A frictionless roller coaster ride is given a certain velocity at the start of the ride. At which point in the diagram is the velocity of the cart the greatest? Assume a frictionless surface. [asy]pair A = (1.7,3.9); pair B = (3.2,2.7); pair C = (5,1.2); pair D = (8,2.7); size(8cm); path boundary = (0,0.5)--(8,0.5)--(8,5)--(0,5)--cycle; path track = (0,3.2)..A..(3,3)..B..(4,1.8)..C..(6,1.5)..(7,2.3)..D; path sky = (0,5)--track--(8,5)--cycle; for (int a=0; a<=8; ++a) { draw((a,0)--(a,5), black+1); } for (int a=0; a<=5; ++a) { draw((0,a)--(8,a), black+1); } for (int a=-100; a<=100; ++a) { draw((0,a)--(8,a+8)); } for (int a=-100; a<=100; ++a) { draw((8,a)--(0,a+8)); } fill(sky,white); draw(track, black+3); clip(boundary); label("$A$", A, dir(120)); label("$B$", B, dir(60)); label("$C$", C, dir(90)); label("$D$", D, dir(135));[/asy] $ \textbf {(A) } \text {A} \qquad \textbf {(B) } \text {B} \qquad \textbf {(C) } \text {C} \qquad \textbf {(D) } \text {D} \\ \textbf {(E) } \text {There is insufficient information to decide} $ [i]Problem proposed by Kimberly Geddes[/i]

An integer $n > 1$ is given. Two players in turns mark points on a circle. First Player uses red color while Second Player uses blue color. The game is over when each player marks $n$ points. Then each player nds the arc of maximal length with ends of his color, which does not contain any other marked points. A player wins if his arc is longer (if the lengths are equal, or both players have no such arcs, the game ends in a draw). Which player has a winning strategy?

Prove that there exists two different permutations $ (a_1,a_2,\dots,a_{2009})$ and $ (b_1,b_2,\dots,b_{2009})$ of $ (1,2,\dots,2009)$ such that \[ \sum_{i\equal{}1}^{2009}i^i a_i \minus{} \sum_{i\equal{}1}^{2009} i^i b_i\] is divisible by $ 2009!$.

$ a_1, a_2, \cdots, a_n$ is a sequence of real numbers, and $ b_1, b_2, \cdots, b_n$ are real numbers that satisfy the condition $ 1 \ge b_1 \ge b_2 \ge \cdots \ge b_n \ge 0$. Prove that there exists a natural number $ k \le n$ that satisifes $ |a_1b_1 \plus{} a_2b_2 \plus{} \cdots \plus{} a_nb_n| \le |a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k|$

A $5 \times 5$ grid is given, called $f_1$: \[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 1 & -1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{array} \] A new grid $f_{n+1}$ is constructed where each cell is equal to the product of its neighboring cells in grid $f_n$. (a) Find the grids $f_6$ and $f_7$. (b) Find the grids $f_{2008}$ and $f_{2009}$. (c) Find $f_{2n}$ and $f_{2n+1}$ for any $n \in \mathbb{N}$. *Note: Neighboring cells are those that share an edge, not just a vertex.*

How many ordered triples $(a, b, c)$ of positive integers satisfy $a \le b  \le c$ and $a
\cdot b\cdot
c = 1000$?

How many positive integers have exactly three proper divisors, each of which is less than 50?

It is known that a certain mechanical balance can measure any object of integer mass anywhere between 1 and 2009 (both included). This balance has $k$ weights of integral values. What is the minimum $k$ for which there exist weights that satisfy this condition?

In a chess tournament, every participant played with each other exactly once, receiving $1$ point for a win, $1/2$ for a draw and $0$ for a loss. [list][b](a)[/b] Is it possible that for every player $P$, the sum of points of the players who were beaten by P is greater than the sum of points of the players who beat $P$? [b](b)[/b] Is it possible that for every player $P$, the first sum is less than the second one?[/list]

Let $0 <a <b$ be two rational numbers. Let $M$ be a set of positive real numbers with the properties: (i) $a \in M$ and $b \in M$; (ii) if $x$ $\in M$ and $y \in M$, then $\sqrt{xy} \in M$. Let $M^*$denote the set of all irrational numbers in $M$. prove that every $c,d$ such that $a <c <d<b$, $M^*$ contains an element $m$ with property $c<m<d$

Suppose that every two opposite edges of a tetrahedron are orthogonal. Show that the midpoints of the six edges lie on a sphere.

