Monthly Archives: December 2014

Fortnightly maths prize for 24 February

No prize-winners this time. Though I know (because they told me) that the year 13 Further Maths students worked out answers, none of them wrote down the answers and handed them in, so no prizes.


If all the people of the world stood shoulder-to-shoulder (with the babies in slings, and the old and sick people held up by the people next to them), how much area would we cover? Continue reading

Fortnightly maths prize for 3 February

The problem: prove that the crescent-moon shaped region, upper left, has the same area as the shaded triangle.


Mariama Bah, Shannon Bradley, and Sharif Quansah all handed in correct solutions and won prizes.

Proof: Let AO be 1 unit. Then the area of the triangle is ½ unit, and AB is √2 units, by Pythagoras.

The area of the semicircle ABC is 2π

The area of the quarter-circle AFBO is π

The area of the segment AFBD is π−½

The area of the semicircle AEB is π

The area of the crescent-shape = area of semicircle AEB minus area of segment AFBD
 =area of triangle ∎