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The 'Impossible' GCSE Maths Question
The diagram is made of three circles, each of radius 4cm.
The centre of the circles are A, B and C, such that ABC is a straight line.
Work out the total area of the two shaded regions.
Give your answers in terms of π.
What's the Problem?
The final question of this year's (2022) first GCSE Mathematics paper from Edexcel has been branded as impossible by many students with even some maths teachers struggling to solve it. There are even stories of some extremely able maths students resorting to using A-level techniques such as polar coordinates and integration to solve it.
We're going to have a look at how to solve this particular question using only GCSE maths techniques and no calculator.
A Video Solution to the 'Impossible' Question
The first thing to spot when solving this question is that, due to the intersecting circles, we can add equilateral triangles to our diagram.
Question With Equilateral Triangles Added
We know that these triangles are equilateral because each side is a radius of one of the circles, hence must all be the same size, 4cm.
As they are equilateral, this then means that all three angles in each triangle are 60°.
We can now add this purple sector to the top. This sector is made up of the top blue area plus a small segment to each side. If we can work out the areas of the sector and segments, then we can solve the question.
The sector is easy enough. Because of the angles on a straight line rule, its angle is 180° − 60° − 60° = 60°. Its straight sides are 4cm as these are radii. The area can therefore be worked out using our sector area formula.
Area = π × 42 × 60/360 = 8π/3 cm2
Now for the segments.
As can be seen in the image above, each segment can be made by taking a 60° circle sector and removing the equilateral triangle.
The sector has the same area as the sector we calculated earlier due to it again being a 60° sector with sides of 4cm. Its area is therefore 8π/3 cm2.
The triangle's area can be calculated using the formula 1/2 ab × sin C.
Triangle area = 1/2 × 4 × 4 × sin 60 = 1/2 × 16 × √3 / 2 = 4√3 cm2.
Note that we should remember that sin 60 = √3 / 2 and not need a calculator for this.
The area of the segment is therefore equal to 8π/3 − 4√3 cm2.
Bringing it All Together
We now have the area of the sector and the two segments and so bringing this together we get:
Area of top blue area = 8π/3 − 2(8π/3 − 4√3) = 8√3 − 8π/3.
The total area is two of these areas and so:
Total area = 2(8√3 − 8π/3) = 16√3 − 16π/3
Should I Worry If I Didn't Answer This?
No. Definitely not. This was the final question on a higher paper designed to go all the way up to grade 9. Grade 9 is the very top grade created to split the very best A* mathematicians from the other A* mathematicians.
To be able to create this split between grade 8 and 9, the examiners need to include questions that can't be learnt by rote or drilled in by hard work and repetition. This question, although only using GCSE maths skills, requires that extra spark to spot how to use those skills.
A student can still very easily get a top grade without having answered this question and anybody not predicted a grade 9 should have no worries whatsoever about having missed this one.
This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.
© 2022 David