help-38

[solid plank]

Yeah it's the same thing!

[native shuttle]

kk

thanks

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[autumn nova]

found h and k no issues for (a) but for b i got y=-2x, but mark scheme is diff ans

[sterile blade]

what did you get for h and k?

[dapper swift]

what work did you do for part b also?

[autumn nova]

h=4/3, k=7/3

[autumn nova]

at first i thought of changing the eq in b into the form y-x = m((x^2)/y) + c

but im having trouble isolating it

[dapper swift]

instead of that try multiplying everything in part b by x^2

[autumn nova]

yeah thats what i did

then i solved but stuck at 8y = (10x^2)/y

[dapper swift]

do you have your working on paper or something?

[autumn nova]

uh yeah but my handwritings kinda indecipherable

[dapper swift]

hmmm send it anyways

[dapper swift]

,rotate

[dapper swift]

ah I think I found a way

rearrange to get $x^2 = \frac{(5 - k)y}{2 + h}$

[solid kiln]

south

[dapper swift]

then sub into part a to get $y^2 = h \frac{(5 - k)y}{2 + h} + y(k + x)$

[solid kiln]

south

[dapper swift]

dividing both sides by y then substituting h and k everywhere gives you the equation of the line

[dapper swift]

yep I found a way

[autumn nova]

hmm ok let me try

[autumn nova]

sorry im still confused could u show how u got to the answer

[dapper swift]

which part are you confused on?

[autumn nova]

well i rearranged to make x^2 the fomula, then subbed it back into the first formula
and i got y^2 = 16/15 +7y/3 + xy which doesn't seem like the correct answe

mark scheme gave y-x=-\frac{2x^{2}}{y}+5

[dapper swift]

oh wait it's a straight line on the graph, sorry

[dapper swift]

okay this bit is not correct

[autumn nova]

mark scheme give this

[dapper swift]

oh really hmmm

[trim jolt]

@autumn nova Has your question been resolved?

[autumn nova]

found a way to solve it

[dapper swift]

omg

[autumn nova]

not reallt sure if its the optimal way to go but at least its some way

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[trim jolt]

Please don't occupy multiple help channels.

[clear cloud]

what would be the condition for A to be "canceled" both sides

like what operation allow you to cancel out the A

and what are the condition for it to be possible

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@lean kraken Has your question been resolved?

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[thorny fulcrum]

Can someone check if my equation is correct please

[split chasm]

is incorrect

how are you getting from:
$$x \cdot (13-x) = 36$$
to
$$x \cdot x = 23$$

[solid kiln]

ℝαμOmeganato5

[thorny fulcrum]

I got 13 the other side and subtracted it

[solid plank]

you didn't get 13x

[split chasm]

is not allowed here

[solid plank]

The equation has to become $13x-x^2=36$

[split chasm]

13 is part of the (13-x) with x being multiplied to it
you can't just take out 13 like that

[solid kiln]

Aria

[thorny fulcrum]

Oh

Ty

.Close

. close

.close

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[young thicket]

Let ABCD be a rhombus with ∠BAD < 90°. The circle passing through D with center A intersects the line CD a second time in point E. Let S be the intersection of the lines BE and AC. Prove that the points A, S,D and E lie on a circle.
Can someone draw me a sketch? My E keeps being on the same with D.

[undone gorge]

Something like this?

[young thicket]

But it’s a rhombus, so if all sides are equal AB and AD would be radius and the circle wouldn’t touch DC unless at D right?

[undone gorge]

Yes AB and AD are radius

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[high raptor]

i tried solving this question using key identity method but apparently i got the wrong answer, i ended up getting: y = C_1 * e^3t + C_1* e^-3t + (tsin(3t)(1-t)) / 3

[high raptor]

is this solvable using key identity or do you use undetermined coeffecient? also it be nice if you could show how you solved it

[knotty locust]

Show your work for how you got that answer

[high raptor]

its a bit messy but ill upload the scratch work

sorry i was just writing all over the place

soln is bottom right, page 1 is top left,

page 2 is top right and page 3 bottom left

but basically i first found the char polynomial and then used that to find the roots, using the roots i wrote the solution of the homogenous solution. after that i looked at the forcing( or the nonhomogeneous function) and checked if the characteristic polynomial of the forcing is already one of the roots , in this case the forcing is = -2tcos3t , so its characteristic polynomial was +- 3i , and the roots i found originally were 3 , -3 . so here the characteristic polynomial of the forcing was not present in the roots i found for the homogenous solution, so then i use the linear operator and i take the derivative of L(e^zt) = p(z)* e^zt , where z is the root , and p is the polynomial characteristic of the original equation, in this case (z^2-9) . i take the derivative of L(e^z) from m + d to m , where m is the multiplicity of the characteristic polynomial of the forcing in this case it was 0 as it was not present in the roots and the degree of the characteristic polynomial here is 1 as we have -2t^1 cos(3t) in our forcing. after doing that i plug in my root from the characteristic polynomial of forcing which is 3i, i do p(3i) and then p'(3i). then i plug them back in my equation then manipulate until i have L( some expression here) = my original forcing, sometimes like this question there is two parts of the expression one is imaginary and one is real, you just choose the one which matches your forcing , here mine was real so i chose the real expression inside L( ) as my solution

this is exactly how my professor told us to solve these but i might ve gotten something wrong

[knotty locust]

The characteristic polynomial

is r^2-9=0

[high raptor]

yup

i just used z

as variable

[knotty locust]

which factors as (r-3)(r+3), so your complementary solution is correct

c1e^(3t)+c2e^(-3t)

[high raptor]

yup

[knotty locust]

So it remains to find the particular solution

[high raptor]

yeah thats what im struggling with

[fresh zephyr]

Hi

[knotty locust]

Try undetermined coefficients

[high raptor]

would you be able to show me how to find the particular using undetermined i have an idea on how to do it but still not completely familiar

[fresh zephyr]

I'm new

[knotty locust]

Have you used the method before?

