geometry-and-trigonometry

[upper karma]

I know no proofs

[tropic stirrup]

It's fun to prove tbh

[dark sparrow]

the pentagon adds up to 3π

[upper karma]

I know nothing...

[tropic stirrup]

Oh also I forgot to make the paper airplane one

[upper karma]

especially proofs

[copper valve]

wait!!!

ann hold on ;o

also wheres the paper airplane theorem 😮

[upper karma]

@tropic stirrup have u seen national mathcounts thingy

Watch this

https://youtu.be/vFTeN17Z4rc

[tropic stirrup]

kek

[upper karma]

All speed based

[tropic stirrup]

Ain't really interested in competition

[upper karma]

Like Koreans like itg

[tropic stirrup]

ew

anyways @copper valve

[upper karma]

?

[copper valve]

oooh!!!

wait i got the star theorem

so the pentagon angles add up to 3pi

[tropic stirrup]

mhm

[copper valve]

and the angles opposite the pentagon (like the ones on the outside) will also add to 3pi

[slim plover]

that's some interesting theory names

[copper valve]

then on each vertex of the pentagon

[slim plover]

tbh

[copper valve]

counting all the angles up will give 2pi

times 5 for each vertex for total of 10pi

[tropic stirrup]

narrows eyes

[copper valve]

that means the other angles will add to 4pi

and sum of all angles of a triangle is pi

theres 5 triangles

so the remaining angles a,b,c,d,e must add to 5pi - 4pi = pi = 180 degrees!!

[tropic stirrup]

my gosh you went around for pretty long but yes you made it x'D

[copper valve]

😄

paper airplane theorem, quadrilateral interior angles add to 2pi, so the last interior angle will be (2pi - (a+b+c))

and the outside angle will be 2pi - (2pi - (a+b+c)) = a+b+c

😮

[upper karma]

Idk how lightening one works

[copper valve]

this one is spooky 😮

[upper karma]

I don't see any obvious reason why

[copper valve]

because it has cool name

[upper karma]

:/

not rlly

I just don't know why

[copper valve]

ya really!!

these are great!

does it hold for arbitrary amount of zigzags @tropic stirrup ?

or only 3

wait is this 3 ;o

however many this is

it works for 2 angles i know that

does it work for 4, 8, any even amount?

[upper karma]

amount of zig zags creating 3 angles on each side ;o

[copper valve]

this is 6 angle version

[upper karma]

I don't think it works with any number

[tropic stirrup]

Yes

[upper karma]

yes what

[tropic stirrup]

This holds for arbitrary amount of zigzags

As long as you keep track of the correct angles, yes

[upper karma]

cooooool

so it would work

with

4 angles?

[tropic stirrup]

Yes

[upper karma]

alrig

[copper valve]

cool!!

[tropic stirrup]

⁉

[upper karma]

?

oh its moved here okay

[dark sparrow]

@gilded star do you know the definition of a parabola that doesn't rely on coordinates?

[gilded star]

It's it just a an open curve?

Wait one sec

[dark sparrow]

no

[gilded star]

My notebook should have it

[dark sparrow]

"open curve" is too vague

[upper karma]

something that has two arms

[tropic stirrup]

ew

[upper karma]

im jk

it is something that

if a line parallel to x-axis touches the thing

it will reflect the thing to the focus

[tropic stirrup]

what

[dark sparrow]

no that's a property

not the definition

[upper karma]

sth like that

oh property okay

[tropic stirrup]

It's a set of points whose distances to a straight line and a point are the same

[upper karma]

ohhh

[tropic stirrup]

👍🏼

[upper karma]

with the directrix

[gilded star]

y^2=4cx

---

y^2=-4cx

Is that it?

[dark sparrow]

no

[upper karma]

PF = PD

[dark sparrow]

those equations can be derived

...i gtg, sorry

[upper karma]

distance from any point to the focus = distance from any point to the directrix

[tropic stirrup]

oops

[gilded star]

Bye!

I've noticed that "foyers" are quite important in conic sections

[tropic stirrup]

foyers?

[gilded star]

That's what they're called in French

[upper karma]

what

are

those

[gilded star]

F,F'

[upper karma]

the foci?

[gilded star]

Not too sure, my entire class is in French, it's like slightly more inside the ellipse (the streched put cirlce) than the summits

[upper karma]

stretched out circle

best definition

[gilded star]

Hey, if it's understood, it's good enough for me

[tropic stirrup]

pfft

Your English seems pretty rad though

[gilded star]

Technically my first language, I'm a linguistic mess

[tropic stirrup]

pfft

Anyways, yeah, those F and F' are called foci, plural of focus

By the way, what's the definition of an ellipse?

and the definition of a hyperbola?

