15,252 reputation
11753
bio website daylateanddollarshort.com/…
location
age
visits member for 4 years
seen 4 hours ago

The mathematician formerly known as "Day Late Don".

Honk if you love the Function Monkey.

See some of my research (and fun stuff) at my blog, "The Bloog!"


1d
comment Inscribed Circles in Triangles
Would the original triangle happen to be a right triangle?
1d
comment Inscribed Circles in Triangles
Your question asks about "the area of the circle". Do you mean "the circles", as in "the total area of all of the circles"?
1d
comment Inscribed Circles in Triangles
Actually, I should've been more clear in my comment: There's always at least one altitude that separates a triangle into two triangles. (For an obtuse triangle, there's only one.)
1d
comment Inscribed Circles in Triangles
Do you mean "altitude" instead of "perpendicular bisector"? A perpendicular bisector doesn't always separate a triangle into two triangles, but an altitude does.
1d
comment Is there any reason why $4-\pi$ is quite close to $\frac{\sqrt{3}}{2}$?
@SamuelYusim: Of course, the approximations are varying degrees of bad here, so the diagram hardly constitutes a rigorous numerical argument. A more-refined diagram would have blue and red chords hugging the circle more-closely. Even so, it's rather striking that the blue and red chords seem to meet so well at this very rough stage.
1d
comment Is there any reason why $4-\pi$ is quite close to $\frac{\sqrt{3}}{2}$?
@SamuelYusim: The blue polygonal arc has $8$ chords of length $1/2$; the big circular arc joining its endpoints therefore has length approximately $4$. That circular arc is clearly a semicircle, plus a smaller arc that's very-nearly subtended by the red chord of length $\sqrt{3}/2$; therefore, the big arc has length also approximated by $\pi + \sqrt{3}/2$. This gives $4 \approx \pi + \sqrt{3}/2$, as desired.
2d
comment Good Reference for Justifying (less well-known fields of) Math?
Perhaps Davis & Hersh's "The Mathematical Experience"?
2d
answered Is there any reason why $4-\pi$ is quite close to $\frac{\sqrt{3}}{2}$?
Aug
18
revised Prove that $\frac{{-\cos(x-y)-\cos(x+y)}}{-\cos(x-y)+\cos(x+y)} = \cot x \cot y$
TeX typo
Aug
17
answered Derivatives of trig polynomials do not increase degree?
Aug
17
answered How can the circular function $\tan(\theta)$ be both a length and a ratio of lengths?
Aug
15
comment Examples of Beautiful Applications of Thale's Theorem
This isn't really a "beautiful application", but ... This result happens to be my primary examples of what I call a "Gilligan's Island" theorem: If you were stuck on a deserted island and really needed to make a right angle (to lay out your huts and whatnot), then Thales' Theorem lets you create one using only vines and sticks.
Aug
15
comment Hypervolume of expanded $n$-simplex
I have added the definition that was given as a comment to my (now-deleted-as-irrelevant) answer. I also TeX-ified the volume expression; please double-check that it's the formula you intended. (I wasn't entirely sure, for instance, just how much stuff was supposed to go under the radical.)
Aug
15
revised Hypervolume of expanded $n$-simplex
Added definition of term, improved math formatting
Aug
15
comment Solving awkward quadratic equation to obtain “nice” solution.
You could post some of your notes about the larger problem as background for this question. (Just add it as an edit under the current stuff, perhaps with a --- separator for clarity.) I'm sure most answerers would appreciate knowing the context of the equation, so that they can decide for themselves whether a "nice" solution is feasible.
Aug
15
comment Solving awkward quadratic equation to obtain “nice” solution.
The value $\rho = 1 - \sin(\pi/n)/n$ is not a root for $n=3$, $4$, $6$, $17$, $55$, or $123$, either. I'd be rather surprised if it were a root for any particular integer $n$.
Aug
15
comment Solving awkward quadratic equation to obtain “nice” solution.
If $\rho = 1-\sin(\pi/n)/n$ were a solution, then you would be able to substitute the value into the equation and get zero. However, what you get is a complicated relation involving $n$ and trigonometric functions of $\pi/n$. This relation may be true for some $n$, but it's not true for all $n$ (for instance, $n=1$ and $n=2$ already fail).
Aug
14
comment Identity with logarithms?
Take the logarithm of both sides.
Aug
14
comment Prove that the length of segment on tangent is constant for $y=\frac a2\ln{\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}}-\sqrt{a^2-x^2}$
FYI: This curve is called the "Tractrix". (The equation given in the Wikipedia article uses a different formulation of the logarithm component; however, you can see that the derivative matches the one given by @ClaudeLeibovici.)
Aug
13
answered Parametrization of curve