competition-math

[glad zinc]

Uh oh

It’s so over 😭

I “proved” it’s not possible for anything other than 3

https://tenor.com/view/facepalm-really-stressed-mad-angry-gif-16109475

Has anyone heard of the ISMTF Olympiad?

[deft wraith]

did you do sine rule way

what are you thinking for silver and gold medal bounadries 👀

[high goblet]

oh well

[high goblet]

nah i just synthetically did it

let AM meet circle BPC at point Q then finish

[high goblet]

probably like 45+ or 50+?

gold will be leaderboard, silver will be just below leaderboard

how did u do?

[vague temple]

I’m not sure but here’s my solution: @hidden oracle

Parameterize the position of each point by a range of [0,1[ with 0 being the beginning and 1 being the full rotation. We fix the first point A at 0. This way, the total volume is just 1x1x1 = 1, so we don’t need to worry about this factor and the winning volume gives us the winning probability. Choose B, C, D such that 0 < B < C < D < 1 or 0 < D < C < B < 1 for a convex quadrilateral. The next part is where I am uncertain. I believe that if I interpreted correctly, these both give the same graph of a right tetrahedron like in the picture, with volume 1/6.

Ok so looks like I indeed messed up the graph so the answer is indeed 2 x 1/6 = 1/3

I was wondering, there shouldn’t be overlap between the graphs but the first one is like x+y+z < 1 that’s why it’s symmetric (wrong)

Hi everyone, other than AoPS by category, where can I train specific AMC 12/AIME topics individually?

[deft wraith]

i think i can get silver then

i got 2-6 (solution seems a bit weird for 6 just did some repeated php magic we will see what i get when it gets marked but hopefully 10- at least)

for 1 i did huge wlog fail, provided construction for n = 3 and proved that you can have it for n >= 4 if you have an increasing sequence of length at least 3 within your circle

im guessing like 2-3 marks max for that lol

[high goblet]

oh damn nice

[deft wraith]

do you lose marks if you state something untrue after

[high goblet]

what was ur soln to 6?

[high goblet]

we mark by the principle "if a subset of the soln is worth 10, then the soln is still worth 10" lol

[deft wraith]

because for this after i tried to prove you could only have increasing sequences anyway (?? not sure what i was thinking here)

[deft wraith]

uh like map each cube to a co-ord 3d grid and do php with like 3d pythagoras

[high goblet]

huh interesting

there's a cute way of doing it with pigeonholes directly

split ur cube into 8 2x2x2 cubes

then you can split each of ur 2x2x2 cubes into 2 bits by a 'chessboard' pattern

and then that basically finishes the problem

[deft wraith]

that's quite nice

[high goblet]

this was the soln george found when he was very drunk lol

[deft wraith]

i did 5 with orders

[high goblet]

how?

[deft wraith]

is this george z

[high goblet]

yeah, IMO 2022

[deft wraith]

well order has to divide 6 so order has to be 1, 2, 3, 6 then prove it doesnt work for 1 or 6 and then you can show for the order equalling 2 it divides one of the expression and for order equalling 3 it divides the other expression and that the only possible orders are 2 and 3

[high goblet]

right ic

well anyway i shall look forwards to marking ur work

[deft wraith]

let me know if you spot it or find any funny solutions

where did people get stuck with 3 is it just a you either spot it or you dont question

[high goblet]

i'd imagine so

[hidden oracle]

nah

you misinterpreted it

[prisma loom]

how did you show ord_p(n)=6 is impossible?

[vague temple]

Ok yeah it was much more involved apparently
https://math.hawaii.edu/home/pdf/putnam/2006.pdf

That's Putnam A6

[hidden oracle]

Oh damn never knew that

[solid kindle]

Any idea how to do the first problem

[weary light]

135

from screen'

to awda

birngq

[misty belfry]

Yeah its 135

anybody know what this function is?

The question says: 17) What is the expanded function term?

[ornate coyote]

math hurts my brain

[shadow spruce]

Pure Mathematics

[gusty verge]

Is it a Green function 🤓🤓

[covert crystal]

https://tenor.com/view/fiji-water-bottle-bottle-bounce-boing-gif-17824457779343478106

[jade widget]

Well…actually🤓👆

[hexed oak]

does anyone know a synthetic proof of the existence of isogonal conjugates?

[snow tartan]

What do people think bmo2 qualification mark will be

[high goblet]

30ish potentially? what r ur thoughts?

i heard ppl found it to be a hard year

[high goblet]

maybe 33-35 for Y13?

[snow tartan]

same

[jade widget]

[hexed oak]

i got a ping here

[jade widget]

idk

[ornate coyote]

huh

[jade widget]

Hi

[jade widget]

[soft vigil]

https://ggim.me/2003/usamo/1

this is cool

[vague temple]

Is there a better way than PIE/casework?
How many distinct five-letter permutations are possible using five of the letters in COMMITTEE?

[pallid dragon]

probably not

it's pretty fast in this case

well i'm not sure i understand the task

[vague temple]

I think casework is slightly simpler

[pallid dragon]

you don't need casework or pie

[vague temple]

Like 2,2,1,1,1 and 2,2,2,1,1

Wait really

How

[pallid dragon]

it has MM or TT or EE and never two pairs

like if a word is MMEEC

that's using 3 letters

but they want 4

oh it was a typo

[vague temple]

Yeah I edited it from a mathcounts question and forgot to change the last part

It was like the same question but 4 letters from MATHCOUNTS

[pallid dragon]

i don't see how pie would work, it's two three cases yeah

one pair or two

or none

[vague temple]

Oh wait it's disjoint I just realized so PIE is not needed

[pallid dragon]

pie would be like, 3(words with MM) − 3(words with MM and TT)

[vague temple]

Ok

[pallid dragon]

but words with MM are hard to count, basically back to square one

[vague temple]

I'm tired I'll take a break and try again

My brain is not processing

[warped anvil]

hmmm

[radiant jasper]

i started to cancel things out from the top, but it turns out that i should be starting from the bottom to cancel things out or else the result would be entirely different?

[ornate coyote]

hmmm?

i think the intended solutioon is to separate 1-7 and 8-14

and then you just continue wiht it

im not sure what you tried to do

[radiant jasper]

Yeah but they started cancelling numbers from 8 to 14

But what i did was going from 14 to 8

So i was wondering how i would figure out to start from 8

[ornate coyote]

cancelling from 14 is the right choice

or the faster one imo

[radiant jasper]

But the result comes out differently

[ornate coyote]

i think you did it wrong

[radiant jasper]

Because cancelling from 14 it’s 1x6x5x …

[radiant jasper]

Rlly?

[ornate coyote]

yeah i dont think you can cancel it like that

[radiant jasper]

Why not?

