I have the test, too, this level is embarrassment; but is more hard than the problem 3. Latin american(olympic team) is bad in geometry $\ldots$ WLOG $EC=\sqrt{2},FD=2,AF=1$ $BD \cap AC \equiv X$ By Menelaus' theorem ($\triangle XCD - M,E,B$) we get $\frac{\sqrt{2}}{EX}=\frac{BD}{BX}$ Since $B$ is the spiral-center $AF \rightarrow XE \Rightarrow \frac{AB}{1}=\frac{BX}{XE} \Rightarrow BD=AB\sqrt{2} \Rightarrow \angle ADB=45^{\circ}$ and $3XE=BX=\frac{3\sqrt{2}}{2}=EC+XE \Rightarrow AX=BX=CX=DX \ldots$