# 6th Benelux Mathematical Olympiad Solutions 

Brugge, May 2-4 2014
![](https://cdn.mathpix.com/cropped/2024_12_15_fb9be07ea97b63ba84aeg-1.jpg?height=405&width=497&top_left_y=104&top_left_x=1448)

## Problem 1

Find the smallest possible value of the expression

$$
\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor,
$$

in which $a, b, c$ and $d$ vary over the set of positive integers.
(Here $\lfloor x\rfloor$ denotes the biggest integer which is smaller than or equal to $x$.)

## Solution

The answer is 9 .
Notice that $\lfloor x\rfloor>x-1$ for all $x \in \mathbb{R}$. Therefore the given expression is strictly greater than

$$
\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}-4,
$$

which can be rewritten as

$$
\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-4 .
$$

Since $t+\frac{1}{t} \geq 2$ for $t>0$, we get that $6 \cdot 2-4=8$ is a strict lower bound for the given expression; since it takes integral values only, we actually get that 9 is a lower bound. It remains to check that 9 can be attained; this happens for $a=b=c=5$ and $d=4$.

## Problem 2

Let $k \geq 1$ be an integer.
We consider $4 k$ chips, $2 k$ of which are red and $2 k$ of which are blue. A sequence of those $4 k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an equal number of consecutive blue chips. For example, we can move from $r \underline{b b b r} \underline{r} \underline{b} b$ to $r \underline{r} \underline{b} b \underline{b b b} b$ where $r$ denotes a red chip and $b$ denotes a blue chip.

Determine the smallest number $n$ (as a function of $k$ ) such that starting from any initial sequence of the $4 k$ chips, we need at most $n$ moves to reach the state in which the first $2 k$ chips are red.

## Solution

The answer is $n=k$.
We will first show that $n \geq k$. Let us count the number $C$ of times a red chip is directly to the right of a blue chip. In the final position this number equals 0 . In the position brbrbr $\cdots b r$ this number equals $2 k$. We claim that any move reduces this number by at most 2 . Denote by $R$ the group of red chips and by $B$ the group of blue chips that are interchanged. Any reduction in $C$ must involve a red chip getting rid of its blue left neighbour or a blue chip getting rid of its red right neighbour. This can only happen with the leftmost chip of $R$ (if its left neighbour is blue) and the rightmost chip of $B$ (if its right neighbour is red), but not with the rightmost chip of $R$ (and its right neighbour) or the leftmost chip of $B$ (and its left neighbour). Hence $C$ is reduced by at most 2 in any move. Therefore the number of moves necessary to change $b r b r b r \cdots b r$ into the final position is at least $\frac{2 k}{2}=k$.
We will now show that $n \leq k$, i.e. that it is always possible to perform at most $k$ moves in order to reach the state in which the first $2 k$ chips are red. Consider the first $2 k$ chips. If at most $k$ of these chips are blue, we can perform one move for each chip, switching it with one of the red chips of the last $2 k$ chips. So then we are done in at most $k$ moves. Now suppose that of the first $2 k$ chips, at least $k+1$ are blue. Then at most $k-1$ chips are red. Hence it is possible to perform at most $k-1$ moves to reach the situation in which the last $2 k$ chips are red. We then perform one final move, switching the first $2 k$ chips and the last $2 k$ chips, ending in the situation in which the first $2 k$ chips are red. Thus, it is always possible to reach the situation in which the first $2 k$ chips are red in at most $k$ steps, hence $n \leq k$.
We have now shown that $n \geq k$ and $n \leq k$, hence $n=k$ as claimed.

## Problem 3

Find all positive integers $n>1$ with the following property:
for each two positive divisors $k, \ell<n$ of $n$, at least one of the numbers $2 k-\ell$ and $2 \ell-k$ is a (not necessarily positive) divisor of $n$ as well.

## Solution

If $n$ is prime, then $n$ has the desired property: if $k, \ell<n$ are positive divisors of a prime $n$, we have $k=\ell=1$, in which case $2 k-\ell=1$ is a divisor of $n$ as well.

Assume now that a composite number $n$ has the desired property. Let $p$ be its smallest prime divisor and let $m=n / p$; then $m \geq p \geq 2$. Choosing $(k, \ell)=(1, m)$, we see that at least one of $2 m-1$ and $m-2$ must divide $n$. However, $m<2 m-1<2 m \leq n$; since $m$ and $n$ are the two biggest positive divisors of $n$, we conclude that $2 m-1$ cannot divide $n$. Therefore $n$ must be divisble by $m-2$. It follows that $m-2 \mid m p-p(m-2)=2 p$, hence $m-2 \in\{1,2, p, 2 p\}$.
We deal with each case separately.

