geometry-and-trigonometry

[upper karma]

that's it?

[dark sparrow]

yeah

[upper karma]

I feel dubious about this formula, honestly

[dark sparrow]

like, assuming you went through the math and found the area of one of the triangles

and then multiplied that by n

that's enough proof

[upper karma]

Is it still possible to do a proof by induction for this?

[dark sparrow]

don't think so

[upper karma]

why tho

[dark sparrow]

going from an n-gon to an n+1-gon shakes shit up quite a bit

i feel like you can fix like

a side, at most

[upper karma]

Well thanks for clarifying things, understood why it works

[slim gorge]

we can also prove that the product of slopes of two perpendicular lines is -1 without trig

[dark sparrow]

hm?

[slim gorge]

i did it while studying analytic geometry

[dark sparrow]

do share

[slim gorge]

it involves a bit of geometry like similar triangles and stuff

so it will be hard to "visualize"

should i go on

[upper karma]

ooh this one uses similar triangles

[slim gorge]

ya

i don't remember the whole proof, but i think i can derive it

[dark sparrow]

i think i get the rough idea, actually

it's easy to prove for lines intersecting at the origin

[slim gorge]

ya

but i did a more general proof

[upper karma]

i accidentally proved it once when I was trying to prove the intersecting chord theorem for circles

I was like

"wtf is dis shit"

and googled the thing

and was laik

"oh kewl beans"

[slim gorge]

lol

what intersecting chord theorem?

[upper karma]

draw two chords that are in the same circle

and the product of their corresponding opposite lengths are the same

well you can google it

[slim gorge]

power of a point?

[upper karma]

don't think so

it uses inscribed angles

and triangle similarity

to prove it

[slim gorge]

oh ya

i remember something like it

[dark sparrow]

@upper karma you mean two chords that intersect each other?

[upper karma]

yeah, that's what the name suggests

well the chord can be a diameter

well the diameter is a chord

nvm

[slim gorge]

yes it is

[vale raven]

"accidentally proved it"

That's amazing xD

[slim gorge]

did u use analytic geometry?

[upper karma]

ikr

yeah

analytic geometry's OP m8

[slim gorge]

ikr

but its bashy sometimes too

[upper karma]

what do you mean

[slim gorge]

like too many calculations and stuff

[upper karma]

yeah man

gad like earlier today

I tried finding the area of a circle

that's separated by a chord such that it's perpendicular if you a draw a radius to it

the whole thing

was a mess

i originally started out with this

[slim gorge]

seperated by a chord
?

[upper karma]

uhm say you have a circle

and draw a random chord such that if you draw a radius to the chord it's perpendicular

so you now have two regions

[dark sparrow]

define radius to the chord

to one of its endpoints?

[slim gorge]

the radius bisects the chord then

[upper karma]

ah yeah

equivalent statement

[dark sparrow]

the radius bisecting any non-diameter chord will be perpendicular to it though

[slim gorge]

ya

[upper karma]

well i forgot the terms and stuff

[slim gorge]

and vice-versa

[upper karma]

:<

anyways I managed to find a function that can help me find the area of one region

[slim gorge]

what region?

[upper karma]

if you draw a chord in the circle

[slim gorge]

can u post a photo of the problem or something?

[upper karma]

that's more easier yeah sure

[slim gorge]

thx

[dark sparrow]

more easier

[upper karma]

it's something like this

https://puu.sh/wwq8l.png

you have two regions

in the pic, there's the bigger one on the left

and the smaller one on the right

blue line's the said chord

[dark sparrow]

ok

so you want the area of a segment

[upper karma]

yeas

as a function of the distance of the chord

let me type out the thing

[slim gorge]

distance from the center?

[dark sparrow]

distance of the chord from the center?

that's cos(θ/2), where θ is the angle spanned by the smaller chord

[upper karma]

i mean a good way to describe this is

imagine the chord

going from right to left

the area on the smaller region

gets bigger and bigger

[slim gorge]

yes...

[dark sparrow]

also are we assuming the radius is 1?

[slim gorge]

or r?

[upper karma]

just r is fine

[slim gorge]

ok

[upper karma]

=tex A(k)=r^2\arccos(\frac{k}{r}) -k\sqrt{r^2-k^2}

[upper karma]

so the r is the radius

[slim gorge]

whoa, what is arccos()?

[upper karma]

arccos inputs length and outputs angles

[dark sparrow]

arccos is the inverse function to cos

[slim gorge]

oh ok

cos^-1?

[upper karma]

oh sorry I use arccos instead of that

[dark sparrow]

yeah

[upper karma]

just the intuitive appeal it has

[slim gorge]

i dont know inverse functions yet :(

[upper karma]

from the word arc

[dark sparrow]

inverse functions = undoing functions @slim gorge

[slim gorge]

lol

ya

[dark sparrow]

like, if f(x) = x - 1 then f^-1(x) = x + 1

[upper karma]

yes

[slim gorge]

i did that

but not inverse trig functions

[upper karma]

inverse trig functions are the same with inverse functions in general

[dark sparrow]

anyway, in terms of the angle θ spanned by the smaller arc

=tex A = \theta \frac{r^2}{2} - \frac{1}{2}r^2 \sin(\theta)

[upper karma]

normal trig functions inputs angles and outputs some length

inverse trig: length to angles

[slim gorge]

ya

ok got it :)

[upper karma]

So like I was unsatisfied of the function that I'd found

simply because

=tex |k|\le r

[slim gorge]

how is that so

[upper karma]

shet

[upper karma]

yes

[dark sparrow]

so what's wrong with that

[slim gorge]

ya

k is the length of the chord right?

[upper karma]

no

[dark sparrow]

no

the distance from it to the center

[upper karma]

yis

[slim gorge]

OH ok

[dark sparrow]

r cos(θ/2)

anyway here's a slightly more formal definition of inverse trig functions

=tex \arcsin(x) = y \iff |y| \leq \frac{\pi}{2} \wedge \sin(y) = x \ \arccos(x) = y \iff 0 \leq y \leq \pi \wedge \cos(y) = x \ \arctan(x) = y \iff |y| < \frac{\pi}{2} \wedge \tan(y) = x

[slim gorge]

what is :

=tex \wedge?

