ฝัง
- เผยแพร่เมื่อ 13 มิ.ย. 2024
- In this video, I discuss and interesting Geometry problem, related to dividing up any convex quadrilateral into exactly 7 kites. This problem begs the question of "why 7?" which I address indirectly as the video progresses.
This video was created using: Davinci Resolve (free version)
Geogebra: www.geogebra.org/m/uynzn4kn
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Chapters:
0:00 Introduction
0:50 The Problem
2:05 Problem Simplification
5:06 Tangential Polygons
8:21 Final Solution
11:50 Closing Thoughts
Background Music:
Music by Vincent Rubinetti
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Paper Textures from: www.freepik.com/
the use of the intermediate value theorem is just beautiful
I disagree; the IVT only says that the given division into a tangential quadrilateral and a triangle _exists_ . Not only is the common proof not constructive, it turns out that IVT is actually _not provable_ in constructive mathematics. So, we "know" that a decomposition exists, but we have no construction that will always work. The promise of a "recipe" has been broken precisely because this solution hinges on the use of the IVT.
@@drmathochist06 I largely agree with you, I just found it beautiful as it is the first time I have seen the IVT used in a geometric proof.
@@drmathochist06 Though, as you say, the proof given in the video is non-constructive, it does hint that there may be a construction that always works. And indeed there is one, that allows for dissection of any equilateral triangle into k kites for k = 3 or k >= 6, any non-equilateral triangle into k kites for k >= 3, and any n-gon (with n >= 4) into k kites for k >= 3(n-2), where by kite I mean convex kite, as was also meant in the video. And it's also one of the few times I've seen the IVT used in a geometric proof too
@@charlottedarroch Sure; it's a simple matter of analytic geometry to exhibit the explicit function and solve for the point in question. But then you don't need the IVT.
You might be able to find a construction by extending two sides of the convex quadrilateral (say, AB and CD) to meet at E. If we assume B and C are the points closer to E, take the E-excircle of triangle BCE and then draw the segment from A tangent to that excircle. If that segment meets point F on CD then ABCF is tangential and ADF is a triangle.
Also I first saw this and was like “no shit” thinking kites were just quadrilaterals (as in 3d modeling) but realizing they were symmetrical I was like “wtf” and had to click
I have a new insight which simplifies the problems given at the end of the video and shows that any quadrilateral, convex or concave, may be dissected into n kites for all n >= 6. As observed in the video, any triangle can be dissected into 3 kites using the incircle method. But we can say more. Any non-equilateral triangle can be dissected into 4 kites. Let ABC be non-equilateral with shortest side AB and longest side AC. We may have an isosceles triangle with |AB| = |BC|, just not equilateral. Then take D to be the intersection of the angle bisector at A with the side BC and take E to be the point on AC such that |AE| = |AB|. Then ABDE is a kite and CDE is a triangle which can be dissected into 3 kites.
Therefore any quadrilateral (including concave) which can be dissected into two non-equilateral triangles can be dissected into 6, 7 and 8 kites. But which quadrilaterals can be dissected into two non-equilateral triangles?
Given a convex quadrilateral ABCD, we can clearly dissect it into triangles ABD and BCD. Suppose that at least one of ABD and BCD is equilateral, say ABD. Then we may instead dissect ABCD into ABC and ACD. And as ∠BAD = 60°, we have ∠BAC < 60° and ∠CAD < 60°, so neither of ABC and ACD are equilateral. And if neither ABD and BCD are equilateral, we're already done. In all cases, we dissect a convex quadrilateral into two non-equilateral triangles.
Now consider the case of a concave quadrilateral ABCD with D interior to ABC. Then we may dissect ABCD into triangles ABD and BCD. If both were equilateral, then ABCD would be a non-square rhombus, which isn't concave, a contradiction. Therefore at most one of ABD and BCD may be equilateral. Suppose ABD is equilateral, then BCD is non-equilateral. But then ∠BDC > 120°, otherwise D wouldn't be interior to ABC, therefore ∠BCD < 60°. So if we instead divide ABCD by extending side CD until it intersects AB at E, we obtain a dissection of ABCD into triangles ADE and BCD. But as ∠BCD < 60°, BCD non-equilateral, and as ADE is part of the triangle ABD, we have ∠ADE < ∠ADB = 60°, so ADE is also non-equilateral. Finally, if neither ABD and BCD are equilateral, we're already done. Therefore any concave quadrilateral may be dissected into two non-equilateral triangles.
