geometry-and-trigonometry

[waxen gorge]

Other way

They switched the whites and blacks

So

Waitttt

They filled in the middle triangle...

ER MER gerd

37+16/64

53/64

[charred spear]

2.16049383

[upper karma]

don't forget that random white triangle off to the side

lol

[urban sonnet]

I need the help.

[thorn talon]

?

[urban sonnet]

Would this be better as a traverse or radial survey?

[rugged moat]

Can someone please help this retarded dimwit by solving this easy(seems like it) question. You have to find the area of the shaded region. Fig: A square of side 6cm with two intersecting quadrants inside

[keen aspen]

Are the points at where the triangles meeting at the midpoint?

[chrome fiber]

are those supposed to be circular?

[rugged moat]

@keen aspen quadrants

@chrome fiber yes

@keen aspen nope, it is a random point

[keen aspen]

Okay find the lengths of the triangle first

[chrome fiber]

there are no triangles in that figure.

[keen aspen]

Uh then idk what it is

[rugged moat]

@keen aspen a quadrant

Better?

A circle divided by 4

[keen aspen]

Not sure what you mean by quadrant .-.

[keen aspen]

ah ok

[rugged moat]

@keen aspen each one is a quadrant

[keen aspen]

can you draw the square in terms of that circle

[rugged moat]

Area of a quadrant is 1/4 * pi *r * r

[keen aspen]

pir^2/4 ye

[rugged moat]

@keen aspen

[keen aspen]

So im assuming one of the curve lines is the outer edge of the circle

[rugged moat]

You there?

[keen aspen]

Ye

[rugged moat]

?

[keen aspen]

Is one of the curved line the outer edge of a circle?

[rugged moat]

It is something like this

[keen aspen]

Yes

[oblique hearth]

this problem isn't trivial ig

if you find the area of both quadrants

and usbtract by overlapping region/2

you get the solution

[rugged moat]

@oblique hearth Doesn't work

Then there is one quadrant whose area is taken as its half

[oblique hearth]

what?

[rugged moat]

Area

Typo

[oblique hearth]

find the area of the overlapping region

divide it by two

[rugged moat]

How do you find the area of the overlapping region?

[oblique hearth]

http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/EMAT6690/Essay2/essay2.html

[rugged moat]

@oblique hearth What part of the page should I read to get the area?

[oblique hearth]

follow the logic of all of it

don't skim it actually understand how they derive it

so u know in the future

also this was the first result on google

[rugged moat]

I read the whole thing, still don't get it @oblique hearth

[oblique hearth]

https://math.stackexchange.com/questions/402858/area-of-intersection-between-two-circles

u will feel much better once u understand it

ok w/e let me show u

this shows it well

so

in our problem we have a equilateral triangle

with side length 6

that would mean the height of it is

=tex \frac{3\sqrt{3}}{2}

[oblique hearth]

and twice the length would be 3sqrt{3}

so we have the cord length of our circular segment

the area of a circular segment is

=tex \sqrt{6^2-\frac{(3\sqrt{3})^2}{4}}

[oblique hearth]

(radius of the circle is 3

this simplifies to

=tex \sqrt{36-\frac{27}{4}}

[oblique hearth]

the circle has symmetry so if you were to find the segment reversed

it would also be the same

like

if that makes sense

but you over measure an area

of one one equilateral triangle

so

(again of side length 6)

so

the area would be

=tex 2\sqrt{36-\frac{27}{4}}-\frac{\sqrt{3}*36}{4}

[oblique hearth]

which is

=tex 3\sqrt{13}-9\sqrt{3}

[rugged moat]

Woah

Atleast tag me

[oblique hearth]

ok sorry haha

anyway we have to divide it by two

[rugged moat]

One sec, let me read the whole thing

[oblique hearth]

and when we do we get the area of half of the segment

ok

[rugged moat]

Where did the equilateral triangle come from?

@oblique hearth

[oblique hearth]

wym

[rugged moat]

Don't we have an isosceles triangle?

[oblique hearth]

it's equilateral because side lengths are 6,6,6

it's the construction of one too

if you connect the centers + one of the top intersections

or bottom

if you go from a center to a intersection you travel 1 radius

because it's on edge of the circle

and both of the radii are equal o the two circles

[rugged moat]

6^2 + 6^2 = 72

√72

[oblique hearth]

what?

[rugged moat]

Pythagoras theorem

[oblique hearth]

that's not how you find the height of an equilateral triangle

[rugged moat]

That is a right triangle?

What triangle are we talking about here?

Can you post an image

[oblique hearth]

verbally communicatnig this is hard becuase i dont have paper and im on a laptop

[rugged moat]

Paint?

[oblique hearth]

i suggest you google it until you get it

convince yourself it's true

[rugged moat]

I will try

@oblique hearth Can you crosscheck this

I used the thing......that....
Ummmmm ........ You compare the equations.........I changed the signs

But my answer was negative

[chrome fiber]

how did you go from 18(2 - pi) to 18(2 + pi)?

and area of a quarter is one-fourth the area of that circle. not 1/2.

