Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$.  (Denmark)  Preamble. Let $X=y \cap z, Y=x \cap z, Z=x \cap y$ and let $\Omega$ denote the circumcircle of the triangle $X Y Z$. Denote by $X_{0}, Y_{0}$, and $Z_{0}$ the second intersection points of $A I, B I$ and $C I$, respectively, with $\omega$. It is known that $Y_{0} Z_{0}$ is the perpendicular bisector of $A I, Z_{0} X_{0}$ is the perpendicular bisector of $B I$, and $X_{0} Y_{0}$ is the perpendicular bisector of $C I$. In particular, the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$ are homothetic, because their corresponding sides are parallel.  The solutions below mostly exploit the following approach. Consider the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$, or some other pair of homothetic triangles $\Delta$ and $\delta$ inscribed into $\Omega$ and $\omega$, respectively. In order to prove that $\Omega$ and $\omega$ are tangent, it suffices to show that the centre $T$ of the homothety taking $\Delta$ to $\delta$ lies on $\omega$ (or $\Omega$ ), or, in other words, to show that $\Delta$ and $\delta$ are perspective (i.e., the lines joining corresponding vertices are concurrent), with their perspector lying on $\omega$ (or $\Omega$ ).  We use directed angles throughout all the solutions.

Solution: Claim 1. The reflections $\ell_{a}, \ell_{b}$ and $\ell_{c}$ of the line $\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\omega$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-46.jpg?height=831&width=916&top_left_y=1338&top_left_x=570)  Proof. Notice that $\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, \ell\right)+\Varangle\left(\ell, \ell_{c}\right)=2 \Varangle(y, \ell)+2 \Varangle(\ell, z)=2 \Varangle(y, z)$. But $y \perp B I$ and $z \perp C I$ implies $\Varangle(y, z)=\Varangle(B I, I C)$, so, since $2 \Varangle(B I, I C)=\Varangle(B A, A C)$, we obtain  $$ \Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle(B A, A C) . $$  Since $A$ is the reflection of $D$ in $x$, $A$ belongs to $\ell_{a}$; similarly, $B$ belongs to $\ell_{b}$. Then (1) shows that the common point $T^{\prime}$ of $\ell_{a}$ and $\ell_{b}$ lies on $\omega$; similarly, the common point $T^{\prime \prime}$ of $\ell_{c}$ and $\ell_{b}$ lies on $\omega$.  If $B \notin \ell_{a}$ and $B \notin \ell_{c}$, then $T^{\prime}$ and $T^{\prime \prime}$ are the second point of intersection of $\ell_{b}$ and $\omega$, hence they coincide. Otherwise, if, say, $B \in \ell_{c}$, then $\ell_{c}=B C$, so $\Varangle(B A, A C)=\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, B C\right)$, which shows that $\ell_{b}$ is tangent at $B$ to $\omega$ and $T^{\prime}=T^{\prime \prime}=B$. So $T^{\prime}$ and $T^{\prime \prime}$ coincide in all the cases, and the conclusion of the claim follows.  Now we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\ell_{b}, D_{c}$ lies on $\ell_{c}$, and  $$ \begin{aligned} \Varangle\left(D_{b} X, X D_{c}\right) & =\Varangle\left(D_{b} X, D X\right)+\Varangle\left(D X, X D_{c}\right)=2 \Varangle(y, D X)+2 \Varangle(D X, z)=2 \Varangle(y, z) \\ & =\Varangle(B A, A C)=\Varangle(B T, T C), \end{aligned} $$  hence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\prime}$ yields $\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D_{c}^{\prime}, D_{c}^{\prime} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right)$. Therefore,  $$ \begin{gathered} \Varangle\left(\ell_{b}, X T\right)=\Varangle\left(D_{b} T, X T\right)=\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right) \\ =90^{\circ}+\Varangle(B I, I C)=\Varangle(B A, A I)=\Varangle\left(B A, A X_{0}\right)=\Varangle\left(B T, T X_{0}\right)=\Varangle\left(\ell_{b}, X_{0} T\right) \end{gathered} $$  so the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows.  Comment 1. After proving Claim 1 one may proceed in another way. As it was shown, the reflections of $\ell$ in the sidelines of $X Y Z$ are concurrent at $T$. Thus $\ell$ is the Steiner line of $T$ with respect to $\triangle X Y Z$ (that is the line containing the reflections $T_{a}, T_{b}, T_{c}$ of $T$ in the sidelines of $X Y Z$ ). The properties of the Steiner line imply that $T$ lies on $\Omega$, and $\ell$ passes through the orthocentre $H$ of the triangle $X Y Z$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-47.jpg?height=811&width=1311&top_left_y=1248&top_left_x=378)  Let $H_{a}, H_{b}$, and $H_{c}$ be the reflections of the point $H$ in the lines $x, y$, and $z$, respectively. Then the triangle $H_{a} H_{b} H_{c}$ is inscribed in $\Omega$ and homothetic to $A B C$ (by an easy angle chasing). Since $H_{a} \in \ell_{a}, H_{b} \in \ell_{b}$, and $H_{c} \in \ell_{c}$, the triangles $H_{a} H_{b} H_{c}$ and $A B C$ form a required pair of triangles $\Delta$ and $\delta$ mentioned in the preamble.  Comment 2. The following observation shows how one may guess the description of the tangency point $T$ from Solution 1.  Let us fix a direction and move the line $\ell$ parallel to this direction with constant speed.  Then the points $D, E$, and $F$ are moving with constant speeds along the lines $A I, B I$, and $C I$, respectively. In this case $x, y$, and $z$ are moving with constant speeds, defining a family of homothetic triangles $X Y Z$ with a common centre of homothety $T$. Notice that the triangle $X_{0} Y_{0} Z_{0}$ belongs to this family (for $\ell$ passing through $I$ ). We may specify the location of $T$ considering the degenerate case when $x, y$, and $z$ are concurrent. In this degenerate case all the lines $x, y, z, \ell, \ell_{a}, \ell_{b}, \ell_{c}$ have a common point. Note that the lines $\ell_{a}, \ell_{b}, \ell_{c}$ remain constant as $\ell$ is moving (keeping its direction). Thus $T$ should be the common point of $\ell_{a}, \ell_{b}$, and $\ell_{c}$, lying on $\omega$.