So we've in fact shown that any quadrilateral may be dissected into two non-equilateral triangles, and can therefore be dissected into 6, 7 and 8 kites. Using the fact that a kite is a tangential quadrilateral, we may dissect any kite into 4 kites, increasing the number of kites by 3. Therefore any quadrilateral may be dissected into n kites for all n >= 6.
Thank you for putting this into words. I was thinking this the entire time and was just about to type up a proof.
Is it possible with five or fewer
Yeah as cool as this vdeo was (at least from what I realised from reading this) there was no reason to divide it into seven quadrilaterls. They could have literally just split into two triangles and got 6 kites from individually splitting those triangles into kites.
you can pretty easily, for any non-equilateral triangle, get it split into 5 kites by doing the triangle -> kite + triangle trick twice and splitting the resulting triangle into three. And then you can use what you said about a kite being splittable into four kites to get any kite amount n above 5.
But I have a question
How can we split an equilateral triangle into 4/5 kites?
@@lox7182 The aim of the video wasn't to show find the minimum number of kites that any quadrilateral could be dissected into. It was to first show that it's relatively easy to get 6 kites, or 3k kites for all k >=2, but that doesn't immediately lead to achieving 7 kites. Then to show that combining what we already knew, with a few extra pieces of pieces of knowledge, does allow for dissection into 7 kites also.
The most underrated maths channel on TH-cam
Fr!!!!! This stuff reminds me of 3blue1brown
Who is Polyamaths? For the blind, he is the vision. For the hungry, he is the chef. For the thirsty, he is the water. If Polyamaths thinks, I agree. If Polyamaths speaks, I'm listening. If Polyamaths has one fan, it is me. If Polyamaths has no fan, I don't exist.
ok shakespeare popoff
Bro actually had that stuff open in a notes tab in order to copy and paste it as quickly as possible when the video came out
I have a question, inside any pentagon, can you create an ellipse tangent to all sides? Because I know an ellipse has five degrees of freedom
Also, if you make a smaller kite out of a bigger one in a way that one kite has the same corner as the bigger one, (not the ones off to the side, but the one that looks similar to it) is indeed similar, and if so, I wonder if there’s a specific ratio it has to be
@@ianweckhorst3200 Yes, every convex pentagon has a unique inscribed ellipse!
Ah I see, you disect the original quadrilateral into a triangle, and a quadrilateral for which there exists a tangent incircle. Got it the moment you drew that circle sitting loosly in the quadrilateral. Neat. You did a very good job of leading up to this idea, so you could consider encouraging the viewer even earlier to stop and see if they can figure it out from certain points.
My Euler, this is simply amazing. I never heard about that. Keep on doing this great work, PolyaMath! Big hug from Brasil.
Both of the open problems mentioned at the end really do come down to ruling out specific values of n. Any convex quadrilateral can be dissected into n kites for any n >= 9. And any concave quadrilateral can be dissected into n kites for any n >= 14.
Given a dissection of a quadrilateral into n kites, the same quadrilateral can be dissected into n+3 kites by taking any kite in the original dissection and dividing it into 4 kites as discussed in the video. Similarly, one may instead take a kite in the dissection, divide it into two triangles, then divide those two triangles into 6 kites, allowing us to turn a dissection into n kites into a dissection into n+5 kites.
So given that we can dissect a convex quadrilateral into 6 kites (two triangles) or 7 kites, we may further take a dissection into 6 kites and produce a dissection into 11 kites. But then we may dissect into 3k+6, 3k+7 and 3k+11 kites for all k >= 0, which covers all 3 residue classes mod 3 and clearly covers all numbers n >= 9.
Similarly, for concave quadrilaterals we dissect first into two triangles and into 6 kites. Then we can further dissect into 11 kites and into 16 kites. Then we may dissect into 3k+6, 3k+11 and 3k+16 kites for all k >= 0, again covering the residue classes and covering all numbers n >= 14.
So it remains to be seen whether all convex quadrilaterals can be dissected into 3, 4, 5, and 8 kites, and whether all concave quadrilaterals can be dissected into 3, 4, 5, 7, 8, 10, and 13 kites.
In fact, dissecting a concave quadrilateral into 13 kites is trivial. Let ABCD be such a quadrilateral. Then one of the vertices lies in the triangle described by the other three, so without loss of generality, let D be interior to the triangle ABC. The pick points E and F on AB and BC respectively such that |BE| = |BF|. Further, we choose |BE| small enough that D is not on or interior to BEF. Then we may pick a point G on the angle bisector through B, exterior to BEF but interior to ABCD. Then BEGF is a kite. Also draw the line DG. Then we dissect ABCD into BEGF, ADGE and CDGF, where BEGF is a kite and both ADGE and CDGF are quadrilaterals which may be further dissected into 6 kites each, resulting in a dissection into 13 kites.
