Skip to main content

Trying to go beyond my comfort level?

Wrapping up year 2022 with a few problems I did in the past week :) 

Edge colouring with $n$ colours $K_n$ 

Prove that if the edges of $K_n$ are coloured with $n$ colours, then some triangle has its edges of different colours. 

Proof: We use induction. For $n=3$ it is true. Now, suppose it is true for $n=1,\dots, l$. We will show it is true for $n=l+1$. Now, consider $k_{l+1}$ with vertex $v_1,\dots,v_{l+1}$. Consider the $k_l$ with vertex $v_2,\dots, v_{l+1}$. Now note that the colours used in that $k_l$ are max $l-1$ colours (since by induction, if $l$ colours then we get a triangle with the property given.)

But since $l+1$ colours are used, at least two of the edges $v_1v_2,v_1v_3,\dots v_{l+1}$ are coloured with $2$ colours which are not used in $k_{l}$. WLOG say $v_1v_2,v_1v_3$, then $v_1v_2v_3$ is a triangle with edges of diff colour. 

Russia 2011 Grade 10 P6

Given is an acute triangle $ABC$. Its heights $BB_1$ and $CC_1$ are extended past points $B_1$ and $C_1$. On these extensions, points $P$ and $Q$ are chosen, such that angle $PAQ$ is right. Let $AF$ be the height of triangle $APQ$. Prove that angle $BFC$ is a right angle.

Proof: Note that $AEC_1Q$ and $AEB_1P$ cyclic. 

Claim: $C_1FB_1C$, $C_1FB_1A$ cyclic

Proof: $$180-\angle C_1EB_1=\angle QEC_1+\angle PEB_1=\angle QPC_1+\angle PAB_1=90-A=\angle C_1CB_1.$$

And we get $F\in (BC_1B_1C)\implies \angle BFC=90$.

Russia 2011 Grade 10 P2

On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$.

Proof: Define $H$ as the orthocenter of $O_1O_2O_3$. Note that $$\angle O_2O_3T=\angle O_2O_1T\implies T\in (O_1O_2O_3)$$ 

Define $M,N$ as midpoints of $AT,DT$. Note that $MN$ is the simson line for $\Delta O_1O_2O_3$ and $T$. Since $H$ is the orthocenter, we get $MN$ bisects $HT$. But we have homothety at $T$ sending $M$ to $A$ and $N$ to $D$. Hence $H\in AD$. And we are done.

Additionally, since $O_2O_3\perp DT,O_1O_2\perp AT$ we have $$HO_1|| DT, HO_3||AT\implies \angle O_1HO_3=\angle DTA.$$

Polynomial in OLY-PD posted by Atul 

Determine the set of all polynomials $P(x)$ with real coefficients such that the set $\{P(n) | n \in \mathbb{Z}\}$ contains all integers, except possibly finitely many of them.

Solved with Sid and Malay!

Proof: We claim that $P(x)=(x+k)/b, k,b\in \Bbb{Z}$ is the only solution. 

By Lagrange interpolation construction, we get coefficients of $P(x)\in \Bbb{Q}$.  If $P$ have degree greater than $2$, $|P(x+1)-P(x)|$ will eventually be greater than $1$ and start skipping infinitely many integers. Hence $P$ is linear.  

Let $P(x)=\frac{ax}{d}+\frac{c}{d}$ with $\gcd(a,c,d)=1$. If $a\ne \pm 1$ then consider prime $p|a$. So $d|ax+c$. So if $p|d\implies p|c$ not possible. So $P(x)\equiv \frac{c}{d}=k\mod p$ for a fixed $k\in \Bbb{Z}_p$ which doesn't cover infinite integers. 

So $P(x)=\frac{x+c}{d}$ is the only solution. 

EGMO 2019 P3

Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$. Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$. Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$ intersect at a point on line $BC$.

Solved with Krutarth and Malay!

