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Problem of the Week
Problem A and Solution


Laila starts with a square piece of paper. Starting at one corner and moving around the square, she labels the corners \(A\), \(B\), \(C\), and \(D\).

Laila folds the paper in half, by folding side \(AB\) onto side \(DC\), to form a rectangle. She opens up the paper and folds it again to form another rectangle by folding side \(AD\) onto side \(BC\). When she opens up the paper this time, she sees two creases in the paper as shown below.

A horizontal dashed line and a vertical dashed line divide the square into four identical smaller squares.

The centre of the square is the point where the two creases intersect. Now, she takes each corner of the square and folds the paper so that each corner touches the centre of the square. Folding all four corners in this way forms another smaller square made up of four triangular regions as shown below.

A square with dashed lines for its sides and that looks like a kite. Its diagonals are horizontal and vertical lines that divide the square into four identical triangles that meet at the centre.

What fraction of the area of the original square is the area of this smaller square? Justify your answer.


The smaller square has an area that is \(\frac{1}{2}\) the area of the original square. A justification of this is given below.

Consider the following image of the smaller square with only one of the four triangular regions shaded.

Underneath the shaded triangle is a region of the original square that is exactly the same size. That is true for all four of the triangles that were formed by having the corners meet at the centre of the original square.

Let’s now shade all four of the triangular regions in the smaller square, and then open up the paper again.

Square ABCD with a smaller square in its interior. The four vertices of the smaller square are at the midpoints of the four sides of square ABCD. The diagonals of the smaller square divide the smaller square into four identical unshaded triangles. Four triangles in the corners of ABCD lie outside the smaller square and are shaded.

We can make the following observations:

It follows that the area of the smaller square is \(\frac{1}{2}\) the area of the original square.