geometry-and-trigonometry

[dark sparrow]

the y axis is four units to the left

and if you go 4 units to the left along r3, how far will you go down?

[upper karma]

uhm

[dark sparrow]

if you go 5 horizontally, you also go 1 vertically

[upper karma]

oh

yes. I see that

[dark sparrow]

that's what it means for the slope of that line to be 1/5

[upper karma]

yes. the inclination of the slope

so..

1/5 + 4?

[dark sparrow]

no

[upper karma]

darn

[dark sparrow]

5 horizontal -> 1 vertical
4 horizontal -> ?

[upper karma]

yes. I get that

slightly less than 1

[dark sparrow]

how far do you go vertically if you only move 1 unit horizontally?

[upper karma]

1/5?

[dark sparrow]

yes

[upper karma]

so

[dark sparrow]

no

1 horizontal -> 0.2 vertical
4 horizontal -> ?

[upper karma]

0.8?

[dark sparrow]

bingo

so the y intercept of r3 is that much below -1

[upper karma]

so -1.8?

[dark sparrow]

💯

[upper karma]

so -9/5

this took me way more to understand than what I should need. why is geometry such a pain sometimes...

[dark sparrow]

i mean

[foggy oxide]

what you were trying to understand

[dark sparrow]

the more boring/algorithmic route would be to write down the line's equation in point-slope form and plug in x = 0

[foggy oxide]

analytic geometry is very different from usual geometry

[dark sparrow]

y = -1 + 0.2(x-4)

[foggy oxide]

and it's essential just like calculus some problems that'd be hard using plane geometry can be simplified with analytic

[upper karma]

yes

I know

I studied calculus too

[foggy oxide]

it's very similiar, isn't it ?

[upper karma]

yeah. minus the integrals

[foggy oxide]

xDD

[upper karma]

anyway were was I

now I am doing the same thing with r4

trying to understand it though

point-slope form... watched a video about it yesterday

but already forgot it

((I mean yesterweek))

[dark sparrow]

y = q + m(x-p) for a line passing through (p,q) with slope m

[foggy oxide]

the interception of points can be solved by a system equation of both lines

[upper karma]

oh

right

[dark sparrow]

or y - q = m(x-p) if you prefer

[upper karma]

hmm

oh my

[foggy oxide]

i like to write, delta y = m delta x

[dark sparrow]

i mean, that's the definition of slope

xP

[foggy oxide]

xD

[upper karma]

yeah

also of derivative

if lim->0

[foggy oxide]

just saying, do you know the condition for perpendicularism ?

it's very important

[upper karma]

eh

[foggy oxide]

you have to know it

[upper karma]

I have to study it again

[foggy oxide]

m_1 * m_2 = -1

[upper karma]

oh

the analytic version of it

[foggy oxide]

basically it's the inverse

yes

[dark sparrow]

the slopes are negative reciprocals

:3

[foggy oxide]

there is a intuitive way to understand this, because you work with the slope in relation to the x axis

[upper karma]

ok

[foggy oxide]

you will enjoy those problems eventually trust me

[upper karma]

I AM TRYING TO

if I didn't enjoy them I wouldn't as heck try to get into university

[foggy oxide]

xDDD

brb

[upper karma]

also thank you guys for the help and for making this server. ALL my friends have literally no idea of maths beyond arithmetics (and if they do they are too busy to help)

[dark sparrow]

lol yw

[upper karma]

well damn me

I just saw it that the Y intercept is the same as the slope

[dark sparrow]

coincidental

[upper karma]

yeah

0 + - 1/5

[dark sparrow]

although... lines have that iff they pass through (-1, 0)

[upper karma]

yep

seeing it

in analytic geometry, what comes after equation of a line?

[foggy oxide]

the equation of circle

after that probably parabola or eliptyc

[upper karma]

yes I know the conic sections

I meant if there is something else

basically speaking

besides distance, medium point , line, conic sections

[foggy oxide]

it can be connected with the notable points of a triangle (centroid, incenter etc...)

and the segments from trapezium too like euler median, harmonic median etc..

[upper karma]

lmao

[foggy oxide]

xDD

@upper karma

xD

a good thing about parallelism it's that you can perceive some triangle similiarities in problems

i always pay attention to informations like "AB is parallel to CD and perpendicular bla bla "

[upper karma]

oh

[foggy oxide]

what are you planning to do at uni ?

[upper karma]

Automation Engineering

I know. I will die a horrible dead

[foggy oxide]

xDDDDDD

that's a cool choice btw

[upper karma]

thank you

if there is no place there I would like Electrical engineering

((or as they call it Engineering of electric energy)

[foggy oxide]

i wanted to do Civil Engineering

idk if that's the correct word for it

i still consider it as option

[upper karma]

Civil?

[foggy oxide]

Ye

[upper karma]

yeah

said to be the easiest engineering

[foggy oxide]

probably

[upper karma]

((of course. 10% less hard. but engineering is already one of if not the HARDEST university course))

[foggy oxide]

in terms of entrance ?

