Ancient astronaut theorist Nutter B. Butter claims that the Caloprians from planet Calop, 30 light years away and at rest with respect to the Earth, wiped out the dinosaurs. The iridium layer in the crust, he claims, indicates spaceships with the fuel necessary to travel at 30% of the speed of light here and back, and that their engines allowed them to instantaneously hop to this speed. He also says that Caloprians can only reproduce on their home planet. Call the minimum life span, in years, of a Caloprian, assuming some had to reach Earth to wipe out the dinosaurs, $T$. Assume that, once a Caloprian reaches Earth, they instantaneously wipe out the dinosaurs. Then, $T$ can be expressed in the form $m\sqrt{n}$, where $n$ is not divisible by the square of a prime. Find $m+n$. [i](B. Dejean, 6 points)[/i]

A point $P$ lies on the internal angle bisector of $\angle BAC$ of a triangle $\triangle ABC$. Point $D$ is the midpoint of $BC$ and $PD$ meets the external angle bisector of $\angle BAC$ at point $E$. If $F$ is the point such that $PAEF$ is a rectangle then prove that $PF$ bisects $\angle BFC$ internally or externally.

Consider the graphs of (1): $ y\equal{}x^2\minus{}\frac{1}{2}x\plus{}2$ and (2) $ y\equal{}x^2\plus{}\frac{1}{2}x\plus{}2$ on the same set of axis. These parabolas are exactly the same shape. Then: $ \textbf{(A)}\ \text{the graphs coincide.} \\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).} \\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).} \\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).} \\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).}$

In triangle ABC, let $P$ and $Q$ be two interior points such that $\angle ABP = \angle QBC$ and $\angle ACP = \angle QCB$. Point $D$ lies on segment $BC$. Prove that $\angle APB + \angle DPC = 180^\circ$ if and only if $\angle AQC + \angle DQB = 180^\circ$.

Each vertex of the 3-dimensional Euclidean space either is coloured red or blue. Prove that within those squares being possible in this space with edge length 1 there is at least one square either with three red vertices or four blue vertices !

Let $\ell$ be a line in the plane, and a point $A \not \in \ell$. Determine the locus of the points $Q$ in the plane, for which there exists a point $P\in \ell$ so that $AQ=PQ$ and $\angle PAQ = 45^{\circ}$. ([i]Dan Schwarz[/i])

Compute the maximum area of a rectangle which can be inscribed in a triangle of area $M$.

Aaron takes a square sheet of paper, with one corner labeled $A$. Point $P$ is chosen at random inside of the square and Aaron folds the paper so that points $A$ and $P$ coincide. He cuts the sheet along the crease and discards the piece containing $A$. Let $p$ be the probability that the remaining piece is a pentagon. Find the integer nearest to $100p$. [i]Proposed by Aaron Lin[/i]

Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side length 1. The sides of the pentagon are extended to form the 10-sided polygon shown in bold at right. Find the ratio of the area of quadrilateral $A_2A_5B_2B_5$ (shaded in the picture to the right) to the area of the entire 10-sided polygon. [asy] size(8cm); defaultpen(fontsize(10pt)); pair A_2=(-0.4382971011,5.15554989475), B_4=(-2.1182971011,-0.0149584477027), B_5=(-4.8365942022,8.3510997895), A_3=(0.6,8.3510997895), B_1=(2.28,13.521608132), A_4=(3.96,8.3510997895), B_2=(9.3965942022,8.3510997895), A_5=(4.9982971011,5.15554989475), B_3=(6.6782971011,-0.0149584477027), A_1=(2.28,3.18059144705); filldraw(A_2--A_5--B_2--B_5--cycle,rgb(.8,.8,.8)); draw(B_1--A_4^^A_4--B_2^^B_2--A_5^^A_5--B_3^^B_3--A_1^^A_1--B_4^^B_4--A_2^^A_2--B_5^^B_5--A_3^^A_3--B_1,linewidth(1.2)); draw(A_1--A_2--A_3--A_4--A_5--cycle); pair O = (A_1+A_2+A_3+A_4+A_5)/5; label("$A_1$",A_1, 2dir(A_1-O)); label("$A_2$",A_2, 2dir(A_2-O)); label("$A_3$",A_3, 2dir(A_3-O)); label("$A_4$",A_4, 2dir(A_4-O)); label("$A_5$",A_5, 2dir(A_5-O)); label("$B_1$",B_1, 2dir(B_1-O)); label("$B_2$",B_2, 2dir(B_2-O)); label("$B_3$",B_3, 2dir(B_3-O)); label("$B_4$",B_4, 2dir(B_4-O)); label("$B_5$",B_5, 2dir(B_5-O)); [/asy]

