Hey:) I have also made an interactive sketch for you guys to play around with. You can interact with the triangle ABC by moving its vertices and see how the location of the Fermat point changes accordingly. Can you find any patterns? Here is the link: editor.p5js.org/psyduck/present/iezcQtpB 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:)
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.
also say we take a wooden board and mark a triangle in it then we drill holes at its vertices say vertex A,B,C Then we take three equal weights say a,b,c and pass the threads so that they are hanging and a's thread passes through A and so on now if we knot the free ends together we will get a physical way of finding the Fermat's point ( _Source- Mark Levi's - The Mathematical mechanic_ )
They ain’t gonna teach you if you are not interested. If you were interested you would’ve found a way to learn this already, and not wait for the teachers to magically pour knowledge into you. Teachers give you basics and help you if you are interested in a topic.
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!
Thank you, glad you liked the video. Yes the Fermat point is also called the Torricelli point, I guess the names are used interchangeably. Good luck with your finals! My finals are starting in a week too:)
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
@@nanamacapagal8342 PPAP stands for pen pineapple apple pen. The lyrics goes like this: I have an apple I have a pen Uh Apple pen I have a pen I have a pineapple Uh Pineapple pen I have an apple pen I have a pineapple pen Uh *Pen pineapple apple pen*
@@Monochrome_math Also the final verse is "Apple peee~n, pineapple peee~n, **uh,** pen pineapple apple pen", and the second line in the second verse is missing the "a".
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. AP+BP+CP+DP=AP'+DP'+CP+DP=AP'+P'D+DP+PC. 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.
Since what we're trying to minimize is the distance to the 4 corners, why would it matter whether the quadrilateral is crossed or not? It doesn't even need to be a quadrilateral, just 4 points on the plane, right?
@@3snoW_ As I stated in my comment: imagine the crossed quadrilateral with Points A, B, C, D. My definition for the Fermat point of a quadrilateral was: "The Fermat point of a quadrilateral lies at the intersection point of the two diagonals. " Let us take an example of a crossed quadrilateral and look at the intersection of the diagonals. (2D Plane) A(0, 0), B(2, 0), C(1, 1) and D(1, -1) If we look at the direction of the two diagonals (AC and BD), which we can find out using vectors: (v stands for vector) v AC = (1-0, 1-0) = (1, 1) and v BD = (1-2, -1-0) = (-1, -1). Since the vectors are multiples of each other, that means the vectors are parallel to each other. Because they are parallel they cannot intersect and therefore have no intersection point, which is contradicting my definition for Fermat point of a concave quadrilateral. However if you do similar calculations as in my previous comment, you come to the conclusion that the Fermat point of a CROSSED quadrilateral lies on the intersection point of the lines AB and CD. Summary: "The Fermat point of a concave quadrilateral lies at the intersection point of the two diagonals. " "The Fermat point of a crossed quadrilateral lies at the intersection point of the two sides AB and CD. "
@@alpe6127 I have a little knowledge of vectors. I did it by Plane Geometry. I firstly experimented a little and then made a claim that that it lies on the intersection of diagonals. Then, I used simple triangular inequalities that sum of any two sides is greater than the third side. Rest was easy. You just have to get the correct (won't write right (90°) as it's confusing) triangles.
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.
returnexitsuccess I was searching for this comment. I think I will have to prove it myself, proving that does lines intersect should not be very hard (at least I hope so).
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!
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.
The challenge problem posed is brilliantly crafted. While yes, it’s obvious that multivariable calculus is the way to go, it seems that there must be another way to do it geometrically, which finding it is the real challenge. Excellent video, keep up the amazing content.
Incredible, Think Twice. Thank you! Fermat's point serves to understand barycentric subdivision, spectral analysis, as well General Economic Equilibrium. Amazing!
I think this is the first time I ever truly grasped that the geometric properties of lines and shapes are functional in character, and so, theorems in geometry are much, much more powerful and interesting this way
The answer to the riddle at the end is the intersection of the diagonals. Proof: 1) chose a random point P. 2) consider the ellipse that has the foci B,D that touches the point P: If we move point P on the ellipse the sum of the lengnths PD and PB will stay constant (by the definition of the ellipse). 3) now, in order to minimise along the ellipse we only need to consider two lengths: PA,PC. The shortest path between two points is a line. Therefore, P (with restrictions to the ellipse) must be on AC. 4) now, we can return to step 1 and ask "what P will give us the ellipse that will have the 'best' minimised P ?" Since in every case P will be on AC (and thus AP+PC -> constant) that best case will be that in which DP+PB is minimised wich is when P is on BD. [|||]
Awesome as always! For the challenge, my guess is that P will be located at the intersection of AC and BD, and for concave quadrilaterals P will be the vertex with >180 degree angle.
