sophie Problem Sets. 
  • I also had my WeMP in Sophie fellowship ( Weekly Math Presentation)
  • We also successfully wrapped up the third season of my math club, called The Philomath club. We had over 20 classes which were so exciting and amazing. And we had over 120+ kids. It was such an amazing experience! 
  • I and my friends are actually working on a secret project! (Hopefully, I can reveal it in my next blog post)
  • I also had my first unit tests and realised I have been quite not into school stuff. For example, I have realised focusing on non-math stuff is hard.
  • I also had my first OMC class on projective geo!! I think it went pretty nice and I had my part 2 on July 3rd! So yeyy that's niceeee!!
  • Tried PSET 5 of Sophie and dang! I realised I am very bad at geometry :(
  • I am yet to start PSET 6 of Sophie and yes I have a lot to work on.
So although I did a fairly okayish amount of problems (not a very nice amount but not less too), very few of the solutions were LaTeXed due to almost every problem I tried was either a math application problem or from a book or from problem sets, each demanding a handwritten write-up.

The Math: 

Here are basically some nice problems which I did and are fairly very cute ✨ ( Oh, and these  problems were done in the past few weeks, thanks to Rohan bhaiya, Atul, Ananda, Sid and Malay for helping me so much 💗)

Problem 1[Ukraine IMO TST P4 2022]:$ABCD$ is a quadrilateral with equal angles $B$ and $D$. Circles $\omega_1$ and $\omega_2$ are such that $\omega_2$ is symmetric to $\omega_1$ with respect to $AC$, moreover $\omega_1$ passes through $B$ and intersects lines $AB, BC$ at points $K$ and $L$ respectively, while $\omega_2$ passes through $D$ and intersects lines $CD, AD$ at points $M$ and $N$ respectively. Prove that lines $KM, LN$ and $AC$ are concurrent. 

Proof: Define everyone's reflection wrt AC.

Note that $D'BAC$ is cyclic. Note that $$\angle BKM'=\angle BD'M'=\angle BD'C=\angle BAC.$$ Note that $$\angle LN'D=\angle BBD'=\angle D'AC$$ $$\implies LM||KM' || K'M|| L'N $$

Now we will show that $KLMN$ is a parallelogram which would prove the concurrency ( This proving concurrency is sort of non trivial but obvious)

Note that $KL=MN$ as $\angle DMN=\angle LBK$

Note that $KM'NL'$ is isosceles trap $\implies M'L'=KN$ but $M'L'=ML$  $$\implies KN=ML$$

And we are done as $KK'MM'$ , $NN'LL'$ are isosceles trapezoid.

Remark: It was just reflection.. 

Problem 2[Well-known lemma]: Let the incircle of triangle $ABC$ touch sides $BC, CA, AB$ at $D, E, F$
respectively. Let $P$ be the foot of the perpendicular from $D$ to $EF .$ Prove that $P D$
bisects $\angle BP C.$

Solution: Define $T=FE\cap BC.$ Note that $(T,D;B,C)=-1.$ Since $\angle TPD=90\implies P D$
bisects $\angle BP C$.

Problem 3: $ABC$ is a triangle and $M,N$ are midpoints of $AB$ and $AC.$ $BN$ and $CM$ meet $(ABC)$ at $X,Y$ and $AD$ is the altitude. Prove that $(XYD)$ is tangent to $(DMN).$

Proof: Note that $$\angle MNB=\angle NBC=\angle XBC=\angle XYC\implies XYMN\text{ is cyclic }.$$

Consider the following circles and their radical axis. $$(ABC), (AMN), (MNXY)\implies AA,XY,MN\text{ are concurrent  at }Z.$$

Moreover, $MN$ is the perpendicular bisector of $AD.$

Hence $ZD=ZA.$ Hence $$ZD^2=ZA^2=ZM\cdot ZN=ZX\cdot ZY\implies ZD\text { tangent to} (DMN),(DXY).$$

Problem 4: $ABC$ is a triangle. Midpoints of $AB$ and $AC$ are $M$ and $N.$ $MH$ and $NH$ meets $(ABC)$ at $P,Q.$ Prove $PQ$ and $MN$ intersect on $A$ tangent.

