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: 603

Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.
Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.
Let $M$ be a point in the triangle $ABC$ such that \[\text{area}(ABM)=2 \cdot \text{area}(ACM)\] Show that the locus of all such points is a straight line.
Let $B = (-1, 0)$ and $C = (1, 0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be [i]nice[/i] if $\text{(i)}$ there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $TQ$ lies entirely in $S$; and $\text{(ii)}$ for any triangle $P_1P_2P_3$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1, 2, 3\}$ for which triangles $ABC$ and $P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$ are similar. Prove that there exist two distinct nice subsets $S$ and $S'$ of the set $\{(x, y) : x \geq 0, y \geq 0\}$ such that if $A \in S$ and $A' \in S'$ are the unique choices of points in $\text{(ii)}$, then the product $BA \cdot BA'$ is a constant independent of the triangle $P_1P_2P_3$.
Let $ABC$ be an acute triangle with $AB = AC$. Let $D$ be the point on $BC$ such that $AD$ is perpendicular to $BC$. Let $O,H,G$ be the circumcenter, orthocenter and centroid of triangle $ABC$ respectively. Suppose that $2 \cdot OD = 23 \cdot HD$. Prove that $G$ lies on the incircle of triangle $ABC$.
The vertices of a quadrilateral lie on the graph of $y = \ln x$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln \frac{91}{90}$. What is the $x$-coordinate of the leftmost vertex? $\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 13$
Let $ABCD$ be a parallelogram. Let $E$ and $F$ be the midpoints of sides $AB$ and $BC$ respectively. The lines $EC$ and $FD$ intersect at $P$ and form four triangles $APB, BPC, CPD, DPA$. If the area of the parallelogram is $100$, what is the maximum area of a triangles among these four triangles?
Consider the curve $\Gamma$ defined by the equation $y^2 = x^3 +bx+b^2$, where $b$ is a nonzero rational constant. Inscribe in the curve $\Gamma$ a triangle whose vertices have rational coordinates.
A cake has the form of an $ n$ x $ n$ square composed of $ n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$. Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $ \mathcal{A}$ by performing a number of switches, defined as follows: A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.
A semicircle of diameter $ 1$ sits at the top of a semicircle of diameter $ 2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune. [asy]unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); filldraw(Circle((0,.866),.5),grey,black); label("1",(0,.866),S); filldraw(Circle((0,0),1),white,black); draw((-.5,.866)--(.5,.866),linetype("4 4")); clip((-1,0)--(1,0)--(1,2)--(-1,2)--cycle); draw((-1,0)--(1,0)); label("2",(0,0),S);[/asy]$ \textbf{(A)}\ \frac {1}{6}\pi \minus{} \frac {\sqrt {3}}{4} \qquad \textbf{(B)}\ \frac {\sqrt {3}}{4} \minus{} \frac {1}{12}\pi \qquad \textbf{(C)}\ \frac {\sqrt {3}}{4} \minus{} \frac {1}{24}\pi\qquad\textbf{(D)}\ \frac {\sqrt {3}}{4} \plus{} \frac {1}{24}\pi$ $ \textbf{(E)}\ \frac {\sqrt {3}}{4} \plus{} \frac {1}{12}\pi$
We consider a game on an infinite chessboard similar to that of solitaire: If two adjacent fields are occupied by pawns and the next field is empty (the three fields lie on a vertical or horizontal line), then we may remove these two pawns and put one of them on the third field. Prove that if in the initial position pawns fill a $3k \times n$ rectangle, then it is impossible to reach a position with only one pawn on the board.
Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.
Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral? [asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy] $\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$
Let $B = (-1, 0)$ and $C = (1, 0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be [i]nice[/i] if $\text{(i)}$ there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $TQ$ lies entirely in $S$; and $\text{(ii)}$ for any triangle $P_1P_2P_3$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1, 2, 3\}$ for which triangles $ABC$ and $P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$ are similar. Prove that there exist two distinct nice subsets $S$ and $S'$ of the set $\{(x, y) : x \geq 0, y \geq 0\}$ such that if $A \in S$ and $A' \in S'$ are the unique choices of points in $\text{(ii)}$, then the product $BA \cdot BA'$ is a constant independent of the triangle $P_1P_2P_3$.
There are 8 points in the plane.we write down the area of each triangle having all vertices amoung these points(totally 56 numbers).Let them be $a_1,a_2,\dots a_{56}$.Prove that there is a choice of plus or minus such that: $$\pm a_1 \pm a_2 \dots \pm a_{56}=0$$
For every pair of reals $0 < a < b < 1$, we define sequences $\{x_n\}_{n \ge 0}$ and $\{y_n\}_{n \ge 0}$ by $x_0 = 0$, $y_0 = 1$, and for each integer $n \ge 1$: \begin{align*} x_n & = (1 - a) x_{n - 1} + a y_{n - 1}, \\ y_n & = (1 - b) x_{n - 1} + b y_{n - 1}. \end{align*} The [i]supermean[/i] of $a$ and $b$ is the limit of $\{x_n\}$ as $n$ approaches infinity. Over all pairs of real numbers $(p, q)$ satisfying $\left (p - \textstyle\frac{1}{2} \right)^2 + \left (q - \textstyle\frac{1}{2} \right)^2 \le \left(\textstyle\frac{1}{10}\right)^2$, the minimum possible value of the supermean of $p$ and $q$ can be expressed as $\textstyle\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $100m + n$. [i]Proposed by Lewis Chen[/i]
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
In $\triangle ABC$ with side lengths $AB=13$, $BC=14$, and $CA=15$, let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}$. There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ$. Then $AQ$ can be written as $\frac{m}{\sqrt{n}}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Find all points on the plane such that the sum of the distances of each of the four lines defined by the unit square of that plane is $4$.
Consider the $ 12$-sided polygon $ ABCDEFGHIJKL$, as shown. Each of its sides has length $ 4$, and each two consecutive sides form a right angle. Suppose that $ \overline{AG}$ and $ \overline{CH}$ meet at $ M$. What is the area of quadrilateral $ ABCM$? [asy]unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W);[/asy]$ \textbf{(A)}\ \frac {44}{3}\qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ \frac {88}{5}\qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ \frac {62}{3}$
Let $n$ be a positive integer. Consider $2n$ distinct lines on the plane, no two of which are parallel. Of the $2n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the plane that lie on at least one red line. Prove that there exists a circle that intersects $\mathcal{B}$ in exactly $2n - 1$ points, and also intersects $\mathcal{R}$ in exactly $2n - 1$ points. [i]Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand[/i]
Consider a rectangle $R$ partitioned into $2016$ smaller rectangles such that the sides of each smaller rectangle is parallel to one of the sides of the original rectangle. Call the corners of each rectangle a vertex. For any segment joining two vertices, call it basic if no other vertex lie on it. (The segments must be part of the partitioning.) Find the maximum/minimum possible number of basic segments over all possible partitions of $R$.
Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$. Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs. [i]Proposed by Ankan Bhattacharya[/i]
Let $n>1$ be an integer. Suppose we are given $2n$ points in the plane such that no three of them are collinear. The points are to be labelled $A_1, A_2, \dots , A_{2n}$ in some order. We then consider the $2n$ angles $\angle A_1A_2A_3, \angle A_2A_3A_4, \dots , \angle A_{2n-2}A_{2n-1}A_{2n}, \angle A_{2n-1}A_{2n}A_1, \angle A_{2n}A_1A_2$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$). Prove that there exists an ordering of the given points such that the resulting $2n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.
Space is dissected into congruent cubes. Is it necessarily true that for each cube there exists another cube so that both cubes have a whole face in common?