PK 2018 Q25
25. Two chords \(PQ\) and \(PR\) are drawn in a circle with diameter \(PS\).
The point \(T\) lies on \(PR\) and \(QT\) is perpendicular to \(PR\). The angle \(QPR = 60°\), \(PQ = 24\) cm, \(RT = 3\) cm. What is the length of the chord \(QS\) in cm?
- A \(\sqrt{3}\)
- B 2
- C 3
- D \(2\sqrt{3}\)
- E \(3\sqrt{2}\)
Pink Kangaroo 2018, Question 25. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(2\sqrt{3}\)
\(PS\) is a diameter, so \(\angle PRS = 90°\) (angle in a semicircle). Let \(U\) be the point on \(QT\) for which \(SU\) is perpendicular to \(QT\). Hence \(RSUT\) is a rectangle, and \(SU = TR = 3\).
In triangle \(TPQ\), we have \(\angle PTQ = 90°\) and \(\angle QPT = 60°\), so \(\angle TQP = 30°\). Also, \(\angle PQS = 90°\) (angle in a semicircle) so \(\angle UQS = 90° - \angle TQP = 90° - 30° = 60°\).
Also \(\sin \angle UQS = \dfrac{US}{QS} = \dfrac{3}{QS}\). So
\[QS = \frac{3}{\sin \angle UQS} = \frac{3}{\frac{1}{2}\sqrt{3}} = 3 \times \frac{2}{\sqrt{3}} = 2\sqrt{3}.\]
Solution © UK Mathematics Trust, from the PK 2018 solutions.