The Integration Premier League has $n$ teams competing. The tournament follows a round-robin system, that is, where every pair of teams play each other exactly once. So every team plays exactly $n-1$ matches. The top $m \leq n$ temas at the end of the tournament qualify for the playoffs. Assume there are no tied matches. Let $A(m,n)$ be the minimum number of matches a team has to win to gurantee selection for the playoffs, regardless of what their run rate is. For example, $A(n,n) = 0$ (everyone qualifies anyway so no need to win!) and $A(1,n) = n-1$ (even if you lose to just one other team, they might defeat everyone and qualify instead of you). Answer the following: $(A)$ FInd the value of $A(2,4),A(2,6)$ and $A(4,10)$ with proof (explain why a smaller value can still lead to the team not qualifying, and show that the respective values themselves are enough). $(B)$ Show that $A(n-1,n) = \frac{n}{2}$ when $n$ even and $ = \frac{n+1}{2}$ when $n$ odd. $(C)$ For bonus marks, try to find a general pattern for $A(m,n)$.

For any positive integer number $k$, the factorial $k!$ is defined as a product of all integers between $1$ and $k$ inclusive: $k!=k\times{(k-1)}\times\dots\times{1}$. Let $s(n)$ denote the sum of the first $n$ factorials, i.e. $$s(n)=\underbrace{n\times{(n-1)}\times\dots\times{1}}_{n!}+\underbrace{(n-1)\times{(n-2)}\times\dots\times{1}}_{(n-1)!}+\cdots +\underbrace{2\times{1}}_{2!}+\underbrace{1}_{1!}$$ Find the remainder when $s(2024)$ is divided by $8$

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in the swamp, and they make the following statements: Brian: "Mike and I are different species." Chris: "LeRoy is a frog." LeRoy: "Chris is a frog." Mike: "Of the four of us, at least two are toads." How many of these amphibians are frogs? $ \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Let $0<a\leq 1$ be a real number and let $a\leq a_{i}\leq\frac{1}{a_{i}}\forall i=\overline{1,1996}$ are real numbers. Prove that for any nonnegative real numbers $k_{i}(i=1,2,...,1996)$ such that $\sum_{i=1}^{1996}k_{i}=1$ we have $(\sum_{i=1}^{1996}k_{i}a_{i})(\sum_{i=1}^{1996}\frac{k_{i}}{a_{i}})\leq (a+\frac{1}{a})^{2}$.

We are given $n$ intervals $[l_1,r_1],[l_2,r_2],[l_3,r_3],\dots, [l_n,r_n]$ in the number line. We can divide the intervals into two sets such that no two intervals in the same set have overlaps. Prove that there are at most $n-1$ pairs of overlapping intervals.

$2002$ cards with numbers $1,2,\ldots ,2002$ written on them are put on a table face up. Two players $A,B$ take turns to pick up a card until all are gone. $A$ goes first. The player who gets the last digit of the sum of his cards larger than his opponent wins. Who has a winning strategy and how should one play to win?

Compute the prime factorization of $1007021035035021007001$. (You should write your answer in the form $p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$ where $p_1,\ldots,p_k$ are distinct prime numbers and $e_1,\ldots,e_k$ are positive integers.)

In a right triangle $ABC$, M is a point on the hypotenuse $BC$ and $P$ and $Q$ the projections of $M$ on $AB$ and $AC$ respectively. Prove that for no such point $M$ do the triangles $BPM, MQC$ and the rectangle $AQMP$ have the same area.

(a) Alenka has two jars, each with $6$ marbles labeled with numbers $1$ through $6$. She draws one marble from each jar at random. Denote by $p_n$ the probability that the sum of the labels of the two drawn marbles is $n$. Compute pn for each $n \in N$. (b) Barbara has two jars, each with $6$ marbles which are labeled with unknown numbers. The sets of labels in the two jars may differ and two marbles in the same jar can have the same label. If she draws one marble from each jar at random, the probability that the sum of the labels of the drawn marbles is $n$ equals the probability $p_n$ in Alenka’s case. Determine the labels of the marbles. Find all solution

