First solution. Between the two moments of meeting, each boy traveled half of the asphalt section and half of the sandy section, and they spent the same amount of time on this. Therefore, each of them spent twice as much time on the entire path, which means they spent the same amount of time overall.

Second solution. Let's draw graphs of the boys' movements along the path: on the horizontal axis, we mark the time \( t \), and on the vertical axis, the position \( y \) of the boy, starting from the beginning of the path. Let \( P_{0}, P_{1}, P_{2} \) be the points corresponding to Petya's start, the moment he switched from the asphalt section to the sandy section, and his finish; let \( V_{0}, V_{1}, V_{2} \) be the analogous points for Vasya. Then the graphs of the boys' movements are the broken lines \( P_{0} P_{1} P_{2} \) and \( V_{0} V_{1} V_{2} \), with the segments \( P_{0} V_{0}, P_{1} V_{1} \), and \( P_{2} V_{2} \) being horizontal (see Fig. 1). According to the condition, the midpoints of the segments \( P_{0} P_{1} \) and \( V_{0} V_{1} \) coincide, from which \( P_{0} V_{0} P_{1} V_{1} \) is a parallelogram. Similarly, \( P_{1} V_{1} P_{2} V_{2} \) is a parallelogram. Therefore, the segments \( P_{0} V_{0}, V_{1} P_{1} \), and \( P_{2} V_{2} \) are parallel and equal. Hence, the time between Petya's and Vasya's finishes is the same as the time between their starts; thus, the answer follows.

A polygon whose area (measured in cm²) is numerically equal to its perimeter (measured in cm) is called good.

Note that the original triangle is good: it is a right triangle with legs of 5 cm and 12 cm, so its area is \( 30 \mathrm{~cm}^{2} \) and numerically coincides with its perimeter, which is \( 5 + 12 + 13 = 30 \) cm.

If some polygon P is divided into good polygons, then the area of P, equal to the sum of the areas of all the polygons in the division, would numerically coincide with the sum of the perimeters of the polygons in the division. But the sum of these perimeters is greater than the perimeter of P (by twice the sum of the lengths of the common parts of the boundaries of the polygons in the division). We obtain that the area of P is greater than its perimeter.

Therefore, no good polygon, including the given triangle, can be divided into several (more than one) good polygons.

One of many possible examples is shown in the figure.

Assume such three numbers exist. Since \( a \) is divisible by \( b + c \), the sum \( a + (b + c) = 2023 \) is also divisible by \( b + c \), which implies that \( b + c \) is odd. Therefore, \( b - c + 1 \) is an even number, and the odd number \( b + c \) cannot be divisible by it.

First solution. Since the polynomial \( \left(x - a_{1}\right)\left(x - a_{2}\right)\left(x- a_{3}\right) - b \) has a leading coefficient of 1 and roots \( c_{1}, c_{2}, c_{3} \), we have \( \left(x - a_{1}\right)\left(x - a_{2}\right)\left(x - a_{3}\right) - b = \left(x - c_{1}\right)\left(x - c_{2}\right)\left(x - c_{3}\right) \). Substituting \( -x \) for \( x \) in the last equation, we get\(\left(-x - a_{1}\right)\left(-x - a_{2}\right)(- \left.-x - a_{3}\right)-b = \left(-x - c_{1}\right)\left(-x - c_{2}\right)\left(-x - c_{3}\right)\), which is equivalent to \( \left(x + a_{1}\right)\left(x + a_{2}\right)\left(x + a_{3}\right) + b = \left(x + c_{1}\right)\left(x + c_{2}\right)\left(x + c_{3}\right) \). From this equality, we find that the three roots of the equation \( b = (x + c_{1})(x + c_{2})(x + c_{3}) \) are the numbers \( -a_{1}, -a_{2}, -a_{3} \).

Second solution. By Vieta's formulas, the following relations hold: \[ \begin{array}{c} c_{1} + c_{2} + c_{3} = a_{1} + a_{2} + a_{3} \\ c_{1} c_{2} + c_{2} c_{3} + c_{3} c_{1} = a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{1} \\ c_{1} c_{2} c_{3} = a_{1} a_{2} a_{3} - b \end{array} \]

These same equations can be rewritten as: \[ \begin{aligned} \left(-a_{1}\right) + \left(-a_{2}\right) + \left(-a_{3}\right) & = \left(-c_{1}\right) + \left(-c_{2}\right) + \left(-c_{3}\right) \\ a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{1} & = c_{1} c_{2} + c_{2} c_{3} + c_{3} c_{1} \\ -a_{1} a_{2} a_{3} & = -c_{1} c_{2} c_{3} - b \end{aligned} \]

From this, it follows that the numbers \( -a_{1}, -a_{2} \) and \( -a_{3} \) are the roots of the equation \( \left(x + c_{1}\right)\left(x + c_{2}\right)\left(x + c_{3}\right) - b = 0 \).

Note that there is exactly one rook in each row and each column.

First, we show that all \( 2n \) rooks could not be in one part. Let \( A, B, C, D \) be the corner cells of the board (in counterclockwise order, see fig. 2). By symmetry, \( A \) and \( C \) must belong to different parts, as well as \( B \) and \( D \). This means that either \( A \) and \( B \), or \( A \) and \( D \) are in one part, and the other two cells are in the other.

Assume for definiteness that \( A \) and \( B \) are in part I. Then all boundary cells between them must also be in part I; indeed, if some such cell \( X \) is in part II, then there is a path from \( X \) to \( C \) in part II, and a path from \( A \) to \( B \) in part I; but these paths must have a common cell, which is impossible.

Thus, the entire row between cells \( A \) and \( B \) lies in part I, meaning there must be at least one rook in it. Similarly, in part II there is also a whole row (between \( C \) and \( D \)), and hence there is a rook. This leads to the required conclusion.

It remains to provide an example where \( 2n - 1 \) rooks are located in one of the parts. One possible example is arranged as follows. Consider the diagonal of the square; cells below it, as well as the lower half of the diagonal itself, will fall into one part; the remaining cells will fall into the second part. Place \( 2n - 1 \) rooks in the cells directly below the diagonal; then they will be in one part. Place the remaining rook at the intersection of the remaining row and column. Figure 2 shows such an example for \( n = 5 \).