[dark sparrow]

"and"

[upper karma]

the thing means and

[slim gorge]

oh ok

couldnt we just use & ?

[upper karma]

well

this one is used in logic pretty frequently so Idk about changing the conventions

[slim gorge]

ok

[upper karma]

anyways

I was unsatisfied with the thing

[dark sparrow]

why tho

[upper karma]

cuz k can be negative

I don't want negative inputs

[dark sparrow]

why not?

[upper karma]

imo it's just

personal preferences

aesthetics and stuff

[dark sparrow]

k < 0 gives you the big sectors

😛

[upper karma]

i knaw

just

uglyness

with negative numbers

😦

[dark sparrow]

why do you hate negative numbers 😦

[slim gorge]

wut

[upper karma]

Iz ugly :((

Anyways

because of this

I tried getting rid of the thing

[slim gorge]

better than fraction tbh

[upper karma]

topkek

So I tried defining k differently

instead of the distance of the chord from the center

[slim gorge]

o k

[upper karma]

I'd defined it as the distance of the chord from the circle

it's kinda hard to explain this in words

but basically

[slim gorge]

i see

[upper karma]

you see that pic I gave

if you draw the radius that bisects the chord

[dark sparrow]

i get what you mean

you basically made a variable for r(1 - cos(θ/2))

[upper karma]

nah

idk what you meant by that actually

I just redefine "k"

[dark sparrow]

θ is the angle spanned by the shorter arc delimited by the chord

[slim gorge]

ok, go on

[upper karma]

https://puu.sh/wwqMS.png

in this diagram

[slim gorge]

oh

[upper karma]

I'd defined k as the distance of the chord from (-1, 3)

but of course the chord starts from the left

instead of the right as depicted in this pic

cuz no negative numbers plx

so from the pic

if you slide the chord

to the left

there's gonna be the regions that will appear

[slim gorge]

hmm

[upper karma]

the left one starts to get smaller and smalelr

and the right one formed by sliding the chord to the left

will get bigger

so if you slide the chord all the way until the chord length=0

you get the area of the circle

so anyways

I tried doing this

[slim gorge]

chord lenght = 0?

[upper karma]

but it was a shitstorm

in the pic

the chord length right now

is 0

[slim gorge]

oh

i see

[upper karma]

if you slide it to the left

[dark sparrow]

honestly...

[upper karma]

it'll start to get bigger until it reached its max at the center

the diameter

I know this is a lot of hassle to get rid of negative numbers for my inputs

but myeh

[dark sparrow]

=tex 2r^2\int_{0}^{\frac{k}{r}} \sqrt{1-(x-1)^2}dx

[slim gorge]

NO CALCULUS PLEASE!!!

[upper karma]

well I suppose I could do it that wayyyyyyyyy

[dark sparrow]

pff ok

[upper karma]

but yeah when I tried to use k in its new definition

I get really long terms

and it's just a mess

and I'm pretty sure I got an error somewhere

I'll just type the thing in

[dark sparrow]

lesson: don't go out of your way to avoid negative numbers

[slim gorge]

lol]

[upper karma]

yeah 😦

[slim gorge]

LIFE LESSON

cant u do it without analytic geo tho

[dark sparrow]

without analytic geometry?

honestly, segment = sector minus triangle

[slim gorge]

ya

[upper karma]

=tex A(k)=\theta\cdot r^2 -x\sqrt{r^2-x^2}

[upper karma]

This is the simplified ver

I'll type the x and theta out

[slim gorge]

theres no k in the RHS

[upper karma]

=tex \theta=\arcsin(\frac{\sqrt{k(2r-k)}}{r})

[upper karma]

=tex x=\sqrt{k(2r-k)}

[slim gorge]

=tex \theta=\arcsin\left(\frac{\sqrt{k(2r-k)}}{r}\right)

[upper karma]

couldve typed out the x first then substitute

gerd

anyways for some reason that I've probably made during finding this thing, the thing is not defined for all values of k

which is from 0 to 2r

So I kinda just took a break from this shit and will try to do it again

tmr

redefining k made me use some other neat theorems like heron's and the intersecting chord theorem

but yeah

I'll stick to negative numbers for now

😦

[slim gorge]

👍

[rotund raven]

Can anyone recommend a good book on elliptic curves which is simple to follow? I'm in year 12 in UK( high school equivalent in USA)

[shut needle]

"Elliptic Curves: Number Theory and Cryptography" by Kenneth Rosen is the easiest one I've seen

Easy while still conveying the material

"The Arithmetic of Elliptic Curves" by Silverman is the classic that everyone eventually reads, but it's slightly harder

[rotund raven]

Thanks I'll have a look

[tawny pewter]

Root welp

[idle escarp]

Hey i need some help with geometry

Please help me!!!!

[dark sparrow]

@idle escarp if you're still here: just post your question

don't ask to ask

just ask

[upper karma]

Why is it that the ratio of the angle of a central arc of a circle to its total angle is equal to the ratio of the area of the of sector that's being subtended by the said central angle to the area of the circle

specifically

=tex \frac{\theta}{2\pi}=\frac{a}{\pi r^2}

[upper karma]

It's a known fact that this is true but I haven't thought about it why this is true

[azure storm]

Because it's homogenous I'd say, plus it has an infinity of symmetry axis.

So if you take a fraction of it, you take a fraction of it's area

[dark sparrow]

yeah

circular symmetry and all

[upper karma]

Is this a consequence from the fact that all circles are similar?

[azure storm]

Wouldn't say so

Take a cube in 3D, if you cut it parallel to a side you get a fraction proportionnal of the volume

[upper karma]

I don't get it

[dark sparrow]

if you cut a cube parallel to one of its faces the pieces will be in the same ratio as the edges that ended up getting cut

[upper karma]

oh so the ratio of one of the pieces to the cube's volume is the same as the ratio of the length of the edge that ended up getting cut to the full length of the edge?

[dark sparrow]

yeah

[upper karma]

I mean this is just an example of what I just said

Is there like a proof?

[vale raven]

=tex a=\int_0^r x\theta \text{ d}x = \frac{r^2\theta}{2} \ \theta \propto a

[vale raven]

...Something along those lines.

