geometry-and-trigonometry

[slim plinth]

3 main questions i like to solve:
log questions, (ax+b)^n=(cx+d)^n, and pythagorean theorem with polynomials

[proven monolith]

Lets try another special triangle, shall we

[tiny acorn]

x^2+(x-31)^2 = (x+1)^2

damp.

[proven monolith]

It is also a quick one

To remember

[slim plinth]

Sure

[proven monolith]

Answer: ||9-40-41||

Interesting reaction I guess

[slim plinth]

Guess what

It is

[slim plinth]

Can you make one where two solutions are real

If not do this

Good luck

[tiny acorn]

x^4+x^6=x^8?

[slim plinth]

yes

Then make it x^8-x^6-x^4=0

[tiny acorn]

dividing both sides by x^4 gives 1+x^2 = x^4

[slim plinth]

factor out x^4

x^4(x^4-x^2-1)=0

[tiny acorn]

1+x^2 = x^4 = (x^2)^2
let u = x^2
we have 1+u = u^2, u is the golden ratio

[slim plinth]

And yes

[tiny acorn]

x^2 = phi

x = sqrt(phi)

[slim plinth]

you will get x=sqrt(phi)

And yay

That’s ur ans

[tiny acorn]

i guess

[slim plinth]

The sqrt(golden ratio)

[tiny acorn]

,w x^4+x^6=x^8

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by multiplexer_12

[tiny acorn]

.... negative sqrt(phi) doesnt count, nor the complex conjugates

[proven monolith]

Me when I have find the velocity of the car -3000km/h on the test

[tiny acorn]

or worse, sqrt(10)*i km/h

[slim plinth]

What

[slim plinth]

What about velocity of the car -3000km/s on the test

[tiny acorn]

its a joke

normally they wouldnt make it so the velocity is -3000 or sqrt(10)i

[spice oar]

how could 0 be a answer tho?

[proven monolith]

Its an invalid answer

[tiny acorn]

1 it would leave the triangle degenerate
2 tfw triangle inequality

[proven monolith]

Also thats because x⁸ - x⁶ - x⁴ -> (x⁴)(x⁴-x²-1)

See the x⁴ there

[tiny acorn]

and the -1 term

that looks sus

[proven monolith]

Hm?

[tiny acorn]

better explaination: 0 is a solution to the resulting polynomial but would make it invalid as the edge length of a non-degenerate right triangle (degenerate triangles are triangles whose all 3 points lay on the same line)

[spice oar]

i didn't knew about degenerate triangles, thank you.

[uneven siren]

Wouldn't that be just a line??

"Degenerate" triangles I mean

[tiny acorn]

uhm somewhat

you can make out the supposed edges

but since all 3 points are collinear some property is lost

hence "degenerate"

[uneven siren]

U cant

I feel like

It's just a line

How can u make out sides

From a line

[tiny acorn]

suppose ABC where a,b,c are collinear, i can still specify AB, AC, BC

but its kinda meaningless

[uneven siren]

Exactly

Like

Whats the point

Waitt

So in that sense

r there "degenerate" version of every polygon?

Square

Hexagon

Pentagon

Etc

[tiny acorn]

probably yes

[uneven siren]

😭

That's just

Like

Cool and useless

And the same time

Like imagine going to my teacher

And showing him a line and saying "that's a degenerate triangle"

Bro would beat me up

Deadahh

[minor heron]

oh ic idk my math teacher said like say we get an extraneous solution on an frq, notate it with the not equals sign and make sure your ap grader knows its extraneous clearly so u dont get marked off for not showing all solutions

[blazing steeple]

I was taught to write a comment and disregard it, for UK exams

[slim plinth]

Can you generate a problem that two solutions both works

[raven laurel]

velkoz

[umbral rivet]

👀

[hot grail]

Can a quaternion be invalid even if it's norm is equal to 1 ?

[vapid kettle]

Invalid?

Norm equal to 1?

[trail matrix]

uh whay?

[hot grail]

I'm making a rendering engine and when I extract a rotation matrix from a quaternion I get a non-invertible matrix for some reason, although the quaternion's is normalized

[vapid kettle]

Probably a good question for software

[meager thunder]

Phlimy why don't you invert the quaternion and convert that to a matrix?

[honest socket]

hi peeps

let's say we have a vector

and this vector ends in point A

this vector is normalized, so it has length of 1

we also have another vector, i call it Offset

X value of Offset means offset from A in direction of the first vector

and Y value of Offset means offset from B in kinda direction upwards from point A

u know what i mean?

[craggy shale]

B also a normalized vector? Linearly independent from A?

Do you mean a change of basis?

[honest socket]

a picture

if offset will be (1, 0) then we just go further in direction of A, i mean we just extend the A by 1

[craggy shale]

And (0,1) we go orthogonal to A?

[honest socket]

so it's like adding, but in orientation of A

yeah

[craggy shale]

Looks like a change of basis

[honest socket]

what is that?

[craggy shale]

So you know normally

If you have the vector (1,0)

Basically you're saying

1 unit to the right

0 units up

Or otherwise

1 unit in the direction of the vector pointing to the right

0 units in the direction of the vector pointing straight up

[honest socket]

ooooh

[craggy shale]

Now then the question is

[honest socket]

it has something common with i j k?

[craggy shale]

Right so (1,0) = 1 i + 0 j

Or more general

(a,b) = a i + b j

Now I'm going to be mean to you

I'm going to say

I've got a vector v

And it is x units in the direction of a vector pointing kinda diagonally upwards

y units in the direction of a vector pointing straight down

=tex \vec{v} = x \begin{bmatrix} 1 \ 1 \end{bmatrix} + y \begin{bmatrix} 0 \ -1 \end{bmatrix}

[craggy shale]

Why would this be better than the x i + y j thing?

We're just using different vectors that tell which direction we're going in

How many units along which vector

We're just changing what vectors we're going along

That's called a change of basis

[honest socket]

so i need to break the vector A into two basis vectors and then multiply and blablabla?

