Prove no three lattice points in the plane form an equilateral triangle.

Sam colors each tile in a 4 by 4 grid white or black. A coloring is called [i]rotationally symmetric[/i] if the grid can be rotated 90, 180, or 270 degrees to achieve the same pattern. Two colorings are called [i]rotationally distinct[/i] if neither can be rotated to match the other. How many rotationally distinct ways are there for Sam to color the grid such that the colorings are [i]not[/i] rotationally symmetric? [i]Proposed by Gabriel Wu[/i]

A polygon is called concave if it has at least one angle strictly greater than $180^{\circ}$. What is the maximum number of symmetries that an 11-sided concave polygon can have? (A [i]symmetry[/i] of a polygon is a way to rotate or reflect the plane that leaves the polygon unchanged.)

Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

Given a real number $\lambda > 1$, let $P$ be a point on the arc $BAC$ of the circumcircle of $\bigtriangleup ABC$. Extend $BP$ and $CP$ to $U$ and $V$ respectively such that $BU = \lambda BA$, $CV = \lambda CA$. Then extend $UV$ to $Q$ such that $UQ = \lambda UV$. Find the locus of point $Q$.

Given $2012$ distinct points $A_1,A_2,\dots,A_{2012}$ on the Cartesian plane. For any permutation $B_1,B_2,\dots,B_{2012}$ of $A_1,A_2,\dots,A_{2012}$ define the [i]shadow[/i] of a point $P$ as follows: [i]Point $P$ is rotated by $180^{\circ}$ around $B_1$ resulting $P_1$, point $P_1$ is rotated by $180^{\circ}$ around $B_2$ resulting $P_2$, ..., point $P_{2011}$ is rotated by $180^{\circ}$ around $B_{2012}$ resulting $P_{2012}$. Then, $P_{2012}$ is called the shadow of $P$ with respect to the permutation $B_1,B_2,\dots,B_{2012}$.[/i] Let $N$ be the number of different shadows of $P$ up to all permutations of $A_1,A_2,\dots,A_{2012}$. Determine the maximum value of $N$. [i]Proposer: Hendrata Dharmawan[/i]

Draw an equilateral triangle with center $O$. Rotate the equilateral triangle $30^\circ, 60^\circ, 90^\circ$ with respect to $O$ so there would be four congruent equilateral triangles on each other. Look at the diagram. If the smallest triangle has area $1$, the area of the original equilateral triangle could be expressed as $p+q\sqrt r$ where $p,q,r$ are positive integers and $r$ is not divisible by a square greater than $1$. Find $p+q+r$.

Let $ ABCDEF$ be a regular hexagon. Let $ G$, $ H$, $ I$, $ J$, $ K$, and $ L$ be the midpoints of sides $ AB$, $ BC$, $ CD$, $ DE$, $ EF$, and $ AF$, respectively. The segments $ AH$, $ BI$, $ CJ$, $ DK$, $ EL$, and $ FG$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ ABCDEF$ be expressed as a fraction $ \frac {m}{n}$ where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$.

On the tetrahedron $ ABCD $ make the notation $ M,N,P,Q, $ for the midpoints of $ AB,CD,AC, $ respectively, $ BD. $ Additionally, we know that $ MN $ is the common perpendicular of $ AB,CD, $ and $ PQ $ is the common perpendicular of $ AC,BD. $ Show that $ AB=CD, BC=DA, AC=BD. $

Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.

There is a complex number $ z$ with imaginary part $ 164$ and a positive integer $ n$ such that \[ \frac {z}{z \plus{} n} \equal{} 4i. \]Find $ n$.

