IMC 2017 Q25
25. The diagram shows a circle with radius 2 and a square with sides of length 2. The centre of the circle lies on the perpendicular bisector of a side of the square, at a distance \(x\) from the side, as shown. The shaded region − inside the square but outside the circle − has area 2.
What is the value of \(x\)?
- A \(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} - 1\)
- B \(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4} - 1\)
- C \(\dfrac{\pi}{3} + \dfrac{1}{2}\)
- D \(\dfrac{\pi}{3} + 1\)
- E \(\dfrac{\pi}{3}\)
Intermediate Mathematical Challenge 2017, Question 25. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A \(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} - 1\)
The diagram shows part of the diagram in the question. Point \(O\) is the centre of the circle and points \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), are as shown. Consider triangle \(BOF\): \(BF\) is equal in length to the side of the square so \(BF = 2\). Also \(OB = OF = 2\) as they are both radii of the circle. So triangle \(BOF\) is equilateral.
Therefore \(\angle BOF = 60\)°, so the area of sector \(OBF = \frac{60}{360} \times \pi \times 2^2 = \frac{2\pi}{3}\).
By Pythagoras' Theorem: \(OA = \sqrt{2^2 - 1^2} = \sqrt{3}\).
So the area of triangle \(OBF = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}\). Therefore the area of the segment bounded by arc \(BF\) and line segment \(BF = \frac{2\pi}{3} - \sqrt{3}\). The area of rectangle \(BFEC = BF \times AD = 2 \times (\sqrt{3} - x)\).
The shaded region has area 2, so the area of the above segment + area of rectangle \(BFEC\) = area of the given square minus 2 = 4 − 2 = 2.
Hence \(\frac{2\pi}{3} - \sqrt{3} + 2\sqrt{3} - 2x = 2\). So \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} - 1\).
Solution © UK Mathematics Trust, from the IMC 2017 solutions. Longer versions with investigations: Solutions and investigations (PDF).