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[oak swift]

Can i just get a little help on how to get started on this question

[loud sundial]

Combine the equations somehow

[oak swift]

That’s what I’ve been trying to do, I’ve been tryna get 2 equations with just 2 variables but idk

Hmm ill try some things

[loud sundial]

oh

um

Do all three at once

[oak swift]

How

[loud sundial]

Don’t try to do substitution or anything like that

Think back to ||standard techniques when you have symmetric systems||

[oak swift]

Symmetric systems?

[loud sundial]

When you have equations that are the same if you interchange the variables

Or that just display some high degree of similarity to each other tbh

[oak swift]

What techniques are used?

[loud sundial]

How would you solve, say,
\begin{align*}
a+b &= n_1 \
b+c &= n_2 \
a+c &= n_3
\end{align*}
For fixed $n_1, n_2, n_3$?

[soft zealot]

Civil Service Pigeon

[loud sundial]

Besides doing a mess of substitution

What’s the fast way to do this

Now think of how that applies to the original question you proposed

[oak swift]

But there’s variables with powers

[loud sundial]

ok

the idea is still the same

Maybe ||trivial inequality|| would’ve been a better hint in hindsight tbh (by virtue of wishful thinking)

But that might give away a lot

So open at your own risk

[oak swift]

Trivial inequality ? I’ve never heard of that

[loud sundial]

Square of a real number is non-negative

aka $x^2 \geq 0$ for all real $x$

[oak swift]

For this one the fastest way is just subtracting the first equation from second then doing simultaneous equations with third right?

[soft zealot]

Civil Service Pigeon

[loud sundial]

It’s quite obvious, hence the name lol

[loud sundial]

There’s a faster way

[oak swift]

Faster?

Really

[loud sundial]

yes

[oak swift]

It’s to do with inequalities?

[loud sundial]

What happened if you add all of these

btw ||that’s what I was trying to get to earlier - adding the three equations||

[oak swift]

2a + 2b + 2c?

[loud sundial]

mhm

Do you see the idea

[oak swift]

I don’t get it

[loud sundial]

nvm we can worry about that later

Just add the three equations in your original question

And try to proceed from there

[oak swift]

I tried to do that originally then thought maybe I could factorise

But I don’t think it worked

Or that I could see

[loud sundial]

Note that you have x^2-2x, y^2-4y, z^2-6z

Which are basically the same as x^2, y^2, z^2 after a shift

if you know about the equation of a sphere, that may help to provide some imitation

[oak swift]

I’ve never done 3d spheres

[loud sundial]

ok nvm don’t worry about that last part then

[oak swift]

Maybe complete the square

[loud sundial]

yes

[oak swift]

Looks like it’ll make the equation look nicer

[loud sundial]

mhm

[oak swift]

That’s nice it cancelled out with the 14 so it’s equal to 0

[loud sundial]

mhm

[oak swift]

Just left with (x-1)^2+(y-2)^2+(z-3)^2=0

So x=1,y=2,z=3

Is one solution

Looks like

But idk if thrrr are other solutions

There*

[loud sundial]

^^

When does equality occur here

[digital sable]

nice sol

[oak swift]

Wdym

Equality?

[loud sundial]

It says that $x^2$ is greater than \textit{or equal} to zero

[soft zealot]

Civil Service Pigeon

[loud sundial]

When is $x^2$ actually equal to zero

[soft zealot]

Civil Service Pigeon

[oak swift]

When x=1

[loud sundial]

🤨

typo?

[oak swift]

I’m lost

For my question ?

[oak swift]

Or this

[loud sundial]

Never mind

[swift tapir]

$1^2 = 1 \neq 0$

[soft zealot]

Mirror

[loud sundial]

When is $t^2$ equal to zero

[soft zealot]

Civil Service Pigeon

[oak swift]

I meant in my question it had x-1

[loud sundial]

I’m asking you this as a general question.

[oak swift]

Oh when t=0

[loud sundial]

yes

Now back to your completed square thing

You have three squares that add to zero

So what is each of those squares forced to be

[oak swift]

Yeah I got that solution

[oak swift]

^

[loud sundial]

My point is that you have three non-negative terms that add to zero

[oak swift]

Oh there can’t be other solution since that’s the only way

[loud sundial]

yes

[oak swift]

To make it 0

Oh shoot

Alr thanks

What was the circle have to do with it

Did*

Oh do u complete square in 3d for a sphere?

That makes sense since u do in 2d

[loud sundial]

When you have the expanded form of the equations of a sphere, it’s common to complete the square to pick off the center and radius

[oak swift]

Yeah

[loud sundial]

Similar to 2d

[oak swift]

Is radius r^3 in 3d

[digital sable]

no

[loud sundial]

$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

[soft zealot]

Civil Service Pigeon

[oak swift]

Why is it squared in 2d

[loud sundial]

Comes from the distance formula and the definition of a circle as a locus

[digital sable]

same reason as 2d

[loud sundial]

By definition, a circle with center (h,k) and radius r is the locus of all points that are a distance of exactly r from (h,k)

So $r=\sqrt{(x-h)^2+(y-k)^2}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

Now square both sides

Similar logic for 3d

Just use the 3d distance formula

[oak swift]

Oh right yeah

Are u in uni?

[loud sundial]

just started so technically yes but spiritually no lol

[digital sable]

good explanation I think

[oak swift]

Lmao is it just proofs for maths?

I heard from somewhere it was

At uni level

[loud sundial]

it’s a lot more proof focused sure

Assuming you had a more typical education

for me high school was actually harder than uni so far lol

But that seems to be a rare experience

[oak swift]

Highschool?

Where u from

[loud sundial]

I assume you’re British

So your secondary school

[oak swift]

Yep

[loud sundial]

The US

[oak swift]

Wait I’m in college

After secondary school

[loud sundial]

Oh true a levels are college

And secondary school is gcse

[oak swift]

Yep

[digital sable]

I think it is typically like this at the start, but if you are consistent then uni never becomes too difficult

being consistent is the hard part 😂

[oak swift]

Are u US?

