geometry-and-trigonometry

[twin prawn]

Let A be the area of the triangle with sides a, b and c

A = a*h_a/2 as we all know, same thing for b and c

So 1/h_a = a/(2A)

But since a < b + c (triangle inequality) we have that

1/h_a < (b + c)/(2A) = 1/h_b + 1/h_c

@thin wasp

[thin wasp]

Yep, thanks

[twin prawn]

🍮

[thin wasp]

The median to a 10cm side of a triangle has length 9cm and is perpendicular to a second median of the triangle. Find the exact value of centimeters of the length of the third median.

We were specifically told NOT to use the law of cosines

I found the sidelengths of the triangle

So what I suppose I could do, is draw a perpendicular down from the centroid to the 10cm base

then calculate the area of the triangle using heron's formula, which would be a pain to do cause I've got multiple surds

and then show that 10*(that perpendicular)/2 = (Area of the triangle/3), and thus calculate the perpendicular length

and then use pythag/similarity

[thin wasp]

Nvm got it ^

[upper karma]

could i have something to prove

[fallen ivy]

of what nature?

[neon fossil]

prove that a dog is topologically a sphere

[fallow edge]

Do spheres have hair?

[zenith ember]

But a dog isn't topologically a sphere, I don't think. It's a torus,

[plain wigeon]

Wow, deep conversation

[neon fossil]

here is the solution if you were curious -_- https://i.warosu.org/data/sci/img/0089/00/1494491021652.png

[forest dove]

Then there's the successor

[fallow edge]

Ahh, math memes

[pliant rivet]

i’m fuckin dying

[tender adder]

oof

[strange crow]

@neon fossil what the hell did that dog do to deserve the change

lol

[coral kiln]

anyone able to help?

<@&286206848099549185>

[upper karma]

[twin prawn]

I’ve got this

In the triangle ABC denote the angles at the respective vertices by a, b and c

PC is tangent to the circle and touches one of the vertices of ABC so by the alternate segment theorem (you might want to look this up) the angle BCP is the same as the angle a

So the angle PCQ has size a + c/2 because CQ bisects c

What’s more, we know that a + b + c = 180 degrees

But also b + BCQ + PQC = 180 degrees, but BCQ = c/2, so we must have that PQC = PCQ = a + c/2

This shows that PQC is isosceles so PC = PQ

@coral kiln

[coral kiln]

ty @twin prawn

[twin prawn]

🍮

[coral kiln]

@twin prawn are u free?

[twin prawn]

What have you tried

[coral kiln]

uh

ye

so the three rules to solve PBCX are opposite angels are sup, exterior angle = opp interior and <XPY = <YBC/<PXY=BCY

can't seem to get any tho

@twin prawn

[upper karma]

@coral kiln you need to prove it by proving <CXP = <CPB

As a chord subtends equal angle anywhere on the circumference...here BC being taken as the chord

And then it is pretty easy to prove

Hint:use property of rhombus's diagonals

[main totem]

There is triangle ABC with all acute angles(Point B is up,A is on the left and C is on the eight ). The length of the height from point B is 7, The length of the height from point C is 9. Then we draw median from A to side BC then marke point M. Then we draw point symetric to point M relative to line AC then name it P.then we draw point symetric to point M relative to line AB then name it Q. Fine the perimeter of quarilater APMQ.
Please help <@&286206848099549185>

[hollow edge]

@main totem sorry for the late reply; I just saw this. Assuming AM is 8, as it is in the diagram (but not your explanation), the answer is 7 + 9 + 8*2.

[main totem]

I got the 8*2 part

But not he 7 and 9 parts yet

[hollow edge]

similarity

@main totem

[main totem]

Wait wait wait

How did we you 7?

[hollow edge]

wasn't that given?

which lengths were given?

[main totem]

But 7 is height of triangle

[hollow edge]

oh 7 as part of the answer?

[main totem]

Yes

[hollow edge]

ok this is confusing with unclear points

[main totem]

Can u please xplain i am dumb

[hollow edge]

ok

so

BDC and MEC are similar triangles

right?

because 90 deg and shared C

[main totem]

Yes

[hollow edge]

what's the ratio of similarity?

[main totem]

1/2

[hollow edge]

that means ME is...

BD is 7

[main totem]

3.5

[hollow edge]

yup

but ME and EP are the same since it's a reflection

[main totem]

@hollow edge thank u very much ur a legend

[hollow edge]

np

glad to be of help

you can do the same with 8

[manic grail]

HELP

I AM DYING FROM GEOMETRY

https://gyazo.com/edf6bc66ae0eb4eee5143b05a78f93d0

[fallen ivy]

lol

[manic grail]

HOW DO I DO THAT THING

[fallen ivy]

angle BDC is 44

[pliant rivet]

and the other is 22 iirc

lemme double check

[fallen ivy]

yeah it will be

[pliant rivet]

and inscribed angles dont change their measure no matter where you put the middle vertex bit

as long as its still on the circle ofc

or rather "in a circle, two inscribed angles with the same intercepted arc are congruent"

[manic grail]

my brain is frozen

i am taking summer geomrtry class

i dont understand anything outside of algebra 1

[past mantle]

Angle bac is half of that arc

[fallen ivy]

We already gave that answer

[past mantle]

...

Welp

I didn't notice

[fallen ivy]

It's okay we still love you <3

[last otter]

https://i.imgur.com/zHpW5Xu.png

hi. Is this shit even true?

I have counterexample. Let B be (0,3) and let E be disjoint union of (0,2), (0,2), (1,3),(1,3),(1,3)

and let p map number t to number t in (0,3)

then p^-1(1.5) has 5 elements but p^-1(0.5) has only 2 elements

this is so annoying

[mint sandal]

pretty sure that's not a covering map

@last otter

[last otter]

how can u be so sure? points in B are evenly covered by intervals (0,2) and (1,3)

oh shit they're not

[mint sandal]

what is the preimage of (1.5, 2.5)

it's disjoint collection of open sets, but not all of them map homeomorphically to (1.5, 2.5)

[last otter]

ty

I can't do sh*t without hints

[mint sandal]

want a hint?

[last otter]

no 😄 I meant this one you gave me

I meant to say that generally I cant prove shit without someone giving me hint

[mint sandal]

alright

[steady widget]

geometry is killing me also, what do i need to do to get this?