Proof: Define $Y=\omega \cap AD$, $Z=\omega\cap AB$, $M=$ midpoint of arc $AB$ ( not containing $C$), $N=$ midpoint of arc $CB$ ( not containing $A$), $K=AD\cap (ABC)$, $S=$ incenter of $\Delta ABK$.


Claim: $YZ||BC$

Proof: Note that $\angle B=\angle YAC=\angle YZA\implies YZ||BC.$


Claim: $C-Y-X$

Proof: Note that $\angle CXA=\angle B=\angle YZA=\angle YXA.$


Claim: $S\in BC$

Proof: Note that $C$ is the midpoint of arc $KA$. By Fact 5, $S\in BC$.


Claim: $AISB$ cyclic

Proof: Note that $M$ is the centre of $AIB$ and $M$ is also the centre of $ASB$. So $MA=MI=MS=MB$.


Claim: $I-Y-S$ collinear

Proof: Note that$$\measuredangle AIY=\measuredangle YZA=\measuredangle SBA.$$

Now using pascal on $MCXNAK$, we get $S=NX\cap MK$. So $XS$ bisects $\angle CBX$. And we are done!

EGMO 2021 P1

The number 2021 is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?


Proof: We begin with the claim. 

Claim: If $2021$ is fabulous then so is $1$.

Proof: If not, then let the least element be $k$. We take cases. 

Case 1: If $k$ is even

Then $$2k+1\leftrightarrow 6k+3\leftrightarrow 3k+1\leftrightarrow \frac{3k}{2}\leftrightarrow \frac{m}{2}$$ is fantabulous.

Case 2:If $k$ is odd

Then $\frac{k-1}{2}$ is fantabulous.

Claim: If $1$ is fantabulous then so is every number.

We do induction. Say $1,2,\dots,n$ is fantabulous. We will show that $n+1$ is also fantabulous. But  this is just similar to our above claim's proof ( we can simply back track stuff stuff)

And yey!

Russia 2015 Grade 11 P7

A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$.

Proof: $ABPQ$ is cyclic ( simple angles.) Define $M_A,M_B$ as arc midpoints. So we have $$\angle APQ=\angle ABQ=\angle ABI=\angle B/2=\angle AM_AM_B\implies M_AM_B||PQ.$$

If $N$ is midpoint of $M_AM_B$ $\implies I-M-N$ collinear. Define $S$ as the midpoint of arc. 

Claim: $IM_BM_AS$ is parallelogram.

Proof: Note that $M_BM_A||SC$ (As $IC\perp SC, IC\perp M_AM_B$). So $M_BS=M_AC=M_AI$. Moreover, we have $\Delta M_BIM_A$ congruent to $\Delta M_BCM_A$. So we have $M_BI=M_BC=M_AS$. 

Done!


Russia 2022 Grade 11 P1

We call the $main$ $divisors$ of a composite number $n$ the two largest of its natural divisors other than $n$. Composite numbers $a$ and $b$ are such that the main divisors of $a$ and $b$ coincide. Prove that $a=b$.

Proof: Let $a=p_1^{\alpha_1}\dots p_k^{\alpha_2}$ and $b=q_1^{\beta_1}\dots q_l^{\beta_l}$.

Note that if $a$ or $b$ is a prime then clearly we are done. So supposed neither $a,b$ is prime.

Since$$\frac{a}{p_1}=\frac{b}{q_1}\implies p_1^{\alpha_1-1}\dots p_k^{\alpha_2}=q_1^{\beta_1-1}\dots q_l^{\beta_l}.$$

Claim: If $\alpha_1-1,\beta-1>0$ then $p_1=q_1$ which forces $a=b$.

Proof: If not wlog $p_1<q_1$ but $p_1|b$. Contradiction as then the biggest divisor would be $\frac{b}{p_1}$ and not $\frac{b}{q_1}$.

