Christmas Streak 50/88: Geometry + Number Theory

$\boxed{\text{Geometry}}$

The congruence of $\triangle \mathrm{ADB}$ and $\triangle \mathrm{AEC}$ is trivial. $\Rightarrow~ \angle\mathrm{ABD}=\angle\mathrm{ACE}.$

Since $\angle\mathrm{AEC}=\angle\mathrm{FEB},$ we see that $\triangle \mathrm{AEC}\sim\triangle\mathrm{FEB}.$

Therefore $\overline{\mathrm{BE}}:\overline{\mathrm{CE}}=\overline{\mathrm{FE}}:\overline{\mathrm{AE}},$ leading to $\overline{\mathrm{BE}}\times\overline{\mathrm{AE}}=\overline{\mathrm{CE}}\times\overline{\mathrm{FE}}=630.$

We know that $\overline{\mathrm{BE}}=x^2+y^2+2-(2x+y^2+1)=x^2-2x+1=(x-1)^2,$ and $\overline{\mathrm{AE}}=2x+y^2+1.$

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$\boxed{\text{Number Theory}}$

$(x-1)^2(2x+y^2+1)=630=2\times3^2\times5\times7$

Since $(x-1)^2$ is a square while also being a divisor of 630, it's either $(x-1)^2=1$ or $(x-1)^2=9.$

$x=2,~y=25$ or $x=4,~y=\sqrt{61}.$

The latter is impossible, so we know that $x=2$ and $y=25.$

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$\boxed{\text{Geometry}}$

Then $\overline{\mathrm{AD}}=630,$ and $\overline{\mathrm{AB}}=631.$

So we know that $\triangle\mathrm{ADB}=\dfrac{1}{2}\times630\times631\times\dfrac{\sqrt{3}}{2}=\dfrac{198765\sqrt{3}}{2}.$

Therefore, $a+b+c=\boxed{198770}.$