Let $ m,n$ be two natural nonzero numbers and sets $ A \equal{} \{ 1,2,...,n\}, B \equal{} \{1,2,...,m\}$. We say that subset $ S$ of Cartesian product $ A \times B$ has property $ (j)$ if $ (a \minus{} x)(b \minus{} y)\le 0$ for each pairs $ (a,b),(x,y) \in S$. Prove that every set $ S$ with propery $ (j)$ has at most $ m \plus{} n \minus{} 1$ elements. [color=#FF0000]The statement was edited, in order to reflect the actual problem asked. The sign of the inequality was inadvertently reversed into $ (a \minus{} x)(b \minus{} y)\ge 0$, and that accounts for the following two posts.[/color]

Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3,4,12, and 13, respectively, and $\measuredangle CBA$ is a right angle. The area of the quadrilateral is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy] $\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$

As shown in the figure, two circles $\Gamma_1$ and $\Gamma_2$ on the plane intersect at two points $A$ and $B$. The two rays passing through $A$, $\ell_1$ and $\ell_2$ intersect $\Gamma_1$ at points $D$ and $E$ respectively, and $\Gamma_2$ at points $F$ and $C$ respectively (where $E$ and $F$ lie on line segments $AC$ and $AD$ respectively, and neither of them coincides with the endpoints). It is known that the three lines $AB$, $CF$ and $DE$ have a common point, the circumscribed circle of $\vartriangle AEF$ intersects $AB$ at point $G$, the straight line $EG$ intersects the circle $\Gamma_1$ at point $P$, the straight line $FG$ intersects the circle $\Gamma_2$ at point $Q$. Let the symmetric points of $C$ and $D$ wrt the straight line $AB$ be $C'$ and $D'$ respectively. If $PD'$ and $QC'$ intersect at point$ J$, prove that $J$ lies on the straight line $AB$. [img]https://cdn.artofproblemsolving.com/attachments/3/7/eb3acdbad52750a6879b4b6955dfdb7de19ed3.png[/img]

In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line. [i]Proposed by Theoklitos Parayiou, Cyprus [/i]

There are 2019 cards in a box. Each card has a number written on one of its sides and a letter on the other side. Amy and Ben play the following game: in the beginning Amy takes all the cards, places them on a line and then she flips as many cards as she wishes. Each time Ben touches a card he has to flip it and its neighboring cards. Ben is allowed to have as many as 2019 touches. Ben wins if all the cards are on the numbers' side, otherwise Amy wins. Determine who has a winning strategy.

Two circles in the plane do not intersect and do not lie inside each other. We choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2'$ intersect. Let $A$ and $B$ be the midpoints of the segments $A_1A_2$ and $B_1B_2$, and $C$ be the intersection point of these segments. Prove that the orthocenter of the triangle $ABC$ belongs to a fixed line that does not depend on the choice of diameters.

[b]p1.[/b] A rectangle $ABCD$ is cut from a piece of paper and folded along a straight line so that the diagonally opposite vertices $A$ and $C$ coincide. Find the length of the resulting crease in terms of the length ($\ell$) and width ($w$) of the rectangle. (Justify your answer.) [b]p2.[/b] The residents of Andromeda use only bills of denominations $\$3 $and $\$5$ . All payments are made exactly, with no change given. What whole-dollar payments are not possible? (Justify your answer.) [b]p3.[/b] A set consists of $21$ objects with (positive) weights $w_1, w_2, w_3, ..., w_{21}$ . Whenever any subset of $10$ objects is selected, then there is a subset consisting of either $10$ or $11$ of the remaining objects such that the two subsets have equal fotal weights. Find all possible weights for the objects. (Justify your answer.) [b]p4.[/b] Let $P(x) = x^3 + x^2 - 1$ and $Q(x) = x^3 - x - 1$ . Given that $r$ and $s$ are two distinct solutions of $P(x) = 0$ , prove that $rs$ is a solution of $Q(x) = 0$ [b]p5.[/b] Given: $\vartriangle ABC$ with points $A_1$ and $A_2$ on $BC$ , $B_1$ and $B_2$ on $CA$, and $C_1$ and $C_2$ on $AB$. $A_1 , A_2, B_1 , B_2$ are on a circle, $B_1 , B_2, C_1 , C_2$ are on a circle, and $C_1 , C_2, A_1 , A_2$ are on a circle. The centers of these circles lie in the interior of the triangle. Prove: All six points $A_1$ , $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are on a circle. [img]https://cdn.artofproblemsolving.com/attachments/7/2/2b99ddf4f258232c910c062e4190d8617af6fa.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].