[dark sparrow]

we're changing the vectors we choose as our "building blocks"

[craggy shale]

Do you want to take this over? Or shall I continue? You know me.

[honest socket]

lol

i'll try ty

if something go wrong i ask you again

[craggy shale]

Sure

[honest socket]

it seems i didn't get anything

😐

@craggy shale halp

[craggy shale]

I'm going to be a little bit lazy.... But I really like this video https://www.youtube.com/watch?v=P2LTAUO1TdA

[honest socket]

k cF

XD

[dark sparrow]

ayyy

[honest socket]

i was googling too, and found khan academia's video but it's so bad, uhh..

[craggy shale]

Khan is pretty bad

I don't like it

Well I like the humanities part of Khan Academy

But the maths part? No thanks

Oh he goes into change of basis matrix also

Do you know matrices?

[honest socket]

yeah

it's like 2d array

[craggy shale]

Yes, basically

And you know how multiplication works with matrices?

[honest socket]

yeah

[craggy shale]

👌 Excellent!

[honest socket]

horizontal by vertical

[craggy shale]

Yep

[honest socket]

i c already what i need

[craggy shale]

That times that plus that times that plus that times that plus ....

[honest socket]

i have 0.7; 0.7 as i, and i need to rotate it by 90 degrees counter clockwise and this will be j

[craggy shale]

So do you know what your change of basis matrix would be?

[honest socket]

not yet, i want to finish the video, maybe i will find out something new

[craggy shale]

Cool

[upper karma]

How do I figure the center of rotation between 2 of points in a 2d plane?

[dark sparrow]

...what?

[upper karma]

uhm sorry for phrasing it poorly

Is it possible to determine the center of rotation of 2 different points in the 2d plane?

[dark sparrow]

what do you mean by center of rotation

[upper karma]

assuming i know the angle it's being rotated

[dark sparrow]

the center of a rotation that sends one point to the other?

[upper karma]

i'll define it as the point where you rotate a given coordinate, (x, y) to another coordinate (a, b) by some angle

[dark sparrow]

you're still wording things way too vaguely and confusingly for me to understand

[upper karma]

i guess an example would be the unit circle and trig

say if i have point (1, 0)

and the point (0, 1)

[dark sparrow]

mhm

[craggy shale]

There's lots of centers you could rotate around to rotate one point into another

[upper karma]

Oh i didnt know that

is there a proof for that?

[craggy shale]

Those two points have to lie on the same circle

But 3 points define a unique circle

[dark sparrow]

any point on the two points' perpendicular bisector is a possible center of rotation

knowing the angle of rotation narrows it down to two points on that; knowing the direction narrows it down to one

[upper karma]

ah

Gee thanks for the help

I was trying to find an analytic way to find a point A', which is the result of rotating the point, A by some arbitrary angle

[dark sparrow]

around what

the origin?

[upper karma]

Nah, an arbitrary point

Writing the circle's equation seems to help

I'll try it out

[upper karma]

Suppose if you have a point (x, y) and you want to rotate it by some angle, θ about an arbitrary center of rotation, (a, b) other than the origin

one way to approach this

[dark sparrow]

...can we call the angle something other than n

θ?

[upper karma]

yes

one can do a translation such that the center of rotation maps to the origin

and subsequently do a rotation about the origin at there

then

translate it back to where it's supposed to be

why does this work?

am i phrasing this poorly

like if you want to rotate a point by some angle θ about an arbitrary center of rotation

[dark sparrow]

yeah

i get what you're saying

:p

[upper karma]

oh kewl

so why does this work tho

[dark sparrow]

imagine that instead of shifting the plane, you shift your coordinate system

in the opposite direction, ofc

[upper karma]

yes

and what

[dark sparrow]

so then your center becomes (0,0) in the new coordinate system

you do your rotation about the origin and the center stays intact

and then you just shift your coordinate system back to where it started

=tex \begin{bmatrix} x \ y \end{bmatrix} \mapsto \begin{bmatrix} (x-a)\cos(\theta) - (y-b)\sin(\theta) + a \ (x-a)\sin(\theta) + (y-b)\cos(\theta) + b \end{bmatrix}

[upper karma]

is this a rotation matrix?

[dark sparrow]

not quite

finding that did involve using a rotation matrix though

i've denoted it with R here

[upper karma]

ah it makes sense now

so basically, what you had wrote down is that

to rotate a point (x, y) by an angle θ about the point (a, b)

one would do a translation such that the center of rotation is at the origin

or

(v-c)

and apply the rotation matrix for it

[dark sparrow]

yeah

[upper karma]

and just translate it back

gee

thanks

[dark sparrow]

ye

[honest socket]

let's say we have a 2d array of dots (pixels), it starts from top-left corner and has width and height 512

we place a point at some pixel and say that this is center of a circle

then we check every point in the array, and if distance between this point and the center is less then radius, we mark this pixel, so that we have a circle of marked pixels

but what, if some dimension of our array is doubled

[dark sparrow]

you don't have to check all of the array, just the square given by x0 - r ≤ x ≤ x0 + r and y0 - r ≤ y ≤ y0 + r

[honest socket]

so, we have the same amount of pixels, but their width is doubled

[dusk snow]

hold on

[honest socket]

yeah i know, bounding box, i do this

[dusk snow]

since when do pixels have a width

[honest socket]

i'm working in some game engine, and there is sprites that are created from texture, they don't change the texture, but i can scale the sprites

so it looks different, but the texture is the same

and then i want to make a hole in a sprite, so i translate world space point into pixel space, and then check pixels in bounding box, you know..

so if i scale the sprite, the hole is scaled also

and i want to fix it

[dusk snow]

ah

[honest socket]

yeah

[dusk snow]

then you change the distance function

so it's (2(x- xc))^2 instead of (x-xc)^2

[dark sparrow]

might be a multiplier on both coordinates

:p

[dusk snow]

well depending on the kind of scaling yeah

either you do a circle equation with the coordinates post-scaling

either you do an ellipse equation with the coordinates pre-scaling

[honest socket]

wait wait

what do u mean

[tranquil yew]

can't you just make the hole smaller initially

[honest socket]

what do u mean?

