Convex Set and an interior point

Question

Given a bounded convex set in $R^2$ with atleast 3 non-collinear points, can we find a point in $R^2$ such that any line passing through it cuts boundary atleast or exactly two points.

Can we prove it logically, I know that any point in triangle formed by 3 non-collinear points should work.

Answer 1

The answer is NO.

For example, the convex set $S= \mathbb R^2$ doesn't have this property.

A non-trivial counterexample is the region below the graph of your favourite concave down function. A vertical line will only meet the boundary once.

One another hand, if your question is

Question: Given a bounded convex set in R2 with at least 3 non-collinear points, can we find a point in R2 such that any line passing through it cuts boundary atleast or exactly two points.

The answer is yes. Note that any such set contains a triangle and hence a disk.

We will prove the following stronger statement.

Lemma If $A \subset \mathbb R^2$ is a bounded set which contains a disk, then any line through any point in this disk will meet the boundary at least twice.

Proof: Fix an arbitrary point $\alpha$ strictly inside. Draw an arbitrary ray from $\alpha$ in any direction. A small interval at the beginning is inside $A$, and towards infinity the ray gets outside $A$ (because $A$ is bounded).

A standard argument show that there is at least a boundary point in between [let me know if you need the details], and this point is not $\alpha$

Now, if we pick any line through $\alpha$, it is made of two such rays in opposite direction. Each of the rays contains a point, which is not $\alpha$, which is on the boundary, and those two points are distinct (since $\alpha$ is the only common point on the two rays).

Answer 2

Let $S \subsetneq \mathbb{R}^2$ be a convex set with at least 3 non-collinear points $A,B,C \in S$. That means $\Delta ABC \subset S$, since $S$ is convex and hence contains the convex hull of any subset.

Let $M$ be the center of mass of $\Delta ABC$, then $M$ is an interior point of $\Delta ABC$. Hence, $\exists r > 0$ such that the open ball $\mathcal{B} = B(M,r) \subsetneq \Delta ABC \subset S$.

Note that since $M$ is an interior point of $\mathcal{B}, S$ and $\Delta ABC$, any line through $M$ is a diameter of $\mathcal{B}$, hence it will cut the boundary of $\mathcal{B}$ in exactly 2 places.

Can you finish this?