A beautiful presentation of a beautiful proof. I enjoyed it immensely. My one suggestion is that you may want to mention that it was first published (as far as I know) in 2012: William Derrick, James Hirstein, "Proof Without Words: Ptolemy’s Theorem", in The College Mathematics Journal, Vol. 43, No. 5 (November 2012), p. 386.
Thank you! I'm glad you liked it, and I appreciate your letting me know the source of this proof. I have added a reference to the article in the video's description.
Thanks for watching! All my videos use Desmos Graphing Calculator plus screen capture, then iMovie to edit the screen capture video. For my first three videos, I controlled the animations by hand - starting sliders, turning expressions on and off, etc. Once the videos got too complicated for that, I wrote a javascript add-on that allows me to program these changes so they run automatically. Here's the Desmos Graph for this video: www.desmos.com/calculator/famrmmf7xs And my animation program is on GitHub: github.com/MathyJaphy/DesmosPlayer (it's got some bugs and quirks, but the instructions are all there if you want to try it).
Thank you! I get that question a lot, so I put the answer in the description. I used Desmos Graphing Calculator to create the animations, and I wrote a plug-in for Chrome that lets me program the animations, which are otherwise controlled manually by moving sliders and displaying/undisplaying the various equations. See the description for links to the code on GitHub if you want to try it. The README file has a link to the Desmos graph I used for this video.
This is really interesting proof. I do have concern on "scaling A" as you can't scale a segment with a factor of another length of segment this with only compass and straightedge. A*B is only interpretable as area, not length. Is there any way to circumvent this issue?
Ok so I looked at the animation again and we can actually scale by "F/A" for second triangle and "B/A" for third triangle instead, we get C+DB/A = EF/A. And some small algebra should lead to the same Ptolemy's theorem.
No, even the Greeks knew how to draw a line segment whose length was equal to the product of the length of any other two line segments. It's the *lengths* of the lines being multiplied, so there's no dimensional problem. Coming from a physics background, I had the same confusion as you for ages.
Hi, and thanks for the comment. I explain what I use to animate my videos in the description, along with links to the software if you want to try it yourself. Basically, it's Desmos Graphing Calculator turned into a programmable animation with a browser plug-in that I wrote which I call DesmosPlayer.
@@MathyJaphy Holy Fuuk you replied to 2 year old video. so can you share the link to the desmos player plugin cause it is removed from Github and thanks
I'm still making videos for the channel, so I notice comments on any of them, new or old! :-). I tested the link in the description before replying, and as far as I can tell, the repository is still there. I see my instructions and my javascript file. I'm not a GitHub expert so I don't know why you can't see it (assuming you entered the link correctly). If you have any idea, let me know and I'll investigate.
@anuamba I may have found the reason that you couldn't get to my GitHub. It looks like a RUclips bug. When you click on the link in the description, it includes the close-parentheses in the URL! I fixed it by removing the period after the close-parentheses. Could you try it now and see if the link works for you?
Any convex quadrilateral whose opposite angles add up to 180° is a cyclic quadrilateral, right?? I mean, I have no doubt that for any cyclic quadrilateral, opposite angles add up to 180°. But is the reciprocal true?, i.e. is it true that any convex quadrilateral whose opposite angles add up to 180° is a cyclic quadrilateral?? (i.e. there is a circle passing through each of the four vertices). How can you prove it?
Excellent question. Yes, the reciprocal is true. By way of contradiction, assume you have a convex, non-cyclic quadrilateral ABCD where angle ABC and angle ADC are supplementary. Draw the circumcircle of triangle ABC, and note that D is not on that circle. If D is outside the circle, find the point on line AD that intersects the circle, and call it E. If D is inside the circle, extend AD until it intersects the circle, and call that point E. Since ABCE is a cyclic quadrilateral, angle AEC is supplementary to angle ABC and must therefore be equal to angle ADC. That's impossible.
You're right. It didn't occur to me until after posting the video that I was not following the usual conventions. I leaned towards capitals for legibility on small screens, but I probably could have made it work with lower case.
