geometry-and-trigonometry

[neat kite]

real?

[low ocean]

[low ocean]

Yeah, Vietnamese type geo

[neat kite]

Ah yeah

[low ocean]

Make a diagram first

[low ocean]

Hmm gimme a min

[low ocean]

Show that MJ//NQ and MJ=NQ

Hence MJNQ is a parallelogram

That makes JQ//MN

[neat kite]

Initially, I thought of proving M is the center of the plane AA'C'C would be the solution and also with N parallel to the BB'D'D plane

So can i conclude that MN is parallel to JQ by that, or is the condition missing?

[low ocean]

you probably can't

[neat kite]

missing conditions ?

Eh nvm i saw it

Thank you !!!

Besides, it's just an assignment to get full marks so I don't think it's worth it. Thanks for your suggestion btw (⁠ㆁ⁠ω⁠ㆁ⁠)

Ima come back with another problem

[low ocean]

np

[neat kite]

im back with a new problem

https://cdn.discordapp.com/attachments/1381951378678747218/1411672396162793643/image0.gif

Given four non-coplanar edges A, B, C, D. Let G1, G2, G3 be the centroids of triangles ABC, ACD, ADB respectively. Prove that (G1G2G3) // (BCD).

i did solve AG1/AM = AG2/AN = 2/3 but idk what to do next

[low ocean]

Idea: construct segments through G1,G2,G3 respectively and parallel to the base if its triangle

You can show that those 3 segments are coplanar

Hence prove the statement

<@&268886789983436800> fishy

[neat kite]

me suck at english

but i kind of get the idea

[low ocean]

I can speak in Vietnamese

If you like

[neat kite]

yes pls

[low ocean]

Qua G1,G2,G3 kẻ 3 đoạn thẳng song song vs đáy của tam giác chứa nó

Từ 3 đoạn đó, ta chứng minh rằng chúng đồng phẳng

Vì 3 đoạn đs song song vs đáy nên hiển nhiên mp chứa 3 đoạn sẽ song song mp đáy

=> đpcm

[neat kite]

tytytytytytyty

ilysm pls dont go bald

thank you !!

em muon dc hoi not cau cuoi a

@low ocean

[low ocean]

Okay

[neat kite]

Cho hình chóp S.ABCD có đáy lần lượt là hbh tâm O. Gọi M, N, P lần lượt là trung điểm của AB, AD, SO. (MNP) lần lượt cắt SB, SC SD tại điểm I, J, K. Tính SI/IB + SJ/JC + SK/KD

https://cdn.discordapp.com/attachments/1381951378678747218/1411672396162793643/image0.gif

[low ocean]

Có hình k

[neat kite]

có ạ

đợi e xíu e vẽ xong h

[low ocean]

Bt tìm giao điểm I,J,K ko

H phải tìm vị trí I,J,K trc

[low ocean]

Gọi H là giao của AC vs MN, HP giao vs SC chính là J

[neat kite]

em khoong

[low ocean]

Căng

H mà bày cho cả cả thì sau lại k bt làm

[neat kite]

em k hiểu tìm phần thiết diện nên h k hiểu cái này

[low ocean]

Tập đi, tốt nhất là làm nhiều bài vào vs đọc lời giải cho kĩ vào

[low ocean]

Kéo dài MN cắt CD tại Q, QJ cắt SD tại K

Tương tự vs bên kia cho điểm I

[neat kite]

em thấy hình ròi

đến đây e tự lm đc

hị hị mai em làm phiền anh tiếp ạ

if you don't mind ofc

[low ocean]

It depends

You can just ask here ig, someone would probably help

[neat kite]

DM lam tutor cho e dko a ʕ⁠´⁠•⁠ ⁠ᴥ⁠•̥⁠`⁠ʔ

[low ocean]

DM rảnh rảnh hỏi bài thì đc

Bận ôn tư duy k tutor đc

[neat kite]

yes sir

[lapis gull]

can someone here understand analytical geometry?

[novel wing]

Peak pfp dude

[novel wing]

Never ever let the synthetic geometers(a cult am in) here you

[low ocean]

And your cause nightmare

[novel wing]

Legend says they invert around point at infinity anyone who dared blaspheme Euclid by complex bashing or coord bashing

[low ocean]

Combinatorics >>>>> Geo no diff no debate

[novel wing]

I like them both ngl

Like this one made me combi-sexual

[sterile beacon]

It's really interesting how any polygon can be described by a system of linear inequalities

That means that if you are programming, say a UI, and you need to figure out if the mouse is overlapping a polygon-shaped UI element, all you need is to check whether the mouse coordinates satisfy the system of linear inequalities that describes the shape of the UI element

No fancy math... just some basic arithmetic

[upper karma]

is this true for concave polygons

its funny this was my first message on this server

asking if this was true for concave polygons

i think no one answered

[sterile beacon]

Ah interesting

I would think that it is

and even for shapes with holes

[upper karma]

it was like 3-4 years ago

lmao

[upper karma]

hmhm

[deep oak]

sure, why wouldnt it be?

[plush brook]

So I was told this involves a quadratic ? I thought about it a little and couldn't really get anywhere

[light fable]

do you know the Pythagorean theorem?

[real sentinel]

Yes it does

[real sentinel]

No

Use the area of the triangle

[light fable]

oh my bad, didn’t read what it’s actually asking for

[plush brook]

Is that the a squared plus b squared equals c squared

[real sentinel]

The area of the triangle is (1/2 x base x height)

[light fable]

yes but nevermind, it’s not what you want here

[real sentinel]

Plug in to get quadratic equation

Solve for x

[plush brook]

So would I have to do 1/2 x x+4 x x-2

[real sentinel]

Be careful, x can't be negative

[real sentinel]

Yes

And equal to a given area from data

[plush brook]

Why not ?

[real sentinel]

Cuz the side (x-2)

Will become negative

And length is never negative

[plush brook]

This feels somewhat wrong

[light fable]

writing the $\times$ symbol between $x$’s in a way that they look identical is bound to cause you undue difficulty

[somber coyote]

ManifoldCuriosity

[plush brook]

What do I use

I see some people do like

Dots, is that valid ?

