Reach for the Summit - M-S6-A2

Note that $C: x^2 + y^2 - 2x - 2y + 1 = 0 \implies C: (x-1)^2 + (y-1)^2 = 1$ is the incircle of $\triangle AOB$ with center at $(1,1)$ and radius $1$ . Then the $\triangle AOB$ with the smallest area is when it is an isosceles triangle with $OA=OB$ and $\angle A = \angle B = 45^\circ$ (see proof below):

Then we have $OA = OB = \sqrt 2 (1+\sqrt 2)$ and the minimum area of $\triangle AOB$ , $S = \dfrac {OA^2}2 = (1+\sqrt 2)^2 = 3 + 2\sqrt 2 \approx 5.828 \implies \lfloor 1000 S \rfloor = \boxed{5828}$ .

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Proof

Let $OB = a$ , $OA = b$ , and $\angle OBA = \theta$ . Then we have:

$\begin{cases} a = \cot \left(\dfrac \theta 2 \right) + 1 \\ b = \cot \left(45^\circ - \dfrac \theta 2 \right) + 1 \end{cases} \quad$ Let $t = \tan \dfrac \theta 2 \implies \begin{cases} a = \dfrac 1t + 1 = \dfrac {1+t}t \\ b = \dfrac {1+t}{1-t} + 1 = \dfrac 2 {1-t} \end{cases}$

The area of $\triangle AOB$ is given by $A = \dfrac {ab}2 = \dfrac {1+t}{t(1-t)}$ to find minimum $A$ , we first find the value of $t$ such that $\dfrac {dA}{dt} = 0$ and $\dfrac {d^2 A}{dt^2} > 0$ .

$\begin{aligned} \frac {dA}{dt} & = \frac {t(1-t) - (1+t)(1-2t)}{t^2(1-t)^2} = \frac {t^2+2t -1}{t^2(1-t)^2} & \small \blue{\text{Putting }\frac {dA}{dt} = 0} \\ t^2 + 2t - 1 & = 0 \\ t^2 + 2t + 1 & = 2 \\ (t+1)^2 & = 2 \\ \implies t & = \sqrt 2 - 1 & \small \blue{\text{Since }t > 0} \end{aligned}$

Since $\dfrac {d^2A}{dt^2} > 0$ , when $t= \sqrt 2 -1$ , $A$ is minimum when $t = \tan \dfrac \theta 2 = \sqrt 2 -1$ . $\implies \tan \theta = \dfrac {2(\sqrt 2-1)}{1-(\sqrt 2-1)^2} = 1 \implies \theta = 45^\circ$ .

The area of $\triangle {OAB}$ will be minimum when $|\overline {OA}|=|\overline {OB}|=2+\sqrt 2$

The minimum area will be

$\dfrac 12\times (2+\sqrt 2) ^2=3+2\sqrt 2\approx 5.828427$

Hence the required answer is $\boxed {5828}$ .