[digital sable]

nah I’m from the uk

[loud sundial]

eh in high school I was used to 50% being good and I was tested on shit like this so

[oak swift]

Ah that’s cool what uni u go

[oak swift]

Highschool?

[loud sundial]

The age that you guys do a levels

[oak swift]

Wtf am I looking at

[digital sable]

???

what type of high school does ts

[oak swift]

Ong??

[loud sundial]

analyzing fractals

[digital sable]

I don’t want to dox myself on this server

[loud sundial]

in essence, repeated geometric patterns

[oak swift]

Ah okay

[digital sable]

was this like
extended essay or smth

[loud sundial]

I was a comp mather so we did stuff like this https://static1.squarespace.com/static/570450471d07c094a39efaed/t/673a4bada1036103a5a7897f/1731873709355/Power2024_Updated_1117.pdf

[oak swift]

What could u possible need this for??

[loud sundial]

this is one of the rare ones where you take it home and have a week

so not as time pressured

I think our submission was like

40 pages or smt?

we actually lost a few points cause there was something I was supposed to fix

but I woke up after the deadline lol

but it was chill

we still did well

[oak swift]

It was a project?

[loud sundial]

fun

[oak swift]

Looks pretty cool

[digital sable]

that is pretty cool though I think
just definitely not normal for hs education 😂

[oak swift]

Yeah

[loud sundial]

yeah the education standards in the us are mad low

but if you go to a specialised hs, then you can do a lot more fun things cause (i) culture and (ii) resources

[oak swift]

Are there imo competitors in this server?

[loud sundial]

yeah

[oak swift]

Who

[loud sundial]

they don't rlly pop out much

they're here tho

[oak swift]

Are there any British ones

[loud sundial]

idk

[oak swift]

It’d be cool to talk

[digital sable]

yes

[loud sundial]

eh you can always send a message out to the void

low chance of a response tho since again

[oak swift]

Void?

[loud sundial]

they don't appear much lol

[loud sundial]

send it out in the open and see what happens

[oak swift]

I’m in an Olympiad server maybe should ask there

[digital sable]

but for smth like this just join the British oly server

[oak swift]

Oh can u send me a link

[loud sundial]

yeah they're a lot more concentrated there

[digital sable]

I’m no longer there but you can just ask on mods

[oak swift]

Alr

[digital sable]

ask joe or someone

joe1

[oak swift]

I got another question which I’ve not looked at yet but I thinks pretty hard

[loud sundial]

[loud sundial]

eh this isn't too bad

[oak swift]

I thought it’d be hard since its from bmo round 2

[loud sundial]

(to spectators: no I am not just trvialising this by quoting the linear independence argument)

[oak swift]

I’ve not had a go at it yet

[oak swift]

No idea what that is

I mean on first thought im thinking to square both sides and get to some factoried couple

Then find multiples and do simultaneous equations

But idk if it would

Work

[loud sundial]

at least some this is relevant

I don't feel like splitting hairs lol

[oak swift]

Hmm im gonna try some stuff

[loud sundial]

@digital sable btw I actually gotta testsolve some questions for another comp (ahem my procrastination) so feel free to take over

[digital sable]

it’s 2am here haha so I’ll be heading soon, I was content with just watching

[loud sundial]

eh a little multitasking never hurt anyone lol

pretty sure I sketched out the soln so it should be chill

we'll see

[oak swift]

Im lost

I tried squaring everything, so that (root a + root b)^2=2009

Im kinda confused how 2009 isn’t a perfect square when the other side of the question is?

Am i bugging

[loud sundial]

I'd ||move one of the square roots to the other side|| first

[oak swift]

Why?

[loud sundial]

it makes 2009 have a more prominent role ... ?

yes I know this is a shit explanation

[oak swift]

Prominent role 😭???

[loud sundial]

more like if you do it like how you did, the square root that's left over doesn't rlly have anything to do with 2009

and 2009 is the thing that's the most restrictive here

something like that ig...?

idk #vibes lol

[oak swift]

Okay

Lmao

So then square it?

[loud sundial]

yeah

[oak swift]

A=2009-2root2009b +b

What so the root has to be a square?

Inside the root

For a to be an integer?

So if i set b=2009 then a=0

Which it works

Ig

Wait it needs to be positive

Nvm

Oh wait the 2009 in the root can be simplified

@loud sundial am i doing this right 😭

[loud sundial]

I like this idea of the root being an integer

[oak swift]

I just need to simplify the root 2009b

2009 has prime factors of 7^2 x41

B=41?

A=1476

@loud sundial why did u think to move the root to the other side, i would never have thought of that ngl

[loud sundial]

there's multiple solutions

[oak swift]

But like still idk, just gotta be some crazy game sense

[oak swift]

Multiple solutions?

What else

[loud sundial]

You arrived at the fact that $\sqrt{2009b}$ has to be an integer, right?

[soft zealot]

Civil Service Pigeon

[oak swift]

Yep

[silver dew]

How

How did you obtain that result I meant

[oak swift]

What

[loud sundial]

so what form must $b$ have

[soft zealot]

Civil Service Pigeon

[oak swift]

What do u mean what form

[loud sundial]

$\sqrt{a}=\sqrt{2009}-\sqrt{b}$, now square both sides and note all the other terms are integers

[soft zealot]

Civil Service Pigeon

[loud sundial]

yk when you say that "if $n$ is an even integer, then $n$ is of the form $2k$ for some integer $k$"

[soft zealot]

Civil Service Pigeon

[loud sundial]

same kinda idea here

[oak swift]

41k?

[loud sundial]

be more specific

[oak swift]

Like b has to be a multiple of 41

[silver dew]

I meant I got different thing from them . 2•2009•sqrt(b) is an integer. which doesn’t imply sqrt(2019b) being an integer at all

[oak swift]

Where u get that from

[silver dew]

This

[scarlet sequoia]

At best you have 2sqrt(2009b) is an integer

Not what you wrote

[loud sundial]

So could $b$ be $82$?

