SAT1000 - P931

Let the measure of the angle between $AB$ and line $l_2$ be $\theta$ . Then we have:

$\begin{cases} \sin \theta = \dfrac 1l & ...(1) \\ \sin(60^\circ - \theta) = \dfrac 2l & ...(2) \end{cases}$

From $(2)$ :

$\begin{aligned} \frac {\sqrt 3}2 \cos \theta - \frac 12 \sin \theta & = \frac 2l = 2 \blue{\sin \theta} & \small \blue{\text{As }(1) : \ \sin \theta = \frac 1a} \\ \frac {\sqrt 3}2 \cos \theta & = \frac 52 \sin \theta \\ \tan \theta & = \frac {\sqrt 3}5 \\ \implies \sin \theta & = \frac {\sqrt 3}{\sqrt{5^2+3}} = \sqrt{\frac 3{28}} \\ (1): \quad \implies l & = \frac 1{\sin \theta} = \sqrt{\frac {28}3} \approx 3.055050463 \\ \implies \lfloor 1000l \rfloor & = \boxed{3055} \end{aligned}$

Place the figure onto the xy-plane. Let the difference between the x-coordinates of A and B be $a$ and let the difference between the x-coordinates of B and C be $b$ . It follows that $a^2 + 1 = l^2$ , $b^2 + 4 = l^2$ , and $(a-b)^2 + 9 = l^2$ .

Using the first two equations, we conclude that $a^{2}=b^{2}+3$ . Substituting this into the third equation and expanding yields $2b^{2}-2b\sqrt{b^{2}+3}+12=b^{2}+4$ which is equivalent to $3b^{4}-4b^{2}-64=0$ for positive $b$ . Applying the quadratic formula and solving for $b$ yields $b = \frac{4\sqrt{3}}{3}$ . Substituting this value for $b$ into $b^2 + 4 = l^2$ gives us $l = \sqrt{\frac{28}{3}}$ . $\lfloor 1000l \rfloor = \boxed{3055}$ .

See this :) what a coincidence!