help-49

[lyric charm]

was anything else given in the problem

it sounds like we may have insufficient data

is this another ex recto one

[cursive swan]

Angle ABC = (360 - 10x - 5x - 7x) = 360 - 22x
angle BAC = angle BCA = (180 - (360 - 22x))/2 = 11x - 90
Also, angle ABD = angle ADB = (180 - 10x)/2 = 90 - 5x
So angle DOA = (11x - 90) + (90 - 5x) = 6x
In triangle AOD, by the angle sum characteristic, 10x + 6x + (90 - 5x) = 180
11x = 90
x = 90/11

[cursive swan]

No. I saw this on Youtube.

Is it correct?

Because the person in the video got x = 10 deg

<@&286206848099549185>

[lyric charm]

video link?

[lyric charm]

im surprised that this kind of angle chasing even works at all here

lemme repro...

[lyric charm]

10x is not an angle in triangle AOD

[cursive swan]

Oh.

[cursive swan]

Lol.

[lyric charm]

this is what plain angle chasing would give us

i will ask once again that you link the video

and timestamp

[cursive swan]

You won't understand it.

https://www.youtube.com/watch?v=sAxfxg2Isys&ab_channel=VedantuOlympiadSchool

[lyric charm]

i can follow the diagram at least

[cursive swan]

Yeah.

Thanks for your help.

.close

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[grim vector]

Ah now the link has changed <@&268886789983436800>

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[molten bay]

Sin|x| differentiability

[molten bay]

d/dx(sin|x|) would be cos|x|.x/|x|

So at x=0 it will be not differentiable

[gaunt nimbus]

Yes

[runic hamlet]

you cant compute the derivative like that at x=0

[molten bay]

??

[runic hamlet]

you have to take the limit

[molten bay]

Then should I have to take limit?

[runic hamlet]

the chain rule doesnt work if the inner function isnt differentiable

[molten bay]

Definition of derivative

[sage helm]

take the limit of that function you got

[sage helm]

this

at x = 0

[molten bay]

F'(0)=f(0+h)-f(0)/h

sinh-0/h

=Sinh/h

and f'(0-h) will be different

[astral terrace]

just plot the graph thats the easiest method

but if you wanna do it mathematically just use the first princ

[astral terrace]

it will be sin|h| not sin(h)

limit of sinh/h exists

sin|h|/h doesn’t

Cz the numerator is positive for 0+h aswell as for 0-h

but denominator changes sign

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[dusty portal]

I can’t integrate $\int_0^1\int_0^1\sqrt{x^2+y^2}dydx$ using a polar subsitution because $[0, 1]\times [0, 1]$ is a square, right?

[grand pond]

;(

[blissful pier]

[\int_0^1\int_0^1\sqrt{x^2+y^2}dydx=\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sec(\theta)}r^2drd\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\csc(\theta)}r^2drd\theta]

[grand pond]

PajamaMamaLlama

[blissful pier]

https://math.stackexchange.com/a/1248640/1049995

you can but to say the substitution is obvious is silly lmao

[dusty portal]

Ok

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[regal barn]

it is differentiable at x=0

[gaunt nimbus]

Solve for?

[tired osprey]

Not differentiable

[humble wolf]

!xy

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@regal barn Has your question been resolved?

[regal barn]

this is the original problem

[humble wolf]

on the right of ur img, it's a comma, indicating that the sentence doesn't end there

[regal barn]

alpha>0

alpha>1

alpha>=0

[humble wolf]

!xy

[midnight plank]

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[regal barn]

alpha>=1

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[chilly adder]

How do I read the left side of this equation? i mean, how do i start writing the terms?

[chilly adder]

I know this is the Vieta's formula for sum of products of r roots for a general polynomial, just curious on how this notation works

[verbal pumice]

Set i_1 = 1 first

then i_2 has to be 2

i_3 is 3

on and on

then incremenet i_r until it hits the max

then increment i_(r-1) and incremenet i_r again

[chilly adder]

ohhhh

[verbal pumice]

try matching the formula up with a small value of n like 3 or 4

[chilly adder]

okay so i choose i_1 to be anything between 1 and n and keep incrementing?

[verbal pumice]

you'll see what I mean

[verbal pumice]

well technically thats fine

but to maintain order you generally should start from 1

[chilly adder]

start with i_1=1 yes

[verbal pumice]

yeah

[chilly adder]

wait so

Let $n=3$ and $r=2$, then

$$\sum\limits_{1\leq i_1<i_2\leq 3}\left(\prod\limits_{j=1}^2\alpha_{i_j}\right)$$

[grand pond]

@chilly adder

[chilly adder]

so if i start with i_1=1, then i_2=2

then for the first term the product becomes $\alpha_{i_1}\alpha_{i_2}=\alpha_1\alpha_2$

then i set i_1=2?

[grand pond]

@chilly adder

[chilly adder]

hhh

if i do that i_2=3
so product is now alpha_2alpha_3

thats it?

there should be an alpha_1alpha_3 term too

[midnight plank]

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[chilly adder]

@verbal pumice my bad if you're busy, i have this last question

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[zenith prism]

Yo stupid question but is
Y/-2
The same as Y × -1/2

[zenith prism]

Okay thanks

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[sage crypt]

Ik this a very easy question

But i need someone to use scaler vectors way strictly
Please dont help if your gonna use coordinate geometry

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[sage crypt]

No

[sage crypt]

.reopen

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[fallow scarab]

did you draw a picture

[shadow schooner]

yeah try drawing a picture

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[merry sail]

Can anyone explain how my teacher went from the first line to second? I can send the whole context if needed

[tough shale]

Do you know the expansion for (a+b)²?

