In a country, there are $100$ towns. Some pairs of towns are joined by roads. The roads do not intersect one another except meeting at towns. It is possible to go from any town to any other town by road. Prove that it is possible to pave some of the roads so that the number of paved roads at each town is odd.

Consider the set $A=\{1,2,3,4,5,6,7,8,9\}$.A partition $\Pi $ of $A$ is collection of disjoint sets whose union is $A$. For example, $\Pi_1=\{\{1,2\},\{3,4,5\},\{6,7,8,9\}\}$ and $\Pi _2 =\{\{1\},\{2,5\},\{3,7\},\{4,5,6,7,8,9\}\}$ can be considered as partitions of $A$. For, each $\Pi$ partition ,we consider the function $\pi$ defined on the elements of$A$. $\pi (x)$ denotes the cardinality of the subset in $\Pi$ which contains $x$. For, example in case of $\Pi_1$ , $\pi_1(1)=\pi_1(2)=2$, $\pi_1(3)=\pi_1(4)=\pi_1 (5)=3$, and $\pi_1(6)=\pi_1(7)=\pi_1(8)=\pi_1(9)=4$. For $\Pi_2$ we have $\pi_2(1)=1$, $\pi_2(2)=\pi_2(5)=2$, $\pi_2(3)=\pi_2(7)=2$ and $\pi_2(4)=\pi_2(6)=\pi_2(8)=\pi_2(9)=4$ Given any two partitions $\Pi$ and $\Pi '$, show that there are two numbers $x$ and $y$ in $A$, such that $\pi (x)= \pi '(x)$ and $ \pi (y)= \pi'(y)$.[[b]Hint:[/b] Consider the case where there is a block of size greater than or equal to 4 in a partition and the alternative case]

Let $ [a]$ be the integer such that $ [a]\le a<[a]\plus{}1$. Find all real numbers $ (a,b,c)$ such that \[ \{a\}\plus{}[b]\plus{}\{c\}\equal{}2.9\\\{b\}\plus{}[c]\plus{}\{a\}\equal{}5.3\\\{c\}\plus{}[a]\plus{}\{b\}\equal{}4.0.\]

Let $\triangle MBT$ be a triangle with $\overline{MB} = 4$ and $\overline{MT} = 7$. Furthermore, let circle $\omega$ be a circle with center $O$ which is tangent to $\overline{MB}$ at $B$ and $\overline{MT}$ at some point on segment $\overline{MT}$. Given $\overline{OM} = 6$ and $\omega$ intersects $ \overline{BT}$ at $I \neq B$, find the length of $\overline{TI}$. [i]Proposed by Chad Yu[/i]

Is there an injective function $f : Z^+ \to Q$ satisfying the equation $f(xy) = f(x) + f(y)$ for all positive integers $x$ and $y$?

Let $\triangle ABC$ be an acute-angled triangle, with circumcenter $O$, circumradius $R$ and orthocenter $H$. Let $A_1$ be a point on $BC$ such that $A_1H+A_1O=R$. Define $B_1$ and $C_1$ similarly. If $\overrightarrow{AA_1} + \overrightarrow{BB_1} + \overrightarrow{CC_1} = \overrightarrow{0}$, prove that $\triangle ABC$ is equilateral.

Find the locus of centers of regular triangles such that three given points $A, B, C$ lie respectively on three lines containing sides of the triangle.

Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$. Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic. [i]Author: Andrey Gavrilyuk, Russia[/i]

Given $x,y,z\in \mathbb{R} ^+$ , that are the solutions to the system of equations : $$x^2+xy+y^2=57$$ $$y^2+yz+z^2=84$$ $$z^2+zx+x^2=111$$ What is the value of $xy+3yz+5zx$? [i](maphybich)[/i]

Let $ABCDE$ be a cyclic pentagon such that $BC = DE$ and $AB$ is parallel to $DE$. Let $X, Y,$ and $Z$ be the midpoints of $BD, CE,$ and $AE$ respectively. Show that $AE$ is tangent to the circumcircle of the triangle $XYZ$. Proposed by [i]Nikola Velov, Macedonia[/i]

Every member of a certain parliament has not more than $3$ enemies. Prove that it is possible to divide it onto two subparliaments so, that everyone will have not more than one enemy in his subparliament. ($A$ is the enemy of $B$ if and only if $B$ is the enemy of $A$.)

Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$? $ \textbf{(A)} A = \dfrac9{16}B \qquad\textbf{(B)} A = \dfrac34B \qquad\textbf{(C)} A=B \qquad\textbf{(D)} A = \dfrac43B \\ \\ \textbf{(E)}A = \dfrac{16}9B $

The function $f(n)$ satisfies $f(0)=0$, $f(n)=n-f \left( f(n-1) \right)$, $n=1,2,3 \cdots$. Find all polynomials $g(x)$ with real coefficient such that \[ f(n)= [ g(n) ], \qquad n=0,1,2 \cdots \] Where $[ g(n) ]$ denote the greatest integer that does not exceed $g(n)$.

Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? [asy] size(200); defaultpen(linewidth(3)); real[] inrad = {40,34,28,21}; real[] outrad = {55,49,37,30}; real[] center; path[][] quad = new path[4][4]; center[0] = 0; for(int i=0;i<=3;i=i+1) { if(i != 0) { center[i] = center[i-1] - inrad[i-1] - inrad[i]+3.5; } quad[0][i] = arc((0,center[i]),inrad[i],0,90)--arc((0,center[i]),outrad[i],90,0)--cycle; quad[1][i] = arc((0,center[i]),inrad[i],90,180)--arc((0,center[i]),outrad[i],180,90)--cycle; quad[2][i] = arc((0,center[i]),inrad[i],180,270)--arc((0,center[i]),outrad[i],270,180)--cycle; quad[3][i] = arc((0,center[i]),inrad[i],270,360)--arc((0,center[i]),outrad[i],360,270)--cycle; draw(circle((0,center[i]),inrad[i])^^circle((0,center[i]),outrad[i])); } void fillring(int i,int j) { if ((j % 2) == 0) { fill(quad[i][j],white); } else { filldraw(quad[i][j],black); } } for(int i=0;i<=3;i=i+1) { for(int j=0;j<=3;j=j+1) { fillring(((2-i) % 4),j); } } for(int k=0;k<=2;k=k+1) { filldraw(circle((0,-228 - 25 * k),3),black); } real r = 130, s = -90; draw((0,57)--(r,57)^^(0,-57)--(r,-57),linewidth(0.7)); draw((2*r/3,56)--(2*r/3,-56),linewidth(0.7),Arrows(size=3)); label("$20$",(2*r/3,-10),E); draw((0,39)--(s,39)^^(0,-39)--(s,-39),linewidth(0.7)); draw((9*s/10,38)--(9*s/10,-38),linewidth(0.7),Arrows(size=3)); label("$18$",(9*s/10,0),W); [/asy] $ \textbf{(A) } 171\qquad \textbf{(B) } 173\qquad \textbf{(C) } 182\qquad \textbf{(D) } 188\qquad \textbf{(E) } 210$

Let $ABC$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto triangle sides $BC$, $CA$, $AB$, respectively. Find the locus of points $P$ such that $AA_1$, $BB_1$, $CC_1$ are concurrent and $\angle PAB + \angle PBC + \angle PCA = 90^{\circ}$.