IMC 2016 Q23
23. A Saxon silver penny, from the reign of Ethelbert II in the eighth century, was sold in 2014 for £78 000. A design on the coin depicts a circle surrounded by four equal arcs, each a quarter of a circle, as shown. The width of the design is 2 cm.
What is the radius of the small circle, in centimetres?
- A \(\frac{1}{2}\)
- B \(2 - \sqrt{2}\)
- C \(\frac{1}{2}\sqrt{2}\)
- D \(5 - 3\sqrt{2}\)
- E \(2\sqrt{2} - 2\)
Intermediate Mathematical Challenge 2016, Question 23. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B \(2 - \sqrt{2}\)
The diagram shows part of the figure, to which have been added \(A\) and \(C\), centres of two of the quarter-circle arcs, \(B\) and \(D\), points of intersection of two arcs, and \(E\), the centre of the small circle. In cm, the radii of each arc and the small circle are \(R\) and \(r\) respectively. Firstly, note that \(\angle BCD\) is a right angle as arc \(BD\) is a quarter of a circle. Therefore, by Pythagoras' Theorem \(R^2 + R^2 = 2^2\) so \(R = \sqrt{2}\). Consider triangle \(ACE\): from the symmetry of the figure we deduce that \(\angle AEC = \frac{1}{4} \times 360\)° = 90°. So, by Pythagoras' Theorem \((R + r)^2 + (R + r)^2 = (2R)^2 = 4R^2\). Therefore \((R + r)^2 = 2R^2 = 2 \times 2 = 4\).
Hence \(R + r = 2\), so \(r = 2 - R = 2 - \sqrt{2}\).
Solution © UK Mathematics Trust, from the IMC 2016 solutions. Longer versions with investigations: Solutions and investigations (PDF).