mn squares of equal size are arranged...

MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 1982Moderate

$mn$ squares of equal size are arranged to from a rectangle of dimension $m$ by $n$ , where $m$ and $n$ are natural numbers. Two squares will be called 'neighbours' if they have exactly one common side. A natural number is written in each square such that the number written in any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the numbers used are equal.

Visualized Solution (Hindi)

Visualizing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m \times n Grid

Consider an $m \times n$ grid of squares.
Each square $(i, j)$ contains a natural number $x_{i, j}$ .
Two squares are neighbors if they share a common side.

The Arithmetic Mean Condition

For any square $x_{i, j}$ , let its $k$ neighbors be $a_{1}, a_{2}, \dots, a_{k}$ .
The condition states: $x_{i, j} = \frac{a _{1} + a _{2} + \dots + a _{k}}{k}$ .
Note: $k$ can be 2, 3, or 4 depending on the position.

Introducing the Maximum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Let $S$ be the set of all numbers in the grid.
Since the set is finite, there exists a maximum value $M = max (S)$ .
Let a square with value $M$ have neighbors $a_{1}, a_{2}, \dots, a_{k}$ .
By definition of maximum, $a_{i} \leq M$ for all $i$ .

The Rigidity of the Average

We have $M = \frac{a _{1} + a _{2} + \dots + a _{k}}{k}$ .
This implies $k M = a_{1} + a_{2} + \dots + a_{k}$ .
Since $a_{i} \leq M$ , if any $a_{i} < M$ , then $\sum a_{i} < k M$ .
To avoid contradiction, we must have $a_{1} = a_{2} = \dots = a_{k} = M$ .

Propagation Across the Grid

By the same logic, the neighbors of $a_{i}$ must also be $M$ .
This property propagates to every square in the connected grid.
Therefore, $x_{i, j} = M$ for all $i, j$ .

Conclusion and Takeaway

Key Takeaway: The Maximum Principle ensures that in an AM-balanced system, the only steady state is uniformity.
Next Challenge: What if the numbers could be negative? Does the existence of a maximum still hold for an infinite grid?

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Visualized Solution (Hindi)

Visualizing the $m \times n$ Grid

The Arithmetic Mean Condition

Introducing the Maximum Value $M$

The Rigidity of the Average

Propagation Across the Grid

Conclusion and Takeaway

Conceptually Similar Problems

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Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

If $p$ be a natural number then prove that $p^{n + 1} + (p + 1)^{2 n - 1}$ is divisible by $p^{2} + p + 1$ for every positive integer $n$ .

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Visualized Solution (Hindi)

Visualizing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m×n Grid

The Arithmetic Mean Condition

Introducing the Maximum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

The Rigidity of the Average

Propagation Across the Grid

Conclusion and Takeaway

Conceptually Similar Problems

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A1​,A2​,…,An​ be the vertices of an n-sided regular polygon such that A1​A2​1​=A1​A3​1​+A1​A4​1​. Find the value of n.

Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p be a natural number then prove that pn+1+(p+1)2n−1 is divisible by p2+p+1 for every positive integer n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If mi​,mi​1​,mi​>0,i=1,2,3,4 are four distinct points on a circle, then show that m1​m2​m3​m4​=1.

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A1​,A2​,…,An​ be the vertices of an n-sided regular polygon such that A1​A2​1​=A1​A3​1​+A1​A4​1​. Find the value of n.

Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p be a natural number then prove that pn+1+(p+1)2n−1 is divisible by p2+p+1 for every positive integer n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If mi​,mi​1​,mi​>0,i=1,2,3,4 are four distinct points on a circle, then show that m1​m2​m3​m4​=1.

Visualizing the $m \times n$ Grid

Introducing the Maximum Value $M$

Let $A_{1}, A_{2}, \dots, A_{n}$ be the vertices of an $n$ -sided regular polygon such that $\frac{1}{A _{1} A _{2}} = \frac{1}{A _{1} A _{3}} + \frac{1}{A _{1} A _{4}}$ . Find the value of $n$ .

If $p$ be a natural number then prove that $p^{n + 1} + (p + 1)^{2 n - 1}$ is divisible by $p^{2} + p + 1$ for every positive integer $n$ .

If $m_{i}, \frac{1}{m _{i}}, m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .

Let $A_{1}, A_{2}, \dots, A_{n}$ be the vertices of an $n$ -sided regular polygon such that $\frac{1}{A _{1} A _{2}} = \frac{1}{A _{1} A _{3}} + \frac{1}{A _{1} A _{4}}$ . Find the value of $n$ .

If $p$ be a natural number then prove that $p^{n + 1} + (p + 1)^{2 n - 1}$ is divisible by $p^{2} + p + 1$ for every positive integer $n$ .

If $m_{i}, \frac{1}{m _{i}}, m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .