prove that $\pi_1 (X,x_0)$ is not abelian. $X$ is like an eight $(8)$ figure. [b]comments:[/b] eight figure is the union of two circles that have one point $x_0$ in common. we call a group $G$ abelian if: $\forall a,b \in G:ab=ba$.

Suppose $A,B,C,D$ are $n \times n$ matrices with entries in a field $F$, satisfying the conditions that $AB^T$ and $CD^T$ are symmetric and $AD^T - BC^T = I$. Here $I$ is the $n \times n$ identity matrix, and if $M$ is an $n \times n$ matrix, $M^T$ is its transpose. Prove that $A^T D - C^T B = I$.

Square $ ABCD$ has sides of length $ 4$, and $ M$ is the midpoint of $ \overline{CD}$. A circle with radius $ 2$ and center $ M$ intersects a circle with raidus $ 4$ and center $ A$ at points $ P$ and $ D$. What is the distance from $ P$ to $ \overline{AD}$? [asy]unitsize(8mm); defaultpen(linewidth(.8pt)); dotfactor=4; draw(Circle((2,0),2)); draw(Circle((0,4),4)); clip(scale(4)*unitsquare); draw(scale(4)*unitsquare); filldraw(Circle((2,0),0.07)); filldraw(Circle((3.2,1.6),0.07)); label("$A$",(0,4),NW); label("$B$",(4,4),NE); label("$C$",(4,0),SE); label("$D$",(0,0),SW); label("$M$",(2,0),S); label("$P$",(3.2,1.6),N);[/asy]$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}$

Let $p$ be a prime such that $p \equiv 3 \pmod 4$. Prove that we can't partition the numbers $a,a+1,a+2,\cdots,a+p-2$,($a \in \mathbb Z$) in two sets such that product of members of the sets be equal.

Tickets for the football game are $\$10$ for students and $\$15$ for non-students. If $3000$ fans attend and pay $\$36250$, how many students went?

Show that the equation ${n \choose k}=m^{l}$ has no integral solution with $l \ge 2$ and $4 \le k \le n-4$.

Find all functions $f:[1,\infty )\rightarrow [1,\infty)$ satisfying the following conditions: (1) $f(x)\le 2(x+1)$; (2) $f(x+1)=\dfrac{1}{x}[(f(x))^2-1]$ .

The perpendicular bisectors of the sides AB and BC of a triangle ABC meet the lines BC and AB at the points X and Z, respectively. The angle bisectors of the angles XAC and ZCA intersect at a point B'. Similarly, define two points C' and A'. Prove that the points A', B', C' lie on one line through the incenter I of triangle ABC. [i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. Darij

Let $\delta (x)$ be the greatest odd divisor of the positive integer $x$. Show that $| \sum_{n=1}^x \delta (n)/n -2x/3| <1,$ for all positive integers $x.$

Vertices $A$ and $C$ of the quadrilateral $ABCD$ are fixed points of the circle $k$ and each of the vertices $B$ and $D$ is moving to one of the arcs of $k$ with ends $A$ and $C$ in such a way that $BC=CD$. Let $M$ be the intersection point of $AC$ and $BD$ and $F$ is the center of the circumscribed circle around $\triangle ABM$. Prove that the locus of $F$ is an arc of a circle. [i]J. Tabov[/i]

Let $T_m$ be a number of non-congruent triangles which perimeter is $m$ and all its sides are positive integers. Prove that: $a)$ $T_{1999} > T_{2000}$ $b)$ $T_{4n+1}=T_{4n-2}+n$, $(n \in \mathbb{N})$

Find maximal numbers of planes, such there are $6$ points and 1) $4$ or more points lies on every plane. 2) No one line passes through $4$ points.

Which number is greater, $\frac{1996^{1995}+1}{1996^{1996}+1}$ or $ \frac{1996^{1996}+1}{1996^{1997}+1}$ ?

