geometry-and-trigonometry

[dark sparrow]

yes obviously.

do you see how to do the problem now?

[west radish]

Sorry for thinking was easyy

[west radish]

Yeahh

[shadow remnant]

minus 1 though

since one of those combinations will be the successful code

[west radish]

Exactly

Thx for helpp

@dark sparrow 👍

[ionic crypt]

Need help.

[obsidian harness]

By similarity, CE/CB = FG/AD

[ionic crypt]

Was it 8?

[obsidian harness]

yes

[ionic crypt]

Help

[obsidian harness]

,calc 180-62-88

[somber coyote]

Result:

30

[obsidian harness]

Yeah so AA

AA is the same thing as AA, naturally if two of the angles are the same, then both triangles will have 180 - x - y as the 3rd angle

[obsidian harness]

also you can't see any side lengths

[obsidian harness]

Ah okay, so similar triangles again

11/x = x/44

short/long in both cases

[ionic crypt]

22?

[obsidian harness]

yeah

,calc sqrt(11 * 44)

[somber coyote]

Result:

22

[obsidian harness]

(the formula comes from them being similar)

[obsidian harness]

again it's really similar

How about this time, you try to set up an equation yourself?

[ionic crypt]

I'm very dumb.

So..

Maybe, Like give me answer.

Wait, is it 9?

[obsidian harness]

Yes, but how do you get there?

[ionic crypt]

Used a math ai.

[obsidian harness]

well don't call yourself that
Everyone has different strengths yk

[obsidian harness]

ah okay so yes this is something you need to do yourself

I can guide you ofc

[obsidian harness]

Like you have the adjacent (the short side) and the hypotenuse in the small and the big triangle

[ionic crypt]

I use ai math, because I'm dumb

I have dislexcia

so..

[fervent coyote]

how are you typing

[obsidian harness]

that doesn't stop you from learning how to problem-solve

[obsidian harness]

the strategy here is to rotate the large right triangle so it's upright like this

[ionic crypt]

is it 7?

[obsidian harness]

stop guessing

The answer is not what matters

The process is what matters

[ionic crypt]

I'll never understand it anyway.

[obsidian harness]

okay I can't help you if you've already given up on yourself

[ionic crypt]

No.

[obsidian harness]

I can't force you and I can't make you want to learn

[ionic crypt]

Just tell me answer.

[obsidian harness]

no

[ionic crypt]

please

[slender flint]

how do you know it's a right traiangle though

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[dark sparrow]

AI is dumber than you.

[ionic crypt]

.

how u know.

[dark sparrow]

• i have seen it firsthand mess up problems that are remotely more complicated than a single operation
• it's very easy to convince an AI into thinking a wrong answer is right

AI cannot be trusted ever

[ionic crypt]

oh.

[dark sparrow]

also maybe stop spouting things like this? it helps nobody, not even you.

[ionic crypt]

now i'll be dumb forever

[dark sparrow]

what did i just say?

[ionic crypt]

without ai.

[ionic crypt]

then help.

[dark sparrow]

you make yourself sound like you want to actively avoid learning.

[ionic crypt]

is it 7?

[dark sparrow]

how did you get x=7?

[ionic crypt]

.

from.

[dark sparrow]

i don't know if that's right or wrong, but i'd like to see your work for it.

[ionic crypt]

gauthmath

i dont what to do on that math problom.

[dark sparrow]

ok, so

you see here a triangle in which one of the angles has been bisected

[quaint sequoia]

Im not trying to interrupt, but I also want to understand this question. And @dark sparrow can you correct me if I’m wrong. Since this triangle has an angle bisector wouldn’t that mean that 6 and 3 - 12 would be equal ?

[dark sparrow]

and this bisector divides its opposite side into two segments

[ionic crypt]

means split in half.

[dark sparrow]

don't drop the x

and no, 6 and 3x-12 aren't equal

[quaint sequoia]

Yes don’t drop it, u set them equal and solve the equation ?

[dark sparrow]

no

they are not equal

[grave pond]

No.

[quaint sequoia]

Oh then idk

[dark sparrow]

@ionic crypt there is a theorem which you should have learned at some point
which says that the ratio of these two segments is the same as the ratio of the other two sides

[ionic crypt]

She understads. so i'm the dumb one here.

[ionic crypt]

yeah.

[dark sparrow]

ok, so you have those segments and sides' lengths marked. can you write down the equation from what i just said?

don't try to solve it

only write down the equation

and don't try to solve it

[dark sparrow]

and stop beating yourself up like this for gods sake

[grave pond]

Hmm, I can see the ratios are in fact equal, but is there a common elementary theorem that tells us so directly?

[dark sparrow]

if you wanna prove the theorem (which is outside the scope of this problem a bit) you can look at the two smaller triangles' areas

and either some basic trig or a proportionality argument that avoids it

[ionic crypt]

https://tenor.com/view/confused-confused-face-confused-look-im-confused-im-so-confused-gif-16067865999736184254

[dark sparrow]

i am talking to tropo here, you can ignore what im saying to him

i gave you something to do

do that

and show me the equation you got

NOT its solution

but the equation itself

[ionic crypt]

is it 7

[dark sparrow]

you didn't do what i asked

i don't give a flying fuck if it's 7 or not, the point isn't to get the right answer and blind ourselves to how we got it

[grave pond]

I asked because it's not clear to me that Alexcia can understand which lengths it is you say are in proportion.

(Though I recognize there are not many options here).

[dark sparrow]

if she does not understand what lengths im talking about, then she should tell me so, and i will make a diagram that makes it clearer what is what

[ionic crypt]

.

[quaint sequoia]

Um if anyone is able to can I please get some help in the algebra channel 😭?

