geometry-and-trigonometry

[amber raven]

But I don't think anyone has read this obscure book

I am stuck on ex 1 in section 2.6 of chapter 2

It says
Show that (AB) = (BC) = (CA) implies angle(CAB) = angle(ABC) = angle(BCA)

I.E. show all three angles of an equilateral triangle are the same

[mint sandal]

what do you have so far?

[sweet heart]

Can someone get me started?

Dunno a bijection for countability. And how to show an open ball in R^d of radius = any positive epsilon contains a rational.

[mint sandal]

cartesian product of countable sets is countable

[sweet heart]

Oh okay

[mint sandal]

you prove that by induction

finite cartesian product*

[sweet heart]

And denseness?

[mint sandal]

every real number is a limit of a sequence of rational numbers

hence the closure is all of R^d

[sweet heart]

Okay, thank you

[tender gull]

Hello, i want some tips to know what matter i have to study, I have nothing in view. Just some matter that is cool

Some ideas?

[mint sandal]

depends on what you already know

what is your background

[tender gull]

pre algebra, algebraI , ALGEBRA 2

Algebra I, Algebra II, pre-algebra

[mint sandal]

then study calculus

and this is probably not the best channel to discuss this

[tender gull]

And what is?

[mint sandal]

probably #math-discussion

[amber raven]

@tender gull if you dislike calculus( I mean who wouldn't, it's just borring) try your hand at introductory group theory. It's really fun, and requires no prerequisites except for imagination

[mint sandal]

calculus is not boring

[amber raven]

It is

[mint sandal]

not as boring as group theory anyway

[amber raven]

Wtf

[tender gull]

ajshkasjhdakjd

no problem

i think i can see the two

[amber raven]

@tender gull do you know what an equivalence relation is?

[tender gull]

No

maybe

wait

[amber raven]

You're gonna learn so much in group theory. It blew my mind when I first encountered it

[tender gull]

Yah, it looks interesting

[amber raven]

Group theory is all about symmetry and structure

Get the book Visual Group Theory if you have the money

[tender gull]

Ok

[mint sandal]

or learn calculus instead

[amber raven]

Yuck

[tender gull]

sjdlaksjdlaskjdals

[amber raven]

I mean your going to have to

Calculus is really important

Calculus changed the world

[mint sandal]

yes

[amber raven]

But they're so much more to math than calculus

High schoolers never get to see the fun stuff

All they do is compute compute compute

A computer could do that!

[tender gull]

@amber raven You have a point

[upper karma]

You have to start somewhere tbh

[mint sandal]

computing is not that bad

it lets you get a feel for stuff

[tender gull]

by other side, i love computing

[amber raven]

I hate it

[mint sandal]

and you wont get that far in calculus if you dont at least intuitively understand whats going on

it's not all about computing

[tender gull]

the less that i saw

[amber raven]

I'd rather imagine and play with concepts and ideas rather compute hard integrals.

[tender gull]

That's true

[mint sandal]

theres much more to calculus than computing hard integrals

[amber raven]

I was lucky they let me skip calculus 2. That torture was avoided

[mint sandal]

and being able to compute something instead of just looking at the theory and thinking about it abstractly is a great + for me

[amber raven]

Except the only things you are able to compute are really really really really special functions.

[mint sandal]

not really

they come up a lot in physics

and computer stuff

well

yeah they're special but useful too

and let you get a feel for stuff as i said before

[amber raven]

Compute integral of e^x * sin(x^5) cos(3/x).

[tender gull]

what is the more open for me now?

[amber raven]

Wtf

Look at the functions

[tender gull]

i saw just a bit of functions and it explode my mind

[amber raven]

They're really special hand pick to use the select tricks that you can use

[mint sandal]

this is just an example aimed to teach you specific techniques which could be useful later on

yes exactly

this is not what calculus is about

this is a practice problem

[tender gull]

yeah

So, let's do this

[amber raven]

Special case tricks

[tender gull]

What are the pre-requisites

[amber raven]

Ok I'll stop

[tender gull]

of calculus

[amber raven]

You have the prerequisites

You should be able to enroll in AP calculus if your school has it

[mint sandal]

or go to khanacademy

i learned calculus from it

[amber raven]

Oh I learned it first from a MOOC

On Coursera

Free course

[mint sandal]

khanacademy is also totally free

[amber raven]

Yep khanacademy is nice

[upper karma]

_free....I think you mean....** Stolen from the tax payers ** _

Also

try

Shelovesmath

Its a great website with a calc section

[amber raven]

The internet is covered to the brim with calculus content

[tender gull]

And it is good?

[amber raven]

It's all the same

[tender gull]

show me why learn group theory

[mint sandal]

yeah im interested too

why would you want to learn group theory at this level

[amber raven]

I learned group theory in highschool and it was really fun and mind blowing. I feel like more highschoolers, enthusiastic highschoolers, should be exposed to this creative stuff

And group theory is fundamentally about symmetry. How can we characterize symmetry? What are the symmeties of an object? How to they play together? How do they compose? A group is essentially a collection of symmetries.

You know what sets are. A set is a collection of elements. A group is a step higher then a set. It's a collection, yes, but not of element but of symmetries. A symmetry for example is rotating a triangle clockwise by 180 degrees. This changes the orientation of the trianle but keeps it occupying the same space. The triangle looks the same before and after you rotated it.

Something about working with symmetries was just so cool and mindblowing. Its a core idea in math and appears every where. It's so easy too.

You don't have to do a rigorous study of group theory. You don't have to learn everything in a first semester of abstract algebra. But I think it's a good idea to familiariaze yourself with some of these ideas. And have fun. If you find it interesting, please be ambitious and try to learn it rigorously and more in depth.