[junior scaffold]

if u have for example sin x = 1/2

and find the values between -2pi and 2pi

[thorn talon]

-2pi?

basically going backwards around the unit circle back to same point

[junior scaffold]

wld the values will be same but negaitve

[thorn talon]

it wouldn't

[junior scaffold]

nvm i get it

[past mantle]

How to find arc length of a circle?

[junior scaffold]

l=theta * r

l is arc length

r is radius

theta is angle

[junior scaffold]

cos^2 pi/6 - sin^2 pi/6

how do u prove tht

is =cos pi/3

[thorn talon]

would evaluating each part count or not?

[junior scaffold]

?

[thorn talon]

probably not then

like exact values of each of those

[wintry fern]

How do i figure out wether somethings acute obtuse or right triangle from 3 numbers

Ex 16,30,34

[upper sedge]

Are those numbers the side lengths of the triangle?

For any 3 numbers $$a,b,c$$ such that $$a\leq c$$ and $$b\leq c$$, the triangle formed by lines of those lengths will be determined by the relationship of the smaller sides to the largest side. We have $$\begin{array}{rcccl}a^2+b^2&\leq&c^2&\to&\text{Obtuse}\a^2+b^2&=&c^2&\to&\text{Right}\a^2+b^2&\geq&c^2&\to&\text{Acute}\\end{array}$$

[upper sedge]

Unless I'm crazy.

And I could be, so consider what I've said with however much salt you want.

[rugged moat]

@chrome fiber I took out the area of the semicircle(by combining the area of 2 quadrants)

Also, I changed the sign by doing that comparison thingy

Like 6- 5 = a + b.... So here you can assume a=6 and b= -5

This was the only way that came to my mind

I may be wrong tho

[wintry fern]

Mathbot helped me

[upper sedge]

@tropic island @abstract arch
Remember the parallelogram stuff we were doing earlier? It's no great accomplishment, but while tutoring earlier I decided to generalize our result. I figured it looked pretty enough to be worth sharing!

For the derivation, recall that $$c$$ and $$d$$ are distances between the point of intersection of the diagonals and the sides, which I'll call $$a$$ and $$b$$.

[upper sedge]

=tex \\begin{array}{c}
\begin{array}{rcl|rcl}
h_a&=&2c&h_b&=&2d\
A&=&h_aa&A&=&h_bb\
A&=&2ac&A&=&2bd\
a&=&\frac{A}{2c}&b&=&\frac{A}{2d}\
\end{array}\\hline\begin{array}{rcl}
P&=&2(a+b)\
P&=&2\left (\frac{A}{2c}+\frac{A}{2d}\right )\
P&=&A\left (\frac{1}{c}+\frac{1}{d})\right )\
P&=&A\frac{c+d}{cd}\
\end{array}\end{array}

[upper sedge]

Now let $$A=72$$, $$c=3$$ and $$d=4$$, and you get $$P=42$$.

[upper sedge]

No mess. No fuss. Just pretty looking math.

[abstract arch]

$$P = A\frac{c+d}{cd} \implies P = 2ac\frac{c+d}{cd}\implies P=2a\frac{c}{d} + 2a = \frac{2ac}{d} + 2a$$

$$P = \frac{A}{d} + 2a$$

Hmm. Well that doesn't work let me look at it

Oh I dropped the,

[upper sedge]

That seems fine.

[abstract arch]

$$P = \frac{A}{d} + \frac{A}{c}$$

[upper sedge]

That also seems nice.

You can get there by distribution from the second to last line in my derivation as well.

I love algebra.

[abstract arch]

Well I thought that would've given me something else but it's just the distributed form of what you got. XD

[upper sedge]

^^

[abstract arch]

Personally though I prefer it in that form, it seems more meaningful

Though your final derivation is probably easier to calculate

[upper sedge]

Yeah, it's much more meaningful as you put it, conceptually.

Yours is essentially exactly what we did earlier.

[abstract arch]

I, don't even remember tbh.

[upper sedge]

It got stuck in my head.

Ended up drawing a parallelogram on the board while the student got up to get coffee.

Scribbled it out on a napkin, and wrote it here when I got home.

[abstract arch]

It just seems more meaningful to say that the sum of the ratios of the Area to the semi-axis(shutup, I'm using the term, Gaussian) of the parallelogram is the perimeter

[upper sedge]

It does, certainly.

[abstract arch]

Yeah I have no idea if you can call those a semi-axis but I'm going to anyway.

Apothem would be better, probably.

[upper sedge]

That seems to be a common thing, giving names to things that already have names.

I don't know what they're called, either.

[abstract arch]

Well, technically apothems are only for regular polygons so that name still doesn't really apply

Could call them semi-altitudes

[upper sedge]

What ever it is that they're to be called, it's half of the height of the parallelogram with respect to a side.

[abstract arch]

hence semi-altitude

the entire height would be an altitude

Interesting though that the formula works out to be in terms of the semi-altitudes rather than the altitudes.

