Constructing an equilateral triangle from an arbitrary triangle by shifting towards an interior point

Question

Suppose $\triangle ABC$ has no angles greater than or equal to $120^{\circ}$ and let $P$ be any arbitrary point inside $\triangle ABC$. Let $\overline{AP}, \overline{BP}, \overline{CP}$ be the line segments connecting the vertices to $P$. Is it always possible to construct an equilateral triangle by shifting the vertices $A,B$ and $C$ along these segments i.e. by moving the vertices inward towards $P$? My intuition says no, but I have yet to come up with a relatively simple counterexample. Any help is appreciated.

Answer 1

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Solution:

This is false, here is a counterexample: $\hspace{40pt}$ The crucial thing is that no angle near $P$ is in $[60^\circ,120^\circ]$ interval.

Formal proof:

Let $\triangle XYZ$ be an equilateral triangle from the problem formulation. This implies that $P \in \triangle XYZ$ and as such angles $\angle XPY$, $\angle YPZ$ and $\angle ZPX$ are all between $[60^\circ,180^\circ]$. Hence, if any of $\angle APB$, $\angle BPC$ or $\angle CPA$ is strictly smaller than $60^\circ$, then no such triangle exists.

I hope this helps $\ddot\smile$