geometry-and-trigonometry

[dusty coral]

thats kinda sad

with 86.6

[upper karma]

🤷 You can screw up at any level.

[copper valve]

not preparing properly doesn't make cheating less serious

[upper karma]

Didn't say it did.

[oak fractal]

I am not in middle school lmfao

I'm in 10th grade

[copper valve]

have i gotten meaner

[upper karma]

Then your inability to use pythagorean theorem should have set off alarm bells.

[dusty coral]

That's sad, then

Why are you in discord in school?

[oak fractal]

cause I needed help

duh

[copper valve]

is it the post (algebraic geometry) stress disorder

[dusty coral]

Ask your teacher.

[upper karma]

In a final tho -_-

Lmao @copper valve

[copper valve]

[upper karma]

The trauma.

[dusty coral]

nah, up till algebra 2 is ez so far in 9th

[upper karma]

I mean..

Is doing simple trig algebra a part of your course rn then @dusty coral

[copper valve]

algebra 2 here is in 2nd year

[dusty coral]

I'm in the highest level for 9th grade.

so idk

[upper karma]

Yeah, I think Americans have Algebra 2 in high schools and Algebra 2 again in college.

But this is the difference between "algebra" and algebra

[dusty coral]

they seem pretty simple to me

[copper valve]

yea :/

u should do abstract algebra!

google: napkin project

[dusty coral]

a B s T r A c T

[copper valve]

it's higher math exposition for high schoolers

[upper karma]

Tbh - if you're worried about any concepts going into an exam. You need to hammer them out. Especially something as foundational as pythagorean theorem.

[dusty coral]

I need help for this one question. Verify $$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$$

[copper valve]

hmmmm~

interesting question ;o

[dusty coral]

I multiplyed both sides of the fraction by $$\cosA-\sinA+1$$

[copper valve]

hmmm

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[copper valve]

i was gonna try writing the right hand side as a single fraction first

[dusty coral]

but no use

[copper valve]

=tex \csc A + \cot A = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1+\cos A}{\sin A}

[copper valve]

hmmm

are u sure this is true

oh maybe it is

lemme think

[dusty coral]

check a value

[copper valve]

yeah im convinced

[dusty coral]

it works

[copper valve]

this is a cool problem

where'd u get it?

[dusty coral]

my book

only problem im stuck on

[copper valve]

nice

what book is it?

[upper karma]

Whew trig identities

[copper valve]

some are really cool tbh

[upper karma]

I still use a reference sheet whenever I need double angle formula, am I a bad person?

Yeah, trig integrals are really neat.

[dusty coral]

Glencoe Algebra 2 CCSS

@upper karma trig is basically common sense algebra

but sometiems you dont know which steps to make

[upper karma]

Pretty much, you have play around with it.

Which if people have been relying purely on plug and chug, becomes very uncomfortable for them.

[dusty coral]

yea, then you get "number sense" and get used to it, however this problem stumps me

Any progress?

[copper valve]

(1+cosA)(cosA+sinA-1)/sinA
= (cos²A+cosAsinA-cosA+cosA+sinA-1)/sinA
= (cos²A-(cos²A+sin²A)+cosAsinA+sinA)/sinA
= (cosA-sinA+1)sinA/sinA = cosA-sinA+1

gotem

[dusty coral]

oof

[upper karma]

👌 Always trust le woog

[dusty coral]

jacobian >>>>>>> everythign

?

[copper valve]

interestingly, this is also equal to cot(A/2)

I leave this as an exercise

😉

[dusty coral]

i just looked at the book answer and they said to divide both sides of the fraction by sin(A)

how did they come up with tat?

[copper valve]

same way u come up with answers to all trig identity problems

just try random stuff pretty much

[upper karma]

Play.

[copper valve]

eventually u get better at guessing what tricks are gonna be more successful than others

[upper karma]

Through process known as "juju"

[dusty coral]

'

[fallow quail]

if i do coordinate geometry with vectors, then the position vector is (i,j) and i get the normal vector by flipping the coordinates of the position vector, and *-1 one of them

does it matter which one i multiply by -1?

if i want to get gradient/equation of the line etc?

like (-j,i) and (j,-i) are both normal vectors

[zenith ember]

Yes.

Your understanding is correct. They are both normal vectors.

And there is no inherent preference of one over the other.

[fallow quail]

thank you

[reef cave]

Can somebody help me with 33?

[fallow edge]

Are you supposed to use a compass for a?

[primal pilot]

Anyone online and help me with a grade 11 functions question

[keen aspen]

sure

[upper karma]

can anyone help me with a tricky situation im in, i recently moved to spain and im forced to go to school here (public) with no language skills. i was doing finnish/american calculations and i noticed they do it kinda different, im dead tired and i have other subjects i need to do as well

[shadow anvil]

@upper karma

I had similar

I moved to spain from england

but I attended an international school

Puhutko espanjaa?

[upper karma]

https://gyazo.com/1fe9ff717ab34647b7730ea44a1d22e8

halp

What’s the formula for the area of a triangle?

._. r u still there?

[unborn kite]

@upper karma ^

[ashen tree]

thoughts?

[zenith ember]

Let's call the intersection of the diagonal E

Then by considering, in turn, the right triangles AEB and DEC, we can create expressions for AB^2 and CD^2

Hm.

Not entirely sure that goes where you want it to...

[ashen tree]

yeah I started with that

it gets pretty mess pretty quickly

not sure if it works out or not

[zenith ember]

Ahh.

I see it now.

I converted it into a parallelogram by adding a copy of it to the right.

BF = CD, and AB = CG

Then CF = BD and CF is perpendicular to AC.

And the result falls out.