[light fable]

it is, or parentheses to separate things getting multiplied is another way

[versed hamlet]

Anyone knows a good book for high school geometry? I suck at identifying patterns on questions and I need to train

[plush brook]

I think I cooked

[light fable]

$\frac{1}{2}(x-2)(x+4)$

[somber coyote]

ManifoldCuriosity

[light fable]

unfortunately the way you used the dots there isn’t good

because by the order of operations, it doesnt equal what you want to have

[plush brook]

Ah okay sorry I'll try again

[light fable]

there are some other issues with how you went down those lines

[plush brook]

Any you can point out to me if that's okay ?

[light fable]

actually it’s just one issue

you essentially did $\frac{1}{2}\cdot A \cdot B = \frac{1}{2}A \cdot \frac{1}{2} B$, like “distributing”

[somber coyote]

ManifoldCuriosity

[light fable]

but that’s a false algebraic “rule”

what is true is $\frac{1}{2} (x-2) = \frac{1}{2}x -1$

[somber coyote]

ManifoldCuriosity

[plush brook]

So I would only multiply the half by either the A or the B, not both ?

[light fable]

but you shouldn’t spread the 1/2 to the other chunk as well

yup

[plush brook]

Okay I'll give this a shot and report back

[light fable]

either one would be fine but not both

[plush brook]

I cooked ?

[light fable]

yuss

wait, uhh

the -2 is wrong

(last one)

[plush brook]

Oh yes shit it's -4

[light fable]

right

so that’s the expression for the area

[plush brook]

So that should be equal to 27.5 ?

[light fable]

yep

[plush brook]

Okay I got the quadratic, I could use a calculator or the formula but I want to try use normal factorisation, if that's possible

I think the AC method would work here ?

[light fable]

I recommend multiplying both sides by 2 to remove the pesky fraction and decimal

[plush brook]

I think I'm on a good streak for getting it right

[light fable]

very close

but you got a 2 on the right where you shouldn’t have

[plush brook]

Isn't that why I moved it to the left hand side

[light fable]

the first line is right, but 0 * 2 = 0

[plush brook]

It would be the middle line but equal to 0

I see, that was a silly slip up

[light fable]

yup

almost there now

[plush brook]

Factorising (not on my own, computer assisted) I get (x-7)(x+9)

And x can't be negative so x is 9

[light fable]

I agree

ermm

no I don’t 😭

[plush brook]

Awwh shucks

[light fable]

I agree that x can’t be negative

but what are the two possible solutions?

[plush brook]

They are -7 and +9

[light fable]

nope!

[plush brook]

Did the computer factorise it wrong ?

[light fable]

the solutions come from x - 7 = 0 and x + 9 = 0

it’s factored correctly

[plush brook]

I don't follow

[light fable]

you got (x - 7)(x + 9) = 0

then the zero-product property tells us either x - 7 = 0 or x + 9 = 0

[plush brook]

Ohhh

OHHH

BECAUSE ONLY ONE OF THEM HAS TO BE ZERO

I SEE IT

[light fable]

woah, I just had some heavy deja vu

[plush brook]

How so ?

[light fable]

I don’t know

[plush brook]

Have you taught someone this before ?

[light fable]

I may have had a similar exchange with someone here before

[plush brook]

Huh, small world

[light fable]

but like your specific phrasing with all caps

really hit me with deja vu

[plush brook]

I apologise (?)

[light fable]

no need, I didn’t mind; just a strange feeling!

[plush brook]

Regardless is my thinking right

[light fable]

yes

[plush brook]

You just invert the signs

[light fable]

yes, very important to remember that

[plush brook]

So the answer for x is 7

[plush brook]

But there's an unlabelled side here, I was told you can't always trust your eyes when judging length, so couldn't that be the shortest ?

[light fable]

it’s good you thought about that

but because that side is opposite from the right angle, it’s the hypotenuse, so it’s actually the longest

[plush brook]

Ohhh Soh Cah Toa

I see this

[light fable]

and between x - 2 and x + 4, we can be sure which is shorter

[plush brook]

So the answer is 7-2 = 5cm ?

[light fable]

bingo

[plush brook]

That was rough, but thank you so much

I apologise if I was difficult at any point

[light fable]

and as a quick check, the other side is then 7 + 4 = 11

and indeed 1/2 * 5 * 11 = 27.5

no need to apologize!

you weren’t difficult

[plush brook]

Much obliged my friend! Thank you and have a good day

[light fable]

sure thing, and likewise

[upper karma]

idk like tell me how to do it with concave polygons

with convex polygons you can take the sides of the polygon as lines and convert them into a linear inequality

but this doesnt work if the plygon is concave

[jaunty jasper]

Can someone help me with 19? I set up the proportion but I don’t think i’m multiplying correctly

[exotic yarrow]

This is correct so far - you multiplied both sides by 2.

[timber hatch]

There is something wrong

[timber hatch]

you are right

[exotic yarrow]

Btw there’s the emoji

[jaunty jasper]

I did this which I believe is correct, x=6

[exotic yarrow]

,w x/(x+2)=(x/2)/(x-2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

You may want to justify why you can divide by x. Other than that, seems good

[jaunty jasper]

division property??

[exotic yarrow]

You can’t divide by zero, so dividing by $x$ assumes that $x \neq 0$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

,w graph x^2-6x

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

,w graph x-6

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

Civil Service Pigeon

[jaunty jasper]

ohhh

[exotic yarrow]

[timber hatch]

Get it

[plush brook]

Yes yes I see

[exotic yarrow]

Is there a reason that you keep pinging me? Do you have a question?

[timber hatch]

Here

[timber hatch]

Sorry.

[slim plinth]

Domain: x≠+/- 2

[deep oak]

You just constrain the linear inequal regions

https://www.desmos.com/calculator/ikbyet9w2p

[jaunty jasper]

Can someone help me with this problem?