[tender gull]

Wow

Wow

[amber raven]

Look at this stuff

Is it enlightening?

Blah.

I hate calculus

Sorry my opinion I'll stop

[tender gull]

Group Theory is related with Geometry topology?

[amber raven]

No it's algebra

But algebra is related to geometry in an intimate way

The set of all "motions" for example translations, rotations, reflections, glide reflections, of the plane is a group

[mint sandal]

yes it is related

a lot of of topology and geometry uses groups

basic example would be the fundamental group, used both in geometry and topology

[amber raven]

I have yet to learn that ^^

Will take intro topology next semester

[mint sandal]

fundamental group is just the group of certain equivalence classes of loops in a space

equivalence relation being one loop being deformable into another

and operation being the path composition

[amber raven]

Loops starting at a common point right?

[mint sandal]

yeah we pick a point

[tender gull]

i want to be a data scientist, do you have some matters to learn related to it?

[mint sandal]

although for a path connected space all fundamental groups are isomorphic

[amber raven]

@tender gull statistics

[mint sandal]

at different points

[tender gull]

==10

[charred spear]

10

[mint sandal]

you should definitely learn calculus

multivariable too

[amber raven]

Yeah learn calculus

[mint sandal]

and linear algebra

[amber raven]

The most useful subjects in math are calculus and linear algebra

[tender gull]

What is your age?

[amber raven]

Me? I'm 18

[mint sandal]

linear algebra is probably one of my favorite subjects

[tender gull]

Man, i'm 14

[amber raven]

Idk about my fav subject

I haven't had many courses

[tender gull]

Man, i want to be a mathematician too , but i'm afraid of salary

[mint sandal]

you're 14

don't think about salary

[tender gull]

sdlamsçdmasçdakd

why?

[mint sandal]

because it doesn't matter right now

just learn as much as you can

[amber raven]

@mint sandal how do I become mod?

[mint sandal]

lol how would i know

[tender gull]

But I have to have a goal

[amber raven]

@mint sandal you're green

[mint sandal]

im not a mod

[tender gull]

i think mods are blue

[mint sandal]

mods have moderator roles

[amber raven]

Yeah

Nvm

[tender gull]

Dude

what I have to be?

what's my vocation?

i'm chained on it

[amber raven]

Computer programmer

[tender gull]

why?

i like math

[mint sandal]

⌨ 🐒

[tender gull]

i like computers and computing

but

mathematician or data scientists

i don't like statistics too much

[amber raven]

Mathematician is reserved for those who are obsessed, cannot live without mathematics

[tender gull]

i'm not like that

I think I know what i gonna do

[amber raven]

That's ok

[tender gull]

I'll do Machine Learning engineering and a university of Math

Boom

dramatic exit

[sweet heart]

Why is the set at the bottom still a cover with removed F?

[neon fossil]

if their intersection is empty then their complement (the union of complements) is full

[mint sandal]

intersection of F is empty means that no element belongs in every element of F, so for every x in [a,b], there's some f in F such that x doesn't belong in f hence x belongs in [a,b]/f

[sweet heart]

Thanks guys

So each x is in some [a,b]\f, which is apparently open?

Not sure why it's open..

[neon fossil]

complement of closed is open

[sweet heart]

Ah oops

Thanks

[sweet heart]

Why is the intersection of the countable subcover empty?

Still the The Heine–Borel theorem from above

[twin prawn]

Call $${[a,b]\backslash F_n}{n\in\mathbb{N}}$$ the countable subcover of $$[a,b]$$, then $$\bigcup{n\in\mathbb{N}}[a,b]\backslash F_n=[a,b]\backslash\left(\bigcap_{n\in\mathbb{N}}F_n\right)=[a,b]$$, showing exactly that $$\bigcap_{n\in\mathbb{N}}F_n=\varnothing$$, i.e. the existence of a countable subfamily of $$\mathcal{F}$$ with empty intersection

[sweet heart]

@twin prawn Oh, thanks a lot!

[twin prawn]

🍮

[upper karma]

Hey people do u have some resources on differential geometry?Thx

[mint sandal]

@upper karma what do you already know about differential geometry. What level?

[upper karma]

@mint sandal I just know calculus and differential equations so I think I dont know nothing =^

=^(

[mint sandal]

Thats the book i used

@upper karma i suggest starting with Tu's introduction to manifolds

[upper karma]

@mint sandal Ty very much

[slim gorge]

hey

im following evan chans book for olympiad geometry

but can't seem to understand directed angles

any help?

<@&286206848099549185>

[mossy vine]

http://puu.sh/Ayp8U/f54ba6db8f.png

<ABC is the angle that starts at A, goes to B, and then to C

technically there are two angles, a minor one, and a major one

but we'll look at the minor one

[slim gorge]

yeah

[mossy vine]

I assume the <ABC = - <CBA is fine

[slim gorge]

okay

[mossy vine]

does that bit make sense?

[slim gorge]

yes it does

[mossy vine]

ok good

well we say that <ABC is positive, because to get the segment AB to the segment BC, we rotate anti-clockwise about B

[slim gorge]

thats it?

[mossy vine]

yes

or at least, that's what I'm used to

I think that's reasonably standard

it doesn't really matter which way you make positive, as long as you stick with it for that question

[upper karma]

ty

[slim gorge]

wait

if <ABC denotes directed angles,

then

in the book,

he mentinoed this theorem for cyclicity

ABCD is cyclic iff <ABC = <ADC

how does that relax the possibility that ABCD maybe be concave?

