help-27

[white elk]

yes

[bronze void]

so 4^3 - 2^3

[white elk]

yup

[bronze void]

i got new role??

[white elk]

oh wait

[solid perch]

active, yes.

and also not 4^4.

[white elk]

the outside one isn't ^4 💀

[bronze void]

is it still just ^3

[white elk]

ye ye

it'll always be ^3

[bronze void]

yea bc its 3d not 4d...

[solid perch]

(hint: 4x4x4. 3 4s here, so...)

[sharp adder]

i sure hope not.. (sorry for jumping oop)

[bronze void]

that would be a problem wouldn't it lol

[white elk]

well tbf it's not too hard to calculate

just hard to visualize

[white elk]

lil comments are okay just not spoiling problems yk

[sharp adder]

if icant visualize it my brain explodes.. :D

[white elk]

D:

[bronze void]

me with vectors this term ong, hate them so much!

[white elk]

vectors keep blowing your mind

but in good ways too!

[fluid stag]

if it isn't R^3 or under just... don't visualize em

[white elk]

there's at least 100 different ways to think about them

[white elk]

no no but I just look at 3D and I just say, "... 4"

and all works out trust

[fluid stag]

I mean to OP

[white elk]

oh sorry

[fluid stag]

otherwise it's, uh, a trip to the asylum when you tell people you can see 10D stuff

[bronze void]

...

[fluid stag]

ok sorry for the random stuff, continue

[white elk]

yeah I am letting myself get distracted again

sorry

[bronze void]

uh so just to recap, for our 4*4*4 we get 4^3 - 2^3 ?

[white elk]

yes

and in general if you come across different kinds of things where you want to think about the boundary (this is more general advice), it's usually quite helpful to think of them as differences of volumes

but I digress

[bronze void]

i.e. thinking about the total # of cubies is hard, but thinking about the ones we dont see is easier

just reframing the question

[white elk]

exactly

[bronze void]

how do u understand it so quickly

[white elk]

it takes a lot of time

[slim rose]

What

[white elk]

but when you do put in the time it feels like magic :D

[bronze void]

mhm

[white elk]

I promise it's very doable for the vast majority of people though

[bronze void]

ty again for the help

ill try to give the next one a proper go

[white elk]

np!

[slim rose]

i dont undstand what this 10d stuff are

and why peoples think this is crezy

[white elk]

well idk about you but I don't see 10 completely independent dimensions

[bronze void]

this cant be that bad

hopefully i get the correct numbers

[white elk]

this is... bizarre

yeah give it a whirl

[slim rose]

who even gived you this question

[bronze void]

its a lot of random math for a competition lol

[slim rose]

you mean dream place

[sharp adder]

i feel like this is sudoku with extra steps /j

[white elk]

yeah lol

who came up with these I gotta know

[bronze void]

some random professer

huh ?

not the same thing

[slim rose]

so what is this

[white elk]

can we not post completely off topic things in a help channel?

[sharp adder]

-# like how do u even come up with conditon b pallandrome what??

[slim rose]

i just want to undstand question become it is weird

[bronze void]

oh for that, thats like 121 for example yea? like racecar -> racecar when we flip the order

or is that something else

[white elk]

it could be 121

but it could also be some other multiple of 11

oh wait you mean the palindrome mb lol

[white elk]

yeah this is right

[bronze void]

kk

[white elk]

reading comprehension kinda short circuited there

[bronze void]

so for a. i got 216

[sharp adder]

okay ithink a good place to start is because both right and left collums are plaindromes we know then that the top right and left squares must be the same as bottom leftand botto mrightsquares

[bronze void]

also doing this without a calculator is pain

[white elk]

can you say what the factorization of 216 is

[bronze void]

6^3

[white elk]

yeah so that's a third power

[sharp adder]

wont work..

[white elk]

it wants something to the power of 6

not base 6

[bronze void]

Oh a sixth power

i understand now

not a base of 6

[white elk]

yeye

[bronze void]

so if the bottom row is a multiple of 11, it should be in the form (a, b, a) right

[white elk]

well multiples of 11 don't have to be of that form

for example 121 is a known multiple of 11 right

but if you add 11 that's still a multiple of 11

[bronze void]

sadge

[white elk]

but that isn't a palindrome

[white elk]

yeah unfortunately you can't skip directly to the answer

[bronze void]

this one req. too much imo

[white elk]

actually I think you were on the right track initially

you wanna start with the top row

[bronze void]

yea, instead of 6 say do 2?

or something like that

2 is too small

[white elk]

yeah it's 64 right

[bronze void]

mhm

[white elk]

so it's not 3 digits, but yeah keep going

[bronze void]

possibly 3^6

works

time to do this by hand rq

[white elk]

yeah it might take a bit lol

[bronze void]

yea

hate the fact calcs are not permitted during this event so i gotta learn all the niche techniques to do addition, multiplication and division 😭

[white elk]

yeah this part is really brute force which is dumb

I don't even think there are good fast ways to do this other than knowing some powers of 3 beforehand

[bronze void]

im doing this rn: 3^6 = (3^3)^2

27*27 = (20+7)(20+7)

[white elk]

oh that's cool

I wouldn't have thought of that

[bronze void]

ive had to do this for other tasks in the past

only reason i know to do this method

[white elk]

well that's also kinda how all people learn things so

you're in good company

[bronze void]

ok so we get 729

[white elk]

yes

[bronze void]

2^6 is too small, and i hypothesize that 4^6 is too large

so 3^6 must be correct

[white elk]

yeah

I would have some powers of two remembered because they come up often

but yeah 4^6 is 4096 which is way bigger

[bronze void]

yea bc thats just (2^2)^6

so 2^12

[white elk]

yup

[white elk]

anyways the top row becomes pretty helpful for the remaining parts

[bronze void]

so if the left and right columns form a palindrome, we get the bottom left to be 7 and bottom right to be 9

[sharp adder]

okay i think ||11 * 69|| might work for bottom row

[sharp adder]

yup!

