#geometry-and-trigonometry | Mathematics | Page 170

hallow trellis Mar 7, 2018, 12:19 AM

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yeah true

distant stag Mar 7, 2018, 12:20 AM

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Complete non-integrability means that in no open subset of the space does the plane field integrate to a submanifold

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This is impossible for vector fields.

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Since that's just ODEs

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You can integrate to submanifolds (lines) with any vector field

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The reason it has to be odd dimensional is because this condition happens to be impossible when it is even dimensional.

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It's just a parity problem.

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When you work out the linear algebra, that's how it happens.

hallow trellis Mar 7, 2018, 12:22 AM

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"plane field" ?

distant stag Mar 7, 2018, 12:22 AM

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Well that's the thing that is completely non-integrable.

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It's like a vector field, except not vectors---planes.

hallow trellis Mar 7, 2018, 12:23 AM

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thonker

distant stag Mar 7, 2018, 12:23 AM

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📎 1200px-VectorField.png

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vector field

hallow trellis Mar 7, 2018, 12:24 AM

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i get that it just kinda sounds like a meme

distant stag Mar 7, 2018, 12:24 AM

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📎 1200px-Standard_contact_structure.png

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plane field

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Why would it be a meme?

hallow trellis Mar 7, 2018, 12:24 AM

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couldn't you convert this to a vector field though by using the normal vectors?

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is the plane field the tangent planes?

distant stag Mar 7, 2018, 12:26 AM

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That involves choosing orientations for the planes.

hallow trellis Mar 7, 2018, 12:26 AM

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i guess what do they represent

distant stag Mar 7, 2018, 12:26 AM

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Which is not always possible.

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And that doesn't change the ability to integrate the plane field or not.

hallow trellis Mar 7, 2018, 12:29 AM

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why is choosing orientation for planes not always possible? does that have to deal with the type of space it is in?

distant stag Mar 7, 2018, 12:30 AM

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Suppose you are looking at a bunch of planes along a circle embedded in the manifold.

hallow trellis Mar 7, 2018, 12:31 AM

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like tangent planes?

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of the circle>

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or something else?

distant stag Mar 7, 2018, 12:32 AM

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A plane field is a map H which assigns to each point of your manifold M a plane.

hallow trellis Mar 7, 2018, 12:32 AM

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okay

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but that still doesn't answer what do the planes represent? are they just randomly chosen?

distant stag Mar 7, 2018, 12:33 AM

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Can I ask what the answer of the analogous question about vector fields is to you?

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Vector fields, plane fields, hyperplane fields, surface fields, etc. They are just things you can make. Do they have to represent something more?

hallow trellis Mar 7, 2018, 12:36 AM

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vector fields represent something though specifically on a surface they represent the change in the surface? like you're adding a context to the planes but I don't understand what that context is do you just assign a plane at each point of the circle? how do you decide what the equation of the plane is equal to

distant stag Mar 7, 2018, 12:37 AM

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The plane field is smooth.

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Like vector fields.

hallow trellis Mar 7, 2018, 12:37 AM

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yeah okay

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so what is the plane field defined as for the circle @distant stag

distant stag Mar 7, 2018, 12:43 AM

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Well I was just going to explain that you could have a plane field which flips along some circle.

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Then the normal vector field is not smooth for any choice of normals.

hallow trellis Mar 7, 2018, 12:45 AM

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so the plane field changes over time?

distant stag Mar 7, 2018, 12:45 AM

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📎 unknown.png

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No, the plane field changes like a vector field does.

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This is what I was going to describe.

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Imagine the red lines as planes.

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So the normal lines to the planes are any perpendicular lines.

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But if you start choosing normals, you can't end this circle without messing it up.

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Kind of like choosing a normal for a mobius band.

hallow trellis Mar 7, 2018, 12:47 AM

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but why is it just a random plane field?

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what is the end goal

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is probably a better question

distant stag Mar 7, 2018, 12:48 AM

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In the case of a contact manifold, it's not a "random plane field"

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It's a completely non-integrable plane field.

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This is just an additional mathematical structure added to the manifold.

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Like adding an operation to a set.

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"What's the end goal of a monoid?"

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I don't know, it's just additional structure.

hallow trellis Mar 7, 2018, 12:49 AM

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okay i see

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so you embed a surface with a plane field attached into a smooth contact manifold that is in odd dimensions.

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this operation allows you to see properties of the mainfold?

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what's the coolest result from this?

distant stag Mar 7, 2018, 12:53 AM

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No, the plane field is additional structure on the manifold. This plane field makes it a contact manifold.

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Then you look at the surfaces in this manifold.

hallow trellis Mar 7, 2018, 12:53 AM

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are completely non integrable plane fields unique for a surface?

distant stag Mar 7, 2018, 12:55 AM

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And you can have multiple different contact structures on a manifold.

hallow trellis Mar 7, 2018, 12:56 AM

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and what are some examples of applications?

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and yeah I would think so since you can embed multiple surfaces into a manifold

distant stag Mar 7, 2018, 12:56 AM

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Well it's not really dependent on the surfaces.

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That's just to do with the characteristic foliations that I work with, personally.

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Contact geometry appears in physics, originated from the theory of differential equations, gives a geometric framework for PDEs, etc.

hallow trellis Mar 7, 2018, 12:59 AM

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what about are they unique per surface?

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and by the characteristic foliations is that just different types of plane fields?

distant stag Mar 7, 2018, 1:01 AM

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The surfaces do not have the plane field

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The contact manifold (the ambient manifold that the surfaces are in) has the plane field.

hallow trellis Mar 7, 2018, 1:01 AM

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then why did you draw the planes on the circle???

distant stag Mar 7, 2018, 1:02 AM

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That was me just restricting your attention to some of the plane field (which lies on the circle)

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That was just an example of why you can't always associate a vector field to a plane field via the normals.

hallow trellis Mar 7, 2018, 1:02 AM

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okay

distant stag Mar 7, 2018, 1:03 AM

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The characterstic foliations are the foliations of the surface that you get when you intersect the plane field with the surfaces.

hallow trellis Mar 7, 2018, 1:05 AM

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okay so are the plane fields on a mainfold unique? or pretty common

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the non integralble ones

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are there special groups or something of plane fields? or some sort of algebra involved

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do you have links to stuff where i can learn more?

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sorry I'm done bugging you @distant stag I hope i didn't cause too much pain 😛

distant stag Mar 7, 2018, 1:13 AM

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No, there are many different contact structures sometimes.

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And there are groups associated with contact manifolds, the contactomorphisms, but that's analogous to a group of diffeomorphisms.

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Probably the best place to learn more is to learn more about manifolds first, since a lot of the stuff requires rigorous definitions.

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But etyner has some notes with applications, let me find them.

hallow trellis Mar 7, 2018, 1:15 AM

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yeah I'm going to eventually get through manifolds

distant stag Mar 7, 2018, 1:16 AM

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http://people.math.gatech.edu/~etnyre/preprints/papers/contlect.pdf

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This one doesn't have the applications, I can't find the one that does.

hallow trellis Mar 7, 2018, 1:18 AM

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so it IS related to the tangent

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📎 Capture.PNG

distant stag Mar 7, 2018, 1:18 AM

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Any plane field is a subbundle of the tangent bundle.

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(well I should say, hyperplane field)

hallow trellis Mar 7, 2018, 1:18 AM

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that was what i was trying to get out of you lol

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i kept asking

distant stag Mar 7, 2018, 1:19 AM

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kept asking what?

hallow trellis Mar 7, 2018, 1:20 AM

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if it was the tangent or something i dont remember i was just confused because what you were saying it sounded like the plane field was completly random except for the fact it was non integrable

distant stag Mar 7, 2018, 1:20 AM

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It effectively is.

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The tangent bundle is how you specify the planes.

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How you define them, that is.

hallow trellis Mar 7, 2018, 1:21 AM

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yeah i asked you how you define them too 😛

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Ive done a bit of the mainfolds so i know what a tangent bundle is paired with M

distant stag Mar 7, 2018, 1:21 AM

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e.g. the tangent bundle to R^3 is R^3xR^3 so it's analogous to specifying a "random" plane in R^3.

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It's just in more general situations, what one means by "hyperplane" becomes generalized to this subbundle definition.

hallow trellis Mar 7, 2018, 1:22 AM

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yeah but is it not random when you actually have a defined manifold though? instead of R^3

distant stag Mar 7, 2018, 1:23 AM

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Well it effectively is still, under choice of coordinates locally.

