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## 'Crazy Shading' printed from http://nrich.maths.org/

The middle sized square passes through the centres of the four circles. Each side of the middle sized square together with the edges of the outer square creates a right angled isosceles triangle with angles of 45$^\circ$. Thus the angles these sides make with the inner square are also 45$^\circ$.

Each side of the middle sized square bisects the area of the circle. The inner half of that circle is made up of two shaded segments with angles of 45$^\circ$ which together are equal in area to the unshaded right angled segment.

Thus the total shaded area is exactly equal to the area of the inner square and hence equal to one quarter of the area of the outer square.

*This problem is taken from the UKMT Mathematical Challenges.*

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