[mossy vine]

I need to think about this

because if D and B are on the same arc of AC then sure that works

but I don't see how it works if they aren't

[slim gorge]

hmm

well there are 1700 ppl on this server

surely SOMEONE would know

[mossy vine]

I remember reading a similar comment in some book I have, I'll go look at it

right ok, i found it

it seems to only be if one of ABC and ADC is looking at the exterior angle

[slim gorge]

hmm?

[mossy vine]

http://puu.sh/AyqdL/7eb6f61181.png7

quite clearly here <ABC ≠ <ADC

[slim gorge]

oh no i was wrong

[mossy vine]

whether you use directed angles or not

[slim gorge]

sorry

[mossy vine]

o

[slim gorge]

i meant

<ABD = <ACD

[mossy vine]

o right ok

[slim gorge]

is it possible for a concave quadrilateral to be cyclic?

i doubt it

anyway, dont deviate rn

[mossy vine]

concave is just that an angle is > 180 degrees?

interior, that is

if so, no it can't be cyclic

idts anyway

[slim gorge]

okay

so

again

this

[mossy vine]

right

if you take a line segment, AB

if you don't use directed angles

[slim gorge]

hmm

[mossy vine]

then <AXB = c, for some c, gives two circular arcs of the same radius ending at AB,

[slim gorge]

wait

ok got it

[mossy vine]

one on each side of AB, ye

the circles made by these arcs intersect at only A and B

so taking X on one arc and Y on the other wouldn't give you a cyclic quadrilateral, even though (undirected) <AXB = <AYB

if we did the same with directed angles

[slim gorge]

wait

can you explain with a picture or something?

this is getting confusing

[mossy vine]

sure

I can try

http://puu.sh/AyqQ5/9d4498402b.png

[slim gorge]

okay

[mossy vine]

if we use directed angles

[slim gorge]

hmm

[mossy vine]

<AXB have and <AYB have opposite signs, so they aren't equal

can you post the statement of theorem 1.9?

[slim gorge]

okay

wait

if they have opposite signs

what are you trying to say

[mossy vine]

one of them is positive, the other is negative

[slim gorge]

yes

then?

[mossy vine]

if you don't use directed angles, they're equal

[slim gorge]

wait

ok

let me try to explain what i think is happening:

in the case where both of X and Y are on the same side, <AXB = <AYB

which is clear

when X and Y are on either side of AB

then one of the ("in-direct") angles is 180* - the other angle

which means both the angles are equal in mod 180*

.

okay, was that what you were trying to say?

[mossy vine]

I don't think so

[slim gorge]

okay

what were you trying to say then?

[mossy vine]

http://puu.sh/Ayrim/51a1f501c5.png

if we consider the quadrilateral AXYB

[slim gorge]

hmm

oh

you were talking about AXYB?

[mossy vine]

I'm not sure what I was talking about tbh, I'm quite tired :^)

let me try and figure out what I'm trying to say

[slim gorge]

lol :P

[mossy vine]

right yeah, we do need to look at AXYB

[slim gorge]

i thought you were talking about AXBY in the latter case

[mossy vine]

yeah I probably was but it doesn't make the point very well :^)

we'll look now at AXYB

[slim gorge]

okay

[mossy vine]

clearly, undirected, the angles <AXB and <AYB are equal

[slim gorge]

hmm

yeah carry on

[mossy vine]

which, if we didn't have the convex requirement, would be saying that it were cyclic, but it can't be

[slim gorge]

(hmm means i got it, carry on)

[mossy vine]

whereas if we look at it with directed angles

we now have <AXB = - <AYB

[slim gorge]

ok true

then?

[mossy vine]

so <AXB is no longer equal to <AYB, which means that it wouldn't be considered cyclic if we dropped the convex property

there's more to be said, such as why this is always true

but I haven't figured that out yet :^)

this gives a sort of reasoning as to why you might be able to drop the convex requirement tho

oh right yeah of course

if we have points A,X,B fixed

then using directed angles, we must have Y on the same arc as X

if we want <AXB = <AYB, that is

[slim gorge]

oh

[mossy vine]

and since they must be on the same circular arc, they're cyclic

[slim gorge]

i dont think i got ALL of it

but i think i will eventually

thx for the help

[mossy vine]

ya

np

as a summary, using directed angles restricts you to just the one arc, which makes it cyclic

whether it's convex or not

with undirected angles, the only way to guarantee that it's on the same arc is by making it convex

[slim gorge]

so you mean concave as in self-intersecting?

[mossy vine]

I believe self-intersecting is neither convex nor concave but idk

but if it's self-intersecting it can't be convex

[slim gorge]

so wdym by "whether its convex or not"?

[mossy vine]

I mean it doesn't matter whether AXYB is convex or not, because using directed angles means that <AXB = <AYB forces X and Y to be on the same arc, and hence AXYB to be cyclic, whether AXYB is convex or not

[slim gorge]

i was just asking what AXYB is if not convex

[mossy vine]

idk what you'd call it other than self-intersecting

[slim gorge]

okay

wait

if its self-intersecting, then is <AXB = <AYB?

[mossy vine]

if you're using undirected angles, yes

if you're using directed angles, no

[slim gorge]

then they are not on the same arc right?

[mossy vine]

yes, they aren't

[slim gorge]

then how is it cyclic?

[mossy vine]

oh, right I see what you're asking now

[slim gorge]

now? xD

[mossy vine]

I thought you were asking about the previous diagram

http://puu.sh/AyrXE/51e74ae065.png

in this case, using directed or undirected angles, <AXB = <AYB

[slim gorge]

i mean

http://puu.sh/Ayrim/51a1f501c5.png

how is it cyclic here?