[upper karma]

are u asking or testing

[gilded star]

An ellispe (and I'm translating from my notebook here) is the area of point where the sum of the distances that has two fixed points, called foci, is constante

I was so wrong, everyone left me 😢

[upper karma]

not the area

the loci

yeah is constant which is also equal to the length of the major axis

wait u actually learn definitions i didnt know any

[gilded star]

We have them written in our notebooks

[upper karma]

ah thats ausome

[gilded star]

The conic sections is the only thing we learnt where our teacher gave us a book to follow with

[upper karma]

BOOK

we got

like 15 sheets of paper

PHOTOCOPIED

[gilded star]

Well ours is more 27 pages

[upper karma]

ah

thats a lot to read

[gritty flare]

nneeed hallff

;/

tanks

@copper valve pls hallppp Q-Q

[copper valve]

hi

[gritty flare]

hi 😄

[copper valve]

well first, to get the center of the circle, we need to equate:
(x-8)²+(y+5)²=0

which means x-8=0 and y+5=0

then to get the radius r of the circle, we have:
(x-8)²+(y+5)²=r²=4

so r²=4

[gritty flare]

so radius is 2

[copper valve]

ya

[gritty flare]

ohh

i get it

so you get the x and y values

and then the radius

thanks a lot

[copper valve]

np!

[gritty flare]

one question

ok

no

i got it

[waxen gorge]

O k

[upper karma]

(8,-5)?

[waxen gorge]

Ya

[upper karma]

What I thought

[queen notch]

hi there

i have a question if possible

[dark sparrow]

[queen notch]

is there an isomorphism between hyperbolic geometry and euclidean geometry?

[copper valve]

what u mean by isomorphism?

[queen notch]

well, a bijection that preserves properties like continuity and differentiability

[copper valve]

there's related concepts, but there's definitely hyperbolic-only phenomena

[queen notch]

like, more precisely a mapping from the euclidean geometry to the conics geometry

[copper valve]

hmm

[queen notch]

and an inverse mapping

[copper valve]

good question :0

I'd guess not, but im not an expert on this. i wanna know too :-0

[queen notch]

can i ask another question

[copper valve]

of course

I'm really interested in mappings of axioms and such

[dark sparrow]

is there an isomorphism between hyperbolic geometry and euclidean geometry?
i'm pretty sure the poincare disk and upper half-plane models both correspond to diffeomorphisms between the euclidean and hyperbolic planes

[queen notch]

let me explain the whole idea i had today

thanks

[copper valve]

woo

[final prairie]

g

[copper valve]

😄

[queen notch]

i was reading about conics geometry and then i tought, is it possible to make machine learning on conics, then, forgive me if there are imprecisions on this, i would like to analyse the difference in computational cost between the 2 cases

[copper valve]

interesting

so like the difference in difficulty​ of computing geometric results in varying non-euclidean spaces?

[queen notch]

yeah

and if they are more optimized

or not?

that is a good question

[copper valve]

good question ^.^

[final prairie]

well should you use the same algorithm for both cases, their complexity will be the same. One might take way more time but the time-complexity would obviously be the same

[copper valve]

i am interested in this stuff too... eeee

pocky lets be buddies

😄

[queen notch]

sure

[copper valve]

i even have pocky with me right now

[queen notch]

could you guys point me to some textbooks that could help

[copper valve]

strawberry flavor

@haughty prawn got any noneuclidean geometry pdfs or algorithmic geometry

[vapid kettle]

algorithmic geometry :0

[copper valve]

or otherstuff that seems relevant

[haughty prawn]

Baby hartshorne goes over noneuclidean geometry for a bit.

I don't know if it's the best resource for it though.

[queen notch]

ok, i'll look into it

[haughty prawn]

It's actually called: Geometry: Euclid and Beyond

[queen notch]

much obliged, this is a long term project but if i can, would it be possible to keep in contact over here

[copper valve]

yes of course!!

[queen notch]

^^

[copper valve]

we need more pocky

[queen notch]

this name is because the one that invited me to discord used to call me that in a game lol

so i said, well why not

[copper valve]

[queen notch]

well, thanks for the lead ann/sqrt(2) and obyeag, i'll continue my studies now

[copper valve]

ok!!

[upper karma]

@copper valve I remember my school used to give those on holidays

lol

[copper valve]

nice 😮

[upper karma]

xd

They don't anymore

[waxen gorge]

Pocky sticks r so good

Strawberry flavor best

[upper karma]

@copper valve do you speak french?

[copper valve]

no

[upper karma]

does anyone know a formula for reflection of a point (x,y) by line y = mx+b

do you know anything about linear algebra

[hidden geode]

like what topic in particular

(oh my bad you mean sem not me)

[upper karma]

a bit

[crude kraken]

(y,x)?

oh, by y=mx+b

hm

=tex \left(\frac{(1-m^2)x_0+2m(y_0-b)}{1+m^2},\frac{2mx_0-(1-m^2)y_0+2b}{1+m^2}\right)

Bah

[queen notch]

you can apply a rotaion matrix on the line so it is coincides with an axis, that will roatate your group of points by an angle, the you just apply like a mirror in optics, the group of points becomes -this new position, then you rotate again the line to the original position, then you get the reflection, someone correct me if i'm wrong.

and sorry for the grammar mistakes

typed really fast

[leaden kindle]

No

Cause you cant lap them over

[dark sparrow]

"lap them over"

[gritty flare]

but there is also not enough information

[dark sparrow]

you mean overlap them

anyway

yes there is

you know all 3 angles for both triangles

[gritty flare]

oh right

😮

[leaden kindle]

Yeah mabad

[gritty flare]

im dumb

[leaden kindle]