A conical surface $C$ is cut by a plane $T$ as shown in the figure on the back of this sheet. Show that $C \cap T$ is an ellipse. You can use as an aid the fact that if you consider the two spheres tangent to $C$ and $T$ as shown in the figure, they intersect $T$ in the bulbs. [asy] // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } // calculate area of space triangle with vertices A, B, and C real trianglearea(triple A, triple B, triple C) { return abs(cross(A - C, B - C)/2); } // calculate incentre of space triangle ABC triple triangleincentre(triple A, triple B, triple C) { return (abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(abs(B - C) + abs(C - A) + abs(A - B)); } // calculate inradius of space triangle ABC real triangleinradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(abs(B - C) + abs(C - A) + abs(A - B)); } // calculate excentre of space triangle ABC triple triangleexcentre(triple A, triple B, triple C) { return (-abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(-abs(B - C) + abs(C - A) + abs(A - B)); } // calculate exradius of space triangle ABC real triangleexradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)); } unitsize(2 cm); pair project (triple A, real t) { return((A.x, A.y*Sin(t) + A.z*Cos(t))); } real alpha, beta, theta, t; real coneradius = 1, coneheight = 3; real a, b, c; real[] m, r; triple A, B, V; triple ellipsecenter, ellipsex, ellipsey; triple[] F, O, P, R, W; path[] ellipse, spherering; theta = 15; V = (0,0,-coneheight); m[1] = sqrt(Cos(theta)^2*coneheight^2 - Sin(theta)^2*coneradius^2)/coneradius; m[2] = -m[1]; alpha = -aTan(Sin(theta)/m[1]); beta = -aTan(Sin(theta)/m[2]) + 180; A = (coneradius*Cos(alpha), coneradius*Sin(alpha), 0); B = (coneradius*Cos(beta), coneradius*Sin(beta), 0); W[1] = interp(V,(coneradius,0,0),0.6); W[2] = interp(V,(-coneradius,0,0),0.4); O[1] = triangleexcentre(V,W[1],W[2]); O[2] = triangleincentre(V,W[1],W[2]); r[1] = triangleexradius(V,W[1],W[2]); r[2] = triangleinradius(V,W[1],W[2]); F[1] = projectionofpointontoline(O[1],W[1],W[2]); F[2] = projectionofpointontoline(O[2],W[1],W[2]); P[1] = O[1] - (0,0,r[1]*coneradius/sqrt(coneradius^2 + coneheight^2)); P[2] = O[2] - (0,0,r[2]*coneradius/sqrt(coneradius^2 + coneheight^2)); spherering[11] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[12] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360); spherering[21] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[22] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360); ellipsecenter = (W[1] + W[2])/2; a = abs(W[1] - ellipsecenter); c = abs(F[1] - ellipsecenter); b = sqrt(a^2 - c^2); ellipsex = (W[1] - W[2])/abs(W[1] - W[2]); ellipsey = (0,1,0); ellipse[1] = project(ellipsecenter + a*ellipsex, theta); for (t = 0; t <= 180; t = t + 5) { ellipse[1] = ellipse[1]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); } ellipse[2] = project(ellipsecenter - a*ellipsex, theta); for (t = 180; t <= 360; t = t + 5) { ellipse[2] = ellipse[2]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); } R[1] = ellipsecenter + 1*ellipsex + ellipsey; R[2] = ellipsecenter - 1.2*ellipsex + ellipsey; R[3] = ellipsecenter - 1*ellipsex - ellipsey; R[4] = ellipsecenter + 1.2*ellipsex - ellipsey; fill(ellipse[1]--ellipse[2]--cycle, gray(0.9)); draw(yscale(Sin(theta))*Circle((0,0),coneradius)); draw(project(V,theta)--project(A,theta)); draw(project(V,theta)--project(B,theta)); draw(Circle(project(O[1],theta),r[1])); draw(Circle(project(O[2],theta),r[2])); draw(spherering[11], dashed); draw(spherering[12]); draw(spherering[21], dashed); draw(spherering[22]); draw(ellipse[1], dashed); draw(ellipse[2]); draw(project(R[1],theta)--interp(project(R[1],theta),project(R[2],theta),0.13)); draw(interp(project(R[1],theta),project(R[2],theta),0.13)--interp(project(R[1],theta),project(R[2],theta),0.76), dashed); draw(interp(project(R[1],theta),project(R[2],theta),0.76)--project(R[2],theta)); draw(project(R[2],theta)--project(R[3],theta)--project(R[4],theta)--project(R[1],theta)); label("$C$", (-1,0.3)); label("$T$", (1.2,-0.8)); dot(project(F[1],theta)); dot(project(F[2],theta)); //dot("$F_1$", project(F[1],theta)); //dot("$F_2$", project(F[2],theta)); //dot("$O_1$", project(O[1],theta)); //dot("$O_2$", project(O[2],theta)); //dot("$P_1$", project(P[1],theta)); //dot("$V$", project(V,theta)); //dot("$W_1$", project(W[1],theta)); //dot("$W_2$", project(W[2],theta)); [/asy]