@ashen tree

[clever moth]

What's topology

[neon fossil]

turning donuts into mugs

[plain wigeon]

morphing autism

[analog dust]

Its stretching dogs and cows to fit the donut

[analog dust]

Dogs aint spheres

And we shouldnt neglect internal organs... then its not even a dog anymore

[upper karma]

Is a coffee cup really a donut?

pours coffee in donut

[analog dust]

Actually mammals are like many holed donuts cuz our skin has holes

I think

I can blow out air through my eyes, so I count up to genus 5 then.

Just read that

Now my eyes hurt

[zenith ember]

The digestive tract is a continuous hole.

[analog dust]

Its a long hoole

[zenith ember]

Doesn't matter, topologically

[austere stratus]

been on a binge of plotting various equations in Vex (Houdini) - polar rose curve

[upper karma]

@shadow anvil i speak it a little but very little, its stressful and i have to google translate home a lot (the topics) i was learning american schooling before and now i have to go back to topics i was in months ago (and some which i havent learned.)

@shadow anvil i have to learn everything in spannish but i speak basic spannish at best, its terrible. (public school) worst thing is that i have to repeat the year anyways (since i only started like 3 weeks ago)

[heavy kernel]

can anyone pls tell me what's the equation for lateral area of prism, pyramid, cones and cylinder?

[neon fossil]

you can literally google "lateral area of prism"

[analog dust]

(Perhaps he lives in china and theres no google there)

[timber hinge]

You can literally Baidu "lateral area of a prism"

[upper karma]

I hate geometry with passion

what kind of geometry?

Some bullshit geometry we learned last semester came in this semesters exam

65 percent of the semesters grade

And they bring up shit from the past semester

sucks fam, but thats how it is, eventually its all gonna make sense

[timber hinge]

Geometry is great

Easily one of my favourites

[tawny pewter]

https://gyazo.com/bbc8792047bfe4a2adabc0e2d5784030

how i do this

[upper karma]

what we learned = basic trig ratios

what came on the exam

Prove cos^2 + sin^2 = 1

thank god I knew what a unit circle was and solved it from there

65 percent

I’ll lose a ton on my averae

Average*

[timber hinge]

I just use ratios for that tbh

(A/H)^2 + (O/H)^2

A^2 + O^2 = H^ from Pythagoras

Then it’ll be H^2/H^2

= 1

Nice and elegant

[zenith ember]

I feel like I may be missing something simple.

So... I've been puzzling over what seems like a simple geometric construction problem.

Given a line and a circle that is tangent to the line, construct a circle (any circle) that is tangent to both the line and the given circle.

Can't see it...

(Note, I do not know if this is actually possible)

But it really seems like it should be.

[upper karma]

besides the obvious one?

[zenith ember]

Yeah.

[analog dust]

I shouldnt have written C1 and C2

[zenith ember]

I don't want the one of equal radius.

[upper karma]

no I mean any circle which is tangent to the point of tangeny of your original circle and the line

[zenith ember]

I want to be able to construct one of arbitrary radius.

[upper karma]

like on the other side of the line

[zenith ember]

Oh, no.

[upper karma]

you can pick any radius you want

[zenith ember]

On the same side.

[upper karma]

ah ok

[zenith ember]

Something like this, but, you know, actually tangent to one another.

[analog dust]

But you want to know like how to construct itm

?

[zenith ember]

Yes.

I can calculate the center position.

That's not interesting. I want to do it purely geometrically.

If that's possible...

[upper karma]

what you mean purely geometrically?

[zenith ember]

Lines and circles.

And intersections.

[upper karma]

ah

[zenith ember]

On the image, A, B, and C are arbitrary.

Everything else should be derivable from those.

[analog dust]

This thing smh, its like, rly close

[zenith ember]

Heh.

[analog dust]

Oh wait

When moving C it changes

[zenith ember]

That's been my issue too.

I just can't figure out a second constraint on the position on the perpendicular at B.

[analog dust]

@zenith ember i found ittt

Except its in terms of a point on the circle

[zenith ember]

Hmmm.

Ahhhh.

Very nice.

[analog dust]

And

For a point on the line

Idk why B is black tho.., you can change it

[zenith ember]

Very nice.

[analog dust]

Indeed

[zenith ember]

No, I see how you did it. Thank you. 😃

[analog dust]

Oke so

Damnit there are no letters

[zenith ember]

I get it.

I just did it on mine. 😃

Don't know why that never occurred to me to do.

[analog dust]

Oke

[zenith ember]

The next trick is to understand -why- that works...

[rough juniper]

I dunno where to even start with this question

Thanks in advance

[upper karma]

you should find the angle measure of ACB first

[rough juniper]

I believe I can use tan-1, right?

tanx = ab/bc

[zenith ember]

Isn't it all similar triangles?

[upper karma]

oh right nvm ^

damn and it's not specified right either fml

[rough juniper]

They don't look similar?

I think it is based on similarity

the next question has a scale and other info

[zenith ember]

Try drawing a horizontal line from E to AB.

[rough juniper]

?

[zenith ember]

Oh, can we assume angle ABC is 90 degrees?

[rough juniper]

Doesn't specify, but we should give it a shot and see if the numbers checkout

[upper karma]

no i don't think so

[zenith ember]

Yeah, so you know the width and hyp. of that triangle. And it is similar to the big triangle.

[rough juniper]

yeah but then how do we get the length of things like EF

[zenith ember]

Not sure we need to.

[rough juniper]

we need it to get EG?

unless AE = EG and GJ = JC

[zenith ember]

They should be.

[upper karma]

if we let EC = x, we can make a proportion with (AE + x)/x

(x+AE)/150 = ?

