IMC 2015 Q24
24. In square \(RSTU\) a quarter-circle arc with centre \(S\) is drawn from \(T\) to \(R\). A point \(P\) on this arc is 1 unit from \(TU\) and 8 units from \(RU\).
What is the length of the side of square \(RSTU\)?
- A 9
- B 10
- C 11
- D 12
- E 13
Intermediate Mathematical Challenge 2015, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 13
In the diagram \(V\) is the point where the perpendicular from \(P\) meets \(TS\). Let the side of the square \(RSTU\) be \(x\). So the radius of the arc from \(R\) to \(T\) is \(x\). Therefore \(SP\) has length \(x\), \(PV\) has length \(x - 8\) and \(VS\) has length \(x - 1\). Applying Pythagoras' Theorem to triangle \(PVS\):
\((x - 8)^2 + (x - 1)^2 = x^2\). So \(x^2 - 16x + 64 + x^2 - 2x + 1 = x^2\).
Therefore \(x^2 - 18x + 65 = 0\), so \((x - 5)(x - 13) = 0\).
Hence \(x = 5\) or \(x = 13\), but \(x > 8\) so the length of the side of the square \(RSTU\) is 13.
Solution © UK Mathematics Trust, from the IMC 2015 solutions. Longer versions with investigations: Solutions and investigations (PDF).