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Because of how a tangent bundle is defined.

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And since any manifold can be embedded in R^N for some suitable N....

hallow trellis Mar 7, 2018, 1:24 AM

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yeah

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but this makes SO much more sense

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now

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thanks for the link

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ill save it for later

upper karma Mar 7, 2018, 9:17 PM

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can someone help me?

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i have afew questions on circles

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i can vc if necessary

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JustAsk

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ok

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📎 image.jpg

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4,5,6,13,15,17,23 then back page 1,2
just need help with those questions
i can go on vc if that makes things easier

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the front is second post

waxen gorge Mar 7, 2018, 10:04 PM

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Why did u send the teachers name

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Now everyone knows where u live

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And the school ur at

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@upper karma

upper karma Mar 7, 2018, 10:24 PM

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lol

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idgaf

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to be honest

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📎 image.jpg

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📎 image.jpg

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4,5,6,13,15,17,23 then back page 1,2
just need help with those questions
i can go on vc if that makes things easier
the front is second post

waxen gorge Mar 7, 2018, 10:37 PM

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Is simple

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25 + x^2 = 4x^2

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25 = 3x^2

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x = 5*sqrt(3)/3

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9 + x^2/4 = 25

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x^2/4 = 16

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x^2 = 64

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x=8

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Idk I haven't done these in a while

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x° = 360- 117/2

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Use power of a point

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x^2 = 18*6

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x= 6*sqrt(3)

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x=215/2

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y = 145/2

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WRONG

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No one can read that

upper karma Mar 7, 2018, 10:52 PM

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lol ok lemme try this

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ty

waxen gorge Mar 7, 2018, 10:56 PM

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I think you missed all the power of a point problems

upper karma Mar 7, 2018, 11:00 PM

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i did this at 3 o clocking thinking i would have schoo

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but i had a day off from snow

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so

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lul

waxen gorge Mar 7, 2018, 11:21 PM

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Lol

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luckyyyy

upper karma Mar 7, 2018, 11:22 PM

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i have a questin

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@waxen gorge

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on number 4

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is it 2x

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or x^2

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nvm

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im retarted

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x*x equals x^2

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nope im lost

waxen gorge Mar 7, 2018, 11:29 PM

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Hi?

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@upper karma the length is 2x

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(2x)^2 = 4x^2

upper karma Mar 7, 2018, 11:31 PM

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but where u get 2x

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u need to multiple x and x?

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bc outside * whole

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wait

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the whole is

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2x

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......

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oh ok

waxen gorge Mar 7, 2018, 11:33 PM

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Lol

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It's just a right triangle

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= Pythagorean theorem

upper karma Mar 7, 2018, 11:34 PM

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oh

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im thinking of segment lengths formed by tanget and secants

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u have time for vc?

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feel like id be easier

waxen gorge Mar 7, 2018, 11:35 PM

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Not RN uwu

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U can ask someone else in general-1 tho

upper karma Mar 7, 2018, 11:53 PM

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@waxen gorge i started to get it

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thanks for the help

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feel like 4 is the only question i need help on

waxen gorge Mar 7, 2018, 11:57 PM

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Mk

shell kestrel Mar 8, 2018, 4:47 AM

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📎 IMG_20180308_174452.jpg

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Is this working correct

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For question

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https://media.discordapp.net/attachments/315464520090779659/421161896813527040/JPEG_20180308_172656.jpg?width=507&height=676

abstract arch Mar 8, 2018, 4:51 AM

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$$-\frac{4}{3}x + \frac{34}{3} = \frac{3}{4}x + 3\newline \frac{25}{3} = x\left(\frac{3}{4} + \frac{4}{3}\right)$$

charred spearBOT Mar 8, 2018, 4:51 AM

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📎 latex.png

abstract arch Mar 8, 2018, 4:51 AM

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$$\frac{25}{3} = \frac{25}{12}x\implies x = 4$$

charred spearBOT Mar 8, 2018, 4:51 AM

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📎 latex.png

abstract arch Mar 8, 2018, 4:52 AM

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Which is the point you have, good.

shell kestrel Mar 8, 2018, 4:53 AM

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what?

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https://media.discordapp.net/attachments/315464520090779659/421167916378226690/unknown.png?width=1442&height=526

abstract arch Mar 8, 2018, 4:54 AM

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Yeah, I don't see a problem with it other than some bad 6's and 4's(hard to distinguish) and you not specifically showing how you got the intersections

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But yeah, it's the right answer and the work seems solid other than that.

shell kestrel Mar 8, 2018, 4:55 AM

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alright sweet, just not sure what your x = 4 calculation achieved, is that just showing the x axis of the intersection?

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and also

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is there some other web application your using for formulating the mathbot

abstract arch Mar 8, 2018, 4:55 AM

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That would be solving for the point of intersection between the line and the perpendicular

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I did that because it wasn't shown in your work and I needed to check it.

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And no, I type it in the chat window.

shell kestrel Mar 8, 2018, 4:56 AM

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jeez, and yeah

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I did the simultaneous equations on the other side

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so is my answer of distance 5 good then?

abstract arch Mar 8, 2018, 4:56 AM

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yeah.

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Although, the position of the line relative to the point is a bit weird.

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Again, not wrong of course.

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Especially since there's no reference axis

shell kestrel Mar 8, 2018, 5:04 AM

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I think its fine for high school

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think_left

abstract arch Mar 8, 2018, 5:05 AM

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implying I'm not in high school

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I'm just saying, if you assume positive y and x are up and right, respectively, as is customary, then the line is on the wrong side of the point. and the point of intersection, (4,6) should be above and to the left of the point (7,2)

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But, it's not wrong, just a little detail that's a bit weird.

shell kestrel Mar 8, 2018, 7:59 AM

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oh yeah

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I didn't draw the points accurately

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I wouldn't normally visualise it but I just wanted to for some reason

upper karma Mar 8, 2018, 10:26 AM

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Any tricks for Trig not taught in school?

upper urchin Mar 8, 2018, 10:53 AM

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e^(ia)=cos (a)+isin (a)

upper karma Mar 8, 2018, 6:32 PM

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everyone knows that

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I'm sure it's taught in school

upper urchin Mar 8, 2018, 6:33 PM

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not usually before the trig stuff

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trig is about 12 yo that is usually 16-18yo

upper karma Mar 8, 2018, 6:33 PM

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yo?

upper urchin Mar 8, 2018, 6:33 PM

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years old

upper karma Mar 8, 2018, 6:33 PM

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oh

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🤷

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it's different depending on the school I'd guess

upper urchin Mar 8, 2018, 6:34 PM

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I would be very surprised if that was taught before trig in any school

upper karma Mar 8, 2018, 6:35 PM

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yeah obviously it wouldn't be

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but maybe in one year

upper urchin Mar 8, 2018, 6:36 PM

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yes but often times there are years separating it and if someone is asking about trig tricks then I'm going to assume they haven't learned that since you can get more or less anything you want from that

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hence a useful trick they at least haven't been taught in schools

upper karma Mar 8, 2018, 6:37 PM

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yeah

upper karma Mar 8, 2018, 7:42 PM

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Exactly

upper karma Mar 8, 2018, 8:02 PM

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I hate memorising trig formulas

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I keep forgetting them

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How do u memorise it

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Use them/learn the intuition behind them.

upper urchin Mar 8, 2018, 8:04 PM

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Using that formula

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Honestly

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Especially for multiple angles

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I guess you could also use vectors/ lines in the unit circle but it boils down to the same thing but less nicely expressed

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Complex geometry is great

upper karma Mar 8, 2018, 8:07 PM

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Haven't learnt vectors yet

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Just a beginner

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Double angle identity is kicking my ass

upper karma Mar 8, 2018, 9:35 PM

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Anyone know if I did this correctly? This is proportion involving line segments.

📎 0CA7CACA-EF8C-4CDC-A323-7B86F4A9A901.jpeg

waxen gorge Mar 8, 2018, 9:42 PM

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Haihai

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Yes

upper karma Mar 8, 2018, 9:52 PM

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Ooh, awesome!

abstract arch Mar 9, 2018, 1:56 AM

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yeah honestly anytime I think I may need double angle formula I'll just derive it real quick from Euler's. I can never remember it either. XD

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Not now that I'm not in the class, anyway.

waxen gorge Mar 9, 2018, 2:51 AM

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It's ez

upper karma Mar 9, 2018, 5:01 AM

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I can't even euler

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I guess I'll learn Euler before doing trig

worldly wave Mar 9, 2018, 10:13 AM

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📎 20180309_150454.jpg

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Guys.