[scarlet sequoia]

But sqrt(2009b) is a rational

[soft zealot]

Civil Service Pigeon

[scarlet sequoia]

And since 2009b is an integer

[silver dew]

I got the question wrong, sorry

[scarlet sequoia]

No choice, 2009b is a perfect square

[oak swift]

Yeah

[loud sundial]

test it

[silver dew]

=sqrt(2009), thought =2009

[scarlet sequoia]

If you want a nudge to a simpler problem, what can you also notice about a?

[oak swift]

Me?

[scarlet sequoia]

Yes

We found b has to be a multiple of 41

What do you think about a

[oak swift]

Idk

[silver dew]

multiple of 41 by a perfect square you meant right

[scarlet sequoia]

Yeah more precisely

[scarlet sequoia]

b and a don't have that different of a role do they

sqrt(a) + sqrt(b) = sqrt(2009) is the same as sqrt(b) + sqrt(a) = sqrt(2009)

[oak swift]

I mean ig yeah you could have swapped a

[scarlet sequoia]

So if we found that something is true for b

[oak swift]

So a also a multiple

Of 41

?

[scarlet sequoia]

times a perfect square, yes

[oak swift]

Yeah

[scarlet sequoia]

$$a = 41m^2$$
$$b= 41n^2$$

[soft zealot]

Raphaelisius Maximus MMIII

[scarlet sequoia]

m and n are some positive integers

Can we find them?

[oak swift]

Wait Why is it times a perfect square

[scarlet sequoia]

We had found sqrt(2009b) is an integer

[oak swift]

Yeah

[scarlet sequoia]

So 2009b is a perfect square

[oak swift]

Yep

[scarlet sequoia]

b is also a multiple of 41

[oak swift]

Isn’t b just a multiple of 41

[scarlet sequoia]

If you combine both informations

You should get that b/41 is also a perfect square

[oak swift]

I don’t see why it has to be multipled against a perfect square

[scarlet sequoia]

Look

2009b = M²

So

[oak swift]

Oh yeah

[scarlet sequoia]

2009(41k) = M²

[oak swift]

Yep

[scarlet sequoia]

So k = (M/(7*41))^2

k = n²

So back to b = 41n²

[scarlet sequoia]

So, what do we do now to find m and n?

[oak swift]

Idk

[silver dew]

^

[oak swift]

?

[silver dew]

Plug in

[scarlet sequoia]

The first thing you know about a and b

Is sqrt(a) + sqrt(b) = ...

[oak swift]

Oh resub

[scarlet sequoia]

So now that you know what a and b look like

[oak swift]

Back into original

So root (41)n+ root (41)m =7 root 41

N+m=7

Yeah?

[scarlet sequoia]

Yes

[oak swift]

7 solutions?

[scarlet sequoia]

Really?

[leaden moon]

add all of the equations together

[oak swift]

Yes?

[silver dew]

No

[oak swift]

Done already

[scarlet sequoia]

Recall what a and b need to be

[oak swift]

Oh 6

[leaden moon]

then complete the squares

[oak swift]

Solutions

[silver dew]

Yes

[oak swift]

Cant be 0

[scarlet sequoia]

@leaden moon

[leaden moon]

why do no one pin this 😭

[oak swift]

How do u think to do that bruh

[oak swift]

Hard question man

[silver dew]

It’s their second question. Original first question solved long time ago

[oak swift]

Have u been here the whole time

[silver dew]

I hang around, when the channels I am interested in aren’t having a discussion

[leaden moon]

try squaring the equation, then distribute the LHS

[oak swift]

Done the question brotato

[leaden moon]

🙏

[oak swift]

I don’t think that’d work

[leaden moon]

how did you do it then

[oak swift]

Move root a to right hand side the square it

[leaden moon]

what root?

[oak swift]

Any

[leaden moon]

there are like 3 roots in here

[oak swift]

Oh root a or b

Sorry

[leaden moon]

kk

[oak swift]

Then square it

[crude cairn]

i only found 2 sets of solutions

[oak swift]

There’s 6

[leaden moon]

i wanted to find that $\sqrt{ab} \in \mathbb N$, that's what i meant

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

or specifically $ab$ is a square number

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

!nosols btw

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[crude cairn]

2√ab = 2009 - a - b is natural

√ab is a multiple of 1/2

simce a, b naturals, thus √ab must be natural too

[leaden moon]

yes

[leaden moon]

i don't think that explanation works

[crude cairn]

oh oops sorry im new to the server

[leaden moon]

something like this should work, idea is still ab is still a square number

[crude cairn]

yeah i lack rigour

[leaden moon]

do you do substitution or smth?

[leaden moon]

how do you imply that $a + b$ must be odd

[soft zealot]

1 divided by 0 equals Infinity

[crude cairn]

first, i concluded √ab is natural

then, 2√ab is even

since 2009 odd, subtract both sides by 2√ab

[leaden moon]

yeah

then you get

ah i see

[crude cairn]

i tried using odd even parity but could only arrive at 2 non-ordered sets of solutions

[lilac oar]

We can show that b must be of the form 41k^2

[final saddle]

@oak swift Has your question been resolved?

[crude cairn]

and from there WLOG can be applied

on a and b

[lilac oar]

Pretty much

[crude cairn]

oh and i found my mistake

the iff shortcut was invalid

[final saddle]

Channel closed

Closed due to timeout

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[icy lintel]

hello, is this system solvable?

[icy lintel]

a,b, and c are constants

[tranquil pine]

Umm I could try, have u started solving it tho, or tried?

[icy lintel]

yea i did

[tranquil pine]

Where'd u get so far

[icy lintel]

i even gave up and gave to AI and it solved it using some invertible matrix theorem

but im wondering if its doable without that

[final saddle]

@icy lintel Has your question been resolved?

[icy lintel]

i isolated x in the 3rd eq and subbed in the second equation then things got messy

im trying again

[high jolt]

is this the result of a lagrange multiplier system?

systems like that tend to have special tricks to solve.

