A non-rectangular trapezoid is called a Pepe Trapezoid if (a) it has integral side lengths, and (b) An ellipse that is not a circle with integral lengths of semi-major axis and semi-minor axis can be inscribed in the trapezoid such that the major axis or minor axis of the ellipse is perpendicular to the bases of the trapezoid. Prove or disprove that there are infinitely many non-similar Pepe trapezoids.

If $ABCD$ is a cyclic quadrilateral; $N_1, N_2, N_3$ and $N_4$ are the nine-point centres of $\vartriangle ABC$, $\vartriangle BCD$, $\vartriangle CDA$ and $\vartriangle DAB$, respectively. A circle with $AD$ as diameter meets $CD$ again at point $E$. Another circle with $AB$ as diameter meets BC again at point F. Prove that $N_1, N_2, N_3$ and $N_4$ are concyclic and their circumcircle is bisected by line $EF$. by @pepemaths

For any given ellipse $\omega$ , let P be a point external to it. $A$ and $B$ are points on $\omega$ such that $P A$ and $P B$ are tangent to $\omega$ . C, D $\omega$ and E are points on $\omega$ such that $AC = CD = DE = EB$. Lines $P C, P D, P E$ meets $\omega$ again at points $F, G, H$, respectively. $M_1, M_2, M_3$ are midpoints of $CF, DG$ and $EH$, respectively. Prove that $P, A, B, M_1, M_2$ and $M_3$ lie on an ellipse. by @pepemaths

A finite number of points are plotted so that the distances between any two are distinct. Each point joins the one closest to it. Find the maximum number of segments that can start from a single point.

Let $I, H, O$ be the incentre, orthocentre and circumcentre of $\vartriangle ABC$ respectively. $D$ is the circumcentre of $\vartriangle AIC$. $H$ is reflected along $BC$ and $AB$ to $E$ and $F$ respectively. Prove that $D, O, F$ are collinear if and only if $DE$ is perpendicular to $EF$. by @pepemaths

A sphere of radius $r$ can be inscribed in a tetrahedron. The distances between the centroid of the tetrahedron and its four faces are $w, x, y$ and $z$. Prove that $wxyz \ge r^4$. by @pepemaths .

Santa Claus decorates his Christmas tree with a decoration which has a shape of a regular $12$-sided polygon. Let the $12$-sided polygon be $A_1A_2A_3...A_{12}$. Suppose $I_1$, $I_2$ and $I_3$ are the incentres of $\vartriangle A_1A_2A_5$, $\vartriangle A_5A_7A_8$ and $\vartriangle A_8A_{11}A_1$ respectively. Prove that $I_1A_8$, $I_2A_1$ and $I_3A_5$ are concurrent.

Let $ABC$ be a non-isosceles acute angle triangle. $A_0$ is the point of intersection of the $A$-symmedian and the circumcircle of $\vartriangle ABC$ (other than $A$). $A_1$ is the mid-point of $AA_0$. $A_2$ is the point of reflection of $A_0$ over the side opposite to $A$ (which is $BC$). $B_0$, $B_1$, $B_2$, $C_0$, $C_1$, $C_2$ are defined similarly. $O_1$, $O_2$, $N$ are the circumcentres of $\vartriangle A_1B_1C_1$, $\vartriangle A_2B_2C_2$ and the triangle formed by the midpoints of the three sides of $\vartriangle ABC$, respectively. $M$ is the midpoint of the centroid and the symmedian point of $\vartriangle ABC$. Prove that $O_1MO_2N$ is a parallelogram.

Let $O$ be a fixed point on a plane. $P_1$, $P_2$, $...$ , $P_{2022}$ are $2022$ variable points on the same plane which do not coincide with $O$. Initially, $P_1$ is an arbitrary point and the line segment $OP_{n+1}$ is $\frac{\pi}{1011}$ anticlockwise to the line segment $OP_n$ for $n = 1$, $2$,$...$, $2021$. In other words, $\angle P_nOP_{n+1} = \frac{\pi}{1011}$ in directed angles. A group of points is said to be "perfect" if $O$ is the point such that the sum of the distances from that point to all the points in the group is the smallest possible. For example, the group of points formed by $P_1$, $P_2$, $P_3$, $P_4$, $P_5$ is said to be "perfect" if $$OP_1 + OP_2 + OP_3 + OP_4 + OP_5 \le XP_1 + XP_2 + XP_3 + XP_4 + XP_5$$for any point $X$ on the plane. (a) Prove that the group of points formed by $P_1$, $P_2$, $...$, $P_{2022}$ is "perfect" initially. (b) You are given a task to remove the points $P_1$, $P_2$, $...$, $P_{2022}$ according the following rules. In each step, you could remove one point, and then rotate another point anticlockwise for $\frac{\pi}{3}$ around point $O$. After each step, the group formed by the remaining points on the plane must remain to be "perfect". Devise an algorithm to remove $1348$ points from the initial $2022$ points and prove that it works.

Let $P$ be a point in the interior of $\vartriangle ABC$. $R_1$, $R_2$, $R_3$ are the circumradii of $\vartriangle PAB$, $\vartriangle PBC$ and $\vartriangle PCA$ respectively. Find the minimum value of $\frac{R_1+R_2+R_3}{AB+BC+CA}$ and prove that it is the minimum.

