The mathematician formerly known as DayLateDon

1d
comment trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?
@JasandPruski: It's a rectangular prism because one pair of opposite faces consists of squares, and the segment joining the centers of those faces (and, therefore also, each segment joining corresponding corners of those faces) is perpendicular to them. The tetrahedron's dihedral angles are not a factor in that deduction.
1d
comment trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?
@JasandPruski: Take your time. By the way: I was inaccurate to say that your rectangular prism observation was inaccurate. :) In general, a tetrahedron's bounding parallelepiped is not a rectangular prism, but for your tetrahedron it is. (In all cases, though, the parallelepiped faces have areas $H$, $J$, $K$.) I need to stop commenting in the middle of the night ... :)
1d
comment trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?
@JasandPruski: "I especially find it strange that: areas [...] can be used in the law of cosines". I find it strange, too, but that's what makes it really interesting! :) Your observation about "the rectangular prism that this tetrahedron resides in" is insightful, but not quite accurate; instead, you want to consider the parallelepiped whose faces have the tetrahedron's edges as diagonals. The face areas of this parallelepiped match the $H$, $J$, $K$ above, although those are not my usual geometric interpretations of pseudo-faces. (Perhaps they should be. They're easier to describe! :)
1d
answered trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?
1d
answered trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles?
1d
answered Number of polyhedron diagonals
2d
comment Wolfram Alpha "x = derivative x"
As others are noting, W|A isn't properly interpreting your intent. See here for examples of how to enter differential equations into W|A. In particular, you should enter x-x'=0 or x=x' to get the exponential solution you expect (with a multiplied constant, of course!).
Aug
28
comment Number of polyhedron diagonals
With $v$ vertices, $\frac{1}{2}v(v-1)$ counts the number of segments joining any two of them; these segments are either diagonals or edges. With $e$ edges, then, $\frac{1}{2}v(v-1) - e$ counts just the diagonals.
Aug
27
revised Relations involving the altitudes and orthocenter of a triangle
Better title; slightly-better TeX
Aug
26
comment Motivations for Hyperbolic Geometry
As one Blue to another: Because it's there! Also, because even some of the "easy" stuff about geometric objects remains unknown, so opportunities for discovery abound. For instance, only recently (2005) did we get a reasonably-attractive formula for the volume of a tetrahedron; and I beileve a number of results in my "Hedronometric Formulas for a Hyperbolic Tetrahedron" are new. This makes exploration all the more satisfying ... and fun!
Aug
26
comment How to derive parametric equations of a curve from its geometric property?
Incidentally, you can parameterize $h$ and $a$ via $h=\cosh t$ and $a = \sinh t$. This will eliminate the distracting square roots in your formula, and also reinforce the idea that $h$ and $a$ (and the relation $h^2-a^2=1$) represent a single degree of freedom, represented by $t$.
Aug
26
comment How to derive parametric equations of a curve from its geometric property?
If $\overrightarrow{OM}$ makes clockwise angle $v$ with the (positive) $x$-axis, then we can write $M = h ( \cos v, \sin v)$. If $\overrightarrow{MP}$ makes clockwise angle $u$ with $\overrightarrow{OM}$ then it makes angle $u+v$ with the $x$-axis, and we have $$\begin{align} P &= M + a (\cos(u+v)), \sin(u+v)) \[4pt] &= (h\cos v + a\cos(u+v), h \sin v+ a \sin(u+v)) \end{align}$$ Your expressions for $x$ and $y$ arise from the substitution $a = \sqrt{h^2-1}$ and the expansions of $\cos(u+v)$ and $\sin(u+v)$. Is that what you want? (BTW: Don't $h$, $u$, and $v$ make three parameters? :)
Aug
26
answered Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus
Aug
24
comment Locus of points on a curve for constant segment lengths squared sum $ OM^2 + MP^2 $
If you're trying to characterize another interesting family of circles, this won't do. Say that $\overrightarrow{OP}$ makes an angle $\theta$ with the $x$-axis; then we can take $M = c \cos f(\theta)$ & $P = c ( \cos f(\theta) + \sin f(\theta) )$ for any sufficiently-well-behaved function $f$. We need $f(0) = \operatorname{acos} (|OM_1|/c)$, to satisfy the initial condition; and $\sin f(\theta) = 0$ at least once, so the traces of $P$ & $M$ "meet" to form a single curve, taking care to ensure that a "U-turn" at that point (if there is one) is differentiable. The result needn't be a circle.
Aug
24
comment I apply the sum-to-product identity for $\sin$, but my result differs from the textbook's
You can tell that there's an error in the book's solution, since $x=\pi/4$ does not satisfy the original equation.
Aug
23
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
@Razorlance: When Sean describes symmetry with respect to "the $x$-axis", he means symmetry with respect to "the line joining $C$ to the origin" (which only happens to align with the $x$-axis in the convenient model). So, an asymmetric shield path in this sense would remain asymmetric upon rotating the room, the path, and the Cap'n. The point here is that shield path is a chain of chords, and it need not be the case (for $n>2$) that this Cap-line meets any of those chords at their midpoints or endpoints the way it does for $n=2$.
Aug
23
revised Line tangent to circle inside an isosceles triangle
Added image
Aug
23
revised Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrent
added 9 characters in body
Aug
23
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
@Razorlance: Thanks! BTW, I use GeoGebra.
Aug
23
revised Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrent
added 1439 characters in body
1 2 3 4 5