the hole has its radius

radius can be different

oh well, i will be back with this tomorrow, i have already closed my ide and that game engine

[honest socket]

either you do an ellipse equation with the coordinates pre-scaling(edited)``` what's faster?

so, all i needed is if (Vector2.Distance(pointOfImpact, new Vector2(x*xScale, y*yScale)) < radius) to add xScale and yScale -_-

[honest socket]

oh no, i was wrong

it works with objects that has width 1 and height 0.5, but doesn't work with object that has width and height 0.5

[honest socket]

k, this works ```float distance = Mathf.Sqrt(Mathf.Pow((xxScale)-(pointOfImpact.xxScale), 2)+Mathf.Pow((yyScale)-(pointOfImpact.yyScale),2));

                if (distance < radius)```

[ashen blaze]

yo

anyone here

[upper karma]

🖐

[ashen blaze]

im actually struggling rn

determine the area of the composite figure

we tried that

but we didnt get the answer (34 sq inches)\

unless we had the wrong measurements

i erased it all

lol

oh i think we had it as 4

alright thank you

[upper karma]

im lost i dont get it 😢

[dark sparrow]

don't get what

[upper karma]

the explanation

[honest socket]

hi, i hope it will be useful

@ashen blaze

[ashen blaze]

oh i see now

thank you so much

[dark sparrow]

it's just a 5 by 8 rectangle minus a 3/4/5 triangle though

:p

[honest socket]

ye, that's even esier

i have a problem

so, let's say i have a pixel array xD

[dark sparrow]

is it that same problem with the hole? xP

[honest socket]

no

another

[dark sparrow]

ok

[honest socket]

and this pixel array is represented as sprite that is streched 2 times by X axis

so it looks like a rectangle

[dark sparrow]

=tex \begin{bmatrix} 2 & 0 \ 0 & 1 \end{bmatrix}

[honest socket]

is this scaling matrix?

[dark sparrow]

yeah 😛

[honest socket]

so. Then I want to cut a rectangular hole in that sprite

when its rotation angle is 0, everything is okay

the hole isn't rectangular, but imagine that it is

but when i rotate the sprite by 90 degrees the hole looks like this

sooo. i do it like this....
at first I rotate the rectangle that will make a hole, then I rotate it by needed angle and calculate bounding box for this rotated rectangle, and then I check every pixel in this bounding box, rotate it by negative angle, and check if this pixel is inside not rotated bounding box, if it's in, then I make a hole

does it make any sense to u?

green pixels indicate area of bounding box of rotated rect, red pixels indicate area of not rotated rect, when they overlap the ara becomes yellow, here is angle = 0

here is angle 45

so, if we draw rectangle in the rotated boudning box, it fits perfectly in its not rotated state

if u know what i mean

but it looks like my rotation function are wrong

or not... the area of not rotated rectangle is wrong

btw it looks okay when the sprite isn't scaled

[honest socket]

dat feel when no one can halp u

🤔

no.

i'll do it myself

i'm just lazy now

lazying

[honest socket]

well, i made some investigation

let's say i set the X scale to 2, and Y to 1. To draw a 100x100 square it has to draw a rectangle with width 50 pixels and height 100 pixels

when I rotate the sprite by 45 degrees, then I take the bounding box and rotate it by -45 degrees, and then I check pixels of this new, rotated rectangle, whether they are inside another bounding box or not

so the problem is, that this rotated rectangle is destorted

here is the sprite isn't scaled

so, the green rectangle is bounding box for another, and black rect is rotated green rectangle

but when i scale it by to along X axis it looks like this

since i use black rect to draw a picture, the picture is distorted as well

also, if i will remove scaling from the last picture, it looks like this

so it rotates good, the problem is in scaling

a few weeks ago i was playing with rotation and found the shear matrix, i was told that i can rotate an image by 3 shear operations, but i didn't get how it works anyway..

but it's good that i remember that shear matrix is exists

my eyes helped me to figure out, that the distorted rectangle looks like sheared

so by experementation with values i have found values like 0.37 and 0.63

here is the sprite is scaled by 2 along X axis and rotated by 45 degrees

but i applied shearing with value 0.37 to X axis, and shearing with value 0.63 to Y axis

so it kinda fixed the problem, but rotation and scaling of the sprite can be different, so i can't searching values for each case

for now i noticed that 0.37 and 0.64 is actually 1

also, it's not perfect, i mean the shape of black rectangle

hm.. when i scale it by 2 along Y axis, the magic numbers are -0.63 for X and -0.37 for Y

here i found some info

for 36.87 degrees 3/5 is 0.6 and -1/3 is -0.33, the values are pretty close to mine

[eager saddle]

How do I find the value of a trigonometric function of theta when given a line and a quadrant?

This is something really simple. ><

Oh wait I think I got it.

Make a triangle with line and X-axis, then find the respective trigonometric function value of the angle you want to find.

[trail matrix]

@eager saddle I learned a nemonic that helps with trig functions. A Spiffy Trig Chart. The first letter of each word stands for All, Sin, Tan, Cos. The position is in what quadrant that function will be positive in

So Tangent is positive in 1st and 3rd quadrants

Sin in 1st and 2nd

Etc

[upper karma]

Or maybe just know what each represents and it's obvious

http://emweb.unl.edu/Math/mathweb/trigonom/Image296.gif

[eager saddle]

@trail matrix I learned that as 'All Students Take Calculus,' but yeah. It's better to know how it works than a mnemonic.

[dark sparrow]

TONS!

it's like, obvious when you think about it in terms of coords

[trail matrix]

@upper karma sometimes mnemonics help

[eager saddle]

How do you find on paper the trigonometric function value of a non-special and non-quadrantal angle?

