radius of black circle
#geometry-and-trigonometry
umbral snow
It wouldn't be too hard to coordinate bash this. You can call the bottom corner of the triangle (0,0). Probably not the most effective way but it's simple to do mechanically
upper karma
yes thats what i did
ie - (2 0) references how the centerpoint of the blue circle is halfway of the bottom side
so I just constructed those equations that restrict the 3 points to those circles
and the 4th equation is supposed to say they all have the same distance from an unknown centerpoint M
which is also the triangle centerpoint (5th equation)
but uh actually trying to solve them didnt work for me
maybe there's something missing or wrong
umbral snow
The big circle is tangent to each circle once each
upper karma
@upper karma perhaps making a big triangle from the points where the colored circles are tangent to the black circle could be helpful?
mossy vine
if you can work out all three points of tangency with the big circle then the (extended) sine rule will give you the radius
sweet heart
Is the left function continuous and the right not?
forest dove
@sweet heart on their domains of definition, both are continuous
sweet heart
Ah okay thanks, I get my mistake I think
lament swallow
A = {(x,y)| y = 0, 0 < x < 1} , is A open or closed?
I say its not open because there doesnt exist a inner point where B(x,r) is a subset of A, because y = 0, and u draw a circle any point in 0 < x < 1, u cant find r > 0 in y-direction
is this correct?
lament swallow
@tropic stirrup
lament swallow
ty 🤗 t!rep girl
loud cedarBOT
🆙 | Poppa has given @tropic stirrup a reputation point!
tropic stirrup
neon fossil
entrance?
serene dirge
What kind of geometry do you want to study?
wind warren
I need a good intro to topology text for undergrads
forest dove
Munkres is supposed to be good, but kinda long-winded
Check these out too: http://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf
@wind warren
clear haven
wind warren
@forest dove thank
upper karma
@wind warren age?
upper karma
I was thinking about this proof of the uniqueness of limits: http://web.mat.bham.ac.uk/R.W.Kaye/seqser/uniqueness.html
it relies on the triangle inequality, when is it possible to prove this without the triangle inequality?
silent canyon
Okay so correct me if I'm wrong
But doesn't the lone definition of limits guarantee uniqueness?
(yes I just forgot the definition)
umbral snow
You can prove that limits are unique in any metric space, using the triangle inequality, which is an axiom of a metric space. That's probably why it's used
I doubt you need it in R
upper karma
right, so specifically I should say I posted it here in topology channel because I am interested in cases where I don't have a metric
silent canyon
You can do it in other sets I just forgot the exact definition
upper karma
since then I know I am not ensured the triangle inequality anymore
silent canyon
Okay I won't comment here and then it turns out I said something horribly wrong, I don't know
umbral snow
If you don't have a metric, how do you define a limit?
upper karma
you need a metric space to have limits?
upper karma
I don't think that's true
umbral snow
I'm not actually sure, lol. Let me read and find out
silent canyon
The horror on my face when i realise that my professor taught that to me a week ago and I f*cking forgot
white harness
With a topological space, you can define limits.
upper karma
Yeah, I can imagine a topological space without a metric and having sets and imagining some kind of limit there
umbral snow
Oh DUH that's a cluster point
silent canyon
To define a cluster point you need a metric
(if I remember correctly)
To define limits you need the term of neighborhood
if you can't measure the distance from x_0, then you can't define a neighborhood
upper karma
I can imagine having a kind of limit, although maybe it's a looser notion of limit in terms of nested subsets or something I'm not sure though
silent canyon
so you can't have a limit
umbral snow
A is a limit point of a set S, if every open set that contains A contains a seperate point of S
silent canyon
cluster point?
umbral snow
Any set may have more than one limit point, of course
umbral snow
That's not really a "limit" in the same sense you were talking about on metric space
upper karma
I'm wanting to know the minimum topological structure needed to ensure that limit points are unique I guess
silent canyon
I'll ask my topology professor lol
upper karma
since I want to throw out the triangle inequality necessary
hopefully what I'm asking makes sense in spirit
since kind of the idea is a metric space is extra structure on top of a topological space
upper karma
I'll look into that more
white harness
You need the separation axioms.
upper karma
yeah that's sort of what I was thinking but I didn't want to say it
I think Hausdorff space is technically the minimum structure needed to have unique limits
but I wanted to see what people said first since I never thought about it before but trying to really unravel what depends on what
to me, something seems suspicious about uniqueness of limits depending on the triangle inequality, but then again maybe this is just bad intuition so I can't really say haha
pliant rivet
ive got a pretty dumb question that ive gotten pretty close to answering but i cant quite get it.
