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?

$f(x)+x f'(t)+1/2 t^2 f''(x)+1/6 t^3 f^(3)(x)+1/24 t^4 f^(4)(x)+\mathcal{O}(x^5)$
Around $x=a$ means $x-a$ not $a$
I fucked up those last messages so much
Disregars them

And again @Nick

@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

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

@Semiclassical I saw that :/
@Huy Don't know, I love triangles :D

4:11 PM
figured you probably had, but thought i'd mention it

:)
@Semiclassical Do you have GeoGebra 5?

and the fact that it's never negative is interesting

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

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

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

4:12 PM
@Sawarnik: Sad to hgear. Why that?

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?"

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

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

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

@Huy Its became boring...

4:15 PM
so yeah, chrome app

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

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?

tried that

:/
:/ :/

@Sawarnik: Any example?

4:17 PM
no big deal, just annouying
and really, not liking mainstream music != not liking music

@Huy Not quite. I forgot them long ago. But you can understand?
@Semiclassical its strange ... :/

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

never heard of that...
@Semiclassical for what?

@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.

you haven't heard of pandora?

4:19 PM
@Huy not all
@Semiclassical no :/

check it out, then: pandora.com

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

@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)

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

@Semiclassical well ask what error huy gets ...

4:22 PM
error?

ask..

@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.

...well, that's shitty

@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?

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.

4:23 PM
ugh, i wish
it's really sillly that they just give you that

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

you might be able to trick it by doing something like
a/0.01

@Semiclassical But by default? :/

...or, hell, just 100a
not a clue
i find it annouying as well

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

4:24 PM
use musicovery, it's different.

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

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

@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.

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

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

4:26 PM
hmm

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 $R=1$)

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

see if you can find it :/
@Semiclassical Ouch..

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

@Chris'ssis okay

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.

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

@Huy That sounds like Honey Singh .. . ugh!
@Semiclassical hmm, in terms of sides then?

@Huy: Romeo, blah blah blah

@Sawarnik: Fortunately, I don't know Honey Singh.
@Nick: That would be "Love Story". :D

@Semi And is sine and cosine acceptable?

4:28 PM
@Nick: youtube.com/watch?v=8xg3vE8Ie_E For you, my dear.

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

@Huy Fortunately indeed..

@Semiclassical link me the triangles problem

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

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

4:29 PM
3

$ABC$ 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? :)

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

@Nick :?

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

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

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

4:30 PM
@UserX ^ah
@Nick Sure, with your obesity?

@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!

@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.

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

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

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

4:38 PM
print screen, paste in microsoft paint, crop appropriately
:)

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

@semi snipping tool!
@UserX i am sure, why?

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

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

@UserX which identity?

4:41 PM
the only thing not shown there explicitly is the circumradius $R$

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

@Sawarnik: that look like the right image?

@Semiclassical l_c doesn't quite look angle bisector :P
but maybe it is :)

is it or is it not?

4:44 PM
checking now. i might've done something dumb

@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

comes in 5 min...

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

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

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

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

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

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)

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

fair enough

back for 5 mins..
@UserX Handwriting's like me :D

4:54 PM
@Sawarnik have you tried basic inequalities for the expression ? :)

@r9m yes, yes, yes!

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

r9m ∈ {expert in geometric inequalities}

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

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

4:55 PM
lol, no worries

I lazily simplified it too. Don't assume my pic is right.
Why are we lazy anyway?

Because we are human and we choose to be.

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

ehh, lazy is really the wrong word
was referring to userX there

@Semiclassical Why? Then it should be the angle bisectors : O :O
How do you produce incenter without bisectors though? :O

4:59 PM
mine was just wrong :P
there's an incircle command
Incircle[point,point,point]

@Semiclassical :O
Wow.
And orthocenter?
Circumcenter?
Centroid?

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

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

for circumcenter, easiest just to do the Circle[point,point,point] form and then take the center
it does include a centroid command

@UserX Oh.
@Semiclassical Well, that was easy.
@Semiclassical Wow.

5:02 PM
yep

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

why its wrong? :O
@r9m lol ... i also wanted to say that you should redirect your old blog to new blog.

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

awesome blog that...where did you have the time?
@Semiclassical And altitudes are a bit annoying to create, isn't it?

No fucking idea, the formulas were right
Plus the algebra

5:05 PM
you just have to make sure of the iff condition

Whatever, I gotta hang out with friends so cya l8r

well then .. bye :)

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

and then create a segment

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

5:06 PM
@UserX: xlr8, be back soon.

@UserX This is confusing.
@Nick never joins triangle discussions.

ok, this should be on target
did I do anything silly this time? :) @Sawarnik

No :)

neat

Probability says no as well.

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

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

gone for 2 min...

@Sawarnik what's wrong with the new address ?

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

@Sawarnik: @Nick would be in the Geometry Room if he wanted to partake in such discussions.
(@Nick is too stupid to do real math)

5:15 PM
@nick, did you see my 'fyi' in the number theory room?

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

@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 :-)

trying to find a link that's relevant

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

back in a bit

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

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? :)

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

@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

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

@r9m INMO :D :P
imp caught!
[I tried though little :(]

ah INMO !! right .. hehe .. but there be official solutions as well right ?

5:28 PM
@r9m: What's RMO and INMO?

^lol

@Nick regional mathematical olympiad

@r9m yeah, but I don't wanna look at it.
I want some original answers :D :D :D

puff ..

@r9m: That's weird, Heron's formula doesn't work for it? That doesn't make sense.

5:30 PM
it will work like butter on bread :)

... mhh?
oh!
ohk

Back again
Here's the linkI had in mind, I saw one of Prof. Garrett's grad students presenting on the stuff in section 0.3

@r9m Actually I didn't try enough..
..it was more of an excuse.

@Sawarnik growls ... try harder IMP :P

And I know enough spectral theory from QM to recognize what they're after

5:34 PM
@r9m ouch .. :|
:P :D

5:53 PM
@r9m I got an answer.
@r9m :D :P :D

@Sawarnik: How do you prove $27xyz\leq(x+x+y)(y+y+z)(z+z+x)$

By using AM-GM for 3 terms of RHS. :P
For example, $x+x+y\geq 3\sqrt[3]{x^2y}$.

... What is AM-GM ? How do I use it?

Why? I obviously know :D :P