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 diagonalsWith $v$ 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 triangleBetter title; slightly-better TeX Aug 26 comment Motivations for Hyperbolic GeometryAs 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'sYou 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 triangleAdded image Aug 23 revised Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrentadded 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 concurrentadded 1439 characters in body