## BxMO 2015

Problems with Solutions

## Problem 1.

## Determine the smallest positive integer $q$ with the following property:

for every integer $m$ with $1 \leqslant m \leqslant 1006$, there exists an integer $n$ such that

$$
\frac{m}{1007} q<n<\frac{m+1}{1008} q .
$$

Solution 1. For $m=1006$, we have

$$
q-q / 1007<n<q-q / 1008
$$

for some integer $n$. If $q \leqslant 1007$, then $q-q / 1007$ and $q-q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q>1007$ and $q-q / 1008 \leqslant q-1$, implying that $n<q-1$, and that $q-q / 1007<q-2$. By rearranging terms, we find $q>2014$, and hence $q \geqslant 2015$.

Let us prove that $q=2015$ works. Indeed, $m q / 1007=2 m+m / 1007<2 m+1$ and $(m+1) q / 1008=2(m+1)-(m+1) / 1008>2 m+1$ for $1 \leqslant m \leqslant 1006$. Hence each pair of inequalities can be satisfied by taking $n=2 m+1$. This completes the proof, and shows that the smallest possible of $q$ is indeed 2015.

Solution 2. For $m=1, \ldots, 1006$, there must exist an integer $N$ divisible by $1007 \cdot 1008$ satisfying the double inequality

$$
1008 m q<N<1007(m+1) q .
$$

Since $N$ is divisible by 1007 and 1008 , we may write

$$
N=1008 m q+1008 k=1007(m+1) q-1007 \ell
$$

where $k, \ell>0$ are integers. Hence

$$
(1007-m) q=1008 k+1007 \ell \geqslant 2015
$$

since $k, \ell>0$. Choosing $m=1006$ in this last inequality, it follows that $q \geqslant 2015$. Conversely, for $q=2015=1008+1007,(*)$ can be satisfied by taking $k=\ell=1007-m$. (Indeed, 1007 and 1008 are coprime, and so integer divisible by 1007 and 1008 is also divisible by their product.) Thus $q=2015$ has the desired property, and we are done.

## Problem 2.

Let $A B C$ be an acute triangle with circumcentre $O$. Let $\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\Gamma_{B}$ again in $X$ and $\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.

The following solutions are valid for the configurations appearing in the diagrams.
Solution 1. Let $O_{B}$ and $O_{C}$ denote the respective centres of $\Gamma_{B}$ and $\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \perp A C$ since $\Gamma_{B}$ is tangent to $A C$ and $O O_{C} \perp A C$ since $O O_{C}$ is the perpendicular bisector of $[A C]$. Hence $A O_{B} \| O O_{C}$, and similarly, $A O_{C} \| O O_{B}$. It follows that $A O_{B} O O_{C}$ is a parallelogram. In particular, $\left|O_{B} X\right|=\left|A O_{B}\right|=\left|O O_{C}\right|$ and $\left|O_{C} Y\right|=\left|O_{C} A\right|=\left|O O_{B}\right|$.

It will therefore suffice to show that the angles $\angle O O_{B} X$ and $\angle Y O_{C} O$ are equal. Noting that $\angle A O_{B} O=\angle A O_{C} O$ since $A O_{B} O O_{C}$ is a parallelogram, this follows by angle chasing:

$$
\begin{aligned}
\angle X O_{B} O & =\angle A O_{B} O-\angle A O_{B} X=\angle A O_{C} O-\left(180^{\circ}-2 \angle O_{B} A X\right) \\
& =\angle A O_{C} O-180^{\circ}+2\left(\angle O_{B} A O_{C}-\angle Y A O_{C}\right) \\
& =\angle A O_{C} O-180^{\circ}+2\left(180^{\circ}-\angle A O_{C} O\right)-\left(180^{\circ}-\angle A O_{C} Y\right) \\
& =\angle A O_{C} Y-\angle A O_{C} O=\angle O O_{C} Y .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-3.jpg?height=966&width=1438&top_left_y=1525&top_left_x=281)

A Variant. As in the main solution, we note that $A O_{B} O O_{C}$ is a parallelogram. Notice that $O_{B}$ and $O_{C}$ lie on the respective perpendicular bisectors of $[A X]$ and $[A Y]$. The following lemma then implies that $O$ lies on the perpendicular bisector of $[X Y]$, completing the proof:

Let $P_{1}, P_{2}, P_{3}$ be three points on a line. Let $O_{1}$ be a point on the perpendicular bisector of $\left[P_{2} P_{3}\right]$, let $O_{2}$ be a point on the perpendicular bisector of $\left[P_{3} P_{1}\right]$ and let $O_{3}$ be the point in the plane such that $O_{1} P_{3} O_{2} O_{3}$ is a parallelogram. Then $O_{3}$ lies on the perpendicular bisector of $\left[P_{1} P_{2}\right]$.
Proof. One can choose coordinates such that $P_{3}=(0,0)$ and $P_{1}=(2,0)$. Then $P_{2}=(2 a, 0), O_{1}=(a, b)$ and $O_{2}=(1, c)$ for some $a, b, c \in \mathbb{R}$. Hence $O_{3}=(a+1, b+c)$ lies on the perpendicular bisector of $\left[P_{1} P_{2}\right]$.

Solution 2. Let $\alpha=\angle B A C$. Observe that $\angle A X B=180^{\circ}-\alpha=\angle C Y A$ by tangential angles. Let $Z$ be the intersection of $B X$ and $C Y$. Thus $Z X Y$ is isosceles with, in particular, $\angle Z X Y=\angle Z Y X=\alpha$. It will thus suffice to show that $O Z$ bisects $\angle X Z Y$.

Now $\angle B Z C=2 \alpha=\angle B O C$ since $O$ is the circumcentre of $A B C$, and hence $B Z O C$ is cyclic. In particular, since $|O B|=|O C|$, it follows that $\angle O Z Y=\angle O B C=90^{\circ}-\alpha$. But $\angle X Z Y=180^{\circ}-2 \alpha$, and so $\angle X Z Y=2 \angle O Z Y$, which shows that $O Z$ bisects $\angle X Z Y$.
![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-4.jpg?height=881&width=901&top_left_y=1408&top_left_x=566)

Solution 3. Consider inversion $\mathscr{I}$ in a circle centred at $A$. Under $\mathscr{I}$,

$$
B \mapsto B^{\prime}, \quad C \mapsto C^{\prime}, \quad O \mapsto O^{\prime}, \quad X \mapsto X^{\prime}, \quad Y \mapsto Y^{\prime},
$$

$\Gamma_{B} \mapsto \gamma_{B}$, a line through $B^{\prime}$ parallel to $A C^{\prime}$,
$\Gamma_{C} \mapsto \gamma_{C}$, a line through $C^{\prime}$ parallel to $A B^{\prime}$,
Notice that $\mathscr{I}$ sends the circumcircle of $A B C$ to the line $B^{\prime} C^{\prime}$, and hence maps $A O$ to a line perpendicular to $B C$. Further, if $[A D]$ is a diameter of this circumcircle and $D \mapsto D^{\prime}$ under $\mathscr{I}$, then $D^{\prime}$ lies on $B C$. Also, $|A D|=2|A O|$ implies $\left|A O^{\prime}\right|=2\left|A D^{\prime}\right|$, and hence $O^{\prime}$ is the image of $A$ under reflection in $B C$. Finally, $\angle O X A=\angle X^{\prime} O^{\prime} A$ and $\angle O Y A=\angle Y^{\prime} O^{\prime} A$, and hence $|O X|=|O Y|$ if and only if $O^{\prime} A$ bissects $\angle X^{\prime} O^{\prime} Y^{\prime}$ externally. We have thus reduced the problem to the following statement:

In triangle $A B C$, let $O$ be the reflection of $A$ over $B C, \gamma_{B}$ be a line parallel to $A C$ through $B$, and $\gamma_{C}$ be a line parallel to $A B$ through $C$. For an arbitrary line $\ell$ passing through $A$, let $X$ and $Y$ be the intersections of $\ell$ with $\gamma_{B}$ and $\gamma_{C}$, respectively. Prove that $O A$ bisects $\angle X O Y$ externally.
![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-5.jpg?height=882&width=643&top_left_y=1196&top_left_x=692)

Let $P=\gamma_{A} \cap \gamma_{B}$. By construction, $P O B C$ is an isosceles trapezoid (and therefore cyclic), $\angle O B X=\angle O B P=\angle O C P=\angle O C Y$. Further, since $X B \| A C$ and $Y C \| A B$, we have $\triangle A B X \sim \triangle Y C A$. Therefore, as $|O B|=|A B|$ and $|A C|=|O C|$ by construction,

$$
\frac{|O B|}{|B X|}=\frac{|A B|}{|B X|}=\frac{|Y C|}{|C A|}=\frac{|Y C|}{|C O|}
$$

It follows that $\triangle O B X \sim \triangle Y C O$. Hence

$$
\frac{|O X|}{|O Y|}=\frac{|X B|}{|O C|}=\frac{|X B|}{|A C|}=\frac{|X A|}{|A Y|}
$$

and so the result follows from the angle bisector theorem.