[shut thorn]

first. What is the total sum of the exterior and interior angles?

[jaunty jasper]

exterior is 360 and don’t I have to know how many sides to know the sum of interiors?

[shut thorn]

you have to find one interior angle

you're given that a full angle

(interior + exterior angle)

where they are in a ratio of 3:1

what will the sum of interior and exterior angle be?

[jaunty jasper]

ohh 180

[shut thorn]

no no

interior and exterior angles add up to 360.

why?

because they complete the entire arc

like uh

wait

can you tell me what the reflex of angle ABC will be?

@jaunty jasper

[jaunty jasper]

the part outside of that angle?

[shut thorn]

ye so isn't the part which makes angle ABC 360° when added up called the reflex angle?

what if I

[jaunty jasper]

it is yes

[shut thorn]

what if I join AC?

is it still a reflex angle or does it become something else entirely?.

[jaunty jasper]

I think it stays a reflex angle

[shut thorn]

but it also become an exterior angle of the triangle doesn't it?

[jaunty jasper]

oh it does?

[shut thorn]

ye

so from this

interior and exterior angles add up to 360

don't they?.

[shut thorn]

now can you find the measure of interior and exterior angles from using this information they've given?.

[jaunty jasper]

yes the exterior is 45 and interior is 135

[shut thorn]

now what is the formula of finding the measure of the angles of a polygon?

[shut thorn]

no wait this is wrong

[jaunty jasper]

oh😭

[shut thorn]

The exterior and interior should add up to 360

[jaunty jasper]

oh

[shut thorn]

because exterior and interior angles are basically reflex angles od each other

[jaunty jasper]

wait it’s exterior is 90 and then interior is 270

[shut thorn]

like in this

[shut thorn]

nice now what's the formula for finding the measure of angles in a polygon?

[jaunty jasper]

is it 360/the measure of one of the exterior angles

[shut thorn]

no

[jaunty jasper]

wait

180(x-2)

[shut thorn]

that's the sum of all interior angles

what if I want to find one?

what should I divide. With so that only one of them is left

[jaunty jasper]

180(x-2)
/x

[shut thorn]

now what should this be equal to

[jaunty jasper]

180(90-2)/ 90 =176

[shut thorn]

no no no where did you get the 90

[jaunty jasper]

i thought 90 was x

[shut thorn]

x is the number of sides of the polygon

and from

[shut thorn]

this

we know that interior angle is 270

so.

[jaunty jasper]

ohhh

[shut thorn]

180(x-2) /x = interior angles

angle*

and what is the interior angle?

[jaunty jasper]

270

[shut thorn]

ye.

s9

so

180(x-2)/x = 270

solve for x

[jaunty jasper]

okay

[shut thorn]

wait wha

[jaunty jasper]

i got 4/5 😭

[shut thorn]

😭

[jaunty jasper]

wait no

[shut thorn]

oh fuuck

exterior angles are supplementary

😭

[jaunty jasper]

i got -4

[shut thorn]

sorry

[jaunty jasper]

oh 😭

[shut thorn]

this would be correct

same process from here

[jaunty jasper]

okay so i slice for x but its equal to 135

slice

solve

[shut thorn]

slice

🔥

[jaunty jasper]

okay so i got like -0.8

which i don’t think it right

[shut thorn]

how rhe hell

do you want me to do it

[jaunty jasper]

please

[shut thorn]

kk

[jaunty jasper]

this is what i did btw

[shut thorn]

mistake in second line

how did that x become 180 in the denominator

the fact that if you work this out even more then it yeilds the correct answer 😭

[jaunty jasper]

because i got rid of the x then I divided by 180

[shut thorn]

ye continue on

[shut thorn]

im also kinda blind 🎉

[jaunty jasper]

wait i just redid it in my calculator and got 8

how did i ever get -0.8

[shut thorn]

no if you work out -2 = -0.25x then you also get 8

[jaunty jasper]

that’s what i redid

[shut thorn]

wait lemme show u how

(-1)(-2) = (-1)(-0.25x)
-> 2 = 0.25x

then

divide by 0.25

[jaunty jasper]

wait i have to make it positive..

[shut thorn]

yeah

[jaunty jasper]

ohh so then 8 is how many sides it has

[shut thorn]

yep

[jaunty jasper]

okok thank you so much

[exotic yarrow]

Alternatively, recall that the exterior angle of a regular $n$-gon is $\left(\frac{360}{n} \right)^{\circ}$. From the ratio, an exterior angle is $180^{\circ} \cdot \frac{1}{1+3}=45^{\circ}$, and thus
$$\frac{360}{n}=45 \implies n=\frac{360}{45}=\boxed{8}.$$

[somber coyote]

Civil Service Pigeon

[jaunty jasper]

where did the 180 x 1/1+3 come from?

[shut thorn]

1x/4x

it's like uh

ratio of exterior angle / total ratio

[sly urchin]

∆ABC inscribed in (O;R), (O';R') is internally tangent to (O) at I (R<R'). Point P ∈ (O). PA, PB, PC ∩ (O') = A_1, B_1, C_1. The reflections of B_1C_1, C_1A_1, A_1B_1 through BC, CA, AB resp form ∆A_2B_2C_2. Prove (A_2B_2C_2) is tangent to (O)

The point of tangency might be easier to find if I draw this on GeoGebra...

[upper karma]

ok but thats kinda cheating

but yeah i was guessing the answer was just that

[rustic lynx]

whats the range of f(x)=sinx-3?

[sly urchin]

|| Is inversion at I helpful here? ||

[upper karma]

hm the reflections suck

[sly urchin]

Usually when I solve problems with circles tangent to each other, I try to find properties of the point of tangency to "redefine" it

I haven't found the property of this one yet though

[upper karma]

yeah thats usually how you solve these

[upper karma]

same lol

[sly urchin]

I've also found that depending on the position, the circles can be either externally or internally tangent

[sage sequoia]

Help I can’t get pass electrons adventures

I require further assistance

[short swift]

does anyone have suggestions how I should take on Geometry at the beginning with learning all the theorems, postulates, and proofs etc?