[high raptor]

yeah

[knotty locust]

Please don't chat here, read #❓how-to-get-help

[high raptor]

couple times but the problems were much easier than this one

[knotty locust]

Okay so I'll tell you what our guess should be, and why, and you go from there

the forcing function is -2tcos(3t) as you said earlier

so it is the product of a linear function and a trig function

the most naive guess would then be

(At+b)(Ccos(3t))

but

[high raptor]

i just want to put this here because this is what my prof provided us

[knotty locust]

we should also involve a sine function in our guess

[knotty locust]

I think this is a very complicated way of thinking about things

[high raptor]

yeah i think too its confusing me because im struggling with the guessing part

[knotty locust]

The guess we should try is y_p=(At+B)(Ccos(3t)+(Dt+E)(Fsin(3t)

this is a product of linear and trig functions

just like our forcing function was

so thats why we guess it

not tooo complicated to think of

much better than that key evaluation thing

[high raptor]

so what would be the guess if it was only a linear function

for example 6e^2t

[knotty locust]

Say it was like y''-9y= 2t

[knotty locust]

This is not a linear function

its exponential

[high raptor]

oops

[knotty locust]

this is linear however

[high raptor]

well exponential too

mb

[knotty locust]

and in this case, we would take our guess to be a cubic polynomial

taking 2 derivatives, we are left with a linear one, which is what our forcing function is

if we guessed anything bigger, like a quartic polynomial, taking 2 derivatives we would be left with a quadratic term which is not in our forcing function

so we wouldnt want to do that

and if we guessed anything smaller, like a quadratic polynomial, taking 2 derivatives we would be left with a constant, which is not a linear function like 2t

so we wouldnt want to do that either

thats why we guess cubic in that case

[knotty locust]

For an exponential function, we just use an exponential function as our guess

because taking derivatives of exponential functions just returns themselves

[high raptor]

like Ae^2t for the example i gave

[knotty locust]

yes

There are tables you can find of like what to guess when for undetermined coefficients, which can be useful, but it is also helpful to think through yourself what the guess should be

[high raptor]

i see

im gonna try to watch some videos and go through the problem with underetermine coeffecient method

my prof doesnt provide soln key using that method

[knotty locust]

That'd be good, good luck!

[high raptor]

so its harder for me to follow

thanks

[knotty locust]

Np

[high raptor]

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[granite bear]

Can someone help me with this? I'm honestly just too lazy to solve it

[shrewd lion]

You have to try sis 😭

[granite bear]

I mean I kinda did

But it just stressed me out more so I gave up

[obsidian zealot]

u need help w all of them?

[granite bear]

Yes please

[shrewd lion]

[shrewd lion]

Well for part A..
Getting from O -> Q is the same as
O -> P -> Q

So OQ = OP + PQ

[granite bear]

Mhm yh I got that, I'm confused about the XY tho

[obsidian zealot]

what’s OX in terms of OP?

[granite bear]

4p

[obsidian zealot]

in terms of OP tho

like without saying p

[granite bear]

Idkkkk

Yhh this question staying blank

[shrewd lion]

I dont care

You’re gna stay up 😭

[granite bear]

Broo I'm half asleep already

I'm not gonna stress

😭😭😭

[shrewd lion]

XY is simple

[shrewd lion]

Okay draw

Er

In 2 colours

How you can get the line xy

[granite bear]

I'm trying to stay awake just bare with me on this

[shrewd lion]

Its 6 am for me

[granite bear]

And it's 5:43pm for me, these people just tryna stress me out after school

[shrewd lion]

U got this sis

[granite bear]

Okay XY= -4p+4q = 4(q-p)

Is that right

[shrewd lion]

Errr

OX + XY = OY
OY - OX = XY
4q-4p = xy
4(q-p) = xy

Yes ure right

[granite bear]

Okay okay greattt

I'm getting there

[shrewd lion]

[granite bear]

So PR is 3q-p?

[shrewd lion]

Where eid u get PR wait

[granite bear]

The second question

XQ is gonna be 12q-4p

[shrewd lion]

Ok so to obtain XR,
It is 1/4 of XQ

Following the saame principle,
OX + XQ = OQ
XQ = 12q-4p
And XR = 3q-p

You’re right

[granite bear]

Yes yes

[shrewd lion]

So how do we get to PR?

[shrewd lion]

Were u referring to XR?

[granite bear]

Yeppp

[shrewd lion]

So any ideas how to get PR?

[granite bear]

-PX + XR

Soo PR is gonna be 3(q-3p)

Causeeee -8p+3q-p = -9p+3q

[shrewd lion]

OP = OX + XP
XP = OP-OX
XP = 12p-4p
XP = 8p

XP+PR = XR
PR = XR-XP

OP + PR = OX + XR
PR = OX + XR - OP
PR = 4p+3q-p-12p
PR = 3q-9p

Good

[shrewd lion]

Ignore im on my bed and i have to do this on disc 😭😭

Do you understand now?