[upper karma]

🤔

[upper karma]

I don't get it 😦

The way I understood this is this

[final prairie]

What's the integral of x dx daniel?

=tex \int{xdx}=\frac{x^2}{2}

[final prairie]

so

=tex \int\theta x dx = \theta\int x dx=\theta\frac{x^2}{2}

[final prairie]

Implying that

[upper karma]

what about the constant?

[final prairie]

+C yes

:p

[dark sparrow]

fix the lower bound at zero

[final prairie]

Yup

And upper bound is r

so sub r subtract form sub 0

and u get

=tex \frac{r^2\theta}{2}

[upper karma]

oh I can see now

=tex \frac{\theta}{2\pi}=\frac{\frac{\theta}{2}}{\pi}=\frac{r^2\cdot\frac{\theta}{2}}{\pi r^2}

[upper karma]

and that just now justifies it

thanks

[final prairie]

Given that A is a fixed point of f, and f(g(A))=g(A):
i) Prove that if f is a non trivial rotation then A is a fixed point of g.
ii) Prove that if g is a glide reflection then f is a reflection or the identity isometry.```

[dark sparrow]

let's see

let's assume A isn't a fixed point of g, so g(A) ≠ A

then f has two fixed points - A and g(A) - while a nonzero rotation can only have one

💯

[final prairie]

Exactly my proof

💯

The second one is similar

If g is a glide reflection then it has no fixed points

[dark sparrow]

proof by contradiction 👌

[final prairie]

so A is a fixed point of f and g(A) is a fixed point of f but g has no fixed points so g(A)≠A thus there are 2 fixed points, A and g(A), meaning there are infinite fixed points, so either a reflection or the identity

💯

[final prairie]

Another one:

Let f be an isometry of the plane, S_l be reflection on l, and A a point in the plane. Given that f(A)=S_l(A), prove:
1) A is a fixed point of S_l circ f
2) If f is a glide then S_l circ f is a reflection

[dark sparrow]

hmm

assume Sl(f(A)) != A

then since Sl is bijective we can apply it to both sides to get

Sl(Sl(f(A))) != Sl(A)

Sl compose Sl = id

so we get f(A) != Sl(A), which contradicts the initial condition

💯

[final prairie]

No need for contradiction

f(A)=S_l(A) // Apply S_l to both sides, ALWAYS allowed since x=y => f(x)=f(y)
S_l(f(A))=S_l(S_l(A))
S_l(f(A))=A

[final prairie]

f is a any glide reflection and S_l is a reflection through l.
A) Prove fºS_l is not the identity isometry
B) Prove that if A is a fixed point of fºS_l then fºS_l is a rotation around A

Nice one ^

[dark sparrow]

hmm

[final prairie]

There fixed

hehe

[dark sparrow]

ok so

[final prairie]

Took me a few mins, solved with contradiction.

[dark sparrow]

if the line of reflection for f is l, then f compose S_l is a translation

not the identity

[final prairie]

It's true but you need to prove that too

I mean it's kinda obvious because then it's T compose S_l and when you compose with S_l they cancel out

[dark sparrow]

yes

[final prairie]

So you stay with T, cool

You solved 1/infinity options now :p

[dark sparrow]

now suppose the line of reflection for f isn't l

err

we can consider the inverse of f compose S_l

[final prairie]

Try contradiction

oki

[dark sparrow]

namely S_l compose f^-1, and f^-1 is also a glide reflection

that shares the line of reflection with f

[final prairie]

Yup

[dark sparrow]

er

fuck

[final prairie]

hehe

.restart();

[dark sparrow]

...i have other things to do right now and i'm also going through some anxiety so my brain isn't functioning properly

sorry

[final prairie]

No worries hehe

Want my solution?

[dark sparrow]

sure

[final prairie]

Suppose for the contradictory that fºS_l=I
Then
f^-1(fºS_l)=f^-1ºI
S_l=f^-1
f=S_l

contradiction

[dark sparrow]

oh

right

of course

functional algebra

is a thing

[final prairie]

hehehe

[dark sparrow]

invertible functions from a set to itself are a group under composition, damnit

[final prairie]

a=b=>f(a)=f(b) is a nifty trick that i forget a lot

Yeah but I didn't really use that

Another nifty trick, fºf is always orientation preserving for every isometry

[upper karma]

so lets take an ABC triangle

all of the angles the triangle has is less than 90 degrees

lets take a P point inside that triangle, how should we position P to minimize the lenght of PA+PB+PC

[final prairie]

Interesting

[upper karma]

to solve this problem i made a coordinate system where the triangle's points were (0,0) (a,0) (b,c)

then i made the f(x,y) function

[dark sparrow]

i'm tempted to say that the point you're looking for is the center of the triangle's circumcircle

[upper karma]

for the distances

[dark sparrow]

but i'm not sure

[upper karma]

ye maybe

[dark sparrow]

gimme a sec

[final prairie]

Makes sense

Circumcircle is the circle inscribed in the triangle right?

[upper karma]

f(x,y)=(x^2+y^2)^(1/2)+((b-x)^2+(c-y)^2)^(1/2)+((a-x)^2+y^2)^(1/2)

so as any other people

i partially derived it

for each variable

then made it to a system of equations

[dark sparrow]

partially differentiated*

[upper karma]

yep sorry

df(x,y)/dx=0 and df(x,y)/dy=0

but they seem hard to solve formally

[dark sparrow]

and that also doesn't guarantee a minimum or maximum point

consider g(x,y) = xy

[upper karma]

ye right

[dark sparrow]

at (0,0), ∂g/∂x and ∂g/∂y are both zero

[upper karma]

but it must be positive

so we can presume that it has a minimum

[dark sparrow]

anyway, i feel that what you're doing here is way harder than it should be

and that there's a purely geometric way to find the desired point

[upper karma]

probably but im interested in this kind of solution

[dark sparrow]

ಠ_ಠ

[dusk snow]

to minimize the sum of the distances, the angles to the 3 points should all be 2pi/3 iirc

i don't remember the coordinates or the construction though

[dark sparrow]

does that give a unique point though

[dusk snow]

it should, unless they are colinear or something

......

.............

[dark sparrow]

errh

[dusk snow]

wait, circles intersect in two points .. right ?

