The mathematician formerly known as DayLateDon

16h
answered Geometry Problem about Cyclic Quadrilateral
20h
comment Geogebra - Move intersection point labels
@user3507584: Be aware that GeoGebra is sometimes finicky about letting you "grab" a label. (I find that it's especially-so when labels are near angle marks.) When that happens, you might have to hide some objects temporarily to get them out of the way. Also, you can change font size via the "Options" menu. In any case, these questions aren't about math, so they aren't really appropriate here. You might want to visit GeoGebra's own forums.
1d
revised What are some rigorous definitions for sine and cosine?
added 147 characters in body; edited title
1d
comment What are some rigorous definitions for sine and cosine?
My answer to a related question shows a link between the geometric definitions and the power series.
1d
comment Explaining $\cos^\infty$
@giorgiomugnaini: I was about to make the same observation about the discontinuity, except I was going to write the equation as $t-\pi/2 = \cos( t - \pi/2)$; that is, the jump happens when $t = \alpha + \pi/2$.
Mar
3
comment Certain Geometry proofs seem not rigorous at all.
There's rigor and then there's rigor. The prime example is the *Principia Mathematica*, which takes about 370 pages of symbolic logic to even suggest that the statement "1+1=2" might be true. :) That said, while I'm personally a big fan of visual, intuitively-appealing proofs, I too find Kiselev's argument lacking. Nevertheless, I suggest further rephrasing your question (and title) to "simply" request a rigorous proof of the Theorem. As it is now, your post is calling for opinions, which isn't really the point of this site.
Mar
3
comment Certain Geometry proofs seem not rigorous at all.
"As an aside, is there a better proof of this theorem or is this as rigorous as synthetic geometry gets about the congruence of arcs in circles?" The result certainly has a more-rigorous proof, but with that aside aside ... What's your question?
Mar
3
comment Proving the Sine Rule with one line.
@Chris: Not to lessen your enthusiasm for abel's contribution, but ... the "SAS" proof is pretty-much the standard proof of the Law of Sines. (I think the one Tim mentions using Thales' Theorem is "better", in that it explains that $a/\sin A$, $b/\sin B$, $c/\sin C$ are all equal to each other because they're equal to the same other thing, namely the diameter of the triangle's circumcircle. But that's beside the point.) Now I'm curious: If you haven't seen either of these proofs, then how was the Law of Sines explained to you in the first place?
Mar
2
revised Why is the tangent of 22.5 degrees not 1/2?
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Mar
1
answered Why is the tangent of 22.5 degrees not 1/2?
Mar
1
revised Concentric Equilateral Triangles
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Feb
28
comment How to prove $x=120^\circ$
+1. Now I remember that I've seen this problem (and this solution) before. Maybe here, but I think elsewhere long ago, as well. My foggy memory is telling me that it's relatively "famous" (so I'm kicking myself for not remembering the trick right off! :).
Feb
28
comment How to prove $x=120^\circ$
@Sawarnik: Yes, it does appear that the $120^\circ$ (and/or $60^\circ$) property is logically equivalent to the intersection-of-circumcircles property.
Feb
28
comment How to prove $x=120^\circ$
@Sawarnik: Similar to imranfat's example. Rotate about $C$ until $D$ lies on $\overline{AC}$ (with $\overline{AC}$ separating $B$ and $E$). Then $\angle APD = 60^\circ$. The thresholds are the circumcircles of the triangles (which gets to your notion of the cyclic quadrilaterals): when, say, $D$ is outside the circumcircle of $\triangle ABC$, then $\angle APD =120^\circ$; inside, $60^\circ$; when $D$ is on the circumcircle ... well ... it looks like $P$ and $D$ coincide (because of cyclic quads!), so the angle as named is undefined.
Feb
28
comment How to prove $x=120^\circ$
If you call the point of intersection $P$, then it isn't always the case that $\angle APD = 120^\circ$; sometimes, the measure is $60^\circ$. Likewise with $\angle BPE$. (The threshold where things change is interesting.) However, it appears that lines $\overleftrightarrow{AP}$ and $\overleftrightarrow{DP}$ always form two $120^{\circ}$ angles (and two $60^\circ$ angles) at $P$. Neat little problem. Where did you find it?
Feb
28
answered Concentric Equilateral Triangles
Feb
28
comment Does there exist a power of 2 which is the concatenation of two powers of 2?
It's perhaps worth nothing that there are triple-concatenations, such as $128$ and (if you allow intervening zeroes) $1024$ and $2048$. I think there was a question here about such things some time ago.
Feb
26
comment Basic trigonometry with popsicle sticks
I'm confused about how exactly you're trying to put the sticks together. My first thought was that you want "vertical" trapezoids, with a horizontal bottom edge that aligns with the $x$-axis, and a slanting upper edge that approximates the curve of the sine wave. But, then, each stick itself doesn't have a length; it has two lengths: the trapezoid's vertical "bases". (Two adjacent sticks share a common base-length.) Is this what you want? Or do you want "horizontal" trapezoids that slope on the left- and right-hand sides? (There, too, you need two lengths per stick.)
Feb
25
revised geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$
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Feb
25
revised geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$
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