Motivation: This is a good example of "If you know one good intersection of a good line and circle, mark the other intersection point too"

Proof: With the above motivation, let's call the other intersection point $P',Q'.$
Note that $$\angle HMN=\angle HP'Q'=\angle Q'PP=\angle NQP\implies MNPQ \text{ is cyclice }.$$ 

Consider the following circles and their radicals.
$$(ABC),(AMN), (MNPQ).$$

We get that $MN, AA, PQ$ concurring

Problem 5 [Pset 5 P6]: Let $D$ and $E$ be points on sides $AB$ and $AC$ of triangle $ABC$ such that
$BDEC$ is cyclic. Let $BE$ intersect $CD$ at $K$ and $AK$ intersect the circumcircle of $ABC$
at $P $. Let $P E$ intersect the circumcircle again at $R.$ Prove that $BR$ bisects $DE.$

Proof: Denote $PD\cap (ABC)=Q.$ By Pascal on $BACQPR,$ we get that $D-CQ\cap BR-E.$ So enough to show that $CQ\cap BR=X$ is the midpoint of $DE.$
Let $AX\cap (ABC)=Y.$ It is enough to show that $(A,Y;B,C)=-1.$
Note that $$\angle ADE=\angle ACB=\angle AYB\implies  BDXY, XYEC \text{ are cyclic }.$$
Moreover, $$\angle YPD=\angle YPQ=\angle YBQ=180-\angle YCQ=180-\angle YEX\implies DEPY\text{ is cyclic}.$$
By Radicals, we get that $$BC\cap DE\cap PY=S.$$

Here I loved this geo, although I couldn't solve it and took a lot of hints.. I really liked it! The idea used were nice! Basically, 
1. pascal should be used as the problem was sort of asymmetric and hence one introduces the point Q.
2. Symmedian and Media are nice with cyclicsss
3. Although now the reduced problem was projective, more elementary observations should be done which is a good reminder of not forgetting elementary stuff. 
4.Rest of the solution was okayish only.. two times projection was nice
5. Additionally, we also have (A,Y;Q,R)=-1
Now, by ceva menelaus, $(S,AP\cap BC; B,C)=-1.$ Projecting it from $A$ to $(ABC)$, we get that $(SA\cap (ABC)=W, P;B,C)=-1.$ Hence $W-P-BB\cap CC=T.$
Anyways, projecting it from $S$ to the $(ABC)$, we get that $$-1=(W,P;B,C)=(A,Y;C,B).$$
And we are done

Problem 6[ Pset 5]: Let the incircle of $ABC$ touch the sides $BC, CA, AB$ at $D, E, F$ respectively
. Suppose $P = EF \cap BC$ and $M$ is the midpoint of $BC$. Let the tangents from $P$ and
$M$ on the incircle different from $BC$ meet at $K$. Show that $AK|| BC$.

Proof: Given $ABC$, $(DEF)$ incirlce. $M$ as the midpoint of $BC$.

Note that $(D,G;B,C)=-1\implies GD=GH.$ Moreover, $(H,G;F,E)=-1$
And it is well known that $AM, DD',EF$ concur. Moreover, defining $Y$ as the $A-$ excircle touchpoint, we get that $MD=MY$ and $A-D-Y$.

Now, we use pascal 2 times.
pascal on $$HHDXXD'$$  gives $A,K, HD'\cap DX$ collinear and then pascal on $$D'D'HDDX$$ which gives $P_{\infty}, HD'\cap DX, A$ collinear.