But if I statt from 8 it works

[ornate coyote]

it doesnt work because if yoou start from 14

lets say 14 pairs with 7, and with 1-6 and 8-13

you cant just set it up with any combination

since when going from 6 x 5 x 4 x 3 x 2 x 1 means that any pairs are possible

but when you start from 8 you wont overcount

4 x 3 x 3 x 2 x 2

[radiant jasper]

Ohh

[radiant jasper]

Thank you

[ornate coyote]

took me a while to word it out

not good at explaining math

[radiant jasper]

Ah

[normal granite]

what's BMO?

[normal granite]

💀 no one does that

It's the other way around because imo qual for US is way harder

[vague temple]

@pallid dragon I forgot, you can use generating functions

Since we have 3 letters appearing 1 time and 2 letters appearing 2 times, we simply find the coefficient of $x^5$ in $5![(1+x)(1+x+\frac{x^2}{2!})]^3$

[gilded halo]

victor

[vague temple]

Hopefully no mistake was made

[ornate blade]

What about China TST

Jk

[vague temple]

Check using casework over number of pairs:
base factor: 5!
Case 0: 6C5 = 6
Case 1: 6C4.3C1/2! = 30/2 = 15 ??
Case 2: 6C3.3C2/(2!)^2 = 15 ??

Wait what

Did I mess something up

Oops

Check using casework over number of pairs:
base factor: 5!
Case 0: 6C5 = 6
Case 1: choose 1 pair of 2 and 5-1x2= 3 remaining from 6 - 1 = 5
3C1.5C3/2! = 15
Case 2: 3C2.4C1/(2!)^2 = 3
Hence 120(6+15+3) = 120(24) = 2880

[normal granite]

hmm never thought abt that, are US and China imo teams pretty even?

[ornate coyote]

i think china usually wins overall

cause yk 4x population

[vague temple]

Bruh of course it was a silly, I wrote 1 instead of 2

Ok both work very nicely then

[radiant jasper]

No

The china tst is much harder than the imo

I’d say getting on chinas imo team is more of an accomplishment than getting gold (you’re getting it anyway)

Like silver medalists and bronze medalists will never get into china’s imo team

[normal granite]

so is usa tst and usamo right

[normal granite]

From their past scores idt it's by a lot right

[radiant jasper]

☠️

No

I’m not a mopper so my opinion doesn’t matter tho

[soft vigil]

usamo is not harder

wtf are you yapping about

[normal granite]

what's the difference between the two

like difficulty*

[ornate coyote]

usamo is like

miles easier than imo

[hearty tendon]

from what ive heard usamo p1>imo1 but imo3>usamo3

[ornate coyote]

debatable

[radiant jasper]

Uh

[soft vigil]

...

...

[soft vigil]

u sure this is harder than imo p1

@ornate coyote you should be able to solve this

[ornate coyote]

dont we just use a short induction

[soft vigil]

ye

[radiant jasper]

bro what is the question even saying T.T

[ornate coyote]

if you draw 4 different segments randomly

what is the probability that you make a triangle that has all 3 points on the circle

so basically you can find out how many sets of 3 points you can pick

to be the vertices of your triangle

and then place the last segment

[hearty tendon]

ok depends on the year

but usamo is not like a lot easier than imo

[radiant jasper]

hmmm

ok that's a weird question

[prisma loom]

This is not true at all

[hearty tendon]

my fault gang

[prisma loom]

Well the first part

The second part is true

[hearty tendon]

i can always kill imo1/4 but i sometimes cant do usamo1/4

so thats what i thought 💀

[prisma loom]

Well there are differences in "problem flavor" between contests

Maybe IMO just suits you better

[hearty tendon]

yeah guess ill have to make it to imo this yr 😔

[vague temple]

Trying out small values, there is a very interesting pattern:
5
75
375
9375
59375
359375
So it seems like we can just stick an odd digit in front. Let’s test this hypothesis:
By induction:
The base cases n=1,2,3,4,5,6 hold.
Now suppose we can append a digit in front. For length n and numbers x_n = 5^n m_n, we set
x_n+1 = d10^n + x_n = d10^n + 5^n m_n
where d is an odd digit.
We need x_n+1 == 0 mod 5^(n+1). Set
d10^n + 5^n m_n == 0 mod 5^(n+1). Equivalently, dividing by 5^n, we simplify to
d2^n + m_n == 0 mod 5
d == -m_n 3^n mod 5
We need to show that this is possible for odd digit d. Notice that for any residue mod 5, within the digits from 0 to 9, we have exactly 10/5 = 2 digits with that residue, alternating between odd and even. This means that there is exactly 1 odd digit that fits each residue mod 5. Hence we pick d to satisfy the and we have constructed x_n+1, so by induction this holds for all positive integer n.

I wonder, is this solution not unique? Can we construct different solutions by not sticking a digit in front? Or can we prove it is unique?

[summer tusk]

does any1 know how to do this 😢

from 27th PMO

[glad zinc]

brit math olym

[shadow spruce]

Pure Mathematics

[dull glade]

Hey, is there anyone competing in PMO?

[soft vigil]

dont think the 26 matters here for parity

just do $\sum_{d|n, d\neq n} d$

[gilded halo]

buboblakistoni

[soft vigil]

$s_{26}(n)$ has the same parity as this

[gilded halo]

buboblakistoni

[jade widget]

[deft wraith]

Find all integers $(n, m)$ such that $(10n + 4)^m + 4$ is a perfect square.

[gilded halo]

al

[deft wraith]

Equating to k^2 and all the usual tricks don't seem to help with this, can anyone a) verify that this has any solutions at all and b) give me some insight into solving the problem?

[deft wraith]

Taking mod 5, doesn't have any solutions, as I thought. Kind of a troll question

[winter sable]

um hi

[vague temple]

Yes this is just the complement of number of perfect squares from 2 to 100 if I am not mistaken

[vague temple]

Powers preserve parity, so s_26(n) == s_1(n) mod 2. If n is prime factored as n = 2^e_0 p1^e1 p2^e2… then the parity of s_1(n) depends on the number of odd factors.
Each even factor added doesn’t change parity, so we can safely ignore them. For odd factors, the sum will be odd if and only if the number of odd factors is odd.
Hence, by cases:
n odd: we subtract 1 to exclude n from its proper odd divisors so
(e1+1)(e2+1)… - 1 is odd iff (e1+1)(e2+1)… is even (if and only if at least one odd prime factor has odd power)
n even: n doesn’t affect parity of s_1(n), so
(e1+1)(e2+1)… must be odd (iff all exponents are even, meaning n is a power of 2 times a perfect square)

[hearty tendon]

does mobius transformation work here LOL

[prisma loom]

s_1 isn't the number of proper divisors, it's the sum of divisors

[compact wing]

anyone doing ritangle here?