[upper sedge]

It is, isn't it.

[abstract arch]

But of course it makes sense when you consider the special case in which c=d, so it is a square

then you have

$$P = \frac{A}{c} + \frac{A}{c} \implies P = \frac{2A}{c} = \frac{4ac}{c}\newline P = 4a$$

[upper sedge]

Well, rhombus.

Wouldn't have to be square, would it?

Man, I just realized that it's been nearly a decade since I sat through a geometry class.

[abstract arch]

Well for one I made a mistake there, because I forgot that c was a semi-altitude. XD

[upper sedge]

I very was distracted, so I didn't notice.

[abstract arch]

this is why terminology is important.

Well I was kinda just sitting here...

"wait the perimeter of a square isn't twice it's side..."

[upper sedge]

xDD

[abstract arch]

"WAIT"

[upper sedge]

The perimeter of a rhombus is also 4a

This is why I wonder if c=d necessarily forces a square.

I don't think it does.

[abstract arch]

Eh maybe it doesn't

Either way I was really just using square as an example

[upper sedge]

ｘD

[abstract arch]

Whether that example was more general than I needed it to be is irrelevant

[upper sedge]

I'm alright either way, honestly. It was a fun thing to look at.

[abstract arch]

Now,

Do it for a trapezoid.

[upper sedge]

xD

Why not?

xD

That's probably insane.

I'll get the whiteboard.

[abstract arch]

""""""Probably"""""

[upper sedge]

Hey, it's an adventure.

It's exploration.

It's joy!

[abstract arch]

I mean gl even setting up an analogous case

[upper sedge]

Right?

[abstract arch]

considering there's only one height for a trapezoid

[upper sedge]

This is why it will be fun. How few assumptions can I make and still get something intelligble?

[abstract arch]

For a general trapezoid, anyway.

[upper sedge]

Right.

The hard part is not turning it into a trig problem.

[abstract arch]

Well first thing you'll need

$$A = \frac{a+b}{2}h$$

[abstract arch]

Calling the semi-altitude c, you get

$$A = \frac{a+b}{2}(2c) = ac + bc$$

[abstract arch]

You sure you want to do this man?

I meaaan

[upper sedge]

I'll play with it when I finish with another question.

Alright.

[abstract arch]

if you let the legs be p and q, then you can say that the area is also going to be equal,(give me a minute to write this one out)

[upper sedge]

Alright.

[abstract arch]

$$\frac{1}{2}\sqrt{p^2 - 4c^2}2c + \frac{1}{2}\sqrt{q^2 - 4c^2}2c + 2ca$$

[abstract arch]

Which is just pythagorean solved for the little tiny base of the two triangles the trapezoid can be broken into, plus the rectangle in the center

$$ac + bc = c\sqrt{p^2 - 4c^2} + c\sqrt{q^2 - 4c^2} + 2ac$$

[abstract arch]

$$a + b = \sqrt{p^2 - 4c^2} + \sqrt{q^2 - 4c^2} + 2a$$

[abstract arch]

$$b = \sqrt{p^2 - 4c^2} + \sqrt{q^2 - 4c^2} + a$$

[upper sedge]

Mmm.... I need to organize some of this.

[abstract arch]

$$A = ac + bc\newline A = 2ac + c\sqrt{p^2 - 4c^2} + c\sqrt{q^2 - 4c^2}$$

[abstract arch]

I'd also like to suggest that for this case we should give ourself the lengths from the intersection of the diagonals to each of the legs

[upper sedge]

That's part of the intuition, isn't it?

[abstract arch]

Because, logically that's the only way you could restrict it enough to get it to work

Otherwise there's multiple trapezoids you could form

Well what I mean is, in the parallelogram you had the length from the centroid to both of the sides, which was only 2 line segments

but because of the shape it was, you knew the lengths from that point to all four sides from that

[upper sedge]

We'd need 3 here, wouldn't we?

[abstract arch]

Yeah, because with the altitude you can restrict the parallel sides,

Well you know Idek if that would be enough

https://upload.wikimedia.org/wikipedia/commons/thumb/1/11/Trapezoid.svg/220px-Trapezoid.svg.png

Imagine the intersection of the diagonals,

now imagine the line going from that point to CB such that it is perpendicular to CB

Well, you could draw a circle with the radius of that line segment,

And so long as that line segment does not intersect or go past a or b, then CB can be rotated to remain perpendicular to the line and still meet a and b, just a and b must change lengths

So we'd need an additional parameter to restrict it to a unique trapezoid

The length of just one of the bases should work, I think.

Because then CB wouldn't be free to rotate, there is a set place it has to link to.

So we'd need the line linking CB, AD, CD, and the length of at least one base.

On top of the area

[upper sedge]

Hmmm... Do we really care about the trapezoid being unique?

We didn't care about the parallelogram being unique.

[abstract arch]

If our information isn't capable of defining a unique trapezoid, then we cannot possibly find the area.