[icy lintel]

yea

[icy lintel]

o okay, is it not possible through substitution?

[high jolt]

no, it is. try $x = ka^2, y = kb^2, dz = kc^2$.

[soft zealot]

Misanthrope

[high jolt]

you will get a neat closed form for (x,y,z,lambda)

but just as an aside, lagrange systems involve more than a typical nonlinear system of equations and usually have patterns such as homogeneity or symmetry you can exploit.

[icy lintel]

whered this come from?

[high jolt]

when you rearrange to solve x, y, z, you see that the variables are proportional to the constants by some constant factor.

[high jolt]

$$x = \frac{y + z}{2\lambda} a^2$$
$$y = \frac{x + z}{2\lambda} b^2$$
$$z = \frac{x + y}{2\lambda} c^2$$

[soft zealot]

Misanthrope

[high jolt]

although i am also slightly cheating because I'm aware the system is represented by a symmetric operator and the equations also imply (x, y, z) is an eigenvector, so I knew x was some constant multiple of a^2 and same for the other variables

also, you can solve for lambda in the three equations and you'll get a symmetric form $\frac{x}{a^2} = \frac{y}{b^2} = \frac{z}{c^2}$

[soft zealot]

Misanthrope

[icy lintel]

tried this but ended getting $\frac{a^2(y+z)}{x} = \frac{b^2(x+z)}{y} = \frac{c^2(x+y)}{z}$

[soft zealot]

moskov

[icy lintel]

i think im getting im getting somewhere by bruteforcing it tho

[final saddle]

@icy lintel Has your question been resolved?

[tranquil pine]

@icy lintel Sry I had to go do smth

[icy lintel]

its fine dude

[tranquil pine]

Any luck?

[icy lintel]

i ended up with a cubic for lambda

im double checking

[tranquil pine]

Hmm alr

[icy lintel]

got $8{\lambda}^3 - 2\lambda(a^2b^2 + b^2c^2 + c^2a^2) - 2a^2b^2c^2 = 0$

[soft zealot]

moskov

[tranquil pine]

But that's not finally tho, no?

Are u to solve for x,y,z

[icy lintel]

$y=\frac{1}{sqrt(\frac{O

oops

$y=\frac{1}{sqrt{\frac{(2\lambda+b^2)^2a^2}{(2\lambda+a^2)^2b^4}+\frac{1}{b^2}+\frac{(2\lambda+b^2)^2c^2}{(2\lambda+b^2)^2b^4}}}$

o

you get the idea

but yea i can get the others similarly

only issue is lambda

[soft zealot]

moskov

[fossil kiln]

$y=\frac{1}{\sqrt{\frac{(2\lambda+b^2)^2a^2}{(2\lambda+a^2)^2b^4}+\frac{1}{b^2}+\frac{(2\lambda+b^2)^2c^2}{(2\lambda+b^2)^2b^4}}}$

[icy lintel]

yea thanks

[tired walrus]

\sqrt{} not \sqrt() btw

[soft zealot]

ch3rry

[tranquil pine]

Noted

[tranquil pine]

So then, find x and z, and make lambda sof

[icy lintel]

x and z have very similar similar answers,
$x=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{(2\lambda+a^2)^2b^2}{(2\lambda+b^2)^2a^4}+\frac{(2\lambda+a^2)^2c^2}{(2\lambda+c^2)^2a^4}}} \quad z=\frac{1}{\sqrt{\frac{(2\lambda+c^2)^2a^2}{(2\lambda+a^2)^2c^4}+\frac{(2\lambda+c^2)^2b^2}{(2\lambda+b^2)^2c^4}+\frac{1}{c^2}}}}$

[icy lintel]

just need lambda

[soft zealot]

moskov
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tranquil pine]

Did u check or again?

[icy lintel]

i mean idk how to solve that cubic

[icy lintel]

<@&286206848099549185> how to do i solve

$8{\lambda}^3 - 2\lambda(a^2b^2 + b^2c^2 + c^2a^2) - 2a^2b^2c^2 = 0$

[soft zealot]

moskov

[hidden marlin]

what if you uhh.

[icy lintel]

a b and c are constants

[hidden marlin]

(2(lambda))³ - 2(lambda)((ab)² + (bc)² + (ca)²) = 2(abc)²

then uh

factor out 2(lambda)?

[icy lintel]

i did but then what?

[hidden marlin]

2(lambda)[(2(lambda))² - (ab)² - (bc)² - (ca)²]

ye I can't do it.

fuu

wait for someone else

[fossil kiln]

you could use the cubic formula but thts hedious

[icy lintel]

ye i havent learned that

[fossil kiln]

[icy lintel]

dude thats insane

[whole halo]

@icy lintel the x, y, z formulas you stated over here dont work

[icy lintel]

i think theres a typo

$y=\frac{1}{\sqrt{\frac{(2\lambda+b^2)^2a^2}{(2\lambda+a^2)^2b^4}+\frac{1}{b^2}+\frac{(2\lambda+b^2)^2c^2}{(2\lambda+c^2)^2b^4}}}$

[soft zealot]

moskov

[icy lintel]

why not the others?

[whole halo]

these are your three equations for x, y, z?

[icy lintel]

yea

[whole halo]

and this is your equation for lambda?

