Median Triangle Areas

For a regular triangle with side length $a$ , the median length will be $m=\frac{\sqrt{3}a}{2}$ . Then the area ratio will be $\frac{m^2}{a^2}\ = \frac{3}{4}$ . And the answer is $7$ .

Let the position vectors of $A, B$ and $C$ be a , b and c respectively. Then those of $D, E$ and $F$ are $\dfrac{1}{2}$ ( b + c ), $\dfrac{1}{2}$ ( c + a ) and $\dfrac{1}{2}$ ( a + b ) respectively. Then area of $\triangle ABC$ is $\dfrac{1}{2}|$ axb + bxc + cxa $|$ , and that of $\triangle DEF$ is $\dfrac{1}{2}|$ ( $\dfrac{1}{2}$ ( b + c )- a ) x ( $\dfrac{1}{2}$ ( a + b )- c ) $|=\dfrac{3}{8}|$ axb + bxc + cxa $|$ . Therefore the ratio of the area of $\triangle ABC$ and the area of $\triangle DEF$ is $\dfrac{1}{2}\times \dfrac{8}{3}=\dfrac{4}{3}$ . So $p=4, q=3$ and $p+q=\boxed 7$

It is easy to determine the answer by examining the specific case of an equilateral triangle. If the sides of the original triangle are all $1$ , then the median (also the altitude) is $\frac{\sqrt{3}}{2}$ , so the ratio of the sides of $\triangle ABC$ to $\triangle DEF$ is $\frac{2}{\sqrt{3}}$ , making the ratio of the areas $\frac{4}{3}$ , so that $p = 4$ , $q = 3$ , and $p + q = \boxed{7}$ .

The general case is more difficult to prove. According to Appolonius' Theorem for medians, $m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$ , $m_b^2 = \frac{2a^2 + 2c^2 - b^2}{4}$ , and $m_c^2 = \frac{2a^2 + 2b^2 - c^2}{4}$ . By Heron's Formula , the area of $\triangle DEF$ is:

$A_{\triangle DEF} = \frac{1}{4}\sqrt{(m_a^2 + m_b^2 + m_c^2)^2 - 2(m_a^4 + m_b^4 + m_c^4)}$

$A_{\triangle DEF} = \frac{1}{4}\sqrt{(\frac{2b^2 + 2c^2 - a^2}{4} + \frac{2a^2 + 2c^2 - b^2}{4} + \frac{2a^2 + 2c^2 - c^2}{4})^2 - 2((\frac{2b^2 + 2c^2 - a^2}{4})^2 + (\frac{2a^2 + 2c^2 - b^2}{4})^2 + (\frac{2a^2 + 2c^2 - c^2}{4})^2)}$

$A_{\triangle DEF} = \frac{1}{4}\sqrt{\frac{9}{16}(a^2 + b^2 + c^2)^2 - 2\frac{9}{16}(a^4 + b^4 + c^4)}$

$A_{\triangle DEF} = \frac{3}{4}\frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}$

$A_{\triangle DEF} = \frac{3}{4}A_{\triangle ABC}$

which means $\frac{A_{\triangle ABC}}{A_{\triangle DEF}} = \frac{4}{3}$ , so $p = 4$ , $q = 3$ , and $p + q = \boxed{7}$ .