All Triangles are Equilateral - Numberphile

I have an elegant solution, unfortunately this YouTube comment box is too narrow to contain it.

I love troll math like this. it reminds me of the pi = 4 thing that was going around the internet a few years ago. it's just fun. wrong, but fun.

In fact, all the following propositions are corrects:

XB = XC
XB* = XC*
BB* = CC*
XBB* = XCC*
AXB* = AXC*
AB* = AC*

I did it again in a drawing software and I realized that if the line C of the triangle is larger than B, then the point C* is located between the points C and A, and not after the point C, like illustrated.
Therefore, in this case the following equations will be:

AB* - BB* = AB
AC* + CC* = AC

Because that, the conclusion "AB=AC" is wrong.
The most that we can affirm is:

AC = AB + BB*

* I'm brazilian and sorry by the poor english.

I FIGURED OUT WHERE YOU WENT WRONG!!
..it's not equilateral

this is learning... an educated man asking amateurs to solve something... I'd love to have my high school classes structured like this

All the congruency is fine since triangles with right angles.
Actual mistake starts at 4.05 when it says that AB* - BB* = AB AND AC* - CC* = AC because one of C* and B* is inside and the other one is outside. It can be obviously proven since ABXC is on one circle (X is the middle of the arc BC), so ABX+ACX = 180, hence one of them is bigger that 90 when other is less(they equal only if AB=AC), hence perpendiculars XC* and XB* can't both be inside or both outside,

Both the points B* and C* cannot lie outside the triangle. The statement AB*=AC* is true, and BB*=CC is true, but for doing AC*-CC* and AB*-BB*, you need to have both the points outside the triangle, which has not been followed here.

The problem here is that the example he used here had generalized angles. I tried recreating the problem using a protractor to draw an exact 30-60-90 triangle. I bisected the 90 degree angle and created the same diagram shown above, however, I found that (corresponding) lines on my diagram, XB* and XC*, did not fall on the outside of the triangle! In fact, line XB* fell onto line AB inside the triangle and line XC* fell on line AC outside the triangle! Using the same geometric rules used in the video, I found that line AB (which included the point B*) was equal to the line AC* (which included the point C). Once this happens, we cannot simply subtract the CC* and BB* from both sides, because that would take away part of a side of the triangle ABC! Therefore, this is FALSE!!! :)

Troll level: MATEMATICIAN

"A proof that all triangles are equilateral - can you see a problem with it?"

AB* only equals AC* if point M is on the line AX.

I was with him up until 3:21. That's when I started going "No no no no NO NO NO NO NO!"

I'll write A° and B° because Google comments use the asterisk to format text into bold text.

The mistake is that the picture shows B and C as always being one the inside, i.e. the nearest points to A compared to B° and C°. However that's not the case. If B is nearer to A than B°, then C° will be nearer to A than C. Paint a picture in GeoGebra if you want to verify it, but it's what will happen.

So everything is correct until the last part, where we don't have AB° = AB + BB° or we don't have AC° = AC + CC°. Which way it is depends on whether AB>AC, or AC>AB (if we knew a priori which one is bigger).

To prove this we have to show that if AB>AB° <=> AC<AC°. Funnily, the thing above is indeed a prove.

Case 1: AB > AB°. We prove it by contradiction. We assume that AC > AC°. And what we will get (as the video has shown above) is that then AB = AC, which is a contradiction. Since AC != AC° anyway, we have that AC < AC°

Case 2: AB < AB°. We do the same thing.

So we have that AB > AB° => AC < AC° and that AB < AB° => AC > AC° which is equivalent to the reciprocal of AB > AB° => AC < AC°, therefore AB > AB° <=> AC < AC°.

PS: Does someone know how to write asterisks in Google comments without screwing up the text formatting?

You cannot infer that AXB* ~= AXC*.  Sure, B* and C* are 90deg, BB* = CC*, and XB* = XC*, but all we know is (AB* & AC*) > (BB* & CC*).  
Also, the A angle bisector intercepts MX once.  We know MX is an angle bisector of B*XC* because BM = CM and BX =CX.  Angle AXC* does not necessarily equal AXB*.  AX ix not necessarily an angle bisector of B*XC* because of its one, not infinite, intersection with MX.  

I think the small lie that makes this work is at 3:28. Saying those two triangles are mirror images presumes AB and AC to be equal, which is more or less the thing we are trying to prove here. Its a statement that sneakily begs the question, putting a whole lot of construction lines on there that are equal and hoping you forget the more important parts...

If you drew this out carefully it'll be really easy to see that if AB is shorter than AC then the point C* will be in between A and C. This means that AB + BB* = AC - CC* and not +. Tricksy

I feel that before he started his explanation, he should have shouted "LEEEEEEROOOOOOOOOY NJEEEEEENKINS!!!!

The diagram is misleading. In a correct construction, C* would actually be inside AC, so the assumption that |AC| = |AC*| - |CC*| is wrong, and |AC| = |AC*| + |CC*|

I think the problem lies on assuming that the point X always exists, because if the angle bisector and the perpendicular bisector never meet it can't work surely.

I work on automated reasoning in highschool Geometry software, so I know very well where this reasoning chain breaks down. This "paradox" pinpoints exactly what is hidden "under the rug" in Geometry education in highschool: In highschool, we are given an illusion of "rigorous" geometry proof because we are following all the axioms (congruences, angle chasing, etc). But little did we know, these axioms depend on very specific topological facts that does not get proven, but only assumed "experimentally" through inspecting the diagram. So if the diagram is incorrect, the topological assumptions are wrong, and then the axiom applications will result in incorrect facts like so.

What topological fact is wrongly assumed in this diagram? It is the fact that both B* and C* are outside of the triangle.

In short, what you thought of Geometry in highschool is not rigorous at all, we are doing it semi-formally by mixing axiom applications and experimentally inspecting the diagram (physically interacting with the geometry embedded in the physical universe). To fix this, we have to have axioms that deal with topological statements and make them very explicit when applying the more traditional axioms.

i love this guy's accent