[dark gust]

Sounds like you'd enjoy a more proof and logic-based book that covers introductory geometry. I'd recommend getting your hands on Kiselev's Geometry

Actually, it's a really old book so you're surely gonna find it as a pdf

[short swift]

I'm just looking to learn Geometry over all to get ahead of my math class next year

[dark gust]

Try some course like highschool geometry on Khan Academy (which is surprisingly good) and I also suggest getting the habit of solving problems. Olympiad problems are gonna get your gears turning like crazy, I've done it before for algebra and after a couple of months I didn't have to worry about most of highschool math ever again.

If at any point you find a problem that requires some skill you haven't learned yet then you'll most likely find a detailed solution on forums and you always have this discord or youtube videos to help you learn particular skills

[slim plinth]

Is this right

70 degree is 180-110

[exotic yarrow]

,w 2x+65=4x+20=y

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

good

[slim plinth]

Thanks

Yea on Geometry I like to calculate algebra stuff

[exotic yarrow]

[dense current]

can anyone help pls?

[shut thorn]

convert angle AOB to degrees

[slim plinth]

I think there is an area formula that has radians

We know it’s pi r^2/4

[shut thorn]

Well I mean-

just use the degree formula st that point.

at

[slim plinth]

Hint hint: find a unit circle

[slim plinth]

Is it just theta pi r

[shut thorn]

OB is given.

[shut thorn]

thtetha/360 pi × r² is the area of a sector

if you convert angle AOB to degrees

[slim plinth]

What’s it in radians

[shut thorn]

tthen you can use it

[shut thorn]

in radians. Ehh wait a min

lemme see

[slim plinth]

Wait

I found it

A= (r² x theta)/2

1/2 x 9^2 x 5pi / 7

[slim plinth]

Okay I assume it’s B

[shut thorn]

kinda wrong there

[slim plinth]

1/2 r^2 theta

[shut thorn]

5pi/7

not just 55po

55pi

fucl

I keep

😭

[slim plinth]

But

It’s still B

Bc Denominator is 7 x 2

The only ans choice that has 14 on the bottom is B

Oh

[shut thorn]

I uh

[slim plinth]

But

[shut thorn]

forgot the cocommand but

[slim plinth]

We can get 405 pi

[shut thorn]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[shut thorn]

ye

[slim plinth]

By 5pi x (9)^2 x 1

Gtg

[opal grove]

Q.10 length of an arc and area of a sector. How do I find the angle???

[indigo slate]

have you tried drawing the clock face out? see where 10 and 35 minutes are

[opal grove]

To measure the angle?

I tried and it was off when I tried to solve it

[quaint rivet]

The angle would be 150 degrees, since the whole clock is 360 degrees and that means 60 minutes will be 360 degrees so 25 minutes (35-10) will be 360/60 × 25 resulting 150 degrees.

[indigo slate]

more like, 60 minutes is 360 degrees

divide both things by 12, and so 5 minutes is 30 degrees

[opal grove]

Hold on I'm tryna visualise this

Okay I think I get it

[opal grove]

So to solve the degree of a clock thingy like this, I just use 360÷60×[minutes between two times] ?

[shut thorn]

Conveniently shaped circular clock beside me:

[opal grove]

I'm gonna steal that clock

[shut thorn]

Nuh

[opal grove]

Yuh

[shut thorn]

I also have a rectangularular coick

For sone reason

[quaint rivet]

You can, If the hour is same then

[opal grove]

WHAT PLS SHOW

[shut thorn]

Why 😭

[opal grove]

I'm intrigued

[opal grove]

Yah

Woah that's cool

[shut thorn]

Byt god damb is it confusuign to look at

Abyways

[shut thorn]

From this

Can ut help

It

[opal grove]

Ja I can see why

I've always had trouble reading analogue clocks which is why I don't have one

[shut thorn]

Whay will the angle between 12:10 and 12:35 be?

[opal grove]

150°C

.

150°*

[shut thorn]

C

🔥

[opal grove]

😭

[indigo slate]

my American clock uses Fahrenheit

[shut thorn]

Do you know rhe formula for calculatung area if sector?

[opal grove]

The formula to calculate sector with these numbers?

[indigo slate]

it's actually just (angle)/(360 degrees) * area of full circle

[shut thorn]

My clock only adds on K?
No need to flex

[shut thorn]

Ye

[quaint rivet]

150/360 × pi × 4^2

4 is the radius i mean hand's length

[indigo slate]

the reason is that 1/2 the central angle is also 1/2 of the entire circle

[shut thorn]

I thunk this is one too many helpers so in Goanna ho

Go

[indigo slate]

so it's directly proportional

[opal grove]

pi×4²×150/360

Yay tyy

[indigo slate]

so the area of a circle sector is directly proportional to the central angle

[shut thorn]

What do you thinj the distance traveled be equal to?

Think of it

Its moving in a circular motion

Did ny network go out

Nop

[opal grove]

So in summary

[shut thorn]

Ye

[opal grove]

Wait let me find my spare calculator bc mine died

[shut thorn]

This is for the angle

[shut thorn]

Oh cmon its just a calculation of 10s and 5s 💔

[opal grove]

What is this meant to mean

Length of arc?

[shut thorn]

Oh eait

Fuck

[opal grove]

What

[shut thorn]

Weing wait wait

I rhiugh you were rekyingto

[opal grove]

😭 my head is gonna implode susmariosep

[shut thorn]

Ye

So how do yoy find the length of an arc?

[opal grove]

Pi×4×150/180=10.47m

:3

[neon wagon]

rn im learing (x+3)²+(y+3)²=3 (and also perfect squares

[shut thorn]

Ye shoul be correct

[shut thorn]

So.. whats the problem?
Having any?

[neon wagon]

no just cofussed a little

[shut thorn]

About

[neon wagon]

like with negatives

[opal grove]

Ok thanks

[shut thorn]

Np

[shut thorn]

?

