competition-math

[topaz mist]

Oh ok XD

But thank you anyway

Unfortunately school is not letting me do those things much homework

[bold hemlock]

Happens

[pearl holly]

Lol

Got a 85/150 on a gauss prac. contest

💀

Its only grade 7

*skill issue for my partHAHAHA

[topaz mist]

What is it ?

[topaz mist]

Nah don't worry, it doesn't mean you're genuienly bad at maths
Something you juste do worse than you thought you would do

[pearl holly]

https://cemc.uwaterloo.ca/resources/past-contests?grade=All&academic_year=All&contest_category=13&block_config_key=past_contest%3A1srMgUG8ZWnN5_Mx6CmP5HeP-aleHwL0jgxyVUuVYE4

These are like introductory math comp. problems

I started grinding on the first one

I'll grind the other one later

[topaz mist]

Oooh nice

Do you think it's doable for me ?

[pearl holly]

because the entire day I have been grinding NT-Prealgebra on alcumus lol

[pearl holly]

We will never know if you don't try

[topaz mist]

True

I'll start someday hopefully

[pearl holly]

Nah bro

You can start now

[bold hemlock]

How is this competition maths?

Or it's just an end of school exam

Or something

But ok

You have like an 1 hour for all the questions

So it kinda makes sence

[topaz mist]

Nah bro, i got freaking spanish first XD

[topaz mist]

Wait they have computer comps ???

Like coding programming and cs stuff ???

[topaz mist]

Imma do that too then, thank you even more for those links

[solar quail]

lol

[pearl holly]

Yeah

Competitive Programming in terms of what people say

:D

[pearl holly]

Honestly, its best to ask them lol

[pearl holly]

One thing about them is that... I do not know where to submit them

because they give you data sets that you to put the inputs in

so yeah

Oh, and incase you didn't know, @topaz mist , they have these Problem Gens

You can have like your own contest-style

I find it fun to just print problems and simulate an actual math contest environment HAHA

Oh, cool there are some other stuffs here, too

I should make like a list of these types of resources

[topaz mist]

That's so cool

[topaz mist]

Could it just be trainings or archives of real life comps ?

[topaz mist]

Ty, I'll look that up elsewhere too

[pearl holly]

I don't think so?

[topaz mist]

Well if you can't submit them use it as self awareness of your current level ig

[pearl holly]

Okay, wait

@topaz mist check your dms

[sleek ivy]

Those are nowhere near the level of amc

[ornate blade]

disagree

AMC is hard for the grade levels they are testing

so students need easier questions to review the foundations and build confidence
CEMC isn't as easy as you claim if you're practicing towards AMC

click on PCF and Euclid

[bold hemlock]

It's nowhere near Polish Junior math olympiad

I mean at least like 99% of questions I guess

Idk how other olympiads are like

[bold hemlock]

Im talking about this

[bold hemlock]

.

AIME questions are easy, aren't they?

Iirc

[bold hemlock]

Ok

They are not

But some old AIME questions are easy

For sure

No

Accualy they are easy

At least some of them are

[sleek ivy]

Uhm

No

[bold hemlock]

Some of them are for sure

Like "Jimmy walks a kilometers per hour he takes a break for b hours, and it takes him 420 hours to get to home, but if he was 10 km/h faster he would get to home after 2 hours"

Seems like an easy system of equations

[ornate blade]

you haven't seen q20 to q25 on the AMC

and you haven't specified which AMC you're talking about

you should compare with AMC 10 to be honest

[swift imp]

early aime yes could be easy
late aime is not easy

[bold hemlock]

K

[ornate blade]

oh wait is the Polish junior one only up to 8th grade? my bad

[bold hemlock]

True

[ornate blade]

https://artofproblemsolving.com/community/c3821951_2024_polish_junior_mo_finals

these are pretty standard for AMC 10 not 8

but then AMC 10 is the first round of qualification

[bold hemlock]

K

[ornate blade]

so you're comparing apples to oranges

[bold hemlock]

It's for 10th graders?