[normal granite]

Is there a questions pdf

When’s round 1 and 2

And what’s the selection like

[vague temple]

Wait true I got confused while writing here, I accidentally overwrote my message while editing another one

[vague temple]

Powers preserve parity, so s_26(n) == s_1(n) mod 2. If n is prime factored as n = 2^e_0 p1^e1 p2^e2… then the parity of s_1(n) depends on the number of odd factors.
Each even factor added doesn’t change parity, so we can safely ignore them. For odd factors, the sum will be odd if and only if the number of odd factors is odd.
By cases:
n odd: we subtract 1 to exclude n from its proper odd divisors so
(e1+1)(e2+1)… - 1 is odd iff (e1+1)(e2+1)… is even (if and only if at least one odd prime factor has odd power)
n even: n doesn’t affect parity of s_1(n), so
(e1+1)(e2+1)… must be odd (iff all exponents are even, meaning n is a power of 2 times an odd perfect square)

Complement should make things easier to count for case n odd, while n even you can probably use direct enumeration.
Total odd integers from 2 to 100: 3 to 99 so 49.
Bad cases: All odd primes have even power, hence it is an odd perfect square.
9,25,49,81
Good: 49 - 4 = 45

Now for n even:
9: 18,36,72
25: 50,100
49: 98
81: none
Hence 6 good cases.
Total: 51

[hoary radish]

Guys if you could help for this problem below, I can pay for it: https://discord.com/channels/268882317391429632/1310148924836806656

[glad zinc]

Google it

[glad zinc]

It’s pretty easy to qualify for the first

Second is much harder to qualify than first (at least for me)

[shadow spruce]

$x+1=2$

[gilded halo]

Aestusy

[thorn rune]

You need solution?

[fading bluff]

I don't know man this seems like Liouville's Theorem i guess

[shrewd girder]

[golden escarp]

[timber wraith]

I did Q5 and this was my approach

[neat ledge]

yo yea im doing it

[deft wraith]

Where is this from?

[vague temple]

Explanation: the first lattice point hit after (0,0) must clear the denominator of 5/8, so we multiply by 8 to get the point (8,5). Hence, the number of reflections is the number of grid squares travelled mod 2.
Vertically, we have 5 == 1 mod 2, so we have 1 vertical reflection which gives the top side.
Horizontally, we have 8 == 0 mod 2, so we have 0 horizontal reflection which gives the left side.
Hence it is the upper left corner.
(In the pictured I scaled each step of (1,5/8) to (5,8) so the intuition was easier with integers, but this is not necessary)

[radiant jasper]

Which contest?

[timber wraith]

27th Philippine Math Olympiad (Qualifying stage)

[wooden thorn]

Y'all can you add me to the IMO

Pls

[ornate coyote]

math is hard

[radiant jasper]

As long as one doesn't try to understand it

[swift imp]

I found British math Olympiad

[vague temple]

I would say definitely no need for so much bashing, just clear the denominator to ensure you land on a lattice point, then reflections applied twice give the identity so we can just consider the numbers mod 2

[vague temple]

Wow, problem 8 is very beautiful

Thanks for sharing

It seems so complicated at first but with induction it gives a contrastingly elegant result

[timber wraith]

i dont have a decent foundation in NT or mod stuff, so it was the thing I thought of doing

but most of the trial and error part can be done mentally

so yeah, it can be thought of as clearing the denominator

[vague temple]

Well you can just notice the pattern that it always goes LRLRLRLR UDUDU so only the parity matters

[deft wraith]

Balkan Maths Olympiad

[golden escarp]

me leavin the class after said Mary gave birth to 2,5 kids

[ornate blade]

TMUA results out

how did you guys do?

[shadow spruce]

How can I solve it in the second way.

[compact wing]

fairss what stage you on?

this has to be a lie 😭

[neat ledge]

Stage 3 wbu

[compact wing]

same

lowkey can’t do genius 3

this shit feels like a lottery

[neat ledge]

I tried number soup

[compact wing]

they cooked this years stage 3

[neat ledge]

I got a solution but it doesn't fit the same digit present rule

[compact wing]

Ohh fairs keep trying there is one

I finished number soup and triangles but can’t do genius 3

[neat ledge]

I feel like I've tried every possible one

[compact wing]

na say ong 😭

[neat ledge]

Yea I'ma try genius 3 now

[compact wing]

yo that’s actually insane

how did you practice

dam

do you need 100% for a 9

mines next year so just tryna get info on this shit

fucking he’ll

hell

dam

that’s actually sick dude well done

[gusty verge]

Jk, But fr good job

I haven't seen the paper, I tutored someone TMUA tho, not sure what they got

[high goblet]

not even full marks smh

anyway gj tho!

u applying ox?

ic ic

which college?

ic

well gl with ur application process!

oxford still haven't sent out interview offers right?

oh fairs

how did the MAT go for u then?

fairs

[summer granite]

stayed 2 hours after school and solved it

[ornate blade]

wow

[ornate blade]

you doing IB by any chance?

[summer granite]

?

whats that

[summer granite]

oh no i dont

[ornate blade]

just asking cause 45/45 is a perfect score in IB

[summer granite]

i got 50/45 im too good

tbh i have no clue what it is

[glad zinc]

think a kid got 59 in my old school, he did 2 extra HLs

[high goblet]

is this the 3rd round?

[summer granite]

Yh stage 3

[high goblet]

ah nice

gj on finishing stage 3!

[compact wing]

you solved genius 3?

[summer granite]

Yh as a team

[compact wing]

oh shit did you do it by code?

[summer granite]

He'll no

That will take ages and I don't really know how to code

[compact wing]

ohh fairs so you guys solved everything already?

[summer granite]

a code for the QRS

[summer granite]

Wdym already it's been 5 days

That's like a whole week

[compact wing]

tf

so you guys solved everything on Friday?

[summer granite]

Yes

[summer granite]

We spent 4 till 12

[compact wing]

4 hours?

[summer granite]

No 8

[compact wing]

oh

tf

dayum just to confirm what school is this?

[summer granite]

St Gregory's

[compact wing]

dam never heard of it

[summer granite]

Yh

Idk what the sixth-form is called lol

[compact wing]

dawg 😭

[summer granite]

It's like St Catherine

[summer granite]

It's like connected to my secondary

[summer granite]

How far are you

[compact wing]

dayum how many people are in your team?

[summer granite]

5

How about you

[compact wing]

like into the competition?

[summer granite]

Yh

[summer granite]

6*

[compact wing]

5 but like 3 of them are mostly offline

[compact wing]

done everything except genius 3

[summer granite]

There was 10 originally but they don't do anything

[compact wing]

guess il at least get an honourable mentions or sum

genuinely can’t figure this shit out

[summer granite]

Think some schools solved it way before

[compact wing]

you get an honourable mention if you manage to do 2 of the puzzles

[summer granite]

The group leader somehow found the max digits for each clue

[compact wing]

we did as well

but can’t figure out from there

guess gotta do it by hand

this shit long tho

might as well give up

[summer granite]

What were you doing

[compact wing]

me and my partner doing it by code :p

that’s how we solved all of the puzzles so far so we thought it would work 😭

[summer granite]

Huh

Even the letter stuff

[compact wing]

yeah

[summer granite]

I guess PQRS too

[compact wing]

yup 😭

[summer granite]

Technically you could have entered stage 3 without pqrs

[compact wing]

oh yeah maybe but

[summer granite]

It's literally in the name

[compact wing]

am assuming there’s a limit on the amount of answers

you can submit

[summer granite]

Ngl for like the first 10 questions on stage 1 we thought it was only 1 try

[compact wing]

dawg 😭

[summer granite]

We were super careful lol

[compact wing]

wait are you guys in year 12 or 13

[summer granite]

12

[compact wing]

dam

[summer granite]

Can 13 participate?

