IMC 2004 Q18
18. In the triangle \(PQR\), there is a right angle at \(Q\) and angle \(QPR\) is 60°. The bisector of the angle \(QPR\) meets \(QR\) at \(S\), as shown.
What is the ratio \(QS : SR\)?
- A 1:1
- B 1:\(\sqrt{2}\)
- C 1:(3 − \(\sqrt{3}\))
- D 1 : \(\sqrt{3}\)
- E 1:2
Intermediate Mathematical Challenge 2004, Question 18. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 1:2
Angle \(PRS\) = 30°, so triangle \(PRS\) is isosceles with \(SP = SR\). Hence \(\dfrac{QS}{SR} = \dfrac{QS}{SP} = \dfrac{1}{2}\) as \(PQS\) is half of an equilateral triangle.
[Alternatively, we can use the angle bisector theorem: \(\dfrac{QS}{SR} = \dfrac{PQ}{PR} = \dfrac{1}{2}\) as \(PQR\) is also half of an equilateral triangle.]
Solution © UK Mathematics Trust, from the IMC 2004 solutions.