There are $9999$ rods with lengths $1, 2, ..., 9998, 9999$. The players Anja and Bernd alternately remove one of the sticks, with Anja starting. The game ends when there are only three bars left. If from those three bars, a not degenerate triangle can be constructed then Anja wins, otherwise Bernd. Who has a winning strategy?

For every non-negative integer $i$, define the number $M(i)$ as follows: write $i$ down as a binary number, so that we have a string of zeroes and ones, if the number of ones in this string is even, then set $M(i) = 0$, otherwise set $M(i) = 1$. (The first terms of the sequence $M(i)$, $i = 0, 1, 2, ...$ are $0, 1, 1, 0, 1, 0, 0, 1,...$ ) (a) Consider the finite sequence $M(O), M(1), . . . , M(1000) $. Prove that there are at least $320$ terms in this sequence which are equal to their neighbour on the right : $M(i) = M(i + 1 )$ . (b) Consider the finite sequence $M(O), M(1), . . . , M(1000000)$ . Prove that the number of terms $M(i)$ such that $M(i) = M(i +7)$ is at least $450000$. (A Kanel)

Triangle $\vartriangle ABC$ has side lengths $AB = AC = 27$ and $BC = 18$. Point $D$ is on $\overline{AB}$ and point $E$ is on $\overline{AC}$ such that $\angle BCD = \angle CBE = \angle BAC$. Compute $DE$.

For how many positive integers $ n$ is $ n^2\minus{}3n\plus{}2$ a prime number? $ \textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \textbf{(D)}\ \text{more than two, but finitely many}\\ \textbf{(E)}\ \text{infinitely many}$

In the diagram, $ABCDEF$ is a regular hexagon with side length 2. Points $E$ and $F$ are on the $x$ axis and points $A$, $B$, $C$, and $D$ lie on a parabola. What is the distance between the two $x$ intercepts of the parabola? [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -3.3215445204635294, xmax = 7.383669550094284, ymin = -4.983460515387094, ymax = 6.688676116382409; pen zzttqq = rgb(0.6,0.2,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((2,0)--(4,0)--(5,1.7320508075688774)--(4,3.4641016151377553)--(2,3.4641016151377557)--(1,1.732050807568879)--cycle, linewidth(1)); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, EndArrow(6), above = true); yaxis(ymin, ymax, EndArrow(6), above = true); draw((2,0)--(4,0), linewidth(1)); draw((4,0)--(5,1.7320508075688774), linewidth(1)); draw((5,1.7320508075688774)--(4,3.4641016151377553), linewidth(1)); draw((4,3.4641016151377553)--(2,3.4641016151377557), linewidth(1)); draw((2,3.4641016151377557)--(1,1.732050807568879), linewidth(1)); draw((1,1.732050807568879)--(2,0), linewidth(1)); real f1 (real x) {return -0.58*x^(2)+3.46*x-1.15;} draw(graph(f1,-3.3115445204635297,7.373669550094284), linewidth(1)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /*yes i used geogebra fight me*/ [/asy]

The hexagon has five $90^o$ angles and one $270^o$ angle (see picture). Use a straight-line ruler to divide it into two equal-sized polygons. [img]https://cdn.artofproblemsolving.com/attachments/d/8/cdd4df68644bb8e04adbe1b265039b82a5382b.png[/img]

(1) Find the possible number of roots for the equation $|x + 1| + |x + 2| + |x + 3| = a$, where $x \in R$ and $a$ is parameter. (2) Let $\{ a_1, a_2, \ldots, a_n \}$ be an arithmetic progression, $n \in \mathbb{N}$, and satisfy the condition \[ \sum^{n}_{i=1}|a_i| = \sum^{n}_{i=1}|a_{i} + 1| = \sum^{n}_{i=1}|a_{i} - 2| = 507. \] Find the maximum value of $n$.

Let A,B,C,D four matrices of order n with complex entries, n>=2 and let k real number such that AC+kBD=I and AD=BC. Prove that CA+kDB=I and DA=CB.

we are given $n$ rectangles in the plane. Prove that between $4n$ right angles formed by these rectangles there are at least $[4\sqrt n]$ distinct right angles.

Prove that $ \forall n > 1, n \in \mathbb{N}$ the equation \[ \sum^n_{k\equal{}1} \frac{x^k}{k!} \plus{} 1 \equal{} 0\] has no rational roots.

Let $ABCD$ be a tetrahedron and $O$ its incenter, and let the line $OD$ be perpendicular to $AD$. Find the angle between the planes $DOB$ and $DOC.$

Let $ABC$ be a triangle with circumradius $17$, inradius $4$, circumcircle $\Gamma$ and $A$-excircle $\Omega$. Suppose the reflection of $\Omega$ over line $BC$ is internally tangent to $\Gamma$. Compute the area of $\triangle ABC$.