@manic grail also same, trying to get into precalc because I know alg 2 and alg 1 so if i learn geometry I can do that

[mint sandal]

@steady widget

angle bisector theorem

[steady widget]

ok ill look into that, thanks

@mint sandal so x=6 and y=4?

[mint sandal]

yah

[steady widget]

nice, thanks for the help

[opal blaze]

Can someone confirm that when they talk about the circles being externally tangent. they do not refer to the angle APB being 90 degrees. (I don't know all the terminology for English mathematics) <@&286206848099549185>

[opal blaze]

cause if it is 90 degrees that means that the solution to this (2+1)^2 = (2+r)^2 + (1+r)^2 should give you the radius...

but I get the incorrect answer...

[hearty barn]

Noe it doesnt mean that

External tangent means that there exists a particular tangent

When when drawn

Meets A at one point and B in the other

Internal tangent means that there exists a particular tangent line

Which when drawn

Meets the two (semi)circles at the SAME point

@opal blaze

[opal blaze]

ok

[torpid perch]

Hello. What is the usual order of describing a triangle. Should the points be given clockwise or counterclockwise. Would be great to get some references as well. Thanks

[zenith ember]

I don't think there is a strong convention

[opal blaze]

@torpid perch You should think of the unit circle. so counterclockwise.

[torpid perch]

ok thanks

[dreamy wraith]

usually counterclockwise is the positive direction in mathematics

[hollow edge]

because x and y are in a counterclockwise direction in that order

[eager pendant]

problem:

I tried extending PQ and LM to meet, and connecting tangency points in hopes of finding something with similar triangles - but I couldn't go from there(edited)
there's a coordinate bash solution here:
but I'm looking for a geometric solution
if someone has a hint, please ping me.

[eager pendant]

<@&286206848099549185>

[hollow edge]

I think this is how you do it

@eager pendant pythagorean with 10 and 26/5

don't forget x2 since it's half of LM

[eager pendant]

yep, got it

thanks!

[hollow edge]

np

go ahead

I don't mind

[ebon gyro]

What is the notation used to define a gradient between two points?
is it ∇f ∇(arrow)f or just ∇?,
like say if I had a line on a grid, and the endpoints marked A and B,
straight line,
scalar function.

[main totem]

Guys i have a little hard problem that i cant solve can anyone The smartest person in this server help me out?
So i need to prove thar triangle BAD = triangle AED
<@&286206848099549185>

BC or AD is not diameter btw

[ebon gyro]

warning not to scale

[mild cargo]

This looks like it can be done pretty sufficiently with congruency theorems and tangent-secant theorems

[main totem]

I ma de a mistake lol

It says prove that they are similart..

Im so dumb i cant eve read right

[mild cargo]

Could still try tangent-secant theorems

I don't have too much time right now, but I will help out later if I can

[main totem]

I dont nu dat

Ok

[mild cargo]

Just try to find relationships between angles or sides.

Triangles are similar if all corresponding sides are equal, a pair of corresponding angles are equal or a corresponding angle and the two sides that share its vertex are equal (however in the last case they would also be congruent, so it doesn't look like it will work here)

[ebon gyro]

What is the notation used to define a gradient between two points?
is it ∇f  ∇(arrow)f or just ∇? or neither???,
like say if I had a line on a grid, and the endpoints marked A and B,
straight line,
scalar function.

<@&286206848099549185>

[mild cargo]

=tex \frac{\Delta x}{\Delta y}

[mild cargo]

The gradient is just the tangent of the angle between the x-axis and the graph of your function for linear functions

[ebon gyro]

It's just that I came across this =
Definition

The gradient of the function f(x,y) = −(cos2x + cos2y)2 depicted as a projected vector field on the bottom plane.
The gradient (or gradient vector field) of a scalar function f(x1, x2, x3,... xn) is denoted ∇f or ∇→f where ∇ (the nabla symbol) denotes the vector differential operator, del. The notation grad f is also commonly used for the gradient. The gradient of f is defined as the unique vector field whose dot product with any unit vector v at each point x is the directional derivative of f along v. That is,

some people use 'm' to denote gradient

=tex m =\frac{\Delta x}{\Delta y}

[ebon gyro]

but is it

=tex \nabla =\frac{\Delta x}{\Delta y}

[mild cargo]

I mean

[ebon gyro]

like if you wanted to be technical

[mild cargo]

You can literally put anything instead of m

Assuming the variable is viable

In the context

[ebon gyro]

I do wonder why people use 'm' to denote gradient

mmmmmmmmmmmmm

[mild cargo]

In vector calculus it is generally accepted to use nabla as the symbol for the gradient

[ebon gyro]

grad f

[mild cargo]

Just as you would denote divergence by $$ \Nabla \cdot $$

[charred spear]

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

[ebon gyro]

ow

[mild cargo]

argh

Don't know the mathbot formulas

But you get the idea

[ebon gyro]

=tex \Nabla
Is the gradient applied to a scalar f(x,y,z)

[charred spear]

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

[ebon gyro]

f*

don't know how to escape

Yeah

cheers for the help.

[mild cargo]

$$ \nabla $$
Is the gradient applied to a scalar f(x,y,z)

[mild cargo]

There we go

[ebon gyro]

😃

also I wanted to correct you @mild cargo
it's
$$ \frac{\Delta y}{\Delta x} $$

[ebon gyro]

not

=tex \frac{\Delta x}{\Delta y}

[ebon gyro]

for gradient

[mild cargo]

Ahh, yes of course!

Thank you for noticing it.

[eager pendant]

Could someone help me with this?

I tried arranging it symmetrically, and I managed to get a solution via a tedious coordinate bash

But I'm not even sure if that's right, and I'm wondering what would be an easier way to do it

[eager pendant]

<@&286206848099549185>

[hollow edge]

um

this might be a possible method

probs not the easiest

use ptolemy's theorem

set the diameter to d

d^3 - 7 d^2 - 2 * 20^2 d = 0

that wasn't as messy as I expected

d (d + 25) (d - 32)

diameter is 32, and radius is 16 @eager pendant

actually I canceled out (d + 7) at one point

so it should be
d (d + 7) (d + 25) (d - 32)

this would be my method

[zenith ember]

Hm.

I find it interesting that you assumed the two 7 length sides were adjacent.

And yet, you get the same answer if you do not make that assumption.