[asy] fill(circle((4,0),4),grey); fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white); fill(circle((7,0),1),white); fill(circle((3,0),3),white); draw((0,0)--(8,0),black+linewidth(1)); draw((6,0)--(6,sqrt(12)),black+linewidth(1)); MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N); [/asy] In this diagram semi-circles are constructed on diameters $\overline{AB}$, $\overline{AC}$, and $\overline{CB}$, so that they are mutually tangent. If $\overline{CD} \bot \overline{AB}$, then the ratio of the shaded area to the area of a circle with $\overline{CD}$ as radius is: $\textbf{(A)}\ 1:2\qquad \textbf{(B)}\ 1:3\qquad \textbf{(C)}\ \sqrt{3}:7\qquad \textbf{(D)}\ 1:4\qquad \textbf{(E)}\ \sqrt{2}:6$

Let $AA _1$ and $CC_1$ be altitudes of an acute triangle $ABC$. Let $I$ and $J$ be the incenters of the triangles $AA_1C$ and $AC_1C$ respectively. The $C_1J$ and $A_1 I$ lines cut into $T$. Prove that lines $AT$ and $TC$ are perpendicular.

Let $a,b,c\in \mathbb{R}$ with $a^2+b^2+c^2=1$ and $\lambda\in \mathbb{R}_{>0}\setminus\{1\}$. Then for each solution $(x,y,z)$ of the system of equations: \[ \begin{cases} x-\lambda y=a,\\ y-\lambda z=b,\\ z-\lambda x=c. \end{cases} \] we have $\displaystyle x^2+y^2+z^2\leqslant \frac1{(\lambda-1)^2}$. [i]Radu Gologan[/i]

Let circle $A$ have radius $9,$ and let circle $B$ have radius $5$ and be internally tangent to circle $A.$ The largest radius $r$ such that there are two circles with radius $r$ that lie inside circle $A,$ are externally tangent to each other, and externally tangent with circle $B$ can be expressed as a fraction $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Let the incircle of triangle $ABC$ touch sides $BC, CA$ and $AB$ at the points $D, E$ and $F$ respectively and let $I$ be the center of that circle. Furthermore, let $P$ be the foot of the perpendicular from point $I$ to line $AD$ and let $M$ be the midpoint of $DE$. If $N$ is the intersection point of $PM$ and $AC$, prove that $DN \parallel EF$.

Let $ ABC$ be a triangle, line $ l$ cuts its sides $ BC,CA,AB$ at $ D,E,F$, respectively. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ AEF,BFD,CDE$, respectively. Prove that the orthocenter of triangle $ O_{1}O_{2}O_{3}$ lies on line $ l$.

Four dice are thrown. What is the probability that the product of the number equals $ 36?$

Let $ABC$ be an equilateral triangle with side length $1.$ Points $D, E,$ and $F$ lie inside triangle $ABC$ such that $A, E, F$ are collinear, $B, F, D$ are collinear, $C, D, E$ are collinear, and triangle $DEF$ is equilateral. Suppose that there exists a unique equilateral triangle $XYZ$ with $X$ on side $\overline{BC},$ $Y$ is on side $\overline{AB},$ and $Z$ is on side $\overline{AC}$ such that $D$ lies on side $\overline{XZ},$ $E$ lies on side $\overline{YZ},$ and $F$ lies on side $\overline{XY}.$ Compute $AZ.$

A convex polygon of sides $\ell_1, \ell_2, ..., \ell_n$ is called [i]ordered [/i] if for all reordering $( \sigma (1), \sigma (2), ..., \sigma (n))$ of the set $(1, 2,..., n)$ there exists a point $P$ inside the polygon such that $d_{\sigma (1)} < _{\sigma (2)} <...< d_{\sigma (n)}$ , where $d_i$ represents the distance between $P$ and side $\ell_i$. Find all the convex ordered polygons.

Derek the Dolphin and Kevin the Frog are playing a game where they take turns taking coins from a stack of $N$ coins, except with one rule: The number of coins someone takes each turn must be a power of $6$. The person who cannot take any more coins loses. If Derek goes first, how many integers $N$ from $1$ to $6^{2021}$ inclusive will guarantee him a win? (Example: If $N = 37$, then a possible sequence of turns is: Derek takes one coin, Kevin takes $36$ coins, and Kevin wins.)

There is a set of $50$ cards. Each card on both sides is colored in one of three colors — red, blue or white, and for each card its two sides are colored in different colors. The cards were laid out on the table. The card [i]lies beautifully[/i] if at least one of two conditions is met: its upper side — red; its underside is blue. It turned out that exactly $25$ cards are lying beautifully. Then all the cards were turned over. Now some of the cards are lying beautifully on the table. How many of them can there be? [i]K.A.Sukhov[/i]