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.
Just to go a step further, the perimeter of the shape described by the wiper arcs is exactly pi times the perimeter of the triangle i.e. (pi * the chord length), whereas the circumference of the circle is (pi * 2 * the radius) and, as noted above, it is easily seen that (2 * the radius) is greater than (the chord length). To check that the perimeter of the SOCW is (pi * the chord length), consider each arc length. If we label the triangle sides A,B,C and their opposite angles a,b,c, then the arc lengths are given by: Aa + Bb + Cc + (B+C)a + (A+C)b + (A+B)c = (A+B+C)(a+b+c) = (chord length)(pi)
To take it one (very small step further), this shape of constant width has the smallest diameter of all shapes of constant width containing all 6 points!
@@Falanwe that suggests it might be a very valuable curve to quickly generate for certain applications, especially manufacturing and construction. Minimization is a huge part of engineering & design.
@@royalninja2823he says that when he was younger, he used to get upset that people only know his game of life, because it was a quick and draft invention. But I also only know his game of life.
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.
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.
I am not sure we were thinking about the same thing but one thing that first crossed my mind seeing all the equal chords was that if i pick any 2 inersecting ones and unite the points they should form up a isosceles trapesoid which is insciptible in a circle. And indeed it is easy to prove the 2 bases are parallels as bases of isosceles opposite triangles and the sides are identical by tringle equalities. Then we just repeat for the other pairs of points and we note that circle of the circle will be somewhere on mediator of bases which also is the mediator of original triangle sides.
Colour proof seems like something I could come up with given enough time and determination (so back to school). The swinging one reminds me of 3B1B exercise of a proof easy to understand, but super hard to come up and write rigorously.
@@Filipnalepa Agreed. It was even similar in the fact of the rotation of a line being a part of it (although in this case it’s a segment rather than an actual line).
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.
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
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!)
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 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!
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!
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).
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.
@@Mathologer I originally came up with the problem myself, but I later found a special case of it on the 1989 All-Russian Math Olympiad. That contest used an equilateral pentagon instead of an equilateral n-gon.
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!
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.
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
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.
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)
The area of Conway’s circle is larger. I think the proof is quite complicated to write down, so I don’t think I can do this. But I can conceptualize the proof like this: There are 3 pairs of fan-shaped sectors that, when combined and subtracting twice the area of the inner triangle, become the area of The Weird Curve. We can choose one pair of large and small fan-shaped sectors opposite each other to prove this, and the other two pairs will be proven similarly. We see that the small fan-shaped sector will always extend beyond the Conway circle, and the large fan-shaped sector will always eat inside the Conway circle. Using the formula to calculate the area of a circular arc, we can prove that the area extending outside the Conway circle of the small fan-shaped sector is always smaller than the area eating inside the Conway circle of the large fan-shaped sector opposite it
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.
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😵💫
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.
Fair enough, it's a super slick one. Have you heard of the BOOK of proofs? Essentially what I asked you to vote on was which of the two proofs should go into the BOOK :) en.wikipedia.org/wiki/Proofs_from_THE_BOOK
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
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
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.
Love the title and the video. Remembering another humble human being. Being 4 years on april the 11th... Thank you Following all kinds of math related channels, this, really gets to my heart. Sometimes the game of life just isnt fair.
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.
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 💭
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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.
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.
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.
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.
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.
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.
*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.
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.
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
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.
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).
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
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.
Regarding the final question: the square of the diameter of the curve of constant width PLUS the square of the diameter of the incircle EQUALS the square of the diameter of Conway's circle. If Conway had lived in another era, he could have been the Little Euler.
My try for a proof: First, if those point are indeed on a single circle the center of this circle should be the center of the inscribed circle of our starting triangle (as it is the center of all bissectrices). Let's call O this point. Next, let's pair the 6 points we are trying to proove are on a circle according to wether they are linked by a straight line on the figure at 2:17. Each of these pairs is distant by the same amount (the sum of the length of our starting triangle's sides). Some not-so-simple trigonometry lets us know for all these pairs adding O makes an isocele triangle, and because O is the center of the inscribed circle all those triangles are similar (they have the same height). Thus we can conclude.
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.