Is it possible to find a set of $100$ (or $200$) points on the boundary of a cube such that this set remains fixed under all rotations which leave the cube fixed ?

Let AOB and COD be angles which can be identified by a rotation of the plane (such that rays OA and OC are identified). A circle is inscribed in each of these angles; these circles intersect at points E and F. Show that angles AOE and DOF are equal.

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: [list=] [*]$L,$ a rotation of $90^{\circ}$ counterclockwise around the origin; [*]$R,$ a rotation of $90^{\circ}$ clockwise around the origin; [*]$H,$ a reflection across the $x$-axis; and [*]$V,$ a reflection across the $y$-axis. [/list] Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.) $\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$

Let a unit square $\mathcal D$ be given in the plane. For any point $X$ in the plane denote $\mathcal D_X$ the image of $\mathcal D$ in rotation with respect to origin $X$ by $+90^\circ.$ Find the locus of all $X$ such that the area of union $\mathcal D\cup\mathcal D_X$ is at most 1.5.

Given a triangle $ ABC$. A circle passes through vertices $ B$ and $ C$ and intersects sides $ AB$ and $ AC$ at points $ D$ and $ E$, respectively. Segments $ CD$ and $ BE$ intersect at point $ O$. Denote the incenters of triangles $ ADE$ and $ ODE$ by $ M$ and $ N$, respectiely. Prove that the midpoint of the smaller arc $ DE$ lies on line $ MN$.

A triangle $ T$ is contained inside a point-symmetrical polygon $ M.$ The triangle $ T'$ is the mirror image of the triangle $ T$ with the reflection at one point $ P$, which inside the triangle $ T$ lies. Prove that at least one of the vertices of the triangle $ T'$ lies in inside or on the boundary of the polygon $ M.$

Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]

If the inradius of a triangle is half of its circumradius, prove that the triangle is equilateral.

Five persons wearing badges with numbers $1, 2, 3, 4, 5$ are seated on $5$ chairs around a circular table. In how many ways can they be seated so that no two persons whose badges have consecutive numbers are seated next to each other? (Two arrangements obtained by rotation around the table are considered different)

Let $ABC$ an acute triangle. (a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$; (b) Determine if exist some points that are centers of $3$ distinct rectangles whose vertices lie on the sides of $ABC$.

Let $w_1, w_2, \dots, w_n$ be complex numbers. A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[ \sum_{k = 1}^n (z_k - w_k) = 0. \] For the numbers $w_1 = 32 + 170i$, $w_2 = -7 + 64i$, $w_3 = -9 +200i$, $w_4 = 1 + 27i$, and $w_5 = -14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the slope of this mean line.

Two circles are internally tangent at $N$. The chords $BA$ and $BC$ of the larger circle are tangent to the smaller circle at $K$ and $M$ respectively. $Q$ and $P$ are midpoint of arcs $AB$ and $BC$ respectively. Circumcircles of triangles $BQK$ and $BPM$ are intersect at $L$. Show that $BPLQ$ is a parallelogram.

Let $A_{1}A_{2} \dots A_{3n}$ be a closed broken line consisting of $3n$ lines segments in the Euclidean plane. Suppose that no three of its vertices are collinear, and for each index $i=1,2,\dots,3n$, the triangle $A_{i}A_{i+1}A_{i+2}$ has counterclockwise orientation and $\angle A_{i}A_{i+1}A_{i+2} = 60º$, using the notation $A_{3n+1} = A_{1}$ and $A_{3n+2} = A_{2}$. Prove that the number of self-intersections of the broken line is at most $\frac{3}{2}n^{2} - 2n + 1$

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles? $\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$