Balkan Mathematical Olympiad Shortlist

The answer is no. Suppose that it is possible. We assert that the previous triangle $A_{n-1}{B}_{n-1}{C}_{n-1}$ has rational coordinates. In fact, the points $A_{n-1}$, $B_{n-1}$, $C_{n-1}$ are the feet of the altitudes of the triangle $A_n{B}_n{C}_n$. Therefore it is enough to show that, if a line segment has its ends with rational coordinates, then the foot of the perpendicular line passing through a point of the plane with rational coordinates has also rational coordinates. This really happens because the coordinates $(x,y)$ of the foot of the perpendicular are the solutions of the system $y=ax+b$, $y=-\frac{1}{a}x+c$ with $a,b,c$ rational. Therefore, every time in the previous step the coordinates must be rational and so, we arrive to the conclusion that the coordinates of the triangle $A_0{B}_0{C}_0$ must be rational. Then from the area formula using coordinates of the vertices we find that the area of the triangle is a rational number. This contradicts the supposition that the area of the triangle is equal to $\sqrt{2}$.