[mossy vine]

it isn't

[slim gorge]

so?

[mossy vine]

because using directed angles, <AXB ≠ <AYB

[slim gorge]

but A,X,B,Y are supposed to be concyclic right?

[mossy vine]

idk wym

[slim gorge]

the points A,X,B,Y are supposed to lie on some circle right?

because (in-direct) <AXB + <AYB = 180*

[mossy vine]

what A,X,B,Y are you talking about?

[slim gorge]

on the last picture

[mossy vine]

the one you sent?

that was an example of why you need the convex criterion if you're using undirected angles

[slim gorge]

okay

what about when you do use directed angles?

[mossy vine]

AXYB in that is not cyclic

when we use directed angles, <AXB = - <AYB

[slim gorge]

but they are still supposed to be concyclic right?

[mossy vine]

what are you talking about

that was a serious question

[mossy vine]

we have <AXB = - <AYB

so <AXB ≠ <AYB

[slim gorge]

so then they are not cyclic?

[mossy vine]

so it's not cyclic

yes

[slim gorge]

but

this guarantees that they must be cyclic right?

because the (in-direct) sum of the angles is 180

cyclic as in they must lie on a circle in some order

[mossy vine]

idk which angles you're summing to get 180

but either way that only applies if the quadrilateral is convex

[slim gorge]

oh ok i get it now

coool

thanks

[mossy vine]

np

there's like one thing I haven't figured out

[slim gorge]

hmm?

[mossy vine]

http://puu.sh/Aysqf/515c89c85a.png

this case

AXYB here

here we have <AYB = -(180-<AXB) = <AXB - 180

[slim gorge]

wait why is <AYB = -(180-<AXB) = <AXB - 180

do you mean direct angles?

[mossy vine]

well if we use undirected angles, we get <AYB = 180-<AXB, which is positive

but <AYB is negative

so using directed angles, <AYB = -(180-<AXB)

[slim gorge]

oh ok

well gn

[mossy vine]

night

[slim gorge]

its 1AM here

cya around

[mossy vine]

cya

[slim gorge]

can someone check my solution?

wait

@upper karma why are in Honorable lol

.

Let ABC be a triangle and let ray AO meet BC at D . Point K is selected
so that KA is tangent to (ABC) and ∠KC = 90◦. Prove that KD is parallel to AB.

.

(ABC) is the circle passing through A,B and C

My solution:

.

Extend the tangent KA and mark some point X on the other side of the tangent.

Then,
<XAB = 0.5 * <AOB
= <ACB
= <ACD
= <AKD (Because ADCK is cyclic)

Therefore, as <XAB = <AKD, AB || KD

.

Note that I used undirected angles, because I don't think there will be any configuration issues.

<@&286206848099549185>

[slim gorge]

Please ping me if someone wants to answer

[upper karma]

I have a question:
Imagine there is 1 rectangle and 1 square
The area of rectangle is the same as the area of a square.
But the perimeter of the square is smaller than the rectangle's perimeter.
To build the square, you had to shorten the length of the rectangle by 12 meters and add 10 meters to the width of the rectangle.
Find the perimeter of the square.

How would I go on about doing that?

[slim gorge]

hmm

okay

let the length of the rectangle be a and the width be b

and let the side length of the square be s

so, ab=s^2 and s = a - 12 = b + 10

just substitute i guess

s + 12 = a = b+22

(s+12)b = s^2 => (s+12)(s-10) = s^2

there you go

because b = s-10,

you get the equation (s+12)(s-10) = s^2. Now just solve for s and multiply by 4

@upper karma

Your answer should be 240 m

[upper karma]

Let me comprehend this., thanks!

[slim gorge]

yeah sure

this should have been in #prealg-and-algebra lol

[upper karma]

oh lol

wait how does s+ 12 = a = b+ 22

the s + 12 = a makes sense since before it was s = a - 12 but the part where b + 22 is doesnt make sense

@slim gorge

[slim gorge]

i added 10 to every equality

we had 3 equations initially:

s = a-12
a-12 = b+10
s = b+10

now add 10 to both sides of each of the equations

[upper karma]

oh

[white harness]

$$\partial C=\left{(x, y, z) \in \mathbb{R}^3|x^2+y^2\leq (1-z)^2, 0\leq z\leq 1\right}$$\$$\omega = z,dx!\wedge!dy\in\Omega^2(\mathbb{R}^3)$$\$$\phi:(r, \theta, z) \mapsto (r\cos\theta, r\sin\theta, z)$$\
$$\int_{\partial C} \omega = \int_{\phi^{-1}(\partial C)} \phi^*\omega$$\$$$$
How to continue this with Lebesgue measure ?

[white harness]

<@&286206848099549185> .

[white harness]

If the question isn't clear then tell me.

[river forge]

helpers @ helping another i like it ;P

[white harness]

It's like when mathematician asking for help for the Riemann hypothesis or ...

[river forge]

see i would help you but im not designated as a helper so.... ;p

[white harness]

<@&286206848099549185> .

[upper karma]

what?

oh ....

[white harness]

$$\partial C=\left{(x, y, z) \in \mathbb{R}^3|x^2+y^2\leq (1-z)^2, 0\leq z\leq 1\right}$$\$$\omega = z,dx!\wedge!dy\in\Omega^2(\mathbb{R}^3)$$\$$\phi:(r, \theta, z) \mapsto (r\cos\theta, r\sin\theta, z)$$\
$$\int_{\partial C} \omega = \int_{\phi^{-1}(\partial C)} \phi^*\omega$$\$$$$
How to continue this with Lebesgue measure ?