So either $\alpha_1-1, \beta-1$ is $0$. Wlog say $\alpha_1=1$ Then the next largest divisor is$$\frac{a}{p_2},\frac{b}{x}\implies p_1|\frac{a}{p_2}\implies p_1|\frac{b}{x}\implies p_1|b.$$

If $p_1\ne q_1$ then$$v_{p_1}(a/p_1)=0\ne v_{p_1}(b/q_1)\implies \frac{a}{p_1}\ne\frac{b}{q_1}.$$So $p_1=q_1$ and hence $a=b$.


Pumac 2019 A1

A finite graph $G$ is drawn on a blackboard. The following operation is permitted: pick any cycle $C$ of $G$, draw a new vertex $v$, connect it to all vertices of $C$, and finally erase all the edges of $C$. Prove that this operation can only be done a finite number of times.

Proof: Note that if $G$ is connected then even after we do the operation the number of edges will be same and the new graph is still connected.. look at e-v, for any n vertex, the connected graph must have atleast $n-1$ edges.. so this will stop eventually.. 

If $G$ is not connected, then consider the components, whenever we do the operation, the connected components still remain connected.. and the number of edges are same.. use the same argument.

ISL 2004 C3 

The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer ${n\ge 4}$, find the least number of edges of a graph that can be obtained by repeated applications of this operation from the complete graph on $n$ vertices (where each pair of vertices are joined by an edge).

Proof: We claim that the answer is $\boxed{n}$.

Note that whenever we perform the operation, the connectedness is preserved. So the no of edges $\ge n-1$.
 
Claim: $n-1$ is not the answer.
Proof: If $n-1$ worked then we have a tree which is bipartite. 
Then we are adding cycles back. 

Subclaim: Whenever we add an edge, no odd cycle is formed.
Proof: Suppose not. Then there must be a first time when the odd cycle was formed. 

Say we have vertices $v_1,v_2,v_3,v_4$ and we are adding the edge $v_1v_4$ and $v_1a_1\dots a_nv_4$ forms an odd cycle. 

But then $a_1a_2a_3a_4a_n\dots a_2$ is also an odd cycle. But we said that $v_1a_1\dots a_nv_4$ was the first odd cycle. 

Construction for $n$ which is done by induction.

Well, we will show for $K_5$, as $K_n$ can be done similarly. 
Imagine, vertex $v$ attached to $k_4$ of $a_1,a_2,a_3,a_4$.  Then consider cycle $v,a_1,a_4,a_2$, delete edge $va_1$. And similarly, we can delete edges $va_2,va_3$. And then $k_4$ gives $4$ edges. 


ISL 2021 G4


Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.

Proof: Define $X=TE\cap AC$ and $Y=TF\cap AC$.

Claim: $FCYD, DXDE$ cyclic
Proof: Note that $$\angle CYD=\angle DBA=\angle ADE=\angle TFD\implies CYFD \text{ cyclic}$$
Similarly $DXAE$ cyclic. 

Claim: $YXEF$ cyclic
Proof: Note that $$\angle TXY=\angle AXE=\angle ADE =\angle TFE.$$

Claim: $YXDT$ cyclic
Proof: Note that $$\measuredangle TYD=\measuredangle FYD= \measuredangle YCD=\measuredangle DAE=\measuredangle DXE.$$

Claim: $(CFDY),(DXAE)$  are tangent at $D$ and $DT$ common tangent
Proof: $$\angle YDT=\angle YXT=\angle TFE=\angle YFD\implies DT\text{ is tangent to }(CFDY).$$
Similarly, $DT$ is tangent to $DXAE$.

Claim: $BD||TK$
Proof: $$\angle BDE=\angle BDA+\angle ADE=\angle BCA+\angle ACD $$ $$=\angle YDF+\angle YDT=\angle EDT=\angle TKE.$$

Claim: $K\in TDXY$
Proof: $$\angle TKE=\angle DCB=\angle TYD.$$

Claim: $KX||BC,KY||BA$
Proof: $$\angle XKE=\angle DTX=\angle DYX=\angle CFD\implies KX||BC.$$

So $\Delta TXK$ similar to $\Delta DCB$. And the concurrency follows by homothety. 

Russia 2018 10.2 


Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic.

