Round 1
The numbers $1,2,3,\dots,2017$ are on the blackboard. Amelie and Boris take turns removing one of those until only two numbers remain on the board. Amelie starts. If the sum of the last two numbers is divisible by $8$, then Amelie wins. Else Boris wins. Who can force a victory?
What is the maximum number of acute interior angles a non-overlapping planar $2017$-gon can have?
Let $M$ be the incenter of the tangential quadrilateral $A_1A_2A_3A_4$. Let line $g_1$ through $A_1$ be perpendicular to $A_1M$; define $g_2,g_3$ and $g_4$ similarly. The lines $g_1,g_2,g_3$ and $g_4$ define another quadrilateral $B_1B_2B_3B_4$ having $B_1$ be the intersection of $g_1$ and $g_2$; similarly $B_2,B_3$ and $B_4$ are intersections of $g_2$ and $g_3$, $g_3$ and $g_4$, resp. $g_4$ and $g_1$. Prove that the diagonals of quadrilateral $B_1B_2B_3B_4$ intersect in point $M$. [asy][asy] import graph; size(15cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-9.773972777861085,xmax=12.231603726660566,ymin=-3.9255487671791487,ymax=7.37238601960895; pair M=(2.,2.), A_4=(-1.6391623316400197,1.2875505916864178), A_1=(3.068893183992864,-0.5728665455336459), A_2=(4.30385937824148,2.2922812065339455), A_3=(2.221541124684679,4.978916319940133), B_4=(-0.9482172571022687,-2.24176848577888), B_1=(4.5873184669543345,0.057960746374459436), B_2=(3.9796042717514277,4.848169684238838), B_3=(-2.4295496490492385,5.324816563638236); draw(circle(M,2.),linewidth(0.8)); draw(A_4--A_1,linewidth(0.8)); draw(A_1--A_2,linewidth(0.8)); draw(A_2--A_3,linewidth(0.8)); draw(A_3--A_4,linewidth(0.8)); draw(M--A_3,linewidth(0.8)+dotted); draw(M--A_2,linewidth(0.8)+dotted); draw(M--A_1,linewidth(0.8)+dotted); draw(M--A_4,linewidth(0.8)+dotted); draw((xmin,-0.07436970390935019*xmin+5.144131675605378)--(xmax,-0.07436970390935019*xmax+5.144131675605378),linewidth(0.8)); draw((xmin,-7.882338401302275*xmin+36.2167572574517)--(xmax,-7.882338401302275*xmax+36.2167572574517),linewidth(0.8)); draw((xmin,0.4154483588930812*xmin-1.847833182441644)--(xmax,0.4154483588930812*xmax-1.847833182441644),linewidth(0.8)); draw((xmin,-5.107958950031516*xmin-7.085223310768749)--(xmax,-5.107958950031516*xmax-7.085223310768749),linewidth(0.8)); dot(M,linewidth(3.pt)+ds); label("$M$",(2.0593440948136896,2.0872038897020024),NE*lsf); dot(A_4,linewidth(3.pt)+ds); label("$A_4$",(-2.6355449660387147,1.085078446888477),NE*lsf); dot(A_1,linewidth(3.pt)+ds); label("$A_1$",(3.1575637581709772,-1.2486383377457595),NE*lsf); dot(A_2,linewidth(3.pt)+ds); label("$A_2$",(4.502882845783654,2.30684782237346),NE*lsf); dot(A_3,linewidth(3.pt)+ds); label("$A_3$",(2.169166061149418,5.203402184478307),NE*lsf); label("$g_3$",(-9.691606303109287,5.354407388189934),NE*lsf); label("$g_2$",(3.0889250292111465,6.727181967386543),NE*lsf); label("$g_1$",(-4.763345563793459,-3.4725331560442676),NE*lsf); label("$g_4$",(-2.663000457622647,6.878187171098171),NE*lsf); dot(B_4,linewidth(3.pt)+ds); label("$B_4$",(-1.5647807942653595,-3.0332452907013523),NE*lsf); dot(B_1,linewidth(3.pt)+ds); label("$B_1$",(4.955898456918535,-0.6583452686912173),NE*lsf); dot(B_2,linewidth(3.pt)+ds); label("$B_2$",(4.104778217816637,5.0661247265586455),NE*lsf); dot(B_3,linewidth(3.pt)+ds); label("$B_3$",(-3.4454819677647146,5.656417795613188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
The sequence $a_0,a_1,a_2,\dots$ is recursively defined by \[ a_0 = 1 \quad \text{and} \quad a_n = a_{n-1} \cdot \left(4-\frac{2}{n} \right) \quad \text{for } n \geq 1. \]Prove for each integer $n \geq 1$: (a) The number $a_n$ is a positive integer. (b) Each prime $p$ with $n < p \leq 2n$ is a divisor of $a_n$. (c) If $n$ is a prime, then $a_n-2$ is divisible by $n$.
Round 2
For which integers $n \geq 4$ is the following procedure possible? Remove one number of the integers $1,2,3,\dots,n+1$ and arrange them in a sequence $a_1,a_2,\dots,a_n$ such that of the $n$ numbers \[ |a_1-a_2|,|a_2-a_3|,\dots,|a_{n-1}-a_n|,|a_n-a_1| \]no two are equal.
In a convex regular $35$-gon $15$ vertices are colored in red. Are there always three red vertices that make an isosceles triangle?
Given is a triangle with side lengths $a,b$ and $c$, incenter $I$ and centroid $S$. Prove: If $a+b=3c$, then $S \neq I$ and line $SI$ is perpendicular to one of the sides of the triangle.
We call a positive integer heinersch if it can be written as the sum of a positive square and positive cube. Prove: There are infinitely many heinersch numbers $h$, such that $h-1$ and $h+1$ are also heinersch.