[faint pasture]

!noadvert

[lime crown]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[dark sparrow]

@ionic crypt should i repeat my instruction

[ionic crypt]

go ahead

[west radish]

BE=2rsinθ
EC=2rsin2θ
BC=2rsinθcosθ
CD=2rsinθcos2θ
CF=2rsin2θcos2θ
DF=2rsinθcos3θ
OD^2 +CD^2 = OF^2 + DF^2 + CD^2 = (r−2rsin2θcos2θ−2rsin2θ)2+(2rsinθcos3θ)2+(2rsinθcos2θ)2
=r2+4r2sin4θcos4θ+4r2sin4θ−4r2sin2θcos2θ−4r2sin2θ+8r2sin4θcos2θ+4r2sin2θcos6θ+4r2sin2θcos4θ
=r2+4r2sin4θcos4θ−4r2sin2θcos2θ−4r2sin2θcos2θ+8r2sin4θcos2θ+4r2sin2θcos6θ+4r2sin2θcos4θ
=r2+4r2sin4θcos4θ−8r2sin2θcos2θ+8r2sin4θcos2θ+4r2sin2θcos6θ+4r2sin2θcos4θ
=r2+4r2sin4θcos4θ−8r2sin2θcos4θ+4r2sin2θcos6θ+4r2sin2θcos4θ
=r2+4r2sin4θcos4θ−4r2sin2θcos4θ+4r2sin2θcos6θ
=r2+4r2sin4θcos4θ−4r2sin2θcos4θ+4r2sin2θcos6θ
=r2−4r2sin2θcos6θ+4r2sin2θcos6θ
=r2

r = 10, 10^2 = 100
OD^2 +CD^2 = 100

[dark sparrow]

there is a theorem which you should have learned at some point
which says that the ratio of these two segments [into which the angle-bisector divides the opposite side]
is the same as the ratio of the *other two sides

write this down as an equation.

don't solve the equation.

@ionic crypt

[ionic crypt]

ok

[dark sparrow]

obviously once you write the equation down you need to show it to me

[ionic crypt]

oh uh.

https://tenor.com/view/byuntear-baby-cry-crying-gif-25019897

[dark sparrow]

what happened?

[grave pond]

Keep gifs in #chill, please.

[fresh meteor]

I don’t understand step 2.
Why is EC=2rsin2θ

[ionic crypt]

my pencil broke.

[dark sparrow]

get a new one?

or a pen?

[ionic crypt]

i dont have one

[fresh meteor]

crazy

[dark sparrow]

you don't have anything else to write with?

[ionic crypt]

nope

[fresh meteor]

Why isit so long tho? i believe there is a requirement for setters to ensure that students can solve it without too much working

[ionic crypt]

just a white bourd

[west radish]

Was in another group

Sorry

Now answring

[dark sparrow]

you have a whiteboard, do you have a marker to write on it?

[ionic crypt]

Dryed out

[fresh meteor]

Use google jamboard

[ionic crypt]

Damn it

[dark sparrow]

like either this is an unfortunate coincidence in which every single writing implement in your house is out of commission,

or you are just trying to find ways to avoid following my instruction

[fresh meteor]

its alr

[grave pond]

Ann it looks a lot like they're trying to find a way to disengage; I think you should let them.

[west radish]

Listen @fresh meteor

[fresh meteor]

Listening

[dark sparrow]

she*

[west radish]

As we know sintheta equals opp by hyp so opp is EC and hyp equals 2r which is BE as we are takung angle 2theta

[ionic crypt]

no helping

[fresh meteor]

What is 2θ?

[west radish]

Lemme its sinsquare theta

My badd

Ill do it again when im freee

Hanan

@bleak salmon

[bleak salmon]

Here

[west radish]

If i thorough the dice 3 times whats the probability that the sum of the ordered pairs would be 7

I want 3 methods

@havoc

@bleak salmon

Do this mann

State your methods guys

[bleak salmon]

Sorry, i was making food for the 17 year old. Give me a sec

[west radish]

😂

[bleak salmon]

Is it a fair dice? Meaning no bias in the probability of dice

[west radish]

Yeaa

Nice thinking though

[bleak salmon]

I would make the list of every possible position of dice that sums up to 7, and then divide that possibility with total possibilities

[west radish]

Like 43. 34. 52 25

And sample sapce would be

[bleak salmon]

Like that, but you said 3 times throwing the dice

So we cannot make a . in the possibility

[grave pond]

The funny thing here is that the problem seems to be underspecified. Once we've drawn EOA, we can pick any point on the upper semicircle as B, and then construct C and D according to the instructions.
To metagame the question a bit, if if has a definite answer at all, it must be that the answer is independent on where we choose to put B.
We can see that if we choose B very close to E, then DO will end up close to 5 cm and CD wil end up close to 0 cm. And this also happens if we choose B very close to A. If we choose B straight above O, CD and DO both end up 5/sqrt(2) cm.
In all of these cases the desired value in cm² is indeed the same ...

[bleak salmon]

Sir Hussain, the possibilities i think would be 223, 331, 421, 511 and times that with 3 as three possibilities

[west radish]

Yea 😂

U can be formal

Hussain it iss

So u mean 6 to the power 3 is the sample space

216

[bleak salmon]

Sorry Sir, im a little used to our usual school environment so I cant just remove a habit Sir Hussain

[west radish]

Alrightttt

[west radish]

Will do it

Its lengthy

[bleak salmon]

Sir Hussain, does the test for tomorrow include integration by parts or not.

@west radish

[west radish]

Yes i told u short trick for ILATE

remember

[bleak salmon]

Ok sir, thx you

[west radish]

And lopital rule will be mixed with it

[bleak salmon]

Classmates were saying the test is day after tomorrow, plz verify it in group Sir

[west radish]

Alrighttt

[neat plover]

i found this in a forum and thought it was fun ;3

[ornate glen]

LOL

[grave pond]

||Draw the circumscribed circle of ABC and use the central angle theorem?||

[lime dune]

i think the easiest method is

||drop an altitude from B||

||30?||

[grave pond]

I get that too, but I don't understand your method.