[merry sail]

yes

[polar star]

im sure you can understand how the second line implies the first line. the trickier part is how the first line implies the second

have you learnt "completing the square"

[merry sail]

yea b/2 squared

[rustic tartan]

$t^2 - 2 \times 2t + 2^2$

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[polar star]

they are using that method to convert it to vertex form

i can link a video that shows how to convert from the standard form to vertex form

https://www.youtube.com/watch?v=pf9LkX8hpTQ

[sage crypt]

Wha can u pls solve it?

[polar star]

https://www.youtube.com/watch?v=MnZXvkYpHOM

[rustic tartan]

Also: $(a-b)^2 = a^2 - 2 \times ab + b^2$

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[polar star]

make another channel

[merry sail]

alright thanks i’ll watch them

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[sage crypt]

Alr

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[dawn pine]

is there any way to solve these types of questions mentally?

37^12 / 23 has a remainder of ....
35^10 / 19 has a remainder of ...

[lethal path]

yes, using modular arithmetic

$37 \equiv 14 \equiv -9 \pmod {23}$

[grand pond]

south

[lethal path]

then repeated squaring, so (-9)^12 = 81^6 mod 23 = (-11)^6 mod 23
= 121^3 mod 23 = 6^3 mod 23 = 36 * 6 mod 23
= 13 * 6 mod 23 = 78 mod 23
= 9 mod 23

that's the closest you can get

for other similar questions you could also use Fermat's little theorem

[dawn pine]

alr i see what u mean

thanks

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[astral terrace]

How could I approach this problem?

[viral dagger]

Q seems the easiest, notice how its just the sum of the squares of P then -10

[astral terrace]

Yeah

[viral dagger]

then for S its kinda the same thing

[astral terrace]

the fourth power comes in

and it is a fifth degree polynomial so it wont be that easy

not even to find the sum of squares ig

how will you solve that?

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[astral terrace]

<@&286206848099549185>

[karmic snow]

What does "zeros alpha1, alpha2, etc." means? @astral terrace

If you know the alpha[] value then this would be trivial no?

[viral dagger]

yes but you cant really find the value for qlpha easily

[viral dagger]

ol i have no idea weather its correct or not but it may be equivalent to
(a^2+b^2+c^2+d^2+e^2)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)^2+4(abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde)(a+b+c+d+e)-4(abcd+abce+abde+acde+bcde)

[karmic snow]

Ok but what are we suppose to use P(x) for?

[viral dagger]

for all the alphas

[viral dagger]

i cant check cause im too lazy to do it manually, chatgpt sucks, wolfram alpha breaks

[astral terrace]

zeroes are roots

But i assume you already know that

man i have an examination rn so i gotta leave now

[karmic snow]

I'm an idiot

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[vague crag]

for this, do i need to show that t(0)=0? if so how?

[lyric charm]

uppercase T not lowercase t, but yes you do need to use that T(0) = 0.

it is not hard if you consider that 0 = 0 + 0

and the defn of linearity

[uneven aspen]

Do you have any information about T? If T is one-to-one, then you can use that.

[lyric charm]

this start is a bit sus though tbh

[vague crag]

oh i see

[vague crag]

nope

[uneven aspen]

Why do you think this is true?

[vague crag]

T is linear and from V to W

[uneven aspen]

Try to find a counter example.

[vague crag]

so it's not always true?

[lyric charm]

i mean i think OP needs to be clear on what exactly his assertions and goals are for this proof

cause right now the beginning's kinda sloppy and the logic is, for lack of a better word, clouded

[uneven aspen]

I claim you can always find a linear transformation that will map any independent set to a dependent one.

@vague crag , try to think of a simple example of a linear transformation T where you end up with a depdendent set.

[lyric charm]

this is a "prove or disprove" question btw right @vague crag

[vague crag]

yeah

[vague crag]

hmm

[lyric charm]

that's a context that you have to give

always without exception

also yes it pays to try and think of the simplest, dumbest linear map that could possibly exist.

why the unamused react?

[vague crag]

wait lemme think

so basically it fails if ker(T) doesn't equal 0 or not injective. So maybe a T where it maps all vectors to the 0 vector?

[uneven aspen]

"So maybe a T where it maps all vectors to the 0 vector?" You tell me, is the set with just the zero vector a dependent set?

[vague crag]

ye

[uneven aspen]

So, in particular, if you apply T to a independent set you'd get a dependent set.

[vague crag]

ya

ty

[uneven aspen]

np

You should repeat the problem assuming T is injective.

[vague crag]

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[spare dawn]

I need help solving this ITF question

tried converting to tan

but in the end this comes

[prime hornet]

wait, this identity doesn't seem true?

[spare dawn]

this is true

[round parcel]

Yes, the inputs for the second arcsine go from 1 - 1 to 1 + 1.

[viral dagger]

sus

[spare dawn]

for further help i used the solution but it confused me even more

[round parcel]

It's false for x = 1, for example.

[polar star]

perhaps the question was to show that there exists only one real solution

[round parcel]

,calc asin(1) + asin(1 - 1)

[grand pond]

Result:

1.5707963267949

[round parcel]

,calc acos(1)

[grand pond]

Result:

0

[spare dawn]

wait guys let me share the solution

maybe you can interpret better

[polar star]

okay

[round parcel]

It gets two solutions in the end, so it's not an identity.

[viral dagger]

thats sesrching for solutions

[spare dawn]

it's not

it's a prove that it equal cos-1 x in this specific on

[round parcel]

They only prove that two solutions work.

They don't prove it's an identity.

[spare dawn]

what happened in 4th step?

[viral dagger]

sin(sin^-1(x))=x

[round parcel]

sin(pi/2 - x) = cos(x) for the right side.

