Expected area of Routh's triangle

Hi everyone! Wish you a very happy New Year! Now onto my first post of the year...

I recently came up a problem in Geometric probability which really had a stunning result that took by surprise. 

Consider a triangle $ABC$. Choose points $D$, $E$ and $F$ uniformly randomly on sides $BC$, $CA$ and $AB$ respectively. The three cevians $AD$, $BE$ and $CF$ form a smaller central triangle, which we call as Routh's triangle, between them as shown in the figure below.

Now what might be the expected value of the ratio of the area of $\triangle PQR$ to that of $\triangle ABC$?

Let the random variable $T$ denote the ratio we are interested in. Then given the positions of $D$, $E$ and $F$, Routh's theorem tells us,

$\displaystyle \frac{\triangle PQR}{\triangle ABC}=\frac{(uvw-(1-u)(1-v)(1-w))^2}{(1-u+uv)(1-v+vw)(1-w+wu)}$

where $DC=u\cdot BC$, $EA=v\cdot CA$ and $FB=w\cdot AB$.

Because $u$, $v$ and $w$ are IID uniform random variables by definition,

$\displaystyle \mathbb{E}(T)=\int_0^1\int_0^1\int_0^1 \frac{\triangle PQR}{\triangle ABC} \,dudvdw$

This integral is very unwieldy despite the symmetry between the variables and I can't see a clear way of solving this triple integral. We can solve this with Wolfram Alpha, but let's see how we can solve this without a CAS.

The trick is to solve this problem is shift our focus on triangles $ARC$, $BPA$ and $CQB$. In fact, over the course of proving Routh's theorem we get as an intermittent result that

$\displaystyle \frac{\triangle{ARC}}{\triangle{ABC}}=\frac{u(1-w)}{1-w+wu}$

and similar results for $\triangle{BPA}$ and $\triangle{CQB}$. Because,

$\triangle{ABC}=\triangle{PQR}+\triangle{ARC}+\triangle{BPA}+\triangle{CQB}$

we can now use the linearity of expectation and idea of symmetry to get

$\displaystyle \mathbb{E}(T)=1-3\int_0^1\int_0^1\frac{u(1-w)}{1-w+wu}\,dudw=1-3\int_0^1\int_0^1\frac{(1-w)(1-u)}{1-wu}\,dudw$

Now using Geometric series and term-by-term integration it is easy

$\displaystyle \int_0^1\int_0^1\frac{1}{1-wu}\,dwdu=\zeta(2)$, $\displaystyle \int_0^1\int_0^1\frac{wu}{1-wu}\,dwdu=\zeta(2)-1$ and

$\displaystyle \int_0^1\int_0^1\frac{u}{1-wu}\,dwdu=\int_0^1\int_0^1\frac{w}{1-wu}\,dwdu=1$

Plugging in all these values, we get the amazing amazing result that $E(T)=10-\pi^2$

Because we've calculated an expected value of a non-negative random variable, we know that this quantity is greater than zero and hence $10>\pi^2$.

Such a beautiful result that is atleast as remarkable as A cute sum of Ramanujan!!

Until then
Yours Aye
Me