[rough juniper]

I got EG = 42.2 GJ = 43.0 JC = 43.0

And where did you get (x+AE)/150

[upper karma]

the length of the longest leg

[rough juniper]

I mean the 150

[upper karma]

the base of the triangle

[rough juniper]

oh

omg this assignment was printed from my printer and I need a ruler for the second question

I hate it whenever teachers do this

[rough juniper]

I'm gonna move my question back to here: https://cdn.discordapp.com/attachments/269573202018041856/445778561488519168/problem.PNG

[keen aspen]

What is angle S?

[sudden stream]

Could I be walked through on this one? http://i.ankatic.ga/8f5be4d1463b0d0c364bc0492745540f.png

[chrome fiber]

Draw the figure

Each chord will form a triangle with the center

The altitude of these triangles is 4

Look at one half of the triangle

For chord AB, one side is 4, the other is (4x - 6)/2 and the hypotenuse is the radius

You should get a similar triangle for chord CD

Use Pythagoras to get two equations, both equal to r^2

Then solve for x

[sudden stream]

@chrome fiber Do you know how to do this one?

or <@&286206848099549185>

[chrome fiber]

What do you think?

[sudden stream]

@chrome fiber I think its < DXA

[chrome fiber]

Why?

[sudden stream]

It has the 8

vs the 6

[chrome fiber]

That's right

[sudden stream]

yeet

[sudden stream]

<@&286206848099549185> Could I get help with this?

[keen aspen]

@sudden stream what do you know about those angles?

Its the average of the two corresponding arc measures right?

[sudden stream]

@keen aspen Yea

[upper karma]

Can anyone help me on number 1 <@&286206848099549185>

[analog dust]

You can find the height of triangle ABC

And brcause thats the same height as triangle DEF

You can find DF

Just by using the area formula for a triangle

[upper karma]

Yes hello

I need help on a problem, i need to learn how to do it for a retake test and it determines if i graduate

[mint sandal]

@upper karma

subdivide it into rectangles

[upper karma]

Thank you

[orchid hill]

can someone help me with these vector questions

[upper urchin]

Use the dot product

[orchid hill]

what

@zenith ember sorry about this but i kinda need this quick this questions seems simple but i dont get could you pls help me

[upper urchin]

Oh actually the easier way would be to say if a is parallel to b then a=cb where c is a scalar

[orchid hill]

a is not parallel to b or i think not

its a+b(x) is paralle to i

[upper urchin]

I just meant arbitrary vectors a and b

Not the specific ones in the question

[orchid hill]

alright

i ve tried to make it into a simultaneoes yet it still doesnt work

[upper urchin]

Well if you have a vector in the I direction, what is its j component?

[orchid hill]

it took out j but made it = to i

like this

2+3(x)=i

5-(x)=i

[upper urchin]

No I mean say you had an arbitrary vector h in the I direction

What would the j component of h be?

[orchid hill]

where

[upper urchin]

say h=a*i where i is the unit vector

then h is parallel to i

so say we have the h=ai+bj

if h is parallel to i

what is b

[upper karma]

What is the shortest distance Nellie Bell Evans can travel to water her horse at the stream and return home if she is 5 km south of the stream that flows due east and she is 8 km west and 6 km north of her cabin?

I don't know how the picture would look like

[golden nymph]

Does anyone know mathematically how I would calculate the Z value for a camera if I want it to capture a full top down view of a square plane? I have the x, y, but how would I calculate z's for orthogonal and perspective top down views?

[upper karma]

@sudden stream My quick guess, (360-(120-84))/2 = 78

[rose geyser]

Is this a good channel to ask a question about Trapezoidal rule?

[upper karma]

For integration? Maybe take to either #calculus or #precalculus

[rose geyser]

No just to understand it better

[upper karma]

Yes.

This isn't geometry or topology.

[rose geyser]

Oh okay

[upper karma]

Well, whatever.

Yes

Trapezium rule generally offers a good approximation for underneath a curbe.

[rose geyser]

Thank you!

[upper karma]

So what’s the area of the base

[acoustic ibex]

I would think 49

[upper karma]

And what’s the area of each of the triangles

[acoustic ibex]

42

[upper karma]

And how many triangles are there

[acoustic ibex]

4

[upper karma]

So the combined area of the triangles is?

[acoustic ibex]

168

Oh I figured it out

Cheers thx mate

I added that by 49 and got 217

[upper karma]

nice job

[acoustic ibex]

I had a formula that said like s=b+1/2PL and confused me

[upper karma]

Yeah that’s why sometimes it’s not good at just use a formula

It’s better to know how to solve it

[acoustic ibex]

Yeah and it would help if my teacher would teach me too so

[upper karma]

....

[acoustic ibex]

Yeah

My teacher just doesn't teach he like gives us notes for us to study

[arctic apex]

I'm dumb and hate this sort of stuff so can you just tell me the answer because I got this type of question wrong on my last quiz (btw this math books makes you go counterclockwise when rotating)

[twin prawn]

dilate by a factor of 2 and then rotate counterclockwise 90 degrees

[arctic apex]

Ty

[upper karma]

yes

crap i did it again

[upper karma]

Can Euclid's 5th postulate be considered as a conjecture?

[timber hinge]

@arctic apex If this is GCSE you will be required to also state the point at which you rotate around (In all honesty it should probs be required anyways)

Its (0,0) here but isnt always

[mint sandal]

Hi

I'm confused about a calculation

how do we deduce the properties in the bottom just from the conditions of being in the lie group?

like i get the 3 equations in the first page

but not what follows next

can someone explain how those conditions follow?

nevermind i got it

[mint sandal]

another question:

how do I verify the formula at the bottom?

small g is the lie algebra of G

[mint sandal]

nevermind

[neon fossil]

nice

[mint sandal]

yes this boiled down to the fact that c^i_jk = - c^i_kj, so that 1/2 c^i_jk (theta^j(X) theta^k(Y) - theta^j(Y)theta^k(X))=1/2 c^i_jk (theta^j(X) theta^k(Y) ) + 1/2 c^i_kj (theta^k(X)theta^j(Y))

still this topic is damn confusing to me :D

like

how do you derive the equations on the bottom

this is probably the most computation intensive chapter i came across in the book so far :D

@neon fossil can you help?

is there a fast way to verify it

or any way really :D

[neon fossil]

you'd just have to plug the situation into the previous equations

[mint sandal]

could you elaborate?

which form of maurer-cartan do i use?

one with structure constants?