[hollow edge]

well

thinking in terms of arc length

they should all have the same circle

since they would "fit" all the segments the same way

and you can abuse the fact that the question implies that it doens't matter

[zenith ember]

True

[eager plume]

false

[eager pendant]

@hollow edge is there a way of doing it without ptolemy?

[hollow edge]

idk

any ideas anyone?

the reason ptolemy is useful here is that it uses the fact that it's an inscribed quadilateral

[eager pendant]

Given triangle ABC with a right angle at B, D is a point on AC such that BD is an altitude to AC and F is a point on BC such that AF is the angle bisector of angle A. If AP=12 and PF=8, find tan(<BAF) and find PD.

I've tried using the angle bisector theorem and made a few fruitless constructions. I know that the three right angled triangles are all similar to each other and that ABF is similar to ADP but I don't know how to start

Diagram for reference:

[zenith ember]

Is this pure geometry? Or can we use trig?

[eager pendant]

I mean, I'd prefer a purely geometric solution but if you can solve it using trig I'd be interested to know how you did it

[zenith ember]

Well, with trig I'd be looking at double-angle formulas

[eager pendant]

Yeah, I know tanx=sinx/(1+cosx)

or at least i think its that

but still, I dont know how I'd go about finding sin BAC

[zenith ember]

Well, in ABF, I know angle B and AF. So I can use Law of Sines to get angle BAF.

angle BAC is then twice that.

[hollow edge]

you can use the similarity ADP to ABF

to get AD:AB

3:5

[eager pendant]

yeah

[hollow edge]

3:4:5 triangles

[zenith ember]

Ah. Nice.

[hollow edge]

you can get AB:BF using the angle bisector theorem

[eager pendant]

yep

[hollow edge]

which is the tangent you're looking for

using that you can get DP from triangle ADP

(last two because of similarities again)

did that work?

[eager pendant]

i dont understand the context of 345 triangles here. like, i can eventually prove that AB:BC:AC = 3:4:5 through a convoluted but i cant see why its relevant here

[hollow edge]

well it's not really important

just a note because it's a special case

3:5 sin always means it's a 3:4:5 triangle

not used here, it's just a note because it's a special case (pythagorean triangle)

[eager pendant]

hm, can you elaborate on

you can get AB:BF using the angle bisector theorem

ik BF/AB=FC/AC

but how are you using this?

[zenith ember]

Hm. I have completely forgotten the angle bisector theorem...

[eager pendant]

given triangle ABC with angle bisector AD, BD/DC=AB/AC

[zenith ember]

Yeah. I'm looking at the wiki page. :)

I think you'll want to apply the angle bisector theorem to ABD.

[eager pendant]

sure, i find that DP/PB = 3/5

and that BD=4x/5, so I've found DP in terms of x

[zenith ember]

One sec. I need a piece of paper. I can't do this in my head. :)

[eager pendant]

@hollow edge Could you elaborate on how you used the angle bisector theorem?

[hollow edge]

ratio of AB:BF is 3 : 3/2 = 2:1

[eager pendant]

How did you know AB:AC=3:5?

(as a side note, thanks to you and samantha for helping. im sorry if my persistent questions are bothering you in any way)

[zenith ember]

It's similar to ABD

ABC is similar to ABD

[eager pendant]

ohhhhhhhhh

I see

[hollow edge]

I was using this from a while ago

oh wait

yeah sorry that wasn't clear

[zenith ember]

What's clear is that I am not actually awake enough to do math.

[eager pendant]

nor am i

So from there, we can solve for x

and find DP

[hollow edge]

well

ADP is similar to ABF

so ratios are all there

super simple

[eager pendant]

@hollow edge ah it's clicked

thank you so much!

[hollow edge]

np

[eager pendant]

thanks @zenith ember too

okay good night

[hollow edge]

gn

[severe folio]

quick question

im writing an essay within n-spheres

and i was wondering how to best give a quick introduction

as im working within a wordcount

could you guys give me a list of things to include so i dont miss anything

[mint sandal]

@severe folio

Essay at what level

Theres a ton of stuff in maths about spheres

What do you want to covee?

[eager pendant]

@hollow edge for what it's worth, I found a solution without ptolemy

[hollow edge]

let's see it

[eager pendant]

lemme do a diagra

[hollow edge]

wat

more explanation pls

[zenith ember]

It's 20-7-20 arcs

Since you know the 7 is parallel to the diameter, you know that from the center to the perpendicular is 3.5

Equalize the height of the perpendicular

And solve.

That's how I solved it, originally.

[hollow edge]

hmm nice

[severe folio]

@mint sandal its for an extended essay within the IB so it should roughly translate to grade 12 - university level

[hollow gazelle]

does anyone know if there is an online calculator for this?

[wide herald]

anyone active

its urgent

[serene yacht]

ye?

[wide herald]

Any good with Compound Area?

[serene yacht]

no just simple area

[wide herald]

like this

https://gyazo.com/4c9edcdec31adf586755e628d5ebbff9

[serene yacht]

lewd

naughty boi

[covert idol]

That's just 2 rectangles added together

[wide herald]

for compound area

at the top of the assignment it says "round to the nearest whole number" so if i got 37, would it be 40

[covert idol]

37 is a whole number

[wide herald]

im retarded, i give up

[covert idol]

The answer is 37 though...

[twin prawn]

A sad story

[last otter]

Hi. Is [0,1) union {5 } (as subspace of R) homeomorphic to disjoint union of [0,1) with single element space?

[serene yacht]

yeah

check that the function which sends [0,1) to itself and {5} to some other singleton is a homeomorphism

[last otter]

Hi. Could anybody help me in questions-3 ?

[slender gyro]

Does anybody understand celestial mechanics? I don't have a question yet but I'd be nice to know

[zenith ember]

Celestial mechanics in the planetary orbit sense?

[slender gyro]

@zenith ember Yeah! Kepler Problem, Hohmann Transer Orbit, Sundman Inequality, etc

[zenith ember]

I kind of do.

[slender gyro]

nice ❤ , it's nice to know.
It's a topic I rarely hear questions about

[fallow edge]

That seems kind of like a physics topic

Idk though

[slender gyro]

It's taught like a mathematics course unfortunately. But yes, it's extremely physicy

[fallow edge]

Unfortunately?