That was my problem when I first posted this video (LOL): ruclips.net/video/8Yo-vrGQoqI/видео.html. But seriously, I assume you mean that upper case letters usually represent vertices, not side lengths. I wasn’t thinking about that when I set up the animation. I guess I preferred the aesthetics of capitals. I’ll be more careful about following convention in the future.
Hey I'm looking for a nice Introductory geometry book with many exercises. I'm starting my major next year and I have been focusing mainly on calc, proofs and linear algebra but I'd like to refresh my knowledge in geometry
Thanks for asking, but I don't have anything to suggest, as I'm not a math professional. I'm sure you can find recommendations on reddit, r/geometry. Good luck with your major! See you in #SoME3, perhaps?
@@SabrinaHoq Yes, you're referring to the part where I scale each triangle by different factors. But I don't know what you mean by "solve". Some commenters have questioned whether it's okay to multiply a length by a length because the units become squared. But I'm not multiplying by a length, just a value that happens to be the same as the length of one of the other sides. You can always scale the sides of a triangle by a constant value without changing its angles. Does that answer your question?
@@SabrinaHoq The lengths of the triangles' sides (A, B, C, D, E and F) are defined by the quadrilateral you start with. A, B, C and D are the lengths of the quadrilateral's sides. E and F are the lengths of the diagonals. For the quadrilateral that I used as a demo in the video, the lengths are approximately (A=1.49, B=1.11, C=3.34, D=1.66, E=1.98, F=1.94). Is this helpful?
Ah, got it. Yeah, that silliness was only for the joke video. I didn’t think I’d use the song again either, but I needed something for the end screen in this particular case.
Yes, indeed! That's what the first 30 seconds of the video is about, only in reverse. I start with Pythagoras, then generalize to the case where opposite sides are _not_ equal. :-)
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
This is one of the best proofs of Ptolemy's Theorem ever.
Agreed
I agree lol
I agreed too
0:20 DAMNN!!! I never saw any geometric proof more elegant than this one. You deserve million views.
Thank you!
Dazzled me for a while!
This channel & this video is so under-rated.
More power to you!
A beautiful presentation of a beautiful proof. I enjoyed it immensely. My one suggestion is that you may want to mention that it was first published (as far as I know) in 2012:
William Derrick, James Hirstein, "Proof Without Words: Ptolemy’s Theorem", in The College Mathematics Journal, Vol. 43, No. 5 (November 2012), p. 386.
Thank you! I'm glad you liked it, and I appreciate your letting me know the source of this proof. I have added a reference to the article in the video's description.
Can you please tell me what software you used for the presentation ? Thanks! Great idea btw
Thanks for watching! All my videos use Desmos Graphing Calculator plus screen capture, then iMovie to edit the screen capture video. For my first three videos, I controlled the animations by hand - starting sliders, turning expressions on and off, etc. Once the videos got too complicated for that, I wrote a javascript add-on that allows me to program these changes so they run automatically. Here's the Desmos Graph for this video: www.desmos.com/calculator/famrmmf7xs And my animation program is on GitHub: github.com/MathyJaphy/DesmosPlayer (it's got some bugs and quirks, but the instructions are all there if you want to try it).
This video is very good, thank you. You deserve more views and likes.
Potpuno se slazhem, ali nema mnogo ljudi koji mogu da vide lepotu u dokazu neke teoreme.
The is literally the best proof I have ever seen of this theorem, like seriously this is pretty amazing
Your channel deserves million views my friend you are a good guy
Beautiful! I love the symmetry of this proof.
Love the animation and explanation.
Beautiful explanation of Ptolemy's Theorem, outstanding!
Thank you!
Finding this video when i was looking for ptolemy astronomy. Even this isnt what I looking for, I'm not disappointed and even thoroughly enjoying it 😊
Thanks for this beautiful, articulate and intuitive proof
🤗
This is the most beautiful prove for this theorem! thank you!
I was the 667th upvote. I saved your video from darkness. You’re welcome.
One of the best background scores ever!
can’t believe he actually used the music. so good!
Beautiful theorem. The pioneering ancient Greeks!
I prefer it with your voice, btw nice work! Keep going :)
Really great and helpful May I know which app is used for animation?