[neon wagon]

but also im in a extra math class bc im not the smartest

[shut thorn]

Np just tell your problem and ill try to help to ny best capabiliy

If youre free that is

[shell birch]

Teach me math

[shell birch]

I did just get done reading a book about the man. Named Euclidean

Who apparently added to this concept

[sly urchin]

Not a problem but... Do you guys like to colors lines in GeoGebra when solving olympiad geo?

I feel like it does make some things pop out more, like circles and some similar triangles

But it could hurt your eyes too

[shell birch]

Teach away

[sly urchin]

Wdym?

[quaint rivet]

Can someone give me a proof problem on circle

[shut thorn]

fine wait wait

,rccw

[shut thorn]

the somewhat unreadable thing on the first line is "AB = 2AC"

[quaint rivet]

Well

[shut thorn]

its kinda easy and kinda hard

wait isnt that called mediocre difficulty

😭

[quaint rivet]

Maybe 🐱

[shut thorn]

maybe

[quaint rivet]

@shut thorn done

[shut thorn]

ok i was so confused. 😭

that r looks so much like a p

[quaint rivet]

😅

[shut thorn]

your r looks like a p and my n and x arent even differentiable

somehow

dont ask how its that bad

i dont know either 😭

[quaint rivet]

I was also thinking that why 4p^2 - 4p^2

[shut thorn]

😭 😭

i also do that sometimes

wait a min

see this

[quaint rivet]

I used to face that too

[shut thorn]

Take a look at rhis and tell me that this equals

sum(x = 1 to n) sin(x)

[shut thorn]

used or use to

[quaint rivet]

Used

[shut thorn]

sure sure

sure

[quaint rivet]

But still may face if I write fast

[shut thorn]

this is me writing slowly

[quaint rivet]

Isn't it an infinite series? Or n upto 360

[shut thorn]

just n

i wanted to ask what it would be equal to tbh

because i got curious of what summing up sin(x) upto n would bee like

[quaint rivet]

You meant this

[shut thorn]

and would it be different then just an integral

[quaint rivet]

Ohh xD

[shut thorn]

ye

[shut thorn]

would this be equal to "int(1 to n) sin(x)dx"?

if so. then why? And if not then why not?

[quaint rivet]

Idk calculus actually 🤷‍♂️

[shut thorn]

ohh right this is calculus... 😭 🙏

[quaint rivet]

But will start after July maybe 🤲🏻

[novel wing]

Well if n = 360 everything cancels out

So we just look at n<360(idk how this helps lmao)

Hmm

[indigo slate]

https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro

you don't need calculus

[shut thorn]

oh

thanks

[shut thorn]

oh yeah it does because sin from 0-180 = -(180-0)

[indigo slate]

yeah, an easier variation on this problem is tan 1 * tan 2 * tan 3 * ... tan 89

[novel wing]

So we just look at n mod 360(if we’re talking in radians)

[indigo slate]

or even easier, cos 1 * cos 2 * cos 3 * ... * cos 180 (lol)

[novel wing]

0 ?

Ye

[shut thorn]

i feel like im much better not knowing this for now..

😭 🙏

[indigo slate]

yeah, it's a dirty trick question

[novel wing]

I once saw an incredible problem regarding sin summation from a Chinese Olympiad

[novel wing]

Some question that would be in a YouTube thumbnail with « genius olympiad(even tho it’s not) problem only 99% can solve)

[quaint rivet]

damn

[sly urchin]

I think I need a clever approach, some new points

[spring vine]

I need help with this

Can anyone help me😭

[exotic yarrow]

Consider how you can write the area of the outermost ring as the difference between the areas of two circles

Now spam that logic

[lament prairie]

I apologize if the answer wasn't clear, as I study mathematics in French, not English

[short swift]

Hi, I am a 8th grader trying to learn Geometry in my free time to advance my classes. It started off going well, but it's getting really hard for me. I don't know how to memorize all of these postulates, properties and theorems at the pace I am trying to go at. I am using two textbooks, Geometry: Seeing, doing, understanding and BIG ideas math Geometry. The big ideas one I keep making mistakes in during the exercises. I am not sure how quite I should approach this, but it isnt going the best right now how I am. Does anyone have any suggestions?

[hardy cypress]

I am a eight grader doing advanced trigonometry, as I was doing geometry in 7th. The strategy I would recommend is to lock in.

[short swift]

YOOO! Can you explain how you did it? I am super interested

locking in is hard lol

[hardy cypress]

I watched YouTube

Very hard strategy

[short swift]

holy yeah

i have two textbooks, but now i regret getting 2 lmao

[shut thorn]

isolatino

isolation

[short swift]

i wanna take twice the notes

[short swift]

im trying to but literally i'm crashing out at small mistakes i make on practice

[shut thorn]

common

i do that too

[short swift]

i keep forgetting names of postulates and like

theorems and rules

[shut thorn]

i remember once i bit down my notebook because i forgot that tanx = sinx/cosx

[short swift]

😂

[shut thorn]

just do more questions which require direct applications of those

[short swift]

i feel like eating my notes

[short swift]

yeah

[shut thorn]

atleast we have something in common

[short swift]

real

[shut thorn]

paper lowkey tastes good

ill eat it

[short swift]

never had it but feels good

[shut thorn]

yk like those wood spoons

which come with ice cream

[short swift]

yeah

mm

[shut thorn]

tastes like that

[short swift]

yum

by the way, while learning geometry, is it okay to have a sheet of paper just will all the postulates or rules on it to reference to in order to find the name?

