Let \(n > 1\) be an integer. In a configuration of an \(n \times n\) board, each of the \(n^{2}\) cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate \(90^{\circ}\) counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of \(n\) , the maximum number of good cells over all possible starting configurations.

Solution:
We will show that the maximum number of good cells over all possible starting configurations is

\[\frac{n^{2}}{4} \quad \text{if} n \text{ is even and}\] \[0 \quad \text{if} n \text{ is odd.}\]

## Odd \(n\)

First, we will prove that there are no good cells if \(n\) is an odd number.

For Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after \(n^{2} \equiv 1 \mod 2\) moves. Thus, it is impossible for Turbo to come back to her initial black cell on her \(n^{2}\) - th move, which is a contradiction. Thus there are no good cells.

## Lower bound for even \(n\)

We will now construct a starting configuration with \(\frac{n^{2}}{4}\) good cells for even \(n\) .

Let \((i,j)\) denote the cell in row \(i\) and column \(j\) . Consider the following cycle

\[\begin{array}{r l} & {(1,1)\rightarrow (1,2)\rightarrow (1,3)\rightarrow \ldots \rightarrow (1,n)}\\ & {\qquad \rightarrow (2,n)\rightarrow (2,n - 1)\rightarrow \ldots \rightarrow (2,2)}\\ & {\qquad \ldots}\\ & {\qquad \rightarrow (2i - 1,2)\rightarrow (2i - 1,3)\rightarrow \ldots \rightarrow (2i - 1,n)}\\ & {\qquad \rightarrow (2i,n)\rightarrow (2i,n - 1)\rightarrow \ldots \rightarrow (2i,2)}\\ & {\qquad \ldots}\\ & {\qquad \rightarrow (n,n)\rightarrow (n,n - 1)\rightarrow \ldots \rightarrow (n,2)}\\ & {\qquad \rightarrow (n,1)\rightarrow (n - 1,1)\rightarrow \ldots \rightarrow (2,1)\rightarrow (1,1).} \end{array} \quad (1,1)\]

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)

Note that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \((1,1)\) is good, we need to find a starting configuration such that Turbo traverses this cycle.

Let \(c_{i}\) be the \((i - 1)\) - th cell on the cycle: so we have \(c_{0} = (1,1)\) , \(c_{2} = (1,2)\) , ..., \(c_{n^{2} - 1} = (2,1)\) . For every \(i\) , we draw an arrow in cell \(c_{i}\) pointing towards cell \(c_{i + 1}\) (or pointing towards \(c_{0}\) if \(i = n^{2} - 1\) ) and then rotate this arrow \(i\) times \(90^{\circ}\) in the clockwise direction. After \(i\) moves, the arrow in \(c_{i}\) will have rotated \(i\) times \(90^{\circ}\) counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle \(c_{0}, c_{1}, c_{2}, \ldots , c_{n^{2} - 1}, c_{0}\) and \((1,1)\) is good.

Every four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since \(n^{2}\) , the length of the cycle, is divisible by 4. Therefore Turbo can also start at any \(c_{i}\) with 4 \(| i\) and follow the same route. Hence the cells \(c_{0}, c_{4}, c_{8}, \ldots , c_{n^{2} - 4}\) are good and there are \(\frac{n^{2}}{4}\) of such cells.

## Upper bound for even \(n\)

We will prove that for even \(n\) and any start configuration there are at most \(\frac{n^{2}}{4}\) good cells.

Let \(a_{0}\) be a good cell. Let \(a_{0},a_{1},a_{2},\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}\) be the sequence of cells that Turbo visits when she starts at \(a_{0}\) . Now suppose there is another good cell \(b_{0}\) and let \(b_{0},b_{1},b_{2},\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}\) be the sequence of cells that Turbo visits when she starts at \(b_{0}\) .

Note that, since \(4\mid n^{2}\) , the arrows are back to their initial configuration after \(n^{2}\) steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again.