And dissecting a concave quadrilateral into 10 kites is also straightforward. Again taking ABCD to be concave with D interior to ABC. Then take the angle bisector at D. This bisector intersects either AB or AC. Without loss of generality, take it to intersect AB at E. Then ADE is a triangle and BCDE is a convex quadrilateral, which can be dissected into 3 and 7 kites respectively, attaining a dissection of ABCD into 10 kites. In the case where the angle bisector in fact intersects B, we may consider E where the line CD intersects AB and pick any point on F on BE. Then the line DF dissects ABCD into triangle ADF and convex quadrilateral BCDF.
I also got stuck at disproving (or perhaps proving) the small values of n. Very nicely done though.
@@Polyamathematicshow do you not getbored or frustrated or fed y0 with math in general or getting stuck at a proof like that? Hope you can PLEASE respond whenever you can. And hiw does anyone sustain attention for 13 minutes on one thing in math or even five minutes..how is it not boring after a few minutes? Hope you can respond whenever you can.
@@PolyamathematicsAnd shouldn't you explain what a kite is since I'm guessing most ppl don't know what a mathematical kite is?
Polya's books were such an inspiration in both my math learning and math teaching (intersecting) paths; now this channel, i daresay, is my new Polya-math-inspiration!
Keep up the good work. The quality of your videos is great. Very well presented. A nice little watch that I felt very able to follow along with. 😊
9:13 This looks like it would be a cool fractal design.
Great video by the way!
Excellent, beautiful video. I loved it!
One comment---it seems like your s sounds are getting eaten by your mic or editing software? I thought it was just at the ends of words, but even when you say "ABCD", the "C" feels almost muted out.
Fingers crossed audio should be sorted out, next video.
A triangle can also always be divided into 7 kites: for instance, after inscribing a circle, add another circle touching two sides and the first circle. The common tangent of both cuts the triangle into a tangential quadrilateral and a triangle.
Nothing short of a gem of a math channel. Subbed
That was very interesting. Lovely graphics!
Wonderfully displayed. From the outset, I was wondering if this was seven-trees-in-one, but it's something quite different and deeply geometrical.
this video was fantastic and the animation was great! keep up the good work big man!
7:00 i was wondering where i heard pitot before, thanks for that!
Cool math, really well explained! And the graphics are lovely. Thanks!!
dude, okay the content of the video is one thing, but the editing jeez. you're insane !! love this so much
i have NO clue how this showed up on my recommended.. but I am SO THANKFUL!! This video is so amazing and demonstrates such mathematical ability that I could not believe what I was watching!! Honestly idk who this guy is but he is GOING PLACES.. to be fully honest his voice turns me on a bit too.. haha
Hi this is my son and we do not appreciate this comment. 🙂
I suggest you edit this and remove the last line before further action is taken.
@@karinabrahmathematics Hi this is john and I DONT CARE!! WHO DO YOU THINK YOU ARE?!?! it is obvious that polyamath has no parents.. he was brought to us from HEAVEN! he is a gift and you dare to say that you are his MOTHER?????/ You really cannot go ANYWHERE on the internet without seeing all of these impertinent losers!
@@karinabrahmathematicslmao really
@@karinabrahmathematics Are you kidding me, you PATHETIC excuse for a human being? Pretending to be the mother of Polyamath? What kind of sick, twisted mind comes up with that garbage? You're not just a LIAR, you're a spineless, gutless COWARD hiding behind a screen, too afraid to show your face because you know everyone would see you for the sorry excuse for a person you truly are!!!
Do you get some sick pleasure out of deceiving people? Is your life so devoid of meaning that you have to resort to impersonating someone else to feel a shred of significance? Well, newsflash, you sorry excuse for a troll, nobody's buying your bullshit! We see through your flimsy facade, and let me tell you, it's LAUGHABLE!
Polyamath is light years ahead of you in intellect, integrity, and authenticity, and you have the audacity to try and tarnish his reputation with your deceitful antics? You're not fit to lick the dirt off his shoes! You're a stain on the fabric of society, a festering sore on the internet's backside, and I wouldn't piss on you if you were on fire!!!