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Is this a correct way of proving this question?

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My teacher said this is how I should do it

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but I have no idea how this proves

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the said condition

eager pendant Mar 9, 2018, 11:45 AM

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Four distinct points are arranged on a plane so that the segments connecting them have lengths a, a, a, a, 2a, and b. Express b in terms of a.

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seeing a, 2a, you could see that b=sqrt(3)*a works

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is there a solution that proves b must equal a*sqrt(3) ?

mint sandal Mar 9, 2018, 12:21 PM

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yes

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i got a solution, but it's kind of messy ;D

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first you prove that there's a equilateral triangle inside

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we know there are 4 a's so there either is an equilateral triangle inside or a rhombus

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but if it's a rhombus, one of the diagonals will have to be 2a, which is impossible if the points are distinct

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so there's an equilateral triangle with sides a

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add a point next to the triangle

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it has to be a distance a from one of the points, because we have 4 a's

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now the maximum distance from other points is 2a, which is when it's colinear with one of the sides of the triangle

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so it has to be colinear

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and the final side you draw on must be b

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with this geometric construction you get that b=sqrt(3)a

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by rule of sines :p

mint sandal Mar 9, 2018, 8:42 PM

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does this proof make sense?

quaint robin Mar 9, 2018, 11:11 PM

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Ehhhh

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This was an old AMC 10 problem

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My solution was form an equilateral triangle with side lengths a, meaning all angles are 60°

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Then make one of the vertices a midpoint so the extended side length would become 2a

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With this you have 5 of the 6 lengths given

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This is a 30-60-90 triangle, simply proven since there are two triangles within the larger triangle, an equilateral triangle and a 120-30-30 triangle

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so the last side, b, is a (sqrt3)

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@mint sandal This means you don't have to use law of sines

slender shore Mar 10, 2018, 1:45 PM

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@upper karma

upper karma Mar 10, 2018, 1:46 PM

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great so

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let me translate a couple things then we can get started

slender shore Mar 10, 2018, 1:46 PM

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before you do anything
did you think about complex numbers? it makes everything way easier in geometry

upper karma Mar 10, 2018, 1:47 PM

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I'm only in 10th grade I've never really gotten a chance to deal with complex numbers so I don't know anything about them

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to be honest I've never really liked math til I started taking geometry

slender shore Mar 10, 2018, 1:47 PM

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alright

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you should consider learning it during the summer or something. it's not that hard

upper urchin Mar 10, 2018, 1:48 PM

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are you talking about this problem?
Four distinct points are arranged on a plane so that the segments connecting them have lengths a, a, a, a, 2a, and b. Express b in terms of a.

upper karma Mar 10, 2018, 1:48 PM

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lemme take a picture of what I'm having problems with

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I'm reading a book called basic mathematics by Serge Lang

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after I'm done with that I'm willing to read anything you recommend

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oh @slender shore nevermind I just looked at it for a while and I get it now 😄

slender shore Mar 10, 2018, 1:49 PM

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lol

upper karma Mar 10, 2018, 1:49 PM

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sorry for wasting your time and thank you

slender shore Mar 10, 2018, 1:49 PM

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what was the problem

upper karma Mar 10, 2018, 1:52 PM

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okay so English ain't my native language, so this is a rough translation, but I didn't understand why with 90 degree triangles, if you divide them into two lesser triangles..

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oh my this is impossible to translate with my current knowledge

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but it is very simple

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nothing interesting

sweet heart Mar 10, 2018, 8:30 PM

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📎 unknown.png

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Does someone know how this thing is obtained in projective geometry?

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Reading Courant, but the pace is very brisk, and it doesn't seem to be explained well

upper urchin Mar 10, 2018, 8:31 PM

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I think it's differential geometry?

sweet heart Mar 10, 2018, 8:31 PM

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Projective

upper urchin Mar 10, 2018, 8:31 PM

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no I mean this is a result you can get using it

sweet heart Mar 10, 2018, 8:32 PM

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Well, perhaps 😄

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But it's apparently supposed to use cross-ratio

upper urchin Mar 10, 2018, 8:32 PM

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where you find a function which has the derivative being normal to each line it passes over

sturdy finch Mar 11, 2018, 3:10 PM

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ABCD rectangle
BC= 6 square root 3
DC= 3 square root 6
AC= ?

pulsar bay Mar 11, 2018, 3:14 PM

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use pythagoras

sturdy finch Mar 11, 2018, 3:14 PM

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hmm

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oki

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thx

pulsar bay Mar 11, 2018, 3:14 PM

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📎 image.jpg

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@sturdy finch

sturdy finch Mar 11, 2018, 3:15 PM

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XD

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thx

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but i didnt want to bother u

pulsar bay Mar 11, 2018, 3:15 PM

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No worries

sturdy finch Mar 11, 2018, 3:16 PM

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:]

upper karma Mar 11, 2018, 8:03 PM

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Anyone know if I did #13 correctly?

📎 E5C8C5E0-7ACC-4310-BDCE-7DA2C7BCA9F7.png

keen aspen Mar 11, 2018, 8:22 PM

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Yes

upper karma Mar 11, 2018, 8:26 PM

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And did I do #20 correctly?

📎 D2EB8F2A-B1B5-484B-B8D0-D58FCE7EB239.png

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Btw, I'm asking if the perimeter is 81.

keen aspen Mar 11, 2018, 9:13 PM

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yes as well

neon fossil Mar 11, 2018, 10:55 PM

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is there a topological manifold with no differentiable structure available?

mint sandal Mar 11, 2018, 10:57 PM

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yes

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https://en.wikipedia.org/wiki/Differentiable_manifold#Relationship_with_topological_manifolds

Differentiable manifold

In mathematics, a differentiable manifold is a type of manifold that is locally similar enough to a linear space to allow one to do calculus. Any manifold can be described by a collection of charts, also known as an atlas. One may then apply idea...

neon fossil Mar 11, 2018, 11:01 PM

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cool

celest swan Mar 12, 2018, 12:56 AM

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«consider p(x, y), a surface that is expanding infinitely in a tridimensional space (x, y, z). If p(x, y) is flat, then p(x, y) is a sphere with an infinite radius, circumference, and area.»```

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I can neither prove it right or wrong... :/

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<@&286206848099549185> can one of you prove or disprove it?

abstract arch Mar 12, 2018, 12:58 AM

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I mean it's just like a limit essentially.

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A similar idea was used to find the area of a circle

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i.e. you assume that the circle represents the limit as n tends towards infinity of an n-sided regular polygon

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This is just another way of thinking about a plane.

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There's also a concept in R^2 in which you can think of a line as a circle with infinite radius

celest swan Mar 12, 2018, 1:04 AM

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Yes but I don’t know about the R^2 stuff

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@abstract arch it’s not really the same thing... because here we search the inclination

abstract arch Mar 12, 2018, 1:06 AM

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What do you mean?

neon fossil Mar 12, 2018, 2:00 AM

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sorry what does this mean?

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expanding infinitely?

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aka not bounded?

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spheres are compact, it can't be a sphere

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if it's flat, it could be R^2

celest swan Mar 12, 2018, 12:06 PM

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Not bound

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Like

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No edge

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Exactly

brittle kraken Mar 12, 2018, 12:35 PM

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Can someone help me with this? Idk what I did wrong

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📎 IMG_20180312_221749.jpg

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#66

upper karma Mar 12, 2018, 12:36 PM

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Dont ping admins for help there is a helper tag

brittle kraken Mar 12, 2018, 12:37 PM

#

Right ty

#

So I ping helpers instead?

#

<@&286206848099549185>

upper karma Mar 12, 2018, 12:44 PM

#

That looks tough :(

pulsar bay Mar 12, 2018, 12:46 PM

#

use sine rule

brittle kraken Mar 12, 2018, 12:48 PM

#

Is there an easier way to solve that aside from using the sine rule? An example given here is used, but I've done everything and I cannot find the answer that I've gotten on the list of choices.