Problem 7[Bulgaria 1997]: Determine all integers $m,n\ge 2$ for which $$\frac{1+m^{3^n}+m^{2\cdot 3^n}}{n}\in \Bbb{Z}$$

Proof: Note that taking $t= m^{3^n}$, we get that $$(1+m^{3^n}+m^{2\cdot 3^n})(m^{3^{n}}-1)\equiv m^{3^{n+1}}-1\pmod n\implies m^{3^{n+1}}\equiv 1\pmod p, p|n$$

Moreover, we note that ord $_{p}(m) \mid 3^{n+1}$. Note that if if ord_$p(m)=3^k$ for $k\le n.$ Then $p|3$ as $m^{3^n} \equiv 1 \pmod p$. And we get $n=3^{\alpha}$.
If ord$_p(m)=3^{n+1}|p-1<n$ But this is weird as $3^{n+1}>>n.$

Hence $$n=3^k.$$  Taking $m\equiv 1\pmod 3$, by LTE we get $$v_3(m^{3^{3^{k}+1}}-1)-v_3(m^{3^{3^k}}-1)= v_3(3^{3^k+1})-v_3(3^{3^k}) =1$$ 

And $m \equiv 2\pmod 3$ doesnt work.

So $n=3$ is the only sol.

Problem 8[Putnam 2021 A5]: Let $A$ be the set of all integers $n$ such that $1 \le n \le 2021$ and $\text{gcd}(n,2021)=1$. For every nonnegative integer $j$, let

$$S(j)=\sum_{n \in A}n^j.$$ 
Determine all values of $j$ such that $S(j)$ is a multiple of $2021$
Proof:  We use the following lemma.
Lemma: Let $p$ be a prime. Consider $\{1,2,\cdots, p-1\}$. Then $$ p-1\nmid x \implies 1^x+2^x+\cdots {p-1}^x\equiv 0\pmod p .$$
Proof: Since $p$ is a prime, it must have a primitive root $g$ and hence $\{1,2,\cdots, p-1\}=\{g,g^2,\cdots, g^{p-1}\}.$
Hence $$ 1^x+2^x+\dots {p-1}^x\equiv g^x+g^{2x}+\cdots g^{(p-1)x}\equiv g^x( g^1+g^2+\cdots + g^{p-1})\equiv g^x(1+2+\cdot p-1 )\equiv 0\pmod p .$$ 
Now, if $42\nmid j.$
Then note that $$S(j)=\sum_{n \in A}n^j=0\pmod {43}.$$
This is because we have $46$ complete residues of $43$ that is $$S(j)=47\times ( 1^j + \cdots + 42^j)- (47\cdot 1+\cdots +47\cdot 43)\equiv 0\pmod {43}.$$

And similarly for $47$.
For $42\mid j$, we get that $S(j)\equiv 1\cdot \phi(2021)\pmod 43$ and same for $46\mid j.$ 

And we are done!

Problem 9[Some USA Problem]: Let $a$, $b$, $c$, $d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

Solution: Answer is 16.
Let $$P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$
So $$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=\prod (x_1j+i)(x_j-i)= P(i)\cdot P(-i)$$
$$=(i^4+ai^3+bi^2+ci+d)((-i)^4+a(-i)^3+b(-i)^2-ci+d)$$ $$=(1-ai-b+ci+d)(1+ai-b-ci+d)=(1-b+d)^2-(ai-ci)^2$$ $$=(1-b+d)^2+(a-c)^2\ge (1-5)^2+0^2= 16$$
And this is achievable for $P(x)=(x-1)^4$

Problem 10[Well known lemma]: $$(a_1^3+a_2^3+\dots +a_n^3)(b_1^3+\dots+b_n^3)(c_1^3+\dots+c_n^3)\ge (a_1b_1c_1+\dots+a_nb_nc_n)^3$$