[neon fossil]

yeah use the one at the end of the previous picture

it's the one which looks most similar

[mint sandal]

it requires a choice of basis and the book isn't explicit about the basis

[neon fossil]

it seems like the alphas and betas and gammas form bases

[mint sandal]

do we take a^i_j - a^j_i and stuff like that as basis?

[neon fossil]

I don't know because the definition isn't shown in what you wrote

[mint sandal]

like a^i_j alone can't be an element of the basis

definition of what?

i can post it

[neon fossil]

of the alphas, betas, gammas

[mint sandal]

it is

like they're a bunch of functions

which together form the canonical 1-form

on the lie group

[neon fossil]

okay they're just defined as the components then

[mint sandal]

yes

like every element of G and g is a matrix

because G is a subgroup of GL

real matrix

hence the matrix form of the canonical form

[neon fossil]

yeah try to use the differences as a basis maybe

the computation is invariant by change of basis

[mint sandal]

yeah

[neon fossil]

so as long as your terms are expressed in terms of what you want

you should get something similar

[mint sandal]

but that seems like a lot of work

[neon fossil]

it does, I don't see a better way

you could go and use mathematica

[mint sandal]

what can mathematica do

[neon fossil]

computations

plug in stuff

[mint sandal]

[mild vale]

Is this the best/fastest way to scale a 2d triangle?

[mild vale]

float scale = 2f;
float mx = (V1.x + V2.x + V3.x) / 3f;
float my = (V1.y + V2.y + V3.y) / 3f;
V1 = new Vector2(mx + (V1.x - mx) * scale, my + (V1.y - my) * scale);
V2 = new Vector2(mx + (V2.x - mx) * scale, my + (V2.y - my) * scale);
V3 = new Vector2(mx + (V3.x - mx) * scale, my + (V3.y - my) * scale);

[surreal bolt]

Do you only deal in triangles? Just saying if you deal with n-sided figures, you might as well write general code for that.

[raven sequoia]

Hello, Im dealing with a non-symmetrical tetrahedron problem here and Im running out of ideas on how to approach the problem. I know the positions of 3 points: A,B,C. The goal is to find the 3d position of point D. In addition I know the angle ADB, BDC and CDA. Any clue what direction I can take this?

[surreal bolt]

hmm do those three angles define the"shape" of the vertex D?

[raven sequoia]

Yep, ended up finding a paper on this exact problem: http://www.mmrc.iss.ac.cn/~xgao/paper/ieee.pdf

[surreal bolt]

okay. here's what i have so far: (It's not everything, but I think it's a very good start)

Trying to find a congruent 3D shape with D' at the origin (and some other restrictions) and then using transformations to rotate and move it into position.

Let's consider vertex D and the angles first.

WLOG:

Let D' be at the origin.
Let A' be on the x-axis or (a, 0, 0).
Let B' be in the xy plane or (b1, b2, 0).
Let C' be a point (c1, c2, c3).

Let alpha be the measure of angle A'D'B'.
Let beta be the measure of angle B'D'C'.
Let gamma be the measure of angle A'D'C'.

b1 = k cos(alpha); b2 = k sin(alpha)

The vector in the direction of D'A' is <1, 0, 0>.
The vector in the direction of D'B' is <cos(alpha), sin(alpha), 0>.
The vector in the direction of D'C' can be determined using beta and gamma, but for now <c1, c2, c3>.

So now we address triangle A'B'C'.

Since we know the "final" positions of A, B, and C, we can determine these lengths exactly: A'B', A'C', B'C'.

[sorry tons of corrections]

[raven sequoia]

in my exact case I can even define
A: (0,0,0)
B: (b,0,0)
C: (0, c, 0)

[surreal bolt]

I understand that 😃 but using D as the origin simplifies finding the "shape" of the figure around vertex D 😃

[raven sequoia]

In the practical use case BAC should always have an angle of 90

makes sense

[surreal bolt]

oh. Am I allowed to use that assumption?

the right angle BAC?

[raven sequoia]

if it helps with the solution: absolutely 😃

[surreal bolt]

Okay, but without that:

We have three points (a, 0, 0), (b cos(alpha), b sin(alpha), 0) and (c(c1), c(c2), c(c3)). This is equivalent to saying we know the 3 lines in R^3 that contain the points A', B', and C'.

Each value of "a" determines up to two possible positions of B' (using distance. We know A', we know the distance from A' to B', so the sphere centered at A' with radius of length AB will intersect line with B' in at most two places) and up to two values of C' (using distance again).

It is left to check if B'C' is the correct distance apart.

On paper, we can calculate the up to two possible positions for B' in terms of "a" and constants. Ditto with the position of point C' (also in terms of "a" and constants). There will be up to 4 values of B'C' again in terms of "a". Set the values of B'C' equal to the length of BC and find what "a" should be?

That should give you the shape of the whole figure. Just rotate and transform to get A'B'C' to ABC.

[raven sequoia]

Ill see if can end up formulating this in to something sensical, thanks!

