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

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Found problems: 85335

1999 Junior Balkan Team Selection Tests - Romania, 4

Tags: geometry
Let be a convex quadrilateral $ ABCD. $ On the semi-straight line extension of $ AB $ in the direction of $ B, $ put $ A_1 $ such that $ AB=BA_1. $ Similarly, define $ B_1,C_1,D_1, $ for the other three sides. [b]a)[/b] If $ E,E_1,F,F_1 $ are the midpoints of $ BC,A_1B_1,AD $ respectively, $ C_1,D_1, $ show that $ EE_1=FF_1. $ [b]b)[/b] Delete everything, but $ A_1,B_1,C_1,D_1. $ Now, find a way to construct the initial quadrilateral. [i]Vasile Pop[/i]

1989 Tournament Of Towns, (204) 2

In the triangle $ABC$ the median $AM$ is drawn. Is it possible that the radius of the circle inscribed in $\vartriangle ABM$ could be twice as large as the radius of the circle inscribed in $\vartriangle ACM$ ? ( D . Fomin , Leningrad)

2004 Baltic Way, 2

Let $ P(x)$ be a polynomial with a non-negative coefficients. Prove that if the inequality $ P\left(\frac {1}{x}\right)P(x)\geq 1$ holds for $ x \equal{} 1$, then this inequality holds for each positive $ x$.

2018 MIG, 18

Tags:
How many paths are there from $A$ to $B$ in the following diagram if only moves downward are allowed? [center][img]https://cdn.artofproblemsolving.com/attachments/f/d/62a14f7959cc0461543b0f76bba51be9786847.png[/img][/center] $\textbf{(A) } 65\qquad\textbf{(B) } 67\qquad\textbf{(C) } 70\qquad\textbf{(D) } 74\qquad\textbf{(E) } 75$

2017 F = ma, 4

Tags:
Several identical cars are standing at a red light on a one-lane road, one behind the other, with negligible (and equal) distance between adjacent cars. When the green light comes up, the first car takes off to the right with constant acceleration. The driver in the second car reacts and does the same 0.2 s later. The third driver starts moving 0.2 s after the second one and so on. All cars accelerate until they reach the speed limit of 45 km/hr, after which they move to the right at a constant speed. Consider the following patterns of cars. Just before the first car starts accelerating to the right, the car pattern will qualitatively look like the pattern in I. After that, the pattern will qualitatively evolve according to which of the following? $\textbf{(A)}\text{first I, then II, and then III}$ $\textbf{(B)}\text{first I, then II, and then IV}$ $\textbf{(C)}\text{first I, and then IV, with neither II nor III as an intermediate stage}$ $\textbf{(D)}\text{first I, and then II}$ $\textbf{(E)}\text{first I, and then III}$

2019 AMC 10, 4

All lines with equation $ax+by=c$ such that $a$, $b$, $c$ form an arithmetic progression pass through a common point. What are the coordinates of that point? $\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

2013 All-Russian Olympiad, 4

On a $55\times 55$ square grid, $500$ unit squares were cut out as well as $400$ L-shaped pieces consisting of 3 unit squares (each piece can be oriented in any way) [refer to the figure]. Prove that at least two of the cut out pieces bordered each other before they were cut out. [asy]size(2.013cm); draw ((0,0)--(0,1)); draw ((0,0)--(1,0)); draw ((0,1)--(.5,1)); draw ((.5,1)--(.5,0)); draw ((0,.5)--(1,.5)); draw ((1,.5)--(1,0)); draw ((1,.5)--(1,0)); [/asy]

1969 IMO Shortlist, 7

$(BUL 1)$ Prove that the equation $\sqrt{x^3 + y^3 + z^3}=1969$ has no integral solutions.

1998 Vietnam National Olympiad, 2

Let be given a tetrahedron whose circumcenter is $O$. Draw diameters $AA_{1},BB_{1},CC_{1},DD_{1}$ of the circumsphere of $ABCD$. Let $A_{0},B_{0},C_{0},D_{0}$ be the centroids of triangle $BCD,CDA,DAB,ABC$. Prove that $A_{0}A_{1},B_{0}B_{1},C_{0}C_{1},D_{0}D_{1}$ are concurrent at a point, say, $F$. Prove that the line through $F$ and a midpoint of a side of $ABCD$ is perpendicular to the opposite side.

2023 UMD Math Competition Part II, 5

Let $0 \le a_1 \le a_2 \le \dots \le a_n \le 1$ be $n$ real numbers with $n \ge 2$. Assume $a_1 + a_2 + \dots + a_n \ge n-1$. Prove that \[ a_2a_3\dots a_n \ge \left( 1 - \frac 1n \right)^{n-1} \]

2008 AIME Problems, 15

A square piece of paper has sides of length $ 100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance $ \sqrt {17}$ from the corner, and they meet on the diagonal at an angle of $ 60^\circ$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $ \sqrt [n]{m}$, where $ m$ and $ n$ are positive integers, $ m < 1000$, and $ m$ is not divisible by the $ n$th power of any prime. Find $ m \plus{} n$. [asy]import math; unitsize(5mm); defaultpen(fontsize(9pt)+Helvetica()+linewidth(0.7)); pair O=(0,0); pair A=(0,sqrt(17)); pair B=(sqrt(17),0); pair C=shift(sqrt(17),0)*(sqrt(34)*dir(75)); pair D=(xpart(C),8); pair E=(8,ypart(C)); draw(O--(0,8)); draw(O--(8,0)); draw(O--C); draw(A--C--B); draw(D--C--E); label("$\sqrt{17}$",(0,2),W); label("$\sqrt{17}$",(2,0),S); label("cut",midpoint(A--C),NNW); label("cut",midpoint(B--C),ESE); label("fold",midpoint(C--D),W); label("fold",midpoint(C--E),S); label("$30^\circ$",shift(-0.6,-0.6)*C,WSW); label("$30^\circ$",shift(-1.2,-1.2)*C,SSE);[/asy]

1964 Polish MO Finals, 1

Prove that the inequality $$ \frac{1}{3} \leq \frac{\tan 3\alpha}{\tan \alpha} \leq 3 $$ is not true for any value of $ \alpha $.