Consider the upper left corner of the board. With standard row and column numbering, the corner cell is \((1,1)\) . This cell has only two neighbours, so both the \(a\) - route and the \(b\) - route must have cells \((2,1),(1,1),(1,2)\) in that order or \((1,2),(1,1),(2,1)\) in that order. Without loss of generality, \(a_{i - 1} =\) \((2,1)\) , \(a_{i} = (1,1)\) and \(a_{i + 1} = (1,2)\) for some \(i\) . Let \(j\) be such that \(b_{j} = (1,1)\) . If \(b_{j - 1} = (2,1) = a_{i - 1}\) then the arrow in cell \((2,1)\) must be pointed in the same direction after \(i - 1\) steps and after \(j - 1\) steps, so \(i\equiv j\) mod 4. But then the arrow in cell \(b_{j} = (1,1) = a_{i}\) must also be pointed in the same direction after \(i\) and after \(j\) steps, so Turbo moves to \(b_{j + 1} = a_{i + 1}\) in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the \(b\) - route is actually identical to \(a_{4t},a_{4t + 1},\ldots ,a_{n^{2}} = a_{0},a_{1},\ldots ,a_{4t - 1},a_{4t}\) for some \(t\) , as any other starting point would have the arrows in the wrong direction initially.

Now suppose instead that \(b_{j + 1} = (2,1) = a_{i - 1}\) . Considering the \(a\) - route, the arrows in the upper left corner after \(i - 1\) steps must be like this:

![](data:image/jpeg;base64,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)

Considering the \(b\) - route instead, the arrows after \(j - 1\) steps must be like this:

![](data:image/jpeg;base64,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)

From the arrows in cell \((1,1)\) we see that \(i\equiv j + 1\) mod 4. However, for the cells \((2,1)\) and \((1,2)\) this gives a contradiction.

We conclude that the only possible good cells are \(a_{4t}\) for \(t = 0,1,\ldots ,\frac{n^{2}}{4} - 1\) , which gives at most \(\frac{n^{2}}{4}\) good cells.


In each cell of a \(2025 \times 2025\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \(r_{i}\) to be the largest value in row \(i\) , and let \(R = r_{1} + r_{2} + \dots + r_{2025}\) . Similarly, define \(c_{i}\) to be the largest value in column \(i\) , and let \(C = c_{1} + c_{2} + \dots + c_{2025}\) .What is the largest possible value of \(\frac{R}{C}\) ?

What is the largest possible value of \(\frac{R}{C}\) ?

Solution:
Answer: \(\frac{2025}{89}\) .

In general, if the table is \(m^{2} \times m^{2}\) , the answer is \(\frac{m^{2}}{2m - 1}\) .

The example is as follows: label rows and columns from 1 to \(m^{2}\) , from top to bottom and left to right. For the first \(m\) columns, write \(\frac{1}{m}\) in all squares whose coordinates have the same residue modulo \(m\) and place 0 everywhere else. For the remaining \(m^{2} - m\) columns, place \(\frac{1}{m^{2}}\) everywhere. Then \(R = m^{2} \cdot \frac{1}{m} = m\) , and \(C = m \cdot \frac{1}{m} + (m^{2} - m) \cdot \frac{1}{m^{2}} = 2 - \frac{1}{m}\) . So the ratio is as claimed.

1/2	0	1/4	1/4

0	1/2	1/4	1/4

1/2	0	1/4	1/4

0	1/2	1/4	1/4

In particular, when \(n := m^{2} = 2025\) , we get \(\frac{2025}{89}\) . Now we need to show that \(\frac{R}{C} \leq \frac{n}{2\sqrt{n} - 1}\) .

For each row, select one cell having the largest value appearing in said row and colour it red. Then, without loss of generality, we may rearrange the columns such that the red cells appear in the first \(k\) columns from the left, and each column contains at least one red cell, for some \(k \leq n\) .

For the \(j^{\mathrm{th}}\) column, for all \(1 \leq j \leq k\) , let \(p_{j}\) and \(n_{j}\) denote the sum and number of red cells in it, respectively. We observe that \(c_{j}\) , the biggest number in the \(j^{\mathrm{th}}\) column, is at least \(\frac{p_{j}}{n_{j}}\) , for all \(1 \leq j \leq k\) . For all other columns, the largest value they contain is at least \(\frac{1}{n}\) , as their sum is 1. Thus, \(C \geq \frac{p_{1}}{n_{1}} + \frac{p_{2}}{n_{2}} + \dots + \frac{p_{k}}{n_{k}} + \frac{n - k}{n}\) .

