Geometry
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 .
Algebra
Prove the inequality which states that if you let $x_1, x_2,..., x_n$ be positive real numbers, with $n \ge 2$; then you have the inequality $$\frac{x_1}{x_2 + x_3 +... + x_k} +\frac{x_2}{x_1 + x_3 + ... + x_k} + ...+ \frac{x_k}{x_1 + x_2 +... + x_{k-1}} \ge \frac{n}{n-1}$$
Let $a_1, a_2, b_2, b_3, c_1, c_2$ be positive numbers such that $a_1b_1-c^2_1 > 0$ and $a_2b_2-c^2_2> 0$. Prove that $$\frac{8}{(a_1 + a_2)(b_1 + b_2)- (c_1 + c_2)^2 } \le \frac{1}{a_1b_1-c^2_1}+\frac{1}{a_2b_2-c^2_2}$$
Find all functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ such that \[ f(x) + f \left( \frac{1}{x} \right) = 1 \ \text{and} \ f(1 + 2x) = \frac{1}{2} f(x) \]for all $x \in \mathbb{Q}^+$.
Given that $x_1, x_2,..., x_k$ are positive reals such that $\sum^k_{i=1}x^{n-1}_i = k - 1$, prove that $$\frac{x^n_1}{x_2 + x_3 +... + x_k} +\frac{x^n_2}{x_1 + x_3 + ... + x_k} + ...+ \frac{x^n_k}{x_1 + x_2 +... + x_{k-1}} \ge 1$$
Discrete
A computer calculates the $n$-th Fibonacci Number ($F_n$, where $F_n = F_{n-1} + F_{n-2}$ and $F_0 = 0$, $F_1 = 1$) using "steps". A step is defined to be a single calculation (this means carrying out $a + b$ when we already know $a$ and $b$ counts as one step, though if we don't know what $a$ or$ b$ are, carrying out $a + b$ takes the number of steps to find $a +$ the number of steps to find $b + 1$, with the $1$ coming from calculating $a + b$). The computer can only have $F_0$ and $F_1$ permanently stored, and it has to calculate everything else all over again when a request to calculate a new Fibonacci number is made. The advantage of this process is that the final addition step (to add one due to calculating $a+b$) gets omitted(by some algorithmic magic). In this algorithm, the steps needed to "call" $F_0$ and $F_1$ (these are the only values for which a call counts as a step) are both $1$. Given that the computer can only carry out simple addition of $2$ numbers at most : $\bullet$ Find an expression for $S(F_n)$, the number of steps taken to find $F_n$ using this approach if we can only have $F_0$ and $F_1$ stored for all $n \ge 2$ where $S(F_0) = 1$ and $S(F_1) = 1$ due to the "calling" feature $\bullet$ Find an expression for gcd$(S(F_n), S(F_m)))$ involving only n,m and the gcd function for $n,m \ge 2$ Now, the computer has the ability to "cache" (store) all previously calculated Fibonacci Numbers, but the final step is not omitted anymore. Assuming that we've already calculated Fk for some $2 \le k \le n - 1$, and the probability of picking an Fk is equally likely for all $k$ : $\bullet$ Show that the expected number of steps taken to calculate $F_n$, $E(S(F_n))$, using the "cache" feature (if we have $F_0$ and $F_1$ stored and there is no calling step) is $\frac{n-1}{2}$. Note : $E(X) =\sum P(X = x_i)x_i$ here where $x_i$ is any possible value $X$ can take. Also note that the computer does not know that $F_2 = F_1$ or that $F_0 = 0$.
Given that $f(x) = x+1$ and $g(x) = 2x$, how many different ways are there of combining $f(x)$ and $g(x)$ (this means doing any number of compositions like $fg(x)$ or $g^3f^2g(x)$ etc) such that the resulting composition is $8x + 8m$ where $m \ge 0$ is an integer? Find a general formula for the number of possibilities in terms of $m$.
Three frogs are initially on the vertices of an equilateral triangle with sides length of $1$. The frogs can jump over each other in the following way: if frog $A$ at point $M$ jumps over frog $B$ at point $N$, then frog $A$ will land on point $O$ such that $MN = ON$ and $M, N, O$ are collinear. By repeated jumping, is it possible that the three frogs eventually move to the vertices of an equilateral triangle with sides length of $10$?
In chess, a knight moves either two squares vertically and one square horizontally, or two squares horizontally and one square vertically in each move. Suppose a knight can visit every squares on a $4\times n$ chessboard exactly once without repeating. Find all the possible values of $n$.
People from $80$ countries participated in the $1$st round of $IGMO$. In order to ensure that participants from any of the $80$ countries can travel to any one of the remaining $79$ countries by at most $2$ flights, the countries have an agreement such that among the $80$ countries, each country's airport connects to at least $n$ other countries' airport. $\bullet$ Find the minimum value of $n$, proving that it is the minimum. $\bullet$ Prove that for this minimum $n$, if there isn't a direct flight between $2$ countries, then there are at least $2$ paths that require $2$ flights between the countries Note: For two airports $A$ and $B$ to be connected, it must be possible to have a flight from $A$ to $B$ and a flight from $B$ to $A$. This adds $1$ to both the number of connections from the airports. Also note that for this problem, each country has only $1$ airport.
Number Theory
For any natural number $n$ and all natural numbers $d$ dividing $2n^2$ show that $n^2 + d$ is not the square of a natural number
Let's define a function $\phi: \mathbb{N} \to \mathbb{N}$, where $0 \notin \mathbb{N}$, as follows \[ \phi(n) = \sum_{k = 1}^n k! \]Let $\mathbb{V}$ be defined as the set of all triplets $(x,y,z) \in \mathbb{N}$ such that $\phi(x) = y^{z + 1}$. For a triplet $x,y,z$ (denoted by $v$) in $\mathbb{V}$, we define \[ f_v(n) = 8 \left \{ \frac{xy}{8} \right \} \lfloor \sqrt{n} \rfloor + \frac{zn}{z + x - y} \]Show that for any $v \in \mathbb{V}$ and $m \in \mathbb{N}$, the sequence \[ m, f_v(m), f_v(f_v(m)), f_v(f_v(f_v(m))), \dots \]contains at least one square of a natural numbers.
Find $a, b, c, d \in N$ such that: $$a + 2^b + 3^c = 3d! + 1$$and exist $p, q$ prime such that : $$a = (p + 1)(2p + 1) = (q + 1)(q - 1)^2$$
Let $k$ be a positive real number such that $\lfloor kn^2 \rfloor$ is perfect square for all $n \in N$. Show that $k$ must be a perfect square.