If you enjoyed this video, click the like button and subscribe with notifications on! Also, if you haven't already, please fill in the Google form here so that I have a better idea of what your math levels are: forms.gle/QJ29hocF9uQAyZyH6 Most geometric explanations are brief because most of you know Euclidean geometry as indicated by the Google form above, but also the video will be less interesting if I explained every step of proving pairs of similar triangles explicitly. Treat those as little exercises if you want!
"Knowing" Euclidean geometry in a poll probably doesn't mean the same thing as "knowing how to construct figures in a plane with compass and straightedge." I think most people see "Euclidean geometry" and think "I took a class on that in 10th grade or whatever, and I remember a bit of it." That's a pretty normal way to think about math in this context. Just as an early example, it is not immediately obvious that the highlighted triangles at 3:00 are similar. In particular, I have to know that the tangent to a circle is perpendicular to the radius at that point, then I have to reason through the proportions implied by the circle inversion, and finally I have to verify that the triangles are indeed similar. My students are pretty good at this, but they would not be happy just seeing a three second visual explanation before I moved on. I think that most viewers of Vsauce's video on Thales' Theorem (with a mispronunciation I cannot get out of my head) would say that they "knew" Euclidean geometry, even though his entire video focused on proving one of its most trivial theorems.
@@EebstertheGreat As I pointed out in this pinned comment, proving some triangles are similar will make the video much less interesting, and actually somewhat distracts us from the main point of the video. If I explained every step here, you and your students might be interested, but not the majority of viewers here, which is why it is gone over pretty quickly to get to the meat. In other words, the approach is to let the viewers see the big picture, rather than stressing on the small details, i.e. the fact that we *can* do inversion of a point is vastly more important than *how* to do it / *why* the method works /in this context/. As for questioning whether people "know" Euclidean geometry, I could assure you that most actually do, because a lot of responses show that they want me to cover topics of differential geometry or complex analysis or topology, or something along those lines, and specifically said that the level of maths on this channel is sometimes too low.
@@mathemaniac IDK. 7% of the comments on a video solving the problem of Apollonius ask you how to solve the problem of Apollonius. I agree that the circle inversion is the key and you don't want to clutter it with unnecessary detail. But I can assure you that just because someone is in grad school does not mean they know Euclidean constructions, because most schools do not even teach those anymore. So it's still a balancing act, and maybe your explanation shouldn't get any longer, hard to say. But I will stand behind my claim that people responding that they "know" a mathematical theory usually don't.
If you can expand the circles and retain an intersection point why not just keep expanding them until the three of them intersect as that is where the middles circle would be "squished" to a radius of 0 and you would find the center of the circle. Then just draw a line to any other circles center and the original radius of the "middle intersecting circle" would be the point in witch it intersects with that circle
Finding your videos after watching 3b1b's channel, I expected this one to be a less interesting knockoff, solely judging by the graphics (my apologies). But to my surprise, after watching just a single video, I completely changed my mind. Your videos are of great quality, really easy to follow, and the voiceover complements this style of videos nicely. I'll surely stay here for longer and hope that you'll steadily grow, improve, and perhaps make a living out of this. Keep it up!
Circular inversion was a turning point in my life in terms of "perspective". Problem of Apolonius was definitely one of the most exciting and beautiful time I have spent in my life.
2:05 - 3:06 was very helpful. In Electrical Engineering, I had to look up a derivation ( for Capacitance between two conductors, with non uniform charge distribution)....no Power System book would show the derivation and had to search in Field Theory book. I never expected to find that relation being used. Thank you very much for this video.👍❤️
Really nice explanation and visualization! Would love to see this extended to 3d to solve the Problem of Apollonius to find a sphere tangent to four given spheres.
Reminds me of a softeng saying I heard; "To implement a difficult change to the system first make the change easy, then make the easy change". Working in inverse-circle-space or using integral transforms to solve ODEs are exactly this, I guess!
I wonder if people like Euclid contemplated the collapsibility of compasses. Any real compass clearly cannot collapse, or you wouldn't even be able to use it to draw a circle. But his axioms do describe a system of construction using only collapsing compasses, because he doesn't explicitly include any translational axiom. (Then he panics almost immediately in his proof of SAS congruence and uses a much stronger one anyway, sometimes called the "method of superposition.") On the other hand, he leaves out plenty of axioms, even the existence of points, so it's hard to say. In practice, Euclid uses non-collapsing compasses frequently. The Compass equivalence theorem is remarkable, and at least in my opinion, the Mohr-Mascheroni theorem is astonishing. It turns out we need way more axioms than the geometers of Euclid's day had imagined, but somehow also fewer. On the one hand, Euclid failed to notice that he didn't include a single axiom relating angles to straight lines, so of course the SAS proof was impossible, but on the other hand, he failed to anticipate that straight lines weren't even necessary in the first place.