Or the perimeter

There would be too much in flux

[upper sedge]

I wonder....

[abstract arch]

The information we had in the parallelogram problem was actually all we needed to define a unique parallelogram

[upper sedge]

I see.

Whiteboard time regardless.

[abstract arch]

I mean even in the formula I had rn

$$A = 2ac + c\sqrt{p^2 - 4c^2} + c\sqrt{q^2 - 4c^2}$$

[abstract arch]

It has the length of a base, and the length of both legs

I'm still trying to figure out how we can incorporate the perpendiculars of the legs to get the length of the legs, but there's no way we'll be able to get a base length

We will have to make that as something we're given in order for the end formula to work

also let me just uuuh

I don't have a whiteboard or anything so I'm visualizing all of this. XD

If we drop an altitude from C and call the point of intersection M, then we can draw a diagonal DM and say angle DMC = theta. Law of cosines would then state

$$a^2 = \overline{DM}^2 + 4c^2 - 4c\overline{DM}\cos(\theta)$$

[abstract arch]

Why am I using law of cosines

This is a right triangle.

derp, that was my whole reason for dropping an alt from C....

$$\tan(\theta) = \frac{a}{2c}\newline \theta = \arctan\left(\frac{a}{2c}\right)$$

[abstract arch]

Ya know I, I feel like I had a plan but

Wait, I think I may have it

$$\overline{DM} = \frac{a}{\sin(\theta)}$$

[abstract arch]

$$\overline{DM} = \frac{a}{\sin\left(\arctan\left(\frac{a}{2c}\right)\right)}$$

[abstract arch]

Let's call angle A alpha

[upper sedge]

Gurl was right. I should sleep. I've done the same 6 steps in circles on the board about 4 time in a row now.

I see that you're digging deep into the trig.

I wonder if you'll be able to come out the other side without it.

At any rate, I'll take a good long look at whatever comes of it in the morning, and see what I can do with it when I have the chance.

[abstract arch]

$$\overline{DM}^2 = p^2 + \left(\frac{2c}{\tan(\alpha)} + a\right)^2 - 2p\left(\frac{2c}{\tan(\alpha)} + a\right)\cos(\alpha)$$

[abstract arch]

$$\frac{a^2}{\sin^2\left(\arctan\left(\frac{a}{2c}\right)\right)} = p^2 + \left(\frac{2c}{\tan(\alpha)} + a\right)^2 - 2p\left(\frac{2c}{\tan(\alpha)} + a\right)\cos(\alpha)$$

[abstract arch]

$$p^2 - 2p\left(\frac{2c}{\tan(\alpha)} + a\right)\cos(\alpha) -\frac{a^2}{\sin^2\left(\arctan\left(\frac{a}{2c}\right)\right)} + \left(\frac{2c}{\tan(\alpha)} + a\right)^2 = 0$$

[abstract arch]

$$p = \frac{2\left(\frac{2c}{\tan(\alpha)} +a\right)\cos(\alpha) \pm \sqrt{4\left(\frac{2c}{\tan(\alpha)} + a\right)^2\cos^2(\alpha) + \frac{4a^2}{\sin^2\left(\arctan\left(\frac{a}{2c}\right)\right)} - 4\left(\frac{2c}{\tan(\alpha)} + a\right)^2}}{2}$$

[abstract arch]

$$p = \left(\frac{2c}{\tan(\alpha)} + a\right)\cos(\alpha) \pm \sqrt{ \left(\frac{2c}{\tan(\alpha)} + a\right)^2\cos^2(\alpha) + \frac{a^2}{\sin^2\left(\arctan\left(\frac{a}{2c}\right)\right)} - \left(\frac{2c}{\tan(\alpha)} + a\right)^2}$$

[abstract arch]

$$p = \left(\frac{2c}{tan(\alpha)} + a\right)\cos(\alpha)\pm \sqrt{- \left(\frac{2c}{tan(\alpha)} + a\right)^2\sin^2(\alpha) + \frac{a^2}{\sin^2\left(\arctan\left(\frac{a}{2c}\right)\right)}}$$

[abstract arch]

@upper sedge yeah Imma just go

[covert osprey]

im confused by how to do this one

[thorn talon]

Well the reference angle is the angle made with the x axis

So in Q1, the reference angle should be equal to theta already right?

And then for the other quadrants

You've probably seen it as like pi - theta, theta - pi, 2pi - theta

Which makes sense if you draw out various angle and lines

[junior scaffold]

ho

how do u do this

[copper valve]

cross multiply

owait

maybe use a double angle formula first

[dire blade]

Could someone explain or send me an link to a good video explainen what cos and sin really is? I find it hard to get a grasp of the idea involved...

explaining*

[waxen gorge]

Hi

Just use khanacademy

[dire blade]

Hmm ill check it out, thanks

[mint sandal]

@dire blade or you could ask here

[waxen gorge]

@mint sandal it's easier to understand from a video or reading than ppl typing random stuff at u

[mint sandal]

Not random stuff

[waxen gorge]

Yes random stuffz

[dire blade]

Ive watched some videos and understand the basic concepts, going to sleep now but im still wondering over the derivative of sin and the unit cirle

but thx anyways

[narrow warren]

@junior scaffold change theta by 2×(theta/2) and apply sin and cos law.