[upper karma]

I mean examns

engineering and science is equally hard to enter

[foggy oxide]

idk, on my country they don't talk like it's thaaat hard

[upper karma]

what country you live in

[foggy oxide]

Brazil ^^

[upper karma]

aaaaaaaaaaaaaa

ou brazil

nao bon... buen... bien...

[foggy oxide]

What uhahuahu

[upper karma]

carajo ese portuguese

[foggy oxide]

i think bien is italian

[upper karma]

meh. I only speak spanish not portuguese

bien is in spanish

[foggy oxide]

xDDDDD

[upper karma]

and french

[foggy oxide]

got it

[upper karma]

I speak spanish, italian and english

and I am italian

[foggy oxide]

That's very nic33

[upper karma]

da

I spent half my life in latin america

[foggy oxide]

my ancestors are from italia

[upper karma]

in the now trashcan that is Venezuela

[foggy oxide]

xDDDDDDDD

[upper karma]

*Buon giorno

😛

[foggy oxide]

whaaaaaaaat?

i didn't said anything

you didn't read anything

[upper karma]

ok ok

[foggy oxide]

no one saw anything

[upper karma]

you are right

[foggy oxide]

NO ONE

[upper karma]

I am blind deaf and mute

[foggy oxide]

it was a illusion ...

rip, i remember my grandpa saying Boun giorno for me

[upper karma]

hehe

[foggy oxide]

xDD

[upper karma]

ok. after dinned I'll do more excercices condering the equation of a line

[foggy oxide]

that's a good idea bro xd

[upper karma]

namely, verify if two lines are parallel and or perpendicular. For every point, determinate the intersection of the lines and draw them. AND; for every point, find the equation and the graphic of the line with the indicated characteristics

[foggy oxide]

well

you will have to find their angle coeficcients, their equaiton too ..

[dark sparrow]

determine*
graph*

:p

[upper karma]

Sorry

[foggy oxide]

and solve their system in order to find the intersection

[upper karma]

Translating from Italian

[dark sparrow]

dw

[foggy oxide]

sqrt2

[upper karma]

Mhm

[dark sparrow]

ah

[foggy oxide]

i was in school, during math class and then the substitute teacher showed a problem and asked to solve it, it was about 1st degree function, (idk how to translate that from portuguese but it's basically this) i solved it with analytic geometry and then i missed a - sign and got the wrong answer (i did it in 30 seconds didn't checked ) then the teacher said that i wasn't supposed to use analytic geometry because "Variation taxe" is different from "Angle coefficient"

and that is the lvl of my teachers

(btw this wasn't exam just a usual math class)

xDDDDD

[upper karma]

Whaaaaaa

Yeah it seems about normal

[foggy oxide]

xDDD

[upper karma]

Analytic geometry ? 👀

yeah

Googled it didnt know it was named that

kek

mhm

the geometry on the cartesian plane

Yeah coordonate geometry

da

[upper karma]

hi everyone. I have a question about the Mercator projection.
I wrote a script to distort a map from equirectangular to Mercator using the formula given in the image below (from the Wikipedia page) and although it gets the distortion right, it scales the whole map vertically by a value that depends on the truncation or aspect ratio, and I can't find how this value is determined.

[upper karma]

hold on, I might be on to something.

[upper karma]

it's ok I found it.

i had to scale it by (1/(2 * pi)) * aspect ratio

[upper karma]

Question : how to determine the truncated angle of a Mercator projection from the aspect ratio of a map ?

[upper karma]

If nobody knows the answer here you might want to ask again at an another discord server (maybe try 'Advanced Mathematics'?)

@upper karma that's the link by the way: https://discordapp.com/invite/5u7JPG

[upper karma]

@Stefan#7792 thanks. I asked there and got some great help. My problem is now solved.

[upper karma]

@upper karma No problem. I would have loved to help you but unfortunately geometry beyond high school syllabus is not my field of expertise.

[dull egret]

i answered him over in advaced and that was like 10th grade math tbh. not difficult

[upper karma]

Oh hahaha shame on me! I'm don't have to use geometry anymore so my estimation was a bit off haha

[dull egret]

and apart from that, his question really came down to "how do i use these equations i found on wikipedia?" (sry chameleon :P)

[blazing pecan]

hey guys i have a small problem calculating the coordiantes of X Y based off an angle and distance, maybe someone can help me with my formula

X + distance * Cos(Math.PI / 180 * 90)

distance = 1 so logically i would assume that X = 1

however i get x = 6.123

[dull egret]

more context?

what are you going out from?

[foggy oxide]

i actually did not understand anything he typed

[blazing pecan]

oh my bad, let me try again

[foggy oxide]

math.PI ?

what is that

why 180 and 90

what is this "distance"

[dull egret]

he s prob copying stuff from a programming language

presumably python

[foggy oxide]

oh

[blazing pecan]

well i wrote the code 😄 im struggling with the algorithm

[dull egret]

what do you wanna compute?

be precise

[foggy oxide]

why 180 and 90

why distance is 1

and if you don't know x how do you get it's value since what you showed is a expression/function not a equation

are u trying to rotate/transform something ?

i'm sorry but those things are not clear for me

[blazing pecan]

compute the X and Y coordinate based of an existing starting point (X,Y), distance = speed (0.5 fixed) * seconds (2 seconds as an example)

[dull egret]

and an angle at which it is going right?

