Mathematics - 2014-10-25 (page 2 of 5)

user112495

4:00 PM

@UserX Wait, but then wouldn't I have $f(0), f'(0), \cdots$ in $f(X+t)$, and $f(t), f'(t), \cdots$ in the other?

UserX

$f(x)+x f'(t)+1/2 t^2 f''(x)+1/6 t^3 f^(3)(x)+1/24 t^4 f^(4)(x)+\mathcal{O}(x^5)$

Around $x=a$ means $x-a$ not $a$

I fucked up those last messages so much

Disregars them

Sawarnik

And again @Nick

Semiclassical

@Sawarnik: re. your triangles problem, the thing I notice playing around in Geogebra is that evidently one has $al_a^2+bl_b^2+cl_c^2\leq 9R\Delta$ with equality iff the triangle is equilateral. that makes the problem a little precise

UserX

It's $f(t) + xf'(t) + \frac{f''(t)}{2!}x^2 +\mathcal{O}(x^3)$

Sawarnik

@Semiclassical I saw that :/

@Huy Don't know, I love triangles :D

Semiclassical

4:11 PM

figured you probably had, but thought i'd mention it

Sawarnik

:)

@Semiclassical Do you have GeoGebra 5?

Semiclassical

and the fact that it's never negative is interesting

user112495

@UserX But because i'm expanding around t=0, don't I need f(0), f'(0) etc. rather than f(t), ...

Sawarnik

@Huy I stopped listening to music few years ago :/

Semiclassical

i use the chrome app, though only because for some reason geogebra no longer works on my laptop

Huy

4:12 PM

@Sawarnik: Sad to hgear. Why that?

Semiclassical

as in, if i double-click the icon it's as though my laptop walks into the back room of a shop to go get something i need, then comes back a few minutes saying "hey man, was there something you wanted?"

Sawarnik

@Semiclassical lol ..it does open late in my comp too :/

Huy

@Sawarnik: I was asking why you stopped listening to music?

Semiclassical

yeah, i dun get it. it used to be working a while back, and then just "NOPE"

Sawarnik

@Huy Its became boring...

Semiclassical

4:15 PM

so yeah, chrome app

Huy

@Sawarnik: If listening to music became boring to you, you did something wrong.

Sawarnik

No way. I still like simple [not sure of the word] music, not the raps and rock that is only here today.

@Semiclassical Well, re-download it?

Semiclassical

tried that

Sawarnik

:/

:/ :/

Huy

@Sawarnik: Any example?

Semiclassical

4:17 PM

no big deal, just annouying

and really, not liking mainstream music != not liking music

Sawarnik

@Huy Not quite. I forgot them long ago. But you can understand?

@Semiclassical its strange ... :/

Semiclassical

pandora's a good way to find stuff (especially if you're lazy like me)

shrug

Sawarnik

never heard of that...

@Semiclassical for what?

Huy

@Sawarnik: I'm trying. I understand if you don't find any additional joy from listening to music, but all kinds of music being boring is very weird to me.

Semiclassical

you haven't heard of pandora?

Sawarnik

4:19 PM

@Huy not all

@Semiclassical no :/

Semiclassical

check it out, then: pandora.com

Huy

@Sawarnik: Today's task for you: Find a song you like and send it to me.

UserX

@user112295 okay, I'll leave this to you because I'm confusing you with my confusion. Last comment; you need the series around $t=0$ and no you don't get $f(0)$ anywhere, when you do taylor around $t=a$ you get $(t-a)$ terms not $a$ terms. What happens when $a=0$?(you get the $t$ terms you want)

Semiclassical

think of a song you like, try typing it into pandora, and see what it generates for you

Sawarnik

@Semiclassical well ask what error huy gets ...

Semiclassical

4:22 PM

error?

Sawarnik

ask..

Huy

@Semiclassical: We are deeply, deeply sorry to say that due to licensing constraints, we can no longer allow access to Pandora for listeners located outside of the U.S., Australia and New Zealand.

Semiclassical

...well, that's shitty

user112495

@UserX I know that you get t^k terms. But don't you have to evaluate the derivatives of f at the value a=0?

