Area Ratio of Two Triangles

By Routh's theorem , if (as shown in the figure of the problem) $\dfrac {AD}{DB} = x$ , $\dfrac {BE}{EC} = y$ , and $\dfrac {CF}{FA} = z$ . Then the area of $\triangle IJK$ is given by:

$\begin{aligned} [IJK] & = \frac {(xyz+1)^2[ABC]}{(xy+y-1)(yz+z+1)(zx+x+1)} \\ \implies \frac {[IJK]}{[ABC]} & = \frac {\left(\left(\frac 23\right)^3-1\right)^2}{\left(\left(\frac 23\right)^2 + \frac 23+1\right)^3} = \frac {\left(-\frac {19}{27}\right)^2}{\left(\frac {19}9\right)^3} = \frac 1{19} \end{aligned}$

Therefore, $p+q = 1+19 = \boxed{20}$ .

Since we are interested in the ratio of areas, it suffices to consider a particular triangle $\triangle {ABC}$ , whose vertices are $A(0,0), B(5,10), C(15,0)$ . Then the position coordinates of $D, E, F$ are $(2,4), (9,6), (9,0)$ respectively. Equations of $\overline {AE},\overline {BF},\overline {CD}$ are $y=\dfrac{2x}{3}, y=\dfrac{5}{2}(9-x), y=\dfrac{4}{13}(15-x)$ respectively. Hence the position coordinates of $I, J, K$ are $(\frac{135}{19},\frac{90}{19}), (\frac{90}{19},\frac{60}{19}),(\frac{155}{19},\frac{40}{19})$ respectively. Hence, area of $\triangle {ABC}$ is $\dfrac{1}{2}\times 15\times 10=75$ , and of $\triangle {IJK}$ is $\dfrac{1}{2}|\frac{135}{19}(\frac{60}{19}-\frac{40}{19})+\frac{90}{19}(\frac{40}{19}-\frac{90}{19})+\frac{155}{19}(\frac{90}{19}-\frac{60}{19})|=\dfrac{75}{19}$ , and the required ratio is $\dfrac{1}{19}$ . Hence $p=1, q=19$ and $p+q=1+19=\boxed {20}$ .

We neglect the -tive sign in the equation.
If each cevian divide its side in the same ratio, the triangle formed in side is an equivalent.

Skewing $\triangle ABC$ to an equilateral triangle will preserve the ratio of areas, so let $a = AD = BE = CF$ and $b = AF = BD = CE$ . Also let $c = AE = BF = CD$ .

$\triangle ABE \sim \triangle AID$ by AA similarity, so $\frac{AI}{AB} = \frac{AD}{AE}$ or $\frac{AI}{a + b} = \frac{a}{c}$ or $AI = \frac{a(a + b)}{c}$ , and $\frac{ID}{BE} = \frac{AD}{AE}$ or $\frac{ID}{a} = \frac{a}{c}$ or $ID = \frac{a^2}{c}$ .

Now $AE = AI + IJ + JE$ , so $c = \frac{a(a + b)}{c} + IJ + \frac{a^2}{c}$ , which solves to $IJ = \frac{c^2 - ab - 2a^2}{c}$ .

The ratio of sides between $\triangle ABC$ and $\triangle IJK$ is $\frac{IJ}{AB} = \frac{c^2 - ab - 2a^2}{c(a + b)}$ , so the ratio of areas is $r = \frac{(c^2 - ab - 2a^2)^2}{c^2(a + b)^2}$ .

By the law of cosines on $\triangle ACD$ , $c^2 = a^2 + (a + b)^2 - 2a(a + b) \cos 60°$ , or $c^2 = a^2 + ab + b^2$ .

Substituting $c^2 = a^2 + ab + b^2$ into the ratio of areas $r = \frac{(c^2 - ab - 2a^2)^2}{c^2(a + b)^2}$ and simplifying gives:

$r = \frac{(b - a)^2}{a^2 + ab + b^2}$ .

In this question, $a = 2$ and $b = 3$ , so the area ratio of $\triangle ABC$ and $\triangle IJK$ is $r = \frac{(b - a)^2}{a^2 + ab + b^2} = \frac{(3 - 2)^2}{2^2 + 2 \cdot 3 + 3^2} = \frac{1}{19}$ , which means $p = 1$ , $q = 19$ , and $p + q = \boxed{20}$ .

See Routh's Theorem