[blazing pecan]

direction = radians, but the user gives me normal angle so I do PI / 180 * angle

exactly

[dull egret]

alright

[X+speed*seconds*cos(direction*PI/180), Y+speed*seconds*sin(direction*PI/180)]

[blazing pecan]

so again if i go with my base example Start(0,0) move at 90 degrees should give me X = 1 (speed = 0.5 * 2 )

[dull egret]

second

let me fix that

there

[blazing pecan]

this is literally what i have now 😄

but it makes no sense

[dull egret]

why?

[blazing pecan]

my X comes out as 6.12303176911188

not as 1

[dull egret]

what is the value of dist and X

show that code snippet

[blazing pecan]

public void Move(int seconds, int direction)
{
var distance = seconds * _speed;
var potentialXResult = X + distance * Math.Cos(AngleToRadians(direction));
X = potentialXResult > Map.X ? Map.X : potentialXResult;

        var potentialYResult = Y + distance * Math.Sin(AngleToRadians(direction));
        Y = potentialYResult > Map.Y ? Map.Y : potentialYResult;
    }

    private double AngleToRadians(int angle)
    {
        return Math.PI / 180 * angle;
    }

_speed = 0.5 fixed

[foggy oxide]

i still don't understand what is the unity of "angle"

[blazing pecan]

so user gives me angle in degrees

my c# doesnt handle degrees so only Radians

[foggy oxide]

so it is radians or degrees ?

[blazing pecan]

radians

[foggy oxide]

ok, now this makes sense

[dull egret]

the only reason x could be that value is cause x or distance are wrong

or what is Map.X

if Map.X is smaller it is gonna be used

[blazing pecan]

these are just boundry conditions

i can remove it

im calculating the current X and Y position of the robot

and i mean in math terms this is what i got ...

distance = 2 * 0.5 = 1

then 0 + 1 * 6.1230317691118863

[foggy oxide]

i got what you're doing but i have no ideia why it is returning this number, i think the problem is in your code too

[blazing pecan]

let me write it out in 1 line

[foggy oxide]

maybe the problem is in the angle thing

because this number looks like it's irrational

i mean

cos(x) can return irrational values

[blazing pecan]

ok now i am super lost

[foggy oxide]

can u show more from this number?

[blazing pecan]

X + distance * Math.Cos(Math.PI / 180 * direction) gives me : 1.224606353822

[foggy oxide]

starting at 0 ?

[blazing pecan]

yh

[foggy oxide]

can u show more digits

from the numbers

try to start

at x = -1 and y=-1

[blazing pecan]

i literally hard coded the values

0 + 2 * 0.5 * Math.Cos(Math.PI / 180 * 90);

i still get 6.12303176911

[foggy oxide]

cos(77 degrees) =~ 0,224951

since the distance is 1

x has to be interpred as 1 to return this value

since cos(x) varies in the interval {-1,1}

(supposing that there is a problem in the angle thing )

maybe there is a problem in both x coordinate thing and in the angle @blazing pecan

that's what i suppose

[blazing pecan]

my Cos = 6.123031769111

for 90 degree angle

[foggy oxide]

whoaaaaaaaaaaaa

wait

WAITWAITWAIT

this is from wolfram

i think this is a deeper problem well

if i press enter it'll show 0

maybe ...

ik

IK

[dull egret]

^-17!!!!!!!!

[foggy oxide]

it's

multyplying

180 per 90

and not

pi

.-.

you're actually multiplying 180 * 90

[dull egret]

ehh no?

[foggy oxide]

.-.

ehh yes

[dull egret]

that cant be, it has to go in order

[blazing pecan]

yh can confirm jewe

i put brackets in now and no difference

[foggy oxide]

so, idk

.-.

[dull egret]

it s just a floating point error. look at the magnitude

[foggy oxide]

is wolfram coded in python too ?

🤔

[dull egret]

no. this is c#

he said that

[foggy oxide]

well ...

it's really a deeper problem

u should go ask some programmer about this

[dull egret]

whatever. it is a floating point error. he s prob missreqding the number

[blazing pecan]

im copying it from the debugger 😄

[foggy oxide]

well it is grotesque because wolfram has the error too

[dull egret]

no wolfram is a floating point error. it is miniscule. you are missreqding the number

[foggy oxide]

nvm

[blazing pecan]

so we're all out of ideas? 😦

[foggy oxide]

well

it's not a geometry problem

but the number wolfram returned is very close to yours (the inittial non zero digits)

[blazing pecan]

soo my radians = 1.5708

and the cos of this should be: -0.00000367

correct?

[foggy oxide]

idk lol, i have to check

type in terms of pi

[blazing pecan]

PI / 180 * 90

[foggy oxide]

well, your radians are not exactly this

should be very close ...

but it's actually 0

cos(pi/2) = 0

maybe

isn't the problem with the pi ?

try to type it approximately

idk

@blazing pecan

did u tried ?

[blazing pecan]

yh not a problem with pie

how ever i am seeing some strange results. ..