Note. The key consideration in the estimate is the following: If four cells \( P, Q, R, S \) are chosen on the boundary, labeled in order, then any path between \( P \) and \( R \) has a common cell with any path between \( Q \) and \( S \).

Using this statement, the estimate can be conducted differently. For example, by symmetry, there are \( 4n \) boundary segments of the board belonging to one part, and \( 4n \) belonging to the other. From the above consideration, each \( 4n \) segments form a group of consecutive segments; hence, among them, there are all segments of some side, and we again obtain that in each part there is one of the boundary lines (vertical or horizontal).

First solution. Assume the contrary: let \( b \geqslant c + 1 \). From the divisibility of \( a b c + 1 \) by \( a b - b + 1 \), it follows that the number \( b(a c - a + 1) = (a b c + 1)-(a b - b + 1) \) is divisible by \( a b - b + 1 \). Since the numbers \( b \) and \( a b - b + 1 = b(a - 1) + 1 \) are coprime, we obtain that \( a c - a + 1 \) is divisible by \( a b - b + 1 \). Clearly, \( a c - a + 1 > 0 \), so either \( a c - a + 1 = a b - b + 1 \), or \( a c - a + 1 \geqslant 2(a b - b + 1) \).

In the first case, we get \( b = a(b - c + 1) \) and, hence, \( b \) is divisible by \( a \), which is impossible by the condition. In the second case, we have

\[ a c - a \geqslant 2 b(a - 1) + 1 > 2 b(a - 1) > 2 c(a - 1) \]

that is, \( a c < 2 c - a < 2 c \). This means that \( a < 2 \), i.e., \( a = 1 \); but this is also impossible by the condition, because then again \( b \) is divisible by \( a \).

Second solution. Again assume the contrary. Since \( a b c + 1 \) is divisible by \( a b - b + 1 \), then \( b c - c + 1 = (a b c + 1) - c(a b - b + 1) \) is also divisible by \( a b - b + 1 \), i.e., \( b c - c + 1 = k(a b - b + 1) \) for some natural \( k \). In other words,

\[ 0 = (b c - c + 1) - k(a b - b + 1) = b(c - k a + k) - (k + c - 1) .( * ) \]

Thus, \( k + c - 1 \) is divisible by \( b \).

By our assumption, \( c < b \). On the other hand, since \( a > 1 \) (otherwise \( b \) is divisible by \( a \)), we have

\[ b c - c + 1 = c(b - 1) + 1 < b^{2} < b(b + 1) \leqslant b(b(a - 1) + 1) \]

from which \( k < b \). Thus, \( 0 < k + c - 1 < 2 b \), and therefore \( k + c - 1 = b \).

Now ( * ) rewrites as \( 0 = b(c - k a + k) - b \), i.e., \( c - k a + k - 1 = 0 \). But then \( k a = k + c - 1 = b \), i.e., \( b \) is divisible by \( a \). This is impossible.

Let the circles with diameters \(AB\) and \(CD\) be \(\omega_{1}\) and \(\omega_{2}\) respectively. Note that point \(S\) lies on segment \(MN\).

Let lines \(CS\) and \(DS\) intersect \(\ell\) at points \(P\) and \(Q\) respectively (see image 3). Since \(CD\) is the diameter of \(\omega_{2}\), we have \(\angle PSQ = \angle CSD = 90^{\circ}\). In the right triangle \(PSQ\), segment \(SM\) is the altitude, so \(\angle MSP = 90^{\circ} - \angle SPM = \angle SQP\). On the other hand, since \(NS = NC\), we have \(\angle SCD = \angle CSN = \angle MSP\). Thus, \(\angle SCD = \angle MSP = \angle SQP\), meaning points \(P, Q, C,\) and \(D\) lie on the same circle \(\gamma'\).

Now let line \(MC\) intersect circles \(\gamma\) and \(\gamma'\) at points \(X\) and \(X'\) respectively (point \(M\) lies on segments \(CX\) and \(CX'\)). Then \(MC \cdot MX = MA \cdot MB = MS^{2}\), since \(M\) is the center of circle \(\omega_{1}\). On the other hand, \(MC \cdot MX' = MP \cdot MQ = MS^{2}\); the last equality again follows from the fact that \(SM\) is the altitude in the right triangle \(PSQ\). Therefore, \(MC \cdot MX = MS^{2} = MC \cdot MX'\), meaning \(X = X'\). But point \(X\) is distinct from \(C\) and \(D\), since \(M\) does not lie on \(CD\); thus, circles \(\gamma\) and \(\gamma'\) have three common points \(C, D, X\), meaning they coincide. Therefore, \(P\) lies on \(\gamma\), which was to be proved.

Image 3

Image 4

Remark 1. The solution could be completed in many different ways. For example, the equalities \(MP \cdot MQ = MS^{2} = MA \cdot MB\) mean that points \(P, Q, A,\) and \(B\) lie on the same circle \(\delta\). Then either circles \(\gamma, \gamma'\), and \(\delta\) coincide, or they are three different circles. In the second case, the radical axes of these three circles must intersect at one point or be parallel; but these radical axes are the lines \(PQ, AB\), and \(CD\), and for them, these statements are not true.

The reasoning above has a flaw: it does not hold in the case when points \(P, Q, A,\) and \(B\) lie on one line. This case is easy to analyze separately (then \(MN\) passes through the center of circle \(\gamma, AB \parallel CD\) and \(AC \perp BD\)).

Remark 2. There are other solutions, conceptually similar to the one given above. For example, one could argue as follows.

Let rays \(CS\) and \(DS\) intersect \(\gamma\) again at points \(P'\) and \(Q'\) (see image 4). Let \(M' = P'Q' \cap MN\). Then \(\angle DQ'P' = \angle DCS = \angle CSN = \angle M'SP'\), from which \(MN \perp P'Q'\). Then \(SM'\) is the altitude in the right triangle, and \(M'P' \cdot M'Q' = M'S^{2}\). On the other hand, if line \(MN\) intersects \(\gamma\) at points \(K\) and \(L\), then \(M'K \cdot M'L = M'P' \cdot M'Q' = M'S^{2}\). However, as can be easily verified, there are only two points \(X\) on segment \(KL\) such that \(XK \cdot XL = XS^{2}\), and these are points \(X = M\) and \(X = N\). Thus, \(M' = M\), which was to be proved.