[urban sonnet]

Anyone good with radial/traverse surveys?

[hazy ledge]

can someone calculate obtuse angle ADC

[upper urchin]

Yes

Can you?

[hazy ledge]

im using sin rule but its not working

(for me)

I did this : 6.5/sin(30) = 11/sin(ADC)

[hazy ledge]

how tf isnt it working it gives ADC=58 degrees but its not right

[mint sandal]

what is the right answer?

[hazy ledge]

122.2

[mint sandal]

sin(180-x) = sin(x)

[tropic stirrup]

AC sin A = CD sin D by sin rule

[mint sandal]

122 = 180-58

[tropic stirrup]

which- yeah pete said it

obviously D is obtuse

[hazy ledge]

thanks guys

[sturdy finch]

was the formula of a
equilateral triangle
b x h

2

???

[mint sandal]

area of equilateral triangle?

[sturdy finch]

even triangle

[mint sandal]

wtf is even triangle

[sturdy finch]

with all the angles 60 grades

XD

[mint sandal]

so yeah

you want to express the area in terms of its side length?

[sturdy finch]

formula

XD

is it

base times height

          2

?

[mint sandal]

this is area

yes

you want to find the are don't you

[sturdy finch]

welp

[mint sandal]

but in terms of what

[sturdy finch]

i know the area

and i need to find the height

thats all

[mint sandal]

GOD

[sturdy finch]

the area is

[mint sandal]

then just say so

be clear when you're asking

[sturdy finch]

radical 3^2

[mint sandal]

i dont care what the area is

[sturdy finch]

XD

[mint sandal]

do you know the formula for the area in terms of its side length?

for equilateral triangle

[sturdy finch]

nop

[mint sandal]

A=ca^2 obviously

ok

[sturdy finch]

we only know

[mint sandal]

try and find it

[sturdy finch]

the area

[mint sandal]

it's easy to derive

very easy

[sturdy finch]

oki

wait

wth

we didnt learn that

ca^2

is the area of an equilateral triangle

[mint sandal]

no it's common sense

this is true because all equilateral triangles are similar

[sturdy finch]

so if a equilateral triangle had its side length of 6

[mint sandal]

scaled up versions of each other

[sturdy finch]

the area would be 36?

[mint sandal]

36 times some constant yeah

[sturdy finch]

ok

[mint sandal]

this is true of any shape

[sturdy finch]

but i need to know the height

XDDD

[mint sandal]

dont rush

you need to learn to do this right

[sturdy finch]

oh ok

[mint sandal]

how do you express the area of an equilateral triangle

A = ah / 2 right?

you know a

it's the side length

[sturdy finch]

ah?

[mint sandal]

you need to find h

the height of the triangle

how do you find it

a*h/2

yeah

[sturdy finch]

wut is a

[mint sandal]

a is side length

[sturdy finch]

oho

[mint sandal]

base length

area of any triangle is a*h/2

base length times height to that base by 2

[sturdy finch]

ohooooo

[mint sandal]

This shouldn't surprise you

[sturdy finch]

so now i got radical 3 cm^2 = ah/2

how do i use the bot

so u see it good

[mint sandal]

wait

dont rush

i told you not to rush

ah/2 is not enough for you

[sturdy finch]

oki

[mint sandal]

wait what do you want to find

[sturdy finch]

height

[mint sandal]

ok

so you need to express a in terms of height

[sturdy finch]

that means???

[mint sandal]

the side length in terms of height

a=f(h)

find the function f

[sturdy finch]

f?

[mint sandal]

yes

[sturdy finch]

wut is f first

[mint sandal]

a function

[sturdy finch]

i didnt learn about those

[mint sandal]

which takes height and returns side length

ok you know what relationships are between variables

[sturdy finch]

hmm im not english so its hard to understad wut u say

[mint sandal]

like A=ah/2 is a relationship which expresses area in terms of base length and height

[sturdy finch]

oh

[mint sandal]

you need to find a formula for side length

you know height of the triangle, which is h

and you need to find the side length a

how do you solve this

[sturdy finch]

a=

height

umm

times

[mint sandal]

hint: use pythagorean theorem

draw some pictures

[sturdy finch]

oki

[sturdy finch]

XD

so

i^2=a1^2 + h^2

that means

a1^2= i^2 - h^2

right?

[mint sandal]

yes

but wait

what is i

[sturdy finch]

hypotenuse

thingy

[mint sandal]

i know

but

[sturdy finch]

length?

[mint sandal]

it's an equilateral triangle

all sides are the same

i is a

[sturdy finch]

yes

but i drew the height

and i formed 2 right triangles

i apllied the theorem

on 1 of the right triangles

[mint sandal]

ok but for it to be useful you need to express i in terms of a1

i=2a1

agree?