[foggy oxide]

try to aproximate the values of cossine

nvm, that wont work i think

whatever go to a specialized chat since this is not a problem with geometry ..

[blazing pecan]

yeah i thought i could get my solutions here

but thanks ! 😃

[foggy oxide]

no, your problem is definitely with some code man

but if u get the answer tell me because i'm curious now

[blazing pecan]

thanks for ur help 😛 i will keep searching for the answer

if i find it i'll let you guys know

[foggy oxide]

ye

[golden geyser]

How would I set up the equation for this question?

[foggy oxide]

are you here

well

3x + 2 = 5x - 78

[golden geyser]

thank you

[foggy oxide]

np

[minor quest]

DE is a fixed hypothenuse?

[foggy oxide]

i won't answer that

but i'll ask you something

what is the measure of segment QD

@minor quest

[minor quest]

21

[foggy oxide]

and QE ?

@minor quest

[minor quest]

uh

[foggy oxide]

remember you're working with a circle

[minor quest]

u can't use pythag here right?

[foggy oxide]

well, nope, you don't know sufficient informations yet to deduce that

but

what is the lenght of QE

?

remember you're working with a circle

[minor quest]

50?

[foggy oxide]

Nope

Look at the image again

QD = 21 right ?

[minor quest]

yes

[foggy oxide]

what fact made you think that QD is 21 ?

you're right

but tell me that

[minor quest]

its a radius

[foggy oxide]

oh

now, try to identify QE

[minor quest]

mentioned

[foggy oxide]

what do you mean by mentioned ?

[minor quest]

wait what do you mean by identify QE?

[foggy oxide]

the lenght of QE

[minor quest]

yea they said r=21 and r is the length of QE?

[foggy oxide]

yes

Now you know what about the triangle QED

?

[minor quest]

theyre isosceles

[foggy oxide]

exactly

[minor quest]

u divide by 2

and do stuff?

[foggy oxide]

not really ...

don't tell him

the formula yet

[minor quest]

ok

[foggy oxide]

i'll help him deduce

don't try to find it on internet

if you don't remember or don't know

[minor quest]

wat

[foggy oxide]

@minor quest

[minor quest]

me?

[foggy oxide]

Now

How many degrees correspond to the internal angles of a circle ?

[minor quest]

@upper karma yea

[foggy oxide]

i mean

[minor quest]

oh wait

well its area of sectore minus the triangle equal area of segment @upper karma

[foggy oxide]

i want to teach him how to find two areas

yes

but now

the question gives a tip

you can use trigonometry

in wich way that would help you

?

[minor quest]

find cosine of theta?

i meant sine

[foggy oxide]

remember what you pointed already, QED is isosceles and you know one angle

well, you have one angle and two sides and you know the triangle is isosceles

[minor quest]

hang on

[foggy oxide]

right

[minor quest]

i was thinking that i'll divide it into two right triangle and then use sin to find the opposite of the right triangle or one half of the base

[foggy oxide]

that's your triangle

[minor quest]

yes

[foggy oxide]

well, you can actually find the other two angles, but don't grip into that

[minor quest]

other two of isosceles?

[foggy oxide]

there are alot of ways to find the are of it

[minor quest]

theyre 65 each?

[foggy oxide]

what is the easiest way for you ?

considering the simmetry, yes but don't grip into that

[minor quest]

hmmm

[foggy oxide]

do you think dividing into 2 right triangles is easier ?

[minor quest]

well thats what came up to my mind first so yes

[foggy oxide]

so

what's x/2 (or x) ?

[minor quest]

about 8.87

[foggy oxide]

i won't check

so make sure you did it right

how do you calculate the area from one of the right triangles ?

(remember you know those right triangles are equal they have same height same hypotenuse and leg )

[minor quest]

u first find the h and then do bh/2

[foggy oxide]

yes

you already know x ( or x/2 idk )

what can you use

to find h ?

[minor quest]

pythag

[foggy oxide]

exactly

go on

[minor quest]

and then find the area of one right triangle then times 2 to get the whole thing

[foggy oxide]

yes !

[minor quest]

thanks for your patience

[foggy oxide]

np, i'm glad that i can help you

it doesnt bother me

[minor quest]

still hard for me to grasp any triangle aside from those that follow b*h right of the bat so it took a while

[foggy oxide]

i like this kind of "Questioning" conversation

well, the important you're understanding it !

[minor quest]

yes

what were you trying to imply with the "you're working with a circle tip"?

[foggy oxide]

with the fact that QD is equal to QE

you did got it already so...

[minor quest]

oh.

[foggy oxide]

well, after finding the area of QED

what do you need to do ?

[minor quest]

find the sector

[foggy oxide]

yes

how do you find the area of a sector ?

what do you have that can be used to find the area of the sector ?

[minor quest]

given deg/360 * pi r^2

[foggy oxide]

pie ?

[minor quest]

yea?

[foggy oxide]

i don't know if you're saying pi or pie

[minor quest]

i meant pi

lol

[foggy oxide]

right, well

more important, do you understand this formula ?