[warm arch]

For math and music nerds

[ornate blade]

yes

10th grade and below

but yeah AMC is a different format, all multiple-choice

https://artofproblemsolving.com/community/c3342708_2023_polish_junior_math_olympiad_finals

[swift imp]

this is cute

[ornate blade]

regardless if you just look at the material covered by each question

[ornate blade]

you'd have to compare this to the AIME, I see

yeah it's hard to compare

[bold hemlock]

For what grade is AIME I?

Oh

It's I

My bad

lol

[ornate blade]

yeah so generally 10th to 12th grade

[bold hemlock]

K

I wonder how hard is Polish MO finals in comparison with USAMO finals, probably trivial lol

I mean of course not all questions

Because Poland isn't bad in olympiads

We have a lot of silver medalists on IMO

And they often fail to do some of the questions on the last round of the olympiad(I mean at the time of the contest)

[ornate blade]

that's normal

well if you can only choose 6 people from each country

the countries with the highest populations will have an advantage

[bold hemlock]

Or the countries that adopt people from other countries

Sorry I had to

And of course I also have to post the memical photo

That is probably like 10 years old or something Idk

[bold hemlock]

This thing is funny

6th question

1/216 people got it

Well accualy he didn't

There was a mistake in his proof

[bold hemlock]

Or it was not enough

Or some other story like that

I haven't seen it

I've just heared the story

[dreamy plank]

"incomplete gamma function"

[maiden panther]

this is the international chemistry olympiad...

[sleek ivy]

ah yes

american olympiad team

[bold hemlock]

Then even better

But some IMO teams were also mostly asian

[soft badger]

send questions

hey guys does anybody has any question regarding any type of maths

[bold hemlock]

You can try

If you want to

[chrome parrot]

All numbers?

Does it mean there aren't infinitely many?

[sleek ivy]

There are infinitely many primes so yea no?

[chrome parrot]

So

n = 1; 3

So n can't be dividable by two, it's not always a prime neither a composite

Ok 5 and 9 don't work either

11 neither

Ig n = 1; 3

[bold hemlock]

Yea

No.

I mean 1 is not a prime

[chrome parrot]

Oh

No

It can't be

[bold hemlock]

And it's not a proof

[chrome parrot]

It's 3 then

[chrome parrot]

Uh

[bold hemlock]

It's probably 3

But I can't solve it

And I don't know anyone who can

So it's probably 3

[chrome parrot]

It's 3, we already solved it

[bold hemlock]

Yes

But we don't know all the numbers that work

[chrome parrot]

Mega Angelo13 and Sebastian Góral solved the problem

Wait

You are the one who created the question

[bold hemlock]

Nah

[chrome parrot]

You just showed it to us?

[bold hemlock]

It's from another server

[bold hemlock]

Yes

But I'm not the author of the question

[chrome parrot]

Problem solved ✅

[bold hemlock]

I mean it was created because some dude chcecked that it fails for a lot of n, to he thought "ok probably only 3 works" so he sent it to some other dude, he couldn't solve it so he sent it to another server, and we couldn't solve it, so it's unsolved lol

[bold hemlock]

Nah

It's "find all p that work"

You have no proof that there is no other p that works

It was checked up to 3000 first prime numbers or something around that

And so far only p=3 works

But there is no proof

At least for now

[chrome parrot]

There is no calculator that can prove if some number p is a prime number above 13

Not in the web

And I am very angry if the number is like 1836829229833664829102873646472992991938473930192747920287373837

Wolfram Alpha can't solve it either

[bold hemlock]

Cool

So it's still unsolved

[high goblet]

heuristically probably, cus sum 1/log(n) log(n^n + 2) converges

um i don't see a good way to prove this

are you sure this is solvable?

[bold hemlock]

No

Lol

[high goblet]

if this is just from exploration & ur unsure if this is open or not, it may be worth posting to #advanced-number-theory and seeing what ppl know abt it

[bold hemlock]

Nah

Because people there asked me "why do I care about it?"

