My friend, mister Paul Yiu has sent to me a very new nice equilateral triangle associated with two [Fermat points](http://mathworld.wolfram.com/FermatPoints.html) and [Kiepert hyperbola](http://mathworld.wolfram.com/KiepertHyperbola.html) of a [reference triangle](http://mathworld.wolfram.com/ReferenceTriangle.html) via email since 4 years ago. But I have no proof. This problem is very nice, so I post at here, I hope that have a solution.

1. Let $ABC$ be a triangle with two Fermat points $F_1$, $F_2$. Circles with center $F_1$ radius $F_1F_2$ meets the Kiepert hyperbola again at three points $M$, $N$, $P$.

Then triangle $MNP$ perspective to ABC at tripped points. This mean: 

- $MA$, $NB$, $PC$ are concurrent
- $MB$, $NC$, $PA$ are concurrent
- $MC$, $NA$, $PB$ are concurrent

Three points above collinear.

[![enter image description here][1]][1]

2. Let $ABC$ be a triangle with two Fermat points $F_1$, $F_2$. Circles with center $F_2$ radius $F_2F_1$ meets the Kiepert hyperbola again at three points $M$, $N$, $P$.

Then triangle $MNP$ perspective to $ABC$ at tripped points. This mean: 

- $MA$, $NB$, $PC$ are concurrent
- $MB$, $NC$, $PA$ are concurrent
- $MC$, $NA$, $PB$ are concurrent

Three points above collinear

**See also:**

* [Morley's trisector theorem](https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem)


  [1]: https://i.sstatic.net/EZ9FU.png