[compact wing]

I think so?

but it doesn’t matter to them I think

[summer granite]

Oh arl

[compact wing]

applications already ended

[summer granite]

Oh

[compact wing]

uni applications

[summer granite]

Fair

I was about to say they can put it in their personal statement

[summer granite]

Only code I made

Took a few hours

[compact wing]

oh fairs

[summer granite]

Apparently we need working out

[compact wing]

yeah

[summer granite]

But I can't find them

It's on WhatsApp

[compact wing]

wdym you can’t find them

[summer granite]

No physical copy

[compact wing]

oh

thought you can just have a digital copy

are you guys sure your gonna be the winning team?

[summer granite]

No

Still a possibility never know

[compact wing]

dam alright

[summer granite]

Btw is rectangle world wide

[compact wing]

oh yeah fuc

never seen an international team win it tho

[summer granite]

Yh we are done for

[summer granite]

Oh

Can't lie I have no working out for modulus graph

Just used desmos

[compact wing]

I mean that’s appropriate use of technology so it should be fine

did you guys already like pre learn further maths and stuff

[summer granite]

Uh no

But I know some stuff

[compact wing]

what did you guys do for like question 26 then

[summer granite]

Already completed a chapter we are not ment to do

[summer granite]

Wdym

[compact wing]

the probability one

one of our mates had to use like a further maths topic to do it

[summer granite]

Ah u use integration

Ik the basics of it

[compact wing]

yeah

ohh fairs

[summer granite]

We made a 1x1 square and did like integration or something like that

[compact wing]

could you give any tips for genius 3?

[summer granite]

My team leader dosent want to disclose that

Sorry

[compact wing]

fairs

it’s cool

just gonna do it by hand now then

fuck coding

[summer granite]

Bro don't code that

How are you even ment to start

[compact wing]

not sure that’s why we take so long ig

[summer granite]

Use chat gpt or something lol

Ai ain't great tho

Plus don't think it's allowed

[compact wing]

don’t work 😭

[summer granite]

crazy

[vague temple]

Have you tried using QwQ-32B? https://huggingface.co/spaces/Qwen/QwQ-32B-preview

Or DeepSeek’s deep think mode with r1-preview at https://chat.deepseek.com

[summer granite]

I mean if they somehow catch you using ai you are disqualified

[vague temple]

Oh it’s banned?

They also stopped allowing AI in USAMTS just recently

People were using it too poorly apparently lol

Oh no

I do wonder why they disallow it, if used correctly it’s extremely powerful

[misty belfry]

aint no way in fuck im doing this

[gusty verge]

Why? It's not that bad

[misty belfry]

u do it ._.

After the final quiz were gonna go into 4D and 5D figures

our professor is really smart

oh nvm i got it

$$\left[\frac{\int_0^1 \int_0^2 \int_0^3 (xyz + \cos(x)) dy dx dz}{\int_1^3 \int_2^5 \int_3^7 (xyz + \tan(x)) dy dx dz}\right] = \frac{6 + 6\sin(1)}{144 + 12\ln(\sec(3)) - 12\ln(\sec(1))} = \boxed{\frac{1 + \sin(1)}{24 + 2\ln(\sec(3)) - 2\ln(\sec(1))}}$$

[gilded halo]

Madchette

[misty belfry]

holy shit that bot is good

[misty belfry]

somebody help me ._.

[prisma loom]

that's nothing

[jade widget]

What’s the question I mean

[timber wraith]

im comprehending the problem till the 3rd statement... Like does it want me to label my vertices in an unordered manner such that it meets the 3rd statement?

my figure's vertices when labeled orderly doesnt meet the 3rd condition, but gets 108 deg as the answer

[prisma loom]

your CE and BD don't intersect?

[manic quarry]

Do you guys have any recommended courses for the Mathematical Olympiad?

[timber wraith]

they do, but constructing a circle that passes through CDGI seems impossible

[prisma loom]

send your drawing

[timber wraith]

u may ignore the circle

[prisma loom]

I think it's because you drew the pentagon to be regular

That probably makes CDGI cyclic impossible

[timber wraith]

yeah

but it says equiangular pentagon. is it any different?

[prisma loom]

Yes

You can have non-regular equiangular pentagons

[timber wraith]

oh

[timber wraith]

oohh

damn

right

i didnt visualize the equiangular pentagon like that

[prisma loom]

It might be easier to start with CDGI and reverse engineer the pentagon

[timber wraith]

yeah

[prisma loom]

But it's gonna be hard to draw either way

[timber wraith]

i was thinking of simply makking a circle that passes through CDGI first

then bruteforce both the drawing and the vertices

but upon observation, wont an equiangular pentagon not be regular if and only if the two non adjacent sides (like the left and right sides) are extended?

coz like if I extend the sorta base or roof, it's not equiangular anymore, no?

[ornate blade]

equiangular polygons can come in all sorts of shapes

[timber wraith]

whaaaa

but for pentagons?

the only way i could think of an equiangular pentagon thats not regular is if 2 non adjacent sides are extended or shrinked

[timber wraith]

trying to make my drawing (or ways i could draw) easier😭

[hexed oak]

anyone have a nice combi?

[winter sable]

i hope everyone remembers me

dx / dy

[vague temple]

Now you can repeat that operation on different pairs of sides

[hearty tendon]

alternatively draw a 108-36-36 triangle and take isoceles snippets off the ends

which i think? spans all the equiangular pentagons

[ornate coyote]

comp math is scary

[deft siren]

My solution:

n x 2^n x 4^n = n x 2^n x 2^2n = n x 2^3n = n x 8^n
8^(2n-72) = 8^n * 8^(n-72)
n = 8^(n-72)
8^72n = 8^n

idk where to go from here tho

n = 0 is really close but idk if "really damn close to 0" is allowed

<@&286206848099549185>

oh nevermind this is so freaking easy

🤣

[ornate coyote]

lol

[cloud basin]

gang

whats the best strat for like amc 10

do u wanna like sweat out problems 1-13 and not worry about the end

or do the whole thing but risk missing some

[cloud basin]

revolutionary statement

[cloud basin]

yeah ....