[icy lintel]

yes

[whole halo]

then the values your formula calculates do not fit the equations

[icy lintel]

lemme check again

[whole halo]

this formula is also not correct
it would lead to y + z = x + z = x + y, so x = y = z, and so a^2 = b^2 = c^2
so it would only be applicable when a^2 = b^2 = c^2 when it should work for any constant a, b, c

[whole halo]

this lambda equation is correct though, lambda needs to fit this equation

[icy lintel]

could you try $y = \frac{\frac{b^2}{2\lambda+b^2} }{\sqrt{ \frac{a^2}{(2\lambda+a^2)^2}+\frac{b^2}{(2\lambda+b^2)^2}+\frac{c^2}{(2\lambda+c^2)^2}}}$
$x =\frac{\frac{a^2}{2\lambda+a^2} }{ \sqrt{\frac{a^2}{(2\lambda+a^2)^2} + \frac{b^2}{(2\lambda+b^2)^2} + \frac{c^2}{(2\lambda+c^2)^2}}}$
$z=\frac{\frac{c^2}{2\lambda+c^2} }{ \sqrt{\frac{a^2}{(2\lambda+a^2)^2} + \frac{b^2}{(2\lambda+b^2)^2}+ \frac{c^2}{(2\lambda+c^2)^2}}}$

[soft zealot]

moskov

[silver dew]

Though that doesn’t look like the original question. Judging by the set up they were trying to find max/min of 2(xy+yz+zx) on that ellipsoid?

[icy lintel]

yeah 8(xy+yz+zx)

surface area of a prism

[icy lintel]

i think these are correct now, i just need to solve that cubic

[fossil kiln]

i gave u the formula

[icy lintel]

i gave it to AI and it made it look like $4\cos^3 \theta - 3\cos \theta = \cos(3\theta)$

[soft zealot]

moskov

[icy lintel]

so theres a better way technically but how would one think of this

[final saddle]

@icy lintel Has your question been resolved?

[swift void]

Have you tried using linear algebra for this?

$\mqty(-2\lambda & a^2 & a^2 \ b^2 & -2\lambda & b^2 \ c^2 & c^2 & -2\lambda)$

[soft zealot]

jelly v20

[vital python]

you need help

[hard river]

Matrix?

[vital python]

yea

.close

[icy lintel]

thanks yall

.close

[final saddle]

Channel closed

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[vital python]

we good?

[dense badge]

maybe some geometric intuition for the og system could be nice: the way i see it, the first 3 equations are planes in R^3 intersecting at least at the origin (0,0,0). Then the last constraint restricts the possible solutions to an ellipsoid (stretched sphere). Hence if the space of solution of the linear system (first 3 eq) is only the origin, your problem is unsolvable because (0,0,0) won’t be on the ellipsoid. If the solution space it’s 1D i.e. a line in R^3 through the origin, then you will have two sol. on the ellipsoid, and if it’s 2D you will have a plane passing by the origin with some sort of ellipse intersecting with the ellipsoïd.

so imo it all depends on the value of a,b,c and lambda that will dictate the dimension of the solution set of the 3 first linear equations

[final saddle]

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[icy lintel]

I was trying to maximize the total surface area of a rectangular prism inscribed in an ellipsoid

[icy lintel]

But I found the solution

It's just that lambda has to be solved via a depressed cubic

Which made things ugly

[final saddle]

@icy lintel Has your question been resolved?

[final saddle]

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[obtuse hedge]

<@&268886789983436800>

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[robust sequoia]

where did I make a mistake I'm crashing out

[tulip coyote]

You'll have to get another channel, this one was taken by a spambot

[proper dagger]

hiya, channel's occupied atm unfortunately (still has not returned to available)

grrr catbit snipe

[robust sequoia]

rip

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[prisma crag]

Task 2. (0–1)
The absolute value of a certain number, reduced by two, is the greatest one-digit prime number.

Which of the numbers listed below satisfies the described condition? Choose the correct answer from those given.

A. 7  B. −9  C. −14  D. 18

[scarlet sequoia]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[prisma crag]

The absolute value, so it must be plus?

Positive number

[scarlet sequoia]

If the original number is positive, its absolute value stays the same number.
If the original number is negative, it becomes positive through abs value

Like |12| = 12

But |-15| = 15

|.| is the notation for abs value just in case

[prisma crag]

Is greatest one digit prime number

so 9…

9*2 is 18

[scarlet sequoia]

9 is the greatest one digit prime number?

[prisma crag]

One digit

Oh

Prime

Numbers

Divides by themself and by two…

[scarlet sequoia]

And by 1*

[prisma crag]

Ya 1

It’s gonna be C. then

7*2 is 14

But it’s absolute number

So it can’t be -14

Right @scarlet sequoia

[scarlet sequoia]

In "reduced by 2" idk if they meant "take away 2" or "divide by 2"

But even so

You take the absolute value of the original number

14 would be the absolute value of your original number

[prisma crag]

two times so it’s division

[scarlet sequoia]

Not necessarily the original number itself

[prisma crag]

|-14| : 2 = 7?

[rapid sky]

Usually, reduced by means subtraction if it's reduced by a nonnegative integer, division if it's a nonnegative fraction, and subtraction of a percentage if it's a nonnegative percentage.

[prisma crag]

So it’s -14

[scarlet sequoia]

Yeah I would agree with that

So it's rather |x| - 2

[scarlet sequoia]

Thank you for the clarification

[rapid sky]

Ahh, I don't know about Polish.

[prisma crag]

It’s ok

[scarlet sequoia]

So this is correct

[prisma crag]

Yeah so it is -14?

[prisma crag]

Right this is so easy?

And this is final high school exam..

Well I know what prime number was but I thought it was natural number

Nvm

[rapid sky]

Yes, you can check by starting with -14, taking the absolute value, dividing by 2, and seeing if the result is the greatest one-digit prime number.

[prisma crag]

And 7 is highest one digit prime number

[rapid sky]

Right.

[prisma crag]

2, 3, 5, 7

Nvm it’s a task from 8th grade exam ! 😂😅

Not high school

That’s why it’s so easy

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[obtuse hedge]

It's this theorem again 😩

[warm python]

Kind of confused here

I don't get why M<1

[obtuse hedge]

Because M = max |f'(x)| and |f'(x)| < 1 (otherwise there could be another fixed point)?

[warm python]

because the supremum of the derivative values is 1

[warm python]

sure but we are only assured a supremum of the derivative values

not a maximum

[silver dew]

sup=max this case

Closed intervals are compact

[warm python]

sure, but then how do we have the limit is 0

[silver dew]

? M^n->0

[warm python]

yea, but if M=1

[signal vector]

Well if max |f'| < 1 then how can sup |f'| >= 1

M <1

[silver dew]

No

max can be achieved at some point y in that closed interval

[signal vector]

Not sure what you mean, we do have max |f'|<1?