Assume the contrary. Let \( S_{m + 1} \) be divisible by \( 2^{s} \), but not by \( 2^{s + 1} \); then \( s \geqslant 2 \). This means that among the numbers \( 1, 2, \ldots, m + 1 \), there is a number \( a \) divisible by \( 2^{s} \). But then the number \( a / 2 \) does not exceed \( m \) and is divisible by \( 2^{s - 1} \); hence, \( S_{m} \) is also divisible by \( 2^{s - 1} \). Therefore, \( S_{m + 1} / S_{m} \) cannot be divisible by a power of two greater than the first.

Note. It can be shown that \( S_{m + 1} > S_{m} \) only when the number \( m + 1 \) is a power of some prime number \( p \); in this case, the ratio \( S_{m + 1} / S_{m} \) will be equal to \( p \).

We will prove that \( d \geqslant 7 \). All numbers on the board are divided into chains of numbers of the form \( a, a + 1, a + 2, \ldots, a + t \) such that numbers from different chains do not differ by exactly 1. Such a division is easy to construct by connecting any two numbers that differ by 1 with a segment and considering the resulting broken lines.

Let there be \( k \) chains, in which there are \( n_{1}, n_{2}, \ldots, n_{k} \) numbers respectively (some chains may consist of a single number). In a chain of \( n_{i} \) numbers, there are exactly \( n_{i}-1 \) pairs of numbers differing by 1. Therefore, the total number of ones in the notebook is equal to \( \left(n_{1}-1\right) + \left(n_{2}-1\right) + \ldots + \left(n_{k}-1\right) = \left(n_{1} + n_{2} + \ldots + n_{k}\right)-k = 99-k \), from which \( k = 99 - 85 = 14 \). This means that in one of the chains there are at least \( 99 / 14 \) numbers, that is, at least 8 numbers. The difference between the largest and smallest numbers in such a chain is at least \( 8 - 1 = 7 \).

It remains to provide an example where \( d = 7 \). Such an example is given, for instance, by the numbers \[ 0 = \frac{0}{14}, \frac{1}{14}, \frac{2}{14}, \ldots, \frac{98}{14} = 7. \]

Indeed, in this example \( d = 7 \), and exactly for the first 85 of these numbers in the set, there is a number that is one unit larger.

Note. The given example is not the only one. All possible optimal examples are arranged such that there is exactly one chain of 8 numbers (from \( a \) to \( a + 7 \)), and also 13 chains, each consisting of 7 numbers; all numbers of these other chains must be located between \( a \) and \( a + 7 \).

First solution. Perform a homothety with center \( A \) and coefficient 2. Under this homothety, points \( M \) and \( N \) map to \( B \) and \( C \) respectively; let points \( K \) and \( P \) map to \( K' \) and \( P' \) respectively (see Fig. 1). Then it is sufficient to prove that quadrilateral \( ABP'K' \) is circumscribed. We will prove that it is circumscribed around the incircle \( \omega \) of triangle \( ABC \). Three sides of the quadrilateral already touch \( \omega \), so it is enough to prove that \( P'K' \) also touches it.

Let \( I \) be the center of \( \omega \). Then \( KK' = AK \), so \( A \) and \( K' \) are symmetric with respect to \( KI \). Further, note that \( \angle P'K'A = \angle PKA = \angle MHA \). But \( MH \) is the median in the right triangle \( AHB \), so \( \angle MHA = \angle MAC \). Thus, \( \angle P'K'A = \angle BAC \). Therefore, lines \( AB \) and \( K'P' \) are also symmetric with respect to \( KI \); since one of them touches \( \omega \), the other does too. This is what needed to be proved.

Remark. The solution above has several variations. For example, similar reasoning can show that in quadrilateral \( AMPK \), the angle bisectors of three angles \( A, M, \) and \( K \) pass through a single point, the midpoint of segment \( AI \). From this, it follows that this midpoint is the center of the desired inscribed circle.

Fig. 1

Fig. 2

Second solution. Let the line \( PK \) intersect line \( AB \) at point \( L \) (see Fig. 2). As in the solution above, we obtain that \( \angle AKL = \angle AHM = \angle LAK \), hence \( LA = LK \).

We will prove that the circles inscribed in triangles \( AKL \) and \( AMN \) coincide (then this will be the inscribed circle of quadrilateral \( AMPK \)). Since both circles are inscribed in angle \( BAC \), it is sufficient to show that they touch line \( AB \) at the same point. As is known, the distances from \( A \) to the points of tangency of these circles with \( AB \) are \( \frac{AL + AK - KL}{2} \) and \( \frac{AM + AN - MN}{2} \) respectively. Thus, we need to prove that \( AL + AK - KL = AM + AN - MN \), or that \( ML - KL = KN - MN \).

Let the semiperimeter of triangle \( ABC \) be \( p \), and let \( a = BC, b = CA, c = AB \). We have \( ML - KL = (AL - AM) - KL = -AM = -\frac{c}{2} \). On the other hand, \( KN - MN = (AN - AK) - MN = \left(\frac{b}{2} - (p - a)\right) - \frac{a}{2} = \frac{a + b}{2} - p = -\frac{c}{2} \), from which the desired equality follows.

Remark. In the second paragraph of the solution, the following known criterion is essentially proven: quadrilateral \( AMPK \) is circumscribed if and only if \( ML - KL = KN - MN \) (where \( N \) and \( L \) are the points of intersection of the extensions of the lateral sides, positioned as in the figure).

