Conway's IRIS and the windscreen wiper theorem 

Conway’s circle has the larger diameter. The “diameter” of the shape of constant width is exactly one of the chords we were discussing. But in Conway’s circle, those same chords are not diameters, so are less than a diameter, so the circle has a greater diameter.

RIP conway, he will be missed

I remember translating this theorem's Wikipedia page to Hebrew a few years ago, I came up with the coloring proof. but when I saw the swivel proof I just couldn't help but smile. I liked it better than my original proof. this proof made me smile so much. this always happens in your videos, in what you call "aha moment" I experience euphoria. your videos just make me smile from ear to ear, every one of them.

I liked the color proof better, because the swivel proof seemed like it relied more on visuals, which could be deceiving.

"You're here for the mysterious iris in the thumbnail, aren't you?" Sorry, but no. I'm here because it's another Mathologer video :)

The "windscreen wiper" method can be used to prove the same result on a sphere, which by extension also proves the 2D case as well. 1)Draw a great circle around a sphere, marking two points at random on this circle. 2) Mark a third point at random, but not on the same great circle, and connect this to the first two points around two new great circles. This creates a spherical triangle. 3)Rotate the sphere so that the "north pole" lies within the triangle in such a way that the northernmost points on the three great circles all lie at the same latitude. 4)Draw the "whiskers" by extending the sides of the triangle as shown in the video, following the great circles in each case. 5)The great circles all have exactly the same radius of curvature at every point, so the windscreen wiper method can be used to rotate and tilt any one of the extended sides around each vertex to bring it into superposition over the adjacent side. Three rotation and tilts brings a side back to its original position, but pointing the opposite direction, just as in the video. This proves that the three extended sides are the same length, and the northernmost points are the exact midpoints of each arc. 6)Starting at the northernmost point of any of the three great circles and measuring the same arc in each direction takes you to two points which lie at the same latitude. Since the three arcs are all the same length, all six end points must lie at the same latitude. Ta dah. 7)Let the radius of the sphere tend to infinity. The surface of the sphere tends to a plane surface, and the original Euclidean result follows directly. Ta dah!

You triggered my math competition PTSD 😂

Iris-free Proof: join the six exterior points consecutively, calling them A' A'' B' B'' C' and C''. Note the numerous isosceles triangles in this diagram and mark the relevant congruent pairs of base angles. Next, observe that A'A''C'C'' is a quadrilateral with supplementary opposite angles. Thus it is a cyclic quadrilateral and A'A''C'C'' (circle 1) are concyclic. The same reasoning shows the sets of points A''B'B''C' (circle 2) and B'B''C'C'' (circle 3) are concyclic. Circle 2 and 3 have three points in common, and so A''B'B''C'C'' are all concyclic, since three non-collinear points determines a circle. This circle has three points in common with circle 1, and so A'A''B'B''C'C'' are all concyclic.

Draw the chord that joins the vertices of the two green segments (top). This gives an isosceles triangle, hence the axis of the chord is a bisector of the top angle of the original triangle. Hence this axis goes through the triangle's incenter, and this incenter is also equidistant from the two origial top points. The same is true for all three pairs of points. Now choose the top right and the most right point: both lie on a (green+blue) segment. Reason just the same as before and you get that these two points also are equidistant from the triangle's incenter. Rinse and repeat, and you get that the incenter is equidistant from all six points.

Puzzle at the end: Conway's circle is larger. The weird curve's diameter is precisely the length of the chords since the chords are perpendicular to the curve's tangent at the intersection (this can be trivially shown from the curve's construction), so the diameter of the circle is necessarily larger (specifically, the square of the diameter of Conway's circle is the square of the incircle's diameter plus the square of the sum of the three sides of the triangle).

Thank you for all the effort you put into these videos. I very much enjoy the learning experience from watching this channel!

colouring proof is the one i came up with after you pointed out the incircle. also, the shape of constant width has width equal to one of the chords of the iris, definitely less than the diameter of the circle.

I like the coloring proof best. Conway left us so many gems.

Also this shows a general method for generating curves of constant width. Given a set of strait lines of general position (no two lines are parallel, no three lines intersect at the same point) you can always draw a constant width curve as complex as you want. I don't remember the algorithm exactly but I suppose you'll manage to reconstruct it.

