Grazing Through A Hexagon

Relevant wiki: 2D Coordinate Geometry - Problem Solving

We know that the projectile's path is in the form of a parabola, hence we shall set up a coordinate system to deal with the problem.

Without loss of generality, let the origin be the midpoint of the side of the hexagon on the ground. Also, we shall assume that the hexagon has length 1. Then, we observe that the projectile's path would have to pass through the points $(-1,\frac{\sqrt{3}}{2}),(1,\frac{\sqrt{3}}{2}),(-0.5,\sqrt{3}),(0.5,\sqrt{3})$ Note that the parabola that depicts the path of the projectile is of the form $y=ax^2+c$ . Now, plugging in each of the points, we obtain $a=-\frac{2}{\sqrt{3}},c=\frac{7}{2\sqrt{3}}$ .

In order to obtain the angle of projection, we must first obtain the gradient of this parabola when $y=0$ . Solving the equation $-\frac{2}{\sqrt{3}}x^2+\frac{7}{2\sqrt{3}}=0$ we obtain the x coordinate of the point of projection to be $x=-\frac{\sqrt{7}}{2}$ Differentiating the parabola, we obtain $\frac{dy}{dx}=-\frac{4}{\sqrt{3}}x$ Subbing in $x=-\frac{\sqrt{7}}{2}$ , we get the gradient to be $2\sqrt{\frac{7}{3}}$ . The angle of projection is thus $\arctan{2\sqrt{\frac{7}{3}}}$ and the answer is $2+7+3=12$ .