First solution. First, we will prove that \( m = 1 \) satisfies the problem's requirements. Note that \( a b + c = a b + c(a + b + c) = (c + a)(c + b) \). Therefore,

\[ \begin{array}{c} \sqrt{\frac{a b}{c + a b}} + \sqrt{\frac{b c}{a + b c}} + \sqrt{\frac{c a}{b + c a}} = \\ = \sqrt{\frac{a b}{(c + a)(c + b)}} + \sqrt{\frac{b c}{(a + b)(a + c)}} + \sqrt{\frac{c a}{(b + c)(b + a)}} = \\ = \frac{\sqrt{a b} \sqrt{a + b} + \sqrt{b c} \sqrt{b + c} + \sqrt{c a} \sqrt{c + a}}{\sqrt{(a + b)(b + c)(c + a)}} \end{array} \]

Thus, it remains to prove the inequality

\[ \sqrt{a b} \sqrt{a + b} + \sqrt{b c} \sqrt{b + c} + \sqrt{c a} \sqrt{c + a} \geqslant \sqrt{(a + b)(b + c)(c + a)} \]

Squaring this inequality, it takes the form

\[ \begin{array}{r} a b(a + b) + b c(b + c) + c a(c + a) + 2 \sqrt{a b^{2} c(a + b)(b + c)} + \\ + 2 \sqrt{b c^{2} a(b + c)(c + a)} + 2 \sqrt{c a^{2} b(c + a)(a + b)} \geqslant \\ \geqslant a^{2} b + a b^{2} + a^{2} c + a c^{2} + b^{2} c + b c^{2} + 2 a b c \end{array} \]

After simplification, the left side remains a sum of roots, and the right side is \( 2 a b c \). But any of the roots is not less than \( a b c \); indeed, for example, \( \sqrt{a b^{2} c(a + b)(b + c)} \geqslant \sqrt{a b^{2} c \cdot a c} = a b c \). From this, the required follows.

It remains to prove that for any \( m > 1 \), the inequality is not always satisfied; it is sufficient to do this for \( 1 < m < 3 \). Let \( m = 1 + 2 t \) for \( 0 < t < 1 \). Set \( a = b = \frac{1-t^{2}}{2} \) and \( c = t^{2} \). Then \( a + b + c = 1 \), however

\[ \sqrt{\frac{a b}{c + a b}} + \sqrt{\frac{b c}{a + b c}} + \sqrt{\frac{c a}{b + c a}} < \sqrt{\frac{a b}{a b}} + \sqrt{\frac{b c}{a}} + \sqrt{\frac{c a}{b}} = 1 + 2 t = m . \]

Second solution. We provide another proof that \( m = 1 \) is suitable. To do this, we will prove that if \( a \) is the largest of the numbers \( a, b, c \), then the inequality

\[ \sqrt{\frac{a b}{c + a b}} + \sqrt{\frac{c a}{b + c a}} \geqslant 1 \]

is even true. Let \( t = 1 / a, \mu = b / c \); note that \( 1 > a \geqslant \frac{1}{3} \), so \( 1 < t \leqslant 3 \). The left side of the above inequality is rewritten as

\[ \begin{aligned} \sqrt{\frac{a b}{c + a b}} + \sqrt{\frac{c a}{b + c a}} & = \sqrt{\frac{1}{1 + c / (a b)}} + \sqrt{\frac{1}{1 + b / (a c)}} = \\ & = \frac{1}{\sqrt{1 + t / \mu}} + \frac{1}{\sqrt{1 + t \mu}} \end{aligned} \]

Thus, it is sufficient for us to prove that

\[ \sqrt{1 + t / \mu} + \sqrt{1 + t \mu} \geqslant \sqrt{(1 + t / \mu)(1 + t \mu)} \]

Squaring this inequality, we obtain

\[ 1 + t / \mu + 1 + t \mu + 2 \sqrt{(1 + t / \mu)(1 + t \mu)} \geqslant 1 + t / \mu + t \mu + t^{2} \]

After canceling similar terms, we find that it is sufficient to prove the inequality

\[ 2 \sqrt{(1 + t / \mu)(1 + t \mu)} \geqslant t^{2}-1 = (t - 1)(t + 1) \]

Finally, this inequality follows from the inequalities \( 2 \geqslant t - 1 \) (since \( t \leqslant 3 \) ) and

\[ (1 + t / \mu)(1 + t \mu) = 1 + t^{2} + t(\mu + 1 / \mu) \geqslant 1 + t^{2} + 2 t = (t + 1)^{2} \]

where we applied the inequality of means.

Clearly, the result of pressing several switches does not depend on the order in which these presses were made—the number of times each bulb is switched does not depend on this order. In particular, we can assume that Petya used each switch no more than once.

The entire cube is divided into 100 layers parallel to the red face. Each switch on a non-red face switches bulbs in one layer, and each switch on the red face switches a bulb in all 100 layers.

After Petya's actions, there will be a layer in which \( d \leqslant k / 100 \) bulbs are lit—let's call one such layer the main layer. Let \( \mathcal{V} \) be the set of \( d \) switches on the red face associated with the lit bulbs in the main layer. We will prove that Vasya can turn off all the bulbs using exactly these switches from the red face.

Let's start a slightly different process, starting from the same initial position. Let \( \mathcal{P} \) be the set of switches from the red face used by Petya, and \( \mathcal{Q} \) be the set of switches he used from the non-red faces associated with the main layer. Let Petya apply \( \mathcal{P} \) and \( \mathcal{Q} \), and then Vasya apply \( \mathcal{V} \). After Petya's actions, the same \( d \) bulbs will be lit in the main layer as before, so after Vasya's actions, all bulbs in the main layer will be turned off. If Vasya now applies in each of the other layers sets of switches from the non-red faces analogous to \( \mathcal{Q} \), then all bulbs will be turned off.

Now let Petya apply all the other switches (from the non-red faces!) that he originally applied, and Vasya apply them once more. All bulbs will still be turned off. In this new process, Petya applied exactly the same switches as in the original, and Vasya used only the switches from the set \( \mathcal{V} \) from the red face (and some from the other faces). Therefore, if in the original process Vasya performs the same actions as he did in the new one, he will achieve the desired result.

Assume the contrary, and let there be different 30-element subsets \( A_{1}, A_{2}, \ldots, A_{50} \) (the sets of participants of each olympiad) such that the intersection of any 30 of them is non-empty, but the intersection of all is empty.

