If a number \(k=abc+ab+ac+bc\)\(\le 10^9-4000\) is good for Petya, then

\(k-2=(a+1)(b+1)(c+1)\)\(- (a+1)-(b+1)-(c+1)\)

is good for Vasya, since \(a+1,\;b+1,\;c+1\gt 100\). Thus, if there are \(p\) good numbers for Petya, we have exhibited \(p\) distinct good numbers for Vasya, all strictly less than \(10^9-4000\).

Moreover, \(10^9-4000=(1000-1)\cdot 1000\cdot(1000+1)\)\(- (1000-1)-1000-(1000+1)\) is also good for Vasya. Therefore, Vasya has at least \(p+1\) good numbers.

Solution. Suppose such a number \(n\) exists. The condition is equivalent to saying that the last two digits of all divisors are pairwise distinct (one-digit numbers are padded with a leading zero). We will call this pair the tail of the number. Note that a tail has the same residues modulo \(4\) and modulo \(5\) as the original number.

Assume \(n\) is divisible by \(5\). Then for any divisor \(d\) not divisible by \(5\) the number \(5d\) is also a divisor and is divisible by \(5\); different \(d\) give different \(5d\). Hence the number of divisors divisible by \(5\) is at least half, i.e. at least \(25\). But the tails of such divisors end in \(0\) or \(5\); there are at most \(20\) possibilities, so two tails coincide — a contradiction. Thus \(n\) is not divisible by \(5\), and the tails of its divisors cannot end with \(0\) or \(5\).

If \(n\) is odd, then all its divisors are odd. There are exactly \(50\) odd tails in total; among them \(10\) end with \(5\) and therefore cannot occur. That leaves only \(40\) options for \(50\) divisors, so two tails coincide — a contradiction.

If \(n\) is divisible by \(2\) but not by \(4\), then its divisors split into pairs \((d,\,2d)\) with \(d\) odd. Every tail of a number \(2d\) is \(2\bmod 4\); among tails not divisible by \(5\) there are only \(20\) such possibilities. There are at least \(25\) numbers of the form \(2d\), so two tails coincide — a contradiction.

Finally, let the highest power of two dividing \(n\) be \(2^r\) with \(r\ge 2\). For any odd divisor \(d\), the numbers \(d,\,2d,\,2^2 d,\,\dots,\,2^r d\) are divisors of \(n\), and these exhaust all divisors. Hence the total number of divisors is a multiple of \(r+1\). Since \(50\) is divisible by \(r+1\), we have \(r+1\in\{5,10,25,50\}\)\(\Rightarrow r\ge 4\).

Therefore \(n\) has at most \(\dfrac{50}{r+1}\le 10\) odd divisors and the same number of even divisors not divisible by \(4\). The remaining divisors (at least \(30\) of them) are multiples of \(4\); their tails are also multiples of \(4\). But there are only \(20\) tails that are multiples of \(4\) and not multiples of \(5\), so two tails coincide — a contradiction.

In all cases we get a contradiction. Therefore such an \(n\) does not exist.

Solution. Suppose there were \(N\) moves in total. Consider the process “backwards”: call the last move the first, the previous (made by the other boy) the second, and so on. Let \(a_0\) be the number of tubers in the sack from which potatoes were moved on the first (i.e. last) move, immediately after that move. Similarly, let \(a_{i-1}\) be the number in the sack from which potatoes were moved on the \(i\)-th move, immediately after that move. Finally, let \(a_N\) be the number in the sack into which potatoes were moved on the \(N\)-th (earliest) move, before that move (so \(a_N=150\)).

Let \(k\le N-3\). Look at the sack that had \(a_k\) tubers after move \(k+1\) (potatoes were just moved from it). Then before move \(k+1\) it had \(300-a_{k+1}\) tubers, before move \(k+2\) it had \(a_{k+2}\), and before move \(k+3\) it had \(300-a_{k+3}\). On move \(k+1\) the number moved from this sack was \((300-a_{k+1})-a_k\), and this must exceed the amount it had before move \(k+3\). Thus \(300-a_{k+3}<300-a_k-a_{k+1}\)\(\Rightarrow a_{k+3}>a_k+a_{k+1}\); since all numbers are integers, \(a_{k+3}\ge a_k+a_{k+1}+1\).

Define \(b_0,b_1,b_2,\dots\) by \(b_0=b_1=b_2=0,\ \ b_{k+3}=b_{k+1}+b_k+1\). Since \(a_0,a_1,a_2\ge 0\), by induction from the inequality it follows that \(a_k\ge b_k\) and \(b_{k+1}\ge b_k\) for all \(k=0,1,\dots\). Hence \(b_N\le a_N=150\).

\(k:\;\;0\;1\;2\;3\;4\;5\;6\;7\;8\;9\;10\;11\;12\;13\;14\;15\;16\;17\;18\;19\;20\)

\(b_k:\;0\;0\;0\;1\;1\;2\;3\;4\;6\;8\;11\;15\;20\;27\;36\;48\;64\;85\;113\;150\;199\).

From \(b_N\le 150\) we get \(N\le 19\).

An instance with 19 moves follows from the above construction. Initially each child has \(b_{19}=150\) tubers. If after the \(k\)-th move (from the start) the mover keeps exactly \(b_{19-k}\) tubers, then on move \(k\) he moves \(300-b_{20-k}-b_{19-k}\) tubers; before any of his earlier moves he has \(300-b_i\) tubers with \(i\ge 17-k\), and \(300-b_i\le 300-b_{17-k}<300-b_{19-k}-b_{20-k}\). Thus the condition is satisfied and the boys can make 19 such moves.

Remove all non-integer numbers from the progression (if there are any). Clearly, after removal, there remains an infinite decreasing arithmetic progression consisting of integers. Let its difference be \( -d \).

Notice that \( q^{23} - p^{23} \) is divisible by \( d \), which means \( d \) is not divisible by \( p \), otherwise \( q^{23} \) would have to be divisible by \( p \), which is not true. On the other hand, \( d \) must be a divisor of the number \( p^{24} - p^{23} = p^{23}(p - 1) \). Since \( p \) and \( d \) are coprime, \( p - 1 \) is divisible by \( d \). Furthermore, \( p^{23} - p \) is divisible by \( p - 1 \) (since it equals \( p(p - 1)(p^{21} + p^{20} + \ldots + 1) \)). Therefore, \( p^{23} - p \) is divisible by \( d \), and since \( p << p^{23} \), we conclude that \( p \) is in our progression. Similarly, \( q \) is in this progression.

Estimation. Divide the \( 2n \times 2n \) square into \( n^{2} \) squares of \( 2 \times 2 \).

Among these squares, no more than \( 2(n - 1)^{2} / 2 = (n - 1)^{2} \) squares have at least 2 cells colored. The remaining \( 2 \times 2 \) squares are at least \( n^{2} - (n - 1)^{2} = 2n - 1 \) in number. From each of them, a three-cell corner can be cut out.

Example. Construct an example by induction for odd \( n \geqslant 1 \). For \( n = 1 \), there are no colored cells, and one corner can be cut out.

For the transition, highlight an outer "frame" of width two cells in the square. In this frame, color all \( 8(n - 2) \) cells adjacent to the inner boundary of the frame (see Fig. 2), and in the square inside the frame, color the cells according to the induction hypothesis. The total number of colored cells is \( 2(n - 3)^{2} + 8(n - 2) = 2(n - 1)^{2} \).