[dark sparrow]

yes

at most two

[dusk snow]

time to open geogebra i guess

[final prairie]

There was a proof to that in my book lol

[vapid kettle]

Nice circles always intersect in exactly two points

[dusk snow]

oh i see the other point i'm dumb

see the set of points M such that AMB is 2pi/3 is part of a circle

so you just draw them and intersect them

and the circle goes through A and B

[lean crest]

well, two identical circles technically intersect in infinitely many points 🤔

[dusk snow]

and if you do that with B and C you get a circle through B and C

and they intersect at B and at the point you want

[glossy minnow]

How would I parametrize 5a^(2)+2a*(3b+11√(2))+5b^(2)+42b*√(2)+106=0

[dark sparrow]

=tex 5a^2 + 2a(3b + 11\sqrt{2}) + 5b^2 + 42b\sqrt{2} + 106 = 0

[dark sparrow]

this?

this gives you a circle

manipulate that expression until you're able to find the circle's center and radius

and then your parameterization is simply a(t) = a0 + r cos(t), b(t) = b0 + r sin(t), where (a0, b0) is the center

@glossy minnow

[glossy minnow]

Thanks

[tawny pewter]

Root

welp

wy=25 the ratio of wx to xy is 3:2 find wx

[hazy field]

@tawny pewter so wx / xy = 3 / 2

wait, is this even solvable unless we know an angle

[dark sparrow]

@tawny pewter is WXY a triangle or something?

also, yes, there's not NEARLY enough information

[tranquil yew]

=tex 25^2 \leq 4j^2+9j^2 \to 13j^2 \geq 625 \to j \geq \sqrt{\frac{625}{13}}

[tranquil yew]

=tex \overline{WX}=3j \geq 3\sqrt{\frac{625}{13}}

[tranquil yew]

So that's the upper bound

[surreal bolt]

No it is a circle for sure. the 3:2 question

locus of points.

[tranquil yew]

what

It's asking for a length

[dark sparrow]

@tawny pewter??

can you clarify what the hell you were asking?

[upper karma]

hi I have a computer geometry math problem, can anybody here help me?

[tranquil yew]

asking to ask the question is frowned upon

[upper karma]

can I post an image to help ask my question?

[tranquil yew]

go for it

[dark sparrow]

yeah

[upper karma]

I have generated the bottom cylinder head of 12 vertices I know where each of these are, I'm trying to make a semi-circlular hood rotated 90 degrees to the cylinder in the axis of vertexes 3 and 9 I have a formula that lets me work out vertex 6 but I don't know how to work out vertices 4,5 7 and 8. The cylinder isn't flat with the ground and may be facing anywhere, can you tell me arbitrarily where vertexes 4 5 7 and 8 will be in relation to the bottom 12 vertices?

[dark sparrow]

okay first off, the plural is vertices 😛

anyway, you can find the coordinates of the center of that circle and thus find the vectors from there to points 3 and 9

and you can also find the vector from the center to point 4

so then the vector from the center to any particular point is v cos(θ) + w sin(θ), where v is the center-9 vector and w is the center-4 vector

[tranquil yew]

Can't you always just align your coordinate system with the two great circles

makes it pretty easy

[dark sparrow]

🤷

[tranquil yew]

like if point 2 was (x,y), point 4 would have to be (y,x)

err

wait

yeah in their respective planes

[dark sparrow]

(x,y,0) to (x,0,y)?

:p

[tranquil yew]

yes

[upper karma]

what is cos(theta) in this instance?

[dark sparrow]

θ is the angle between the center-9 vector and the vector from the center to a point on the semicircular hood

[upper karma]

is this the angle between the center point and the point I'm trying to get so point 4 would be cos(150) ?

[dark sparrow]

v cos(150°) + w sin(150°)

[upper karma]

so each point are 30 degrees apart right?

[dark sparrow]

are they evenly spaced?

[upper karma]

yes

like a clock

[dark sparrow]

then yes they are 30 degrees apart

[upper karma]

There's a lot of confusion about computers using Vector3, they are not the same as your mathematical vectors now are they?

[dark sparrow]

why wouldn't they be? lol

[upper karma]

the vertexes I have are stored as Vector3's which are essentially just a 1D array of 3 floating point values

[dark sparrow]

yeah so?

you can define vector addition and scaling by a constant just fine

like, [a, b, c] + [d, e, f] = [a+d, b+e, c+f] and k * [a, b, c] = [ka, kb, kc]

[upper karma]

oh so vertex[4] = abs(vertex[9]-vertex[3])*cos(150)+ (!?center4vector)*sin(150

you said that you can also find the vector from the center to point 4, how?

[dark sparrow]

okay first off

no it's not abs(vertex[9]-vertex[3])

if anything, it's (v9-v3)/2

that's the center-9 vector

also it looks like you already have the center-6* vector

sorry

i meant 6, not 4

[upper karma]

xyz position

[dark sparrow]

anyway i was referring to the normal

[upper karma]

which is a vector 3 sure

so would it be vertex[4] = ((v9-v3)/2) x cos(150) + ((v6-(v9-v3)/2)/2) x sin(150) ?

and for vertex[5] is it ((v9-v3)/2) x cos(120) + ((v6-(v9-v3)/2)/2) x sin(120) ?

and for vertex[7] = ((v9-v3)/2) x cos(60) + ((v6-(v9-v3)/2)/2) x sin(60) ?

vertex[9] = ((v9-v3)/2) x cos(30) + ((v6-(v9-v3)/2)/2) x sin(30) ?

is that all right?

or have I messed up?

[dark sparrow]

errh

add (a+b+c+d)/4.0 to all of these

[upper karma]

just add it on at the end?

[dark sparrow]

yes

[upper karma]

This is my final bit of code you think it'll work:

//j4
_output_vertices.push_back( (((_output_vertices[branch+9]-_output_vertices[branch+3])/2.0)*cos(150.0)) + ( (((_output_vertices[branch]+_output_vertices[branch+3]+_output_vertices[branch+6]+_output_vertices[branch+9])/4.0)+(Plane(_output_vertices[branch],_output_vertices[branch+3],_output_vertices[branch+6]).normal)-((_output_vertices[branch+9]-_output_vertices[branch+3])/2.0)/2.0)*sin(150.0) ) + ((_output_vertices[branch]+_output_vertices[branch+3]+_output_vertices[branch+6]+_output_vertices[branch+9])/4.0) );

[glossy minnow]

Referring what?