[granite bear]

Yes I do thankyou

[shrewd lion]

Good

For part C

Collinears what do we know

[granite bear]

What's vector method again

[shrewd lion]

No clue 😂 i’m assuming they want you to use vectors to prove it

[granite bear]

Imma just leave this question for my teacher to work

So I can stress her out

[shrewd lion]

😂😂

U got this girl

No stress ya:)

[granite bear]

You're right the teacher will take the stress I wont

[shrewd lion]

[granite bear]

I give up, I'm going to nap

[shrewd lion]

Tskk

[granite bear]

Thankyou for all your help

[shrewd lion]

:> if you need anymore help my dms are open

Or just come back here

Goodluck:)

[granite bear]

Thankyou,awww ur sooo sweeet

💖💖💖

[shrewd lion]

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[slim pawn]

Let ABC be a triangle with ∠BAC = 70°, ∠ABC = 50°, and M be the midpoint of side AC. Let P and Q be points on lines BC and BA, respectively, such that the circle passing through P, M, B is tangent to BA and the circle passing through Q, M, B is tangent to BC. What is the measure of the angle ∠PQB?

(A) 60°
(B) 70°
(C) 80°
(D) 90°
(E) 100°

[slim pawn]

I already know that the arc BP and the arc BQ measures 100°

[wet lintel]

||similar triangles||?

[slim pawn]

Not necessarily

[wet lintel]

your drawing indicates so

[slim pawn]

My drawing is definitely not reliable

[wet lintel]

so change it up fr

[slim pawn]

I mean, it is just for having a better understanding of the question, but the statement doesn't show any drawing

Just ignore it if you prefer

[simple panther]

Wtf did I just get?

It's supposed to be tangent to BA right?

[slim pawn]

Yes but you forgot the point p

[simple panther]

But it's not possible

[slim pawn]

The statement says it is

[simple panther]

But from the diagram, if we draw a circle tangent to BA passing through M, it doesn't intersect BC

To give P

[slim pawn]

We should focus on the statement instead of the diagram, it is possible to intersect BC, and this intersection is going to be p

[simple panther]

K, I'll try it

I tried it in desmos geometry too. I just can't wrap my head around this diagram

If it's given in the question, then it should be possible to sketch

[trim jolt]

@slim pawn Has your question been resolved?

[wet lintel]

let me draw it

[wet lintel]

yeah PQ is not necessarily parallel to AC

[slim pawn]

yeah

[wet lintel]

i give up

this is a very sus question

[slim pawn]

np

probably the thing isn't with graphing it, if it were, the statement would give the diagram

And if you had to make this diagram, you would have to try a lot of possible locations for p and q, this olympiad would not expect you to do that with such a short time

[cobalt karma]

I remember seeing a problem like this that was ridiculously confusing, and I still haven't been able to wrap my head around it to this day. Olympiad problems in regards to geometry are notoriously difficult, and they certainly require remarkable types of people to be able to solve in such short lengths of time.

Here's a noteworthy question that might take you step in the right direction: where is this problem from? Is it from a textbook, and from what country? This is important, because different cultures always have different methods of solving. The problem I was referring to was in Vietnamese, which relied on the knowledge of excircles and the like—something that I was never taught in American schooling.

If you can give us the source of the problem, we might be able to get closer to solving. I'll try cracking it for now. Haven't contemplated long enough to illustrate a diagram yet.

[slim pawn]

Thank you very much, I can certainly do that!

[slim pawn]

This problem is from a Brazilian Olympiad called OMpD (Olimpíada Matemáticos por Diversão). It is not very well known, even because there is no solution for this problem on the internet, but this Olympiad is associated with very large Olympiads, such as the OBM (Olimpíada Brasileira de Matemática) which is the national mathematics Olympiad.

[slim pawn]

You can find this problem here in this site: https://matematicopordiversao.wordpress.com/5a-olimpiada-matematicos-por-diversao-ompd-2024/
You don't need to open it.

This test is from 2024, the first phase. Even below is the original problem

[simple panther]

When I try to keep the angle 90, it doesn't intersect to get P, when I try to get it intersect, the angle is not 90. Similarly with Q

Perhaps, we can extend the lines BC and BA?

[slim pawn]

which angle are you talking about

[simple panther]

Angle of the radius with AB for P

And radius with BC for Q

Angle should be 90 for tangents right?

[slim pawn]

Hmm, I understand what you are talking about

I am going to start with the old tests because they are probably going to be easier for me

Anyways, thanks a lot guys

[slim pawn]

I really don't have idea what is wrong here, I even thought that I had translated the problem badly, but it seems to be unlikely

Alright, thanks a lot

.close

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[simple panther]

Ohhh

The lines have <--> above them

So they can be extended

[slim pawn]

.reopen

[trim jolt]

✅

[cobalt karma]

Of course they can. Euclid Postulate 2.

You can extend any line infinitely. It's a given.

[simple panther]

I didn't see the original, so I didn't know they can be extended here

I thought we're working with line segments

[slim pawn]

Yeah, I forgot to say that, I am really sorry

[cobalt karma]

No matter what problem it is, you are allowed to extend the line segment infinitely.

[cobalt karma]

This is also true.

[cobalt karma]

I have faith in your diagram, as it is what I have drawn as well, even when adjusting the orders of the points. The angle of PQB should be always be somewhere near the angle BAC, and they might even be similar—accredited to ooaa—but you have to find a way to prove this. I notice that two of the answer choices takes this notion into account.

Last time I solved a problem like this, it took about half a day. Olympiad problems are not so generous as to ever give out the answer, of course! 🤣 If you're willing to leave this thread open, or re-open it, I'll get back to you if I make any breakthroughs.

[slim pawn]

I send this same question in #geometry-and-trigonometry and #competition-math , thanks a lot for trying to help!

Thank you guys, I have to go now. Have a great day!