Consider the two triangles $ ABC$ and $ PQR$ shown below. In triangle $ ABC, \angle ADB \equal{} \angle BDC \equal{} \angle CDA \equal{} 120^\circ$. Prove that $ x\equal{}u\plus{}v\plus{}w$. [asy]unitsize(7mm); defaultpen(linewidth(.7pt)+fontsize(10pt)); pair C=(0,0), B=4*dir(5); pair A=intersectionpoints(Circle(C,5), Circle(B,6))[0]; pair Oc=scale(sqrt(3)/3)*rotate(30)*(B-A)+A; pair Ob=scale(sqrt(3)/3)*rotate(30)*(A-C)+C; pair D=intersectionpoints(Circle(Ob,length(Ob-C)), Circle(Oc,length(Oc-B)))[1]; real s=length(A-D)+length(B-D)+length(C-D); pair P=(6,0), Q=P+(s,0), R=rotate(60)*(s,0)+P; pair M=intersectionpoints(Circle(P,length(B-C)), Circle(Q,length(A-C)))[0]; draw(A--B--C--A--D--B); draw(D--C); label("$B$",B,SE); label("$C$",C,SW); label("$A$",A,N); label("$D$",D,NE); label("$a$",midpoint(B--C),S); label("$b$",midpoint(A--C),WNW); label("$c$",midpoint(A--B),NE); label("$u$",midpoint(A--D),E); label("$v$",midpoint(B--D),N); label("$w$",midpoint(C--D),NNW); draw(P--Q--R--P--M--Q); draw(M--R); label("$P$",P,SW); label("$Q$",Q,SE); label("$R$",R,N); label("$M$",M,NW); label("$x$",midpoint(P--R),NW); label("$x$",midpoint(P--Q),S); label("$x$",midpoint(Q--R),NE); label("$c$",midpoint(R--M),ESE); label("$a$",midpoint(P--M),NW); label("$b$",midpoint(Q--M),NE);[/asy]

Given two (identical) polygonal domains in the Euclidean plane, it is not possible in general to superpose the two using only translations and rotations. Prove that this can however be achieved by splitting one of the domains into a finite number of polygonal subdomains which then fit together, via translations and rotations in the plane, to recover the other domain.

A game is played by $n$ girls ($n \geq 2$), everybody having a ball. Each of the $\binom{n}{2}$ pairs of players, is an arbitrary order, exchange the balls they have at the moment. The game is called nice [b]nice[/b] if at the end nobody has her own ball and it is called [b]tiresome[/b] if at the end everybody has her initial ball. Determine the values of $n$ for which there exists a nice game and those for which there exists a tiresome game.

The parabola $\mathcal P$ given by equation $y=x^2$ is rotated some acute angle $\theta$ clockwise about the origin such that it hits both the $x$ and $y$ axes at two distinct points. Suppose the length of the segment $\mathcal P$ cuts the $x$-axis is $1$. What is the length of the segment $\mathcal P$ cuts the $y$-axis?

Let $L$ be the line with slope $\tfrac{5}{12}$ that contains the point $A=(24,-1)$, and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$. The original coordinate axes are erased, and line $L$ is made the $x$-axis, and line $M$ the $y$-axis. In the new coordinate system, point $A$ is on the positive $x$-axis, and point $B$ is on the positive $y$-axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha+\beta$.

Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

Given a regular $ 2n$-gon, show that each of its sides and diagonals can be assigned in such a way that the sum of the obtained vectors equals zero.

Let $M$ and $N$ be two regular polygonic area.Define $K(M,N)$ as the midpoints of segments $[AB]$ such that $A$ belong to $M$ and $B$ belong to $N$. Find all situations of $M$ and $N$ such that $K(M,N)$ is a regualr polygonic area too.

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) [asy]import three; import math; size(180); defaultpen(linewidth(.8pt)); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7));[/asy]$ \textbf{(A)}\ 210 \qquad \textbf{(B)}\ 560 \qquad \textbf{(C)}\ 840 \qquad \textbf{(D)}\ 1260 \qquad \textbf{(E)}\ 1680$

$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.

Let $L$ be a figure made of $3$ squares, a right isosceles triangle and a quarter circle (all unit sized) as shown below: [img]https://wiki-images.artofproblemsolving.com//f/f9/Weirwiueripo.png[/img] Prove that any $18$ points in the plane can be covered with copies of $L$, which don't overlap (copies of $L$ may be rotated or flipped)

Points $ A \equal{} (3,9), B \equal{} (1,1), C \equal{} (5,3),$ and $ D \equal{} (a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ ABCD$. The quadrilateral formed by joining the midpoints of $ \overline{AB}, \overline{BC}, \overline{CD},$ and $ \overline{DA}$ is a square. What is the sum of the coordinates of point $ D$? $ \textbf{(A)} \ 7 \qquad \textbf{(B)} \ 9 \qquad \textbf{(C)} \ 10 \qquad \textbf{(D)} \ 12 \qquad \textbf{(E)} \ 16$

Suppose a square of sidelengh $l$ is inside an unit square and does not contain its centre. Show that $l\le 1/2.$ [i]Marius Cavachi[/i]

Consider a sphere and a plane $\pi$. For a variable point $M \in \pi$, exterior to the sphere, one considers the circular cone with vertex in $M$ and tangent to the sphere. Find the locus of the centers of all circles which appear as tangent points between the sphere and the cone. [i]Octavian Stanasila[/i]