A salesman and a customer altogether have $1999$ rubles in coins and bills of $1, 5, 10, 50, 100, 500 , 1000$ rubles. The customer has enough money to buy a Cat in the Bag which costs the integer number of rubles. Prove that the customer can buy the Cat and get the correct change.

Let ABC be a triangle with$|AB|< |AC|. $ Let $k$ be the circumcircle of $\triangle ABC$ and let $O$ be the center of $k$. Point $M$ is the midpoint of the arc $BC $ of $k$ not containing $A$. Let $D $ be the second intersection of the perpendicular line from $M$ to $AB$ with $ k$ and $E$ be the second intersection of the perpendicular line from $M$ to $AC $ with $k$. Points $X $and $Y $ are the intersections of $CD$ and $BE$ with $OM$ respectively. Denote by $k_b$ and $k_c$ circumcircles of triangles $BDX$ and $CEY$ respectively. Let $G$ and $H$ be the second intersections of $k_b$ and $k_c $ with $AB$ and $AC$ respectively. Denote by ka the circumcircle of triangle $AGH.$ Prove that $O$ is the circumcenter of $\triangle O_aO_bO_c, $where $O_a, O_b, O_c $ are the centers of $k_a, k_b, k_c$ respectively.

The pentagon $ABCDE$ has an inscribed circle and the diagonals $AD$ and $CE$ intersect in its centre $O$. Prove that the segment $BO$ and the side $DE$ are perpendicular. (Folklore)

A triangle is given on a sphere of radius $1$, the sides of which are arcs of three different circles of radius $1$ centered in the center of a sphere having less than $\pi$ in length and an area equal to a quarter of the area of the sphere. Prove that four copies of such a triangle can cover the entire sphere. A. Zaslavsky

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? [asy] // diagram by SirCalcsALot size(250); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real pos = 2.5; pair s1 = (-10,-2.19); pair s2 = (15,2.19); pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); fill(circle(origin + s1, 1), grey1); for (int i = 0; i < 6; ++i) { draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25)); } fill(circle(origin, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i),1), grey2); draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25)); } fill(circle(origin+s2, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i)+s2,1), grey2); fill(circle(2*pos*dir(60*i)+s2,1), grey1); fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1); draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25)); } [/asy] $\textbf{(A)}\ 35 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 39 \qquad \textbf{(D)}\ 43 \qquad \textbf{(E)}\ 49$

In an acute triangle $\triangle ABC$, the midpoint of $BC$ is $M$. Perpendicular lines $BE$ and $CF$ are drawn respectively on $AC$ from $B$ and on $AB$ from $C$ such that $E$ and $F$ lie on $AC$ and $AB$ respectively. The midpoint of $EF$ is $N.$ $MN$ intersects $AB$ at $K.$ Prove that, the four points $B,K,E,M$ lie on the same circle.