[sharp adder]

(also spoiler cause that is the asnwer to the question u jsut asked

[bronze void]

well i had a look and idk how to get it so time to figure it out

[white elk]

I got ||11*67 which I won't even comment on the funny coincidence that is||

[bronze void]

bc whats the point in the answer if you dont know how to get it

[white elk]

yes

[sharp adder]

but that only gives us ||737|| and we need a start of 7 and end with 9

that is funny though lol

[white elk]

||oh wait did my arithmetic wrong oops||

[bronze void]

so for bottom row, 11*n = 7k9 where k is just some diget 1-9

[white elk]

yeah

[sharp adder]

|| turns out it was th e other funny number lol...||

[white elk]

it would help if you knew some divisibility by 11 rules

[bronze void]

i dont no

[white elk]

but I guess you can brute force it a little

[bronze void]

i guess 11^j to get close to 7k9

[white elk]

okay let me know if you wanna go through that

[sharp adder]

okay a small tip i can give is that just calculate at what number do u multiply by eleven to get a value higher than 700 (this is pretty simple to do without calc)

and then u know that each time u multiple by one more

u will add +1 to the last

do that until it ends with a 9

[white elk]

ah yeah that's a good idea

[sharp adder]

||(doing 11 x 70 will obbvoiusly get you 770 and since u need the last value to be 9 if u go down one u get the asnwer)||

[sharp adder]

thisi s probably the best way to answer ti but i dont wanna spoil it^

[white elk]

if you know divisibility by 11 I think it's simpler but it's about the same yeah

[bronze void]

hmm

idk if this process is correct:

[sharp adder]

||well i would argue t hat its faster since 770 - 11 is bassicly the ownly arthemtic u need to do but ia gree it would help altough i am not a sucker for rules tbh i have bad memory||

[rugged gorge]

Nice, new way of thinking about it.

[white elk]

hmm okay

[bronze void]

do i now just add 11?

739?

[white elk]

yes

exactly

[sharp adder]

yes add 11 until the number ends with 9

[white elk]

or actually wait the number is different I should've checked again

[bronze void]

so 759

[white elk]

but do this

[sharp adder]

YUP!

[white elk]

yes

perfect

[sharp adder]

you got it well done!!

[bronze void]

my method worked!

[white elk]

eyyy

[sharp adder]

yup!

[sharp adder]

although i would reccomend u look a tthis might be usefull in the future ;3

[white elk]

also there's a nice forbidden jutsu for divisibility by 11 that's similar to divisibility by 3

but you do an alternating sum on the digits

[sharp adder]

ooo tell me?

[bronze void]

thats like, too useful

its like a cheat code 😭

[white elk]

you start from the one's place and add and subtract: 9 - 5 + 7

[white elk]

this is probably better general technique though

[bronze void]

so final part now, how do I know what the side numbers should be? all im given is that they are palindromes

[sharp adder]

wait i never knew this this isactually really usefull thanks-

[white elk]

you're actually done

[sharp adder]

i dont think that matters given that the q only asked for bottom..

[bronze void]

i am?

[white elk]

you only needed the bottom row

[sharp adder]

well it doenst matter u can put any number really

[bronze void]

Oh it does

[white elk]

there's some nice reasons for it but you need to know some number theory

[sharp adder]

:o time to go wiki diving hehe!!

[glossy dew]

is there not more than one final answer

[bronze void]

this a number theory question ?

[sharp adder]

i belive not 6^2 is too smal and 6^4 is too big

[glossy dew]

what

[white elk]

:D https://brilliant.org/wiki/proof-of-divisibility-rules/

[glossy dew]

125, 216, 512 and 729 are 3 digit cubes

[bronze void]

oh guys

759 isnt the answer

[sharp adder]

the question wants 6 digit cube- right?

[bronze void]

its 175

[sharp adder]

ah nvm..

[white elk]

hmm

[white elk]

wait what

[bronze void]

oh wait nvm

[sharp adder]

the top row has to form a 6th power no??

[bronze void]

i looked at the wrong answer

...

[white elk]

oh okay

almost had a heart attack

but it's fine

[sharp adder]

wait so is it correct or not iam doubting myself 😭

[glossy dew]

never mind

[sharp adder]

same-

[bronze void]

125 is answer for this question

[white elk]

oh I didn't even do it yet dang

[bronze void]

i forgot which one i was on

so i gave answer of 5 for 4

[white elk]

for question 4?

[bronze void]

yea

[white elk]

it was 759 I believe

[bronze void]

wait i may be confusing lemme restate: Question 4 ans is 759, Q5 is 175

[white elk]

well I haven't done 5

[sharp adder]

okay this is pretty simple if i udnerstood it right

[bronze void]

so there is at least 7*33 pages

[sharp adder]

what did u try doing so far kb?

[bronze void]

oh nvm thats way too many

[sharp adder]

why not?

[bronze void]

well i know the answer is 175, and i got 231 doing this approach

[sharp adder]

(if i understood the q correctly every page only has 1 integer correct
so page 1(1) and 2 is (2) and 3 (3) until 9 i assume?

[sharp adder]

oh true..

[white elk]

is that what it was?

[sharp adder]

what does it mean be consectvie integers..

[white elk]

I think it just said consecutive

like 10, 11 12

[bronze void]

1, 2, 3, ,,, 10, 11, 12, ,,, ... i think no?

[white elk]

like one comes after the other

[white elk]

how did you get 7*33?

