IMC 2019 Q21
21. The diagram shows a right-angled triangle \(PQR\). The point \(S\) is the midpoint of the side \(QR\) and \(\tan \angle QPR = \tfrac{3}{2}\).
What is the value of \(\sin \angle QPS\)?
- A \(\dfrac{1}{\sqrt{3}}\)
- B \(\dfrac{1}{\sqrt{2}}\)
- C \(\dfrac{1}{2}\)
- D \(\dfrac{3}{5}\)
- E \(\dfrac{4}{5}\)
Intermediate Mathematical Challenge 2019, Question 21. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{3}{5}\)
Let \(PQ\) have length \(4x\). Then, as \(\tan \angle QPR = \dfrac{3}{2}\), \(QR\) has length \(6x\). Therefore, as \(S\) is the midpoint of side \(QR\), \(QS\) has length \(3x\). By Pythagoras’ Theorem in triangle \(PQS\), \(PS^2 = PQ^2 + QS^2 = (4x)^2 + (3x)^2 = 16x^2 + 9x^2 = 25x^2\). Hence \(PS = \sqrt{25x^2} = 5x\).
So \(\sin \angle QPS = \dfrac{QS}{PS} = \dfrac{3x}{5x} = \dfrac{3}{5}\).
Solution © UK Mathematics Trust, from the IMC 2019 solutions. Longer versions with investigations: Solutions and investigations (PDF).