geometry-and-trigonometry

[dark sparrow]

advertising your own discord is against the rules here lmao

got any more creative insults to hurl my way?

ah. mans got bopped.

[next grotto]

i wud suggest to not rely on desmos this much

like only use it on polynomials of powers more than or equal to 4

[short panther]

You have no idea on how much I have accidentally messed up things😭 if you accidentally mess up a single mistake then you’ve basically cooked yourself

If a graphing calculator can be used it will be used!

[proper matrix]

Can someone explain to me why the answer for A is 50 minutes and not 52…? Please!!!

[dark sparrow]

they are rounding to the nearest 10 minutes @proper matrix

[proper matrix]

But when I did .871 (60) my answer was 52.2631272 and wouldn’t the nearest minute just be 52…?

[dark sparrow]

no

not the nearest minute

nearest 10 minutes

[proper matrix]

OH!!!! 😭 I was so confused when they added that part in…. That was never explained in class so I just thought it was the solid number before the decimal. Thank you!

[sudden solstice]

Can anyone help me understand Trigonometry??? I don’t know how to do it at all and I think I got like a 30 on my quiz I am learning con tan and sin and I don’t know when I am supposed to have the -1 with the con tan and sin I am also doing it with angles, triangles and ratios

[lime dune]

!da2a

[lime crown]

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/

[lime dune]

show us a question you’re struggling with

[split bluff]

what's the answer?

[trail tendon]

y = (Do) sin( (Your)x + (Math) ) + (Problem)

jkjk what have you tried 💀

[next jewel]

im on this unit in algebra 2 right now

currently this one seems a bit farther than what we have covered so far but i immediately notice that the peaks when y = 0 are definitely displaced when they should be on 2 pi and 0 radians right? so there has to be a horizontal shift

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[mystic umbra]

notice how the function's crests are at 0

what does it tell you?

[split bluff]

idk

[mystic umbra]

😔 this guy

do you even know what function translation and scaling is?

[next jewel]

in the regular sine graph, the crests lie at y = 1

[proper matrix]

Can some one explain how to do y=-1 +2 tan(x) is the real problem just y=tan(x)?

[signal vapor]

They will be different

One is tan-¹ (y) while the other is tan-¹ ((y+1)/2)

[proper matrix]

Oh okay thank you! 🙋🏻‍♀️

[dark sparrow]

missing brackets

[sudden solstice]

Does anyone know where I can get physical science help and also I need help learning trigonometry like sin cos tan and cos-1 sin-1 and tan-1 idk how to know to put the -1 and not cause I got a 30 on my math quiz

[sudden solstice]

I am struggling doing the SOH CAH TOA things and solving for angles missing angles and sides and missing sides with triangles

[strange lark]

But what kind of questions/concepts? There are laws of sine, laws of cosine, SSA triangles, and it may involve other properties you probably forget in middle school.

[silent plank]

inverse trig functions are usually accessible on hand held scientific calculators by pressing
shift/2nd followed by the respective trig function

[strange lark]

1/sin(x) and sin‐¹(x) are two distinct concept. 1/sin(x) is simply the reciprocal of sine (cosecant), while the other one is inverse function of sine. (you get an angle for the respective input value of sine)

[next jewel]

rip i forgot that lol

[waxen temple]

mathmathmath

[pine thunder]

Why is cross product of two vectors defined the way it is? I mean the right hand rule for the unit vector into the ab sin(theta) seems extremely random. Is there some logic behind this.

[lime dune]

right hand rule is simply arbitrary convention

you'd need to learn some physics to see why cross product is defined the way it is

[proper matrix]

So I feel like these aren’t supposed to connect this way. Am I missing a step?

[lime dune]

take note of the vertical asymptotes, your graph shouldn't be intersecting them

[proper matrix]

i understand now!!! thank you!

[lime dune]

np

[proper matrix]

The radical 3 is throwing me off can someone help explain the first value and then I can get the rest after…?

[knotty wasp]

why is cosine related to x values and sin related to y values

[lime dune]

we rewrite as $3\cdot\frac{1}{\sqrt{3}}$

[somber coyote]

elrichardo1337

[lime dune]

but we can write 3 as $\sqrt{3}\cdot\sqrt{3}$

[somber coyote]

elrichardo1337

[lime dune]

so one of the sqrt3s cancels

and you're left with $\sqrt{3}$

[somber coyote]

elrichardo1337

[proper matrix]

Because when you use cosine you get the value of x/1 which is why it’s only x and the same goes for sin because when you get y/1 it ends up just being y

[lime dune]

^ take the unit circle for the above

[proper matrix]

🙏 thank you your a life saver!

[lime dune]

now if you have a ray from the origin that makes a counterclockwise angle $\theta$ with the positive $x$ axis

[somber coyote]

elrichardo1337

[lime dune]

it intersects the unit circle at the point $(\cos\theta,\sin\theta)$

[somber coyote]

elrichardo1337

[proper matrix]

Because when you use cosine you get the value of x/1 which is why it’s only x and the same goes for sin because when you get y/1 it ends up just being y.

[knotty wasp]

then why is the equation for polar coordinates r*cos theta=x

and not just

cos theta=x

[cunning lion]

x = cos(theta) is only for the unit circle, x = r cos(theta) applies to all circles centered at the origin

[plain storm]

How much should I try and memorize vs intuitively understand with trig? Ik it's a lot more memorization based than other math fields but I'd like to keep memorizing to minimum

Thanks that should be pretty useful

[proper matrix]

Should I not use my quarter period for this or would it even matter if I picked different values…?

[maiden brook]

how much trig are u doing? and geometric trig or what?

[fathom bone]

Hi I wantsome help:
Can someone share the proof/ how to prove similarity of triangles.
i.e if angles are equal the sdes are proportional.

closes thing I found was
Basic Proportionality Theorem for Triangles
But that's not exactly "it" as it talks about splitting the line ratio.
Not side to side ratio ig.

Can anyone point me in the right direction/share link etc

[mystic umbra]

can you be more specific?

[fathom bone]

If triangle AbC is similar to triangle DEF.
AB/DE = BD/EF = AC/DF

How to prove the ratios are the same essentially

Given similar triangles have all angles equal

[mystic umbra]

if you know the angles, the proportionality comes from sine law

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

[somber coyote]

parabolicinsanity

[fathom bone]

Ohh ok.
Thank you!
I'll take a look at that

[mystic umbra]

so if $\frac{a_1}{\sin A}=\frac{b_1}{\sin B} \land \frac{a_2}{\sin A}=\frac{b_2}{\sin B} \implies \frac{a_1}{a_2}=\frac{b_1}{b_2}$

[somber coyote]

parabolicinsanity

[fathom bone]

Another question:
For calculation of pi:
For circle properties.
Was it just noticed
:
pi = circumference/diameter consistently

Or is there some other way of calculating pi
Or proving circumference = 2pi.r

Most explanations seem to link these.

two together circularly making a chicked-egg situation in my mind ig.
Trying to find a proof for this ig.

Would someone be able to help me out.

Ty in advance!

[worn stag]

There are other ways of defining pi yeah

[cunning lion]

well if we have [ \pi = \frac{C}{d} ]
then we also know that $d = 2r$ so you could equivalently write [ \pi = \frac{C}{2r} ]
and multiplying both sides by $2r$ we have [ 2\pi r = C ]

[somber coyote]

cloud

[cunning lion]

so really those two expressions relating pi are equivalent

[fathom bone]

Yeah so,
is there like a proof for this,
Or was it just an observation of property?

[cunning lion]

i mean i just did that right here

[fathom bone]

I meant proving pi = C/D essentially

[cunning lion]

that's the definition of pi

[mystic umbra]

you can't "prove" a definition 😔

[cunning lion]

if you're wondering why that ratio is the same for all circles you can have a look at this: https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-circle-basics/v/seeing-circle-similarity-through-translation-and-dilation

[fathom bone]

So like :
Say I want to prove
Circumference of circle = 2 . pi. r

How would I do this?
If we go by what you posted:
pi = c/d
And that way
c = 2.pi.r
Which is fine but how do we get pi.

