Cake cutting with shortest cut

We can see that the polygons formed must be a triangle and a quadrilateral, with the triangle having smaller area. We can see if that there existed a minimal cut with $\frac{[\text{Triangle}]}{[\text{Quadrilateral}]} = 2$ , then there exists a smaller cut parallel to this cut with $\frac{[\text{Triangle}]}{[\text{Quadrilateral}]} = \frac{1}{2}$ . Thus, it remains to find the smallest line segment which creates a triangle with $\frac{1}{3}$ of the area of the original triangle.

We have three cases, one where the line intersects the sides with lengths $13$ and $14$ , $14$ and $15$ , and $15$ and $13$ . We do the one where it intersects the sides with lengths $15$ and $14$ .

Let the intersection of the two sides be called $A$ . Then, call the intersection of the segment and the side with length $15$ point $B$ , and the other intersection point with the side with length $14$ is to be called $C$ . Let $AB = 15a$ . We can then see that because we must have the area of this triangle equal to a third of the original, we must have

$\huge \frac{\frac{(AC)(15a)(\sin{ \angle BAC})}{2}}{\frac{(14)(15)(\sin{ \angle BAC})}{2}} = \frac{1}{3}$ , giving us $AC = \frac{14}{3a}$ .

Now, we use law of cosines to get the length of $BC$ .

$BC^2 = (15a)^2 + (\frac{14}{3a})^2 - 2(15)(\frac{14}{3})(\cos{\angle BAC}) = (15a)^2 + (\frac{14}{3a})^2 - 2(15)(\frac{14}{3})(\frac{3}{5})$

Where $(\cos{\angle BAC}) = (\frac{3}{5})$ can be obtained from the original triangle.

This gives us

$BC^2 = (15a)^2 + (\frac{14}{3a})^2 - 84 \geq 2(15a)(\frac{14}{3a}) - 84 = 56$

Where the last inequality was obtained by AM-GM .

But then, equality holds when $15a = \frac{14}{3a}$ , i.e. when the two sides of the triangle are equal.

This actually brings up a simple fact: Over all triangles with a given angle and area, the one that minimizes the length of the side opposite the given angle is the isosceles triangle with the given angle as the vertex angle.

This allows us to acquire an alternative method for looking for the shortest segment satisfying the givens: look for the segment created with the the triangles with the given angles, with area $\frac{1}{3}$ of the original, and with two sides including the angle equal.

This gives us the minimum value of the segment for this case to be $\sqrt{56} = 2\sqrt{14}$ . Doing the two other cases analogously yields the possible minimum values of $8$ and $\sqrt{\frac{140}{3}}$ . Of these, the smallest is $2 \sqrt{14}$ , so our answer is $2 + 14 = \boxed{16}$ .