Thanks to the editor for adding those extra graphics to clarify the initial statement of the problem.

I actually got scared by the photo that suddenly appeared. One of a more impressive video in my recent memory.

In the end it's supposed to be 3(b²-c²)² instead of 3(b²-c²), it solves the dimension problem and also the inner root now equals 0 when a=b+c or OABC is ciclic (equivalent by Ptolemy's theorem)

I solved it by choosing a different coordinate system, with the origin in A, and the x-axis aligned with AC. For side length s, the coordinates of C are (s, 0) and B is at (0.5*s, 0.5*sqrt(3)*s). Then, the circles' center O is in (x,y). The squared distances from the centers to the triangle vertices are AO^2 = a^2 = x^2 + y^2; BO^2 = b^2 = (x-0.5*s)^2 + (y-0.5*sqrt(3)*s)^2; CO^2 = c^2 = (x-s)^2 + y^2; three equations with unknowns x, y, s. Solve for s. No sines or cosines required, just algebraic manipulation :-).

One obvious application of this: planets orbiting the sun and determining when they are equidistant from each other. Of course orbits are elliptical in general, but this is a good starting point to visualize the problem.

5:26

b^2-c^2 should be (b^2-c^2)^2 in the solution for x.

Something is wrong in the solution for x, the units don't match up in the middle term under the deepest root: all terms need to be quartic in dimension to make sense.

It should have been mentioned that the necessary condition for the existence of such a triangle is: a=0 iff |b-c|

Not one of Michael Penn's finer moments, since the dimension check is basically instantaneous. As several people have pointed out, -3*(b^2-c^2) should be -3*(b^2-c^2)^2 which not only gets the units right but also changes the inner square root expression to the symmetric form (and one would certainly expect symmetry) sqrt ( 6*(a^2*b^2 +b^2*c^2 +c^2*a^2) -3*(a^4 +b^4 + c^4) ) The argument in this square root can be of either sign, and it's plausible that the negative sign bounces out combinations of a,b,c where construction of an equilateral triangle is impossible.

Interestingly, if b=2c and a=3c, i.e. the circles have integer radii 1, 2, & 3 c, then x = sqrt(7) c.

Way to go with the AOC frame when mentioning angle AOC!

Around 10 minutes, when you substituted in for cos²(θ), you dropped most of the "1 -" leading it: this threw off the rest of your calculation

5:27 photo of AOC - hilarious.

Subliminal AOC. Love it.

13:39

i enjoy these gnarly problems !

An interesting result happens when a = 3, b = 2 and c = 1. (I am using the corrected version of the equation, S = sqrt((a^2+b^2+c^2+/-sqrt(6a^2(b^2+c^2)-3(b^2-c^2)^2-3a^4))/2).) The discriminant becomes 0 and either way we get S = root7. I can make this a triangle that does not surround the center, but I cannot find a case that surrounds the circle's center.

I wonder if this can be solve using vectors, by that I meant describing one of the side length of the triangle with linear combinations of the vector AO, BO and CO?

what's with the photo that suddenly appeared (AOC? )??? ok , i got it...

is there a 3 dim. regular tetrahedron (a tetrahedron in which all four faces are equilateral triangles) equivalent eqn. ??

A "famous" geometry problem solved by a good mathematician Mike.

As umpteen people have already pointed out: the expression under the inner root sign should be 6a^2.(b^2 + c^2) - 3(b^2 - c^2)*^2* - 3a^4, which simplifies to the much more sensible-looking: 3(a^2 + b^2 + c^2)^2 - 6(a^4 + b^4 + c^4) Which brings me to what I really wanted to talk about: Why does Penn impose the condition a > b > c ? Surely, if we plonk an equilateral triangle down in the plane, somewhere, we know everything about it if we know the position vectors, *a*, *b*, *c*, of its vertices [note: we can generate Penn's concentric circles by spinning the vectors]. The length of the side of our triangle shouldn't depend on what we call these vectors. Thus our result for the length of the triangle's side in terms of a = |*a*|, etc should be invariant under a permutation of a, b, c; as, in fact, it is. We might even try for a solution approach that makes use of this symmetry, or that at least avoids making "choicees" in respect of *a*, *b*, and *c*. One such approach might be to press into service the result for the area of a triangle: area ABC = ½ | det [ (1,*a*), (1,*b*), (1,*c*) ] |, plus some defining conditions for an equilateral triangle: (*b* - *a*).(*c* - *a*) = ½x (*c* - *b*).(*a* - *b*) = ½x (*a* - *c*).(*b* - *c*) = ½x I'll spare you the details

I don’t under how you got the result at 13:00, I tried recreating it, but for the (b^2 - c^2) part I cannot get it in terms of a^2 and bc except with a square root as (b^2 - c^2) = sqrt((a^2 + 2bc)(a^2 - 2bc))

@MathTheBeautiful also used to flash images of Messi when he said the word "messy" - this is so 2010's

Am I going insane or can you just construct a right angled triangle OAC and use Pythagoras to get x = sqrt(a^2-c^2)

clearly, by inspection, the line AC is tangent to the circle with radius c, and then the radius is easy to get /j

Pfew that was painful algebra... Isn't there a more "geometrical" solution to this problem?

I think that he is about to marry the subliminal woman !

clearly, by inspection, the like AC is tangent to the circle with radius c, and then the radius is easy to get /j

5:27 WTF. Not just me?

What's with the subliminal messages?

@"Prof" Penn: You seriously need to do a better job of minimizing/eliminating your calculation/notation/transcription errors in the live-action working-out of the problems segments of your videos; because, in almost every video of yours that I click/tap on there seems to be at least one such error missed by you in-situ and which you don't correct for in post-production editing before posting said content. Perhaps, one way to minimize/eliminate those errors would be to shift your content production focus away from quantity in favor of quality. Ergo, instead of posting new videos at an average of one-a-day* you should allow however much minimum necessary post-production time is needed for the editing of each video to make sure that all in-situ missed during production are corrected in post-production before publishing them.

Good place to stop

First

5:26 Bro don't involve politics

Please don't flash random pictures in the middle of your videos. Thanks.

I hope this was your last video i put dislike on

clearly, by inspection, the line AC is tangent to the circle with radius c, and then the radius is easy to get /j

What's with the subliminal messages?