[spare dawn]

aah got it

thanks guys

[round parcel]

No problem.

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[twilit lodge]

anyone know why part b might be wrong

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[twilit lodge]

<@&286206848099549185>

[karmic snow]

Isn't there a formula for p?

[twilit lodge]

Umm i honestly just plug everything in ti84 for these problems

But let me check I don’t think you can solve without technology

Yea no I looked it up and it’s pretty much only done using technology and I seem to have followed the steps correctly

It’s probably just an error

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[inland patio]

I'm reading a proof of the fact that a set $\mathcal C$ of commuting linear operators are simultaneously diagonalizable (the other direction is also true and a bit easier). They proceed by induction on the dimension of the whole space. When the whole space has dimension $1$, they say that every linear operator is diagonalizable by the standard basis. This phrasing confuses me, as the whole space is just an arbitrary vector space. It does not necessarily have "a standard basis", or?

[grand pond]

psie

[azure oracle]

ummm... why wouldnt theree be a standard basis?\

[inland patio]

.close

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[twilit field]

I was thinking of using induction

But that doesn't seem to work

[dawn dagger]

you can prob prove it also direct but whats the step you stuck

[twilit field]

To show it's increasing, we perform induction.
\
$a_{n} - a_{n-1} ≥0$
\
so

Now inducting $a_{n+1} - a_n = 3 - \frac{1}{a_n} -a_n$
\

[gaunt jetty]

Showing that it is increasing and a_n < 3 for all n should be two things you handle separately

[grand pond]

What a wonderful world !

[twilit field]

I'm doing that

[gaunt jetty]

Well, it's more logical to perform induction on the a_n < 3 for all n case

[twilit field]

let $a_n<3$

[fresh sparrow]

no, strict

<

[grand pond]

What a wonderful world !

[fresh sparrow]

aw it's a heart <3

[twilit field]

We then have $a_{n+1} = 3 -\frac{1}{a_n}$. but $\frac{1}{a_n}>\frac{1}{3}$

[grand pond]

What a wonderful world !

[fresh sparrow]

indeed

[twilit field]

so $\frac{-1}{a_n}<\frac{-1}{3}$

[grand pond]

What a wonderful world !

[twilit field]

Adding 3 to both sides

$3 -\frac{1}{a_n} < \frac{8}{3}<3$

[vocal inlet]

pretty sure it's 8/3

but still works

[grand pond]

What a wonderful world !

[vocal inlet]

yup

[twilit field]

Now to prove it's increasing

for that I have to show 0 isn't possible

Not too sure

😔

[vocal inlet]

just... induction

suppose a_n ≥ a_n-1 for some n and prove a_n+1 ≥ a_n

[twilit field]

I did try that earlier

[twilit field]

here

[dawn dagger]

Apply definition of a_n

[vocal inlet]

yeah

[twilit field]

I have...

[vocal inlet]

you'll get 3 - 1/an -3 + 1/a_n-1

[twilit field]

yes, so 1/a_{n-1} -1/a-n

[vocal inlet]

yeah

[twilit field]

but I already know that that's positive ( by inductive hypothesis), I suppose?

[vocal inlet]

yes

[twilit field]

ohh

thanks

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[last slate]

why -1.175?

shouldnt it be positive

cuz the z sign is > and the value next to z is negative

.close

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[zenith mirage]

can someone help me qith this question? like the formulas to use and making sure i do i correctly (but also letting me do the qork instead of doing it for me type of thing?

[zenith mirage]

please ping me as qell

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[zenith mirage]

i quiteliterally asked for the formulas so that eould indicate i dont knoe the formula to even start?

[polar star]

https://www.youtube.com/watch?v=H6gXSV8OHW8

https://www.youtube.com/watch?v=Ln97Hd7AiDc

the first one is simple and compounded yearly

the second one is continuous compounding

[zenith mirage]

thank you

couldyou type the forumlas outso i dont go through the video looking for it?

i feel itd be easier to do that than

looking for videos but

actually nevermind

.close

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[polar star]

$SI = P \times r \times t
A = P(1 + r)^t
A = Pe^{rt}$

wawit

$
[
SI = P \times r \times t
]
[
A = P(1 + r)^t
]
[
A = Pe^{rt}
]$

[grand pond]

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

[zenith mirage]

I have close the channel

[polar star]

oh

[zenith mirage]

Please leave me alone.

[polar star]

um okay

[midnight plank]

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[small zenith]

i am having trouble getting my head around this

[small zenith]

how did $csc^2 x$ disappear

[grand pond]

Lavandula

[small zenith]

i am not familiar with cscx cotx integration

[slender walrus]

do you know the derivative of cot(x)?

[small zenith]

not rly

[gaunt jetty]

They did a $u$-substitution by taking $u = \6\cot x -1$ before applying integration by parts. Then you can refer to what ramonov said

[grand pond]

Aero ݁˖ ❀ ⋆｡˚

[small zenith]

i only know tan(x) derivative is sec^2 x tho

[gaunt jetty]

It's not particularly hard to derive. Do you know the quotient rule for derivatives by any chance?

[small zenith]

yes

[gaunt jetty]

So, knowing that [
y = \6\cot x = \4{\6\cos x}{\6\sin x},
]
use the quotient rule to find $dy/dx$

[grand pond]

Aero ݁˖ ❀ ⋆｡˚

[small zenith]

• csc^2 (x)

Why did a bullet pt replace that

[gaunt jetty]

you can put a \ before the - to disable that

Anyways you're right

That's it

Do you see how and why it cancelled now then?