Like, how do I find sin(91°)?

The calculator spits out .998... something in degree mode, but where did that come from?

[lean crest]

you can use the taylor series of sin(x) or cos(x)

but u gotta work with radians

i.e the cos of an angle x is approximately 1-x^2/2

[grave echo]

OMG

@lean crest UR HERE TOO

WHAT A SURPRISING COINCIDENCE

omg so many people here

i thought i entered a server with random peeps

[lean crest]

who r u

[grave echo]

oi wtf

incept

[tranquil yew]

@eager saddle It turns out pretty much all of the functions we commonly use can be represented by some infinite polynomial

And polynomials are easy to calculate

so you can calculate any elementary function to any digit of precision you want by just using more and more terms of that infinite polynomial that represents it

Just to clarify, each elementary function has a distinct infinite polynomial that represents it

[dark sparrow]

@eager saddle convert your angle to radians and then use the Taylor series for sine

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! ...

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - x^10/10! ...

[honest socket]

it seems i don't understand how vector's scaling work -_-

uh, nvm

oooh, okay, now i understand it a little bit

once again, in my case i have a square 1x1 which i have to display on a sprite 2x1, since sprite's ratio is 2x1, i need to make my square 0.5x1, otherwise it will look stretched along X axis

when I rotate the sprite by 90 degrees, and then try to rotate the square by -90 degrees, i rotate a rectangle 0.5x1 which placed on the sprite, and since the rotated square already is 1x0.5, and the sprite is scaled by 2 along X axis, it produce this

where green rectangle is original 0.5x1 square, and black rectangle is rotated version of the first rectangle

if i cancel scaling of the sprite it look like this

so basically we have two similar rectangles, but the black is rotated by -90 degree

and since the sprite is rotated by 90 degree, its X axis is vertical now, so that's why the black rectangle looks stretched vertically on the first image

my task is to make the black rectangle on the first image look like the green one

now it's obvious that i need to scale it by 0.5 along its X axis and scale it by 2 alongs its Y axis

but I want to make it work for an arbitrary angle

with 15 degrees rotation it looks like this

on a sprite with 2x1

and like this without scaling

so yeah, scaling for the green rect is okay since its X axis is aligned with X axis of the sprite, but for the black rect X axis isn't aligned with X axis of the sprite

can I change and use basis vectors for scaling?

i mean this

[honest socket]

hm...

so, for 45 degrees my X basis vector should be 0.7; 0.7, and Y basis vector is -0.7; 0.7, right?

[dark sparrow]

not quite

replace 0.7 with sqrt(2)/2

:p

[honest socket]

oh, i know i know, it's not so important for now

cause i don't need it to has exact values, it just should look good

oka, so here is what i have without changing basis vectors and scaling

the degree is 45

now i use changed basis vectors and scale it by 0.5 along X axis

looks pretty good, it's what i need

but now i want to scale it by 2 along Y axis

i drew axes

on the first image X is (0.7;0.7) and Y is (-0.7; 0.7)

so i scaled the black rectangle along X axis by 2

and now i want to scale it along blue axis by 2

so it should be (0; 1), yeah?

oh god, i'm dying.....

i guess i poke the right thing

[tranquil yew]

Well the exact values might help you

do you know what might have the answer of sqrt(2)/2

oh wait im 4 hours late

[hybrid cypress]

How do you find the measure of an arc, given that you have an angle of 90 degrees and circumference of 180cm

[dark sparrow]

what fraction of a full circle is 90°?

[dark sparrow]

answer my question, please

what fraction of a full circle is 90°?

[hybrid cypress]

1/4?

[dark sparrow]

yes

what is 1/4 of your circle's circumference?

[hybrid cypress]

45

[dark sparrow]

that's your answer

[hybrid cypress]

ooh ty

how about finding the circumference of a circle if you dont have

radius or diameter?

[dark sparrow]

what do you have?

[hybrid cypress]

arc length of 90cm and an angle of 81 degrees

[dark sparrow]

what fraction of a full circle is 81°?

[hybrid cypress]

roughly 1/4

[dark sparrow]

i don't want any approximations

@upper karma please don't

i'd rather help lordyonetime by myself

@hybrid cypress how many degrees are there in a circle?

[hybrid cypress]

360

[dark sparrow]

what fraction of a full circle is 81°?

@hybrid cypress

...welp

[hybrid cypress]

sorry idk is it 2/5?

[dark sparrow]

how did you get that?

[hybrid cypress]

idk honestly wait

i think thats wrong anyways

[dark sparrow]

it is, yes

[hybrid cypress]

its like 4 2/5 of the circlr

[dark sparrow]

🤦

you chop up a circle into 360 pieces (a.k.a. degrees) and take 81 of them

what's the fraction that represents that?

@hybrid cypress

[hybrid cypress]

BUttt

dont you take

360/81

or is it 360/279

[dark sparrow]

🤦

it's the other way around

it's 81/360

[hybrid cypress]

o god im laughing so sorry

[dark sparrow]

i'm surprised at how you managed to answer the same question correctly when the angle in question was 90°

do i need to go over how to simplify that fraction down to 9/40?

[hybrid cypress]

no

[dark sparrow]

good

so now

90 cm is 9/40 of what?

@hybrid cypress

[hybrid cypress]

wai what

[dark sparrow]

you said your arc's length was 90 cm, right?

9/40 of a circle is 90 cm
how much is the whole circle?

[hybrid cypress]

yhe whole circle is 360

[dark sparrow]

NO!

how are you getting that?

[hybrid cypress]

idk im sorry 😱 i cant see my question anymore

[dark sparrow]

your question was "an 81° arc of a circle is 90 cm long. what is the circumference of the whole circle?"

[hybrid cypress]

yesyes

[dark sparrow]

we have established that the arc is 9/40 of the whole circle

[hybrid cypress]

Ok

[dark sparrow]

it is also 90 cm long

90 is 9/40 of what?

[hybrid cypress]

IM SCARED UM 400?