#

http://prntscr.com/iq3dsm

Lightshot

Screenshot

Captured with Lightshot

pulsar bay Mar 12, 2018, 12:52 PM

#

i mean this is a different problem to do with similar triangles

#

their ratios are the same

#

either that or im looking at the wrong q

#

hold on

#

sorry was looking at 67

#

ratios pretty much the only way to do it

#

although you are given a bunch of answers

#

so you can find a ratio on the right triangle and substitute the values in

#

to find the same ratio

brittle kraken Mar 12, 2018, 12:57 PM

#

Ah I see.

#

Wait were you looking at 66 this time?

#

Or the example?

pulsar bay Mar 12, 2018, 1:09 PM

#

66

#

The one above the angles

#

And sides

past mantle Mar 12, 2018, 9:30 PM

#

im cofused

#

📎 Capture.PNG

upper urchin Mar 12, 2018, 9:35 PM

#

what's the question?

upper sedge Mar 12, 2018, 9:36 PM

#

probably get x

#

Name all of the angles, and write down everything you know about them.

upper urchin Mar 12, 2018, 9:37 PM

#

probably

#

but you never know

flat geyser Mar 12, 2018, 9:37 PM

#

180-10x-2x^2 = 0

upper sedge Mar 12, 2018, 9:47 PM

#

Oh, maybe it wants the arc length from A to B. That's a fair amount more work from here, isn't it?

#

I'll take a closer look when I get home.

keen aspen Mar 13, 2018, 12:03 AM

#

Yes arclength

abstract arch Mar 13, 2018, 12:03 AM

#

Since angle D and Angle C both subtend the same arc(and have a vertex on the circumference), AB, they must be equal

keen aspen Mar 13, 2018, 12:03 AM

#

^

abstract arch Mar 13, 2018, 12:04 AM

#

$$2x^2 = 10x \implies 2x = 10 \implies x =5$$, since $$x \neq 0$$

charred spearBOT Mar 13, 2018, 12:04 AM

#

📎 latex.png

abstract arch Mar 13, 2018, 12:05 AM

#

Now, since an angle with a vertex on the circumference is half the angle of the central angle it subtends, we can say

#

$$\theta = 10^\circ$$

charred spearBOT Mar 13, 2018, 12:06 AM

#

📎 latex.png

abstract arch Mar 13, 2018, 12:06 AM

#

So, a- wait we don't have a radius

#

uh... I'll just use r

#

So arc length is given by

#

$$\frac{\theta}{180}\times \pi r$$

#

Where theta is in degrees

#

Wait what wrong formula

past mantle Mar 13, 2018, 12:07 AM

#

oh

keen aspen Mar 13, 2018, 12:07 AM

#

Yeah

charred spearBOT Mar 13, 2018, 12:07 AM

#

📎 latex.png

past mantle Mar 13, 2018, 12:07 AM

#

are we using inscribed angles? we just learn't that

keen aspen Mar 13, 2018, 12:08 AM

#

Anyways, since the angle is on the edge of the circumference of that circle, the indicated arc corresponding to that is (theta)*2

past mantle Mar 13, 2018, 12:08 AM

#

we dont use theta for some reason

keen aspen Mar 13, 2018, 12:08 AM

#

Angle measure

abstract arch Mar 13, 2018, 12:09 AM

#

just call it whatever you want, theta is irrelevant. it's just a letter

past mantle Mar 13, 2018, 12:09 AM

#

ik ik

abstract arch Mar 13, 2018, 12:10 AM

#

$$\therefore AB = \frac{10^\circ}{180^\circ}\times\pi r = \frac{\pi}{18}r$$

charred spearBOT Mar 13, 2018, 12:11 AM

#

📎 latex.png

abstract arch Mar 13, 2018, 12:11 AM

#

Can't figure out how to do the arc sign.

keen aspen Mar 13, 2018, 12:11 AM

#

Where did you get 10 from

#

You are doing it in a complicated way

#

You have the two angles equal to each other so using algebra you can say that x is 5, where each angle measure is 50 degrees

#

since the inscribed angle is made with the two chords and the circumference of the circle, all we need to do is 50*2

#

So the arc measure is just 100 degrees

#

You are finding the arc length, however the measure is 100 degrees

abstract arch Mar 13, 2018, 12:17 AM

#

Oh yeaaaaah. forgot to say get the actual angle after I got x... whoopsies

#

=tex \newcommand{\overarc}[1]{\ooalign{\hfil\scalebox{0.9}{$#1$}\hfil\cr\hfil\raisebox{8pt}{\rotatebox{90}{\scalebox{1}{$)$}}}\hfil\cr}} \begin{align} 2x^2 &= 10x \implies x = 5 \text{, Since } x \neq 0\ \theta &= 10(5) = 50^\circ\ \therefore m\overarc{AB} &= 2(50) = 100^\circ\end{align}

charred spearBOT Mar 13, 2018, 12:20 AM

#

📎 latex.png

keen aspen Mar 13, 2018, 12:20 AM

#

👍🏻

abstract arch Mar 13, 2018, 12:20 AM

#

I had to find a command definition to make that arc symbol.

#

Appreciate it

charred spearBOT Mar 13, 2018, 12:23 AM

#

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

upper sedge Mar 13, 2018, 12:51 AM

#

Oh cool, they got to it.

#

Thanks guys.

keen aspen Mar 13, 2018, 2:06 AM

#

ok

brittle kraken Mar 14, 2018, 1:33 AM

#

📎 IMG_20180314_113323.jpg

#

What am I doing wrong so far @ #74?

keen aspen Mar 14, 2018, 2:58 AM

#

Just remember that the sum of 2 sides must always be greater than the length of the third

#

So when you add 10 and 6 you get 16

#

well, the third length cannot be any thing greater than 16

#

So that leaves with just B.

#

and the 4 suffices because you can do 4.01+6 = 10.01 which is greater than 10

#

So we can say that the possible lengths for that triangle would be 4<x<16

brittle kraken Mar 14, 2018, 4:19 AM

#

I see, thanks

#

📎 IMG_20180314_142031.jpg

#

And can somebody teach me how to do 77?

thorn talon Mar 14, 2018, 4:22 AM

#

Well

#

I would just check each

upper sedge Mar 14, 2018, 4:22 AM

#

I'm afraid that's entirely upside down.

thorn talon Mar 14, 2018, 4:22 AM

#

For the first one, all sides are given

#

So if all the matching sides are in an equal ratio

#

Then you have similarity

brittle kraken Mar 14, 2018, 4:26 AM

#

I'm really bad at geometey so spare me the embarrassment

#

Ye... My phone is weird sometimes

thorn talon Mar 14, 2018, 4:27 AM

#

Match smallest sides with smallest

#

Medium with medium

#

Largest with largest

analog dust Mar 14, 2018, 10:22 PM

#

Use trig to find the other sides if you need them

#

@brittle kraken

#

Or just use the definitions of similar triangles (if in the drawing, they did code same angles with same things)

upper karma Mar 15, 2018, 1:47 AM

#

Is anyone here good at circles

keen aspen Mar 15, 2018, 2:41 AM

#

JustAsk

upper karma Mar 15, 2018, 2:53 AM

#

PJS

#

Questions-5

#

he asked there

#

tried to help but failed, cant seem to recall the way to solve those

#

So if you could help him id be also thankful 😃

keen aspen Mar 15, 2018, 3:35 AM

#

👍🏼

steady hull Mar 15, 2018, 4:12 PM

#

how would one take a 3d cross section of a 4d tetrahedron?

#

<@&286206848099549185>

upper urchin Mar 15, 2018, 4:13 PM

#

do you mean a 4-simplex?

steady hull Mar 15, 2018, 4:13 PM

#

im dum pls take it slow

#

I mean a 5 cell

upper urchin Mar 15, 2018, 4:14 PM

#

yes a 4-simplex is a 5-cell

steady hull Mar 15, 2018, 4:14 PM

#

ok

#

so how to take the cross section?