Sawarnik

true

@Semiclassical But do you know how to improve accuracy of the algebra toolbar? It just gives 2 digits now .. I want more by default.

Semiclassical

4:23 PM

ugh, i wish

it's really sillly that they just give you that

user112495

@UserX I get it now. I was getting confused with something else. Thanks!

Semiclassical

you might be able to trick it by doing something like

a/0.01

Sawarnik

@Semiclassical But by default? :/

Semiclassical

...or, hell, just 100a

not a clue

i find it annouying as well

Sawarnik

I can get magnitudes better approximation by CAS but otherwise...

Nick

4:24 PM

use musicovery, it's different.

Semiclassical

though i mostly just use geogebra to either 1) check if a Math.SE problem makes sense, 2) give a Math.SE problem a better picture

Huy

@Nick: I meant 1989 earlier. Do you like her music?

Sawarnik

@Semiclassical Your idea's is not bad though :D .. nice shortcut :D

Me too. And accuracy is more important for that...
But I do much some other stuff time to time as well.

Semiclassical

trying to think of a good approach to that triangles problem, hmm

Chris's sis

@robjohn you have to see an amazing question ...

Sawarnik

4:26 PM

hmm

Semiclassical

one thought is to do it in complex numbers, and have the vertices all be on the circle $|z|=1$ (obviously it doesn't matter if we scale things to make $R=1$)

Nick

@Huy: There was that one song about some breakup. That's all I remember from her.

Sawarnik

see if you can find it :/

@Semiclassical Ouch..

Huy

@Nick: Are you sure? That doesn't sound like her music. :P

robjohn

@Chris'ssis okay

UserX

4:27 PM

@user112495 I'm glad you got it. Note that if you considered those series trivial and concluded like you did in your first proof, when asked how are they equal you would have trouble.

Semiclassical

eh, the only thing that seems problematic to me is getting $l_a,l_b,l_c$ out of that

Sawarnik

@Huy That sounds like Honey Singh .. . ugh!

@Semiclassical hmm, in terms of sides then?

Nick

@Huy: Romeo, blah blah blah

Huy

@Sawarnik: Fortunately, I don't know Honey Singh.

@Nick: That would be "Love Story". :D

Sawarnik

@Semi And is sine and cosine acceptable?

Huy

4:28 PM

@Nick: youtube.com/watch?v=8xg3vE8Ie_E For you, my dear.

Semiclassical

maybe. if there's a nice expression for the incircle in complex coordinates then getting $l_a,l_b,l_c$ seems like it'd work

Sawarnik

@Huy Fortunately indeed..

UserX

@Semiclassical link me the triangles problem

Semiclassical

@Sawarnik: probably, though i tend to favor avoiding sine/cosine if i'm sticking to a complex algebra approach

Nick

@Sawarnik: Yo Yo Honey Singh sings with singhams about singhs singing bout yoyos, dawg!

Semiclassical

4:29 PM

3

Q: Proving a triangle equilateral given condition $al_a^2+bl_b^2+cl_c^2=9R\Delta$

$ABC$ is a triangle, with $l_a$, $l_b$, $l_c$ as angle bisectors, $R$ as circumradius and $\Delta$ as area, such that: $$al_a^2+bl_b^2+cl_c^2=9R\Delta$$ Is it true that $ABC$ is equilateral? I am not sure how to approach this problem. Ideas, anyone? :)

Chris's sis

@robjohn $$\lim_{n\to\infty} \left(\sum_{k=0}^n \binom{n}{k}k! n^{-k}-\sqrt{\frac{\pi n}{2}}\right)$$

Sawarnik

@Nick :?

UserX

@Semiclassical can you get me a pic? I don't know geometry terminology in english

Nick

@Sawarnik: Yes, Did I leave you speechless? I do that with all the ladies.

Semiclassical

sure, i've got something pretty close to that in geogebra right now

Sawarnik

4:30 PM

@UserX ^ah

@Nick Sure, with your obesity?

Nick

@Sawarnik: Yup, I'm so fat that when I sit around the house, I sit around the house.

