PK 2017 Q24
24. The points \(G\) and \(I\) are on the circle with centre \(H\), and \(FI\) is tangent to the circle at \(I\). The distances \(FG\) and \(HI\) are integers, and \(FI = FG + 6\). The point \(G\) lies on the straight line through \(F\) and \(H\). How many possible values are there for \(HI\)?
- A 0
- B 2
- C 4
- D 6
- E 8
Pink Kangaroo 2017, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 6
Let \(x\) be the length \(FG\) and let \(r\) be the radius. Then \(FI = x + 6\) and \(GH = HI = r\).
Angle \(FIH\) is a right angle (the tangent and radius are perpendicular) so \(FI^2 + HI^2 + FH^2\), which gives \((x + 6)^2 + r^2 = (x + r)^2\). Expanding this gives \(x^2 + 12x + 36 + r^2 = x^2 + 2rx + r^2\), which simplifies to \(12x + 36 = 2rx\). Halving this gives \(6x + 18 = rx\), which rearranges to \(r = 6 + \tfrac{18}{x}\). Since \(r\) is an integer, \(x\) must be a (positive) factor of 18, namely 1, 2, 3, 6, 9, 18; each of these six factors give a different value of \(r\) (or \(HI\)) as required.
Solution © UK Mathematics Trust, from the PK 2017 solutions.