Wow, totally mesmerized by the beauty of the proof, as well as all the content on this channel is elegant mathematics, highly refined , interesting and well animated. Trust me , your channel is at par with 3b1b , keep doing the hardwork. 🤗
I am very bad at geometry, sometimes it is not so easy at a first glance at it seems. You as a very good teacher know much more, than you show. In your head, there is no problem I have a problem, you have already in your head many solutions and that is the problem for me.
Actually with four corners the solution is even more obvious. It has to be the point where the diagonals meet, as any other point would make the paths from A to C or from B to D longer. The same idea works for any even number of corners.
When ABCD is convex, PA + PC >= AC and PB + PD >= BD. So P should be the intersection point of AC and BD. When ABCD is not convex, P should be the vertex where the inner angle is no less than 180 degrees.
I think the Fermat Point of a quadrilateral is the intersection of the 2 diagonals if it exists otherwise it is the vertex with an angle greater than tau/2.
Great animation and clear explanation. Like! PS: there is a physical solution. Consider three equal masses on long ropes. Connect these three ropes in one point and place this point inside the triangle somewhere. Let all the ropes to pass through a single vertex or the triangle and then hang in the constant gravitational field. The energy of the system then will go to the minimum, as the ropes moves and the masses go down. At the energy minimum point, the sum of lengths of the ropes inside the triangle will be minimum, as the rest of the rope length pulled by gravity. At the same time, we know from force balance that three equal forces are in balance when the angle between them is 120°, so we immediately obtain the solution of 120°.
To use this physical solution, you need to know the exact location of point P. Is it correct? Otherwise it will not come to balance position. I think this is a physical explanation of the problem and you can't find the location of P, using this method.
@@rasoulkhoshravan5912 well, yes and no. In order to use physical solution you do not need to know the location of point P at all. You just assume it is somewhere inside the triangle and then let the energy to minimize so that the system reaches equilibrium. Once you obtained the equilibrium, you know that the angle between PA and PB is equal to the angle between PB and PC, and PC and PA, and equal to 2*pi/3. Now you need to find where is the point P is. It is actually very easy. Consider a circle of unit radius and two points on this circle X=(sqrt(3/4), 0.5) and Y=(-sqrt(3/4), 0.5). Put the third point Z anywhere between X and Y on the smaller arc of the circle. The angle XZY is always 2*pi/3. Now let's say the triangle side AB is XY, and the point P is Z. You just need to construct a circle on the side AB, so that AB maps to XY. The intersection of two such circles constructed on AB and AC will yield to the point P.
What a surprise I got. Subscribed well within two minutes of watching the video. Never before did I bump into this channel. Great content, congratulations! What program do you use for these videos?
Generalize with axes of n-Fermat segments with n+1 Fermat points. If there's *only* and only one orthocentre (intersection of all side's axes) than you found Fermat point of polygon. For a quadrilater split poligon in two triangles and do the same you did before. If two found points are aligned with two opposite vertexes than you have a Fermat point for 4gon
For the quadrilateral question challenge, I only found the solution for convex quadrilaterals. Answer: The point is the intersection of two diagonals. Solution: Construct a convex quadrilateral, ABCD and denote the intersection of AC and BD as P. Now, we need to prove that P is the point that satisfy: PA+PB+PC+PD is a minimum. Now we will divide into three cases. Case 1: P' is a point on line AC but is not a point on line BD. By the Triangle Inequality, we have: DP'+P'B>DB=DP+PB and we also have: AP'+P'C=AC=AP+PC Add these together and we get: P'A+P'B+P'C+P'D>PA+PB+PC+PD so P' is not the desired point. Case 2: P" is a point on line BD but is not a point on line AC. By the same reasoning as Case 1, we get: P”A+P"C>AC=PA+PC P"B+P"D=BD=PB+PD Add these together and we will get: P"A+P"B+P"C+P"D>PA+PB+PC+PD so P" is also not the desired point. Case 3: P* is not a point on either line AC or line BD. By using Triangle Inequality, we get: P*B+P*D>BD=PB+PD P*A+P*C>AC=PA+PC Add these together and we get: P*A+P*B+P*C+P*D>PA+PB+PC+PD so P* is also not the desired point. Conclusion: We will get a minimum of PA+PB+PC+PD if and only if P is the intersection of two diagonals of a convex quadrilateral.
