Prove that for all positive real numbers $x, m, a, s$, $$6 + 6^{x+m} + 6^{x+m+a} + 6^{x+m+a+s} >\frac12(6^{x+1}) +\frac12(6^{m+1}) +\frac13(6^{a+1}) + 6^s$$

Santa and an invisible elf play a hide and seek game in the Euclidean plane. Firstly, the elf chooses $3$ points, $A_1$, $A_2$ and $A_3$. These points are known to Santa. Also, we define $A_s = A_{s-3}$ for all $s \ge 4$. Then the elf chooses a point $P_0$ such that the distant between $P_0$ and $A_1$ is $100$. $P_0$ is his original position, and it is not known to Santa. In the beginning of round $n$, Santa chooses a number $\theta_n$ between $45$ to $90$, and then the elf will move to point $P_n$, which is defined as the point where $P_{n-1}$ is rotated $\theta^o_n$ anti-clockwise about $A_n$. The point is not known to Santa since the elf is invisible. Santa will then choose an area to scan using an elf detector. The detector can scan a circular area of radius of $1$. If the invisible elf is within the area of scanning (inside the circle or on the edge of the circle) of the detector, then Santa wins. Does there exist a winning strategy to ensure that Santa can win within $2022$ rounds? Prove your claim. Note : Assume the invisible elf is a point, i.e. he has no area.

Santa decorates his Christmas tree with a triangular decoration. Suppose the triangular decoration can be represented by $\vartriangle ABC$. Let $\omega$ be its incircle and $\omega_A$, $\omega_B$, $\omega_C$ be its $A$-, $B$-, $C$-excircles respectively. Let $J_A$, $J_B$, $J_C$ be the A-, $B$-, $C$-excentres of $\vartriangle ABC$ respectively. $X$ is the radical centre of $\omega$, $\omega_B$, $\omega_C$. $Y$ is the radical centre of $\omega$, $\omega_C$, $\omega_A$. $Z$ is the radical centre of $\omega$, $\omega_A$, $\omega_B$. Prove that $XJ_A$, $Y J_B$, $ZJ_C$ are concurrent

Because of inflation, Santa can’t afford buying gifts for all children this year. He decided to divide his ordered list of good children into those who will get a gift and those who won’t. To be as fair as possible, he came up with the following (seemingly random) rule: ”Each child has its own number on my list. If $n$ is a positive integer which satisfies $$\tau (nk) \le k \cdot \tau (n)$$for all positive integers $k \ge 2$, then the $n^{th}$ child from my list will get a gift. ” Characterize all positive integers $n$ such that the $n^{th}$ child from the list gets a gift this year. Note: $\tau (n)$ denotes the number of positive divisors of $n$.

Santa draws a Christmas tree in the following way. He first draws an acute-angled triangle $\vartriangle ABC$. He then lets $M$ be the mid-point of $BC$, $A'$ be the point of reflection of $A$ over $BC$, $D$ be the point of intersection of line segment $AM$ and the circumcircle of $\vartriangle A'BC$, $E$ and $F$ be points on $AB$ and $AC$ respectively such that $D$, $E$, $F$ are collinear. Prove that $\frac{AE}{ED \cdot DB} = \frac{AF}{FD\cdot DC}$ . Note: Line segments $AE$, $ED$, $DB$, $BC$, $CD$, $DF$, $FA$ form the shape of a Christmas tree!

After delivering all the Christmas presents, Santa finally have some leisure time to do Maths, which is his favourite hobby. Santa proposes two new sequences: Christmas sequence and Santa sequence. Numbers in the Christmas sequence are known as Christmas numbers. The Christmas sequence is defined as: $C_0 = 0, C_1 = 1, C_{n+1} = 2022C_n + C_{n-1}$ for $n \ge1$. The Santa sequence is defined as: $S_0 = 2, S_1 = 2022, S_{n+1} = 2022S_n + S_{n-1}$ for $n \ge1$. Santa finds $4043$ children and labels them from $1$ to $4043$. He asks the $n^{th}$ child to express $$C_1S_{2023} + C_2S_{2024} + ... + C_{2022}S_{4044}$$as a sum of $n$ non-zero Christmas numbers. Those who can do so can get an extra gift, which is a cute Christmas frog. Amongst the $4043$ children, who can potentially get an extra gift? Prove your claim.