TITLE: Triangle tangent to 3 Parabolas, finding the common area
QUESTION [5 upvotes]: Triangle $ABC$, $AB=4$, $BC=15$, $AC=13$.
Two sides are tangents to the respective Parabolas.
We have to find the area shaded.
My approach-

I tried finding the area of the quadratures(Archimedes) formed but it
doesn't help as- I have to find the area between the 2 intersecting
Parabolas also which I can't find.

REPLY [6 votes]: Bullard's paper is fine but uses coordinates: I'll try to give a more geometric proof.
I'll rely on a couple of well-known properties of parabolas, the first one (also recalled by Bullard) states that:

If from a point $C$ tangents $CA$ and $CB$ are issued to a parabola
($A$ and $B$ being the tangency points), and $F$ is the midpoint of
$AB$, then the intersection point of $CF$ with the parabola is the
midpoint $I$ of $CF$, and the tangent to the parabola at $I$ is
parallel to $AB$.

This entails that the tangent at $I$ passes through the midpoints $D$ and $E$ of $BC$ and $AC$. Segment $IF$ is called a diameter of the parabola and half-chord $AF$ (parallel to the tangent at $I$) is the corresponding ordinate. The second property of parabolas is the following:

The diameter is proportional to the square of the corresponding
ordinate: $IF \propto AF^2$.

Let parabola $AIB$ intersect median $AD$ at $V$, and let $VM$ parallel to $AB$ be the ordinate corresponding to diameter $IM$. From the second property it follows that
$$
{IM\over IF}={VM^2\over BF^2}.
$$
If $O$ is the centroid of $ABC$ (see figure below), then
$IO={1\over6}CF={1\over3}IF$, hence the above equation can be written as:
$$
\tag{1}
{IM\over IO}=3{VM^2\over BF^2}.
$$
On the other hand we have from triangle similarity $VM:ID=OM:OI$, that is:
$$
\tag{2}
2{VM\over BF}=1-{IM\over IO}.
$$
Combining $(1)$ and $(2)$ we finally get
$$
\tag{3}
{VM\over BF}={IM\over IO}={1\over3}.
$$
Note that we can repeat the same reasoning for the intersection $W$ of the same parabola with the other median $BE$, obtaining the same result. Hence points $VMW$ are aligned, $M$ is the midpoint of $VW$ and $DV/DO=EW/EO=1/3$.
But of course we would obtain the same ratios for the other two parabolas tangent to the sides of the triangle: it follows that every couple of parabolas meet at their intersection with a median of the triangle. The area common to the three parabolas is then formed by triangle $VWZ$ and by three parabolic segments ($VIW$ and the anologous for the other two parabolas).
Triangle $VWZ$ is similar to $ABC$, with similarity ratio $1/3$. Hence its area is
$$
{\cal A}_{VWZ}={1\over9}{\cal A}_{ABC}.
$$
From $(3)$, recalling that $MO/MZ=1/3$ we also obtain $IM/MZ=1/6$ and consequently ${\cal A}_{VWI}={1\over6}{\cal A}_{VWZ}={1\over54}{\cal A}_{ABC}$.
But the area of parabolic segment $VWI$ is $4/3$ that of triangle $VWI$, hence
$$
{\cal A}_{\text{par. seg.}}={2\over81}{\cal A}_{ABC}.
$$
The same result applies to the other two parabolic segments, hence summing all pieces together the requested area is:
$$
{1\over9}{\cal A}_{ABC}+3{2\over 81}{\cal A}_{ABC}={5\over27}{\cal A}_{ABC}.
$$