My approach at coming up with a proof in advance: For each of the long lines, both ends need to be on the circle, and so, the center of the circle must be orthogonally above the center of the line. Due to the pythagorean theorem, this distance must be constant for all three lines, as the radius of the outer circle is equal for each one. But this means that the middle points of each line all have the same distance to the total center, meaning that there is a smaller circle that is tangent to each of the sides of the triangle, in other words, the incircle, which shares the same center as the original outer circle that we started with. After coming to this point, we can start arguing in the other direction so that instead of exploring the concepts related to our wanted conclusion, we start with what we already know about these. With this, our goal becomes: Prove that for a triangle's incircle, when you take any tangent point of said circle and split the triangle's edge into two halves at that point, going in the direction and adding the triangle's sidelength adjacent to the other direction, you get the same result as if you chose the opposite direction at the start. But this is easy to show given the knowledge that the line from a triangle's corner to its incircle always bisects that corner's angle, since with this, you know that adjacent edge halfs are always equal in length, letting you rearrange the pieces of the triangle's edges to form each of the long lines, making it obvious how the incircle's tangent point lies in the center of the line. Given that the long line has a length of L, you know that the distance of the end of the long line to the center of the incircle of radius r is R=sqrt(r²+(L/2)²), which is constant for each of the points at the end of the long lines. A lot of my explanation relies on having a picture of the geometry in my mind, so it probably isn't very understandable from the comment (unless the video happens to have the same explanation, which wouldn't completely surprise me, as the incircle probably is what is meant with the "iris"), so this is mostly just me assuring myself of being ready to present my proof whenever conway comes up to me with a pen and paper in hand. Edit: After watching the first two proofs, I can say that what I described was exactly the "coloring proof". Naturally, that's the one that I liked best. Edit 2: The swiping proof is definitely more interesting, even if not as easy to show to be correct. The shape at the end has a smaller diameter than the circle, because its diameter is equal to one of the long lines, while the diameter of the circle is larger than that, according to my earlier calculation of R.
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!
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.
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.
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!
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.
Tough call... The swivelling proof is pleasing, but relies on the fact that the number of sides is odd, meaning it's less likely to generalize to other shapes. So the colouring proof seems more likely to be more useful.
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.
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.
Exactly :)
Just to go a step further, the perimeter of the shape described by the wiper arcs is exactly pi times the perimeter of the triangle i.e. (pi * the chord length), whereas the circumference of the circle is (pi * 2 * the radius) and, as noted above, it is easily seen that (2 * the radius) is greater than (the chord length).
To check that the perimeter of the SOCW is (pi * the chord length), consider each arc length. If we label the triangle sides A,B,C and their opposite angles a,b,c, then the arc lengths are given by: Aa + Bb + Cc + (B+C)a + (A+C)b + (A+B)c = (A+B+C)(a+b+c) = (chord length)(pi)
Yes, you are right. My intuition was wrong. I love how these videos get people talking-any ideas on getting 15-16 year olds talking like this?
To take it one (very small step further), this shape of constant width has the smallest diameter of all shapes of constant width containing all 6 points!
@@Falanwe that suggests it might be a very valuable curve to quickly generate for certain applications, especially manufacturing and construction. Minimization is a huge part of engineering & design.
RIP conway, he will be missed
Very much so :(
He *is* missed, sorely. I'm disappointed I didn't get to know everything he'd done before he died, only knowing about his Game of Life.
@@royalninja2823he says that when he was younger, he used to get upset that people only know his game of life, because it was a quick and draft invention.
But I also only know his game of life.
Apparently the last thing he saw before he died was 3 black squares.
“Houston, they got me in a glider pattern. I’m just a dead cell now. Report: dead cell now only.”
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 make you smile, that's great ! :)
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.
I am not sure we were thinking about the same thing but one thing that first crossed my mind seeing all the equal chords was that if i pick any 2 inersecting ones and unite the points they should form up a isosceles trapesoid which is insciptible in a circle. And indeed it is easy to prove the 2 bases are parallels as bases of isosceles opposite triangles and the sides are identical by tringle equalities. Then we just repeat for the other pairs of points and we note that circle of the circle will be somewhere on mediator of bases which also is the mediator of original triangle sides.
I liked the color proof better, because the swivel proof seemed like it relied more on visuals, which could be deceiving.
same
Colour proof seems like something I could come up with given enough time and determination (so back to school). The swinging one reminds me of 3B1B exercise of a proof easy to understand, but super hard to come up and write rigorously.
@@Filipnalepa Agreed. It was even similar in the fact of the rotation of a line being a part of it (although in this case it’s a segment rather than an actual line).
Thank you for all the effort you put into these videos. I very much enjoy the learning experience from watching this channel!
You triggered my math competition PTSD 😂
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.
My first thoughts as well! Go team bisectors!
yes, this is the proof i got also.
I did it with mirrors, but I found (and posted) essentially the same proof. But you beat me to it, congrats.
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
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!)
That's great :)
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.
Correct. I guess you are ready for when you meet Conway in the afterlife :)
"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 :)
Of course that's also a reason that gets my seal of approval :)
Yes, if I have to choose, I watch the latest Mathologer video before the latest 3B1B.