[small zenith]

ohh

r there any others

i only know derivatives of sinx, cosx and tanx

regarding the ones angle-related

[gaunt jetty]

I mean I wouldn't say you need to memorise them, as they're clearly easy to derive if you know the derivatives of cos and sin

But it's generally good to know like, cos, sin, tan, sec, csc, cot, arcsin, arccos, arctan

[small zenith]

would u happen to have like a table of these?

lemme try google

[gaunt jetty]

Yeah just Google it

[small zenith]

alright got it thank you

.close

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[twilit field]

So first I use the root test to find that $0≤ \sqrt[n]{\abs{c_n}}< \frac{1}{4}$

Now checking $c_n (-4)^n$ for absolute convergence

$

[grand pond]

What a wonderful world !

[twilit field]

$\sqrt[n]{ \abs{ 4^n c_n} } = 4 \sqrt[n] {c_n}$

[grand pond]

What a wonderful world !

What a wonderful world !

[twilit field]

Which is between 0 and 1/4 so this converges

now for $c_n (-2)^n$

[grand pond]

What a wonderful world !

[twilit field]

this is betewen 0 and 1/2 with the root test , so it converges

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

Does this work

[visual tiger]

uh

first of all

how did you find that series of c_n(-4)^n is absolutely convergent?

[visual tiger]

and also why is this true

[midnight plank]

@twilit field Has your question been resolved?

[tribal temple]

(and ideally, that isn't the way to go about it either!)

[quiet hinge]

So what converges?

[midnight plank]

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[mint moth]

can someone explain what happened at this point? i understand everything else

[mint moth]

especially in the denominator, why did sin get squared?

[thorny plinth]

its not supposed to be squared

it was a mistake

[mint moth]

so whats the real solution

cuz i get

how 1 - sin^2x turned into cos^2x

but even so, how does sinxcosx turn into just sin x

and the cos^2x goes where?

[thorny plinth]

just remove the square in that one step

[thorny plinth]

they cancelled the common factor from the numerator and denominator

[mint moth]

so cos^2x gets cancelled out once

by the cosx in the denominator

leaving just

cosx over sinx?

[thorny plinth]

yes

[mint moth]

okay thank u

[thorny plinth]

[mint moth]

.close

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[cursive swan]

(Source: IOQM 2024)

[dreamy lichen]

I'd just try to label the sides and apply similarity and perhaps pythagoras

[cursive swan]

It is known that hypotenuse >= the corresponding height / 2

AC >= 24

Because we can draw the right angled triangle in semi circle of radius AC/2 and then get that the maximum value of h as the sides vary is R = AC/2

[lyric charm]

i think you can make a rather elegant solution to this if you consider that triangles ADB and BDC are similar and in so doing get a relationship between AD and DC

[cursive swan]

AD * DC = 144

AD + DC = integer
min(AD + DC) = 24
AC = 24 is not possible due to Fermat's last theorem.
AC = 25 is possible.
Done?

[exotic stratus]

In this case wouldn't the triangles just be congruent if they were similar?

[lyric charm]

AC = 24 is not possible due to Fermat's last theorem.
you're joking right

what's fermat-wiles gotta do w/ this

[cursive swan]

If AD = DC, then AB = BC.

Nvm

Fermat's last theorem doesn't work for n = 2

[lyric charm]

you could have just said "they can't both be 12 bc then the perimeter would be 24 + 24sqrt(2)" or something

[cursive swan]

What?

How?

[dreamy lichen]

i dont think they'd necessarily be congruent

[cursive swan]

Oh. I didn't know! Thanks!

[lyric charm]

was that sarcastic or genuine?

[cursive swan]

genuine

What to do after AD * BC = 144?

And min(AD + BC) = 24?

[dreamy lichen]

This essentially works, doesnt it? If you change the reasoning from fermat's last theorem to what Ann said

[exotic stratus]

Oh yea there is 1 counter case which is when $\angle A = \angle DBC$, otherwise they're congruent I think

[grand pond]

@exotic stratus

[dreamy lichen]

they are almost never congruent

[cursive swan]

Lol.

[exotic stratus]

If $\angle A = \angle C$ then they're congruent by ASA rule

[grand pond]

@exotic stratus

[cursive swan]

Mhm. We're lucky.

[cursive swan]

What do you mean by this though?

[cursive swan]

And this

[dreamy lichen]

you can have sth like this

[exotic stratus]

I was just thinking since $\bar{BD}$ is common between $\triangle ABD$ and $\triangle BDC$, $\angle D = 90^{\circ}$ (both of them) if $\triangle ADB \sim \triangle BDC$ it may imply $\triangle ADB \cong \triangle BDC$

[grand pond]

@exotic stratus

[cursive swan]

May?

[dreamy lichen]

BD is common, but it corresponds to different kind of side in each case

[cursive swan]

Yes

[exotic stratus]

It was a thought

[exotic stratus]

Yea that's the conclusion I ended up with

[dreamy lichen]

e.g. here BD is the longer leg of smaller triangle and smaller leg of larger triangle

[exotic stratus]

I found the contradiction lol no worries

[cursive swan]

Okay.

Thanks y'all.

.clsoe

.close

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[cosmic epoch]

im studying SVD right now, and noticed that A*A^T has the same eigenvalues as A^T*A--aside from the multiplicities of their zeroes and typically you choose to find the eigenvalues of A^{T}A, but that matrix may be larger than the former.
so id like to pick the smaller square matrix of the two to speed up computations since i have an exam on this topic.

im just curious whether this is valid in all cases. im aware of the proof for eigen(AA^T) = eigen(A^TA). i just need to ask if theres any picky considerations i should be aware of with using this strategy

[midnight plank]

@cosmic epoch Has your question been resolved?

