Fermat Points?

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\(\triangle ABC\) is rotated about point \(A\) \(60^{\circ}\) clockwise to form \(\triangle AB'C''\). Then, \(\triangle ABC\) is rotated about point \(B\) \(60^{\circ}\) clockwise to form \(\triangle A''BC'\). Finally, \(\triangle ABC\) is rotated about point \(C\) \(60^{\circ}\) clockwise to form \(\triangle A'B''C\).

Problem 1: Prove that: \[AC'=BA'=CB'\] and \[AB''=BC''=CA''\]

Problem 2: Prove that \(AC',BA',CB'\) are concurrent, and so are \(AB'',BC'',CA''\).


Side Note I think I have, by making those problems up, just rediscovered the first and second Fermat Point. Now that I think about it, what I did is basically the same as attaching equilateral triangles.

If you extend my questions a little bit (the first question to be exact), you will find that they prove that the first Fermat point guarantees the shortest distance between the point and each of the triangle's vertices. You may also want to try that out. Explore! In fact, this question was inspired by me trying to find a generalized Pompeiu's Theorem.

Note by Daniel Liu
4 years, 3 months ago

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Wow. I am so terrible with geometry proofs that I'm not even going to look at this. But how did you make the picture? I notice that everything is exactly aligned the way it's supposed to be. :O

Finn Hulse - 4 years, 3 months ago

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Asymptote Vector Graphics Language.

Daniel Liu - 4 years, 3 months ago

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Aha. Oh yeah that's right! Nevermind. Dude did you see my solution to your other proof problem?

Finn Hulse - 4 years, 3 months ago

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