Skip to main content

Problems done in July

 July was LIOG month. Just LIOG was done, and then we had MMC ( till July 14). SO from the last two weeks, I was doing LIOG only. And I did some non-geo too! Also, I am not adding all the problems I did. A few only in fact. One can go through my apps post ( Jelena_ivanchic)

Here are a few problems and solutions/walkthroughs:

Problem: Let $A_1$ be the intersection of tangent at $A$ to the circumcircle of a triangle $\Delta ABC$ with sideline $BC$. Similarly define $B_1,C_1$. Show, that $A_1,B_1$ and $C_1$ are collinear

Walkthrough: 

  • Use this $\Delta ACA_1\sim \Delta BAA_1.$
  • So we get$$\frac{AC}{BA}=\frac{CA_1}{AA_1}=\frac{AA_1}{BA_1}$$$$ \implies \frac{CA_1}{AA_1}\cdot \frac{AA_1}{BA_1} =\frac{AC^2}{BA^2}$$$$\implies  \frac{AC^2}{BA^2}=\frac{CA_1}{BA_1}. $$Now, apply menelaus on $ABC.$

Problem: Let $\Gamma$ be a circle and let $B$ be a point on a line that is tangent to $\Gamma$ at the point $A$. The line segment $AB$ is rotated about the center of the circle through some angle to the line segment $A’B’$. Prove that $AA’$ passes through the midpoint of $BB’$.

Walkthrough:

Solved with PC and Krutarth.

  • Let $D$ be the midpoint of $BB',$ we get $\angle XAB=\angle XDB,$ hence $XADB$ cyclic.
  • Now by similarity (spiral but let's not bring fancy words), we get $XBB'\sim XAA'$ ( actually isosceles but we don't need that fact).
  • So $\angle  XBB'=\angle  XAA',$ but $\angle XBB'=\angle XAF,$ where $F$ is a point on ray $DA.$



Problem: Given an infinite sequence of numbers $a_1, a_2, a_3,...$ . For each positive integer $k$ there exists a positive integer $t = t(k)$ such that $a_k = a_{k+t} = a_{k+2t} =...$. Is this sequence necessarily periodic? That is, does a positive integer $T$ exist such that $a_k = a_{k+T}$ for each positive integer k?

Walkthrough: No, it’s not necessary.

Here’s the construction:

Let $X$ denote the place of the sequence, for example in sequence $1,2,3,4,\dots $ $1st $ place is $1.$

Then let$$1\rightarrow v_2(x)=0 $$$$2\rightarrow v_2(X)=1 $$$$3\rightarrow v_2(X)=2 $$$$4\rightarrow v_2(X)=3 $$$$5\rightarrow v_2(X)=4$$$$\vdots $$

So the sequence looks like$$1213121412131215 \dots.$$Clearly, it’s not periodic.

Problem: Find all pairs of natural numbers $(m,n)$ bigger than $1$ for which $2^m+3^n$ is the square of the whole number.

Walkthrough: Note that $\mod 3$ gives $m$ is even and $\mod 4$ gives $n$ is even. We get$$2^{2k}= (X^2-3^l)(X^2+3^l)\implies X^2-3^l= 2^a,~~ X^2+3^l=2^b \implies 2^a| 2^b \implies 2^a|2\cdot 3^l| \implies a=0 \text{or} 1$$So $X^2-3^l=1, 2$ but squares are $0,1\mod 3.$

So$$X^2-3^l=1 \mod 3$$Now we will try to solve$$X^2-3^l=1$$

Note that$$X^2-1=(X-1)(X+1)=3^l\implies X-1=3^w,X+1=3^v\implies 3^w|2\implies w=0 \implies X-1=1\implies 2.$$

Clearly, $X=2$ doesn't satisfy. So we don't have any solutions.

Problem: Let $BD$ be a bisector of triangle $ABC$. Points $I_a$, $I_c$ are the incenters of triangles $ABD$, $CBD$ respectively. The line $I_aI_c$ meets $AC$ in point $Q$. Prove that $\angle DBQ = 90^\circ$.

Walkthrough: 

  • Let $F:= I_aI_c \cap BD,~~ J:=BI_a\cap AC,~~ K:=BI_c\cap AC.$
  • Now, we have$$ \angle I_aDI_c=90,~~\angle FDI_a=\angle I_aDA.$$Then we have$$-1=(E,F;I_A,I_C)$$projecting through $B$ on $AC.$ We get$$ -1=(E,F;I_A,I_C)=(E,D;J,K).$$We already have$$\angle JBD=\angle KBD\implies \angle DBE=90.$$

 


Problem: Let $ABC$ be a triangle with incenter $I$, let $,I_{1}, I_{2}, I_{3}$ be the incenters of triangles $BIC, CIA, AIB$ respectively. Prove that lines $AI_{1}, BI_{2}, CI_{3}$ are concurrent.