[surreal bolt]

Okay I'm cooking up an example to see if it works. Don't keep your hopes up 😃

[raven sequoia]

Im looking at other sources aswell, worst case scenarion I think it might be viable for me to implement an iterative approach to finding the most reasonable value of a given your approach

[junior scaffold]

How does it go from intergration of cos^2 x to that

[mint sandal]

cos2x = cos^2 x - sin^2 x = 2 cos^2 x - 1 so cos^2 x = 1/2( cos2x + 1)

@junior scaffold

[junior scaffold]

ohh, ty

[eager pendant]

a semicircle with centre O is inscribed in ABCD and AO=OB

prove (AO)^2=AD times BC

<@&286206848099549185>

ive tried drawing liens from the corners of the quadrilateral to O
a

and looking for other stuff, like congruences

i suspect i need to do something with the point of tangents

but i dont know what

[analog dust]

Are you working with vectors or lengths?

[eager pendant]

idk what a vector is

so probably the latter

[analog dust]

So like youre saying distance between points O and A right?

[eager pendant]

yes

[analog dust]

Doesnt quite match up...

Unless i did the figure wrongly

wrongly

[eager pendant]

why are you doing stuff in geogebra

[analog dust]

Cuz why not

[eager pendant]

you're proving AO^2=AD*BC ?

[analog dust]

No.. im showing that here.. theyre not equal

Look on the left

OA² ≠ AD * BC

4 ≠ 3.97

[eager pendant]

(i can read)

i mean, its definitely true

[analog dust]

Eh

How did you do the figure thing?

Draw it

Or whatever you did with it..

[eager pendant]

wdym

[analog dust]

Hmm okay

[eager pendant]

nvm i solved it

[mint sandal]

how did you solve it

[eager pendant]

ive colour coded those ones, they're obviously congruent

p, q, r are tangent points

and assume that all red lines connect p,q,r with O

so then <AOP=<BOR, <POD=<QOD, <QOC=<ROC

so <DOA=90-<ROC

=<BCO

so DOA and COB are similiar

and from there we're basically done

[mint sandal]

👌

[analog dust]

Damnit i just finished drawing it

Geogebra on phone is not nice

[mint sandal]

ok...

in the same context as my yesterday's posts

I'm still trying to get through that chapter...

i pretty much worked it out with the maurer-cartan form btw

like it wasn't that difficult once you choose the basis

there are essentially just 2 types of elements, and both "switch" places of two basis elements, switch one sign and equate everything to zero

so just looking at how they act on some basis elements made things much easier

now another question in the same chapter:

In screenshot i posted, I get everything except the last sentence

how do I convince myself that it maps that tangent vector into b^i X_i?

i feel like there's something really simple lurking behind it but can't get my head around it..

O(M) is just the principal bundle of orthogonal frames on M

M is a sphere, and f is defined here:

@neon fossil 🤠

@copper valve too :D

[copper valve]

oooh

not now tho I'm meeting with my prof ;o

[mint sandal]

ok!

hi btw

[neon fossil]

sorry brain fried today

ask killingform

[mint sandal]

already did :(

[mint sandal]

nevermind i figured it out

actually it literally is really simple

my head is just not right...

[sweet heart]

Why is completeness not topological (i.e. why is it not preserved by homeomorphism)?

[mint sandal]

do you want a counterexample?

like why would it be preserved by homeomorphism?

the two spaces could have two totally different metrics

[sweet heart]

Ah wait, I need to find a Cauchy sequence that doesn't converge after homeomorphism I think

[mint sandal]

like take a homeomorphism between R^2 and a unit open disk

oh

[sweet heart]

I think.....

[mint sandal]

such that it's a cauchy sequence in both spaces?

and converges in one and diverges in another?

[amber raven]

@sweet heart what textbook do you use?

[sweet heart]

It's from measure, topology and fractal geometry by Gerald Edgar

[amber raven]

For what course?

[sweet heart]

I self-study

[amber raven]

Oh cool

[mint sandal]

well idk if this helps but like D^2 is homeomorphic to R^2 but R^2 is complete while D^2 is not

[mint sandal]

by D^2 i mean x^2 + y^2 < 1

open disk

that seems like a sufficient answer to me

[sweet heart]

Okay, yeah, I'll try 😃

Thanks, Cheburashka

[mint sandal]

lolol

np

you're the first one to say that

[sweet heart]

Where you from if you know Cheburashka?

[mint sandal]

eastern europe

[sweet heart]

Me too

[mint sandal]

👌

[amber raven]

How is D^2 homeomorphic to R^2

@mint sandal

[mint sandal]

is it not?

[sweet heart]

D2 = {(x, y) ∈ R2 : x2 + y2 ≤ 1} ?

[amber raven]

Oh I see

Tangent

[mint sandal]

not x2 + y2 ≤ 1, but x2 + y2 < 1

one is compact other is not

in fact, the unit open disk is diffeomorphic

not only homeomorphic to R^2

it's true higher dimensions too

let me find a proof

[amber raven]

Is Tangent a homeomorphism from (-1,1) to R?

[mint sandal]

arctangent

yeah i think so

well it's the same

as long as you restrict the domain

tan or arctan

[amber raven]

Then we can just do a tangent fuction in every direction of the plane

To construct a homeomorphism from D^2 to R^2

[mint sandal]

the diffeomorphism for general spaces is provided by x → x/sqrt(1-x^2)

well not the diffeomorphism

but one of them

what do you mean by a tangent function in every direction of the plane?

oh i think i can see it

hmm that might work too!

[amber raven]

Restricting our function along a line through the origin gives the tangent

X/sqrt(1-x^2) makes me think of tanget too for some reason..

