IMC 2017 Q22
22. The diagram shows an arc \(PQ\) of a circle with centre \(O\) and radius 8. Angle \(QOP\) is a right angle, the point \(M\) is the midpoint of \(OP\) and \(N\) lies on the arc \(PQ\) so that \(MN\) is perpendicular to \(OP\).
Which of the following is closest to the length of the perimeter of triangle \(PNM\)?
- A 17
- B 18
- C 19
- D 20
- E 21
Intermediate Mathematical Challenge 2017, Question 22. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 19
Consider triangles \(MON\) and \(MPN\). Note that \(MO = MP = 4\) because \(M\) is the midpoint of \(OP\); \(\angle OMN = \angle PMN = 90\)° because \(MN\) is perpendicular to \(OP\); side \(NM\) is common to both triangles. So triangles \(MON\) and \(MPN\) are congruent (SAS). Therefore \(PN = ON = 8\) because \(ON\) is a radius of the circle. By Pythagoras' Theorem in triangle \(OMN\), \(ON^2 = OM^2 + MN^2\), so \(8^2 = 4^2 + MN^2\). Therefore \(MN^2 = 8^2 - 4^2 = 48\). So \(MN = \sqrt{48}\).
The perimeter of triangle \(PNM\) is \(PN + NM + MP = 8 + \sqrt{48} + 4 = 12 + \sqrt{48}\).
Now \(6.5 < \sqrt{48} < 7\), since \(\sqrt{42.25} < \sqrt{48} < \sqrt{49}\). So \(\sqrt{48}\) is closer in value to 7 than it is to 6. So \(12 + \sqrt{48}\) is nearer to 19 than it is to 18.
Solution © UK Mathematics Trust, from the IMC 2017 solutions. Longer versions with investigations: Solutions and investigations (PDF).