[dark sparrow]

yes

what exactly are you scared of?

[hybrid cypress]

you youre scary

BUT HELPFUL

youre like

a teacher

[dark sparrow]

i apologize for scaring you :/

[hybrid cypress]

oh no dont im just super slow

ty foe your help i appreciate it

instead of just giving me the answer 👌

[dark sparrow]

yeah, that's something i'm never doing to anyone

and i mean, i am planning to become a tutor once some irl stuff is sorted out, fwiw

[hybrid cypress]

that's good, and yay i hope you do

good luck with your future, hope everything works out

[tough saffron]

Oh nice

I could use a tutor for this

[tranquil yew]

I would refrain from using the face palm emoji

[honest socket]

hi

what's new ? xD

[weak hemlock]

whast wrong with the face palm emoji

oh alright lol, just seemed out of nowhere. i guess it was deleted or something

[sour sorrel]

the comment was lost on me as well, but now it makes perfect sense XD

[spiral lintel]

hey guys I need help with this one prob

A circle is tangent to lines y=4/3x -12, y=3/4x +29/4, and y=4/3x +14/3. Find the equation of the cirlce

[dark sparrow]

drawing those lines would help

try locating its center @spiral lintel

[spiral lintel]

okay I'll do the hard ones later

can you go to alg section, there's another question that didnt make sense to me

[tranquil yew]

@spiral lintel do you know any properties about tangents and their relationship to radiuses of a circle?

[dark sparrow]

radii

the plural is radii

xP

[upper karma]

They are also perpendicular to the radius

The one that touches them

Good point lol

There exists a perpendicular radius :p

[slim gorge]

hey guys

[surreal bolt]

hi

[slim gorge]

check out this prob:

In rectangle ABCD, AB = 8 and BC = 20. Let P be a point on AD such that angle BPC = 90 degrees. If r_1, r_2, r_3 are the radii of the incircles of triangles APB, BPC and CPD respectively, what is r_1 + r_2 + r_3?

took me a couple hrs to decode as im not good at geo

[surreal bolt]

is the answer a scalar?

[slim gorge]

um... sorry, whats a scalar?

only magnitude?

[surreal bolt]

oops I meant a constant

[slim gorge]

lol, ya it is

[dark sparrow]

think i can solve it, hold on

[surreal bolt]

you could theoretically just draw it out

[slim gorge]

i did

[surreal bolt]

even using analytic geometry 😃

[slim gorge]

loll

[surreal bolt]

okay I am on it. Geogebra

[slim gorge]

just take a pen and paper

[upper karma]

Does a point on AD even exist such that it's 90°?

[slim gorge]

yes it does

@dark sparrow , if i can solve it, then why cant u too?

[upper karma]

Ohh I misread the lengths

[slim gorge]

after i found the "trick", it seemed like i am just stupid

[surreal bolt]

kk found the sides of all three right triangles

[slim gorge]

u dont have to

just in terms of vars

[slim gorge]

um... you dont have to do ANY of that

no trig

[dark sparrow]

there's no trig involved here

[slim gorge]

its an olympiad sample prob

oh sorry, i thought i saw cos

[dark sparrow]

there is no trig involved here

=tex \frac{24}{3+\sqrt{5}} + \frac{40}{5+3\sqrt{5}} = 6(3-\sqrt{5}) + \frac{4}{5}(3\sqrt{5}-5) = 18 - 6\sqrt{5} + \frac{12}{5}\sqrt{5} - 4 = 14 - \frac{18}{5}\sqrt{5}

[upper karma]

I got 20 + 12 sqrt(5)

[surreal bolt]

I got a headache 😃

[slim gorge]

i wont tell the answer

[dark sparrow]

arithmetic

[slim gorge]

😃

can this be a featured prob?

@dark sparrow

[dark sparrow]

it's already been discussed so prob not

[upper karma]

Is my answer correct or not?

[slim gorge]

no

should i give a hint?

[upper karma]

Oh those are incircles

Not normal circles

[slim gorge]

lol

[surreal bolt]

Triangle dimensions:
8, 14, 2rt(65)
8, 6, 10
10, 2rt(65), rt(360)

r = (ab)/(a + b + c)

Too lazy? 😃

[slim gorge]

or:

this is a BIG hint

should i continue?

[upper karma]

8

[slim gorge]

BRAVO

[upper karma]

Gg

[slim gorge]

this took me hours to do

[upper karma]

I didn't use any tricks, I just used incircle radius = area / semiperimeter

[slim gorge]

to "figure out" to be more accurate

[upper karma]

The summed those

[slim gorge]

i used r = (a+b-c)/2 for right triangles

incircle radius = inradius

[upper karma]

Yeah so its a similar solution

What's the trick then?

[slim gorge]

thats why i put it in ""

😉

[upper karma]

Oh

So it's not a real trick :P

[slim gorge]

ikr

i expected somethin more

that exam is not easy to clear

[upper karma]

I thought there was a trick like the way the radii line up and make it perfect for something to happen

[slim gorge]

me too

like the sum of the indiameters = 20 and stuff

so sum = 10

i just made the word indiameters :p

[upper karma]

I read that as india meters

[slim gorge]

http://olympiads.hbcse.tifr.res.in/regional-mathematics-olympiad-2016-rmo-2016-registration/

me too

oh wait i read it as in diameters

[honest socket]

hey peeps, do u know what is pentagon?

i mean geometric figure

[upper karma]

it's a five sided polygon

[honest socket]

yeah

[upper karma]

What do you wanna know about it

[honest socket]

so, let's say we have an ordered list of points

and sometimes they form a pentagon

i know that in this case there is only 5 points

how to check whether it's a pentagon or not?

[dark sparrow]

do you want a regular pentagon?

[upper karma]

Regular meaning all the 5 sides have equal length

[dark sparrow]

with all sides equal and all angles 108° / 3π/5?

@upper karma regular polygons have to have all equal angles as well

[upper karma]

Isn't that a consequence of them having equal sides?

