It is known from Blaschke and later Sas that any convex region $P$ with area $1$ admits an inscribed triangle with area at least $\frac{3\sqrt{3}}{4\pi}$. What if we require the triangle to have a given boundary point of P as vertex? The optimal constant should be between $\frac{1}{4}$ and $\frac{1}{3}$, but I can't tighten these bounds by too much. 



Largest inscribed triangle with a given vertex
The answer is indeed $1/\pi$, equality achieved in the case of $P$ a semicircle and the point its center. 


For an arbitrary convex region $P$ and a point $X$ on its boundary, consider the convex hull of $P$ and $2X - P$, and call it $S$. Then any $S$ is clearly symmetric with respect to $X$. Any triangle in $P$ having $X$ as a vertex with area $a$, induces a parallelogram in $S$ with center $X$ and area $4a$. Also we have the trivial inequality $\operatorname{area}(S) \ge 2 \operatorname{area}(P)$. So it suffices to show that, given a centrally symmetric convex region $S$ with respect to $O$, there exists a parallelogram in $S$ with center $O$ and area at least $(2/\pi) \operatorname{area}(S)$. 


This can be proved by essentially the same technique used by Sas. After scaling and rotation, let $S$ be contained in the unit disc, with $(1, 0), (-1, 0) \in S$. Parametrize the boundary of $S$ by $\pm (\cos \theta, f(\theta) \sin \theta)$, where $f : (0, \pi) \to [0, 1]$. Then the area of $S$ is
$$ \operatorname{area}(S) = 2 \int\_{0}^{\pi} f(\theta) \sin^2 \theta.$$
The area of the parallelogram with vertices $\pm (\cos \theta, f(\theta) \sin \theta)$ and $\pm (-\sin \theta, f(\theta + \pi/2) \cos \theta)$ is
$$\operatorname{area}(\mathrm{Par}\_\theta) = 2 \Bigl( f(\theta) \sin^2 \theta + f\Bigl( \theta + \frac{\pi}{2} \Bigr) \sin^2 \Bigl( \theta + \frac{\pi}{2} \Bigr) \Bigr).$$
Averaging this area over $0 \le \theta \le \pi / 2$ shows that there exists some $\theta$ such that
$$\operatorname{area}(\mathrm{Par}\_\theta) \ge \frac{2}{\pi} \operatorname{area}(S). $$