[ornate glen]

find a lazy method

[lime dune]

||drop an altitude BX to AC, call the length BX=s||

now by various trig ratios $AB=2s,$ $AX=s\sqrt{3},$ $XC=s(2+\sqrt{3}),$ $BC=s(\sqrt{6}+\sqrt{2})$

[somber coyote]

elrichardo1337

[lime dune]

in particular $XM=s$ as $M$ is the midpoint of $\overline{AC}$

[somber coyote]

elrichardo1337

[lime dune]

but that means $\angle XBM=45^\circ$

[somber coyote]

elrichardo1337

[lime dune]

since $\angle ABX=60^\circ,$

[somber coyote]

elrichardo1337

[lime dune]

the answer is $x=135^\circ-60^\circ-45^\circ=30^\circ$

[somber coyote]

elrichardo1337

[grave pond]

Hmm, mine feels simpler than that.
O is the circumcenter of ABC. ||By the central angle theorem we have angle COB=60°, angle BOA=30°, angle COA=90°. Since COB is isosceles with apex angle 60°, it is equilateral. Thus B and M both have the same perpendicular distance to AO, namely half of that of C. So BM is parallel to AO, perpendicular to OC, and bisects angle OBC.||

[stray willow]

hola me puden ayudar

[lime dune]

oh damn that’s pretty nice

I just saw “angle with convenient trig ratio value” and went unga bunga

[knotty pasture]

Is anyone able to help me study for my geometry exam tomorrow. I only need help understanding arcs and tangents

[lime dune]

you can help us start off by telling us a specific question you need help with

I’ll grab one of my old note sheets I wrote up in HS

[knotty pasture]

Do just type out the question or send a picture

[lime dune]

whatever’s more convenient for you

[knotty pasture]

Ok give me a moment

Two concentric circles have radii 8mm and 12mm. What is the length of a chord of the larger circle that is tangent to the smaller circle?

[lime dune]

first step is to draw the diagram, what do you notice?

[knotty pasture]

They share a center

[stray willow]

using the pythagorean theorem

[lime dune]

is there a radius you can draw in to reveal more info?

[knotty pasture]

Yes

[lime dune]

what points does it connect?

[knotty pasture]

The center to the edges of the circles

[lime dune]

try drawing one from the center to an endpoint of the chord

and another one from the center to the point where the chord is tangent to the inner circle?

[knotty pasture]

Ok, done.

[lime dune]

what do you notice about the triangle we just formed?

[knotty pasture]

30-60-90

[lime dune]

not quite

but it is a right triangle

so what can we do with that?

[knotty pasture]

The Pythagorean theroem

[lime dune]

mhm

we know the hypotenuse (12) and one leg (8)

so we can solve for the other leg

[knotty pasture]

4 x square root of 5?

[lime dune]

yeah

now that’s half the length of the chord

so our answer is?

[knotty pasture]

8 x square root of 5

[lime dune]

nice

[knotty pasture]

Thanks

[lime dune]

np

generally if you have a problem with tangents

right triangles will more often than not be involved

especially if you are trying to find lengths

[knotty pasture]

This was the only test I did bad on 😅

[lime dune]

rip

[sacred nimbus]

how to solve this?

[knotty pasture]

Geometric mean

Do you know that?

[sacred nimbus]

3 x 24 than square root?

[knotty pasture]

Yes

[ornate glen]

man

[gritty topaz]

abcd is paralellogram

answer: ||50°||

i dont have a geometric solution

difficulty: 2nd stage olympiad or harder (i guess)

[dark sparrow]

well, basic angle chasing fills in these angles i think

i may have fucked up the arithmetic somewhere

[gritty topaz]

no these are correct

[round wolf]

we have to find the value of X right

[gritty topaz]

yes

[dark sparrow]

i am thinking that maybe we're supposed to make an isosceles or maybe right triangle happen somewhere

but i don't see where

[gritty topaz]

i tried using (2, 46, 62);(8, 12, 50) ceva as lemma but couldn't get anything out of it

[round wolf]

i think 64+14 = the angles in the D part

[dark sparrow]

that's correct, but also doesn't help much i think

[round wolf]

true

[sudden shell]

I need to find the minor arc AB

[terse cypress]

can anyone help me with this?

[fresh meteor]

Do u mean DO approaches 10cm instead?

[fresh meteor]

Do u know cosine rule/sine rule

[grave pond]

Yeah, somehow I had gotten into my head that 10 was the diameter instead of the radius ...

[fresh meteor]

I see, then it all makes sense now…

This is an olympiad qn btw, usually they want us to do some imagination and perhaps move stuff around or draw extra stuff which makes it challenging

[north kindle]

Im not sure where to start with b, I tried extending YQ to be the same length as XZ but it doesn’t seem to help very much

[covert dew]

uhh

isosceles something

prolly HL stuff

vertical angles?

idk

im too lazy to actually read the question

[north kindle]

The only isocoles triangle is PRQ

And I’ve tried messing with it a bit but can’t get any good ideas

[covert dew]

oh

qwz and rxz both right triangles

so u jus need like

one angle

oh wait

i think i found solution

[north kindle]

I’m not sure what angles have to do with it

[covert dew]

if <R is X

then <PQR is like 180-2x, then by vertical angles so is <ZQW

and uh

thats it

nvm

[north kindle]

We aren’t talking about angles

[covert dew]

wait

XZ and YQ are parallel

[covert dew]

that's how similarit works?

bro

[north kindle]

???

[covert dew]

like

for example

AA similarity theorem???

[north kindle]

I know similarity

[covert dew]

SAS?

AAS?

[north kindle]

But wtf does it have to do with this

[covert dew]

cuz

if 2 angles in any triangle are similar

congruent*

the triangles

are similar

[north kindle]

I know what similarity is

[covert dew]

yeah

if you find

2 congr angles

[north kindle]

Use your words to explain how it’s supposed to help with this problem

[covert dew]

then u can prove similarity

wait

which problem'

lol

A or B

[north kindle]

Reading my message might help

[covert dew]

im stupid

my bad

@north kindle

try drawing altitude from q to xz

idk if that helps

[north kindle]

It might actually, I was thinking of something like that

[covert dew]

oh

@north kindle

yo it solves it

i got answer

[north kindle]

Good for you

[covert dew]

...

im jus tryna help bruh

XYQ and whatever you choose to call intersection point of XZ and altitude is a rectangle

[north kindle]

I think all I need to show is that angle XZR = angle RZW

[north kindle]

I think I got it

[maiden brook]

||two tangents theorem||

[sudden shell]

So triangle ABC is isosceles

so side BC is 7 so side AB is also 7? @maiden brook

[maiden brook]

no?

where do u get that from

[sudden shell]

no BE = BF

[maiden brook]

yeah

and what else?

[sudden shell]

FC = GC

[maiden brook]

yeah so u can get the rest right?

[sudden shell]

Ab = 16

[maiden brook]

yes

[maiden brook]

what can u say about triangle FOH

[fresh meteor]

Is there a way to prove that position of B doesn’t matter?

[grave pond]

Most likely, but I've not spent enough effort to find it.