[honest socket]

yes, regular

ofc i can check all 5 angles, but maybe there is a faster way to check whether it's a pentagon or not?

oh, i c, i can also check distance between the points

it must be faster

i think

[upper karma]

Are you saying that the 5 points given can be anywhere on the pentagon?

Assuming it does lie on the pentagon

[honest socket]

no, there are only 5 points, and they form a pentagon

so one corner - one point

[upper karma]

oh so you're saying the 5 points are on the vertex of the pentagon?

[dark sparrow]

@honest socket i don't think there's a faster way honestly

[upper karma]

@honest socket Maybe try inscribing this one into a circle

And determine an equation for the circle

and plug in those points into the equation

[honest socket]

🤔 i need to code it, so it's better to keep it simple

[upper karma]

Simple as in it uses less memory?

[honest socket]

yeah, and time

cpu time

[upper karma]

Ah well I'm ignorant about this but isn't using less memory implies less time being spent?

No

Usually it's a compromise between the two

If you use more memory you can store some things to save on time

Or you can take more time to do something but end up using less memory

Ah okay

[dark sparrow]

@honest socket figure out the lengths first. if they're all the same, then take dot products between the vectors representing adjacent sides. if they're all the same up to sign, you have a regular pentagon

[honest socket]

okay, ty

[dark sparrow]

@ember kraken you know right triangles, right?

[ember kraken]

Yes

90 degrees triangle

[dark sparrow]

a triangle with a 90° angle, yeah

[slim gorge]

SOH CAH TOA is my jam

[dark sparrow]

ಠ_ಠ

okay so basically

sometimes you know the angles in a right triangle, but want to know its sides

the trigonometric functions serve precisely this purpose

[slim gorge]

how do u write in bold?

[dark sparrow]

here's a triangle

@slim gorge **like this**

anyway

[slim gorge]

ok

[dark sparrow]

in that triangle,

[ember kraken]

Guys

[dark sparrow]

=tex \sin(\alpha) = \frac{a}{c}, \cos(\alpha) = \frac{b}{c}

[ember kraken]

I need to take a bath can you leave explanation here please?

I mean just write explanation I will read after I finish my bath

[dark sparrow]

ok

so just to expand on that a bit, sine tells you what fraction/percentage/whatever you want to call it the opposite side to the angle is when compared to the hypotenuse
and cosine tells you the same, but for the adjacent side to the angle

(this means sin and cos are always between 0 and 1 for an angle between 0 and 90°)

[slim gorge]

its easy to prove too

[dark sparrow]

the legs are always shorter than the hypotenuse

[slim gorge]

so u dont even have to prove it for acute angles

so to summarize

=tex \sin(\alpha) = \frac{a}{c} \ \cos(\alpha) = \frac{b}{c} \ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{a/c}{b/c} = \frac{a}{b}

[dark sparrow]

double backslashes for newlines

[slim gorge]

i did put double

[dark sparrow]

you didn't seem to

[slim gorge]

=tex \sin(\alpha) = \frac{a}{c} \ \cos(\alpha) = \frac{b}{c} \ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{a/c}{b/c} = \frac{a}{b}

[slim gorge]

oh sry

[dark sparrow]

oh yeah, tan stands for tangent and is the ratio of the two legs

[slim gorge]

cotan(x) = 1/tan(x) right?

[dark sparrow]

cotan
ಠ_ಠ

but yes

[ember kraken]

Hi im back

[upper karma]

You can also check out on the unit circle definition of the trig ratios

cosine being the x coordinate and sine being the y coordinate

tangent being the slope

[dark sparrow]

tangent has a geometric representation too though 😛

[upper karma]

so geometrically, what is tangent

ow

A line being tangent to the unit circle

[upper karma]

If you take a circle of radius 1 and put it in the center then its circumference is 2pi, so we can define the angle by a point on the circle, and 2pi is a full angle (360 degrees)

[slim gorge]

oh

thats why tan(90 degrees) is undefined right?

[upper karma]

Ye

Cuz it'll never touch the tan line

They're parallel

[tranquil yew]

@upper karma Geometrically, tangent is the slope of the hypotenuse of the triangle when it's legs are aligned with the axes

[upper karma]

what do you mean by "its legs are align with the axes"

Well I interpret it as the y coordinate of the point of intersection between the line when you extend the hypotenuse of the triangle outside of the unit circle and the line x=1

[tranquil yew]

As in like the legs are parallel to the axes

That's the same thing daniel

slope is rise over run

if you make the run 1

then the rise will be the slope

[upper karma]

Well I didn't really get what you meant initially

[tranquil yew]

The slope of the line from the origin to a point on the unit circle is given by tan(theta)

[upper karma]

But the thing is

That's not a geometric definition

Like

It gives a quantity

If you define it as the slope of the hypotenuse, that is

I think the picture given above is the geometric definition

Slope is just a measure of how steep the line is

[tranquil yew]

Exactlyu

so tangent is also

it is a geometric definition

it ties the value of tangent to a geometric value

[upper karma]

Yes

[slim gorge]

Hey

Can we use the power of a point theorem without a circle?

[dark sparrow]

power of a point?

[slim gorge]

Um... Ya

[dark sparrow]

what's that

okay nvm

from what i saw on wiki it does require a circle xP

[slim gorge]

Ok...

Thx

[upper karma]

What do you mean without circle?

[slim gorge]

I mean that no circle is involved

[surreal bolt]

tan(x) = opposite / adjacent. If the adjacent leg is 1, then tan(x) is equal to the opposite leg. Note that in the picture the two triangles are similar. SO, the central angle is the same and the tangent of that central angle is the same. BUT note for the bigger triangle the adjacent side = 1 so the vertical leg is equal to the tangent.

[violet nest]

Anyone remember the sine rule I was talking about?