[mint sandal]

@amber raven

https://math.stackexchange.com/questions/1213847/are-open-sets-in-rn-homeomorphic-to-rn

this person did exactly what you suggested

[sweet heart]

https://math.stackexchange.com/a/296511/204029

Oh, here's an easy one

[neon fossil]

yeah the morphisms of metric spaces are not continuous functions but lipschitz ones

[sweet heart]

How do we show this?

[mint sandal]

it converges to 0

and 0 is not in (0,1)

and limits in R are unique

if they exist

because R is haussdorf

[sweet heart]

I see... Thank you

[simple steeple]

Hi there if anyone is familiar with the sweepline algorithm I could use some help in interpreting some definitions

the bottom of page 3 in this document references the ith leftmost endpoint of a vertex given an input of S[1..n] http://jeffe.cs.illinois.edu/teaching/373/notes/x06-sweepline.pdf

For each segment endpoint, we spend O(log n) time updating the binary tree, plus O(1) time performing pairwise intersection tests—at most two at each left endpoint and at most one at each right endpoint. Thus, the entire sweep requires O(n log n) time in the worst case. Since we also spent O(n log n) time sorting the endpoints, the overall running time is O(n log n).
Here’s a slightly more formal description of the algorithm. The input S[1 .. n] is an array of line segments. The sorting phase in the first line produces two auxiliary arrays:
• label[i] is the label of the ith leftmost endpoint. I’ll use indices into the input array S as the labels, so the ith vertex is an endpoint of S[label[i]].
• isleft[i] is True if the ith leftmost endpoint is a left endpoint and False if it’s a right endpoint.

My issue is that I have no clue what i is a reference for in this document.

[mint sandal]

ok we have the structure equation:

which can also be written as this:

theta is the canonical form, big theta is torsion and omega is the connection form

now much later in the book, to derive a certain formula, they apply the exterior derivative

and if they had written this as $$0=- d\omega \wedge \theta + \omega \wedge \Theta + d\Theta$$ i would understand

[mint sandal]

but this is not the covariant derivative

so how can they just substitute $$d\omega$$ with $$\Omega$$ and $$d\theta$$ with $$\Theta$$?

[mint sandal]

they hadn't applied the horizontal projection operator

[mint sandal]

nevermind

that just follows from the structure equations lol

you just have to apply them again

[prime junco]

How do I write trigonometry in ratios

Nvm just found out lol

[sweet heart]

B > 0 here?

[mint sandal]

yes

[sweet heart]

Okay, thanks, not sure why the don't specify

[mint sandal]

metric is a positive valued function

nonnegative*

[sweet heart]

I know, but they usually state a constant is greater than zero

If B was 0, that would mean the function maps everything from S to some 1 point in T though?

[mint sandal]

yes

i think

[sweet heart]

hah okay 😃

[mint sandal]

d(x,y) = 0 implies x=y

[sweet heart]

I wish they'd just write whether B>0 or maybe >=0

[mint sandal]

the only assumption needed here is that B is a real number as far as i can see

[sweet heart]

0

[forest dove]

I mean sure but it's one of those things where if there is a constant which is a real number, then there is a constant > 0

Since if there is a constant it's ≥ 0, and then you can choose any larger number and it's still valid

[sweet heart]

Oh

[icy forge]

hi

[raven relic]

Oh I can fuck around here and meme about lagrange's

[lusty escarp]

yes

[fallow edge]

This is the definition of sine, cosine, etc.

In a right triangle

[raven relic]

For sin, you can think of unit circle/approximate with exponents, approximate for limx->0 as x

[lusty escarp]

@icy forge Sin Cos and Tan are just functions that are used in triangles

[raven relic]

or just as aw avy function with period 2 Pi

[icy forge]

ok

[raven relic]

I think this image is clear :l

[icy forge]

can you prove something then?

[lusty escarp]

what

[icy forge]

how is SinX+CosX=1

[lusty escarp]

no

thats incorrect

[icy forge]

i forgot the powers

[upper karma]

sin^2x + cos^2x

[lusty escarp]

ok

so the best prove is by defenition

Sin is defined to be O/H

Opposite / Hypotenuse

[icy forge]

yh

[lusty escarp]

Cos is defined to be A/H

Adjecent/ Hypotenuse

[icy forge]

yh, SOH, CAH, TOA

[lusty escarp]

We know that pythagoras states

[lusty escarp]

A^2+o^2 = h^2

[fallow edge]

Unit circle

[lusty escarp]

dont use that pls

[icy forge]

yh

I would rather listen to one person, if 2 people try to teach me at once i dunno who to follwo.

[upper karma]

I’m here too

^^

[icy forge]

Oh i get it

[raven relic]

What do you mean

pythag

[icy forge]

the image helped somewhat

[raven relic]

On that unit circle is so clear

[lusty escarp]

=tex \sin^2(x) + \cos^2(x) = (\frac{o}{h})^2+(\frac{a}{h})^2 = \frac {o^2+a^2}{h^2} = \frac {h^2}{h^2} = 1

[lusty escarp]

Thats how I proved it

[icy forge]

❤ ty

[lusty escarp]

there are 6 Trigonometric Functions

[icy forge]

i only know 2 lol

[lusty escarp]

2 of which are never mostly used

there is

Sine, Cosine, Tangent, Cotangent, Secant, Cosecant

Every other function has Co

[icy forge]

sin^x/cos^x=tan^x

[lusty escarp]

yes

[upper karma]

Which functions aren’t used?

[lusty escarp]

Secant and Cosecant

[upper karma]

Oh

[raven relic]

why not :l

[lusty escarp]

they arent tought in my school

[upper karma]

why is cotan used

[lusty escarp]

:/

[raven relic]

You use them in calc?

[lusty escarp]

yes

[icy forge]

What does secant mean, is it like SOH CAH TOA, where it means something?