And I don't have a reason

I mean I'm just curious

It isn't important to any mathematical reaserch I'm aware of, or something

[high goblet]

basically if the question is solvable, then there'll be like factoring trick or like some modular arithmetic trick that you can do

but there doesn't feel like much you can do, so i'd imagine the problem's open

cus basically every problem regarding "are there infinitely many primes of the form X" is unsolved

[bold hemlock]

Probably

But Idk

[high goblet]

why do we care that there are infinitely many primes?

why do we care about infinitely many 1 mod n primes?

ultimately like there is no real practical application

just say like u were curious what we currently know about primes p s.t. f(p) is prime

[bold hemlock]

I'm just saying what people told me on #advanced-number-theory

[bold hemlock]

Modular arithmetic does not work

I mean in some cases the smallest divisor is like over 100000

But it might have some factorisation

I mean it seems to have some factorisation

Because it seems to be divisible by something of size near to p^2

But it probably does not work

I don't know any polynomial that would work for a lot of cases

Or anything like that

[high goblet]

yh i doubt it

you can probably use sieve methods to bound the number of primes with this property <= x

actually ur bound's probably gonna be kinda awful

[high goblet]

anyway plugged this into wolframalpha and got an answer of ~2 so yeah we probably expect finitely many primes of ur form

[bold hemlock]

Also where does the sum come from

I mean Ik that prime number theorem

But where exacly

It's like "assuming p^p+2 is a random number"

Or something like that

And then we sum the probabilities of a number of some number of that form being prime?

[high goblet]

model each number n as having a 1/log(n) chance of being prime

then find the expected number of primes

[bold hemlock]

Makes sence

Also

It has to be of form 6k+5

Because modulo

[bold hemlock]

Shouldn't it be a sum of 1/ln(ln(n)(n^n+2))?

I mean

n^n+2 because the form of the number

and ln(n) because prime number theorem

?

[high goblet]

log(n) for probability of n being prime, log(n^n+2) for probability of n^n + 2 being prime, assume the events are indepedent & multiply them together

ultimately it doesn't really matter too much cus this is just a heuristic that says there's probably only finitely many of them

[bold hemlock]

true

Also we should sum from like n=10000

Or something

I mean it was checked for first 3000 primes

Ye wolfram is not enough

LOL

But we probably can something that "statisticly around 0,01 of a number is prime" lol

I mean it probably is lower than that

But like what's the point of heuristic arguments like that?

We know that n^n+2 is not random

LOL

[radiant jasper]

im in grade 9

yall

[high goblet]

ad hoc heuristic arguments are why mathematicians believe i.e. there are infinitely many mersenne primes but not infinitely many fermat primes

[radiant jasper]

pleasure to meet you mathematicians

[bold hemlock]

K

Cool

[still geode]

ok

[acoustic nova]

is that (r^r) + 2

[ornate blade]

the vast majority of us are students or other people interested in maths

and not professional researchers

[radiant jasper]

i love maths

but im buns

clazz is always boring bc its js stuff i know

so ion like class

tbh

the teacher gives us friggin packets

its unbearable

[ornate blade]

that sucks

maths can be fun

[radiant jasper]

ya its so buns

i love math but whatever we're learning in class sucks

[iron terrace]

what are you learning

[radiant jasper]

we are doing scale factor

it sucks

i hate it

[trim vortex]

does anyone have resources on how I can prepare for the AMC12

[radiant jasper]

the actual calculation is nice, the concept it easy to grasp. its just the explaining

[iron terrace]

oh like dilations and stuff?

[radiant jasper]

im in ninth grade 😭

[iron terrace]

why are you still learnign that in 9th grade

that's like 7th grade stuff

no offense ofc

[radiant jasper]

i dont know, the curriculum here in canada is different

winnipeg to be specific

i do logs

💀

im joking

but i attempted to learn logarithms, i was trying some exponential equations that utilise in natural logs and whatever

yeah i know im annoying, math js excites me

[sleek ravine]

the mistake here is thinking people learn things in a specific grade

[sleek ravine]

ur not (:>

[iron terrace]

i wasn't trying to be mean

[ornate blade]

middle school topics vary a lot in terms of timing

people here come from all over the world yeah

[ornate blade]

they might mean this by 'scale factor'

[radiant jasper]

yes we use, SF = diagram / actual

[ornate blade]

ah true so on a map

then are you asked to find the real-life area of something on the map?