It remains to understand how many corners can be cut out in this example. Any corner from uncolored cells lies entirely either in the frame or in the inner square (such by assumption \( 2(n - 2) - 1 = 2n - 5 \)). No more than 4 corners can be cut out from the frame - each such corner must contain at least 2 cells of one of the corner \( 2 \times 2 \) squares, and two corners intersecting with one square cannot be cut out. Thus, the total number of corners is no more than \( (2n - 5) + 4 = 2n - 1 \).

Assume such a construction exists. Project the tetrahedron onto a plane \( \alpha \), perpendicular to the line \( \ell \). The projection of the cylinder will be some circle \( \omega \). Denote the projections of the vertices of the tetrahedron by \( A, B, C, D \); they will all be distinct (otherwise, one of the lines containing the sides of the tetrahedron would be parallel to \( \ell \), and such a line cannot have exactly one common point with the cylinder). Each of the lines connecting the points \( A, B, C, D \) must have exactly one common point with the circle \( \omega \), i.e., be tangent to this circle. Moreover, the points \( A, B, C, D \) cannot lie on a single line (since the vertices of the tetrahedron do not lie in one plane). Thus, either three of them lie on one line, without loss of generality \( B, C, D \), or no three of these points lie on one line. In any case, the lines \( AB, AC, AD \) are pairwise distinct, yet they all are tangent to the circle \( \omega \) and pass through the point \( A \), which is a contradiction.

We will show that the triplet \( (a, b, c) \) is mysterious if and only if \( a b c = 1 \). From this, the required result in the problem will immediately follow. Assume that \( a b c < 1 \). Then \( \sqrt{a^{2} + \frac{1}{a^{2} c^{2}} + 2 a b} > \sqrt{a^{2} + b^{2} + 2 a b} = a + b \), similarly \( \sqrt{b^{2} + \frac{1}{b^{2} a^{2}} + 2 b c} > b + c \) and \( \sqrt{c^{2} + \frac{1}{c^{2} b^{2}} + 2 c a} > c + a \). Therefore, in this case, the left side of the equation in the condition is greater than the right side. Reasoning similarly, in the case \( a b c > 1 \), we have that the right side is greater than the left, and in the case \( a b c = 1 \), equality is achieved, which is what was required.

Solution. Applying Menelaus’ theorem to triangle \(ETZ\) with transversals \(AXB\) and \(CYD\), we obtain \( \frac{AZ}{AE}\cdot\frac{BE}{BT}\cdot\frac{XT}{XZ}=1 \) and \( \frac{CE}{CZ}\cdot\frac{DT}{DE}\cdot\frac{YZ}{YT}=1 \). From \(AZ=CE\) and \(BE=DT\) it follows that \(AE=CZ\) and \(BT=DE\). Substituting yields \( \frac{XT}{XZ}=\frac{YZ}{YT} \), i.e. the points \(X\) and \(Y\) are symmetric with respect to the midpoint \(S\) of segment \(ZT\).

From the condition we get that the rays \(AD\) and \(BC\) meet at a point \(F\) at a right angle. Hence, in right triangle \(XFY\) the median \(FS\) equals half of the hypotenuse \(XY\).

Let \(M\) and \(N\) be the midpoints of \(AD\) and \(BC\), and let \(O\) be the center of the circumcircle \((ABCD)\). Then \(O\) is the intersection of the perpendicular bisectors of \(AC\) and \(BD\), which coincide with the perpendicular bisectors of \(EZ\) and \(ET\), respectively. Thus \(O\) is also the center of the circumcircle \((ETZ)\), and \(OE\) is its radius. It suffices to prove \(OE=FS\). We show that \(OEFS\) is a parallelogram.

Since \(EN\) is a median in triangle \(EBC\), while \(MS\) connects the midpoints of opposite sides of quadrilateral \(AZTD\), we have \( \overrightarrow{EN}=\frac{\overrightarrow{EB}+\overrightarrow{EC}}{2}=\frac{\overrightarrow{DT}+\overrightarrow{AZ}}{2}=\overrightarrow{MS} \).

In right triangle \(FBC\), the projections of the median vector \( \overrightarrow{NF} \) onto lines \(BF\) and \(CF\) are \( \overrightarrow{BF}/2 \) and \( \overrightarrow{CF}/2 \). Since \(O\) and \(M\) are the centers of \((ABCD)\) and \((ADF)\), their projections onto the same lines hit the midpoints of \(AB,CD\) and of \(AF,DF\), respectively. Therefore, for \( \overrightarrow{OM}=\overrightarrow{AM}-\overrightarrow{AO}=\overrightarrow{DM}-\overrightarrow{DO} \) the projections onto \(BF\) and \(CF\) are \( (\overrightarrow{AF}-\overrightarrow{AB})/2=\overrightarrow{BF}/2 \) and \( (\overrightarrow{DF}-\overrightarrow{DC})/2=\overrightarrow{CF}/2 \). Hence \( \overrightarrow{NF}=\overrightarrow{OM} \).

Consequently, \( \overrightarrow{OS}=\overrightarrow{OM}+\overrightarrow{MS} \)\(=\overrightarrow{NF}+\overrightarrow{EN} \)\(=\overrightarrow{EF} \), so \(OEFS\) is a parallelogram. Therefore \(OE=FS=\frac{XY}{2}\), and \(XY\) equals the diameter of the circumcircle of \(\triangle ETZ\).

We will show that for \( k = 2n \) (and certainly for \( k << 2n \)), Leon cannot uniquely determine the pair \( P, Q \). Suppose he named \( x_{1} < x_{2} < \ldots < x_{2n} \). Let \( A = (x - x_{1})(x - x_{2}) \ldots (x - x_{n}) \) and \( B = (x - x_{n+1})(x - x_{n+2}) \ldots (x - x_{2n}) \), so that \( A(x_{i}) = 0 \) for \( i = 1,2, \ldots, n \) and \( B(x_{i}) = 0 \) for \( i = n+1, n+2, \ldots, 2n \). Then if Ilya thought of \( P_{1} = A + 2B \) and \( Q_{1} = -A - 2B \), the \( i \)-th column of the table will contain the numbers \( \pm 2B(x_{i}) \) for \( i = 1,2, \ldots, n \) and the numbers \( \pm A(x_{i}) \) for \( i = n+1, n+2, \ldots, 2n \). But the same table is suitable for the pair \( P_{2} = A - 2B \) and \( Q_{2} = -A + 2B \), which Sasha could have thought of.

On the other hand, we will show that for \( k = 2n + 1 \), Ilya's table can satisfy at most one pair of polynomials \( P, Q \). Assume the contrary, and there are two such pairs: \( P_{1}, Q_{1} \) and \( P_{2}, Q_{2} \). Then \( P_{2} \) coincides with \( P_{1} \) or \( Q_{1} \) for at least \( n+1 \) different values of the argument, say, with \( P_{1} \). But then \( P_{1} \) and \( P_{2} \) are the same polynomials (since their difference is a polynomial of degree at most \( n \), having at least \( n+1 \) different roots). From the table, we then obtain that the values of \( Q_{1} \) and \( Q_{2} \) coincide at \( 2n+1 \) points, and thus \( Q_{1} = Q_{2} \).

Let \(\omega\) be the circle \((ETZ)\) and \(d\) its diameter. Since \(BE = DT\), it follows that \(BT = BE + ET = DT + ET = DE\). From the condition \(\angle A + \angle D = 90^{\circ}\), it follows that rays \(AB\) and \(DC\) intersect at some point \(F\) at a right angle (see image 3).