Let among the sets \( A_{1}, A_{2}, \ldots, A_{50} \), there are two sets \( B \) and \( C \) having \( k \leqslant 28 \) common elements \( x_{1}, x_{2}, \ldots, x_{k} \). For each element \( x_{i} \) among the sets \( A_{1}, A_{2}, \ldots, A_{50} \), find a subset \( D_{i} \) that does not contain \( x_{i} \) (such a subset \( D_{i} \) can be found, otherwise \( x_{i} \) is a common element of the sets \( A_{1}, A_{2}, \ldots, A_{50} \)). (Note that among the subsets \( D_{i} \), there may be duplicates.) Then the intersection of no more than 30 subsets \( B, C, D_{1}, \ldots, D_{k} \) is empty. This contradicts our assumption (one can add a few more subsets to these to make it 30 subsets, and the intersection remains empty).

Thus, such two sets \( B \) and \( C \) do not exist. Then the intersection of any two of the sets \( A_{1}, A_{2}, \ldots, A_{50} \) contains exactly 29 elements. Let \( A_{1} \cap A_{2} = \left\{y_{1}, y_{2}, \ldots, y_{29}\right\} \), so that \( A_{1} = \left\{z_{1}, y_{1}, y_{2}, \ldots, y_{29}\right\}, A_{2} = \left\{z_{2}, y_{1}, y_{2}, \ldots, y_{29}\right\} \). Find a subset (let's say, for definiteness, this subset is \( A_{3} \)) that does not contain \( y_{1} \). Since \( \left|A_{1} \cap A_{3}\right| = \left|A_{2} \cap A_{3}\right| = 29 \), \( A_{3} \) must contain all elements \( z_{1}, z_{2}, y_{2}, y_{3}, \ldots, y_{29} \). These elements are 30 (all of them are distinct), so \( A_{3} = \left\{z_{1}, z_{2}, y_{2}, y_{3}, \ldots, y_{29}\right\} \). Consider any subset \( A_{i} \) from the subsets \( A_{4}, \ldots, A_{50} \). Assume that \( A_{i} \) contains an element outside the 31-element set \( K = \left\{z_{1}, z_{2}, y_{1}, y_{2}, \ldots, y_{29}\right\} = A_{1} \cup A_{2} \cup A_{3} \). Then \( A_{i} \) must intersect with each of the subsets \( A_{1}, A_{2}, A_{3} \) by the same 29-element subset of the set \( K \). But \( \left|A_{1} \cap A_{2} \cap A_{3}\right| = 28 \), so such a 29-element subset does not exist—a contradiction. Hence, we conclude that all sets \( A_{1}, A_{2}, \ldots, A_{50} \) are subsets of the set \( K \). But in the set \( K \), the number of 30-element subsets is \( 31 < 50 \). We obtain a contradiction, completing the solution to the problem.

From the condition, it follows that \( (a b c + 1)-(a b - b + 1) = a b c - a b + b = b(a c - a + 1) \) is divisible by \( a b - b + 1 \). Notice that \( b \) and \( a b - b + 1 = (a - 1) b + 1 \) are coprime, hence we obtain that \( a c - a + 1 \) is divisible by \( a b - b + 1 \).

Next, we notice that \( 0 < a c - a + 1 < 2(a b - b + 1) \). Indeed, \( 2(a b - b + 1) = (2 a - 2) b + 2 > (2 a - 2) c + 2 = a c + (a - 2) c + 2 \geqslant a c + 2 > a c - a + 1 \). Therefore, the divisibility of \( a c - a + 1 \) by \( a b - b + 1 \) is only possible in the case of equality \( a c - a + 1 = a b - b + 1 \).

We have \( a(c - 1) = a c - a = a b - b = (a - 1) b \). We see that \( (a - 1) b \) is divisible by \( a \), but since \( a - 1 \) and \( a \) are coprime, it follows that \( b \) is divisible by \( a \), which is what we needed to prove.

Note that \( PQ \parallel CD \), so \( PQ \) is the midline of the right triangle \( AHD \). Thus, \( PQ \) intersects the hypotenuse \( AH \) at its midpoint \( M \), so \( MA = MD = MH \) (see image 6).

Image 6

We have \( \angle MDH = \angle MHD \), and since \( MH \perp BC \) and \( HD \perp CD \), we also have \( \angle MHD = \angle BCD \). We obtain the equality \( \angle MDH = \angle BCD \), from which it follows that the line \( MD \) is tangent to the circle (\( BCD \)) at point \( D \). Hence, \( MD^{2} = MP \cdot MQ \) (by the power of a point theorem).

Further, \( MA^{2} = MP \cdot MQ \). Thus, triangles \( AMP \) and \( QMA \) are similar (angle \( AMQ \) is common and \( MA / MP = MQ / MA \)). From this, \( \angle MQA = \angle MAP \), therefore \( \angle MPA + \angle MQA = \angle MPA + \angle MAP = \angle HMQ = 90^{\circ} - \angle MHD = \angle CBD \). Thus, \( \angle APB + \angle AQB = \angle CBD \), and since \( \angle APB + \angle AQB = (\angle MPA + \angle MQA) + (\angle MPB + \angle MQB) \), to complete the solution, it remains to verify that \( \angle MPB + \angle MQB = 180^{\circ} - \angle CBD \).

For definiteness, we further assume that \( P \) lies between \( M \) and \( Q \). We have \( \angle MPB + \angle MQB = 180^{\circ} - (\angle BPQ - \angle PQB) \). Since \( PQ \parallel CD \), the arcs \( PD \) and \( CQ \) are equal, and hence the inscribed angles subtending them are equal. Then \( \angle BPQ - \angle PQB = \angle BDQ - \angle PCB = (\angle BDC - \angle QDC) - (\angle DCB - \angle DCP) = \angle BDC - \angle DCB = 90^{\circ} - \angle DCB = \angle CBD \), which completes the proof.

Let \( \overline{x_{i} y_{i} z_{i}} \) be the decimal representation of the three-digit number \( a_{i} \). Substituting into the left side of the equation \( x = 1000 \) gives \( a_{9} \cdot 1000^{9} + a_{8} \cdot 1000^{8} + \ldots + a_{1} \cdot 1000 + a_{0} = \overline{x_{9} y_{9} z_{9} \underbrace{0000 \ldots 0}_{27 \text { zeros }}} + \overline{x_{8} y_{8} z_{8} \underbrace{0 \ldots 0}_{24 \text { zeros }}} + \ldots + \overline{x_{1} y_{1} z_{1} 000} + \overline{x_{0} y_{0} z_{0}} = \overline{x_{9} y_{9} z_{9} x_{8} y_{8} z_{8} \ldots x_{0} y_{0} z_{0}} \). Thus, after substituting the 30-digit number \( \overline{x_{9} y_{9} z_{9} x_{8} y_{8} z_{8} \ldots x_{0} y_{0} z_{0}} \) for the asterisk, the equation will have the root 1000.