So was pi just an observation of circle properties defined as such

[mystic umbra]

yeah

[fathom bone]

Ohh ok.
Thank you both!

[cunning lion]

you start with one formula (the definition of pi, which is pi = C/d), then use that to prove the thing you want

[fathom bone]

I've been trying to go through basic math formulas (working my way up) and get proofs/build up theory/understanding for their implementation/how trhey were derived,

Since I feel it gives better clarity overall than just remembering the end formulas,
That and at times it becomes circular
x lay because y law
y law because x law doesn't make sense in that sense.

So was just trying to figure out w.r.t the circle here and similarity of triangles Q before.

Though ig if that's how it's defined as a baseline.
It makes sense or can be said to be observed property that the pi ended up same for all circles ig

That understanding is not incorrect right?

[lapis moon]

Spivak's Calculus gives a rigorous definition of pi

having all the desired properties seen in elementary math

[fathom bone]

Thank you!
I'll give that a look.
Any chapter/section I can jump to /find it under?

[lapis moon]

chap 15/16

[fathom bone]

Ty!

[unkempt topaz]

yes, the first definition of pi is the ratio of circle's circumference to diameter

you can derive the area of a circle in terms of a circle's circumference and radius too, but people tend to prefer using a radius to define a circle rather than the circumference, which is why its written with the constant pi

thinking about it, you can define the constant 'e' in at least 3 ways, although I prefer to think there was 1 original definition and the other two were derived

[obsidian harness]

what have you tried?

[rapid hearth]

what do you mean? like how did i solve?

if ur asking that i knew the parabola opened upwards because the focus is above the directix so then 4p(y - k ) = (x - h)^2 then i plug in my values which is 4(y + 2)= (x+6)^2

[obsidian harness]

oh like why did you post the image then

I thought you were asking for help

[rapid hearth]

no i just wanna know if im right cause i gotta show how i did it tmr in front of the class for review before the text

test*

it would be kinda awkward if i did all that and got it wrong

[obsidian harness]

yes, so the vertex is equidistant to the directrix and the focus

so it does have vertex (-6, -2) and that's the only option that matches

[rapid hearth]

thanks for your help 🙂 glad im correct

[obsidian harness]

,w directrix y = -3 focus (-6, -1) parabola

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

-28 - 12 x - x^2 + 4y = 0 which is the same as that

[rapid hearth]

ah ok i see

[obsidian harness]

"click here to refine your query online"

WA the website is better

[rapid hearth]

ive never seen this website it looks cool

[obsidian harness]

indeed

well it's so annoying cause it doesn't understand around half of queries

[rapid hearth]

thats like every app i feel like they usually get half wrong or more

[rapid zealot]

what queries you got?

[rapid zealot]

just check which points satisy the equation bj checking the options

is A the answer?

[daring plank]

If you know the psychology behind the exercises

You can instantly eliminate two of the options

[unique cradle]

The first thing to look at is the center. Find the center point. Then convert that into the formula.

(x-h)^2 + (y-k)^2 = r^2

Your center is (h,k) and your radius is r.

So for this problem, your center is (3, 4) and your radius is 3.

So your formula would be (x-3)^2 + (y-4)^2 = 9

[unique cradle]

If you need more help or explanation with parabolas, circles, ellipses, hyperbolas, etc, let me know. I teach precalculus and just taught these concepts two weeks ago to my seniors.

[upper karma]

Hi how do I put log into a calculator

I want to do Log 45

[pale sentinel]

Depends on the calculator

Most scientific calculators have a dedicated button for it

[undone sundial]

never knew how hard geometry would be for me

idek if its just me but algebra 1 was a lot easier?

[upper karma]

Just figured, since log isn't too common on simple calculators, sucksto suck I guess!!

[undone sundial]

can anybody relate ^^^

[pale sentinel]

Are you using a finance calculator?

i.e. just a simple one with +-x/, numbers and some memory functions?

[upper karma]

yes

Also tried the phone calculator

[pale sentinel]

Yeah, you're not going to get far doing (school) maths with a calculator like that

[upper karma]

but thing is that calculator only gives logbase10

[pale sentinel]

There is a formula for changing logs, if you can remember it

Taking d = 10 (for log) or d = e (for ln) you can calculate the log of any base from this

[upper karma]

This concept is stupid

its mixing exponents and Squareroots together

[pale sentinel]

It's a core property of logarithms

Further, square roots are just when your exponent is a half

And further still, logarithms are intrinsically linked with exponents almost by definition lol

[upper karma]

cuberoot fourthroot, squareroot doesnt matter its all the same dumb idea with a little more steps

ridiculous concepts!!!!!!

no hate tho thanks for the help lol

these concepts are just killing me

[next jewel]

rooooooots

dont worry im rooting for all of you to do good 👍

[next jewel]

for some calculators like ti-84, going into math if u scroll down far enough there should be an option called log base which will let u choose the base

[warm obsidian]

hi guys

question about law of cosines

so, ik the equation has a minus sign in it

and so first I tested it out using the minus sign, and then I tested out the equation using a plus sign instead of a minus sign

and the two angles combined formed 180 degrees

so if u use a plus sign instead of a minus sign, are u basically not finding the inner angle of the triangle?

but instead finding the "outer angle" / supplement of the triangle's angle?

plz ping, thanks!

[next jewel]

@warm obsidian hey this seems pretty creative! when u say u tested using a plus sign in the law of cosines, did u do (a^2 + b^2 + c^2)/2ab to evaluate cos (C)?

[warm obsidian]

yes

I evaluated them both the same way

[next jewel]

u evaluated angle c the same when u added c^2 and subtracted c^2?

[warm obsidian]

yes

and both of those added up to 180

[next jewel]

hmm i see

[warm obsidian]

yeah so I just wanted to make sure that was a valid way of finding the supplement of a triangle's angle

[next jewel]

i tried what u said with a non-right triangle with sides 6, 7, and 8 where a=8, b=6, and c=7. angle c = 57.9 but when u try adding c^2 like u said u end up doing cos^-1(1.55208) which is undefined

so ur idea might or might not work conditionally but im not so sure abt using it generally 👍

specially because cos^-1(x) has a pretty strict domain

[warm obsidian]

hm lemme try

[warm obsidian]

hey bro I think u might have messed something up

cuz I just tried it and I got 57.91 with the minus and 122.09 with the plus

[next jewel]

what were ur steps

[warm obsidian]

so literally do the same thing as u would with the minus

[mystic umbra]

@warm obsidian if you have + instead of -, you're just finding the resultant of the vector addition

something like this

[warm obsidian]

oh

[mystic umbra]

$R=\sqrt{A^2+B^2+2AB\cos{C}}$

[somber coyote]

parabolicinsanity

[mystic umbra]

I hope that helped

[warm obsidian]

I was talking about the angle specifically

it seems like doing that finds the supplement of the triangle's angle

[mystic umbra]

making the angle negative?

[warm obsidian]

instead of - 2ABcos(theta) u have + 2ABcos(theta)

[mystic umbra]

yeah and that is the resultant 😔

of vector addition

but yes

it is the supplement

as $-\cos{(A)}=\cos{(\pi-A)}$

[somber coyote]

parabolicinsanity

[warm obsidian]

I see

so it is a valid way of finding the supplement then?

[warm obsidian]

also isn't the resultant found with just the a^2 and b^2 and the angle found using arctan?