[silver dew]

SoM=|f’(y)|<1

[warm python]

I'm asking how do w have this

[silver dew]

M<1

[signal vector]

Oh, well because of this I guess

[warm python]

I don't really follow

[signal vector]

Continuous function on compact set achieves maximum

[warm python]

sure

[signal vector]

|f'| is continuous function on compact set

[silver dew]

Yeah you agreed yourself twice, so what is the problem

[signal vector]

So |f'| achieve maximum at some point y

And |f'|<1 for all points, in particular y

[silver dew]

^

[signal vector]

So M <1

[warm python]

right, yes, but that doesn't mean the supremum is not 1

[silver dew]

sup=max<1

[signal vector]

If max of a set is known then sup = max

[warm python]

okay, so here the set has a max

[silver dew]

max(A)=sup(A) if former exists by definition

[signal vector]

That's.. what this is saying
#help-36 message

[warm python]

Noted

thanks

both you and @silver dew

.close

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[torn dome]

This question is from the graph theory chapter so I guess we have to use the handshake theorem

[torn dome]

But idk how to apply it here

[trail mango]

what does the handshake lemma say?

[torn dome]

sum of degrees = 2 * edges

[trail mango]

ok also without that

a handshake between a and b increases each of a’s and b’s handshake count by 1 (so 2 total)

and a sum of 2s is even

[torn dome]

can we do this using graphs

[trail mango]

sure

[trail mango]

every handshake corresponds to 1 edge

so, same idea as what i said

honestly

not even same idea, it’s simpler than that

[torn dome]

ok so every handshake makes one edge, what now

[trail mango]

so, sum of degrees is even (it is equal to 2 * # of edges)

and sum of degrees is what the problem is asking you to prove is even

interpreting the situation as a graph where each person is a vertex and they are adjacent if they shook hands

[torn dome]

degree of vertex = num of people a person has shaken hands with , right

[trail mango]

yes

[torn dome]

cool

ty

.close

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[burnt maple]

Is integral (1/dc) = 1/c ?

[warm python]

what

[round ore]

no, the term you are talking about is undefined

[warm python]

$\int \frac{1}{dx}$?

[soft zealot]

waimas

[blissful turtle]

i don't think that works.

[round ore]

yes i believe thats what he meant

[round ore]

yes it doesn't work

[burnt maple]

💀

[blissful turtle]

$\int \frac{dx}{x}$?

[soft zealot]

Blob

[blissful turtle]

u sure your problem sn't something like this?

[burnt maple]

No

It was a physics question and this was used

[final tangle]

can you show the original question

and/or where this is coming up

[burnt maple]

Wait a sec

[blissful turtle]

watch it be some kind of Feynman type shiii

[burnt maple]

This

[tired walrus]

how did "1/dc" come in

[burnt maple]

The solution

[blissful turtle]

that's.... wrong.

[tired walrus]

....

yeah what is going on there

[echo mason]

The reason why integral of 1/dc is equal to 1/C is because you are adding small capacitances in series.

This is a physics thing.

You need to provide context

They are not saying 1/dC is equal to C.

[blissful turtle]

are you sure d is not distance in this case?

WAIT.

d is distance

[echo mason]

It's just that you are adding dC capacitances in series.

[dreamy warren]

we’re not integrating $1/dc$ the correct relation is $\dfrac{1}{C} = \int \dfrac{dx}{\varepsilon_0 K(x), A'}$

[soft zealot]

anflo

[burnt maple]

Thanks

.close

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[lilac bison]

this will be the integral right? cause my professor has solved it with washer method

[dreamy warren]

oh

yes

[blissful turtle]

are you rotating around x =3 or y = 4?

[lilac bison]

the question is very vague about that

[blissful turtle]

YEAH

[lilac bison]

i assume y = 4

[blissful turtle]

that's what i thought was the problem.

[blissful turtle]

then the sqrt(bleh) - 3 should actually be 4 - sqrt(bleh)

[dreamy warren]

correct me if im wrong but the axis of rotation is x=3 not y=4

[trail mango]

anflo you’re so awesome

[blissful turtle]

no idea. problem doesn't say.

[lilac bison]

no idea

[blissful turtle]

what in the compliments is going on in a help room

[dreamy warren]

it’s the vertical red line in the diagram

[dreamy warren]

omg thank you girll 🫶

[lilac bison]

oh wait yeah its prolly x = 3

okay lets say question says about x =3, even in that case we wont use washer right?

[blissful turtle]

nope.

[lilac bison]

damn knew these notes werent reliable

[blissful turtle]

there's no hole for the washer.

in this case, it's disk

[lilac bison]

ye

thanks for the help y'all

.close

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[blissful turtle]

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[sharp wing]

where do i even start with this

[sharp wing]

imma be writing it down real quick but i can see ur messages

[obtuse hedge]

How annoying

[sharp wing]

yo cool when did u turn japanese

[obtuse hedge]

But remember how tangent line is defined with the derivative

[sharp wing]

yeah

[obtuse hedge]

The derivative here has 2 unknowns

Tangent and normal give you two equations

Now you can solve for those two unknowns

Do you know definition of normal line

[sharp wing]

y = mx + b?

[hasty mist]

coolemplud

[obtuse hedge]

For Christmas haha

[sharp wing]

oh lol

[obtuse hedge]

Thats a line

The normal line is perpendicular to the tangent line

So take the slope of the tangent, and make it the perpendicular slope

[sharp wing]

i remember how to do that

[obtuse hedge]

Let me know if you need any guidance on any of the steps I described

[robust mulch]

i think you can get away with not having one of those pieces of information

[sharp wing]

$$ slope is m = \frac{1}{23}$$

[obtuse hedge]

Oo a cheating

[soft zealot]

sherman

[hasty mist]

slopism

[sharp wing]

lol i dont know whats wrong with it

[obtuse hedge]

You put the text inside the $

[hasty mist]

slop is $m = frac1{23}$

[soft zealot]

Oléagineux Distillièr IX

[sharp wing]

ohhh makes sense

[obtuse hedge]

Slop

[sharp wing]

???? rainbow??

wait is perpendicular slope also negative?