Thanks for the video! I felt compelled to try to find a proof, so here it is. The center of the big circle must be the intersection of the three lines cutting the angles of the triangle in two. This is also the center of a small circle, inscribed in the triangle. Consider a segment of length equal to the sum of the three sizes of the triangle. If we glue the middle of this segment to side of the small circle, and rotate this construction, the extremities of the segment draw the big circle (I skipped a few details that were a pain to write concisely : ) Thank you for making me feel worthy! I hope I got it right

Wonderful proof. The incentre is slowly becoming my favourite triangle centre! (overtaking the orthocentre as my previous favourite)

Very interesting! This reminds me of a result I came across recently. Take a regular polygon positioned anywhere in a circle. Extend the sides to form two sets of segments with lengths x_1,x_2,...,x_n and y_1,y_2,...,y_n, so that each y_i is counterclockwise from x_i. Then applying power of point to each vertex, we get x_k(L+y_(k+1))=y_k(L+x_(k-1)), where L is the side length of the polygon. Adding all these equations and cancelling the x_ky_(k+1) terms, we get x_1+x_2+...+x_n=y_1+y_2+...+y_n. So the two groups of segments have equal sums.

Kudos! To think that such simple principles could go unnoticed for so long. Which is why I love this channel, come to think of it. Not only is the content absolutely amazing, I always feel like I've learned something new. (Almost as if taking a glimpse into the realm of "sacred geometry" or something!)

I liked the colored line proof the best. The other one seems like something someone would show you and it would not turn out to not be true, but just looks like it. Love your channel!

Weird curve's diameter= L1+L2+L3 which is a chord of Conway's circle not containing its center < diameter of Conway's circle. BTW, (Conway's Radius)^2 = r^2 + p^2, here r is the incenter's radius and p is both the triangle's semiperimeter and the weird curve's "radius" (there's no center). You can get r = H/p, using Heron's Formula (H)

I've only watched 45 seconds of this video, so far. And I have to say... "Wow!" A circle... Wow. Greetings from Dallas, Texas!

The iris was interesting, but I'm here for Conway!

This was heartachingly beautiful. Thank you.

To prove this, consider two intersecting chords. If you establish that the product of the segments created by their intersection point is equal for both chords, then they are points on a circle. This is evident from the equation r(b+g)=r(g+b), indicating they lie on a circle. To further prove that this circle shares the same center as the inner circle, examine, for example, a red isosceles triangle. The center of the larger circle lies on the bisector of these red arms. This bisector also serves as the bisector of two sides of the circle where the center of the inner circle lies, confirming they share the same center.

Came for a fractal that looks like an iris, stayed for the beautiful visual geometric proofs

Labeling the endpoints CW from top left: A, B, C, D, E, F Any circle through A and B is centered somewhere on the angle bisector of the top angle Any circle through B and C is centered somewhere on the angle bisector of the left angle (similar for C-D, D-E, E-F, F-A) QED: all the points lie on a circle centered at the intersection of the angle bisectors. (The intersection of the angle bisectors is the center of the incircle.) Not proved here, but true: the angle bisectors intersect at a single point.

2:12 Consider the vertex of the triangle at the bottom-left. There are two lines going through it, both with red on one side and green+blue on the other side. Thus, the angle bisector of these two lines consists of all points that are equidistant from the red endpoints, _and_ (separately) equidistant from the green+blue endpoints. But applying the same reasoning to the other vertices, this leads us to the intersection of all three angle bisectors of the triangle, namely the incenter. We have six equalities between the distances from the incenter to the endpoints, which combine to show that they are all the same! q.e.d.

focus on one intersection of two supposed chords and notice that it has the same power wrt one pair of endpoints as the other (the colors show it) and this means the ends of those two chords lie on the same circle, now if we show that an endpoint of horizontally oriented chord lies on the smame circle as the circle formed by endpoints of other two chords, we are done. Label P,Q;R,S;U,V the endpoints in the positive angular direction starting with horizontal red endpoint and A the red vertex of triangle, B blue vertex and C green vertex. Since PBV is isosceles, angle

My favourite proof is the chord version. Even after playing around with this in Geogebra, I spent extra time with that. Conway discovered so many things in math, he truly was a genius.

Hello Mathologer! You again created an amazing video about an interesting geometric theorem and made the proof visible for us. Magic! - Thanks, Greetings and best wishes!

Construct the angle bisectors of the triangle. They meet at the incenter of the triangle. Draw a circle around the incenter through one of the endpoints, that are supposed to lie on a circle. Choose an angle bisector and do a reflection operation across it. The reflection maps the circle onto itself and two pairs of endpoints to each other. All endpoints can be mapped to each other by a sequence of reflections across angle bisectors. In conclusion, all these points lie on a circle. QED

Swivel , just because I didn't originally catch the insight into the incircle... and the swivel has reflective properties that feels like decoding a secret message through a mirror - very cool. Colouring is ok. It's nice.