Proof: By CS, $$ (a_1^3+a_2^3+\dots +a_n^3)(b_1^3+\dots+b_n^3)\ge ((a_1\cdot b_1)^{3/2}+\dots+(a_n\cdot b_n)^{3/2})^2$$
And hence $$(a_1^3+a_2^3+\dots +a_n^3)(b_1^3+\dots+b_n^3)(c_1^3+\dots+c_n^3)\ge ((a_1\cdot b_1)^{3/2}+\dots+(a_n\cdot b_n)^{3/2})^2(c_1^3+\dots+c_n^3) $$
Note that $$((a_1\cdot b_1)^{3/2}+\dots+(a_n\cdot b_n)^{3/2})^{2/3}(c_1^3+\dots+c_n^3)^{1/3}\ge (a_1b_1c_1+\dots+a_nb_nc_n)$$
And hence $$((a_1\cdot b_1)^{3/2}+\dots+(a_n\cdot b_n)^{3/2})^{2}(c_1^3+\dots+c_n^3)\ge (a_1b_1c_1+\dots+a_nb_nc_n)^3$$
Hence $$(a_1^3+a_2^3+\dots +a_n^3)(b_1^3+\dots+b_n^3)(c_1^3+\dots+c_n^3)\ge (a_1b_1c_1+\dots+a_nb_nc_n)^3$$

P.S. Malay said it's just a generalized holder

Problem 11[USAMO 2004]: Let $a, b, c > 0$. Prove that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a + b + c)^3$.

Proof: Note that $$(a^2-1)(a^3-1)\ge 0\implies a^5-a^2-a^3+1\ge 0\implies a^5-a^2+3\ge a^3+1+1.$$
So, if we show, $$(a^3+1+1)(b^3+1+1)(c^3+1+1)\ge (a+b+c)^3$$
we are done.

Which is true by using the famous lemma $$(a_1^3+a_2^3+a_3^3)^{1/3}(b_1^3+b_2^{3}+b_3^{3})^{1/3}(c_1^3+c_2^3+c_3^3)^{1/3}\ge (a_1b_1c_1+a_2b_2c_2+a_3b_3c_3)$$

Then we get that $$(a^3+1+1)(1+b^3+1)(1+1+c^3)\ge (a+b+c)^3$$

The equality happens when $a=b=c=1.$

Problem 12[Some ISL]: Prove that for all integers $n$, there exists a set of $n$ integers, call it $S$, such that the sum of any subset of $S$ is a perfect power. 

Proof: Let's look at the set of the form $\{d,2d,\dots, 1992 d\}$
Then, the sum of the elements of a subset is of the form $kd, k\in \{ 1,\dots , \frac{1992\cdot 1993}{2}\}$.

In general, to show that $\exists d$ s.t $kd$ is a perfect power for $ 1\le k\le N.$

We just simply spam CRT. I feel the CRT idea is just so so nice.

Let $$p_1^{a_{1k}}\dots p_r^{a_rk}=k$$ and let $$d= p_1^{x_1}\dots p_r^{x_r}$$ then we just want $$\gcd( a_{1k}+x_1,\dots , a_{rk}+x_r)\ge 2$$
So consider 

$$x_j\equiv -a_{j1} \pmod{p_1}$$
$$x_j\equiv -a_{j2} \pmod{p_2}$$
$$\vdots$$
$$x_j\equiv -a_{jm} \pmod{p_N}$$

Note that $p_N$ denotes the $N$th prime.

By CRT, such $x_j$ exists. And by this we get that $$p_i\mid \gcd( a_{1i}+x_1,\dots , a_{ri}+x_r)$$

Problem 13[China TST]: Let $a$ be a fixed positive integer. Prove that the equation $n! = a^{b} - a^{c}$ has a finite number of solutions $(n, b,c).$


Proof: Let $p\nmid a$. As $\nu_p(a^n-1)\le \nu_p(a^{p-1}-1)+\nu_p(n)$, we get that  $\nu_p(n!)=\nu_p(a^b-a^c)=\nu_p(a^{b-c}-1) \le \nu_p(a^{p-1}-1)+\nu_p(b-c)$.

Using the inequality,  $\lfloor\frac{n}{p}\rfloor \ge n/p-1$ we get that $\nu_p(a^{p-1}-1)+\nu_p(b-c)\ge n/p-1$.

