IMC 2018 Q16
16. The diagram shows a triangle with edges of length 3, 4 and 6.
A circle of radius 1 is drawn at each vertex of the triangle.
What is the total shaded area?
- A \(2\pi\)
- B \(\dfrac{9\pi}{4}\)
- C \(\dfrac{5\pi}{2}\)
- D \(\dfrac{11\pi}{4}\)
- E \(3\pi\)
Intermediate Mathematical Challenge 2018, Question 16. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C \(\dfrac{5\pi}{2}\)
The sum of the interior angles of a triangle is 180°. So the three unshaded sectors inside the triangle have a total area equal to one half of the area of a circle of radius 1. Therefore the total shaded area is equal to the area of two and a half circles of radius 1, so the required area equals \(\dfrac{5}{2} \times \pi \times 1^2 = \dfrac{5\pi}{2}\).
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).