You even know all three sides

But you can do this without trig

[dark sparrow]

no trig required

neither is knowledge of the other two sides of either triangle

[leaden kindle]

Yeah true

[gritty flare]

180 minus those angles

ye sry im dumb sometimes

[narrow sleet]

You need to just practise

[gritty flare]

o.o

[dark sparrow]

you're shown the dimensions for one triangular face

you can calculate its area

[gritty flare]

i dont know the surface area formula

[dark sparrow]

just add up the areas of all the faces!

i thought that was obvious

[gritty flare]

but

o.o

oh

ohh

oh shit

[queen notch]

can someone help me please

[upper karma]

[queen notch]

the problem is the following i have this triangle

[upper karma]

what do u need help with

isnt this gurls problem

[queen notch]

well, let me elaborate

consider now this tiangle please

i want to relate both as the following the point p in which the lines i defined intersect was taken arbitrarily

the problem is the following

the important part is that

there exists a p

in which no matter

what kind of position it is inside the triangle

i'll be able to draw lines in which i intersect p with the vertexes

and i claim more, there will be always a way in which i am not considering the line PA i am estabilishing another point W in which i claim it belongs to the line below and it is perpendicular to p

now, following from the first image i have

[tropic stirrup]

...no

no, you don't generalize before you prove something

you must first prove that A, P, W is on a line before constructing such diagram

[queen notch]

APW is not a line

PW is a different line than AP

[tropic stirrup]

Then you shouldn't draw a diagram that makes it look like A, P, W are on a line

[queen notch]

it does not change the angles equations i drew

[tropic stirrup]

It does.

You stated that 40 + 180 - epsilon + beta + 30 = 180, correct? I'm assuming that is from triangle AWC.

[queen notch]

let me rewrite something, that should have been erased i understand why you were angry now

[tropic stirrup]

.....

[queen notch]

epsilon = 90 that we both agree right

[tropic stirrup]

by the way, the answer is that AB is equal to BC, not AC, so whatever you're trying to prove, it's wrong

yeah

Hey hey hey before you do anything

Do it with this diagram.

[queen notch]

and you do agree that the angle 150 + 110 = the upper angle of the lower triangle

ok, i'll do it

[tropic stirrup]

Yeah

[queen notch]

but the condition PW being perpendiculat to BC must hold

[tropic stirrup]

Yes, I am aware.

[queen notch]

this is my last attempt, if it is incorrect i stopped caring at any rate

[waxen gorge]

Looks right...

[runic jackal]

what is y/sin(110-i)

[queen notch]

i decomposed it in 90+20 because that way i would have the same angle

it is just the substitution by a+i = 80

[runic jackal]

and how did u get 90+20

[queen notch]

110 = 90 + 20

[runic jackal]

yeah how do u get 110 then

[queen notch]

a+i = 80

[runic jackal]

oh okay

[queen notch]

sin(a+30)

[runic jackal]

looks right

[queen notch]

thanks i spent hours in it

lol

and sorry that my initial approach was wrong

but as the saying goes i think this is a place to learn right

[runic jackal]

yep

that was a great solution

[queen notch]

thanks, that comment made my day, seriously

[runic jackal]

ahaha

[queen notch]

i'll go and have lunch now, i'm starving

[runic jackal]

cya

[gritty flare]

Gurl

@tropic stirrup You are korean or chinese?

or Japanese even

[mossy vine]

korean iirc

[gritty flare]

i need koreans help

😐

[queen notch]

there is a theorem that states that the area of of 2 times that solid is equal to

where

[dark sparrow]

no that's overcomplicated

[queen notch]

in that case i apologize

[dark sparrow]

V = CF * (BC * AC)/2

[waxen gorge]

@queen notch -_-

[queen notch]

calm down, i'll try to help but i might not be the best to aid

give me some time, and by that i mean a lot of time, is that possible, but i promise i won't give up until this is solved

[dark sparrow]

i just gave you the formula you can use to work everything out

plug in what you know into the formula

84 = 12 * (4 * AC)/2

84 = 24 * AC

[waxen gorge]

3.5

[dark sparrow]

it happens to be a non-integer

...so what?

[gritty flare]

how to square a circl

[waxen gorge]

U wot

[upper karma]

O

[waxen gorge]

?

[mossy vine]

so you know how the equation of the circle is x^2 + y^2 = r^2

for simplicity we'll look at the case x^2 + y^2 = 1

[waxen gorge]

how would u square it tho

[mossy vine]

now consider x^n + y^n = 1

this uniformly converges to a square as n tends to infinity, thus squaring the circle

[waxen gorge]

Ohhh

[upper karma]

U draw square of side a = √2pi* r

?

[mossy vine]

@waxen gorge squaring the circle is, given a circle, constructing with straightedge and compass a square with the same area

[upper karma]

So im right

[mossy vine]

no

it's impossible :^)

[upper karma]

Oo

But

[mossy vine]

that would have the same area

but it's not a straightedge and compass construction

[gritty flare]

want me to show you?