[sturdy finch]

yup

cuz its an equilateral triangle

[mint sandal]

put it in

🥛

[sturdy finch]

a1^2=2a1-h

[mint sandal]

get expression for the area in terms of height

WRONG

[sturdy finch]

ummm

a1^2=2a1-h^2

XD

wait

a1^2=2a1^2-h^2

[mint sandal]

2 gets squared

becomes 4

[sturdy finch]

a1=4(a1^2)- h^2

????

wait

i think im wrong

is it

a1=4a-h^2?

@mint sandal

!!

[mint sandal]

a1 = h/sqrt(3)

[sturdy finch]

where is that 3 from?

[mint sandal]

wait

=tex 3a_1^2 = h^2

[sturdy finch]

hmmm

true

but

2
---??
1

[mint sandal]

what

???????

[sturdy finch]

3a^2 right?

equals

h^2

[mint sandal]

haha yes

[sturdy finch]

oho

[mint sandal]

so what is a_1 equal to?

in terms of h

=tex a_1 = \frac{h}{\sqrt{3}}

[sturdy finch]

but

that 3l

doesnt seem right

3a*

[mint sandal]

=tex a= 2a_1 = \frac{2}{\sqrt{3}} h

[sturdy finch]

wait

u got that 3a^2 by decreasing

4a with a^2?

[mint sandal]

yeas

[sturdy finch]

but u cant do that

thats wut my teacher said

[mint sandal]

4a^2

[sturdy finch]

u have to let it like that

[mint sandal]

4a^2 - a^2 = 3a^2

quick mafs

[sturdy finch]

why 4a^2

and not 4a

[mint sandal]

(2a)^2 = 4a^2

[sturdy finch]

oooh

nvm

[mint sandal]

eeeeh

[sturdy finch]

u took it from the height

[mint sandal]

komo

[sturdy finch]

im so dumb

[mint sandal]

🥛

ok so

=tex a= 2a_1 = \frac{2}{\sqrt{3}} h

[mint sandal]

agree?

[sturdy finch]

2a1

wuts that

XD

[mint sandal]

2*a_1

it's twice as long

[sturdy finch]

oh

[mint sandal]

because height divides base in half

=tex A = \frac{ah}{2} = \frac{\frac{2}{\sqrt{3}}h*h}{2}

[sturdy finch]

this problems arent for my homework doe

[sturdy finch]

i need to exercice for a tournament

[mint sandal]

oh god

what tournament

[sturdy finch]

its called Comper

[mint sandal]

ok

[sturdy finch]

Olympics

[mint sandal]

you should do simpler problems first

[sturdy finch]

thingy

[mint sandal]

and build up from there

[sturdy finch]

this is wut my teacher gave me

;[

[mint sandal]

=tex A = \frac{ah}{2} = \frac{\frac{2}{\sqrt{3}}h*h}{2}

[mint sandal]

do you know where to go from here?

[sturdy finch]

i still dont get wut u did with the height

like

2/radical 3

XD

is the radical 3 from the area

XD

i think u are so mad at me

angry

XD

because of me

not at me

[mint sandal]

=tex a_1^2 = (2a_1)^2 - h^2 = 4a_1^2 - h^2

[mint sandal]

=tex 3a_1^2 = h^2

[sturdy finch]

but

[mint sandal]

=tex a_1^2 = \frac{h^2 }{3}

[sturdy finch]

=tex a_1^2

[sturdy finch]

doesnt that mean

a^2

cuz its divided by 1?

[mint sandal]

what

[upper karma]

thats not a^2/1

i dont think

[sturdy finch]

right?

[mint sandal]

what

[sturdy finch]

ima just let it like this

h = 3l^2

but by curiosity

wut was the final answer

even if i wont write it

2 radical 3?

[mint sandal]

final answer doesnt matter

only how you solve it matters

other things are boring

[sturdy finch]

yea

i know

but in this paper

the teacher gave me

it has more answers

like

a. radical 3
b. 2 radical 3
c. 3 radical 3
d 4 radical 3

is it b?

[mint sandal]

i dont know

i didnt calculate

[sturdy finch]

oh ok

[mint sandal]

if it's a multiple choice question

your best bet is to draw the situation

and use a ruler 📏

[sturdy finch]

o

[mint sandal]

to determine lengths and areas

and see which answer is the closest to truth

[sturdy finch]

yea but the choice thingies are with radical

XDDD

[mint sandal]

there are ways to manualy calculate radicals

[sturdy finch]

ill just say 2 radical 3

because the area would be

ah/2

and the area is radical 3

wait no

id just say radical 3

because

lets say a is 2

2 radical 3/2 = radical 3

XDDDD

lol

this problems

are just way too hard for 7th grade

[mint sandal]

i showed you how to solve it

[sturdy finch]

look at this

[mint sandal]

which country are you from

[sturdy finch]

romania

7th grade

look at this problem

the maximum value of p is?

p= 3 - x - x^2

choices are

3

0

3,5

3,25

id say 3

0

excuse me

cuz its minus

and u cant get more than 3

because its

3- x - x^2

right?