Arguing similarly, we find that the point \(C'\), diametrically opposite \(E\) on the circle \((CED)\), lies on the circle \(\omega\), and also \(C'Y \perp ZT\). Thus, \(B'C'\) is a chord of the circle \(\omega\), and \(X\) and \(Y\) are the projections of points \(B'\) and \(C'\) on line \(ZT\), so \(XY \leq B'C' \leq d\), which was required.

Note 1. In fact, \(B'C'\) is the diameter of the circle \((ETZ)\), which is easy to establish by counting angles, but this is not required for the solution. The equality \(XY = d\) is achieved if and only if the original quadrilateral is inscribed.

Note 2. We provide a plan for another approach to the problem. Using the notations from the solution above, and introducing new ones: \(x = BF, y = AB, z = CF, t = DC, k = \frac{DE}{EB}\), \(m = \frac{AE}{FC}, p = ZT, \alpha = \angle AED\). From Menelaus' theorem for \(\triangle EZT\) and line \(AYB; \triangle EZT\) and line \(CYD\), we find: \(XZ = YT = p \cdot \frac{1}{kl - 1}\). By the sine theorem for triangle \(ZET: d = \frac{p}{\sin \alpha}\), due to the above \(XY = XZ + YZ + ZT = p \cdot \frac{kl + 1}{kl - 1}\). Thus, it is sufficient to prove that \(\sin \alpha \leq \frac{kl - 1}{kl + 1}(\star)\). From Menelaus' theorem for \(\triangle AFC\) and line \(BED; \triangle BFD\) and line \(AEC\), it is easy to see that \(k = \frac{t(x + y)}{zy}, l = \frac{y(z + t)}{xt}\), hence \(kl = \frac{(x + y)(z + t)}{xz}\) and \(\frac{kl - 1}{kl + 1} = \frac{(x + y)(z + t) - xz}{(x + y)(z + t) + xz}\).

Let \(\angle FAC = \beta, \angle FDB = \gamma\), then \(\alpha = 90^{\circ} + \beta + \gamma\). Thus, \(\sin \alpha = \cos \gamma \cos \beta - \sin \gamma \sin \beta = \frac{(x + y)(z + t) - xz}{\sqrt{\left(x^{2} + (z + t)^{2}\right)\left(z^{2} + (y + t)^{2}\right)}}\), the last equality is obtained from the right triangles \(AFC\) and \(BFD\). It remains to note that \(\sqrt{\left(x^{2} + (z + t)^{2}\right)\left(z^{2} + (y + t)^{2}\right)} \geq xz + (z + t)(y + t)\) by the Cauchy-Schwarz inequality, obtaining the required inequality.

Let's divide all positions into two halves of 100 columns each - the "left" and the "right". Suppose that in each row where there are two suns in one half (let's call them sunny), the wind blew off the sheets from one of such pairs of suns, and in each non-sunny row - there are exactly \( 101^{2} \) such rows - the wind revealed the positions of all suns (in a non-sunny row there are no more than two suns, so the wind could do so). Then Yuri will realize that the rows where two suns are open in one half are definitely sunny, and therefore, the non-sunny rows are exactly those \( 101^{2} \) rows in which the wind did not open two suns in one half. Each of them has all suns open, so the images covered by sheets in these rows are moons.

First solution. Without loss of generality, we can assume that the rays \(AB\) and \(DC; CB\) and \(DA\) intersect. Let the segments \(AC\) and \(BD\) intersect at point \(G\), and also let \(A'B'C'D'\) be the quadrilateral formed by the lines \(\ell_{a}, \ell_{b}, \ell_{c}, \ell_{d}\) (see Fig. 4). Also, denote by \(X\) the intersection of \(AB\) and \(CD'\), and by \(Y\) the intersection of \(CD\) and \(AB'\).

Fig. 4

Let \(\angle B'A B = \alpha\). From the inscribed quadrilateral \(A'B'C'D'\) and the conditions \(AX \parallel \ell_{d}, CY \parallel \ell_{b}\), we have: \(\alpha = \angle B'A X = 180^{\circ} - \angle A'B'C' = \angle C'D'X = \angle YCA'\). Thus, firstly, the points \(A, D', X, C'\) lie on one circle, denote it \(\gamma_{1}\); secondly, the points \(C, Y, A', B'\) lie on one circle, denote it \(\gamma_{2}\); thirdly, the points \(A, X, C, Y\) lie on one circle, denote it \(\gamma_{0}\). Note that point \(B\) is the radical center of the circles \(\gamma, \gamma_{0}, \gamma_{1}\) (since it lies on the lines \(AX\) and \(C'D'\)); point \(D\) is the radical center of the circles \(\gamma, \gamma_{0}, \gamma_{2}\) (since it lies on the lines \(CY\) and \(A'B'\)). Thus, \(BD\) is the radical axis of the circles \(\gamma_{0}\) and \(\gamma\); \(AC\) is the radical axis of the circles \(\gamma_{0}\) and \(\omega\); \(EF\) is the radical axis of the circles \(\omega\) and \(\gamma\), therefore these three lines intersect at one point, which was to be proved.

Second solution. We introduce the notations as in the first solution. For a point \(P\) on the plane, denote by \(f(P)\) the difference of the powers of point \(P\) with respect to the circles \(\omega\) and \(\gamma\). Since \(EF\) is the radical axis of the circles \(\omega\) and \(\gamma\), it suffices to prove that \(f(G) = 0\). Moreover, it is easy to see that \(f(A) = AC' \cdot AB'\) and \(f(C) = -CD' \cdot CA'\).

Fig. 5

Note that the function \(f\) is linear, that is, for a point \(P\) on the segment \(QR\), the equality \(f(P) = \frac{PR \cdot f(Q) + PQ \cdot f(R)}{QR}(\star)\) holds. We will prove this statement later. For now, applying it to the points \(A, G, C\), we obtain that \(f(G) = \frac{AG \cdot f(C) + CG \cdot f(A)}{AC}\).

Thus, it suffices to prove that \(\frac{f(A)}{-f(C)} = \frac{AG}{CG}(\star \star)\). Note that \(\frac{AG}{GC} = \frac{d(A, BD)}{d(C, BD)} = \frac{S_{ABD}}{S_{BCD}} = \frac{AB \cdot AD}{CB \cdot CD}\) (the last equality follows from the fact that \(\angle BAD + \angle BCD = 180^{\circ}\); by \(d(P, \ell)\) we denote the distance from point \(D\) to the line \(\ell\)). Therefore, the equality (\(\star \star\)) is rewritten as: \(\frac{AC' \cdot AB'}{CD' \cdot CA'} = \frac{AB \cdot AD}{CB \cdot CD}\).

From the inscribed quadrilateral \(ABCD\) and the given parallelisms of the lines in the condition, we have the angle equalities: \(\angle CA'D = 180^{\circ} - \angle ADB' = \angle BAD = 180^{\circ} - \angle BCD = \angle CBC' = \angle AC'B; \angle ABC' = \angle CDA'\) and \(\angle BCD' = \angle B'AD\). Thus, \(\triangle ABC'\) and \(\triangle CDA'\), as well as \(\triangle DAB'\) and \(\triangle BCD'\), are similar by two angles. From the similarity, we obtain the ratio equalities \(\frac{AC'}{CA'} = \frac{AB}{CD}\) and \(\frac{AB'}{CD'} = \frac{AD}{BC}\), we just need to multiply these equalities.