I know "Numberphile" might be a complicated word around these parts.. but i will never forget watching that interview Brady did with him [Conway] when he's looking out the window and says "i wish i knew whyyy..." .. in reference to the strange universe-implying numbers in group theory - why the monster and no more?.. why any of the sporadjcs?.. especially when symmetry seems to play such a big role in fundamental physics

Very neat video. That little twist at the end resembled somewhat the rotary wheel of a Wankel motor. Yes, I'm that old...

7:39 the coloring proof I suppose. The static picture that you can look at and check over feels more comforting somehow.

I'm pausing to try to prove it, and I think I got it! It's hard to explain without pictures, but the incircle meets each side of the triangle, splitting each into two segments. We can say red=a+b, blue=b+c, and green=c+a. The line extended from the red side has a length of a+blue = a+b+c below its tangent point to the incircle. And it has a length of b+green = b+c+a above the tangent point. So it is tangent to the incircle at its midpoint. The same goes for the other two extended lines. Since each line is the same length, and they're all balancing on the incircle at their midpoints, they are the same thing, just rotated around the incircle's center. Thus their endpoints all lie on a circle.

I have proven in my head... It just creates bunch of isosceles triangles, so it is easy to prove then that for every quadrilateral that you can make out of 4 consequtive points opposite angles add to 180... So all of them of them are cyclic. And as all adjecent ones share 3 points and those 3 points define a circle all six are on a same circle.

from triangle inequality the diameter of wiper curve is larger because if you take triangle with base of diameter of wiper and center of circle then 2r_circle < d_wiper

Personally I like to think of mathematics as inhabited by amazing creatures that move around interacting and spawning new creatures. So I was immediately attracted to the wiper approach.

Remember I was young and there was Conway's fan club on the next street: they listen punk rock, drink beer and has Conway's iris emblem on the jackets.

I liked the color proof a lot more, it felt a lot more clear, straightforward and obvious to me!

That was a good one. Again I always watch this on a Sunday afternoon. Not a mathematician but my theory is that they are the same.

That is the most fantastic Mathologer I've ever seen 🤩. My mind is spinning ! I will surely dream of this - hoping it won't transform in a nightmare😵‍💫

*Proof of Circle* : Given any two lines, they lie on a circle because the power of the intersection point is the same. This is the product of the lengths of the parts of any chord on which the point lines. If the lengths are a,b,c then the power of one of the points, evaluated through one line is a*(b+c), and evaluated through the other line is a*(c+b). This makes the 4 points cyclic because given 3 of the points, we can define a circle, 2 of the points are on a single line and that gives the power of the point. We can then determine what point gives the correct power for the other line, and since the point at which the circle and the other line intersect gives that same power (and power is linear with point position along the line), this intersection point is the same as the 4-th point. Now we find the center of the circle formed. Note that the two lines, lets name them ABC and ACB, are mirror reflections along the angle bisector since the lengths of the sections formed by the point are the same, a and (b+c). Thus the center lies along the angle bisector. Note that the center also lies along the perpendicular bisector of the two chords ABC and ACB. The distance along the perpendicular bisector is a) identical, can be found bu evaluating position of perpendicular along the lines then noticing the equal angle, hence congruence. Evaluating another pair of lines and checking the lengths, we see that the center point is the incircle center, and the distance along the perpendiculars is all the same (by congruence of triangles), and is the incircle radius, which are both known to be unique. Also, the diameter of the circle has to be larger because two of the opposite points form the chords aka whiskers of the triangle, which is smaller than the circle's diameter.

Answer to the puzzle: The circle has a larger diameter. Because the weird curve's diameter is exactly the same as the sum of the 3 sides of the triangle (can be seen because it meets the lines perpendicularly, by its construction). The Conway circle is larger as it also passes through the same two points on any of those lines, but has a center which is not on the line.

The last curve is making nodes and antinodes with the circle.

5:41: The two yellow segments are congruent because they are formed by two tangent lines where they intersect the circle and each other. If you connect the two points where the tangents intersect the circle, the newly made angles are interior angles that intercept the same arc. If you use the formula { measure of angle = 1/2 measure of arc in degrees} you can set the angles equal. This makes the triangle isosceles, and the two yellow segments of the triangle congruent.