[mint sandal]

looks to me like you're integrating a 2-form over a 3d chain

[white harness]

It isn't correct.

$$\psi:(\theta, z)\mapsto ((1-z)\cos\theta, (1-z)\sin\theta, z)$$\ $$\psi^*\omega = z(1-z)(\sin^2(\theta)-cos^2(\theta))d\theta!\wedge!dz$$\ $$d\psi(\theta, z) = \left(\begin{matrix}-(1-z)\sin\theta & (1-z)\cos\theta & 0\ -\cos\theta & -\sin\theta & 1\end{matrix}\right)$$

[white harness]

The final result is $$\frac{\pi}{3}$$.

[mint sandal]

psi is not surjective onto \partial C

[white harness]

Why ?

[mint sandal]

according to your definition of \partial C, it includes the origin, while psi doesn't map to the origin

origin = (0,0,0)

perhaps you made a typo

[white harness]

psi of (theta, 1) is (0, 0, 1) and psi of (theta, 0) is (cos theta, sin theta, 0).

[mint sandal]

yes

in the first line, perhaps you meant C not \partial C?

[white harness]

C is the cone, \partial C is the boundary.

But the base is missing.

[mint sandal]

ok, because you defined \partial C as the cone

[white harness]

Using Stokes theorem, the result is also pi/3.

[mint sandal]

so what's the problem?

[white harness]

$$\int\limits_{{(x, y, z) \in \mathbb{R}^3 | x^2+y^2\leq 1, z = 0 }} \omega$$.

[mint sandal]

well omega vanishes when z=0

right?

so this integral should be 0

[white harness]

But with a translation the result is different.

Is $$\omega!\wedge!\omega = 0$$ ?

[upper karma]

[mint sandal]

\omega!\wedge!\omega is a 4-form in a 3d space

that should tell you enough

[white harness]

Yes. For higher dimension is it always zero ?

https://www.claymath.org/sites/default/files/yangmills.pdf

Page 2.

[mint sandal]

for higher dimensions d>n, if you have n+1 vectors, 1 of them is going to be a linear combination of n others

a form being alternating is equivalent to saying that if an argument repeats anywhere, then the value of the form is 0

hence the value of a higher dimensional form is always 0

[white harness]

Yes.

[upper karma]

I need help solving this thing

im honestly like clueless

[zenith ember]

Solving what thing?

[upper karma]

wait i should probably put this in math help

[main totem]

how can one find the minimum and maximum distance from two circumference's? if 1 circle has a radius of 2 and the other has a radius of 3, and the distance between their centers is 8

[zenith ember]

I don't understand what is being asked, exactly.

[main totem]

oh sorry i have bad english

wait

um

[zenith ember]

I get the geometric configuration, but what minimum and maximum distances are being looked for?

The minimum and maximum distances between two points on the two circles?

[main totem]

yes

[zenith ember]

Ok.

[main totem]

a point on their circumference

[zenith ember]

Right.

Well, if you draw a diagram of it.

It should be pretty clear that the points for minimum and maximum will all be on the line joining the centers of the circles.

[main totem]

but i need to prove it :/

i thought of like graphing it

with the centers at the x axis

[zenith ember]

Hm.

[upper karma]

find the distance between them?

[zenith ember]

Well, you can write the distance equation for -any- two points on the two circles.

[main totem]

what equation

[zenith ember]

It's the distance equation. Pythagoras.

[main totem]

can you pls gib

[zenith ember]

Do you know calculus?

[main totem]

no

[zenith ember]

Ok.

[upper karma]

=tex d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }

[zenith ember]

Nevermind that approach then.

It would require you to perform a derivative.

[upper karma]

well he can learn calc

[zenith ember]

That seems a little overkill.

Do you know trigonometry?

[main totem]

ill try to graph it and try to prove that the minimum and max distances lie on the x axis through inequalities

[zenith ember]

Ok.

[main totem]

if in triangle ABC the median from angle B lands on AC. is AC-AB/2 less than the length of the median?

[mossy vine]

is that (AC-AB)/2 or AC - (AB/2)

[main totem]

@mossy vine (AC-AB)/2

[white harness]

Vectors ?

[mossy vine]

reverse triangle inequality imo

BD > |AB-1/2AC|

where D is the midpoint of AC

although now that I think about it

it follows directly from the normal triangle inequality (the reverse triangle inequality does too tbf :^))

you have AD+BD > AB

BD > AB-AD

BD > AB - 1/2AC

since AB > 0, AB > 1/2 AB

so BD > 1/2 (AB-AC)

[main totem]

is it true that BM will always be less than AM?

or am i just imagining things again

[fallen ivy]

depends on angles

order theorem

[main totem]

also

how can i prove that (AC-AB)/2 is less than AM ?

i tried doing AC < AB + BC

[fallen ivy]

AB + AM > BM
AC + AM > MC
AB + AC + 2AM > BC
AM > (BC - AB - AC)/2

not quite what you want

but close

[main totem]

ugh im so dumb

i ask too much questions

i promise this will be the last one

So im given two points (A,B) and a line. How do i find a point (K) on the line where AK+BK will be the minimum sum

i minimum like compared to other points on the line

[alpine latch]

so there’s an easy case here and a harder one

if A, B are on opposite sides of the line, it should be farily straightforward to see the solution

think about that case first, and when you’ve done that think about whether you can use the learned knowledge to figure out the case when both are on the same side of the line

that bit is significantly harder, it requires (imo) a clever idea

or some calculus but like don’t

the geometric solution is very pretty

(there’s also the cases where one or both are on the line, the latter case is degenerate in that there’s multiple solutions; and the case where one’s on the line should be easy too)

[upper karma]

:o I’m actually really curious about what the pretty solution is

[main totem]

@alpine latch i forgot. it is given that the points lie on the same side

[alpine latch]

still think about the other case first

[main totem]

hmm

[alpine latch]

and my first hint is that you need the solution for the other case for this one

in some way

[mossy vine]

ahaha

I think I just saw it

that is very pretty

[alpine latch]

pm me if you want

[main totem]

i think i know the solution for the first one

[alpine latch]

which is?