First solution. Since \( AM \) is the bisector of angle \( LAN \), segments \( LM \) and \( MN \) are equal as chords subtending equal arcs (see fig. 3). Now it is enough to prove that \( CM = LM \) (then \( CM = LM = MN \), which means \( CNL \) is a right triangle, and \( NM \) is its median drawn from the right angle).

Since \( \angle BKA = \angle NAK = \angle BAK \), triangle \( ABK \) is isosceles (symmetric with respect to the perpendicular bisector to \( AK \)). Mark a point \( X \) on side \( BK \) such that \( LX \parallel AK \). From the symmetry of triangle \( ABK \), we have \( KX = AL \). Then we have \( KX = CK \) and \( MK \parallel LX \), which means \( MK \) is the midline of triangle \( CLX \), hence \( CM = LM \), which completes the solution.

Fig. 3

Fig. 4

Second solution. Let \( \angle BAD = 2\alpha \).

Note that \( DM \) is the bisector of angle \( ADC \). Indeed, extend \( AK \) to intersect \( CD \) at point \( Y \) (see fig. 4). Then, using the similarities \( AML \sim YMC \) and \( AYD \sim KYC \), we have \( AM / MY = AL / YC = CK / YC = AD / DY \). From the obtained equality \( AM / MY = AD / DY \), it follows that \( DM \) is the bisector of triangle \( ADY \). Hence \( \angle MDC = \angle ADC / 2 = (180^{\circ} - 2\alpha) / 2 = 90^{\circ} - \alpha \).

From the inscribed property of \( ALMN \), we have \( \angle CMN = \angle LAD \), and from the parallelism \( AB \parallel CD \), it follows \( \angle LAD = \angle CDN \). Therefore, \( \angle CMN = \angle CDN \), which means quadrilateral \( CMDN \) is inscribed. Hence \( \angle MNC = \angle MDC = 90^{\circ} - \alpha \).

From the inscribed property \( \angle LNM = \angle LAM = \alpha \). Then \( \angle LNC = \angle MNC + \angle LNM = 90^{\circ} \), which is what was required to prove.

Note. In solution 2, when proving that \( DM \) is the bisector of angle \( ADC \), it was not used that \( AM \) is the bisector of angle \( A \). Also, when proving the inscribed property of \( CMDN \), the equality \( AL = CK \) was not used.

Number the poles from 1 to \( n \) along the road and assume the distance between neighboring poles is 1. A pair of poles of the same color, between which there are no other poles of the same color, will be called a good pair.

Estimation. Suppose the \( n \) poles are painted such that the condition of the problem is satisfied. Let \( n_{i} \) be the number of poles of the \( i \)-th color (we further assume that \( n_{i} \geqslant 1 \), i.e., all colors are present, otherwise \( n \) can be increased by adding a pole of a new color at the end). Let \( a_{i} \) and \( b_{i} \) be the numbers of the first and last poles of the \( i \)-th color.

In total, we have \( t = \left(n_{1}-1\right) + \left(n_{2}-1\right) + \ldots + \left(n_{k}-1\right) = \left(n_{1} + n_{2} + \ldots + n_{k}\right)-k = n - k \) good pairs of poles. Since all distances between poles in good pairs are different, the smallest of these distances is at least 1, the next is at least 2, and so on. Thus, for the sum \( S \) of distances in all good pairs, we get the estimate \( S \geqslant 1 + 2 + \ldots + t = t(t + 1) / 2 \).

On the other hand, the sum of all distances for the \( i \)-th color is equal to \( b_{i}-a_{i} \). Therefore, \( S = \left(b_{1} + \ldots + b_{k}\right)-\left(a_{1} + \ldots + a_{k}\right) \). The sum \( b_{1} + \ldots + b_{k} \) does not exceed the sum of the \( k \) largest among the numbers \( 1,2, \ldots, n \), and the sum \( a_{1} + \ldots + a_{k} \) is not less than the sum of the \( k \) smallest among the numbers \( 1,2, \ldots, n \), so \( S \leqslant(n + (n-1) + \ldots + (n - k + 1))-(1 + 2 + \ldots + k) = k(n - k) = k t \).

Thus, \( t(t + 1) / 2 \leqslant S \leqslant k t \), from which \( t \leqslant 2 k - 1 \) and \( n = k + t \leqslant 3 k - 1 \).

Example. For instance, the coloring

\[ 1,2, \ldots, k - 1, k, k, k - 1, \ldots, 2,1,2,3, \ldots, k - 1, k . \]

Here, for color 1, there is a single good pair, and the distance between the poles in it is \( 2 k - 1 \). For all other colors, there are two good pairs, with color 2 having distances \( 2 k - 3 \) and 2, color 3 having distances \( 2 k - 5 \) and 4, and so on, for color \( k \) - distances 1 and \( 2 k - 2 \).

Note. There are other, more complex examples. For instance, the first \( k \) poles can be painted in colors \( 1,2, \ldots, k \), and then the pole with number \( k + s \), where \( s = 2^{p}(2 q - 1) \), can be painted in color \( q \) (for example, for \( k = 8 \) the coloring will look like: \( 1,2,3,4,5,6,7,8,1,1,2,1,3,2,4,1,5,3,6,2,7,4,8 \)). An inductive description of suitable examples is possible.

We will prove that \( (x - y) \sqrt{3 x^{2} + y^{2}} \geqslant (x - y)(x + y) = x^{2} - y^{2} \). If \( x \geqslant y \), then \( x - y \geqslant 0 \) and \( \sqrt{3 x^{2} + y^{2}} \geqslant \sqrt{x^{2} + 2 x y + y^{2}} = x + y \). If \( x \leqslant y \), then \( x - y \leqslant 0 \) and \( \sqrt{3 x^{2} + y^{2}} \leqslant \sqrt{x^{2} + 2 x y + y^{2}} = x + y \).