[radiant jasper]

yes

[ornate blade]

damn okay you're doing IGCSE confirmed

[radiant jasper]

i dont know what that is lolzers

[ornate blade]

oh interesting

that's what 9th and 10th graders study in the UK + many international schools

[ornate blade]

yes so the very important thing is (area ratio) = (length ratio)^2

[radiant jasper]

yez

[ornate blade]

also like keep the ratio order the same, so large/small for both, or small/large for both

yeah

[ornate blade]

this is actually related

(volume ratio) = (length ratio)^3

like you know a cube has three lengths being multiplied right

length * width * height

so all 3 get multiplied or divided by the length ratio

that's why it's ^3

[radiant jasper]

indeed indeed

cubic

[ornate blade]

nice

[ornate blade]

hey so if you want extra practice material by any chance

for your maths in school

check out: https://mrmannmaths.wordpress.com/home/igcse-extended-maths-0580/igcse-extended-revision-by-topic/

[radiant jasper]

hell yah

ty

LMAOOO

we arent learning this

its too advanced for the ppl in my classroom

[ornate blade]

https://rocketrevise.com/igcse-cambridge/cie-igcse-maths/

[ornate blade]

yeah it's definitely challenging for this grade level

think you might enjoy it though if you have any free time

[radiant jasper]

yes lolzz, i am looking forward towards it

this one girl in my class is doing piece-wise functions

[ornate blade]

wow

yeah that's like precalc

[radiant jasper]

yes

i do pre cal

early

im taking IB

[ornate blade]

all of this maths you learn in school comes really handy for contest maths also

if you don't know a broad range of topics for your year level

you won't even get half the questions right

[radiant jasper]

so trueeee

[ornate blade]

to be fair competition problems are hard in their own way

but people ask how do I get better at the AMC? what should I study?

a lot of maths is freely accessible online

[radiant jasper]

whats AMC?

[ornate blade]

that's why I point students towards the UK curriculum

IGCSE then A level mathematics

[ornate blade]

it's the American one

[radiant jasper]

ohhh

[ornate blade]

first round of olympiads

[radiant jasper]

hey, i think you should give me a hard math problem fs

[ornate blade]

enjoy suffering

[radiant jasper]

what continent are you in?

[ornate blade]

Oceania

[maiden panther]

if you are excited about math i think you will enjoy AMC

[radiant jasper]

thats so far 😭

[ornate blade]

🇦🇺 ✅

🇳🇿 ❌ btw

[radiant jasper]

its like australaisa i believe i learnt that

[ornate blade]

yeah but that's outdated

we should be together with all the Pacific islands

only makes sense

[radiant jasper]

yes

[ornate blade]

Aus + NZ + all these islands = Oceania

then that makes us a real continent

a real continent hidden under the sea for large parts

(Zealandia is the 8th largest continent, cause a lot got sunk over and only New Zealand and a few other islands remain)

[radiant jasper]

ohhhhhh

i had this friend

who moved to asia

he was talking about the continents but he didnt include oceania 😭

thats js amazing

[radiant jasper]

i am so determined to solve ts

i am setting up my parameters

the center of each smaller circle lies on the circumference of the larger circle

this looks like that one maths problem i did

but its different bc it had

this means that the distance from the center of the large

r

circle to the

center of any smalelr circle is simply gna be R (radius of the larger circle)

and each smaller circle has a radius of r (each smaller circle)

do i need to

oh wait nvm

bro how do i get the relationship between R and r 😭

@ornate blade sorry for pimg

[radiant jasper]

ya man i give up im chopped

whats the answer

can we play hot or cold

is the answer between 5 and 9

wait

idk

😭

R = r + r(sqrt(2)/2)

R^2 = r^2(1 + sqrt(2)/2)^2

squaring sides

ya i aint doing ts

fck it

[topaz mist]

I want to know the answer too now 😭

What's weird is that there are 4 small circles that are useless

[bold hemlock]

Looks like fun

[pearl holly]

Sup

How do you guys increase your stamina in solving problems?