[main totem]

you just connect the lines

[alpine latch]

yep

the points you mean

[main totem]

yes

but how can i use it for the other case

wait

If you draw points symetric to the given ones on the other side of the line

then connect the points

will it be right ?

or am i dreaming again

[alpine latch]

ding ding ding!

we have a winner!

[main totem]

😃

[alpine latch]

the reason it works is that if B’ is the reflection of B and P a point on the line

then B’P = BP

[main totem]

wait

[alpine latch]

so you can reason with the mirror image instead

[main totem]

but how do we know it will cross at the same point on the line

[alpine latch]

uh, they do because… uh

[main totem]

oof

[alpine latch]

symmetry reasons

^^

[mossy vine]

B'K = BK whichever point K you choose

[alpine latch]

I guess they do because otherwise the proof wouldn’t work ^^

[upper karma]

Aw I think I’m the only one who didn’t solve it myself

[alpine latch]

and the proof works without that fact

[upper karma]

:-(

[alpine latch]

I actually didn’t either

[mossy vine]

so the minimum of BK + AK is just the minimum of B'K + AK

[upper karma]

oh

[alpine latch]

I just remembered it from when I was shown

[upper karma]

damn now I wish I didn’t know the solution

[main totem]

wait a minute

does anyone have a picture of points symetric relative to a line?

oof

nevermind

i thought i could draw a parallelogram

[alpine latch]

you can draw a trapezoid

[main totem]

oh yeah

[alpine latch]

with the parellel lines orthogonal to the line

and the line splits it in halves

[main totem]

and trapezoid diagonals cross at 1 point

[alpine latch]

and the diagonals meet in the middle

cause it’s a symmetric trapezoid

but the way more pragmaticargument is: the proof that P is that unique point doesn’t care whether you reflected A or B, so it must give the same result either way

[main totem]

i literally strained my brain for 4 hours trying to solve this problem and this was the fridging solution

[alpine latch]

it’s a hard puzzle

without any hints at least

it’s a pretty solution though

you have to admit

[mossy vine]

another argument for it being the same point is that the reflection doesn't affect the sum, and in the reflection the point is obviously unique

[main totem]

im about to go to a pretty hard entance exam. its the best international school in israel, for advanced minds

i feel like im gonna fail

and embarrass myself

Exam results: Score 2/50

i want to try but

i dont want to get that kind of score

[upper karma]

[main totem]

:(

[upper karma]

:p

Wut grade r u?

[main totem]

bout to go to grade 9

[upper karma]

oo I see

[main totem]

if i pass the entrance

[upper karma]

[main totem]

Staph with the lul emojis

[upper karma]

[main totem]

kills himself

[covert idol]

[main totem]

kills himself in hell

[twin prawn]

🍮

[main totem]

We pick a point inside a triangle, then we draw lines from the vertices to the point. how do i prove that the sum of those lines is less than the perimeter?

[narrow slate]

Tringle ineq

[upper karma]

ok i am new to this server

i am going into 8th grade and i would be in geometry (one year ahead) but i think i am have capabilites to skip that and go into Algebra II(two years ahead)

could someone give me an outline of geometry

Basically study of shapes and their areas, along with the graphs of functions

Oh wait maybe that’s algebra

I forget

what about proofs?

Oh you start with two column proofs

Basically column 1 is your statement and column to is your reason

And you go from statement to statement until you arrive at what you want to prove

so like converting the quadratic formula to the quadratic equation?

or vice versa

to prove that they are equal?

Errr no that’s algebra

geometry is basically mostly triangles, squares, etc

Lots and lots of triangles

what about them

like knowing how to measure angles?

yeah

k

will it be easy to skip?

what will be on a geometry standardized test

[fallen ivy]

Won't be easy to skip and I wouldn't recommend you do

[mossy vine]

geometry in high school basically comes down to points, lines, shapes and various properties of them

[upper karma]

Ya

[mossy vine]

as well as being able to provide a logical proof

[upper karma]

I wonder if there’s modern geometry

[mossy vine]

there is, but generally not euclidean

[fallen ivy]

It's an entirely different way of doing mathematics than you're used to, if the teacher's good it's proof-based

[upper karma]

@mossy vine do you think i could skip

[mossy vine]

I have no idea

[upper karma]

are there properties of it that will make algebra ii hard?

[fallen ivy]

Differential geometry 😩
Algebraic geometry 😩

[upper karma]

if i don't know

._.

[fallen ivy]

Hm, I think more make Calculus hard than algebra 2 hard

[mossy vine]

I've never seen you do maths, and idk the american curriculum very well

[upper karma]

bro

It’s pretty shite : ^)

the american curriculum is sh!

lol

: ^)

[fallen ivy]

Geometry is more disconnected from the rest in America

[upper karma]

i qualified for the national history bee without any studying

[fallen ivy]

But I think it's an important class

Simply since it's a first exposure to proofs

Admittedly two-column proofs, but you take what you can get

[mossy vine]

from what I've seen I don't think it's taught very well

[fallen ivy]

It's usually taught horribly

[upper karma]

In fact I haven’t done a proof in school since geometry class :I

[forest dove]

Oh geometry? Okay let's say we have some insert AG object here satisfying conditions ... (left as an exercise)

[upper karma]

._.

ok @fallen ivy

[fallen ivy]

I had a very good geometry teacher

[upper karma]

are proofs the most important part of geo?