.close

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[worn sandal]

can i get some1 to look over my greens theorem problem code

[worn sandal]

its on wolfram cloud

[ionic pendant]

why are you calculating the "gradient"? green's theorem uses curl

[worn sandal]

because I used it to get the P and Q stuff

as components of the vector

is that a invalid method

oop i see where i went wrong

how is this @ionic pendant

[ionic pendant]

seems reasonable

[worn sandal]

yay

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[worn sandal]

ty

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[frank bough]

i need help

[frank bough]

just help me, please

[trim musk]

just ask

[frank bough]

i need help

[trim musk]

!da2a

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[frank bough]

oh wait i gor the anser

[zinc ginkgo]

!done

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[frank bough]

.close

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[lament shoal]

can someone help me with this

find solutions for x, y ∈ ℕ given xy - x = y+6
What if x, y ∈ ℤ

[cedar adder]

$xy - x - y = 6$

[solid kiln]

SELVATOR

[cedar adder]

$y(x - 1) - x = 6$

[solid kiln]

SELVATOR

[cedar adder]

$y = \frac{6 + x}{x-1}$

[solid kiln]

SELVATOR

[cedar adder]

Good?

[unkempt oriole]

good

[cedar adder]

Do u know how to complete?

[unkempt oriole]

yes

[cedar adder]

How?

[unkempt oriole]

we need to factorise it first

In a different way

xy-x-y=6

Add 1 both the sides

xy-x-y+1=7

x(y-1)-(y-1)=7

(x-1)(y-1)=7

And now it can be easily solved

[cedar adder]

Solve it

[unkempt oriole]

ok

Then we will break 7 into its factors

It can be only written as 1×7 or 7×1

So,
(x-1)(y-1) =1×7 or = 7×1

[cedar adder]

I got u

[unkempt oriole]

Can u solve it from here?

[cedar adder]

x = 8
y = 2

[unkempt oriole]

yes, and?

[cedar adder]

What else?

[unkempt oriole]

x=2 and y=8

[cedar adder]

Inverse

[unkempt oriole]

There are two pairs of solutions

Yes

[cedar adder]

Anyways, i thought u were owner of ticket. And maybe u think i'm owner of ticket

But he is @lament shoal

[unkempt oriole]

Ik that

but if x and y were integers

There wouldve been more solutions

[unkempt oriole]

Can u find them?

[cedar adder]

[cedar adder]

Wwould we use this?

[unkempt oriole]

No

Lemme give u a hint

Think of two integers that on multiplication gives 7

Except 1×7 and 7×1

[cedar adder]

3.5 × 2

1.75 × 4

[unkempt oriole]

No

I said integers

[cedar adder]

Some thing like 1,2,3,4 ?

[unkempt oriole]

Nope

[remote stone]

lol he peaced out

[cedar adder]

What are integers?

[unkempt oriole]

umh

[cedar adder]

Gimme an example

[unkempt oriole]

Ok

[cedar adder]

10/3?

[unkempt oriole]

-10, -200 , 1, 17, 26

No

[cedar adder]

Owh

[unkempt oriole]

Not fractions

[cedar adder]

N but with negative numbers

[unkempt oriole]

and 0 too

Yea

[cedar adder]

Isnt 0 inside N?

-1 × -7

[unkempt oriole]

0 is not a natural number

[unkempt oriole]

Yayyy

-1×-7 and -7×-1

We can compare it again with (x-1)(y-1)

[cedar adder]

-8 × -2 = 16

[unkempt oriole]

no

compare it sillyy

😔

[cedar adder]

I feel like I'm trying to find a solution to some integration.

[unkempt oriole]

Loll

x-1=-1 and y-1 =-7

[cedar adder]

Let's say x = -7 and y = -1
(-7 - 1) (-1 -1 ) =16

[unkempt oriole]

And the inverse of it

[cedar adder]

Owh

[unkempt oriole]

Why is x=-7?

And y=-1?

[cedar adder]

(x-1) = -1
(y-1) = -7

x = 0...

[unkempt oriole]

Correct

Yea

[cedar adder]

...

[unkempt oriole]

and y= -6

👍

which one

I don't understand that language

i can try but im kinda dumb

[cedar adder]

U were good with numbers.

ANYWAYS

That's his channel

@lament shoal

[unkempt oriole]

I can't solve it

Tell me how to do it

[cedar adder]

Bruh...

[unkempt oriole]

Who cares

[cedar adder]

Do u see that expression at (a)

[unkempt oriole]

Yea

[cedar adder]

Substitute the position of f(x)

And enter the integration

[unkempt oriole]

Wdym enter the integration

oh nvm

[cedar adder]

It would be: $\int f(x) = 2\int f'(x) - \int f"(x)$

[unkempt oriole]

Then?

[solid kiln]

SELVATOR

[cedar adder]

What is $\int f'(x)$ dx ?

[solid kiln]

SELVATOR

[unkempt oriole]

Wait

[cedar adder]

Do u really have to calcul it?

[unkempt oriole]

-xe^x

[cedar adder]

Bruh

[unkempt oriole]

Lemme check

[cedar adder]

Just focus

[unkempt oriole]

Yes it is -xe^x

Am i wrong

[cedar adder]

Did u calcul f'(x)?