I just solved a really hard problem and my stamina got drained

It was this problem in NT/Prealg thata drained it haha

[pallid tundra]

ah an alcumus question

not often that we get those

that sounds odd, are you sure you didn't just level up or smth

and/or are you possibly confusing this for your "accuracy" stat

[pearl holly]

What do you mean by this?
Yes, but those type of problem I get are in the hard difficulty that's why its a bit challenging

[hushed charm]

why specifically uk?

is it harder or something like that then other curriculums?

[quick frost]

Let radius = r. ||Radius of smaller circle is cr where c = 1/sqrt(2). Area of intersection = area of two sectors - area of two triangles. Angles subtended at the centres = 4 arcsin(c/2), 2 arccos(c/2) = 4 t, pi - 2t, say. So area of intersection = r^2/2 (4 t + (pi-2t) c^2) - r^2 sin(2t) = r^2 (2t + (pi/4-t/2) - 2 sqrt(7)/8) = (3t/2+ pi/4 - sqrt(7)/4) r^2. So pi r^2 - 4(that) = (pi - 6t - pi + sqrt(7)) = (sqrt(7) - 6t) r^2 = 48pi||. I give up (and this is probably terribly wrong too).

[radiant jasper]

yea, i gave up pretty quick

[spark minnow]

If anyone has any ways on how to self study calculus please let me know thanks

[pallid tundra]

wrong channel, and why haven’t you looked for any yourself first?

!da2a

[scarlet root]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[deft wraith]

You need very low technology to do well, especially compared to say AMC/AIME/USAMO

[hushed charm]

wdym low technology

[deft wraith]

Doesn’t need lots of theory to solve the questions

[hushed charm]

sure

[chrome parrot]

Competition math, huh? Find every x:
√x + x = x^2

Which numbers solve this equation?
x^3 - 2x = x

[primal cosmos]

guys im 8th grader what should i use to study for amc 12 i can only get 40

[pallid tundra]

far too basic to be a competition problem

there’s no depth or sophistication needed to solve this at all

[maiden panther]

18

[radiant jasper]

did u check it 😭

gauth ai vs poe.com vs chatgpt

vs flying_fly

[maiden panther]

who do you trust more

[radiant jasper]

idk humans are better

lol 😭

ai is under dev

so its hard to deicde with 4 diff answers

[hushed charm]

how did u find this asnwer

[maiden panther]

it's $\sqrt{\frac{48\pi}{\sqrt7 - 6\sin^{-1}\left(\frac{\sqrt2}{4}\right)}}$

[gilded halo]

flying_fly

[maiden panther]

more or less what this guy said

[pallid tundra]

!nogpt

[scarlet root]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[pallid tundra]

if you need an online calculator to check your work just fucking use wolframalpha or smth not this ai slop nonsense 😭😭

[vestal grove]

is anyone here

[ornate saffron]

yes

no

[ornate blade]

solid foundation

[ornate blade]

not necessarily, but it's challenging in a way that is also accessible
also during a year you'd study across algebra and functions, geometry, trig and vectors, stats, and calculus in later year levels

it's probably the best organised curriculum among all Anglophone countries
well not many people know about the Irish Leaving Certificate, or New Zealand's NCEA which are also solid

those may be a bit harder to find materials that are not provided by the exam board

[pearl holly]

If I'll be honest, I don't even use WolframAlpha since I do not understand how it works. I often use ai to help me understand mathy-stuff and when it comes to the answer/solving part, I just let it explain the solution to me :PPP

[pallid tundra]

you're doing yourself a MASSIVE disservice.