Walkthrough: 

  • Note that $\angle ABI_3= B/2=\angle I_1BC.$ Similarly we get $\angle I_1CB=C/2=\angle I_2CA,~~\angle BAI_3=\angle CAI_2.$
  • Now apply jacobi and we are done


Apart from that CAMP started! We had Hanabi night, where we got 20/25, which I feel is quite good at least for beginners. 

And then LIMIT results came out. In subjective I got 21/120. I was sad since I thought I could have got more marks. I might send my sols here too, and maybe someone can open grade here :O.

Sad note: My mom decreased my net time to 9pm :(, earlier it was 11pm.

Anyways, hope you enjoyed this post!

Sunaina💜



Labels:

Comments

Post a Comment

Popular posts from this blog

My experiences at EGMO, IMOTC and PROMYS experience

Yes, I know. This post should have been posted like 2 months ago. Okay okay, sorry. But yeah, I was just waiting for everything to be over and I was lazy. ( sorry ) You know, the transitioning period from high school to college is very weird. I will join CMI( Chennai Mathematical  Institue) for bsc maths and cs degree. And I am very scared. Like very very scared. No, not about making new friends and all. I don't care about that part because I know a decent amount of CMI people already.  What I am scared of is whether I will be able to handle the coursework and get good grades T_T Anyways, here's my EGMO PDC, EGMO, IMOTC and PROMYS experience. Yes, a lot of stuff. My EGMO experience is a lot and I wrote a lot of details, IMOTC and PROMYS is just a few paras. Oh to those, who don't know me or are reading for the first time. I am Sunaina Pati. I was IND2 at EGMO 2023 which was held in Slovenia. I was also invited to the IMOTC or International Mathematical Olympiad Training Cam...
11 comments
Read more

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$.  Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
7 comments
Read more

IMO Shortlist 2022 C1

  Today we shall try IMO Shortlist $2022$ C1. A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and$$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$ We claim that the answer is $\boxed{506}$. $506$ is the upper bound. Just consider the sequence $$+1,-1,-1,+1,+1,-1,-1,+1\dots,-1,-1,+1,+1,-1.$$ Here $1, -1, -1, 1$ is repeated $505$ times and $1,-1$ is concatted to it. Now,our sequence would be $a_1,a_3,a_4,a_5,a_7,\dots$ which on summing would give $506$. And clearly, this would give the upper bound. Now, we show that $506$ is attainable by every sequence. WLOG there are at least $1011$ positive numbers in the sequence. Then we choose $+1$ whenever we can. Let the sequence be $c_1,b_1,\dots, c_n,b_n$ where $c_i$ are ...
1 comment
Read more

IMO Shortlist 2021 C1

 I am planning to do at least one ISL every day so that I do not lose my Olympiad touch (and also they are fun to think about!). Today, I tried the 2021 IMO shortlist C1.  (2021 ISL C1) Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a,b,c,d \in S$ with $\gcd(a,b) \neq \gcd(c,d)$. Prove that there exist three pairwise distinct $x,y,z \in S$ such that $\gcd(x,y)=\gcd(y,z) \neq \gcd(z,x)$. Suppose not. Then any $3$ elements $x,y,z\in S$ will be $(x,y)=(y,z)=(x,z)$ or $(x,y)\ne (y,z)\ne (x,z)$. There exists an infinite set $T$ such that $\forall x,y\in T,(x,y)=d,$ where $d$ is constant. Fix a random element $a$. Note that $(x,a)|a$. So $(x,a)\le a$.Since there are infinite elements and finite many possibilities for the gcd (atmost $a$). So $\exists$ set $T$ which is infinite such that $\forall b_1,b_2\in T$ $$(a,b_1)=(a,b_2)=d.$$ Note that if $(b_1,b_2)\ne d$ then we get a contradiction as we get a set satisfying the proble...
1 comment
Read more