2 ways

[dark sparrow]

go ahead, we'll try to find the flaws

[upper karma]

Sphere to tetrahedron 🤔

[gritty flare]

=tex \bigcirc^2

[dark sparrow]

😒

[gritty flare]

yep

[mossy vine]

I already gave a meme way to do it alexoid

[gritty flare]

oh btw

if u set circles radius to be equal to 1

and squares side to be sqrt of pi

[upper karma]

@gritty flare i think i solved cuz i get expression similar to answer im very happy 🙂

[gritty flare]

their areas should be pi

[mossy vine]

it's impossible because you can only construct "algebraic" numbers using a straightedge and compass, and pi is transcendental

[gritty flare]

:3

cough

sry

i just coughed

hmph

way to ruin my fun

=tex \bigcirc^2

[gritty flare]

sticking with this then

[mossy vine]

I prefer the uniform convergence method :^)

[gritty flare]

:/

wats that

[mossy vine]

let S_n be the set of solutions (x,y) to x^n + y^n = 1

the circle is S_2

as n tends to infinity, S_n uniformly converges* to a square

uniform convergence probably isn't the correct term

but idk what it would be

it's a similar principle though

someone who knows better than I do tell me what it would be :^)

[queen notch]

can someone point where is the mistake i got seven for AB in that problem using my method

i'll elaborate

[waxen gorge]

U didn't do x2

@queen notch

[queen notch]

did the figure load? the connection is slow

just a correction it is 12*y2

= 0

and 4*x1 = 0

sorry it is hard to type in paint

[upper karma]

U get 48z1 from determinant

How do u get 84*2

Oh wait

What this determinant is equal to

[queen notch]

the are of the figure shown above

[upper karma]

Yes

Its volume is [u v w]

So (u×v).w

(48z1)=but whats its volume

[queen notch]

84 was the volume enunciated in the problem

[upper karma]

Z1=21/12

[queen notch]

=calc (842)/(124)

[charred spear]

3.5

[queen notch]

bingo

i messed up in my simplifications

sorry for the mistake and thanks for the help

[upper karma]

Isnt z1=1.9

[queen notch]

but the solid in question was not a parallelepiped it was a prism

[upper karma]

O

[queen notch]

that is why i added the 2

[upper karma]

I see

[queen notch]

thanks martin, i could not have done it without you

[upper karma]

Nah u done urself

😀

[queen notch]

thanks, that is the third comment that made me won my day today

xD

[waxen gorge]

Lol

[tropic stirrup]

@queen notch, sorry but you did a mistake

[vague sierra]

Any smart geometry people here

I need help with my homework.

[upper karma]

@vague sierra

[vague sierra]

oh ok

https://gyazo.com/232258385d624c78bd33478e0328b31b

I want to say its 3

But it's a equailiterial traingle

Not sure if it could be an isoceles

It can't be 1 because that's over 180

2 Does not have 2 congruent angles

4. Doesnt have to congruent angles

[mossy vine]

an isosceles triangle has 2 angles that are the same

at least 2*

[vague sierra]

Correct.

So it's three?

[upper karma]

Yep, go for the safest bet.

[mossy vine]

it has to be 3

none of the others are

and 3 is isosceles, it just also happens to be equilateral

[vague sierra]

Alright dope, thank you.

https://gyazo.com/a6ddd47cc8a511160b8782ce93456e7e

For this question would I just do 8x + 19 = 83

And then I'd get 31 as the answer

Nevermind it's right.

https://gyazo.com/c4d72092335dbf514df836f3fca298bc

I'm not good with the sides question would the answer be 1

[mossy vine]

yes

because 1 + 2 = 3

and then you have a straight line instead of a triangle

[copper valve]

you're not wrong

🤔

[mossy vine]

because 2 of the sides sum to the other one

[copper valve]

good point

[mossy vine]

triangle inequality :^)

[queen notch]

here we go again

now to complete the proof that was indeed flawd look at this

=wolf sin(20+x)/cos(20-x)

[charred spear]

Timeouts
Reduce, BranchPoint, ComplexMap, RiemannSurface, ODE, Sequence, DomainAndRange, GlobalExtrema, InterestingDefiniteIntegrals, InterestingMultiDimensionalDefiniteIntegrals, MathematicalFunctionData

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(20%2Bx)%2Fcos(20-x)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[tropic stirrup]

ok?

[queen notch]

from the domain of the function we get that x < 20 in n = 0

[tropic stirrup]

....what

domain?

[queen notch]

the possible domain of the function

[tropic stirrup]

possible domain. Can the domain be broader?

[queen notch]

sorry i need to organizar this but i did get it this time

i'll be back in some minutes

we have that the angle x is lesser than 110 in fact

then we cosider the another sin law angle

=wolf sin(50) = sin(20+x) x < 110

[charred spear]

Timeouts
Reduce

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(50)+%3D+sin(20%2Bx)+x+<+110
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[queen notch]

= wolf sin(5) = sin(20+x); x < 110

[charred spear]

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(5)+%3D+sin(20%2Bx)%3B+x+<+110
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[queen notch]

= wolf sin(50) = sin(20+x); x < 110

[charred spear]

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(50)+%3D+sin(20%2Bx)%3B+x+<+110
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[tropic stirrup]

....what?

sin(50) = sin(20+x) where did this even come from?