Returning to the proof of the linearity of the function \(f\). We introduce Cartesian coordinates such that the centers of the circles \(\omega\) and \(\gamma\) lie on the x-axis, let their coordinates be \((x_{1}, 0)\) and \((x_{2}, 0)\), and the radii of the circles be \(R_{1}\) and \(R_{2}\). Then for an arbitrary point \(P\) with coordinates \((x, y)\), by the definition of the power of a point, we obtain that \(f(P) = (x - x_{1})^{2} + y^{2} - R_{1}^{2} - (x - x_{2})^{2} - y^{2} + R_{2}^{2} = ax + b\), where \(a\) and \(b\) are two constants. If point \(P\) lies on the segment \(QR\) and \(x_{q}, x_{r}\) are the coordinates of points \(Q\) and \(R\) on the x-axis, then \(x = \frac{PR \cdot x_{q} + PQ \cdot x_{r}}{QR}\), from which (\(\star\)) immediately follows.

Remark 1. A linear function in arbitrarily introduced Cartesian coordinates has the form \(f(x, y) = ax + by + c\). If it is not constant, then the solution of the equation \(f(x, y) = 0\) will be a line. For example, in this way (reasoning as in the given solution), it is proved that the radical axis of two circles is a line perpendicular to the line of centers.

Remark 2. We present a plan for solving the problem using complex numbers. Let the circle \(\omega\) be the standard unit circle, and the coordinates of the points \(A, B, C, D\) be denoted as \(a, b, c, d\), respectively. Consider the lines (more precisely, the equations of the lines) \(\ell_{ac}(z) = ac \bar{z} + z - a - c, \ell_{bd}(z) = bd \bar{z} + z - b - d\) - these are the corresponding diagonals \(AC, BD\). The line \(\ell_{a}\) through point \(A\) parallel to \(BC\) has the equation \(\ell_{a}(z) = bc \bar{z} + z - bc/a - a\), similarly for the other three lines from the condition. The circle \(\gamma\), passing through the points of their intersection, has the equation \(F(z) := \ell_{a}(z)\ell_{c}(z) - \ell_{b}(z)\ell_{d}(z)\): on one hand, the intersection points satisfy the condition \(F = 0\), on the other hand, the coefficients at \(z^{2}\) and \(\bar{z}^{2}\) cancel out, so it is indeed a circle. The radical axis of the circles \(\gamma\) and \(\omega\) has the equation \(\ell_{ef}(z) := F(z) - T(z \bar{z} - 1)\), where \(T\) is a suitable coefficient to make it a line (it will not be important for us later). We will prove that for some coefficients \(\tau, \theta\), the identity \(\ell_{ef}(z) + \tau \ell_{ac}(z) + \theta \ell_{bd}(z) = 0(\mathrm{A})\) holds, from which the required will follow. The identity (A) is sufficient to check when \(z\) is one of the vertices of the quadrilateral \(ABCD\), because they do not lie on one line. Substituting, for example, \(z = a\), we obtain the identity \(0 = F(a) + \theta \ell_{bd}(a) = -\ell_{b}(a)\ell_{d}(a) + \theta \ell_{bd}(a)\). This gives the value of \(\theta\), and it must be consistent with the value obtained for \(z = c\), that is, the identity \(\ell_{b}(a)\ell_{d}(a)/\ell_{bd}(a) = \ell_{b}(c)\ell_{d}(c)/\ell_{bd}(c)\) must hold, and the second similar one. All these values are calculated directly (and decomposed into factors, for example, \(\ell_{bd}(a) = bd/a + a - b - d = (a - b)(a - d)/d\)), after which it is not difficult to prove the required identity.

We will measure the height of the snowdrift in decimeters. We will also assume that the side of one cell is 1 dm, meaning that 1 dm³ of snow falls on a cell each night.

We will prove that after the hundredth morning there will be a snowdrift of at least 112 dm. Suppose there is no such snowdrift. Since the janitor completely shovels the snow from some row on the hundredth morning, there is no snow in ten cells of the square. In each of the remaining 90 cells, by our assumption, there is no more than 111 dm³ of snow, meaning there is no more than 9990 dm³ of snow in total. However, over 100 nights, a total of 10 000 dm³ of snow fell. Contradiction.

We will show how the janitor can act so that after the hundredth morning each snowdrift has a height of no more than 112 dm (i.e., no more than 112 dm³ of snow in each cell).

Solution 1. For the first 11 days the janitor shovels snow from the second column to the first; for the next 11 days from the third to the second; then 11 days from the fourth to the third; and so on. After 99 days there will be no snow in the tenth column. Let’s calculate how much snow is in column \(i\le 9\) after 99 days. On the evening of the \(11(i-1)\)\(-\)th day there was no snow in column \(i\), and in column \(i+1\) there was \(11(i-1)\) dm³ of snow in each cell. The next evening there will be \(11(i-1)+2\) dm³ of snow in each cell of column \(i\). Then for ten more days the amount of snow in each cell of column \(i\) will increase by \(2\), and then for \(11(9-i)\) days by \(1\). In total, after 99 days, there will be \(11(i-1)+22+11(9-i)=110\) dm³ of snow in each cell of column \(i\). On the hundredth night, another 1 dm³ will fall in each cell, and on the hundredth morning the janitor will shovel snow from the tenth column to the ninth. Thus, there will be no more than 112 dm³ of snow in each cell.

Solution 2. Let the janitor shovel snow from the 2nd column to the 1st, from the 3rd to the 2nd, …, from the 10th to the 9th. Then on the evening of the 9th day there will be 10 dm³ of snow in each cell of the first nine columns, and there will be no snow in the tenth column. Then the janitor performs a similar process in reverse order: from the 9th to the 10th, from the 8th to the 9th, …, from the 2nd to the 1st. Then on the evening of the 18th day, there will be 20 dm³ of snow in the cells of the last nine columns, and there will be no snow in the first column. Repeat such shifts (each lasting 9 days) nine more times, and after 99 days, we will have 110 dm³ of snow in the cells of nine columns and one extreme column empty. On the hundredth morning, we shovel snow from this extreme column to the neighboring one and obtain at most 112 dm³ of snow in each cell.

Solution 1. Let \(H'\) and \(X'\) be the points symmetric to \(H\) and \(X\) with respect to the midpoint of side \(BC\) (see figure).

Then \(HXH'X'\) is a parallelogram. Since \(\angle BX'C=\angle BH'C=\angle BHC=180^\circ-\angle BAC\), the points \(X'\) and \(H'\) lie on the circumcircle \(\Omega\). Moreover, since \(H'B\parallel CH\perp AB\), the point \(H'\) is diametrically opposite to \(A\) on \(\Omega\); therefore \(AH'\) passes through \(O\). Noting that \(XO\parallel H'X'\), we obtain \(\angle AYX'=180^\circ-\angle AH'X'=180^\circ-\angle AOX=\angle AYX\); hence \(Y\), \(X\), and \(X'\) are collinear.

Solution 2. Since \(\angle BHC=180^\circ-\angle ABC\), the circumcircle of triangle \(BHC\) is symmetric to \(\Omega\) with respect to \(BC\); let \(O'\) be its center and \(M\) the midpoint of \(BC\). Then \(M\) is also the midpoint of \(OO'\).

It is known that \(AH=2\,OM\) (this can be proved, for example, via a homothety centered at the centroid of triangle \(ABC\) with ratio \(-2\)). Therefore \(OO'=2\,OM=AH\). Since \(OO'\parallel BC\parallel AH\), the quadrilateral \(AHO'O\) is a parallelogram.