[brittle grotto]

hmmm

if $\lambda$ is an eigenvalue of $A^TA$, then $A^TA=\lambda x$

[grand pond]

00100000

[brittle grotto]

then, $AA^TAx=A(\lambda x)$

[grand pond]

00100000

[brittle grotto]

so $AA^T(Ax)=\lambda (Ax)$ for eigenvector $x\neq0$ of $\lambda$

so, you're right

[grand pond]

00100000

[brittle grotto]

which actually kind of surprised me. definitely would mess around with multiplicities tho, since the matrices aren't necessarily of the same size

[midnight plank]

@cosmic epoch Has your question been resolved?

[cosmic epoch]

the multiplicities can be found easily right? via knowing the rank
if eigen(AA^T) = l1, l2, l3, .... where AA^T is (mxm)
then eigen(A^TA) = l1, l2, l3, ..., 0, 0, 0, ... until we hit the rank of the matrix

[paper inlet]

use integration

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[molten bay]

Lol

When i posted it was empty

[midnight plank]

@cosmic epoch Has your question been resolved?

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[tidal turret]

can I get some help

[dawn dagger]

1. looks like you will have to solve a quadratic in k

[tidal turret]

wdym?¿

[dusty portal]

Okay, so can you extend the distance formula to 3D? Also, can you set up the formula?

[tidal turret]

oh this is for 2D, I am working in 3D

[tidal turret]

how?

[dusty portal]

Try to think of a rectangular prism.

[dusty portal]

How would you find the long diagonal connecting the opposite points?

[tidal turret]

pythagoras ig, idk

[dusty portal]

Perfect. So try it out. And if you feel more comfortable talking in Spanish, that’s alright.

[spice mason]

[dawn dagger]

time will reveal

[tidal turret]

I prefer english so the ppl can understand us hehexd

[dusty portal]

Alright then, just apply the formula.

[dawn dagger]

2. looks like you have to construct a right triangle at B

[tidal turret]

1. Determine all values of ( k \in \mathbb{R} ) such that
   [ d(A, B) = 5, \text{ where } A = (-1, 0, k) \text{ and } B = (3, k, 3).]

2. Given
   [ A = (1, -2, 3), \quad B = (2, 3, 5) \text{ and } P = (4, a, -1), \quad \text{find } a \in \mathbb{R} \text{ such that the triangle } ABP \text{ is right-angled at } B.]

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[dawn dagger]

so AB and BP must be orthogonal if yk

[dusty portal]

||Dot product||

[tidal turret]

interesting, let me finish 1) first

[dawn dagger]

hopefuly it wont finish you

[tidal turret]

lmfao

[dawn dagger]

grammar finished me

[dusty portal]

Woah woah

[dawn dagger]

apparently i understand spanish

[tidal turret]

,, A = (x_1, y_1, z_1) \ B = (x_2, y_2, z_2) \ d(A,B) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

[dusty portal]

I also have a question about translation but we can do this after

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[dawn dagger]

the values are missing

[dawn dagger]

that's literally here

i see, you took care of k already

no you skipped to 2.

usually you start with 1

i see

[tidal turret]

A = (-1,0,k)
B = (3,k,3)

,, A = (x_1, y_1, z_1) \ B = (x_2, y_2, z_2) \ d(A,B) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\ A = (-1, 0, k) \ B = (3, k, 3) \ d(A,B) = \sqrt{(3 + 1 )^2 + (k - 0)^2 + (3 - k)^2}

[dawn dagger]

and d is also equal to something

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,w sqrt(16 + k^2 + (3-k)^2)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

what?

[dawn dagger]

you have to set it equal to 5

square both sides

[tidal turret]

oh ok

[dawn dagger]

and solve the quadratic in k dear

[tidal turret]

,, 5 = \sqrt{(3 + 1 )^2 + (k - 0)^2 + (3 - k)^2}

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

(sqrt(x))^2 = |x|?

no

its just x

[dawn dagger]

no

[tidal turret]

ok

[dawn dagger]

the inside is always nonnegative here

[tidal turret]

25 = 16 + k^2 + (3-k)^2

,calc 25-16

[grand pond]

Result:

9

[dawn dagger]

time revealed

[tidal turret]

0 = k^2 + (3-k)^2 - 9

,w expand (3-k)^2

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

0 = k^2 + (k^2 + 6k + 9) - 9

0 = 2k^2 + 6k

zero product property

0 = k(2k + 6)

1. k = 0
2. 2k = -6 ==> k = -3

k = {0, -3}

[dawn dagger]

mistake

[dawn dagger]

-6k

[tidal turret]

my god

[dawn dagger]

only thing that changes is -3 turns to 3

just edit your msgs and you're good

[tidal turret]

0 = k^2 + (k^2 - 6k + 9) - 9
0 = 2k^2 - 6k
0 = k(2k - 6)

1. k = 0
2. 2k = 6 ==> k = 3

[dawn dagger]

there you go

[tidal turret]

k = {0,3}

what about 2)

[dawn dagger]

what about it

[tidal turret]

AB . P = 0

or what?

[dawn dagger]

AB.BP = 0

something along that yea

[tidal turret]

ok

AB = B - A = (2,3,5) - (1,-2,3) = (1,5,2)

BP = P - B = (4,a,-1) - (2,3,5) = (2, a - 3, -6)

,w (1,5,2).(2,a-3,-6) = 0

[grand pond]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[dawn dagger]

embodiment of lazy 🦥

[tidal turret]

I dont really understand the key idea

AB.BP = 0

[dawn dagger]

this is why you should draw a picture

draw the 3 points A,B and P

you want at B a right angle

well B can only be connected to A and P

so you get two segements

that need to be perpendicular at B

[tidal turret]

A = (1,-2,3)
B = (2,3,5)
P = (4,a,-1)

the right angle is at B

[dawn dagger]

it does not need to be perfect but yea

[tidal turret]

wdym?