[bronze void]

digit 7 occurs 33 times

[sharp adder]

-# nvm i think 7*33 we both misunderstood the question

OH i get the question..

(i think) ((nvm i dont))

[white elk]

well yes but that's a symbol and you want the number of pages

I guess it's kinda the same if each page was given by a letter and multiplying B * 33

[sharp adder]

wait it begins at one and counts up-

i belive

[white elk]

yeah

[sharp adder]

so i tihkn it just asks

[white elk]

just ye ol book

[sharp adder]

how high do u have to count to encounter the number 7 33 times

pretty cool question!

[sharp adder]

^^

[bronze void]

from 1-10 we get 7 1 time

[white elk]

yeah I would go by milestones

[bronze void]

from 11-20 we get 7 1 time

so total 2

from 21-30 we get 7 1 time

total 3

and so on

[white elk]

until?

[bronze void]

until from x->y we get 7 a total of 33 times

[sharp adder]

huh wait thats gonna be way higher than 175 wha-

[white elk]

no it's right

[bronze void]

from 1-69 we get 7, 6 times

[white elk]

lol

irony

but yes

[bronze void]

from 70-79 we get 7, 3 times ?

[white elk]

well try writing the numbers out for a second

[bronze void]

70, 77

oh wait

[white elk]

all of them

[bronze void]

nvm

[white elk]

yeah

that's where the bulk of 7s come from

[bronze void]

70, 71, ... 77, 79, 79

so 10 sevens

[white elk]

well there are 10 numbers

and each of them has a 7

[sharp adder]

(actual yes uare on the rihgt path u got this connue)

[white elk]

but there's another straggler on the 77

[sharp adder]

its 11 sevens then

[white elk]

yes

[bronze void]

from 70-79 we have 9 numbers no?

oh wait nvm

im dumb

yes its 11 total 7's

[white elk]

yeah I also hate off by ones 💀

[sharp adder]

70 71 72 73 74 75 76 77 78 79 (double 7 at 77)

[white elk]

okay so now we gotta go to the next mile marker

[sharp adder]

lets go to 100

[white elk]

(and the 6 from before)

[sharp adder]

from 80 to 100 u have how many sevens?

[bronze void]

2

[white elk]

mhm

so from 1-100 how many have we used up

[sharp adder]

okay now from 100 to 169 how muc do u get

[bronze void]

6 more

[sharp adder]

its gonna be 6 + 11 + 2 + (the other 6)

[sharp adder]

okay now try to do 170 and count up

[bronze void]

hmm

at 175 i get 31

[sharp adder]

nope-

[sharp adder]

u forget the 6 from 100 to 169

[bronze void]

no? from 1-169 we get 25

[sharp adder]

wai huh-

wait wha am i missing

[white elk]

yeah I think there was some bad bookkeeping along the way

[sharp adder]

wait wait 1-69 has sevens sevens lol

[bronze void]

it does

[sharp adder]

so it should be

7 + 11 + 2 + 7

[bronze void]

at 169, 27

which is our missing 2

[sharp adder]

Yup!

well done!

(sorry for my many confusion n this question )

[white elk]

oh wait

yeah

[bronze void]

half the challenge is decrypting these

[white elk]

lol

[sharp adder]

yup-

[bronze void]

oh...

idk how to do this one

like at all

[white elk]

I hate off by ones

[white elk]

oh okay if need be I can try to give a walk through

oh it isn't too bad

[sharp adder]

aproach ti the other way its raally easy-

[white elk]

think about it for a second first and if you're having trouble I can explain

[bronze void]

how many dont?

[sharp adder]

no no power instead of square

[bronze void]

i.e. the lowest number when squared is greater than 10000

[white elk]

mmm it's pretty hard to count non-sqares

[sharp adder]

if uwant a hint ican give u one

[white elk]

general powers are harder

[white elk]

this is good information

[bronze void]

yea, and subtract 10 from that

9*

bc we dont want anything below 100

[white elk]

yeah there you go

another off by one almost getting in the way again

[white elk]

yes

[sharp adder]

okay it should help to say that.. you double then number.. every time-

[bronze void]

ah so if 10^2 = 100 and 100^2 = 10, 000

[white elk]

those are powers of 2

[bronze void]

then its just 100-10 = 90

[white elk]

not squares

[sharp adder]

oh wait yeah oops ( i am stupid)

[glossy dew]

+1 to that

[white elk]

well by getting rid of 10 you got rid of 10^2 too

[bronze void]

OH asi asdhio asdiohasd h io asd

[white elk]

lol

[sharp adder]

-# i think iam gonna take abreak for a bitgoodluck to you all!

[bronze void]

^ ^

[white elk]

we just talked about the off by one

💀

it's fine

[bronze void]

so 91

[white elk]

yes

[bronze void]

lowk thats so much easier

than i anticipated

[white elk]

yeah sometimes it looks hard but it just takes a second to realize it's easy

[bronze void]

i wanna see how far i can get with this

[white elk]

ooh I like these yeah

fun geo

[bronze void]

is it not just 3*pi cm^2?

[white elk]

not the circles

[bronze void]

oh wait nvm

yea

[white elk]

like in between the circles

[bronze void]

the "free" area

i got this

easy (i hope)

[white elk]

I'll be back in like 5 minutes btw

[sharp adder]

this is funn but my brain is fried rn so iam jsut gonna watch usolve it

[bronze void]

kk

[white elk]

(I'm back whenever you're ready)

[bronze void]

mhm ive made pretty good progress

[white elk]

oh sick

[bronze void]

area = A_triangle - 3pi

and A_triangle = 2h

but im thinking about how to find my height

[white elk]

sorry where is the 3pi from? from all 3 circles?

[bronze void]

Area of circle is pi r^2 and we got 3 circles of radius 1

so 3pi(1)^2

so 3pi

[white elk]

can you draw a picture of the triangle that encloses the three circles

[bronze void]

"rough" sketch

ignore the 2

[white elk]

okay so when you're subtracting the area of the circles from the big triangle

you also get little bits of area in the corners

so you're not just counting the region you want to

[bronze void]

is it not the whatever the total area of the triangle is - area of my circles?

that gives all the unshaded regions

[white elk]

yes but the question only asks for the part that is surrounded by all the coins

but you're also counting some of the outside, such as the corners of the triangle

[bronze void]

oh

so its like uhh:

[white elk]

yes

[glossy dew]

join the three centers

[white elk]

||don't give answers man, he's trying to figure it out||

[glossy dew]

what?

[bronze void]

idek what you mean by join the three centres so its ok

[white elk]

oh okay lol

[glossy dew]

the circles have centers

[white elk]

no please don't elaborate lol

[glossy dew]

okay

[bronze void]

idk how thats relevant

im so dumb

[white elk]

no no take your time

this is way easier if you've seen it before

[bronze void]

i know the "answer/approach" has been given already but i dont understand it, could I get a hint to continue bc im lost

ive never calculated an area like this before

[sharp adder]

you got this kb!

[white elk]

yeah sure I'm not sure I would know if it was my first time

[white elk]

so in geometry you want to leverage on what you know

and you know what the radius is

so now you should make constructions based on that

[bronze void]

so at the intersection where they meet, the line from the centres will be 2

[white elk]

based on the radius rather than on the outside of the circle that is

[white elk]

yes exactly

and what shape does it make when you connect all the circle centers

[bronze void]

a triangle

[glossy dew]

bro..

[white elk]

wow learning takes time who woulda thunk it

[sharp adder]

(please only post ur question in one channel)

[white elk]

wrong channel

[glossy dew]

i literally said the same thing with less info

[white elk]

yes, more specifically equilateral

[bronze void]

but that was without like context...

[white elk]

and yet he didn't understand

so

[glossy dew]

just

go on

[bronze void]

im sorry for not understanding

[white elk]

no it's fine lol

I was saying that because it takes context to know the purpose of a statement

anyways

look at what smaller regions make up that equilateral triangle

[bronze void]

?

like the three circle things and the centre shape thing

[white elk]

yeah

okay first how would you find the area of the triangle

we can handle the circle parts later

[bronze void]

1/2 bh

[white elk]

mhm

[bronze void]

Oh

provided its 90deg

but this isnt

so we can split it down the middle to make it 90deg

[white elk]

it is not 90º no

[white elk]

yes

[bronze void]

oh so its our 1, 2, sqrt3 right triangle

thats a funny result

[white elk]

yeah lol

[bronze void]

so its area is just sqrt3

for the whole triangle

[white elk]

yes

okay now we have to find the circle areas

have you seen those kinds of shapes before?

[bronze void]

like the uhh segments?

is that the word for it

arcs

[white elk]

sectors for area

yeah

[bronze void]

yes that one

[white elk]

do you remember a formula?

[bronze void]

uhh l*theta?

[white elk]

if not we can go through it

[bronze void]

or is that arc length

[white elk]

that's for the length of the arc

yeah

[bronze void]

hmm i do not recall any formulas no

[white elk]

okay that's fine

well it should depend on the angle in the sector (call it theta for now)

[sharp adder]

-# btw u technicaly can derive that formula if uever forget it

[white elk]

what portion of the circle does that sector make up?

in terms of theta

[bronze void]

pi r^2 * theta/2pi

[white elk]

yes

[bronze void]

?

yes??

[white elk]

yup

that's the area

[bronze void]

i guessed that

lol

[white elk]

nice

[bronze void]

so simplify and we get theta/2

r^2

r^2 theta/2

[white elk]

mhm and r is one so you technically don't need to worry about it

what's theta?

[bronze void]

uhh

pi/6

or pi/3

[white elk]

why π/6?

[bronze void]

bc our triangle

thing

[white elk]

yeah this is right

[bronze void]

the segments (for the bottom two) are at the pi/6 side

and the top one is at pi/3

oh but that would be 2 * pi/3

since we had to bisect it

[white elk]

oh you can get rid of that height line if you still have it

we only needed it to find the area of the triangle

it's easier to handle the sector without it

[bronze void]

so we get A_circles = pi/12 + pi/3

i believe

[white elk]

sorry where is π/12 from?

[glossy dew]

can you see we need to remove three pi/3 sectors from the equilateral triangle in the middle

[bronze void]

uhh 2 lots of pi/3 angle, and stuff

oh yea its equilateral

[white elk]

I think your diagram's a little messed up

[white elk]

this is better

[bronze void]

oh so the bottom two aren't pi/6 then

[glossy dew]

its very symmetric

[white elk]

yeah they're all π/3 radians

[bronze void]

then how did we get the thingy

lemme make a rough sketch

oh nvm

i just remembered by triangle wrong

[white elk]

all good

[bronze void]

so its pi/2 (the area)

[white elk]

yeah

[sharp adder]

||also just realized this can be done with bassilcly no math lol||

[bronze void]

so its just sqrt3 - pi/2

[white elk]

so now it's the area of the big triangle minus those

yup

[bronze void]

is our funky shape

[white elk]

wait wdym

[bronze void]

the next few are super boring, so let's just do final one and ill head out

[white elk]

okay

I was about to say I'm getting pretty tired

[white elk]

ah this is a classic

[sharp adder]

||well really only need to know the area of traingle
since each circle here is 1/6 of a whole cricle we know that in total
they take up 3/6 which is bassicly 1/2 of a circle so we only really need math to calculate the area the circle
since everything else is pretty intutive and an area of traingle is just half of taht of sqaure so ig the only rule
u really need to do is the area of circle rule no?||

[sharp adder]

bassicly whaqt i mean by no bassicly no math is bassicly not alot of rules / only 1 rule bassicly -

[sharp adder]

-# although i could be wrong here idk

[thorn cypress]

if you close this channel and start a new one for this question then this question will be pinned so people coming in to help would know what the question is (becuase it might get buried after a bunch of discussion)

[white elk]

nah it's alright I've been with them a while

[sharp adder]

-# does this makes sense nano or am i compeltety lost?

[white elk]

yeah I guess if you compound everything it does become nicer

but it's essentially the same amount of work

[fallen lava]

good morning

[white elk]

ig you just don't need the sector area explicitly

[sharp adder]

it jsut become (area of traingle) - pie/2

[sharp adder]

tru tru

[white elk]

ik bro I just pulled an all nighter

[bronze void]

but that insight only comes about after figuring out the question in whole no? unless your like insane at math

[fallen lava]

well i didnt

[white elk]

oof

[sharp adder]

this is how i thought of the question the first time i am pretty bad at rules so i dont really memorize rules alot and just rely on uhh i think that makes snese- probably not the most helpfull thing dont be like me

[white elk]

literally leonhard euler

[white elk]

well I think it's easier when you see the diagram

[sharp adder]

yeah this ^^

[white elk]

thank you blurple

[bronze void]

how can a function be neither even nor odd...

[white elk]

anyways you should work on the problem

[bronze void]

like that doesnt make sense

[glossy dew]

a random function

its not like a number

[fallen lava]

honestly im thinking decimal points cuz i remember learning odd/even means if modulus 2 = 0

[glossy dew]

x^2 is even and x^3 is odd for example

[white elk]

it's not symmetric about the x axis at all

no nice relations

[fallen lava]

sin(x!!!!!!)

[bronze void]

oh even odd like that

that makes sense

ive been wokring with uhh one-to-one and inverse functions n stuff recently so thats probs why i got it confused

[white elk]

mm yeah even is kinda the opposite of one-to-one

this problem's pretty tricky by the way

so let me know if you want help

[frozen aurora]

not really

[glossy dew]

try using f(x) = E(x) + O(x)

[bronze void]

it sounds easy but like idk how to like "prove" it

[glossy dew]

and the other two properties given to you to form another equation

[frozen aurora]

let f(x) = E(x) + O(x). what is f(-x) then?

[glossy dew]

ok better

[white elk]

so the idea is that no matter what f I get I can always write an E and O in terms of it

[bronze void]

E(x) - O(x)

[frozen aurora]

great! now solve for E in terms of f and O in terms of f

[white elk]

oh actually this is a more generative way to do it

[bronze void]

E(x) = f(-x) +O(x)

[frozen aurora]

what is f(x) + f(-x) equal to

[bronze void]

2E(x)

[frozen aurora]

yup

so E(x) = ?

[bronze void]

f(-x) + O(x)

[frozen aurora]

no

[bronze void]

?

oh

[white elk]

let's go back to the begining for a second

[bronze void]

nvm i got it

[white elk]

oh great

what was it

[bronze void]

E(x) = 1/2 (f(-x) + O(x))

i think right?

[frozen aurora]

yup!

nice! do you know how to do O(x)?

[white elk]

well I think you made a typo

[frozen aurora]

i'll let you try and think about it

[frozen aurora]

oh yeah lol

[bronze void]

huh

what typo

[frozen aurora]

sorry it's supposed to be 1/2 (f(-x) + f(x))

you want to solve for E in terms of just f

[bronze void]

oh yea mb

i was lookin at wrong one

[frozen aurora]

ok cool

then try figuring out a similar strategy for finding O(x)

[bronze void]

mhm okk gimme a min

can i not just use same system but solve for O(x) instead?

or is there more to it

[glossy dew]

yes

[white elk]

yup (not more to it, same system)

[bronze void]

mhm kk

so instead of adding, we subtract and get O(x) = 1/2 (f(x) - f(-x))

[white elk]

mhm

and now show that f = e + o and you're done

[frozen aurora]

correct

so given f you just found two functions E and O s.t. f = E + O

so you are done

[white elk]

ig that's all then?

[bronze void]

ayy f(x) ≡ f(x)

thats a good result

[white elk]

yeah it's quite neat

comes up when doing integrals a lot actually

[bronze void]

so regardless of what transformation we take for f, we just get f

[white elk]

yup

[bronze void]

whats funny about this???

did i use symbols incorrectly

anyway... Tysm for the help everyone! Much appreciated

[white elk]

yeah ofc!

[frozen aurora]

i mean f(x) = f(x) all the time lol it's just funny when written like that

but all the symbols are correct

[bronze void]

at least i didnt get 1=0

(which may has happened before)...

[white elk]

we do like when things are true

tautology when pitology walks in

[bronze void]

you know is bad when they start saying words that sound made-up 😭

[white elk]

well one of them is 💀

[bronze void]

so overall, i should work on framing the questions in a way that makes them easier to digest and work on basic number theory for divisibility and multiplication stuff

[white elk]

yeah math is all about taking a hard looking problem and chopping it into smaller digestible ones using information you have

[white elk]

don't worry too much about the number theory though

I think it's only useful if you're going to learn the whole shebang of modular arithmetic

which is cool but based on the competition you're doing, I'm not sure it's necessary

[bronze void]

we do need to know mod arithmetic for this as well

😭

[white elk]

oh. okay then I would learn that

💀

[bronze void]

loll

again, tysm for the help

.close

[devout snow]

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[devout snow]

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[proven anchor]

a group is cyclic and its order is 8

[proven anchor]

so i am looking for subgroups which are made of 'a^2'

[thorn cypress]

can you post the actual question if you have one?

[devout snow]

@proven anchor Has your question been resolved?

[devout snow]

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[zenith vault]

Let N1,...,Nm be independent random indices in {1,2,3}; construct the random product of Fresnel matrices Mm = Am...A1 and compute the asymptotic limit of (1/m) log ||Mm||.

[zenith vault]

how to do this?

Let N1,...,Nm be independent random indices in {1,2,3}; define Ak = [[1/tk, rk/tk], [rk/tk, 1/tk]], with rk = (Nk - Nk+1)/(Nk + Nk+1) and tk = 2Nk/(Nk + Nk+1).

these are the matrices

[devout snow]

@zenith vault Has your question been resolved?

[devout snow]

@zenith vault Has your question been resolved?

[gloomy aurora]

Hm you could always post in an advanced channel if you find no one responds to your question @zenith vault

I feel this can go to #advanced-algebra

[devout snow]

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[heavy current]

. this question isn’t suited for that channel, I fear

. that question probably does better in a channel like #advanced-probability, judging by its look

[devout snow]

Please don't occupy multiple help channels.

[neon folio]

[sand quarry]

<@&268886789983436800>

[neon folio]

mods can you just ban him

its the 3rd mod ping on the same guy

[devout snow]

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[faint gorge]

like they are not tryign everyday

[fathom verge]

Only senior mods have hammer powers; no need to be like this tho, I handled it the best I could

[neon folio]

ok

[twilit comet]

thanks mat !

[neon folio]

oh lance please

i didnt't know only smods can ban

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[heavy current]

<@&268886789983436800>

[devout snow]

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[thick lotus]

How can I help

[summer summit]

this is a closed channel.

[devout snow]

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[barren dune]

My answer in blue is still right?

[sand dove]

well "costly and time consuming" doesn't really reflect what the answer in red means

The only way to test a firework is... to use it

So sure, it's costly, but because you would have to consume all fireworks

which is like "I'm suspicious of this kind of firework, I'll buy all of them to check my hypothesis"

[devout snow]

@barren dune Has your question been resolved?

[near jolt]

is the red what the grader wrote?

if so, then i believe ur answer is right

[upper schooner]

(it's what the mark scheme probably gives as the preferred answer - I'm trying to see whether there's a note to accept OP's answer)

[near jolt]

maybe they wanted u to add the point that it cannot be reused to support ur "costly" reason. But like cmon, comment in red is kinda pedantic if this doesn't get full mark.

[devout snow]

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[near jolt]

like even if it were not destroyed after use, who in their right mind buys everything to test

[devout snow]

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[vocal hull]

If $b^2p-2b+p=0$ \
show that $b=\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}$.

[grand warren]

what is a

[vocal hull]

Oh sorry it's p not a

[woven radish]

Weinnion

[grand warren]

ah ok, use the formula for quadratic equations i think

[vocal hull]

That wouldn't work.

I guess so

[grand warren]

it works well

[vocal hull]

That is an algebraic question including ratio proportion methods

But i don't know the ratio proportion methods

[grand warren]

idk too 💔

what is p tho?

[vocal hull]

Okh i will solve this with Gemini

[versed glacier]

try simplifying both sides individually

[untold ravine]

bruh

[vocal hull]

P is a variable bruh

[grand warren]

?

[versed glacier]

maybe find the value of p and substitute it into the second equation

[vocal hull]

I KNOW THE ANSWER

[versed glacier]

[untold ravine]

its a proof there is no answer

[green crypt]

i will like to help you

[finite fable]

-1 and 2-p/p

upon further solving

wait

[finite fable]

disregard this bs

idk how i ended up there

lmao

[vocal hull]

Given instruments,
$b^2p-2b+p=0$ \
$b^2p+p=2b$ \
$p(b^2+1)=2b$ \
$p=\frac{2b}{b^2+1}$ \
$\frac{1}{p}=\frac{b^2+1}{2b}$ \ [Invertendo]
$\frac{1+p}{1-p}=\frac{b^2+1+2b}{b^2+1-2b}$ [Componendo-Dividendo] \
$\frac{1+p}{1-p}=\frac{(b+1)^2}{(b-1)^2}$ [It becomes $(a+b)^2 formula because 1 can be written as $1^2$] \
$\frac{\sqrt{1+p}}{\sqrt{1-p}}=\frac{b+1}{b-1}$ \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{b+1+b-1}{b+1-b+1}$ [Componendo-Dividendo] \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{2b}{2}$ \
$\frac\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=b$

Shown with ease

[versed glacier]

just solve for p and substitute it's value into the second equation after simplifying that's it 🫪🥀

[vocal hull]

Still not solved

[versed glacier]

you can clearly see LHS=RHS

[vocal hull]

Given instruments,
$b^2p-2b+p=0$ \
$b^2p+p=2b$ \
$p(b^2+1)=2b$ \
$p=\frac{2b}{b^2+1}$ \
$\frac{1}{p}=\frac{b^2+1}{2b}$ \ [Invertendo]
$\frac{1+p}{1-p}=\frac{b^2+1+2b}{b^2+1-2b}$ [Componendo-Dividendo] \
$\frac{1+p}{1-p}=\frac{(b+1)^2}{(b-1)^2}$ [It becomes $(a+b)^2 formula because 1 can be written as $1^2$] \
$\frac{\sqrt{1+p}}{\sqrt{1-p}}=\frac{b+1}{b-1}$ \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{b+1+b-1}{b+1-b+1}$ [Componendo-Dividendo] \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{2b}{2}$ \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=b$

Shown with ease

[woven radish]

Weinnion

Given instruments,
$b^2p-2b+p=0$ \\
$b^2p+p=2b$ \\
$p(b^2+1)=2b$ \\
$p=\frac{2b}{b^2+1}$ \\
$\frac{1}{p}=\frac{b^2+1}{2b}$ \\ [Invertendo]
$\frac{1+p}{1-p}=\frac{b^2+1+2b}{b^2+1-2b}$ [Componendo-Dividendo] \\
$\frac{1+p}{1-p}=\frac{(b+1)^2}{(b-1)^2}$  [It becomes $(a+b)^2 formula because 1 can be written as $1^2$] \\
$\frac{\sqrt{1+p}}{\sqrt{1-p}}=\frac{b+1}{b-1}$ \\
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{b+1+b-1}{b+1-b+1}$ [Componendo-Dividendo] \\
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{2b}{2}$ \\
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=b$

Shown with ease
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   I
l.54 ...ac{1}{p}=\frac{b^2+1}{2b}$ \\ [Invertendo]
                                                  
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

[vocal hull]

Given instruments,

$b^2p-2b+p=0$ \
$b^2p+p=2b$ \
$p(b^2+1)=2b$ \
$p=\frac{2b}{b^2+1}$ \
$\frac{1}{p}=\frac{b^2+1}{2b}$ \quad \text{[Invertendo]} \
$\frac{1+p}{1-p}=\frac{b^2+1+2b}{b^2+1-2b}$ \quad \text{[Componendo-Dividendo]} \
$\frac{1+p}{1-p}=\frac{(b+1)^2}{(b-1)^2}$ \
$\frac{\sqrt{1+p}}{\sqrt{1-p}}=\frac{b+1}{b-1}$ \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{b+1+b-1}{b+1-b+1}$ \quad \text{[Componendo-Dividendo]} \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=\frac{2b}{2}$ \
$\frac{\sqrt{1+p}+\sqrt{1-p}}{\sqrt{1+p}-\sqrt{1-p}}=b$

[woven radish]

Weinnion

[vocal hull]

Now see

This

[supple knot]

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[vocal hull]

.close

[devout snow]

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[vocal hull]

This has no relation with the quadratic formula. The question i asked is a ratio algebraic equations

[finite fable]

you can solve it using quadratic tho

lemme check

i got it

@vocal hull

can you rationalize the question?

[vocal hull]

What?

I already did it by using Ratio Proportional Methods

[finite fable]

We got b as 1± root(1-p²) /p

[devout snow]

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[finite fable]

@vocal hull reopen the channel

[finite fable]

um

nvm

[finite fable]

so. we also got b as 1+root(1-p^2) / p from

since our b satisfies the b from quadratic. its proved ig

[supple knot]

!done

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[finite fable]

@vocal hull is it done gang?

[devout snow]

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[devout snow]

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[true rock]

Hey, I’m actually kind of confused on these questions

As I have a Unit Test tmrw, These questions tend to trick me and idk what to do

[drifting epoch]

KM should equal LN i think

and if KN and LM both equal 12

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[drifting epoch]

wait does MN and LK also equal 12v

[true rock]

Hi

Im back

[drifting epoch]

wait wtf did i just help someone when it was closed 😭

[drifting epoch]

^

[true rock]

Yep

[true rock]

Why 12v

[drifting epoch]

typo

i was asking a question

if MN and LK = 12 as well

it’s just a square

do the diagonal should js be 12sqrt(2)

[true rock]

Wait i dont understand tho

Like KM and LN is 12 bc its given

[true rock]

Can u explain why we do this

[drifting epoch]

does MN and LK also = 12

[true rock]

They should because they’re congruent parts right?

[drifting epoch]

KM = LN = sqrt(KN^2 + LK^2)

KN is 12 so replace it

[true rock]

Its KN

[drifting epoch]

what?

[true rock]

Wait what

[drifting epoch]

both of them are 12

you can replace KN and LK with 12

[true rock]

KN and LM =12 yeah

[drifting epoch]

cause MN = LK

i’m doing a horrible job explaining this

[true rock]

Ohhh i see so if ur saying that the result of it being replaced would be MN=LM=12

[drifting epoch]

yeah

and if replace it it’s just 12^2 twice

so 144 + 144

under the root so it’s sqrt(288

which is just 12sqrt(2)

[true rock]

Why should we square jt

[drifting epoch]

pythagorean theron

theorem

it’s a square

[dapper tiger]

you need right angles to use pythagorean theorem

[true rock]

Ohhh

[drifting epoch]

it should be one tho

a parallelogram in a circle is a square

wait now imma google this

[dapper tiger]

obviously not

[true rock]

Wair

[dapper tiger]

KM is a radius, like KN

[drifting epoch]

well a rectangle

[dapper tiger]

you have two equilateral triangles

[drifting epoch]

which is still 90 defeee angles

[dapper tiger]

still not

[drifting epoch]

ignore what i said then @true rock

[true rock]

180?

[dapper tiger]

https://cdn.discordapp.com/attachments/903486480075354182/1499511186058903734/IMG_5515.jpg?ex=69f5101e&is=69f3be9e&hm=6c5be7c51710ab44fa60d915235e2cc7a22cd9e78896fd790d8327fa72bdc7b1&

[drifting epoch]

if it adds up to 180 then each angle must be 90

[true rock]

60

[drifting epoch]

if the opposite angles of the parallelogram are equal

dude i could be talking nonsense

melo gonna explain this better

[dapper tiger]

KM is a radius so KM = KN = 12, the angles of an equilateral triangle are 3 60°

KLM and KNM are the equilateral triangles

3 sides are 12

for both

so your angle at K is 60+60 = 120

the shape KNML is a rhombus (but not square)

with one diagonal that is KM = 12 and the other you can find with trig

[true rock]

Wait

Isnt the triangle make up 180

?