The swivel approach resonated better with me.

The curve of constant width at the end was neat. Also, Conway's game of life is a simplification of a game invented by John Von Neumann. Von Neumann was studying the idea because he was thinking about self-replicating machines.

Triangle area = inradius times semiperimeter (just consider the 6 small triangles defined by the incircle). The 6 Conway points are thus simply the points where the circle of squared radius = squared inradius + squared semiperimeter about the incenter intersects a (produced) side of the triangle.

Draw the lines between neighboring points. Notice that each one forms an isosceles triangle with (alternating) the inner and the outer angles of the original triangle. Choose any four consecutive points to form a quadrilateral. This is a cyclic quadrilateral as two opposing angles will sum to 360 degrees minus the sum of the angles of the original triangle, i.e. 180 degrees. Since I can repeat the argument for any group of four neighboring points, all these possible cyclic quadrilaterals must have the same circumscribing circle.

The swivel proof is more visually striking and easier to follow, but I have no idea how I'd even start formalizing this so I don't have to wonder whether it's just a little bit off and I just don't see it. The color section proof in comparison seems pretty easy to formalize without worrying about my visual acuity. After you said your bit about the swivel proof: Okay, thinking about it a bit more it's pretty obvious that the swivel proof is also pretty easy to formalize.

Nice video. My proof of the first part: it is enough to show that the quadrilateral formed by the a and c whiskers is cyclic. For this, we use the 3 isosceles triangles (two whisker triangles and one triangle with side length a+c). Then opposite angles are complementary.

I personally find the proof with the two colors clearer than the proof wit the swivelling. Maybe it's just me, but I find saying that the sum of some numbers is equal to the sum of some other numbers more obvious than saying that swivelling some line segments around makes them coincide.

My proof: label the 3 triangle lengths a,b,c, and their opposite angles A, B, C. connect 4 of the outer points, two from each of two pairs of the extensions (say two of extensions b, and two of extensions c, to form a quadrilateral. Draw from where the b's meet to the broad edge of the new quadrilateral a line with angle A. do the same from where the c's meet, again with angle A. Since each of those origination points have an angle C and A, (or B and A) and lie on a line, the interior angles are B and C respectively. There fore B, a, C must give you a congruent triangle to the original triangle. There fore you now have a couple more isoceles triangles. Now go to opposite corners. You can quickly find that, you'll get, for instance, angle 90 - C/2 opposite angles 90 -A/2 and 90 - B/2 on the other. These add up to 180, and so are supplementary, and so the new quadrilateral is an inscribed quadrilateral. Since there was nothing special about which of the three quadrilateral we drew, all three are inscribed quadrilaterals. You can draw more quadrilaterals in the same way, to find a bunch more of these half angles, and since 3 points define a circle, you can group them so that eventually you find that all 6 points are parts of a whole bunch of inscribed quadrilaterals that have to be on the same circle.

Each line extending from the edges of the triangle is the same length, since its length is the sum of the lengths of all three edges. The lines are also the same distance from the incenter. If you construct a line from the incenter to the closest point, and a line from the incenter to the ends of the lines, it will form a right triangle. Then you can calculate the distance from the ends of the lines to the incenter. If the closest point on the line exactly divides it in half, then the distance from the incenter to every point is the same.

I would like to humbly submit my vote for the color proof. In any case, this is a beautiful pattern with circles and their symmetries.

I like the coloured proof more than the swivel one, because whilst both have formal versions in algebra and linear algebra (I think) respectively, the algebraic proof is much easier to explain simply and its what I'd be more likely to come up with on my own

I like the swivel proof and the color proof equally. The Action Lab made a good video about shapes of constant width, he also constructs one. Fun fact: if you make Conway's iris with a regular triangle, the radius of the outer circle is roughly 5.29 times (twice root 7 times) bigger than the radius of the inner circle (which was a fun algebraic challenge to compute).

Conway's circle has the large diameter. Diameter of the shape of constant width is the cord formed by extending the triangle. Cord < circle diameter

Each pair of the lines forms a “pair of scissors”, symmetrical wrt an angle bisector. The three angle bisectors meet at incenter. Every pair of neighboring points are equal distant from the incenter. due to symmetry. So all 6 points are equal distance from the incenter. Hence they are all on the same circle.

Coloring proof!

7:38 the colored one is exactly the proof that i was able to get when paused at 2:12. I like this proof more as it shows, that for triangle with sides a, b and c, if we split a in two segments by the touching point of inscribed circle, then difference of these segments is equal to difference of remaining sides of the triangle.