## Problem 3.

Does there exist a prime number whose decimal representation is of the form $3811 \cdots 11$ (that is, consisting of the digits 3 and 8 in that order followed by one or more digits 1 )?

Solution. Write

$$
a(n)=38 \underbrace{11 \cdots 11}_{n \text { digits } 1} .
$$

There are three cases to consider, depending on the remainder of $n$ upon division by three.

- If $n=3 k+1 \equiv 1(\bmod 3)$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by 3 , and hence so is $a(n)$.
- If $n=3 k+2 \equiv 2(\bmod 3)$, then note that $a(2)=3811=3700+111$ is divisible by 37 . By induction, as $a(3 k+2)=1000 a(3 k-1)+111$, it follows that $a(3 k+2)$ is divisible by 37 for each $k \geqslant 0$.
- If $n=3 k \equiv 0(\bmod 3)$, observe that

$$
9 a(3 k)=342 \underbrace{99 \ldots 99}_{n \text { digits } 9}=\left(7 \cdot 10^{k}\right)^{3}-1
$$

which is properly divisible by $7 \cdot 10^{k}-1$, a number that is larger than 9 . Hence $a(3 k)$ admits a non-trivial factor and so is not prime.

## Problem 4.

An arithmetic progression is a set of the form $\{a, a+d, \ldots, a+k d\}$, where $a, d, k$ are positive integers and $k \geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.

Let $n$ be a positive integer. For each partition of the set $\{1,2, \ldots, 3 n\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value $S$ that can attain?
(A partition of a set $A$ is a collection of disjoint subsets of $A$ whose union is $A$.)

Solution. The maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\{1, n+1,2 n+1\}, \ldots,\{n, 2 n, 3 n\}$, each of difference $n$.

Suppose indeed that the set has been partioned into $N$ progressions, of respective lengths $\ell_{i}$, and differences $d_{i}$, for $1 \leqslant i \leqslant N$. Since $\ell_{i} \geqslant 3$,

$$
2 \sum_{i=1}^{N} d_{i} \leqslant \sum_{i=1}^{N}\left(\ell_{i}-1\right) d_{i}=\sum_{i=1}^{N} a_{i}-\sum_{i=1}^{N} b_{i}
$$

where $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now

$$
\begin{aligned}
& \sum_{i=1}^{N} b_{i} \geqslant 1+2+\cdots+N=N(N+1) / 2 \\
& \sum_{i=1}^{N} a_{i} \leqslant(3 n-N+1)+\cdots+3 n=N(6 n-N+1) / 2
\end{aligned}
$$

and thus

$$
2 \sum_{i=1}^{N} d_{i} \leqslant N(3 n-N) \leqslant 2 n^{2}
$$

as $N(3 n-N)$ is an increasing of $N$ on the interval $[0,3 n / 2]$ and since $N \leqslant n$. This completes the proof.

Remark. The maximising partition is in fact unique: from the above, it is immediate that all maximal partitions must have $n_{i}=3$ and hence $n=N$. It follows that the minimal and maximal elements of the arithmetic progressions of a maximal partition are precisely $1,2, \ldots, n$ and $2 n+1,2 n+2, \ldots, 3 n$, respectively. Consider the progression $\{n+1-d, n+1, n+1+d\}$ of difference $d$. Then $n+1-d \geqslant 1$ and $n+1+d \geqslant 2 n+1$,
which implies $d=n$, and hence $\{1, n+1,2 n+1\}$ is an element of any maximal partition. By induction, it follows similarly that $\{k, n+k, 2 n+k\}$ is an element of any maximal partition for $1 \leqslant k \leqslant n$, since integers allowing for smaller or larger differences have already been used up. This proves the partition $\{1, n+1,2 n+1\}, \ldots,\{n, 2 n, 3 n\}$ is the unique maximising partition, as claimed.