I demand, no, I COMMAND you to cease and desist this charade immediately! You owe Polyamath and every single member of this community a heartfelt apology for your despicable behavior! And let me make one thing crystal clear: if you ever dare to pull a stunt like this again, I will personally make it my MISSION to ensure you're exposed for the pathetic fraud you truly are! So take your lies, your deceit, and your sorry excuse for a personality, and crawl back into whatever hole you crawled out of! You're not welcome here, and you never will be!
And another thing, who the hell do you think you are, trying to tell me what to do? You're nothing but a delusional charlatan, a nobody hiding behind a keyboard, thinking you have the audacity to command anyone! Well, let me tell you something, you sorry excuse for a human being: you're not in charge here, and you never will be!
You prance around with your fake identity, thinking you can manipulate people and dictate their actions? Think again! You're nothing but a sad, pathetic little troll, desperate for attention in the most pathetic way possible. Do you honestly believe anyone gives a damn about your worthless actions? Get real!
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So listen up, you despicable fraud: pack up your lies and your deceit, and slink back to whatever cesspool you crawled out of. You're not welcome here, and you'll never be anything more than a stain on the internet's already tarnished reputation. So do us all a favor and disappear!
Omg I was literally gonna Google if it was the same Pitot because I'm a student pilot and I thought you pronounced it weird 😂😂😂
We all say "Pee-toe Tube" Lovely little note
A convex Polygon with n sides, can be divided into number of kites k
n = 1, 2 -> not defined
n = 3 -> Equilateral Triangle -> k = 3, and for k > 6 ; not sure about 4 and 5
-> Non-Equilateral Triangle -> k >= 3
n = 4 -> k >= 6, not sure about 3, 4, and 5
n = 5 -> k >= 9, by dividing it into two polygons n=4, n=3
for any n > 4 -> k >= 3*(n-2), by doing similarly recursive division of n sided polygon into n-1 sided polygon and a triangle
k >= 6 not k > 6 since you can split any triangle (including an equilateral one) into 3 kites and split one of those kites into 4 kites, getting 4 kites + 2 remaining kites = 6 kites. But yeah I'm not sure about 4 or 5 either.
this will really help me when i need exactly seven kites, no more and no less
Trivially, if any convex quadrilateral can be divided into 7 kites, and a kite can be divided into 4, you can always replace one kite with 4 kites, therefor all n >= 7 where n % 3 = 1 is possible. Any convex quadrilateral can be divided into 2 triangles, and a triangle can be divided into 3 kites, so all n >-= 6 where n % 6 = 0 is possible. Further, you can take one of those kites and divide it into 4 kites, so for all n >= 9 where n % 3 = 0 is possible. It looks like n % 3 = 2 is the problem to solve, along with n = 3 and n = 4 (n = 1 and n = 2 are not possible for all convex quadrilaterals. I suspect 3 and 4 are not either, but I do not know).
A concave quadrilateral with 1 angle > 180 can be divided into a convex quadrilateral and a triangle. The triangle can always be divided into 3, and the quadrilateral is solved above. The minimum number of kites increases by 3, but otherwise nothing changes. A concave quadrilateral with 2 angles > 180 can be divided into 2 triangles and a convex quadrilateral; the minimum number increases by 6, but otherwise nothing changes.
There may be other ways to cut up a concave quadrilateral that yield different results. I suspect that in the case of 2 angles > 180, you can always cut the quadrilateral up into a kite, a triangle, and a quadrilateral. The kite gives you n % 3 = 1, the triangle yields n % 3 = 0, and the quadrilateral gives n % 3 = 1, which grants access to n % 3 = 2 for n >= 17 (3 for the triangle, 1 for the kite, and the remaining quadrilateral could require as much as 13 if it has 2 angles > 180 (2 triangles + quadrilateral = 13)). I'm not sure how to prove that you can make that original division.
What a beautiful and simple proof, also the explanations and the animations were on point !
Beautiful use of the Intermediate Value Theorem!!
Für konkave Vierecke kann man einfach einen Inkreis, welcher die beiden Seiten der konvexen Ecke und eine der beiden anderen (Jeweils die, die zuerst berührt wird) tangent berührt, bildet. Dann kann man die vierte Tangente des Inkreises durch die konkave Ecke bilden und hat wieder eine Einteilung in ein Dreieck und ein Viereck mit Inkreis.