[pearl holly]

Let me clarify

when it comes to the answer/solving part, I just let it explain the solution to me

It's where I either got a correct/incorrect ans to a problem

[pallid tundra]

it doesn't actually know what it's doing

it's just "hmm what do i think is the next word" :clueless:

[pearl holly]

At this point, I don't even know what PhD people do with o1 to help them with these sort of stuff

Can we come to an agreement that everybody has their own form of preferences?

Whether ai or not haha

[ornate blade]

you're very mistaken if you think that any PhD students use AI to explain the maths to them

they consult books and papers!
they ask their fellow students, or arrange to meet with their supervisor!

they always have sources based on real experience and knowledge, and not something generated by a predictive text algorithm which can't tell you where it got that exact info from
(AI is famous for hallucinating and generating made-up references)

[pearl holly]

Ahh, I see

Then its best to stick with learning from solution given problems

[ornate blade]

exactly

there's always real people and materials that real people have written

[pallid tundra]

gpt is HILARIOUSLY bad at anything even slightly beyond the surface level

[ornate blade]

all my profs and lecturers said that GPT sucks balls at writing essays

it doesn't really know how to answer a question and tailor its knowledge, or to write an extended response

it just rambles about the general ideas that the question touches upon

which is why AI search is good if you want to get a surface level understanding, a summary

but it sucks for anything any deeper

[swift imp]

Just ask this discord server for help is better than ai

[pearl holly]

Alright, thanks!

Alright chat, I need clarifications

Let p = prime

What is the largest prime factor of p!

Would it be p itself?

Oh, nvm, I have made the correct assumption

[pallid tundra]

yea lol

[acoustic nova]

bruh

[hushed charm]

Besides that. gpt does not have any kind of creativity or problem-solving skills, he is just going always to opt for solutions way harder than should be and brute forcing operations with massive numbers

Like this question, I remember asking for chatgpt how to solve this problem and he always tried to compute 2024³, dividing stuff by 2024³

Let x and y be real numbers such that x + y = 2024 and xy = 8. If

1/x³ + 1/y³ = p/q,

where p and q are positive integers that are coprime, what is the remainder of p - q when divided by 5?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

[summer roost]

I remember asking chatgpt a simple inequality question

And it gave me some calculus solution

Which I cannot even verify

Here is the question

Prove that

[gilded halo]

Triaengle

[summer roost]

o3 is really good at closed ended questions but not in open-ended and proofs I tried several competition proving problems and it gave the wrong answer for most questions

[sleek ivy]

its 1 right?

[obsidian herald]

!showwork

[scarlet root]

Show your work, and if possible, explain where you are stuck.

[obsidian herald]

hint: ||use binomial expansion, then manipulate the terms a bit to get the term in the expansion of another binomial||

[summer roost]

I tried that but I am a bit lost

I need to find where I solved a question, I will open a help channel

[tough patrol]

could we try induction?

[obsidian herald]

idk but i'm sure that that won't be simpler than a direct proof

[hushed charm]

Actually it is 4

[sleek ivy]

Huh

Prolly cuz i was too lazy to check if the fraction was in simplest form

[umbral dock]

Idk I need help

[tepid surge]

There's a cool way to solve this due to polynomial symmetry

we know that $\frac{1}{x^3} + \frac{1}{y^3} = \frac{x^3 + y^3}{(xy)^3}$

[gilded halo]

hiidostuff

[tepid surge]

well maybe youve already solved it

idk

then we have that $x^3 + y^3 = (x + y)^3 - 3xy^2 - 3x^2y = (x + y)^3 - 3xy(x + y)$

[gilded halo]

hiidostuff

[hushed charm]

Yeah, that is how I solved it

[tepid surge]

nice

[hushed charm]

Is this called polynomial symmetry?