[indigo slate]

yes

geometry questions test your ability to think and make connections, not to memorise

[short swift]

🙏 oh bet

Im going to connect my brain to my paper

[shut thorn]

nah connect those teeth to it

[short swift]

i am actually going to 😭

[shut thorn]

it tastes damn fine

[short swift]

I just sold in my real math class by forgetting to fully solve a question

i knew how to i just didnt see the instructions

[short swift]

lemme try rq

[shut thorn]

take a small piece

[short swift]

kinda like bread

[shut thorn]

so you dont choke

[shut thorn]

i think we need help dont we

[short swift]

yeah

its kinda like a bland ritz cracker

[shut thorn]

tf is a ritz cracker

[short swift]

YOOO YOU DONT KNOW?

these things

[shut thorn]

oh

[short swift]

but bland

and not crunchy

nvm bad analogy lol

[shut thorn]

forgot what theyre called here

[short swift]

ight imma go lock in

im behind like 2 days in my geometry

[shut thorn]

fire

[short swift]

more like 3-4 actually 😭

[spring vine]

Thanks you

[short swift]

@hardy cypress yo quick question, how much time every day did you spend learning?

[opal grove]

I lwk forgot how to do these things. How would I find the perimeter and area of this?

[zealous berry]

for area, it'd be A = deg/360 pi(r^2)

so then it would be 270/360 cuz of the 90 degree cut taken from it as indicated

[opal grove]

270°?

[zealous berry]

which is 3/4

[zealous berry]

yea

[opal grove]

Ok

[zealous berry]

3/4pi(8mm^2) would be the calculation for ur area

[opal grove]

What about the perimeter?

[shut thorn]

Same way

Use the perimeter formula though

[opal grove]

Wait, would that be sector?

[zealous berry]

yea

[opal grove]

Ohh okay thanks

Same thing with a quarter circle?

Just 90°?

[zealous berry]

well you would just plug in the circumfrence formula

which is 2pi(8mm) here

that times 3/4 cuz a quater is taken off

then add the two straight sides twice as the length of the radius

[opal grove]

Could I use the sector perimeter instead?

[zealous berry]

sure

[opal grove]

Aight thanks

[umbral shoal]

If , cot∅+tan∅=sec∅+cosec∅
Then Find the value of ∅

[shut thorn]

convert cot, tan, sec and cosec to their sin and cos parts

also man just use cotA or something.

I don't wanna write cot(phi) everytime 😭

[plucky notch]

The value of ∅ is 0

[umbral shoal]

cosA/sinA+sinA/cosA=1/cosA+1/sinA
cos²A+sin²A/sinAcosA=sinA+cosA/sinAcosA
1/sinAcosA=sinA+cosA/sinAcosA
So what would you do now? Two things you can suppose sinAcosA= 0 or suppose it's not 0

[umbral shoal]

Prove it

[plucky notch]

We got sin ∅+cos ∅=1 right?

[umbral shoal]

If we assume sin∅cos∅≠0 then yes

But

If you say

∅= 0

Then sin∅cos∅ would be 0 × 1 = 0

So

[plucky notch]

Theres no valid solution

[shut thorn]

I think so too?

[plucky notch]

L.H.S≠ R.H.S for any real solution

So check the question before asking

[plucky notch]

Yes

[quaint rivet]

0 and 90?

[umbral shoal]

Are you sure

[umbral shoal]

How? Can you prove it

[quaint rivet]

Ignore the bad handwriting

But one more thing, tan and cot cannot contain 90 degrees or 0 degree theta 🥀

[umbral shoal]

One thing one thing

You considered sin∅cos∅≠0

At that line sin∅cos∅ ÷ sin∅cos∅ = sin∅+cos∅

[quaint rivet]

yes ig

[umbral shoal]

You are already proved incorrect

[quaint rivet]

Gg

[umbral shoal]

Haha actually

Wanna know the real answer?

[quaint rivet]

What?

[umbral shoal]

You got it right it's an invalid question made by me to confuse others

🫡

It looks easy

But

[quaint rivet]

[umbral shoal]

It's not

Because I think questions like this So Just spreading curiousity

[umbral shoal]

Still if someone wants to solve it they can

And I will prove them incorrect

[hardy cypress]

Ur right, there is no answer that is valid

[lament prairie]

U are welcome 🤗

[spring vine]

Teacher forgot to check if the answer was right today

I will have to see next week 😭

[lament prairie]

Ohh i hope it’s right 👍🏻 😅

Where are u from??

[tardy field]

,w cot(x) + tan(x) = sec(x) + csc(x)

[somber coyote]

Results provided by WolframAlpha

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

Requested by altanis_

[spring vine]

I'm from Vietnam

But I'm in Finland

Thanks for asking

[lament prairie]

Ohh that's nice, It's a beautiful country

[lament prairie]

I did nothing 😊✋🏻

[spring vine]

Thx

[low ocean]

[lament prairie]

I'm just being honest 🩷, btw I'm from Morocco

[spring vine]

cool

[lament prairie]

Thanks, I've just subscribed to ur YouTube channel

[spring vine]

aw thx

[cosmic cedar]

whats the question mate

[long geyser]

How long is this course supposed to take? https://www.khanacademy.org/math/geometry

[cunning lion]

about one school year i should think

[sly urchin]

Why are (HBC) and (HPQ) tangent?

[winter shadow]

gng what was the formula for law of cosines again?

@everyone

[indigo slate]

just Google it

[winter shadow]

alr bro

[umbral shoal]

Actually no I just figured out This question has a solution

[umbral shoal]

∅ will be 0 or 90 your answer is right but you failed to explain me that if you consider sin∅cos∅ ≠ 0 how you wrote sin∅cos∅ ÷ sin∅cos∅ = 1 which I figured out

anyone else want to give the explanation of how ∅ will be 0 or 90 without considering sin∅cos∅ ≠ 0

Btw the question was
If , cot∅+tan∅=sec∅+cosec∅
Then Find the value of ∅

[dark sparrow]

please don't use the empty set symbol as a stand-in for theta 😭

[umbral shoal]

I can't find anything else

[dark sparrow]

just write "theta" or Google the Greek alphabet for copy and paste

[umbral shoal]

k

[desert nimbus]

its the guy from geometry dash

[sly urchin]

What...

[feral rapids]

Does anyone knows?

[shut thorn]

Translate?

[upper echo]

Draw perpendiculars from D to AB and AC. Then 11=(x-5)+7.

[feral rapids]

Thanks Mate!