Let $$C[0,1]$$ be the set of continuous, real-valued functions on the interval $$[0,1]$$ and let $$d^:C[0,1]\times C[0,1] \rightarrow \mathbb{R}$$ be the distance function defined by $$d^(f,g) = \int_0^1 |f(x)-g(x)|dx$$, for $$f,g \in C[0,1]$$. Define $$I(f) = \int_0^1 f(x) dx$$. Prove that $$I: (C[0,1], d^*) \rightarrow (\mathbb{R},d)$$ is continuous.
charred spearBOT
pliant rivet
ive gotten 2 pretty decent attempts but the first makes a logical mistake along the way and the second has the abs outside the integral instead of inside
here's my better attempt:
Let $$x\in \mathbb{R}$$ be arbitrary. Then for some $$\epsilon>0$$ we have that
\begin{equation*}
I^{-1}(B(x,\epsilon)) = \left{f\in C[0,1],\Big|,x - \epsilon < \int_0^1!f(t),dt < x + \epsilon\right}
\end{equation*}
Subtracting from the inequality, we have that
\begin{equation*}
-\epsilon < \int_0^1!f(t),dt - x < \epsilon \Longrightarrow \left|\int_0^1!f(t),dt -x\right| < \epsilon
\end{equation*}
charred spearBOT
white harness
$$|I(f)-I(g)| \leq \int\limits_0^1 |f-g|$$.
charred spearBOT
pliant rivet
ah okay i see
thanks!
i had something similar to that my first attempt but i jumped the gun and moved my abs around
timber hinge
=tex SMN + LMS = LMN \ (2 + 9x) + 50 = (12x + 16)
charred spearBOT
timber hinge
@pseudo grove
pseudo grove
Ty dad
timber hinge
Yw son
onyx beacon
180
onyx beacon
yes alright thank you
i knew it was an equation
so would i combine the two expressions and equal it to 180 or is it just the one in bold
timber hinge
Both
Since both make up 180
Then once you have the value of x from said equation you can use it to calculate the angle in bold
onyx beacon
😩
shadow prairie
Very nice
gritty flare
@earnest storm come here
earnest storm
amber raven
Find a non housdorf space such that each point has an open nhbd homeomorphic to U^n = n dimensional disk for n = 1 or 2
<@&286206848099549185>
Line with 2 origins is non housdorff
umbral snow
I'm dumb when it comes to topology, but aren't hausdorff spaces preserved under homomorphism?
And Uⁿ is hausdorff, no?
umbral snow
OH but the neighborhoods have to be hausdorff
Derp
The topology itself doesn't have to be
amber raven
Yeah
umbral snow
Well here I am not helping
amber raven
I was thinking about half plane with a tail
umbral snow
But yeah line with two origins is non-hausdorff. Everywhere other than zero it looks like the interval
amber raven
Yeah
But we wanna mix 1 and 2 dimensional topology to create a space that is nonhousdorff
I don't wanna cheat and look it up XD
umbral snow
I'll think about it too. I'm not sure atm
There's got to be some properties that make this easier
Is this disc open or closed?
