Hamilton 2017 Q5
5. In the trapezium \(ABCD\), the lines \(AB\) and \(DC\) are parallel, \(BC = AD\), \(DC = 2AD\) and \(AB = 3AD\).
The angle bisectors of \(\angle DAB\) and \(\angle CBA\) intersect at the point \(E\).
What fraction of the area of the trapezium \(ABCD\) is the area of the triangle \(ABE\)?
Hamilton Olympiad 2017, Question 5. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: \(\tfrac{3}{5}\)
Let \(BC = AD = k\), so that \(DC = 2k\) and \(AB = 3k\), and let the point \(X\) lie on \(AB\) so that \(XBCD\) is a parallelogram, as shown in the diagram on the left below. It follows that \(DX = k\) and \(XB = 2k\) (opposite sides of a parallelogram), so that \(AX = k\).
Hence triangle \(AXD\) has three equal sides—it is therefore an equilateral triangle. In particular, this means that angle \(DAX\) is equal to 60°.
As a consequence, the trapezium \(ABCD\) is actually made up from five equilateral triangles, as shown in the diagram on the right above.
Now the triangle \(ABE\) comprises one equilateral triangle and two half-rhombuses. The area of the two half-rhombuses is equal to the area of two equilateral triangles.
Therefore the area of the triangle \(ABE\) is \(\tfrac{3}{5}\) of the area of the trapezium \(ABCD\).
Solution © UK Mathematics Trust, from the Hamilton Olympiad 2017 solutions.