Irish Mathematical Olympiad

The 16 solutions of $x^2 + y^2 = 65$ in integers are
$$(\pm 8, \pm 1), (\pm 1, \pm 8), (\pm 4, \pm 7), (\pm 7, \pm 4).$$

These form the vertices of a hexadecagon. Calculating the perimeter of this hexadecagon gives rise to the lower bound for $\pi$ as follows. Let $P_1 = (8, -1)$, $P_2 = (8, 1)$, $P_3 = (7, 4)$, $P_4 = (4, 7)$, $P_5 = (1, 8)$, $P_6 = (-1, 8)$ etc. and denote the lengths of the line segments by $a_i = |P_i P_{i+1}|$, $i \ge 1$. The perimeter of the hexadecagon is then equal to $4a_1 + 8a_2 + 4a_3$ (see diagram).

From the coordinates of the $P_i$ we easily obtain $a_1 = 2$, $a_2 = \sqrt{10}$ and $a_3 = 3\sqrt{2}$. This shows that the perimeter of the hexadecagon is equal to
$$8 + 8\sqrt{10} + 12\sqrt{2}.$$
Because the circumcircle of the hexadecagon has radius $\sqrt{65}$, we obtain the inequality $8 + 8\sqrt{10} + 12\sqrt{2} < 2\pi\sqrt{65}$ hence
$$\frac{4 + 4\sqrt{10} + 6\sqrt{2}}{\sqrt{65}} < \pi.$$
The upper bound is obtained by observing that the circle with equation $x^2+y^2 = 121/2$ lies inside the hexadecagon. Indeed, because $\frac{11}{\sqrt{2}} < \frac{5}{2}\sqrt{10} < 8$, this circle is tangent to $P_3P_4$ but does not intersect the line segments $P_1P_2$ and $P_2P_3$. Hence, its circumference is bounded above by the perimeter of the hexadecagon. This gives the inequality $11\pi\sqrt{2} < 8 + 8\sqrt{10} + 12\sqrt{2}$, hence
$$\pi < \frac{4}{11} (3 + \sqrt{2} + 2\sqrt{5}).$$