An artist's discovery of the golden ratio.

[@udic01]

4:44 There is a convention for notating 2 similar/identical triangles. You have to write both of them in the same similarity/identity order. Meaning ∆BFE~∆GCE and that way it is easy for whoever reads the proof, to identify the matching parts of both the triangles.

[@robertveith6383]

You're missing notation/characters. Either write "Triangle BFE ~ Triangle GCE" or use that triangle symbol in front of each group of letters.

[@udic01]

@@robertveith6383 you are right. I was writing The comment on my computer and didn't have the ∆ symbol. I didnt want to write triangle because it would have been less clear in my opinion. I will fix it in a minute.

[@Grizzly01]

@@udic01 If you're on Windows 10, press and hold X then press ; and you'll get a whole heap of mathematical symbols you can use, alongside the usual emojis and stuff.

[@comic4relief]

@@udic01 check out Character Map

[@mathflipped]

Golden ratio is, for lack of a better word, perfect! Great job, Michael.

[@davidcroft95]

You could stop at 6:30, that's literally the definition of the golden ratio

[@laurentpaget5388]

Thank you for sharing this puzzle. With the theorem of intersecting chords, the solution can also be found. Let set DE=BE=1 (equilateral triangle) and EF=GD=y. The unknown x is DE/EF=1/y. According to this theorem, GE.EF=BF.CF (1) and GE=GD+DE => (y+1).y=1*1=1 y+1=1/y As x=1/y and 1/x=y 1/x+1=x or x^2=x+1 x^2-x-1=0 Solving for x>o (ratio of distances), x is the golden ratio. x=(1+5^½)/2

[@Vladimir_Pavlov]

Theorem: If two chords of a circle intersect, then the product of segments of one chord is equal to the product of segments of the other. BE=EC=DE =a EF=GD=b a*a=(a+b)*b DE/EF=a/b (a/b)^2-(a/b)-1=0 a/b=(1+ √5)/2

[@koenth2359]

I started with a triangle with vertices A=(-1, 0) B=(1,0), C=(0, √3). We can easily check AB=AC=BC=2. Point D is ½(A+C) = (-½,½√3) and point E is ½(B+C) = (½,½√3). Therefore DE= 1. The circle's centre is that of the triangle: (A+B+C)/3 = (0, √3/3) and the circle has radius 2/3 √3. So the equation of the circle is x^2+(y- √3/3)^2=4/3. Using the y-coordinate of D and E, this equation gives x^2 = 5/4 so F = (½√5, ½√3). Therefore FE/DE = ½√5 - ½ and its reciprocal is ½√5 + ½ = φ.

[@goodplacetostop2973]

0:03 Question in the thumbnail, _gives the answer in the title and 5 seconds in the video_ 😂 7:24 Good Place To Stop

[@comic4relief]

I wonder what is the measure of arc BG. Using a protractor, the measure of angle BCG looks like about a hair past 32 degrees. This would make the subtended arc BG 64 or 65 degrees. The arctan of 1.618 is 64.76. Curious.

[@comic4relief]

Again we see the golden ratio having a simple and pure sort of quality, demonstrated by just circle and equilateral triangle.

[@jfcrow1]

If DE is not at the midpoints ie moving up down equilateral triangle. Does Ratio still hold? DE and EF being colinear.

[@luisaleman9512]

No. Only at the midpoints the ratio will be equal to the golden ratio.

[@jfcrow1]

@@luisaleman9512 Thanks

[@RexxSchneider]

The first thing that occurred to me was that I'd be interested in the function f(x) describing how the the ratio of DE/EF changed as the lengths BD and BE were moved to a fraction x ( 0

[@manucitomx]

I love Geometry. Thank you, professor! Also, Happy Thanksgiving!🦃

[@karolakkolo123]

Thank you for uploading everyday. I don't know how you do it, but it motivates me to keep going with math everyday. Thanks!

[@comic4relief]

Interesting.

[@kashnigahbaruda]

I love x = 1/ (1+1/x)

[@lambdcalculus]

It's enough to show DE/EF = DF/DE, by the definition of the golden ratio. By power of a point applied at E, we get that: GE*EF = BE*EC but GE = DF, and BE = EC = DE, so we get: DF * EF = DE^2 DF/DE = DE/EF, as we wanted to show.