[maiden brook]

can u link to the original problem I am having problems getting to it for whatever reason

[sudden shell]

isosceles?

[maiden brook]

reread the given information

[maiden brook]

one sec I might be getting this

[sudden shell]

i need help

[fresh meteor]

oh

[maiden brook]

nvm coord geo got messy

that probably works

but let’s see if there’s anything else I notice

[maiden brook]

ok i was helped a bit with this

but u get CD^2 = AD * BD

and r^2 - DO^2 = AD * BD

when u add u get CD^2 + OD^2 = r^2

@fresh meteor

[fresh meteor]

Is there a reason for this

Which theorem did this come from

[maiden brook]

similar triangles

[wooden wind]

@maiden brook you look like you know a lot about triangles, can u help me

[maiden brook]

whats the problem pls?

[wooden wind]

its in the help16

[fresh meteor]

what about this one

isit also similar triangles

[maiden brook]

power of a point

[fresh meteor]

whats that

[maiden brook]

look it up

[fresh meteor]

isit case 1

[maiden brook]

yes

[fresh meteor]

We let D be the point

then AD x BD = (r-DO)(r+DO)

[maiden brook]

^

[rigid sleet]

dude

angle PQR and ZQW are equal

[north kindle]

Yes

I did in fact notice that

[weary palm]

How to do it😭 I am failing math p2 tmrw

[cunning quail]

before doing a question directly lets write what we see

tn = tp

tq is parallel to pr

3.1

so angle qtp = x

npq = x

ATQ

TQP = X

ATQ

[weary palm]

Damn

I like ur solving

[weary palm]

How about 3.2

[cunning quail]

well

basics is

the sum of opposite angles of a cyclic quadrilateral is 180 degrees

so we have to prove the sum is 180

opposite angles

[weary palm]

Ah

So how

😂

Where is the center

[cunning quail]

bro

its not

a cyclic quadrilateral

as cyclic quadrilateral vertisec should go be on the circle

but is not

on to the 3.3

[weary palm]

😂😂🙏

Blud ima be calculating my % if I leave all euclidean geometry tmrw

☠

[knotty pasture]

Is anyone able to explain a problem to me

[ornate glen]

wat problem

[knotty pasture]

This was on my math exam I just took and I had no idea how to solve for x

[ornate glen]

thats a triangle correct?

is it about sin cos tan?

[knotty pasture]

On the exam it wasn’t a triangle

[ornate glen]

wat was it

[knotty pasture]

Just those lines

[ornate glen]

<@&286206848099549185>

[sturdy stone]

Use power of a point

9*16=x^2

[knotty pasture]

Would you do the same thing for this problem

[sturdy stone]

Yeah that's also pop

[knotty pasture]

It seems so simple now can’t believe I forgot that

[gritty topaz]

solution

1. Constructed the ADE triangle on BC side as BCF triangle. Since ABFE and CDEF are paralellograms, the angle distrubution would be as showm. We're looking for the angle BCF on BECF quadrilateral.
2. Let the center of circumcircle of the triangle BCE be G. GBE=GEB=44°, GBC=GCB=20°, GEF=30°. We choose a point H on EF such that BGH=16°. Let the center of circumcircle of the triangle EGH be K. GKH is an equilateral triangle, BGH=EGK=16°, BGH and EGK triangles are congruent triangles, BH=GH
3. We choose a point L on BC such that FHL=66° and we constructed the lemma. We know that BH=LH=FL from the lemma. GHL=40°
4. We reflected the point H with respect to GL and marked that as M. Since we know that GH=GM and BGH=CGM=16°, we can say that BGH and CGM are congruent triangles, therefore LM=GM=CM, GCM=16°, MCL=MLC=4°. CFLM is an isosceles trapezoid, CLM=FCL=a=4°, x=a+46=50°

what a beast

[gritty topaz]

there isn't

you just try until you find

this is one solution to one problem

another similar rating problem and its solution

keep in mind there are numerous solutions for these

[wheat matrix]

how do you find the ratio of the perimeter of similar triangles when only the area is given

[dark sparrow]

area scale factor = (length scale factor)^2

do you mean that both areas are given?

@wheat matrix

[wheat matrix]

yeah

[dark sparrow]

ok then what i said stands

[wheat matrix]

area of the first triangle is 64m, the second is 36m, find the ratio of the perimeter

[dark sparrow]

not meters

area is not measured in meters

[wheat matrix]

m^2

[dark sparrow]

yes

anyway

area scale factor = (length scale factor)^2

[wheat matrix]

thanks

[neon kayak]

How he concluded cos(alpha)/-a = sin(alpha)/-b = p/c = 1/±sqrt(a²+b²)?

[nocturne remnant]

If we rewrite the second equation to (cosα)x+(sinα)y+(-p)=0, since it describes the same line as the first equation, we will have (cosα):(sinα):(-p)=a : b : c

So cosα/(-a)=sinα/(-b)=p/c

And then using the pythagorean trig identity we have the fourth term being equal too

[neon kayak]

How can I show that (cosα):(sinα):(-p) = a : b : c implies cosα/(-a)=sinα/(-b)=p/c? Rigorously speaking

[nocturne remnant]

By definition

[neon kayak]

But isn't the definition a:b = c:d iff a/b=c/d? Then a: b : c = d:e:f is something like a/b/c = d/e/f

[shut bronze]

which in this case would be cosα/a = sinα/b = -p/c

multiplying everything by -1 gives exactly what you want right

[neon kayak]

I wrote it wrong, sorry

[nocturne remnant]

Not exactly; it would be more like a/c=b/d (which is equivalent to yours) for two sets of values but for three sets of values it’s a/d=b/e=c/f

a/b/c doesn’t make much sense

[primal veldt]

i am a noob at geometry

how are u guys doing good math

geometry

i should seriously write my math notes if i want to participate haha

[obsidian harness]

practice and also we probably have more experience with maths than you do

[primal veldt]

ah

revitalize the math in me by looking at the experiences and creating problems again hahaha

geometry was so issuablefooorr me

cuz i didn't like the redundant math u have to do

i like thinking outside of the box here but

is there a change of path in geo ?like in geo logyy ? or some

[gritty topaz]

prove theorems by yourself

[primal veldt]

ah so lame but no hate im too lazy

[gritty topaz]

nothing else is as valuable as doing this

[primal veldt]

imma deduce is via theorems we alr know

?

