IMC 2014 Q23
23. A sector of a disc is removed by making two straight cuts from the circumference to the centre. The perimeter of the sector has the same length as the circumference of the original disc. What fraction of the area of the disc is removed?
- A \(\dfrac{\pi - 1}{\pi}\)
- B \(\dfrac{1}{\pi}\)
- C \(\dfrac{\pi}{360}\)
- D \(\dfrac{1}{3}\)
- E \(\dfrac{1}{2}\)
Intermediate Mathematical Challenge 2014, Question 23. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A \(\dfrac{\pi - 1}{\pi}\)
Let the radius of the circle be \(r\) and let the angle of the sector be \(a\)°.
Then the perimeter of the sector is \(2r + \tfrac{a}{360} \times 2\pi r\).
This equals \(2\pi r\), the circumference of the original circle.
So \(2r + \tfrac{a}{360} \times 2\pi r = 2\pi r\).
Therefore the fraction of the area of the disc removed is
\(\tfrac{a}{360} = \tfrac{2\pi r - 2r}{2\pi r} = \tfrac{\pi - 1}{\pi}\).
Solution © UK Mathematics Trust, from the IMC 2014 solutions. Longer versions with investigations: Solutions and investigations (PDF).