umbral snow
@amber raven
Then it has to be homeomorphic to Rⁿ too. Not sure if that helps though
upper karma
Help me please /\ /\ /\ 👆 👆 👆 👆
surreal star
Someone pleaee help me
🙏🙏🙏
<@&286206848099549185>
Pleaaasseee
Breh
Halp
Me
surreal star
Yes
deft ridge
basically, you know how DBC is 90 right (100-10)
steady hull
ok lol or he can do it
deft ridge
XFE = EFY + XFY = 70 degrees 20 minutes, + 19 degrees, 40 minutes
surreal star
It means 70 1/3 deg
deft ridge
20 min + 40 min = 60 min = 1 degree
which is 90 degrees
so the 2 equal each other
sorry
surreal star
Everything tbh
deft ridge
ok so you know what you're trying to find right
or rather, what you're trying to prove right
surreal star
Yea if dbc is congruent to xfe
deft ridge
mhm and so the way to prove congruence is to provoe that they're the same angle
surreal star
Ye
deft ridge
and so you find the degree of each angle and if they're equal to each other, you know it's congruent
so how do you find DBC
deft ridge
ABC - ADB
surreal star
90 deg
surreal star
K
exotic patrol
I need help on my math homework
and i cant figure this one out
Someone help me :CCC
exotic patrol
,...
shell fable
I’m lost
little osprey
...
upper karma
...
little osprey
@exotic patrol Well I would help you if you were still online.
grand atlas
AB/BF = AC/CG
12/BF = 24/14
12/BF = 12/7
BF = 7
neon fossil
hahaha
upper karma
FUUUUU
clear haven
ono
neon fossil
clear haven
dont really know who to get mad at tbh
neon fossil
you have been dabbed on
upper karma
¯_(ツ)_/¯
neon fossil
upper karma
Help Me!!!
neon fossil
oh it's 10
neon fossil
it's 40
neon fossil
upper karma
HELP ME! HELPF ME!
neon fossil
oh that's 30
upper karma
HELP!!!
neon fossil
false true false false true true
runic cobalt
mystic shadow
all the angles on a straight line add up to 180°
so you will make two equations using this fact
exotic patrol
help.
upper karma
same
hearty barn
@upper karma for triangles to be similar, the ratio of sides must be equal
In particular, If the triangles MPR and ADN are similar
upper karma
So it's not similar
hearty barn
=tex \frac{MP}{PR}=\frac{AN}{AD}
charred spearBOT
runic elbow
I need to solve for POS but I don’t know how to solve these types of problems? Can someone help?
shadow anvil
What other information is given to you
@runic elbow
So do I
but it's also simultaneous so there must be some relation between them
charred spearBOT
rugged moat
You have 3 equations and 2 variables, solve them
runic elbow
That’s the problem I don’t know how to solve for them from there
rugged moat
$$3x-4y=x-y$$
charred spearBOT
charred spearBOT
rugged moat
So using the second equation, we get $$x=2y-10$$
charred spearBOT
charred spearBOT
rugged moat
=pup Solve 3x-4y=x-y=y-10
rugged moat
Yup
uncut zephyr
upper karma
I did matrix transformation, and the result vector is supposed to be 20, 10, 10, 1
However, my result is
-1642983424
0
0
0
written in C
I am modeling it off of http://www.opengl-tutorial.org/beginners-tutorials/tutorial-3-matrices/
What am I doing wrong?
white harness
struct vec4 transform(struct mat4 a, struct vec4 b) seems incorrect. If c1 is a column vector and b is a column vector., then the multiplication is incorrect.
@upper karma .
white harness
Segments or lines ?
white harness
First solve it for lines intersection then check if the point is in both segments.
upper karma
first for lines?
for lines is just the cross product right?
i mean, if they dont intersect the cross product will be 0
white harness
But you want the coordinates of the point.
white harness
The determinant should solve that question.
upper karma
but i think i need the point
white harness
Use the lines method.
upper karma
what method?
white harness
Compute the lines equation.
upper karma
imposible.
just think of it
how many points does a line/segment have?
and how many points does a computer have?
white harness
Let A, B, C, D be points. Find intersection of [AC] and [BD]. If ABCD is a convex quadrilateral then there is an unique solution.
upper karma
b
hybrid tartan
I have a 15x20 rectangle and I'm trying to find the area of the shape such that any two points inside of it are at most 30 units away from X.
hybrid tartan
rectangle*
fixed
wait lemme label the vertices
I started by drawing the outline of the shape (in red).