Let \(T\) lie on the ray \(O'X\) with \(O'T=2\,O'X\). Then \(XT=O'X=O'H=AO\). Furthermore, from isosceles triangle \(O'XH\) we get \(\angle TXO=\angle O'XH=\angle O'HX=\angle AOX\), so triangles \(TXO\) and \(AOX\) are congruent; hence \(\angle TOX=\angle AXO\).

Since \(XM\) is the midline in triangle \(O'TO\), we have \(\angle MXO=\angle TOX=\angle AXO\), i.e. \(XO\) is the bisector of \(\angle AXM\). In the circumcircle \((AXOY)\) we have \(OA=OY\), so \(O\) is the midpoint of arc \(AXY\); therefore \(XO\) is the external bisector of \(\angle AXY\). It follows that \(\angle AXM\) and \(\angle YXM\) are supplementary, hence \(X\), \(Y\), and \(M\) are collinear.

Let's construct an example of 8 quadratic trinomials that satisfy the problem’s condition: \(\,f_1(x)=-x^2+2\,\), \(\,f_2(x)=3x^2-2\,\), \(\,f_3(x)=-4x^2+3\,\).

\(\,f_4(x)=2x^2-3\,\), \(\,f_5(x)=-4x^2+x+4\,\), \(\,f_6(x)=4x^2+x-4\,\).

\(\,f_7(x)=-5x^2-x+5\,\), \(\,f_8(x)=5x^2-x-5\,\).

These polynomials are constructed so that their values at the points \(x=-1,0,1\) correspond to the following table:

The trinomials in this example do not have a common root (not even \(f_1(x)\) and \(f_2(x)\)). It remains to show that they satisfy the condition. Obviously, no two of these trinomials sum to zero.

Suppose some pair of these quadratic trinomials is chosen. If the pair \(\,(f_{2k-1}(x),\,f_{2k}(x))\,\) was chosen, where \(k=1,2,3,4\), then all polynomials can be split into pairs \(\,(f_1(x),f_2(x))\,\), \(\,(f_3(x),f_4(x))\,\), \(\,(f_5(x),f_6(x))\,\), \(\,(f_7(x),f_8(x))\,\); each of these sums has the root \(0\).

Otherwise, it is easy to verify that the value of the sum of the two chosen trinomials is zero either at \(x_0=-1\) or at \(x_0=1\) (possibly both). Choose such \(x_0\). The remaining trinomials at \(x_0\) take the values \(-1\) and \(1\) exactly three times each, and they can be split into pairs so that at \(x_0\) the sums of all four pairs equal zero, i.e., \(x_0\) is their common root.

We prove that in the analogous problem for a row of \(2n\) children, the maximum possible number of good pairs is \((n+1)^2-3\).

Number the children \(1,2,\dots,2n\) in descending order of height. If we arrange them as \(n+1,n+2,\dots,2n,1,2,\dots,n\), then every pair \((i,j)\) with \(i\le n\lt j\) is good; there are \(n^2\) such pairs. In addition, every pair \((i,i+1)\) is good; there are \(2n-1\) such pairs. The pair \((n,n+1)\) is counted twice, so the total number of good pairs is \(n^2+(2n-1)-1=(n+1)^2-3\).

It remains to show there cannot be more than \((n+1)^2-3\) good pairs. Proceed by induction on \(n\). For \(n=1\) the statement is trivial, since there is only one pair.

Now let \(n\gt 1\). Choose a good pair \((a,b)\) with \(\lvert a-b\rvert\) maximal; assume \(a\lt b\) and child \(a\) stands to the left of \(b\). Call a child \(c\) excellent if they form good pairs with both \(a\) and \(b\).

Lemma. There are at most two excellent children.

Proof. If \(c\) is excellent, then by the choice of \((a,b)\) we have \(c-a\le b-a\) and \(b-c\le b-a\), hence \(a\lt c\lt b\). Such a child \(c\) cannot stand between \(a\) and \(b\); otherwise \((a,b)\) would not be good. Therefore any excellent child stands either to the left of \(a\) or to the right of \(b\).

Suppose there are two excellent children \(c_1\lt c_2\) to the left of \(a\); then \(a\lt c_1\lt c_2\lt b\). The child \(c_1\) cannot stand between \(a\) and \(c_2\), otherwise \((a,c_2)\) would not be good; hence \(c_1\) is left of \(c_2\). Then \(c_2\) is between \(c_1\) and \(b\), and \((c_1,b)\) is not good — a contradiction. Thus at most one excellent child is left of \(a\), and similarly at most one right of \(b\). This proves the lemma.

To finish the induction, remove \(a\) and \(b\). All good pairs not containing \(a\) or \(b\) remain good; by the induction hypothesis there are at most \(n^2-3\) of them. Pairs involving \(a\) or \(b\): \((a,b)\); \((a,c)\) and \((b,c)\) for any excellent \(c\); and for each remaining \(c\), at most one of \((a,c)\) or \((b,c)\). In total this is at most \(1+(2n-2)+2=2n+1\) pairs, so altogether at most \((n^2-3)+(2n+1)=(n+1)^2-3\), as required.

Note. The pair \(a,b\) can be chosen in several ways: take a good pair farthest apart, or take \(a=1\) and the largest \(b\) with \((1,b)\) good.

Assume that the planks are laid out on a number line. Let \( 1 = 1 \text{ cm} \).

Choose \( 0 < \varepsilon < 0.01 \) such that the difference in lengths of any pair of adjacent planks is greater than \( 10 \varepsilon \). Mark an infinite arithmetic progression in both directions on the line with a difference of \( \varepsilon \) so that the ends of the planks are not marked. The grasshopper will jump only on the marked points, and the lengths of its jumps will be from the set \( \{\varepsilon, \ell, 2 \ell, 4 \ell, 8 \ell, 16 \ell, 32 \ell, 64 \ell\} \), where \( \ell = N \varepsilon \), and we choose a natural \( N \) such that \( \ell < 2 \) and \( 64 \ell > 101 \).

The grasshopper's strategy will be as follows: jump to the right on the green plank by \( \varepsilon \) as long as possible, and then jump over the next red plank with the shortest possible jump length (such a length will be found since the length of the longest jump is greater than \( 100 + \varepsilon \)). So, suppose a jump of length \( 2 d \) is made from a green segment over the next red segment \( [a, b] \). We need to ensure that after this jump, the grasshopper ends up in the next green segment \( [b, c] \). Assume this is not the case, and the grasshopper jumps from point \( a - x \), where \( 0 < x < \varepsilon \), to point \( a - x + 2 d > c \). We see that \( 2 d > (c - b) + (b - a) > 2 \), which means that the set of jump lengths of the grasshopper includes the length \( d \). Further, by the choice of \( \varepsilon \), we have \( (c - b) > (a - b) + 10 \varepsilon \), so we can estimate \( 2 d > (c - b) + (b - a) > 2(b - a) + 10 \varepsilon \). We see that \( d > (b - a) + \varepsilon \), which means the grasshopper could have jumped over the red segment \( [a, b] \) with a shorter jump than \( 2 d \). Contradiction.

Solution 1. Let \(\angle ADC = x\). From the isosceles triangles \(DME\) and \(DMF\) (or from the fact that \(M\) is the center of the circle \((DEF)\)), we have \(\angle EMF = 360^{\circ} - 2x\) (see fig. 4). To solve the problem, it remains to understand that \(\angle EBF\) is equal to this.

