Intersection of nested triangles is non-empty Let $T_0$ be the interior of a triangle in $\mathbb{R}^2$ with vertices $A,B,C.$ Let $T_1$ be the interior of the triangle whose vertices are the midpoints of the sides if $T_0$ and $T_2$ be the interior of the triangle whose vertices are the midpoints of the sides if $T_1$ and so on. Then find $$\bigcap_{i=0}^\infty T_i$$
My guess is that the intersection will contain only the centroid. Since the diameter of the triangle goes to zero so the intersection will contain only the centroid. But I am unable to give a concrete proof. 
 A: What you've written basically gets you there already. It's easy to see that the centroid $x$ lies in $\bigcap T_i$ since it clearly lies in each $T_i$. Then as you note $\mbox{diam}(T_i) \to 0$ as $i \to \infty$. If there are two points $x,y \in \bigcap T_i$ then $x,y \in T_j$ for each $j$ so $\mbox{diam}(T_j) \geq |x-y| > 0$ for each $j$, giving a contradiction so $\bigcap T_i = \{x\}$
A: Since $T_{i+1}$ has the same centroid as $T_i$, they all have the same centroid, so they will all contain it. Now, notice that the triangle $T_{i+1}$ is a $\frac{1}{2}$ homothetic copy of $T_i$. Therefore, the diameters of $T_i$ decrease to $0$. If we have $P$, $Q$ in $T_i$ for all $i$, we must have $|PQ|\le \operatorname{diam}T_i$ for all $i$. Therefore $|PQ|=0$, $P=Q$.  
A: By Thales theorem, all the triangles that you create have the same centroid, coinciding with the centroid of $ABC$. So the intersection contains the centroid, too.
The intersection is also closed (for the Euclidean topology) and convex, because every triangle is closed and convex.
So the intersection is a closed, convex set in $\mathbb{R}^2$ of diameter $0$ (because the diameter of the triangles shrinks to $0$). Such a set is necessarily a point, and we are done.