[trim bramble]

1-15

[cloud basin]

scores a 105 if u get all those right is that really enough tho?

id say 115 would be ideal 120 ofc is the real goal but still

[trim bramble]

1-15 make sure

then find easy 16-25 to do

[jolly solstice]

what have you got so far

[vague temple]

(a) 3n == 1 mod 2^k
3n + m2^k = 1
mod 2:
n == 1 hence n is odd
(b)
By Euler’s totient. mod 2^k:
3^phi(2^k) == 1
phi(2^k) = 2^k(1-1/2) = 2^(k-1)
3^2^(k-1) == 1
Hence 3^2^(k-2) is an inverse.
By Hensel’s lifting lemma:
1 is the modular inverse of 3 mod 2^1, so we use the refinement from 2^k to 2^(k+1):
n_(k+1) = n_k - (2 - n_k mod 2^k) mod 2^k

For (c) and (d) I am stuck

[jolly solstice]

I think for (d) they accidentally swapped n and k

otherwise just pick n=2 and you can't find a k that satisfies it

[vague temple]

Apparently for (c) and (d) it is supposed to be odd

[jolly solstice]

I suppose the fact that you have hensel's lemma that might be enough to just knock them out

[jolly solstice]

I think that's not enough to fix (c)

[vague temple]

Really

[jolly solstice]

well it becomes trivial if that's the fix

k=1

then 1^2 * 9 = 1 mod 2

for all n = 1 mod 2

[vague temple]

Wait let me find the corrected version

[jolly solstice]

flipping n and k makes the question make sense to me on both (c) and (d) I'm pretty sure that's the intended meaning

for every positive integer k show there's an n such that 3n=-1 mod 2^k

that would lead to a more natural hensel lifting

that's for d, similarly for c, for each positive integer k, show there's an n such that n^2 is an inverse of 9 mod 2^k - and this would follow from part (b) more naturally as well

see what I mean?

[vague temple]

Oh yeah

[jolly solstice]

my mind autocorrected this to them swapping and that's why I made this comment about hensel's lemma

hopefully that doesn't sound totally crazy lmao

imo I think there's an even easier method but I'll keep it secret for now

[vague temple]

It’s supposed to lead to this

I was told it’s a troll question that’s why the pointage is so high

[jolly solstice]

smh

[vague temple]

k > 2

That’s the correction

[jolly solstice]

it's the collatz conjecture

[vague temple]

For n >= 5

[terse canyon]

well

it should be

[vague temple]

Oh

Unsolved problem? Really, it looks so simple

[terse canyon]

8^2(n-36) because you need to square the 8

so then n=2n-72

n=72

oh oops u got it

bro that is NOT competition

go to #help-32

[jolly solstice]

I don't think that's right, I have already lost faith in the person making this question to not want to continue with it after seeing it's a collatz conjecture question lol

[vague temple]

Ok

[terse canyon]

100 points is mad wt

[jolly solstice]

since (d) isn't my interpretation according to this "correction" I'll just give my solution to my version where for all k show that there is an n such that 3n=-1 mod 2^k

-1/3 = 1/(1-4) = 1+4+4^2+4^3+4^4+... all higher powers of 2 get killed eventually mod 2^k so what's left is the positive integer

this is the geometric series in the 2-adics

[vague temple]

Thanks for the help… I have a question, I learnt Hensel’s lifting lemma as geometric series expansion, but then I was told it’s also Newton’s method. I don’t understand, and the Wikipedia page is incomprehensible. Can you help me?

[jolly solstice]

yeah it's the same as Newton's method from calculus as far as how you use it

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

[gilded halo]

Merosity

[jolly solstice]

but you'll have to make sure it satisfies some conditions first

[vague temple]

What is the f here

[jolly solstice]

$f(x_0) =0 \mod p$ and $f'(x_0) \ne 0\mod p$

[gilded halo]

Merosity

[jolly solstice]

a polynomial

[vague temple]

Ok

[jolly solstice]

with rational coefficients and denominators not divisible by p

that's just to make sure things stay well defined, there are other versions of Hensel's lemma, but this is a common version to see, we can kind of walk through how to derive it if you haven't seen that

[vague temple]

Ok

[ornate coyote]

not the freaky math

[timber wraith]

the problem's driving my illustrations nuts

[timber wraith]

, rotate

[timber wraith]

i got this rough illustration that seem to make sense and got 108deg as the answer (D.) But apparently it must be 72deg (B.)?

[vague temple]

It seems like you can’t construct it using a regular pentagon, it really has to be a deformed equiangular pentagon otherwise CDGI cannot be concyclic

So start with the circle and try to construct the pentagon along it

[timber wraith]

why does <FGH not look = 72deg there either😭

[vague temple]

IDK my drawing is horrible

Angles are not 108 degrees I didn’t measure

[vague temple]

Calculation comes from EFGC and BHGD being cyclic, I am not sure about the parallel lines, I was just looking at my (inaccurate) drawing so don’t trust it

Cyclic is true since <BHD = <BGD = 36 degrees

Hence EGF + BGH = ECF + BDH = 180 - 144 = 36 degrees

FGH = CGB + EGF + BGH = 36 + 36 = 72

[jade widget]

[jade widget]

Have u guy receive the amc score?

My friend said her friend got it

[zealous fulcrum]

Hello everyone

I am in high school and would like to start competition math and maybe start a club with some people

What branches of mathematics are usually relevant in competition math that I should study?

I am currently studying calculus, but I'm not sure how meaningful calculus is in competition math

[radiant jasper]

Algebra, Geometry, Number Theory, Combinatorics

[iron terrace]

does the imo have college stuff in it

[deft wraith]

i mean in theory you can kind of do most (if not all (?)) IMO questions with "elementary methods" although some of the "elementary methods" might not be taught in your country until college i.e you are not explicitly taught the euclidean algorithm in the uk unless you take a specific further math module which most people do. read: https://web.evanchen.cc/handouts/Syllabus/Syllabus.pdf

[iron terrace]

oh thank you so much; that's very helpful

[ornate coyote]

imagine making imo

could never be me

[plush abyss]

Guys how to solve 3 cos x +2=5

[pallid dragon]

you imagine the unit circle

[ornate coyote]

cosx=1

so x=0

[lavish scaffold]

x can also be 2k*pi radians, where k is any integer

[minor matrix]

Cos(x+2kpi)

man trigonometry was a long time ago

i nearly forgot everything

[ornate blade]

@iron terrace https://www.mathcounts.org/sites/default/files/u5328/Solutions December2014 Mini.pdf

gahhhh this is the solution, q6

[barren sparrow]

can someone explain this question

[hybrid vessel]

Hii

[radiant jasper]

Hi! First calculate the number of unique arrangements without restrictions

then find the number of arrangements that dont satisfy the restriction and take that away from the number of unique ones

[vague temple]