[mint sandal]

it's more than 3

[sturdy finch]

how

wth

[mint sandal]

because -x can be positive

[sturdy finch]

oh

[mint sandal]

and x grows faster than x^2 in [0,1]

but there are rigorous ways to find the maximum

just find the vertex of the parabola

[sturdy finch]

i will just say 0 with a pencil

cuz i didnt learn any method

for this

at school

😄

[mint sandal]

0 is wrong

i told you it's bigger than 3

[sturdy finch]

but the teacher

didnt teeach us

any way

to find it

:/

[mint sandal]

you can figure it out yourself

teachers dont teach a lot of stuff

doesn't mean you can't solve it

[sturdy finch]

so teachers are almost useless

[mint sandal]

wrong

[sturdy finch]

so how should i find this out??

[mint sandal]

complete the square

[sturdy finch]

gimme the way u think ill understand fastest

[mint sandal]

wait no

ok

lemme think how to explain it

3-x-x^2?

[sturdy finch]

yes

[hazy ledge]

just assume x is 1

[mint sandal]

do you know how to solve quadratic equations?

[sturdy finch]

nop

XD

[mint sandal]

@hazy ledge ?

[sturdy finch]

if i assume that

[mint sandal]

ok look

velfy

[sturdy finch]

i wont get any of the choices

i got

ok

[mint sandal]

p=-3-x-x^2

the maximum of this function is maximum of -x-x^2 plus 3 right?

[sturdy finch]

not -3

3

[hazy ledge]

actually nvm i am wrong , if x is postive then its always bigger than 3

[mint sandal]

so only left is to find the maximum of -x-x^2

which is the minimum of x+x^2

=x(1+x)

[sturdy finch]

where did the 3 went

?

ik ur right

[mint sandal]

i subtracted it

[sturdy finch]

but i want explanation

with wut

[mint sandal]

you just shift the function down

maximum also shifts down

max(p-3) = max(p) - 3

it makes your equation simpler

so you only need to find the maximum of -x-x^2

[sturdy finch]

wut is that type of solving called

[mint sandal]

it's called thinking

[sturdy finch]

XD

oki

[mint sandal]

so we need to find the minimum of x(x+1)

[sturdy finch]

but if u transformed

x^2 in 1+x

doesnt that mean x is 1?

[mint sandal]

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

this is just factorization

[sturdy finch]

im 7th grade

😭

so 0 is wrong u said?

[mint sandal]

yes 0 is wrong

[sturdy finch]

ok excluding that

[mint sandal]

the answer is greater than 3

[sturdy finch]

i will excule 3 also

we are left with 2

[mint sandal]

2 is wrong

i said the answer is more than 3

[sturdy finch]

bro

2 choices

XD

3.5

and 3.25

[mint sandal]

i know

[sturdy finch]

3.5

it says maximum value

so the biggest number?

XDDDD

[mint sandal]

no

[sturdy finch]

then 3.25

?

[mint sandal]

function might not attain 3.5 anywhere

[sturdy finch]

oh

[mint sandal]

we dont know the answer yet

that's why we're solving it

[sturdy finch]

this arent fucking grade 7th problems

x cant be more than

-3

or -2

but

if it would be lets say

3- (-3) - (-3^2)

ot would give us

3+3- 9

-3

:/

its wrong

[mint sandal]

were not looking for x

we're looking at p

[sturdy finch]

i think ill just have to try all posibilites

yes

to find p

we need x

[mint sandal]

this is an extra problem

he's going to go to a math contest

[sturdy finch]

problem for a olympiad

yep

my teacher gave me something to solve

but these arent for my fu**ing

grade

[mint sandal]

@sturdy finch use khanacademy

they will teach you a lot of useful stuff

[sturdy finch]

in 1 day?!?!@?!??!?!?

[mint sandal]

yes

it's doable

[sturdy finch]

u mean

[mint sandal]

with 2x speed

[sturdy finch]

till tomorow

[mint sandal]

is the contest tomorrow?

[sturdy finch]

nop

i need to solve all these

till tommorow

[mint sandal]

ok but you obviously can't

[sturdy finch]

i had more time

like 4 days

but i forgot

XDD

welp true

this sheet of paper

say

[mint sandal]

velfy

[sturdy finch]

its for 7th grade

???

[mint sandal]

do you by any chance play sa-mp

[sturdy finch]

i did

[mint sandal]

oh

[sturdy finch]

a long time ago

why?

is there someone with my name?

