IMC 2011 Q18
18. The diagram contains six equilateral triangles with sides of length 2 and a regular hexagon with sides of length 1.
What fraction of the whole shape is shaded?
- A \(\dfrac{1}{8}\)
- B \(\dfrac{1}{7}\)
- C \(\dfrac{1}{6}\)
- D \(\dfrac{1}{5}\)
- E \(\dfrac{1}{4}\)
Intermediate Mathematical Challenge 2011, Question 18. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{1}{5}\)
As the diagram shows, each equilateral triangle may be divided into four equilateral triangles of side 1, whilst the hexagon may be divided into six equilateral triangles of side 1.
Therefore the fraction of the whole shape which is shaded is
\(\dfrac{6}{6 \times 4 + 6} = \dfrac{6}{30} = \dfrac{1}{5}\).
Solution © UK Mathematics Trust, from the IMC 2011 solutions. Longer versions with investigations: Solutions and investigations (PDF).