Under a homothety with center \(D\) and coefficient \(1/2\), points \(E, F, B\) will map to \(E', F', B'\) - the midpoints of segments \(DE, DF\), and \(DB\), respectively. Instead of \(\angle EBF\), we find \(\angle E'B'F'\), noting that \(E'\) and \(F'\) are projections of \(M\) onto \(AD\) and \(CD\), and \(B'\) is the center of the parallelogram, or the midpoint of \(AC\), thus \(B'\) is the projection of \(M\) onto \(AC\). We see that \(M, E', A, B'\) lie on a circle with diameter \(MA\). Hence \(\angle DE'B' = \angle AMB' = \angle AMC / 2 = \angle ABC / 2 = x / 2\). Similarly, \(\angle DF'B' = x / 2\). From the quadrilateral \(E'B'F'D\), we see that \(\angle E'B'F' = 360^{\circ} - x - x / 2 - x / 2 = 360^{\circ} - 2x\), which is what was required.

Fig. 5

Solution 2. It is sufficient to prove the equality of angles \(\angle ABE = \angle CBF\) (i.e., the isogonality of \(BE\) and \(BF\) with respect to \(AB, BC\)). Indeed, then \(M\) will lie on the external bisector of angle \(EBF\) and on the perpendicular bisector to \(EF\), and thus will coincide with the midpoint of the arc \((EBF)\).

The equality of angles \(\angle ABE = \angle CBF\) is equivalent to the similarity \(ABE \sim CBF\). We will prove this similarity.

Mark a point \(X\) on the ray \(AB\) beyond point \(B\) such that \(BX = BC\), and a point \(Y\) on the ray \(CB\) beyond point \(B\) such that \(BY = BA\). It is easy to see that triangles \(BMC\) and \(BMX\) are equal by two sides and the angle between them. Then \(MX = MC = MA\). Consider the perpendicular bisector to \(DF\), then it is perpendicular to the parallel line \(AX\), and since \(MA = MX\), it is also the perpendicular bisector to \(AX\). Thus, trapezoid \(ADFX\) is isosceles, and since \(ABCD\) is a parallelogram, \(CXBF\) is also an isosceles trapezoid, with \(CB = BX = XF\) and \(\angle XBC = 180^{\circ} - \angle ABC\). We obtain a similar result for trapezoid \(AYBE\). We see that \(AYBE \sim CXBF\), from which follows the required \(ABE \sim CBF\).

Remark. The required similarity \(ABE \sim CBF\) in solution 2 can also be proven differently by establishing the equality \(AE \cdot BC = CF \cdot AB\). This can be done, for example, by calculating in sines through the elements of triangle \(ABC\), expressing the projection of \(AM\) onto \(AD\), then expressing segment \(AE\), and so on.

There are also other computational solutions, including those using complex numbers.

Solution 1.

Number the white segments from left to right as \( w_{1}, w_{2}, \ldots, w_{y_{1}} \), and the black ones as \( b_{1}, b_{2}, \ldots, b_{y_{2}} \). For each black segment \( b_{j} \), define its strength \( S\left(b_{j}\right) \) as the number of indices \( i \leqslant y_{1} - 1 \) such that \( b_{j} \) intersects both \( w_{i} \) and \( w_{i + 1} \). If two black segments intersect with some pair \( (w_{i}, w_{i + 1}) \), they share a point, which is impossible by the condition. Therefore, each such pair is counted in the strength of at most one black segment, hence,

\[ \sum_{j = 1}^{y_{2}} S\left(b_{j}\right) \leqslant y_{1} - 1 \]

Consider the following \( y_{1} \) groups, each consisting of \( x_{1} \) white segments: for \( 0 \leqslant i \leqslant y_{1} - x_{1} \), group \( G_{i} \) consists of segments \( w_{i + 1}, w_{i + 2}, \ldots, w_{i + x_{1}} \), and for \( y_{1} - x_{1} + 1 \leqslant i \leqslant y_{1} - 1 \), group \( G_{i} \) consists of segments \( w_{i + 1}, w_{i + 2}, \ldots, w_{y_{1}} \), as well as \( w_{1}, w_{2}, \ldots, w_{i + x_{1} - y_{1}} \) (in other words, each group consists of \( x_{1} \) consecutive segments in cyclic order). For group \( G_{i} \), let \( N\left(G_{i}\right) \) denote the number of black segments that do not intersect any of the segments in \( G_{i} \). By the condition, \( N\left(G_{i}\right) \leqslant x_{2} - 1 \); therefore,

\[ \Sigma : = \sum_{i = 0}^{y_{1} - 1} N\left(G_{i}\right) \leqslant y_{1}\left(x_{2} - 1\right) . \]

On the other hand, each black segment \( b_{j} \) intersects with at most \( 1 + S\left(b_{j}\right) \) white segments, and all these white segments are consecutive. Then the number of groups containing at least one of these white segments does not exceed \( 1 + S\left(b_{j}\right) + (x_{1} - 1) = S\left(b_{j}\right) + x_{1} \). Therefore, segment \( b_{j} \) is counted in at least \( y_{1} - (S\left(b_{j}\right) + x_{1}) \) numbers of the form \( N\left(G_{i}\right) \). Therefore,

\[ \begin{array}{c} \Sigma \geqslant \sum_{j = 1}^{y_{2}}\left(y_{1} - S\left(b_{j}\right) - x_{1}\right) = y_{2}\left(y_{1} - x_{1}\right) - \sum_{j = 1}^{y_{2}} S\left(b_{j}\right) \geqslant \\ \geqslant y_{2}\left(y_{1} - x_{1}\right) - \left(y_{1} - 1\right) \end{array} \]

From the two obtained estimates for \( \Sigma \), it follows that \( y_{1}\left(x_{2} - 1\right) \geqslant y_{2}\left(y_{1} - x_{1}\right) - \left(y_{1} - 1\right) \Longleftrightarrow\left(y_{1} - x_{1}\right)\left(y_{2} - x_{2}\right) \leqslant x_{1} x_{2} - 1 \), which is what was required to prove.

Solution 2.

Assume that the statement of the problem is false for some \( x_{1}, x_{2}, y_{1}, y_{2} \): \( \left(y_{1} - x_{1}\right)\left(y_{2} - x_{2}\right) \geqslant x_{1} x_{2} \), and under this condition, the sum \( y_{1} + y_{2} \) is the smallest possible.

Without loss of generality, then \( y_{1} - x_{1} \geqslant x_{1} \). Take the \( x_{1} \)-th white segment from the left \( W \) and the \( \left(y_{2} - x_{2}\right) \)-th black segment from the left \( B \). One of them has its right endpoint to the left.

1) Suppose the right endpoint of \( W \) is to the left (or the endpoints coincide).

Then the rightmost \( x_{2} \) black segments do not intersect with the leftmost \( x_{1} \) white segments. Contradiction.

2) Suppose the right endpoint of \( B \) is to the left. Remove all white segments to the left of \( W \) (including it) and all black segments to the left of \( B \) (including it). The remaining white segments (at least \( x_{1} \) of them) do not intersect with the removed \( y_{2} - x_{2} \) black segments; hence it follows that \( y_{2} - x_{2} < x_{2} \).