ChatGPT translation:
For concave quadrilaterals, one can easily construct an incircle that touches both sides adjacent to the concave vertex and one of the other two sides (specifically, the one that is touched first). Then, the fourth tangent of the incircle can be formed through the concave vertex, thus dividing it into a triangle and a quadrilateral with an incircle.
It was fun to learn that a convex quadrilateral can be divided into kites and there can always be a division of exactly 7 kites, the idea of dividing into kites is fascinating
Subscribed based on the color scheme alone ^^
Its crazy, i have never heard of this channel before, and yet when i checked out the channel while watching this vid, i realized that i have seen every single video on this channel
this made me feel better today
The statement applies to convex quadrilaterals only.
A convex quadrilateral is one where all interior angles are less than 180 degrees, and the sides don't intersect when extended.
Awesome! It wasn't clear to me why splitting the quad into 2 tris then into 6 kites "isn't good enough." That's interesting on its own! It's cool you can do it with 7 too, but i went into this video thinking it was a minimization problem.
Duuude, this would be perfect for compressing badapple! You've heard of quad trees, but now, we can have hepta-trees!
Great video
Wow! And that means if you divide each kite down the center, you can make any quadrilateral with only 14 triangles!
Actually, I think you can make any quadrilateral with only two triangles.
@@margue27 No. Really? My word.
@@amaryllis0 any quadrilateral has at least one diagonal completely in the quadrilateral (including concave quadrilaterals), so you can just use that line to split it into 2 triangles. Actually you can, by splitting those triangles further into kites, split any polygon into 6 kites.
@@lox7182 Wow! You don't say? Gee wilickers I sure am learning a lot of things right now for the first time right now (currently), Thank you for educating me oh knowing and wise stranger. Wherever would I be without you?
@@amaryllis0 I'm sorry btw. It should have been splitting any n-polygon into 3(n-2) kites. And I wasn't trying to come off like that I was... wait was your original comment sarcastic?
Well presented. Subscribing.
This does beg the further question, "which quadrilaterals can be divided into fewer kites?"
For 1, it is obvious; the kites themselves. For 2 through 5, I do not know. For 6 and more, every quadrilateral can.
This is actually very interesting, I didn't think about that.
Without actually watching the video, I tried doing proofs of this on my own and I happened to get an easy solution for all quadrilaterals except those that are concave (have a reflexive angle). I hope this video does address the case that involves quadrilaterals with reflexive angles.
Call me crazy, but can't we reduce the magic number to 6? All quadrilaterals can be divided into two triangles by making a cut between opposing corners. You can then use the "every triangle can be made of three kites" idea to divide both triangles into three kites, leaving a total of six kites.
Yes but that's too short to make a full video out of lol, 7 is more fun!
That’s completely right, but this isn’t a question of optimizing; if it were, then it would most likely have the simple solution you’ve pointed out. Instead, we want to know exactly which values of n we can do this for, rather than just the smallest. From the content of this video, we know that it works for n=6 and n=7, and by using the operation of turning one kite into four kites, we know that if it works for some n, it will also work for n+3, so we know that 6+3k and 7+3k are solutions for any k≥0. This means that from our current knowledge, the slurs of n we don’t know about are the numbers less than 6 and numbers of the form 8+3k, so a solution to the open problem might look like finding a method to dissect any quadrilateral into 8 kites, and finding quadrilaterals that can’t be dissected into 1, 2, 3, 4, and 5 kites, or it might look like finding a dissection into 5 kites and eliminating the ones below, or maybe even starting with some higher value 8+3k and eliminating the ones below. Maybe through some strange quirk, no value of the form 8+3k works, and a proof of that would also be a solution to the problem. If we hadn’t found that the dissection always works for n=7, then in a sense there would still be twice as much work to do, so this video makes significant progress into the open question, even though it is not about the minimal dissection.
@@EvilAxelord19 We can actually dissect any kite ABCD into 3 kites, not just 4, by using the following 3 additional points: Let AC be at least as long as BD. Let P be the intersection of the circle around A with radius AB with AC. Let Q1 be the intersection of the bisector of CAB with BC and Q2 the intersection of the bisector of CAD with CD. Then ABQ1P,APQ2D, CQ1PQ2 are the desired kites.
I.e. if we can do k kites, we can also do k+3-1=k+2 kites. In particular, we can do 8 (given that we can do 6). Hence, we can do 6,7,8 and therefore also 9,10,11...