[slender gyro]

My opinion is not too well founded, I still need to study more. But the notation seems to feels somewhat unnatural, not very well chosen. A lot of intuition that could come from physics, an extra simplicity gained from using physics concepts seems to be somewhat disregarded. For example conservation of energy is proved for the gravitational case rather than as a property of conservative forces. Don't trust me though. I might be wrong

[fallow edge]

I see

I thought it was nearly impossible to predict long-term orbits due to the three body problem

But now I hear there’s a whole file devoted to it

Field*

[slender gyro]

I mean sure. But you can always add extra restrictions. Like a 2 body problem. Or for example moon-earth-sun is a three body problem and I think the mass of the moon is considered approximately 0 and that helps out. (unless I misunderstood my teacher. Because it does seem weird to consider the mass of the moon to be 0 when the earth is not so, so radically heavier than the moon)

[fallow edge]

Hm

That seems very weird to me

Like purposely avoiding the problem

[slender gyro]

Well yeah. She told me a very interesting thing today. If you have a satellite around some body (say the earth). Orbiting in an elipse. And you want the satellite to orbit in some other different ellipse. How can you change from one to the other?

[fallow edge]

Hmm

Wdym by “different”?

[slender gyro]

If the problem is too hard you just have to avoid it. All you can do for the hard cases is get approximations. And you do have to develop some fancy math for those approximations, like perturbation theory and stuff

[fallen ivy]

Numerical analysis

[slender gyro]

Hm. Just imagine one ellipse on the solar system, and another ellipse on the solar system. I'm trying to find a video

[fallen ivy]

They're homotopic to each other so they're already the same, topological answer

[slender gyro]

Are you saying perturbation theory is basically numerical analysis or no?

lel

[fallen ivy]

No but they're related to each other and have cross-polination

If I recall

[slender gyro]

Well, that makes sense

@fallow edge I wish I could find a better example but this is one

https://www.youtube.com/watch?v=aKXNbf6NCKU

I just find it to be a super elegant solution. I thought it was cool

[zenith ember]

If you really want an intuition for that stuff... play Kerbal Space Program.

https://xkcd.com/1356/ <-- It's funny, because it's true.

xkcd: Orbital Mechanics

To be fair, my job at NASA was working on robots and didn't actually involve any orbital mechanics. The small positive slope over that period is because it turns out that if you hang around at NASA, you get in a lot of conversations about space.

[fallow edge]

I do actually play that

I’ve heard that it has fixed spheres of influence though, so it’s definitely not a perfect simulation

[zenith ember]

No, it's not.

But it'll teach you Hohmann transfers

And other orbital maneuvers.

[fallow edge]

True

[manic grail]

https://gyazo.com/e2bfde95d1eb13138615466b4a92d9a2

can someone help me with this plz

[rugged bridge]

Is there a simple equation for lines in 3d space?

Like for planes there's y=mx+c

Or do you just use parametric equations for 3d

[wanton forge]

z can be defined with x or y too

it makes a sort of parametric equation too

[rugged bridge]

Ah

Is it possible for non parallel lines to not intersect

[neon fossil]

yeah

consider (t,0,0) and (0,t,1)

[zenith ember]

In 3d, it is far more likely that any two given lines do not intersect than that they do.

[jolly wigeon]

In the diagram A is the point (-1,3) and B is the point (3,1). the line L1 passes through A and is parallel to OB the line L2 passes through B and is perpendicular to AB the lines L1 and L2 meets at C. Find C

[real mulch]

First find L1 and L2.

Can you do that?

@jolly wigeon

[jolly wigeon]

ye

[slender gyro]

A light weight satellite orbiting around earth elliptically MUST have earth as on of is focci? Right?

It's just keplers laws for two bodies

[zenith ember]

Yes.

[slender gyro]

Thanks mate!

[opal blaze]

<@&286206848099549185> How can you find the area of a regular pentagon. without using trigonometry to solve for angles with no exact value? (or rather no calculator)

[keen aspen]

(1/2)a*p

Where a and p are apothem and perimeter respectively

If you know the length of one of the sides, multiply it by 5 to get your perimeter, and with that its pretty easy to find the apothem with triangles

But that involves trig :/

[hollow edge]

well, you could do stuff with golden ratios

[upper karma]

could you do something with the length of the points to the center (or centroid)?

[hollow edge]

if you can find that without trig, sure

[zenith ember]

You can get the exact value for sin(72) and cos(72) without a calculator.

It's just tricksy

[grim lion]

@zenith ember http://puu.sh/AZLBY/96148487fa.jpg

I'm not sure how I could do this. it's a triangle that has 150 degrees in the inside

how would I know what the side angles are?

I feel like I'm forgetting some key formula

@hollow edge

[thorn valley]

the question is whether the angle is bigger than 30 or not

[grim lion]

yeah

not entirely sure. it appears to be smaller

but

I can't prove it

i mean

d is the only alteernative imo

but it's strange logic

since triangle has 150

[thorn valley]

visually, you can draw a 30 degree line from the very beginning

[grim lion]

u cannot have any angl 30 or greeateer

cause if one side is 30 or more
then 150 + 30 + third angle = 180

impossible

so 25 is the only one that would work

but it sounds janky as fuck

lolz

[thorn valley]

yeah i think you got it

the answer must be within the range 0~30 degrees

and D is the only one

[grim lion]

this is such a janky question

I was expecting something more

😦

[thorn valley]

lol

happens a lot in math where the answer ends up being a lot easier than expected

[cinder crag]

i hate questions like that

[gloomy parcel]

how do i do this

[thorn valley]

this should be in questions 1~5

but you find factors that appear twice

[gloomy parcel]

oh..

okay

[thorn valley]

like divide by 2 see where that takes you

[pallid comet]

@grim lion haven't done geometry in a while, but you could've tried making a parallelogram using the parallelogram law

[static bronze]

how the goemetrical shapes have an equation. like circle,parabola,ellipse.

[upper sedge]

What do you mean?

[hollow edge]

they are the set of points that satisfy an equation @static bronze

[static bronze]

i mean how can they formulated the equations from the geometry of shapes on what,which bases they are deriving the equations.

[hollow edge]

the definition of a circle is "the set of points that are a fixed distance away from the center"

the equation for distance is sqrt(x^2 + y^2)

let's say the center is the origin

and the "fixed distance" is r

then the set of points that distance away can be represented as "the set of points satisfying the equation sqrt(x^2 + y^2) = r"

[upper karma]

What is the difference between a vector and a line segment?

[umbral snow]

You can use a vector to represent a line segment. Not all vectors are line segments

Why. Any question in mind?

