The Fermat Point of a Triangle | Geometric construction + Proof |

Learn more theorems in Euclidean geometry and their applications at:
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Summary:

The Fermat point of a triangle ABC is a point P such that the sum of distances PA+PB+PC is a minimum.

To find the Fermat point of a triangle ABC:
1. Construct equilateral triangles on each side of ABC
2. Connect vertices A,B and C to the opposite and outermost vertex of equilateral triangle.
3. The point at which the three lines intersect is a Fermat point of triangle ABC.

In the case where one of the angles of triangle ABC is greater than 120 degrees the Fermat point will be located at the obtuse-angled vertex of ABC.

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Music by:

[ThinkTwiceLtu]

I should also mention that the sketch doesn't work as well if opened on a mobile so I recommend using PC for better experience.

Thanks for watching and until next time:)

[MuPrimeMath]

I usually don't read geometric proofs because the lines and labels everywhere confuse me, but this animation makes it super clear! Thank you!

[plominecraft]

I have watched the first 15 secons of video and i have to say your editing is gorgeous af.

keep it up man.

[mathemaniac]

Brilliant video! I learnt about this proof a couple of years ago, and the animation makes it better! This is also (part of) the reason why soap bubbles always meet at 120 degrees to each other.

[JorgetePanete]

7 minutes and 20 seconds of beautiful interesting content

i love it

[AdityaKumar-ijok]

1:03 *that's really a pro gamer move*

seriously no High school geometry teaches the power of such transformations

[DiegoMathemagician]

Super neat! I didn't know it was called Fermat's point; I have always called it Torricelli's point. Apparently, Fermat sent him a letter stating the problem and Torricelli solved it.
Anyway, great video, the explanation was super clear and the animation smoother than a differentiable function hehe. I will take the challenge next Friday because next week I will be doing my finals. Cheers!

[alpe]

I tackled the quadrilateral challenge and wanted to present my solution:
First I considered the case of a square. Then I chose an arbitrary point P, which is inside the square (for simplicity).
Afterwards, I connected the vertices A, B, C and D with point P.
Following that, I rotated the whole square with all its length 90° with the preserving point A.
I named the new points B', C' and D' respectively (and in my case B' was exactly on top of D, which is why I refer to it as D).
Since we rotated the square the distances are preserved, which means that AP = AP' and BP = B'P' =DP'.
The Fermat point is defined to be the sum of distances from the vertices to the point P, so that the sum is a minimum -> AP+BP+CP+DP.
We see that the first distance AP' ends where the following distance P'D starts (for all of the distances).
Since we want the minimum distant we pay our attention to the starting and ending point, which are A and C.
Therefore P must lie on AC and by symmetry we P must also lie on BD.

The Fermat point of a quadrilateral lies at the intersection point of the two diagonals.

Additional notes:
I don't think this solution holds true for crossed quadrilaterals.
Imagine that there is a crossed quadrilateral, where AC and BD are parallel to each other.
Then there is no intersection point and therefore there is not Fermat point.

I am not sure if my solutions holds true for concave quadrilaterals.

[desco]

In a convex quadrilateral the Torricelli-Fermat point is just the intersection of the diagonals. To prove it let ABCD be a convex quadrilateral and P a point and consider the triangles APC and BPD. By the triangle inequality we have PA+PC≥AC and PB+PD≥BD, thus PA+PB+PC+PD≥AC+BD. The equality, which also is the minimum value we are looking for, holds if and only if the two triangles degenerate into two segments, meaning P is on both AC and BD, so the Torricelli-Fermat point is the intersection of the diagonals

[Asdayasman]

1:30 PPAP



I HAVE A POINT
IT IS EQUIDISTANT
UH
EQUIDISTANT POINT

[mikikaboom]

This is really beautiful. How creative one must be to invent it...

[returnexitsuccess]

Great animations! If I'm not mistake then the one thing missing is just proving that this point actually exists i.e. that the three lines you draw actually do always meet in one point.

[AnonEMoose-mrjm]

This is honestly one of my favorite channels. Mathematical visualization plus mellow lo-fi beats. Its great!

[carlosrocha]

This is a pretty cool visual demonstration. Congrats =D

[snomad]

lo-fi math? never thought about them together but I must say, where have you been this entire time?! I definitely love this style of videos, keep it up!

[Invalid]

Beautiful video, beautiful proof, chill music
-> 11/10

[balajisriram]

You pause the visualizations for some time to help us understand on whats going on, and that is simply amazing!!! Thanks

[victorscarpes]

Watching the video I was like: LET ME USE A COORDINATE SYSTEM AND MINIMIZE A FUNCTION THAT TAKES A 2D VECTOR AS THE INPUT!"
I guess geometry also works. Gorgeous animations, by the way.

[yugiohsc]

Someone’s been working on project Euler

[Alexnt]

Your videos have an amazing edition. You are the most underrated channel on YT.