I got broken compass PTSD from highschool. The lead would be unusably dull, the spike would be wobbly or missing, and the joint would be so loose my arcs and circles would turn out as spirals.
Hi Mr Salomone, thanks for your fantastic and really effective videos. I'd really like to have something like this when studying on my own elliptic curves, modular forms or reciprocity laws. I've just a small question, if possible. You told the straight edge is not a ruler, and this makes a lot of sense (also in projective geometry you can use some similar construction without the need of a metric at all, I guess) but I cannot figure out how you was able to draw the "1" segment within your proof of multiplicative closure with similar triangles. You have not a ruler and you have not used the compass neither, so how we were able to deduce the "1" segment out of the 2 original numbers, without measuring them? Trying to refine better: is there a "fast" way to get a "unit" segment from a couple constructible numbers using just the compass and the straight edge? My feeling you have to have a couple of them in order to establish a ratio between them, and then deduce the unit from that ratio, somehow. But if the ratio between the 2 is not "rational", how you can deduce the unit from it? EDIT: thinking of it deeply: of course there are infinite units...choosing one them just means rescaling the lengths upon it. So probably you just chose one unit segment and put it on the left of the first segment, am I right? Hope my question will not sound too stupid. Best and thanks Ricky
Yeah, I realize I could have been more up front about that, but you're right: we couldn't have constructed segments of length k1 and k2 to begin with unless we started (somewhere on the page, unseen) with a segment of length 1. So wherever that segment was, go set your compass to its length and copy it into place. Good question!
Very good, educational video! Very understandable and good use of graphics. You made a mistake early on though. The large cube has a volume of 3√3 V, not 3V.
Small point but at time 8:08 you show the difference (in red) as k1 - k2 but I think it should be k2 - k1. Unless you're allowing for negative numbers (negative lengths?) which maybe you are...
Very nice video, what software did you use? I can imagine using some kind of presentation software but how do you get a mixture of that with hand-drawings and clean, accurate constructions?
The multiplication argument was really difficult to follow. Probably because geometry has been a while, but I'm not seeing the ratios occur in either triangle
The demonstration for the product k1 k2 = P was confusing to me. k1 and 1 are not sides of the pink triangle. We need to include the two small right triangles on the left as well to show how the ratio works.
When I try to use the similarity of the triangles, by comparing the horizontal edge to the altitudes I get (1+k1+k2+p)/(1+k1+k2) = (1+k1)/1 which can simplify to p = (1+k1+k2)(k1) which implies (1+k1)k1 = 0 if p = k1*k2. This seems problematic, I wonder if I'm misunderstanding what you labeled as p.
Wait, I think I misunderstood that k1 and k2 are laid on top of each other, not put end-to-end. Edit, no, I'm extremely confused about which segment is labeled with what length.
multiplication as explained cannot be understood. Better: 1. draw a line length 1, mark it, and extend until it is length a, at the zero point, make a line right angle up and perpendicular with length b. 2. complete the triangle by adding 3rd side, hypotenuse. 3. From end of a, put a line parallel to the hypotenuse to close the larger similar triangle. 4. Call the vertical length of the larger triangle p. Now you can see b:1 as p:a or b/1=p/a or ba=p.
I LOVE that you know what straightedge music is!!!!
I didn't know what it was until just now when I came across your comment and looked it up.
I got broken compass PTSD from highschool. The lead would be unusably dull, the spike would be wobbly or missing, and the joint would be so loose my arcs and circles would turn out as spirals.
Hi Mr Salomone,
thanks for your fantastic and really effective videos. I'd really like to have something like this
when studying on my own elliptic curves, modular forms or reciprocity laws.
I've just a small question, if possible. You told the straight edge is not a ruler, and this makes a lot
of sense (also in projective geometry you can use some similar construction without the need
of a metric at all, I guess) but I cannot figure out how you was able to draw the "1" segment
within your proof of multiplicative closure with similar triangles.
You have not a ruler and you have not used the compass neither, so how we were able
to deduce the "1" segment out of the 2 original numbers, without measuring them?
Trying to refine better: is there a "fast" way to get a "unit" segment from a couple constructible numbers
using just the compass and the straight edge?
My feeling you have to have a couple of them in order to establish a ratio between them, and then
deduce the unit from that ratio, somehow. But if the ratio between the 2 is not "rational", how
you can deduce the unit from it?
EDIT: thinking of it deeply: of course there are infinite units...choosing one them just means
rescaling the lengths upon it. So probably you just chose one unit segment and put
it on the left of the first segment, am I right?
Hope my question will not sound too stupid.
Best and thanks
Ricky
Yeah, I realize I could have been more up front about that, but you're right: we couldn't have constructed segments of length k1 and k2 to begin with unless we started (somewhere on the page, unseen) with a segment of length 1. So wherever that segment was, go set your compass to its length and copy it into place. Good question!
Thanks Mr Salomone, for your quick answer.
Very good, educational video! Very understandable and good use of graphics. You made a mistake early on though. The large cube has a volume of 3√3 V, not 3V.
Small point but at time 8:08 you show the difference (in red) as k1 - k2 but I think it should be k2 - k1. Unless you're allowing for negative numbers (negative lengths?) which maybe you are...
Very helpful video.
Very nice video, what software did you use? I can imagine using some kind of presentation software but how do you get a mixture of that with hand-drawings and clean, accurate constructions?
The multiplication argument was really difficult to follow. Probably because geometry has been a while, but I'm not seeing the ratios occur in either triangle
The ratio 1/k2 from small triangle is equal to the ratio k1/p from large triangle.
3:40
Did you mean: side length sqrt(2A)?
The area is 2A, so the side length is the square root...
Ivo Kroon that’s what I thought too
thanks a lot sir 🙏🏻
The demonstration for the product k1 k2 = P was confusing to me. k1 and 1 are not sides of the pink triangle. We need to include the two small right triangles on the left as well to show how the ratio works.
When I try to use the similarity of the triangles, by comparing the horizontal edge to the altitudes I get (1+k1+k2+p)/(1+k1+k2) = (1+k1)/1 which can simplify to p = (1+k1+k2)(k1) which implies (1+k1)k1 = 0 if p = k1*k2. This seems problematic, I wonder if I'm misunderstanding what you labeled as p.
Wait, I think I misunderstood that k1 and k2 are laid on top of each other, not put end-to-end. Edit, no, I'm extremely confused about which segment is labeled with what length.
I got p as k1^2 + k1*k2 if p represents the length of the orange line.
multiplication as explained cannot be understood. Better: 1. draw a line length 1, mark it, and extend until it is length a, at the zero point, make a line right angle up and perpendicular with length b. 2. complete the triangle by adding 3rd side, hypotenuse. 3. From end of a, put a line parallel to the hypotenuse to close the larger similar triangle. 4. Call the vertical length of the larger triangle p. Now you can see b:1 as p:a or b/1=p/a or ba=p.
The volume with a length of a long diagonal is 3*sqrt3 V?
4:50
but k1 and 1 ain't the sides of the smaller triangles........
Oh man, this video was triangularwesome! 👍📐