- If $m-2=1$ we have $m=3$. As $p \leq m$, we have $p \in\{2,3\}$, hence $n=6$ or $n=9$.
- If $m-2=2$ we have $m=4$, hence $n$ is even and $p=2$, so $n=8$.
- If $m-2=p$, we may assume that $p>2$ (since we already discussed the case $m-2=2$ ). We have $m=p+2$, hence $n=p(p+2)$. Applying the condition in the problem to the pair $(k, \ell)=(1, p)$, we get that $p-2$ or $2 p-1$ divides $n$. If $p-2$ divides $n$, we must have $p=3$ since $p$ is the smallest prime divisor of $n$. If $2 p-1$ divides $n$, we must have $p+2=2 p-1$, since $p+2 \leq 2 p-1<p(p+2)$ and $p+2$ and $p(p+2)$ are the biggest positive divisors of $n$. Therefore $p=3$ in both cases, and we get $n=15$ as a candidate.
- If finally $m-2=2 p$ we have $m=2 p+2$, so $n$ is even. Hence $p=2$ and $n=12$.

We conclude that if a composite number $n$ satisfies the condition in the problem statement, then $n \in\{6,8,9,12,15\}$. Choosing $(k, \ell)=(1,2)$ shows that $n=8$ does not work; choosing $(k, \ell)=(3,6)$ shows that 12 is not a solution. It is easy to check that $n=6, n=9$ and $n=15$ have the desired property by checking the condition for all possible pairs $(k, \ell)$.
We conclude that the solutions are given by the prime numbers, $n=6, n=9$ and $n=15$.

## Problem 4

Let $A B C D$ be a square. Consider a variable point $P$ inside the square for which $\angle B A P \geq 60^{\circ}$. Let $Q$ be the intersection of the line $A D$ and the perpendicular to $B P$ in $P$. Let $R$ be the intersection of the line $B Q$ and the perpendicular to $B P$ from $C$.
(a) Prove that $|B P| \geq|B R|$.
(b) For which point(s) $P$ does the inequality in (a) become an equality?

## Solution

![](https://cdn.mathpix.com/cropped/2024_12_15_fb9be07ea97b63ba84aeg-4.jpg?height=447&width=927&top_left_y=982&top_left_x=551)

We claim that $\triangle A B P$ and $\triangle R C B$ are similar triangles. Indeed, if we denote the intersection of $B P$ and $C R$ by $S$, then $\angle R C B=\angle S C B=90^{\circ}-\angle S B C=90^{\circ}-\angle P B C=\angle A B P$. Moreover, the right angles in $P$ and $A$ imply that $A$ and $P$ lie on the circle with diameter $[B Q]$, so either $A B P Q$ or $A Q B P$ is a cyclic, convex quadrilateral. In either case, $A$ and $Q$ lie on the same side of $B P$, so $\angle P A B=\angle P Q B$. Since $C R$ and $P Q$ are perpendicular to $B P$, these lines are parallel and hence $\angle P Q B=\angle C R B$. Together with $\angle P A B=\angle P Q B$ and $\angle R C B=\angle A B P$, this implies the claim that $\triangle A B P \sim \triangle R C B$.
This similarity yields the equality $|A P| /|B R|=|B P| /|B C|$. Since $\angle B A P \geq 60^{\circ}$, we get

$$
\begin{aligned}
0 \leq & (|A B|-|A P|)^{2}=|A B|^{2}+|A P|^{2}-2 \cdot|A B| \cdot|A P| \\
& =|B P|^{2}+2(\cos \angle B A P-1) \cdot|A B| \cdot|A P| \\
& \leq|B P|^{2}-|A B| \cdot|A P|=|B P|^{2}-|B R| \cdot|B P| .
\end{aligned}
$$

This implies that $|B P| \geq|B R|$, as desired.
In order for equality to occur, one needs equality in each of the inequalities considered above: $|A B|=|A P|$ and $\angle B A P=60^{\circ}$. Hence there is exactly one point $P$ for which we have equality; this is the unique point inside the square such that $\triangle A B P$ is equilateral.