[quick heart]

I am a specialist maths student but my school never taught me certain topics in algebra. I do online school and am mostly self taught, can someone explain what is meant by chords (I think its to do w circles) bc my teacher said that it would be on the test but I didn't get to ask him

[steep escarp]

on a circle?

a chord is a line segment that connects two points on its circumference from what i remember

[upper karma]

its just [insert the whole solution]

and insert clown emoji

are you asking because you found that in a solution or are you trying to find a solution through this claim

because iirc thats not how you solve this problem

i mean its not the best sol

ok now i will hibernate

[long geyser]

Why can't you subtract cos(55) from both sides?

Nvm the reason is that it's not the same answer. What makes dividing by cos(theta) better than subtracting both sides by it? Is it not still something that you can algebraically do?

[sly urchin]

I saw it in a livesolve

[sly urchin]

Considering the person drew the diagram, solved the problem then explained it within less than half an hour live, I would argue the sol was quite good

[upper karma]

then whats the context

[sly urchin]

Power of a point of some sort, but please say it i'm not too familiar

[upper karma]

is that the first step of the sol

[sly urchin]

It's just his quick observation, he inverted at A right after that

[sly urchin]

The video didn't really explain as it was "obvious" by PoP, but I need some elaboration

[upper karma]

then i misremembered something

[quick heart]

ohhh okay thanks

[tropic widget]

NO SCHOOL
NO SCHOOL
NO SCHOOL
NO LEARNING
NO LEARNING
NO LEARNING
LEAVE THIS SERVER

[upper karma]

mirror emoji

[tropic widget]

bye

[sly urchin]

Please explain, i'm new to PoP, i might not even know the theorem

[silent plank]

you can, its just not meaningful in solving for ED

[steep escarp]

ofc!

[long geyser]

Tysm. Why though?

[upper karma]

assuming (HBC), (HPQ) are tangent at H, then radax on (HBC), (HPQ), (ABC) implies BC, PQ, HH concur. so HR is the tangent to HBC at H and this line is trivially parallel to EF

so why did he not immediately finish

[silent plank]

that'll lead to
$$\frac{4}{ED} - \cos(55\deg) = 0$$

[somber coyote]

ραμOmeganato5

[sly urchin]

"Why are the circles tangent?" Is my question

[upper karma]

(HBC) and (HPQ) tangent is easily equivalent to the statement of the problem

[upper karma]

im not trying to prove that, i wanted to show you this is literally equivalent to the problem basically

[sly urchin]

Wdym radax?

[long geyser]

I meant like this

[upper karma]

the radical axes of those circles concur

so i kinda doubt its that easy to see theyre tangent

[long geyser]

Ohhh because LHS will be ED*0?

[upper karma]

like, i rememebr it wasnt, but maybe im remembering it wrong, ill check

[silent plank]

no

different reason

[long geyser]

Oh

[silent plank]

that's not valid because what you effectively did was divide the left side by cos(55°)

[sly urchin]

Maybe i misunderstood the idea then

Maybe that was just working backwards state the problem in an equivalent way

[long geyser]

So it's supposed to be ed-cos=4-cos then?

[upper karma]

maybe he was just saying that after proving it, the finish would be quick by pop and that kind of stuff

cause the hard part in the sol of that problem was indeed inversion at A and doing smart stuff

it would be weird if he did all that after claiming that tangency

[long geyser]

except ed doesn't equal 4

[silent plank]

if you were to subtract cos(55°) from both sides there, you'd get
ED*cos(55°) - cos(55°) = 4- cos(55°)
you could factorise the left side to get
cos(55°)(ED - 1) = 4 - cos(55°)

which doesn't really help you at all in solving for ED

[long geyser]

I see, tysm.

[silent plank]

take trig out of this

[long geyser]

All of the terms in the LHS are subtracted by cos(55°)?

[silent plank]

this is the same principle as solving a linear equation like
4x = 7

subtract the same value from each side once would be a valid operation

if you have multiple terms, you don't subtract that many times

there was only one term on each side anyway

[upper karma]

i remembered this wrong, but its just because that condition can be rephrased in many ways and youre not necessarily focussing specifically on that

[long geyser]

Right. Since cos(55) = some constant, why wouldn't subtracting it from both sides to get ED=4-cos(55) be the simplification? I don't know how to factor a trig ratio

[silent plank]

subtraction doesn't mean erase a component that looks the same

note that cos(55°) is being multiplied to ED

this is the same principle as solving a linear equation like
4x = 7

4x - 4 isn't x

[long geyser]

Because of PEMDAS? (ED*cos(55))-cos(55) isn't the same as ed(cos(55)-cos(55))?

[silent plank]

if you have
4🍎
subtracting number 4 (non apple, non fruits) doesn't give you one 🍎

[long geyser]

Yep

[silent plank]

yes, also violating pedmas like that would result in 0

[long geyser]

I see, tysm!! I think my problem was that I was doing it all in my head which made it difficult to see the error, I intuitively knew to divide both sides but I wanted to find out if subtracting could yield the same result.

[silent plank]

and i used "erase" because you essentially rubbed out cos(55°) and not subtract in any way

[long geyser]

Exactly

[long geyser]

Yeah, in my mind that was normal/ok to do

i can't believe i can make mistakes like that after all of the practice i've had with algebra

That's ok though, I've learned why it doesn't work now

[cosmic cedar]

hii

[long geyser]

I constructed this in the Desmos geometry tool, are there any special properties about this?

[shut thorn]

Well-
First thing I see is that there's a circle 😭🙏

[long geyser]

[shut thorn]

Cyclic quadrilateral

And the cyclic quadrilaterals diagonals (the radii) are being bisected and are also equal. Which means that it is a rectangle

[long geyser]

It’s a square though, no?

Because all sides are equal

[shut thorn]

Ye but how the hell will you prove them equal

[long geyser]

Hmmmm

[long geyser]

Through similar triangles no?

I mean congruent

[shut thorn]

Ye but no congruency as much as I try

Wait lemme uh

For it to be q square. One of these should be true. The angles are being bisected by the diagonals OR the adjacent sides are equal

Since it is not given that the diagonals bisect at 90. We cannot assume that it is already a rhombus

[long geyser]

I see

[long geyser]

I constructed this to where it's perpendicular though. So it's given, no?

