NZMO Round Two

Solution:
Since $AD = DC$ and $\angle ADC = 90^{\circ}$, we can fit four copies of quadrilateral $ABCD$ around vertex $D$ as shown in the diagram.

The outer shape is a quadrilateral because $\angle DAB + \angle BCD = 180^{\circ}$. Moreover it is a rectangle because $\angle ABC = 90^{\circ}$. In fact it is a square with side-length $2021$ because of rotational symmetry and $AB + BC = 2021$. Also $D$ is the centre of the square because it is the centre of the rotational symmetry. So $BD$ is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus
$$BD = \frac{1}{2}\left(2021\sqrt{2}\right) = \frac{2021}{\sqrt{2}}.$$

Solution:
First let $x = AD = DC$ and $a = BD$ and $y = AB$ and $z = BC$. Now initially we can apply Pythagoras in triangles $CDA$ and $ABC$ to get $x^{2} + x^{2} = AC^{2}$ and $y^{2} + z^{2} = AC^{2}$ respectively. Putting this together gives us
$$x^{2} = \frac{y^{2} + z^{2}}{2}.$$
Now note that the opposite angles $\angle ABC$ and $\angle CDA$ (in quad $ABCD$) are supplementary. Therefore $ABCD$ is a cyclic quadrilateral. Equal chords subtend equal arcs (and chords $AD = DC$ are equal) so $\angle ABD = \angle DBC$. Furthermore, since $\angle ABC$ is a right angle, this means that $\angle ABD = \angle DBC = 45^{\circ}$.

For any three points $P$, $Q$ and $R$, let $|PQR|$ denote the area of triangle $PQR$. Now consider the total area of quadrilateral $ABCD$ calculated in two ways:
$$|ABC| + |CDA| = |ABD| + |DBC|.$$
We calculate the areas of the right-angled triangles using the $\triangle = \frac{b h}{2}$ formula, and we calculate the area of the $45^{\circ}$-angled triangles using the $\triangle = \frac{1}{2} ab\sin C$ formula.
$$\frac{yz}{2} + \frac{x^{2}}{2} = \frac{1}{2} ay\sin (45^{\circ}) + \frac{1}{2} az\sin (45^{\circ})$$
At this point we can substitute $x^{2} = \frac{1}{2} (y^{2} + z^{2})$ into this equation, and rearrange:
$$\frac{yz}{2} + \frac{x^{2}}{2} = \frac{ay\sin(45^{\circ})}{2} + \frac{az\sin(45^{\circ})}{2}$$
$$\frac{yz}{2} + \frac{y^{2} + z^{2}}{4} = \frac{ay}{2\sqrt{2}} + \frac{az}{2\sqrt{2}}$$
$$2yz + y^{2} + z^{2} = \frac{ay + az}{2\sqrt{2}}$$
$$\frac{(y + z)^{2}}{4} = \frac{a(y + z)}{2\sqrt{2}}$$
$$\frac{y + z}{\sqrt{2}} = a.$$
Finally since $y + z = 2021$ this gives our final answer of $BD = a = \frac{2021}{\sqrt{2}}$.