The question is posed as follows:
Let $O(0,0), A(2,0), and B(1, \frac1{\sqrt3})$ be the vertices of a triangle. Let $R$ be the region consisting of of all those points $P$ inside $\triangle OAB$ which satisfy $$d(P, OA) \leq \min\{d(P, OB), d(P,AB)\}$$ where $d$ denotes the distance from the point to the corresponding line. Sketch the region $R$ and find its area.
My solution is as follows:
Firstly, we note that $\triangle OAB$ is an isosceles one with $|OA| = |BA|$. So, our region will be symmertric with respect to the median through $A$.
Now, we divide $R$ into infinitesimal regions (strips) along the $X$ axis. The line $OB$ is one such strip itself as every point on it is at a distance 0 from $OB$ and is at most also at a distance $0$ from $OA$ or $AB$. Taking the next strip, right "above" $OB$, this is essentially a slightly "shorter" strip as we must remove the point of its intersection with either side where the distance would be $0$ vs $\epsilon > 0$ to the base. Similarly, the strips will form some sort of shape that will reduce in length until we reach a point where (by symmetry), all the sides are equidistant. The shape will be a triangle, as of course we have seen the base is $OB$ (a line) and as we move up, the sides get closer to the median of this region linearly.
We may find the third vertex, say, $C$ of $R$ by the fact that it is the incenter of $\triangle OAB$, or by our earlier reasoning (or by any other valid method) that $C(c_x, c_y)$ is equidistant from $OA$ and $OB$ to get $C \equiv (1, 2-\sqrt 3)$. It is now trivial to draw the graph and find the area.
$\mathbf{My\ problem}$: is that it is not at all mathematically rigorous as a proof and so I would appreciate help in making it as such. (or even another, slicker solution would be cool).
This is tagged with contest-math as it appeared in JEE 1997(2). JEE is an Indian entrance examination for the system of Indian Institutes of Technology for undergraduate courses.
Edit: There is a good alternate answer to this question due to @RossMillikan and his comments under this question. I would like to see if it is possible to flesh out my own answer to make it viable also. Thank you to all participants.
Put the isosceles triangle in the usual orientation, sitting on the base. If you think about points near one of the base angles of the triangles, they will be closer to the base than one of the equal sides if they are below the bisector of the base angle. The area closer to be base is inside the inner triangle in the figure below.