[mystic umbra]

yes but it's kinda redundant when you can just use the identity i just gave

[mystic umbra]

the arctan angle gives the angle

or inclination

but the magnitude is given by the formula i provided, exactly what you're doing

[warm obsidian]

oh

[next jewel]

damn then i have no clue what i did wrong in trying to check this lol

nvm seems like it works

[warm obsidian]

no yeah ik this was probably a weird thing to ask, it's just that I'm reading an engineering-related book, and they basically had that there

[mystic umbra]

weird

[warm obsidian]

it was the law of cosines except it had a plus instead of a minus

[mystic umbra]

basically what you're doing is inverting a triangle and that gives the resultant

[warm obsidian]

so yeah I asked ChatGPT about it, and it told me that that finds the supplement angle

but it's ChatGPT, so I wanted to make sure it was accurate lol

[mystic umbra]

$\vec{A}+\vec{B}+\vec{C}=0$

[somber coyote]

parabolicinsanity

[mystic umbra]

$\vec{C}=-(\vec{A}+\vec{B})$

[somber coyote]

parabolicinsanity

[mystic umbra]

since $\vec{R}=\vec{A}+\vec{B}$

[somber coyote]

parabolicinsanity

[mystic umbra]

$\vec{R}=-\vec{C}$

[somber coyote]

parabolicinsanity

[mystic umbra]

now if we invert C, which means we take the supplement of the angle of C in the cosine law

we get R

$\vec{R}=\vec{C}_{anti-parallel}$

[somber coyote]

parabolicinsanity

[mystic umbra]

which is basically what you did

hope that made sense

infact, that's one way to prove the cosine law

via vector addition

[warm obsidian]

interesting

[warm obsidian]

I assume this identity holds true for sin as well?

[mystic umbra]

no

[warm obsidian]

oh shit

[mystic umbra]

$\sin{(\theta)}=\sin{(\pi - \theta)}$

[somber coyote]

parabolicinsanity

[mystic umbra]

refer to CAST rule for specifics i guess

[warm obsidian]

interesting

well, thanks!

[mystic umbra]

happy mathing

[sly venture]

This one anyone

[obsidian harness]

probably just use area = 1/2 ab sin C

also the statement is equivalent to proving $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$, if that helps

[somber coyote]

south

[obsidian hornet]

From sin area formula easy to see that opposite triangles area in sum gives quadrilateral area

[sly venture]

Yeah I got that thnx

[sly venture]

Thnku mate

[obsidian hornet]

No problem

[sly venture]

Tell me something is the condition for riemann sum limited

[obsidian hornet]

I don’t actually do calculus, but i think that something like continuity or boundness with finite discontinuities

[sly venture]

Ohk np
Thanks

[obsidian harness]

the proper condition is that the upper and lower sums converge to the same value

search up 'Darboux sums'

[sly venture]

Ik that

[sly venture]

I was just asking coz I was doing a question and I was stuck should use it or not but they used it anyway

[upper karma]

@deft summit <--- this guy is very good at this stuff hell solve it

[deft summit]

I'm not and I gave this problem an attempt and got nowhere any hints are much appreciated

[upper karma]

(hint: you can do this...!!!)

[lyric quail]

heyy

[daring plank]

Alright

So bring that question back up

[lyric quail]

Anna notices: In the sink, the water doesn't drain away so quickly.
The metal plate over the drain has 6 small holes, each with a diameter of 1 cm.
The opening of the siphon has a diameter of 4.4 cm.

Anna is wondering whether even half of the available area is used to let the water drain away.

6 small circ: 4,71

Half of big circ: 7,6

[daring plank]

So what it's asking is if the area of half of the bigger circle is as large as the area of all 6 other smaller circles

We know that the area of a circle is πr²

And that the radius is half of the diameter

So what this becomes is π(4.4/2)² for the larger circle

[lyric quail]

7,6>4,71 so the 6 circ fit inside the big one. ans is yes?

or did I misunderstand smth

[daring plank]

It's not asking if the 6 circles fit in the big one
It's asking if half of the area of the big circle is equal to the area of all the smaller circles combined

[lyric quail]

wait like 6A_s=1/2A_B

?

[daring plank]

So yes the area of the smaller circles combined is 4.71

[lyric quail]

A_small and A_big

[daring plank]

Yes

[lyric quail]

so ans is no

?

[daring plank]

But the area of the larger circle is 15.2
Not 7.6
You forgot to half the diameter of the larger circle to get the radius

I think?

So

Wait no you did it right

[lyric quail]

the q considers half of the area

[daring plank]

Half of the larger circle is 7.6

[lyric quail]

yep

[daring plank]

So the answer is no
Half of the available area isn't used to drain the water

[lyric quail]

what av. area exactly?

like the area of the big circ?

[daring plank]

Yeah

[lyric quail]

hmm okayy

[daring plank]

The big circle is the siphon

A siphon is the end of a faucet

[lyric quail]

okay so half of the siphon is 7,6

[daring plank]

And the 6 small holes are 4.71

[lyric quail]

and the 6 circ are 4.71 together

[daring plank]

Approximately

[lyric quail]

ye

so what is the q asking again?

[daring plank]

If half of the area of the big circle is the same as the area of the small circles

[lyric quail]

hmm okay

I'm so fkd rn

idk why

everyone says something different

[daring plank]

Anna should replace her siphon that shit's a scam

[lyric quail]

Are the 6 holes together smaller than half the area of the drain – or larger

is that right?

thats the q?

[daring plank]

Mmm

[lyric quail]

that what chat gpt say

says

[daring plank]

Anna's wondering if even half of the siphon is used to drain

So

Yeah

[lyric quail]

@austere silo

[austere silo]

So you've calculated the area of the small circles and of the big circle?

[lyric quail]

yep

6 circ: 4.71

half of big: 7.6

[austere silo]

And that was using pi*r^2

[daring plank]

Mhm

[lyric quail]

yep

[daring plank]

We halved the diameter to get the radius

[lyric quail]

We have no problem with the math. is just the q

[austere silo]

Then yeah, I think that implies that the water is draining in less than half of the area of the big circle

But also, ask your teacher.

[lyric quail]

one q to you two: is the q well ,,asked''?

[daring plank]

It's pretty weirdly worded
But that's the fun in word problems

[austere silo]

When the problem doesn't include a specific question or instruction, I would guess that there's a specific quesiton or instruction that's missing.

[daring plank]

What could be missing?

It's asking to compare the area of the smaller circles combined and the bigger circle
We got their areas
And we compared them

[austere silo]

The question "Can water drain through more than half the area" or someth like that

A direct question

Not just "Anna is wondering"

"Anna is wondering" isn't a question or instruction, it's a statement.

[lyric quail]

is there anyway to turn the q into comparing circles?

just cuz I'm an idiot

^^

[daring plank]

It's implying the question

[austere silo]

If it wasn't a word problem, then it would ask "Is the sum of the areas of the smaller circles greater than, less than, or equal to half of the area of the larger circle?"

[daring plank]

Well yeah

But
This is a word problem

This is the fun in word problems

You have to think about what it's really asking mathematically

Otherwise they wouldn't give you word problems in schools

[lyric quail]

my problem is my language weakness😭

[austere silo]

As a word problem, it should say "Can water drain through more than half the area?"

I've also seen stuff like "Anna thinks water can't drain through more than half the area. Is Anna right?"

[daring plank]

Yes

It's implying that you answer the question

You have to find that

You help solve what Anna is wondering about

[austere silo]

It's bad form to have a problem that doesn't directly ask a question or provide an instruction.

[daring plank]

I find it thought provoking and fun

[daring plank]

Don't worry, I'm not native to the English language either

I'm Romanian

[austere silo]

Or even "Anna wonders if water can drain through more than half the area. How could she find that out?" And the answer would be "She should use pi*r^2 to find the area of the circles, and compare 6 times the area of the smaller circles to half the area of the larger circle"

[lyric quail]

if the 6 circs are bigger than half of the bigger circ would be a hole right?

[daring plank]

What would be a hole

[lyric quail]

the half of the bigger circ

ye

its a hole then

[austere silo]

The circles are each a smaller hole.

Or I guess, 6 small holes

[daring plank]

I'd consider the question solved, right?

[lyric quail]

ai's solution

[daring plank]

That's correct too

[lime dune]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[fallen portal]

can you simplify 64 * ((3 * sqrt3) / 2) + 288

[austere silo]

64 is divisible by 2, and then you have two integer coefficients on sqrt3.

[undone sundial]

I hate proofs

[mystic umbra]

maths is mostly proof based so better stop complaining and start proving

[undone sundial]

I used to be so good at math until geometry.

idk if it's me or my teacher

[mystic umbra]

probably the latter because euclidean geometry is trivial for most maths people usually...

in general maths is easy if you put your head into it, like it makes intuitive sense once you look into it proper

ex: orthocentre, centroid, and circumcentre are collinear with a 2:1 ratio

it might seem weird or daunting but it makes sense once you look into the proof for it (via homotheities and triangle similarity, congruency)

[golden crystal]

mmm :(

[azure helm]

are all squares rectangles and are all rectangles squares?

how to prove it?

