It Shouldn't be this Hard

Let the $x$ -coordinate of $B$ be $n>0$ , and since the ratio of the $y$ -coordinate and the $x$ -coordinate of the coordinate $C$ is $\tan(60^\circ) = \sqrt3$ , we can denote the coordinate of $C$ to be $\left(m, m\sqrt3\right)$ , where $0<m\leqslant 3$ .

Define $D$ as the square of the distance $BC$ , then using the distance formula , $D = (m\sqrt3)^2 + (m-n)^2 = 3m^2 + (m-n)^2.$

Since $P(3,3)$ passes through the straight line $BC$ , the gradients of the straight line $CP, PB, BC$ are all identical: $\dfrac{m\sqrt3 - 3}{m-3} = \dfrac3{3-n} = \dfrac{m\sqrt3}{m-n}.$

Setting $n$ as the subject gives $n = \dfrac{3m}{m\sqrt3 - 3} \left(m \sqrt3 - 1\right) .$

Thus, upon substitution, we can express $D$ as a single variable function, $D= \dfrac{6m^2 (2m^2 - 3(1 + \sqrt3) m + 9)}{(m \sqrt3 - 3)^2} .$

At the critical point, $\dfrac {d}{dm} D = 0 \quad \implies \quad m \cdot \dfrac{(24 \sqrt3) m^3 + m^2 (-198 - 18\sqrt3) + m(162 + 162\sqrt3) - 324}{3 \sqrt3 m^3 - 27 m^2 + 27 \sqrt3 m - 27} = 0$

Simplifying this gives $m^3 (4\sqrt3) + m^2 (-33 - 3\sqrt3) + m(27 + 27\sqrt3) - 54 = 0$ . Using Cardano's method , we get $m = \dfrac{11 + \sqrt3}{4\sqrt3} + \dfrac q{\sqrt[6]{27648}} - \dfrac{7\sqrt3 - 8}{\sqrt[6]{6912} \cdot q} \approx 2.504653757,$ where $q = \sqrt[3]{391 - 195 \sqrt3 + \sqrt{258876 - 147744 \sqrt3}}$ .

Checking the endpoints of the domain $0 < m \leqslant 3$ , the critical point of $D(2.504653757)$ is smaller than $D(0)$ and $D(3)$ . Hence, this critical point must be a minimum value.

We have $\min(D) \approx 21.39856894117064819658652$ with coordinates $B \approx (4.1104861276 , 0), \quad \quad C \approx (2.504653757949, 4.338187564).$

The answer is $\left \lfloor 10^5 \cdot \sqrt{21.398568941170648196586523\ldots} \right \rfloor = \boxed{462585}.$

---

Addendum:

1. Even if we allow $n< 0$ , that is, the $x$ -coordinate of $B$ is negative, then the straight line $BC$ can never pass through $(3,3)$ . A simple graphing shows that the straight $AC$ and $BC$ does not intersect at the coordinate $C$ .

2. The maximum value of the length $BC$ is $3\sqrt3$ , and it occurs when $ABC$ is a right triangle at $\angle B$ , with $m=n = 3$ .

Let $\angle B = \theta$ . By sine rule, $\dfrac {CP}{AP} = \dfrac {\sin \angle CAP}{\sin \angle ACP} = \dfrac {\sin 15^\circ}{\sin (120^\circ - \theta)} = \dfrac {\sin 15^\circ}{\sin (60^\circ + \theta)}$ . And $\dfrac {BP}{AD} = \dfrac {\sin \angle PAB}{\sin \angle PBA} = \dfrac {\sin 45^\circ}{\sin \theta}$ . $\implies \dfrac {CP}{BP} = \dfrac {\sin 15^\circ \sin \theta}{\sin 45^\circ \sin (60^\circ + \theta)}$ .