[dawn dagger]

the drawing

now imagine you connect red with blue and blue with green

you then get vectors in space floating around, but now you want them to be perpendicular, so they make a right triangle

thats the idea behind

[tidal turret]

wait but

P = (4,a,-1) is a line

P € L

[dawn dagger]

nah you are making stuff up now

[tidal turret]

L : X = a(0,1,0) + (4,0,-1)

[dawn dagger]

a is not a parameter but a constant

[tidal turret]

oh ok

[dawn dagger]

it's an arbitrary point you move around in space

a=5 happens to be the value where you get the right triangle

[tidal turret]

ok

I appreciate the help provided as always

sorry if I say garbage

.solved

[midnight plank]

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[marble trellis]

Is there a way to find the answer by jus looking at it

Knowing what the origional graph looks like

Nvm forget thay

Can i get help wid dis

[midnight plank]

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[twilit field]

Trying to find $\pdv{z}{s}$and $\pdv{x}{t}$

[twilit field]

We start with $\pdv{z}{s}$

[grand pond]

What a wonderful world !

[twilit field]

$\pdv{z}{r} \cdot \pdv{r}{s}+ \pdv{z}{\theta} \cdot \pdv{\theta}{s}$$

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

So that gives us $e^rcos(\theta) \cdot t -e^r \sin(\theta) \cdot \frac{ s}{\sqrt{s^2+t^2}}$

[grand pond]

What a wonderful world !

[twilit field]

is this right

[grand pond]

What a wonderful world !

[last slate]

is that the final answer? i thought its gonna continue further

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

It is yea

.close

[midnight plank]

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[twilit field]

Now I know I can just expand this

but is there no other way

$\pdv{z}{r} = f_x(x,y) \cos(\theta) + f_y(x,y) \sin(\theta)$

[grand pond]

What a wonderful world !

[twilit field]

$\pdv{z}{\theta} = -f_x(x,y) \cdot r \sin(\theta) + f_y(x,y) r \cos(\theta)$

[grand pond]

What a wonderful world !

[boreal plaza]

Guys, I can't solve this question, I didn't understand about centroids and everything, and this is one of the last questions I need to finish from the list that I have to submit today. Please help me, I'm in high school and I entered this math olympiad to challenge myself because I enjoy it, but I think it's too much for me with so little time to complete a list at this level along with other questions.I know that these concepts will be used, because I was recommended to read a very dense material about it, but I understood very little, especially about barycentric coordinates, because of that, I can't move forward to apply Ceva's Theorem.

ADVANCED LEVEL

[cursive swan]

!occupied

[midnight plank]

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[twilit field]

!occupied

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[cursive swan]

Open a new channel.

[boreal plaza]

THANKS

[twilit field]

okay, so what I did works

.close

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[midnight plank]

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[cursive swan]

Prove that a tangent is perpendicular to the line joing the point of tangency to the centre.

[cursive swan]

angle CDB is always 90 deg.
As D gets closer to the point of tangency B, it basically falls on the tangent! And that will be when it is coincident with B. And that will be when angle DCB = 0 deg. So angle CBD will be 180-(0 + 90) = 90.
Is this a correct ||proof||?

[lyric charm]

was there any reason for spoiler-tagging the word "proof"?

anyway, no. i'd say it's an ok thing for intuition but i would not consider it a rigorous proof at all.

[cursive swan]

Okay.

Thank you.

[cursive swan]

Yes, to shock the helper.

.close

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[lyric charm]

........

oh yeah because shocking the helper is such a good thing to set up on purpose.

(/s)

[cursive swan]

/s = sarcasm?

[exotic stratus]

Yes /s = $\sum_{i = 1}^{sarcasm} i$

[grand pond]

@exotic stratus

[lyric charm]

https://tenor.com/view/jarvis-delete-gif-12630970875383007652

[lyric charm]

yes

[lyric charm]

this joke was supremely unfunny.

[surreal moon]

||un||funny

[lyric charm]

can you please not?

[midnight plank]

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[graceful ferry]

$\int(lnx)^2dx$

[grand pond]

prograce

[graceful ferry]

what method

[deep oriole]

ibp i reckon

[graceful ferry]

What is u and what is v' ?

[silver bluff]

differentiate (ln(x))^2 and integrate 1

after 1 iteration you will realise

[worthy spruce]

Or u sub x=e^y if you fancy. Still have to use ibp, but it makes it a bit easier

[graceful ferry]

.solved

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[graceful ferry]

$\int_{0}^{1}\frac{ln(1+x)}{x^{\frac{3}{2}}}dx = \lim_{\epsilon \to 0^+}\int_{0-\epsilon}^{1}\frac{ln(1+x)}{x^{\frac{3}{2}}}$ ?

[graceful ferry]

help

[lyric charm]

missing } just before the last dollar

[grand pond]

prograce

[lyric charm]

lower limit should be just epsilon not -epsilon

also missing dx also missing \ln command

[graceful ferry]

ok

I have to find indefinite integral can u help ?

also \ln for what

[gaunt jetty]

Classic integration by parts

[graceful ferry]

I thought I could use taylor series for ln(1+x) and then calculate integral for polynomials

[graceful ferry]

o ok

[gaunt jetty]

\[
\int_0^1 \frac{\ln(1+x)}{x^{\frac32}}\,dx = \lim_{\epsilon \to 0^+}\int_\epsilon^1\frac{\ln(1+x)}{x^{\frac32}} \, dx
\]

[grand pond]

Aero ݁˖ ❀ ⋆｡˚

[gaunt jetty]

This is what she meant I believe

[gaunt jetty]

Overcomplicated. No need

[lyric charm]

$\ln(x)$ and not $ln(x)$

[grand pond]

Ann

[lyric charm]

btw do you need to calculate the integral or just test it for convergence?

