IMC 2015 Q25

Points \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) on the perimeter of the triangle are as shown. Let \(AD\) have length \(x\) so that \(DB\) has length \(3x\). Let the perpendicular from \(C\) to \(AB\) have length \(4h\). So, by similar triangles, the perpendicular from \(E\) to \(DB\) has length \(h\). The area of triangle \(ABC\) is \(\tfrac{1}{2} \times 4x \times 4h = 8xh\). The area of triangle \(DBE\) is \(\tfrac{1}{2} \times 3x \times h = \tfrac{3}{2}xh\). So the area of triangle \(DBE\) is \(\tfrac{3}{16}\) of the area of triangle \(ABC\).
Similarly, by drawing perpendiculars to \(CB\) from \(A\) and from \(F\), it may be shown that the area of triangle \(FEC\) is \(\tfrac{3}{16}\) of the area of triangle \(ABC\).
So the fraction of the area of the triangle that is shaded is \(1 - \tfrac{3}{16} - \tfrac{3}{16} = \tfrac{5}{8}\).