[unkempt oriole]

Yes

Oh i needed integrate it

Idk how to do that

[cedar adder]

[trail rover]

think

[cedar adder]

Integration by parts

But

We wont use it

[unkempt oriole]

Idk integration

I just know basic formulas

Im just 41🥺

[cedar adder]

Owh

Anyways

[unkempt oriole]

yes

[cedar adder]

$\int f'(x)$ dx = f(x)

[solid kiln]

SELVATOR

[unkempt oriole]

Yes that's correct

[trail rover]

• C ✌🏻✌🏻

[unkempt oriole]

You r doing great

[cedar adder]

$\int f"(x)$ dx = f'(x)

[solid kiln]

SELVATOR

[unkempt oriole]

Dammnnn

[cedar adder]

Idk why, but we dont add it for now

[unkempt oriole]

Yes yes

[cedar adder]

$\int f(x)$ dx = $[2f(x)] - [f'(x)]$

[solid kiln]

SELVATOR

[cedar adder]

[unkempt oriole]

How?

Where's the proof

[cedar adder]

.

[cedar adder]

.

[cedar adder]

.

[unkempt oriole]

Where did greatest integer function come from?

[cedar adder]

Do you learn how to integrate at 14 years old?

[unkempt oriole]

Just the basics

like integration of sinx

[cedar adder]

At 14, I probably didn't even know f(x)

[unkempt oriole]

and i know the u substitution method

[cedar adder]

[unkempt oriole]

How old is u

[cedar adder]

More than 14

[unkempt oriole]

52?

[cedar adder]

Call me grandpa!

[unkempt oriole]

:) ok granpa

[cedar adder]

Well

Good grandson

[unkempt oriole]

:D

Where is @lament shoal

[cedar adder]

Do u know how to solve ax² + x + c = 0?

[unkempt oriole]

Lemme think

Yea use the quadratic formula

Or maybe by completing the square

[cedar adder]

U are more than 14

Why r u manipulating?

[unkempt oriole]

Prove it

[cedar adder]

U know harmonic formula

[unkempt oriole]

Yes ik

[cedar adder]

[unkempt oriole]

F=mrw²

[cedar adder]

Unless you're Chinese, that's proof enough.

[unkempt oriole]

Not chinese

But asian

[cedar adder]

Proved

[cedar adder]

Not enough

[unkempt oriole]

😬

[cedar adder]

What is golden ratio in maths?

[unkempt oriole]

idk the ratio

But ik the series

[cedar adder]

Does it converge?

[unkempt oriole]

112358

the ratio?

Or the series?

[cedar adder]

$\phi$

[solid kiln]

SELVATOR

[unkempt oriole]

What

[cedar adder]

This

[unkempt oriole]

theta?

wdym

[cedar adder]

[unkempt oriole]

the series don't converge

well i was preparing for maths and physics Olympiad

That's why ik that stuff

the question that jutin asked was of diophantine equation, and it is an important topic in mathematical Olympiads

[cedar adder]

I understand, maybe you switched roles with him.

[unkempt oriole]

look at his name

i didn't wanted to tell him my name

who wants to tell their real age to daddy dumpling?

[cedar adder]

@unkempt oriole

The court has found you guilty.

Or try to create 6 different stories

[unkempt oriole]

Actually im 18

[cedar adder]

U are 22

Don't judge!

[unkempt oriole]

im just a jobless homeless guy

i don't even have money to but clothes

i don't have any job

i live under a tree

Im 49

[cedar adder]

Nah bro, u have to move on

[cedar adder]

Before you create a new law

[unkempt oriole]

Move on from being a poor?

[cedar adder]

Move on from under the tree

[unkempt oriole]

Ic

[lament shoal]

.closed

[unkempt oriole]

dudee

Where were u

[lament shoal]

pooping

[unkempt oriole]

💀

[lament shoal]

you know i ate like 50 kilograms of chips

[unkempt oriole]

Hell no

Let's not talk abt that

[lament shoal]

wait i got it

.close

[trim jolt]

Channel closed

Closed by @lament shoal

Use .reopen if this was a mistake.

[unkempt oriole]

So the solution to the problem that u asked are - (2,8),(8,2),(0,-6),(-6,0)

[lament shoal]

yea dw i solved it in the bathroom

[unkempt oriole]

💀

[cedar adder]

U are in ur first university year

[unkempt oriole]

That's too mucch

[cedar adder]

Engineering or maybe physics scientist

[unkempt oriole]

That's just permutations and combinations

[cedar adder]

I dont think computer sciences, ur brain too smooth to be one of them

[unkempt oriole]

Wottt?

No

im 13

I mean 15

What did i told u my age was?

14 im 14

[cedar adder]

I would say high school. But u published ur question in february

So university

[unkempt oriole]

So im 18?

Or 17?

[cedar adder]

Range: 17 ~ 19.88

[unkempt oriole]

well no

im older

[cedar adder]

Nah, i proved it this time

[unkempt oriole]

But idk integration

[proper kernel]

nice
I already typed a solution though so you can read it if you want

xy - x = y + 6
xy - x - y = 6
xy - x - y + 1 = 7
(x - 1)(y - 1) = 7

the factors of 7 are 1 and 7
as a result, these are the only 4 ways that integers can multiply to 7:
1 * 7 = 7
7 * 1 = 7
-1 * -7 = 7
-7 * -1 = 7
now you can rewrite these 4 ways to look slightly different:
(2 - 1) * (8 - 1) = 7
(8 - 1) * (2 - 1) = 7
(0 - 1) * (-6 - 1) = 7
(-6 - 1) * (0 - 1) = 7
this tells you that (x, y) = (2, 8), (8, 2), (0, -6), (-6, 0) are the 4 solutions, which come from the only 4 ways that integers can multiply to 7

the original problem has x, y ∈ ℕ which wouldve had the same steps, but you remove the (0, -6) and (-6, 0) solutions since those use negative numbers

[unkempt oriole]

I love the dedication man

[cedar adder]

Yes, i believe u

U know advanced integration

[unkempt oriole]

i wont lie im 16

[unkempt oriole]

No

[cedar adder]

I believe u, twice

[unkempt oriole]

how old is u

[cedar adder]

Goggle it!