Thank you! I get that question a lot, so I put the answer in the description. I used Desmos Graphing Calculator to create the animations, and I wrote a plug-in for Chrome that lets me program the animations, which are otherwise controlled manually by moving sliders and displaying/undisplaying the various equations. See the description for links to the code on GitHub if you want to try it. The README file has a link to the Desmos graph I used for this video.
Thanks a lot sir
Your explanations are awesome!
This is really interesting proof. I do have concern on "scaling A" as you can't scale a segment with a factor of another length of segment this with only compass and straightedge. A*B is only interpretable as area, not length. Is there any way to circumvent this issue?
Ok so I looked at the animation again and we can actually scale by "F/A" for second triangle and "B/A" for third triangle instead, we get C+DB/A = EF/A. And some small algebra should lead to the same Ptolemy's theorem.
Granted, but there's nothing that prevents scaling a triangle by a unitless quantity that happens to be equivalent to one of the side lengths.
if you admit that (1,1) and (0,0) can be constructed A*B can also be constructed.
No, even the Greeks knew how to draw a line segment whose length was equal to the product of the length of any other two line segments. It's the *lengths* of the lines being multiplied, so there's no dimensional problem. Coming from a physics background, I had the same confusion as you for ages.
@@QuantumHistorian if you are given a unit length, then yes. Is it possible without using unit length?
👍Good stuff! Everything in this channel is very worthwhile.
Wow! Amazing work!
Look at the triangle show, it's really easy to understand if we follow his teachings really great.
Fantastic as always!
I would love more videos from you. also some longer and more in depth!
No.1 proof bro , I love it so much 🤟
This proof deserves handreds of thousands of likes.
Amazing,astonishing,unbelievable beautiful
The animation is what helped me understand it most. How do you do it?
Hi, and thanks for the comment. I explain what I use to animate my videos in the description, along with links to the software if you want to try it yourself. Basically, it's Desmos Graphing Calculator turned into a programmable animation with a browser plug-in that I wrote which I call DesmosPlayer.
@@MathyJaphyBut the player plugin is deleted from github. Can you share it?
@@MathyJaphy Holy Fuuk you replied to 2 year old video.
so can you share the link to the desmos player plugin cause it is removed from Github and thanks
I'm still making videos for the channel, so I notice comments on any of them, new or old! :-).
I tested the link in the description before replying, and as far as I can tell, the repository is still there. I see my instructions and my javascript file. I'm not a GitHub expert so I don't know why you can't see it (assuming you entered the link correctly). If you have any idea, let me know and I'll investigate.
@anuamba I may have found the reason that you couldn't get to my GitHub. It looks like a RUclips bug. When you click on the link in the description, it includes the close-parentheses in the URL! I fixed it by removing the period after the close-parentheses. Could you try it now and see if the link works for you?
I found an amazing Chanel, if only there were more videos to see..
Any convex quadrilateral whose opposite angles add up to 180° is a cyclic quadrilateral, right??
I mean, I have no doubt that for any cyclic quadrilateral, opposite angles add up to 180°. But is the reciprocal true?, i.e. is it true that any convex quadrilateral whose opposite angles add up to 180° is a cyclic quadrilateral?? (i.e. there is a circle passing through each of the four vertices). How can you prove it?
Excellent question. Yes, the reciprocal is true. By way of contradiction, assume you have a convex, non-cyclic quadrilateral ABCD where angle ABC and angle ADC are supplementary. Draw the circumcircle of triangle ABC, and note that D is not on that circle. If D is outside the circle, find the point on line AD that intersects the circle, and call it E. If D is inside the circle, extend AD until it intersects the circle, and call that point E. Since ABCE is a cyclic quadrilateral, angle AEC is supplementary to angle ABC and must therefore be equal to angle ADC. That's impossible.
@@MathyJaphy Wow, nice! Thank you!
*You meant: " If D is outside the circle, find the point on line AD that intersects the circle, and call it E."
Thanks for pointing out my typo. I’ve corrected it.
You are doing amazing great
absolutely wonderful
I would have used small letters a, b, c, d, e, f for the measures of the line segments. Capital letters are conventionally used to denominate points.
You're right. It didn't occur to me until after posting the video that I was not following the usual conventions. I leaned towards capitals for legibility on small screens, but I probably could have made it work with lower case.
this is crazyyyyyyyyy
v cool! (but could small letters be used instead?)