[upper karma]

It means 1/cos

[lusty escarp]

Secant is opposite of cosine

[raven relic]

^

[lusty escarp]

Cosecant is opposite of Sine

[upper karma]

So hypotenuse/adjacent

[lusty escarp]

so its

[raven relic]

Opposite is a misnomer, reciprocal

[upper karma]

they’re reciprocals

[icy forge]

oh inverse functions thanks.

[raven relic]

Eh no

That's cos-1 or inverse cos

xD

[lusty escarp]

SOH CAH TOA CAO SHA CHO

[fallow edge]

Can I just say something?

[lusty escarp]

this is full

;3

yes?

[raven relic]

Best to memorize the first three tho and stick with them and know when to use reciprocal function

[icy forge]

damn

[raven relic]

You never need to actually use reciprocal function, just have to recognize them normally cus csc x=1/sin x

and you just can write 1/sin x

[fallow edge]

The definition of cosecant is 1/sin, not the opposite

[raven relic]

^

He corrected himself tho

somewhere

[lusty escarp]

Which is the reciprocal

[fallow edge]

Oh nvm

[raven relic]

opposite=-sin x

[fallow edge]

I thought the opposite was the negativr

[raven relic]

technically

^

[fallow edge]

Yeah

[lusty escarp]

welp

[fallow edge]

Anyway, I’ll go out

[lusty escarp]

https://www.khanacademy.org/ I recommend using this website

to learn maths

it helps

[fallow edge]

i love that website

[raven relic]

Anyway if I wanted to meme I could give someone.... how do you prove the lagrange thereom's maximum must be parallel to the curve....

Khan is nice I don't like some of the older stuff

[lusty escarp]

ye

the quality isnt the best for old ones

[icy forge]

How do we get trig functions in the real world?

[raven relic]

Umm circles

Waves

useful in calc

[lusty escarp]

most probably Architecture

[raven relic]

Physics

Lots in physics

harmonic motion

[lusty escarp]

Really mostly used in Physics

[raven relic]

Dampening, electricity

[lusty escarp]

yep

I mean Math is just a foundation to bigger things

[icy forge]

Yh, i need to learn it for calculus.

[raven relic]

Pure math

[lusty escarp]

Saph has this pic

I need to find it

[icy forge]

Philosphy is even furuther than maths

[raven relic]

Math is hard to make cash fam

[lusty escarp]

going left to right

are the most pure

[icy forge]

Wasn't newton a philosphy dude?

[lusty escarp]

Physician

Physicist

how youd like to say

[icy forge]

I thought he did philosphy in the past, and all maths science was under philosphy

[lusty escarp]

thats incorrect

Philosophy is mostly based on Science

Biology and Chemistry

[icy forge]

science is based on philosphy,

[lusty escarp]

which are based around physics and human bodies

which is mostly Math

[icy forge]

from my p.o.v science is just stories we use to understand our surroudings, and maths is a language which helps out.

[lusty escarp]

math is a language and foundation to all that is created

[icy forge]

I dunno much

But there is a problems, which was prooven to not be solvable, but was also prooven to not be not solvable.

[lusty escarp]

tbh

[icy forge]

wait ill send you a link, i dont know much about it.

[lusty escarp]

Lets break the case

[icy forge]

Maybe youll understand the steps i didnt

[raven relic]

wrong chat probs....

[lusty escarp]

Phylosophy is just on whole different level

[keen aspen]

Not related to the appropriate channel

[lusty escarp]

its like math + phyloso = Eart

Earth

[icy forge]

Ill private msg it to someone, and maybe they can explain it to me?

[lusty escarp]

wht?

[fallow edge]

Explain what?

[raven relic]

/eshrugs

[junior scaffold]

How do u do this

[zenith ember]

Draw line segments from the center to A and B. Then draw the perpendicular to AB that passes through the center.

Because it is a chord of a circle this perpendicular must bisect the chord.

Which gives you a couple of congruent right triangles.

That's a good start.

[raven relic]

Woah that looks like an AMC problm but not really

and what samantha said

Make an isoleces triangle from oa to ob

you know of sidelength 8

than calcualte the area of sector of circle

subtract triangle area

Or just plan the math before you do it

[upper karma]

Yay! John M. Lee is publishing a new edition for his Riemannian Geometry: An Introduction to Curvature that he's calling Introduction to Riemannian Manifolds.
https://www.springer.com/gp/book/9783319917542

[sweet heart]

It's not continuous, right? But how do I show it?

Can only think of this

But then epsilon is changing

[twin prawn]

What are S, T and varrho? Metric spaces and metric?

[sweet heart]

That's all I have above. Any S, T

[twin prawn]

You can find a very easy counter-example

Consider a function f from the real numbers to itself given by x + 1 if x >=0 and x if x<0

Clearly the function is inverse Lipschitz with constant A = 1 but it's also clearly discontinuous

Case closed @sweet heart

oh and I equip the real numbers with the standard metric but that usually goes without saying

🍮

[sweet heart]

Hey sorry if it's a stupid question

And thanks

[twin prawn]

Nah don't be sorry, questions are always to be had and I'm glad to help you out

You're welcome

[hybrid plinth]

Once I fix two points on the circle, call them A and B, then any point x on the "same" side of the chord AB, where x is holding A and B will have the same inscribed angle? Am I understanding this theorem right?

[zenith ember]

I think so.

[upper karma]

yeah

[lusty escarp]

@zenith ember Miss Teacher I need your halp again

I created this shit but

[lusty escarp]

<@&286206848099549185>

[fallow edge]

Is the line coming out of A perpendicular with BD?

[lusty escarp]

Yes

sry didnt write that

[lusty escarp]

its basically another similarity problem

I just cant figure it out

Have alrdy been working on it for past 2 hours

Samantha (the new mod) helped me understand it

but this problem

_<

[fallow edge]

What do you know so far?