[unkempt oriole]

r u famous? 😏

[cedar adder]

Yes

[unkempt oriole]

Ur name?

[cedar adder]

Selvator

[unkempt oriole]

Pikachu?

Oh ok

U r a painting?

[cedar adder]

Im ur grandfather

[unkempt oriole]

[cedar adder]

[unkempt oriole]

Nicee

[cedar adder]

.close

[unkempt oriole]

It's already closed

[cedar adder]

But u aren't yet

[unkempt oriole]

💀

ok close me

[trim jolt]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[grizzled zenith]

I understand how the unwise solution is reached and why it is an unwise solution but I have no idea how I would even approach getting to a wise solution.

[knotty locust]

Why is it unwise

[grizzled zenith]

Because if n is large then you will have a very large number of values to multiply together which would take a very long time

[knotty locust]

Is there a way to avoid that?

[grizzled zenith]

presumably. You could separate it into two sums and have a^2 for when n=m and then another sum for when n=/=m but that hardly speeds up the calculation

[knotty locust]

yeah

maybe do it for like

sum n=1 to 5

a_n

See if you can do any better

just with the small example

[willow urchin]

wiseness has nothing to do with computing time

[knotty locust]

That is based

[grizzled zenith]

Ah ok, i see what you mean, theres repeated multiples, so you could somehow create a summation when you have 2a_n a_m for when n<m so you only have to do each multiplication once and then multiply by 2.

[willow urchin]

theres a general law for multiplying sums, can u think of it?

[grizzled zenith]

Like this?

I feel like you need to be so creative to come up with that, thank you!

[willow urchin]

$\br{\sum_na_n}\br{\sum_mb_m}=~?$

[solid kiln]

ロケットジャンプ

[grizzled zenith]

Yeah I feel like it would equation this no?

[willow urchin]

very different

[grizzled zenith]

oh no

because the a and b are different functions?

[willow urchin]

possibly different terms yes

and n,m may have different ranges

here ill give it

$\br{\sum_na_n}\br{\sum_mb_m}=\sum_n\sum_ma_nb_m$

[solid kiln]

ロケットジャンプ

[grizzled zenith]

Yes, is it not more complicated than that though?

[proud flower]

note that $\left(\sum_na_n\right)^2 \neq \sum_n\sum_n (a_n)^2$

[grizzled zenith]

Because the question is asking for the same summation squared and writing it like that for a very large n would take a very large computing time

[solid kiln]

richard feynman

[willow urchin]

again wiseness has nothing to do with computing time

[proud flower]

the computing time is constant across these different representations of the same sum. the "wiseness" here is referring to the fact that it is hard to work with the product of two sums

[willow urchin]

here wiseness is about allowing unambiguous algebra

[grizzled zenith]

Ah ok! Thank you

So you need to differentiate between a_n and a_m instead of saying a_n and a_n

because you need all terms multiplied together?

[willow urchin]

yes and whenever we have multiple sums its good practice to use different indices

ofc there are exceptions where u plan to do some algebra that doesnt care too much about this practice

[proud flower]

to be explicit, if this were a finite sum:

$\left(\sum^n_{i=1}a_i\right)^2 = \sum^n_{i=0}\sum^n_{j=0}a_ia_j$

[solid kiln]

richard feynman

[grizzled zenith]

Yes ok, thank you so much, have a wonderful day!

[willow urchin]

np 🙂

[grizzled zenith]

.close

[trim jolt]

Channel closed

Closed by @grizzled zenith

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[urban copper]

let me translate it

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

lowkey

A = (4,6)

B = (2,-1)

wdym?

like a 2D vector can be divided into two components, x and y component (physics)

y component of (4,6) is

(4,6) . sin(theta)

I think

idk if its math related because is for physics intro class ,idk

I just need to decompose a vector in its y component

idk if its dot product related

nop

answers will be released the 26th

yes one moment

yea

mb

ok, good

tan(x) = 4/6

do I use arctan or no?

arctan is restricted to first quadrant tho

in this case will work

but idk if its bullet proof

,w range arctan

[solid kiln]

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[urban copper]

whatever, lets not overcomplicate things for now

,w arctan(4/6)

[solid kiln]

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[urban copper]

,w arctan(4/6) in degrees

[solid kiln]

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[urban copper]

sort of

not too much

yeah

is geometric definition of dot prod or something

didnt knew it was going to be helpful, but idk

but doing the way we did using trig was better tbh

the vector that is in the fourth quadrant is a little harder

yep finding (2,-1) angle formed with y axis

oops

you are right I am stupid

only A vector is needed

yep just ||A-B||

||(2,7)||

sqrt(4 + 49)

sqrt(53)

no?

wdym?

sometimes only one vector is given

using trig it would be translating the fourth quafrant vector to a vector in first quadrant and use symmetry of y axis I think

or something along those lines I think

let me show you, one sec

no only (4,6) was asked

well

the inner angle of the triangle that connects to y axis is 63,4°

the outer one is 243,4°

oops

63,4° + ? = 180

? = 180-63,4

? = 116,6°

!show

[trim jolt]

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

[urban copper]

which number crunching u did?

to get same angle?

yep that angle is 116,6

yep exactly

,w (2,-1)*(0,-1)

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[urban copper]

arcos(sqrt(5))

[solid kiln]

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[urban copper]

,w arccos(1/sqrt(5)) in degrees

[solid kiln]

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[urban copper]

nice ^^

I prefer the dot prod way now

physics is just hard

it isnt like math that you can be certain of things

you have to interpret statements and shit

good bie and thanks

.solved

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[wraith hinge]

Namaste Bhai/behna

Yahan par apne prashna ko angrezi mein translate kar ke daliye

Apko maddad mil jaye gi

[trim jolt]

@lean kraken Has your question been resolved?