[upper karma]

I'm learning how to be a game developer.

And there's something about vectors.

Now what's the difference between magnitude and length?

[umbral snow]

Vectors have an addition, you can take any two, add them together, and get a new vector

Line segments can't do that

[upper karma]

So you cannot add two lengths and get one length?

[clear haven]

yeah but thats a scalar quantity

1 dimensional

not good for game development haha

if you go north 1 meter and go east one meter after that, the total distance you have traveled would be 2 meters, but as a vector you would take the shortest path

[unborn bane]

Hi can I get a rigorous definition of a pseudo vector?

[sour furnace]

lol

http://mathworld.wolfram.com/Pseudovector.html

@atomic vigil

[raven citrus]

I need a couple of hard geometry finals practice tests pls send if u guys found any

[rugged moat]

What level? (Uni, high school, middle school) etc.

[steep temple]

I have a 40m sphere. On an 'infinite' graph, what are the possibilities of a 1x1x1 brick being in the sphere. I hope I've explained this clear enough. I need a function/formula for always getting the brick in the sphere. If anyone could help, that'd be great.

^ Imagine it in the same space as kind of like 3B1B's Putnam problem

[upper karma]

Point Z is on side PR of triangle PQR such that m<PZQ = m<PQZ, and m<PQR - m<PRQ = 42. Find m<RQZ

[raven citrus]

High school

@rugged moat high school

[naive scarab]

quick sanity check: for the last sentence, we would need the regular value to be in F(U), not just R^m (so we need a non-trivial regular value).

[dense holly]

consider the contrapositive with n=1,m=2 and F(x)=(x,0). then (0,1) is a regular value, contradicting their conclusion that there are no such regular values

you're correct

[rugged moat]

Hmm... Middle school, Lemme check if I have any

(The maths taught is quite advanced compared to most countries)

Nope.... I think I threw all of them out

[cinder crag]

can someone give me a quick rundown of vectors?

[white harness]

The vectors live in a vector space.

[cinder crag]

okay

but like what EXACTLY are they

the stuff online doesnt really help me grasp the whole concept

[dense holly]

look at the wiki page for euclidean vectors

that's probably what you want, but vectors are more general than that

real answer is that first you have a vector space, then vectors are elements of the set the vector space is defined over

[cinder crag]

ok...

how would you show vectors on paper?

and what are they used for?

like ik its used for physics and geometry but

[real mulch]

Vectors are general things.

Things in a vector space are called vectors.

[dense holly]

do you know if you want to know about euclidean vectors (like in hs physics) or general vectors like in math?

[real mulch]

And everything can be a vector space that satisfies certain criteria.

[cinder crag]

eucledian vectors

and wdym by "a certain criteria"?

[dense holly]

OK, those you can draw as arrows, add them by connecting tip to tail, and multiply with scalars by stretching them

[cinder crag]

okay

and are they like representations?

[real mulch]

https://en.wikipedia.org/wiki/Vector_space - look for "axioms"

To qualify as a vector space, the set V and the operations of addition and multiplication must adhere to a number of requirements called axioms.

[cinder crag]

o

[dense holly]

the point of euclidean vectors is to have a direction and a length/magnitude

[zenith ember]

Don't go all formal right out of the gate...

[dense holly]

yeah those axioms are satisfied by euclidean vectors, but they are also much more general

[cinder crag]

so

i guess you could say theyre kind of like representations?

[dense holly]

and euclidean vectors have a dot product as well anyway

[cinder crag]

in physics

[dense holly]

representations?

[cinder crag]

of things moving

like

lets say you throw a ball

and you wanted to represent that on a 3d graph

would you use a vector?

[zenith ember]

Yes

[dense holly]

uh kinda?

[cinder crag]

and if you want to calculate the amount of force needed for it to higher you could mess with the vector?

[dense holly]

you would use a vector to represent the velocity of the ball at any single point in time

or the force acting on it

[cinder crag]

okay

can vectors be curved?

like parabola shaped

[zenith ember]

No

[cinder crag]

hmm ok

[zenith ember]

Not euclidean vectors

[dense holly]

you represent the trajectory as a sequence of vectors, not a single one

[zenith ember]

They have magnitude and direction.

That is all.

[dense holly]

although you don't actually use vectors for position really

[zenith ember]

Eh. Sometimes

[cinder crag]

ok

[dense holly]

more like a function from time to position, which is just undecorated R^3

(cause adding positions is meaningless)

[cinder crag]

okay

[zenith ember]

I mean, vectors in 3d graphics are used for positioning objects.

[cinder crag]

thats what i need it for actually

im learning some programming and im trying to use vectors

[zenith ember]

This is a big topic...

[cinder crag]

yeah i just wish there was some place that would simplify it

[dense holly]

what is and isn't a vector is a mathematical distinction, which goes out the window in CG

[zenith ember]

And I think it's worth learning properly.

[cinder crag]

okay

[dense holly]

http://www.opengl-tutorial.org/beginners-tutorials/tutorial-3-matrices/

this is good for a taste

[cinder crag]

alright ill check it out

thanks for the help

i guess sometimes advanced is too advanced

[zenith ember]

There is also a 3blue1brown series on the essence of linear algebra. I haven't watched it, but I've heard others say it is good.

That may be helpful.

[dense holly]

oh yeah actually 3b1b is excellent for this

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[fallow edge]

^^^

[cinder crag]

alright thanks

[zenith ember]

Just watched that first video. Pretty good intro.

[hollow edge]

yeah I love that series

[naive scarab]

Anyone here familiar with the discontinuous action of a group on a manifold? I'm trying to work out how this induces a differential structure on the quotient manifold and I'm having some trouble with a technical point.

so this is the relevant part of do carmo, my question is with the very last claim. I can see why this is true:

=tex \text{for each } p_2 \in X_2(W) \text{ there exists a unique } g \in G \text{ such that } \pi_1^{-1}\circ \pi_2(p_2) = gp_2.

[naive scarab]

I'm having trouble proving that's there's one g which makes this equality hold on all of X_2(W). Anyone have any ideas?

[white harness]

If p_1 and p_2 are equivalent then there is g such that p_2 = gp_1.

[naive scarab]

yeah I get that. I want to say pi_1^{-1} \circ pi_2(p) = g(p), but how do I know it's the same g in the entire domain?

