Solutions and Comments

7.

Let
$S=\frac{{1}^{2}}{1\xb73}+\frac{{2}^{2}}{3\xb75}+\frac{{3}^{2}}{5\xb77}+\dots +\frac{{500}^{2}}{999\xb71001}\hspace{1em}.$
Find the value of
$S$.
Solution. [Miranda Holmes] Each term of
$S$ is of the
form
$\frac{{i}^{2}}{(2i1)(2i+1)}=\frac{1}{4}(\frac{i}{2i1}+\frac{i}{2i+1})\hspace{1em}.$
Thus
$S=\frac{1}{4}\sum _{i=1}^{500}(\frac{i}{2i1}+\frac{i}{2i+1})\hspace{1em}.$
$4S=\sum _{i=1}^{500}(\frac{i}{2i1}+\frac{i}{2i+1})=\frac{1}{1}+\frac{1}{3}+\frac{2}{3}+\frac{2}{5}+\frac{3}{5}+\frac{3}{7}+\dots +\frac{500}{999}+\frac{500}{1001}\hspace{1em}.$
Since
$\frac{i}{2i+1}+\frac{i+1}{2i+1}=1\hspace{1em},$
we can combine all such terms to get
$4S=1+499+\frac{500}{1001}=\frac{501000}{1001}\hspace{1em}.$
Thus,
$S=\frac{125250}{1001}\hspace{1em}.$

8.

The sequences
$\{{a}_{n}\}$ and
$\{{b}_{n}\}$ are such
that, for every positive integer
$n$,
${a}_{n}>0\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}{b}_{n}>0\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}{a}_{n+1}={a}_{n}+\frac{1}{{b}_{n}}\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}{b}_{n+1}={b}_{n}+\frac{1}{{a}_{n}}\hspace{1em}.$
Prove that
${a}_{50}+{b}_{50}>20$.
Solution. Note that
$x+(1/x)\ge 2$ for every positive
real
$x$. Consider the sequence
${c}_{n}=({a}_{n}+{b}_{n}){}^{2}$.
Since
${c}_{2}=({a}_{2}+{b}_{2}){}^{2}=({a}_{1}+\frac{1}{{b}_{1}}+{b}_{1}+\frac{1}{{a}_{1}}{)}^{2}=[({a}_{1}+\frac{1}{{a}_{1}})+({b}_{1}+\frac{1}{{b}_{1}}){]}^{2}\hspace{1em},$
we find that
${c}_{2}\ge (2+2){}^{2}=16$.
For each positive integer
$n$,
$\begin{array}{cc}{c}_{n+1}\hfill & =({a}_{n+1}+{b}_{n+1}){}^{2}=({a}_{n}+\frac{1}{{b}_{n}}+{b}_{n}+\frac{1}{{a}_{n}}{)}^{2}\hfill \\ \multicolumn{0}{c}{}& =[({a}_{n}^{2}+{b}_{n}^{2}+2{a}_{n}{b}_{n})+(\frac{1}{{a}_{n}^{2}}+\frac{1}{{b}_{n}^{2}}+2\xb7\frac{1}{{a}_{n}}\xb7\frac{1}{{b}_{n}})\hfill \\ \multicolumn{0}{c}{}& \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+2\xb7{a}_{n}\xb7\frac{1}{{a}_{n}}+2\xb7{b}_{n}\xb7\frac{1}{{b}_{n}}+2(\frac{{a}_{n}}{{b}_{n}}+\frac{{b}_{n}}{{a}_{n}})\hfill \\ \multicolumn{0}{c}{}& ={c}_{n}+(\frac{1}{{b}_{n}}+\frac{1}{{a}_{n}}{)}^{2}+2+2+2\xb7(\frac{{a}_{n}}{{b}_{n}}+\frac{{b}_{n}}{{a}_{n}})>{c}_{n}+8\hspace{1em},\hfill \end{array}$
Thus,
${c}_{n+1}>{c}_{n}+8\hspace{1em}.$ It follows that
${c}_{50}>{c}_{49}+8>{c}_{48}+2\xb78>\dots >{c}_{2}+48\xb78\ge 16+48\xb78=400\hspace{1em}.$
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.
Comment. Before giving the solution to the problem, we
present a result that you should be aware of; pay also close
attention to the proof.
Proposition. There are six points in the plane or in
space, no
three of them collinear. Each of the segments between two of
them is coloured in one of two colours. There exists a
triangle whose vertices are three of the given points and
whose sides are of the same colour.
Proof. Let
$A$ be one of the points. There are five
segments joining
$A$ to the other points. Since they
have one of two colours, by the Pigeonhole Principle, at
least three of them must have the same colour. Wolog,
suppose that
$\mathrm{AB}$,
$\mathrm{AC}$,
$\mathrm{AD}$ are coloured the same.
If any of the segments
$\mathrm{BD}$,
$\mathrm{BC}$,
$\mathrm{CD}$ has this colour,
then we will have a triangle in this colour. Otherwise,
$\mathrm{BCD}$ must be a triangle all of whose edges have the
other colour.
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 onecolour 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.
Solution. Consider the six points and all triangles whose
vertices are any three of them. Colour the
(uniquely determined) largest side of each
triangle black, and colour the remaining edges red.
There must be a triangle all of whose edges are the same
colour. This colour cannot be red. (Why?) So there must
be a triangle all of whose edges are black; its smallest
edge must be the largest edge of some other triangle.

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.
Solution. [Miranda Holmes] If a circle of diameter 1 can
be placed, it means that there must be a point in the rectangle
such that every point of every square is more than 1/2 units
away from it to the centre of the circle. The maximum area A
around each square in which the centre of the circle cannot be
located is the area of the figure
$F$ formed by
(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
$456455>0$, 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.
Solution. [Miranda Holmes] Since the lines divide the square into
two quadrilaterals, they cut opposite sides of the square.
Let the vertices of the square be
$A$,
$B$,
$C$,
$D$ (counterclockwise),
and let one of the lines intersect
$\mathrm{AB}$ at
$M$ and
$\mathrm{CD}$ at
$N$.
We can represent these points in an appropriate coordinate plane
as
$A(0,0)$,
$B(1,0)$,
$C(1,1)$,
$D(0,1)$,
$M(m,0)$,
$N(n,1)$.
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 100sided 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
$a+b=c+d$.
Solution. Since the given polygon is regular, it can be inscribed
in a circle. There are exactly 50 diagonals of the polygon which
pass through the centre of the circle. As they are diameters,
they are of equal length. Consider the positive differences of
two vertices which are endpoints of the same diagonal. Since the
mark numbers are from among
$1,2,\dots ,49$, the range of the
differences is between 0 and 48. So, we have 49 possible values
for 50 differences. Hence, there are at least two diagonals
with the same difference. Without loss of generality, denote
these diagonals as as
$\mathrm{AC}$ and
$\mathrm{BD}$ and suppose that
$a\ge c$ and
$d\ge b$. Then
$ac=db$, so that
$a+b=c+d$. The quadrilateral
$\mathrm{ABCD}$ has two diagonals
of equal length and with the same midpoint, so it is a rectangle,
which satisfies all of the required conditions.