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

In $\triangle ABC$, point $M$ is the middle point of $AC$. $MD//AB$ and meet the tangent of $A$ to $\odot(ABC)$ at point $D$. Point $E$ is in $AD$ and point $A$ is the middle point of $DE$. $\{P\}=\odot(ABE)\cap AC,\{Q\}=\odot(ADP)\cap DM$. Prove that $\angle QCB=\angle BAC$. [url=https://imgtu.com/i/4pZ7Zj][img]https://z3.ax1x.com/2021/09/12/4pZ7Zj.jpg[/img][/url]
Let $k$ be a positive integer. The organising commitee of a tennis tournament is to schedule the matches for $2k$ players so that every two players play once, each day exactly one match is played, and each player arrives to the tournament site the day of his first match, and departs the day of his last match. For every day a player is present on the tournament, the committee has to pay $1$ coin to the hotel. The organisers want to design the schedule so as to minimise the total cost of all players' stays. Determine this minimum cost.
Let $a_1,a_2, \cdots$ be a sequence of integers that satisfies: $a_1=1$ and $a_{n+1}=a_n+a_{\lfloor \sqrt{n} \rfloor} , \forall n\geq 1 $. Prove that for all positive $k$, there is $m \geq 1$ such that $k \mid a_m$.
Mark on a cellular paper four nodes forming a convex quadrilateral with the sidelengths equal to four different primes. [i](Proposed by A.Zaslavsky)[/i]
Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. He knows that the sides of the carpet are integral numbers of feet and that his two storerooms have the same (unknown) length, but widths of 38 feet and 50 feet respectively. What are the carpet dimensions?
The negation of the statement "No slow learners attend this school" is: $ \textbf{(A)}\ \text{All slow learners attend this school} \\ \textbf{(B)}\ \text{All slow learners do not attend this school} \\ \textbf{(C)}\ \text{Some slow learners attend this school} \\ \textbf{(D)}\ \text{Some slow learners do not attend this school} \\ \textbf{(E)}\ \text{No slow learners do not attend this school}$
A complete graph is in a plane such that no three of its vertices are collinear. The edges of the graph, which are straight segments connecting the vertices, are colored with two colors. Prove that there is a non-self-intersecting spanning tree consisting of edges of the same color.
Decide whether any rectangle that can be covered by 25 circles of radius 2 can also be covered by 100 circles of radius 1.
Let $\omega_1$ and $\omega_2$ be two circles that intersect at two points: $A$ and $B$. Let $C$ and $E$ be on $\omega_1$, and $D$ and $F$ be on $\omega_2$ such that $CD$ and $EF$ meet at $B$ and the three lines $CE$, $DF$, and $AB$ concur at a point $P$ that is closer to $B$ than $A$. Let $\Omega$ denote the circumcircle of $\triangle DEF$. Now, let the line through $A$ perpendicular to $AB$ hit $EB$ at $G$, $GD$ hit $\Omega$ at $J$, and $DA$ hit $\Omega$ again at $I$. A point $Q$ on $IE$ satisfies that $CQ=JQ$. If $QJ=36$, $EI=21$, and $CI=16$, then the radius of $\Omega$ can be written as $\frac{a\sqrt{b}}{c}$ where $a$, $b$, and $c$ are positive integers, $b$ is not divisible by the square of a prime, and $\gcd(a, c) = 1$. Find $a+b+c$. [i]Proposed by Kevin Zhao[/i]
Two circles $\omega_A$ and $\omega_B$ have centers at points $A$ and $B$ respectively and intersect at points $P$ and $Q$ in such a way that $A$, $B$, $P$, and $Q$ all lie on a common circle $\omega$. The tangent to $\omega$ at $P$ intersects $\omega_A$ and $\omega_B$ again at points $X$ and $Y$ respectively. Suppose $AB = 17$ and $XY = 20$. Compute the sum of the radii of $\omega_A$ and $\omega_B$.
