Yiu's equilateral triangle-triplet points

Question

In more than 2300 years since Euclid's Elements appear, there were only two equilateral triangles become famous: The Morely equilateral triangle and the Napoleon equilateral triangle. In more than 20000 points of Encyclopedia of Triangle Centers don't has any triangle perspective triplet to $ABC$.

But 4 years ago, Professor Paul Yiu had sent to me a very new nice equilateral triangle associated with two Fermat points and Kiepert hyperbola of a reference triangle. But I did not have a proof. I posed at here and 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 $MNP$ is equilateral and perspective to $ABC$ at triplet points. This mean:

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

Three points above collinear.

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 $MNP$ is equilateral and perspective to $ABC$ at triplet 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

Answer 1

To prove that $MNP$ is equilateral note that hyperbola through $ABCF_1F_2$ is orthogonal i.e. goes through the orthocenter of $MNP$ and also goesthrough its circumcenter $F_1$. So this hyperbola is Feuerbach hyperbola for $MNP$. Also note that the midpoint $M$ of $F_1F_2$ is the center of hyperbola, so is the Feuerbach point of $MNP$. But $|F_1M| =$ circumradius of $PMN$, so $MNP$ is equilateral. $\Box$