[trail tendon]

look! A square!

[azure helm]

how to prove the latter is false and the former holds for every square

[trail tendon]

what's the definition of a rectangle and whats the definition of a square

[azure helm]

unsure

[deft bluff]

@azure helm
Well, let's look at it carefully then.

Do you think this image represents a square?

If not, why?

[azure helm]

the image represents lowkey a rectangle

but I dont think the interior angles are 90 90 90 90

[deft bluff]

Yep, but why not a square?

[azure helm]

because is slanted

[azure helm]

the length of every side is not equal

[deft bluff]

I'm pretty sure Viper made his diagram assuming that the angles are all 90, it was just shakily drawn

[deft bluff]

Aha - and there we have our definition

[azure helm]

ok

[deft bluff]

A square is a four sided shape where:

• All the angles are 90
• All the side lengths are equal

[azure helm]

a polygon that has 90 90 90 90 interior angles and every side is of equal length

[deft bluff]

Exactly.

Now let's define what a rectangle is

[deft bluff]

Based on this, what do you think a good definition for a rectangle is?

[azure helm]

just a irregular polygon with every interior angle being 90 deg and two sides being equal to each other and two sides being equal to each other but not all of them are equal, not the four sides are equal to eacher other

[deft bluff]

Yeah.
Although, the way we usually define retangles is we don't have that irregularity condition. That is, a rectangle is just a four sided polygon with the interior angles as right angles.

[mystic umbra]

all squares are rectangles, the connverse is not true

why so serious about basic euclidean geometry

[deft bluff]

And this falls under what you are asking:

1. All squares are rectangles
   A rectangle is a four sided shape with all 90 degree angles in it. Squares have all 90 degree interior angles, so a square is a rectangle!
2. All rectangles are not squares
   A square has a specific condition - that all sides are the same. Not all rectangles have equal sides, and so, not all rectangles are squares.

[mystic umbra]

weird for an undergrad to be asking a problem about basic euclidean geometry 😢

yes, i am discriminating

because you should know as much 😭

[obsidian hornet]

I tried to solve the problem from problem board and stuck. O,I-incenter and circumcenter of ABC, A'-reflection of A relatively OI. Need to proof that 3 point lye on one line. I found some inscribed quadrilaterals and chased some angles, but it leads to nowhere. Appreciate for help in advance. P.S. Additionally noticed: black circle is trident circle for triangle XYA', tried to proof that D has equal powers above black and red circle(coz Pow(D,Black)=(p-b)(p-c) and this seems like a good condition for doing this, but i don’t know anything about Z point for counting Pow(D,Red)), centre of the red circle lies on green circle.

[whole smelt]

hi

[visual flume]

help pls

[mystic umbra]

#competition-math people could help you better

[mystic umbra]

feeks like extended sine law

$\frac{a}{\sin A}=2R$

[somber coyote]

parabolicinsanity

[mystic umbra]

try visualising first

could help

[visual flume]

ty

[fresh lava]

am i allowed to ask a question that technically slips into physics here? about breaking a velocity vector into components

and its real world application in bumper car collisions

[lime dune]

i think you could probably fakesolve by setting both B and D on the perpendicular bisector of AC?

[mystic umbra]

yeah, it's maths obviously

you can

technically breaking down a vector into its components is projecting the vector onto the axes or something

[fresh lava]

alrighty then

i have a question regarding bumper car collisions:
if a bumper car collides with another stationary one from the back at an angle, can you break the velocity vector of the moving car into two parts: the normal component and the tangential component? The normal component is the line between the two centers of the bumper car, and the tangential is perpendicular to that. Is it true that the normal component is the one that is responsible for the deformation of the stationary bumper car (as its the direction of the force) and the tangential is responsible only for the sliding of the bumper cars-- no deformation?
is that true?

[mystic umbra]

hmmm try visualising it first

would help if it's just particles really, don't worry much about the shapes

[fresh lava]

i think it would look something like this.
green is the velocity vector
blue is the normal component
yellow is tangential component
and so the normal is responsible for the deformation of the bumper and the tangential just for them sliding together with no deformation. is that true?

[mystic umbra]

ehhh, you vould've drawn just points because the problem isnt really focused on the geometry of the vehicles themselves

but also, denote the angle (usually theta) as well

[strange lark]

M is midpoint of line AB. I feel like ∠BOM is a right angle, but I couldn't tell why.

[pale sentinel]

There's nothing there to suggest that

What's the full question, if there's any more text? (even if it's in Chinese (though do your best to translate if you can))

[strange lark]

On the coordinate plane, O is the origin, A is (6,0), OB=2√6. If M is the midpoint of segment AB, and tan ∠AOM=√2, what's the value of tan ∠AOB?

[grave pond]

At least you can throw algebra at it. You can write the coordinates of M and then B as functions of the x-coordinate of M, and the solve for the x that gives OB the right length.

There may or may not be a clever synthetic approach instead.'

[pale sentinel]

That's the right idea, as far as I can make out

[grave pond]

Memes go in #chill.

[pale sentinel]

[contextually] drop an altitude from M to OA, call this point D.

tan AOM then equals MD / OA

[grave pond]

Actually, the locus of possible B will be a straight line through (-6,0) and a slope of sqrt(2).

[pale sentinel]

But then that's also how you'd calculate the gradient of OM

[grave pond]

So we can just solve the (SSA) triangle between O, B, and (-6,0).

[grave pond]

THe slope of OM was explicitly given in the problem, if I read it right.

[pale sentinel]

yeah, via tan AOM

[strange lark]

B is in the second quadrant.

[pale sentinel]

So - let M be the point (x,y).

1. since OM has gradient tan AOM = sqrt(2), what relation can you give between x and y?
2. Because M is the midpoint of AB, where is B?
3. Since OB has length 2sqrt(6), what other relation can you give between x and y?
4. solve for x and y.
5. Where then is B?
6. Drop an altitude from B, say it lands on a point D on the x-axis
7. find tan (DOB), then find tan(AOB).

I've yet to calculate anything, but this would be my approach here

[grave pond]

What I was getting at: call the point (-6,0) E; then triangles AOM and AEB are similar, so tan AEB = tan OEB = sqrt(2) too. Now only O, B, E matters.

[pale sentinel]

Ah that's not a bad idea

I'm just wondering though how that assists with angle AOB

[grave pond]

We're now just looking at solving triangle EOB. This is now a standard SSA problem, so there will be two solutions (unless the problem ends up impossible or degenerate).

[strange lark]

The question explicitly stated that B is in the second quadrant.

[grave pond]

Ah, perhaps that will exclude one of the SSA solutions.

[pale sentinel]

It will

I think the other SSA solution has B in the 3rd quadrant

Because it's symmetric

[grave pond]

It's not symmetric -- the slope of OM was sqrt(2) rather than -sqrt(2).

[pale sentinel]

Ah

Well, nearly

Extend OM past O to superficially get a new M'

Extend AM' appropriately to a point B' such that OB' has length 2sqrt(6)

[pale sentinel]

oh wait then this approach will also get you two solutions (eh not quite?)

[grave pond]

But then B' gets an x-coordinate left of -6, so |OB| can never be as small as 2sqrt(6).

[pale sentinel]

Ah igu

Anyways, you've got two lines of attack then - Troposphere's more geometric approach or my more algebraic approach

Combine with B having to be in the second quadrant and you should be golden

[pale sentinel]

(I never really thought about it, but given what it means as well, "troposphere" is a rad name to have ngl)

[high wadi]

hey there I want to ask if someone could explain why x = r cosθ = y/tanθ as shown in my notes in the photo

I have been slumped for a solid 20 minutes on why it is like that, any help is greatly appreciated!

[pale sentinel]

From cosθ = x/r we have x = r cos θ

From tanθ = y/x we have x = y/tan θ

Since these are both referring to x, r cos θ = y / tanθ

[high wadi]

is it just solving for x in the case of x/r = cosθ and etc

[keen marsh]

AB is parallel to CD, I have to find the length of BD

can someone help me?