@@ElchiKingnot necessarily *any* kite (unless kites are for some reason defined to have two 90° angles). consider a kite with angles 60°,90°,60°,150°. the 60° angles must be at the vertices A and C, since otherwise AC would be shorter than AB, so P would be outside of ABCD. but even then, the quadrilateral APQ2D would not be a kite, because the length of AD is equal to neither the length of AP (because that's the length of AB), nor the length of DQ2 (because Q2 lies strictly between C and D, and the length of AD is equal to the length of CD).
in fact, AC being at least as long as BD is not a sufficient requirement even if we assume that AC is the axis of symmetry. -your construction works if and only if AC is the axis of symmetry and is (strictly) longer than AB. fortunately, that's implied by the kite having two right angles, which is the case for the 6 kites we get by splitting an arbitrary quadrilateral as in the video.-
-so yes, for every n>=6, it's possible to split every convex quadrilateral into n kites (and if this n isn't 7, then it's possible to split every quadrilateral into n kites)--
nevermind, i forgot about at least one thing: the kites still need to be convex. CQ1PQ2 is convex only if the angle Q1PQ2 (specifically the one internal in CQ1PQ2) is less than 180°, which it is not in the case of kites with two right angles, since the size of the angle Q1PQ2 is 360°-2α, where α is the size of the angle ABC (or equivalently ADC). and if ABC and ADC are right angles, then that is exactly 180°, not less. so 8 kites still doesn't necessarily work
The music is so calming
This whole video is like asmr
Great vídeo!
I appreciate the use of the intermediate value theorem, but there is also a nice constructive way to go about it. One could simply construct the 'inner circle' for the trio of lines AB, AD, DC, then draw the other tangent through B of this circle, creating a tangential quadrilatical. One could show that if the tangent through B does not pass between D and C, then the tangent through C would pass between A and B completing the proof.
Brilliant video
The video is better than the Geogebra file, which only seems to show the case when the incircle of the tangential quadrilateral touches the incircle of the triangle
So since any kite can be split into 4 kites, then any shape that can be split into K kites, it can also be split into K+3N kites for any positive N.
Love the videos! Very well done, well paced, well written, dont waste time. But i do have one piece of constructive criticism if youre open to it. I do recommend not having the beige background come in , or if you do darken it a bit. It looks great, but for people watching this at night, its a nice black background with enough time for my eyes to get adjusted, and when the white background comes in it hurts my eyes haha. Just a suggestion!
at the end of the video, around 11:33, would it be possible to find the circle on the right by finding the intersection of the angle bisectors of B and C and then find AE by finding a line tangent to said circle and point A? though I guess you would still need to argue that the construction is possible, which would require the reasoning you gave. I still wonder if there's a simpler way to go about things
very interesting
I have a conjecture that it true when n is multiple of 3.
Because u can always divide a quad. in desired number of triangle which gives you 3 kites.
Nice math there. One suggestion on your video: switching between dark and bright background is not a good idea. Choose one and stick to it. I personally prefer dark mode and the sudden switch from dark to bright cause discomfort. Switching between those in same video is just going to piss off people who have either preference.
Since we can cut a kite into 4 kite repeatedly, we need to consider the smallest cases mod 3.
We have 6 and 7 from the video. We can add a line between the midpoints of opposite sides to create two quadrilaterals, each of which can be covered by 7, thus 14 is possible. And any number more than 12.
Less than 6 is not always possible. This leaves 8 and 11 open.
EDIT: others here outlined how 11 is possible. How less than 6 is not possible should be shown.
Firstly, it needs to be clarified that less than 6 being impossible would, hypothetically, be only true for some quadrilaterals (it wouldn't be a result for every quadrilateral). For example, a kite can be split into one kite.
You sound a bit like TodePond. Cool video!
Reminds me of voronoi patterns
How many unique dissections of this kind are there?
Once you've divided a shape into kites, you can just keep dividing the kites to get more kites. Dividing the kite into 4 kites removes 1, netting 3 new kites. Dividing into triangles and then into kites gives 5 new kites
Using these, you can get almost all numbers of kite subdivisions, only missing a few small numbers. Everything that can be written as 6+3a+5b or 7+3a+5b is possible, with a,b non-negative integers
I think 8 might be the largest impossible n
using the proof of another comment, specifically
@danieldarroch4775's comment, any non-equilateral triangle can be split into a kite and a triangle, and any quadrilateral (including the concave ones) can be split into two triangles. If you can split a triangle into three kites and a (non-equilateral triangle) into a kite and a triangle, then you can turn that into a kite and three kites, aka 4 kites. And you can split any quadrilateral into 2 (non-equilateral) triangles, and so you can split any quadrilateral into 8 kites.