[fallen ivy]

So my experience of it is a bit different

Mm, yeah

Should be at least

like all math

[upper karma]

what is on a geometry standardize test

[mossy vine]

proofs are generally the most important part of maths

[upper karma]

true

[fallen ivy]

^

It's just high school mathematics doesn't reflect that

On a standardized test it depends

If it's an exam given by the teacher at your school, probably proofs

[upper karma]

i wouldn't think proofs

because i think it would be multiple choice

but who know

[fallen ivy]

If it's a standardized test then it would be multiple choice based on properties of geometric objects

[upper karma]

If u want I can test ur geometry skills : ^)

sure

pm me

[fallen ivy]

And perhaps combining those properties

[upper karma]

my geometry tests were all proofs

[fallen ivy]

To get new ones

[upper karma]

And hard as FUK

rip

[fallen ivy]

So we're mine

[upper karma]

i will study two column proofs sometime

no I’ll send them here for the collective audience to try

oh ok xd

[fallen ivy]

Two column is dum

[alpine latch]

two column proofs

what is that?

never heard of that concept

[fallen ivy]

A horrid invention

[upper karma]

but i need to learn it

right?

[fallen ivy]

Tbh no

No one needs two-column proofs

[alpine latch]

well if it’s gonna be on your exams you better learn it

but I’ve never even heard of them thus far

[upper karma]

sascha bear

[fallen ivy]

It won't be on exams

[upper karma]

i am going into 8th grade

[mossy vine]

I wonder how hard these horrible geometry questions will be :^)

[upper karma]

If in a triangle with side lengths 30,60,90, the side opposite the 60 degree angle is 2, then what is the length of the hypotenuse

and i want to skip geometry which is a 9th grade class

horrible -_-

[fallen ivy]

Lol that's easy

[alpine latch]

most proofs I know are just texts with the occasional propositional logic thrown in

or an equation

[upper karma]

It’s easy on purpose -.-

[alpine latch]

it’s like 50% prose at least

[fallen ivy]

And not even a proof based question

[mossy vine]

it's v. easy to do purely euclidean as well

[upper karma]

bruh

[fallen ivy]

Unless you do it right

[upper karma]

Oh ur right

alright let me think

[fallen ivy]

Trig :^)

[mossy vine]

nah

[fallen ivy]

Yeah do it purely Euclidean for the meme

[upper karma]

prove that the median to the base in an isosceles triangle is also the altitude to that side

[fallen ivy]

That's a good one

[upper karma]

thanks : U)

[alpine latch]

isn’t that almost by definition

[upper karma]

Wut no

[fallen ivy]

Yeah Sascha that's how it should be

[upper karma]

wat

[mossy vine]

just reflect the triangle to get an equilateral one :^)

[upper karma]

ummmm

[fallen ivy]

How it should be as in to the proofs being prose

[mossy vine]

to do the first one euclidean

[upper karma]

Tbh I think I forgot how to prove this : ^)

[alpine latch]

what can I assume as given knowledge for this proof though?

[upper karma]

it would be whatever you have to multiply to get root 3 to become 2*2

[alpine latch]

can I assume I know that isosceles are symmetric?

[upper karma]

right?

i forgot the numbers, is it root 4?

[fallen ivy]

I would say you can assume the ITT

[upper karma]

you have an isosceles triangle ABC with congruent sides AB and AC. Point D is the midpoint of BC. Prove that AD is an altitude to BC

[alpine latch]

I don’t know what that means

the ITT that is

[upper karma]

I don’t either

[fallen ivy]

That angle congruency for isosceles triangles

[upper karma]

small numbre

did i get the first one right

Yeag

[fallen ivy]

We had acronyms for ours

[upper karma]

No it’s not root 4

ik, but what i said first

Uh

I’m not sure I understand what you’re saying

it would be whatever you have to multiply to get root 3 to become 2*2

[mossy vine]

x sqrt(3) = 4 is what he's saying I think

[upper karma]

almost

sqrt3*2=x

answer 2x

right?

[fallen ivy]

Anyway the median one is more important IMO

Draw a diagram and reason about it

Just remembered the proof we used myself

(Doesn't actually need ITT)

Mudkip no

Cubing should solve

[upper karma]

what is an altitude

[fallen ivy]

Angry

[alpine latch]

my thing here is to construct an isosceles I would use that thing that we’re to prove

[fallen ivy]

Angery

[alpine latch]

it’s such a fundamental thingy that I don’t even know where to start

[upper karma]

nvm i know what an altitude is

[alpine latch]

why are easy things always so hard to prove

[fallen ivy]

Start by drawing a geometrical diagram

[upper karma]

k

[fallen ivy]

I can dm you a solution Sascha

[upper karma]

Oh I remember how to prove it now

[alpine latch]

sure

[upper karma]

ikr

[alpine latch]

go ahead

[mossy vine]

because you have so little to prove them with

[upper karma]

You prove that the two triangles are congruent, then the two corresponding angles are supplementary, so they’re each 90

Wait what proof are you guys talking about?

small numbre

can you just dm me so that if i don't understand something u can explain

[fallen ivy]

Numbre

Why did you do that

:(

You needed to wait for cubing to solve

[upper karma]

i don't understand proofs

how can i prove that, too me it is just draw a line and don't be blind :/

america, explain

[fallen ivy]

Alright

Start out with the diagram

[upper karma]

k

[fallen ivy]

Now mark your congruent sides

[upper karma]

k

[fallen ivy]

We usually do this as a 1 tick along two lines for those two lines being congruent

[upper karma]

yeah that's what i did

[fallen ivy]

Or two tick marks

For the other pair

Now you have two triangles

[mossy vine]

I see how to do this without ITT but that's kinda long-winded in comparison

[fallen ivy]

What properties do these two triangles share?