[late flume]

What is that

[lean kraken]

Sure

Just a second

[late flume]

Sanskrit !!!

[trim lichen]

no, more likely Hindi.

[trim lichen]

you will need to translate though

for those of us that don't speak hindi

[strong delta]

i speak hindi idk these words wth

[trim lichen]

hm

[wanton bluff]

it asks locus i think

[trim lichen]

is it a different language then or do you just not know the math words

[lean kraken]

.close

[trim jolt]

Channel closed

Closed by @lean kraken

Use .reopen if this was a mistake.

[trim lichen]

suit yourself...

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[limpid siren]

Prove that the angle bisector of an angle of a triangle and the exterior angle bisectors of the other two angles of the triangle are concurrent.

[limpid siren]

😄

Let ABC be a triangle. BF is the angle bisector of angle ABC. AF and CF are the exterior angle bisectors of the other two angles, that is, they are the angle bisectors of angles DAC and ACE.

[queen burrow]

use coordinate geometry, but that is very lengthy and time taking

[limpid siren]

Coordinate geoemtry (black magic) is prohibited.

Pure Euclidean Geometry.

[queen burrow]

yeah makes sense lol

[trim jolt]

@limpid siren Has your question been resolved?

[trim lichen]

characterize each bisector as consisting of precisely those points which are equidistant from the sides of the angle it bisects.

this alone almost kills the problem

[main sigil]

And it also exposes one interesting property of the point of intersection

[limpid siren]

I have proved it IG.

Wait.

I have drawn BD as the angle bisector of angle ABC, and AD as the angle bisector of the exterior angle A of triangle ABC. These two intersect at D. I have joined C and D. It will be enough to prove that CD is the angle bisector of the exterior angle C of triangle ABC. I have also drawn all the interior angle bisectors of the angles of the triangle ABC.

angle FAD = angle DAC = x (let)
angle BAE = angle EAC = (180 - 2x)/2 = 90 - x
So angle EAD = 90 deg
Now let angle ACE = y = angle ECB
angle BAC + angle ABC = (180 - 2y)
angle BAE + angle ABE = (90 - y)
angle AEB = (180) - (90 - y) = 90 + y
angle AED = 180 - (90 + y) = 90 - y

Well angle BAC = 180 - 2x and angle BCA = 2y
So angle ABC = 180 - (180 - 2x + 2y) = 2x - 2y
angle AEC = 90 deg + (2x - 2y)
Angle AED = 90 - y [found earlier, check]
angle DEC = 90 + (2x - 2y) - (90 - y) = 2x - 2y + y = 2x - y
angle DAC = x

<@&286206848099549185>

I am in a panicky situation.

Assuming that the theorem holds true:
If DC was the exterior angle bisector of angle ACB of triangle ABC, angle ECD has to be 90. Since angle EAD is 90 too, quad. AECD must be cyclic. So angle DEC = angle DAC. This means x = 2x - y
Or x = y

Which is absurd.

[limpid siren]

All of this was proved without assuming that the theorem holds

[limpid siren]

With the assumption that the theorem holds, this was written.

So I have somehow logically arrived at the conclusion that x = y

This is absurd because this implies that angle ABC is 0 deg (trust me).

[limpid siren]

<@&286206848099549185>

Fast.

[trim jolt]

@limpid siren Has your question been resolved?

[limpid siren]

<@&286206848099549185>

Exactly 30 minutes have passed since I pinged the helpers.

So I rightfully ping the helpers again.

[trim jolt]

@limpid siren Has your question been resolved?

[stray sable]

Hey, is the set of all real numbers a subset of itself? If not, is there any notation that may hint to it being true?

[limpid siren]

!occupied

[trim jolt]

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).

[limpid siren]

<@&286206848099549185>

[versed trail]

?

[limpid siren]

?

Are you mental?

[versed trail]

?

[versed trail]

Yes, every real number is a subset of itself

$\mathbb{R} \subseteq \mathbb{R}$

[solid kiln]

kanisan

[versed trail]

What’s your problem @limpid siren

[limpid siren]

This is my channel

[versed trail]

Deadass I know

What you need help with

[limpid siren]

Everything is written above.

[versed trail]

Ok

[summer tapir]

well, DEC = DAC is correct

[limpid siren]

yeah?

what's wrong

There must be some mistake

[summer tapir]

no

it is correct

[limpid siren]

So x = y is correct?

[summer tapir]

no

DEC is not 2x-y

[limpid siren]

Oh no.

I made a mistake. I knew it.

Wait

[summer tapir]

AEC = 90 + x - y

is the actual thing

[limpid siren]

Let angle FAD be x degrees = angle CAD. Let angle ECA = angle EBC = y degrees
angle BAC = 180 - 2x
angle ABC = 2x - 2y
angle AEC = 90 + angle ABC/2 = 90 + x - y
angle AEB = 90 + (2y)/2 = 90 + y
angle AED = 90 - y
angle DEC = 90 + x - y - (90 - y) = x = angle DAC, which shows that quad. AECD is cyclic. Since angle EAD = 90 deg, angle ECD = 90 deg.
angle ACD = 90 deg - y
angle DCG = 180 - (90 - y + 2y) = 180 - (90 + y) = 90 - y = angle ACD

Proved

[limpid siren]

Thank you

[summer tapir]

welcome

[summer tapir]

also this is a much better method

[limpid siren]

What I did was misremembering my unclear memory of angle AEB being 90 + angle ACB (which is not the case, it is 90 + angleACB/2)

[limpid siren]

I don't understand that

[summer tapir]

consider an angle, and consider any point on its angle bisector, srop two perpendiculars from that point onto the lines of the angle

[trim lichen]

the bisector of an angle is the locus of a point equidistant from both of the angle's sides

[limpid siren]

oh

wait

okAY

[limpid siren]

.close

[trim jolt]

Channel closed

Closed by @limpid siren

Use .reopen if this was a mistake.