[upper karma]

DARN RADIANS

[dark sparrow]

btw

you can enclose code blocks in triple backticks

like this

[upper karma]

''' this better '''

better

yay I can code in discord now, thanks root 2

btw ur code didn't work

[dark sparrow]

my code?

[upper karma]

I even converted the degrees to radians

root 2 something's not quite right

[dark sparrow]

it was up to you to implement it lol

[upper karma]

well my interpretation of your math something's not working out

I mean obviously I've made some new vertices and I've followed your math as faithfully as possible but the end result doesn't match up what you described

_output_vertices.push_back( (((_output_vertices[branch+9]-_output_vertices[branch+3])/2.0)*cos(2.617994)) + ( (((_output_vertices[branch]+_output_vertices[branch+3]+_output_vertices[branch+6]+_output_vertices[branch+9])/4.0) + (Plane(_output_vertices[branch],_output_vertices[branch+3],_output_vertices[branch+6]).normal) - ((_output_vertices[branch+9]-_output_vertices[branch+3])/2.0)/2.0)*sin(2.617994) ) + ((_output_vertices[branch]+_output_vertices[branch+3]+_output_vertices[branch+6]+_output_vertices[branch+9])/4.0) );

thanks for the backticks help

I think I've probably made a schoolboy error somewhere

perhaps you can help me?

you're obviously a lot cleverer than me @dark sparrow there must have been something lost in translation

anyway I see you're busy with your game so I won't disturb you

[dark sparrow]

not a game but i am busy with other things

also i really don't want to debug code right now

[upper karma]

Is it possible, dare I say you might possibly be wrong?

[dark sparrow]

i'm pretty sure i am right but not getting my point across

[upper karma]

lost in translation then ok perhaps later when you are less busy

[upper karma]

@dark sparrow Hi I managed to generate 5 vertexes along the line but I'm still a little confused about how you make a vertex go "up" along the normal, how do you do this?

Basically I used : (1.0*((v2-v1)/6.0))+v1 and (2.0*((v2-v1)/6.0))+v1 and (3.0*((v2-v1)/6.0))+v1 ...

[surreal bolt]

kk I can help I think.

so is the "bottom" circle centered at (0, 0, 0)?

[upper karma]

@surreal bolt no the cylinder may be facing ANYWHERE

so now I have 5 vertices splitting the cylinder

[surreal bolt]

okay. Gotcha.

So it can be centered anywhere and facing any direction.

[upper karma]

basically say there were a laser beam along a normal and I wanted to pop a vertex distance 1.0 say along that normal how would I do that?

basic basic vertex math

vector math

[surreal bolt]

yeah. I am making sure. So you have a circle in a plane and you have the center of the circle and the normal to the plane.

[upper karma]

yeah the circles in a plane but it may be facing anywhere

[surreal bolt]

no problem.

So do you want the semicircle to always come out of the base circle at points 3 and 9 on the base?

[upper karma]

ok this is where it gets confusing... the old tree's top cylinder had 13 vertexes, because of something known as UV mapping and its like a flat piece of paper rolled into a pipe but you don't see the edge

even though it has 12 sides

the 0th and 13th vertex are in the same place

this group is called [branch]

[surreal bolt]

ah good

[upper karma]

so [branch+0] has the same xyz as [branch+12]

now I create a new set of 13 vertexes along one half of the cylinder 7 recycle the xyz of one half of the old [branch] vertexes

but vertices 4 5 6 7 and 8 are in new positions to split the branch into two

Here's the same branch in wireframe mode from behind

this way these new 5 vertexes in an arc will allow the creation of two new 12 faced cylinders

Does that make sense?

[surreal bolt]

no. But hold up re-reading

[upper karma]

here's a little diagram if that helps explain things easier

[surreal bolt]

cool I was going off this other diagram.

Okay so just to make sure you want a semicircle, you have the radius, and could find the center and have a vector that goes through the center and 90-degree mark of the semicircle you want to draw?

[upper karma]

well I figured that the distance from the midway point between 2 or 15 or 4 is the same

likewise the distance between 1 and 11 and 5/18 is the same

and the distance between 0/12, 6/19, 13/25 are the same

[surreal bolt]

hmm in the circle?

[upper karma]

I already got the hood, I just want to pop it out perpendicular to the cylinder

formulaically

[surreal bolt]

Yeah that works too.

[upper karma]

ok tell me very very slowly how do I create a vertex B that's a distance of D away along the normal N of vertex A ?

like pretend you're explaining this to a five year old with down syndrome

[surreal bolt]

Usually you would normalize the N vector and multiply by D and add to A

so you have a direction vector.

But it would be really nice if its length were 1.

[upper karma]

well I'm going to use distance_to which gets a floating point value of between two points

[surreal bolt]

kk hold up. 😃 You have this N vector.

And you want to know if I go D distance along this N vector what would be the distance_to?

[upper karma]

bingo

no

[surreal bolt]

well it is an intermediate step.

[upper karma]

what would be the xyz position

I go D distance along N vector from vertex A.xyz what is B.xyz?

b.xyz = ?

[surreal bolt]

kk what are the components of the N vector?

(na, nb, nc)?

[upper karma]

they are 3 floating point values

i don't know what they are going to be

but they are derived from the cylinder so they face outwards

[surreal bolt]

okay so pick your poison I'll call them (Nx, Ny, Nz). Is that okay?

[upper karma]

fine by me

[surreal bolt]

kk so what is the length of N at this point?

[upper karma]

Call it D

[surreal bolt]

=tex |N| = \sqrt{ (Nx)^2 + (Ny)^2 + (Nz)^2 }

[upper karma]

good old fashioned pythagorus?

[surreal bolt]

yessir 😃

=tex \frac{N}{|N|}D + A

[surreal bolt]

Is D a vector?

or a scalar?

[upper karma]

??

[surreal bolt]

Like D is just the distance between A and B?

