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5h
comment Continous family of $n$-gons
@coffeemath's answer settles this quite elegantly. I'll mention that in this answer, I describe how any $n$-gon decomposes into a "sum" of regular (convex, starry, & multiply-traced) $n$-gons. Any $n$-gon, then, can be continuously transformed into a regular convex $n$-gon by continuously reducing the contribution of the non-convex components. I've wondered ---but have never investigated--- how (or whether) one might maintain convexity throughout such a process if the initial polygon is convex. If do-able, we can transform $A$ to regular to $B$.
9h
revised Regular/Right hyperbolas through three points
edited title
13h
comment Using oblique projection can you always rotate a triangle to look like an equilateral triangle?
Related (duplicate?): "Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection?"
13h
comment Regular/Right hyperbolas through three points
I suspect that the task intends $A$ and $B$ to be distinct points, in order that you get three equations in three unknowns ($x_0$, $y_0$, $a$). Allowing $\alpha=\beta$ will create a "one-parameter" family of solutions; and, if you're supposed to consider those, then you're probably also supposed to consider $\alpha = \beta = \gamma =0$: the two-parameter family of all regular hyperbola that pass through the origin.
13h
comment Regular/Right hyperbolas through three points
You're assuming not only are you assuming that "regular" means "right", but also that the hyperbola's transverse axis is parallel to the $x$-axis; to cover $90^\circ$ rotations of these, you simply replace your $a^2$ parameter with an un-squared $a$ that you allow to be negative. (Rotated hyperbolas complicate matters considerably.) Note that you've derived the relation $x_0 = \frac{1}{2}(\alpha+\beta)$; if you think about it, this is "obvious": points $A$ and $B$ (if distinct) must be symmetrically arranged about an axis of the hyperbola.
14h
comment Regular/Right hyperbolas through three points
Removed hyperbolic-geometry tag, since the question is not about non-Euclidean space.
14h
revised Regular/Right hyperbolas through three points
edited tags
1d
answered If $A+B+C=π$, prove that
1d
answered How to prove that $x^2 + 3y^2 = 1$ is contained inside of the unit ball?
1d
comment If $A+B+C=π$, prove that
Changing the $\cos^2C$ on the left-hand side to $\sin^2C$ gives a valid identity.
1d
comment If $A+B+C=π$, prove that
The statement as given is false. It's a near miss of this identity $$1 -\cos^2 A - \cos^2 B - \cos^2 C-2\cos A\cos B\cos C= 0$$
2d
revised Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
added 49 characters in body; edited title
2d
comment Ellipse set with one fixed focus, co-tangential at origin
You've put the foci on a some line through the origin, so that the ellipse (if non-degenerate) can only meet the origin at one of its vertices. For the ellipse to be tangential to the $x$-axis at a vertex, the ellipse's own major axis must coincide with the $y$-axis; in that case, $a=0$, and finding the equation is straightforward.
2d
comment Complex derivative numerically using real $h$ and imaginary $h i$?
Consider $f(z) := z$. The derivative should be $1$, but the formula shown gives $1+i$.
2d
revised Solving $\sec(3\alpha+30^\circ)=\csc(7\alpha-40^\circ)$
added 24 characters in body; edited title
Feb
6
reviewed Reject suggested edit on triangle construction given side, angle and median
Feb
6
answered Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix
Feb
5
comment Interesting cube subdivisions: what is going on here, and what are these polytopes?
On George Hart's "Conway Notation for Polyhedra" page, your construction corresponds to the "ambo" operation. Thus, in Conway notation, your polyhedra are "C" (cube), "aC", "aaC", "aaaC", etc.
Feb
5
revised Prove that $16\cos^5A-20\cos^3A+5\cos A=\cos5A$
Better title
Feb
4
revised Perimeter of a teardrop (made by two adjacent circles)
added 36 characters in body
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