[minor quest]

pi r^2 is the area of circle but 50 deg is specific so u take area of circle times the specific deg then you get the desired sector

[foggy oxide]

yes, i like to think like that for the sector formula 360/(given angle) = (total area)/(piece, what you want)

now that you know how to calculate the sector

and you already have QED area

you just have to do what you said earlier

[minor quest]

how do you cross out a text?

[foggy oxide]

~~ text ~~

[minor quest]

~~ cool ~~

[foggy oxide]

~~ <= this

text

[minor quest]

hmm copy and paste didn't take it in

[foggy oxide]

@upper karma do you need something man ?

[minor quest]

yea i saw him lurking the whole time lol

[foggy oxide]

xD

@minor quest Think about a different solution for finding the x (base of the isosceles)

it's a good thing

thinking about different solutions for the same problem

this helps you seeing more solutions in the new ones

[minor quest]

there is a different way?

[foggy oxide]

yes

using trigonometry too

[upper karma]

@foggy oxide naah. I was just about to reply to an old message but eventually thought it wasn' timportant. Thanks for asking though

[foggy oxide]

@upper karma no problem man, if you feel like i can help you some way just ask it here xD

[minor quest]

im still not sure if DE is the only possible hypothenuse

[foggy oxide]

suppose that the triangle wasn't isosceles, how would you find it ? @minor quest

[upper karma]

@foggy oxide thanks 😋

[minor quest]

well it'll be easier if its a triangle with a distinct hypothenuse

[foggy oxide]

you have a angle and two sides

well hypotenuse is meant to be the bigger side for a right triangle

if it was scalene (all sides are different and all angles too ) it wouldn't have a hypotenuse

[minor quest]

oh

[foggy oxide]

legs and hypotenuse are expressions for right triangles

that's a very important fact to know

and now you know it 😉

[minor quest]

then how else would u solve for this

[foggy oxide]

you can use cossine law

did u learned about it already ?

[minor quest]

no not yet

[foggy oxide]

well it's very useful

but, forget about it, you will learn it in the correct time

brb

i'll take a shower if you need a help just ask here

[minor quest]

ok

[minor quest]

@foggy oxide

[foggy oxide]

there is a known formula for the area of a parallelogram when u have height and base (if im not mistaken)

[minor quest]

its perimeter @foggy oxide

right?

[foggy oxide]

oh

sorry, didn't read it propely

if there was a angle xDD

i think you can do it

with the diagonals

true

but well

you gave him the done answer

[minor quest]

ima go shower be back in like 10-20 mins

[foggy oxide]

@minor quest

he meant that

[minor quest]

wat?

2(a+b) for the parallelogram?

[foggy oxide]

well i cannot drawn a proper pallalogram

[minor quest]

for the parallelogram or the rectangle?

@upper karma

[foggy oxide]

i still drawing a pallalelogram xDDDDDD

not only the perimeter but the area too, well

why did u deleted it

it was a good pic

why ?

xD

xDD

rip

fuckn gif

[minor quest]

Bruh

[foggy oxide]

wtf

[minor quest]

U got this

[foggy oxide]

well, it still doesn't work for me

even if i click

lol

xD

i think there is not enough information to answer the question u sent @minor quest

[minor quest]

Well what is the best answer out of that question then?

[foggy oxide]

so

no one

also

(the last option )

you can deduce that

^^

well if you had only 1 angle you could calculate everything about this parallelogram

well, i'm going to sleep

cya

[surreal bolt]

um isn't the answer none of the above?

[upper karma]

why am I so stupid

can anyone help me find the equation of this line

https://cdn.kastatic.org/ka-perseus-graphie/a6f83535387b3fd7b8c60d7eef60d76cfc221e32.png

just. tell me if y=-1/11x - 96/11 is correct. I don't care anymore

[dark sparrow]

it's not that

the line's slope is -1/3

[upper karma]

aaaaaaaaaaaaaaaaa

...

[dark sparrow]

y = -(x+16)/3

[upper karma]

this geometry can go to hell

yeah. on point C I mistook x for y

[foggy oxide]

a one a two - Hoje às 01:06
um isn't the answer none of the above?

@a one a two#9669

not really

because we cannot prove that the answer is not no one of the above

if i use the simmetry i could find the perimeter actually

but it's not said that the img is drawn to scale

[surreal bolt]

oh is someone quoting me?

[upper karma]

Yeah he was

[surreal bolt]

is it good news or bad?

[upper karma]

Good news

by that I mean a mathematical nightmare

[surreal bolt]

oh i hope so

[upper karma]

So uhm

https://cdn.kastatic.org/ka-perseus-graphie/a6f83535387b3fd7b8c60d7eef60d76cfc221e32.png

The problem consisted in finding B knowing that BA is 1 fifth of AC

But how do you find a point knowing how distant another point is to it?

(Or is the linear equation a better way of doing it)

[dark sparrow]

is B on AC?

[vapid kettle]

@surreal bolt

[upper karma]

@dark sparrow Sorry. Yes it is

[dark sparrow]

okay, so

to go from A to C, you'll need to move 15 units horizontally

to go from A to B, you'll need to move 1/5 as far

1. how far horizontally do you need to move to go from A to B?
2. what is the x coordinate of B?
3. thus, what is B?