2024 Polish MO Finals, 5

We are given an integer $n \ge 2024$ and a sequence $a_1,a_2,\dots,a_{n^2}$ of real numbers satisfying \[\vert a_k-a_{k-1}\vert \le \frac{1}{k} \quad \text{and} \quad \vert a_1+a_2+\dots+a_k\vert \le 1\] for $k=2,3,\dots,n^2$. Show that $\vert a_{n(n-1)}\vert \le \frac{2}{n}$. [i]Note: Proving $\vert a_{n(n-1)}\vert \le \frac{75}{n}$ will be rewarded partial points.[/i]

2009 Postal Coaching, 1

A circle $\Gamma$ and a line $\ell$ which does not intersect $\Gamma$ are given. Suppose $P, Q,R, S$ are variable points on circle $\Gamma$ such that the points $A = PQ\cap RS$ and $B = PS \cap QR$ lie on $\ell$. Prove that the circle on $AB$ as a diameter passes through two fixed points.

2024 Baltic Way, 18

An infinite sequence $a_1, a_2,\ldots$ of positive integers is such that $a_n \geq 2$ and $a_{n+2}$ divides $a_{n+1} + a_n$ for all $n \geq 1$. Prove that there exists a prime which divides infinitely many terms of the sequence.

the 12th XMO, Problem 3

Let $a_0=0,a_1\in\mathbb Z_+.$ For integer $n\geq 2,a_n$ is the smallest positive integer satisfy that for $\forall 0\leq i\neq j\leq n-1,a_n\nmid (a_i-a_j).$ (1) If $a_1=2023,$ calculate $a_{10000}.$ (2) If $a_t\leq\frac{a_1}2,$ find the maximum value of $\frac t{a_1}.$

2016 USAMTS Problems, 5:

Tags:
Consider the set $S = \{ q + \frac{1}{q}, \text{ where } q \text{ ranges over all positive rational numbers} \}$. (a) Let $N$ be a positive integer. Show that $N$ is the sum of two elements of $S$ if and only if $N$ is the product of two elements of $S$. (b) Show that there exist infinitely many positive integers $N$ that cannot be written as the sum of two elements of $S$. (c)Show that there exist infinitely many positive integers $N$ that can be written as the sum of two elements of $S$.

2023 BMT, 7

Tags: algebra
Nikhil constructs a list of all polynomial pairs $(a(x), b(x))$ with real coefficients such that $a(x)$ has higher degree than $b(x)$ and $a(x)^2 + b(x)^2 = x^{10} + 1$. Danielle takes Nikhil’s list and adds all polynomial pairs that satisfy the same conditions but have complex coefficients. If Nikhil’s original list had $N$ pairs and Danielle added $D$ pairs, compute $D - N$.

2005 Thailand Mathematical Olympiad, 14

A function $f : Z \to Z$ is given so that $f(m + n) = f(m) + f(n) + 2mn - 2548$ for all positive integers $m, n$. Given that $f(2548) = -2548$, find the value of $f(2)$.

2003 Tuymaada Olympiad, 4

Given are polynomial $f(x)$ with non-negative integral coefficients and positive integer $a.$ The sequence $\{a_{n}\}$ is defined by $a_{1}=a,$ $a_{n+1}=f(a_{n}).$ It is known that the set of primes dividing at least one of the terms of this sequence is finite. Prove that $f(x)=cx^{k}$ for some non-negative integral $c$ and $k.$ [i]Proposed by F. Petrov[/i] [hide="For those of you who liked this problem."] Check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62259]this thread[/url] out.[/hide]

1989 China National Olympiad, 5

Given $1989$ points in the space, any three of them are not collinear. We divide these points into $30$ groups such that the numbers of points in these groups are different from each other. Consider those triangles whose vertices are points belong to three different groups among the $30$. Determine the numbers of points of each group such that the number of such triangles attains a maximum.

2018 239 Open Mathematical Olympiad, 8-9.3

Is it possible to divide all non-empty subsets of a set of 10 elements into triples so that in each triple, two of the subsets do not intersect and in their union give the third? [i]Proposed by Vladislav Frank[/i]

2021 Kyiv City MO Round 1, 9.2

Roma wrote on the board each of the numbers $2018, 2019, 2020$, $100$ times each. Let us denote by $S(n)$ the sum of digits of positive integer $n$. In one action, Roma can choose any positive integer $k$ and instead of any three numbers $a, b, c$ written on the board write the numbers $2S(a + b) + k, 2S(b + c) + k$ and $2S(c + a) + k$. Can Roma after several such actions make $299$ numbers on the board equal, and the last one differing from them by $1$? [i]Proposed by Oleksii Masalitin[/i]

2019 LIMIT Category A, Problem 7

The digit in unit place of $1!+2!+\ldots+99!$ is $\textbf{(A)}~3$ $\textbf{(B)}~0$ $\textbf{(C)}~1$ $\textbf{(D)}~7$

1988 Irish Math Olympiad, 12

Tags: trigonometry
Prove that if $n$ is a positive integer ,then \[cos^4\frac{\pi}{2n+1}+cos^4\frac{2\pi}{2n+1}+\cdots+cos^4\frac{n\pi}{2n+1}=\frac{6n-5}{16}.\]