We can also observe that \(R = p_{1} + p_{2} + \dots + p_{k}\) .

Therefore we have to show that:

\[p_{1} + p_{2} + \dots +p_{k} \leq \frac{n}{2\sqrt{n} - 1} \cdot \left(\frac{p_{1}}{n_{1}} + \frac{p_{2}}{n_{2}} + \dots + \frac{p_{k}}{n_{k}} + \frac{n - k}{n}\right). \quad (*)\]

By construction, \(n_{1} + n_{2} + \dots + n_{k} = n\) , and, as the numbers in every column are nonnegative, we see that \(p_{j} \leq 1\) for every \(j\) . Also, since each number in a red cell is at least \(\frac{1}{n}\) , we also have \(p_{j} \geq \frac{n_{j}}{n}\) .

Since our inequality is linear in each \(p_{j}\) , it suffices to prove it when each variable equals one of its two critical values. By relabeling, we may assume that \(p_{j} = \frac{n_{j}}{n}\) for \(1 \leq j \leq t\) , and \(p_{j} = 1\) for \(t + 1 \leq j \leq k\) , for an integer \(0 \leq t \leq k\) .

First, if \(t = k\) , we observe that \(p_{1} + p_{2} + \dots +p_{k} = \frac{n_{1}}{n} + \frac{n_{2}}{n} + \dots + \frac{n_{k}}{n} = 1\) , and that \(\frac{p_{1}}{n_{1}} + \frac{p_{2}}{n_{2}} + \dots + \frac{p_{k}}{n_{k}} = \frac{k}{n}\) , so the inequality becomes \(1 \leq \frac{n}{2\sqrt{n} - 1}\) , which is true.

From now on we may assume that \(t < k\) . We need to show that:

\[\frac{n_{1} + \dots + n_{t}}{n} +k - t \leq \frac{n}{2\sqrt{n} - 1} \cdot \left(\frac{t}{n} + \frac{1}{n_{t + 1}} + \dots + \frac{1}{n_{k}} + \frac{n - k}{n}\right).\]

By Cauchy- Schwarz inequality we have that:

\[\frac{1}{n_{t + 1}} + \dots + \frac{1}{n_{k}} \geq \frac{(k - t)^{2}}{n_{t + 1} + \dots + n_{k}} = \frac{(k - t)^{2}}{n - (n_{1} + \dots + n_{t})}. \quad (CS)\]

Let \(n_{1} + \dots +n_{t} = n\cdot q\) , where \(0\leq q< 1\) . Thus, it is now enough to show that:

\[q + k - t\leq \frac{n}{2\sqrt{n - 1}}\left(\frac{t}{n} +\frac{(k - t)^{2}}{n - nq} +\frac{n - k}{n}\right).\]

Let \(k - t = \ell \geq 1\) . The inequality becomes:

\[q + \ell \leq \frac{1}{2\sqrt{n} - 1}\cdot \left(n - \ell +\frac{\ell^{2}}{1 - q}\right).\]

Rearranging this we get:

\[n + q + \frac{\ell^{2}}{1 - q}\geq 2(q + \ell)\sqrt{n}.\]

If \(q = 0\) the inequality is trivially true by \(A M - G M\) . Suppose now that \(0< q< 1\) . Then, by Cauchy- Schwarz, we have:

\[q + \frac{\ell^{2}}{1 - q} = \frac{q^{2}}{q} +\frac{\ell^{2}}{1 - q}\geq (q + \ell)^{2}.\]

It is therefore enough to show that \(n + (q + \ell)^{2}\geq 2(q + \ell)\sqrt{n}\) , which is true by \(A M - G M\) , completing the proof.