I loved everything about the video except the end, it is good u provide an exercise to viewers but you've got to include the solution as well! Just give a brief pause and show it, if someone is interested they'll simply pause it. If they are just casually viewing they will still want to see it and will feel robbed if it's not provided. And of course anyone trying to solve the exercise needs the answer! Not everyone is able to prove things to themselves
3:42 by the definition at the beggining of the video, the intersection of CB',BA',AC' line segments isn't the only fermat point, every element of their union is a fermat point.
the fermat point of a quadrilateral is the point at which all four angles connecting the points are 90 degrees, cause doing another angle would result a non-straight line, and that's always longer than a straight line. unless the quadrilateral is concave, in which the non straight line is ok if the point's in the concave quadrilateral.
for the quadrilaterall, let A,B,C,D in order be the points of the quadrilateral and x the candidate for the "fermat" point. we want to minimize |x-A|+|x-B|+|x-C|+|x-D|. We take the gradient with respect to x and get (x-A)/|x-A|+....+(x-D)/|x-D|=0 for the minimum (assuming it exists). So let a,b,c,d be the vectors that are in the same direction as x-A,...,x-D but with unit length. This means we want a+b+c+d=0, since vector adition is tip to tail this means that we have a quadrilateral with lengths of the same size, ergo they consititue a rombus ergo it is a parallelopiped, so a+c=0, b+d=0. This means that x is between A and C as well as B and D. Meaninf that x is the intersection of AC and BD.
This was really a nice theorem. I really enjoyed it and its clever way of proof. When I was in high school, I didn't hear about this theorem. Is this thought in schools in other countries?
Don't call this theorem, it's a beautiful problem. Calling it *theorem* spoils the fun... And this is generally for Math Olympiads so less chances of teaching this in schools...
Well, the quadrilateral challenge was much easier. Apply triangle inequality on both the diagonals, and you should get the minimum where both the diagonals meet.
Q: isnt this a special case where you said to choose and arbitrary point P and rotate it 60 degrees around A. Wouldn’t it only work because angle BAP was 30 degrees?
No. Notice that when he constructs the equilateral triangle, the path from B' to C is not straight (because the arbitrarily chosen point is not the correct Fermat Point of the triangle), he then shows in the animation how moving the selected point P around changes the resulting path. The originally selected P doesn't have any bearing on the proof, it's just an example to demonstrate the rotation, after that it's about how to use the rotation to find the optimal P.
So you can find the Fermat point of a triangle with ruler and compass by constructing a line 60 degrees off from one angle with the same length as its side and draw a line from that point to the opposite corner. Do that for two sides and the intersection is the point. All without choosing an arbitrary point
Hey so, I have came up with the solution for the problem u gave us at the end (well atleast for quadeilateral only). I tried searching for generalised versions on internet b4 but couldn't find them. Is it possible I am the first person to come up with? Even if there are solutions, are those the same as mine? If I make a video, would it get stolen and I would lose my credit, so maybe I should publish a paper abt it? What are ur views all the readers who read my entire comment.
Thank you! Beautiful except for the difficult to see deep purple color! If only we had this method of wonderful visualization in the mid-1960s when I was a young teen! You young ones don't know how lucky you are! However you should know I anticipated this and other things would be realized!
So this problem asked by Fermat and solved by Toricelli. You should notice that what was formula for minimum value which depends on sides of triangle. (Also there is maximum value too)
Surprisingly, I'm looking around and now here to see the shortest distance among four points. I was surprised once again because my problem is mentioned on this problem. But huge problem is, my problem is for 1st grade student of highschool
Hey:) I have also made an interactive sketch for you guys to play around with. You can interact with the triangle ABC by moving its vertices and see how the location of the Fermat point changes accordingly. Can you find any patterns? Here is the link: editor.p5js.org/psyduck/present/iezcQtpB
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:)
Move A through BC and it breaks the visualisation
I love p5.js! Also I really love the videos. How do you get the animations in your videos?
Eyyy p5. js!