Solution: As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\omega$. Thus, it suffices to prove that $\Varangle\left(T X_{0}, T Y_{0}\right)=$ $\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, or, equivalently, $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$.  Recall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\frac{1}{2} \overrightarrow{I D}$. Define the shifting vectors $\vec{y}=\frac{1}{2} \overrightarrow{I E}, \vec{z}=\frac{1}{2} \overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$.  Claim 2. $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}, \overrightarrow{I V}=2 \overrightarrow{Y_{0} Y}, \overrightarrow{I W}=2 \overrightarrow{Z_{0} Z}$.  Proof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal.  The projection of $\overrightarrow{X_{0} X}$ onto $I B$ equals $\vec{y}$, while the projection of $\overrightarrow{I U}$ onto $I B$ is $\overrightarrow{I E}=2 \vec{y}$. The projections onto the other axis $I C$ are $\vec{z}$ and $\overrightarrow{I F}=2 \vec{z}$. Then $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}$ follows.  Notice that the line $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle(I U, I V)=$ $\Varangle(W U, W V)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, and we are done.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-48.jpg?height=888&width=1106&top_left_y=1161&top_left_x=475)


Solution: Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above).  Notice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre.  Construct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\Omega$.  Consider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\frac{1}{2}$. Furthermore, let $h=h_{2} \circ h_{1}$. The transform $h$ can be a homothety or a translation, and  $$ h(T)=h_{2}\left(h_{1}(T)\right)=h_{2}(S)=T $$  hence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\Omega$, and this completes the proof.


Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $s n$ and of $s k$ are equal.  (Ukraine)  Answer: All pairs $(n, k)$ such that $n \nmid k$ and $k \nmid n$.