[graceful ferry]

This one test convergence

I feel like now you mentioned it that what I'm doing trying to calculate the integral is a bit useless and there's another way?

I'm a bit stuck now too

[visual tiger]

if you only want to test for convergence

just try to compare ln(1+x) with an easier function to work with

only after we established convergence can we start manipulating the integral to find its value

[graceful ferry]

Like x?

[visual tiger]

like x

[graceful ferry]

then its just 1/x^1/2 and since the power is less than 1 then it converges

since integral is between 0 and 1

right?

.solved

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[modern shard]

did I do something wrong? or answer is given wrong?

aha nvm found it

it would be root w

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[cursive swan]

Find all integer solutions to (5^a - 2^b)^2 = 1

[opaque mural]

minus or positive 1

the part in the ( )

[nimble leaf]

So every power of 2 which is either 1 above or 1 below a power of 5

Trivial examples are a=0 b=1 and a=1 b=2

[cursive swan]

Minus. Clearly, it can't be positive.

[nimble leaf]

Why not?

[opaque mural]

modules could help u find a solution

[midnight plank]

@cursive swan Has your question been resolved?

[lyric charm]

5^a - 2^b = -1 has a lot less solutions than 5^a - 2^b = 1, or so one would think.

wow ok the strikethrough completely obliterated the minuses.

[fast estuary]

5^1 - 2^2 = 1

1^2 = 1

Not sure if maybe there's a misunderstanding about what was meant here

[cursive swan]

The question is:

Find all integer solutions to (5^a - 2^b)^2 = 1

[fast estuary]

I understand the question

Someone offered the observation that this means the quantity in parentheses must equal 1 or -1

[cursive swan]

Yes, that is, |(5^a - 2^b)| = 1.

Because there could be some pairs that satisfy 2^x - 5^y = 1, where x > y. That is why I squared it.

I found no solutions for 0 <= a, b < 100.

[midnight plank]

@cursive swan Has your question been resolved?

[dreamy lichen]

(a, b) = (0, 1)
(a, b) = (1, 2)

those are 2 solutions with 0 <= a, b < 100

[cursive swan]

I mean there are not more than (0, 1) and (1, 2)

[dreamy lichen]

For even x:
2^(2x) - 1 = 5^y
(2^x + 1)(2^x - 1) = 5^y
this would mean that both 2^x + 1 and 2^x - 1 are powers of 5, which is clearly impossible

5^(2y) - 1 = 2^x
(5^y - 1)(5^y + 1) = 2^x would mean both are powers of 2, which is also impossible

Oh wait

2^n - 1 will always be divisible by 3

or maybe im stupid

i honestly dont know

yeah, im stupid

it will only be divisible by 3 for even n

or maybe

no

5^n - 1 is divisible by 6 for even n

and always by 4

[cursive swan]

yes, by 24

[dreamy lichen]

just brainstorming

i honestly have no idea on how to approach this

if only 4 wasnt a power of 2

for odd n, the other factor will be
5^n + 5^(n-1) + .. + 1

5^n + 1 will often be diviisble by 3

by 6

ohhh

namely when n is even

oh..

damn it

[dreamy lichen]

well

it was proven in 2002

no elementary proof for us ig then

[cursive swan]

Thanks y'all.

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[west bay]

Help

[west bay]

Is the answer for the first one 16? How do i do the rest?

[midnight plank]

@west bay Has your question been resolved?

[fathom onyx]

I'm getting 16 for a as well

But the other ones are just elaborate lines, they don't contain regions of points as such

So I'm getting 0 for those

No wait mb I'm massively misreading that

Do each of these with cases, depending on which quadrant the point (x,y) is in

These are still collections of (geometric 1x1) squares, but you need to determine which squares

[midnight plank]

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[inland patio]

Let <x,y>=xAy* be an inner product on C^2, with A={{1, i} , {-i, 2}}. Why does the conjugation overline become a conjugate transpose?

[grand pond]

psie

[sharp coral]

it's a 1x1 matrix which makes it automatically symmetric

[inland patio]

right, I agree that it's a 1 x 1 matrix, but shouldn't the inner product be a scalar?

[sharp coral]

in linear algebra we tend to implicitly use the isomorphism between 1x1 matrices and scalars by treating them interchangeably

[inland patio]

ok, that makes sense

[midnight plank]

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[opaque elk]

What's the actual difference between the pigeonhole principle and it's generalized form?

[opaque elk]

ah nvm

.exit

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[inland patio]

Let $x_1,x_2\in\mathbb C$. Consider the expression $$|x_1|^2+2\Re(ix_1\overline{x_2})+2|x_2|^2.$$ How can I convince myself this is positive for $x_1$ or $x_2$ not equal to $0$? If one them is zero and the other isn't, it's obviously $>0$. However, if both of them are not equal to $0$, I don't see how to motivate that the expression is $>0$.

[grand pond]

psie

[sharp coral]

the inner product is only guaranteed positive if you take the inner product of a vector and itself

[inland patio]

Yes, in my example we had $$A=\begin{pmatrix}1&i\ -i&2\end{pmatrix},$$and we were considering $\langle x,y\rangle=xAy^\ast$. Now I'm trying to show that $\langle x,x\rangle >0$ if $x\neq 0$. If $x=(x_1,x_2)$, then $xAx^\ast$ reduces to the expression above.