Starting from X, we can travel in XY until we get to Y. Now we are 15 units away from Y.
Then, we can continue traveling in YZ until we are 30 units away from Z. Then, we will have travelled some distance x in YZ, which I calculated to be sqrt(15) using the Pythagorean Theorem.
We can also do the same process with W and Z instead of Y and Z, which gives us x = sqrt(10).
The rest of the outline would then be part of a circle's outline - more specifically, the one with its radius equal to the sides of the square in blue.
For the horizontal side, we get a length of 20-sqrt(15), and for the vertical side, we get a length of 15-sqrt(10).
Obviously these must be equivalent, so we have 15-sqrt(10)=20-sqrt(15).
Substract 15 from both sides, and we get sqrt(10)=5-sqrt(15).
However, sqrt(10) ~ 3.16, while 5-sqrt(15) ~ 1.13.
What did I do wrong?
hard gale
so your goal is to get the area in orange right?
white harness
$$\pi30^2\cdot\frac{\frac{\pi}{2}-2\arccos\left(\frac{15}{30}\right)}{2\pi}$$.
charred spearBOT
Rendering failed. Check your code. You can edit your existing message if needed.
white harness
$$\pi30^2\cdot\frac{\frac{\pi}{2}-2\arccos\left(\frac{15}{30}\right)}{2\pi}$$.
charred spearBOT
hard gale
(the angles are not the same since it's a 15x20 square) @white harness
white harness
$$\pi30^2\cdot\frac{\frac{\pi}{2}-\arccos\left(\frac{15}{30}\right)-\arccos\left(\frac{20}{30}\right)}{2\pi}$$.
charred spearBOT
hybrid tartan
No, I'm trying to get this area:
hybrid tartan
but i ended up getting sqrt(10)=5-sqrt(15), which is wrong
covert verge
hallow hornet
can someone help me with topology
eager pendant
hallow hornet
alright so
if i were to have an set U that is a subset of X in a topology T of X generated by some basis
i could show that U is open
by showing that for any element in U
there is some basis element containing x
which is contained in U
right
?
amber raven
I think so
Yes
Prove that U is the union of all basis element that contain a point of U and are contained in U
I love topology, lessgetit
Topology ❤️
Makes me gay
@hallow hornet
hallow hornet
ok thank ❤
balmy vigil
im going to kill myself whenever i get to geometry arent i
crimson isle
If we have angle X, and the length of A, how would we determine the length of ?
errant ether
the tangent of x is the ratio of the side opposite of x to the one adjacent to it
what is that in this case?
unique tulip
Tan(x)= ?/A
mystic shadow
@unique tulip you have that backwards, tan(x)=?/A
crimson isle
I vaguely remember that.
crimson isle
Thanks guys, I took geometry like 2 years ago, it didn't stick well.
unique tulip
someone HELP me pls🙏🙏🙏🙏
shell fable
Can someone help me with number 5
It doesn’t need to be exactly right
I just need an idea of what to put down
I’m fucking lost lol
sly narwhal
Any three angles of a triangle measure up to 180 degrees.
clear haven
animorphs were the shit
narrow sleet
I wanna ask Is a cylinder a prism?
marsh beacon
@narrow sleet no?
narrow sleet
Why?
lunar wind
@clear haven hell yeah Tobias was such a g
distant sonnet
narrow sleet
Most of uz?
upper karma
Circle-based prism, why not 🤷
oak steeple
who doesn't?
hard gale
@steady sleet doesn't
sudden helm
I don't know if this belongs here but how to draw orthogonal trapezoid plane in room. (Hope I explained clearly)
this is rectangle for example
tribal wind
you didn't
sudden helm
didnt explain clearly? @tribal wind
tribal wind
right
sudden helm
okay, like imagine that you are looking at the object (trapezoid) from other point of view. not from up but like diagonally
tribal wind
difference in what?
nimble bay
circumference
tribal wind
oh it's calculus right?
real light
Can anyone help me with a problem? >~>
tribal wind
NO
real light
Oh...
little osprey
@real light Just post it.
real light
Ah! I got help already
little osprey
My disappointment is immeasurable and my day is ruined.
polar mason
amber raven
Non- measurable sets
white harness
Axiom of choice.
amber raven
You have to choose your dissappointment from uncountably many boxes
uncut shoal
are the boxes distinguishable?
runic elbow
Can someone help me with this. I’m really not sure how to prove JL Congruent MO and if it’s side angle side
little osprey
Uhhhh
I mean it's basically given you everything.