We can use complementary counting. Let’s name our events:
A: no 2 S’s consecutive
B: no 2 C’s consecutive
Now, our desired answer counts the number of possibilities within:
A and B
which is logically equivalent to
not (not A or not B)
So now our events are (give it new names for convenience):
C = not A: at least 2 S’s consecutive
D = not B: at least (exactly) 2 C’s consecutive because there are only 2 C’s
The total without restrictions is 7!/(3!2!1!1!) = 420.
Let’s find the count of (C or D) = (C or (D \ C)) = ((C \ D) or D). In this case, since D is simpler, we can probably restrict it more easily, so let’s go with (C or (D \ C)). (You can also use PIE here if you prefer)
C: we need at least 2 S’s consecutive. The keyword “at least” hints at complementary counting. So let’s take the complement.
Total: 420
Bad (no S’s consecutive): first, place all the other letters c,c,e,u with x4!/2! = 12 for ordering. Then place the remaining S’s in 5 spots: .c.c.e.u. (5 choose 3) = 10. So bad = 12x10 = 120.
C = 420 - 120 = 300
(D \ C): exactly 2 C’s consecutive and no S’s consecutive. A similar approach: place cc as a single entity with e and u: cc, e, u so x3! = 6 for ordering. Then .cc.e.u. so for S’s (4 choose 3) = 4.
D\C = 6x4 = 24.
Now the size of (D or C) = 300 + 24 = 324.
Taking the complement, the answer is 420 - 324 = 96.

[winter sable]

?

[vague temple]

Alternatively, using constructive counting:
place 3 S’s, we need to fill in the middle gaps to be non-consecutive.
.s.s.s.
Let t be a placeholder for e or u (no restrictions) and we will multiply by 2! at the end for permutations.
We will place the c’s first since they have restrictions.
Cases:
Distinct slots
cscs.s x2 for reflection
cs.scs x2 (same calculation as first so we’ll group these)
cs.s.sc
scscs
Same slot:
ctcs.s.s not possible
sctcsts x2
Case 1: cs.scs x4
Fill the gap: .c.st.s.c.s.
then 6 spots for other t
4x6=24
Case 2: cs.s.sc
Fill the gaps, 1 possibility
Case 3: scscs
.s.c.s.c.s.
t in separate slots: 6 choose 2 = 15
t in same slot: 6
(Alternatively, by stars and bars (2+5 choose 5) = (7 choose 2))
6+15=21
So total for distinct slots: 24+1+21=46

Same slot:
Case sctcsts x2 for reflection
2x1=2

Total: 2(46+2)= 96

[iron terrace]

oh thank you

[barren sparrow]

Got it thanks @vague temple @radiant jasper

[jade widget]

[timber wraith]

For Q14, i cant wrap my head around the idea how ai can form a cyclic Quadrilateral w/ circumcenter O

[timber wraith]

got this

[fervent ember]

Hi, which books and websites should I use to prepare for competitions as an A-Level Student

[high goblet]

which competitions r u thinking of?

[vague temple]

I wonder, how did you get that first equation 180 = 54 + 2x + (36 - y)? I don’t see it
I got the same answer using ADC = AOC = 108

[timber wraith]

sum of interior angles of triangle BAC

i substituted "y" to <ACB coz i was lazy

since <ACO = 36

[vague temple]

Isn’t that the second equation

[timber wraith]

the 2nd eq is the sum of interior angles of triangle DAC. since <ADC = 54 + x (exterior angle theorem)

[vague temple]

Then shouldn’t it read 180 = 54 + 2x + y

I’m not sure how you got that, 36 - y is <OCD

[timber wraith]

oh wait right

yeah i missed that

how did u get <ADC = 108?

[vague temple]

Since AODC is cyclic, ADC and AOC subtend the same arc, so they are equal. Hence AOC = 108 = ADC

I think the problem choices are too easy to guess, it’s very likely to be some multiple of 15 or 18 for these problems since you usually multiply by integer multiples or take complements in 90 or 180 so 18 is the only plausible answer since you start with 54

[fervent ember]

idek i j wanna get better at maths

[high goblet]

oh fair

i grinded out like all of the UKMT maths challenges, that made me quite a bit better at maths

in general for maths, i think you should try focusing on properly understanding some of the content taught in schools

like in school, sometimes they can be quite bad at teaching with emphasis focused on memorisation

[fervent ember]

which website has like hard a level content

except for madasmaths

or is madasmaths the best

[high goblet]

ngl i didn't really do much a level maths revision lol

i just kinda grinded olympiads

[high goblet]

but yeah i have heard madasmaths is supposed to be good

[jaunty orchid]

when aime cutoff come out

[prisma loom]

when they are done grading and such

[jaunty orchid]

usually its 12/1 isnt it

[prisma loom]

they'll probably be out soon

check aops forums

[fervent ember]

u reckon i shld j do that

[iron terrace]

does the AMC 8 include number theory & counting and probability?

[thorn rune]

Yes

But basic

[iron terrace]

ah okay

thanks

[tacit knoll]

idek where to put this anymore, this was from a past paper and prob gonna be in my test soon

yr 12 paswt paper

trig ig

[radiant jasper]

did you do the british olympiad training camp

[jade widget]

How to prepare for aime?

Just solving problems?

[vague temple]

(a) probably just divide the factor out and use quadratic formula
to take square root just set sqrt(z)=a+ib so let z = (a+ib)^2, solve for coefficients
(b)
(i) multiply both sides by (w-1), it becomes the third roots of one
(ii) z2 = wz1
(iii) rotate PR by third rotation so z4 = w(z3-z1) = w(w^2z1-z1)=(1-w)z1

[swift imp]

yes

.

just solve problems

that are slightly above ur skill level

[tacit knoll]

Thanks

[ornate coyote]

i do many mocks

should i do jmo practice problems while im still prepping for the aime

[jade widget]

[radiant jasper]

Can some plzz solve this

[high goblet]

well i mean if ur goal is to just get good at a level maths then probs not

[high goblet]

yh

[radiant jasper]

Anyone?

[swift imp]

Depends on how well u do on aime

If u just started learning aime stuff probably not but if ur consistently getting like 10,11,12 + it might be worth looking at jmo and oly

[dull glade]

I've my maths Olympiad in some days. Can someone guide me where to study and what to practice

This is my syllabus

[radiant jasper]

bruh anyone

[ivory ember]

distance formula + collinearity moment

[fervent ember]

how would you get better at like tmua

[high goblet]

grind out tmua, grind out MAT, grind out olys etc.

basically just grind maths ig

i think trying to get good at SMC can be quite beneficial for that

[radiant jasper]

bro orz

[high goblet]

lol ty

[radiant jasper]

did u make the team

[high goblet]

almost

i got to like the final stage (like the top ~10s)

but i didn't quite make the team

but i am on quite a few of the leaderboards

[radiant jasper]

Can i have 10m sb coins

[high goblet]

no

[radiant jasper]

Buh

[radiant jasper]

Oh

Did u try this years usamo problems

[high goblet]

nah i don't really look at USAMO

i have enough of my own work to do lol

[radiant jasper]

If you have time try to look at it

[ornate coyote]

💀

[swift imp]

@high goblet do u like classical music

[radiant jasper]

If its a high level competition, focus mainly on developing number theory skills and advanced geometry

[jade widget]

[high goblet]

yh, why?

