Inspired by Calvin Lin: Any part 1

Here is the solution: $AE\times AF = 3\times GE\times GF$

$\implies 2AE\times 2AF = 6\times GE\times 2GF$

$\implies AC \times AB = 6\times GE\times GC$ ...(1)

But we know that the medians divide a triangle into six $equal$ smaller triangles. So,

$Area of \Delta ABC = 6\times Area of \Delta GEC$

$\implies AC\times AB\times \sin(\angle A) = 6\times GE\times GC\times \sin(\angle EGC)$

Using (1), $\implies \sin(\angle A) = \sin(\angle EGC)$

$\implies \angle A = \angle EGC$ (** justification for this step: at last)

This makes quadrilateral $AEGF$ cyclic. Applying Ptolemy's theorem,

$AE\times GF + AF\times GE = AG \times EF$

$\implies 2AE\times 3GF + 2AF\times 3GE = 2\times \frac{3}{2}\times AG \times 2EF$

$\implies AC\times CF + AB\times BE = 2\times AD \times BC$

$\implies$ the answer is $2\times BC = 20.$

Pending justification: If $180 - \angle A = \angle EGC$ , this would mean:

$\angle A = \angle EGF = \angle BGC$ ,

That is $BC$ subtends equal angles at $A$ as well as at $G$ $\implies$ $G$ lies on the circumcircle i.e. outside of the triangle (since only the vertices of a triangle are the triangle's points on this circle, the other points of the circle are necessarily outside)

But this is impossible: Think of the centroid's role as the centre of mass, hundreds of other reasonings!

Let the medians at $A,B,C$ have lengths $m_a,m_b,m_c$ . Then Apollonius' Theorem tells us that $m_a \; = \; \tfrac12\sqrt{2b^2 + 2c^2 - a^2}$ and similarly for $m_b,m_c$ . The condition $AE \times AF = 3 GE \times GF$ can be read as $\begin{array}{rcl} \tfrac12b \times \tfrac12c & = & 3 \times \tfrac13m_b \times \tfrac13m_c \\ m_bm_c & = & \tfrac34bc \\ \sqrt{(2a^2 + 2c^2 - b^2)(2a^2 + 2b^2 - c^2)} & = & 3bc \end{array}$ Squaring and simplifying, we obtain $(a^2 + b^2 + c^2)(2a^2 - b^2 - c^2) \; = \; 0$ so that $b^2 + c^2 = 2a^2$ , and therefore $m_a = \tfrac{\sqrt{3}}{2}a$ .

Now $\begin{array}{rcl} (AB \times BE + AC \times CF)^2 & = & (cm_b + bm_c)^2 \; = \; c^2m_b^2 + 2bcm_bm_c + b^2m_c^2 \\ & = & \tfrac14c^2(2a^2 + 2c^2 - b^2) + \tfrac32b^2c^2 + \tfrac14b^2(2a^2 + 2b^2 - c^2) \\ & = & \tfrac12(a^2+b^2+c^2)(b^2+c^2) \; = \; 3a^4 \end{array}$ Thus $\frac{AB \times BE + AC \times CF}{AD} \; = \; \frac{\sqrt{3}a^2}{m_a} \; = \; 2a \; = \; \boxed{20}$ It is worth noting that the triangle can be equilateral, with $a=b=c=10$ and $m_a = m_b = m_c = 5\sqrt{3}$ .