[primal veldt]

aghhhh

is it side a is (some math symbol) to side b

[gritty topaz]

proving the angle bisector theorem saves you from solving 1000s of problems related and trying to remember what the theorem was

for example

[obsidian harness]

you sound really really unhinged

[primal veldt]

hahaha is that the case

🗣️

[gritty topaz]

quite

[primal veldt]

okay so is this as fundamental as this gets

bisector theorem and applying the memory of it unto every systematic thing possible

(any attention to detail from examplar math can help)

angle bisector theorem ... ay ..

[gritty topaz]

learn what xy plane is
learn what coordinates mean
learn what lines, line segments and rays are
learn what angles are
learn what triangles are
learn what similarity is
learn to find the length of segments

i don't think there's anything else you can't derive from these informations

[primal veldt]

can i start reading this too ?

https://en.wikipedia.org/wiki/Geometry

to get the information ?

[primal veldt]

this is so complex i cant

i cant believe it

i cant comprehend what this is first

[gritty topaz]

socrates was more patient than me i guess

what a waste of time

[primal veldt]

time to get that knowledge now

as speak

the definitions of the xy plane the defitions of the coordinate in a given diagram figure wtv the lines line segments n ray r

triangles must be a great

the length of segments n more

my gout time to play ball to the numbers and easy shapes

[neon kayak]

I see, thanks then

[primal veldt]

im a noob ig hahaha

[primal veldt]

how connected are these as a result ?

[lime dune]

just pick up some book and start reading

[astral shadow]

Wouldn't it be x^2= 9(9+16)

[sturdy stone]

Yes

True

[upper karma]

Can you derive that there exists (or doesn't exist) eight points on the plane, no three on the same line, no four on a circle, with integer pairwise distances?

[river raptor]

Can someone check this?

[gritty topaz]

eventually

even if i cant show it, at least i'll be able to understand it when someone shows the solution

if you get what i mean

[autumn venture]

Talking about rhombuses, is opposite sides parallel a corollary from all sides being equal? Or am I going about learning the shapes wrong?

[dark sparrow]

all sides equal does imply opposite sides parallel, and you should be able to sketch a proof of that

[stray willow]

can you help me I have to calculate x

[grave pond]

Law of sines.

[stray willow]

I just have to use Pythagorean theorem

[grave pond]

That'll be trickier less systematic. Do you know the side ratios in a 30°-60°-90° triangle?

[stray willow]

yes

[grave pond]

Make such a triangle by drawing the altitude from B.

[stray willow]

If I already have two triangles of 30, 60, 90, I have the hypotenuse, but I don't know how to use the theory of that triangle to find the legs because if I have the legs, I would also have those of the other triangle 45, 45, 90 and I would only find the hypotenuse

[dark sparrow]

have you done as tropo told you?

[stray willow]

yes

[dark sparrow]

ok, can you show the picture with the altitude that you've drawn?

@stray willow

[stray willow]

I'm just not sure of my answer plus in 12 squared minus k squared

[dark sparrow]

this is nonsense

actually even BD^2 = 12^2 + k^2 is already bullshit, but also that is not equal to 11^2 in any world.

basically all of your work is garbage and unsalvageable @stray willow

well, ok. the angles are correct.

but everything else is useless.

tropo asked you:

Do you know the side ratios in a 30°-60°-90° triangle?
and you answered:
yes
he was implying that you should use those known side ratios to find AD and BD.

[warm drift]

this makes me exited to take geometry next year

[obsidian harness]

ayyyyyyyy geometry is super fun and not like algebra

but the two often come together like in this question or others

[dark sparrow]

!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.

[obsidian harness]

to be fair they have shown working but like yeah

[upper karma]

are all regular/irregular polygons simple?

[grave pond]

Regular polygons are simple.

"Irregular" just means "not regular", right? So all non-simple polygons are examples of irregular polygons that are not simple.

[upper karma]

oh wait so

this thing is wrong right?

for irregular polygons

consider self intersecting ones as well then i guess since you classed irregular as everything else

[dark sparrow]

the exterior angle sum is 360° for all convex polygons, regular or otherwise, and the interior angle sum is 180(n-2)° for all non-self-intersecting polygons, regular or otherwise.

[grave pond]

This may be true, but kindly refrain from stating it so harshly that it sounds like a personal criticism. We want to encourage helpees to show work -- getting that kind of response just teaches them that they had better clam up and say nothing, because otherwise they'll be attacked for their mistakes.

[obsidian harness]

sane mods sane mods

[upper karma]

oh i see, so the better "distinction" is between convex and concave polygons?

[dark sparrow]

myeh

maybe??

[upper karma]

there doesn't seem to be any "properties" that you can split between regular and irregular other than everything that isn't equiangular or same side length

well at least there seems to be more you can do with convex and concave

[upper karma]

wait building on this, why does this not have an exterior angle sum of 360

yeah this is concave

but what sets it apart?

[dark sparrow]

try to draw an external angle at the vertex where the internal angle is reflex

[upper karma]

yeah it's acute

unlike the others

[obsidian harness]

the sum of the exterior angles in a convex polygon relies on the fact that if you start at one vertex, you turn 360 degrees in total to get back to where you started

But this relies you turning in only one direction

In your shape you're turning one way and then the other

[upper karma]

but why are we using directed angles?

considering you say "direction"

[obsidian harness]

We aren't actually using directed angles

[upper karma]

i mean you just said turning one direction

why don't we just look at the magnitude

[obsidian harness]

Another way to look at the magnitude is if you imagine yourself walking along the edge of the convex polygon

And following all the lines

[upper karma]

are these the exterior angles?

[obsidian harness]

So the sum of all the exterior angles is how much you are turning

[obsidian harness]

nah

[upper karma]

isn't this the interior angle?

for that specific reflexive angle

[obsidian harness]

yes the red one is

[upper karma]

red one is what? exterior?

[obsidian harness]

But the other green angles you have drawn apart from 360 - the red one are not the exterior angles

[upper karma]

okay probably is exterior cuz exterior -> outside

[obsidian harness]

the red would be interior

[obsidian harness]

good use of English but that's not how it's defined in maths

[upper karma]

wait huh

[obsidian harness]

exterior angle + interior angle = 180

that's just a definition

oh wait that's only for convex angles

[upper karma]

yeah

okay wait

i have the wrong definition of an exterior angle lol

i have to extend the lines

[obsidian harness]

yes

[upper karma]

so how do i get the exterior angle

of the concave thing

it's not this right?