If you just move one point of the triangle, for example A, around, does P trace an arc of a circle? I'm pretty sure it does but I'm not sure.
@@MagicGonads duh 😂
I usually don't read geometric proofs because the lines and labels everywhere confuse me, but this animation makes it super clear! Thank you!
Happy to hear that:)
For me it is otherwise...
It's best to repeat the constructions oneself on a blank paper while you read. I agree that nothing beats the animations though. :)
I have watched the first 15 secons of video and i have to say your editing is gorgeous af.
keep it up man.
Thanks!!
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.
also say we take a wooden board and mark a triangle in it
then we drill holes at its vertices say vertex A,B,C
Then we take three equal weights say a,b,c and pass the threads so that they are hanging and a's thread passes through A and so on
now if we knot the free ends together we will get a physical way of finding the Fermat's point
( _Source- Mark Levi's - The Mathematical mechanic_ )
Pack 7 bubbles and then you get a hexagon! I think this works.
@@hamiltonianpathondodecahed5236
Mark Levi, he is a legend. His work in mathematics with help of physics feels so satisfying and intuitive.
@@52.yusrilihsanadinatanegar79 0
7 minutes and 20 seconds of beautiful interesting content
i love it
Almost 53 seconds of which was sponsored. 6 minutes 27 seconds of beautiful interesting content
Fibonacci is seen. That's it.
1:03 *that's really a pro gamer move*
seriously no High school geometry teaches the power of such transformations
They ain’t gonna teach you if you are not interested. If you were interested you would’ve found a way to learn this already, and not wait for the teachers to magically pour knowledge into you. Teachers give you basics and help you if you are interested in a topic.
You inspire my idle thoughts daily
That's a really nice line I've never heard that before, did you come up with it yourself? :)
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!
Thank you, glad you liked the video. Yes the Fermat point is also called the Torricelli point, I guess the names are used interchangeably. Good luck with your finals! My finals are starting in a week too:)
@@ThinkTwiceLtu Good luck & have fun as well
I call it the Fermat-Torricelli point.
@Bob Trenwith yes
I would have called it a "Steiner Point", as it is the solves the Steiner tree problem for a triangle.
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
excellent simple proof. no need for geometric transformations.
@@riadsouissi
But to *claim* this, it would require intense experimentations...
Now question is, how to prove it holds for concave quadrilaterals as well?
This is really beautiful. How creative one must be to invent it...
1:30 PPAP
I HAVE A POINT
IT IS EQUIDISTANT
UH
EQUIDISTANT POINT
uhh what
@@nanamacapagal8342 PPAP stands for pen pineapple apple pen. The lyrics goes like this:
I have an apple
I have a pen
Uh
Apple pen
I have a pen
I have a pineapple
Uh
Pineapple pen
I have an apple pen
I have a pineapple pen
Uh
*Pen pineapple apple pen*
@@Monochrome_math no no i get the joke, it's just that i wasn't expecting it to pop up here
@@nanamacapagal8342 :() Oh.
@@Monochrome_math Also the final verse is
"Apple peee~n, pineapple peee~n, **uh,** pen pineapple apple pen", and the second line in the second verse is missing the "a".
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.
AP+BP+CP+DP=AP'+DP'+CP+DP=AP'+P'D+DP+PC. 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.
great work
keep it up dude
Since what we're trying to minimize is the distance to the 4 corners, why would it matter whether the quadrilateral is crossed or not? It doesn't even need to be a quadrilateral, just 4 points on the plane, right?
@@3snoW_ As I stated in my comment: imagine the crossed quadrilateral with Points A, B, C, D. My definition for the Fermat point of a quadrilateral was:
"The Fermat point of a quadrilateral lies at the intersection point of the two diagonals.
"
Let us take an example of a crossed quadrilateral and look at the intersection of the diagonals. (2D Plane)
A(0, 0), B(2, 0), C(1, 1) and D(1, -1)
If we look at the direction of the two diagonals (AC and BD), which we can find out using vectors:
(v stands for vector) v AC = (1-0, 1-0) = (1, 1) and v BD = (1-2, -1-0) = (-1, -1).
Since the vectors are multiples of each other, that means the vectors are parallel to each other.
Because they are parallel they cannot intersect and therefore have no intersection point, which is contradicting my definition for Fermat point of a concave quadrilateral.
However if you do similar calculations as in my previous comment, you come to the conclusion that the Fermat point of a CROSSED quadrilateral lies on the intersection point of the lines AB and CD.