[dense holly]

how does one act with an entire group G on a point p?

is the result the set of points {gp | g in G}?

also what is pi | U in this context?

[naive scarab]

so we have a group G, acting on a manifold M. We first require that the map taking p to gp is smooth for each g in G

we're considering a discontinuous action, which means that for each point p in M, there is a neighborhood U such that g(U) is disjoint from U

this basically makes the action faithful

then we quotient out by the orbit of the action, so p1 is equivalent to p2 if p1 = gp2 for some g in G

we call the set of equivalence classes induced by this relation M/G

pi is just the canonical projection from M to M/G, so pi | U is the restriction of pi to an open set U in M

[forest dove]

Okay yeah I do see the concern, do Carmo does it for a point and just says "It follows easily"

[naive scarab]

always a trap lol

[forest dove]

So we already know that pi_i is continuous, yeah?

I'm wondering if that coupled with the fact that the action of G is discontinuous does it

So like, you sorta say okay, assume that for a different point also in that neighborhood g' is what does it for you, then somehow take tiny neighborhoods where g and g' send it to disjoint stuff

[naive scarab]

hmm

actually the pi_i are open mappings right

[forest dove]

I think so

[naive scarab]

and we also know g(U) and h(U) should be disjoint if g is not equal to h

[forest dove]

Err, for some small enough U, yeah

[eternal orchid]

g maps a n'hood U of p_1 diffeomorphically to a n'hood V of p_2 (excuse my differing notation).

If for some other point q in U we have hq in V, then g^(-1)hq is in U. So by discontinuity, h=g.

[naive scarab]

oh I see it

[eternal orchid]

(essentially what you guys were saying above, just slightly different phrasing)

[naive scarab]

yeah ok makes sense

thanks guys

[forest dove]

Well that's a convenient way to show that RP^n is a manifold

[eternal orchid]

Np, good practice for me. I didn't take any good formal manifolds courses in undergrad so I mostly taught myself this stuff, and often skipped the somewhat technical/dreary stuff like this.

[forest dove]

Sorta same, my difftop prof didn't really want to get too technical so everything was in R^n, etc

And he'll be teaching big boy difftop next year, wonder how that'll go

[eternal orchid]

Yeah my first geometry course was classical diffgeom of curves and surfaces in the style of do carmos book with that title pretty much, so more emphasis on computations of curvature etc. abstract manifold stuff was all on me to learn when I needed it.

I remember it being a headache to get used to the way of thinking

but after "getting" the way charts and invariant objects work its nice, and you don't get bogged down anymore.

[forest dove]

Makes sense

[steady widget]

wait so are the distance between these two points 6, or sqrt(37)? because using the formula its sqrt(37) but logically 6 makes better sense.

but 0^2=1

o wait fuck

yeah just realized

jesus i can be stupid sometimes, thanks @ebon vapor

[forest dove]

= p

[upper karma]

So obviously a Straw has one hole, but if you took a straw and poked a needle through it (two layers of plastic) then how many holes would a straw have?

[jagged oracle]

3

[upper karma]

Really?

Why wouldn’t it be 2?

[shadow anvil]

Why would it be 2 is a better question

Make it easier for you to imagine

[upper karma]

Because the one hole goes all the way through, similar to the one it started out with

[shadow anvil]

Shrink the straw into a donut

It's still the same p much right?

1 hole going through

now the holes made by the pin will create two holes on opposite sides of the donut

Unlike the original hole they're in no way connected

[upper karma]

Ok

[shadow anvil]

If that makes sense?

[upper karma]

So then does that mean that a t-shirt has 3 holes then? If you had a fabric that you could stretch and bend to your hearts desire then could you not form this 3 holes straw into a shape resembling a t-shirt?

[shadow anvil]

Urm yeah

I guess so

in a topological mind set

If you're ignoring all of the weird creases and stuff

and you make the tshirt into a tube

with the two arm holes being holes in the sides of the tube

[upper karma]

So then topologically speaking a t-shirt only has 3 holes then

[shadow anvil]

I guess so

[upper karma]

So then this would have 7 holes then

[jagged oracle]

do the other holes go through the shirt completely?

[upper karma]

Yea

[shadow anvil]

Then yes

[upper karma]

Ok thank you

Another question, if somebody got shot with a bullet, but it doesn’t go all the way through, is it still considered a hole

[shadow anvil]

What's a hole

Something that you can pass through

[upper karma]

Because I got into a debate about this with my dad

[shadow anvil]

Or

More exactly

In topology

"A hole in a mathematical object is a topological structure which prevents the object from being continuously shrunk to a point. When dealing with topological spaces, a disconnectivity is interpreted as a hole in the space."

So no

It's more of a crater type thing

[upper karma]

Ok thanks

That’s what I thought

[shadow anvil]

What did your dad say on the matter?

Just out of curiosity

[upper karma]

Well he said that a straw has two holes, so if the bullet goes in you that’s one, and if it goes all the way through you you’d had two

[shadow anvil]

no

The straw is hollow

[upper karma]

Exactly

[shadow anvil]

your body is not

hopefully atleast

[jagged oracle]

technically the human body is genus 1 cuz the the digestive system runs continuously to two openings

"technically"

[shadow anvil]

idk about you

but my asshole is shut most of the time

[upper karma]

But it can be continuously deformed

All life on earth, begins as the sacred topology: the torus.

Although now That I think about it this might not be 7

[shadow anvil]

Why not

The only thing that would change that

Is if the holes don't pass out on the other side

[upper karma]

You don’t know if the hole is the back is two small holes aligned with the front, or it could just be one big hole in the back

[shadow anvil]

Maybe

If this is a question given to you though

From some school or whatever

You'd hope they wouldn't leave ambiguities like that

[upper karma]

No, i saw it online

[shadow anvil]

ah oke

[upper karma]

And t got me thinking

It*

[sacred pilot]

or the inside of the shirt is white

below the top

[upper karma]

Could be

[sacred pilot]

lmao

[upper karma]

Anyway @shadow anvil and @jagged oracle thanks guys

[jagged oracle]

np

[shadow anvil]

A friend of mine just actually raised an interesting point

If you consider the shirt to be homeomorphic to a sphere

Then it has 4 holes

(A normal shirt not the one with holes)

[jagged oracle]

i suppose it depends whether you consider a straw to be a sphere with 2 holes or a torus

[upper karma]

Is it thicc

[jagged oracle]

i can see the logic for both

[shadow anvil]

Yeah

I guess you could take either route

What would be correct though

I guess torus tbh given it's a shirt

It's gonna have thickness

https://www.reddit.com/r/math/comments/5rped8/topology_how_many_holes_in_a_tshirt/

Just found this

Or rather my friend did

[jagged oracle]

ah, so really we need to find how many holes are in a t-shirt because it's made of cloth

[shadow anvil]

lmao

many

But are they actually holes

The fabric isn't a continuous sheet

They're actually just ropes winding

and tangled

but now

This is scaling way out of control and nobody really cares 😂

So I'm going to walk away

[upper karma]

So then taking into account the thickness of the cloth strands, what would the collection of cloth be homeomorphic to? Can we ignore the holes in the shirt at this micro-scale?