#

I sort of understand the concept of 4d

#

its sort of like a bunch of 3d figures layered together

#

ugg

#

y can't humans be like 10d creatures

#

so we can see 9d and understand everything below

#

can anyone help me?

upper urchin Mar 15, 2018, 4:24 PM

#

Have you tried Wikipedia? There's a whole section on projection of a 4-simplex

steady hull Mar 15, 2018, 4:24 PM

#

yea

#

I read it all

#

its tough

#

im 13 😦

#

and first exposure

#

to higher dimensions

naive scarab Mar 15, 2018, 4:25 PM

#

so it depends on how good your description of the 4d object is

#

and it also depends how you want to project it

steady hull Mar 15, 2018, 4:25 PM

#

5 cell

#

thats all the questions says

naive scarab Mar 15, 2018, 4:26 PM

#

ok, and how are you asked to take the cross section? cause there are a bunch you could take

steady hull Mar 15, 2018, 4:27 PM

#

well

#

it gives us a 3d version

#

where its a tetrahedron ABCD

naive scarab Mar 15, 2018, 4:27 PM

#

My point is that there are infinitely many cross sections you could take

#

so which one

steady hull Mar 15, 2018, 4:28 PM

#

and a square is drawn parallel to side AB and CD

naive scarab Mar 15, 2018, 4:28 PM

#

just like for a 3d object there are infinitely many 2d planes through it

steady hull Mar 15, 2018, 4:28 PM

#

yea

#

equidistand

#

and parrallel to the sides

#

I should screenshot

naive scarab Mar 15, 2018, 4:32 PM

#

where'd you get this question from?

steady hull Mar 15, 2018, 4:32 PM

#

y?

#

the internet

#

trying to learn this stuff

naive scarab Mar 15, 2018, 4:33 PM

#

oh it's just an interesting question

#

and it's hard to come across this in normal math education

steady hull Mar 15, 2018, 4:34 PM

#

I dont lol

#

im bored of school

#

look for other stuff to do

naive scarab Mar 15, 2018, 4:34 PM

#

cool 😃

steady hull Mar 15, 2018, 4:35 PM

#

common core in az never covers ths

naive scarab Mar 15, 2018, 4:35 PM

#

lol I can imagine

steady hull Mar 15, 2018, 4:36 PM

#

so at first I thought that it was a cube

#

bc its up one dimension

#

but that would be too ez

#

so them maybe it could be a tetrahedron

#

bc 4d is a lot of 3d figures

#

and now I am confused

naive scarab Mar 15, 2018, 4:36 PM

#

I mean it's bad to just guess in this way, there should be a way of thinking it through

steady hull Mar 15, 2018, 4:37 PM

#

yea

#

I guess its more of a conceptual prpblem

#

so no ideas from u?

naive scarab Mar 15, 2018, 4:39 PM

#

I'm thinking lol, but it is quite hard

steady hull Mar 15, 2018, 4:40 PM

#

its really interesting bc the problem set also has a question about knot theory

naive scarab Mar 15, 2018, 4:40 PM

#

a good math problem will keep you stuck for hours, and this is probably one such example. Don't let that discourage you.

steady hull Mar 15, 2018, 4:40 PM

#

k

naive scarab Mar 15, 2018, 4:41 PM

#

so here's how I would do it if I was being pretty serious

#

step 1: find points A,B,C,D,E in 4d space which are the vertices of a 5-cell

#

do you know how to calculate distance in 4d?

steady hull Mar 15, 2018, 4:41 PM

#

well its a unit 5 cell

#

yea

naive scarab Mar 15, 2018, 4:41 PM

#

ok, so in principle you can do that

#

step 2: write down the equation that would be satisfied by points in this hyperplane

#

step 3: try and simplify that to something you can understand

#

there is probably a way to get a visual intuition, or maybe there's a clever trick, but I feel like what I've described would be the only way to really prove it

steady hull Mar 15, 2018, 4:44 PM

#

ok

hearty pilot Mar 16, 2018, 12:06 AM

#

How do I get the inverse function of 2^x?

upper urchin Mar 16, 2018, 12:21 AM

#

Use logs

umbral snow Mar 16, 2018, 12:21 AM

#

log2(x) is the inverse by definition

analog dust Mar 16, 2018, 5:08 PM

#

Anyone got fun and medium hard geometry problems?

dull cedar Mar 16, 2018, 8:06 PM

#

What is the perimeter of gAY

upper urchin Mar 16, 2018, 8:17 PM

#

There is a set of straight lines in a plane such that each line intersects exactly ten others. Which of the following could not be the number of lines in that set?
A 11
B 12
C 15
D 16
E 20

#

somewhat interesting reasoning question

mossy vine Mar 16, 2018, 10:49 PM

#

oh huh this years SMC

#

I remember enjoying that question

upper urchin Mar 17, 2018, 12:16 AM

#

I was just looking through past papers Iol

#

A lot easier than i remember

upper urchin Mar 17, 2018, 12:31 PM

#

try drawing a line from D to F

#

hang on

#

actually I'm not sure you can work that out with the given infomation

#

FED is 75 yes

keen aspen Mar 17, 2018, 4:58 PM

#

BCE is 28, so the corresponding arc is 2*28, so 56

#

if that's 56, that makes angle A, half that which is 28

#

So BAC is 28

#

Since BAC is 28, and BAC forms an isosceles triangle, anle B and C are both (180-28)/2, so 76

#

That makes arc ABC 76(2)+56 so 208

#

Angle ADC is half of that so 104

#

Now , arc BC is 56, and when you make a triangle out of COB, both CO and BO are gonna be the same length

#

So you can say that angle COB is 56

#

So to sum up things.....

#

a) 28 b) 76 c) 104 d) 56

past mantle Mar 17, 2018, 5:06 PM

#

im confused

#

📎 Capture.PNG

keen aspen Mar 17, 2018, 5:35 PM

#

Hmm well for angle EJG its just 180-89 bc they are supplements

#

so 91

#

Using the Ange of Intersecting Chords, theorem, 91=1/2(137+x) where x is eg

#

So using algebra, we can say that mEG is 45

#

Using this theorem again for mDE, we do 89=1/2(61+x) where x is DE

#

using algebra, we can say that mDE is 117

#

See if you can do 29 and 30 by yourself

zenith ember Mar 17, 2018, 7:01 PM

#

Anyone here play Euclidea? I've been trying to figure out how to find the intersection of angle bisectors in an arbitrary triangle with only 6 operations.

#

The hint says it's "Circle, Circle, Circle, Line, Circle, Line" but I still can't see it.

upper urchin Mar 17, 2018, 7:02 PM

#

what are the 6 operations?

zenith ember Mar 17, 2018, 7:03 PM

#

Drawing a circle is an operation. Drawing a Line is an operation.

#

It's a geometric construction game.

#

Things like, "Given a line and a point not on the line, construct a line parallel to the given line that passes through the given point."

eager pendant Mar 17, 2018, 10:59 PM

#

@zenith ember which level is this?

zenith ember Mar 17, 2018, 11:00 PM

#

Beta-2, the E challenge.

eager pendant Mar 17, 2018, 11:11 PM

#

@zenith ember so i just redownloaded it and finished beta

#

are you sure its level 2?

zenith ember Mar 17, 2018, 11:12 PM

#

2.2: Intersection of Angle Bisectors. Yeah. looking at it right now.

#

It's been added since Euclidea was first released.

keen aspen Mar 18, 2018, 4:49 AM

#

@upper karma to?

dusty timber Mar 18, 2018, 7:10 AM

#

Hey

#

📎 JPEG_20180318_021010.jpg

#

How do I approach this?

shadow anvil Mar 19, 2018, 1:08 AM

#

@dusty timber Still stuck?

dusty timber Mar 19, 2018, 1:08 AM

#

Yeah I stopped looking at it after a while

shadow anvil Mar 19, 2018, 1:09 AM

#

Just realised this is like 18 hours late

#

but oh well

#

Well

#

First

#

It's safe to say that triangle is isosceles

upper urchin Mar 19, 2018, 1:10 AM

#

scalene

shadow anvil Mar 19, 2018, 1:10 AM

#

wat

upper urchin Mar 19, 2018, 1:10 AM

#

it says it's a scalene triangle

shadow anvil Mar 19, 2018, 1:10 AM

#

oh

#

literally says

#

rip

#

didn't even read that ngl

#

Well ok

upper urchin Mar 19, 2018, 1:11 AM

#

reading is good

shadow anvil Mar 19, 2018, 1:11 AM

#

We can find that bottom angle pretty easy

#

Just through sin rule

#

=tex \frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}

charred spearBOT Mar 19, 2018, 1:12 AM

#

📎 latex.png

shadow anvil Mar 19, 2018, 1:13 AM

#

So the sin of an angle divided by its opposite side

#

is equal to the sin of another angle divided by its opposite side

#

and so on

#

So then once we have that angle

#

we can find the length of the final side

zenith ember Mar 19, 2018, 1:16 AM

#

Wasn't there a "No trig" specification?

shadow anvil Mar 19, 2018, 1:16 AM

#

Was there?

zenith ember Mar 19, 2018, 1:16 AM

#

No, nevermind.