[upper karma]

Pokys query
"= wolf sin(50) = sin(20+x); x < 110"

[queen notch]

from the other side, the equality is this x/sin(20+i) = y/(cos(i-20) = z/sin(50)

but i made a mistake, the idea is now to try to compare the domains of the functions and reach a single solution, i'll try to imrpove it

the ideal would be then that 2(pin - 8pi + 15) an angles that is greater than 0 and less than 180

forms

which occurs in x = 15

sorry 30 = x

it is wrong i know, i'll be back to the drawing board

[tropic stirrup]

Okay why do you keep changing y and z into x?

they're different variables

[sullen flint]

no idea what to do

AC is 4+x perhaps ?

[runic jackal]

oh

if angle DAB is 45 degrees

u should know what triangle ABD is

u would be able to find what AD is

for DC, u can use sine rule and get that

sine rule everything and u will get all sides

[sullen flint]

but i didnt learn sin yet

how to do it without it

[dark sparrow]

have you learned the 30°-60°-90° triangle?

@sullen flint

[sullen flint]

is it like if angle is 30° then side that its facing will be double ?

i dont know

[sullen flint]

nope didnt learn that

[dark sparrow]

huh?! idk how the hell they're expecting you to do that problem you posted then lol

like BCD is the same triangle as the one i showed

just 4 times larger

[runic jackal]

lol

maybe imagine it as an equilateral triangle divided into half

[tropic stirrup]

That's where it came from tbh

[runic jackal]

yeah lol

[dark sparrow]

yeah

[sullen flint]

i still dont understand how to find edge lengths

[queen notch]

wow, i never used this kind of brainpower in a math problem before lol, i'm not sure there are not flaws this time but i'll post what i think to be something now, i'll draw a picture so it will take some minutes

[queen notch]

v

a part was cut, where i mentioned all the sin law inside the triangle i apologize for that

[waxen gorge]

that font 😮

use like cambria math or something

[queen notch]

sorry it is the standard font so i did not even consider changing it

i'll be more careful next time

[upper karma]

what was the question again?

[queen notch]

the question was

a point forms angles of 40 10 30 20 with certain vertex

the ones noted

exactly in those position

prove that if that occurs the triangle is isoceles

[upper karma]

what is K again?

[queen notch]

K is the line inside the triangle

the only one that is missing sorry

[upper karma]

oh ok

[queen notch]

i was sure i added the letter

but here is a new picture in case of doubt

[upper karma]

I cant spot the error, but

= wolf sin(20)/cos(20)

[charred spear]

Assumptions
Assuming trigonometric arguments in degrees. Use 🇦 radians instead

Query made by @upper karma
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(20)%2Fcos(20)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[upper karma]

And you have that A/C=sin(20)/cos(20)

If A=C then A/C=1 but this is not the case

So there is something weird going on

[queen notch]

true i see the point

[waxen gorge]

=pup sin(20°)/cos(20°)

[charred spear]

Query made by @waxen gorge
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(20°)%2Fcos(20°)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[queen notch]

can i retry tomorrow?

[upper karma]

Sure, Im gonna see if I can spot where the problem is

[queen notch]

thanks

[tropic stirrup]

A =/= C.

I've already said this a few times and you're just ignoring

A = B is what you have to prove

[queen notch]

well man i'm trying, you have no idea how many attempts i did at this problem now lol, but i'l keep at it for a while, i'll just have coffee and restart working onto it

[waxen gorge]

Doesn't that mean ur diagram is wrong

cuz law of sines

sin(110)/a = sin(100)/b

So a =/ b

[past mantle]

What are the 6 properties of parallelograms?

[waxen gorge]

Opposite sides congruent

Opposite angles congruent

Consecutive angles are supplementary

If one angle is right all others are right

The diagonals of a parallelogram bisect each other

[past mantle]

And after u list them, how to prove?

[runic jackal]

hmm what do u mean by proving

[waxen gorge]

Each diagonal separates into 2 congurent triangles

[runic jackal]

u mean proving properties?

[waxen gorge]

Ok so start with the opposite angles congruent

If u draw a circle around the shape

And u know angle measure is arc degrees/2

Then we can easily prove opposite angles add up to 180

[tropic stirrup]

thonk

[waxen gorge]

Then we go a step further

And then we have the two other angles next to each other also add to 180

[past mantle]

We haven't talked about arcs yet

So that. Kind of proof is obsolete

[waxen gorge]

Meaning that if a + b = 180 and b+c= 180... Ohhh then lmme use a simpler method

[runic jackal]

why would u draw a circle around sth that u have proven

[copper valve]

circles are cool

[tropic stirrup]

squares are not cool

:(

[waxen gorge]

Have u learned sss, etc

[tropic stirrup]

hisss

[past mantle]

Were in the chapter of polygons and property's and proofs of them

[tropic stirrup]

sss sas asa rhs n stuff

[waxen gorge]

^

hl, ll

[runic jackal]

^^

[past mantle]

SSS sas Asa

[tropic stirrup]

Sassy

[waxen gorge]

Ok that's all u need

[tropic stirrup]

What the hell

[copper valve]

stop spamming multiple channels

[waxen gorge]

U baka

[copper valve]

is this a test question?