[shadow anvil]

Both

Torus and sphere

I'd say torus though

Is the winner

just because thickness

but still

Either works

[upper karma]

Yeah that’s what I think as well

[dense holly]

easiest way for me to think about T-shirt problems is to take the waist hole and stretch it out all the way to infinity, so the shirt is now a flat plane with some holes in it, maybe with shirt arms reaching up to some of them

if you do this, a T-shirt clearly has n-1 holes where n is the number of different openings (ie, equivalence classes of paths you could take from over the shirt plane to under it which are homeomorphic without intersecting the shirt)

[steep temple]

Think I asked this a couple days ago but here we go again

On an infinite 2D graph with a circle plotted on it, points are randomly dotted against the whole graph. How can I make it so they are always dotted inside the circle?

[errant ether]

consider only points with radius less than that of the circle?

how are you generating points

[steep temple]

Solved ^

[thin wasp]

Let PA and PB be tangents to a given circle with P outside the circle. Draw a line through A parallel to PB such that it intersects the circle again at C. PC intersects the circle once more at D. Prove that AD extended bisects PB.

(prove PM=MB here)

[thin wasp]

<@&286206848099549185>

[hollow edge]

this is cool

[rugged moat]

Angle PDM= Angle CDM

Or

Angle PAM=Angle BAM

Prove one of them, and you are done

[eager pendant]

@rugged moat why? angle bisector isn't the same as a median

[hard gale]

jazza well in an isoceles triangle yes

[rugged moat]

I read the question wrong

Rip

My bad

[thin wasp]

@hard gale I think he meant in the diagram here

Also, PAB is isosceles but the line comes from the vertex A not p

Actually, what happens if you extend AC to D such that DAPB is a parallelogram

or rhombus in this case

And then extend DB and AM to meet at X

How could we prove that PMA is congruent to BMX? That would imply PM=MB as desired.

Please ping me if you can help

[hollow edge]

PAM BAM
nice

sry I went out for something for a bit

I'm guessing you use the fact that PAM = ACD

I got to AB = BC

[tiny sphinx]

Equivalent to proving angle APB = acos(1/4)

[hollow edge]

why

no

the answer doesn't depend on the angle

you could construct this with any angle for APB

[rugged moat]

On that note, Are there imaginary angles?

Like in the form of a+bi

[tiny sphinx]

AP . BP = ||AP|| ||BP|| cos(PAB) = 2Lcos(PAB). 2Lcos(PAB) = 1/2 L => cos(PAB) = 1/4 ?

[hollow edge]

no @rugged moat I don't think so

[rugged moat]

y tho

[hollow edge]

angles have a literal meaning

the arc length along a unit circle

length (or ratios of lengths) are real

@tiny sphinx I don't follow

define L

PB?

[tiny sphinx]

Dot the vector A - P onto the vector B - P.
||A - P|| = L
||B - P|| = L
(A - P) . (B - P) = 2Lcos(PAB)
2Lcos(PAB) = L/2 => cos(PAB) = 1/4 => PAB = acos(1/4)

[hollow edge]

PM?

[rugged moat]

Can an arc length be imaginary?

[hollow edge]

length isn't imaginary

[shadow anvil]

no lmao

[rugged moat]

y tho

[shadow anvil]

"How far is it to the shops"

"4+3ikm"

Like what does that even mean

[rugged moat]

Sure, I can't comprehend it

[hollow edge]

t!wiki Euclidian distance

[rugged moat]

But it should mean something

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Euclidean_distance **

[hollow edge]

the idea of "length"

[rugged moat]

We can just say that it is not a concept definable by Euclidean geometry?

[hollow edge]

that's like saying "2.3 stair steps" should mean something

[rugged moat]

Maybe

I think I have come up with my goal in life

To prove that imaginary angles and lengths exists

[hollow edge]

invent a new geometry space

you did it

[rugged moat]

My autocorrect is ducking correcting my completely right spellings

Reeeeeeeeee

[hollow edge]

congrats now your life is meaningless

[rugged moat]

[hollow edge]

@tiny sphinx where tf did you get 2L

[rugged moat]

[rugged moat]

[hollow edge]

you dot vectors, not scalars

[tiny sphinx]

I think my mistake is assuming the line is perpendicular

[hollow edge]

||A - P|| is a scalar

[tiny sphinx]

That's a typo

[hollow edge]

where did 2L come from though

[tiny sphinx]

Dot the vector A - P onto the vector B - P.
||A - P|| = L
||B - P|| = L
(A - P) . (B - P) = L^2cos(PAB)
L^2cos(PAB) = L/2 => cos(PAB) = 1/2L => PAB = acos(1/2L)

This is wrong though

[hollow edge]

shouldn't that be L^2

[tiny sphinx]

No

Oh

Yes

lol

[hollow edge]

where'd you get L/2

[tiny sphinx]

Want the midpoint

I wanted the point (A - P) . (B - P) along BP to be the midpoint

[hollow edge]

how can you say that the dot product is equal to the length of PM?

[tiny sphinx]

Only if AM is perpendicular to AC, right?

[hollow edge]

still no

wrong dimensionality

the dot product is 2 dimensional, length is 1

[tiny sphinx]

I don't know what you're saying, but I'm convinced I made a mistake anyway

[hollow edge]

I mean the equality depends on what "unit" you use for length

[tiny sphinx]

Equality of what?

[hollow edge]

it doesn't hold if you scale the entire thing

of your equation

that seems to be randomly put together

[rugged moat]

g/s works?

[tiny sphinx]

I don't follow what you're saying

Nevermind

[rugged moat]

Or should I use V/m

[tiny sphinx]

All I can intuit is that APB is an isosceles triangle, but I can't even prove that.