Thanks for an interesting solution using the inversion at a circle. I had encountered this type of inversion only for proofs of Ptolemy's Theorem so far, and I'm happy to see other applications. Could you recommend more applications of this method if inversion?
Thanks for the appreciation! As far as I know, the other applications are only Pappus and Steiner chains, but I am sure there are others that I just can't name off the top of my head.
The inversion has many interesting/recreational properties. For instance, the lemniscate of Bernoulli can be constructed as a circle inversion of a hyperbola. Wikipedia also shows its utility in a few other nontrivial proofs. Sphere inversions also exist in higher dimensions. It's not exactly a font of power though, and I can't think of any times I have seen it come up naturally except in some proofs of Ptolemy's Theorem.
@@EebstertheGreat - thank you. Could you tell me which Wikipedia page shows the utility? I know en.wikipedia.org/wiki/Inversive_geometry that says a statement about the Euler line could be proven this way, do you know any others?
Sorry for not explaining enough because otherwise the video becomes less interesting... Join the centres of two of the circles (O_1 and O_2) using a line, which will intersect the circles at two points, say A and B, so that O_1, A, B, O_2 appear in that order. Find the midpoint of A and B and call it M. Then AM = MB, and so you can inflate all the three circles by the same amount AM. This is done by constructing another circle centred at O_1 with radius O_1M, and another centred at O_2 with radius O_2M, and for the third circle, just draw a circle centred at a point C on the circumference with AM as the radius (using the non-collapsing compass), and draw a line joining O_3 (the centre of the third circle), and C. The line will intersect the newly drawn circle (centred at C with AM radius) at two points. Let's say the point further from O_3 is called D, then draw a circle with centre O_3 and radius O_3D. This will inflate the 3 circles by the same amount. I just think it ruins the lovely animation for inflation if I explain all these construction steps.
@@mathemaniac I think the video would have gotten more educational and thus more valuable. I have a lot of interest in complete explanations. There is certainly a way to show these transformations being done in a different area of the screen.
@@xBZZZZyt Draw a circle around each of your endpoints with radius equal to the length of the line segment in question. Those two circles will intersect above and below the line. Join the two intersection points and you'll get a perpendicular that crosses the original line segment exactly at the midpoint. Note that the actual radius of the two endpoint circles doesn't matter, as long as its larger than half the length of the line segment. But they have to be exactly equal or your perpendicular won't cross the midpoint. Using the full length of the line segment is just the easiest way to ensure equal radii without requiring a "non-collapsible" compass.
You should do a video showing Algebraic solutions, or how coordinates can make solving some geometric problems almost trivial. As a bonus it would be nice to have it being solved by some Matrix algebra software.
Really good rendition of this phenomenal problem :) I wish you explained HOW we can construct the circle tangent to two parallel lines and another circle. Also would this be possible with a collapsing compass? I don't see how this is simple.
Thanks for the compliment! I sort of already explained the construction of the tangent circle to two parallel lines and another circle bit in the middle of explaining the entire process. All moves that can be done with non-collapsing compasses can also be done with the collapsing ones with just a few more steps.
@@ShefsofProblemSolving 8:06 You draw the middle line with basic steps (I count 7 steps). Then with a non-collapsing compass you take half the distance between the 2 original lines. This distance X is the radius of the tangent circle we want to construct. Draw a circle C with radius X and center where the green circle and midline intersect (there are two choices). The other intersection point of the circle C with the midline is the center of the tangent circle and you have its radius so it's done
Tiny thing I just noticed, after seeing it mentioned in another old mathologer video... the "non-collapsing" condition isn't really actually a new condition. The trick is that you can translate any line segment to an endpoint of your choice, by constructing the diagonals of a parallelogram. So if you have the midpoint of a circle, and any point on its boundary, then you can clone that circle to any other center point of your choice. And if you don't you can still sometimes get them (as long as you have ANY point inside the circle), but the non-collapsing compass wouldn't be able to handle those circles either. So yea, your "noncollapsing compass and straightedge" constructions are just ordinary compass and straightedge constructions. Which is great!