[lusty escarp]

uh

That Perpendicular line coming from A is

[fallow edge]

Do you see the two similar triangles?

[lusty escarp]

of course

[fallow edge]

And you know what the scale factor is?

[lusty escarp]

No

That is unknown to me

[fallow edge]

Look at the line BD

[lusty escarp]

and?

[fallow edge]

It’s cut into three equal sections, right?

[lusty escarp]

its a hypotenuse

yes

[fallow edge]

So if one similar triangle encompasses one of those sections, and the other similar triangle encompasses two of those sections, what is the scale factor?

[lusty escarp]

I literally have no idea what you're talking about

[fallow edge]

One sec

Do you see the 2 triangles now?

[lusty escarp]

yes

oh

so its 2

LOL

k = 2 then

but look

you would just says

so AD = 38 sqrt(2)

[fallow edge]

Well, what’s the scale factor to get from 1/3 to sqrt(2)/3

[lusty escarp]

I think i did proportions wrong

but i got AD = 38 sqrt(2)

but answer was incorrect

[fallow edge]

I got AD = 38

[lusty escarp]

yes

thats the correct answer

_<

[fallow edge]

So, the scale factor is sqrt(2), right?

[lusty escarp]

isnt the scale factor 2?

how is it sqrt(2)

2/3 : 1/3 = 2

[fallow edge]

No, because the 1/3 side corresponds to the sqrt(2)/3 side

[lusty escarp]

what?

im confused now

ok lets call O the point the triangles meet

wait

ill draw it

[lusty escarp]

BD times that

[fallow edge]

If you rotate the smaller triangle around point O, the 1/2 side corresponds to the sqrt(2)/3

All of the measurements have BD times them, so we can leave it out

[lusty escarp]

wait

like how..

pls tell me

[mint sandal]

pixel geometry 😍

[lusty escarp]

i seperated them

[fallow edge]

In the picture, imagine rotating the smaller triangle 90 degrees

[lusty escarp]

yes

and the side which is 1/3

would correspond to side which is 2/3

ok lemme draw the triangle ABD only

[lusty escarp]

my eyes lied to me

O.o

fuck

lol

[fallow edge]

Do you see?

[lusty escarp]

YES

like after a visual proof

[fallow edge]

Good!

[lusty escarp]

like my eyes are shit right now

_<

i dont like geometryy

this summer im gonna visit Khan academy again

:3

will have to buy whole stack of A2 papers

or A4

which ones cheaper

[fallow edge]

Well, anyway, after you determine the scale factor is sqrt(2), 19xsqrt(2)xsqrt(2)=?

[lusty escarp]

ye

thats ez

good thing i proved what the perpendicular line is

lol

so i did half the problem

actually more than half apparently

😐

[fallow edge]

Yes

[lusty escarp]

thank you so much

[fallow edge]

Np

[lusty escarp]

like my book

has COMPLETELY DIFFERENT PROOF

ill send you books proof yesterday

and you explain to me

lol

[fallow edge]

Ok lol

[lusty escarp]

it was involving x

2 3

x^2

LIKE WUT

but thank you so much

,3

❤

[fallow edge]

👍

[sweet heart]

How can we count the balls?

[mint sandal]

what do you mean

are you asking why the collection B is countable?

@sweet heart

We count them as B_1/1, B_1/2, B_1/3 and so on..

[sweet heart]

Yea that's the question

Not like this?

[upper karma]

This is a way to count through the rationals (which since we can count them, are countable).

[sweet heart]

So correct/incorrect for this case?

[upper karma]

Which case?

[mint sandal]

countable = in bijection with the natural numbers

[sweet heart]

I was thinking if D was countable infinite

[mint sandal]

or finite

[upper karma]

Where D is a countable set and open balls are indexed by reciprocals of natural (which are also countable0.

[sweet heart]

Yeah, so I'm trying to find bijection

[mint sandal]

n <-> B_1/n

here's the bijection

[sweet heart]

So I don't need to think about "a"s from D?

[mint sandal]

oh right you do

yeah you can do it like with the rationals

just use any bijection N <-> NxN

then compose it with (n1, n2) → B_1/n1(a(n2)

[upper karma]

Cantor bijection is a fun one.

[sweet heart]

Okay, thanks

[stuck jewel]

how is a concave sphere ven possible? i googled it and its a thing

[fluid falcon]

I don’t know how to start this geometry question

Question ii

@timber hinge ^ I have no clue how to start

[timber hinge]

Wouldnt it just be 30? You have 60 degrees in that corner and it must sum to 90 since its a right angle?

[fluid falcon]

But if you look at the second question, how does that work

Cos doesn’t exterior angle equal to sum of two opposite interior angles?

@timber hinge

[timber hinge]

ABE is an equilateral triangle, all interior angles must be 60 degrees. Thus DAE has to be 30 degrees.

As for part iii im not sure how to approach it

I dont believe that DB and AC are straight lines either

[fluid falcon]

Hmm okay know anyone that may be able to help me start off the question?

For iii

[timber hinge]

Something like this

[fluid falcon]

Hmm alright

Thanks for your help

[lusty escarp]

60+75 isnt 180

how is that possible

@fluid falcon It must be a trick question

there is no way that angle is 75 degrees

[raw flume]

No, angle DEA is 75 degrees

You can remark that triangle DEA is isocele (because AE = AB = AD)

And knowing this and the value of angle DAE, it's pretty simple

[upper karma]

https://gyazo.com/34b75a090618e0313a87c143481e4f01

What is d1 and d2?

I don't get this concept please can someone explain it to me

[keen aspen]

Diagonal lengths

[amber raven]

I need help with Ewald's Geometry