IMC 2013 Q17
17. The diagram shows four equal discs and a square. Each disc touches its two neighbouring discs. Each corner of the square is positioned at the centre of a disc. The side length of the square is \(2/\pi\). What is the length of the perimeter of the figure?
- A 3
- B 4
- C \(\dfrac{3\pi}{2}\)
- D 6
- E \(2\pi\)
Intermediate Mathematical Challenge 2013, Question 17. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 6
The radius of each disc in the figure is equal to half the side-length of the square, i.e. \(\tfrac{1}{\pi}\). Because the corners of a square are right-angled, the square hides exactly one quarter of each disc.
So three-quarters of the perimeter of each disc lies on the perimeter of the figure. Therefore the length of the perimeter is \(4 \times \tfrac{3}{4} \times 2\pi \times \tfrac{1}{\pi} = 6\).
Solution © UK Mathematics Trust, from the IMC 2013 solutions. Longer versions with investigations: Solutions and investigations (PDF).