Hence $\nu_p(b-c)\ge n/p-k$ for some $k$ independent of $n$.

Hence $$n^n \geq n! = a^{c} (a^{b-c} - 1 ) \geq a^{\log_p{b-c} + c} \geq a^{p^{\frac{n}{p} - k} }$$

Taking logs of each side gives that
$$ n \ln{n} \geq p^{\frac{n}{p} - k} \ln{a} $$

But note that $n\ln{n} \le n^2$ for sufficiently large $n$ and $p^{\frac{n}{p} - k} \ln{a}=p^{\frac{n}{p} }c.$ And it is well know that exponent function grows faster than polynomial function. So $ n \ln{n} \geq p^{\frac{n}{p} - k} \ln{a} $ will have finaite solutions.

OMC Blog: 

  • The OMC blog team is really hardworking and is working a lot for the OMC blog. 
  • I am kind of enjoying my supervisor position and it's fun to be a dictator ( lol)
  • We are currently working on the IMO interview which is going to be held soon after IMO. ( I guess during the unofficial TC, we are not sure..)

Unofficial IMOTC:

Continuing the tradition, here we proudly introduce the Unofficial TC 2022, also known as Unofficial Training Camp 2022. Because of the IMOTC not being conducted due to the COVID crisis, here we are back again to conduct the unofficial online version of the IMOTC called the Unofficial TC 2022! We hope through this camp, the INMO awardees ( mostly the juniors) can socialise with other INMO awardees and (obviously) learn new maths! The camp's teachers are professors, International Olympiads medalists, senior INMO awardees and of course the camp's students too!

  • When? August 1 - August 31

  • Who are invited? The INMO 2022 awardees and additionally, a few non-INMO awardees will also be selected through the application process described on the Apply page.

  • Where? Online

  • Describe the structure of the classes? We are planning to conduct two classes every day except Wednesday, Saturday and Sunday. The duration of the classes would be 1.5 hours but sometimes they might extend till 2hrs. Additionally, problem sets, handouts and lecture notes would also be provided. Moreover, some classes might be PSS i.e Problem Solving Session and some might be completely college math-related. Some might be guest lectures! Some might even be just seniors giving advice sessions!

  • Will it be too rigorous for me? Although we will be having a lot of classes, our main aim would be to just have fun! We want people like us to socialize with each other, have fun, play various games, listen to songs and enjoy doing math together. We hope that this unofficial TC could at least give a taste of how fun IMOTC and maths are!

  • Who will be teaching? We will soon have a team page, but for now, we have 25+ instructors who have confirmed their positions.

  • Do you have a syllabus? We don't have a fixed syllabus and the classes would depend on students' requests and vary from instructor to instructor. However, if interested, please visit the lectures page.

Note that Unofficial IMOTC is completely independent of HBCSE.


I am just so so so excited for the unofficial TC! It's my first time organising something this big! Moreover, the confirmed lectures are already looking so so so exciting!

Thanks to Atul, Gunjan and Pranav for organising this and obviously big thanks to Aatman dada and Rohan Bhaiya for guiding us! I hope this 1-month long camp runs smoothly and nicely. 

The courses are just so so interesting and tempting! Please do check it out! Also, I am solely responsible for the website, so please do check it out! I worked really hard for it! 

Link: Eat, Love, Kill: 

What a drama! I started watching this drama expecting some romance since I prefer romantic dramas and I knew these two actors are very well known! But little did I know that this drama was actually more of a mystery! 

As the Mydramalist describes: It is a fantasy mystery drama about a man and woman who share the same emotional state. Eun Gye Hoon is a chef who sets up a restaurant in the town where his twin sister went missing 20 years ago. He finds himself randomly experiencing emotions one day, spontaneously crying and laughing, and it turns out that they are the emotions of a woman named Noh Da Hyun.