#

That was something else.

shadow anvil Mar 19, 2018, 1:16 AM

#

Dude

#

YOu made me panick

#

That would've been the second thing here I just didn't read

zenith ember Mar 19, 2018, 1:17 AM

#

Sorry.

shadow anvil Mar 19, 2018, 1:18 AM

#

Is there even a way to approach without trig?

zenith ember Mar 19, 2018, 1:18 AM

#

You know, I'd completely forgotten about that triangle property. The Sin A/A = Sin B/B

#

I dunno.

#

I didn't think about it that much. Probably not.

shadow anvil Mar 19, 2018, 1:18 AM

#

And it's not really that common I don't think so

#

The sin thing that is

zenith ember Mar 19, 2018, 1:19 AM

#

Trying to soothe my ego? Not necessary. It's probably been a couple of decades since I last used it.

shadow anvil Mar 19, 2018, 1:19 AM

#

lmao

keen aspen Mar 19, 2018, 1:38 AM

#

Law of sines are common :v

#

more common than the law of cosine

shadow anvil Mar 19, 2018, 1:39 AM

#

Neither is common smh

keen aspen Mar 19, 2018, 1:39 AM

#

In trig class yes

zenith ember Mar 19, 2018, 1:45 AM

#

I haven't taken trig in ages.

keen aspen Mar 19, 2018, 1:50 AM

#

I'm taking it now

#

I love it

#

I got a 100 on my last trig test

analog dust Mar 19, 2018, 9:04 PM

#

Prove that the angle AKD = 90°

📎 JPEG_20180319_220350.jpg

#

Btw DE is not always a diameter

#

And this is just for people who want to do it.. Thought it was a quite interesting problen

prime urchin Mar 20, 2018, 12:45 AM

#

not really anything tho I have a quiz tomorrow which goes over Circumference & area, arc length + sector area, and degrees to radians, radians to degrees

#

can someone please help>

#

?*

keen aspen Mar 20, 2018, 2:20 AM

#

Sure, just dm me for help @prime urchin

prime urchin Mar 20, 2018, 2:20 AM

#

alright

eager pendant Mar 20, 2018, 6:29 AM

#

@analog dust how isi the 2nd circle defined?

zenith ember Mar 20, 2018, 6:48 AM

#

I think it's based off the perpendicular.

#

Centered at the midpoint, radius = half the length of the perpendicular

analog dust Mar 20, 2018, 9:09 AM

#

@eager pendant The diameter is AH

charred spearBOT Mar 20, 2018, 9:28 AM

#

Information on wolfram has been send to you privately.

royal marsh Mar 20, 2018, 9:41 AM

#

The equation of a line segment AB are A (2, -5) and B (10, x), the midpoint of AB is M (m,3). Find the values of x and m and the equation of the line AB.?

#

@violet nest @magic arrow @shadow anvil

violet nest Mar 20, 2018, 9:51 AM

#

do you know how to do this?

shadow anvil Mar 20, 2018, 9:52 AM

#

Which bits are you confused by

royal marsh Mar 20, 2018, 9:52 AM

#

I got no clue on how to do this

#

ik the general formulas for midpoint

#

and gradient

#

and what not

#

but im not sure how to find x and m

shadow anvil Mar 20, 2018, 9:53 AM

#

Well the mid point

#

Is going to be half way

#

in both x and y directions

#

So you know AB starts at 2 in x

#

and half way from A to B in x is 10

#

Does that make sense?

royal marsh Mar 20, 2018, 9:55 AM

#

not really 😂

shadow anvil Mar 20, 2018, 9:55 AM

#

Ok

#

So

#

We've got a point

#

2,-5

#

A

royal marsh Mar 20, 2018, 9:55 AM

#

yes

shadow anvil Mar 20, 2018, 9:55 AM

#

And we've got a second point

#

B(m, 3)

royal marsh Mar 20, 2018, 9:55 AM

#

with a pronumeral

#

yes

shadow anvil Mar 20, 2018, 9:55 AM

#

Now we're going to look at a to b

#

in the x plane

#

So we're only looking at x

#

So AB in x is 2 to m

#

the middle of that liine is at 10

#

So the distance from 2 to 10

#

and 10 to m

#

is the same

royal marsh Mar 20, 2018, 9:56 AM

#

ok

#

so 8

shadow anvil Mar 20, 2018, 9:56 AM

#

Yep

#

So if the distance from 10

#

to m

#

is eight

#

what is m

royal marsh Mar 20, 2018, 9:57 AM

#

umm

#

18

shadow anvil Mar 20, 2018, 9:57 AM

#

yep

#

So did you understand what we just did?

royal marsh Mar 20, 2018, 9:57 AM

#

yeh kind of if i imagined it visually

shadow anvil Mar 20, 2018, 9:57 AM

#

Imagining it visually is perfcet

#

It's kinda awkward to explain it through text

royal marsh Mar 20, 2018, 9:57 AM

#

yeh

shadow anvil Mar 20, 2018, 9:57 AM

#

That's why I suggest drawing it

#

and marking out the x and y components seperately

#

But now you do the same as we just did

#

with with y instead

royal marsh Mar 20, 2018, 9:58 AM

#

The equation of a line segment AB are A (2, -5) and B (10, x), the midpoint of AB is M (m,3). Find the values of x and m and the equation of the line AB. SO we know M is (18,3)

#

so the distance between -5 and 3?

shadow anvil Mar 20, 2018, 11:09 AM

#

yes

violet nest Mar 20, 2018, 11:41 AM

#

@royal marsh you online?

#

Lemme show you an easy way

#

=tex \text{Midpoint between 2 points} = \left(\frac{x_1+x_2}{2} , \frac{y_1 + y_2}{2} \right)

charred spearBOT Mar 20, 2018, 11:42 AM

#

📎 latex.png

violet nest Mar 20, 2018, 11:43 AM

#

so the midpoint is (m, 3) so

#

=tex (m, 3) = \left(\frac{2+10}{2} , \frac{x -5}{2} \right)

royal marsh Mar 20, 2018, 11:44 AM

#

yeh im here

charred spearBOT Mar 20, 2018, 11:44 AM

#

📎 latex.png

violet nest Mar 20, 2018, 11:44 AM

#

just solve for m now

#

and x

#

m = 6

#

and 6 = x-5, x = 11

#

ez pz

royal marsh Mar 20, 2018, 11:44 AM

#

haha thanks a lot

#

this helped

analog dust Mar 20, 2018, 12:29 PM

#

A triangle ABC is inscribed in a circle with center O. The altitude from A intersects BC in H.
The circle with diameter AH intersects AB in D and AC in E.
Show that DE is perpendicular to OA

#

I posted the drawing yesterday

broken sierra Mar 20, 2018, 6:51 PM

#

Can someone help me understand radians and when they should be used over degree

chrome fiber Mar 20, 2018, 7:27 PM

#

A radian is an angle subtended at the center of a circle by an arc the length of which is equal to the radius of the circle

#

And you should usually use radians when your calculations involve pi

#

Degrees and radians are the same thing, in essence

analog dust Mar 20, 2018, 7:43 PM

#

They mesure ANglE

zenith ember Mar 20, 2018, 9:21 PM

#

They're different the way meters and yards are different.

brittle kraken Mar 21, 2018, 1:26 PM

#

Can you guys review my answers for the following? (#77, #80, #86, #87, #88, #91)

#

📎 IMG_20180321_232321.jpg 📎 IMG_20180321_232426.jpg 📎 IMG_20180321_232413.jpg

#

📎 IMG_20180321_232454.jpg 📎 IMG_20180321_232506.jpg

#

Apologies for the poor camera pictures, since my camera acts weird as of late.

zinc zodiac Mar 22, 2018, 11:19 AM

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I have a math question which I don't know how to solve. This is just a short test for a certificate in construction. Could anyone help? Message me and I'll send the question.

mint sandal Mar 22, 2018, 11:20 AM

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just post it here

zinc zodiac Mar 22, 2018, 11:23 AM

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sure

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https://gyazo.com/b6e1737150e723eadb6093925b36170b

Gyazo

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You can't work it out algebraically as far as I know. I used a ruler and it's not to scale (although it looks it is). The problem is clearly the top left length.

umbral snow Mar 22, 2018, 11:28 AM

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Not enough info.

mint sandal Mar 22, 2018, 11:32 AM

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yeah you could shift the left side horizontally and all parameters stay the same while perimeter changes

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so it's impossible

zinc zodiac Mar 22, 2018, 12:54 PM

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okay what about algebraically?

upper urchin Mar 22, 2018, 12:55 PM

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just say the unknown length is x

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otherwise you can't solve it

chrome fiber Mar 22, 2018, 1:11 PM

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Why not?

