This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 85335

2007 Nicolae Coculescu, 2

Tags: inequalities
Let be $ a,b,c\in (0,\infty)$ such that $ 3abc\equal{}1$ . Show that : $ \frac{3a^5}{3a^5\plus{}2bc}\plus{}\frac{3b^5}{3b^5\plus{}2ca}\plus{}\frac{3c^5}{3c^5\plus{}2ab}\ge 1$

2010 Contests, 3

Tags:
Find the number of $4$-digit numbers (in base $10$) having non-zero digits and which are divisible by $4$ but not by $8$.

2001 Mexico National Olympiad, 4

For positive integers $n, m$ define $f(n,m)$ as follows. Write a list of $ 2001$ numbers $a_i$, where $a_1 = m$, and $a_{k+1}$ is the residue of $a_k^2$ $mod \, n$ (for $k = 1, 2,..., 2000$). Then put $f(n,m) = a_1-a_2 + a_3 -a_4 + a_5- ... + a_{2001}$. For which $n \ge 5$ can we find m such that $2 \le m \le n/2$ and $f(m,n) > 0$?

2016 PUMaC Number Theory A, 7

Compute the number of positive integers $n$ between $2017$ and $2017^2$ such that $n^n \equiv 1$ (mod $2017$). ($2017$ is prime.)

2013 Tournament of Towns, 7

On a table, there are $11$ piles of ten stones each. Pete and Basil play the following game. In turns they take $1, 2$ or $3$ stones at a time: Pete takes stones from any single pile while Basil takes stones from different piles but no more than one from each. Pete moves fi rst. The player who cannot move, loses. Which of the players, Pete or Basil, has a winning strategy?

2004 Bosnia and Herzegovina Junior BMO TST, 4

Let $ABCD$ be a parallelogram. On the ray $(DB$ a point $E$ is given such that the ray $(AB$ is the angle bisector of $\angle CAE$. Let $F$ be the intersection of $CE$ and $AB$. Prove that $\frac{AB}{BF} - \frac{AC}{AE} = 1$

2018 Nepal National Olympiad, 4a

Tags: geometry
[b]Problem Section #4 a) There is a $6 * 6$ grid, each square filled with a grasshopper. After the bell rings, each grasshopper jumps to an adjacent square (A square that shares a side). What is the maximum number of empty squares possible?

2008 F = Ma, 25

Tags:
Two satellites are launched at a distance $R$ from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed $v_\text{0}$ and enters a circular orbit. The second satellite, however, is launched at a speed $\frac{1}{2}v_\text{0}$. What is the minimum distance between the second satellite and the planet over the course of its orbit? (a) $\frac{1}{\sqrt{2}}R$ (b) $\frac{1}{2}R$ (c) $\frac{1}{3}R$ (d) $\frac{1}{4}R$ (e) $\frac{1}{7}R$

2021 Science ON all problems, 4

The numbers $\frac 32$, $\frac 43$ and $\frac 65$ are intially written on the blackboard. A move consists of erasing one of the numbers from the blackboard, call it $a$, and replacing it with $bc-b-c+2$, where $b,c$ are the other two numbers currently written on the blackboard. Is it possible that $\frac{1000}{999}$ would eventually appear on the blackboard? What about $\frac{113}{108}$? [i] (Andrei Bâra)[/i]

1996 Mexico National Olympiad, 1

Let $P$ and $Q$ be the points on the diagonal $BD$ of a quadrilateral $ABCD$ such that $BP = PQ = QD$. Let $AP$ and $BC$ meet at $E$, and let $AQ$ meet $DC$ at $F$. (a) Prove that if $ABCD$ is a parallelogram, then $E$ and $F$ are the midpoints of the corresponding sides. (b) Prove the converse of (a).

CNCM Online Round 1, 7

Tags:
Three cats--TheInnocentKitten, TheNeutralKitten, and TheGuiltyKitten labelled $P_1, P_2,$ and $P_3$ respectively with $P_{n+3} = P_{n}$--are playing a game with three rounds as follows: [list=1] [*] Each round has three turns. For round $r \in \{1,2,3\}$ and turn $t \in \{1,2,3\}$ in that round, player $P_{t+1-r}$ picks a non-negative integer. The turns in each round occur in increasing order of $t$, and the rounds occur in increasing order of $r$. $\newline \newline$ [*] [b]Motivations:[/b] Every player focuses primarily on maximizing the sum of their own choices and secondarily on minimizing the total of the other players’ sums. TheNeutralKitten and TheGuiltyKitten have the additional tertiary priority of minimizing TheInnocentKitten’s sum. $\newline \newline$ [*] For round $2$, player $P_{2}$ has no choice but to pick the number equal to what player $P_{1}$ chose in round $1$. Likewise, for round $3$, player $P_{3}$ must pick the number equal to what player $P_{2}$ chose in round $2$. $\newline \newline$ [*] If not all three players choose their numbers such that the values they chose in rounds 1,2,3 form an arithmetic progression in that order by the end of the game, all players' sums are set to $-1$ regardless of what they have chosen. $\newline \newline$ [*] If the sum of the choices in any given round is greater than $100$, all choices that round are set to $0$ at the end of that round. That is, rules $2$, $3$, and $4$ act as if each player chose $0$ that round. $\newline \newline$ [*] All players play optimally as per their motivations. Furthermore, all players know that all other players will play optimally (and so on.) [/list] Let $A$ and $B$ be TheInnocentKitten's sum and TheGuiltyKitten's sum respectively. Compute $1000A + B$ when all players play optimally. Proposed by Harry Chen (Extile)

1984 AIME Problems, 12

A function $f$ is defined for all real numbers and satisfies \[f(2 + x) = f(2 - x)\qquad\text{and}\qquad f(7 + x) = f(7 - x)\] for all real $x$. If $x = 0$ is a root of $f(x) = 0$, what is the least number of roots $f(x) = 0$ must have in the interval $-1000 \le x \le 1000$?