See, I cannot be insulted :D

To whomever it may concern, I'm 5'2'' and 195 pounds and I am proud of it!

robjohn

@Chris'ssis Interesting... I may not get to look at this until later this afternoon. We have plans this morning that might keep me away.

Chris's sis

@robjohn Look at it whenever you have time, no hurry with my questions. I'd like to add something like that to my book.

Nick

Now, let me enjoy Peter Capaldi's expected portrayal of the Doctor :D

Semiclassical

urk. can't figure out how to export images from chrome's geogebra app

Mike Miller

4:38 PM

print screen, paste in microsoft paint, crop appropriately

:)

UserX

@Sawarnik are you sure it's not $9R^2$?

Sawarnik

@semi snipping tool!

@UserX i am sure, why?

Semiclassical

@mikemiller lol, just finished doing that (though copied image instead of print screen)

UserX

@Sawarnik because 9R^2 is a known identity nothing more

Sawarnik

@UserX which identity?

Semiclassical

4:41 PM

the only thing not shown there explicitly is the circumradius $R$

UserX

$9R^2=a^2+b^2+c^2$

Semiclassical

@Sawarnik: that look like the right image?

Sawarnik

@Semiclassical l_c doesn't quite look angle bisector :P

but maybe it is :)

Semiclassical

hmm

Sawarnik

is it or is it not?

Semiclassical

4:44 PM

checking now. i might've done something dumb

UserX

@Sawarnik I'm too bored to LaTeX so I'll send you a handwritten image in chat, I collapsed the problem with no trigonometry to proving a three variable equality

Sawarnik

comes in 5 min...

Semiclassical

i constructed it by having Geogebra produce the incenter of the triangle, and then clicked the intersections of the incircle with the triangle

maybe i should've done 'intersections' at the command line instead

yeah, they're not angle bisectors. annouying

that's still correct enough to recognize the geometry, but not enough to include in the question

UserX

@Sawarnik s28.postimg.org/3paxw132k/20141025_194557.jpg (clarifications; $\beta=b, \gamma=c$) mixed them up because I'm greek :P

Semiclassical

why not just do the math here? robjohn's chatjax link takes care of symbolism

UserX

4:51 PM

Because I'm bored and this isn't a complete answer. Otherwise I would have posted it to his question.

Plus it was done so fast I can assume I have a mistake in there

Semiclassical

ah, i more meant "why make us read hand-written photographs of math when we can read mathjax instead" :)

UserX

Because of the magic of handwriting(i'm bored and my mobile doesn't render mathjax so it's time-consuming checking for tex mistakes if you can't see them rendered)

r9m

@Sawarnik baccha u were looking for me ? :) :P

Semiclassical

fair enough

Sawarnik

back for 5 mins..

@UserX Handwriting's like me :D

r9m

4:54 PM

@Sawarnik have you tried basic inequalities for the expression ? :)

Sawarnik

@r9m yes, yes, yes!

Semiclassical

nope :) i'm not really working on the problem itself

Sawarnik

r9m ∈ {expert in geometric inequalities}

Semiclassical

right now i'm just lazily putting together the right picture in geogebra

r9m

@Semiclassical I wanted to tag @Sawarnik .. sorry :)

Semiclassical

4:55 PM

lol, no worries

UserX

I lazily simplified it too. Don't assume my pic is right.

Why are we lazy anyway?

Nick

Because we are human and we choose to be.

Sawarnik

@UserX Who told you this? Its wrong, in my opinion.

Semiclassical

ehh, lazy is really the wrong word

was referring to userX there

Sawarnik

@Semiclassical Why? Then it should be the angle bisectors : O :O

How do you produce incenter without bisectors though? :O

Semiclassical

4:59 PM

mine was just wrong :P

there's an incircle command

Incircle[point,point,point]

Sawarnik

@Semiclassical :O

Wow.

And orthocenter?

Circumcenter?

Centroid?