Determine the greatest number of figures congruent to [img]https://cdn.artofproblemsolving.com/attachments/c/6/343f9197bcebf6794460ed1a74ba83ec18a377.png[/img] that can be placed in a $9 \times 9$ grid (without overlapping), such that each figure covers exactly $4$ unit squares. The figures can be rotated and flipped over. For example, the picture below shows that at least $3$ such figures can be placed in a $4 \times4$ grid. [img]https://cdn.artofproblemsolving.com/attachments/1/e/d38fc34b650a1333742bb206c29985c94146aa.png[/img]

Let $n$ be a positive integer. Ana and Banana play a game with $2n$ lamps numbered $1$ to $2n$ from left to right. Initially, all lamps numbered $1$ through $n$ are on, and all lamps numbered $n+1$ through $2n$ are off. They play with the following rules, where they alternate turns with Ana going first: [list] [*] On Ana's turn, she can choose two adjacent lamps $i$ and $i+1$, where lamp $i$ is on and lamp $i+1$ is off, and toggle both. [*] On Banana's turn, she can choose two adjacent lamps which are either both on or both off, and toggle both. [/list] Players must move on their turn if they are able to, and if at any point a player is not able to move on her turn, then the game ends. Determine all $n$ for which Banana can turn off all the lamps before the game ends, regardless of the moves that Ana makes. [i]Andrew Wen[/i]

If $x$, $y$, and $z$ are integers satisfying $xyz+4(x+y+z)=2(xy+xz+yz)+7$, list all possibilities for the ordered triple $(x, y, z)$.

Let $n$ be a positive integer and let $P$ be the set of monic polynomials of degree $n$ with complex coefficients. Find the value of \[ \min_{p \in P} \left \{ \max_{|z| = 1} |p(z)| \right \} \]

A block of mass $m_\text{1}$ is on top of a block of mass $m_\text{2}$. The lower block is on a horizontal surface, and a rope can pull horizontally on the lower block. The coefficient of kinetic friction for all surfaces is $\mu$. What is the resulting acceleration of the lower block if a force $F$ is applied to the rope? Assume that $F$ is sufficiently large so that the top block slips on the lower block. [asy] size(200); import roundedpath; draw((0,0)--(30,0),linewidth(3)); path A=(7,0.5)--(17,0.5)--(17,5.5)--(7,5.5)--cycle; filldraw(roundedpath(A,1),lightgray); path B=(10,6)--(15,6)--(15,9)--(10,9)--cycle; filldraw(roundedpath(B,1),lightgray); label("1",(12.5,6),1.5*N); label("2",(12,0.5),3*N); draw((17,3)--(27,3),EndArrow(size=13)); label(scale(1.2)*"$F$",(22,3),2*N); [/asy] (A) $a_\text{2}=(F-\mu g(2m_\text{1}+m_\text{2}))/m_\text{2}$ (B) $a_\text{2}=(F-\mu g(m_\text{1}+m_\text{2}))/m_\text{2}$ (C) $a_\text{2}=(F-\mu g(m_\text{1}+2m_\text{2}))/m_\text{2}$ (D) $a_\text{2}=(F+\mu g(m_\text{1}+m_\text{2}))/m_\text{2}$ (E) $a_\text{2}=(F-\mu g(m_\text{2}-m_\text{1}))/m_\text{2}$

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

$(GBR 5)$ Let us define $u_0 = 0, u_1 = 1$ and for $n\ge 0, u_{n+2} = au_{n+1}+bu_n, a$ and $b$ being positive integers. Express $u_n$ as a polynomial in $a$ and $b.$ Prove the result. Given that $b$ is prime, prove that $b$ divides $a(u_b -1).$

Say an integer polynomial is $\textit{primitive}$ if the greatest common divisor of its coefficients is $1$. For example, $2x^2+3x+6$ is primitive because $\gcd(2,3,6)=1$. Let $f(x)=a_2x^2+a_1x+a_0$ and $g(x) = b_2x^2+b_1x+b_0$, with $a_i,b_i\in\{1,2,3,4,5\}$ for $i=0,1,2$. If $N$ is the number of pairs of polynomials $(f(x),g(x))$ such that $h(x) = f(x)g(x)$ is primitive, find the last three digits of $N$.

Let $ A,B,C,D,E$ be five distinct points, such that no three of them lie on the same line. Prove that \[ AB\plus{}BC\plus{}CA \plus{} DE < AD \plus{} AE \plus{} BD\plus{}BE \plus{} CD\plus{}CE .\]

Given an acute triangle $ABC$ with $AB <AC$. Points $P$ and $Q$ lie on the angle bisector of $\angle BAC$ so that $BP$ and $CQ$ are perpendicular on that angle bisector. Suppose that point $E, F$ lie respectively at sides $AB$ and $AC$ respectively, in such a way that $AEPF$ is a kite. Prove that the lines $BC, PF$, and $QE$ intersect at one point.

There are $2017$ mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when Luciano stops drawing.

Let be two sequences $ \left( a_n \right)_{n\ge 0} , \left( b_n \right)_{n\ge 0} $ satisfying the following system: $$ \left\{ \begin{matrix} a_0>0,& \quad a_{n+1} =a_ne^{-a_n} , &\quad\forall n\ge 0 \\ b_{0}\in (0,1) ,& \quad b_{n+1} =b_n\cos \sqrt{b_n} ,& \quad\forall n\ge 0 \end{matrix} \right. $$ Calculate $ \lim_{n\to\infty} \frac{a_n}{b_n} . $ [i]Florian Dumitrel[/i]

Find all integers $n$, $n>1$, with the following property: for all $k$, $0\le k < n$, there exists a multiple of $n$ whose digits sum leaves a remainder of $k$ when divided by $n$.