[limpid siren]

Note: I haven't yet been able to do it using the method told by Ann.

[trim jolt]

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[kindred jungle]

can someone please help w this

ping for reply pls ty

[simple panther]

There is nothing stopping you from directly substituting x = 0

Just do it

[kindred jungle]

we have to use epsilon delta to prove it

[simple panther]

Oh, yeah

Sorry

Didn't read it

[kindred jungle]

lol allg

[willow urchin]

@kindred jungle might not be able to stick around but heres the process of doing epsilon delta scratchwork

[kindred jungle]

mhm

[willow urchin]

to do scratchwork to find $\delta$ for
$$\lim_{x\to a}f(x)=L$$
\begin{enumerate}
\item let $0<|x-a|<\delta$ (with $\delta$ to be found later)
\item start at the expression $|f(x)-L|$
\item write a chain of inequalities (only involving larger expressions) and equalities. you should eventually end at an expression that is independent of $x$ and dependent on $\delta$
\item make a choice of $\delta$ in terms of $\ep$ which makes the last expression in the chain $<\ep$ (or $=\ep$ if the chain already contains $<$)
\end{enumerate}
once you finish these steps, write the proof using your choice of $\delta$, simply reformatting the scratchwork as the proof

[solid kiln]

ロケットジャンプ

[kindred jungle]

yeah so heres the problem im having

i got delta = min(1, epsilon/4)

which should be right

BUT

this is what my professor did

which i js dont understand at all

[willow urchin]

thats an eyesore

[kindred jungle]

fr 😭

here it is more zoomed in

[willow urchin]

ur delta looks off

if u follow my scratchwork process u should get

$|f(x)-L|\le3\delta^2+\delta$

[solid kiln]

ロケットジャンプ

[kindred jungle]

huhh

sorry i kinda rushed but

how is it so different 😭

[willow urchin]

i suggest u try doing scratchwork again using my process

[kindred jungle]

what do i do after |x||3x+1| < epsilon?

[willow urchin]

pls take a few mins to read it thru and do each step 🙂

[willow urchin]

here

[kindred jungle]

yeah im on step 3

started w |3x^2 + x| < e

factorised LHS

|x||3x+1| < e

idk what to do w the |3x+1|

ohhh wait

i have an idea

let me check smth

[kindred jungle]

ok i got to here

[willow urchin]

we not writing e anywhere yet 🙂 read 3 carefully

[kindred jungle]

i didnt know i could do that but now i do lol

oh what

ohhh i dont do e until later

ok hang on

ok yep i got to |f(x) - L| < 3d^2 + d

now i choose 3d^2 + d = e?

and solve for d here?

[willow urchin]

i would do algebra like this

$|3x^2+x|\le3|x|^2+|x|<3\delta^2+\delta$

bc triangle ineq

[kindred jungle]

triangle inequality

mhm

ohhh then you js sub in x for delta

[willow urchin]

ya

[solid kiln]

ロケットジャンプ

[kindred jungle]

mhm

but how do i choose like

how do i choose delta = smth in terms of epsilon?

[willow urchin]

we gotta pick smart conditions on delta

[kindred jungle]

oh yeah also does triangle inequality work w 3 variables?

i.e. |a+b+c| <= |a| + |b| + |c|

[willow urchin]

do u agree we should try to make each term <e/2?

[kindred jungle]

whys that?

[willow urchin]

and yes

well if so then ..+..<e/2+e/2

[kindred jungle]

ohhh so 3d^2 < e/2 and d < e/2

so 3d^2 + d < e

[willow urchin]

ya

[kindred jungle]

okayyy

[willow urchin]

now we get two conditions on d

theyre both d<… can u say em?

[kindred jungle]

d < sqrt(e/6) and d < e/2

is sqrt(e/6) bigger or is e/2 bigger

[willow urchin]

we dont know

[kindred jungle]

oh

[willow urchin]

and we dont need to care

[kindred jungle]

ohh

[willow urchin]

how can we pick d to satisfy both?

[kindred jungle]

im not suree

czu we dont know whats bigger

[willow urchin]

actually we can take nonstrict ineqs here

[kindred jungle]

ie if we pick d = e/2 and sqrt(e/6) is bigger then

[willow urchin]

bc we already have strict ineq earlier

[kindred jungle]

oh what

[willow urchin]

we already have strict ineq here

so the later ineqs can be nonstrict

[kindred jungle]

whys that?

oh because < implies <=?

[willow urchin]

no

if theres < anywhere in the chain then |f(x)-L|<e

doesnt matter how many = or <= there are

[kindred jungle]

oh lol

yeah fair enough

[willow urchin]

ok heres a general problem

if we want d<=a and d<=b how can we pick d?

[kindred jungle]

d <= a+b?

or d < a+b?

[willow urchin]

no like define d

[kindred jungle]

OH

um

im not sure 😭

d = a-b

?

[willow urchin]

no

[kindred jungle]

mmmm