Remark. Another way to show \(n + q + \frac{\ell^{2}}{1 - q}\geq 2(q + \ell)\sqrt{n}\) is to split it into two \(A M - G M\) 's, as:

\[\left(n(1 - q) + \frac{\ell^{2}}{1 - q}\right) + (nq + q)\geq 2\ell \sqrt{n} +2q\sqrt{n}.\]


Solution:
We prove the main inequality \((\ast)\) in a slightly different manner. Instead of the strong lower bound \(p_{j}\geq \frac{n_{j}}{n}\) , we use the weaker, simpler and more immediate lower bound \(p_{j}\geq 0\) (thus proving the inequality in a larger regime).

As in Solution 1, suppose \(p_{j} = 0\) for \(1\leq j\leq t\) and \(p_{j} = 1\) for \(t + 1\leq j\leq k\) , with \(\ell = k - t\) . We also denote by \(m = n_{t + 1} + \dots +n_{k}\) and note that \(m\leq n - t\) , since \(n_{j}\geq 1\) for each \(i\leq t\) . We need to prove that:

\[\ell \leq \frac{n}{2\sqrt{n} - 1}\cdot \left(\frac{1}{n_{t + 1}} +\dots +\frac{1}{n_{k}} +\frac{n - k}{n}\right).\]

Rearranging and using the same Cauchy- Schwarz (CS) as in Solution 1, we see it suffices to show that:

\[(2\sqrt{n} -1)\ell \leq n - k + \frac{n\ell^{2}}{m},\]

or equivalently, that:

\[2\sqrt{n}\ell \leq n - t + \frac{n\ell^{2}}{m}.\]

But since \(n - t\geq m\) this immediately follows from \(2\sqrt{n}\ell \leq m + \frac{n\ell^{2}}{m}\) , which is a simple application of \(A M - G M\) .

(Note that in Solution 1 the case \(t = k\) , which is equivalent to \(m = \ell = 0\) , was dealt with separately, to avoid the appearance of \(\frac{0}{0}\) terms such as \(\frac{n\ell^{2}}{m}\) . It is easy to verify that the corresponding term should in fact be 0 and the final \(A M - G M\) replaced with \(0\leq 0\) , and all transitions are valid. Alternatively, the case can be argued directly by simply noting that \((\ast)\) evaluates to \(0\leq \frac{n - k}{2\sqrt{n} - 1}\) , which is obvious.)


Solution:
1". This is an alternative way of getting the upper bound on \(\frac{R}{C}\) from

\[\frac{R}{C}\leq \frac{p_{1} + p_{2} + \ldots + p_{k}}{\frac{p_{1}}{n_{1}} + \frac{p_{2}}{n_{2}} + \ldots + \frac{p_{k}}{n_{k}} + \frac{n - k}{n}}.\]

Using the fact that \(\sum_{j = 1}^{k}n_{j} = n\) , we can rewrite the above right hand side as follows:

\[\frac{\sum_{j = 1}^{k}p_{j}}{\sum_{j = 1}^{k}\left(\frac{p_{j}}{n_{j}} +\frac{n_{j} - 1}{n}\right)}.\]

We notice that this is a quotient of affine functions in the \(p_{j}\) 's, for which the denominator does not vanish over the set defined by \(0 \leq p_{j} \leq 1\) . Therefore the maximum of this function is attained when a certain number of \(p_{j}\) 's are 1 and the others are 0. Without loss of generality we may assume that the first \(t\) are equal to 1 and the other \(k - t\) are 0 for some \(0 \leq t \leq k\) . Then one has that the previous expression is at most

\[\frac{t}{\sum_{1 \leq j \leq t} \left(\frac{1}{n_{j}} + \frac{n_{j} - 1}{n}\right) + \sum_{t < j \leq k} \frac{n_{j} - 1}{n}}.\]

We now lower bound the denominator by observing that the second sum is non negative, while each term of the first sum can be bounded by \(AM - GM\) as follows:

\[\frac{1}{n_{j}} + \frac{n_{j} - 1}{n} \geq \frac{2}{\sqrt{n}} - \frac{1}{n}.\]

We therefore have

\[\frac{R}{C} \leq \max_{1 \leq t \leq k} \frac{t}{\sum_{1 \leq j \leq t} \left(\frac{2}{\sqrt{n}} - \frac{1}{n}\right)} = \frac{n}{2\sqrt{n} - 1},\]

which finishes the proof.