Solution: As usual, the number of divisors of a positive integer $n$ is denoted by $d(n)$. If $n=\prod_{i} p_{i}^{\alpha_{i}}$ is the prime factorisation of $n$, then $d(n)=\prod_{i}\left(\alpha_{i}+1\right)$.  We start by showing that one cannot find any suitable number $s$ if $k \mid n$ or $n \mid k$ (and $k \neq n$ ). Suppose that $n \mid k$, and choose any positive integer $s$. Then the set of divisors of $s n$ is a proper subset of that of $s k$, hence $d(s n)<d(s k)$. Therefore, the pair $(n, k)$ does not satisfy the problem requirements. The case $k \mid n$ is similar.  Now assume that $n \nmid k$ and $k \nmid n$. Let $p_{1}, \ldots, p_{t}$ be all primes dividing $n k$, and consider the prime factorisations  $$ n=\prod_{i=1}^{t} p_{i}^{\alpha_{i}} \quad \text { and } \quad k=\prod_{i=1}^{t} p_{i}^{\beta_{i}} $$  It is reasonable to search for the number $s$ having the form  $$ s=\prod_{i=1}^{t} p_{i}^{\gamma_{i}} $$  The (nonnegative integer) exponents $\gamma_{i}$ should be chosen so as to satisfy  $$ \frac{d(s n)}{d(s k)}=\prod_{i=1}^{t} \frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=1 $$  First of all, if $\alpha_{i}=\beta_{i}$ for some $i$, then, regardless of the value of $\gamma_{i}$, the corresponding factor in (1) equals 1 and does not affect the product. So we may assume that there is no such index $i$. For the other factors in (1), the following lemma is useful.  Lemma. Let $\alpha>\beta$ be nonnegative integers. Then, for every integer $M \geqslant \beta+1$, there exists a nonnegative integer $\gamma$ such that  $$ \frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M}=\frac{M+1}{M} . $$  Proof.  $$ \frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M} \Longleftrightarrow \frac{\alpha-\beta}{\beta+\gamma+1}=\frac{1}{M} \Longleftrightarrow \gamma=M(\alpha-\beta)-(\beta+1) \geqslant 0 . $$  Now we can finish the solution. Without loss of generality, there exists an index $u$ such that $\alpha_{i}>\beta_{i}$ for $i=1,2, \ldots, u$, and $\alpha_{i}<\beta_{i}$ for $i=u+1, \ldots, t$. The conditions $n \nmid k$ and $k \nmid n$ mean that $1 \leqslant u \leqslant t-1$.  Choose an integer $X$ greater than all the $\alpha_{i}$ and $\beta_{i}$. By the lemma, we can define the numbers $\gamma_{i}$ so as to satisfy  $$ \begin{array}{ll} \frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=\frac{u X+i}{u X+i-1} & \text { for } i=1,2, \ldots, u, \text { and } \\ \frac{\beta_{u+i}+\gamma_{u+i}+1}{\alpha_{u+i}+\gamma_{u+i}+1}=\frac{(t-u) X+i}{(t-u) X+i-1} & \text { for } i=1,2, \ldots, t-u . \end{array} $$  Then we will have  $$ \frac{d(s n)}{d(s k)}=\prod_{i=1}^{u} \frac{u X+i}{u X+i-1} \cdot \prod_{i=1}^{t-u} \frac{(t-u) X+i-1}{(t-u) X+i}=\frac{u(X+1)}{u X} \cdot \frac{(t-u) X}{(t-u)(X+1)}=1, $$  as required.  Comment. The lemma can be used in various ways, in order to provide a suitable value of $s$. In particular, one may apply induction on the number $t$ of prime factors, using identities like  $$ \frac{n}{n-1}=\frac{n^{2}}{n^{2}-1} \cdot \frac{n+1}{n} $$


Four positive integers $x, y, z$, and $t$ satisfy the relations  $$ x y-z t=x+y=z+t . $$  Is it possible that both $x y$ and $z t$ are perfect squares?  (Russia)  Answer: No.