Summary:
"The Fermat point of a concave quadrilateral lies at the intersection point of the two diagonals.
"
"The Fermat point of a crossed quadrilateral lies at the intersection point of the two sides AB and CD.
"
@@alpe6127
I have a little knowledge of vectors. I did it by Plane Geometry. I firstly experimented a little and then made a claim that that it lies on the intersection of diagonals. Then, I used simple triangular inequalities that sum of any two sides is greater than the third side. Rest was easy. You just have to get the correct (won't write right (90°) as it's confusing) triangles.
This is honestly one of my favorite channels. Mathematical visualization plus mellow lo-fi beats. Its great!
Clearly your best video so far
Beautiful video, beautiful proof, chill music
-> 11/10
:)
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.
returnexitsuccess I was searching for this comment. I think I will have to prove it myself, proving that does lines intersect should not be very hard (at least I hope so).
You pause the visualizations for some time to help us understand on whats going on, and that is simply amazing!!! Thanks
This is a pretty cool visual demonstration. Congrats =D
Very good video. Loved the pauses to let you think.
Thank you:)
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!
Your videos have an amazing edition. You are the most underrated channel on YT.
Thank you for the support:)
I'm a simple man. I see ThinkTwice post a video, I click on it.
Pure Bliss!
Thanks for the support:)
Each video is better than the last. Love your channel!!!
Thank you:)
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.
Okay this was satisfying to watch
Very Rich animations. Also the style is very professional. I love it
Thank you:)
absolutely loved it!
:)
The challenge problem posed is brilliantly crafted. While yes, it’s obvious that multivariable calculus is the way to go, it seems that there must be another way to do it geometrically, which finding it is the real challenge. Excellent video, keep up the amazing content.
For quadrilateral, Multivariable calculus is overkill.
This channel is too underrated keep the beauty of visual proofs alive :)
Shared your video in my while department......best one
The music and editing is so good... I love it!
Amazing prof, thank you so much 🙏
Someone’s been working on project Euler
Incredible, Think Twice. Thank you!
Fermat's point serves to understand barycentric subdivision, spectral analysis, as well General Economic Equilibrium. Amazing!
Thanks for watching:)
Bro whoever has made this is a geometry god
Seeing is believing. So true.
Extremely beautiful maths
Great work, I love your content!
Thank you:)
That’s so cool! Great video
Thanks 😊
I think this is the first time I ever truly grasped that the geometric properties of lines and shapes are functional in character, and so, theorems in geometry are much, much more powerful and interesting this way
You've really outdone yourself with this video! Well done making this proof beautiful and accessible!
This was your best video ever. Keep up the good work and keep being awesome.
Thanks for the support!
Think Twice is here again to rock our minds. Missed you dude!!
This is so beautiful!
Brilliant!
Absolutely dope animations 🤩🔥
Thanks:)
Fabulous animation
Fantastic content and exposition, Thank You!
Brilliant video!
Notification squad reporting in
Man, this is awesome!
:)
glad you got a sponsorship
just discovered this channel. So wonderful! Thank you :D :D
The answer to the riddle at the end is the intersection of the diagonals.
Proof:
1) chose a random point P.
2) consider the ellipse that has the foci B,D that touches the point P:
If we move point P on the ellipse the sum of the lengnths PD and PB will stay constant (by the definition of the ellipse).
3) now, in order to minimise along the ellipse we only need to consider two lengths: PA,PC.
The shortest path between two points is a line.
Therefore, P (with restrictions to the ellipse) must be on AC.
4) now, we can return to step 1 and ask "what P will give us the ellipse that will have the 'best' minimised P ?"
Since in every case P will be on AC (and thus AP+PC -> constant) that best case will be that in which DP+PB is minimised wich is when P is on BD.
[|||]
beautiful transitions =)
Thanks:))
Great content!
Just brilliant!
Nice one. Keep it up 👍
Chillest math videos on the interwebs 🔥🔥🔥
😌😌
wow, interesting)! Subscribed!)
Thanks!
Brilliant video buddy :) u smashed it!
Thank you:)
Beautiful !
Awesome as always! For the challenge, my guess is that P will be located at the intersection of AC and BD, and for concave quadrilaterals P will be the vertex with >180 degree angle.
