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

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.

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If I'm not mistaken, that is equivalent to saying that if you draw any three rays from a point $P$, you can pick a point on each ray such that the three points form an equilateral triangle. I'm not sure how to translate your 120 degree condition though. –  Keshav Srinivasan Sep 17 '13 at 21:17
@confused Not really, your triangle can be very, very, very, ..., very small. –  dtldarek Sep 17 '13 at 21:37

Hint:





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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 $\ddot\smile$

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The hint is, that once you make an equilateral triangle small enough so that it fits on each ray (on both sides) near $P$, then you can lock two of its vertices on two rays (so it could "slide"). This "sliding" is continuous, so if the triangle starts with its third vertex on one side of the third ray and ends on the other, there has to be a position you are looking for. The only way it could fail is that your triangle would start and end with its third vertex on the same side (or slip under). However, that could only happen if... –  dtldarek Sep 17 '13 at 22:00
@confused If you are still stuck, I've added a solution. –  dtldarek Sep 17 '13 at 22:15
@confused That's why I included the black triangle in the picture. When transforming green to brown (dark red) you will have to go below the third ray and the trajectory of the third vertex won't intersect. Or so I thought, I didn't make it formal, so there might be something weird going on. –  dtldarek Sep 17 '13 at 23:32
@confused I've added a formal proof (it is rather trivial). –  dtldarek Sep 17 '13 at 23:48