[obtuse hedge]

Yes

But keep in mind those are at different points

[sharp wing]

slope is$$ m = -\frac{1}{23}$$

[soft zealot]

sherman

[obtuse hedge]

That tangent line is at x = -1 abd that normal line is at x = -4

[sharp wing]

so now what do with this perp slope

[obtuse hedge]

Nothing with the perp slope of the tangent line

[sharp wing]

whyd u make me do that then 💀

[obtuse hedge]

But the perp slope of the normal line gives you the slope of the tangent line at that point

Since normal is perpendicular to tangent

[sharp wing]

i see

[obtuse hedge]

Recalling slope of the tangent = derivative, you can set equal and solve

[robust mulch]

??? why are we using -1/23 slope

[obtuse hedge]

But yes now that I look at it there is a way to cheat

[obtuse hedge]

We're not

[robust mulch]

ok

[sharp wing]

yo lets get the cheat

[obtuse hedge]

You have these equations and x-values

So you can just plug them in

Combined with the point already given

You have 3 points on a parabola

That completely defines the parabola

Will that make it easy? Not sure

I'd do it the intended way xd

[obtuse hedge]

(If a,b,c are not nice numbers, this won't work)

[sharp wing]

i forgor the intended way 😭

[obtuse hedge]

Okay let's go through it again

Derivative is what we'll rely on

What's the derivative of our function

[sharp wing]

imma check my notes real quick i can still see ur messages

[sharp wing]

derivative?

[spark jasper]

can we get a,b,c with just 2 conditions?

[obtuse hedge]

That thing that they defined with the limits that we went through last time

[obtuse hedge]

Not sure

[sharp wing]

I found some old notes example five

We just have to reverse engineer that somehow

[obtuse hedge]

Good you have constant multiple rule

Is there a rule above it called the power rule

[spark jasper]

use f(x) = f'(x0)(x-x0)+f(x0) its better

[sharp wing]

i know how to do power rule, its easy

[obtuse hedge]

So you should be able to tell me the derivative of f(x)

Of y

They noted it y

[sharp wing]

which y tho

[obtuse hedge]

The original, the function we are trying to figure out

[sharp wing]

quick question, what is the diff between normal y and tangent y

[obtuse hedge]

y and x are just variables, the difference is which is the parent function and which is a tangent line to that function

https://tenor.com/view/secant-line-converging-to-tangent-line-animation-gif-18716967

Here the black is the original function

Green is a tangent line
Particularly, y = x^2 and at x = 2, there is a tangent line y = 4x - 2

[sharp wing]

but i mean there are two y equations here right

what is the difference between first y and tangent y equation

[obtuse hedge]

This

The tangent line y = 23x + 13 is tangent to the quadratic

It looks like that green line in the gif

Where the quadratic is the black function

[sharp wing]

mhm

[obtuse hedge]

Recall the definition of tangent is just that it touches the function at one point

[sharp wing]

yeah i recall

[obtuse hedge]

Otherwise they're just equations, and we use y = f(x) for equations

The meaning is in the words, which tells us who is tangent to whom

So the "difference" is in their relationship, but otherwise they are just graphs

[sharp wing]

i see

[obtuse hedge]

Similar to how a mother and child are just people, what differentiates them is the fact that one is a child to the other

And what brings them all together in this case is the derivative which we will use

[sharp wing]

u are saying y = - 1x/65 - 9234/65 are mother and child to y = 23x +13?

[obtuse hedge]

y = ax^2 + bx + c is our parent, y = 23x + 13 is tangent to it

[sharp wing]

what is this then: y = - 1x/65 - 9234/65 ?

[obtuse hedge]

y = -1x/65 - 9234/65 is normal to it

[sharp wing]

whats a normal

[obtuse hedge]

Meaning perpendicular to the tangent line at x = -4

[sharp wing]

oh ok

https://tenor.com/view/brain-brain-meme-big-brain-big-brain-meme-big-brain-time-gif-2441110471562975014

it all clicks now

[obtuse hedge]

Do you want to try it or work through it together

[sharp wing]

quick repost

latex didnt load in one sec

hmm imma think for a min

alright first from what i analyze is that $$ y = ax^2 + bx + c = f(3) = -30 $$

[soft zealot]

sherman

[obtuse hedge]

Sure, that's a choice you can make 👍 but is that correct?

Right now you have written $y = f(x) = f(3) = -30$

[sharp wing]

yeah its correct

[soft zealot]

クリスマスempire93

[obtuse hedge]

You never plugged in 3

[sharp wing]

i mean there aint no use pluggin 3 cuz it would still be y = 9a +3b +c

[obtuse hedge]

Right, 9a+3b+c = -30 is what we have based on the first thing they give us

What are you going to do next

[sharp wing]

to find slope of tangent line we had to take derivative of quadratic and plug in the point, but now we have slope and we are tryna find the quadratic

just dont know how to find quadratic tho

[obtuse hedge]

So what is the derivative

[sharp wing]

derivative of the quadratic?

[obtuse hedge]

Yes

[sharp wing]

hmm, is it y' = 12?

[obtuse hedge]

No, not quite

[sharp wing]

oh wait silly me

y = 2ax + b

[obtuse hedge]

Right

And you said that we plug in the point to get the slope of the tangent line

$f'(-1)$ = slope of the tangent line

[soft zealot]

クリスマスempire93

[obtuse hedge]

What is the slope of the tangent line we are given

[sharp wing]

23

[obtuse hedge]

Right, and what is the derivative when we plug in -1

[sharp wing]

-2a+b?