Semiclassical

dunno about orthocenter and centroid, but wouldn't shock me

UserX

@Sawarnik you have to prove that this equality holds iff $a=b=c$ iff the triangle is equilablabla

Semiclassical

for circumcenter, easiest just to do the Circle[point,point,point] form and then take the center

it does include a centroid command

Sawarnik

@UserX Oh.

@Semiclassical Well, that was easy.

@Semiclassical Wow.

Semiclassical

5:02 PM

yep

UserX

@Sawarnik it's completely wrong disregard it. I W|A the result. I might try a complex number approach later

Sawarnik

why its wrong? :O

@r9m lol ... i also wanted to say that you should redirect your old blog to new blog.

Semiclassical

i don't see a command for orthocenter off the bat, though obv. one can just use the altitudes

Sawarnik

awesome blog that...where did you have the time?

@Semiclassical And altitudes are a bit annoying to create, isn't it?

UserX

No fucking idea, the formulas were right

Plus the algebra

Sawarnik

5:05 PM

you just have to make sure of the iff condition

UserX

Whatever, I gotta hang out with friends so cya l8r

Sawarnik

well then .. bye :)

Semiclassical

not really, just create a perpendicular line using the side and the opposite vertex

Sawarnik

and then create a segment

UserX

My approach was wrong, not your conjecture that it is equiblabla

Nick

5:06 PM

@UserX: xlr8, be back soon.

Sawarnik

@UserX This is confusing.

@Nick never joins triangle discussions.

Semiclassical

ok, this should be on target

did I do anything silly this time? :) @Sawarnik

Sawarnik

No :)

Semiclassical

neat

Sawarnik

Probability says no as well.

Semiclassical

5:11 PM

for my part, what i'd really like would be an expression for the difference in the LHS and RHS in terms of the three angles

r9m

@Sawarnik from the time I saved when I successfully escaped the IMP :P

Sawarnik

gone for 2 min...

r9m

@Sawarnik what's wrong with the new address ?

Semiclassical

the equality would presumably follow from that, but would also indicate the extent to which the equality fails when the angles are different

Nick

@Sawarnik: @Nick would be in the Geometry Room if he wanted to partake in such discussions.

(@Nick is too stupid to do real math)

Semiclassical

5:15 PM

@nick, did you see my 'fyi' in the number theory room?

Nick

@Semiclassical: Yes... now, I forgot what it was. Let me go check again.

r9m

@Sawarnik by Stewart(or otherwise ?) we have $l_a^2 = bc - \dfrac{a^2bc}{(b+c)^2}$ and similar other terms .. $\sum\limits_{cyc} al_a^2 = 3abc - abc\sum\limits_{cyc} \dfrac{a^2}{(b+c)^2} = 9R\Delta = \frac{9}{4}abc$ .. tell me if you recognize the equality obtained :-)

Semiclassical

trying to find a link that's relevant

r9m

$\displaystyle \sum\limits_{cyc} \dfrac{a^2}{(b+c)^2} \ge \frac{3}{4}$ (by CS and Nesbitt / otherwise .. ?) equality holds iff $a=b=c$

Semiclassical

back in a bit

Sawarnik

5:21 PM

back again..

@r9m nothing, i was telling to redirect the old one to the new one .. your blogs have strange names always though :D

@r9m Well, post it as an answer?

@Semiclassical see, i told r9m's a genius.

0

Q: Inequality of area of two triangles

Let $ABC$ be a triangle with sides $a,b,c$ and $A_1B_1C_1$ be another triangle with sides $a+\frac{b}2$, $b+\frac{c}2$, $c+\frac{a}2$. Prove that: $$\frac94[ABC]=[A_1B_1C_1]$$ I tried using Heron's formula, but it is not getting me to the answer. Can anyone help? :)

geometry triangle

r9m

@Sawarnik is that good enough I can post it as an answer .. or are you looking for some purely geometric proof ?

@Sawarnik WHAAAAA ?? -_-

Sawarnik

@r9m Good enough .. and a geometric answer would be better :)

@r9m Truth.

Another question that I have no idea :O @r9m @semi

@r9m :P :P

r9m

@Sawarnik .. its a RMO question (as far as I can remember) .. its a must try on your own problem ! ^^

Sawarnik