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Use similar triangles?

hearty barn Mar 22, 2018, 1:21 PM

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Nay wont work

chrome fiber Mar 22, 2018, 1:23 PM

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Why not?

hearty barn Mar 22, 2018, 1:25 PM

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@chrome fiber which triangles are similar?

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Wait lemme label the corners

chrome fiber Mar 22, 2018, 1:26 PM

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The one with 140mm, 100mm sides and the one with the unknown length and 80mm side

hearty barn Mar 22, 2018, 1:26 PM

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📎 JPEG_20180322_185557.jpg

chrome fiber Mar 22, 2018, 1:26 PM

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ABC ~ CDE

hearty barn Mar 22, 2018, 1:26 PM

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Oh fook

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Nay

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Nope

chrome fiber Mar 22, 2018, 1:27 PM

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?

hearty barn Mar 22, 2018, 1:27 PM

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If angle BCA = x

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Then angle DCE is 90 - x

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Oh damn

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Sorry

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Yep u right

chrome fiber Mar 22, 2018, 1:28 PM

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The two lines are parallel

hearty barn Mar 22, 2018, 1:28 PM

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The triangles ARE SIMILARR

chrome fiber Mar 22, 2018, 1:28 PM

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Yeah

hearty barn Mar 22, 2018, 1:28 PM

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Shall i post the solution plzzz

chrome fiber Mar 22, 2018, 1:29 PM

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No

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Let them try it on their own

hearty barn Mar 22, 2018, 1:29 PM

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xD

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Wokei

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Damn u have good eyes xD

chrome fiber Mar 22, 2018, 1:30 PM

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It was pretty obvious, they gave you three sides

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So you could just make two triangles and go from there

mint sandal Mar 22, 2018, 1:44 PM

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wait so you're saying the perimeter problem is solvable?

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@chrome fiber how are the triangles similar again?

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there's no guarantee that the three points lie on the same line

hearty barn Mar 22, 2018, 2:43 PM

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shinshen chck it again

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@mint sandal

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Even i dont feel they are similar...but therz solid proof there is

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The angles of the triangles are same

mint sandal Mar 22, 2018, 2:44 PM

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but i provided a proof that it's impossible to solve 🤔

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@hearty barn show me the proof

hearty barn Mar 22, 2018, 2:44 PM

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📎 JPEG_20180322_185557.jpg

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ABC is similar to CDE

mint sandal Mar 22, 2018, 2:45 PM

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wait why

hearty barn Mar 22, 2018, 2:45 PM

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All the angles are equal 🙏

mint sandal Mar 22, 2018, 2:45 PM

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are they? 🤔

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i don't think that's necessarily true

hearty barn Mar 22, 2018, 2:46 PM

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PM me

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Even i cant believe xD

mint sandal Mar 22, 2018, 2:46 PM

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probably because it's false...

hearty barn Mar 22, 2018, 2:48 PM

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Noww

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@chrome fiber

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I have solid proof to disprove ur proof

mint sandal Mar 22, 2018, 2:49 PM

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👏

ruby swallow Mar 22, 2018, 3:01 PM

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@hearty barn you have no proof that they are all 90

mint sandal Mar 22, 2018, 3:15 PM

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@ruby swallow its assumed

chrome fiber Mar 22, 2018, 3:15 PM

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Yeah, that's a reasonable assumption though

mint sandal Mar 22, 2018, 3:15 PM

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Hi june

chrome fiber Mar 22, 2018, 3:15 PM

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Hey

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But yes

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I'm pretty sure it's okay to assume the points are collinear

mint sandal Mar 22, 2018, 3:16 PM

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Why

chrome fiber Mar 22, 2018, 3:16 PM

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You're not given any other piece of info

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Two sides of two right triangles, one unknown

mint sandal Mar 22, 2018, 3:18 PM

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Well then you might as well just use a ruler

chrome fiber Mar 22, 2018, 3:19 PM

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?

umbral snow Mar 22, 2018, 3:20 PM

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Wat? I can shrink AB

mint sandal Mar 22, 2018, 3:25 PM

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@umbral snow yeah thats the first thing i said when i saw the problem

hearty barn Mar 22, 2018, 3:28 PM

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I thought exactly the same @umbral snow @mint sandal when jun said abt similar trianglea

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*triangles

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📎 Screenshot_2018-03-22-21-00-45.png

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that awkward realisation

mint sandal Mar 22, 2018, 3:33 PM

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😅

hearty barn Mar 22, 2018, 3:36 PM

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This gotta be de meme of de month atleast

chrome fiber Mar 22, 2018, 3:41 PM

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That's the most you can do with what's been given

hearty barn Mar 22, 2018, 3:42 PM

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👍

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Orr we can assume the shape as a circle and use 2pi r -_-

mint sandal Mar 22, 2018, 3:44 PM

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@hearty barn great method to solve it

hearty barn Mar 22, 2018, 3:45 PM

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👊

mint sandal Mar 22, 2018, 3:51 PM

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ouch

hearty barn Mar 22, 2018, 3:53 PM

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panda_dead sorry xD

mint sandal Mar 22, 2018, 3:55 PM

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its ok

chrome fiber Mar 22, 2018, 5:32 PM

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You're making it seem like it's a far-fetched assumption when it's not

mint sandal Mar 22, 2018, 10:10 PM

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📎 unknown.png

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why is the form on M skew-symmetric?

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halp

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I don't think it was assumed that the form on P was skew-symmetric

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@copper valve

olive tinsel Mar 22, 2018, 10:17 PM

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What is G

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What is V

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What is E

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What is P

copper valve Mar 22, 2018, 10:20 PM

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GWcfcThonk

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is skew symmetric the same as alternating

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o it is

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i have no idea tbh. i do remember skimming ahead in calc on manifolds

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where it brings up exterior algebras and wedge products, which look really similar, and mentions multilinear skew symmetric k-forms

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looks super similar GWcfcThonk
idk enough of this stuff, just noticed similarity XD

mint sandal Mar 22, 2018, 10:58 PM

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@olive tinsel G is the group acting on the bundle, V is the standard fibre which is also a vector space on which G acts by representation, E is the associated bundle, P is the principal bundle

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skew symmetric is the same as alternating, yeah

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basically if any two vectors repeat as arguments, the output is 0

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or equivalently switching 2 arguments switches the sign

mint sandal Mar 22, 2018, 11:17 PM

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eh i guess ill just move on

upper karma Mar 23, 2018, 12:35 AM

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Pentagon ABCDE has three right angles at A, B and C, as shown. The five sides have integer lengths with AB = BC = 10 units. What is the largest possible perimeter of this pentagon, in units?

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📎 unknown.png

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this is the picture

zenith ember Mar 23, 2018, 12:39 AM

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Hm.

upper karma Mar 23, 2018, 12:40 AM

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I drew the diagonal but I don't think that helps

zenith ember Mar 23, 2018, 12:40 AM

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If all of the sides have integer lengths, that limits what ED can be.

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I think what you want to do is draw a 'ghost' point where the 4th corner of the square would be.

upper karma Mar 23, 2018, 12:41 AM

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ok

zenith ember Mar 23, 2018, 12:41 AM

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Call that point F.

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Then consider the triangle DEF.

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If all of the sides of ABCDE have integer lengths, then DEF must also have sides of integer length.

upper karma Mar 23, 2018, 12:42 AM

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yeah

zenith ember Mar 23, 2018, 12:42 AM

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And there are a limited number of right triangles with sides of integer lengths.

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Where the 'legs' are less than 10.

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Well, less than 9. AE can't be 0.