Proof: Define $AD\cap MN=D'$. Define $L$ as the midpoint of $BC$, $X$ as the midpoint of $MN$, $E=$ parallel through $A$ to $BC$ intersecting at $(ABC)$, $X'$ as the midpoint of $AE$.

First note that $\Delta LMN$ is congruent to $\Delta AMN$. So $MD'=KN\implies X\text{ is midpoint of }D'K.$ Note that $LX'$ is the perpendicular bisector of $AE$. Also $ADLX'$ is a rectangle. 
So $$D-X-X'\implies D-K-E.$$

Now, $$\angle KQC=\angle EQC=\angle EAC=\angle ANK\implies Q\in(CNK).$$


Indian TST D1 P2 (Practice Test)


Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers $$(a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1$$ are all powers of $5$


Proof: Suppose there exists such $a_1,\dots,a_n$.
Let the minimum out of all powers (WLOG) be $(a_1)^{2018}+a_2=5^{\alpha}$. Since it is the minimum, we have $$ (a_1)^{2018}\equiv -a_2\mod 5^{\alpha}$$
 $$ (a_1)^{2018^2}\equiv -a_3\mod 5^{\alpha}$$
$$\vdots$$
$$ (a_1)^{2018^{2017}}\equiv -a_{2018}\mod 5^{\alpha}$$
$$ (a_1)^{2018^{2018}}\equiv -a_1\mod 5^{\alpha}$$
$$\implies a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}.$$


Claim: $5\nmid a_i\forall i\in [2018]$

Proof: If $5|a_k\implies 5|a_{k+1}\implies 5|a_{k+2}\implies \dots \implies 5|a_{k-1}$
WLOG $v_5(a_1)$ be the lowest. Then $$v_5((a_{2018})^{2018}+a_1)=v_5(a_1).$$ Which is not possible.

So by our, above claim, $$a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}\implies a_1\equiv 4\mod 5\implies a_i\equiv 4\mod 5\forall i\in [2018].$$

So we have $$(a_1)^{2018^{2018}-1}+1\equiv 0\mod 5^{\alpha}\implies (a_1)^{2018}+a_2|(a_1)^{2018^{2018}-1}+1.$$

Now, by LTE, $$v_5((a_1)^{2018^{2018}-1}+1)=v_5(a_1+1)+v_5(2018^{2018}-1)=v_5(a_1+1).$$

Since $(a_1)^{2018}+a_2$ is a power of $5$, we get that $$(a_1)^{2018}+a_2|a_1+1.$$ But $(a_1)^{2018}+a_2>a_1+1$ as $a_i\equiv 4\mod 5$.

The Euler Circuit Theorem


A connected graph contains a closed Euler circuit iff every vertex has an even degree

Thankyou Rohan Bhaiya, Malay and Sid!

Proof: We know that the graph $G$ will contain a cycle $C.$ Now delete the cycle $C.$ Note that the new graph will still have even degrees. 
If the new graph still has non-zero degrees then we claim that it will have a cycle. Consider any vertex $u$ with non zero degree. So it must be adjacent to $a_1$ and $a_2$. Since degrees are even $a_1$ must be adjacent to another vertex $a_3$ and similarly $a_2$ too. Continue this process. This process must terminate sometime, and this process can terminate only when we get a cycle.  

Delete it and so on. Repeat the process. Note that this will continue till our graph becomes a cycle.

Now, we merge all these cycles. 
Say we have a cycle, $V_1\rightarrow V_2\rightarrow \dots \rightarrow V_k\rightarrow V_1$ and we have another cycle, $V_i\rightarrow D_1\rightarrow \dots \rightarrow D_k\rightarrow V_i.$

Then consider this circuit, 
$$V_1\rightarrow V_2\rightarrow \dots V_{i-1}\rightarrow V_i\rightarrow D_1\rightarrow \dots \rightarrow D_k\rightarrow V_i\rightarrow \dots V_k\rightarrow V_1$$

Note that to merge the circuit and the cycle, we must have a vertex in common in both the existing circuit and the cycle to be merged. 

