The mathematician formerly known as DayLateDon

8h
answered Limit of two variable
1d
comment Broken line parallel with angle bisector
My formula $(\star)$ can be greatly improved by replacing $d$, $e$, $f$ with their representations in terms of edge lengths, as well as $a = |BD|+|DC|$, $b = |CE|+|EA|$, $c = |AF|+|FB|$: $$|AF| |BD|+ |BD| |CE|+|CE| |AF| = |EA||FB|+|FB||DC|+|DC| |EA| \qquad(\star\star)$$
1d
comment Broken line parallel with angle bisector
I have no time right now to post a coordinate proof of the following Ceva-like theorem: Take $D$, $E$, $F$ on respective edges $BC$, $CA$, $AB$. Let $$a = |BC| \qquad b = |CA| \qquad c = |AB|$$ and $$d = \frac{|BD|}{|DC|} \qquad e = \frac{|CE|}{|EA|} \qquad f = \frac{|AF|}{|FB|}$$ Then lines through $D$, $E$, $F$, parallel to respective angle bisectors at $A$, $B$, $C$, meet at a common point if and only if $$ b c ( 1 + d )( 1 - e f ) + c a ( 1 + e )( 1 - f d ) + a b ( 1 + f )( 1 - d e ) = 0 \qquad (\star)$$ In (my interpretation of) the problem given, $d = e = f = 1$, so that $(\star)$ holds.
1d
comment Broken line parallel with angle bisector
A GeoGebra sketch seems to show that, taking $A_1$, $B_1$, $C_1$ as midpoints of respective sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and taking lines $p$, $q$, $r$ parallel to the respective angle bisectors at $A$, $B$, $C$, the lines meet at a point. Interesting problem. That said, the "broken line" terminology is unfamiliar to me. I can see where the phrase might mean "angle", but it's rather unexpected that the "centre" of such a thing would be the midpoint of the side opposite that angle; it would seem easier (to me) to just name the side. I'm somewhat intrigued by this.
2d
comment How to explain to a 14 years old that $\sqrt{(-3)^2}$ isn't $-3$?
Tell him to hang onto his hat. In a couple of years at school (or minutes at a search engine), he'll learn that (complex) exponentiation is multi-valued. This tends to put confusion over the sign of a square root, and over when/whether powers and roots simply cancel, into some perspective, because this is only the beginning. One realizes that many?/most?/all? elementary definitions and rules we learn eventually need some tweaking. At some point, the square-roots-cancel-squares rule is bound to break; it may not be too hard to accept that that threshold is reached with the problem at hand.
2d
answered Recurrence relation for right-angled triangles stuck-together
Mar
27
comment Updates to the site
Blue always gets my attention. :)
Mar
27
comment Simplifying identity cos*cos+sin*cos
This answer may help your understanding.
Mar
26
comment Textbook geometry problem
Writing $a$, $b$, $c$, $d$ for the radii of $\bigcirc A$, $\bigcirc B$, $\bigcirc C$, $\bigcirc D$, it happens that $ac = bd$. (The radius of $\bigcirc X$ is irrelevant!) I got this result by a tedious coordinate argument; I'll seek an elegant one later.
Mar
26
comment Recurrence relation for right-angled triangles stuck-together
Are you familiar with rotation matrices?
Mar
24
answered Straightedge-only construction of segment of length $\sqrt{7}$, given regular hexagon with unit sides
Mar
24
revised Sequences and series of $\tan^n x$
TeXification
Mar
24
revised Straightedge-only construction of segment of length $\sqrt{7}$, given regular hexagon with unit sides
Better title. Minor revision of terminology.
Mar
24
comment Straightedge-only construction of segment of length $\sqrt{7}$, given regular hexagon with unit sides
How do you propose to construct the vertical blue and red lines using only an unmarked straightedge? (With a small adjustment, the basic idea can be salvaged.)
Mar
23
comment Geometric proof of this property of the ellipse
Just "thinking out loud". It's interesting to me that the segments $\overline{AX^\prime}$ and $\overline{PF}$ are congruent, so I'm curious about whether there's a nice way to demonstrate this fact without resorting to calculating their actual lengths. The congruence simply strikes me as being more than an algebraic coincidence. Maybe there's a cool application of the reflection property at work here ... or something. I may or may not have time to investigate this myself, so I thought I'd post the idea in case it might pique someone else's curiosity.
Mar
23
comment Geometric proof of this property of the ellipse
Hmmm ... Draw a circle of radius $|OF|$ about $O$ and say that it meets $OQ$ at $X$; and drop a perpendicular from $X$ to $X^\prime$ on the ellipse's major axis. Then $|OX^\prime| = c \cos\theta$, so that (writing $A$ for the ellipse's far-right vertex) $|AX^\prime| = a - c\cos\theta = a ( 1 - e\cos\theta) = |PF|$. Computations aside, I wonder: Is there a geometrically-obvious reason why $\overline{AX^\prime}\cong\overline{PF}$? I'm not seeing one (yet).
Mar
23
awarded Nice Answer
Mar
23
comment Series for computing trigonometric ratios
In this answer, I discuss the link between the geometric interpretations and the power series for sine, cosine, tangent, and secant. (Cotangent and cosecant have eluded me thus far.)
Mar
23
comment How to prove that $2\cos(x)\sin(y)=\sin(x+y)-\sin(x-y)$ with Euler's formula?
@MPW: That OP describes the process as "really complicated" suggested to me that perhaps there wasn't a keen-enough awareness of the exponential forms of the trig functions, and/or their usefulness to this situation. I can imagine OP having trouble teasing-out the identity via an argument like David's (which is quite nice, although not necessarily obvious), rather than realizing that a straightforward (obvious?) binomial product gets the job done. In any case, I intended the comment to be more helpful than "Just answer your own question"; my apologies if I have fallen short of this goal.
Mar
23
revised How to prove that $2\cos(x)\sin(y)=\sin(x+y)-\sin(x-y)$ with Euler's formula?
Texification, tag clean-up
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