Triangle-Ception I

Let's start with determining where $AD$ should intersect side $BC$ . Since triangle $ABD$ should have $\frac{1}{4}$ the total area of the triangle, we can deduce using $Area=\frac{1}{2}Base \times Height$ that point $D$ should be placed $\frac{1}{4}$ of the way down side $BC$ from vertex $B$ .

The same reasoning can be used to to determine where segment $DE$ should intersect side $AC$ . We have a triangle $ACD$ for which we need to divide into three triangles of equal area, so point $E$ should be $\frac{1}{3}$ of the way down side $AC$ from vertex $C$ .

Last, we can use once again the same reasoning from above to determine where segment $EF$ should intersect segment $AD$ . We need to split the remaining triangle $ADE$ into two triangles of equal area, so point $F$ should lie on the midpoint of $AD$ .

Now that we know our ratios, lets put some numbers to the segments. Because our side lengths for the equilateral triangle are $12$ ,

$BD=3, CD=9, CE=4, AE=8$ .

Using Law of Cosines :

$DE^{2}=4^{2}+9^{2}-(2)(4)(9) \cos (60^{o})$ $DE=\sqrt{61}$

Using Law of Cosines again, $AD^{2}=3^{2}+12^{2}-(2)(3)(12) \cos (60^{o})$

$AD=\sqrt{117}$

Since $F$ lies on the midpoint of $AD$ ,

$AF=DF=\frac{\sqrt{117}}{2}$

We now have all the information for triangle $ADE$ to use Stewart's Theorem , which yields us $(\frac{\sqrt{117}}{2})^{2} \times \sqrt{117}+EF^{2} \times \sqrt{117}=\frac{\sqrt{117}}{2} \times \sqrt{61}^{2}+\frac{\sqrt{117}}{2} \times 8^{2}$ .

Solving for $EF^{2}$ gives us

$EF^{2}=\frac{133}{4}$

which certainly cannot be reduced. Our answer is then

$\large 133+4=\boxed{137}$

Find DC=9,DB=3.EC=4,EA=8,AD^2=117, ED^2=61, using cosine theorem and area division in four equal parts.

Now use Pythagorean’s relation to find EF^2=133/4 and answer as 133+4=137