Ok so after proving congruence
runic elbow
Yeah but what’s the proof called
little osprey
just say JL=MO(congruent sides on congruent triangles)
lavish thistle
little osprey
@lavish thistle He's not asking for the proof, he wants deduction
little osprey
Dw.
runic elbow
What’s the proof for JL=MO
little osprey
JL=MO(matching sides of congruent triangles)
umbral heron
CPCTC?
little osprey
Wtf is CPCTC.
umbral heron
Congruent Parts of Corresponding Triangles are Congruent?
runic elbow
Ok
little osprey
Is it a universally accepted term?
runic elbow
We were going to learn what that is tomorrow but he put it on the homework
little osprey
Then why would you keep it in a universal math's chat.
@runic elbow You're one step ahead.
You're probably going to learn more than just how to deduce.
little osprey
Where.
umbral heron
in school and practice presumably
little osprey
What?
No.
Look at a trapezium for example
The line segments are parallel
But they are not congruent.
If you mean
lines, 'no fixed end points'
Then technically yes.
runic elbow
What’s the name of the proof that shows line pq is equal to line st (<p and <t are equal)
Can I assume jk and kn are equal?
clear haven
isnt it abbreviated like congruent parts of congruent triangles are congruent
or something
warm tartan
Well if JMK = KLM
And you know for sure JKM=NKL
Then the last angles must also be equal because the angles must add up to 180 degrees
But i think you would need those lines saying that the sides are equal
clear haven
and the second set of triangles i dont think you can assume as congruent at most similar
runic elbow
Thanks
uncut shoal
in german you'd call those Z-angles (Z-Winkel)
my preferred proof would be via parallels and opposite angles. but you have to extend the lines in the picture for that
steady sleet
#❓how-to-get-help
a) min 15 min rule: wait 15 minutes before pinging helpers
b) please don't delete the ping
upper karma
ok
white harness
Sneaky ping.
cold plaza
So what field of math deals with higher-dimensional geometries (like analogs of circle, sphere in 4d and higher dimensions, analogs of planes and all that)? Is it 'topology'? For instance, what does 3x^2/sin(zs) = 32^(xys)/ln(k/w) even mean geometrically (or spatiallly) in this 6-d space? wtf is it? like a bended sphere or some shit?
tribal wind
@cold plaza you're looking for differential geometry, but things like topology/algebra can deal with higher dimensions too
cold plaza
@tribal wind thanks
wanton prawn
hey guys. I need to find the angel alpha dependent on the radee r and R as well as finding the distance a dependent on the radee r and R . I cant really find a triangle .
<@&286206848099549185>
uncut shoal
the tangents form similar right triangles
small r being rotated is a ruse
you can now use your trusty Vierstreckensatz 😉
wanton prawn
@uncut shoal ty buddy! Vierstreckensatz was the thing I needed!
mystic ermine
yeah i tried
clear haven
remember it's squared too
charred spearBOT
clear haven
same difference as the first one
jaunty horizon
well no that is a shit method
mystic ermine
jaunty horizon
c^2
mystic ermine
as far as ive gone
jaunty horizon
You forgot to square c
clear haven
it should be (3x)^2 and (5x)^2
hard gale
and you didn't square the 3x and 5x
jaunty horizon
lol ye
mystic ermine
okay so 64+9x=25x
clear haven
yes
jaunty horizon
no
mystic ermine
is it just an equation?
hard gale
9x^2 : 25x^2
white harness
Yes.
