– puzzle ©Alex Bellos (The Guardian)
Here are the questions each followed by two methods of solution. Sometimes the simple action of drawing in a few extra lines reveals the solution clearly.
1. A point inside a square is connected to its four vertices. What fraction of the square is shaded?

Solution: 1/2

By segmenting the square like this, you can see that half of each segment is shaded.

Pass a line through the point that is parallel to the base. The illustration shows what happens when you shear the top and bottom triangles along the line: the process of shearing, which keeps base and height the same, does not change the area.

2. A shaded region is constructed within an equilateral triangle using the midpoint A of a side, and the midpoint between AB. What fraction is shaded?

Solution: 1/4

Draw the line from the top vertex of the shaded triangle to the top vertex of the equilateral triangle. The largest white triangle has an area that is half the area of the equilateral triangle, since it has the same base but half the height. The remaining two triangles are congruent, so they each have area a quarter of the equilateral triangle.

Draw in a rectangle as shown. The area of the shaded triangle is half the area of that rectangle, since the shaded area outside the rectangle is equal to the area bounded by the dashed line. The area of the rectangle (half of the base x half of the height of the equilateral triangle) is half the area of the equilateral triangle.

3. In this regular hexagon, three diagonals have been drawn to form this parallelogram. What fraction of the hexagon is shaded?

Solution: 1/3

By rearranging the two shaded triangles, yo can see that the parallelogram has the same area as the shaded rectangle. A hexagon is made from six equilateral triangles. In the image, one half of the hexagon has been dissected into six halves of equilateral triangles. The rectangle has the area of four of these halves, in other words it has the area of two equilateral triangles, which is a third of the total hexagon.
The parallelogram has a side measuring half a diagonal of the shaded rhombus, and a height equal to twice the other diagonal of the rhombus. You can also see that both figures have the same area using two shearings. Three of these rhombuses form the hexagon, which gives the answer of 1/3.
4. A square is shown with its inscribed and circumscribed circle. What fraction of the larger circle is shaded?

Solution: 1/2

You needed Pythagoras for this one! If the big circle has radius R, and the small circle radius r , then a right angled triangle is drawn with hypotenuse R and the two other sides r . By Pythagoras, R = r √2, so r( squared ) = R(squared)/2. Since the area of a circle is π times the radius squared, the shaded area must be half of the big circle’s area. (Apologies these captions don’t let me right superscripts for squared).

r ^{2} = R^{2} /2

The area of the grey square can be calculated in two ways. Using the small radius r it is (2 r )(squared) = 4 r (squared). Using the big radius R, which is half a diagonal, it is 2 x R(squared). These terms are equal so again r( squared ) = R(squared)/2.

5. This last puzzle is about angles. Four identically-sized squares are arranged in an L. What’s the angle marked in blue?

Solution: 45 degrees

This for me is the nicest proof: no words needed!
Or you can use trigonometry, by considering arctan 2 and arctan 1/3. This is ugly and also impossible to format properly in a caption so the rest will be left to the reader!