Now, consider any vertex from the existing circuit, say $a$ and any vertex from the cycle say $b$. Since the graph is connected, there must be a path from $a$ to $b$. Say $a\rightarrow a_1\dots \rightarrow a_n\rightarrow b$. 

Say $a,\dots,a_k$ is in the circuit. Then add $a_{k+1}$  and so on ( There must be a cycle with edge $a_ka_{k+1}$. 

And we can repeat this process.

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

How to prepare for RMO?

"Let's wait for this exam to get over".. *Proceeds to wait for 2 whole fricking years!  I always wanted to write a book recommendation list, because I have been asked so many times! But then I was always like "Let's wait for this exam to get over" and so on. Why? You see it's pretty embarrassing to write a "How to prepare for RMO/INMO" post and then proceed to "fail" i.e not qualifying.  Okay okay, you might be thinking, "Sunaina you qualified like in 10th grade itself, you will obviously qualify in 11th and 12th grade." No. It's not that easy. Plus you are talking to a very underconfident girl. I have always underestimated myself. And I think that's the worst thing one can do itself. Am I confident about myself now? Definitely not but I am learning not to self-depreciate myself little by little. Okay, I shall write more about it in the next post describing my experience in 3 different camps and 1 program.  So, I got...
7 comments
Read more

Introduction

  Hey Everyone!! This is my first Blog post. So let me give a brief introduction about myself. I am Sunaina Pati. I love solving Olympiad math problems,  learning crazy astronomical facts , playing hanabi and anti-chess, listening to Kpop , love making diagrams in Geogebra and  teaching other people maths 😊 . I love geometry , number theory and Combinatorics . I am starting this blog to keep myself a bit motivated in doing studies 😎 . Right now, I am planning to write walkthroughs on some of the best problems I tried over the week which can refer for hints 'cause solutions contain some major spoilers and one learns a lot while solving the problem on his own rather than seeing solutions . Also, there will be some reviews about Kpop songs, study techniques, my day to day lifestyles,exam reviews and ofc some non-sense surprises 😂.  I am planning to  try  posting every week on Sundays or Saturdays ( most probably) ! Though there is no guarantee about when I ...
21 comments
Read more

How to prepare for INMO

Since INMO is coming up, it would be nice to write a post about it! A lot of people have been asking me for tips. To people who are visiting this site for the first time, hi! I am Sunaina Pati, an undergrad student at Chennai Mathematical Institute. I was an INMO awardee in 2021,2022,2023. I am also very grateful to be part of the India EGMO 2023 delegation. Thanks to them I got a silver medal!  I think the title of the post might be clickbait for some. What I want to convey is how I would have prepared for INMO if I were to give it again. Anyway, so here are a few tips for people! Practice, practice, practice!! I can not emphasize how important this is. This is the only way you can realise which areas ( that is combinatorics, geometry, number theory, algebra) are your strength and where you need to work on. Try the problems as much as you want, and make sure you use all the ideas you can possibly think of before looking at a hint. So rather than fixing time as a measure to dec...
9 comments
Read more

TOP 10 problems of Week#1

This week was full Geo and NT 😊 . Do try all problems first!! And if you guys get any nice solutions , do post in the comments section! Here are the walkthroughs of this week's top 5 geo problems! 5th position (PUMac 2009 G8) : Consider $\Delta ABC$ and a point $M$ in its interior so that $\angle MAB = 10^{\circ}, \angle MBA = 20^{\circ}, \angle MCA =30^{\circ}$ and $\angle MAC = 40^{\circ}$. What is $\angle MBC$?  Walkthrough: a. Take $D$ as a point on $CM$ such that $\angle DAC=30^{\circ}$, and define $BD\cap AC=E$ . So $\Delta DAC$ is isosceles . b. Show M is the incentre of $\Delta ABD$ c. Show $\angle EDC=60^{\circ}$ d. Show $\Delta BAC$ is isosceles . e. So $\boxed{\angle MBC=60^{\circ}}$ 4th position (IMO SL 2000 G4):  Let $ A_1A_2 \ldots A_n$ be a convex polygon, $ n \geq 4.$ Prove that $ A_1A_2 \ldots A_n$ is cyclic if and only if to each vertex $ A_j$ one can assign a pair $ (b_j, c_j)$ of real numbers, $ j = 1, 2, \ldots, n,$ so that $ A_iA_j = b_jc_i - b_ic_j$ fo...
10 comments
Read more