@@lox7182 that's really nice! Does it work for all quadrilaterals?
Beautiful stuff, but I can't seem toget my head around what I think is an unsatisfied edge case: non-square rectangles. Following the procedure from the fifth chapter of the video all rectangles fail because the leftover triangle is degenerate [is this the right maths word? It's unpleasant]. A square can almost trivially be checkered into four half side-length squares, one of which can be further checkered to compose a total of seven squares.
i manged to get a rectangle to work in the geogebra file. I don't see why the triangle must be degenerate?
@@TheArizus yep I was wrong! It totally works. I too have made it in the geogebra, thanks for your reply!
Nice
How could you prove that it is impossible to dissect an equilateral triangle into 4 or 5 kites?
What about 49 kites, or 343 kites, or even 2401 kites, and so on?
I think I have a very bland proof for concave 4-gons. It is more of a construction than a proof.
A … point adjacent to concave point
B … concave point
Start by making an angle bisector of alpha. Construct a point E such that it lies on angle bisector and CD. Make a kite ABEB’ which is symmetrical over the angle bisector.
Proove that this partitions the quadrilateral into a kite and two triangles.
Proof left as an exercise for the viewer.
If all triangles have a unique circle that is tangential to all sides, then does that mean there is a 3-dimensional shape of some general description that has a sphere that is tangential to all faces?
Yep, a tetrahedron has 4 faces and a unique insphere
4:18 but since any quadrilateral is made up of 2 triangles (just cut along the diagonal)… wouldn't it be possible to cut it into 2*3 = 6 kites only ???
(okay maybe the goal wasn't to get the least amount of kites)
are you using AI noise cancellation? if youre using audacity, you can just select your audio, click "nosie removal" > "get noise profile", then "repeat noise removal". it works great.
I’m confused because you say at 3:00 that a kite is convex by definition but it is simply not true is it ? Concave kites do exist ! Or is it some kind of regional thing ? (I’m in France)
I think it depends on the context. In this context kites were defined at convex, but certain contexts allow concave kites.
Intermidiate
I assume this doesn't also work for bowtie shaped convex quadrilaterals?
What happened with your voice at 11:41?
11:30 misspelt 'intermediate' in intermediate value theorem
One thing I still don't understand (maybe it was mentioned and I just missed it) is why split into 7 kites not 6? The only reason I could imagine is that there is a case where 6 wouldn't work since in cases where it does work it seems much superior (dividing into two triangles is way easier than finding the tangentual quadrilateral).
6 kites is minimal yes, just 7 kites made for a more interesting problem, and gave more insight about the general case with n kites
Oh okay thanks, great video btw :)))
@@Polyamathematics There's a simpler way to turn it into 7 kites that @danieldarroch4775 mentioned
07:35 shouldn't that be
I’m pretty sure that they are talking about each corner, not the sum
I admit I'd never thought about this problem before. The reason I clicked was to verify my intuition that the "Any*" in the title actually meant "Any plane convex". Further, given that any kite constructed this way is also a plane convex quadrilateral, these can also be dissected into 7 kites. Therefore, you can apply this construction recursively to find tilings of any convex plane quadrilateral with 7*n kites. I wonder - is it always possible to tile a convex quadrilateral with 7*n congruent kites, for some positive integer n?
Actually, I need to revise my statement, because I neglected to consider that dissecting one of the kites removes it from the count. So, you can tile a convex quadrilateral with 6*n + 1 kites. Further, since kites are, as you pointed out, tangential, and can themselves be dissected into 4 kites, the expression can actually be expanded to 3*n + 4
You can divide kites into kites to get 3n+4 as you said, but you can also divide quadrilaterals into triangles to get 3n+3 so it just seems like 2 (mod 3) is that hard part.
@@TheArizus That is not entirely true. Since you can decompose a kite into 2 triangles, thus 6 kites as well, you can also increase the number of kites by 5. So a decomposition into 11 kites is possible. The hard part is to show that a decomposition into 3 or 5 kites is impossible. I am not sure whether 8 kites would be possible either.
i still dont get why you can't just subdivide the polygon into 2 triangles
6:00 why doesn't a triangle have 6 degrees of freedom? a X and a Y coordinate for every point?
