Maclaurin 2017 Q1
1. The diagram shows a semicircle of radius \(r\) inside a right-angled triangle. The shorter edges of the triangle are tangents to the semicircle, and have lengths \(a\) and \(b\). The diameter of the semicircle lies on the hypotenuse of the triangle.
Prove that
Maclaurin Olympiad 2017, Question 1. © UK Mathematics Trust. Original paper on ukmt.org.uk
Label the points as shown above, and join \(O\) to \(P\), \(Q\) and \(C\). Then \(OP\) is perpendicular to \(BC\) and \(OQ\) is perpendicular to \(CA\), since each of \(OP\) and \(OQ\) is a radius and each of \(BC\) and \(CA\) is a tangent.
Method 1
The area of triangle \(AOC\) is \(\tfrac{1}{2}rb\), that of triangle \(BOC\) is \(\tfrac{1}{2}ra\) and that of triangle \(ABC\) is \(\tfrac{1}{2}ab\).
Hence, adding, we have
Dividing each term by \(\tfrac{1}{2}abr\), we obtain
Method 2
Since \(OPCQ\) is a square, \(AQ = b - r\). The triangles \(AOQ\) and \(ABC\) are similar (AA), so that
Multiplying each side by \(b(b - r)\), we get
and now, dividing each term by \(abr\), we obtain
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2017 solutions.