Let $BC=a$ . Then $\dfrac {CP}{BP} = \dfrac {a\sin \theta - 3}3 = \dfrac {\sin 15^\circ \sin \theta}{\sin 45^\circ \sin (60^\circ + \theta)}$ and:

$\begin{aligned} a & = \frac {3 \sin 15^\circ \sin \theta+ 3 \sin 45^\circ \sin (60^\circ + \theta)}{\sin 45^\circ \sin \theta \sin (60^\circ + \theta)} \\ & = \frac {3(\cos (\theta - 15^\circ)-\cos(\theta + 105^\circ))}{\sin 45^\circ (\cos 60^\circ - \cos (2\theta + 60^\circ))} \\ & = \frac {3\sin 60^\circ \sin (\theta + 45^\circ)}{\sin 45^\circ(\cos 60^\circ - \cos (2\theta + 60^\circ))} \\ & = \frac {6\sqrt 6 \sin (\theta + 45^\circ)}{1 - 2\cos (2\theta + 60^\circ)} \\ \frac {da}{d\theta} & = \frac {6\sqrt 6 (\cos (\theta + 45^\circ)(1 - 2\cos (2\theta + 60^\circ)) - 4 \sin (\theta + 45^\circ)\sin (2\theta + 60^\circ))}{(1 - 2\cos (2\theta + 60^\circ))^2} \end{aligned}$

As there is no maximum value of $a$ , to find $\min (a)$ , we find the $\theta$ such that $\dfrac {da}{d\theta} = 0$ .

$\begin{aligned} \cos (\theta + 45^\circ)(1 - 2\cos (2\theta + 60^\circ)) & = 4 \sin (\theta + 45^\circ)\sin (2\theta + 60^\circ) \\ \frac {1 - 2\cos (2\theta + 60^\circ)}{4\sin (2\theta + 60^\circ)} & = \tan (\theta + 45^\circ) \\ \frac {1-\cos 2\theta + \sqrt 3\sin 2\theta}{2\sqrt 3\cos 2\theta + 2\sin 2\theta} & = \frac {1+\tan \theta}{1-\tan \theta} & \small \blue{\text{Let }t = \tan \theta} \\ \frac {1+t^2 - 1 + t^2 + 2\sqrt 3t}{2\sqrt 3 - 2\sqrt 3 t^2 + 4t} & = \frac {1+t}{1-t} \\ \frac {\sqrt 3t + t^2}{\sqrt 3 + 2t - \sqrt 3 t^2 } & = \frac {1+t}{1-t} \\ (\sqrt 3-1)t^3 - t^2 - 2t - \sqrt 3 & = 0 \end{aligned}$

Solving the cubic equation, we have $t \approx 2.701519565$ and $\min (a) = \dfrac {3\sqrt 2(1+t)\sqrt{1+t^2}}{\sqrt 3t +t^2} \approx 4.625858725$ . Therefore $\lfloor 10^5 \min(a)\rfloor = \boxed{462585}$ .

Draw $AP$ , which by the distance formula a length of $AP = \sqrt{(3 - 0)^2 + (3 - 0)^2} = 3\sqrt{2}$ .

Since by the slopes formula $\tan \angle PAB = \frac{3 - 0}{3 - 0}$ , $\angle APB = 45°$ , this makes $\angle PAC = \angle CAB - \angle PAB = 60° - 45° = 15°$ .

Let $\theta = \angle APB$ . Then from the straight line $\angle APC = 180° - \theta$ and by the angle sums of $\triangle APC$ and $\triangle APB$ , $\angle PBA = 135° - \theta$ and $\angle PCA = \theta - 15°$ .

By the law of sines on $\triangle APC$ , $CP = \cfrac{3\sqrt{2} \sin 15°}{\sin (\theta - 15°)}$ , and by the law of sines on $\triangle APB$ , $PB = \cfrac{3\sqrt{2} \sin 45°}{\sin (135° - \theta)}$ , so $BC = \cfrac{3\sqrt{2} \sin 15°}{\sin (\theta - 15°)} + \cfrac{3\sqrt{2} \sin 45°}{\sin (135° - \theta)}$ .

The minimum will occur when $\cfrac{dBC}{d\theta} = \cfrac{3\sqrt{2} \sin 15° \cos (\theta - 15°)}{- \sin^2 (\theta - 15°)} + \cfrac{3\sqrt{2} \sin 45° \cos (135° - \theta)}{\sin^2 (135° - \theta)} = 0$ , which solves numerically to $\theta \approx 65.3126396562°$ .

Substituting $\theta$ value into the equation for $BC$ gives $BC \approx 4.625858725$ , so $\lfloor 10^5 \cdot BC \rfloor = \boxed{462585}$ .