[dull glade]

I can send the sample paper and can then you tell if it's high level or not?
And how can I develop skills on it? Any YouTube channel or playlist recommended

[jade widget]

[swift imp]

slayy

[prisma loom]

This is dumb advice

Why only focus on half of the potential problems?

In math olympiads, being good at everything is pretty much always better than being very good at a single topic and mediocre at everything else

[novel karma]

fr

[ornate coyote]

DAMN IT JIMMY

[lavish scaffold]

I find that number theory is quite flexible and can be helpful for a large range of problems

[radiant jasper]

I might want to get involved in competition math, how difficult is it

For reference I have fairly deep conceptual understandings of algebra and geometry and functions in general, and I learn quickly

[ornate coyote]

depends

on like what level

[radiant jasper]

Like what level of competition math?

[ornate coyote]

yeah

there are simpler ones like amc10-aime early questions

[radiant jasper]

Idrk I don't know the levels

[ornate coyote]

then there are more difficult ones like the MOs

[ornate coyote]

what math courses have you taken

[radiant jasper]

I've taken up to algebra 2 but I have a fairly deep understanding of precal and know differential calculus

[ornate coyote]

try taking like an amc10 mock test

or like an aime

[radiant jasper]

admits

[ornate coyote]

huh

i think comp math gets much more freaky than normal math classes tho so like you need to read up on combinatronics and some number theory stuff

[radiant jasper]

I hate number theory

💀

Combinatronjcs tho 🤤

[acoustic nova]

do the problem i sent

[dull glade]

What would you guys rate it on difficulty scale? Is it high level or average or eas

[radiant jasper]

Where

[ornate coyote]

very easy

id say around amc8 level

[dull glade]

Nice. Can you tell me how can I practice on similar but a bit difficult problems? Any YouTube playlist, or website with mock tests etc.

[ornate coyote]

artofproblemsolving

google amc8 previous tests

and you will find like 20 tests that are around that level

or you can use the aops free training platforms

like alcumus or mathcounts trainer

aops is the goat for all things math

[dull glade]

If it's amc8, practicing higher levels will make them more easy, right?

[dull glade]

Okkk. Thanks man

[ornate coyote]

yes

[dull glade]

This is my first competition and without anyone to guide so kinda unfamiliar with everything and nervous

[ornate coyote]

mostly the later amc8 problems are on the higher level

compared to the paper you sent me

[dull glade]

Okk. What would you ultimately recommend practicing? Sticking to amc8? Or something else

[ornate coyote]

i think amc8 is the best choice

because it has a few more difficult problems

and there are so many to practice on

[dull glade]

Okay.

[ornate coyote]

then as you get better and better you can do like amc10, amc12 and aime

[dull glade]

Ok

Thanks man.

[high goblet]

me completely ignoring every single geometry problem and somehow almost making the UK team

[shadow spruce]

Is there people who are good at a specific topic and not good at generalising

Geometry is basically Pure Mathematics, pure talent required. So does Number Theory.

[prisma loom]

bro what

Oly geo is extremely niche and pretty much completely divorced from all other math topics

I know lots of people who are extremely good at geo but bad at everything else

and being good at number theory won't help you in a combinatorics problem about snails on a grid

[deft wraith]

I would argue combinatorics is the topic that “talent” comes into the play the most

But I mean there’s people who are talented with geometry, algebra etc so

Anyway I agree with above

[radiant jasper]

how?

talent means nothing

its about how hard you work

also im a geo main

[prisma loom]

Talent isn't the most important thing but it can have an impact

[acoustic nova]

no there is talent

a lot of talent

in math

especially higher math

[radiant jasper]

we're talking about comp math

if youre talented and youre motivated then youre obviously going to be really good

but if youre just an average person but youre also motivated you can still be really good

it just matters how much you try

[prisma loom]

100% true

There are countless people who were "bad at math" before they started doing competitions

It's really just a matter of motivation

[acoustic nova]

yeah I guess that’s true. not something I like as much (I’m not trying to claim I’m very talented it’s just something I think)

[radiant jasper]

YES

LMAO

[vapid herald]

What is the socioeconomic status?

[radiant jasper]

If you're from a rich neighborhood, youv have access to better education

[sturdy viper]

When is cutoff released

[swift imp]

The math contest education is far more different compared to normal education

But poor people don’t know as much about contests as rich people

[radiant jasper]

And have less access to good tutors bc the cost

But internet helping that

[swift imp]

Yaaaa

But u don’t need tutors as much anymore

[radiant jasper]

Thankfully

[swift imp]

Imo tutoring isn’t as useful

[radiant jasper]

Yt clutching

[swift imp]

As ppl think

[swift imp]

AoPS forums:

[ornate coyote]

tutors arent really that useful

cause literally everything is on aops

and there are lots of free sources and you probably can buy a few books

[acoustic nova]

How do u do this: a^2 = 3^b + 37 solve in integers

[ornate blade]

disagree, tutors provide the face-to-face interaction

otherwise AoPS online classes would have zero demand given the reasons you stated

not everyone does well when self-studying

[acoustic nova]

as if AoPS classes aren't like as much as the opposite of face-to-face. screen to screen not even face to face over screen
but anyway i get what u r saying

[ornate blade]

exactly so it's even better

[hearty tendon]

combo talent is real

💀

[marble grove]

guy i got a 58!!!

i think

maybe it was a 52.5 or sum

or a 50.5

idk

[swift imp]

The issue with self study is that no one knows how to self study properly

Once they do it goes a lot further tho

[marble grove]

there isn't a place to start.

since aops covers so much its rlly hard to just find out what you know and what you don't and where to start

[swift imp]

That’s the hardest part

For me personally I’ve gotten a. Lot further self studying

But it’s also because I was more passionate and didn’t have like people forcing me to do stuff I didn’t want to

But it’s probably a case by case thing

Finding what to start with is hard but in general just do more problems a little above ur skill level

[marble grove]

idk my skill level bruh

like some times i can go up to question 19 but also don't know how to do question 6

[swift imp]

Different categories of math?

Like

Are they different type of questions

[deft wraith]

i mean there exists talent that allows people to start at a higher level or learn quicker

i think thats undeniable

but yes for sure its not "only" talent and saying that discredits all the hard work that talented people put in

[vague temple]

Looking at the graph of y = x^2 and y = 3^x + 37 in the real numbers, we see there are 2 intersections so at most 2 solutions, one positive and one negative.

Now plug in for candidate solutions (estimating the graphs, it should not take too long for exponential to catch up to quadratic):
Positive b:
3^b = 1, 3, 9, 27, 81, ...
We see 27 + 37 = 64 is a perfect square, so (+-8)^2 = (3)^3 + 37 giving (a,b) = (+-8,3).
Negative b:
3^b will be fractional, so a^2 cannot be an integer so a neither.
Hence, (+-8,3) are the only integer solutions.

