Is the information sufficient 8?

The problem came from this.

Consider an equilateral triangle $ENA$ , with a point $P$ located inside that triangle such that, $\overline{EP} = 19 \ , \ \overline{PN} = 180 \ , \ \overline{PA} = 181 \ \cdot$

Now, reflect point $P$ with respect to line segment $EN$ . Let's name the point of reflection $U \ \cdot$

So, we can conclude that: $\overline{EU} = 19 \ , \ \overline{UN} = 180 \ \cdot$

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Next, reflect point $P$ again, this time respect to line segment $EA$ . Let's name the point of reflection $S \ \cdot$

Again, we can conclude that: $\overline{ES} = 19 \ , \ \overline{SA} = 181 \ \cdot$

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Lastly, reflect point $P$ with respect to line segment $AN$ . Let's name the point of reflection $H \ \cdot$

So, we can conclude that $\overline{HN} = 180 \ , \ \overline{HA} = 181 \ \cdot$

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It is left to the readers how $m \left(\angle UES\right) = m\left(\angle UNH\right) = m\left(\angle HAS\right) = 120^{\circ} \cdot$

Connecting the 6 points will give us the Hexagon, named EUNHA S .

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As $\triangle EUS$ is isosceles, $\implies \angle EUS = 30^{\circ}$

As $\triangle UNH$ is isosceles, $\implies \angle NUH = 30^{\circ}$

As $\triangle SUH$ is right (right-angled at $U$ ), $\implies \angle EUN = 30^{\circ} + 30^{\circ} + 90^{\circ} = 150^{\circ} \ \cdot$

As the quadrilateral $EUNP$ is a kite, $\implies \angle EPN = 150^{\circ} \ \cdot$

and therefore,

$\left( m \left(\overline{EA}\right) \right)^2 = \left( m \left(\overline{EN}\right) \right)^2 = 19^2 + 180^2 - 2(19)(180)( \cos (150^{\circ})) = 32761 + 3420\sqrt{3} = a + b\sqrt{c} \\ \implies a = 32761 \ , \ b = 3420 \ , \ c = 3 \\ \implies \left\lfloor \sqrt{a + b + c} \right \rfloor = \left\lfloor \sqrt{32761 + 3420 + 3} \right \rfloor = \boxed{190}$

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This is what is it look like. :)