Let \( x_{1}^{\prime} = x_{1}, y_{1}^{\prime} = y_{1} - x_{1} \geqslant x_{1}^{\prime}, x_{2}^{\prime} = x_{2} - \left(y_{2} - x_{2}\right) \) and \( y_{2}^{\prime} = y_{2} - \left(y_{2} - x_{2}\right) = x_{2} \geqslant x_{2}^{\prime} \); then there are \( y_{1}^{\prime} \) white and \( y_{2}^{\prime} \) black segments left. Consider any \( x_{1}^{\prime} \) remaining white and \( x_{2}^{\prime} \) remaining black segments. If there are no intersecting ones among them, then by adding all the removed black segments to them, we obtain a set of \( x_{1} = x_{1}^{\prime} \) white and \( x_{2} = x_{2}^{\prime} + \left(y_{2} - x_{2}\right) \) black segments of the original set, among which there are no intersecting ones; this is impossible. Thus, the remaining set satisfies the condition (for the new numbers \( x_{1}^{\prime}, y_{1}^{\prime}, x_{2}^{\prime} \) and \( y_{2}^{\prime} \)), and it contains fewer segments than the original, therefore

\[ \begin{array}{l} 0 < x_{1}^{\prime} x_{2}^{\prime} - \left(y_{1}^{\prime} - x_{1}^{\prime}\right)\left(y_{2}^{\prime} - x_{2}^{\prime}\right) = \\ = x_{1}\left(2 x_{2} - y_{2}\right) - \left(y_{1} - 2 x_{1}\right)\left(y_{2} - x_{2}\right) = \\ \quad = x_{1} x_{2} - \left(y_{1} - x_{1}\right)\left(y_{2} - x_{2}\right) \leqslant 0 \end{array} \]

Contradiction.

Note. It is worth noting that for \( x_{1} = x_{2} = 1 \), the statement of the problem turns into the two-color (one-dimensional) Helly's theorem.

We will measure the height of the snowdrift in decimeters. We will also assume that the side of one cell is 1 dm, that is, 1 dm \(^{3}\) of snow falls on the cell each night.

We will prove that after the hundredth morning, there will be a snowdrift of at least 112 dm. Suppose there is no such snowdrift. Since the janitor completely shoveled the snow from some row on the hundredth morning, there is no snow in ten cells of the square. In each of the remaining 90 cells, according to our assumption, there is no more than 111 dm \(^{3}\) of snow, that is, a total of no more than 9990 dm \(^{3}\) of snow. However, over 100 nights, a total of 10000 dm \(^{3}\) of snow fell. Contradiction.

We will show how the janitor can act so that after the hundredth morning, each snowdrift has a height of no more than 112 dm (that is, no more than 112 dm \(^{3}\) of snow in each cell).

Method 1. For the first 11 days, the janitor shovels snow from the second column to the first, for the next 11 days, the janitor shovels snow from the third column to the second, then 11 days from the fourth to the third, and so on. After 99 days, there will be no snow in the tenth column. Let's calculate how much snow is in column \( i \leqslant 9 \) after 99 days. On the evening of the \( 11(i - 1) \)-th day, there was no snow in column number \( i \), and in column \( i + 1 \), there was \( 11(i - 1) \) dm \(^{3}\) of snow in each cell. The next evening, there will be \( 11(i - 1) + 2 \) dm \(^{3}\) of snow in each cell of column \( i \). Then for ten more days, the amount of snow in each cell of column \( i \) will increase by 2, and then for \( 11(9-i) \) days by 1. In total, after 99 days, there will be \( 11(i - 1) + 22 + 11(9-i) = 110 \) dm \(^{3}\) of snow in each cell of column \( i \). On the hundredth night, another 1 dm \(^{3}\) will fall in each cell. And on the hundredth morning, the janitor will shovel snow from the tenth column to the ninth. Thus, there will be no more than 112 dm \(^{3}\) of snow in each cell.

Method 2. Let the janitor shovel snow from the 2nd column to the 1st, from the 3rd to the 2nd, ..., from the 10th to the 9th. Then on the evening of the ninth day, there will be 10 dm \(^{3}\) of snow in each cell of the first nine columns, and there will be no snow in the tenth column. Then the janitor performs a similar process in reverse order: from the 9th to the 10th, from the 8th to the 9th, ..., from the 2nd to the 1st. Then on the evening of the 18th day, there will be 20 dm \(^{3}\) of snow in the cells of the last nine columns, and there will be no snow in the first column. Repeat such shifts (each lasting 9 days) 9 more times, and after 99 days, we will have 110 dm \(^{3}\) of snow in the cells of nine columns and one extreme column empty. On the hundredth morning, we shovel snow from this extreme to the neighboring one and get no more than 112 dm \(^{3}\) of snow in each cell.

Let the line \( AH \) intersect the circle \( \omega \) again at point \( D \). Then the line drawn through \( O \) as per the condition is the perpendicular bisector of the chord \( AD \), and let it intersect the sides \( AB \) and \( AC \) at points \( X \) and \( Y \), while the line through \( H \) intersects them at points \( Z \) and \( T \). Since \( XY \parallel AC \), the circle ( \( AXY \) ) is tangent to the circle \( \omega \) at point \( A \). By symmetry with respect to \( XY \), the circle \( \omega \) maps onto itself, and the circle ( \( AXY \) ) maps onto the circle ( \( DXY \) ), so it is also tangent to the circle \( \omega \).

Since \( ZH \perp AO \), we have \( \angle AZH = 90^{\circ} - \angle OAB = \angle ACB = \)\( = \angle ADB \).

Fig. 6

Therefore, the quadrilateral \( BZHD \) is cyclic. Then \( \angle BZD = \angle BHD = 90^{\circ} - \angle HCB = \angle ACB \). Hence, as mentioned above, \( \angle XYD = \angle AYX = \angle ACB = \angle BZD \), so point \( Z \) lies on the circle ( \( DXY \) ). Similarly, point \( T \) also lies on this circle, which proves the required statement.

Remark. One can reason slightly differently: establish (using similar angle equalities) that points \( X, Y, Z, T \) lie on one circle, and also that the circles ( \( DXY \) ) and ( \( DZT \) ) are tangent to the circle \( \omega \) at point \( D \). However, these three circles cannot be different, because in such a case their radical axes do not intersect at one point, which is easy to verify. Therefore, all these circles coincide, which is what we needed to prove.

We call a road that both mayors want to build a reliable road. In each component, consider the cheapest road \( AB \). Then it is reliable. Suppose there is another reliable road \( CD \) in this component (clearly, cities \( C, D \) are different from \( A, B \)) and consider the path along the roads from one of the cities \( A, B \) to one of the cities \( C, D \) - without loss of generality, it has the form \( A X_{1} X_{2} \ldots X_{k} C \) (cities \( X_{1}, \ldots, X_{k} \) are different from \( A, B, C, D \), possibly \( k = 0 \)). Then the mayor of city \( X_{1} \) wants to build the road \( AX_{1} \) (the mayor of city \( A \) wants to build \( AB \)), the mayor of city \( X_{2} \) wants to build the road \( X_{1}X_{2} \) (the mayor of city \( X_{1} \) wants to build \( X_{1}A \)), and so on, the mayor of city \( C \) wants to build the road \( CX_{k} \), not \( CD \) - a contradiction.

Thus, in each component, there is exactly one reliable road. For each of the \( n(n - 1) / 2 \) pairs of cities \( A, B \), consider the random variable \( \xi_{AB} \), which equals 1 if \( AB \) is a reliable road, and 0 otherwise. From the proven, it follows that \( M \) is the sum of \( \xi_{AB} \) over all \( n(n - 1) / 2 \) pairs \( \{A, B\} \). For given \( A, B \), the event \( \xi_{AB} = 1 \) means that the road \( AB \) is the cheapest of the \( 2n - 3 \) roads leading from \( A \) or \( B \), so the probability of such an event is \( 1 / (2n - 3) \) (due to the symmetry of the price distribution, these roads are equally likely, so each of them is the cheapest with probability \( 1 / (2n - 3) \)). Therefore, the expected value of \( \xi_{AB} \) is \( 1 / (2n - 3) \), and the expected value of the random variable \( M \) is the sum of these expectations over all pairs, that is, \( \frac{n(n - 1)}{2(2n - 3)} \).