[waxen gorge]

Use vietas

Now go away

Ok back to the quadrilateral

Parallelogram*

[past mantle]

On a test of proofs, I was writing Asa but I wrote as(s)

[waxen gorge]

-_-

Ok whatever so u split it into two triangles

[past mantle]

Xd

[waxen gorge]

Now if u assume the two opposite sides are congruent

[past mantle]

Oh, I see

[waxen gorge]

Actually

U don't need to

But

For this proof u have to know opposite angles add to 180

So we have to prove that first

[past mantle]

In 2 units (after similar shapes) we get into trig :D

[waxen gorge]

Do u just want me to send vids?

Instead of typing all this out

[past mantle]

Nah

[waxen gorge]

Uh... K it would be a lot simpler but sureee

[languid plank]

is there a way to construct a segment that has a length equal to the sum of two rectangular areas

using ruler and compass

[dark sparrow]

length
sum of two areas

unlikely

[umbral snow]

The area of a rectangle is just b×h, which is easily constructable. Where is the difficulty?

[dark sparrow]

you can't really have an equality between a length and an area

unless it's numerical equality, but that's dependent on a measurement system

[umbral snow]

Oh I see what you mean. I assumed numerical equality.

[queen notch]

here goes, last try i think everything is in order now, you be the judges of that:

[queen notch]

just 2 corrections that won't affect the result, writing in paint is horrible lol, where i state that 4sin(20)cos(20) = sin(20)cos (alpha)+
cos(20)sin(alpha)/(sin(110)cos(alpha)-sin(alpha)cos(110)), there is a bracket mistake 4sin(20)cos(20) = (sin(20)cos(alpha) +
cos(20)sin(alpha))/(sin(110)cos(alpha)-sin(alpha)cos(110)), and in the line K/C/K/A = sin(20)/sin(150)/sin(30)/sin(110)

instead of what is written

[tropic stirrup]

K/C/K/A bit is wrong

[queen notch]

in the calculations i considered the corrected fact

it was a transcription error

[tropic stirrup]

then please correct that

[tropic stirrup]

wrong

A/C = sin(110)sin(20)**/**4

[queen notch]

=wolf sin(20)sin(110)/(sin(30)sin(150))

[charred spear]

Assumptions
Assuming trigonometric arguments in degrees. Use 🇦 radians instead

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(20)sin(110)%2F(sin(30)sin(50))
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[queen notch]

=wolf sin(20)sin(110)/(sin(150)sin(30)

[charred spear]

Assumptions
Assuming trigonometric arguments in degrees. Use 🇦 radians instead

Query made by @queen notch
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(20)sin(110)%2F(sin(150)sin(30)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[queen notch]

and there this time you are wrong

[tropic stirrup]

Seriously....

Next time you write anything get the constant out in the front okey

like

4sin(110)sin(20)

[queen notch]

i'll keep that information for future record

[tropic stirrup]

.....what is this

[queen notch]

taking alpha = 80 - beta, we have that beta + 30 = 110 - alpha

[tropic stirrup]

So you're saying \
$$\sin (110^{\circ}-\alpha)=\sin\alpha\cos 110^{\circ} - \cos\alpha\sin 110^{\circ}$$

[queen notch]

sin(110)cos(a) - cos(110)sin(a0)

[tropic stirrup]

why do you do this to me 🤦🏼

[queen notch]

is there another mistake?

[tropic stirrup]

Tell me how you got from \
$$\tan \alpha = -\tan 20^{\circ}\frac{4\cos^2 20^{\circ} - 1}{4\sin^2 20^{\circ} - 1}$$ \
to $$\arctan \alpha=60^{\circ}$$

[queen notch]

oh in that case i used a calculator, no way to do that on hand

[tropic stirrup]

nope

[queen notch]

ann helped me on that part

[tropic stirrup]

calculator.....

[copper valve]

lol

[tropic stirrup]

this whole compilation of problems disallows any use of calculators

[queen notch]

well, if you say so

but just for record, the calculation is on the bot-test section

[tropic stirrup]

no you need to prove the calculation with hand!

Everything, without calculator, prove

[queen notch]

i won't do that, that is clearly a number

[tropic stirrup]

No calculator allowed means you can't use a calculator to prove anything

For example if a question asked to find the value of $$\sin 20^{\circ}\cos 10^{\circ} + \sin 10^{\circ}\cos 20^{\circ},$$ would you use a calculator

[copper valve]

gurl!!

spill the beans

how do u solve it

[tropic stirrup]

I don't know

[copper valve]

WAT

[tropic stirrup]

But @queen notch, it's not like I'm not respecting your work. You've come pretty far, and it's quite impressive you made it till that thing

The last step seems quite hellish though

[queen notch]

the 2 only things i have not considered in this problem are a change of coordinates and inscription in a circunference

i have a feeling the problem is much easier through inscription on a circunference

[tropic stirrup]

The hell is an inscription on a circumference?