Some NMTC sub-junior level Problems

Well.. Many people don't know but I was a part of STEM's Horizon ( Now, I have left them due to boards etc.) BTW STEM's Horizons is really great! And anyone interested in Olympiad math should join it! More info about it in below (make sure to check it out!). So here are some problems I sent to them. They are fairly easy, and most of them are repetitive ideas. But they are my first sets of problems ( I know UMO was there but still..) The solutions will be posted in another blog posts. You guys can type out sols in the comments sections too :) Problems:- 1. What is maximum possible number dividing  $x^2+x+1$ and $x^5 +x^4 +x^3 + 3x^2 +2 x +4$ for all $x\in \Bbb{N}$ 2. Let $P$ be the sum of all $x$ and $y$ satisfying $45^x-2^x=2021^y.$ What is the last two digits of $p^2+p+1.$ 3. What is the greatest value of $r$ such that $3^r$ is factor of $10^{2022}-8^{674}$. 4. Find all possible tuples $ (x,y,l)$ such that $\frac{x}{100}=\frac{20}{y}=\frac{5}{l}.$ 5. Consider the following...
13 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

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...
11 comments
Read more

TOP 10 problems of Week#3

 This week was full algebra biased!! 😄  This is my first time trying inequalities, so this pure beginners level.   Do try all the problems first!! And if you guys get any nice solutions , do post in the comments section!  These problem uses only  Power mean Inequality   and  Titu's lemma . The First few problems happen to be  not problems, but tricks(?) which are extensively used..  Here are the walkthroughs of this week's top 10 Inequality problems! 10th position:  Prove that for any real $a>0$ , $a+\frac{1}{a}\ge 2$ Walkthrough: a.  only AM-GM 9th position: Prove that for any real  $a>0$ , $\frac{a}{1+a^2}\le \frac {a}{2a}$. Walkthrough: a. Only AM-GM  b. Use AM-GM to show that   $1+a^2\ge 2a $  8th position:  Prove that for any real $x,y>0$ ,$\frac{1}{x+y}\le \frac{1}{4x}+\frac{1}{4y}$ Walkthrough: a. AM-HM inequality (cute 💖)  7th position :  Prove that for any real positiv...
14 comments
Read more

Problems done in August

  Welcome back! So today I will be sharing a few problems which I did last week and some ISLs. Easy ones I guess. Happy September 2021!  Problem[APMO 2018 P1]: Let $ABC$ be a triangle with orthocenter $H$ and let $M$ and $N$ denote the midpoints of ${AB}$ and ${AC}$. Assume $H$ lies inside quadrilateral $BMNC$, and the circumcircles of $\triangle BMH$ and $\triangle CNH$ are tangent. The line through $H$ parallel to ${BC}$ intersects $(BMH)$ and $(CNH)$ again at $K$, $L$ respectively. Let $F = {MK} \cap {NL}$, and let $J$ denote the incenter of $\triangle MHN$. Prove that $FJ = FA$. Proof:  By angle chase, we get $\angle FKL=\angle FLK.$    Hence $KL||MN\implies FM=FN.$   And we get $\angle MFN=2A\implies F$ is circumcentre if $(AMN)\implies FA=FM=FN.$   And we get $\angle MHN=180-2A$    Hence $MFHN$ is cyclic.    By fact 5, $ME=FJ=FN\implies FJ=FA.$ Problem[Shortlist 2007 G3]: Let $ABCD$ be a trapezoid whose diagonals meet at $P$....
17 comments
Read more