Question

A square of n units by n units is divided into \[{n^2}\] squares each of area 1 sq. unit. Let \[\dfrac{{{n^2}\left( {n + k} \right)}}{m}\]be the number of ways in which 4 points (out of \[{\left( {n + 1} \right)^2}\]vertices of the squares) can be chosen so that they form the vertices of a square. Find \[k + m\]?

Answer 1

 Hint: No. of squares in a grid of \[n \times n\] is given by = \[{1^2}\;\;\; + {2^2} + {3^2} + ...... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
For example if \[n = 4\]
No. of squares of 1 sq. unit \[ = {4^2} = 16\]

No. of squares of 2 sq. unit \[ = {3^2} = 9\]
No. of squares of 3 sq. unit \[ = {2^2} = 4\]
No. of squares of 4 sq. unit \[ = {1^2} = 1\]
∴ Total no. of squares \[ = 1 + 4 + 9 + 16 = 30\

Complete step-by-step answer:
Given number of ways in which 4 points (out of \[{\left( {n + 1} \right)^2}\]vertices of the squares) can be chosen so that they form the vertices of a square = \[\dfrac{{{n^2}\left( {n + k} \right)}}{m}\] Eqn (i)
Now we know that total no. of squares in a grid =\[{1^2}\;\;\; + {2^2} + {3^2} + ...... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
Let us take \[N1 = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
Now suppose n is even,
And if we select 4 points from the \[n \times n\]grid
Then,
No. of squares of area \[\;\dfrac{{{n^2}}}{2}\] square units \[ = {1^2}\]
No. of squares of area \[\dfrac{{{{(n - 2)}^2}}}{2}\;\] square units \[ = {3^2}\]


No. of squares of area \[\dfrac{{{2^2}}}{2}\;\] square units \[ = {(n - 1)^2}\]
Then total no. of squares that can be made (apart from N1) if we select 4 points will be:
\[N2 = {1^2} + {3^2} + {5^2} + ... + {(n - 1)^2} = \;\dfrac{{n(n - 1)(n + 1)}}{6}\]




Similarly, if n is odd then,
No. of squares of area \[\;\dfrac{{{{(n - 1)}^2}}}{2}\] square units \[ = {2^2}\]
No. of squares of area \[\dfrac{{{{(n - 3)}^2}}}{2}\;\] square units \[ = {4^2}\]
.
.
.
No. of squares of area \[\dfrac{{{2^2}}}{2}\;\] square units \[ = {(n - 1)^2}\]
Then again total no. of squares that can be made (apart from N1) if we select 4 points will be:
\[N2 = {2^2} + {4^2} + {6^2} + ... + {(n - 1)^2} = \;\dfrac{{n(n - 1)(n + 1)}}{6}\]




∴Value of N2 doesn’t change whether n is even or odd,
Hence
\[\begin{gathered}
  Total{\text{ }}no.{\text{ }}of{\text{ }}squares{\text{ }}formed{\text{ }}which{\text{ }}can{\text{ }}be{\text{ }}obtained{\text{ }}by{\text{ }}taking\;4\;points{\text{ }}out{\text{ }}of\;{(n + 1)^2}\;points\; = N1 + N2 = \\
   \Rightarrow \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n - 1)(n + 1)}}{6}\; = \;\dfrac{{{n^2}(n + 1)}}{2} \\
\end{gathered} \]
Now equation above with eqn (i)
We get \[k = 1\] and \[m = 2\]
Hence \[k + m = 3\], which is the required answer.
Note: In this question, we are selecting only 4 points. More no. of squares can be formed due to intersection of lines if we select 3 points for example,