geometry-and-trigonometry

[sacred canopy]

sorry

[timber kraken]

your fine I should be sorry 😭

[shut thorn]

E C H O L O C A T I O N

Mb vro i fell asleep

😭

[crisp zenith]

How to do trig prove qs

[gray vessel]

Manipulate algebraically until the expression is simplified on both sides

And make sure the equality stands

[graceful ocean]

convert ratios into sin and cos, it works 90% of the time

[unkempt kestrel]

rate my geo notes (triangle parts)

there is a quite stupid mistake in the proof of the first area thingy but the diagram should make the plan clear enough

[real wing]

Bro that book looks cute

[timber kraken]

big fat notebook series

it contains pre-algebra and algebra, geometry, and algebra 2 along with books that explain and practice books

[bold nymph]

Hey everyone I have a acuplacer test for college so I can take my other classes but it’s the only thing holding me back from being able to take my other classes and I was wondering if anyone had any resources and etc on these topics

[sacred canopy]

I have some pdfs on linear equations (affin geometry), linear aplications and geometry

however i dont have them in english, maybe you can translate them , idk how

[bold nymph]

Why language is it in ?

[sacred canopy]

spanish

[bold nymph]

Ripp maybe I can just translate it

But thank you !

[sacred canopy]

its nothing

so you want them?

nevermind

i just remembered a better source

https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

you should be fine with that

[bold nymph]

Thank you sm 🙌

[forest atlas]

😂😂😂

[somber coyote]

Nerdyasianguy

[sly urchin]

Nvm got it

[timber kraken]

so basically idk

[past path]

Just to be sure this is just inscribed angles right?

[obsidian harness]

and then to finish, once you have angle ADC ||just use the alternate segment theorem||

[vagrant sparrow]

in a level further maths vectors, is there a way to like understand it intuitively rather than memorising a method for the like 8 different types of questions they can ask?

[obsidian harness]

from another server:

the point of practice is not to memorize solutions, but the understand them enough so you know when to use which approach/technique
after solving a problem:
check your work
make the most of every problem and really understand each problem
identify keywords or indications on which approaches to use
record patterns you observe repeatedly in a notebook, refer to it when stuck
if you were not able to get the answer in one go
-- identify where and why you went astray
-- ensure you understand why certain steps were taken (eg in probability, a multiplication could be independent events or conditional events. make sure you can identify it when unprompted by the question)
if you were able to get the answer in one go
-- redo problem but instead change what to solve for. eg you are given a, b, and you find c; try problem with b, c and solve for a
if the question appears frequently, each time you do it, aim for shorter time and higher accuracy

[sly urchin]

How can I look up a triangle's center name just from its construction?

[silent plank]

Wdym

[cosmic igloo]

You can identify a triangle center by matching its construction to the known families of classical concurrence points.

[sullen path]

Looks amazing!!

Did you scan the image? all looks like digital but also in paper way, i don't know how to explain it in a fluid way hehe

I say it looks great, everything, the order, info

[river wren]

can anybody help explain binomial expansion/theorem to me

[crude peak]

Hi can someone help me with this please?

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[crude peak]

I think it’s 40 but the more I think about the more I think it’s wrong

[past hearth]

Try to use the concept of cyclic quadrilaterals to solve for y (if the question supplies that the quadrilateral's points are all on the circle)

[crude peak]

Okay I’ll try it thank you

[crude peak]

I’m sorry I’m not getting it 🙁

[past hearth]

If AD is a staright line, then CBD is 40 degrees

try to find other angles and then use the fact that opposite angles in cyclic quadrilaterals are supplementary to find y

[glass slate]

Then y is 35 degrees.

[unkempt kestrel]

yup

[unkempt kestrel]

from my notebook

[sullen path]

How? i didn't understand

[unkempt kestrel]

idk i just scanned the pages out of my notebook with my phone lol

the last page got a lil tilted

[sullen path]

I did reading wrong, my bad, i understand now hehe, very good job btw

[sullen path]

Stills fine

[unkempt kestrel]

fill in what you know whenever you get an angle inside one of the isosceles triangles, then use circle properties to set up an equation

after getting the angles that the arrows are coming from in this, you have enough info to solve

[crude peak]

Thank you everyone:)

[sly urchin]

What's the easiest way to construct the Anti-Steiner point?

[outer forge]

use the fact that the point at infinity perpendicular to the steiner line is the isogonal conjugate of the antisteiner point

which is easy to prove with angle chasing

basically this means you can draw the line perp to steiner through A, reflect it over the bisector of <A, and intersect the new line with (ABC) to find the antisteiner

[sly urchin]

Thank you

[obsidian harness]

waiting for the person who actually studied Oly geo instead of half-assing their way through be like....

[outer forge]

?

[obsidian harness]

it's a compliment, take it

[ivory spruce]

hi

[inland summit]

Hmmm only 140° is known

Oh wait line A is straight until D

Ooh nvm i get it

[sly urchin]

Acute-angled scalene triangle ABC. Incenter I, circumcenter O, orthocenter H, Euler point E, Spieker point S, Nagel point N. Prove OI//ES//HN

[rugged mortar]

A line is drawn in a trapezoid with bases 4 and 8, that is parallel to the bases of the trapezoid and divides the trapezoid into two congruent areas. What is the length of this line?

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[soft river]

I'm looking for a study partner for probabilities any one in interested

[neon prairie]

Please don't spam this.

[soft river]

ok sorry

[neon prairie]

No worries.

[languid steppe]

60?

[exotic yarrow]

Arc ABC is 60 degrees. But x is half of that. [https://www.mathwarehouse.com/geometry/circle/angle-tangent-and-chord.php]

[past path]

yes, but x = 30

[slim plinth]

Idk this I need help

I thought it’s 3/5

[shut thorn]

TThe thing with scaling is that the triangles which you're scaling will always be similar So if you wanna find the factor by which it was shrunken. Then just divide all the sides by each other such as blue AB / orange AB = k

from the graph

[short swift]

does anyone have any suggestions as to how I should memorize all of the area formulas? I learned them a couple years back but should know them now that I am learning Geometry. (2d things like trapezoids, triangles, etc..)

[shut thorn]

blue AB = -6 -(-1) -> -6 + 1

[shut thorn]

do more questions on them?

[short swift]

well yeah, but how many types of formulas are there for area of 2d figures

[shut thorn]

-> -5. And similarly orange AB = -6-(-4) which is -2. So the answer is -2/-5 which is 2/5? Is that correct, do double check because I, not confident myself.

[shut thorn]

I know like five formulas which are always on my fingertips (rectangle, circle, ||gm, rhombus ,triangle)

[short swift]

what is the rhombus and ||gm fomrulas?

[shut thorn]

||gm: area = base × perpendicular height. Perimeter = 2(l+b)

for rhombus

AArea = (1/2) × d1 × d2 where d1 and d2 are the lengths of the diagonals

perimeter = 4b

[short swift]

thank you! I'll learn those ASAP

[shut thorn]

np.

[shut thorn]

oh so you just did a calculation error. Nothing else

[tiny musk]

can someone help me on this i'm new to trig and all i really know is SOHCAHTOA

[shut thorn]

what can you not get?

a specific part or the entire incorrect proof?

[vapid remnant]

Is it just me or this question is just not possible with the given context?

[shut thorn]

isn't this

I'm having deja vu

brb

[shut thorn]

it is possible.

byt in the figure

they marked the wrong angles

it says angle ECD not angle ECB

[shut thorn]

also yeah this question is in my book. Just with different angles.

[crimson nymph]

they didn't mark ECD and ECB is 31

[shut thorn]

No ECD is given 20 in question

read it

at the top

[crimson nymph]

yes

[shut thorn]

ye it's possible.

The diagram makes it seem impossible but of you read the question then it's possible.

[exotic marlin]

Look at angle C and theta carefully and find out what might be the relation between them and what Aubrey has wrote

[vapid remnant]

I trusted both the diagram and the question and still think it's impossible
In the answersheet they assumed that angle CDE= angle BAC

We can easily get CDE

But not BAC unless that assumption is explicitly stated

This is where I'm standing at

[shut thorn]

alright so from here

do you know the theorem "angle in same segments" ?

[vapid remnant]

Uhhh I don't remember the proper name but if you tell me what it does maybe I'll know it

[vapid remnant]

By the name I think it's talking about segments of the same size which there are none here

[shut thorn]

you see how uh

BC is an arc?

[vapid remnant]

BC is the diameter

Both triangles are not congruent you can tell by just looking

[shut thorn]

it is not given the diameter.

there's no center given for the circle so we cannot determine if it's the diameter or not

[vapid remnant]

Yeah but it is definitely not an arc

[shut thorn]

why not?

oh I know your thingy

you like

[vapid remnant]

Alr I see where this is going

[shut thorn]

make a circle around this figure

[vapid remnant]

Dude look at this

This is much neater

[shut thorn]

ye

so from here

do you see that B And C are point on the circumference?

[vapid remnant]

Yes

[shut thorn]

so don't they enclose an arc?

[vapid remnant]

If BC was an arc angles BCD and CBA wouldn't exist

[shut thorn]

segments

I FORGOT THE NAME

AAAAAAAAA

no wait

[vapid remnant]

BC is a segment yes....

But doesn't really take you anywhere from this

[shut thorn]

ye so can you tell me which two angles share the same side?

God damn I'm so bad at expaexplaining

can you tell

which two angles

[vapid remnant]

BCD and CBA

[shut thorn]

share the same segment (BC)

[vapid remnant]

Share BC

[shut thorn]

ye

so

[vapid remnant]

As there "segment"

[shut thorn]

if you find BCD.tha

that would be the answer

[vapid remnant]

Which I have found....

31+20

[shut thorn]

yep

[vapid remnant]

Which is 51

[shut thorn]

no no

[vapid remnant]

BCD+BCA=51

[shut thorn]

The figure isn't correct wait

also jesus I can't explain.

wait brb

[vapid remnant]

I'll come back when you have a proper explanation or even an answer
Ima do other questions

[shut thorn]

this red part here stretching from B to C is called an arc. An arc is commonly denoted straight up by "arc BC". So by saying arc BC. We mean the part of the circle enclosed by those two points.

If we wanna say which angles have the common arc BC then we say that Angle BAC and angle BDC have same arc because they're both made by the same points (B and C) and two different points anywhere on the circumference of the circle.

[sly urchin]

What's the fastest way to show that the incenter of a triangle is the Nagel point of its medial triangle?

[shut thorn]

from this if you calculate angle BDC then you get angle BAC too. Do you know how to do that @vapid remnant ?

tell me if it's a bit unclear

I'll try to clear it ,ore

mire

MORE

[shut thorn]

in books it is denoted by a symbol which I can't type on a keyboard.

let me try to fund it.

find

[slim plinth]

It’s 5/2

Why

[shut thorn]

The scaling factor?

[slim plinth]

The person states that “You’re on the right track…Keep in mind that if the image is larger than the pre-image, the scale factor must be greater than 1.”

[shut thorn]

Oh

Oh you know what mistake i did?

[slim plinth]

What

[shut thorn]

It said that it was scaled up

So the okd triangle (supposedly the big one) is the smaller one

[slim plinth]

I’ll give you the formula

[shut thorn]

The thin still applies

[shut thorn]

Ye there

The old image was the smaller one

I thought it was the bigger one

[slim plinth]

Then “Not quite…Remember that triangle ABC is the pre-image 😊”

[slim plinth]

I did is I counted 3 grids from the ABC to the larger triangle

[shut thorn]

Sonething cinfusing is that both of them are called ABC

Can you atleast say orange or green

😭

[shut thorn]

Oh

[slim plinth]

Orange

To green

[slim plinth]

To separate it they made the green one A’ B’ C’

[shut thorn]

Ye its from -6 - (-4) [side AB if i remember the figure correctky]

[shut thorn]

Hm

Ok so the image was made bigger?

[slim plinth]

Then when I searched these up it’s 2.5 as a scale factor

Or 5/2

[shut thorn]

Tbh the question didnt specify which one was the after image. Or im just dumb as sho

Oh my god

[slim plinth]

Idk

[shut thorn]

Dialation

DIALTUON

[slim plinth]

Yes

[shut thorn]

AAAAA

Ye

Dilation

The image was made bigger

[slim plinth]

It said “center of dilation”

(-6,2) is the point

[shut thorn]

That doesnt matter

You just need to find the scale

Ok wait

[slim plinth]

I assume it’s the distance from that point to A (Orange) and from (-6.2) to A’(green)

It looks like this right now

[shut thorn]

Because the blue image is the after image. One way i could think of putting it like is this.
If we do it normally and find the ratio if one side between the similar trianggles with it being AB/ A'B'. Which is equal is equal to 2/5. But since 2/5 is smaller than 1. If we apply it to the orange one. It would become smaller instead of bigger. But if we apply it to the bigger triangle then it becomes the smaller triangle. So to make the smaller triangle the bigger one. You have to reciprocal the scaling factor. And hence. You get 5/2

[shut thorn]

Center of dialation diesnt matter tbh

[slim plinth]

Okay

Is it telling me the x axis or the y axis

[shut thorn]

Read this and uh tell me if theres something you canr understand

Ill try my best

[slim plinth]

Thanks

[shut thorn]

Its just a problem on similar triangles but because its being dialated instead of shrunken diwn. The scaling factor would be the reciprocal of the normal factor you would get

[slim plinth]

Oh

[shut thorn]

Oh and no offense but you shoukd try and practice on the better counting of unit boxes in a graph because you counted -6 to -4 as -3

[shut thorn]

That big paragraph i wrote should give a good reasoning in why thats the case

Anyways

[slim plinth]

Ima go to algebra little bit cuz im interested and mainly in Algebra

[shut thorn]

Fire

[shut thorn]

I found a place where they used the arc symbol. Yoy mightve seen it in NCERT too

I also have the NCERT but i feel too lazy to open it rn 👍

[rugged mortar]

Can someone help me with this? I'm kinda not sure what to do here, honestly.

I mean, a bot told me something, but it doesn't give much help

[low ocean]

You should make a diagram that would attract more helper

[timber kraken]

I ain't reading allat

ixl thinks their funny

we'll see who's laughing after well...

[shut thorn]

after wha

also vro this is just adding 😭

to get all three sides of the big triangles equal

you add the smaller equal sides

[timber kraken]

yea but who feels like reading all of that

its confusing enough when they don't let me work out the proof myself and its more confusing when they give me some confusing model too

I got it correct anyway its just so tedious it makes me wanna rip all of my blood vessels outta my body

or split an atom on my head so it goes kaboom

[shut thorn]

its lowkey not tedious

[timber kraken]

for you

this is giving me anxiety

[shut thorn]

listen to music them

then

[timber kraken]

what type of music

im listening to hamilton and I break into dance every time I hear something favorable

[shut thorn]

oh welp

god damn my chance of pinging the mods

😭

[timber kraken]

its ok you'll have your chance one day

[shut thorn]

ehh something like

idk

field of memories?

[timber kraken]

k

[shut thorn]

theres a song called grieving and it might be the best thing ive heard in a while 😭 🙏

[timber kraken]

i like music thats very loud

like not loud because you can adjust the volume by loud I mean lots of energy

[shut thorn]

flamewall might be the best type of that music genere ive heard then

[timber kraken]

W

I might check this out

[shut thorn]

do

[timber kraken]

btw mods if you're reading this, this is totally related to geometry

uhm

$sqrt(67+67)$

[shut thorn]

we can dash too :)

[somber coyote]

varq

[shut thorn]

😭

[timber kraken]

thats not sqrt

[timber kraken]

wym by dash?

[shut thorn]

geoemtry dash.

https://tenor.com/view/geometry-dash-flamewall-gd-gif-5918391188933860766

[timber kraken]

ohhh

so im just gonna upload a proof im doing to keep the channel on topic

how’s my work look

[shut thorn]

oh my god i got so confused because usually our book doesnt give soem angles equal by just drawing them onto the figure 😭

[timber kraken]

oh lol

this isn't a traditional workbook, I bought it off of amazon.

[shut thorn]

jij

huh

[timber kraken]

like its not a workbook you'd use in a classroom

something you'd use at home.

[shut thorn]

i just use my book tbh. It already has a gazillion questions

[winged oyster]

is it (cos, sin) or (sin, cos)

i think its cos, sin but idk

[cunning lion]

if you need to remember you can do SOH CAH TOA in the first quadrant

[cerulean lintel]

$2root(43-23)$

[somber coyote]

emiya

[cerulean lintel]

How do you do square root

With the bot

What do i say

$5x+4y=7(f(x)-6+-8.5$

[somber coyote]

emiya

[jaunty jasper]

can someone help with this problem?

[safe agate]

Similarity of triangles, let's say ADE and ABK

[jaunty jasper]

I have this but how do I get DE?

[safe agate]

EF is given and AK is median so DE...

[jaunty jasper]

wait the median carries through all the triangles?? 😭

[safe agate]

Yes

[shut thorn]

assuming DF GJ and BC are parrallel. Then try to use similarity of triangles between AD and AB for triangle ADF and ABC

[verbal aspen]

I have an algebra quiz about trig tomorrow. And it's asking for me to find the trigonometric ratio. What is a trigonometric ratio????

[dark gust]

Trigonometric ratios are the relationships established in the trigonometric functions

so in this case you've got tan A

remember tan(x) = opposite/adjacent

[obsidian harness]

SOH CAH TOA

[obsidian harness]

okay so you're on right-angled trig, and you have these new functions sin, cos, tan

you want to know how sin, cos, tan are even defined

you literally take the right two sides in the given right-angled triangle, and divide them for the ratio

so for example, if you wanted to know sin A

SOH CAH TOA
sin = opp/hyp

the side opposite to angle A is a
the side which is the hypotenuse (longest side) is c

so sin A = a/c

[verbal aspen]

HOw can I easily find the opposite and adjacent in a triangle? I'm so confused with that

[obsidian harness]

can you see that side a is opposite to angle A?

[verbal aspen]

Yes

[obsidian harness]

similarly, the hypotenuse is always opposite the right angle

[verbal aspen]

Ohhhhh

Tangent is opp/adjacent right? So a is opposite and b is adjacent? Is tan A a/b?

[obsidian harness]

and then there's only 3 sides in a triangle

the hypotenuse is c
the opposite side to angle A is a
the adjacent side to angle A is ....

[obsidian harness]

yep!

[verbal aspen]

I just need to remember this image then I'll think I'll be fine

Tysm for the help

[obsidian harness]

no worries! if you just try to remember SOHCAHTOA

sock-a-toe

you'll be good

[verbal aspen]

Ok! I'll make sure to keep that in mind

[woven remnant]

hi

I suck at algebra

But somehow so far I’m understanding trig

🙏🏼

I hope I don’t encounter a lot of algebra topics that I don’t remember during trig

I still am trying to study algebra a

[obsidian harness]

are you comfortable with expanding brackets (the distributive law) and solving equations in general?

you don't need to know much of the algebra 2 stuff to be understanding trig

[woven remnant]

I think so

Like 5 ( 3+2) =25? Is that what u mean by expanding the brackets

[woven remnant]

Good I took algebra 2 freshman year and even then I was a horrible student

I wish I would’ve never skipped a year in math but oh well

[obsidian harness]

yes, so you should absolutely be able to expand polynomial brackets

like (x^2 - 2x + 4)(2x - 3)

it's the same principle as 5 * (3 + 2) = 5 * 3 + 5 * 2, just with more steps and the variable x

[woven remnant]

I’d probably get this wrong

[obsidian harness]

the thing about maths is that it's like a brick wall

topics build on each other so you can't have shaky foundations to progress further

[woven remnant]

😔

2x^3 -4x - 12 is what I’d get the answer but I think that’s wrong

[woven remnant]

I need something to measure my progress so illl try to use khan academy

since there’s like a mastery check and stuff

[obsidian harness]

yes

I also have this great self-help resource

https://adventurousandrewmaths.blogspot.com/2022/05/aaslaahlaihl-foundations-of-algebra.html

[woven remnant]

I wanna take ap calc next year

bc I need to take calc in college and don’t want to

So I need to practice

[obsidian harness]

then work hard so that you're at the right level for calc AB

[woven remnant]

yeah I wanna take ab but everybody’s telling me to take bc but I only need ab credit

[obsidian harness]

calc AB is around half precalculus so you'll be fine

[obsidian harness]

don't listen to those people

[woven remnant]

Dc pre calc was frying me

[obsidian harness]

there's humility in knowing your own level and knowing what's realistic for you

[woven remnant]

yeah

I also don’t want to work harder for no reason

I understand I’d learn but I don’t want to be frustrated to take math

I’m currently taking dual credit pre calculus

And I feel I’m doing way better now that I have a teacher

With trigonometry

[obsidian harness]

ah okay

I just think it's important not to rush through all this new material, cause maths isn't a race

[woven remnant]

Yeah I don’t really have time to rush anyways so that’s fine for me

I just wanna pass trig

And be ready for ap calc ab and ap stats by August 2026

I’m taking 5 college courses so I don’t usually have spare time

[obsidian harness]

wait you're already taking 5 college courses?

[woven remnant]

yeah

So when I graduate. High school I will have an associate degree

But I know I won’t use all the courses for my major but it’s wtv

[obsidian harness]

damn

how you finding that, stress-wise?

[woven remnant]

It kills me

especially last semester I had no math teacher

[obsidian harness]

why not drop a few courses at college then?

[woven remnant]

I can’t but I mean it’s fine

College is just stressful

[obsidian harness]

you can't?

[woven remnant]

No other way around it

I made it easier for me this semester tho by ensuring I have instructors

[obsidian harness]

no, there's academic advisors you can speak to and counsellors

[woven remnant]

At my High School

[obsidian harness]

I'm just curious why you're taking a heavy workload despite the mental and physical costs

[woven remnant]

So I can graduate college sooner

I wanna start making money

[obsidian harness]

why do you want to graduate college sooner?

[woven remnant]

so I can get a job

[obsidian harness]

I don't think that's necessarily an advantage

[woven remnant]

😭

[obsidian harness]

you can still work when you're at high school

[woven remnant]

I also see it’s a good opportunity so I took advantage

It’s money saved

$6000

[obsidian harness]

you scared of relying on your parents financially or something?

[woven remnant]

and at least 1 year saved from getting my bachelors degree

oh yeah most definitely

I have a job but it doesn’t pay much, so they would help me alot probably unless I get a scholarship

and I’m not rich so id qualify for full rides

not full rides free tuition

[obsidian harness]

okay let me try and unpack all of that info

[obsidian harness]

what sort of career are you trying to get into after uni?

[woven remnant]

accounting

I used to want to be a teacher but this year showed me to not choose that path

[obsidian harness]

nice, well I don't know much about accounting though

[woven remnant]

I don’t know much either, I really just know that it’s a good degree and that they work with the expenditures a business makes

[obsidian harness]

I'm not sure what the college requirements to graduate with an accounting degree are

[woven remnant]

And it’s able to have multiple career choices

[obsidian harness]

most likely you'd just want to aim for AP calc BC credit and that's it for maths

[woven remnant]

A college im considering needs me to take CALC 1

Not calculus 2

So I can take ap calculus ab

[obsidian harness]

cause a lot of STEM degrees don't exactly lead to a specific job either

[woven remnant]

I hope it’s much easier

[obsidian harness]

if you take chem for example, you could find a connection somewhere and end up being a lab technician yes

equally likely however is that you'll end up in a job which doesn't use any of your chem knowledge

[woven remnant]

Yeah

[woven remnant]

Technically they need me to take business calculus

[obsidian harness]

but yeah no matter what they tell you, you're better off having a college degree than going without

[woven remnant]

But I’d rather just try to satisfy the credit during my senior year since this year I will finish a lot of my credit

[woven remnant]

Yeah I really just want to get a degree cuz I feel like education is a privilege

even if I don’t get a job from a college degree I won’t be homeless or anything

[obsidian harness]

from that perspective it'd be great if you could take AP calc this year, but it's no biggie if you end up taking it freshman year college either

[woven remnant]

I don’t wanna take it in college tho

[obsidian harness]

exactly it insulates you from the worst shocks of the economy

[woven remnant]

professors are much more stricter then a high school teacher

[obsidian harness]

of course to be 100% insulated you'll need to have been born in a rich family or something

[queen venture]

$$ -4a^2k^2 - 8a^4 = k^4 - 2ahk^2 $$

[somber coyote]

fysch

[queen venture]

how to find vertex of this parabola

i got this locus

[obsidian harness]

do you mind if we continue our discussion here haha

I know this isn't the right place

[obsidian harness]

is that so from your experience?

[woven remnant]

well, maybe not much more but

In my case

[obsidian harness]

I mean, they certainly aren't going to be holding your hand, but from my experience if you just show up to office hours with some questions prepared, they're happy to help

[woven remnant]

high school side of my school can just turn in work whenever they feel like it

and it seems like they get babysat

[obsidian harness]

oh okay so from that angle

[woven remnant]

As opposed to college

[obsidian harness]

yeah absolutely it's sad to see HS turning into that

[woven remnant]

yeah I noticed this with my cc class, he had horrible RMP reviews but I think it was just student error

[obsidian harness]

yeah it can be just salty students who haven't gotten out of that HS mindset

[woven remnant]

Yeah I didn’t even really know how different college was from hs

it’s like a light switch

And nobody really told me about it

[obsidian harness]

mhm

[woven remnant]

Still scared for university 😭

[queen venture]

yall know the cmd to complete the square ?

[obsidian harness]

I think you're pretty prepared if you're already aiming for an associate's degree already

the other 2 years of uni aren't necessarily different

[woven remnant]

just major coursework I believe for me

[obsidian harness]

just a bit more in depth into the degree content you're studying

yeah and maybe a bit more coursework

especially like a final year project or something

[woven remnant]

wow

Did u take ap stats

[obsidian harness]

no I didn't, I'm not American

[woven remnant]

So you didn’t take ap calc ab either??

[obsidian harness]

I did take calc BC though

yeah I mean people cover similar stuff around the world

[woven remnant]

Also I know the ap calc teacher already

He helped me with my college algebra

[obsidian harness]

oh that's good

[woven remnant]

And he teaches ap pre calculus and ap calc ab and calc bc

He said he wants me to take his class

[obsidian harness]

I think it's good if you can take it in a high school environment

[woven remnant]

but it’s up to me which class he wanted me to start with ap pre calculus during my 2nd semester of junior year so right now but it would’ve been too much on my plate

[obsidian harness]

yeah you're doing trig this semester so yeah

[woven remnant]

Yeah I really wanna take advantage of my college credit I can earn so hopefully I can not spend too much time at my college

[woven remnant]

He also said ap pre calculus is doing trig this part of the year so I’m not too worried about that for calculus ab

[obsidian harness]

I think that's understandable, and yeah doing community college first then transferring in is a sound strategy

[woven remnant]

Yeah and it also helps my class rank

I’m 23 out of 1098 students in my grade level

[obsidian harness]

oh your school still does class ranks in your entire grade

[woven remnant]

Since I think ap and college classes are weighed the same

[woven remnant]

No just my grade

[obsidian harness]

that's a bit quaint, but I get the idea

[woven remnant]

I made an error

[obsidian harness]

oh right okay

I feel though AP doesn't give many subjects justice

you're just trying to speedrun and cram the content

[woven remnant]

yeah I’ve taken two ap classes and only liked one

Because of the teacher

I don’t really like the structure and the way it works

[obsidian harness]

conversely with college you're actually learning in that sort of more independent environment

[woven remnant]

and how u can pass the class with a 97 or something and not get college credit

[obsidian harness]

many students don't either

[woven remnant]

and it’s still in the high school format, and at times it’s even more difficult then dual enrollment which does give college credit

[obsidian harness]

that's insane that AP is harder than the actual dual enrollment course

[woven remnant]

I think this is just some courses

Like I think Ap calculus ab and Bc are easier then their college counterparts

[obsidian harness]

oh that absolutely makes sense

they definitely cover more at college

[woven remnant]

but apush and ap physics would probably be more difficult then their college counterpart

[verbal aspen]

Can anyone tell me if I'm right
sin A = a/c
cos A = b/c
tan A = a/b

[exotic yarrow]

good

[verbal aspen]

Yayyy thank youy

[exotic yarrow]

[verbal aspen]

So is the cosecant c/a, the secant c/b and the cotangent b/a?

[obsidian harness]

yep!

there you go, that's all 6 trig ratios

[verbal aspen]

That means my notes are right? Yay

[obsidian harness]

oh, a is opposite to angle A

opposite and adjacent only make sense relative to a specific angle

so sin A = ...
cos A = ....

and so on

[verbal aspen]

Oh ok

But how about finding the sec, csc and cot in a graphing calculator?

[obsidian harness]

oh there are buttons for those

if you don't see any buttons, remember the reciprocal rule

so csc(20) = 1/sin(20)

sec(50) = 1/cos(50)

[obsidian harness]

they're hidden under some menu though, so not buttons exactly

sin, cos, tan are the buttons

[verbal aspen]

Let me try finding

[verbal aspen]

How do I change the graphic calculator from radian to degree?

Oh wait I did it I just pressed enter

[versed iron]

is this correct for the quadrantal angle?
Quadrant I = 0-90 degree angles
Quadrant II = 91-180 degree angles
Quadrant III = 181- 270 degree angles
Quadrant IV = 271-360 degree angles

[barren hound]

yes

[crimson nymph]

lower bounds are not correct

90.001 lies in 2nd quadrant

[queen venture]

use [] or ()

like [0,pi/2]

then (pi/2, pi]

[slim plinth]

What’s a?

[sly urchin]

The information given is insufficient

[slim plinth]

I thought we can make a slope intercept form

[silent plank]

there's nothing fixing the position of a

a could be anywhere between 0 and 1

[slim plinth]

I can assume that it’s around 1 or 2

[spiral lodge]

No, you can't

At least if this is the only information given

[karmic hazel]

It doesn't even solve by the distance formula

[slim plinth]

We can try

But then we have 2 variables

[timber kraken]

are the squares equal to one unit

[slim plinth]

Assuming it is

Cuz x=3

[timber kraken]

slope of the line is one than

[slim plinth]

Yes

[timber kraken]

Just need the y intercept

[slim plinth]

Use slope formula

Or point slope

[timber kraken]

$3-a/3-0$

[somber coyote]

varq

[timber kraken]

WITH A FRACTION YOU DUMB

wait lemme cook

$frac{3-a}{3-0}$

[somber coyote]

varq

[timber kraken]

$/frac{3-a}{3-0}$

[somber coyote]

varq

[slim plinth]

The b on “y=mx+b” as to be negative

[timber kraken]

yea

oh smart

[tardy field]

$3 - \frac{a}{3} = 0$

[somber coyote]

Altanis

[tardy field]

So close 🥀 🥀

[queen venture]

no

[crimson nymph]

the question is incomplete

they say the two area are equal

[slim plinth]

Hard

[slim plinth]

Scoob

Look

[silent plank]

ok thats some new info that we didn't have access to

[slim plinth]

Yep

[slim plinth]

What’s that stated

[silent plank]

?

[slim plinth]

That said “Unit five square slant …”

[silent plank]

not sure what you're asking atm

[slim plinth]

Something

[silent plank]

can you speak in full sentences/phrases

single word responses mean nothing to me

[crimson nymph]

ok'

[slim plinth]

That’s what the chalk board said

Rest of them I don’t understand

I only read “unit five square slant ? the area of the square to half”

[silent plank]

the image has 5 unit squares
the slanted line splits those into two parts
the unshaded part and the shaded yellow part
these areas are equal

[slim plinth]

Yes

Now I wonder if we can use these info to solve x

Cuz (a,0) a is in the x-axis

[crimson nymph]

do you see how

[slim plinth]

Yes

It’s 5 shaded boxes

And some are half

[crimson nymph]

uh ok

then

[slim plinth]

I thought of making it like a puzzle moving pieces around

[crimson nymph]

gl diagram no to scale

ahh moment

[slim plinth]

I assume that’s not 1-1 scale

Like grids

[crimson nymph]

anyways let's not assume shaded portions are x% shaded

[crimson nymph]

do you see the right triangle

what is its height

[slim plinth]

3

u^2

[crimson nymph]

right and base?

[slim plinth]

We don’t know cuz

[crimson nymph]

you can write in terms of a

[slim plinth]

It’s (a) units

[crimson nymph]

m.. no

[slim plinth]

2.5 u

[crimson nymph]

red is 3

[slim plinth]

2 and an half

[crimson nymph]

yellow is red - blue

what is blue?

[slim plinth]

a

[crimson nymph]

right

so red - blue =

[slim plinth]

2-a

Ohhh

[crimson nymph]

uh red is 3

[slim plinth]

Oh so it’s 3-a*

[crimson nymph]

yes

so your base is 3-a, now what is area

[slim plinth]

We can use A=bh/2

So it’s (3)(3-a)/2

[crimson nymph]

yes

[slim plinth]

Distribute 3 to 3-a to get 9-3a/2

[crimson nymph]

subtract small square

you will have area of shaded

[slim plinth]

Is it the black one

That’s 1 u

[crimson nymph]

yes

[slim plinth]

It’s 7-3a/2

[crimson nymph]

yes

now area of shaded is half

how many squares are there

[slim plinth]

5/2

[crimson nymph]

yes

you have a

[slim plinth]

5/2=7-3a/2

[slim plinth]

I got a=2/3

[tropic monolith]

anyone have tips to remember trig exact values

[silent plank]

look up trig exact values hand

[summer pebble]

I am trying to find the volume enclosed by a moving boundary with velocity 'v', anyone can help me find some resources

[spiral lodge]

#geometry-and-trigonometry message

[topaz goblet]

Provide some more information lol

[west kettle]

If you have to remember a lot you could watch

https://youtube.com/shorts/77vlRweCtGE?si=_D2LRnfZFwOIwmVt

after understanding how the six trig ratios worked together through triangle congruence, I was able to prove so many formulas in just a few minutes

[summer pebble]

Ok.

[summer pebble]

Given a time-dependent region Omega(t) whose boundary moves with velocity v, how can the enclosed volume be computed, or how can its rate of change be expressed in terms of v?

I came across this equation. Does this work?

[topaz goblet]

Shouldn't the integral be replaced with a surface integral

Try posting this in the calculus section not here

[summer pebble]

Okay

[upper karma]

oki im here dude

uhmm

[low ocean]

<@&268886789983436800>

[sly urchin]

Can someone please send me the inversion sol of IMO2011 P6?

[outer forge]

why do you want to invert on that problem

there are a few inversion sols on aops but i dont think it makes the problem any easier than just doing the normal sol, which i think is also more instructive since a lot of problems of that kind are solved with that trick.

https://artofproblemsolving.com/community/c6h418983p8689639

this one looks decent

[sly urchin]

Thank you

[sly urchin]

I saw the complex sols quite a few times but when looking at at some sol pages, it mentions there being an inversion sol but i couldn't find it

So curiosity got me

[outer forge]

oh ok

[sly urchin]

I have this so far:
EF is A-antiparallel wrt ∆ABC (E ∈ AC, F ∈ AB) so the symmedian of ∆AEF is the median of ∆ABC. Similarly, it follows that AX, BY, CZ are concurrent at centroid G of ∆ABC

I need AG ⊥ YZ but I don't know how

[outer forge]

angles give you ||Y, D, foot of C onto AM (lets call it U) are collinear|| and same on the other side (lets call the other point V).

now if you add ||CU int BG = I and BV int CG = J|| notice that ||CI//BJ perp AG||

try to prove YZ is also parallel to those 2 with proj

[sly urchin]

By "same on the other side" you mean Z,V,D?

[outer forge]

aka ||try to show (GY,IB) = (GZ,CJ)||

[outer forge]

yes

i hope i didnt mess up the letters cause i drew a lot of diagrams with other triangles as reference and i didnt name a single point lmao

[sly urchin]

What was your intuition on this problem btw? Why did you think about those intersections?

My first thoughts on this were antiparallels through X,Y,Z or the Lemoine L of ∆ABC and its pedal triangle

But those didn't get me very far

[outer forge]

first i tried to see what was the best way to define Y and Z, and what seemed good enough to me was defining Y as ||the point on BG such that <YDB = <GAC||

this makes you try to use the equal angles to find a cyclic quad, which you do by introducing the point U

then i took DUV as reference triangle and tried defining the whole problem in terms of that because it looked a bit simpler, idk if this actually helps, you dont really need it

[sly urchin]

Oh this is because ∆ABC ∪ G ~ ∆ DBF ∪ {G_DBF}

Didn't pop out to me

[sly urchin]

Yeah but I really enjoy seeing different people's approaches, it's fascinating to see how you do it

[sly urchin]

I don't think I know how, do you proj through D and then V?

Wait Idk what i'm talking about

[outer forge]

look at each one individually, for example if you look at (GY, IB) then what are the best points to project this from

like, ideally you'd want a point K such that most of KG, KY, KI, KB are already in your diagram

if not all

here ||you have a point for which all those lines are already there for you|| and that point is ||suspence....||||U||

||if you keep in mind you want to project this one from U and similarly the other one from V, its not hard to see where you want to project them exactly||

[sly urchin]

There isn't UB in the diagram though, right?

Wait did i follow your points correctly

[outer forge]

ok mb then just stick to the "almost all of them"

[sly urchin]

No worries

I'll put this here to make it easier

Do you keep B as it is and denote M as the midpoint of BC

U(GYIB)= U(MDBC)

And Maclaurin?

[outer forge]

wdym

do the same thing on the other side

and youre done

[sly urchin]

I just got enlightened

[outer forge]

anyway since you technically also had other good enough points to project through, ill tell you why i think U is the better one: if youre trying to solve this and you got to this point, you realise that points Y and Z kinda suck because Y = BG int DU is not so easy to control, so what you want to do with proj is projecting through points that allow you to undo the constructions: projecting from D or U would allow you to delete Y and line UD, so that you delete the bad points. So U and D should be your first tries, and here U seems a lot better

usually this is a very useful idea in proj, you're always trying to undo the constructions and deleting the bad points

and proj is one of the best tools to do this

[sly urchin]

Thanks again, you've helped me a lot, both with the ideas and even more so with the explanations

[novel wing]

WoW there are Olympians here 🥹

Olympians are so oppressed lmao

Whenever I mention math olympiads some dude says « Olympiad math is not real math, it’s memorisation »

[low ocean]

It's both memorization and solving skill

It has a lot of memorization

[upper karma]

you never actively try to memorize stuff

you actually dont even need to know that much stuff

[low ocean]

Eh but we do that by solving problems, it's a fun way tho

[upper karma]

compared to uni math etc

[low ocean]

Hmmm true

Tho if you memorize a whole bunch of small and unique problems that would be an advantage

[novel wing]

Well not if the problem is entirely new

[long geyser]

Is it ok to assume lengths like the green color even when it's not given? And if so, then can you assume that for more lengths like in the blue or do you have to prove that? Maybe you can only assume for one segment's length and make it something neat like 1 for example and work around that?

[shut thorn]

shouldnt AE = AD mean that angle AED = angle ADE

[long geyser]

I suppose, but that's not my question, I'm just giving an example of a random triangle, I don't mean this triangle but just any shape in geometry or math at all.

I hope that clarifies my question

[shut thorn]

hm

the thing with assuming one thing is that if youre going to use it then your entire thing would be based on it

[long geyser]

I'm just assuming lengths, not proving anything, so AE being 3 is no problem here. Except for the fact that it might not match reality, so I'm wondering what are the limits on saying what has what quantity?

[long geyser]

True, however can you also like scale things by a k factor to match the original maybe?

And so whatever shape you are trying to interpret is similar to the real one if that makes sense?

[shut thorn]

ye then it would be the same as the original. We are basically asumming that a length x = 2k and not just 2

[long geyser]

Can I say this is 10 and this is 3 without being given that? I know it's a really dumb question, but I'm not sure.

[shut thorn]

no but you could say that they are 10k where k is any variable

10k is a variable Whereas 10 is a constant

[long geyser]

I think the reason I can't do that is because it's not necessarily similar, because if we only choose one side then we can calculate the other sides and stuff with proportions to the real triangle.

[shut thorn]

one reason is that you are fixing the value of a variable where it isnt even given

[long geyser]

Can you say this?

[shut thorn]

i think so

[long geyser]

Tysm for the help!

[shut thorn]

np

yep heres an example.

no wait

shi lemme make the figure first

fuck

i wanna sleep

night

[long geyser]

Sorry I didn't catch your messages, goodnight!!! ❤️ and thanks again!!

[lime crown]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[unkempt kestrel]

does anyone know about this bot

can i ask about its origins rq

[shut thorn]

alright my brain is working somewhat now so here's one question I could think of wich uses this idea. "Prove that the ratio of the perimeter of similar triangles is the same as the ratio of their sides" Hint: ||take p/q = some factor k. Where p and q are sides of the similar triangles||

if you want me to then I can also provide the solution

[split osprey]

What program is this

[split osprey]

What program is this

Anyways
I have a argument with someone that we don't need to add 90 to the angle arccos(7/9). First Image part d is the question, height is 7. The second image is what he made and the third image is for the purple angle arccos(7/9)

[nimble belfry]

How is it 7/9

[split osprey]

7 = adjacent
9 = hypotenuse

[nimble belfry]

O wait wrong question

Yeah u do add 90

[timber kraken]

properties of isocelese triangle!

[long geyser]

Sure! This is interesting

[long geyser]

Khan Academy

[shut thorn]

Solution?

[long geyser]

Yea

[shut thorn]

Wait

[long geyser]

Okay

Also happy early birthday

[shut thorn]

Thankd

[long geyser]

Ofc

[shut thorn]

Forget the bullet like marks on the page

Theyre just there because.. y e s

[long geyser]

Np

[shut thorn]

I think ive also eaten a part of one page 😭🙏

Like its just missing

Really small part but yes

[long geyser]

XD

[tribal compass]

Where did you prove what was asked ?

I guess the proof is not complete if i am correct

[shut thorn]

Gdo damnit i forgot to write what was to be proved

[tribal compass]

xd

[shut thorn]

😭

[tribal compass]

Np

[shut thorn]

Tis

No wait i did write it

[long geyser]

Cool proof, tysm!

[shut thorn]

Np

[tribal compass]

@shut thorn i guess you should show k equal to p/q in the end

Cause that expression is the ratio of perimeters

[shut thorn]

Its from similarity of triangles

Idk why i wrote AB/PQ as p/q in the first lune

Then just refused to use it

[tribal compass]

And we should explicitly state the equality we were asked

[tribal compass]

Yeah, there was even no need for the variable k

You couldve used p/q in place of that every time

[shut thorn]

The k was somewhat needed to show that AB is just PQ but scaled down by a factor of k

The question was ti show that the perimeter is in the same ratio as the sides of two similar triangles

And you can only do that if you show that the sudes of the similar triangles are equal IF muktipkied by some factor k

Multipkied

MULTIPLIED

[tribal compass]

No we could have also wrote that the AB is the scaled vsrsion of PQ where the scaling factor is p/q

[shut thorn]

Took it as k for simplicity

Ye that works too

P/q

Being the scaling factor

[tribal compass]

Ye thats fine

[tribal compass]

Yeah if the side(s) are equal, the triangles are congruent

[shut thorn]

if the sides are equal then the k factor making the sides of the similar triangles is equal to 1

[tribal compass]

Can anyone check my attempt on this question ?

[crimson nymph]

looks good

[tribal compass]

Ok thanks

[upper karma]

@sly urchin why did you delete it :(

[sly urchin]

Oh i didn't think there was anyone interested

And thought to open a help channel

Sorry

I'll resend it then

∆ABC, incircle (I), orthocenter H. Tangent of (I) parallel to AB cuts BC, CA at L, K. Tangent of (I) parallel to AC cuts BC, AB at G, J. KJ cuts HB, HC at M, N. (I) is tangent to BC at D.

Prove that HD, LM, GN are concurrent

[upper karma]

i solved it

[sly urchin]

Man you're insane

I was trying Menelaus but didn't get far

[upper karma]

its actually really easy to get rid of H, M, N with proj

[sly urchin]

So you try to prove the cross ratio is the same when projected through T?

I need to get better at proj, it's my weakest part of geo

[upper karma]

my idea was to show that GN int HD is the same point as LM int HD, so you can try to show (H, JK int HD; NG int HD, D) = (H, JK int HD; ML int HD, D)

[exotic marlin]

Yeah, seems fine to me

[upper karma]

projecting the first one from N and the second one from M, you get rid of H, M, N and you're left with normal incenter config points

you still need a bit of work but you took away the annoying points so now its doable

[sly urchin]

I'll try that now, thank you

You're so good

[sly urchin]

I'm guessing you're projecting the points onto BC, thus i should name JK ∩ BC?

[upper karma]

yes

[sly urchin]

S = HD ∩ KJ, R = KJ ∩ BC, T = HD ∩ ML, T' = HD ∩NG

[exotic marlin]

It can be proven using Desargues' Theorem

[sly urchin]

I try to prove M(HSTD)=N(HST'D) or equivalently M(BRLD)=N(CRGD)

[sly urchin]

Please elaborate

[exotic marlin]

We can consider the triangle HMN and GDL (although GDL is a degenerate triangle) but the theorem still works. So if we let the correspondence of vertices like this:
H-D , N-G, M-L
Then we want to prove that the corresponding vertices of the two triangles of interest are concurrent

Acc to Desargues' Theorem, if their the intersection points of the corresponding sides of those two triangles are collinear, it just be true that the lines mentioned are concurrent

[sly urchin]

Oh I didn't know Desargues applied to lines (degenerate triangles)

Can you fact check that?

[upper karma]

desargues doesnt work with degenerate triangles

[exotic marlin]

Sure ?

[upper karma]

yes

[sly urchin]

If it worked like that, i could prove every 2 triangles are similar lol

[upper karma]

yeah

[sly urchin]

So back to this

I tried simply using the definition of cross ratios

And the fact that DB/DC=AB/AC

[shut thorn]

i thought i knew the question too well

anyways im gone

because this will go out of the topic

[upper karma]

this is not true

[sly urchin]

Oh shoot

[upper karma]

it would be if D was AI int BC

[sly urchin]

AD is not the bisector

[upper karma]

yeah

[sly urchin]

My bad

[exotic marlin]

Oh yes sorry, double checked it. Big palmface

[sly urchin]

So what I need is (BRLD)=(CRGD)

I notice that these 2 have 2 points in common

So my thought is just to apply the definition

Oh wait

Menelaus?

Wait no

Those points are collinear

What am I talking about lol

[sly urchin]

This is equivalent to BL/BD . CD/CG . RG/RL = 1

Or I guess a better way to write it would be LB/LR . DC/DB . GR/GC = 1

[sly urchin]

@upper karma am i on the correct track?

Or should I have continued projecting

[upper karma]

you can reduce the calculations a little bit if you project once more, but you can also continue on that path

[sly urchin]

Hm... "calculations"

[upper karma]

lol i mean its still just ratios

not heavy calculations

btw idk if you noticed already but you know <GIC and <BIL

[low ocean]

<@&268886789983436800>

also in pre-calc

[sly urchin]

No I didn't notice that, in fact i'm trying to prove it

IG and IL don't seem very nice

[upper karma]

theyre actually not bad at all once you notice it

[sly urchin]

Wait is I the midpoint of JK?

[upper karma]

yes

[sly urchin]

I sense a mixtilinear incircle lurking around

There might be more right angles at play than i thought

I "see" that BIL=CIG=90° but haven't proven it

This should be obvious though

I need ∠GIL=90° - ∠A/2

GDI and GIC are similar...

I can't believe i still can't prove GIC=90°

<@&268886789983436800>

[sharp monolith]

Funny that they post that the same second I show up

[sly urchin]

Oh yeah

That's exactly what this is

[upper karma]

oh btw i found a way to do another projection that makes this a lot faster, if you want it its ||project from I to (BIC)||

[sly urchin]

Uhmm... Project which cross ratio?

[upper karma]

i mean the ones at the start

[upper karma]

those

[sly urchin]

Also why (BIC)? The only nice property about that circle which I know is OB=OI=OC (O being the center of (BIC))

IR is a tangent of (O) but what to do with the points G,D,L?

They don't seem to project nicely onto (O)

[upper karma]

for G you can use the 90 degree angle you have

and same for L

[sly urchin]

There might be properties of cross ratios that I am unaware of

[upper karma]

so like ||G will go to a point G' such that G'B perp BC||

[upper karma]

nah i mean you have <GIC = 90

[sly urchin]

Yes

[upper karma]

then you can just do the same thing you were doing before

like, use the definition of cross ratio

you know that the same definition with the ratios of segments also works on circles

and everythings so much better from here

as youll see

[sly urchin]

I don't get this part

Do you project through ∞_AH

[upper karma]

no its through I

G goes to IG int (BIC)

[sly urchin]

G'C is the diameter of (O)

So G'BC=90°

Okay i see it

And i guess i call ID ∩ (O) = D'

I'll show you my idea after this I guess

I(BRLD)=I(BRL'D')
I(CRGD)=I(CRG'D')

Doesn't seem very nice

[sly urchin]

Sorry carbonara, I just don't understand

Btw in while trying, I noticed a few things

The A-mixtilinear incircle wrt ∆ABC is tangent to AB, AC at J,K

Similarly to the B and C -Mixtilinear incircle and BC

[novel wing]

Damn noticed ur good in geo

[sly urchin]

This observation makes R a point with many propterties

[novel wing]

Try AGO 3rd edition P6

[sly urchin]

Hell nah i'm not

[sly urchin]

I'm sucking at this

[novel wing]

P3

[novel wing]

Dude ur d7+

Ur awesome

[sly urchin]

Nuh uh, i'm struggling with this easy problem even after many many hints

[sly urchin]

So we try to prove (BLRD)=(CGRD) instead to make use of R and D more

Which is equiv to:
RB/RL : DB/DL = RC/RG : DC/DG

<=> RB/RC . RD/RL = DB/BL . DG/DC

If I let XI ∩ BC = P then (BCRP)=-1

Or RB/RC=PB/PC=XB/XC=JB/KC

I feel like a sol should "click" anytime soon

[sly urchin]

We need L'B/L'R : D'B/D'R = G'C/G'R : D'C/D'R

But L'B=G'C so we only need RG'/RL'=D'B/D'C

I need sleep, it's about 2AM here

Maybe i'll wake up with a simple sol

[sly urchin]

This Mixtilinear approach seems promising btw

[timber kraken]

geometry 🥲

Lowk you can see the reflection AHH

ok that’s better

[upper karma]

why are you using R here

once you project (BRLD) through I onto (BIC), R goes to I so youre looking at (BIL'D')

i see why you were struggling lmao, idk if you got confused but the cross ratio is that one

[upper karma]

with this fix you only need IG'/IL' = D'B/D'C and you should be able to see why this is true

[upper karma]

anyway i'll also say how i would have proceeded from here since you should still be able to know how to continue even when your path is suboptimal (edit: it turns out its not that bad).
||using ratio lemma,
LB/LR = sinLIB/sinLIR * IB/IR = IB/IRsin(90+ICB) = IB/IRsinDIC
similarly GR/GC = IRsinDIB/IC||
so ||LB/LR * GR/GC = IBsinDIB / ICsinDIC = DB/DC||
which is not that bad.

Actually, if youre ok with seeing a bit of trig like that then heres how you can do it by just using the definition of cross ratio with sines:
||(BR,LD) = (sinBIL/sinLIR) / (sinDIB/sinRID) = (1/sinDIC) / (sinDIB/sinRID) = sinRID/(sinDIC*sinDIB)
this is symmetric so its equal to (CR,GD)||

(so ig sorry for making it seem harder than what it was)

tbh at first i didnt see this so maybe youll think this is even better than what i suggested, but hopefully youll agree projecting onto (BIC) also makes the final equality of ratio much easier to see, maybe if you want to avoid trig at all costs (although its not like i bashed with trig or anything, those sols are still elegant imo)

the way i initially did it was ||rotate (IB,IR;IL,ID) by 90 degrees to get (IL,IA;IB,I(inf_BC)) and project this onto BC so that if Z = AI int BC the cross ratio is equal to LB/BZ, so you want to prove LB/BZ = GC/CZ||. Then you can ||rewrite as LB/GC = BZ/CZ, but BL = BI^2/BD and CG = CI^2/CD, so you only need (BI/CI)^2 = BD/DC * BZ/CZ, but this is known to hold whenever ID and IZ are isogonal in <BIC, which they are so were done.||

[pallid totem]

What in the world is that

[sly urchin]

Uh oh, yeah I'll blame my lack of sleep

[sly urchin]

Oh nice, last part is Steiner's theorem or sth right?

[upper karma]

oh yes i didnt know the name but yes its that

[tall dove]

I got a nice question - If a circle is inside or outside another circle, prove that the points on them with least distance between, and the centres of both circles lie on same line.(Easy but classic)

[neat kite]

is this the right place to ask about spatial geometry

i lowkey stuck

[crimson nymph]

any type that is pre university

[neat kite]

idk if it's the part of the curriculum at you all university thou, i might just drop the problem

Consider a quadrilateral prism ABCD.A'B'C'D'. Let M and N be the centers of the lateral faces AA'C'C and BB'D'D, respectively. Prove that MN is parallel to the plane ABCD.

[low ocean]

Okay

As soon as I read the question I know where you from