Question

A line of symmetry is defined as a line that divides a figure into two parts in a way such that each part is a mirror image of the other part about that line.
The given figure consists of 16 unit squares arranged as shown. In addition to the three black squares, what is the minimum number of squares that must be coloured black, such that both $PQ$ and $MN$ form lines of symmetry? (The figure is representative)

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

When a figure must be symmetric about two lines, take the “orbit” of each marked cell under both reflections (and their composition). The minimal coloring is the union of these orbits.

Answer 1

The Correct Option is C

Label the $4\times4$ grid cells by coordinates $(x,y)$ with $x=1,2,3,4$ from left to right and $y=1,2,3,4$ from bottom to top.
From the figure, the initially black squares are $(3,4)$ (top row, third column), $(1,3)$ (second row, first column), and $(3,1)$ (bottom row, third column).
Symmetry conditions:
- Reflection about the vertical midline $PQ$ maps $(x,y)\mapsto(5-x,y)$.
- Reflection about the diagonal $MN$ (the line $y=x$) maps $(x,y)\mapsto(y,x)$.
Close the set under both reflections (and hence under their composition).
Start with $S=\{(3,4),(1,3),(3,1)\}$.
Apply $PQ$: $(3,4)\!\to\!(2,4)$, $(1,3)\!\to\!(4,3)$, $(3,1)\!\to\!(2,1)$, so add $\{(2,4),(4,3),(2,1)\}$.
Apply $MN$: $(2,4)\!\to\!(4,2)$ and $(2,1)\!\to\!(1,2)$; the images of $(3,4),(1,3),(3,1),(4,3)$ are already in $S$. Add $\{(4,2),(1,2)\}$.
Applying the reflections again yields no new cells (closure reached).
Final symmetric set:
$\{(3,4),(2,4),(4,3),(1,3),(4,2),(1,2),(3,1),(2,1)\}$ — a total of $8$ squares.
Since $3$ were already black, the minimum additional squares needed $=8-3=5$.
\[ \boxed{5} \]