5 matchsticks, placed horizontally and vertically, divide the square into 2 equal parts as shown above. The 2 parts are equal in shape and size. Another obvious solution would be 1 vertical and 4 horizontal matchsticks based on the same principle.
But there is another clever solution! In what other way can you place the 5 matchsticks inside the square to divide it exactly into 2 parts equal in shape and size?
AC is exactly the length of 3 matchsticks and BC exactly 4 matchsticks. Applying the Pythagoras's theorem you can proof that AB is exactly the length of 5 matchsticks:
(AB)² = (BC)² + (AC)²
(AB)² = 16 + 9
(AB)² = √25
AB = 5
Rectangle A is one third of the size of rectangle B with 6 and 14 matchsticks used in A and B respectively.
Transfer 1 matchstick from B to A which means you now have 7 matchsticks in A and 13 matchsticks in B. Create 2 new shapes, not necessarily rectangles, with shape A still one third of the size of shape B.
You can place the broken matchstick with head on any of the open places as long as it does not form a row (horizontal, vertical or diagonal).
Below is one example: Place your broken matchstick in the middle top row as shown above. The other player can now only play position 3. (1 or 2 would make him/her a loser). Thereafter your move could be 1 or 2.
The matchsticks cover an area of 3 square matchsticks. Unfortunately 2 of the matchsticks have not been used. Move 4 matchsticks to still cover an area of 3 square matchsticks, but now all the matchsticks must be used.