[quick sandal]

hello! does anyone wanna be a study buddy for geomtry / trig?

or if anyone just wants to volunteer their help to me😭

[obsidian harness]

Similar triangles

[mystic umbra]

ratio of the sum of the two lengths of the sides and respectively 12.3 14.8 is constant

due to similarity, yeah

[thick schooner]

Any angles?

[pale sentinel]

This is just algebraic manipulation at that point

[pale sentinel]

You don't need to know the values of any angles

If we let the top vertex be called E for the sake of nomenclature, then
angle E = angle E (reflexive property [meaning they're the same angle])
angle EAB = angle ECD (corresponding angles because AB || CD)
angle EBA = angle EDC ( " )
So triangle EAB is similar to ECD

[thick schooner]

Ik, but you could do it another way

Math is a full of paths (if done correctly and applied correctly) which lead to one destination, the correct answer

[pale sentinel]

I mean, sure

But you don't even have sufficient information to calculate any angles

Functionally you have to spot the similar triangles to make any progress here

[upper karma]

guys forget this

you know you can derivate sine(x) just with geometry

[upper karma]

yo this easy (12.3/15.5 = 14.8/BD)

its the thale thingy

you can prove it considering areas of triangles and stuff

[echo sigil]

Does anyone know why there is AM/AC multiplied? Isn’t it just PC/MP from sinuses theorem?

[obsidian hornet]

No, u just using additional angels: CP/sinCAP=AC/sinAPC ; MP/sinMAP=AM/sinAPC => sinCAP/sinMAP*MP/CP=AM/AC=> sinCAP/sinMAP=AM PC/MP AC

[echo sigil]

Got it,thanks!

[obsidian hornet]

No problem 👍

[hardy bobcat]

This gave me a headache

[jade sierra]

pls help

due in a hour

[zealous pike]

Is this solvable with using the area of an equilateral triangle? That being the case, is the area of that region

$\frac{\pi}{6} - \frac{\sqrt{3}}{4}$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

?

The first fraction being the area of the whole sector from the vertex to the extremities of the chord, and the second fraction being the area of the triangle that's formed by connecting the chord extremity points to the vertex

My reasoning is that, if there is a radius of 1, then it is reasonable to consider that the chord is equal to the radius, therefore, by connecting the extremities of that chord to the vertex we get a triangle with sides of 1, which is an equilateral triangle

We have that the area of an equilateral triangle is $\frac{\sqrt{3}s^2}{4}$ where s = 1 in this context

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

if we subctract the area of the whole sector by the area of that triangle, we get the area of the shaded area

[barren venture]

I think your approach and result is correct

[obtuse karma]

I have an trigonomety examn

help

im in spain 4t eso

[spiral lodge]

We don't help during the test (of course...)

But if you want to check your answers afterwards, sure

[rotund hemlock]

Where are the numbers yo 💔

[whole smelt]

Can someone help

#help-30 message

[azure helm]

can I get some hints

[obsidian harness]

you could 'cheat' by letting P be point A

then G3 would be the midpoint of AB, G2 would be the midpoint of AC, and G1 would be the centroid of ABC

the question assumes the area is an invariant

but yeah for a proper approach try barycentric coordinates

[azure helm]

WDYM?

[obsidian harness]

that the area is the same regardless of where P is

[azure helm]

this section is on : Triangle Medians and Altitudes

[obsidian hornet]

You can just use homothety with midtriangle of ABC

[obsidian hornet]

||center is P and k=2/3||

[obsidian harness]

Smart

[hoary totem]

id try similarity

full solution using similarity:
||say M3 is the midpoint of AB and M2 the midpoint of AC and M1 the midpoint of BC.||
||take triangles APB and APC. then G3 lies 2/3 of the way on line PM3 and G2 lies 2/3 of the way on line PM2.||
||since the line M3M2 is parallel to BC, and cuz G3 and G2 lie at the same fraction along those lines, G3G2 is also parallel to BC. by the same argument G3G1 is parallel to AC and G1G2 is parallel to AB.||
||thus G1G2G3 is similar to ABC and obviously M1M2M3 is also similar to ABC so G1G2G3 is also similar to M1M2M3.||
||G1G2G3 is similar to M1M2M3 by a factor of 2/3 (cuz the G points lie 2/3 the way between P and M) and M1M2M3 is similar to ABC by a factor of 1/2 so G1G2G3 is similar to ABC by a factor of 2/3 * 1/2 = 1/3.||
||the question asks for area so square 1/3 to get the area factor which would be 1/9. if ABC has an area of 18 then G1G2G3 has an area of 18 * 1/9 = 2||

@obsidian harness this works right? (sorry for ping)

[azure helm]

did you came back from military service btw?

[azure helm]

I will check it later, ty.

[hazy sinew]

whats honothety

[mystic umbra]

its a way to describing similar triangles which preserves different properties

[obsidian hornet]

Homothety with center O and coefficient k is a transformation of the plane in which each point X is associated with a point X^ such that OX/OX^=k, parallel lines turn into parallel lines, circles into circles (or points as a special case), etc. Basic properties: the center, the image, and the prototype lie on the same straight line; if the composition of three homotheties is an identical transformation, then their centers lie on the same straight line (example: Monge's theorem)

[hoary totem]

no

[pale sentinel]

I think they misinterpreted the 전역 in your bio

(Though in fairness I'm not sure what the rest of that bio is supposed to mean - the 三 is throwing me)

[hoary totem]

三 is simply the rank insignia of corporal (상등병)

[pale sentinel]

Ah okay

[hoary totem]

"corporal - 2025/12 discharge"

[pale sentinel]

I knew it was something I was lacking context for lol

[hoary totem]

u speak korean?

[pale sentinel]

Not much, I can read it and understand a bit of grammar

But it's really rusty

[pale sentinel]

Oh as in the insignia here, I was thinking the literal character 三 i.e. "three")

Though I guess yh it's the third rank from the bottom so

I'm much better with Japanese though so ironically I have better understanding from reading the hanja there 🤣

[hoary totem]

lmao

hanja is fun

[pale sentinel]

ikr

[echo sigil]

Does anybody know why CQDX and BQXA are cyclic from Miquel point?I don’t get it

[azure helm]

i didnt read the bio but i think he said he was going to do miltary service like 2 years ago

maybe i am mistaken, sorry if that's the case

[pale sentinel]

Ah

That tracks somewhat

If I'm reading it correctly, he's out at the end of the year...?

[hoary totem]

it hasnt even been a year lol

[obsidian hornet]

Miquel point is point of intersections four circumcircles in complete quadrilateral(this four circles are dual to inversion version of quadrilateral in point Q), so, circumcircles of CDX and BXA are passing from Q by definition of Miquel point

[cerulean cipher]

Identify a circle

Cos(66)=sin(14)

$\pi$

[somber coyote]

∑

[crimson rune]

Hey is anyone here ever partcipated in IMO?

[mystic umbra]

ask in #competition-math its more suitable for your inquiry

[crimson rune]

Ok

[warped edge]

Any help in surfaces

[lean light]

What shape

[copper surge]

Find xc anyone

Igcse level maths

[pale sentinel]

What's special about C here?

[lean light]

I can see an equatorial triangle but i don’t know if it’s mathematically right

Could be 3 sides = -3a / a + -3b/b

[pale sentinel]

But what does the original thing say about C?