I wanted to see little creatures crawling around the screen...
@@tulliusexmisc2191 Same here.
I like the coloring proof best. Conway left us so many gems.
This was heartachingly beautiful. Thank you.
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!
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!
Sounds right :)
If this is correct, then something similar might be expected also to work in hyperbolic surfaces of constant curvature.
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).
I've only watched 45 seconds of this video, so far. And I have to say... "Wow!" A circle... Wow. Greetings from Dallas, Texas!
Greetings from Melbourne in Australia :)
Spoiler alert:
Irving here...
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.
Where did you come across this?
@@Mathologer I originally came up with the problem myself, but I later found a special case of it on the 1989 All-Russian Math Olympiad. That contest used an equilateral pentagon instead of an equilateral n-gon.
@@MathFromAlphaToOmega I see, thanks for sharing this with me :)
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!
Glad you liked it :)
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.
Came for a fractal that looks like an iris, stayed for the beautiful visual geometric proofs
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
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.
Yes....
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)
Correct :) Here is a little a little trickier question than. Which is larger the AREA of Conway's circle or that of the curve of constant width?
The area of Conway’s circle is larger. I think the proof is quite complicated to write down, so I don’t think I can do this. But I can conceptualize the proof like this:
There are 3 pairs of fan-shaped sectors that, when combined and subtracting twice the area of the inner triangle, become the area of The Weird Curve.
We can choose one pair of large and small fan-shaped sectors opposite each other to prove this, and the other two pairs will be proven similarly.
We see that the small fan-shaped sector will always extend beyond the Conway circle, and the large fan-shaped sector will always eat inside the Conway circle.
Using the formula to calculate the area of a circular arc, we can prove that the area extending outside the Conway circle of the small fan-shaped sector is always smaller than the area eating inside the Conway circle of the large fan-shaped sector opposite it
Wonderful proof. The incentre is slowly becoming my favourite triangle centre! (overtaking the orthocentre as my previous favourite)
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.
Very neat video. That little twist at the end resembled somewhat the rotary wheel of a Wankel motor. Yes, I'm that old...
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😵💫
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.
I guess you are ready for an encounter with Conway in the afterlife :)
I really enjoyed this one, John Conway did so much cool stuff.
RIP
7:39 the coloring proof I suppose. The static picture that you can look at and check over feels more comforting somehow.
I liked the color proof a lot more, it felt a lot more clear, straightforward and obvious to me!
Fair enough, it's a super slick one. Have you heard of the BOOK of proofs? Essentially what I asked you to vote on was which of the two proofs should go into the BOOK :)
en.wikipedia.org/wiki/Proofs_from_THE_BOOK
Oooh, that's awesome, thanks for the link and for the response! I learned something today :D
The last curve is making nodes and antinodes with the circle.
A beautifully done video & presentation! Thank you for the mention too! 😉
Thanks again for your help answering my questions :)
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!
Glad you liked it!
7:40 I liked the colouring proof best. It was less obvious; and so, a nice surprise 🙂.
I like the coloring one!
It reminds me of the wobbly table fix proof.
Thanks so much for the video!
The iris was interesting, but I'm here for Conway!
Definitely the better reason :)
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
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
Great as always
The swivel approach resonated better with me.
👍🏻 Another great T-shirt. Of course the lecture too. 😉
High quality math video. The ending score makes it perfect.
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.
Love the title and the video.
Remembering another humble human being.
Being 4 years on april the 11th...
Thank you
Following all kinds of math related channels, this, really gets to my heart. Sometimes the game of life just isnt fair.
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.
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 💭
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 liked the swivel proof best, but that’s probably because when I tried to prove it I found the color proof
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.
Beautiful
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.
Correct :)
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.
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.
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.
*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.
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.
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
13:50 FINALLY I remembered from decades ago Martin Gardner's column on equal-width shapes, including this as a method of generating them.
Martin Gardner sure covered a LOT of ground in his writings :)
I love both of the (perpendicular) bisection proofs SO MUCH.
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.
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).
Love this. Thank you.
Glad you enjoyed it!
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
Coloring proof!
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.
Seriously? Not a joke? :) Which Conway?
Conway's circle diameter > curve's diameter, one more interesting thing to observe is that, perimeter of curve = pi * perimeter of triangle.
Whoa proof not immediately obvious to me for that
@@pingnick Google Barbier's theorem
@@Mathologer 🤯
Regarding the final question:
the square of the diameter of the curve of constant width PLUS the square of the diameter of the incircle EQUALS the square of the diameter of Conway's circle.
If Conway had lived in another era, he could have been the Little Euler.