[upper karma]

Hmm

[foggy oxide]

what is the symbol we use when two triangles are similiar, i forgot it

[upper karma]

You need to move 3?

(15/5)

[foggy oxide]

~

is that ?

it is the ~

right

alright

[dark sparrow]

@upper karma yes

[vapid kettle]

did somebody say ~?

[upper karma]

Ok

[dark sparrow]

what point on the line is 3 away horizontally from A?

[vapid kettle]

~

[dark sparrow]

~

[upper karma]

5 -7?

[dark sparrow]

💯

[upper karma]

Hon hon hon

Thank you

Saw. Did you need the distance formula?

Or just glance at it

[dark sparrow]

the distance formula is not needed

[foggy oxide]

~~ offtopic : what is the joke about ~ ?~~

[vapid kettle]

~

[foggy oxide]

😭

LOL

😄

LOOOOL

the ~~ reduces the size when used on a emoji

i didn't noticed that b4

maybe because it's not big when it's accompaigned of some letter/word

[upper karma]

~ 😄 ~

Hm

[foggy oxide]

can you guys tell me what is the fun behind the ~ joke ?

i'm really curious

[upper karma]

Because online

Is a way of expressing sensuality

[foggy oxide]

.-.

[upper karma]

Or cuteness

[foggy oxide]

.--.

[upper karma]

Like "Hey~"

Sorry I mean

"Hey baby~"

That's moar like it

[vapid kettle]

I can confirm that isn't what it means ~

[upper karma]

It is in roleplay

BRING IT ON

[vapid kettle]

if curious you can ask @final prairie

[upper karma]

damit

Now I want to know

What you invented

[final prairie]

Y u do dis to me

:(

[upper karma]

Tell it

[final prairie]

I wish i knew

He is just teasing me

:p

[upper karma]

Darn

X3

[foggy oxide]

lol there was a question about finding the radius from the incircle of a isosceles

we are in #geometry-and-trigonometry let's talk about geometry

lol

[upper karma]

Ok

[foggy oxide]

ok

how would you find the radius of the incircle ?

[upper karma]

Hmm

The radius of the circle minus the radius of the outside circle

[foggy oxide]

i made a scheme for it

then found a veryyy

easier

sol for it

[dark sparrow]

lol there was a question about finding the radius from the incircle of a isosceles
given what

[foggy oxide]

height and base

it wasn't really about a triangle and circle

[dark sparrow]

area divided by half-perimeter

or bh/p

[foggy oxide]

lol, mine was a fucking mess

well, this one is the "easier" i found

[dark sparrow]

=tex r = \frac{bh}{b + 4\sqrt{4h^2 + b^2}}

[foggy oxide]

mine was like that

:

QBR ~ QDA ~ BAP <=> QM \\ BC

LOL a fucking mess

(i considered BQ = BP and DO = QW )

i don't remember the theorem that is about BQ=BP (it's about tangent i only know that)

[minor quest]

@foggy oxide u here?

[foggy oxide]

yes

[minor quest]

I use Pythagorean to find GI right? @foggy oxide

[foggy oxide]

i'm back

sorry

for making you wait

i was talking with a friend

well

are you here ?

[fringe basin]

just simple pythagoras right?

the line GI is totally unnecessary

[minor quest]

well u can too....

[foggy oxide]

well, i was thinking it wanted to find the GI

lol

didn't read it propely

well, yes man, i mean it's simply pyt

[minor quest]

ok..

[foggy oxide]

by the way

do you want to know how to calculate GI

?

the GI segment is the relative height to hypotenuse

[minor quest]

what?

[foggy oxide]

GI is a height from the triangle

[minor quest]

i was just trying to do proportion to find Gi

[foggy oxide]

do you want to know how to calculate it ?

it's useful

Well, i mean, you don't need to calculate GI

[minor quest]

sure

[foggy oxide]

it's just a question of sometime you need it idk

[minor quest]

well can you pm how?

[foggy oxide]

here

i'll show here

i'm drawing it

are you familiarized with the notation BÂC ?

For example this notation means the angle at the A corner starting from B and ending at C

so, the red angle

You know those facts that i pointed out because if triangle BDC is right triangle and angle BCD is in the triangle the the angle DBC is the red one

[fringe basin]

is the reason that they are similar triangles because they share a side?

[foggy oxide]

not only that

the 3 internal angles are the same

[fringe basin]

could you explain how please?

[foggy oxide]

here : You know those facts that i pointed out because if triangle BDC is right triangle and angle BCD is in the triangle the the angle DBC is the red one

[fringe basin]

like that kinda intuitively makes sense

[foggy oxide]

a height is a perpendicular segment

that means it makes 90 degree in relation to AC (in that case it's AC the hypotenuse that's why it's called the height relative to the hypotenuse)

imagine the red angle were 30 deg

[fringe basin]

oh yeah if you look at B DBC and ABD have to add to 90

[foggy oxide]

right?