Solution:
\(1^{\prime \prime}\) . We follow the same notation as above. First, we apply Cauchy- Schwarz as follows :

\[\left(\sum_{i = 1}^{k} \frac{p_{i}}{n_{i}}\right) \left(\sum_{i = 1}^{k} p_{i} n_{i}\right) \geq \left(\sum_{i = 1}^{k} p_{i}\right)^{2} = R^{2}.\]

We now write \(z_{i} = 1 - p_{i}\) for all \(1 \leq i \leq k\) , and observe that all \(z_{i}\) are positive. Moreover, we have that \(\sum_{i = 1}^{k} z_{i} = k - R\) , and \(\sum_{i = 1}^{k} p_{i} n_{i} = n - \sum_{i = 1}^{k} p_{i} z_{i}\) . Thus, from our last inequality we get

\[\frac{p_{1}}{n_{1}} + \dots + \frac{p_{k}}{n_{k}} \geq \frac{R^{2}}{n - \sum_{i = 1}^{k} n_{i} z_{i}}.\]

As before we have \(C \geq \frac{p_{1}}{n_{1}} + \dots + \frac{p_{k}}{n_{k}} + \frac{n - k}{n}\) , and so

\[C \geq \frac{R^{2}}{n - \sum_{i = 1}^{k} n_{i} z_{i}} + \frac{n - k}{n} = \frac{R^{2}}{n - \sum_{i = 1}^{k} n_{i} z_{i}} + \frac{n - R - \sum_{i = 1}^{k} z_{i}}{n}.\]

Putting everything together we get

\[\frac{R}{C} \leq \frac{R}{\frac{R^{2}}{n - \sum_{i = 1}^{k} n_{i} z_{i}} + \frac{n - R - \sum_{i = 1}^{k} z_{i}}{n}} = \frac{n}{\frac{R n}{n - \sum_{i = 1}^{k} n_{i} z_{i}} + \frac{n - \sum_{i = 1}^{k} z_{i}}{R} - 1}.\]

Applying \(AM - GM\) to the denominator we get \(\frac{R}{C} \leq \frac{n}{2 \sqrt{n \frac{n - \sum_{i = 1}^{k} z_{i}}{n - \sum_{i = 1}^{k} n_{i} z_{i}} - 1}}\) , which finishes the proof by noting that \(\frac{n - \sum_{i = 1}^{k} z_{i}}{n - \sum_{i = 1}^{k} n_{i} z_{i}} \geq 1\) .


Solution:
This is an alternative approach that goes via an intermediary quantity in order to establish to the upper bound on \(R / C\) .

Let \(x_{i j}\) be the entry in row \(i\) and column \(j\) . Let \(n = 2025\) . The key idea is to analyze the expression:

\[T:= \sum_{i,j}x_{i j}\max \left(x_{i j},\frac{1}{n}\right).\]

On one hand, since \(c_{j}\geq 1 / n\) , we have \(x_{i j}\max (x_{i j},\frac{1}{n})\leq x_{i j}c_{j}\) for every \((i,j)\) . So

\[T\leq \sum_{j}\sum_{i}x_{i j}c_{j} = \sum_{j}c_{j} = C.\]

On the other hand, let \(j_{i}\) be one of the indices for which \(r_{i} = x_{i j_{i}}\) . We therefore have:

\[T=\sum_{i}\left(x_{i j_{i}}\max \left(x_{i j_{i}},\frac{1}{n}\right)+\sum_{j\neq j_{i}}x_{i j}\max \left(x_{i j},\frac{1}{n}\right)\right)\] \[\geq\sum_{i}\left(r_{i}^{2}+\frac{1}{n}\sum_{j\neq j_{i}}x_{i j}\right)\] \[=\sum_{i}\left(r_{i}^{2}-\frac{1}{n}r_{i}+\frac{1}{n}\right)\geq\sum_{i}\left(\frac{2}{\sqrt{n}}r_{i}-\frac{1}{n}r_{i}\right)=\left(\frac{2}{\sqrt{n}}-\frac{1}{n}\right)R.\]

This gives the claimed result.