[upper karma]

ok and what about distance D?

scalar

yes

single float

[surreal bolt]

yeah then that equation is correct (but worry when the denominator is zero)

Um looking up how to Tex vectors.

[upper karma]

its extremely rare that a normal is 0,0,0 I've tried putting that in and all my meshes look black

[surreal bolt]

okay well here goes my final edit and equation hopefully.

=tex \frac{\vec{N}}{|\vec{N}|}d + A

[surreal bolt]

Everything with 3 components is capitalized.

[upper karma]

implying vector3s rather than scalar1's

[surreal bolt]

well technically A is a displacement vector from the origin.

[upper karma]

yep

[surreal bolt]

so I could put a vector symbol on that as well.

[upper karma]

vertex xyz is stored as vector3

[surreal bolt]

try it out.

[upper karma]

Thank you for your help sir!!!

[surreal bolt]

come back if it doesn't work. :). I make mistakes.

otherwise, you are welcome.

[upper karma]

@a one a two#9669 are you still about?

@a one a two#9669 I've nearly got it, its just that the vertices are off by one if that makes sense

[terse marten]

Nice

[surreal bolt]

sorta back lol should be sleeping

what is going on?

[upper karma]

yeah one of my vertexes just won't go to the right place

@surreal bolt https://paste.ee/p/huBpD this is the code I use to generate just taht little tip at the end

you don't want to know the rest of the tree trust me

its like thousands of lines of code

@surreal bolt so yeah I moved all the vertexes along one and all but the top one seems right its the vertex j8 if that makes sense

I couldn't see anything that was obviously wrong, the only thing I could think of was the normal is being derived from a reflex angle triangle

[surreal bolt]

@upper karma hmm text tag. are ya out there? Just PM me so we don't annoy the Geometry channel.

[upper karma]

@upper karma why godot and not unity?

[upper karma]

@upper karma do I seriously have to dignify that with an answer?

in short because unity blows

I'm seriously asking so yeah

Because from what I've seen Godot is very similar to Unity

I'll give you 10 reasons Godot is better:

1. MIT license

2. Open source

3. Friendly dev team, very helpful

4. no licensing hell of gibmedats

5. Global illumination

6. No invasive app forced upon your system in order to do any development

7. Transparency, you know what you're building

What invasive app?

8. Poolvectors allocating memory to heap efficiently

9. low memory consumption

10. Its free to have the full features, unlike unity's fremium methodology

godot has no invasive app

How does unity?

Brb

[ember kraken]

Where the plot goes?

[fallen ivy]

Don't ask to ask, just ask ^^

message about S&W?

and I've heard all those words but I wouldn't be able to answer

you can, but it might be better to dm?

It's an awesome show definitely

[upper karma]

can anyone help me with a direction problem?

I'm using some code to draw these semicircles, thank you last time for your help but one's the correct way around the other is wrong, is there a way I can easily change the direction of the one that's facing inwards?

This is the code that I use to generate the semicircles: https://paste.ee/p/cG6Ie

[surreal bolt]

yes

the current normal vector is (x, y, z) make it point (-x, -y, -z)

do you also need to know how to detect this?

[upper karma]

sure

@surreal bolt my hunch is to use L1 and L2 normal vectors to get a sense of which way is "outbound"

But I don't know how to combine L2 with N2 so it flips direction

And combine L1 with N1 in the same way so it stays the same

[surreal bolt]

ah you need to use dot product (or basically law of cosines) to find out what the angle is between consecutive generations of branches. If the angle isn't less than 90, you are facing the wrong way.

[upper karma]

formula?

I tried changing the x coords of the branch2's circle and the hood faced the correct way but then the normal of the branch that spawned out of hood2 was all wrong

so the answer lies with some way of "detecting" which way is going away from the tree

[surreal bolt]

I'm a bit too tired. 😦 Catch me tomorrow?

[upper karma]

no problem!

[surreal bolt]

kk

[upper karma]

Is there a way to "flip" axel A 180 degrees based on the direction of L1?

[surreal bolt]

yeah use -L1 ... I think.

[terse marten]

Hrmmm...

[upper karma]

@upper karma thanks for the godot recommendation!

[upper karma]

@upper karma no problem, I'm having a lot of fun making stuff in C++ with it, trying to get my head around 3d vector math, its not a bad way to learn

sounds great

[upper karma]

Say you have two random triangles in 3D space. I am using the sum of (C-A).(B-A) My gut tells me that if you flip B and C around it gives you the same result. I mean i know obviously that if you move A into a different corner the normals will change but can you just confirm for me that my hunch is indeed correct that if the two corners are labelled B instead of C and C instead of B the results are the same?

I need to know because if I'm right I can save myself a lot of bother coding, thanks in advance

and no I am not sitting in an exam hall and no I am not trying to cheat on a test.

i.e are vector dot products order sensitive or not?

[hoary spoke]

dot products are not order sensitive

but maybe don't mix up whatever you're subtracting

[granite seal]

https://cdn.discordapp.com/attachments/269861536867155968/335038923854184448/Screen_Shot_2017-07-13_at_10.44.48_pm.png

can someone tell me how to solve the last one?

what would the formula be?

[dark sparrow]

it's a square minus four quarter circles

[granite seal]

ye but how can I find the answer?

12.6mm would be the entire side

[dark sparrow]

yes

so what'd the area of the square be?

[granite seal]

ye minus quarter circles

[dark sparrow]

can you find the area of the square?
can you find the total area of the quarter circles that have been cut out of it?

[granite seal]

I can find the entire square area but how to find that circle area

[dark sparrow]

what are the radii of all of the circles?

[granite seal]

😮

wow got it I guess

I got a way to find the area of the circle. firsst find their radii that is 4.2/2 = 2.1 then use Pi*r^2 that is pi * 2.1^2 then we could just do 12.6^2 (complete area of the box) -pi * 2.1^2 (circle's area)

is this the right way?

[dark sparrow]

no!

their radii are 4.2 mm!!!

not 2.1!

[granite seal]

oh

[dark sparrow]

why did you decide they were 2.1?

[granite seal]

because I split the entire box into 4 small boxes and took one

[dark sparrow]

what?

