Maclaurin 2020 Q2
2. In the square \(ABCD\), the bisector of \(\angle CAD\) meets \(CD\) at \(P\) and the bisector of \(\angle ABD\) meets \(AC\) at \(Q\).
What is the ratio of the area of triangle \(ACP\) to the area of triangle \(BQA\)?
Maclaurin Olympiad 2020, Question 2. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: 2 : 1
We know that, since \(AC\) is the diagonal of a square, \(\angle PCA = 45^\circ = \angle QAB\). Also \(\angle DAC = 45^\circ = \angle DBA\). Since \(AP\) and \(BQ\) are angle bisectors, \(\angle PAC = 22.5^\circ = \angle QBA\).
Hence the triangles \(CPA\) and \(AQB\) have equal angles, so they are similar. Now \(CA = \sqrt{2}\) and \(AB = 1\), so the sides are in the ratio \(\sqrt{2} : 1\), and the areas are in the ratio \(2 : 1\).
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2020 solutions.