IMC 2013 Q10
10. \(ABCDE\) is a regular pentagon and \(BCF\) is an equilateral triangle such that \(F\) is inside \(ABCDE\).
What is the size of \(\angle FAB\)?
- A 48°
- B 63°
- C 66°
- D 69°
- E 72°
Intermediate Mathematical Challenge 2013, Question 10. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 66°
The sum of the interior angles of a pentagon is 540° so \(\angle ABC = 540^\circ \div 5 = 108^\circ\). Each interior angle of an equilateral triangle is 60°, so \(\angle FBC = 60^\circ\).
Therefore \(\angle ABF = 108^\circ - 60^\circ = 48^\circ\). As \(ABCDE\) is a regular pentagon, \(BC = AB\). However, \(BC = FB\) since triangle \(BFC\) is equilateral.
So triangle \(ABF\) is isosceles with \(FB = AB\).
Therefore \(\angle FAB = \angle AFB = (180^\circ - 48^\circ) \div 2 = 66^\circ\).
Solution © UK Mathematics Trust, from the IMC 2013 solutions. Longer versions with investigations: Solutions and investigations (PDF).