- 7.
- Let

Find the value of $S$.

Thus

Since

we can combine all such terms to get

Thus,

- 8.
- The sequences $\{{a}_{n}\}$ and $\{{b}_{n}\}$ are such that, for every positive integer $n$,

Prove that ${a}_{50}+{b}_{50}>20$.

we find that ${c}_{2}\ge (2+2){}^{2}=16$.

For each positive integer
$n$,

Thus, ${c}_{n+1}>{c}_{n}+8\hspace{1em}.$ It follows that

Since ${a}_{50}+{b}_{50}>0$, it follows that ${a}_{50}+{b}_{50}>20$.

- 9.
- There are six points in the plane, no three of them collinear. Any three of them are vertices of a triangle whose sides are of different length. Prove that there exists a triangle whose smallest side is the largest side of another triangle.

Note 1: The minimum number of points with such a property is
6. If there are five points, it is possible to colour the
segments between any two of them so that a triangle with
edges of a single colour does not exist. For example, for
a regular pentagon, we can colour all the sides with one
colour and all the diagonals with the other.

Note 2: A triangle of one-colour always exists when we have
17 points in the plane (no three collinear) and three colours
are used for the segments. This can be given a similar proof.
>From any point, at least six of the segments emanating from
it have the same colour. Now look at the six points
terminating these segments.

Now we can solve the problem.

- 10.
- In a rectangle, whose sides are 20 and 25 units of length, are placed 120 squares of side 1 unit of length. Prove that a circle of diameter 1 unit can be placed in the rectangle, so that it has no common points with the squares.

(a) the square;

(b) four rectangles of dimensions
$1\times \frac{1}{2}$ external
to the sides of the square;

(c) four quarters of circles with radius 1/2 units
external to the square with and centres at
the vertices of the square.

Hence,
$A=1+\frac{1}{2}\xb71\xb74+(\frac{1}{2}){}^{2}\pi =3+\frac{\pi}{4}$. As there are
120 squares, the sum of all such areas within the
rectangle does not exceed
$120\xb7(3+\frac{\pi}{4})<455$.

As the circle should be placed inside of the rectangle, its centre
cannot be less than 1/2 units away from the rectangle's sides,
*i.e.*, it can be only in the rectangle with sides 19 and 24
units of length, whose sides are parallel to the rectangle's sides
on the distance 1/2 units from them. The area of this rectangle
is
$19\times 24=456$. But
$456-455>0$, so at least one point
is not covered by any of the 120 figures
$F$ described above.. This
point can be the centre of a circle of diameter 1 lying within the
rectangle and having no point in common with any of the squares.

- 11.
- Each of nine lines divides a square into two quadrilaterals, such that the ratio of their area is 2:3. Prove that at least three of these lines are concurrent.

Let
$[\mathrm{AMND}]:[\mathrm{MBCN}]=2:3$. Then
$[\mathrm{AMND}]=(m+n)/2=2/5$, because
the area of the whole square is 1. The midpoint of
$\mathrm{MN}$ is the point
$S(\frac{1}{2}(m+n),\frac{1}{2})=S(\frac{2}{5},\frac{1}{2})$, which
does not depend on the points of intersection of
$M$ and
$N$ and, hence,
is the same for all such lines. So, each line which divides the square
into two quadrilaterals in this way, must go through the point
$S$.
Because of the symmetry, there are three other possible points
in the square
$(\frac{3}{5},\frac{1}{2})$,
$(\frac{1}{2},\frac{2}{5})$,
$(\frac{1}{2},\frac{3}{5})$, and each of the
given 9 points must pass through one of them. Applying the
Pigeonhole Principle for 9 lines and 4 points, we find that at
least three of the lines must pass through the same point, and
because of that, they are concurrent.

- 12.
- Each vertex of a regular 100-sided polygon is marked with a number chosen from among the natural numbers $1,2,3,\dots ,49$. Prove that there are four vertices (which we can denote as $A$, $B$, $C$, $D$ with respective numbers $a$, $b$, $c$, $d$) such that $\mathrm{ABCD}$ is a rectangle, the points $A$ and $B$ are two adjacent vertices of the rectangle and $a+b=c+d$.