[fringe basin]

and that's what the two angles which aren't 90 in a right triangle have to add to

[foggy oxide]

if you have 30 deg and 90 deg on triangle BDA

yes

they are complementar

i think it's complementar the correct term for that

so you can use triangle sim to solve it

[fringe basin]

complementary?

[foggy oxide]

yes, i think

it's correct "complementary"

@minor quest with those facts, the 3 angles from those triangles being the same you can use the relations of proportion

[minor quest]

well that is kind of what i was taught

[foggy oxide]

:D

[short sky]

am I doing this right

[dark sparrow]

blurry

[dark sparrow]

yup, looks alright

[short sky]

was wondering how you deal with theta > pi though

I mean pi/2

[dark sparrow]

i usually prove the law with coordinates

place C at the origin, B at (a, 0), A at (b cos(x), b sin(x)) and just compute AB^2 (a.k.a. c^2) directly, and out pops the formula

c^2 = (a - b cos(x))^2 + (b sin(x))^2
= a^2 - 2ab cos(x) + b^2 cos^2(x) + b^2 sin^2(x)

= a^2 - 2ab cos(x) + b^2

💯

@short sky

[short sky]

ohh

I guess it's equivalent

a = a_1 + a_2 isn't really needed if you have the angle

[vestal cape]

I know how to do Q3 a-c, e-j but I got stuck at d. and k

[thorn talon]

Its the same x coordinate

So it's a vertical line

There's no gradient to work with, so special case

It just has to take the form x = number

[vestal cape]

Thank you!

[upper karma]

Move this to #chill but not here please

[minor quest]

@foggy oxide senpai!

[vocal plinth]

Guys, Can someone help me because I am retaking the Geometry Regents tomorrow?

[foggy oxide]

@minor quest Hello

[minor quest]

Nice hello edit lol

[foggy oxide]

@vocal plinth What do you need, i can help u tomorrow, i'm going to sleep in some minutes

[minor quest]

Oh...

Nvm then

[foggy oxide]

no i can help you very quickly

i'll just ask some things

just as advice

it's not necessary but u can draw it

it'd be better to understand

[minor quest]

i did it

nvm

[foggy oxide]

i already memorized the problem

[minor quest]

😉

[foggy oxide]

you solved it ?

[minor quest]

proud of me senpai

[foggy oxide]

xDD, good boi

[minor quest]

lol gn

[foggy oxide]

cya

[gray umbra]

Can someone correct my work? It's simple geo but I really don't want to screw this up, thanks in advance

[foggy oxide]

i think everything is correct

[bleak lintel]

Talking about vectors and the distance from a point p (x1,y1) to a vector A

How do you do the top part?

as there's only this formula for me, no examples of how to use it

is it just the square root of u.x1 + v*y1 + w?

drawing that goes with it

or is it this

[dark sparrow]

you know what |x| means, right?

[bleak lintel]

the lenght of x?

or absolute value

my god im stupid

you just add those shits up, get the absolute value and then divide them with that square root?

[dark sparrow]

yes

[bleak lintel]

ffs xD

so much confusion for no reason

thanks x3

[upper karma]

hehe, i do the same thing a lot

[bleak lintel]

NOW I see why you take the absolute value

my god

that took ages

because a length cant be negative

xD

[foggy oxide]

@bleak lintel

do you know how to deduce this formula ?

it's not reaaaaaally hard, and it's very important to understand

just involves some concepts that can trick you but aren't that complicated

[dark sparrow]

something something linear algebra

[foggy oxide]

oh xD

[bleak lintel]

Well, its just the derpy me

nothing to special

the full conversion is in my pdf so thats alright

[foggy oxide]

:D that's good

[foggy oxide]

?

How is this related to geometry, btw, how is this related to this server at all, (except from the fact that both servers are from discord)

[dark sparrow]

deleted

[fringe basin]

I take back what I said about the unit circle btw

but I think how it looks in american middle and high school textbooks is horrible

with all the angles around the outside and stuff

[upper karma]

What 😄

[dark sparrow]

with the multiples of π/6 and π/4 labeled?

[bleak lintel]

how do you convert ax²+bx+c again to a* (bx+c)² +D

not sure, kinda forgot how

[dull egret]

complete the square

also, you dont need the b infront of the x

[bleak lintel]

it's just there for my OCD reasons

[dull egret]

alright then..

[dark sparrow]

then there's no unique conversion

[bleak lintel]

I'll use the -b/2a

version

way easier

[fringe basin]

also yeah so the way inverse function works

is that a point (1,3) on f(x) will be (3,1) on f^-1(x)?

[dark sparrow]

yeah

[fringe basin]

okay

image=range?

[dark sparrow]

range, yeah

as in, image of the whole domain

[fringe basin]

more technical language yeah

okay so if I'm rubbish at visualising them now

[dark sparrow]

@upper karma codomain R

output space

:3

[upper karma]

whats your definition of domain / codomain ?

anyone

I think i actually got it

i guess i call it 'arrival set' usually when talking

codomain that is

[dark sparrow]

🛬

f: 🛫 -> 🛬

[upper karma]

Usually 'ensemble d'arrivée' which is literraly the definition for codomain = codomaine

Definition set usually

Image ^^

[fringe basin]

okay yeah I think I've worked it out

using facts about where stationary point is, domain and range of original function

and also knowing they have to cross on line y=x

if they ever equal each other

[dark sparrow]

https://www.desmos.com/calculator/muqpfwmigo

made a thing

if you let k bounce back and forth between 0 and 1 you get an animation of the graph flipping back and forth across the y=x line

[fringe basin]

so wait how do you make it do inverse?