[obtuse hedge]

System of equations so far: $$\begin{cases}9a + 3b + c = -30 \ -2a + b = 23\end{cases}$$

[soft zealot]

クリスマスempire93

[obtuse hedge]

Do you see how I got there

[sharp wing]

yeah

[obtuse hedge]

Similar to a physics problem, we want to exhaust all the information we're given, even if it's a path that we know won't give us the full answer

Normally like you said we have a point and we find slope

Now we are given slope, but we have unknowns, so that's still useful to us

Okay, what do you plan to do next

[sharp wing]

can we solve those system of equations?

[obtuse hedge]

In general, you need at least as many equations as variables to solve for all of them

So we'll need 1 more equation to solve for 3 variables

Thankfully, they give us one more piece of information

[sharp wing]

oh the perpendicular line

[obtuse hedge]

Yeah!

What can we do with that 👀

[sharp wing]

is the y of the perpendicular line equal to the y of the quadratic?

[obtuse hedge]

Technically, yes, but I hope you don't plan to plug in a value to that ugly y with all those fractions when we can use the fact that the slope of a normal line is related to the slope of the tangent line 👀

[sharp wing]

alright u said normal is perpendicular to tangent right?

[obtuse hedge]

Yep

[sharp wing]

we can do y = 65x +b?

[obtuse hedge]

We don't actually need a line, all we end up using is slope, right

So yes the tangent line will be in that form

But what's more important is slope of the tangent line at x = -4 is 65

So now we can build it up like we did the other one

[sharp wing]

how?

[obtuse hedge]

Remember how we worked with the tangent line slope before

[sharp wing]

hmm

-2a + b = 65?

[obtuse hedge]

Close, but not quite

[sharp wing]

oh

-8a + b = 65

[obtuse hedge]

Perfect!

$$\begin{cases}9a + 3b + c = -30 \ -2a + b = 23 \ -8a + b = 65\end{cases}$$

[soft zealot]

クリスマスempire93

[sharp wing]

hmm we have a problem

actually

nvm

well that worked

def gotta practice that over and over again with software

[obtuse hedge]

I think I mentioned it before, but we see it here, one of the big important things of getting into calculus is remembering all the meanings of the derivative

In this case, we relied heavily on the fact that

The derivative at $x = a$ is the slope of the tangent line at $x = a$

[soft zealot]

クリスマスempire93

[obtuse hedge]

In other words, if $y = mx + b$ is tangent to $f(x)$ at $x = a$, then $f'(a) = m$

[soft zealot]

クリスマスempire93

[obtuse hedge]

This is why I favor renaming the parent function f(x)

[sharp wing]

it will always come together and click eventually

[obtuse hedge]

Ahmed did it here as well, very common

[sharp wing]

right now im in a hot mess

[obtuse hedge]

Yes, I just want to give it to you in writing so that whenever you come across it you have a definition to look at

[sharp wing]

cool thanks!

[final saddle]

@sharp wing Has your question been resolved?

[final saddle]

@sharp wing Has your question been resolved?

[sharp wing]

.close

[final saddle]

Channel closed

Closed by @sharp wing

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[boreal smelt]

I can't come up with a substitution to simplify this

[magic sparrow]

(3x-4)/(x-1)=3(x/(x-1))-4/(x-1)

[rancid zenith]

Well, if the question is well defined then you can get away with computing the integral of the identity.

[boreal smelt]

how?

[rancid zenith]

f=identity satisfies that equation.

[boreal smelt]

I don't get it

[rancid zenith]

Question:
Let f be a function satisfying [...] Then compute integral of f over 3/2 to 5/2

f(x)=x is a function satisfying [...], so let's just compute the integral of x over 3/2 to 5/2. (Not saying that this is correct or how it should be solved, but just explaining my point)

[boreal smelt]

ohh so you are assuming a function?

ohh

f(x)=x satifies it nicely

but how did you see through it?

[rancid zenith]

Perhaps austin has an "actual" method compared to this fake one.

[boreal smelt]

yea i would like to know about that too @magic sparrow

[rancid zenith]

It's just something you do habitually. Either way this "method" gives you 2 as an answer.

[boreal smelt]

ohh

[magic sparrow]

yes

so

you can solve it just with a substitution

[old quarry]

x-->x+1
then x-->1/x
then x-->x-1
u get f(x) + f(4-x) = 4 and use kings rule on the integral

[boreal smelt]

actually by some substitutions I got f(x+2) + f(2-x) = 4

[old quarry]

now do x-->x-2

cause when we use kings rule on integral we will get 4-x not 2-x

[boreal smelt]

trueeee

thank you so much

[old quarry]

np

[boreal smelt]

well i have another question

of differential eqn

[old quarry]

ohk send

[old quarry]

take 1-y^2 = t

[boreal smelt]

ok

lemme try

btw I would like to know your thought process until I solve it

how did you think of this sub

[boreal smelt]

i got $\frac{dt}{dx} = \frac{t^2}{x^3} - \frac{t}{x} + \frac{t^2}{x}$

[soft zealot]

Prathmesh

[old quarry]

hm

that dosent look very nice

let me try something

[boreal smelt]

ok

[full jungle]

i thought of y^2 = t

[dreamy warren]

because 1-y^2 appears and its derivative matches the ydy term

[old quarry]

nono wait

[dreamy warren]

after simplifying the DE becomes $\dfrac{dt}{dx} + \dfrac{1}{x}t = \left(\dfrac{1}{x^3} + \dfrac{1}{x}\right)t^2$

[soft zealot]

anflo

[dreamy warren]

which is a bernoulli equation

[boreal smelt]

ohhh

[dreamy warren]

using the standard substitution v = 1/t reduces it to a linear DE

[old quarry]

oh ye mb

[boreal smelt]

in bernoullis we use the substitution v=t^(1-n) right?

[old quarry]

yes

[dreamy warren]

n is 2 over here

[boreal smelt]

yes

[dreamy warren]

i think you can do it now

[boreal smelt]

i got $v^2\frac{dv}{dx}-xv=-(\frac{1+x^2}{x})$