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That should be enough to get you to a solution. Do you agree?

upper karma Mar 23, 2018, 12:44 AM

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I will see what I can get with that

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Thank you

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If I don't get it I'll ask

zenith ember Mar 23, 2018, 12:44 AM

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Ok.

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good luck!

upper karma Mar 23, 2018, 12:49 AM

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@zenith ember I don't know what to do

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:/

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WAIT

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it's a 3-4-5 triangle

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right?

zenith ember Mar 23, 2018, 12:50 AM

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That's the only one I know of that would fit.

upper karma Mar 23, 2018, 12:50 AM

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ok

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thank you

chrome fiber Mar 23, 2018, 12:56 AM

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6-8-10 would fit too

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But that would not give you the maximum perimeter

zenith ember Mar 23, 2018, 1:01 AM

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Ah. Good point.

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I had discarded that one because of the 10, thinking that wouldn't fit.

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But the 10 is the hypotenuse, not the legs, so it would be fine.

void dirge Mar 23, 2018, 2:00 AM

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copy pasting this here because I'm kinda desperate

I need some serious help right now. I already know how to do this question in calculus, but that seems kinda cheap, and I want to do itwith geometry or algebra, so I'm asking here.

I need to prove two things - that the tangent of theta is greater than theta, and that the sine of theta plus the tangent of theta all over two is greater than theta. Theta is in radians, and is in the range (0, pi/2).

I was thinking of using the unit circle, but got stuck because I couldnt compare a curve to a line

charred spearBOT Mar 23, 2018, 4:25 AM

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📎 latex.png

void dirge Mar 23, 2018, 4:18 PM

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Yeah

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Basically need to prove that

past mantle Mar 23, 2018, 10:32 PM

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📎 Capture.PNG

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i dont understand

waxen gorge Mar 23, 2018, 10:50 PM

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Hi

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just add them lol

past mantle Mar 23, 2018, 10:54 PM

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really?

waxen gorge Mar 24, 2018, 4:25 AM

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If you travel <1,1>

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Then travel <2,2>

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U travel a total of

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<1+2,1+2>

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= <3,3>

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And then walk another <3,3>

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Gives <3+3,3+3>

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= <6,6>

stoic steeple Mar 25, 2018, 12:06 AM

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📎 JPEG_20180325_053624.jpg

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Anyone here?

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I need help

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2

thorn talon Mar 25, 2018, 12:08 AM

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well, 5,6,7 are the sides of the original triangle

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what must the shortest side of the new triangle be?

stoic steeple Mar 25, 2018, 12:09 AM

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4,5,6 are the sides of the original triangle

thorn talon Mar 25, 2018, 12:09 AM

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oh whoops

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true

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4,5,6

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so what must the sides of the new triangle be?

stoic steeple Mar 25, 2018, 12:10 AM

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2/3 of 4 5 and 6

thorn talon Mar 25, 2018, 12:10 AM

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yes

stoic steeple Mar 25, 2018, 12:10 AM

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I know that but

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U have to do it somewhat like this:

📎 JPEG_20180325_054058.jpg

thorn talon Mar 25, 2018, 12:12 AM

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hmm

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ok

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so

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well, all angles must be the same

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so you can draw the small triangle first

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then extend it

hearty barn Mar 25, 2018, 12:05 PM

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📎 JPEG_20180319_220350.jpg

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A triangle ABC is inscribed in a circle with center O. The altitude from A intersects BC in H.
The circle with diameter AH intersects AB in D and AC in E.
Show that DE is perpendicular to OA

shadow anvil Mar 25, 2018, 12:06 PM

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I swear that's been posted like 4 times now

hearty barn Mar 25, 2018, 12:06 PM

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xD its the second time

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N i havent got any solutions

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@stoic steeple NCERT?

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xD all the best

upper karma Mar 25, 2018, 12:14 PM

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Oh I see

stoic steeple Mar 25, 2018, 12:24 PM

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@hearty barn yea

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@hearty barn is Ur exam on 28th??

hearty barn Mar 25, 2018, 12:28 PM

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Nay mine is done

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M in class 12

crimson harness Mar 25, 2018, 3:58 PM

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Hi, I would like to ask. Given parametric curves in space R^3, how to find points one on each curve, such that the distance between them is minimal. If curves are given by r1=F(t): t is element R and r2=G(s); s is element R. I've also got derivatives of function and I want to find proper system of equation

neon fossil Mar 25, 2018, 4:04 PM

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f(u,v) probably

crimson harness Mar 25, 2018, 4:05 PM

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yes

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f(s,t)=||r(u)-s(v)|| should be minimized

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So what u did here is: set equations for distance between two points

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and then calculated the minimum of the function using derivatives?

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smart, thank you for help!

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I think this would be right approach

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Would it be wrong what I tought? I wanted to set perpendicular vector on both tangents

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since this would be the shortest distance. So I wrote down vectortangent1 = 0 vectortangent2 = 0, where vector is defined as Point1-Point2

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Can I ask why exactly?

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aha

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ok, thank you for your time and help!

analog dust Mar 25, 2018, 5:42 PM

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@hearty barn Why u post my post

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Oh its u

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But still

stoic steeple Mar 26, 2018, 12:32 AM

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@hearty barn PCM? nice my last exam is on 28th

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I hate 10th so much

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Legit I am studying stuff which is useless af

hearty barn Mar 26, 2018, 1:43 AM

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@analog dust i dint understand french

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@stoic steeple Been there

analog dust Mar 26, 2018, 10:34 AM

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@hearty barn Why u no parler le french ? ¿

hearty barn Mar 26, 2018, 10:49 AM

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@analog dust Messa not french!!!

upper karma Mar 27, 2018, 12:33 AM

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Can someone help me out with these?

📎 image.jpg

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📎 image.jpg

keen aspen Mar 27, 2018, 12:57 AM

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What do you need help with

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We aren't here to do your homework problems

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The least you could do is attempt it

upper karma Mar 27, 2018, 8:24 PM

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I am here to do homework problems

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Dont speak for all of us pls

keen aspen Mar 28, 2018, 1:44 AM

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Lol okay then

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Not beneficial for both you and him

upper karma Mar 28, 2018, 2:39 AM

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Q2 is so hard

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I'm dying

river forge Mar 29, 2018, 12:41 AM

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lul

steady sky Apr 1, 2018, 10:41 AM

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I don't know where my mistake is:

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📎 cylinder.png

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To calculate the Volumina of a cylinder we usually do

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=tex V_1 = (r^2 \pi) \cdot h

charred spearBOT Apr 1, 2018, 10:42 AM

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📎 latex.png

steady sky Apr 1, 2018, 10:43 AM

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"Basically taking the ground area and dragging it from 0 to h"

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I thought to myself: "What if we take the Rectangle $$r\cdot h$$ and just rotate it by $$2 \cdot r \cdot \pi$$ ?"

charred spearBOT Apr 1, 2018, 10:45 AM

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📎 latex.png

steady sky Apr 1, 2018, 10:45 AM

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like this

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=tex V_2 = rh \cdot 2r\pi = 2r^2\pi \cdot h

charred spearBOT Apr 1, 2018, 10:46 AM

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📎 latex.png

steady sky Apr 1, 2018, 10:47 AM

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I don't know what's wrong about my idea 🤔 😦

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it definitely shows the wrong result, but no idea where my mistake is (most probably it's the rotating part 🤔 )

steady sky Apr 1, 2018, 11:06 AM

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this not a April fools' question , just sayin'. Been like 15 years since I did my last serious April fools.

knotty shell Apr 1, 2018, 12:57 PM

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@steady sky Not sure how to do that in carthesian coordinates but if you use sperical coordinates it works this way

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Doing something times $$2 \pi r$$ doesn't rotate over that part I think

charred spearBOT Apr 1, 2018, 12:57 PM

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📎 latex.png

knotty shell Apr 1, 2018, 12:59 PM

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For example if you go to cylindrical coordinates you get:

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$$\iiint_C r ,dr,d\theta,dz$$

charred spearBOT Apr 1, 2018, 1:00 PM

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📎 latex.png

knotty shell Apr 1, 2018, 1:01 PM

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And if you take the right borders for the integrals you get the volume of that cylinder

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You get an extra factor of r in that integral since the jacobian matrix gives a factor of r when you calculate it

steady sky Apr 1, 2018, 6:01 PM

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@rapid palm Ahh, I needed to use the correct coordinate system for my problem