1996 Putnam, 3

Tags:
Let $S_n$ be the set of all permutations of $(1,2,\ldots,n)$. Then find : \[ \max_{\sigma \in S_n} \left(\sum_{i=1}^{n} \sigma(i)\sigma(i+1)\right) \] where $\sigma(n+1)=\sigma(1)$.

Kyiv City MO Seniors 2003+ geometry, 2018.11.4.1

In the quadrilateral $ABCD$, the diagonal $AC$ is the bisector $\angle BAD$ and $\angle ADC = \angle ACB$. The points $X, \, \, Y$ are the feet of the perpendiculars drawn from the point $A$ on the lines $BC, \, \, CD$, respectively. Prove that the orthocenter $\Delta AXY$ lies on the line $BD$.

2023 IMC, 10

For every positive integer $n$, let $f(n)$, $g(n)$ be the minimal positive integers such that \[1+\frac{1}{1!}+\frac{1}{2!}+\dots +\frac{1}{n!}=\frac{f(n)}{g(n)}.\] Determine whether there exists a positive integer $n$ for which $g(n)>n^{0.999n}$.

1985 AMC 12/AHSME, 30

Let $ \lfloor x \rfloor$ be the greatest integer less than or equal to $ x$. Then the number of real solutions to $ 4x^2 \minus{} 40 \lfloor x \rfloor \plus{} 51 \equal{} 0$ is $ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

1991 IMO Shortlist, 2

$ ABC$ is an acute-angled triangle. $ M$ is the midpoint of $ BC$ and $ P$ is the point on $ AM$ such that $ MB \equal{} MP$. $ H$ is the foot of the perpendicular from $ P$ to $ BC$. The lines through $ H$ perpendicular to $ PB$, $ PC$ meet $ AB, AC$ respectively at $ Q, R$. Show that $ BC$ is tangent to the circle through $ Q, H, R$ at $ H$. [i]Original Formulation: [/i] For an acute triangle $ ABC, M$ is the midpoint of the segment $ BC, P$ is a point on the segment $ AM$ such that $ PM \equal{} BM, H$ is the foot of the perpendicular line from $ P$ to $ BC, Q$ is the point of intersection of segment $ AB$ and the line passing through $ H$ that is perpendicular to $ PB,$ and finally, $ R$ is the point of intersection of the segment $ AC$ and the line passing through $ H$ that is perpendicular to $ PC.$ Show that the circumcircle of $ QHR$ is tangent to the side $ BC$ at point $ H.$

1997 AMC 12/AHSME, 14

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The number of geese in a flock increases so that the difference between the populations in year $ n \plus{} 2$ and year $ n$ is directly proportional to the population in year $ n \plus{} 1$. If the populations in the years $ 1994$, $ 1995$, and $ 1997$ were $ 39$, $ 60$, and $ 123$, respectively, then the population in $ 1996$ was $ \textbf{(A)}\ 81\qquad \textbf{(B)}\ 84\qquad \textbf{(C)}\ 87\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ 102$

2003 JHMMC 8, 13

Tags:
A problem author for a math competition was looking through a tentative exam when he realized that he could not use one of his proposed problems. Frustrated, he decided to take a nap instead, and slept from $10:47\text{ AM}$ to $7:32\text{ PM}$. For how many minutes did he sleep?

2000 Hungary-Israel Binational, 1

Let $A$ and $B$ be two subsets of $S = \{1, 2, . . . , 2000\}$ with $|A| \cdot |B| \geq 3999$. For a set $X$ , let $X-X$ denotes the set $\{s-t | s, t \in X, s \not = t\}$. Prove that $(A-A) \cap (B-B)$ is nonempty.

1992 APMO, 2

In a circle $C$ with centre $O$ and radius $r$, let $C_1$, $C_2$ be two circles with centres $O_1$, $O_2$ and radii $r_1$, $r_2$ respectively, so that each circle $C_i$ is internally tangent to $C$ at $A_i$ and so that $C_1$, $C_2$ are externally tangent to each other at $A$. Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.

2007 Princeton University Math Competition, 10

Let $f_n$ be the Fibonacci numbers, defined by $f_0 = 1$, $f_1 = 1$, and $f_n = f_{n-1}+f_{n-2}$. For each $i$, $1 \le i \le 200$, we calculate the greatest common divisor $g_i$ of $f_i$ and $f_{2007}$. What is the sum of the distinct values of $g_i$?

1994 Baltic Way, 2

Let $a_1,a_2,\ldots ,a_9$ be any non-negative numbers such that $a_1=a_9=0$ and at least one of the numbers is non-zero. Prove that for some $i$, $2\le i\le 8$, the inequality $a_{i-1}+a_{i+1}<2a_i$ holds. Will the statement remain true if we change the number $2$ in the last inequality to $1.9$?

2020 Iran MO (3rd Round), 2

For each $n$ find the number of ways one can put the numbers $\{1,2,3,...,n\}$ numbers on the circle, such that if for any $4$ numbers $a,b,c,d$ where $n|a+b-c-d$. The segments joining $a,b$ and $c,d$ do not meet inside the circle. (Two ways are said to be identical , if one can be obtained from rotaiting the other)

2010 Tuymaada Olympiad, 2

We have a number $n$ for which we can find 5 consecutive numbers, none of which is divisible by $n$, but their product is. Show that we can find 4 consecutive numbers, none of which is divisible by $n$, but their product is.