[obsidian harness]

ah it would be like this

[upper karma]

so they're using directed angles

i've never taken geometry

[obsidian harness]

https://math.stackexchange.com/questions/4758493/how-do-we-define-exterior-angle-of-a-concave-polygon-whose-interior-is-reflex

[upper karma]

well i guess i have to learn this lol

[obsidian harness]

yes it only works out like that for some reason

that you have to use directed angles

honestly you don't come across directed angles much until trigonometry

[upper karma]

i've already done trig

never seen directed angles

[obsidian harness]

So most people (myself included) have never wondered about this

[upper karma]

i think that's only a competition math term

lol

[obsidian harness]

really, you have never seen sin(-30 deg) or something?

[upper karma]

yeah but like u can use the unit circle for those i guess

[obsidian harness]

oh right I see your point

yeah never in this context

[upper karma]

and also now that i think about it

using the periodicity of sine

you technically never deal with negatives at all

also also just sin(-x) = -sin(x)

anyway okay yeah

[obsidian harness]

yeah ofc ofc

[upper karma]

so something like this?

[obsidian harness]

yes!

[upper karma]

wait i think my diagram is a bit wrong

[upper karma]

okay thanks but like wait

how does this relate to the interior angle

seems like a random construction

[obsidian harness]

Ehhh the interior angle minus that green angle

because it's directed

equals 180 degrees

[upper karma]

what's the point of drawing that orange

[obsidian harness]

You're left with the angle on a straight line

[obsidian harness]

so that we can generalise our definition of exterior angles to concave polygons

[upper karma]

hmm okay thanks

but i don't think i'll remember this lol

maybe i should read up more on this

[grave pond]

For what it's worth, I never knew that definition of "exterior angle" either. The concept it describes is meaningful and sometimes useful, but I'd say don't sweat over remembering the name.

[upper karma]

right makes sense

[twin crag]

does this theorem work in reverse

if the ratios are arranged in that way, then the blue line is the angle bisector of the angle highlighted?

sorry i havent done geo in a while

[shut bronze]

yes

the converse of the angle bisector theorem is true

and I believe it's actually easier to prove as well

[winged aurora]

yep, and u can also write it as a/x=b/y

[wide kettle]

How to solve

[twin crag]

i dont need to prove it lol

i just got a random image off of google

[obsidian harness]

Taylor series gives $\frac{2x + (2x)^3 / 3 + 8x + (8x)^3 / 3- 10x - (10x)^3 / 3}{(2x) - (2x)^3 / 6 - 2x - (2x)^3 / 3}$

[somber coyote]

south

[obsidian harness]

So 40 or B

[wide kettle]

wtf

you did it so quickly 😭

@obsidian harness Can you teach me the method

Of Taylor series

[abstract finch]

@wide kettle wait lemme try it

I remember there was also some other method to solve it

If you want to go with the Taylor series then you'll have to remember the expansion of tanx, that will be a lot of memorization

[wide kettle]

I tried changing tan into sin/cos, i did get the answer but there must be a simpler way

Cuz i spent forever on that method

[wide kettle]

I don't even understand Taylor series 😅

[trail tendon]

same, bro. same.

💀

[abstract finch]

It's just an expansion of a couple of functions used while solving limit related questions

You memorize them and while solving the question you just expand it and put in the value of x

Btw another method is LHospital method which only works if you know differentiation and the question is of the form 0/0

[wide kettle]

I only know like sinx becomes cosx, tanx becomes sec²x, like only the basic stuff

[abstract finch]

This question is of 0/0 form meaning you get 0/0 after putting the value of x at first. LHospital rule says that if it's 0/0 then you have to differentiate both the numerator and denominator and then put the value of x and then you'll get the answer

[abstract finch]

Cool, go ahead and differentiate it

[trail tendon]

or infinity/infinity

[abstract finch]

Yeah that too

Lol

[wide kettle]

i thought thats equal to 1

[trail tendon]

do you know like chain rule as well?

[trail tendon]

nope

[abstract finch]

Nahh it's not

[trail tendon]

infinity * 5 = infinity, infinity/infinity = 5?

infinity/infinity is indeterminate

0/0 is also indeterminate

[wide kettle]

not in trigonometric limits

only in algebra limits

[trail tendon]

no like derivative chain rule

derivative of sin(5x) is 5cos(5x)

for example

[abstract finch]

Yeah you'll need chain rule in this question but not the u/v rule, you'll have to differentiate the numerator and denominator separately

[trail tendon]

you mean not quotient?

yes

[dark sparrow]

btw why is all this talk happening in #geometry-and-trigonometry

[wide kettle]

so u'v+v'u?

[subtle cape]

U/v rule will not be used

[trail tendon]

great question

[dark sparrow]

this needs to move to #calculus

[trail tendon]

nope, thats the product rule, you dont need that

[wide kettle]

o

like

dy/du *du/dx and stuff?

[trail tendon]

or not 💀

[trail tendon]

yes

[abstract finch]

It's not (p/q)' it's simply p'/q'

[wide kettle]

But its trigonometry

[dark sparrow]

i mean it's only trigonometry insofar as the limit is of a trigonometric function

[trail tendon]

rip that idea

[trail tendon]

it includes trigonometry but its really calc

my internet ._.

[abstract finch]

It's calculus guys

I was taught this in the class of calculus and not of trigonometry

[subtle cape]

Ya I know calculas is the big concept

And it is also difficult

[abstract finch]

It is but with enough practice we can do its questions very fast and save time to spend on other sections

[wide kettle]

im at high school so i just clicked the pre university and theres no #calculus channel sry

[subtle cape]

Yaa

I am in 12 currently

Science students

I need someone to help me in calculus part 😭

[wide kettle]

Physics?

[trail tendon]

what part

[subtle cape]

From the starting

[subtle cape]

No a part 😭

[subtle cape]

Ya it's easy as compare to maths

[subtle cape]

Can you help me

[trail tendon]

wait what??

💀

[trail tendon]

with what?