By summing the proven inequality \( (x - y) \sqrt{3 x^{2} + y^{2}} \geqslant x^{2} - y^{2} \) with similar inequalities \( (y - z) \sqrt{3 y^{2} + z^{2}} \geqslant y^{2} - z^{2} \) and \( (z - x) \sqrt{3 z^{2} + x^{2}} \geqslant z^{2} - x^{2} \), we obtain the required result.

First solution. We will prove that for any pair \((x, y)\) written on the board, the number \(2x - y\) is divisible by 7.

Indeed, for the pair \((1,2)\), the number \(2 \cdot 1 - 2 = 0\) is divisible by 7.

Suppose for the pair \((a, b)\) the number \(2a - b\) is divisible by 7. Then for the pair \((-a,-b)\), the number \(2 \cdot (-a) - (-b) = -(2a - b)\) is divisible by 7, and for the pair \((-b, a + b)\), the number \(2 \cdot (-b) - (a + b) = -a - 3b = 3(2a - b) - 7a\) is divisible by 7.

Suppose for the pairs \((a, b), (c, d)\) the numbers \(2a - b, 2c - d\) are divisible by 7. Then for the pair \((a + c, b + d)\), the number \(2(a + c) - (b + d) = (2a - b) + (2c - d)\) is divisible by 7.

Since for the pair \((2022,2023)\), the number \(2 \cdot 2022 - 2023 = 2021\) is not divisible by 7, this pair cannot appear on the board.

Second solution. We will add a third number \(c = -a - b\) to each pair \((a, b)\) on the board. Then the sum of the numbers in each triplet will be zero, and the rules for adding new pairs will be as follows: if the triplet \((a, b, c)\) is on the board, then the triplets \((-a,-b,-c)\) and \((-b,-c,-a)\) can be added, and if the triplets \((a, b, c)\) and \((d, e, f)\) are on the board, then the triplet \((a + d, b + e, c + f)\) can be added - we will call this triplet the sum of the triplets \((a, b, c)\) and \((d, e, f)\). Also, for the triplet \((a, b, c)\) and an integer \(k\), we denote by \(k \cdot (a, b, c)\) the triplet \((ka, kb, kc)\).

We will prove that all triplets appearing on the board have the form

\[(a, b, c) = k \cdot (1,2,-3) + \ell \cdot (2,-3,1) + m \cdot (-3,1,2)\]

with integers \(k, \ell\), and \(m\). Initially, this is true: \((1,2,-3) = 1 \cdot (1,2,-3) + 0 \cdot (2,-3,1) + 0 \cdot (-3,1,2)\). Now it is enough to show that from triplets of the form \((*)\), only such triplets are obtained. For the operation of taking the sum of triplets, this is obvious. For the other operations, it is also not difficult to verify: if \((a, b, c)\) has the form \((*)\), then

\[ \begin{array}{l} (-a,-b,-c) = (-k) \cdot (1,2,-3) + (-\ell) \cdot (2,-3,1) + (-m) \cdot (-3,1,2), \\ (-b,-c,-a) = (-m) \cdot (1,2,-3) + (-k) \cdot (2,-3,1) + (-\ell) \cdot (-3,1,2). \end{array} \]

The statement is proven.

Now assume that the triplet \((2022,2023,-4045)\) appeared on the board, i.e., it has the form \((*)\). Then we have

\[2022 = k + 2\ell - 3m \quad \text{and} \quad 2023 = 2k - 3\ell + m.\]

Expressing from the first equation \(k = 2022 - 2\ell + 3m\) and substituting into the second, we get \(2023 = 2 \cdot 2022 - 7\ell + 7m\), i.e., \(7(m-\ell) = 2023 - 2 \cdot 2022 = -2021\). However, this is impossible since 2021 is not divisible by 7.

Note. It can be shown that the specified operations yield all triplets of the form \((*)\). Also, note that \((-3,1,2) = -(1,2,-3) - (2,-3,1)\), so in the formula \((*)\), the third term can be omitted.

A similar solution can be obtained without adding a third number to the pair (although it will look less natural). Namely, it can be proven that all pairs appearing on the board in the original process have the form

\[(a, b) = k \cdot (1,2) + \ell \cdot (2,-3)\]

with integers \(k\) and \(\ell\). Note here that if the pair \((a, b)\) has the form \((***)\), then

\[(-b, a + b) = \ell \cdot (1,2) + (\ell - k) \cdot (2,-3).\]

Let \( P \) be a point on the ray \( HE \) such that \( PB \perp BC \) (see Fig. 7). We will prove that the points \( C, O \), and \( P \) are collinear.

Indeed, by Menelaus' theorem for triangle \( ADE \) and line \( CMB \), we have \( \frac{EC}{CD} \cdot \frac{DM}{MA} \cdot \frac{AB}{BE} = 1 \). Since the lines \( PB, AH \), and \( OM \) are parallel to each other (as they are all perpendicular to line \( BC \)), we have \( AB / BE = HP / PE \), and also \( DM / MA = DO / OH \). Therefore, \( \frac{EC}{CD} \cdot \frac{DO}{OH} \cdot \frac{HP}{PE} = 1 \), which implies that the points \( C, O \), and \( P \) are collinear by Menelaus' theorem for triangle \( EDH \). Thus, point \( P \) is diametrically opposite to point \( C \) on the circle \( \omega \). Similarly, if \( Q \) is the intersection point of the perpendicular to line \( BC \) passing through point \( C \) and line \( HF \), then point \( Q \) is diametrically opposite to point \( B \). From this, it follows that \( \angle EXC = \angle PXC = 90^{\circ} \), and similarly, \( \angle FYB = \angle QYB = 90^{\circ} \).

Denote by \( H', T_b \), and \( T_c \) the intersection points of line \( AH \) with lines \( PQ, BY \), and \( CX \), respectively (see Fig. 8). Note that triangles \( HXT_c \) and \( HH'P \) are similar as right triangles with vertical acute angles. Therefore, \( HT_c / HX = HP / HH' \), or \( HT_c = HX \cdot HP / HH' = HB \cdot HC / HH' \). Similarly, \( HT_b = HB \cdot HC / HH' \). Thus, the lines \( BY \) and \( CX \) intersect line \( AH \) at the same point, which was to be proved.