My try for a proof:
First, if those point are indeed on a single circle the center of this circle should be the center of the inscribed circle of our starting triangle (as it is the center of all bissectrices). Let's call O this point.
Next, let's pair the 6 points we are trying to proove are on a circle according to wether they are linked by a straight line on the figure at 2:17. Each of these pairs is distant by the same amount (the sum of the length of our starting triangle's sides). Some not-so-simple trigonometry lets us know for all these pairs adding O makes an isocele triangle, and because O is the center of the inscribed circle all those triangles are similar (they have the same height). Thus we can conclude.
To assist your engagement; I liked the coloring better, the idea of the semi-sides each composing the half the chords is pleasing.
Thank you for assisting :)
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.
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.
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.
Any nice geogebra file to share ? :)
Definitely like the color one better. It keeps in mind the lengths of the original, arbitrary triangle.
I like the swivel proof because it was nice, and the coloring proof seems like the more obvious sort of approach to take.
So did I, for the same reason.
Sliding the ends of a chord around the outer circle sweeps the iris (twice). The chord's center draws the inner circle (pupil).
misread the title as Conway's IRS. i was very curious to learn about John Horton Conway's tax collection system.
I am sure Conway had something interesting to say about the IRS, whatever he said was interesting :)
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
My approach at coming up with a proof in advance:
For each of the long lines, both ends need to be on the circle, and so, the center of the circle must be orthogonally above the center of the line. Due to the pythagorean theorem, this distance must be constant for all three lines, as the radius of the outer circle is equal for each one.
But this means that the middle points of each line all have the same distance to the total center, meaning that there is a smaller circle that is tangent to each of the sides of the triangle, in other words, the incircle, which shares the same center as the original outer circle that we started with.
After coming to this point, we can start arguing in the other direction so that instead of exploring the concepts related to our wanted conclusion, we start with what we already know about these. With this, our goal becomes:
Prove that for a triangle's incircle, when you take any tangent point of said circle and split the triangle's edge into two halves at that point, going in the direction and adding the triangle's sidelength adjacent to the other direction, you get the same result as if you chose the opposite direction at the start. But this is easy to show given the knowledge that the line from a triangle's corner to its incircle always bisects that corner's angle, since with this, you know that adjacent edge halfs are always equal in length, letting you rearrange the pieces of the triangle's edges to form each of the long lines, making it obvious how the incircle's tangent point lies in the center of the line.
Given that the long line has a length of L, you know that the distance of the end of the long line to the center of the incircle of radius r is R=sqrt(r²+(L/2)²), which is constant for each of the points at the end of the long lines.
A lot of my explanation relies on having a picture of the geometry in my mind, so it probably isn't very understandable from the comment (unless the video happens to have the same explanation, which wouldn't completely surprise me, as the incircle probably is what is meant with the "iris"), so this is mostly just me assuring myself of being ready to present my proof whenever conway comes up to me with a pen and paper in hand.
Edit: After watching the first two proofs, I can say that what I described was exactly the "coloring proof". Naturally, that's the one that I liked best.
Edit 2: The swiping proof is definitely more interesting, even if not as easy to show to be correct. The shape at the end has a smaller diameter than the circle, because its diameter is equal to one of the long lines, while the diameter of the circle is larger than that, according to my earlier calculation of R.
Looks like you are ready for on the spot challenges from mathematical geniuses :)
I like the coloring proof because swiveling is hard to imagine
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!
Correct :)
I would guess same diameter ;) Nice content, Nice video name, thank you for the chronological facts too.
And my guess is false ;)
It is right when the original triangle is equilateral! Then the "funny" circle coincides with the second, outer circle.
I guessed quite a few of the swivelling proof steps just from feeling. But I find the colouring proof more elegant.
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.
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.
I liked Both Versions of the Proof, the coloured one is a Little Bit more intuitive. Thanks again for a Great Video ❤
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!
The swiveling proof is very appeasing
Always cool T-shirts
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 liked the swivel proof better, but I may have been biased by hearing it first. The color-splitting proof is also quite neat.
I cant believe something that basic is discovered only few years ago.
Conway was really a genuis
Once you have a close look you'll find that there are really a LOT of very simple, beautiful results that were missed for thousands of years :)
Thank you for such a nice and informative video. I liked The colour proof more than the swivel proof
Using property of chords that product of 2 segments is constant
So use it ...yeh
Tough call... The swivelling proof is pleasing, but relies on the fact that the number of sides is odd, meaning it's less likely to generalize to other shapes. So the colouring proof seems more likely to be more useful.
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.
I like the color proof best. Seems like something I might have been able to come up with myself. Whereas I never would for the swivel proof.
The swivelling proof is better because it opens the idea to so many other uses in geometry.