On square $ABCD$, point $E$ lies on side $\overline{AD}$ and point $F$ lies on side $\overline{BC}$, so that $BE=EF=FD=30$. Find the area of square $ABCD$.

Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by $10$ percent. Over the weekend, Lou advertises the sale: "Ten percent off the listed price. Sale starts Monday." How much does a pair of shoes cost on Monday that cost $40$ dollars on Thursday? $\textbf{(A)}\ 36 \qquad \textbf{(B)}\ 39.60 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 40.40 \qquad \textbf{(E)}\ 44$

Given sequences $a_n=\frac{1}{n}{\sqrt[n] {_{2n}P_n}},\ b_n=\frac{1}{n^2}{\sqrt[n] {_{4n}P_{2n}}}$ and $c_n=\sqrt[n]{\frac{_{8n}P_{4n}}{_{6n}P_{4n}}}$, find $\lim_{n\to\infty} a_n,\ \lim_{n\to\infty} b_n$and $\lim_{n\to\infty} c_n.$

Let $k\geq 2$ be a natural number. Suppose that $a_1, a_2, \dots a_{2021}$ is a monotone decreasing sequence of non-negative numbers such that \[\sum_{i=n}^{2021}a_i\leq ka_n\] for all $n=1,2,\dots 2021$. Prove that $a_{2021}\leq 4(1-\frac{1}{k})^{2021}a_1$.

A right circular cone has a base with radius 600 and height $200\sqrt{7}$. A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}$. Find the least distance that the fly could have crawled.

For any real number $a$ and positive integer $k$, define \[ {a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}. \]What is \[{-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}?\] $ \textbf{(A)}\ -199\qquad\textbf{(B)}\ -197\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ 197\qquad\textbf{(E)}\ 199 $

The probability that event $A$ occurs is $\frac{3}{4}$; the probability that event $B$ occurs is $\frac{2}{3}$. Let $p$ be the probability that both $A$ and $B$ occur. The smallest interval necessarily containing $p$ is the interval $ \textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad\textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad\textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad\textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad\textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big]$

The ``Manhattan distance" between two cells is the shortest distance between those cells when traveling up, down, left, or right, as if one were traveling along city blocks rather than as the crow flies. Place numbers from $1$-$6$ in some cells so the following criteria are satisfied: $1.$ A cell contains at most one number. Cells can be left empty. $2.$ For each cell containing a number $N$ in the grid, exactly two other cells containing $N$ are at a Manhattan distance of $N.$ $3.$ For each cell containing a number $N$ in the grid, no other cells containing $N$ are at a Manhattan distance less than $N.$ [asy] //credits to fruitmonster97 for the diagram unitsize(1.25cm); //The gridlines for(int i=-3;i<4;++i){ draw((i,3)--(i,-3),lightgray+linewidth(1)); } for(int j=-3;j<4;++j){ if (j==0){ draw((4,j)--(-4,j),lightgray+linewidth(1)); }else{ draw((3,j)--(-3,j),lightgray+linewidth(1)); } } //The outline draw((-4,-1)--(-4,1)--(-3,1)--(-3,3)--(3,3)--(3,1)--(4,1)--(4,-1)--(3,-1)--(3,-3)--(-3,-3)--(-3,-1)--cycle); //The numbers label("$1$",(-0.5,0.5)); label("$1$",(0.5,-0.5)); label("$2$",(-1.5,1.5)); label("$2$",(-2.5,-1.5)); label("$3$",(2.5,1.5)); label("$3$",(-3.5,0.5)); label("$4$",(3.5,-0.5)); label("$4$",(1.5,-2.5)); label("$4$",(-1.5,2.5)); label("$5$",(-2.5,1.5)); label("$5$",(2.5,-1.5)); label("$6$",(1.5,-1.5)); [/asy]

Show that the following equation has finitely many solutions $(t,A,x,y,z)$ in positive integers $$\sqrt{t(1-A^{-2})(1-x^{-2})(1-y^{-2})(1-z^{-2})}=(1+x^{-1})(1+y^{-1})(1+z^{-1})$$

We have a blue triangle. In every move, we divide the blue triangle by angle bisector to $2$ triangles and color one triangle in red. Prove, that after some moves we color more than half of the original triangle in red.

Given a positive integer $n \geq 2$, whose canonical prime factorization is $n = p_1^{\alpha_1}p_2^{\alpha_2} \ldots p_k^{\alpha_k}$, we define the following functions: $$\varphi(n) = n\bigg(1 -\frac{1}{p_1}\bigg) \bigg(1 -\frac{1}{p_2}\bigg) \ldots \bigg(1 -\frac {1}{p_k}\bigg) ; \overline{\varphi}(n) = n\bigg(1 +\frac{1}{p_1}\bigg) \bigg(1 +\frac{1}{p_2}\bigg) \ldots \bigg(1 + \frac{1}{p_k}\bigg)$$ Consider all positive integers $n$ such that $\overline{\varphi}(n)$ is a multiple of $n + \varphi(n) $. (a) Prove that $n$ is even. (b) Determine all positive integers $n$ that satisfy this property.

