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In more than 2300 years since Euclid's Elements appear, there were only two famous 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

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

My friend, 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 via email for 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 $MNP$ 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

In more than 2300 years since Euclid's Elements appear, there were only two famous 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