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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.

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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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