[queen notch]

i mean as methods of resolving the problem

[tropic stirrup]

Also refrain from using non-English words in your proof, really throws people off sometimes

[queen notch]

one second i'll look for a proper definition for that

inscribed is the word

and circumscribed i think

[tropic stirrup]

nono I mean in your mspaint proof

[queen notch]

i understood, but that was from the previous comment, a delay in answer, I'll try to improve so thanks for correcting my proofs, and also for the reminder of not using non-english words

i hope i can be of use at some particular problems in here in the future, this one was hard, really hard

and sorry i made you angry it was never my intent in the start of things

[upper karma]

He's just angry sometimes.

[tropic stirrup]

No no I'm not angry x'D

I might sound like I am, but trust me, I'm never angry about a person who wants to truly solve something and do something

That time when I swore at you was um... well I had really bad things happening that day -- to be precise someone close to me died -- so yeah, exceptions occur sometimes ^^;;

[haughty prawn]

kek

[queen notch]

i understand

wanna know something, i'll go and study some complex analysis now, it might be easier than geometry at some points, this is a joke by the way.

i mean, my statement is a joke

[narrow sleet]

....

Oh ok???

[queen notch]

like, complex analysis, easier, the irony, i'll stop here, thought it would make people laugh

[narrow sleet]

Never studied complex analysis before, so I don't know

[copper valve]

me neither :0

[upper karma]

It's apparently purty

[queen notch]

it was just a game of words, i don't really mean what i said

every subject is difficult in its own set of problems

[upper karma]

Too tru

[queen notch]

at any rate, i once again give thanks to @Gurl for posting the problem and for helping me amongst all the others that took their time to read the solutions i posted and comment so that i could improve myself and the solution as well, even though it is not yet complete, I'm a bit hungry, gonna eat something, so see you all.

[copper valve]

okay~ see you later~

[tropic stirrup]

No problemo 👌🏼

[plucky perch]

If two angles are supplementary to the same angle ( or to congruent angles), then these angles are congruent.

what does this look like

[runic jackal]

supplementary angles are like

they sum up to 180

[plucky perch]

what if i have like 120 and 60 though

aren't they 180 , but not congruent

i don't get the to the same angle part

[runic jackal]

ohhh

lets say i have a and b

and "b" is the "same angle"

hey if a is supplementary to b, a = 180-b

if c is also supplementary to b, c = 180-b

u can see a and c are congruent

:D

[barren moat]

a=c in conclusion

bc

a=180-b=c

[dark sparrow]

@plucky perch if a+b = 180 and c+b = 180 then a=c

[vast pasture]

okay guys

i need help

[narrow sleet]

Angle B = Angle A

[vast pasture]

yea

[narrow sleet]

And u can use that to find the value of x

[vast pasture]

I don't understand how to get angle e

O I think i understand

[sour wharf]

hey

so does anyone know why in trigonometry people use "theta" for angles? was it like a random thing or was there a certain reason behind it?

[waxen gorge]

No particular reason

Kinda just chosen by culture u could say

[sour wharf]

oh okay

[upper karma]

@vast pasture did u get it

[narrow sleet]

I think he gets it

[upper karma]

pretty easy problem since you're given x

[vast pasture]

yea i got it

[upper karma]

👌

[gritty flare]

i like theta more then alpha and beta

but in Identities sometimes you need to use alpha beta

[dark sparrow]

no

the names of any variables in an identity are completely arbitrary

[gritty flare]

well

😄 its better to use alpha beta in like cos(alpha + beta)

[sterile mason]

Is there any formula to find the median of a right triangle?

[gritty flare]

i think this is it

this is for solving side a

[dark sparrow]

a right triangle has 3 medians though

so @sterile mason you'll have to clarify what you mean by "the median"

[gritty flare]

just median

1/3

[sterile mason]

the median of the hypothenuse

[dark sparrow]

yeah, that's always half the hypotenuse

in a right triangle

[gritty flare]

from C to to c

[sterile mason]

👍

[surreal hawk]

Thanks man

[gritty flare]

it was 61?

wow

😐

i can do things

wait

oh not again

why so many dans

@vast pasture angle is 61

[dark sparrow]

there's apparently 5 people named Dan, up to capitalization

[gritty flare]

2 of them have same capitalization

[runic jackal]

AHAHAH wait did u get the wrong dan

[gritty flare]

YEP

Q-Q

u see i remembered the name dan nothing else

[runic jackal]

welp refer from the avatar

is one way

[gritty flare]

i didnt remember it

[runic jackal]

at least remember how the avatar looked like

great

[rose glacier]

So, stupid question time, but, what is the between quaternion and axis-angle rotation representation?

Pratically, both seem the same to me, as both store a direction to rotate around and an angle on how much to rotate around it, so why do game engines prefer quaternions over axis-angle since there's no difference?

[copper valve]

less operations required to calculate

making it more efficient for certain things like 3d graphics in video games 😛

[rose glacier]

Yes but why?

[copper valve]

i have a book that describes the difference

unfortunately its not with me atm 😦

[rose glacier]

Oh

What's the name of the book?

I'd love to check it on amazon

Aw crap

[copper valve]

https://press.princeton.edu/titles/6429.html

[cursive nexus]

It's just faster to multiply quaternions

[copper valve]

quaternions and rotation sequences

its more efficient for certain algorithms

[rose glacier]

I see

[copper valve]

im not quite sure how much better they are though. i didnt to the end of the book haha

[rose glacier]

Thanks for indication me the book, I'll definitely be checking that out :D

[copper valve]

np!