This is my proof, I think

I discovered it when I was at high school, there were a crap ton of class work and I was too bored with using the sine and cosine rule, so I thought this would be a nice change

But then I noticed after rearranging, asinC and csinA part

[rare talon]

It will still require a circle, whether stated on the problem or we've constructing a circle when the problem need it

[slim gorge]

so, given ANY four distinct points A, B, C and D that make the quadrilateral ABCD, and AC and BD intersect at X, is
AX * CX = BX * DX ?

[dark sparrow]

i feel like that need not be the case at all

or maybe not

[slim gorge]

me too, but sometimes it feels very true

[dark sparrow]

hmm

[slim gorge]

every quadrilateral isn't cyclic is it?

[dusk snow]

some are, some aren't

[slim gorge]

so no?

[dark sparrow]

yeah, not every quadrilateral can be inscribed in a circle

:p

[upper karma]

it's not hard to prove either

[slim gorge]

ya

we can make a circle that goes through 3 point

[upper karma]

exactly

[slim gorge]

add a point outside the circle, done

or even inside, who cares

btw, in which field of mathematics (or geometry) are cyclic quadrilaterals used the most?

[upper karma]

https://math.stackexchange.com/questions/1859652/where-do-you-see-cyclic-quadrilaterals-in-real-life

[hybrid cypress]

Does anyone know how I'd solve for AE?

[dark sparrow]

...are you not able to take screenshots?

[hybrid cypress]

I am but thats more work

since I use my phone for discord

[dark sparrow]

redrew your pic for clarity

so for starters, @hybrid cypress, what can you tell me about this whole picture?

[hybrid cypress]

The circles are equal

[dark sparrow]

no they aren't

if they were, their shared tangents would be parallel

which they clearly aren't

[hybrid cypress]

Oh

Can we solve for ED

[dark sparrow]

you certainly can!

[hybrid cypress]

Using Pythagorean Theorem

[dark sparrow]

but can you tell me what GD equals first, and why?

[hybrid cypress]

7?

[dark sparrow]

yes

can you tell me why GD = 7?

[hybrid cypress]

Because GF equals 7

[dark sparrow]

why are GD and GF equal, then?

😛

[hybrid cypress]

bc they have common external tangents?

[dark sparrow]

no

try again

[hybrid cypress]

hm

I just assumed bc it looks like GF

has a radius of 7

so the other

[dark sparrow]

radius
aha!

that's the key word

GF and GD are radii of the same circle

so they are equal by definition

[hybrid cypress]

Ohh

[dark sparrow]

okay, so now

what is ED?

[hybrid cypress]

24

[dark sparrow]

good

so now, the thing i wanted to hear from you about the whole picture

[hybrid cypress]

So

[dark sparrow]

is that it is symmetric across the line EB

[hybrid cypress]

Is AE 68

[dark sparrow]

why is AE equal to 68?

[hybrid cypress]

Because we know that ED is 24 and DC is 44

[dark sparrow]

that gives you EC, though

not EA

why are EC and EA equal?

[hybrid cypress]

bc theyre common ex tangents? 😓

[dark sparrow]

that works, sure.

[hybrid cypress]

What makes them equal otherwise?

[dark sparrow]

well, you could also tell me they're equal because they're symmetric across EB

but saying they're tangents to the same circle from the same point works

[hybrid cypress]

Ah gotcha

[dark sparrow]

yknow, in geometry, and for that matter in math in general, you really ought to know why the things you say or learn are true

not just that they're true

[hybrid cypress]

I know. You have to know how you're getting the answers you do and why

I'm just not in the program rn

But I'll get better at understanding why things are in math

oh gosh lol sorry

btw what do you go by?

[dark sparrow]

what do you mean

you can call me sqrt(2) or Anto, i'll respond to either

[hybrid cypress]

aw thats cute

Anto is nice

I have another question

[dark sparrow]

shoot

[hybrid cypress]

except its with internal tangents this time

[dark sparrow]

ok

[hybrid cypress]

Can I show you?

[dark sparrow]

ofc

post it here

[hybrid cypress]

Sorry the qualitys bad

[dark sparrow]

is it implied that E lies on HB?

[hybrid cypress]

I believe so but theres no point

[dark sparrow]

it's not marked, hence my request for a clarification

[hybrid cypress]

Im not sure myself

I think it is

because we have points ED and EF

or line*

[dark sparrow]

i mean, it must lie on HB, since those tangents to the circles are symmetric to each other across the line joining the circles' centers

[hybrid cypress]

Yeah

[dark sparrow]

actually, the whole pic has twofold rotational symmetry around E!

and i hope it's obvious that EI = EG and EA = EC

and due to that rotational symmetry, EI = EC, so EA = EC = EI = EG

can you find the length of any of these lines?

[hybrid cypress]

👀💥

Idk because I think Id need

to find

either HF or DB

first?

[dark sparrow]

HF is easy to find

think about it

[hybrid cypress]

Im thinking its 7

[dark sparrow]

and why is that? 😛

[hybrid cypress]

bc of the radii afain

again*

[dark sparrow]

yup!

[hybrid cypress]

Ok gimme a sec

Wait

can EG be a decimal?

[dark sparrow]

i assume you're trying to take the square root of a number that isn't a perfect square?

[hybrid cypress]

yeah hold on let me double checj

Ohh im so slow

I just did this

[hybrid cypress]

Hey Anto I have one last problem

[dark sparrow]

post it

[dark sparrow]

what can you say about triangles DEC and DAB?

[hybrid cypress]

Theyre symmetrical at D?

[dark sparrow]

nope

there is a transformation with its own name that turns one into the other though

but anyway

they are similar

[hybrid cypress]

ok bc its a transformation in size

[dark sparrow]

dilation is the somewhat-fancy name for it

or scaling, if you wish

[hybrid cypress]

ahh

[dark sparrow]

can you figure out the scaling factor?

[hybrid cypress]

Yes, by figuring out the lengths of either the smaller or larger figure first?

We already know the larger ones

AD = 20, DB = 25 and AB = 15

[dark sparrow]

you also know AE = 16

what is AD?