Episode 11 Review: Very nice episode! In fact, this whole drama is just so goood. Full of mystery and a tiny bit of romance! The chemistry is in point and the acting is just so good that you feel you are in the neighbourhood. Finally, the killer was revealed. Curious about how Dahyun's family is related and how Dahyun's memory got erased. Sad that they broke up but hmm.. I am sad that JinGuen got killed ( like he did die ig)? Another weird mystery I am curious about is, the neighbourhood ladies seem sus and the couple who now leaves in the house where the girls were kidnapped. Danggg still so much mystery! Normally I am not into mystery dramas ( rather I prefer cliche romantic ones) but this drama is so unique. I highly recommend people to watch it! Can't wait for the next episode!

Another thing which I really liked was the chemistry between Ji WonTak and Hwang-Min-Jo is also really nice! I still wonder how they broke up and they too seem to have some mystery.











Webtoons:

  • Basically, I have again started reading webtoons and they don't consume much time.
  • Currently, I have subscribed to like three of them and my favourite is " Wished you were dead"
  • I like the storyline and it is a bit ( like a very tiny bit) similar to the Remarried Empress storyline. But the "Wished you were dead" storyline is quite unique. I agree, the main plot is sort of similar to "Remarried Empress" but it is still unique ( Aaaaa Irdk how to explain this without giving spoilers )
  • And the artwork is just so so so good!





TxT minisode 2: Thursday Child

The songs are so good. My favourite song is definitely Opening Sequence. Actually, everything song is just so good!!!! 

Now talking about the Good boy gone bad era. I feel the GBGB era for TXT is just so musically good. Talking about aesthetics, I prefer the innocent cute era, but dang I think Taehyung just nailed the GBGB era.



Taehyun nailed it! 




My Plans: 

  • My July-August plan is basically me doing a lot of work. My current goal is to get get a great score in the half-yearly examination.I am planning to post every day a few ISL problems from tomorrow. Let's see how many I can do!
  • Kind of very ambitious, so don't expect much!

    Well, that's it for this month's blog! See you all tomorrow or very soon!

     Sunaina💚
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Bio is Love..

Adios, everyone! Boards preparation at its peak :(  However, I am not able to study how I used to. Every time I try to study for boards, I just keep thinking much about a topic, stare at the book, jam a song or just start doing procrastination by bookmarking random cute problems in HSO. It's been more than a year I have studied like with a focus on a book. My lappy is being a big distraction tbh. So after INMO score come out, I will just give my lappy for repair and say papa to bring it back home after June 2.  Milk and Mocha I literally am taking 2 days to complete 1 bio chapter, some times even 3. The rate of my "slowness" is probably because I am like every 15 minutes checking discord to see if the INMO scores are out or not. So HBCSE, thank you for keeping me anxious.  Funfact:- we must be grateful that there is an organisation that is conducting these national Olys. There are some countries where no Olys are being conducted. ( Same dialogue which mumma uses, but in p
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Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral. Solution:  Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral. Problem:  A convex quadrilateral $
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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 r
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Just spam combo problems cause why not

This post is mainly for Rohan Bhaiya. He gave me/EGMO contestants a lot and lots of problems. Here are solutions to a very few of them.  To Rohan Bhaiya: I just wrote the sketch/proofs here cause why not :P. I did a few more extra problems so yeah.  I sort of sorted the problems into different sub-areas, but it's just better to try all of them! I did try some more combo problems outside this but I tried them in my tablet and worked there itself. So latexing was tough. Algorithms  "Just find the algorithm" they said and they died.  References:  Algorithms Pset by Abhay Bestrapalli Algorithms by Cody Johnson Problem1: Suppose the positive integer $n$ is odd. First Al writes the numbers $1, 2,\dots, 2n$ on the blackboard. Then he picks any two numbers $a, b$ erases them, and writes, instead, $|a - b|$. Prove that an odd number will remain at the end.  Proof: Well, we go $\mod 2$. Note that $$|a-b|\equiv a+b\mod 2\implies \text{ the final number is }1+2+\dots 2n\equiv n(2n+1
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IMO 2023 P2

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang
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