hybrid tartan
yeah
mystic ermine
but i still dont have any of the sides
jaunty horizon
$$64+9x^2=25x^2$$ yay
charred spearBOT
clear haven
solve for x and then plug x back into the triangle sides
jaunty horizon
Solve for x and then plug in
clear haven
then you'll have all of the sides you need
clear haven
np
mystic ermine
wait so i know 64 has the value of 16x's and to get the value of x i can just divide 64 with 16 right
hard gale
it's x²
mystic ermine
yeah get the value of x²
jaunty horizon
what
mystic ermine
okay i solved it thanks for the help!
clear haven
?
tribal wind
yes
mystic ermine
anyone can help me with thius
how do i calculate this i know its something with like parrerals or something
wild hamlet
I don't read german, can you translate it to me?
white harness
u=(2, 2), v=(4, -1).
mystic ermine
ohhhhhh
wild hamlet
u = -2v i think
mystic ermine
let me calculate this real quick
wild hamlet
hm
mystic ermine
okay am im disabled
wild hamlet
o yeah, the two w's in problem 1 and 2 are different from each other. makes sense
mystic ermine
i cant make this work
i feel asleep in the lecture when my teacher was going through this
wild hamlet
sqrt((2)^2 + (2)^2) * sqrt((4)^2 + (-1)^2) i think
there's some courses on khan academy and also 3b1b on youtube has some stuff on it
mystic ermine
well im looking at another example in my book
white harness
w=2u+v, w=(6, 3), length \sqrt{6^2+3^2}.
w=u-v, w=(-2, 3), length \sqrt{(-2)^2+3^2}.
mystic ermine
2u + 3v = (7,8)
i found that example
if thats correct wouldent this be 2u + v = (2,2) x (4,1) = (8,2) ?
is that right?
white harness
2(2, 2)+(4, -1)=(6, 3).
mystic ermine
where do u get the -1 from?
white harness
It's 'going down'.
mystic ermine
still confused why is it going down?
white harness
You need a basis to write those coordinates.
pliant rivet
yo this came up on a problem set and i just want to make sure i have the definition right: a metrizable space would be some underlying set endowed with the topology generated by an arbitrary metric, right? so i'd have to do this with the axioms of a metric instead of just saying "let X have the discrete metric" and proving it that way
pliant rivet
ya actually that's a dumb question i got it lol
twin prawn
What do you know about angle 1 + angle 2?
upper karma
sorry i had to go i know that there supp
rotund orbit
@upper karma cesyou know what supplementary means right?
assume angle1 = x
angle1 = angle4 right?
so angle1 = angle4 = x
now express angle 2 in terms of angle1 (or x) and express angle3 in terms of angle4 (or x)
use the definition of supplementary angles
analog dust
1+2=3+4 <=> 2=3 you're welcome
upper karma
@upper karma if supplementary angles congruent, then the angles are congruent
upper karma
tfw shadow ping
timber hinge
upper karma
my 1D mind
rugged moat
You cannot truly experience anything really
upper karma
ay who you callin stoopid
rugged moat
Eyes are 2D
timber hinge
Negative dimensions when
upper karma
@upper karma Such is life
sick veldt
hey could someone help me with a geometry question
twin prawn
upper karma
Hello
May someone help me study geometry
I just need the definitions to all the postulates
guys
imagine i have a set of points
all with x >= 0 and y >= 0
i can sort them from the origin going to polar form and see the angles
but how can i sort them if instead of being the origin is a generic point a,b?
hard gale
so you know the (x,y) coords of the points in the set right?
upper karma
imagine they are sorted like this
this is the easy way
but i wanna sort them like this
hard gale
yeah 1sec on my whiteboard
so a point (x,y) in the red basis has coordinates (x-a,y-b) in the blue basis
@upper karma
upper karma
okey
so this
could work?
Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
float angle = atan(y/x);
if(x < 0 && y >= 0) angle = PI/2 - angle;
else if(x < 0 && y < 0) angle += PI/2;
else if(x >= 0 && y < 0) angle *= -1;
return new Point2D(mod, angle);
}
ok if u dont know programing
my point center (x,y)
is somewhere
hard gale
so this refers to the point you're ordering right?
hard gale
seems right for the logic yeah
hard gale
well idk for the syntax :/
upper karma
no, the syntax is not the problem
are the operations
aaah
i see my errors :P
i think
if(x < 0 && y >= 0) angle = PI/2 - angle;
else if(x < 0 && y < 0) angle += PI/2;
else if(x >= 0 && y < 0) angle *= -1;
there are my errors
since i am not working on (0,0)
i need to update the edges
hard gale
ah yeah duh me
hard gale
yeah replace all the xs by center.x (same for ys)
hard gale
idk
upper karma
i know that if they are on 1st quadrant
angle is angle
if it is in second quadrant?
hard gale
i just supposed the atan function was broken on that language
hard gale
yes ik
if(x < 0 && y >= 0) angle = PI/2 - angle; isn't it pi-angle here?
(with the center changed ofc)
upper karma
PI is pi
hard gale
i guess you just have to add PI whenever at least one of the coordinates is negative
hard gale
i guess you just have to add PI whenever at least one of the coordinates is negative try this
hard gale
what's your language? (i want to see how the arctan function behaves on there)
upper karma
java
oh
maybe thats the problem
i am adding pi
and maybe i dont need to
aaaaaaaah
it returns values between -Pi/2, Pi/2
actually, what is that?
a full loop from 180º to 180º but passing first over 0º?
hard gale
you mean the graph of arctan?
hard gale
for you just have to add PI whenever at least one of the coordinates is negative
upper karma
x >= 0 and y >= 0
x < 0 and y >= 0
x < 0 and y < 0
x >= 0 and y < 0```ah i'm dumb
upper karma
instead of 0 will be center.x and center.y
hard gale
x >= 0 and y >= 0, x >= 0 and y < 0 : for those two cases atan returns the good value
hard gale
for the two others, just add PI
hard gale
tan(x+pi)=tan(x) this
hard gale
yea
hard gale
i am adding pi indeed
upper karma
in both cases
hard gale
yas always add
upper karma
why? on the black one u need to minus
hard gale
the angle which will have the same tan as black on Q2 is my blue on Q4
honestly just try
upper karma
aaaaah
true
:P
so
x < center.x && y >= center.y
x < center.x && y < center.y
for those i add pi?
upper karma
...
it isnt working
again...
mmm
i dont know why
but it fails in some cases :P
hard gale
ie when the point is right on the y-axis
cause you'll get some infinity shit going on
but does it work for some at least?
upper karma
it does i think when they all are on the left side
or on the right side
if they are in between...
hard gale
yea das ist the problem
well just do a case for numbers that are right on the y-axis (i mean at x==center.x ofc)
like this basically
upper karma
hard gale
and maybe handle the case when the point is exactly at the center
upper karma
this is how it is sorting them
upper karma
:)
Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
float angle = atan(y/x);
if(x < center.x) angle += PI/2;
return new Point2D(mod, angle);
}
so how should i change this? u.u
hard gale
yes you should add pi
hard gale
what does it do? (the graph output i mean)
hard gale
and you should add 2PI whenever it is in Q4 now that i think about it
hard gale
how worse? (yes i love that graph dope)
upper karma
hard gale
😮 shits i gotta sleep now
hard gale
One last thing : the error may be elsewhere
upper karma
@hard gale
i dont think so
look
i dont know where is the error
the mess up when they are on diff quadrants
white harness
What is the issue ?
upper karma
i am trying to sort some points by their angle
from another point which is the reference
and this last one case
isnt working
and i am sorting them on clockwise
and this is the code i am using to get the angle
Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
float angle = atan(y/x);
if(x < center.x && y >= center.y) angle += HALF_PI;
else if(x < center.x && y < center.y) angle += TWO_PI;
return new Point2D(mod, angle);
}
this is the point i wanna sort, and center is the red point on the giffs
white harness
hard gale
X<center x and Y<center y is not quadrant 4
upper karma
this is the atan i am using
no, it is 3rd
but my results from atan is -pi/2, pi/2
hard gale
So did you actually try the combo "add pi if point in Q2 or Q3 + add 2pi if point in Q4"? If you did I'm really gonna sleep
white harness
#include <cmath>
#include <iostream>
int main()
{
float x, y;
std::cin >> x >> y;
std::cout << std::atan(y / x) << std::endl;
return 0;
}
``` Toy program.Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
if(x < 0 && y > 0) angle += PI;
else if(x < 0 && y < 0) angle += PI;
else if(x > 0 && y < 0) angle += TWO_PI;
return new Point2D(mod, angle);
}
``` This should work.and what if the point is on the edges?
