Three Cubes

Can you work out the dimensions of the three cubes?

Farhan's Poor Square

From the measurements and the clue given find the area of the square that is not covered by the triangle and the circle.

Parallel Parking

Scientist Bryan Rickett has a vision of the future - and it is one in which self-parking cars prowl the tarmac plains, hunting down suitable parking spots and manoeuvring elegantly into them.

Get Cross

Stage: 4 Challenge Level:

Congratulations for your solutions to Hyeyoun Chung, age 17, St Paul's Girls' School, London; Royce Ferguson; Ang Zhi Ping, age 16; Yatir Halevi, age 17, Maccabim and Reut High-School, Israel; and Joe Nielson, Rowan Maclennon-Ryde and Elizabeth Brewster from Madras College, St Andrew's, Scotland.

The radius of the circle OA can be found by using the right triangle formed by $AB = (1 + (\sqrt 2)/2)$ and $BO = (\sqrt 2)/2)$. Using Pythagoras' theorem, the radius is found to be $OA = \sqrt(2 +\sqrt 2)$ units and the area of the circle to be $\pi (2 + \sqrt 2)$.

Now, connect the centre of the circle to the 8 points on it's circumference where the white meets red. This divides the white into circular sectors and quadrilaterals.

Using the cosine rule to find $\angle AOD$ we have $AC=CD=1$ and so $AD = \sqrt 2$ and hence

$$\cos \angle AOD = {{OA^2 + OD^2 -AD^2}\over 2OA.OD} = {1 \over \sqrt 2}.$$

Hence $\angle AOD = 45$ degrees.

The area of the triangle $AOD$ is

$${1\over 2} OA \times OD \times \sin \angle AOD = {\sqrt 2 \over 4}(2 + \sqrt 2).$$

To find the area of the minor segment $AD$ we subtract the area of triangle $AOD$ from the area of sector $AOD$ which gives

$${\pi \over 8}(2 + \sqrt 2) - {\sqrt 2 \over 4}(2 + \sqrt 2).$$

To get the total red shaded area we now add the area of triangle $ACD$ and multiply by $4$ which gives:

$$4[{\pi \over 8}(2 + \sqrt 2) - {\sqrt 2 \over 4}(2 + \sqrt 2)+ {1\over 2}]={\pi \over 2}(2 + \sqrt 2)- 2\sqrt 2 \approx 2.535 \rm {\ sq. \ units} .$$