[hybrid cypress]

oh! youre right

AD = 20, DB = 25, DE = 4 and AB = 15

im stumped

[dark sparrow]

yup

can you figure out DC?

[hybrid cypress]

Well yes but how?

I dont have EC

[dark sparrow]

DEC and DAB are similar, remember?

DE = 4 and DA = 20, so the scaling factor between them is 5

one triangle is exactly 5 times as big as the other

[hybrid cypress]

oh i see

So that makes CB = 20

[dark sparrow]

👍

[hybrid cypress]

Alrighhhht

tysm Anto

thanks for all the help

[dim plover]

Could someone point out which step of the following process is losing me answers?
Q: Find the exact solutions of the equation in the interval [0, 2pi)
Eq: 2tan(2x) - 2cot(x) = 0
Steps:
2tan(2x) = 2cot(x)
2((2tan(x))/1-tan^2(x)) = (2/tan(x))
4tan(x) = (2(1-tan^2(x)))/tan(x)
4tan(x) = (2 - 2tan^2(x)) / tan(x)
4tan^2(x) = 2 - 2tan^2(x)
6tan^2(x) = 2
tan^2(x) = 1/3
tan(x) = +-1/sqrt(3)

[dark sparrow]

all your steps are correct so far

it's weird that you didn't halve both sides in 2tan(2x) = 2cot(x) tho xP

@dim plover

anyway yeah you haven't lost anything, dw

[dim plover]

Hmm, sorry wasn't paying attention

I should have lost something though since i only get 4 of the 6 answers

If I turn the tangent and cotangent into their sine and cosine components I get all six

Any ideas?

[dark sparrow]

hang on, yeah, that's true

gimme a sec

ah

you're losing pi/2 and 3pi/2 because you're involving tan(x) in your manipulations, which is undefined at those points

[dim plover]

Oh, then really whats the point of the double angle formula for tangent :/ if I can just end up losing things

Doesn't that sort of imply that you should avoid working directly with tangent manipulations then?

[dark sparrow]

it does make sense to be careful with them

they're safe if you know cos(x) != 0 though

[dim plover]

So preventatively I would have just had to plug in the values where cosine or sine are 0 to avoid this?

(since we also had a cotangent)

I guess then... The only reason this was an issue was because of the double angle, does that mean I always need to check for this when working with the double angle of tangent?

[dark sparrow]

yeah

[dim plover]

Hmm, alrightie, glad I wasn't doing something blatantly wrong in the math

Thanks for the help

[dark sparrow]

=tex \vec{r}(t) = \left< \cos(t)e^{p\sin(t)}, \sin(t)e^{p\cos(t)} \right>

[dark sparrow]

TIL the curve given by this parametric eq has a cusp precisely when p = sqrt(2)

[upper karma]

it'd be something else if the cusp looked like the radical

[final prairie]

Interesting

Proof? :p

[dark sparrow]

cusps correspond to r'(t) = 0

differentiate this thing and see for yourself that t = π/4 is the point of the cusp

[final prairie]

Interesting

Imma visualize that

[final prairie]

@dark sparrow Does there exist another p such that r(t) has a cusp?

[dark sparrow]

don't think so

maybe -sqrt(2)

[final prairie]

Nope, apparently

maybe sqrt(2)+πu for all u in Z

Nope apparently nope

Oh wait -sqrt(2) does have a cusp

I just graphed t from 0 to 1 not 2π

Fishh

[lean crest]

this feels so oddly satisfying

[final prairie]

yeah hehe

[umbral rivet]

👀

[torpid galleon]

No one is responding qq

[azure storm]

What have you tried so far?

[upper karma]

There are 2 main ways you can approach this

You can find a point on the line that's closest to Q, and then find distance between 2 points

Or you can use some formula if you were taught any for this

[azure storm]

I'd minimize the general distance

[upper karma]

I think point and then distance is easier @azure storm

[torpid galleon]

Something like this

[upper karma]

Okay so

You see how the closest path to the line from point is a perpendicular line?

[torpid galleon]

Yeah

Oups

Sent itbagain

[upper karma]

So, you could take a perpendicular line that goes through Q, and you know that the nearest point to Q is where those 2 lines intersect

[azure storm]

This only works for lines though

[torpid galleon]

And further

?

[dark sparrow]

Q ends up above that line, btw

😛

[torpid galleon]

Oh

[dark sparrow]

anyway

if you draw a line that is perpendicular to h and goes through Q, it'll meet h at the point you're looking for

the slopes of any two perpendicular lines multiply to -1

[torpid galleon]

How come

[dark sparrow]

er, gimme a moment

okay, so basically if you have a line with slope m1 and a line with slope m2, then the (signed) angles they make with the x-axis are arctan(m1) and arctan(m2)

and the angle between them is arctan(m1)-arctan(m2)

so denoting that angle with θ, we get

=tex \tan(\theta) = \frac{m_1 - m_2}{1 + m_1m_2}

[dark sparrow]

and if θ = π/2, then the left hand side is undefined, which means that there must be a division by zero error on the right hand side

thus m1m2 + 1 = 0, and m1m2 = -1

or more rigorously you could calculate cot(θ), which would then be zero for θ = π/2

@torpid galleon

[upper karma]

I tried finding a function that inputs the number of sides of a regular polygon and that it can outputs me the area of any regular n-sided polygon

=tex A(n)=\frac{nx^2}{4\tan(\frac{\pi}{n})}

[upper karma]

where x is the length of one of the side of the polygon

and n is the number of sides

of course

this is just

a conjecture

I want to prove this by induction

but how do I start?

I've never dealt a proof by induction that involves the trig functions

[dark sparrow]

p sure you don't need induction for that

[upper karma]

So how do I prove this tho

[dark sparrow]

split your n-gon into n isosceles triangles

[upper karma]

I actually used that reasoning to arrive to this formula

[dark sparrow]

the top angle in each will be 2π/n

oh

then you've proven it

[upper karma]

what