Consider a 4-point configuration in the plane such that every 3 points can be covered by a strip of a unit width. Prove that: 1) the four points can be covered by a strip of length at most $\sqrt2$ and 2)if no strip of length less that $\sqrt2$ covers all the four points, then the points are vertices of a square of length $\sqrt2$
Let $\vartriangle ABC$ be a triangle with side lengths $AB = 13$, $BC = 14$, and $CA = 15$. The angle bisector of $\angle BAC$, the angle bisector of $\angle ABC$, and the angle bisector of $\angle ACB$ intersect the circumcircle of $\vartriangle ABC$ again at points $D$, $E$ and $F$, respectively. Compute the area of hexagon $AF BDCE$.
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.
Do there exist two infinite sets of non-negative integers such that every non-negative integer can be uniquely represented in the form $a + b$ with $a$ in $A$ and $b$ in $B$?
Determine all functions $f : R \to R$ satisfying $f (f(x)f(y) + f(y)f(z) + f(z)f(x))= f(x) + f(y) + f(z)$ for all real numbers $x, y, z$.
A flea jumps in a straight numbered line. It jumps first from point $0$ to point $1$. Afterwards, if its last jump was from $A$ to $B$, then the next jump is from $B$ to one of the points $B + (B - A) - 1$, $B + (B - A)$, $B + (B-A) + 1$. Prove that if the flea arrived twice at the point $n$, $n$ positive integer, then it performed at least $\lceil 2\sqrt n\rceil$ jumps.
The plane is partitioned into regions by a finite number of lines no three of which are concurrent. Two regions are called "neighbors" if the intersection of their boundaries is a segment, or half-line or a line (a point is not a segment). An integer is to be assigned to each region in such a way that: i) the product of the integers assigned to any two neighbors is less than their sum; ii) for each of the given lines, and each of the half-planes determined by it, the sum of the integers, assigned to all of the regions lying on this half-plane equal to zero. Prove that this is possible if and only if not all of the lines are parallel.
Josh writes the numbers $1,2,3,\dots,99,100$. He marks out $1$, skips the next number $(2)$, marks out $3$, and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$, skips the next number $(4)$, marks out $6$, skips $8$, marks out $10$, and so on to the end. Josh continues in this manner until only one number remains. What is that number? $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$
For how many values of $ a$ is it true that the line $ y \equal{} x \plus{} a$ passes through the vertex of the parabola $ y \equal{} x^2 \plus{} a^2$? $ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ \text{infinitely many}$
The cells of a square $100 \times 100$ table are colored in one of two colors, black or white. A coloring is called admissible if for any row or column, the number $b$ of black colored cells satisfies $50 \le b \le 60$. It is permitted to recolor a cell if by doing so the resulting configuration is still admissible. Prove that one can transition from any admissible coloring of the board to any other using a sequence of valid recoloring operations.
A convex polyhedron has $2n$ faces ($n\ge 3$), and all faces are triangles. What is the largest number of vertices at which converges exactly $3$ edges at a such a polyhedron ?
There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n.$ Each pebble is coloured in one of $n$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied: [list] [*]The total weights of both piles are the same. [*] Each pile contains two pebbles of each colour. [/list] [i]Proposed by Milan Haiman, Hungary and Carl Schildkraut, USA[/i]
In eight years Henry will be three times the age that Sally was last year. Twenty five years ago their ages added to $83$. How old is Henry now?
(a) A point $A$ is marked inside a sphere. Three perpendicular lines drawn through $A$ intersect the sphere at six points. Prove that the centre of gravity of these six points does not depend on the choice of such three lines. (b) An icosahedron with the centre $A$ is placed inside a sphere (its centre does not necessarily coincide with the centre of the sphere). The rays going from $A$ to the vertices of the icosahedron mark $12$ points on the sphere. Then the icosahedron is rotated about its centre. New rays mark new $12$ points on the sphere. Let $O$ and $N$ be the centres of mass of old and new points respectively. Prove that $O = N$.
Solve in $R$ equation $[x] \cdot \{x\} = 2001 x$, where$ [ .]$ and $\{ .\}$ represent respectively the floor and the integer functions.