[low leaf]

How

[dark sparrow]

pic's sideways

also, standard question:

what have you tried so far and where did you get stuck?

[low leaf]

Whoops sorry lol

[dark sparrow]

what have you tried so far and where did you get stuck?

[low leaf]

I dunno if I'm doing this right lol but I don't know what to do with the 14 or what the second part is supposed to look like

Is that just the line that connects them?

[dark sparrow]

yes

[low leaf]

Oh ok lol

Thank you!

[dark sparrow]

i assume you can take it from here

[low leaf]

I made it complicated in my head lol yeah thanks!

[cursive vigil]

UUghhhh

ABCDE is a regular pentagon and BCF is an equilateral triangle such that

F

is inside ABCDE

What is the size of angle FAB?

[dark sparrow]

what have you tried so far and where did you get stuck?

[cursive vigil]

My teacher told me to like

Form triangles from the sides

from 1 point only

I just dont see any equilateral triangle anywhere

forming

I see iscoceles

everywhere.

[dark sparrow]

FAB is the angle you're after, right?

what can you say about the triangle FAB?

@cursive vigil

[cursive vigil]

OH

ok i did not think the triangle was meant to be like that

i thought it had to link to each vertex

if each full interior angle of the pentagon is 108 degrees

and equilateral triangle is 60 degrees

108-60 is 48

so angle ABF is 48

AFB is 48 too

[dark sparrow]

no

ABF is 48°, but AFB isn't

[cursive vigil]

oh

is it scalene?

[dark sparrow]

no it's not

AFB is not scalene

[cursive vigil]

ok so if ABF is 48 degrees

180 - 48 = 132 degrees

so 132/2

is 66

[dark sparrow]

indeed

[cursive vigil]

thank you

very much

[restive void]

my goal here is to find a

but the way my teacher explained it is really confusing for me

<@&286206848099549185>

[nocturne estuary]

those are similar triangles

[restive void]

yeah, so i know i need to set up proportion

[steady sleet]

Or you could just find the bottom side-length using Pythagorean

[restive void]

but i dont know what proportion

[steady sleet]

^

[restive void]

cant you use a proportion as well?

[nocturne estuary]

you know the standard notation for numbers, like 1x10^-12 is pico and stuff?

whats 1x10^-48

[restive void]

i havent learned that

so

uh

well

so 6/12?

i think itd be better if i redraw it

but im stil really confused

jesus i suck at math lol

would it be 12/a = 6/a?

or wait

wrong

12/a = a/6

or something

yes

so a/6 = 6/12

?

okay a = 3

math is so hard for me

science is my strong point lol

[upper karma]

Hey everyone its my first day here

[copper valve]

welcome!

dang o_o

you beat me to it

[upper karma]

I'm really struggling in math models, next week is the end of the semester and I have a %22

[waxen gorge]

With what concepts are you struggling

[upper karma]

So right now we are working on linear functions

I'm really bad at graphing

[runic jackal]

off topic: ur avatar is really adorable

[copper valve]

yup!

[upper karma]

Thank you!

Please i need help really bad and my mom is making fun of me for having ADD pls

[narrow sleet]

Give us a question then we can help you to go through it

[upper karma]

It sounds really confusing but, i have to evaluate f(x) for the given values of x and then i have to graph the ordered pairs. I get that part but i dont know what to do when i dont have a number

Would you like to see?

[thorn talon]

sure

[runic jackal]

yeah just post the question here

[thorn talon]

f(x) = x + 1?

or what?

[upper karma]

Question 50

[thorn talon]

well

by filling in the table

you are basically just finding points of the line

for x = -2

you basically do like

f(-2) = (-2) + 1

f(-2) = -1

so (-2, -1) is a point on the line

you understand that part?

[upper karma]

I understand that part

Actually what happens to the 1?

[thorn talon]

where?

[upper karma]

The +1

[thorn talon]

f(-2) = (-2) + 1 ?

[upper karma]

Yeah

[thorn talon]

-2 + 1 = -1

i just simplified it i guess for the next line

[upper karma]

help

[mossy vine]

the circumcentre is the same distance from all three points

[upper karma]

So

[mossy vine]

how do you find the line of points the same distance from the two points?

[upper karma]

Man i'm so confused right now

🤔

[mossy vine]

perpendicular bisector

[upper karma]

oh

[vast pasture]

I dont think im right

[mossy vine]

so the point of concurrency of the perpendicular bisectors will be same distance from all three points

[upper karma]

no

you aren't

[vast pasture]

yea ;(

[mossy vine]

so it'll be the circumcentre

[upper karma]

Subtraction property of equality

[vast pasture]

but there's addition

[upper karma]

no

you're subtracting BC from both sides

[vast pasture]

o

[upper karma]

Okay i'm on the same question but different row

[thorn talon]

?

[tawny pewter]

how do i solve

interior angles

for a trapezoid when i have no leads

i only have a triangle next to it that is 60 60 60

zzz

[upper karma]

can u post the problem

[tawny pewter]

ok

[upper karma]

ok

what is the question