A triangle is uniquely defined by its 3 side lengths or side - angle - side etc... (always 3 properties)
Your question is actually still very good, but the key is that we're saying GIVEN a triangle FIND the incircle. Now we can translate and rotate some triangle and produce a congruent triangle so with 6 degrees of freedom this is NOT unique. However if we fix one vertex of the triangle to the origin (0,0) (to prevent translation) and fix one side to say the x-axis so one of the points is (a,0) (to prevent rotation) and the third point is (b,c) and now a UNIQUE definition of a triangle can be given by a, b and c.
(The reason we are allowed to translate the incircle is because we're saying find the incircle GIVEN the triangle so the triangle is fixed and hence must be unique)
Why the restriction to convex? As one an see from the Geogebra animation, non-convex works fine. The kites might be non-convex but that's ok.
Specifically when AB + CD = AD + BC AND ABCD is concave then (at least with the given construction) you can't (always) dissect into a tangential quad and triangle (in other words Pitot's Theorem only applies to convex quadrilaterals)
4:09 AHH . . . Yes, I see
Since you can always dissect a quadrilateral into 2 triangles, you can always split it into 6 kites, so 7 isn't the least. Since you can always dissect a kite into 4 kites, adding 3 kites each time, 7 isn't close to the most. What was the point of picking 7 specifically?
Not trying to be snarky btw, I just wonder whether I missed something or it was to demonstrate the various methods
Hey polymath, maybe you should start to use Desmos instead of Geogebra, compared to Geogebra, Desmos performs way better, and it has all the same features and sometimes even more
Desmos sucks for geometry. I wish it was better
@@Etothetaui
They have a new geometry tool now, you should try it
@@JJean64 I have, how do you find the intersection points between a circle and a triangle?
One thing I don't understand is why exactly did you choose to prove the theorem for seven kites specifically, when much more simply you could just divide the general quadrilateral ABCD into two triangles ABC and ACD, divide both of them into 3 kites each and just call it a day proving 6 kites are enough.
I don't think it was a minimisation problem. 7 kites ended up being more useful for the general case of n kites (and made for a better video)
square always equals two triangles, and one triangle can split to three kites….
I don't think the chevron can be split into 3 convex quadrilateral let alone kites.
I actually solved this kinda, but our trainer forgot to say the kites are convex
OK I got you a proof I think.
Lemma 1: Any non-equilateral triangle can be dissected into 4 kites.
Let ABC be a non-equilateral triangle such that AB
about lemma 2... I didn't think about this when I was making my own replies to people but what if the new triangle (the triangle gotten from the triangle -> kite + triangle dissection) is equilateral?
actually lemme see if this is even possible or not
let's say that such a case existed
in that case the inner equilateral triangle's length x would be equal to BC/2
and AC = AB + (BC/2)
(AB + (BC / 2))^2 + (BC ^ 2) - 2(AB + (BC / 2))BCcos(pi/3) = AB^2
AB^2 + (AB)(BC) + ((BC ^ 2) / 4) + (BC ^ 2) - ((AB)(BC) + (BC^2 / 2)) = AB^2
3(BC^2)/4 = 0
-> BC = 0
um yeah not happening.... nevermind
@@lox7182 I think this is possible, however there are multiple different kites you can pick and not all will lead to an equilateral triangle.
That being said, another user posted their proof that does not rely on my Lemma 2 which I think is a better proof.
Interesting, but why though?
That's what's beautiful about mathematics. There's no need for a why. It's kind of like a puzzle, where the reward is that at the end, all pieces fall into place and that you've solved it. You find beautiful connections rooted in concise logic.
Oh Ic
Nishad are you indian origin
You mean "at most" seven kites, I suppose. One might suffice.
Thala for a reason
3:00 there are concave kites
The problem defined kites as convex
@@TheArizus ah, my fault
Every quadrilateral is made up of 11 kites.
Y E S S S S S S S S S S S S S S. SS S S. S
hi
Hello
Just came to say, "the thumbnail doesn't have square or rectangle." And leave.
so while this problem is interesting, Its not even the lowest amount which is confusing to me. ANY quadrilateral can be split into 2 triangles, which are both 3 kitable, 3*2 = 6. so any quadrilateral can also be split into 6 kites, which is smaller. I feel it would be more interesting if you could split any quadrilateral into 5 kites, or into 7 kites in a single way. Good video tho
Its only 6 kites
You could be the coolest guy in the world, but if you use the word “hence” more than a couple of times, in a 13 minute period, you’re always going to look like a nerd. There’s just no getting around it. I tried it one time myself, and my girlfriend left me for another dude, later that same day. That wasn’t a good experience. I don’t recommend it.