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures? [asy] //kante314 size(11cm); filldraw((0,0)--(29,0)--(29,29)--(0,29)--cycle,mediumgray); draw((36,29/2)--(54,29/2),EndArrow(size=7)); draw((36,29/2)--(52.5,29/2),linewidth(1.5)); filldraw((61,22)--(63,22)--(63,6)--cycle,mediumgray); fill((63,6+1*17/16)--(80,6+1*17/16)--(80,6+2*17/16)--(63,6+2*17/16)--cycle,lightgray); fill((63,6+3*17/16)--(80,6+3*17/16)--(80,6+4*17/16)--(63,6+4*17/16)--cycle,lightgray); fill((63,6+5*17/16)--(80,6+5*17/16)--(80,6+6*17/16)--(63,6+6*17/16)--cycle,lightgray); fill((63,6+7*17/16)--(80,6+7*17/16)--(80,6+8*17/16)--(63,6+8*17/16)--cycle,lightgray); fill((63,6+9*17/16)--(80,6+9*17/16)--(80,6+10*17/16)--(63,6+10*17/16)--cycle,lightgray); fill((63,6+11*17/16)--(80,6+11*17/16)--(80,6+12*17/16)--(63,6+12*17/16)--cycle,lightgray); fill((63,6+13*17/16)--(80,6+13*17/16)--(80,6+14*17/16)--(63,6+14*17/16)--cycle,lightgray); fill((63,6+15*17/16)--(80,6+15*17/16)--(80,6+16*17/16)--(63,6+16*17/16)--cycle,lightgray); draw((63,6)--(63,23)--(68,23)--(69,12)--(80,6)--cycle); filldraw((69,12)--(69,27)--(67,28)--cycle,mediumgray); filldraw((69,12)--(69,29)--(80,23)--(80,6)--cycle,white); fill((69,12+1*15/13)--(80,6+1*15/13)--(80,6+2*15/13)--(69,12+2*15/13)--cycle,lightgray); fill((69,12+3*15/13)--(80,6+3*15/13)--(80,6+4*15/13)--(69,12+4*15/13)--cycle,lightgray); fill((69,12+5*15/13)--(80,6+5*15/13)--(80,6+6*15/13)--(69,12+6*15/13)--cycle,lightgray); fill((69,12+7*15/13)--(80,6+7*15/13)--(80,6+8*15/13)--(69,12+8*15/13)--cycle,lightgray); fill((69,12+9*15/13)--(80,6+9*15/13)--(80,6+10*15/13)--(69,12+10*15/13)--cycle,lightgray); fill((69,12+11*15/13)--(80,6+11*15/13)--(80,6+12*15/13)--(69,12+12*15/13)--cycle,lightgray); fill((69,12+13*15/13)--(80,6+13*15/13)--(80,6+14*15/13)--(69,12+14*15/13)--cycle,lightgray); draw((69,12)--(69,29)--(80,23)--(80,6)--cycle); draw((87,29/2)--(105,29/2),EndArrow(size=7)); draw((87,29/2)--(102.5,29/2),linewidth(1.5)); fill((112,6+1*17/16)--(129,6+1*17/16)--(129,6+2*17/16)--(112,6+2*17/16)--cycle,lightgray); fill((112,6+3*17/16)--(129,6+3*17/16)--(129,6+4*17/16)--(112,6+4*17/16)--cycle,lightgray); fill((112,6+5*17/16)--(129,6+5*17/16)--(129,6+6*17/16)--(112,6+6*17/16)--cycle,lightgray); fill((112,6+7*17/16)--(129,6+7*17/16)--(129,6+8*17/16)--(112,6+8*17/16)--cycle,lightgray); fill((112,6+9*17/16)--(129,6+9*17/16)--(129,6+10*17/16)--(112,6+10*17/16)--cycle,lightgray); fill((112,6+11*17/16)--(129,6+11*17/16)--(129,6+12*17/16)--(112,6+12*17/16)--cycle,lightgray); fill((112,6+13*17/16)--(129,6+13*17/16)--(129,6+14*17/16)--(112,6+14*17/16)--cycle,lightgray); fill((112,6+15*17/16)--(129,6+15*17/16)--(129,6+16*17/16)--(112,6+16*17/16)--cycle,lightgray); draw((112,6)--(129,6)--(129,23)--(112,23)--cycle); draw((112+17/2,6)--(129,6+17/2),dashed+linewidth(.3)); draw((111.7,6.7)--(111.7,23.3)--(128.3,23.3),linewidth(1)); draw((111.75,6.6)--(111.75,6.3)); draw((128.4,23.25)--(128.7,23.25)); [/asy] [asy] //kante314 size(11cm); label(scale(.85)*"\textbf{(A)}", (2,55)); filldraw((7,31)--(13,31)--(19.5,37)--(26,31)--(32,31)--(32,37)--(26,43.5)--(32,50)--(32,56)--(26,56)--(19.5,50)--(13,56)--(7,56)--(7,50)--(13,43.5)--(7,37)--cycle,mediumgray); label(scale(.85)*"\textbf{(B)}", (44,55)); filldraw((49,31)--(55,31)--(61.5,37)--(68,31)--(74,31)--(74,37)--(74,50)--(74,56)--(68,56)--(61.5,50)--(55,56)--(49,56)--(49,50)--(49,37)--cycle,mediumgray); label(scale(.85)*"\textbf{(C)}", (86,55)); filldraw((91,31)--(116,31)--(116,56)--(91,56)--cycle,mediumgray); filldraw((91+25/4,31+25/4)--(116-25/4,31+25/4)--(116-25/4,56-25/4)--(91+25/4,56-25/4)--cycle,white); label(scale(.85)*"\textbf{(D)}", (2,24)); filldraw((7,0)--(32,0)--(32,25)--(7,25)--cycle,mediumgray); filldraw((7+25/4,25/2)--(32-25/4,25/2)--(7+25/2,25-25/4)--cycle,white); label(scale(.85)*"\textbf{(E)}", (44,24)); filldraw((49,0)--(74,0)--(74,25)--(49,25)--cycle,mediumgray); filldraw((49+25/4,25/2)--(49+25/2,25/4)--(74-25/4,25/2)--(49+25/2,25-25/4)--cycle,white); [/asy]