Steve is piling $m\geq 1$ indistinguishable stones on the squares of an $n\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform [i]stone moves[/i], defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l\leq n$, such that $i<j$ and $k<l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively, or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively. Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves. How many different non-equivalent ways can Steve pile the stones on the grid?

Prove that if \( a \), \( b \), and \( c \) are positive real numbers, then the following inequality holds: \[ \frac{a + 3c}{a + b} + \frac{c + 3a}{b + c} + \frac{4b}{c + a} \geq 6. \]

Suppose that points $A$ and $B$ lie on circle $\Omega$, and suppose that points $C$ and $D$ are the trisection points of major arc $AB$, with $C$ closer to $B$ than $A$. Let $E$ be the intersection of line $AB$ with the line tangent to $\Omega$ at $C$. Suppose that $DC = 8$ and $DB = 11$. If $DE = a\sqrt{b}$ for integers $a$ and $b$ with $b$ squarefree, find $a + b$.

In the plane, consider a point $ X$ and a polygon $ \mathcal{F}$ (which is not necessarily convex). Let $ p$ denote the perimeter of $ \mathcal{F}$, let $ d$ be the sum of the distances from the point $ X$ to the vertices of $ \mathcal{F}$, and let $ h$ be the sum of the distances from the point $ X$ to the sidelines of $ \mathcal{F}$. Prove that $ d^2 \minus{} h^2\geq\frac {p^2}{4}.$

Let $y(x)$ be a continuously differentiable real-valued function of a real variable $x$. Show that if $y'(x)^2 +y(x)^3 \to 0$ as $x\to \infty,$ then $y(x)$ and $y'(x) \to 0$ as $x \to \infty.$

Let $ M \equal{} \lbrace 2, 3, 4, \ldots\, 1000 \rbrace$. Find the smallest $ n \in \mathbb{N}$ such that any $ n$-element subset of $ M$ contains 3 pairwise disjoint 4-element subsets $ S, T, U$ such that [b]I.[/b] For any 2 elements in $ S$, the larger number is a multiple of the smaller number. The same applies for $ T$ and $ U$. [b]II.[/b] For any $ s \in S$ and $ t \in T$, $ (s,t) \equal{} 1$. [b]III.[/b] For any $ s \in S$ and $ u \in U$, $ (s,u) > 1$.

The numbers $1$ to $12$ are arranged in a sequence. The number of ways this can be done equals $12 \times11 \times 10\times ...\times 1$. We impose the condition that in the sequence there should be exactly one number that is smaller than the number directly preceding it. How many of the $12 \times11 \times 10\times ...\times 1$ sequences satisfy this condition?

Let $f$ be a polynomial such that, for all real number $x$, $f(-x^2-x-1) = x^4 + 2x^3 + 2022x^2 + 2021x + 2019$. Compute $f(2018)$.

We are given a polyhedron with at least $5$ vertices, such that exactly $3$ edges meet in each of the vertices. Prove that we can assign a rational number to every vertex of the given polyhedron such that the following conditions are met: $(i)$ At least one of the numbers assigned to the vertices is equal to $2020$. $(ii)$ For every polygonal face, the product of the numbers assigned to the vertices of that face is equal to $1$.

Let $x, y$ be positive integers such that $\frac{x^2}{y}+\frac{y^2}{x}$ is an integer. Prove that $y|x^2$.

Every point of the plane is painted with one of three colors. Can we always find two points a distance $1$ cm apart which are of the same color?

A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains $N$ astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength $g$. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength $g'$ . What is the ratio $g'/g$ in terms of $N$?[asy] import roundedpath; size(300); path a = roundedpath((0,-0.3)--(4,-0.3)--(4,-1)--(5,-1)--(5,0),0.1); draw(scale(+1,-1)*a); draw(scale(+1,+1)*a); draw(scale(-1,-1)*a); draw(scale(-1,+1)*a); filldraw(circle((0,0),1),white,black); filldraw(box((-2,-0.27),(2,0.27)),white,white); draw(arc((0,0),1.5,+35,+150),dashed,Arrow); draw(arc((0,0),1.5,-150,-35),dashed,Arrow);[/asy] $ \textbf{(A)}\ 2N/(N-1) $ $ \textbf{(B)}\ N/(N-1) $ $ \textbf{(C)}\ \sqrt{(N-1)/N} $ $ \textbf{(D)}\ \sqrt{N/(N-1)} $ $ \textbf{(E)}\ \text{none of the above} $

Find the sum of the positive integer solutions to the equation $\left\lfloor\sqrt[3]{x}\right\rfloor+\left\lfloor\sqrt[4]{x}\right\rfloor=4.$

In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$. Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$. [i]Proposed by Lewis Chen[/i]