Wow, totally mesmerized by the beauty of the proof, as well as all the content on this channel is elegant mathematics, highly refined , interesting and well animated. Trust me , your channel is at par with 3b1b , keep doing the hardwork. 🤗
Wow; that's a neat derivation
I am very bad at geometry, sometimes it is not so easy at a first glance at it seems.
You as a very good teacher know much more, than you show. In your head, there is no problem
I have a problem, you have already in your head many solutions and that is the problem for me.
Inspiring geometry lesson.
Actually with four corners the solution is even more obvious. It has to be the point where the diagonals meet, as any other point would make the paths from A to C or from B to D longer. The same idea works for any even number of corners.
Thanks, the video is super cool.... Can you please make a video on the Feuerbach's Theorem too
When ABCD is convex, PA + PC >= AC and PB + PD >= BD. So P should be the intersection point of AC and BD.
When ABCD is not convex, P should be the vertex where the inner angle is no less than 180 degrees.
I think the Fermat Point of a quadrilateral is the intersection of the 2 diagonals if it exists otherwise it is the vertex with an angle greater than tau/2.
Good student 👦📖🎒
Interesting use of 'dynamic' geometry graphics.
Great animation and clear explanation. Like!
PS: there is a physical solution. Consider three equal masses on long ropes. Connect these three ropes in one point and place this point inside the triangle somewhere. Let all the ropes to pass through a single vertex or the triangle and then hang in the constant gravitational field. The energy of the system then will go to the minimum, as the ropes moves and the masses go down. At the energy minimum point, the sum of lengths of the ropes inside the triangle will be minimum, as the rest of the rope length pulled by gravity. At the same time, we know from force balance that three equal forces are in balance when the angle between them is 120°, so we immediately obtain the solution of 120°.
To use this physical solution, you need to know the exact location of point P. Is it correct? Otherwise it will not come to balance position. I think this is a physical explanation of the problem and you can't find the location of P, using this method.
@@rasoulkhoshravan5912 well, yes and no. In order to use physical solution you do not need to know the location of point P at all. You just assume it is somewhere inside the triangle and then let the energy to minimize so that the system reaches equilibrium. Once you obtained the equilibrium, you know that the angle between PA and PB is equal to the angle between PB and PC, and PC and PA, and equal to 2*pi/3. Now you need to find where is the point P is. It is actually very easy. Consider a circle of unit radius and two points on this circle X=(sqrt(3/4), 0.5) and Y=(-sqrt(3/4), 0.5). Put the third point Z anywhere between X and Y on the smaller arc of the circle. The angle XZY is always 2*pi/3. Now let's say the triangle side AB is XY, and the point P is Z. You just need to construct a circle on the side AB, so that AB maps to XY. The intersection of two such circles constructed on AB and AC will yield to the point P.
What a surprise I got. Subscribed well within two minutes of watching the video. Never before did I bump into this channel. Great content, congratulations! What program do you use for these videos?
Generalize with axes of n-Fermat segments with n+1 Fermat points. If there's *only* and only one orthocentre (intersection of all side's axes) than you found Fermat point of polygon.
For a quadrilater split poligon in two triangles and do the same you did before. If two found points are aligned with two opposite vertexes than you have a Fermat point for 4gon
For the quadrilateral question challenge, I only found the solution for convex quadrilaterals.
Answer: The point is the intersection of two diagonals.
Solution:
Construct a convex quadrilateral, ABCD and denote the intersection of AC and BD as P.
Now, we need to prove that P is the point that satisfy: PA+PB+PC+PD is a minimum.
Now we will divide into three cases.
Case 1: P' is a point on line AC but is not a point on line BD.
By the Triangle Inequality, we have:
DP'+P'B>DB=DP+PB
and we also have:
AP'+P'C=AC=AP+PC
Add these together and we get:
P'A+P'B+P'C+P'D>PA+PB+PC+PD
so P' is not the desired point.
Case 2: P" is a point on line BD but is not a point on line AC.
By the same reasoning as Case 1,
we get:
P”A+P"C>AC=PA+PC
P"B+P"D=BD=PB+PD
Add these together and we will get:
P"A+P"B+P"C+P"D>PA+PB+PC+PD
so P" is also not the desired point.
Case 3: P* is not a point on either line AC or line BD.
By using Triangle Inequality, we get:
P*B+P*D>BD=PB+PD
P*A+P*C>AC=PA+PC
Add these together and we get:
P*A+P*B+P*C+P*D>PA+PB+PC+PD
so P* is also not the desired point.