[dapper tiger]

the sum of the 3 angles of a given triangle is 180

[true rock]

Yes

[dapper tiger]

that's why equilateral triangles, which can only have 3 times the same angle

have 60° angles

60*3 = 180

[true rock]

Mhm

Then to get K

We double the 60?

[dapper tiger]

the angle LKN is angle LKM + angle MKN

so 60+60 = 120

[true rock]

So both those angles = 60

And we use the 2 angles in triangle to get the 120

[dapper tiger]

yes bc they are angles of equilateral triangles

LKM and MKN

[true rock]

Because of that we add

[drifting epoch]

this problem js made me realize how much of an idiot i am

[true rock]

Its ok

Geometry is shit

[drifting epoch]

no idk why i thought it was a parallelogram it’s a kite 😭😭

[dapper tiger]

well we add because it's a decomposition, like you have to understand visually that angle LKN = angle LKM + angle MKN, draw the segment KM it will become more visual

[dapper tiger]

no it's a rhombus, so it's indeed a parallelogram

your mistake is thinking that parallelogram necessarily have 90° angle

which is wrong

a rhombus is nor a rectangle nor a square

[true rock]

Alright then wait KM is = KN right

I forgot the rule it applies

[dapper tiger]

yes because radiuses have same length

[true rock]

Ohh

[true rock]

Wb LM then

We use trig?

Law of Sines or Law of Cosines???,

[dapper tiger]

yes because now you know the angle at K, which is 120, so the question becomes very simple: KLN is an isosceles triangle with small sides 12, and angles are 120 at K, so the others are 30

from this point there are at least 20 ways to do it

law of sines would be particularly effective

it's still doable without it tho

[true rock]

c squared=b squared + a squared + sinC?

[dapper tiger]

that's not the law of sines

[true rock]

Isnt it easier to use trig

[dapper tiger]

I mean law of sines is already kinda trig but sure let's do without it

[true rock]

Alr

[dapper tiger]

draw the segment [KM], it intersects [LN] at some point, let's call it O
since KNML is a rhombus, they intersect at a right angle, ie KLO is a right triangle, and by symmetry LN is twice LO

LO is just cos(30°)*KL

[true rock]

Im drawing it

[dapper tiger]

so sqrt(3)/2 * 12

multiplied by 2, LN = sqrt(3) * 12

like, your problem is really just finding the big side

in a isosceles triangle

[true rock]

We either use cosine or tangent

[dapper tiger]

go for it, I gave a possible way to do it, there are a lot

in a 120-30-30 isosceles, you should always find that the big side is sqrt(3) times the small

[true rock]

I see

[dapper tiger]

that's also why the law of sines would have been very very good

you know all angles and a side

the law of sines is really made for those cases

you can always do without in isosceles tho

[true rock]

I used it just now

20.78

[dapper tiger]

yep

it's sqrt(3)*12

[true rock]

I see

I’ve got it

.close

[drifting epoch]

i got it too 💀

.close

[devout snow]

Channel closed

Closed by @drifting epoch

Use .reopen if this was a mistake.

[drifting epoch]

i think it’s mine so i have to close

yeah

[devout snow]

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[balmy crescent]

Can someone help me understand why this is undefined? Thanks

[misty crest]

what is sqrt(2) - 2

[balmy crescent]

Ah it's negative!

For some reason I thought sqrt 2 was bigger than 2

[lyric hornet]

click the wrench symbol in the top right

[balmy crescent]

im sorry

[lyric hornet]

and toggle "Complex Mode" in the popout

Desmos will evaluate it in C

[balmy crescent]

Thanks!

I was trying to figure out the side length of a triangle with base 1 and hyp sqrt(2)/2

.close

[devout snow]

Channel closed

Closed by @balmy crescent

Use .reopen if this was a mistake.

[misty crest]

well yea this is a contradiction since sqrt(2)/2 is about 0.7 < 1

[lyric hornet]

for a right angle triangle this is impossible

[balmy crescent]

Ah my bad the base should've been 1/2

you're right

[lyric hornet]

yes that is a possible right triangle

[balmy crescent]

I was looking for this

[lyric hornet]

that would not quite be the base length

recall that (b=\sqrt{c^2-a^2})

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[lyric hornet]

the squaring is important

and assuming principal root and all that

[crystal dawn]

.reopen

[devout snow]

✅ Original question: #help-27 message

[balmy crescent]

Omg you're right

[crystal dawn]

I'd also not recommend smashing everything into one fraction from the get-go

[balmy crescent]

b=sqrt((2sqrt(2)-1)/4)

[balmy crescent]

I got ahead of myself

that's true

[lyric hornet]

[b\neq\sqrt{\frac{2\sqrt{2}-1}{4}}]

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[lyric hornet]

try taking this formula, and putting in your known c and a directly

instead of having your brain try to simplify the arithmetic immediately

that, most of the time, causes slips

[balmy crescent]

0.5?

[balmy crescent]

I see

[lyric hornet]

yes!

[balmy crescent]

Cool!

[lyric hornet]

and observe that

,calc (1/2)^2+(1/2)^2

[woven radish]

Result:

0.5

[lyric hornet]

wait

nvm I am being silly

[balmy crescent]

1/4+1/4=2/4,
2/4=2/4,
1/2=1/2

[lyric hornet]

,calc (sqrt(2)/2)^2

[woven radish]

Result:

0.5

[devout snow]

@balmy crescent Has your question been resolved?

Channel closed

Closed by @balmy crescent

Use .reopen if this was a mistake.

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Remember:
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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crystal lava]

I need help finding X

[strange adder]

which angle is 236-15x can u clear the diagram

[crystal lava]

i think it is this one

[last parrot]

x^2 = 236 -15x

They are all subtending same arc

Then just solve for quadratic equation