white harness
Edge of ?
white harness
You can fix that.
white harness
[0;2\pi[.
upper karma
no
i mean
the variable
u dont initialize it
u just say angle + two_pi
but u dont do anything
white harness
It is angle += TWO_PI;;
white harness
angle = atan(y / x);
white harness
Why ?
white harness
Plot it.
upper karma
look
well
ofc it will never work
lol
i need to change the output of atan
cuz i want
my angle from 0 to 2Pi
otherwise
if there are points on different semiplanes
it will never work
or will it?
wait...
do i need to return the abs value?????????
white harness
Of ?
white harness
The variable angle is between 0 and 2pi.
upper karma
@hard gale -pi/2, pi/2 isnt what u draw T.T
@white harness i know, but it returns something else
on the Q2, it returns -angle
white harness
If x < 0 and y > 0 then atan(y/x) < 0, if x < 0 and y < 0 then atan(y/x) > 0, if x > 0 and y < 0 then atan(y/x) < 0. By adding the right value, angle has to be positive.
upper karma
for fuck sake
ok look
Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
float angle = atan(y/x);
return new Point2D(mod, angle);
}
just with that code, to return atan only
there is the point
- the angle on degrees
white harness
It returns new Point2D(mod, angle).
upper karma
the thing is
for this enviroment
the origin is top left
not bot left
so possitives y goes down
and possitive x goes right
idk if this affects
but i think so
cuz the only possitive angles
are
white harness
You should have printed x, y, not this.x nor this.y.
white harness
Exactly, print x, y the difference.
white harness
Yes.
upper karma
that is a minus
right?
between point and center
now they are sorted properly
but this is luck
dont think it works
look
white harness
It isn't right.
upper karma
the first point color green
seems to be ok
is a bit on the left from the center
-15 pixels
and far away down
200 pixels
and angle in degrees is
idk
actually
that angle is this
white harness
What I did it is for angle in radians.
upper karma
yes i know it is for radians, i am just printing this
sortByAngle[index].polar(sortCenter).y*360/TWO_PI
the value itself is in radians
but i am printing the degrees value
but any way, look that picture
i am doing this wrong hahaha
white harness
Not right. 86 + 180 is different from -86 modulus 360.
upper karma
i said
y >= center.y
was positve
but here
the positive y value
is down
not up
the points above my center
has negative y
i think this has been my issue
fck
i wanna sort them this way
but
as how is the coordinates system implemented
it sorts them this way
but i will try to sort them on the second way first
and if i got it working, i move to the other one
white harness
Maybe (x, -y).
upper karma
okey i manually wrote some points
so now i wish u can help me better
acutally they are sorted XD
-90 to 90
from the lower to the higher
so the sorting algorithm works xD
now i need to guess the angles
white harness
The base isn't direct.
white harness
It depends on how you are dealing with x = 0 or y = 0.
upper karma
no
the problem is
asjbfhakhbv
idk
i know why
rofl
i have been always doing
x < center.x
but x is the distance between the point and x
not the actual x of the point
that should be x + center.x < center.x
so i move the point from the center
ah nop
not that i think
no, i fixed that
now i have a diff problem
why the left point
has 90 degrees?
it should be 180
white harness
0 again.
Point2D polar(Point2D center){
float mod = center.dist1(this);
float x = this.x - center.x;
float y = this.y - center.y;
float angle = atan(y/x);
if(x < 0 && y >= 0)
angle += Math.PI;
else if(x <= 0 && y < 0)
angle += Math.PI;
else
angle += 2. * Math.PI;
return new Point2D(mod, angle);
}
```.