[subtle cape]

Calculus

I do know some of the basic of it

[trail tendon]

thats...vague

[trail tendon]

what do you know

[subtle cape]

About the rules

Little limit

[spiral lodge]

Too vague

[subtle cape]

Yaa

So can anyone help me with it

[trail tendon]

do you know what a limit is?

you can go to khan academy to learn btw

[abstract finch]

What topics are in the syllabus? If you know limits then continuity and differentiability will be easy. Differentiation is also easy. Integration is where things get a little tricky but I don't know if y'all have integration in your syllabus.

[subtle cape]

Yaa I have

[trail tendon]

you have what?

[abstract finch]

Khan academy is the best place btw

[subtle cape]

On YouTube

[abstract finch]

Just go there and lock your room for a couple of days and you'll be able to get calculus done.

[subtle cape]

Oo

[abstract finch]

If you procrastinate then you'll never get it done

[subtle cape]

Hmm

[abstract finch]

Also close discord, it's a distraction

[subtle cape]

Ya

But if I have some doubts then

I only use discord for study

Only have 2 servers

On which I am there

[abstract finch]

Just don't come here whenever you have a doubt. That breaks your study flow. Write all your doubts on a piece of paper while studying and come here after you have like 20-30 doubts

[subtle cape]

Ook

Thank you

[abstract finch]

Good. Now go study and make your parents proud

[subtle cape]

Yaa

[abstract finch]

Damn, i became a motivational speaker lol

[trail tendon]

Yaaah

XD

[wide kettle]

Motivate me too pls

😁🙏

[abstract finch]

I can say a lot of things but I'll say study. Build discipline because motivation is temporary. Make a daily schedule and stick to it.
In a year or two you'll be thanking your past self for studying hard. If not then you'll regret wasting your time.

[wide kettle]

o

that hits me hard cuz im a procrastinator

[abstract finch]

Go study man. You're still procrastinating

[wide kettle]

ok ty for advice

[grand sail]

Does someone know why is volume of cone 1/3×(r^2×3.14xh)

[sharp ruin]

You can find that using calculus

Although there must be geometric explanation too
@.dlardo._58635

[grand sail]

Sorry but i do not know calculus can you explain how i can do that by calculus (just a bit do not go depply)

[sharp ruin]

You can think of very thin discs

Infinitely many of which (obviously of decreasing radii) stack over each other to form a cone

You can now have the volume of the disc in terms of it's small thickness 'dx'

Have a function which represents volume in terms of the radii or the height (whichever makes the integration easier for you)

And then integrate it accordingly to get your result

Did you get that?

Obviously you'll need some hold of calculus to understand this completely but I've tried to keep this simple

[grand sail]

Thanks, that is going to be useful but why i cant use formula for triangle and change it with radius. I tried and i get 1/2 not 1/3.

[sharp ruin]

'cause it doesn't work that way

[trail tendon]

could you be more specific?

[sharp ruin]

I mean you'll know in integration that a single power x yields a 1/2

Which give you the area

Further integration of which yields the 1/3

[trail tendon]

even I have no idea what you're talking about, and i know integration 💀

[sharp ruin]

Man I'm just stating the reverse power rule

[trail tendon]

the integral of x is not 1/2

[sharp ruin]

Integral of x gives x^/2 right?

[trail tendon]

1/2 * x^2 yeha

[sharp ruin]

He questioned about the 1/2 factor so I explained that accordingly

That might not be correct (completely) but works until he learns calculus

[trail tendon]

there is no 1/2 factor

[sharp ruin]

He said why can't I use the 1/2 .base .height and transform it

[trail tendon]

because you just got the area of a triangle instead of the volume?

[trail tendon]

if you rotate it i mean you sort of can

[sharp ruin]

But that will require some calculus won't that?

Maybe idk

[trail tendon]

well sure

but it is the sum tho

[sharp ruin]

Sum of triangles=> infinitely many of them => integration

[trail tendon]

like if you have a triangle with base, height, and infinitely small width, and you add an infinite amount of them you get volume

[trail tendon]

yes but people can understand sum of triangles

maybe not integration

i mean like if you just whip out the name "integration" or integral symbol

no idea what going on

but if you say its like the sum

[grand sail]

If is understandable

[trail tendon]

of triangles with infinitely small width

[grand sail]

I know that is not correct i tried experiment

[sharp ruin]

That would just become a Sigma series eventually leading to integration
But yeah for easier explaeination, your way is better

[trail tendon]

you can explain infinite series and not use sigma notation

i mean math isnt that hard, they just put weird symbols and greek letters and make it seem complicated lol

[sharp ruin]

By that you get it for a cylinder

[trail tendon]

in general

[sharp ruin]

Yeah but that's just to make things compact

[trail tendon]

half of a cyllinder

[sharp ruin]

Yeah

Sorry didn't notice that

[upper karma]

[trail tendon]

honestly though

the hardest part is creating math up on ur own honestly

which is like what the challenge problems do

but if they actually teach you methods to solve problems and change the numbers

it aint hard to understand

[upper karma]

Go check channels beyond "early university" thread

[trail tendon]

just dont make mistake

[trail tendon]

i prob wouldn't understand, greek letters/symbols 😂

[upper karma]

yeah but there they can't teach you methods?

[trail tendon]

what do they teach you then?

[upper karma]

I mean actual math doesn't have numbers beyond 10

so what is there to solve and change numbers in

[upper karma]

idk lol

The impression is that school math is easy because you just repeat the same algorithm over and over again

I don't think this generalizes to the math you encounter later cuz u have to actually think instead of just using pattern recognition

and there's no algorithm to learn cuz i don't think they care about doing any problems algorithmically relevant then cuz a CAS can do it for them

[trail tendon]

nah but at a certain point they have to teach you how to come up with math

its only hard if they don't teach you how to do it

[trail tendon]

the human brain is trained to think in a pattern recognition way lol

like even coming up with stuff

is still pattern recognition, you just have to use a bit more creativity

[keen canopy]

whats the area of a tringular prism?

anyone??

[maiden brook]

not true

[trail tendon]

hmmmm

there is math that exists that is hard, but as a whole, i don't think so?

[cunning lion]

we use greek letters because we ran out of latin letters

[trail tendon]

☠️

i mean the new math that is being discovered right now may be "difficult" but it hasn't even been invented yet

the point is tho

that i feel like a lot of math is taught a lot slower than it could be

it's a lot easier than it appears, for the most part

once you understand it, i think at least for me its very common that i find another way that i could have been taught it much quicker

like if they just explained it in a different way, or something, ect.