[grand pond]

psie

[sharp coral]

well in the worst case scenario you would have 2Re(ix1 x2bar) = -2|x1||x2|

[inland patio]

how do you know that 2Re(ix1 bar(x2))< -|x1|^2-2|x2|^2? I don't see it.

[sharp coral]

well it's greater than that

and that's the only part which can be negative

so we need to convince ourselves that the parts that must be positive will necessarily be bigger than the parts that could be negative

[inland patio]

right, and I can't see why 2Re(ix1 bar(x2))< -|x1|^2-2|x2|^2 is false

[sharp coral]

well at this point we can just replace the magnitudes entirely with real numbers

so we want to convince ourselves that x^2 - 2xy + 2y^2 > 0 for real x,y (equality if both are 0)

[inland patio]

so if |x1|=x and |x2|=y, how do you know Re(ix1bar(x2))=xy? That is what I gather from your reasoning.

[zealous schooner]

Re(ix1 x2bar) >= -xy

[sharp coral]

the magnitude of ix1 bar(x2) is xy, so -xy <= Re(that) <= xy

[inland patio]

ok, I think I'm beginning to understand, thanks both of you

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[green mesa]

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[green mesa]

.reopen

[midnight plank]

✅

[green mesa]

It's been a little while since I've done anything with torque. To verify, do I set the rotational axes at the ends of the ladders, calculate the torque generated by the person and the ladder, then calculate from that the force the wall/floor has to exert to keep it in equilibrium?

For this, I don't really know what to do

[midnight plank]

@green mesa Has your question been resolved?

[green mesa]

<@&286206848099549185>

[vast ginkgo]

the rotational axis in this case is the point on the ladder on the pavement

(since it's kept in place)

[green mesa]

so what do I do to find the force on the ladder from the wall?

[vast ginkgo]

well you can see the wall's only purpose here is to balance the torque for the ladder by pushing back on it

[green mesa]

wait

I meant how do you find the force from the pavement

I got the wall, rotational axis floor, torque equals torque from wall

but what about the floor?

[vast ginkgo]

there's only one horizontal force here

the system is in mechanical equilibrium

[green mesa]

So the force from the floor...

Is gravity?

[vast ginkgo]

no

well

gravity would be one of the forces

acting on the ladder at the floor

what's the other

(horizontal one)

[green mesa]

normal force from the wall

so the torque from the normal force equals the torque from the weight of the ladder and person. But how do I calculate the force from the floor?

oh

[green mesa]

friction

[vast ginkgo]

yes

what is friction cancelling out in this case

(the only other horizontal force)

[green mesa]

not normal force fromt he wall?

[vast ginkgo]

it is the normal force from the wall

that's what the frictional force will be equal to

[green mesa]

Oh

so find one, find the other?

[vast ginkgo]

yes

but remember the floor has gravitational normal force being acted on it as well

[green mesa]

Right

So how do I answer it?

[vast ginkgo]

you have the wall force right

[green mesa]

Not yet

[vast ginkgo]

oh

[green mesa]

I haven't calculated anything yet

I've been doing other homework, and I'm tired of the numebrs lol

[vast ginkgo]

fair

[green mesa]

I'm taking a break and just planning this one

[vast ginkgo]

well id find the wall force first

[green mesa]

okay
It's kind of confusing, though, because it implies that there's only one force acting on the ladder from the floor

maybe I should ignore the weight?

[vast ginkgo]

there's 2

1 net force

[green mesa]

So do you think I should find the magnitude of the net force?

[vast ginkgo]

yes

[green mesa]

okay

and how do I calculate the weight? I assume it's not mass times gravity because of the wall

[vast ginkgo]

unless the question specifically asked for like the magnitude of the friction

[vast ginkgo]

the wall doesn't matter

[green mesa]

okay, so just normal weight calculation

[vast ginkgo]

the weight force is always equal to an object's mass times gravity

pointed straight down

[green mesa]

that should do it for the first one, thank you very much for your help

I'll go back and look at this later

Any tips for number 2?

I haven't got an idea of how to do that one

[vast ginkgo]

well it definitely will involve torque

do you know what the "pivot point" in this case would be

[green mesa]

It's just one?

[vast ginkgo]

yes

[green mesa]

mm

the center of mass?

[vast ginkgo]

yep

[green mesa]

okay
What can I calculate from the angle of the strings?

[vast ginkgo]

well theres tension

from the mass

[green mesa]

right, but I can't calculate it exactly

but the tension applies torque?

[vast ginkgo]

all forces can

[green mesa]

So I can use the angle of the torque somehow to determine distance?

The torque would have to be equal, right?

[vast ginkgo]

torque would have to be equal yes

about the center of mass

[green mesa]

and the forces are equal

the tension, I mean

[vast ginkgo]

well not necessarily

[green mesa]

But in this case?

[vast ginkgo]

the sum of both tension y components will equal the mass

i mean weight

[green mesa]

right

and the x components would cancel?

[vast ginkgo]

the sum of both tension x components will equal 0

yes

and remember the torque exerted by each tension force is just the y component times r (dist from center of mass)

from what it looks like you never actually have to solve for tension

youre just using it to relate equations to actually solve for x

[green mesa]

would it also work if you did the magnitude of the tension times sin of the angle between the cables and the beam, and the distance from the center of mass?

Is that the same as the y component times sin 90 times distance?

[vast ginkgo]

yes

T_1 and T_2 would represent the magnitudes of the ropes

(for example)

you can use mechanical equilibrium to solve for x from component forms of the tensions

(eg. T_1 * sin(theta))

you want to rearrange for two expressions equal to T_1/T_2 (or the other way around)

then you'll be left with x as the only unknown

which you can solve for

(based on how i believe the problem would work out)