The mathematician formerly known as DayLateDon

50m
comment Triangle inscribed inside a circle: prove that abc = 4 * area * R
Inscribed Angle Theorem?
10h
comment Function on plane with incenter
Here's the corresponding problem for tetrahedra: math.stackexchange.com/questions/857273/…
1d
comment System of quadratic equations for a tetrahedron
By the Spherical Law of Cosines, if you know the face angles at the apex, then you can find the dihedral angles at the apex, and vice-versa.
1d
awarded Yearling
1d
awarded Yearling
1d
comment Why does base*height work?
"Because of my level, ...". It would help if you explained exactly what your level is.
2d
comment A strange trigonometric equation
Using complex numbers, Euler's Formula lets you write $$\sin x = \frac{1}{2ip}(p^2-1) \quad \cos x = \frac{1}{2p}(p^2+1) \quad \cos 2x = \frac{1}{2p^2}(p^4+1)$$ with $p = e^{ix}$. Expanding terms, collecting, and factoring gives $$\frac{1}{2p^8} ( p^8 + 1 ) ( 3 p^8 + 7 p^4 + 3 ) = 0$$ Solving this eqn is straightforward. However, you mention needing this "done in class"; the expand/collect/factor steps here are really too much to do by hand, so this approach is worse than your attempt and what other answers show.
2d
comment A strange trigonometric equation
Conversion to complex exponentials is usually helpful in cases like this. Here, the exponential form of $16\sin^{10} x + 16\cos^{10}x - 29 \cos^4 2x$ decomposes into straightforward factors.
Jul
27
revised Help needed verifying a trigonometric identity
De-huge-ifying, and other TeX clean-up.
Jul
27
comment An equilateral triangle formed using points of tangency
@rah4927: The "measure" (which is not to say the length) of an arc is equal to the measure of the central angle that subtends ("looks at") it; that's definitional. (The length of an arc is proportional to its measure, with proportionality factor $r$ ... or $\pi r/180$ if the measure is in degrees.) Since the measure of an inscribed angle is proportional (always with factor $1/2$) to the measure of the central angle that subtends the same arc, we conclude that arcs (in measure or length) are also proportional to these inscribed angle measures.
Jul
27
comment Is there a way to calculate the area of this intersection of four disks without using an integral?
Looks like you're making good progress learning the TeX markup! Just a note: Avoid re-posting your answer with every little edit. Use the preview area (under the editor) to see how things look as you type. The StackExchange software is pretty good at remembering the state of your last text entry if you happen to accidentally close your browser window or something, so you don't have to keep "saving" your progress by repeatedly posting. (TeX Tip: Wrap the "words" in your equations with \text{}.)
Jul
27
comment Surprising necessary condition for a "shift-invariant" determinant
How many of the $12,870$ matrices with exactly $8$ zeros have determinant $1$?
Jul
27
comment Trigonometric Identities Need Help
Duplicate? math.stackexchange.com/q/879158/409
Jul
26
awarded Necromancer
Jul
26
answered problem about length of perpendicular chords
Jul
26
comment problem about length of perpendicular chords
Hypotenuse-Leg, all around.
Jul
26
comment problem about length of perpendicular chords
The perpendicular from $O$ to $\overline{KL}$ must also be perpendicular to $\overline{MN}$. (Why?) Let $P$ and $Q$ be the points where the perpendicular crosses these segments. Show that $\overline{OP}\cong\overline{OQ}$. (Hint: The perpendicular from $O$ to $\overline{AB}$ meets the segment at its midpoint, which is also the midpoint of $\overline{DE}$. (Why?)) Then $\triangle OPK$, $\triangle OPL$, $\triangle OQM$ and $\triangle OQN$ are all congruent, and the result follows.
Jul
26
comment Strategies to work with system of trigonometric inequality
If you take $a=b$, and $p_i = q_i = r$, your inequality reduces to $$0 \neq a e^{-2ir} \cos^3r \sin^3 r (e^{ir} - \cos r) = aie^{-2ir}\cos^3r\sin^4r$$ This is satisfied by any $a\neq 0$ and $r \neq n\pi/2$ (for integer $n$). But that's really over-thinking things. As @RossMillikan notes, almost-any values of the individual variables will make your expression non-zero.
Jul
25
revised An equilateral triangle formed using points of tangency
added 603 characters in body
Jul
25
comment An equilateral triangle formed using points of tangency
+1. I was just going to post an identical demonstration (which I'll probably go ahead and do). I hadn't identified $D$ as trisector of the large arcs, though; now it seems obvious. :)
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