a famous geometry problem

[@LCDL6]

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)

[@yeech]

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

[@user-yt198]

5:26

[@pablojesusmolinaconcha4504]

Came here for the same, still have no answer

[@firefly618]

AOC is apparently the initials of an American politician called Alexandria Ocasio-Cortez. (I just took a screenshot of that frame and pasted it in Google Image search)

[@bourhinorc1421]

​@@pablojesusmolinaconcha4504just type AOC in google and see who pops, just a "funny put on random image related to the stuff that is said" similar to when some someone in a video says "Geez...but still!" And the editor adds a multiple pictures of goose and then a piece of steel

[@scottmiller2591]

Jump scare.

[@BridgeBum]

​@@pablojesusmolinaconcha4504American (famous) politician who is known by her initials AOC.

[@martenjanderuiter5474]

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 :-).

[@insouciantFox]

Algebraic geometry is a heck of a drug huh

[@robert-skibelo]

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

[@ihavepotatopie]

Especially the graphic depicting triangle AOC.

[@Tiqerboy]

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.

[@Simon-fg8iz]

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.

[@angrybhalo1671]

You're right, that 3(b²-c²) isn't matching up in dimensions

[@wolliwolfsen291]

@@angrybhalo1671Yes, and the result for x in the case of the pythagorean triple can‘t follow from the equation with the yellow arrow. Obviously he fixed it in his scratchbook and not on the chalk board

[@bozorgmehrmehr6772]

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

[@57thorns]

Yes, I think he skipped over that part. The general solution is more interesting after all. (Even if there might have been a typo on the board.)

[@etaoinshrdlu7247]

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.

[@edwardlulofs444]

For a professor, this guy makes very few mistakes. In my 10 years of taking college classes, there were many who were not even required to give evaluations. For example, the U of Calif. Two profs especially stand out. After class all of students would have to get together afterwards and discuss the lecture. We were trying to figure out what was even said. For one class, we all met to try and figure out what the homework assignment was even asking for. It was so out of date, links and references no longer existed. We sat near each other and studied the assignment sheet. Every 5 minutes or so, someone would say “Oh, I figured out what sentence 3 in the first problem means.” This would go for an hour or more. In another, but graduate class, the answer was obvious but required 10 pages of hand written algebra, calculus, limits, infinite sums, matrices, etc. one little mistake and we would discard the 10 pages and start over. The graduate advanced quantum mechanics required an infinite number of integrations over all paths. By contrast, Dr Penn sometimes makes a small typo which is always discovered and explained in the comments.

[@vanosaur]

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

[@bozorgmehrmehr6772]

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

[@billmorgan6082]

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

[@mireyajones810]

5:27 photo of AOC - hilarious.

[@stephenhamer8192]

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

[@stephenhamer8192]

_Query_: how can we be sure that 3(a^2 + b^2 + c^2)^2 - 6(a^4 + b^4 + c^4) is non-negative?

[@57thorns]

The a>b>c comes from the initial presentation of the problem. Since we can label the three points in any way we want before starting the calculations, we can always assure that this is true without losing any generality.

[@LillianRyanUhl]

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

[@naturalphilosiphers9848]

I was doodling yesterday and came up with this exact problem for myself, now I learn it’s internet famous lol

[@Qermaq]

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.

[@goodplacetostop2973]

13:39

[@muenstercheese]

thank you for your service

[@charleyhoward4594]

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

[@tscoffey1]

Subliminal AOC. Love it.

[@roberttelarket4934]

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

[@freddupont3597]

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

[@charleyhoward4594]

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

[@CTJ2619]

i enjoy these gnarly problems !

[@edwardhudson815]

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)

[@edwardhudson815]

*edit Ok this is just when AC is tangent to OC, which it looks to be in the picture

[@freddupont3597]

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

[@MyNordlys]

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

[@Qermaq]

It's not a subliminal message. Those actually don't work anyway. It's a joke. AOC is what that woman, a well-known American politician, is known as. Not surprised many outside the US didn't get it, but it's not subliminal messaging.