Let $\vartriangle ABC$ be a triangle. Its incircle $\omega$ touches $ BC$, $CA$, $AB$ at points $A_0$, $B_0$, $C_0$ respectively. $AA_0$ meets $\omega$ at $A_1$. $A_2$ is the point of reflection of $A_1$ over $B_0C_0$. $B_1$, $B_2$, $C_1$, $C_2$ are dened similarly. Prove that the circumcentre of $\vartriangle A_2B_2C_2$ lies on the Euler line of $\vartriangle A_0B_0C_0$.

For $\vartriangle ABC$, let $D$ and $E$ be points on line segments $AB$ and $AC$ respectively such that $DE$ is parallel to $AB$. Suppose that $BD$ and $CE$ meet at $F$, and the tangents to the circumcircles of $\vartriangle ADF$ and $\vartriangle AEF$ at $D$ and $E$ respectively meet at $X$. Prove that there exist points $P, Q, R, S, T$ on lines $AF$, $AX$, $AB$, $BC$, $CA$ respectively such that (1) $PR + QR = PS + QS = PT + QT$ and (2) $AS, BT$ and $CR$ are concurrent.

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

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!

Let $\vartriangle ABC$ be a scalene (i.e. non-isosceles) triangle with circumcentre $O$. Let $J_A$, $J_B$, $J_C$ be the $A$-, $B$-, $C$-excentre of $\vartriangle ABC$ respectively. $OJ_A$ intersects $BC$ at $X$, $OJ_B$ intersects $CA$ at $Y$ , $OJ_C$ intersects $AB$ at $Z$. Prove that $AX$, $BY$ , $CZ$ and the Euler line of $\vartriangle ABC$ are concurrent. Note: The $A$-excentre of $\vartriangle ABC$ is the centre of the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond C. The $B$-, $C$-excentre of $\vartriangle ABC$ are defined similarly. The Euler line of a triangle is the line which passes through the orthocentre and the circumcentre of the triangle.

Let $\vartriangle ABC$ be an acute-angled scalene (i.e. non-isosceles) triangle. $D$ and $E$ are the foot of $B$-altitude and C-altitude of $\vartriangle ABC$ respectively. $M$ is the midpoint of $BC$. The circumcircles of $\vartriangle BEM$ and $\vartriangle CDM$ meet at $F$ (other than $M$). $X$ and $Y$ are the points of projection of $F$ on $BD$ and $CE$ respectively. Prove that $AF$ and $XY$ meet at the circumcircle of $\vartriangle DEM$.

Pepe has a large garden for frogs. He plays a game with Lintz the pig in the garden. The rules are as follows: 1. A positive real number $s$ is randomly generated by a computer system, called FROG (short form for “Fast, Random, Outstanding Generator of numbers”). 2. Lintz then draws a circle $K_1$ in Pepe’s garden (the garden is assumed to be an Euclidean plane). 3. Pepe then draws another circle $K_2$ in his garden. 4. After that, Lintz colours a finite number of arcs of $K_1$ in yellow, such that the total length of yellow arcs is more than two-third of the circumference of $K_1$. He also colours a finite number of arcs of $K_2$ in purple, such that the total length of purple arcs is more than two-third of the circumference of $K_2$. 5. Pepe puts three frogs, $A_1$, $B_1$, $C_1$ on the yellow arcs of $K_1$. He also puts three frogs, $A_2$, $B_2$, $C_2$ on the purple arcs of $K_2$. (Assume that the frogs are points, i.e. they have no areas.) Define $m_{A_1}$ , $m_{B_1}$ , $m_{C_1}$ , $h_{A_1}$ , $h_{B_1}$ , $h_{C_1}$ as the lengths of the $A$-median, $B$-median, $C$-median, $A$-altitude, $B$-altitude and $C$-altitude of $\vartriangle A_1B_1C_1$ respectively. $m_{A_2}$ , $m_{B_2}$ , $m_{C_2}$ , $h_{A_2}$ , $h_{B_2}$ , $h_{C_2}$ are defined similarly for $\vartriangle A_2B_2C_2$. Pepe wins if $$m^2_{A_2} (m^2_{B_1} + m^2_{C_1}- m^2_{A_1}) + m^2_{B_2} (m^2_{C_1} + m^2_{A_1}- m^2_{B_1}) + m^2_{C_2} (m^2_{A_1} + m^2_{B_1}- m^2_{C_1})$$$$+ \frac{1}{h^2_{A_2}}\left( \frac{1}{h^2_{B_1}}+\frac{1}{h^2_{C_1}}-\frac{1}{h^2_{A_1}}\right) + \frac{1}{h^2_{B_2}}\left( \frac{1}{h^2_{C_1}}+\frac{1}{h^2_{A_1}}-\frac{1}{h^2_{B_1}}\right) +\frac{1}{h^2_{C_2}}\left( \frac{1}{h^2_{A_1}}+\frac{1}{h^2_{B_1}}-\frac{1}{h^2_{C_1}}\right)= s$$Otherwise, Lintz wins. Find the range of $s$ where Pepe has a winning strategy and the range of $s$ where Lintz has a winning strategy. Prove your claim. PSI added last equation as an attachment also because it had too many indices, in case I mistyped any of them.