I like the colour proof a lot more! The swivels matching points is not very intuitive - it looks good, but I have a hard time imagining why that would be the case. The colours make the case immediately that it would be true of any triangle.

7:40 I liked the colouring proof best. It was less obvious; and so, a nice surprise 🙂.

I guessed quite a few of the swivelling proof steps just from feeling. But I find the colouring proof more elegant.

Coloring proof. In my head I started with the rotations proof but I don't like them as "proofs" since it just leads to statements like "these two points are the same" which I can't visually differentiate. I could write code to simulate it, but in the end, that is just solving it algebraically and presenting the output visually. The coloring proof actually shows the detail that the rotational proof needs in order to justify doing its rotations. For me it really came down to realizing that the center of the chord must be the tangent to the inscribed circle. Then its obvious that the distance of any point depends only on the radius of the inscribed circle and the length of the triangle perimeter. Which, of course justifies the rotational symmetry that I started with.

To assist your engagement; I liked the coloring better, the idea of the semi-sides each composing the half the chords is pleasing.

The coloring proof was the one I came up with. It was obvious that if the points were all on a circle then it had to be concentric with the inscribed circle of the triangle. Since the six pieces obtained when cutting the sides at the tangent points to the inscribed circle would be pairwise identical I could show that the cords were indeed bisected.

The proof is visible in the thumbnail picture. The whole line segments are of the same length which is the sum of the side lengths of the triangle. Looking at a pair of them, the shorter outer partitions are the opposite side length of the triangle (the partition has to be shorter because one side length of a triangle can never be longer than the other two), the longer ones have to be the sum of the other two side lengths. So its possible to construct a circle that goes through the four end points using the angle bisector and the normal line at the half point (of the whole length). We can do this with each pair and each pair shares a leg, so it becomes obvious each of the line segments is a secant of the same length of one similar outer circle. If we rotate them, they form the inner circle. Edit: This took me a month, perhaps because it's so obvious.

I like the swivel proof because it was nice, and the coloring proof seems like the more obvious sort of approach to take.

All a+b+c have same length, and all have same distance to center of the circle within the triangle. So rotating them around this center gives the same outer points on the circle(, its actually three times the same full line, just with different angles).

The length of each strut of an iris is equal to the perimeter of the triangle. Thus, the midpoint of each strut is at the semi-perimeter! There must be some connection with the various formula involving the semi-perimeter, such as Heron's Formula!

I prefer the swiveling proof, while I would go for the coloring one. My intuition told me that I would find equal lengths if I go for the segment lengths; however, I find the transformation proofs more elegant and more beautiful. Beautiful music, BTW!

2:22 I know I'm late to the party, but here's my attempt: Let's name the vertices of the original triangle X(top one), Y(right one) and Z(left one), and the vertices on top of X are named A(left one) and B(right one). Similar to C, D near Y and E, F near Z. First, we can easily show that the 4-gon ABDE forms a isosceles trapezoid, hence A,B,D,E are cocyclic. -(1) Similarly, the 4-gons CDFA and EFBC are both isosceles trapezoids, hence C,D,F,A and E,F,B,C are both cocyclic. Now consider the 4-gon ABCD. By checking the angles, we'll find out that "angle A + angle C=180°", so the 4-gon ABCD is cyclic. -(2) Similarly, the 4-gon ABEF is also cyclic. So {A,B,C,D}, {A,B,D,E}, {A,B,E,F} are 3 sets of cocyclic vertices which has 2 common vertices A and B, therefore they are all cocyclic.

My attempt: in one of the vertices (call it A) of the triangle, draw the bissecting line. Let A₁ and A₂ be the extensions. Every point X on the bissecting line makes an isoceles triangle XA₁A₂. Do the same on the other vertices B and C. The intersecction between the bissecting lines shows us that A₁, A₂, B₁, B₂, C₁ and C₂ are all in the same circunference. :)

I liked Both Versions of the Proof, the coloured one is a Little Bit more intuitive. Thanks again for a Great Video ❤

I love both of the (perpendicular) bisection proofs SO MUCH.

The colour proof is the one I would most likely find on my own. The swivel proof I think is more elegant and more appealing

Definitely like the color one better. It keeps in mind the lengths of the original, arbitrary triangle.

