Maclaurin 2018 Q6
6. The diagram shows a triangle \(ABC\) and points \(T\), \(U\) on the edge \(AB\), points \(P\), \(Q\) on \(BC\), and \(R\), \(S\) on \(CA\), where:
Prove that
Maclaurin Olympiad 2018, Question 6. © UK Mathematics Trust. Original paper on ukmt.org.uk
Let \(TU = PQ = RS = k\), \(AB = c\), \(BC = a\) and \(CA = b\).
Triangles \(\triangle RSY\) and \(\triangle CAB\) are similar, since \(RS\) is parallel to \(CA\), \(SY\) is parallel to \(AB\) and \(YR\) is parallel to \(BC\), so \(SY = \dfrac{ck}{b}\).
Similarly \(YP = \dfrac{ck}{a}\). Now
and so \(\dfrac{1}{c} = \dfrac{ab}{(ab + bc + ca)\,k}\).
If three such expressions are added, the factor \(ab + bc + ca\) cancels and we obtain the desired result.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2018 solutions.