[@wyboo2019]

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

[@ethanbartiromo2888]

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))

[@Flikus97]

There's a typo, it should be (b^2-c^2)^2. As other comments said that makes sense even for dimensional reasons!

[@dang-x3n0t1ct]

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?

[@TheKudo555]

this is pretty much what he did, except formulating it in geometrical terms. For example: \vec OA = \vec OB + \vec BA (forgoing "\vec"s :) OA^2 = OB^2 + BA^2 + 2 ||OB|| ||BA|| cos(OB, BA) a^2 = b^2 + x^2 + 2ax cos(\pi/3 - \theta) and so on for other equalities

[@brabhamfreaman166]

5:27 WTF. Not just me?

[@shruggzdastr8-facedclown]

@"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.

[@shruggzdastr8-facedclown]

(*by my perception)

[@wyboo2019]

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

[@copernicus633]

Not necessarily true. That is only an appearance of the diagram, not a logical consequence of the construction.

[@avhuf]

What's with the subliminal messages?

[@bobh6728]

The angle was her name.

[@jimpim6454]

Alexandria orcasio Cortez, famous American crazy person

[@bilalabbad7954]

Good place to stop

[@joeeeee8738]

5:26 Bro don't involve politics

[@d3mux_man]

You know who is she?

[@bobbaum8974]

de.m.wikipedia.org/wiki/Alexandria_Ocasio-Cortez Nickname AOC ​@@d3mux_man

[@PotatoBTD6]

​@@d3mux_man Google says she's Alexandria Ocasio-Cortez. But I'm from Spain, I don't know anything about her.

[@skoosharama]

@@PotatoBTD6 : She's one of a handful of self-identified socialists (members of the Democratic Socialists of America) elected to the United States Congress in 2019, and it was made a very big deal in American media, especially on right-wing media. She was also one of the youngest elected members of Congress at the time, pretty good-looking, and very active on social media, so she's kind of become a celebrity as well as a politician. Easily recognized among many Americans, no matter their political leanings. The flash of her photo is just a joke: Ocasio-Cortez is widely known by her initials AOC, and Michael mentioned triangle AOC at that moment.

[@PotatoBTD6]

@@skoosharama Thank you. I figured it was some kind of joke.

[@alchun3340]

First

[@9adam4]

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

[@aug3842]

it wasn’t random, he said ‘triangle AOC’ then as a joke flashed a picture of the politician commonly known as ‘AOC’

[@sugarfrosted2005]

​@@aug3842 Given the avatar I'm pretty sure op knows op's just throwing a minor political temper tantrum

[@9adam4]

@@aug3842 I'm aware it wasn't literally random - I've made the same joke myself when tutoring geometry. When you have a circle and a triangle labeled ABC, angle AOC comes up pretty often. I just found the sudden flashed image disconcerting.

[@9adam4]

@sugarfrosted2005 I didn't assume there was anything political about about flashing that image; I'm not sure why you'd assume my request is political.

[@orlandoatrahasis5172]

I hope this was your last video i put dislike on

[@forcelifeforce]

Explain *why* you put a "dislike" on the video. Your post is not meaningful to the rest of us without explaining it.

[@Qermaq]

I hope your post to explain is your next post. If you don't say what you don't like you win nothing.

[@wyboo2019]

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

[@dandrestor]

What's with the subliminal messages?

[@TobiasFord-u3q]

Dude exactly wtf was that

[@TobiasFord-u3q]

His voice is breaking there are random funking women wtf

[@ScienceTalkwithJimMassa]

@@TobiasFord-u3q It was a joke. There's a politician in the US who's name is Alexandria Ocasio Cortez. She is commonly known as AOC. When Michael mentioned triangle AOC, her photo was flashed. That's all. Just a little fun.