[granite seal]

uh wait nvm >_> now that I see your answer is right I guess let me try it

ok yeh that's right. Thanks a lot! :>

[stray mesa]

i need some help with sperical coordinates

wrong chat

[foggy oxide]

@granite seal when you're working with this kind of geometry i recommend you to try splitting the figures in a way you know how to calculate the pieces

that's always a good strategy xD

[granite seal]

Okay, thank you guys ^_^

[queen wigeon]

I've got this neat area puzzle I made a few whiles ago. As you can see, the requested area is that of the shaded triangle. I can elaborate on any questions you have about the image. (Being the one who accidentally made and solved it, I can say it's harder than it looks.) There IS an algebraic representation for its exact area, but I'll accept a decimal approximation, too.

[devout oak]

(4-b)^2=3^2+a^2
a^2=(2-c)^2+d^2
b^2=d^2+c^2

Solve for a, multiply by 3, divide by 2

[upper karma]

Cant we find top angle with al kashi or smth like that and then use trig@in yellow triangle ?

[dusk snow]

yeah but it's kinda weird using transcendental functions two times to get an algebraic answer anyway

[upper karma]

True

[azure storm]

there is probably a smart way to extend a line

maybe to the top right corner

but i'm not good at either trig or geometry, so ....

[dusk snow]

the al kashi method is the easiest though

for me

[azure storm]

yep it is, but your algebraic solution will be ugly

made this problem on geogebra

i know the answer LOL

[queen wigeon]

oh woops!!! computers are always one step ahead

[azure storm]

ok i might have an idea of solution but it involves solving trig equations

ok my solution is too hard nvm

[queen wigeon]

yeowch

I can give a hint that leads to my method of finding the algebraic expression, in dms or something

yeah i'm pretty sure some simple trig could suffice if done right

.....that is way off the mark!

oh wiat

wait hold on, I misread the tilde as a negative sign

not AS way off.

i'm a sillyhead

you're close

[dusk snow]

well if you have a negative area you have a problem lol

[queen wigeon]

but not on target

sure sure

BAM!

[dusk snow]

2.489775 ?

[queen wigeon]

👏 ya got it

[dusk snow]

9sqrt(15)/14 ?

[queen wigeon]

god job y'all two

oh

you got the algebraic answer!

congratulations

[azure storm]

proof?

[dusk snow]

trig

[azure storm]

2 AlKashi in a row?

[dusk snow]

m gonna try and translate it geometrically

[azure storm]

trig is fine 😛

[queen wigeon]

i don't know a lot about this problem so i'd like to see how trig ties into all this

[upper karma]

I have angles using Law of Cosines but I'm not finished 😛

I'm a slow worker

[queen wigeon]

yeah law of cosines is actually a really good method for finding the solution wow 😮

[upper karma]

That's what I did

Cuz you get top angle easily

[queen wigeon]

it turns out i'm the weird one when it comes to how i solved it, huh.

anyways

So how I solved it (no nonsense I swear), I used Heron's Formula (or at least, a close relative so to say) to find some fake positions for every point on the 2-3-4 triangle, and with some linear algebra and geometric-algebraic identities (particularly about perpendicular lines), I found the positions of the points on the shaded triangle, and tied it all together with pythagorean theorem and the 1/2bh equation.

I'm self-driven in my mathematical endeavors :p

however

given you could find it that easily with some simple tirg

i will concede the point that my solution to it is absolutely outrageous

anyways i hope you all enjoyed that puzzle

i like to give Heron's formula some love, it's really cool for both area and constructing triangles ❤

[dusk snow]

I think you can almost divine out the formula by knowing when it has to be zero

[upper karma]

Finally done. 2.4898

told you I was slow

[queen wigeon]

don't worry, you did good

[dusk snow]

actually the ratio between the inner triangle and the big triangle is a rational fraction of the sides

[fiery moon]

Is it possible to construct a 75 degree angle in only 3 lines given the available tools?

[azure storm]

yes, because all levels of this games are solvable 😛

[fiery moon]

.< been trying to find a solution for hours

[azure storm]

solutions are on youtube, but its better to find by yourself 😛

just looked at it, and its not an easy one... But its kinda common "trick" past like the 4th "chapter"

[cunning zinc]

what is that

[azure storm]

euclidea iirc

[cunning zinc]

interesting, never heard of it

[uneven zephyr]

http://prntscr.com/fvtfxe Can somebody help me

I have the length XY as 2 * sqrt(33), and I'm quite sure this question becomes a similar triangles one, but I can't find it :/

ty

[final prairie]

Interesting

Assume X is at (0, 0, 0) and find the coordinates of Y

@uneven zephyr ^

[uneven zephyr]

Yeah idk

I'm pretty sure it's similar shapes.

[final prairie]

Wait

I misunderstood the question

completely

sorry lol

[uneven zephyr]

Like I have the triangle, and I make a smaller one with the same angles

[final prairie]

I understand that now

mhm interesting

[cunning zinc]

it's space!

[uneven zephyr]

we're not meant to use calculators btw

[final prairie]

Let's start with finding where XY intersects the bottom face of the 2 cube

[uneven zephyr]

wait is it ratios?

like 1/3 of the part in the 3 cube should be equal to 1/10 of the whole line right?

[final prairie]

I don't think

[uneven zephyr]

that would give the answer A

[final prairie]

Find the equation of XY and the plane of the top face of 3

[cunning zinc]

oh I misread the question still lol

I thought it was from X to the cube wiith side length 3

[final prairie]

It's confusing

[cunning zinc]

it's just the part in the cube

[final prairie]

The newline at the word 3 is confusing hehe

PQE should u use 3d line equations? did u learn that stuff?

How come?

[uneven zephyr]

Yeah that's what I said @sour briar

Yeah I'm pretty sure that's right, ty. I have one more question

http://prntscr.com/fvtl5u

I have been working on this for like an hour and got nowhere.

I was thinking it may be the altitude but that didn't make sense.

[final prairie]

:o mhm

[uneven zephyr]

I may have something

nvm actually :/

yes

I know the sum of the areas of the triangles

[final prairie]

30+x+y=40+x+(5-y)

30+y=45-y

y=15-y

y=7.5

My guess

[uneven zephyr]

wait x's cancel so you have to divide the hypotenuse to make the perimeters the same

[final prairie]

I'll explain steps

30+x+y is the perimeter of first triangle

[uneven zephyr]

Dude that's not one of the options though?