It just seems that if it's not, PA and PB wont be tangent

[hollow edge]

it is

you can prove it by using the center

and proving that two triangles are congruent

PAO and PBO

when O is the center

therefore AP = BP

[tiny sphinx]

How do you prove they're congruent?

I see that ABO has two sides of length r

Oh

I see

PAO and PBO both have sides of length r and... k

And I guess we can assume PO bisects APB

So SSA

Wait

Why can I assume that

[zenith ember]

Tangent lines make right angles to radii

[tiny sphinx]

Oh, so I get those two right angles

It implies that PAO and PBO are also right angles

So the triangles both have right angles and have sides of length r and PO

SSA

[zenith ember]

Don't think SSA, because in general that's similarity, not congruence.

But for right triangles, we can use Pythagoras to find the missing side.

[tiny sphinx]

If two triangles are similar and have two sides of the same length, then doesn't the last side have to be the same length?

[zenith ember]

No

Hm

Maybe SSA doesn't even prove similarity, in general

[hollow edge]

well

[tiny sphinx]

https://en.wikipedia.org/wiki/Congruence_(geometry)#Determining_congruence

It says SSA applies to definition of congruence

[hollow edge]

SSA proves similarity for right angles

no

[tiny sphinx]

Oh, no it doesn't

[hollow edge]

you need additional information

[tiny sphinx]

"The SSA condition (Side-Side-Angle) which specifies two sides and a non-included angle (also known as ASS, or Angle-Side-Side) does not by itself prove congruence."

[zenith ember]

Right. That was my point

[hollow edge]

RHS similarity congruence or something

it was called

[zenith ember]

I remember it due to the ASS ordering.

[tiny sphinx]

But since the angle in question is a right angle, we know the sides are equal as you said due to the pythagorean theorem

[zenith ember]

Yes

[tiny sphinx]

Alrighty

You guys haven't solved this yet, right?

[hollow edge]

I'm working on it

[tiny sphinx]

Is OAC equal to OCA?

Oooh

It is

Here's all I have so far

[hollow edge]

here's my progress

im kinda multitasking

can't focus completely

[tiny sphinx]

How do you know AB and BC are equal?

[hollow edge]

angles

PAB = ACB

PBA = CAB

PAB = ACB because of circle geometry

[tiny sphinx]

I can't figure out the details of "because of circle geometry"

Oh, I screwed up

[tiny sphinx]

Hm, I got that AB and AC are equal by a different way

[tiny sphinx]

Oops. Lavendar is wrong

[tiny sphinx]

I think I'm done for now. Let me know if you figure it out

Kinda just procrastinated not figuring much out

[thorny depot]

After drawing AB, CB, and DB and angle chasing a lot with the inscribed triangles and all, we've got a lot of similar triangles

which i think combining that w/ ptolemys might do the trick?

eg, we have that MDB and BPC are similar

and PDM, CDA

AMB, BMD, CBD

[tiny sphinx]

How do you get that MDB and BPC are similar?

[thorny depot]

One sec, I'll go back and retrace my steps

[tiny sphinx]

I do see that ABC is similar to PAB

[thorny depot]

oh oops, i think i typoed that one

ugh why is my handwriting bad lol

ok total similar trianlges i see: PDM~CDA, ABM~BDM~CDB, PBD~BAD~PCB

and thena lso the PAB to BAC

[tiny sphinx]

Looks right, but I haven't proved all of those

Anyway, I think your strategy is right

You can use the quad ACBD to apply Ptolemy's theorem

[thorny depot]

yeah

i just don't wanna bash out the similar triangles w/ ptolemys

but something there should end up cancelling

[tiny sphinx]

I'm sure at some point you can get 2MB = PB

[thorny depot]

i just wonder if there's a cleaner way

yeah its just a matter of getting there

i wonder if it would be possible to just snow from teh similars that pm = mb but idk if that'll be possible

[tiny sphinx]

I didn't know about this theorem. I wouldn't have figured it out

[thorny depot]

probably will need ptolemys?

ya is a useful theorem for cyclics

[tiny sphinx]

Oh, yeah, maybe you can show PM = MB too

I'm sleepy

I'm going to check back for the spoiler when I wake up 😛

[thin wasp]

By circle geometry I assume he means the alternate segment theorem

<@&286206848099549185> Could someone help? My question is posted above

[bright shard]

maybe tangent-secant theorem? idk if things will simplify down though

i think it'll work out if you can prove that CD = 2AD (or PD = 2MD), but i can't see how to prove that

[thin wasp]

update: I got it

angle chasing from a construction

[hollow edge]

@thin wasp what was your approach exactly?

[tiny sphinx]

I have a geometry problem. I found a paper for computing the soft shadow of a sphere occluder and sphere light analytically here: https://sourceforge.net/p/stellarium/mailman/attachment/[e-mail]/2/

To determine how much the light is occluded from point p, the method projects the spheres onto a hemisphere of radius 1 around p, scaling the spheres by their distance from p. From here, it treats the spheres as circles on a plane and computes the overlap.

I am interested in finding the area overlapped by the intersection of two spheres (purple volume) and a sphere light. How can I extend the above method to solve this problem? I'd like to know alternative easier methods too.

[opal blaze]

Am I missing something here

[mint sandal]

?

[opal blaze]

h is the same in both triangles but the opposite is not the same

oh

nvm

[mint sandal]

What are you trying to do

[opal blaze]

h needs to be equal to g

[mint sandal]

Why is there a circle

@opal blaze youre trying to solve the exercise?

[opal blaze]

yeah

[mint sandal]

Why a circle?

[opal blaze]

because they need to have the same length and a circle is shows you all the points a certain distance apart from each other.

[mint sandal]

I have a very nice solution for this exercise

Label the vertices ABC

Say AB=BC

[opal blaze]

hmm

I can

't get this to make sense

[mint sandal]

Then triangle CAB is congruent to triangle ACB

Hence angle CAB is equal to angle ACB

[opal blaze]

they are not congruent if you only know that 2 sides are equal

@mint sandal

[mint sandal]

3 sides are equal

AC=CA, AB=CB, BC=BA

[opal blaze]

It says 2 sides in the exercise?

[mint sandal]

...

Im talking about 2 triangles

CAB and ACB

They are equal

Because 2 sides are equal

[opal blaze]

ohh