!original

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lean light]

No i read all

Draw error

The text says parallel

But does parallel mean sides are equal

[pale sentinel]

oh wait I saw that

No, parallel doesn't mean that

Parallel in this case allows us to say one of those vectors is a scalar multiple of the other

[somber coyote]

Waes (Wires)

Waes (Wires)

[pale sentinel]

How else could we describe XC though? @copper surge

[copper surge]

Here is the solution

[copper surge]

Yh

[copper surge]

Sm ppl r getting different answers

Wht i sent might be wrong

[hoary totem]

heres a purely geometric solution

we will draw a line parallel to OA through B and label the intersection points as in the diagram

angle OAB = angle ABD and angle AXO = angle YXB so triangles AOX and YXB are similar by AA similiarity

the ratio of similarity is 3:2 because AX : XB is 3:2 so AO : YB is also 3:2
AO = DB so DB : YB = 3:2
so DY : DB = 1 : 3 so DY : YB = 1 : 2

now, angle DCY = angle YOB and angle DYC = angle OYB so triangles YDC and YBO are similar by AA similarity. the ratio is 1:2 because DY : YB = 1:2

so DC : OB = 1:2
since OB = AD = b, we have DC = 0.5b and AC = AD + DC, therefore AC = 1.5b

you found earlier that AX = -0.6a + 0.6 b
since AC = AX + XC
XC = AC - AX = 1.5b - (-0.6a + 0.6b)
= 0.6a + 0.9b

the solution you posted was correct

(ignore the P in the diagram lol)

@copper surge

[copper surge]

Thank you

[viscid hawk]

I love geometry

[hoary totem]

same

kind of

lol

[viscid hawk]

I feel like It what I’m best at that’s probably why I like it so much lol

[hoary totem]

do u know any fun problems

[viscid hawk]

Well I love pi so I like a lot of problems with pi so that’s fun to me what about you?

[hoary totem]

i like a lot of problems regarding spherical geometry - tho i can only do basic stuff

[viscid hawk]

Yeah spherical geometry is very interesting I like trigonometry a lot as well

[hoary totem]

yea me too

[viscid hawk]

Honestly I love mathematics but I hate chemistry

[hoary totem]

lol who likes chemistyr

[viscid hawk]

I know right like I’m not gonna use it any where to be honest any way

[hoary totem]

lol

[toxic jacinth]

you have to consider the traingle OXB and triangle CXA to be similar

[hoary totem]

oh thats much faster lol

[mystic umbra]

wait till you learn calculus and have to get to differential geometry

[viscid hawk]

Actually I’m just a freak because I like calculus as well

[hoary totem]

i have a fun problem

[mystic umbra]

a lot of the geometry you know right now follows directly from

[hoary totem]

do u wanna try solving it

[mystic umbra]

the elements

[viscid hawk]

I like any mathematics to be honest

[mystic umbra]

perchance if its not too hard

[hoary totem]

the red ellipse is an ellipse with semi-major axis 13 and semi-minor axis 5.
let points F and F' be the foci of the ellipse, and point P be a point on the ellipse.
find the exact value of cos(PFF') such that the radius of the circumcircle of triangle FPF' (the blue circle) is minimum.

answer: ||(13 - sqrt119)/24||

@viscid hawk perhaps u should also give it a try

[viscid hawk]

Okay my interest is peaked

[hoary totem]

its actually not that hard at all lol

[mystic umbra]

just feels like its tedious or something

[hoary totem]

not at all

[mystic umbra]

i was searching for this

[viscid hawk]

Haha I feel like I’m overthinking this problem to much

[viscid hawk]

Haha sorry but I could never hate math trust me

[hoary totem]

why what is ur line of thinking?

tbh i gave u the first hint already lol - drawing coordinate axes

but ig thats not that big of a hint

[viscid hawk]

Ok so my thought process right now is to find the radius of the blue circle which should be 16 but I’m starting to think that finding the radius of the blue circle is futile and doesn’t quite matter

[hoary totem]

howd u get 16

[hoary totem]

tbh this may be a throwoff

not a hint lmao

[viscid hawk]

Hmm well let see if I go based of your model presented and look at the red eclipse it passes through both 5 and -5 so then if we take that it is slightly larger sliding it half way down the circle yet it would not be able to fit the middle of it was a orb and not a circle so then you take the graph and get around 15 both ways to the middle but their is also no way both could be 15 simply because then the blue circle would pass through -25 so statistically speaking the only other option is 16

[hoary totem]

the diagram does not represent the state where the radius of the circumcircle is minimum

i just chose a random position of P for that diagram

try without coordinate axes

the red ellipse is an ellipse with semi-major axis 13 and semi-minor axis 5.
let points F and F' be the foci of the ellipse, and point P be a point on the left half of the ellipse.
find the exact value of cos(PFF') such that the radius of the circumcircle of triangle FPF' (the blue circle) is minimum.

[viscid hawk]

Okay then I was just overthinking that lol

[hoary totem]

yeah probably

btw finding the location of P that would make the circumcircle smallest is crucial

but hint it doesnt involve coordinates

maybe u havent learned about ellipses yet, which would make this difficult

[viscid hawk]

Wouldn’t it just be 0

[hoary totem]

no

the circumcircle cannot have radius 0

[viscid hawk]

Yes that’s right I’m thinking about the wrong thing

[hoary totem]

the cosine of PFF' can be 0 but thats not the correct answer

[viscid hawk]

I should be writing it down so I could actually see my madness lined out for me

[hoary totem]

oh i should give a constraint

P is on the left side of the ellipse

[viscid hawk]

Okay thank you

[hoary totem]

so P cant be on the right half of the ellipse

this is cuz if u allow that u get 2 correct answers

[viscid hawk]

Wait is this Fregier’s theorem

Or am I just imagining it

[hoary totem]

nope

u dont need such fancy math for this

[viscid hawk]

Haha I really need to stop reading so much into this problem

[hoary totem]

do u want me to give u the solution?

[viscid hawk]

No

Now I’m very peaked and laughing at myself

[hoary totem]

the solution is just a bit of geometry then a bit of algebra

its quite easy

[mystic umbra]

geometry is not my strong suit 😔

i just solve differential equations most of the times

[hoary totem]

lmao

[mystic umbra]

i just recently learned about laplace transform and i've been using it like a monkey that just discovered fire

[hoary totem]

i once knew what that meant - i forgot all the math i learned

[mystic umbra]

perhaps its a sign to open up a book

and start learning

[hoary totem]

well ill learn once im back at university

[mystic umbra]

that's kinda bad 💀

[hoary totem]

ive been out of college for like a year now

[hoary totem]

tbh not hard u should be able to solve it if u know a thing or two about ellipses and circumcircles

[mystic umbra]

i guess i'm not built for ts right now, might have to go over geometry again 😔

[hoary totem]

lmao

[viscid hawk]

Hmm what is the solution?

[hoary totem]

the radius of the circumcenter is the distance from the center of the circumcenter to point F
also, the center of the circumcenter must lie on the perpendicular bisector of FF'
so lets draw the perpendicular bisector of FF'

the circumcenter (blue X) must lie on the blue line (the perpendicular bisector of FF')
its now evident that for the radius to be minimum, the center must lie at the midway point of FF'

as in this diagram

if a triangle's circumcenter lies on one of its sides, that means that that triangle is a right triangle with that side as the hypotenuse

so FPF' is right, with FF' as the hypotenuse

that means cos(PFF') = FP/FF'

[mystic umbra]

so the extremity of the latus rectum 😔

i had a feeling

but doens't that make the cosine of it 0? 😔

[hoary totem]

calling the midway point O, we have FO = 12 cuz the ellipse's semimajor axis is 13 and semiminor axis 5

[mystic umbra]

oh wait

no its

something else

nvm

angle FPF' is 90 instead

[hoary totem]

so FF' = 24

[viscid hawk]

Oh okay so I was doing something right

[hoary totem]

now calling FP = x

we have PF' = 26 - x

cuz FP + PF' = 2 * semimajor axis

so pythagoras

x^2 + (26 - x)^2 = 24^2

this gives x = FP = 13 - sqrt(119)
(it also gives x = 13 + sqrt119, which is why i restricted P to be on the left half only)

therefore cos(PFF') = FP/FF' = (13 - sqrt119)/24

[viscid hawk]

Okay that makes sense

[hoary totem]

theres another way to figure out that that angle is right

the diameter of the circumcircle is just the term in sine rule

so FF/sin(FPF') = diameter of circumcircle

to minimize that, sin(FPF') must be maximum

so FPF' must be right

the rest of the question continues identically

[viscid hawk]

Well that was fun thank you

[hoary totem]

np

[hoary totem]

its close to 0, (13 - sqrt119)/24 = 0.08713699522 lol

[hoary totem]

@viscid hawk wanna give another easy problem a try?

let A be a point in the first quadrant on the ellipse x²/64 + y²/16 = 1

let the foci of the ellipse be F' and F.

lets B be the center of a circle C with its center on the negative y axis thats tangent to both AF and AF'.

if the area of quadrilateral AFBF' is 72, what is the radius of circle C?

answer ||9||

[sly patrol]

How do I do b?

