How $A$ can be divided to $n+1$ sets $a = \cup_{i=0}^n A_i$ such that $diam(A_i) \lt 1$ for $i=0,...,n.$

Question

Given: Let $ A \subset \mathbb R^n$ be a closed convex set with smooth boundary and diameter $diam(A) = 1$.

Question: How $A$ can be divided to $n+1$ sets $a = \cup_{i=0}^n A_i$ such that $diam(A_i) \lt 1$ for $i=0,...,n.$

I'm still not sure how to approach this, any hints/approaches wouch be highly appreciated.

Answer 1

This is not possible for $n$ large enough: Borsuk's conjecture

Answer 2

For n = 2, let p, the geometric center of A, be placed at (0,0).
A is a subset of a closed disk D with radius 1/2 centered at p.

Draw a line u from p at a 45 degree angle from the x-axis.
Draw a line v from p at a -45 degree angle from the x-axis.

The wedge between u and v is one part.
For the other two parts divided the disk without the wedge by the x-axis.

Use the conditions on A to prove this construction divides A into three not empty convex subsets.