Saudi Arabia Mathematical Competitions 2012

Denote by $a$, $b$, $c$ the side lengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that $$2m_b/3=MD=AD=CD=b/2,$$ so $m_b=3b/4$, which means that $$(3b/4{)}^2=m_b^2=(a^2+c^2)/2-b^2/4.$$ This is equivalent to $13b^2=8(a^2+c^2)$.

Next, apply the Menelaus theorem to get $$EC/EB=4=FA/FB$$ and deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrilateral $AFED$ is therefore a trapezoid; it is cyclic if and only if $AF=DE$.

We now express the two lengths in terms of $a$, $b$ and $c$. Recall that $FA/FB=4$ to obtain $AF=4c/5$. Next, apply Stewart's theorem in triangle $BCD$ to get $D{E}^2=b^2/2-4a^2/25$. By the preceding, the quadrilateral $AFED$ is cyclic if and only if $25b^2-8a^2=32c^2$. Recall that $13b^2=8(a^2+c^2)$ to express $b$ and $c$ in terms of $a$: $$b=2a\sqrt{2}/3\text{and}c=2a/3.$$ Finally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that $$EN=a/2-a/5=3a/10,$$ and that the triangles $ANC$ and $PNE$ are similar. Then we obtain $$NP=3m_a/5,$$ so $$NA\cdot NP=3m_a^2/5=3(2(b^2+c^2)-a^2)/20=a^2/4=NB\cdot NC.$$ The conclusion follows.