I guess since we need to inflate the circles until two touch, I would first determine which circles are to be touching. To do this, I would trace a line between ever two centers, and whichever line has the least "outside distance" (the part of the line that is outside the two circles of the two centers it is connecting), will belong to the circles that will be touching. Now, the middle of that "outside distance" will be the the point of tangency of the two inflated circles. you can easily determine the mid point of a line using straightedge and compass. Therefore, you can redraw the two circles by placing the compass at each center, extend it to the established tangency point, and draw a circle. For the third circle, the increase in radius would be equal to half the "outside distance": trace a line passing through its center and use the point of intersection of that line with the circle to add the required distance to the radius using the compass (measure the distance using the compass, place its point on the point of intersection, draw a circle and wherever that circle intersects the line on the outside, this is where the new radius should reach).
@@michaeltamer274 The "outside distance" idea doesn't really work because we don't have a way of measuring distance. That said, the same thing can be accomplished by just doing the construction for all three pairs of circles and picking the one where its the two "inner" extended circles that are tangent to each other (the other two lines should have one "inner" and one "outer", barring exactly equal distances in which case just pick one). The rest of your construction looks fine (albeit a little hard to parse out - I almost added another correction before realize I'd just misunderstood you and was building the same construction after all).
Do you mean the applications of circle inversions? I am not aware of any other applications other than those that are already covered by other big channels, but some people also suggested some Euclidean geometry, so I will consider that. However, the next video will be group theory as suggested by the poll on my community post, so if I decided to make a video on Euclidean geometry, it will have to be after the next one.
@@mathemaniac Another application of circle inversion: In school I learnt to use circle inversion as a means to construct affine mappings of the form f(z) = (a*z+b/c*z+d) with complex a,b,c,d, in the complex plane. For this, you need to know how to construct 1/z (for complex z), and this inversion can be done geometrically, too, hence you are doing circle inversion. This was in 1987, and pupils were invited to bring their home computers to school to demonstrate how self-written software can do these transformations on screen.
You can just use the fact that if two circles are tangent, than if you connect their mid points the tangent point (the point the circles meet) will be on this line too
Man, there are a few places when I had to stop the vid and check the date to be sure it was NOT 4-1-21. Or candid camera. I do enjoy looking at the pretty pictures tho. Thanks!
Sorry for not explaining enough because otherwise the video becomes less interesting... Join the centres of two of the circles (O_1 and O_2) using a line, which will intersect the circles at two points, say A and B, so that O_1, A, B, O_2 appear in that order. Find the midpoint of A and B and call it M. Then AM = MB, and so you can inflate all the three circles by the same amount AM. This is done by constructing another circle centred at O_1 with radius O_1M, and another centred at O_2 with radius O_2M, and for the third circle, just draw a circle centred at a point C on the circumference with AM as the radius (using the non-collapsing compass), and draw a line joining O_3 (the centre of the third circle), and C. The line will intersect the newly drawn circle (centred at C with AM radius) at two points. Let's say the point further from O_3 is called D, then draw a circle with centre O_3 and radius O_3D. This will inflate the 3 circles by the same amount. I just think it ruins the lovely animation for inflation if I explain all these construction steps.
Animations and talk are fun 'n' all, but.... In this topic, specifically, it'd've been nice to make the transformations using the limited tools. Ya know: without the digital magic. Human behavior relative to tools: such's intuitively replicable magic.
1) Take your point A on the line OA and draw a circle with radius R around it. The size of R doesn't matter (well, as long as its >0 of course.. just extend the line as needed). That circle will intersect the line at two additional points P and P' that are both equidistant from and on opposing sides of the original point. 2) From each of the new points P and P' draw additional circles with radius |P-P'| = 2R. These two new circles will intersect above and below line OA at new points Q and Q'. 3) Use your straightedge to connect Q and Q' into line QQ'. Because the construction was done symmetrically across OA (mirrored vertically with respect to the video's diagram), we can be guaranteed that QQ' will be perpendicular to OA. Likewise, because the construction was done symmetrically along OA (horizontally) and centered on point A, we can be guaranteed that the new perpendicular will pass through A. 4) Extend your new perpendicular line until it intersects the original circle from the video (the one centered on O) to find point B.
Didn't know enough about the history to know that reference, but a good one after learning the reference, even though it was Archimedes, not Apollonius haha :)
it's called a "centimeter" not a "cee-em". What is it with you people and pronouncing the abbreviation, do you also say "one eye-en" when someone writes "1in"?
cm for centimeter takes two initials of words contained in "CentiMeter", while "in" only abbrevaites "inch", This is quite different types of unit name usages.
The original picture of the problem didn't indicate we would actually know the center points of any of the circles. This solution is therefore suspicious.
