Expectation of overlapping triangles

Question

Let E denotes a triangle PQR with PQ = QR = 2 unit and angle Q = 90 degree . L , M and N are mid-points of PQ , QR and RP respectively . Let A denotes triangle PLN , B denotes triangle LQM , C denotes triangle NMR and X denotes triangle LMN . If A , B and C can move freely inside E but must keep parallel with E in moving . Find the expected area of the overlapped portion of A and B .

Answer 1

Let's use a coordinate system in which the coordinates of $P$, $Q$, and $R$ are (respectively) $(0,0)$, $(2,0)$, and $(2,2)$. Now let $\triangle A(s,t)$ be a triangle congruent to triangle $\triangle PLN$, with its three sides parallel to the sides of $\triangle PQR$ but "moved" so that its vertices are at coordinates $(s,t)$, $(s+1,t)$, and $(s+1,t+1)$; and let $\triangle B(u,v)$ be a triangle congruent to triangle $\triangle LQM$, with its three sides parallel to the sides of $\triangle PQR$ but "moved" so that its vertices are at coordinates $(u,v)$, $(u+1,v)$, and $(u+1,v+1)$.

To place triangles $\triangle A(s,t)$ and $\triangle B(u,v)$ at random locations, we have to specify what random distribution they should follow. Notice that $(s,t)$ and $(u,v)$ must each be somewhere inside or on the original location of triangle $\triangle A$; for any measurable subregion of that triangle, let the probability that $(s,t)$ lies within that subregion be proportional to the area of the subregion, and similarly for $(u,v)$.

To find possible values for $s$ and $t$, you can first take $0\leq s \leq 1$ and then take $0 \leq t \leq s$. Likewise, for $u$ and $v$ you can first take $0\leq u \leq 1$ and then take $0 \leq v \leq u$. So if $f(s,t,u,v)$ is the area of the overlapping portions of triangles $\triangle A(s,t)$ and $\triangle B(u,v)$, the expected area of the overlapping portions of the triangles is $$ E(f(s,t,u,v)) = \frac{\displaystyle\int_{s=0}^1 \displaystyle\int_{t=0}^s \displaystyle\int_{u=0}^1 \displaystyle\int_{v=0}^u f(s,t,u,v) \, dv\, du\, dt\, ds} {\displaystyle\int_{s=0}^1 \displaystyle\int_{t=0}^s \displaystyle\int_{u=0}^1 \displaystyle\int_{v=0}^u 1 \, dv\, du\, dt\, ds}. $$ In order to write this in a different way, let $V$ be the set of all coordinates $(s,t,u,v)$ such that $0\leq s\leq 1$, $0\leq t\leq s$, $0\leq u\leq 1$, and $0\leq v\leq u$, and let $\mu$ be a suitable measure function over $(s,t,u,v)$-space. In particular, this means $$\mu(V) = \int_{s=0}^1 \int_{t=0}^s \int_{u=0}^1 \int_{v=0}^u 1 \,dv\,du\,dt\,ds = \frac14,$$ that is, $\mu(V)$ gives the four-dimensional "volume" of $V$. Let $\newcommand{X}{\mathbf X} \X$ be a random set of coordinates within $V$. Then the expected area of the overlapping portions of the triangles can be written $E(f(\X))$, and $$ E(f(\X)) = \frac{\displaystyle\int_{\X\in V} f(\X) \, d\mu(V)}{\mu(V)}. $$

The advantage of writing the expected value this way (aside from a shorter formula) is that this integral, which integrates a bounded function over a finite region, is sufficiently "nice" that we can relatively freely choose the order in which we "add up" the $f(\X)\, d\mu(V)$ values. We are not obliged to choose one of the variables $s,t,u$, and $v$ for each of four nested integrals.

Now consider the different ways in which $\triangle A(s,t)$ and $\triangle B(u,v)$ can overlap. It is possible for the two triangles to be identical, or for at least one side of $\triangle A(s,t)$ to be collinear with one side of $\triangle B(u,v)$, but those cases have zero probability. With probability $1$, one of the vertices of either $\triangle A(s,t)$ or $\triangle B(u,v)$ will lie inside the other triangle; so $V$ can be partitioned in six parts, each part obtained by choosing one of the two triangles, then choosing one of the three vertices of that triangles to lie inside the other triangle.

Consider just the portion of $V$ in which the lower-left vertex of $\triangle B(u,v)$ lies within $\triangle A(s,t)$. Give the name $V_1$ to this part of $V$.

Now let $p = 1 + s - u$. Then $p$ is the length of either leg of the right triangle formed by the overlapping portions of $\triangle A(s,t)$ and $\triangle B(u,v)$; that is, the area of overlap is $\frac 12 p^2$. By considering all possible pairs of triangles with a given value of $p$, it should be clear that these are just the pairs of triangles for which the coordinates $(s,t)$ satisfy $0 \leq s \leq p$ and $0 \leq t \leq s$. Moreover, if we let $q = v - t$, then we can integrate over all of $V_1$ by letting $0\leq p \leq 1$, $0\leq q\leq 1-p$, $0 \leq s \leq p$, and $0 \leq t \leq s$: \begin{align} \int_{\X\in V_1} f(\X) \,d\mu(V_1) &= \int_{p=0}^1 \int_{s=0}^p \int_{t=0}^s \int_{q=0}^{1-p} \frac12 p^2 \,dq\,dt\,ds\,dp \\ &= \int_{p=0}^1 \int_{s=0}^p \int_{t=0}^s (1-p) \frac12 p^2 \,dt\,ds\,dp \\ &= \int_{p=0}^1 \left(\frac12 p^2\right) (1-p) \frac12 p^2 \,dp \\ &= \frac14 \int_{p=0}^1 (p^4 - p^5) \,dp \\ &= \frac14 \left(\frac15 - \frac16\right) = \frac{1}{120}. \\ \end{align}

We will get the same value by integrating over each of the other five subregions of $V$. This is obvious for the three subregions in which an acute-angled vertex of one triangle lies inside the other triangle; for the two subregions in which a right-angled vertex of one triangle is inside the other triangle, we can either apply an area-preserving linear transformation to the plane so that $\triangle PQR$ is mapped to an equilateral triangle, in which case the complete symmetry of all six subregions of the partition is obvious, or we can consider the partition in which the right-angled vertex of $\triangle B(u,v)$ is inside $\triangle A(s,t)$ and assign $p$ and $q$ so that the integral of $f(\X)$ works out the same as the calculation for $V_1$. Either way, we find that $$ \int_{\X\in V} f(\X) \,d\mu(V) = 6 \int_{\X\in V_1} f(\X) \,d\mu(V_1) =\frac{1}{20} $$ and therefore $$ E(f(\X)) = \frac{1/20}{1/4} = \frac15. $$