[upper karma]

Wait wut

two congruent sides?

isocecles?

still want me to DM u the solution

oh nvm wolfs got it

[fallen ivy]

They're not isosceles

[upper karma]

what

@mossy vine what’s ITT

[fallen ivy]

But they have some sides congruent

Look at the diagram and the two triangles formed

[upper karma]

you have an isosceles triangle ABC with congruent sides AB and AC. Point D is the midpoint of BC. Prove that AD is an altitude to BC

You can prove this using coordinate geometry : ^)

this problem, right?

[mossy vine]

idk what it stands for, Isosceles ____ Theorem I think

[upper karma]

yeah

oh I see

[mossy vine]

Isosceles Triangles Theorem

I'm dumb

[fallen ivy]

Yep

Anyway

[upper karma]

it is isoscles @fallen ivy

Ngl I completely forgot how to prove that the angles in a triangle sum to 180

[fallen ivy]

The triangle there is

You said the triangles it splits into are

Which isn't truey

[upper karma]

ohhh

that's what you meant ok

they both have right angles

[mossy vine]

kinda hard to prove that @upper karma if you want to do it properly

[fallen ivy]

You don't know that cubing

That's what you want to prove

What do you explicitly know

[upper karma]

Isn’t it external angles

they are congruent

they are supplementary

[fallen ivy]

The triangles are congruent?

[upper karma]

right?

Also how the fuk do u prove that the angles in an n-gon sum to (n-2)(180)/n

[fallen ivy]

Angles are supplementary, not angles

*not triangles

Mistype

[upper karma]

:/

[mossy vine]

that's a corollary

[upper karma]

huh?

[fallen ivy]

Split up the n-gon into triangles

Triangulation

[upper karma]

the vertices that meet at angle d are supplementary

[mossy vine]

^

[upper karma]

Oh I see

[fallen ivy]

@upper karma ok good

[upper karma]

oh nice that’s pretty elegant

[fallen ivy]

What can you derive from the fact that the triangles are congruent?

[upper karma]

they have same angle measurement

and same side lenght

[fallen ivy]

Yep

So if they have same angle measurement and the angles are supplementary

Write an equation from that

[upper karma]

ok

Wait how do you prove SAS again

do you even need to prove something like that

angle BDA+angle CDA = 180

[mossy vine]

SAS is normally taken as an axiom I think

[fallen ivy]

Yeah

[upper karma]

Oh I see

how do you prove that a line can be drawn between any two points?

[mossy vine]

then the other triangle congruences can be derived from it

another axiom :^)

[fallen ivy]

I believe there are ways to prove it (???) but they're not suitable for high school courses

[upper karma]

:I

yeah probably

[mossy vine]

ya

[upper karma]

Proving SAS isn’t hard though right?

so that is kind of a generic proof?

Using trigonometry

[mossy vine]

that'll probably be circular

[fallen ivy]

^

[upper karma]

oh god flashbacks to circular

[fallen ivy]

Anyway @upper karma

angle BDA+angle CDA = 180

[upper karma]

yes

[mossy vine]

trigonometry is just similarity of right-angled triangles

[fallen ivy]

The two angles are the same

Call them both x

x + x = 180

[upper karma]

k

2x

[fallen ivy]

What is x

[upper karma]

90

oh

wow

But if you prove trigonometry constants without using right angles triangles, then use trigonometry to prove SAS

[fallen ivy]

Proof done

[upper karma]

that was easy

knowing that, will pretty much any geometric proof in high school derive from this

[alpine latch]

But if you prove trigonometry constants without using right angles triangles, then use trigonometry to prove SAS

we defined π to be the first positive root of a certain power series so…

[upper karma]

err no

@alpine latch

[fallen ivy]

Congruency properties of triangles then yes basically :p

[upper karma]

are proofs not hard, but require a lot of thinking?

[fallen ivy]

Hmm

Depends on the proof

[upper karma]

No proofs can be hard : ^)

could i have another example

[mossy vine]

sure

[fallen ivy]

And requiring a lot of thinking makes something hard by defn

[upper karma]

and see if i can do it without help

[alpine latch]

proofs are mostly creativity + experience

[upper karma]

Oh god

[alpine latch]

in terms of difficulty

[upper karma]

Have mercy @mossy vine :I

[fallen ivy]

I wish I could dig up my prof's midterms

He's doing a masters in math rn

[upper karma]

oh how about this

[fallen ivy]

And he teaches geometry at our high school

[upper karma]

cool

[mossy vine]

let ABC be a triangle, and let L, M, N be the midpoints of BC, CA and AB respectively. Prove that <LAC = <ABM if, and only if, <ANC = <ALB

[upper karma]

In a right angled triangle ABC with hypotenuse BC, a point D is on BC such that AD = DB = DC. Prove that ABC is isosceles

mine is easier :I

I think at least

[mossy vine]

it is

[upper karma]

k ill do numbre's first

[mossy vine]

ya mine's kinda tricky

probably ~BMO1 or BMO2 level

[upper karma]

wait what

Oh wtf

one sec