Conclusion:
We will get a minimum of PA+PB+PC+PD if and only if P is the intersection of two diagonals of a convex quadrilateral.
2:46 ooh I see where this is going
Essa foi a demonstração mais foda que eu já vi na minha vida
I loved everything about the video except the end, it is good u provide an exercise to viewers but you've got to include the solution as well! Just give a brief pause and show it, if someone is interested they'll simply pause it. If they are just casually viewing they will still want to see it and will feel robbed if it's not provided. And of course anyone trying to solve the exercise needs the answer! Not everyone is able to prove things to themselves
3:42 by the definition at the beggining of the video, the intersection of CB',BA',AC' line segments isn't the only fermat point, every element of their union is a fermat point.
the fermat point of a quadrilateral is the point at which all four angles connecting the points are 90 degrees, cause doing another angle would result a non-straight line, and that's always longer than a straight line.
unless the quadrilateral is concave, in which the non straight line is ok if the point's in the concave quadrilateral.
for the quadrilaterall, let A,B,C,D in order be the points of the quadrilateral and x the candidate for the "fermat" point. we want to minimize |x-A|+|x-B|+|x-C|+|x-D|. We take the gradient with respect to x and get (x-A)/|x-A|+....+(x-D)/|x-D|=0 for the minimum (assuming it exists).
So let a,b,c,d be the vectors that are in the same direction as x-A,...,x-D but with unit length. This means we want a+b+c+d=0, since vector adition is tip to tail this means that we have a quadrilateral with lengths of the same size, ergo they consititue a rombus ergo it is a parallelopiped, so a+c=0, b+d=0. This means that x is between A and C as well as B and D. Meaninf that x is the intersection of AC and BD.
Very good
great channel
This was really a nice theorem. I really enjoyed it and its clever way of proof. When I was in high school, I didn't hear about this theorem. Is this thought in schools in other countries?
Don't call this theorem, it's a beautiful problem. Calling it *theorem* spoils the fun...
And this is generally for Math Olympiads so less chances of teaching this in schools...
How did you make this? It’s very nice and clean btw
can you make more such videos on eucledian geometry? It will be a great help
Well, the quadrilateral challenge was much easier. Apply triangle inequality on both the diagonals, and you should get the minimum where both the diagonals meet.
Great Stuff!
Thanks 😊
Please , make an video on Steiner point of a triangle.
Masterful!
Bravo!
Q: isnt this a special case where you said to choose and arbitrary point P and rotate it 60 degrees around A. Wouldn’t it only work because angle BAP was 30 degrees?
No. Notice that when he constructs the equilateral triangle, the path from B' to C is not straight (because the arbitrarily chosen point is not the correct Fermat Point of the triangle), he then shows in the animation how moving the selected point P around changes the resulting path. The originally selected P doesn't have any bearing on the proof, it's just an example to demonstrate the rotation, after that it's about how to use the rotation to find the optimal P.
So you can find the Fermat point of a triangle with ruler and compass by constructing a line 60 degrees off from one angle with the same length as its side and draw a line from that point to the opposite corner. Do that for two sides and the intersection is the point. All without choosing an arbitrary point
I wonder how much time it took to make this video .. btw great vids bro , keep it up 👌👌👍👍
Hey so, I have came up with the solution for the problem u gave us at the end (well atleast for quadeilateral only). I tried searching for generalised versions on internet b4 but couldn't find them. Is it possible I am the first person to come up with? Even if there are solutions, are those the same as mine? If I make a video, would it get stolen and I would lose my credit, so maybe I should publish a paper abt it? What are ur views all the readers who read my entire comment.
Thank you! Beautiful except for the difficult to see deep purple color! If only we had this method of wonderful visualization in the mid-1960s when I was a young teen! You young ones don't know how lucky you are! However you should know I anticipated this and other things would be realized!
math and lo-fi, great idea
So this problem asked by Fermat and solved by Toricelli. You should notice that what was formula for minimum value which depends on sides of triangle. (Also there is maximum value too)
Please use colours with higher contrast. I'm colorblind and got a hard time reading the variables
Surprisingly, I'm looking around and now here to see the shortest distance among four points. I was surprised once again because my problem is mentioned on this problem. But huge problem is, my problem is for 1st grade student of highschool
is fermat also the centroid of the triangle