I like the swiveling proof more, because it uses less colors at the same time. It also seems more directed. In the color proof, the result seems to just fall out, but in the swivel proof It takes less work to understand why what we're doing will result in the proof. Unlike some of the other commenters, I felt like I could trust both of the proofs equally. Maybe that's because I imagined the swivels as reflections?

I just watched the first minute and came up with a proof as suggested. Here it is: First notice that the proof in the other direction is very easy: Given a circle and three lines of equal length that lie on the circle you will always get this colour pattern. (1) This is because when looking at two lines at a time there is a symmetry axis between them so the line segments have equal length. For the desired direction it is to show that given three lines of equal length that overlap with the specified colour pattern their endpoints lie on a circle. First consider two of those lines. Their four endpoints lie on a circle because we know that for three points there is a circle that goes through them and the fourth is also on it because of symmetry. Now it needs to be shown that the third line is also on the circle. For any given angle of the third line we know that a line exists that is on the circle and follows the colour pattern because of (1). But there is only one line of this angle that satisfies the colour pattern so it must be the one that lies on the circle. Why is there only one? You have two degrees of freedom to move the line: Parallel to itself and orthogonally to itself. The position along the orthogonal direction is fixed by the colour condition. With this fixed the position along the parallel direction is also fixed by the colour condition. qed

I really enjoyed this one, John Conway did so much cool stuff. RIP

This looks like an optical illusion, because the lines don't really appear to be of equal length!! Fascinating stuff!! Even though the professor maintains that they are all of equal length, or so I thought 💭 I heard him say!! Hmmm!!! Food for thought 💭

Wooo I realized this is a new Mathologer video after watching it! 🙂 Great names of the theorems and this is the best recreational math channel!

The thumbnail helps a bit. Each side is extended into a line with the length of the perimeter. And they are all the same distance away from the incenter of the triangle. I believe that this proves all six points are the same distance from the incenter

The swivelling proof is better because it opens the idea to so many other uses in geometry.

Since the wiper segments pivot from the same point when they wipe, the traced path from when they start to end wiping is on a circle with the pivot point as the midpoint of the circle. This means that the "weird curve" shape is made of overlapping circles!

I think the swivel proof was more visually appealing but the coloring proof was more clearly a proof that that holds for any configuration and not just this one. It was more obvious with the way it was colored that this is always the case.

Diameter of the circle is larger. If you superimpose the circle over the shape, you can form a triangle of the shape's width and two radii of the circle. Due to triangle inequality, the two radii, and therefore the circle's diameter, must be longer.

Sliding the ends of a chord around the outer circle sweeps the iris (twice). The chord's center draws the inner circle (pupil).

I cant believe something that basic is discovered only few years ago. Conway was really a genuis

13:54 I JUST THOUGHT ABOUT THOSE SHAPES AND THE WAYS TO MAKE THEM THIS WEEK!!!!!!!

I like the swiveling proof better. Once you've observed that the two lengths from a corner to the points of intersection are equal, the rest follows with very little explanation. The colouring proof has its own appeal, but it feels a little clunkier.

The three diametral chords of the weird curve are off-center chords of the Conway's circle, which must therefore have a larger diameter.

Physics tells me that the circle is the extrem in area of the setup, but gut feeling leads me to the areas being identical. Btw. thx so much, this looks like solving an engineering problem of mine as well 😂

Beautiful! The last animation makes me think of a rotary engine! Constant diameter shape seems like another way to design a rotary engine where the outer part could be simpler than the one actually used in rotary engines.

High quality math video. The ending score makes it perfect.

One proof for 2:20 : By 3 point there is one and only one circle. We will divide our 6 points into two groups of 3 points. I'll add names : starting on the horizontal line on the left we name the point A. Then we turn clockwise and name D,B,E,C and F. Our two groups are A,B,C and D,E,F. Then we can show that AB = FE, BC = DF and AC = DE using the colored length. Let be C1 the only circle defined by A,B,C and C2 defined by D,E,F. The triangles ABC and DEF have the same lenghts, hence C1 and C2 have the same radius. Does C1 and C2 share the same center ? If they are, we are done. Let's focus and the green angle between D and B. Let be Δ the bisector. The axial symetry applied on [DF] gives [BC], thus C1 become C2. Another axial symetry based on another bisector will also send C1 and C2. Hence the center of both C1 and C2 is the intersection of the 3 bisectors. ㅁ

I proved this myself, quite nice, I particularly like the formulas for the green red and yellow lengths at 7:22 : calling the sides of the original triangle a, b, c, those lengths are (-a + b + c)/2, (a - b + c)/2, (a + b - c)/2