PK 2016 Q10
10. A square is split into nine identical squares, each with sides of length one unit. Circles are inscribed in two of these squares, as shown. What is the shortest distance between the two circles?
- A \(2\sqrt{2} - 1\)
- B \(\sqrt{2} + 1\)
- C \(2\sqrt{2}\)
- D 2
- E 3
Pink Kangaroo 2016, Question 10. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A \(2\sqrt{2} - 1\)
Each square has side-length 1 unit, so by Pythagoras' Theorem the diagonals have length \(\sqrt{1^2 + 1^2} = \sqrt{2}\). The distance between the two circles consists of a whole diagonal and two part diagonals (from the corner of a square to the circle). This is the same as the length of two whole diagonals, minus a diameter, that is \(2\sqrt{2} - 1\).
Solution © UK Mathematics Trust, from the PK 2016 solutions.