Reach for the Summit - M-S5-A1

Position coordinates of $C$ are $\left (\dfrac {35}{9},0\right )$

Equation of perpendicular bisector of $\overline {MN}$ is $x=1$ , of $\overline {AM}$ is

$y-2=-\dfrac 34 (x-\frac 32)$

Solving we get the position coordinates of the circumcentre of $\triangle {AMN}$ as $\left (1,\dfrac {19}{8}\right )$

Equation of perpendicular bisector of $\overline {BC}$ is $x=-\dfrac 59$ ,

of $\overline {AB}$ is $y-2=-2(x+1)$

Solving we get the position coordinates of the circumcentre of $\triangle {ABC}$ as

$\left (-\dfrac 59,\dfrac {10}{9}\right )$

So, $x_1=1,y_1=\dfrac {19}{8},x_2=-\dfrac 59,y_2=\dfrac {10}{9}$

Hence the required answer is $\boxed {8527}$ .

Since $A(3,4)$ , we note that $AM = 5$ . As $BM=AM=5$ , $\triangle AMB$ is isosceles and $\angle ABM = \angle BAM = \angle CAN = \theta$ . We also note that $\triangle ABC$ and $\triangle CAN$ are similar. Then we have:

$\frac {AC}{NC} = \frac {AB}{AN} = \frac {BC}{AC}$

Then $\dfrac {AC}{NC} = \dfrac {AB}{AN} = \dfrac {\sqrt{8^2+4^2}}{\sqrt{1^2+4^2}} = \sqrt{\dfrac {80}{17}}$ . Let $NC = a$ ; then $AC = \sqrt{\dfrac {80}{17}}a$ . And also:

$\begin{aligned} \frac {BC}{AC} & = \frac {AB}{AN} \\ BC & = \frac {AB}{AN} \times AC \\ 7+a & = \sqrt{\frac {80}{17}} \times \sqrt{\frac {80}{17}} a = \frac {80}{17}a \\ 119 + 17a & = 80a \\ \implies a & = \frac {119}{63} \end{aligned}$

We can find the coordinates of circumcenter by first finding the circumradius using the formula $2r = \dfrac a{\sin A}$ , where $a$ is the side length opposite $\angle A$ in a triangle.

For $\triangle AMN$ , $r_1 = \dfrac {AN}{2 \sin \angle AMN} = \dfrac {\sqrt{17}}{2 \times \frac 45} = \dfrac {5\sqrt{17}}8$ . Noting that $x_1$ is the midpoint of $MN$ or $x_1 = 1$ , then

$\begin{aligned} x_1^2 + y_1^2 & = r_1^2 \\ 1^2 + y_1^2 & = \frac {25 \times 17}{64} \\ \implies y_1 & = \frac {19}8 \end{aligned}$

Similarly for $\triangle ABC$ , $r_2 = \dfrac {AC}{2 \sin \angle ABC} = \dfrac {\sqrt{\frac {80}{17}}\times \frac {119}{63}}{2\times \frac 1{\sqrt 5}} = \dfrac {10\sqrt{17}}9$ . $x_2 = \dfrac {-5+2+a}2 = - \dfrac 59$ , and $\\ y_2 = \sqrt{\left(\dfrac {10 \sqrt{17}}9 \right)^2-\left(-\dfrac 59+5\right)^2} = \dfrac {10}9$ .

Therefore $\lfloor 10^6(x_1 + 2y_1 + 3x_2 + 4y_2)\rfloor = \boxed{8527}$ .