[obsidian harness]

start from the definition of the distance formula

$PQ^2 = (ap^2 - aq^2)^2 + (2ap - 2aq)^2$

[somber coyote]

south

[sly patrol]

Sorry I meant a

I got up to this but can’t factorise it into smth useful

[deft summit]

hey

whats the minimum number of non coplanar vectors whose sum will be 0

[pale sentinel]

I'm guessing at least 4?

The sum of three vectors can always be representative of a triangle if their sum is 0, so you need a fourth

[grave pond]

Yeah: If you have three vectors whose sum is 0, one is in the span of the two others, so they are coplanar.

[deft summit]

yes thank you

[grave pond]

On the other hand, 4 is easily possible.

[deft summit]

my friend was refusing to believe this

[rancid spindle]

Try

[toxic jacinth]

could you not use more lighter pencil

[viscid hawk]

Ohh sorry. I fell asleep last night

[obsidian harness]

ah, okay if I try myself using point-slope form

$$y - 2ap = -p(x - ap^2)$$
$$2aq - 2ap = -p(aq^2 - ap^2)$$
$$2a(q - p) = -pa(q - p)(q + p)$$

since $a \ne 0$ by defn. of parabola and $p \ne q$:

$$2 = -p(q + p)$$

[somber coyote]

south

[obsidian harness]

should be straightforward from this step

[obsidian harness]

yeah should be just $2ap = 2ap^2 + c$

[somber coyote]

south

[obsidian harness]

the p^3 appeared by mistake and that's why everything after that is incorrect

[sly patrol]

Ik what happened I did negative instead of positive for one of them

[magic barn]

What's the meaning of hyperbolic trig function

And are they useful

[cunning lion]

we define the hyperbolic sine function
[ \sinh(x) = \frac{e^{x} - e^{-x}}{2} ]
and the hyperbolic cosine function
[ \cosh(x) = \frac{e^{x} + e^{-x}}{2} ]
other hyperbolic functions (e.g hyperbolic tangent, secant, etc.) are defined by quotients of $\sinh$ and $\cosh$ analogously to circular trig

[somber coyote]

cloud

[cunning lion]

they're not nearly as ubiquietous as the circular trig functions but they do have uses

[trail tendon]

https://www.youtube.com/watch?v=65yw9klPklk dis u?

[pale sentinel]

ubiquitous*

Sorry, I had to lol

But yh that's the gist of it

[sly patrol]

Can someone help with part b?

[hoary totem]

no worries

[lime dune]

!status

[lime crown]

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

[visual flume]

2

lol

[lime dune]

this just looks like ||angle bisector theorem|| spam

[hoary totem]

not really spam i dont think?

but yes it is the ||angle bisector theorem||

oh no it is spam

[upper karma]

in 3 dimensions
?

6 in 3d?

[magic barn]

Explain

[cunning lion]

one example is that a hanging rope/chain will naturally form a shape called a caternary, which is given by the graph of the hyperbolic cosine function

they also give a parameterization of the "unit hyperbola" x^2 - y^2 = 1

[upper karma]

oh i get it

4 in 3d

[upper karma]

I was adding one more that now i think about it wasnt nesscarutfy... but base vecs all would be six?

[hoary totem]

man what a nice problem
semicircle with radius 1
P is on the semicircle
angle PAB is θ
H is a perpendicular line to AB drawn at B
line AP intersects H at Q
draw a tangent to the semicircle at P, the tangent intersects H at R

if the area of triangle PQR (shaded) is S, find the limit of S/θ³ as θ goes to 0+

||answer is 1||

[pale sentinel]

Yh the question needs to be in at least 3 dimensions, else all vectors are necessarily coplanar

[potent schooner]

can someone help me with thos

[signal vapor]

Hint ||divide the hexagons into 6 equilateral triangles||

[odd pivot]

Am i correct in this? Not sure if shifting element from one side to other is correct

[signal vapor]

In the 6th line it is better to just remove the negative sign

The ans is correct

[odd pivot]

how does the remove of - would look like compared with 3rd line when shifting elements? Or is just straight negation

[signal vapor]

Multiplying both sides by -1

[odd pivot]

oh, thank you

[upper karma]

yeah thats trivial, i skidadled, but dont the 4 vectors reduce to more simple vectors in six, does the question ask for any vector ? like 4 for 2d.

[pale sentinel]

@upper karma The question only asks for the number of non-coplanar vectors with zero sum

[upper karma]

yes but what kinda vectors?

[pale sentinel]

wdym what kind?

I'd presume geometric R^n-type ones

[upper karma]

i mean the ones where the other two coords are zero must make up the 4, but then those are six

i do have a proof of the non coplanar thingy that i got to know from someone but it feels goofy

he considered the 3 basic vectors but 1 resultant vector to get 4

[pale sentinel]

Yeah, so e.g. i, j, k and -(i+j+k)

[upper karma]

ye

[pale sentinel]

These would be 4 non-coplanar vectors in R^3

Whose sum is 0

[upper karma]

ye

[pale sentinel]

I didn't need a 4th dimension, let alone a 6th

[upper karma]

but the other type

i mean

a 3d system has 2 2d planes by the origin

since the sum of two vects are required for a zero sum

in 3d total 6 vects are rquired

i think this only is for basic unit vectors

[upper karma]

wdym thats not what i meant

[pale sentinel]

okay I see

[pale sentinel]

But what does this mean?

[upper karma]

like the x, y plane

and the

yz

[pale sentinel]

Given a point, there are infinitely many lines that go through them; thus there are infinitely many planes that go through them

[upper karma]

srry

[pale sentinel]

"coplanar" means there exists a 2D plane that goes through them (it doesn't have to be the coordinate ones)

[upper karma]

not the origin

[upper karma]

i dont mean that

i know what coplanar is

do this

the four vectors you spoke of, reduce them. then six vectors are there

[pale sentinel]

Are you suggesting you need a +i, -i, +j, -j, +k and -k vector, so 6 in all, so that they're not coplanar and that their sum is still 0?

[upper karma]

ye

[pale sentinel]

Because then that's not answering the question

6 is, surprisingly, not smaller than 4

[upper karma]

thats wht i m saying

what typa vectors

[pale sentinel]

So we haven't actually reduced the number of vectors

[upper karma]

i mean reduce in the other sense

[pale sentinel]

Are you suggesting "basis" vectors?

[upper karma]

like components of vectors

[pale sentinel]

Those are either components of vectors or basis vectors

[upper karma]

because

[pale sentinel]

Either way, since neither is stated, the question implies general vectors

[upper karma]

the one guy i talk to

[pale sentinel]

e.g. a = (1,4,5) is one vector, not 3

[upper karma]

he take 3 basis vects and one resultant

[pale sentinel]

Yh, so that's 4 vectors

[upper karma]

someone

[pale sentinel]

3 of the vectors are your basis vectors in the space

[upper karma]

ohhhh

i get it noiw

[pale sentinel]

and your fourth is the negative of their resultant

[upper karma]

generally

[pale sentinel]

(idk how you ended up with 6 from that 🤣)

[upper karma]

i get

[upper karma]

dude

vectors in the opp direction exist

dont they?

[pale sentinel]

Yh but if say "i" is a basis vector, we don't then take "-i" as another basis vector

It's already a scalar multiple of one, namely of "i"

[upper karma]

but there not same

only the same in magnitude right?

[pale sentinel]

Sure

[upper karma]

because then

[pale sentinel]

But since one is a scalar multiple of another, we don't need a second basis vector to describe -i

[upper karma]

then there would be 3 vectors

the resultant ones?

