IMC 2018 Q25

In the diagram shown, \(F\) is the foot of the perpendicular from \(D\) to \(AB\), \(G\) is the foot of the perpendicular from \(D\) to \(AC\) and \(H\) is the foot of the perpendicular from \(E\) to \(AB\). So \(DF = DG = EH = \sqrt{3}\). Consider triangles \(AFD\) and \(AGD\). Edge \(AD\) is common to both triangles, \(DF = DG\) and angles \(AFD\) and \(AGD\) are both right angles. So the triangles are congruent (RHS). Therefore \(\angle DAF = \angle DAG = 60° \div 2 = 30°\).
So in triangle \(AFD\), \(\tan 30° = \dfrac{DF}{AF}\).
Therefore \(AF = \sqrt{3} \div \left(1/\sqrt{3}\right) = \sqrt{3} \times \sqrt{3} = 3\). The same argument shows that \(BH = 3\). It is clear that \(FH = DE\), so \(AB - DE = AF + BF = 3 + 3 = 6\).
Therefore the difference between the lengths of the edges of the two triangles equals 6.