That was my problem when I first posted this video (LOL): ruclips.net/video/8Yo-vrGQoqI/видео.html.
But seriously, I assume you mean that upper case letters usually represent vertices, not side lengths. I wasn’t thinking about that when I set up the animation. I guess I preferred the aesthetics of capitals. I’ll be more careful about following convention in the future.
It was hard to believe that it could be this easy...
Hey I'm looking for a nice Introductory geometry book with many exercises.
I'm starting my major next year and I have been focusing mainly on calc, proofs and linear algebra but I'd like to refresh my knowledge in geometry
Thanks for asking, but I don't have anything to suggest, as I'm not a math professional. I'm sure you can find recommendations on reddit, r/geometry. Good luck with your major! See you in #SoME3, perhaps?
Loved the way of describing
Such a circle is known as Circumcircle. It's a circle that passes through all vertices (corner points) of a polygon.
I know. I chose to avoid using that technical term in my script. It sounds just a little bit funny. :-). Thanks anyway for the comment!
very nice pictorial explanation
Hermosa demostración.
Amazing video made with only four minutes!
Which software you use for animation
Desmos graphing calculator.
Excellent proof ! Vow !!
Excellent video. Thanks a lot.
I know the answer to the last question (3:54): VERY!
Can you explain me how can you solve the factor of 'A','B' or 'F'..
I would be happy to, but I don't understand the question. What do you mean by "solve the factor of..."?
@@MathyJaphy it would be helpful for my project if you could provide the measurement of the sides of the triangle..
@@SabrinaHoq Yes, you're referring to the part where I scale each triangle by different factors. But I don't know what you mean by "solve". Some commenters have questioned whether it's okay to multiply a length by a length because the units become squared. But I'm not multiplying by a length, just a value that happens to be the same as the length of one of the other sides. You can always scale the sides of a triangle by a constant value without changing its angles. Does that answer your question?
@@MathyJaphy Can I use the same constant value every time,??
@@SabrinaHoq The lengths of the triangles' sides (A, B, C, D, E and F) are defined by the quadrilateral you start with. A, B, C and D are the lengths of the quadrilateral's sides. E and F are the lengths of the diagonals. For the quadrilateral that I used as a demo in the video, the lengths are approximately (A=1.49, B=1.11, C=3.34, D=1.66, E=1.98, F=1.94). Is this helpful?
Nice!
Very Nice!
Thanks sir
Thank you ❤️
Beautiful ❤️
So…. I’m seeing a pattern of video content just leaving my sight into the edges of my screen.
Hey, where are new videos? It's good content, keep work
I thought you were saying it's a sick quadrilateral, I was like maybe corona.
So, did you disinfect your hands? 😆
Subscribed!
Wow, que satisfactorio es entender algo
Very cool.
LETS SHOUT FOR PTOLEMY!
It was awesome!
Just wow!
so nice,cool !
Very nice!
PS love the outro.
Ketiva v chatima tova sister. I also found a good channel of a friend called bethalgebra just new
Indeed beautiful.
🤗
Nice!!
Insanely tought. Dictionary is licked to thank you😷
Thanks for the comment! I don't understand it, but I'll take it as a compliment. :-)
The video is great but would be better with voice over in the second part
I am from 🇮🇳India.
🤨
Proud to be the 69th subscriber
Your joke is in a bad taste.
NICE
Post more
Soon… 😃
I cannot believe that you are not a mathematician.
subbed
Using technology makes math more fun!!😎
best 😉😉😉😉
Boing! ???
Translation…? :-)
@@MathyJaphy The "Boing!" at the end's missing? 🙃
Ah, got it. Yeah, that silliness was only for the joke video. I didn’t think I’d use the song again either, but I needed something for the end screen in this particular case.
Damnn crazy broo❤
Yup, your welcome :)
Thanks, I'm a big fan! Got any other theorems for me to animate? :-)
One peculiar thing is if all the opposite sides are equal then you'll get the Pythagorean Theorem.
Yes, indeed! That's what the first 30 seconds of the video is about, only in reverse. I start with Pythagoras, then generalize to the case where opposite sides are _not_ equal. :-)