[maiden brook]

nah I kinda disagree

[trail tendon]

hmm

why?

[maiden brook]

it’s because generally very little application to things is shown so there are very hard problems in much of what is covered that are never given

[trail tendon]

what makes the problems hard?

[maiden brook]

connecting things in a way u have never thought about before, really requiring understanding of how things work

[trail tendon]

so if you take a new perspective, and if you actually understand how it works, then its actually not hard, correct?

is the hard part actually learning how it works?

or learning how to take a new perspective? or learning techniques of which perspective to take maybe?

I mean I guess the farther away the method gets from actually completing the math, the harder it is, to an extent

but if you practice it and are able to do it, it becomes easy again

so was it actually hard ._.

again, this might not apply to some crazy math, but yeh

i could be wrong tho idk

[maiden brook]

what about this famous problem https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_6

where a totally new technique was made because of it

[faint pasture]

IMO is IMO though

You cant just bust up an IMO problem to prove that math is hard

[vernal pilot]

This is a fallacy. Math which requires new math is of course going to be famous. But for every difficult math problem there are a hundred relatively easy problem.

[abstract finch]

You need to have some good level of understanding of number theory to solve such problems

But that's imo, no need to do that

Stick with what's relevant

[north kindle]

I don’t know how to go about this question

This is the chapter about similar triangles so presumably there is something with one somewhere but I am unable to find it

[upper karma]

this can't be solved, probably the question is wrong

[north kindle]

I somehow doubt you’ve proven its unsolvability in 5 minutes

[twin flint]

if the ratios were equal it would be possible

[north kindle]

It should be possible

I haven’t ran into an impossible question in any of these books yet

[twin flint]

there's a first to everything, but let me see if i can solve it

[dusky locust]

there's only two triangles in the diagram, if that helps

(this might sound sarcastic but i'm being dead serious)

[north kindle]

I know

But I can’t see how to use them

[twin flint]

one or more values are wrong

[dusky locust]

have you calculated AE and AD?

[north kindle]

Yes

[dusky locust]

alright cool

compare that with the values of AB and AC

[twin flint]

the ratios arents equal

[dusky locust]

really?

[twin flint]

6/30 != 9/20

[dusky locust]

that is.. not how similar triangles works

[lime dune]

if there aren't any similar triangles

try forcing there to be one

[dusky locust]

pretty valid way of putting it

[north kindle]

I’m trying but I can’t find anything

[lime dune]

oh wait LMAO

ok i see it

nothing extra needs to be drawn in

[north kindle]

Oh wait I see I think

6/20 = 9/30 = 3/10, and they share angle EAD

[lime dune]

yea

[north kindle]

So CAB is similar to DAE

[dusky locust]

yessirrrr

[lime dune]

it's kinda disguised bc the small triangle is "flipped"

[dusky locust]

also yessirrrrr

[lime dune]

took me a second to see it too lol

[north kindle]

So ED is 130/3 I think

[obsidian harness]

6 30
9 20

6/20 = 9/30, interesting

[lime dune]

130/3 sounds right to me

[obsidian harness]

yeah so I guess the thing that makes you realise that is that there isn't any other info in the question

Yep so 13 * 10/3 exactly

[north kindle]

Nice

Thx

[lime dune]

npnp

[north kindle]

Is the 6 referring to WZ or YZ

[faint pasture]

YZ id guess

no way to know for sure

[upper karma]

XY = 4
YZ = 6
all the triangles are similar to each other, so:
4/WY = WY/6
WY² = 24
WY = sqrt(24) = 2 sqrt(3)
2 sqrt(3)/6 = 6/YV
2 YV sqrt(3) = 36
YV sqrt(3) = 18
YV = 18/sqrt(3) = 6 sqrt(3)

[obsidian harness]

btw !nosols

!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.

[north kindle]

I didn’t ask dude

[upper karma]

oh 😅

[gritty topaz]

xy×yz=wy²

i'm pretty sure they would write 6 above WZ if it was meant for it

[gritty topaz]

do you know the law of cosines

a²=b²+c²-2*a*b*cos(x)

[gritty topaz]

???

[north kindle]

No, that has nothing to do with this

[gritty topaz]

what would you do if the triangles weren't similar 🤔

[north kindle]

I don’t know, but they are

[gritty topaz]

trust me that has something to do with this

[north kindle]

So I don’t care

[gritty topaz]

fair enough

[grave pond]

I'd end up at the law of cosines too.

Though I suppose you could avoid mentioning it explicitly by drawing some auxiliary right triangles and reproducing a proof of the law rather than the law itself.

[gritty topaz]

that's basically using the law of cosines, just trying to set the angle to 90

although i think it looks simpler to the eye

[grave pond]

I think the main sticking point here is that "use the law of cosines" can sound like you need to calculate some cosines and/or arccosines -- whereas in this problem you don't ever need to know the numerical value of the angles, since you can shortcut it by cos ∠CBD = -cos ∠CBA.

[gritty topaz]

sure that's where i was getting at

[upper karma]

how do I prove it?

[obsidian harness]

https://math.stackexchange.com/questions/805821/how-can-i-find-the-following-product-tan-20-circ-cdot-tan-40-circ-cdot

I recommend using https://approach0.xyz/search/ to search through questions

https://math.stackexchange.com/questions/2080657/hi-i-have-a-problem-with-a-trigonomic-excercise-about-tan?noredirect=1

[upper karma]

umm.. I did till (co20 - cos 60)sin 80 / (co60 + cos 20)cos 80

How do I continue after this

[obsidian harness]

Well, cos(x + y) - cos(x - y) = -2 sin x sin y

(cos x cos y cancels)

Or equivalently cos(x - y) - cos(x + y) = 2 sin x sin y

And similarly cos(x - y) + cos(x + y) = 2 cos x cos y

ah wait sorry that's reversing your steps

[upper karma]

ah yes

I got more confused

[obsidian harness]

Sorry that answer isn't correct

,w (cos 120 deg + cos 40 deg) - (2 cos 40 deg - 1)

[somber coyote]

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Requested by sooouth.

[obsidian harness]

it's not 0

[upper karma]

The what?!