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....
3 comments
Read more

Symmetric Polynomials #week 6

Well... I haven't seen much symmetric polynomials in Olympiads, but still I am learning, because I found them cute. And I am basically using this blog as my notes :P What are symmetric polynomials?  One can understand this with  examples. If we are considering over 3 variables, $x_1,x_2,x_3$ then  $$\sum_{sym}x_1^2\cdot x_2^3\cdot x_3=x_1^2\cdot x_2^3\cdot x_3+x_1^2\cdot x_3^3\cdot x_2+x_2^2\cdot x_1^3\cdot x_3+x_2^2\cdot x_3^3\cdot x_1+x_3^2\cdot x_1^3\cdot x_2.$$ See? $3!$ terms! Let's take one more example with again over 3 variables, $x_1,x_2,x_3$ then $$\sum_{sym}x_1^2\cdot x_2^2= x_1^2\cdot x_2^2+x_1^2\cdot x_3^2+x_2^2\cdot x_1^2+x_2^2\cdot x_3^2+x_3^2\cdot x_1^2+x_3^2\cdot x_2^2$$ Wait.. why 2 times ? So basically what happens in symmetrictric sums, is we go through all $n!$ possible permutations. So, here we have $a^2\cdot b^2\cdot c^0$ as like the "general" form type, right? Now, list down all the $3!=6$ permutations of $x_1,x_2,x_3$, and put them in the gene...
3 comments
Read more

Number Theory Revise part 1

I thought to revise David Burton and try out some problems which I didn't do. It's been almost 2 years since I touched that book so let's see! Also, this set of problems/notes is quite weird since it's actually a memory lane. You will get to know on your own! I started with proving a problem, remembered another problem and then another and so on! It was quite fun cause all these questions were the ones I really wanted to solve! And this is part1 or else the post would be too long. Problem1: Prove that for $n\ge 1$  $$\binom{n}{r}<\binom{n}{r+1}$$ iff $0\le r\le \frac{n-1}{2}$ Proof: We show that $\binom{n}{r}<\binom{n}{r+1}$ for $0\le r\le \frac{n-1}{2}$ and use the fact that $$\binom{n}{n-r}=\binom{n}{r}$$ Note that $\binom{n}{r}= \frac{n!}{r!(n-r)!}, \binom{n}{r+1}=\frac{n!}{(r+1)!(n-r-1)!}$  Comparing, it's enough to show that $$\frac{1}{n-r}<\frac{1}{r+1}\text{ or show } n-r>r+1$$ which is true as $0\le r\le \frac{n-1}{2}$ Problem2: Show that the exp...
8 comments
Read more

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...
2 comments
Read more

Reflecting on past

INMO Scores are out!! I am now a two times INMO awardee :) I got 16|0|1, so 17 in total! Yes, 16 in P1 T_T. I was thinking I would lose marks because of the way I wrote.  Lemme tell ya'll what happened that day but first I should share a few thoughts I had before the exam. My thoughts Honestly, my preparation for INMO was bad. In fact, I should say I didn't work hard at all. As I have said earlier, I had lost all my hopes for INMO and Olympiads as a whole after EGMO TSTs happened.  Art by Jelena Janic EGMO TSTs i.e European Girl's Mathematical Olympiad Team selection Tests 2022.  Literally my thoughts after EGMO TSTs I feel very ashamed to share but I got 1 mark in my EGMO TSTs. Tests in which I literally gave my whole life. I did so many ISLs ( like SO MANY), I mocked EGMO 2021 TST where my score was 28/42 and I perfected Day 2. 1 mark in the TST just showed my true potential. There are way better people than me in olys. A friend even said to me, "If I wouldn't...
7 comments
Read more

INMO Scores and Results

Heya! INMO Results are out! Well, I am now a 3 times IMOTCer :D. Very excited to meet every one of you! My INMO score was exactly 26 with a distribution of 17|0|0|0|0|9, which was a fair grading cause after problem 1, I tried problem 6 next. I was hoping for some partials in problem 4 but didn't get any.  I am so so so excited to meet everyone! Can't believe my olympiad journey is going to end soon..  I thought to continue the improvement table I made last year! ( I would still have to add my EGMO performance and also IMO TST performance too) 2018-2019[ grade 8]:  Cleared PRMO, Cleared RMO[ State rank 4], Wrote INMO 2019-2020[ grade 9]:  Cleared PRMO, Cleared RMO[ State topper], Wrote INMO ( but flopped it) 2020-2021[grade 10]:  Cleared IOQM, Cleared INMO [ Through Girl's Quota] 2021-2022[grade 11]:  Wrote EGMO 2022 TST[ Rank 8], Qualified for IOQM part B directly, Cleared IOQM-B ( i.e INMO) [Through general quota],  2022-2023 [grade 12]:  Wrote E...
Post a Comment
Read more