[@TheBlueArrow27]

Can someone explain why it's obvious DE is the same length as BE?

[@Qermaq]

BD, BE and DE form an equilateral triangle.

[@altmike]

When you join the midpoints of the sides of a triangle, that line segment will be parallel to the base. This implies that the angles in the small top triangle will be the same measure as the angles in the big triangle, and therefore since the big triangle is equilateral, the small one is too. To prove the mid-segment theorem is not too tricky, but more than I should include in this comment.

[@RexxSchneider]

When you have a triangle with an angle of 60° and it also has the two adjacent sides equal, it's an equilateral triangle. We know that ∠ABC = 60° because we were told ∆ABC was equilateral. We constructed BD and BE to be the mid-points of AB and BC. We know that AB=BC because ∆ABC is equilateral. So we know that BD = BE. That means ∆BDE has two equal sides that enclose an angle of 60° so it has to be equilateral All of its sides are therefore the same length and DE = BE.

[@michaelempeigne3519]

what are you putting m with angle for ? are you multiplying by a value for m ?

[@mikeschieffer2644]

m stands for the measure of the angle. Angles can be congruent, while measures of angles can be equal.

[@karolakkolo123]

@@mikeschieffer2644 I still don't like that notation tbh, ever since it was introduced to me in high school geometry class. I think it looks better without the m. Sort of like when we have zero vectors and zero matrices, they are also just denoted by a boldface zero, but we can easily see which one of these is used based on the context

[@robertveith6383]

@@karolakkolo123 You must indicate the measure. If not with "m," then write it out as "the measure of ..."

[@karolakkolo123]

@@robertveith6383 I understand that, and I understand the difference between the angle and the measure of the angle, but tbh, whether the angle or the measure is used can be easily deduced from the context, in my opinion making the distinction in notation unnecessary

[@Grizzly01]

@@robertveith6383 I'm not sure he should have in this case, as he wasn't stating the measure of any angles at all. He should have written ∠BFG ≡ ∠BCG (or ∠BFG ≌ ∠BCG if you prefer).

[@raystinger6261]

I think the intersecting chords theorem is much easier for this problem. Chords GF and BC intersect at point E. By the intersecting chords theroem we get: GE*EF = BE*EC [equation 1] However, we get: BE=EC=DE and GE = GD + DE and GD = EF => GE = DE + EF So, from equation 1, we get: (DE + EF)*EF = DE² =>DE² - EF*DE - EF² = 0 Dividing this whole equation by EF² (granted that EF isn't 0): (DE/EF)² - (DE/EF) - 1 = 0 Solving this quadratic, we get: DE/EF = (1 +- sqrt(5))/2

[@jacemandt]

He really did use the intersecting chords theorem here, but instead of citing it directly, he proved it along the way, maybe for people unfamiliar with the theorem. The similar triangle argument proves precisely that theorem.

[@Tom-cq2ui]

Neat!

[@CM63_France]

Hi, Great! I didn't know that way of finding phi. For fun: 3:07 : "ok, great".

[@takyc7883]

I think its much easier to use trigonometry to find the distances in terms of the radius, and then it all cancels out nicely into the golden ratio

[@takyc7883]

less pretty and less rigorous but hey ho it does the job

[@channelsixtysix066]

Every time I hear or read "Golden Ratio" I turn off.

[@Nah_Bohdi]

🥺

[@jorgecologico6289]

Angles BEG is equal to BCA but BFG is not equal to BEG

[@robertveith6383]

That should be the *measure* of angle BEG is equal to the *measure* of angle BCA, but the *the measure of angle* BFG is not equal to *the measure of angle* BEG. The latter is true. Why did you bring it up? Are you claiming Michael said it and/or wrote it as such?

[@jorgecologico6289]

@@robertveith6383 BFG is not equal to BCG

[@Qermaq]

@@jorgecologico6289 3:08 he explains this.

[@Grizzly01]

@@jorgecologico6289 ∠BFG ≌ ∠BCG as can be proven by the inscribed angle theorem. I realise you made a typo in your first comment (you typed E when you meant to type C), which actually renders it correct, but your subsequent comment is wrong.

[@cicik57]

illuminati signs even here :(

[@michaelempeigne3519]

can hardly hear @michaelpenn in video