[dark sparrow]

as in, desmos?

just have it graph x = f(y)

the stuff with k is just a clever lil bit of linear algebra to make it animated

[upper karma]

any travelers of the nth dimension here?

how does one map higher dimensional coordinates to lower dimensional ones?

projections

is one way

or do you mean bijective maps?

don't know what i mean!

well i kinda do

i have to leave for a bit, but i might be back

but if you describe what you're interested in and/or why, that might help!

the hypothetical is suppose we live in some higher dimensional space

but you need to specify a 3d coordinate within that space

to kind of shortcut 3d space

wormholes etc

as in suppose one had acquired a wormhole generator/stargate/device for interdimensional travel

but they're a cubelander

so they need to feed this device a string of coordinates in n dimensions

to specify a unique 3d point

how does that.... work?

and in n space not 3 or 4 or whatever

or even drop the special case of 3 dimensions, that's just what i'm most familiar with. but the question is more general, so i can work on one case and then induce

or even what particular topic would this be. i'll definitely look into projections and bijective maps

thanks in advance!

or in a mandelbrot set, how does one specify one particular "location" in this recursive infinite thing?

[upper karma]

i'm here

so the idea is projections

which are idempotent maps

so f(x) = f(f(x))

where f maps from n dimensional space

to m dimensional space

where m < n

and it's a linear map

so f(x + y ) = f(x) + f(y)

and f(a*x) = a * f(x)

when a is a scalar

and x and y are n-dimensional vectors

woah ok

a little too abstract for me

could we work like a small concrete example?

so for lets say we live in a 6 dimensional world, i assume 6 dimensions would mean each point in that space needs 6 unique values, yea?

• for background, i just finished up my second semester of calculus and am about to take vector calc and differental equations. but i've been fascinated by math in a imaginative way, mathematical fiction and imagination more than technical rigor and abstraction

so here n=6 and m=3

is a point in a space a scalar or a vector?

i asssume scalar since there's no directional aspect?

[quasi lion]

a scalar is a one-dimension vector

[upper karma]

i was only taught vectors in physcis and so associate them with directions and forces, is that not the case in pure math?

[quasi lion]

lol no

[upper karma]

enlighten me oh wise ones

[quasi lion]

What do you want to know?

[upper karma]

vector

what makes a vector different than a scalar in pure math

or

well as you say

[quasi lion]

a scalar is a one-dimension vector

[upper karma]

a scalar is a 1d vector

then define vector

[quasi lion]

It's just multiple numbers that you put in a single mathematical object

[upper karma]

wiki says "an element in a real coordinate space"

[quasi lion]

You have 2 bananas and 6 oranges, you have (2,6) in the plane (bananas)x(oranges)

(with the origin being 0 banana and 0 orange)

[upper karma]

is it assumed that those dimensions are orthogonal?

or can they not be

[quasi lion]

yes they need to be orthogonal

because else the vector would be equal to another vector and the maths wouldn't fit

[upper karma]

ah ok so thats why we have the unti vectors

to orient the whole thing

unit*

[quasi lion]

It's slightly more complicated but I get what you mean

[upper karma]

i really do want to understand this stuff

[quasi lion]

the space is generated by the span of the vectors that directs the space

[upper karma]

i have my primitive ideas of space

i live in euclids world

but i know, at least i believe that

at some point things become

less grounded

and so need more clear definitions

[quasi lion]

How old are you?

[upper karma]

23

well

by who's count?

😉

[quasi lion]

What do you want to know?

[upper karma]

zeno dialogue [at the fountain]

z; Notice the fountain…
b; What of it?
z; Particularly the transformations of each glimmering droplet
through the air. Can you identify a single one for any extended time?
b; I’m afraid not.
z; Notice too, how it seems the fountain
never resides wholly in one instant, but is rather perpetually in flux.
b; This is indeed true… why is it so?
z; It is said it would not be fitting the majesty
of the Creator to create but once,
rather to create, and then perpetually recreate!

[quasi lion]

I mean "vectors" is too general.

[upper karma]

i want to understand "space"

[quasi lion]

I can't answer that

[upper karma]

higher dimensional geometry

[quasi lion]

Oh

[upper karma]

what are the rules once i get passed 3 dimensions

[quasi lion]

Higher dimensional geometry isn't really fun, it's just adding an extra dimension

Hold on

[upper karma]

how do i deal with things in places i cant imagine

[quasi lion]

https://www.youtube.com/watch?v=4URVJ3D8e8k

This video might help you

Especially if you are physicist

[upper karma]

thanks

ahhhh

yesssss

cool

[quasi lion]

Honestly, I don't think I can define "space"

[upper karma]

the void or something

i've literally spent hours contemplating infinitesimals