[prisma loom]

what?

if you look at the intersections of the graphs x^2 and 3^x+37 you're only getting solutions where a=b

[jolly solstice]

maybe you could find integer points on the elliptic curves y^2=x^3+37, y^2=3x^2+37, and y^2=9x^2+37 then throw out the solutions to x that aren't powers of 3.

[ornate blade]

clever idea

[jolly solstice]

idk how easy that is, I don't think nagel lutz applies to the other two but I haven't thought about it... also just noticed I put x^2 where I meant x^3 in those

[acoustic nova]

but that’s wrong

[swift imp]

Imo very little is actually talent

Like basically negligible because I know of many people who just spent many hours in math to get better talent doesn’t get you problem solving skills

[pallid dragon]

talent is when you don't need effort, but also you don't need time

time and effort substitute for each other

so yeah, the majority of people had to spend one of those for sure

so talent is "very little" statistically, you can just assume you don't have it, so you're like everyone

[high goblet]

where is this problem from?

[acoustic nova]

Me

[high goblet]

oh

the answer is if u've cooked up a random problem by urself, then it's highly possible there is no elementary soln

(see like Fermat's Last Theorem or Catalan conjecture lol)

[high goblet]

^ sounds like an reasonable approach

idk how to find integer points on elliptic curves, but you can plug those elliptic curves into a database and find all integer points

[jade widget]

[vague temple]

You can just replace 3^b by x^3?

[vague temple]

True, my bad

[vague temple]

Using modular arithmetic, I have narrowed down the solutions to:
a = 18n + 8
for some integer n
b = 2m + 1
for some integer m

According to WolframAlpha:

[vague temple]

You posted this?
https://math.stackexchange.com/questions/5007345/a2-3b-37

[low fiber]

can someone help me make my amc 10 study schedule

[swift imp]

Why do you want a study schedule

It's better to explore rather than trying to stick ti a plan

[vague temple]

Agreed, if you have no experience then just first list your goals and then try different approaches and record your experiences, then you can decide on which approach to proceed with

[low fiber]

because i feel im not learning enough through just exploring the AoPS book

[swift imp]

Don’t explore books at first

Do problems

[low fiber]

ik the kind of stuffs rhat on amc. I bombed it cause i didn’t study

[swift imp]

Then you don't know it

[swift imp]

I don't think books are good

For problems

Just find problems in contest collections that look interesting and do them until you have an idea of what contest is like

[ornate coyote]

i get mine secondhand from a moper i know

[swift imp]

I have aops books from piracy 💀

[ornate coyote]

i read books when i dont have access to the internet

and sometimes without power

so i like having the books

[radiant jasper]

what does ap calculus course cover?

[valid socket]

single variable calc

bc covers up to polar and series

and parametric

[radiant jasper]

I'm currently learning korean curriculum so I'm just confused

like exponent functions and log functions are in pre calc I assume?

like the trignometry function

and then it's ap calc ab?

[swift imp]

Exponent functions and log functions are in algebra courses in the usa, and trigonometry is usually covered in precalc

AP Calc AB and AP Calc BC are both taken

They are both year long courses

[radiant jasper]

oh because in korean high school last year

we learn calculus

wait

lemme share a picture

[swift imp]

AP Calc AB is a separate course from BC, they share many similarities though, Calc BC just covers more

[radiant jasper]

are stuff like that taught in ab?

[swift imp]

Stuff like what?

[radiant jasper]

like things in the image i sent you

[swift imp]

Area under the curve?

That is taught in AB and BC

[radiant jasper]

that's finding the area covered by a function and its inverse function

[swift imp]

Maybe

I don't really remember my AP Calc course

[radiant jasper]

ah

[valid socket]

it’s possible to show up on a question but highly unlikely

[radiant jasper]

because I'm currently studying in canada

[valid socket]

ap calc is very basic imo

[radiant jasper]

and the BC curriculum here is so bad

[swift imp]

Yea the AP curriculums arent that good imo

[valid socket]

covers less than stewart’s calc did

[radiant jasper]

no not ap like regular math course

[swift imp]

oh

[radiant jasper]

like I'm taking pre-calc 11 and they're still working on quadratic function

like the basics of it...

[valid socket]

NA education

[swift imp]

Yea

[hearty tendon]

LMAO u korean

[swift imp]

Stay ahead

Just learn from aops problems lol

[radiant jasper]

hi

[hearty tendon]

hi

[swift imp]

(That's how I learnt math 😭 )

[radiant jasper]

r u korean?

[hearty tendon]

yeah 😭

[radiant jasper]

yeah that helps

[hearty tendon]

i was scrolling and i saw that

[radiant jasper]

what grade r u in??

have you taken ap calc ab?

[swift imp]

I just learnt from doing competitive math lol

[hearty tendon]

11 and yeah

[radiant jasper]

oh so ur in ap calc course rn i assume?

[hearty tendon]

nah im taking dual enrollment

[radiant jasper]

yeah but I also wanna start working on calc because gpa is more important

[radiant jasper]

what's that

[hearty tendon]

college stuff

[swift imp]

GPA is not that much more important imo

[radiant jasper]

oh

[swift imp]

Doing well in math contests can get you a lot further than gpa

[valid socket]

and it shouldn’t be challenging to balance

[radiant jasper]

but do you know how ap courses work?

[hearty tendon]

yeah but if u cant do well in gpa dont expext to do well in comps lmao

[radiant jasper]

well i know for sure that im not good enough to make it to usamo..

[hearty tendon]

💀

[swift imp]

well erm

[hearty tendon]

yeah

[swift imp]

my gpa isnt the top but its ok ig

[radiant jasper]

okok good

do you know the korean math curriculum?

[hearty tendon]

are u good?

[swift imp]

But like I do math contests a lot

[hearty tendon]

at math

comps

[swift imp]

Yea

[hearty tendon]

sorta

[swift imp]

that's all i focus on 💀

[hearty tendon]

did u amo

[radiant jasper]

lol my friend qualified to aime with 0.5% score but got 90% in school math

[swift imp]

hm maybe

[radiant jasper]

are you literate in korean by any chance??

[swift imp]

I don't want to answer that

[hearty tendon]

yeah

[swift imp]

Doing well in school math $\neq$ doing well in contest math

[radiant jasper]

can u dm

[gilded halo]

stasya

[hearty tendon]

i mean i did and gpa is free man

😭

[swift imp]

IN math yea

[radiant jasper]

yes

[hearty tendon]

if u got a q hmu

[swift imp]

I just

[radiant jasper]

math = free gpa booster

[swift imp]

dont know how to write essays LMAO

[hearty tendon]

oh thats fair

[swift imp]

i usually write them in like 30 minutes and the teacher gives me like a week or 2

And I just turn in my first draft 💀

[hearty tendon]

lwk same but i bs a lot

what grade are u

[swift imp]

senior

[hearty tendon]

o cool

i tried gridning mop once and bombed jmo

😭

[radiant jasper]

i sent ")

[swift imp]

why jmo?

amc 10 kinda bad ngl