Canadian Mathematical Olympiad

Solution:
Let the circle be tangent to $\overline{P Q}$, $\overline{A B}$, $\overline{A D}$ at $T$, $U$, and $V$, respectively. Let $p = P T = P U$ and $q = Q T = Q V$. Let $a = A U = A V$ and $b = B U = D V$. Then the side length of the rhombus is $a + b$.

Let $\theta = \angle B A D$, so $\angle A B C = \angle A D C = 180^{\circ} - \theta$. Then (using the notation $[XYZ]$ for the area of a triangle of vertices $X, Y, Z$)
$$\begin{aligned} & [A P Q] = \frac{1}{2} \cdot A P \cdot A Q \cdot \sin \theta = \frac{1}{2}(a-p)(a-q) \sin \theta \\ & [B C P] = \frac{1}{2} \cdot B P \cdot B C \cdot \sin (180^{\circ} - \theta) = \frac{1}{2}(b+p)(a+b) \sin \theta \\ & [C D Q] = \frac{1}{2} \cdot D Q \cdot C D \cdot \sin (180^{\circ} - \theta) = \frac{1}{2}(b+q)(a+b) \sin \theta \end{aligned}$$
So
$$\begin{aligned} [C P Q] & = [A B C D] - [A P Q] - [B C P] - [C D Q] \\ & = (a+b)^2 \sin \theta - \frac{1}{2}(a-p)(a-q) \sin \theta - \frac{1}{2}(b+p)(a+b) \sin \theta - \frac{1}{2}(b+q)(a+b) \sin \theta \\ & = \frac{1}{2}\left(a^2 + 2 a b - b p - b q - p q\right) \sin \theta \end{aligned}$$
Let $O$ be the center of the circle, and let $r$ be the radius of the circle. Let $x = \angle T O P = \angle U O P$ and $y = \angle T O Q = \angle V O Q$. Then $\tan x = \frac{p}{r}$ and $\tan y = \frac{q}{r}$.

Note that $\angle U O V = 2x + 2y$, so $\angle A O U = x + y$. Also, $\angle A O B = 90^{\circ}$, so $\angle O B U = x + y$. Therefore,
$$\tan (x + y) = \frac{a}{r} = \frac{r}{b}$$
so $r^2 = a b$. But
$$\frac{r}{b} = \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\frac{p}{r} + \frac{q}{r}}{1 - \frac{p}{r} \cdot \frac{q}{r}} = \frac{r(p+q)}{r^2 - p q} = \frac{r(p+q)}{a b - p q}.$$
Hence, $a b - p q = b p + b q$, so $b p + b q + p q = a b$. Therefore,
$$[C P Q] = \frac{1}{2}\left(a^2 + 2 a b - b p - b q - p q\right) \sin \theta = \frac{1}{2}\left(a^2 + a b\right) \sin \theta$$
which is constant.

Solution:
Alternate Solution: Let $O$ be the center of the circle and $r$ its radius. Then $[C P Q] = [C D Q P B] - [C D Q] - [C B P]$, where $[\ldots]$ denotes area of the polygon with given vertices. Note that $[C D Q P B]$ is half $r$ times the perimeter of $C D Q P B$. Note that the heights of $C D Q$ and $C B P$ are $2 r$ so $[C D Q] = r \cdot D Q$ and $[C B P] = r \cdot P B$. Using the fact that $Q T = Q V$ and $P U = P T$, it now follows that $[C P Q] = [O V D C B U] - [C D V] - [C B U]$, which is independent of $P$ and $Q$.