Must be A,B,C,D or E

[final prairie]

no it's not the final answer

[uneven zephyr]

oh right

what's y then?

wait it divides into an isocoles triangle 1 sec

[final prairie]

y is the length from one of the veritcies to the point where the fence intersects

[uneven zephyr]

That should be 20

I think

30 + 50-x = 40 + x ?

where x is that length

and then you create an isocoles triangle with two sides of length 30?

and the other side is the length of the fence

You divvy it up into 30/30/x and 20/40/x triangles?

that works because 30 and 20 still sum to 50

Is there a ratio for an isosceles triangle?

[final prairie]

Interesting guess

[uneven zephyr]

hmm not sure

give different perimeters:

oone is 58.57, the other is 61.4

You work out point of intersection by:

30 + 50 -x = 40 + x

40 = 2x

x -> 20

http://prntscr.com/fvtqi5

but idk where to go from there

[dusk snow]

that triangle doesn't look isoceles at all o.o

[uneven zephyr]

This should be solvable without a calculator

How do you work out the angle without sin/cos/tan?

hmm

[upper karma]

Can't you just use law of cosine to solve the left angle?

=tex \frac{\sin\theta}{50}

[upper karma]

You can't unlesss you have an angle , right?

[uneven zephyr]

I'm working on something

kind of substitution/pythagoras

[upper karma]

I think it's easier using law of cosines

why don't you do it then, it's not much to calculate

you can solve without a calculator it's just power of 2

oh

then what are you trying to do with law of sines? XD

No it's just all about the fractions

then it's just

=tex \frac{\sin\theta}{50} = \frac{\sin\alpha}{40} = \frac{\sin\beta}{30}

[uneven zephyr]

(4 * sqrt(60x)^2 is 16x60x right?

ok ty

it is

[upper karma]

well whole point of law of sines is to have atleast one angle so you can show ratio and solve the other two angles lol

[glossy minnow]

Ooh geometry problem

Gimme

[upper karma]

:DDD

[glossy minnow]

GIMME

[upper karma]

I've used it too much now, I wish I had your fortune

@glossy minnow whats the area of a cone

[glossy minnow]

Lateral area?

[uneven zephyr]

Man this problem is crazy.

[upper karma]

Lateral?

[uneven zephyr]

It's gonna have a really simple solution I can tell

[upper karma]

What problem?

lmao

it's basically same thing but written in two different ways, cosine law is basically pythagoras theorem pimped and law of sines is about ratio

[uneven zephyr]

Who wants the solution?

[upper karma]

what's the problem even

[glossy minnow]

Of what problem?

[uneven zephyr]

http://prntscr.com/fvu15t

[glossy minnow]

Does it involve law of sines?

[uneven zephyr]

No

[glossy minnow]

Wait don't gimme

I wanna solve it

[uneven zephyr]

We forgot ratios existed when I posted this image:
http://prntscr.com/fvtqi5

[glossy minnow]

Wait a second

This is a geometry theorem

[uneven zephyr]

http://prntscr.com/fvu21e

AX / AC = AY / AD

Then you can create the triangle CXY and use pythagora's theorem

super nasty question

[glossy minnow]

Wait how did you get that ratio?

XY and CD aren't parallel

[uneven zephyr]

No they are I'm just bad at drawing @glossy minnow

[glossy minnow]

What is x

[uneven zephyr]

http://prntscr.com/fvu3m0 Here is the full solution.

It is a pretty nice solution

Just pretty difficult to discover lol

[upper karma]

I just thought I'd give a huge shoutout to all the mathematics team for all your help you've given me and I thought I'd share with you the fruits of my labor, I have successfully developed a 3D game procedural tree generator, no 2 trees are the same, its all done with some very clever vector3d geometry math

[sour sorrel]

oh nice, pretty.

[upper karma]

well I'm sure you math pros that taught me will think its very simple math really

[vapid kettle]

looks cool to me ~

not sure I'd qualify as the "pros"

[sour sorrel]

kinda reminds me of morrowind; ashy kinda landscape.

[haughty prawn]

Looks dope, I can't program for my life.

[sour sorrel]

^

[stiff heart]

visually, multiplying by i turns a point 90 degrees in the complex plane

what does raising an integer to the power of I do to it visually

[queen wigeon]

that's superbly done! i hope you can make good use out of it

[stiff heart]

Sigh

[haughty prawn]

Lemme try to help, one sec

[upper karma]

you could figure it out from this

=tex k^i = e^{i 2\pi \frac{\ln (k)}{2\pi}}

[upper karma]

if you squint hard enough, it might mean something

[haughty prawn]

Tbh, what he said ^

[upper karma]

I had a thought on this a long while back of like splitting fractions up

n^1, n^{-1}, n^i, n^{-i} etc

you can sorta imagine maybe making a fraction with 4 parts instead of 2

or infinitely many parts just n^{w} for w = e^{i 2pi k} basically but

I don't know how useful that is

[fallen ivy]

it's actually not that difficult, your formula obscurs the meaning

[upper karma]

enlighten us

[fallen ivy]

2^i = e^(i * ln 2) = (e^i)^(ln 2)

[upper karma]

I'm just trying to figure it out, not claiming to have any answers here

[fallen ivy]

or to be more succinct, a ln 2 radian turn around the unit circle

[upper karma]

seems good to me I guess

[fallen ivy]

in general you'll get for positive x that x^i = a ln x turn around the unit circle

[haughty prawn]

I wasn't sure whether to assume they knew euler's formula, but all is well that ends well. 👌

[upper karma]

actually kinda surprised that raising an integer to the i always makes |n^i|^2= 1

[fallen ivy]

well, not quite

[upper karma]

😉

[fallen ivy]

only positive integers

[upper karma]

and negative

[fallen ivy]

for negative integers you get e^(-pi) * (|n|^i)

[upper karma]

=tex (-2)^{-i} (-2)^i = (-2)^{-i+i} = (-2)^0 = 1

[fallen ivy]

?

what's special about that :P

[upper karma]

this is a counter example to your false claim