Consider a rectangle of size \( 5 \times 15 \), half of whose area is 37.5. To satisfy the condition, it is necessary to cut out a rectangle with an area of 37 or 38 from this rectangle. There are only three such rectangles: \( 1 \times 37, 1 \times 38 \), and \( 2 \times 19 \). Note that the longer side of each of these rectangles is at least 19. On the other hand, the diagonal of the original rectangle is \( \sqrt{250} \), but \( \sqrt{250} < \sqrt{256} = 16 < 19 \), so none of these rectangles can be cut out from the \( 5 \times 15 \) rectangle.

Let the second intersection point of the angle bisector of angle \( ABC \) with the circumcircle of triangle \( ABC \) be \( F \) (see image 5). Then point \( F \) is the midpoint of arc \( AC \), so \( OF \) is the perpendicular bisector of chord \( AC \). Since the inscribed angle is half of the central angle subtending the same arc, \( \angle FOC = 2 \angle FBC \). On the other hand, since \( BD = DC \), \( \angle DCB = \angle CBD \), and thus \( \angle CDF = \angle DCB + \angle DBC = 2 \angle DBC = 2 \angle FBC \) as the external angle to triangle \( BCD \). Therefore, \( \angle FOC = \angle FDC \), so point \( F \) lies on the circumcircle of triangle \( COD \). Similarly, we find that \( \angle AOF = 2 \angle ABF = \angle AEF \), and point \( F \) also lies on the circumcircle of triangle \( AOE \). Thus, points \( P \) and \( Q \) are the centers of the circumcircles of triangles \( AOF \) and \( COF \), and these triangles are symmetric with respect to \( OF \). It follows that points \( P \) and \( Q \) are also symmetric with respect to \( OF \). Consequently, either points \( P \) and \( Q \) lie on line \( AC \), or \( P, Q, A, C \) are the vertices of an isosceles trapezoid, and therefore lie on the same circle.

Image 5

Let \( d = \frac{1}{a}-\frac{1}{b}-\frac{1}{c} \). Now notice that

\[ \left|\frac{b}{a}-\frac{b}{c}\right| + \left|\frac{c}{a}-\frac{c}{b}\right| + |b c + 1| = |b d + 1| + |c d + 1| + |b c + 1| . \]

If \( d = 0 \), then two of these terms are equal to 1, and thus the sum is at least 2. Otherwise, the numbers \( a, b, d \) are non-zero. Therefore, some two of them have the same sign, and then their product is positive, making the corresponding term greater than 1. Since the other two terms are non-negative, the total sum is greater than 1.

We will prove by induction on \(n\) that in any connected graph containing \(2n\) vertices, it is possible to color them in red and blue such that the number of edges with endpoints of different colors (we will call such edges multicolored) exceeds the number of edges with endpoints of the same color (we will call such edges monochromatic) by at least \(n\) - from this the statement of the problem will follow. The base case \(n = 1\) is trivial, let's prove the induction step.

Assume that in a graph with \(2n\) vertices, there is a pair of vertices connected by an edge, upon the removal of which the graph remains connected; denote these vertices by \(u\) and \(v\). Color the remaining vertices such that the number of multicolored edges is at least \(n - 1\) more than the number of monochromatic edges - this can be done by the induction hypothesis. Note that the vertices \(u\) and \(v\) can now be colored in such a way that the difference between the numbers of multicolored and monochromatic edges increases. Indeed, without loss of generality, assume that if the vertices \(u\) and \(v\) do not both have even degrees, then vertex \(u\) has an odd degree. Then color vertex \(u\) in the color that is in the minority among its neighbors (in case of equality, color in any color), and then color vertex \(v\) in the same way. Obviously, with each coloring, the required difference did not decrease, and at least with one coloring, the corresponding vertex had an odd number of colored neighbors, i.e., the difference increased with this coloring. Since before coloring the vertices \(u\) and \(v\), the difference between the number of multicolored and monochromatic edges was at least \(n - 1\), after this coloring it became at least \(n\).

On the other hand, if there is a pair of pendant vertices in the graph, then obviously, upon their removal, the graph still remains connected, and by the same algorithm, the entire remaining graph can be colored, and then these pendant vertices, in such a way that the difference between the numbers of multicolored and monochromatic edges will be at least \(n\). We will prove that in any connected graph with at least three vertices, either there are two adjacent vertices, upon the removal of which the graph remains connected, or there are two pendant vertices.

Indeed, consider an arbitrary spanning tree of this graph and suspend it by any non-pendant vertex. Let \(v\) be the most distant pendant vertex from the root of this tree, and \(u\) be the ancestor of this vertex. Denote the descendants of this ancestor by \(v_{1}, \ldots, v_{k}\). Note that the vertices \(v_{1}, \ldots, v_{k}\) are pendant in the considered spanning tree. Consider several cases.

Case 1. Among the vertices \(v_{1}, \ldots, v_{k}\), there is a pair of vertices connected by an edge in the original graph. Then, upon the removal of these two vertices, the spanning tree (and hence the original graph) retains connectivity.

Case 2. Among the vertices \(v_{1}, \ldots, v_{k}\), there is a pair of vertices that are pendant in the original graph. Thus, in the original graph, there are at least two pendant vertices.

Case 3. Among the vertices \(v_{1}, \ldots, v_{k}\), there is at most one vertex that is pendant in the original graph. Without loss of generality, assume that if such a vertex exists, it is vertex \(v_{1}\). Then reattach each of the vertices \(v_{2}, \ldots, v_{k}\) to any of its neighbors other than \(u\): since these vertices are not pendant in the original graph, such a neighbor always exists. After all reattachments, the vertices \(u\) and \(v_{1}\) can be removed from the graph, and the spanning tree will remain connected, and hence the graph itself.

Since at least one of the cases takes place, and in each of them, there is either a pair of adjacent vertices in the graph, upon the removal of which the graph remains connected, or a pair of pendant vertices, the induction step is proven.

Note. The inequality from the problem is exact: in particular, in a complete graph on \(2n\) vertices, the corresponding difference cannot be strictly greater than \(n\).