All comparisons in this solution are made modulo \( p \). If \( a \) and \( b \) are integers, with \( b \) not divisible by \( p \), then \( a / b \) denotes the unique remainder \( c \) modulo \( p \) for which \( a \equiv bc \).

Let the numbers \( 1^{0}, 2^{1}, 3^{2}, \ldots, k^{k - 1} \) yield exactly \( d \) distinct remainders when divided by \( p \). Let \( t = [p / 4] \). Then expressions of the form \( f(a) = (2a)^{2a - 1} / \left(a^{a - 1}\right)^{2} = 2^{2a - 1}a \) for \( 1 \leqslant a \leqslant t \) yield at most \( d^{2} \) distinct remainders.

We call a pair of natural numbers \( (a, b) \) such that \( 1 \leqslant a < b \leqslant t \) exceptional if \( f(a) \equiv f(b) \). We will show that for each \( \delta = 1, 2, \ldots, t - 1 \), there is at most one exceptional pair \( (a, b) \) where \( b - a = \delta \). Indeed, if \( (a, b) \) is such a pair, then from \( 2^{2a - 1}a = 2^{2b - 1}b \) it follows that \( a / b \equiv 2^{2\delta} \), hence \( b = a + \delta \equiv 2^{2\delta}b + \delta \), or \( b(1 - 2^{2\delta}) \equiv \delta \). Such a remainder \( b \) is at most unique (since \( \delta \neq 0 \)), and \( a \) can be recovered from it.

Thus, there are at most \( t - 1 \) exceptional pairs; let their number be \( S \). Let the numbers \( f(1), f(2), \ldots, f(t) \) yield exactly \( \ell \) distinct remainders modulo \( p \), occurring \( a_{1}, a_{2}, \ldots, a_{\ell} \) times respectively. Then \( \sum_{i = 1}^{\ell} a_{i} = t \) and \( S = \sum_{i = 1}^{\ell} C_{a_{i}}^{2} \). The following chain of inequalities holds:

\[ t - 1 \geqslant S = \frac{1}{2}\left(\sum_{i = 1}^{\ell} a_{i}^{2} - \sum_{i = 1}^{\ell} a_{i}\right) \geqslant \frac{t^{2} / \ell - t}{2}\]

from which \( \ell \geqslant t^{2} / (3t - 2) > t / 3 \). Recalling that \( \ell \leqslant d^{2} \), we obtain the estimate \( d > \sqrt{t / 3} \geqslant [\sqrt{p / 12}] \).

Thus, as the desired constant \( c \), one can take, for example, the number \( 1 / 24 \): for primes \( p < 12 \), the inequality \( d > \frac{\sqrt{p}}{24} \) is trivial, and for \( p > 12 \), it follows from the inequality \( [x] \geqslant x / 2 \) for \( x > 1 \).

We will expand the set of situations in which Taya wins (i.e., can obtain \( n \) equal numbers).

(1) Suppose we have \( n \) odd numbers. Then in one operation, we can obtain \( n \) twos.

(2) Suppose no sum of two neighboring numbers is a power of two. Then in one operation, we can obtain situation (1).

(3) Suppose the arithmetic mean \( s \) of all numbers is not a power of two. We will show that we can reach situation (2). We use the following lemma, the proof of which is given at the end of the solution.

Lemma. Let \( a_{1}, a_{2}, \ldots, a_{n} \) be real numbers, \( s \) their arithmetic mean. In one move, we change the set \( a_{1}, a_{2}, \ldots, a_{n} \) to \( \frac{a_{1} + a_{2}}{2}, \frac{a_{2} + a_{3}}{2}, \ldots, \frac{a_{n} + a_{1}}{2} \). Then for any \( \varepsilon > 0 \), after several moves, all numbers will lie in the interval \( (s - \varepsilon, s + \varepsilon) \).

Clearly, \( s > 1 \). Choose \( \varepsilon > 0 \) such that the interval \( (s - \varepsilon, s + \varepsilon) \) is entirely between consecutive powers of two: \( 2^{t - 1} < s - \varepsilon < s + \varepsilon < 2^{t} \) for some natural \( t \). We will perform the operation of replacing a pair of neighbors with their sum many times. Then, according to the lemma, there will be an \( N \) such that after \( N \) operations, all numbers will lie in the interval \( \left(2^{N}(s - \varepsilon), 2^{N}(s + \varepsilon)\right) \), and hence in the interval between consecutive powers of two \( 2^{N} \cdot 2^{t - 1} \) and \( 2^{N} \cdot 2^{t} \). Therefore, after \( N - 1 \) operations, condition (2) was satisfied.

(4) Suppose all numbers are at least 2. If we are not in situation (2), there is a pair of neighbors \( a, b \) whose sum is \( 2^{t} \), where \( t \geqslant 2 \) is natural. We will try to perform the following operation arbitrarily, only replacing \( a \) and \( b \) with the number 2. Suppose in such an attempt we did not reach situation (3), i.e., we obtained a situation where the arithmetic mean \( s \) is a power of two. Then we will make another attempt, in which all pairs are changed the same way, only \( a \) and \( b \) are replaced with 4. Compared to the first attempt, \( s \) increased by \( \frac{2}{n} \), so we will find ourselves in situation (3).

(5) Let the initial set of numbers be arbitrary. Then after one operation, we have situation (4).

Proof of the lemma. We will make re-designations, let \( s + x_{0}, s + x_{1}, \ldots, s + x_{n} \) be the given numbers, so that \( x_{0} + \ldots + x_{n} = 0 \). Let \( M = \max \left\{\left|x_{0}\right|, \ldots,\left|x_{n}\right|\right\} \). Clearly, after a move, \( M \) does not increase. It is enough to understand that after some number of \( k \) moves, this maximum deviation will become no more than \( \lambda M \) for some fixed \( 0 < \lambda < 1 \). Below we will see that we can set \( k = n \) and \( \lambda = \frac{2^{n}-n - 1}{2^{n}} \).

After \( n \) moves, we will have the set \( s + y_{0}, s + y_{1}, \ldots, s + y_{n} \), where \( y_{0} = \frac{1}{2^{n}}\left(x_{0} + C_{n}^{1} x_{1} + C_{n}^{2} x_{2} + \ldots + C_{n}^{n - 1} x_{n - 1} + x_{n}\right) \) and so on.

Since \( x_{0} + x_{1} + \ldots + x_{n} = 0 \), we have \( x_{0} + C_{n}^{1} x_{1} + C_{n}^{2} x_{2} + \ldots + + C_{n}^{n - 1} x_{n - 1} + x_{n} = \left(C_{n}^{1}-1\right) x_{1} + \left(C_{n}^{2}-1\right) x_{2} + \ldots + \left(C_{n}^{n - 1}-1\right) x_{n - 1} \). Hence \( \left|y_{0}\right| \leqslant \frac{1}{2^{n}}\left(\left(C_{n}^{1}-1\right) + \ldots + \left(C_{n}^{n - 1}-1\right)\right) M = \frac{2^{n}-n - 1}{2^{n}} M \). Similarly, all \( \left|y_{i}\right| \leqslant \frac{2^{n}-n - 1}{2^{n}} M \).