
3. We are given a sheet of paper in the form of a rectangle $x \times y$, where $x$ and $y$ are integer numbers larger than 1. Let us draw a lattice of $x \cdot y$ unite squares on the sheet. Rolling up the rectangle and gluing it along its opposite sides we shape a lateral surface of a circular cylinder. Join each two distinct vertices of the marked unit squares on the surface by a segment. How many of all these segments are passing through an interior point of the cylinder? In the case $x>y$ decide when this number of "internal" segments is larger - for the cylinder with bases of perimeter $x$, or $y$ ?

(Vojtech Bálint)


Solution. We will compute the requested number $P$ of all internal segments for the cylinder formed by gluing the rectangle $x \times y$ along the opposite sides of length $y$.

This cylinder has two bases of perimeter $x$ and its lateral sides are of length $y$. We will use an obvious formula $P=P_{0}-P_{1}-P_{2}$, where $P_{0}$ denotes the total number of segments, while $P_{1}$ and $P_{2}$ denote the numbers of segments which lie on the lateral surface or on one of the two bases, respectively.

Note that the vertices of the unit squares are situated on the surface of the cylinder in such a way that exactly $y+1$ of them lie on the same of the $x$ lateral sides and, in the same time, exactly $x$ vertices lie on the same boundary circle of the two bases. These facts lead immediately to the following formulæ

$$
\begin{aligned}
& P_{0}=\left(\begin{array}{c}
x(y+1) \\
2
\end{array}\right)=\frac{x(y+1)(x y+x-1)}{2}, \\
& P_{1}=x \cdot\left(\begin{array}{c}
y+1 \\
2
\end{array}\right)=\frac{x(y+1) y}{2}, \\
& P_{2}=2 \cdot\left(\begin{array}{l}
x \\
2
\end{array}\right)=x(x-1) .
\end{aligned}
$$

Consequently,

$$
\begin{aligned}
P & =P_{0}-P_{1}-P_{2}=\frac{x(y+1)(x y+x-1)}{2}-\frac{x(y+1) y}{2}-x(x-1) \\
& =\frac{x(x-1)\left(y^{2}+2 y-1\right)}{2} .
\end{aligned}
$$

In view of symmetry, the number $Q$ of internal segments for the other cylinder (with base's perimeter $y$ and lateral sides of length $x$ ) is given by

$$
Q=\frac{y(y-1)\left(x^{2}+2 x-1\right)}{2} .
$$

To decide which of the inequalities $P>Q$ or $Q>P$ holds in the case when $x>y$, we factorize the difference $P-Q$ (since $P=Q$ if $x=y$, the polynomial $P-Q$ must be divisible by $x-y$ ):

$$
\begin{aligned}
2(P-Q)= & \left(x^{2}-x\right)\left(y^{2}+2 y-1\right)-\left(y^{2}-y\right)\left(x^{2}+2 x-1\right) \\
= & \left(x^{2} y^{2}-x y^{2}+2 x^{2} y-2 x y-x^{2}+x\right) \\
& -\left(x^{2} y^{2}-x^{2} y+2 x y^{2}-2 x y-y^{2}+y\right) \\
= & 3 x y(x-y)-(x-y)(x+y)+(x-y) \\
= & (x-y)(3 x y-x-y+1) .
\end{aligned}
$$

Thus $x>y$ implies that $P>Q$ if we show that the same condition $x>y$ implies that $3 x y-x-y+1>0$. The last is almost evident: it follows from $y \geqslant 2$ that $3 x y \geqslant 6 x$ and hence

$$
3 x y-x-y+1 \geqslant 5 x-y+1>4 x+1>0 \text {. }
$$

Answer. In the case when $x>y$, the number of internal segments is larger for the cylinder with bases which perimeter has length $x$.

Remark. Let us describe a shorter way of determining the number $P$. The orthogonal projection of each internal segment to the fixed base of the cylinder is one of the $\frac{1}{2} x(x-1)$ segments connecting $x$ vertices on the boundary circle. Each of these projections is common for exactly $(y+1)^{2}-2=y^{2}+2 y-1$ internal segments, because $y+1$ is the number of vertices on the same lateral side and all the segments connecting two distinct lateral sides are internal, with exception of the two segments lying on the bases of the cylinder. Hence,

$$
P=\frac{x(x-1)\left(y^{2}+2 y-1\right)}{2}
$$

## Second Round of the 63rd Czech and Slovak Mathematical Olympiad (January 14th, 2014)

![](https://cdn.mathpix.com/cropped/2024_04_17_207c009ec429634e3f25g-16.jpg?height=226&width=534&top_left_y=475&top_left_x=812)