[mint sandal]

just that there are a lot of romanian servers on sa-mp

seems like the whole nation is playing that game

[sturdy finch]

XD

i play mostly roblox and brawlhalla

im animating

things

for roblox games

like punching running etc

[hazy ledge]

i play roblox c;

[mint sandal]

i play roblox too c;

[sturdy finch]

hehe

my name is Velfy

cuz its my original name

i put it everywhere

[hazy ledge]

my name is masterkoga

cuz its my orginal name

[sturdy finch]

i created this name

so no one can steal it

[hazy ledge]

i put it everywhere

[sturdy finch]

as long as they dont know bout it

the value of x is somewhere between 2 and 3

i think

if x = radical 5n+3

does that mean x is a natural number?

or an irational number

?

or something else

ye

the v thingy

yea

i think yea

yep

but not with brackets

nop

it says

its a natural number

so its 0 or more

right?

ohoo

so then n is a natural number too

why??

it will give .blah blah

irational number?

ok thx

o

look at this

i know the answer already

x + 1/x=2

x^3 +1/X^3 =?

its 2

right?

or 1

i think its 2

[past mantle]

What is the equation of a tangent line?

[keen aspen]

y=mx+b :v

idk, don't know calc

but I know a tangent line is linear

[past mantle]

well, we just leared about circles and the ways lines intersect circles

[upper karma]

Are you specifically wondering about tangent lines on a circle?

[past mantle]

yes, and 2 circles

[upper urchin]

You differentiate implicitly then solve simultaneously

[mild cargo]

Hey g8ys

At the nth stage of the sierpinski carpet is it true that

=tex \left ( \frac{8}{9} \right )^n

[mild cargo]

of the original pattern remains?

<@&286206848099549185>

[graceful totem]

Yep

1/9 of each square is removed at each step

[mild cargo]

Thanks 👍

[hasty mural]

Please help

[upper urchin]

use the fact that any triangle in a semicircle must be right angled

https://www.petervis.com/mathematics/Triangle_Inscribed_in_a_Semicircle/Triangle_Inscribed_in_a_Semicircle.html

this thing

[vital oxide]

welp pls

[mint sandal]

@Nam#0784 explain what you've done so far and what you're struggling with

[vital oxide]

what do they mean by shorter diagonal

[upper urchin]

there are two diagonals from each of the opposite corners

[mint sandal]

there are 2 diagonals in a parallelogram

[upper urchin]

find the shorter one

[mint sandal]

yes

[past mantle]

so, in geometry, were almost at probability, i want to understand everything before it happens. what are all of the symbols used?

[distant stag]

mu for measure, Sigma for the sigma-algebra, ...

[hallow trellis]

@distant stag ????????? who are you responding to

[distant stag]

mr. jon

[hallow trellis]

but what kind of geometry can totally change the answer. and different naming conventions and different languages/countries????? idk it just seems weird to answer a question that you dont even know the answer to without more contex

[neon fossil]

D for derivative

T for tangent space

[hallow trellis]

you guys are awful

[distant stag]

using D for derivative

[hallow trellis]

I was going to mention that but I didn't feel like it was worth it

[neon fossil]

maybe you have a small d

HAAAAAAAAAAAAAAA perfect setup and execution

:dab:

[distant stag]

not using pushforward

[neon fossil]

😦

[distant stag]

Let's talk some hot sweaty geometry here, guys.

[hallow trellis]

how do you apply the langarian on infinite dimensional frechet space

[distant stag]

You already know where to find that answer

Also that's BORING

No one likes to take integrals here.

[hallow trellis]

@distant stag talk about your geometry then

[distant stag]

I have not done my contact geometry routine?

[hallow trellis]

yeah but everytime you do it I dont understand the point or what exactly is happening

you just post a picture of a bunch of tangent planes on a sphere

[distant stag]

THAT'S THE COOL PART

[hallow trellis]

EXPLAIN HOW

[distant stag]

The rough idea is that you can reduce some of the structure of the ambient manifold down to the surfaces it contains.

So you learn stuff about a manifold by looking at embedded surfaces.

Kind of like studying subgroups or representations to study groups.

[hallow trellis]

what do you mean by the structure of the ambient mainfold

so basically you're breaking down a manifold into all of it's surfaces?

[distant stag]

Not quite.

You have some contact manifold (who cares what exactly that is, for the time being) M.

Now you want to study the additional contact structure of M.

So how do you study this?

Well what I am doing is studying contact manifolds by the surfaces which embed in it.

[hallow trellis]

what does it mean for a surface to embed in it? I've never learned embedding

[distant stag]

You can imagine it as a topological embedding.

So a homeomorphism onto its image.

Though that is technically wrong.

Because here we have smooth manifolds.

But it's good enough intuition.

Re: blind mathematicians, one of the mathematicians who pioneered the stuff I am doing right now Is E. Giroux who happens to be blind.

[hallow trellis]

wrong channel

but cool!

so what properties do you get by embedding a surface on it? do you have to find specific surfaces that work or what?

what does complete non-integrability mean?

[distant stag]

It's the right channel, this is geometry.

[hallow trellis]

why does it have to be odd dimensional?

no i mean the Re blind mathematicians

we were talking about it in pure math

[distant stag]

Yeah, I figured I would put it here since it relates to this.