Points $E$ and $F$ lie on diagonal $\overline{AC}$ of square $ABCD$ with side length $24$, such that $AE = CF = 3\sqrt2$. An ellipse with foci at $E$ and $F$ is tangent to the sides of the square. Find the sum of the distances from any point on the ellipse to the two foci.

In the following base-$10$ equation, each of the letter represents a unique digit: $AM\cdot PM=ZZZ$. Find the sum of $A+M+P+Z$. $\text{(A) }15\qquad\text{(B) }17\qquad\text{(C) }19\qquad\text{(D) }20\qquad\text{(E) }21$

The diagram below shows a square of area $36$ separated into two rectangles and a smaller square. One of the rectangles has an area of $12$. What is the smallest rectangle's area? [asy] size(70); draw((0,0)--(2,0)--(2,6)--(0,6)--cycle); draw((2,2)--(6,2)--(6,6)--(2,6)--cycle); draw((2,2)--(6,2)--(6,0)--(2,0)--cycle); label("$12$",(1,3)); label("$?$",(4,4)); label("$?$",(4,1)); [/asy] $\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }16\qquad\textbf{(E) }\text{Not Enough Information}$

Suppose you have $27$ identical unit cubes colored such that $3$ faces adjacent to a vertex are red and the other $3$ are colored blue. Suppose further that you assemble these $27$ cubes randomly into a larger cube with $3$ cubes to an edge (in particular, the orientation of each cube is random). The probability that the entire cube is one solid color can be written as $\frac{1}{2^n}$ for some positive integer $n$. Find $n$.

The area of a convex pentagon $ABCDE$ is $S$, and the circumradii of the triangles $ABC$, $BCD$, $CDE$, $DEA$, $EAB$ are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$. Prove the inequality \[ R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\geq {4\over 5\sin^2 108^\circ}S^2. \]

Let $A$ be a non-empty subset of the set of all positive integers $N^*$. If any sufficient big positive integer can be expressed as the sum of $2$ elements in $A$(The two integers do not have to be different), then we call that $A$ is a divalent radical. For $x \geq 1$, let $A(x)$ be the set of all elements in $A$ that do not exceed $x$, prove that there exist a divalent radical $A$ and a constant number $C$ so that for every $x \geq 1$, there is always $\left| A(x) \right| \leq C \sqrt{x}$.

The sequence $ (a_n)$ is defined by $ a_1\equal{}a_2\equal{}a_3\equal{}1$ and $ a_{n\plus{}1}a_{n\minus{}2}\minus{}a_n a_{n\minus{}1}\equal{}2$ for all $ n \ge 3.$ Prove that $ a_n$ is a positive integer for all $ n \ge 1$.