Solution: Arguing indirectly, assume that $x y=a^{2}$ and $z t=c^{2}$ with $a, c>0$.  Suppose that the number $x+y=z+t$ is odd. Then $x$ and $y$ have opposite parity, as well as $z$ and $t$. This means that both $x y$ and $z t$ are even, as well as $x y-z t=x+y$; a contradiction. Thus, $x+y$ is even, so the number $s=\frac{x+y}{2}=\frac{z+t}{2}$ is a positive integer.  Next, we set $b=\frac{|x-y|}{2}, d=\frac{|z-t|}{2}$. Now the problem conditions yield  $$ s^{2}=a^{2}+b^{2}=c^{2}+d^{2} $$  and  $$ 2 s=a^{2}-c^{2}=d^{2}-b^{2} $$  (the last equality in (2) follows from (1)). We readily get from (2) that $a, d>0$.  In the sequel we will use only the relations (1) and (2), along with the fact that $a, d, s$ are positive integers, while $b$ and $c$ are nonnegative integers, at most one of which may be zero. Since both relations are symmetric with respect to the simultaneous swappings $a \leftrightarrow d$ and $b \leftrightarrow c$, we assume, without loss of generality, that $b \geqslant c$ (and hence $b>0$ ). Therefore, $d^{2}=2 s+b^{2}>c^{2}$, whence  $$ d^{2}>\frac{c^{2}+d^{2}}{2}=\frac{s^{2}}{2} $$  On the other hand, since $d^{2}-b^{2}$ is even by (2), the numbers $b$ and $d$ have the same parity, so $0<b \leqslant d-2$. Therefore,  $$ 2 s=d^{2}-b^{2} \geqslant d^{2}-(d-2)^{2}=4(d-1), \quad \text { i.e., } \quad d \leqslant \frac{s}{2}+1 $$  Combining (3) and (4) we obtain  $$ 2 s^{2}<4 d^{2} \leqslant 4\left(\frac{s}{2}+1\right)^{2}, \quad \text { or } \quad(s-2)^{2}<8 $$  which yields $s \leqslant 4$.  Finally, an easy check shows that each number of the form $s^{2}$ with $1 \leqslant s \leqslant 4$ has a unique representation as a sum of two squares, namely $s^{2}=s^{2}+0^{2}$. Thus, (1) along with $a, d>0$ imply $b=c=0$, which is impossible.


Solution: We start with a complete description of all 4-tuples $(x, y, z, t)$ of positive integers satisfying (*). As in the solution above, we notice that the numbers  $$ s=\frac{x+y}{2}=\frac{z+t}{2}, \quad p=\frac{x-y}{2}, \quad \text { and } \quad q=\frac{z-t}{2} $$  are integers (we may, and will, assume that $p, q \geqslant 0$ ). We have  $$ 2 s=x y-z t=(s+p)(s-p)-(s+q)(s-q)=q^{2}-p^{2} $$  so $p$ and $q$ have the same parity, and $q>p$.  Set now $k=\frac{q-p}{2}, \ell=\frac{q+p}{2}$. Then we have $s=\frac{q^{2}-p^{2}}{2}=2 k \ell$ and hence  $$ \begin{array}{rlrl} x & =s+p=2 k \ell-k+\ell, & y & =s-p=2 k \ell+k-\ell, \\ z & =s+q=2 k \ell+k+\ell, & t=s-q=2 k \ell-k-\ell . \end{array} $$  Recall here that $\ell \geqslant k>0$ and, moreover, $(k, \ell) \neq(1,1)$, since otherwise $t=0$.  Assume now that both $x y$ and $z t$ are squares. Then $x y z t$ is also a square. On the other hand, we have  $$ \begin{aligned} x y z t=(2 k \ell-k+\ell) & (2 k \ell+k-\ell)(2 k \ell+k+\ell)(2 k \ell-k-\ell) \\ & =\left(4 k^{2} \ell^{2}-(k-\ell)^{2}\right)\left(4 k^{2} \ell^{2}-(k+\ell)^{2}\right)=\left(4 k^{2} \ell^{2}-k^{2}-\ell^{2}\right)^{2}-4 k^{2} \ell^{2} \end{aligned} $$  Denote $D=4 k^{2} \ell^{2}-k^{2}-\ell^{2}>0$. From (6) we get $D^{2}>x y z t$. On the other hand,  $$ \begin{array}{r} (D-1)^{2}=D^{2}-2\left(4 k^{2} \ell^{2}-k^{2}-\ell^{2}\right)+1=\left(D^{2}-4 k^{2} \ell^{2}\right)-\left(2 k^{2}-1\right)\left(2 \ell^{2}-1\right)+2 \\ =x y z t-\left(2 k^{2}-1\right)\left(2 \ell^{2}-1\right)+2<x y z t \end{array} $$  since $\ell \geqslant 2$ and $k \geqslant 1$. Thus $(D-1)^{2}<x y z t<D^{2}$, and $x y z t$ cannot be a perfect square; a contradiction.  Comment. The first part of Solution 2 shows that all 4-tuples of positive integers $x \geqslant y, z \geqslant t$ satisfying (*) have the form (5), where $\ell \geqslant k>0$ and $\ell \geqslant 2$. The converse is also true: every pair of positive integers $\ell \geqslant k>0$, except for the pair $k=\ell=1$, generates via (5) a 4-tuple of positive integers satisfying $(*)$.


Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5 n$. Suppose that the sequence  $$ \frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}} $$  forms an arithmetic progression. Prove that the terms of the sequence are equal.  (Thailand)

Solution: Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\Delta=\frac{c}{d}$, where $d>0$ and $c, d$ are coprime.  We will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \leqslant i \leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\left(b_{i}\right)<v_{p}(d)$. $\left(v_{p}(x)\right.$ stands for the exponent of $p$ in the prime factorisation of $x$.)  Claim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$.  Proof. Let $\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $\left.p^{\alpha}\right)$ such that $i \not \equiv j(\bmod p)$. Then the least common denominator of $\frac{a_{i}}{b_{i}}$ and $\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\alpha}$. But this is impossible because in their difference, $(i-j) \Delta=\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\alpha}$ divides the denominator $d$.  Claim 2. $d$ has no prime divisors greater than 5 .  Proof. Suppose that $p \geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \ldots, n$, at most $\left\lceil\frac{n}{p}\right\rceil<\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\frac{p-1}{p} n-1$ of $b_{1}, \ldots, b_{n}$. Since these denominators are distinct,  $$ 5 n \geqslant \max \left\{b_{i}: p \mid b_{i}\right\} \geqslant\left(\frac{p-1}{p} n-1\right) p=(p-1)(n-1)-1 \geqslant 6(n-1)-1>5 n, $$  a contradiction.  Claim 3. For every $0 \leqslant k \leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \ldots, b_{k+30}$, at least $\varphi(30)=8$ are divisible by $d$.  Proof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \cdot(3-1) \cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \mid b_{i}$ for every uncovered index $i$.  Claim 4. $|\Delta|<\frac{20}{n-2}$ and $d>\frac{n-2}{20}$.  Proof. From the sequence (1), remove all fractions with $b_{n}<\frac{n}{2}$, There remain at least $\frac{n}{2}$ fractions, and they cannot exceed $\frac{5 n}{n / 2}=10$. So we have at least $\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\frac{10}{n / 2-1}=\frac{20}{n-2}$.  The second inequality follows from $\frac{1}{d} \leqslant \frac{|c|}{d}=|\Delta|$.  Now we have everything to get the final contradiction. By Claim 3, we have $d \mid b_{i}$ for at least $\left\lfloor\frac{n}{30}\right\rfloor \cdot 8$ indices $i$. By Claim 4 , we have $d \geqslant \frac{n-2}{20}$. Therefore,  $$ 5 n \geqslant \max \left\{b_{i}: d \mid b_{i}\right\} \geqslant\left(\left\lfloor\frac{n}{30}\right\rfloor \cdot 8\right) \cdot d>\left(\frac{n}{30}-1\right) \cdot 8 \cdot \frac{n-2}{20}>5 n . $$  Comment 1. It is possible that all terms in (1) are equal, for example with $a_{i}=2 i-1$ and $b_{i}=4 i-2$ we have $\frac{a_{i}}{b_{i}}=\frac{1}{2}$.  Comment 2. The bound $5 n$ in the statement is far from sharp; the solution above can be modified to work for $9 n$. For large $n$, the bound $5 n$ can be replaced by $n^{\frac{3}{2}-\varepsilon}$.  The activities of the Problem Selection Committee were supported by  DEDEMAN  LA FANTANA vine la tine