IMC 2010 Q19
19. A snail is at one corner of the top face of a cube with side length 1 m. The snail can crawl at a speed of 1 m per hour. What proportion of the cube's surface is made up of points which the snail could reach within one hour?
- A \(\dfrac{\pi}{16}\)
- B \(\dfrac{\pi}{8}\)
- C \(\dfrac{1}{4}\)
- D \(\dfrac{1}{2}\)
- E \(\dfrac{\sqrt{3}}{4}\)
Intermediate Mathematical Challenge 2010, Question 19. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B \(\dfrac{\pi}{8}\)
In one hour, the snail can reach points within 1 m of the corner at which it starts. So it can reach some of the points on the three faces which meet at that corner, but none of the points on the other three faces. On each of the three reachable faces, the points which the snail can reach form a quarter of a circle of radius 1 m.
So the required fraction is \(\dfrac{3 \times \frac{1}{4}\pi \times 1 \times 1}{6 \times 1 \times 1} = \dfrac{\pi}{8}\).
Solution © UK Mathematics Trust, from the IMC 2010 solutions.