, 2023

Since $AB$ and $BC$ are both radii of the circle, then $AB = BC$. Since $ABC$ is a quarter-circle centred at $B$, then $\angle ABC = 90\degree$. Thus, $\triangle ABC$ is isosceles and right-angled, which means that $\$\angle$
BAC = $\angle$ BCA = $45\degree$. We add some additional labels to the diagram: [[IMAGE0]] We note that the angles between the straight lines at

P$,$Q$,$R$, and$U are all right angles. Since

$\angle$ PAD = $\angle$ BAC =
$45\degree$$, this means that$$\angle PAD = \angle ADP = \angle QDE = \angle DEQ = \angle REF = \angle EFR = \angle UFG = \angle UGF = 45\degree$ This in turn means that each of

$\triangle$ APD$,$$\triangle$ DQE$,$$\triangle$ ERF$, and$$\triangle$ FUG is right-angled and isosceles. Since the side length of each square is 10, then

BT = 10$and$TQ = TR + RQ = 20$. Since$$\angle$ BTQ = $90\degree$$, then by the Pythagorean Theorem,$$BQ = \sqrt{BT^2 + TQ^2} = \sqrt{10^2 + 20^2} = \sqrt{500}$$We note that$$\sqrt{500} = \sqrt{100 \times}$ 5} = $\sqrt{100} \times \sqrt{5} = 10\sqrt{5}\$.$

Since $BQ$ is a radius of the circle, then $BQ = BA = BC =
$10\sqrt{5}$$. Since$BP = TQ = 20$, then$AP = BA - BP = $10\sqrt{5}$ - 20$. Thus,$PD = AP = $10\sqrt{5}$ -
20$. Since$PQ = 10$, then$DQ = PQ - PD = 10 - $(10\sqrt{5}$ - 20) = 30 - $10\sqrt{5}\$.$

Thus, $QE = DQ = 30 - 10\sqrt{5}$.

Since $PQ = QR$ and $DQ = QE$, then $PD = ER = 10\sqrt{5} - 20$. Using similar reasoning, $ER = RF =
$10\sqrt{5}$ - 20$and$UF = UG = $30-10\sqrt{5}\$.$

The total area, $\mathcal{A}$, of the shaded regions equals the sum of the areas of $\triangle DQE$, $\triangle ERF$ and $\triangle FUG$. Therefore, $$\begin{align*} \mathcal{A} & = \tfrac{1}{2} \times DQ \times QE + \tfrac{1}{2} \times ER \times RF + \tfrac{1}{2} UF \times UG \\ & = \tfrac{1}{2}(30-10\sqrt{5})^2 + \tfrac{1}{2} (10\sqrt{5}-20)^2 + \tfrac{1}{2}(30-10\sqrt{5})^2 \\ & = (30-10\sqrt{5})^2 + \tfrac{1}{2} (10\sqrt{5}-20)^2 \\ & = (30^2 - 2 \times 30 \times 10\sqrt{5} + (10\sqrt{5})^2) + \tfrac{1}{2}((10\sqrt{5})^2 - 2 \times 10\sqrt{5} \times 20 + 20^2) \\ & = (900 - 600\sqrt{5} + 500) + \tfrac{1}{2}(500 - 400\sqrt{5} + 400) \\ & = 1400 - 600\sqrt{5} + 450 - 200\sqrt{5} \\ & = 1850 - 800\sqrt{5} \\ & \approx 61.14\end{align*}$$ and so the integer closest to $\mathcal{A}$ is 61.