[pale sentinel]

There are three basis vectors: i, j and k

[upper karma]

no 2

yeah

[pale sentinel]

Their resultant is their sum, i+j+k

[upper karma]

yeah

[pale sentinel]

That is one extra vector

Hence 4 in total

[upper karma]

i know that

wht about the zero thingy

that involves a negative basis vector

-i, -j, -k

[pale sentinel]

We take the negative of that resultant vector instead

So we have four vectors: i, j, k and -i-j-k

[upper karma]

yeah so 6 vectors

[pale sentinel]

four

[upper karma]

why

why not

-i, -j, -k?

[pale sentinel]

Because if we can do it with fewer vectors, we should

This is a question about minimising

[upper karma]

can i ask one more

wait

[pale sentinel]

Ask on the server, do not DM me

[upper karma]

what difference does it make

ok fine

but do answer

[pale sentinel]

It's kind of standard? You revolve a circle around the x-axis and calculate a volume of revolution

https://math.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/single-variable-calculus/volume/

[upper karma]

i integrated with cylinders instead of rectangles with the help of x^2 + y^2 = z^2(2d)

but im having trouble

i have done it once

but now have forgoten

[hoary totem]

ok wtf

this is a parabola with focus F(0, 1)

if u construct a triangle such that FH = PH

where P is on the parabola and PH is perpendicular to the x axis

the area of the incircle of this triangle is phi pi

phi being the golden ratio

the proof is probably very obvious lemme try

[hoary totem]

crap im out of time

ill try tomorrow

coords of P = (x, y)
HP = FH
-> x^2/4 = sqrt(1 + x^2)
x^4/16 = 1 + x^2
-> x^2 = 8 + 4sqrt(5)
-> x = sqrt(8 + 4sqrt(5))
-> y = 2 + sqrt(5)

found this out at least

[odd pivot]

Sorry for pinging, does it matter if i a or b is unknown? like if instead of a=14 it would be b=14?

[upper karma]

the assumption that a in some sense is equal to b in some sense should hold for all values that hold that assumption

[odd pivot]

Does it matter with that eqation if both sides are equal or not?

c^2 = a^2 + b^2

speaking in the sense of numbers

i know the length of side will be different so the triangle will look differently

[upper karma]

ohh

yes

only certain number combinations like the pythsagorean triplets hold

[odd pivot]

Can this be final solution or it should be
minus root of a squared equal minus root of squared 21?
-√a^2 = -√21

this number supposed to be on xy axis on x axis second quadrant

it supposed to be negative but i dont want to trust guide and pull numbers out of butt

[upper karma]

No the two minus 1s on both sides go away

[upper karma]

A minus in a square root would be hard to imagine

[odd pivot]

well its in 2nd quadrant on xy axis so it has to be negative

[upper karma]

?

I wasn’t following with your question

Well it can be either +root21 or -root 21

The a part

This guy will helwp

@pale sentinel help him

[odd pivot]

but is that negative part is just because its in the 2nd quadrant and on x axis? or there is more math

[upper karma]

It’s the other way around

If it’s negative x coord is negative and if positive it positive

The root 21

[odd pivot]

oh so its basically quadrant being like [x, y] and 2nd quadrants its like [-x,y] and i probably just need to multiple it by -1 or 1

[upper karma]

No For sine(-x)=-sin(x)
For cos(-x) = cos(x)

!noping

[lime crown]

Please do not ping individual helpers unprompted.

[upper karma]

!yesping

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[upper karma]

!display

!math

!show

[lime crown]

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

[upper karma]

!explain

[odd pivot]

Why he multiple by root of 21 and not negative root 21? does it change the sign by multiplying both sides

[regal wedge]

he multiples by sqrt(21)/sqrt(21) which rationalizes the denominator. This is the same as multiplying by 1. Multiplying by -1 would give a different solution

[odd pivot]

so its something like this?

[regal wedge]

yea but generally you just multiple by sqrt(21)/sqrt(21) not -sqrt(21)/-sqrt(21)

[odd pivot]

well im learning and i think i need to know that, cuz else its just looks wrong. Ok so multiplying just negates the signs

well to be fair why did he used the - root 21 in the first place than and didint explained

[tepid oyster]

i’m assuming it’s because of ASTC

[odd pivot]

so csc, sec, cot is just fliped sin, cos, tan

[tepid oyster]

yes that’s true but i don’t think that’s the answer to your question

sorry i’m trying to send an image but wifi is kicking my ass lol

[tepid oyster]

ok sorry the image isn’t sending i’ll *try to explain it myself— do you know what quadrant he was working in ?

[odd pivot]

2nd

topleft

[tepid oyster]

ok that’s why

in the second quadrant the tangent will be negative

second and fourth tan is negative

google “ASTC rule trigonometry” and a graph will show up with which quadrants are positive and negative in each function (sorry i can’t send images)

[tepid oyster]

also just in case— in the 2nd quadrant cosine and secant are also negative as well as tan and cotan which is why he also made it negative in the first equation

[odd pivot]

Oh, i found it, i thought sine sign depends if its negative x or positive x but the 4th quadrant debunked my logic

[tepid oyster]

awesome sauce !! i can try to dm my notes from precalc if you needed a more visual aid ??

[odd pivot]

I got this

[upper karma]

What

[odd pivot]

so it seems that these things have they own axis

[tepid oyster]

that’s not exactly what i was talking about sorry— i’ll try to send the image again

awesome sauce !! my internet is working better :))

the sign of the functions remain constant through the inverse functions as well

as well as their corresponding arc-functions

[odd pivot]

ive seen this image, so these values are negative because in which quadrant they are and the -x or x and -y ot y changes the sign of these operations

[tepid oyster]

the sign of the coefficients don’t affect the functions sign the quadrants do :))

[odd pivot]

like the above image i showed sin is on y axis so it positive in 2nd quadrant

[tepid oyster]

specifically for reference angles you shouldn’t be focusing too much on axis the axis comes in more when you start graphing them !!

[odd pivot]

well knowing that sine value is of y on axis helps me to quickly know if sine is negative or positive in quadrant.
I think thats the case. If y is positive the sine is positive if y is negative the sine is negative

[tepid oyster]

if it works for you that’s great— i wasn’t taught that way so im not sure that there is a connection but i am also still a student and i don’t think im qualified to say whether there is or isn’t a correlation

(just a side note i’m not completely understanding your perspective but just in case— the quadrant has priority over the equation in the way that if your equation from the unit circle gives you -1 but the function is positive in the quadrant your working in, disregard the sign of the equation it’s positive 1 in the quadrant)

[odd pivot]

i thought of like coords in quadrant like in 2nd quadrant = [-x, y] which can be [-1, 1] and if sin would be positive i could multiple by -1 and get negative. Im also not sure if im right. Still reading on and watching guides

[tepid oyster]

ooh ok i understand what your doing now, yea that works !! :))

[visual flume]

should i js assume that the first pair of congruence goes to dm and me

[obsidian hornet]

||AX-median, 2/3AX=1/2AX+2 => AX=12 and AG=8||

[visual flume]

help plsssss ahhhhhhhh

[visual flume]

yeah ty i did it earlier

but this problem is so annoying i swear

[obsidian hornet]

||using median length formula: M2=1/4(2a^2+2b^2-c^2) getting that 1/36x^2=1/4(18+32-x^2)=> x=3sqrt5||

[final ferry]

In rectangle ABCD, AB = 4 and BC = 3. Find the side length of rhombus AXY Z, where X is on
AB, Y is on BC, and Z is on BD.

Anyone who can solve this I will give u 1 million

[visual flume]

tysm

[hoary totem]

perimeter = 2 + sqrt5 + 2 + sqrt5 + 3 + sqrt5 = 7 + 3sqrt5

area of triangle = (2 + sqrt5) sqrt(8 + 4sqrt5) / 2
= r * (7 + 3sqrt5)/2
r = (2 + sqrt5) sqrt(8 + 4sqrt5) / ((7 + 3sqrt5)
r² = (9 + 4sqrt5) (8 + 4sqrt5) / (94 + 42sqrt5)
= (152 + 68sqrt5) / (94 + 42sqrt5)
= 2 (38 + 17sqrt5) / (47 + 21sqrt5)
rationalize
= 2(38 + 17sqrt5)(47 - 21sqrt5)/(47² - 21² * 5)
= (1 + sqrt5)/2
therefore area of incircle = φπ