[shut thorn]

Oh so the diagonals are perpendicular.?

[long geyser]

Yep

[shut thorn]

Yep

[long geyser]

which is why i was confused

ah

Okay then!!

[shut thorn]

You didn't tell me vro 😭

[long geyser]

XD my bad!!

[shut thorn]

🤸‍♂️
🦽🏌️‍♂️

I may be a bit angry

[long geyser]

Sorry I thought like you somehow knew, but I forgot you should never assume something from a diagram if it's not given.

[short swift]

would anyone by chance have a free pdf answer key (full) to Geometry: Seeing, doing understanding third edition?

[hot gazelle]

Let ABCD be a rectangle. Let H be the projection of A onto the diagonal BD (AH \perp BD at H). Let M be the midpoint of HD.
​A line passing through M and perpendicular to AM intersects the line AB at N.
​The line CM intersects the line AB at K.
​A line passing through K and parallel to AD intersects BD at P.
​Prove that: The points N, C, and P are collinear.

My teacher's suggestion is to prove AM intersects BC at the mid point of BC

I still dont know wat to do

🙁

[sly urchin]

Wdym

[hot gazelle]

I meant NM

[shut thorn]

I think he meant BP?

oh

[hot gazelle]

Sorry :<

[gusty grotto]

Why do i get different results of (1-i)^3 when i try to solve complex number using polar form and formula

[cunning lion]

you shouldn't

show your work

[gusty grotto]

Got these 2

Maybe i did the polar form one wrong

[gusty grotto]

^

Photomath got different results too

[tranquil iron]

Are drafting pencils good for math?(especially graphs)

[gusty grotto]

I use pen

[cunning lion]

when raising the polar form to a power you must multiply the argument by that power

[gusty grotto]

Yh i already did

[cunning lion]

15pi/4 is the same argument as the original number

[spring needle]

guys how do i get better at proofing triangles, what r some tips

[dark sparrow]

the verb is "to prove" not "to proof"

but also what do you mean by "proving triangles" exactly? @spring needle

[tired bone]

how do i calculate angle from a vector 2 or vector 3 can anyone explain

[spiral lodge]

!original

[tired bone]

?

[spiral lodge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tired bone]

i want to apply this logic to the code i am working with where i want to get the vector position of the mouse then calculate the angle i need to rotate the object to look towards the mouse position so i want to calculate vector to angle

[spiral lodge]

Mmh I see

Try asking in #computing-software , they probably have some more precise suggestion

[tired bone]

oh ok

[mellow moth]

if you take AB to be 2
how is the length of the middle of ABCD to T sqrt(2)?

like this i mean

how is that line going to T sqrt(2)

[upper echo]

||Triangles AHD and ABC are similar, so if L is an intersection of MN and BC then ∠BLA=∠HMA (ABLM is cyclic). This implies that AL is a median of triangle ABC (just as AM is a median of AHD). Then the properties of a trapezoid KBCP give that N,C,P are collinear.||

[shut thorn]

ok thats cool and all but why do you have so many tabs open vro

😭

FIFTEEN TABS

[long geyser]

It's only like 2 if you group similar ones

[shut thorn]

still nine..

[short swift]

My laptop is worse...

One of my accounts has like 40 tabs lmao

[shut thorn]

im scared of opening more than 4 tabs at once vro 🥀

i should lowkey switch to microsoft edge then

[short swift]

Edge isn't that good lol

Google is fine

[rustic lynx]

Bruh can anyone help with domain range and graph of functions

[shut thorn]

any specific question youre stuck on?

[viscid vapor]

uhmm guys i need help with something?

[neat gale]

?

What?

[viscid vapor]

not or so proof or an equation but like a friend who's a major in maths keeps talking about how tri-angles are not polygons and now i feel like a dummy cuz this feels like it goes against common sense

[neat gale]

So

You think that tri andlges are a polygon?

[viscid vapor]

are they not?

[neat gale]

Yes, triangles are polygons. A triangle is defined as a polygon with three sides, three vertices, and three angles. It is the simplest type of polygon in geometry, and the sum of the interior angles of a triangle is always 180 degrees.

there.

Your friend is wrong

[distant crag]

8th one.

Anyone

Online?

[cunning lion]

,rccw

[upper echo]

Draw the diagram and use geometric interpretation for the sum of several vectors. Then the answer is more or less obvious.

[spring needle]

congruency

[dark sparrow]

ok right

• remember the congruence tests
• gather data about equal angles or equal sides
• recall the angle sum of any triangle is 180° (helps with angles)

[spring needle]

yeah i know all this but

i keep proofing too much

[dark sparrow]

proving*

also what do you like, mean exactly

"proving too much" is a logical fallacy but i want to know if that's what you're talking about or not

@spring needle can you show an example where you proved too much?

[somber kite]

hey

[mossy aspen]

anyone please

its due 12 am

TOT

[cyan scaffold]

draw it out

[mossy aspen]

i did and i keep getting it wrong

[cyan scaffold]

what did u draw?

[cyan scaffold]

well where is point A and point B?

label them

[cyan scaffold]

yeah so

youre trying to find B-A

what do you know right now?

[mossy aspen]

i did tan (2)=181.7/x

which is like 5202.6

and that minus 787

which is 4415.6

and its wrong

[cyan scaffold]

ok so

wait are u in degrees or radians

[mossy aspen]

degrees

[cyan scaffold]

ok

just making sure

[mossy aspen]

ok

[cyan scaffold]

ok so im getting a different number for the height

im not getting 181.7

can u check ur work

[mossy aspen]

ok

[cyan scaffold]

oh shoot im wrong mb

wait hol on

[mossy aspen]

ok

[cyan scaffold]

i think

youre just rounding wrong

does it say to round between intermediate steps?

[mossy aspen]

no

ok lemme try again

ahhh its still wrong

i got 4416.1