Given 98 points in a circle. Mary and Joseph play alternatively in the next way: - Each one draw a segment joining two points that have not been joined before. The game ends when the 98 points have been used as end points of a segments at least once. The winner is the person that draw the last segment. If Joseph starts the game, who can assure that is going to win the game.

The circumference inscribed on the triangle $ABC$ is tangent to the sides $BC$, $CA$ and $AB$ on the points $D$, $E$ and $F$, respectively. $AD$ intersect the circumference on the point $Q$. Show that the line $EQ$ meet the segment $AF$ at its midpoint if and only if $AC=BC$.

Click for solution Let $M$ be the intersection of $EQ$ and $AB$. We have \begin{eqnarray*} & & AM = MF \\ & \iff & AM^2 = MF^2 \\ & \iff & AM^2 = MQ \cdot ME \\ & \iff & \Delta AMQ \sim \Delta EMA \\ & \iff & \angle AEM = \angle QAM \\ & \iff & \angle EDA = \angle DAB \\ & \iff & DE \parallel AB \\ & \iff & \angle EDC = \angle ABC \\ & \iff & 90^{\circ} - \frac{1}{2} \angle ACB = \angle ABC \\ & \iff & \angle ACB + 2 \angle ABC = 180^{\circ} \\ & \iff & \angle ABC = \angle BAC \\ & \iff & AC = BC. \blacksquare \end{eqnarray*}

Find the minimum natural number $n$ with the following property: between any collection of $n$ distinct natural numbers in the set $\{1,2, \dots,999\}$ it is possible to choose four different $a,\ b,\ c,\ d$ such that: $a + 2b + 3c = d$.

There are representants from $n$ different countries sit around a circular table ($n\geq2$), in such way that if two representants are from the same country, then, their neighbors to the right are not from the same country. Find, for every $n$, the maximal number of people that can be sit around the table.

Find the maximal possible value of $n$ such that there exist points $P_1,P_2,P_3,\ldots,P_n$ in the plane and real numbers $r_1,r_2,\ldots,r_n$ such that the distance between any two different points $P_i$ and $P_j$ is $r_i+r_j$.

Let $\lambda$ the positive root of the equation $t^2-1998t-1=0$. It is defined the sequence $x_0,x_1,x_2,\ldots,x_n,\ldots$ by $x_0=1,\ x_{n+1}=\lfloor\lambda{x_n}\rfloor\mbox{ for }n=1,2\ldots$ Find the remainder of the division of $x_{1998}$ by $1998$. Note: $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$.

Click for solution We have $\frac{x_{n + 1}}{\lambda} < x_{n} < \frac{x_{n + 1} + 1}{\lambda}$, so $\left[\frac{x_{n + 1}}{\lambda}\right] = x_n - 1$. As $\lambda = 1998 + \frac{1}{\lambda}$, $x_{n + 1} = \left[\lambda x_n\right] = 1998x_n + \left[\frac{x_n}{\lambda}\right] = 1998x_n + x_{n - 1} - 1$, so $x_{n + 1} \equiv x_{n - 1} - 1 \pmod{1998}$. It follows that $x_{1998} \equiv 1000 \pmod{1998}$.