[tepid surge]

well

the type of polynomial tricks youre using is due to the fact that everything expands to a symmetric polynomial

a symmetric polynomial is a polynomial of x_1, x_2, ... x_n that remains the same no matter how you permute the variables

so for example, x^3 + y^3 = y^3 + x^3

turns out that every symmetric polynomial of x,y can be expressed in terms of simpler elements of x and y

that being (x + y) and xy

[hushed charm]

awesome

[iron terrace]

fr

[pallid tundra]

shameless self promotion time

https://drive.google.com/file/d/1f_QI_U3iip--oyRnggvLf-Wt2W9yvquL/view problems from a contest we just hosted at usc

[soft vigil]

im getting A

is that right

[soft vigil]

or just reverse it

1/x+1/y=2024/8 and 1/xy=1/8

[hushed charm]

I checked again, the correct answer is indeed E

I can show if you want

[hollow rose]

Anyone here done the euclid?

[tepid surge]

its 4

each term in the numerator is divisible by 8 so p/q is an integer

then (2024)^3 / 8 = (1012)^3 which is 3 mod 5

then 3(8)(2024)/8 = 3(2024) which is congruent to 2 mod 5

wait i mightve screwed up

[maiden panther]

it's 129553457/8

[tepid surge]

no wait it is A

[soft vigil]

Im getting A

[maiden panther]

how?

[tepid surge]

the numerator is 1 mod 5

and p/q is an integer

therefore its A

[maiden panther]

bruh

[soft vigil]

Hmm

I get denominator is 3 mod 5

Specifically it is 8

[tepid surge]

thats irrelevant

[soft vigil]

What

Is it not p-q

[tepid surge]

i should say, when you reduce the rational expression, the numerator is 1 mod 5

[soft vigil]

Huh

I get that p/q=11232021/8

Numerator 3 mod 5

Denominator 3 mod 5

[tepid surge]

nah

it cant be odd on top

[soft vigil]

Why

[tepid surge]

you should get $\frac{(2024)^3 - 3(8)(2024)}{8}$

[gilded halo]

hiidostuff

[soft vigil]

Wait a sec

I might be wrong

[tepid surge]

you are

the numerator must be even

[soft vigil]

Oops yeag

Accidentally squared

[tepid surge]

its ok tho

from here the problem is pretty easily solvable

firstly, reduce the expression

[soft vigil]

Yes of course

[tepid surge]

to get $\frac{(1012)^3 - 3(2024)}{1}$

[gilded halo]

hiidostuff

[soft vigil]

The problem itself is not hard

[tepid surge]

then 1012^3 is 3 mod 5

3(2024) is 2 mod 5

so 1 mod 5 on top of 1 mod 5

0 mod 5

[maiden panther]

no

divide by 512

[chrome parrot]

Which numbers work for
2^n - 3 = prime number?
(finite answer)

[pearl holly]

Maybe im wrong, my attempt is just $n > 2$ where n is prime

[gilded halo]

Shawn

[pearl holly]

frick my punctuation is bad

[chrome parrot]

Try so, there is a finite amount of numbers

[pearl holly]

What are we looking here for? What do you mean by finite amount of numbers?

Is there some kind of constraint?

like 3 < n < 9999?

[chrome parrot]

There are several numbers which, when used, you get a prime number

1 < n < 100

In that area

[pearl holly]

Oh, then, its just {3,5}

Oh, wait

nv,

I just brute forced it and disregarded the prime numbers for n

My bad

[pearl holly]

Since if $2^n - 3 = prime$, then we check the numbers that're less than 100 for $2^n$.
Simple brute force should do the trick.

n=2:
$2^2 - 3 = 1$ (Not prime)

n=3:
$2^3 - 3 = 5$

n=4:
$2^4 - 3 = 13$

n=5:
$2^5 - 3 = 29$

n=6:
$2^6 - 3 = 61$

Therefore, the numbers for n are: {3,4,5,6}

[gilded halo]

Shawn

[pearl holly]

Why did my brain think that the bold text thing whould work for TeXit

[chrome parrot]

Not only

[pearl holly]

What do you mean?

[chrome parrot]

I mean these are not all the numbers to make a prime number

[hushed charm]

$\frac{1}{x^3} + \frac{1}{y^3} = \frac{p}{q} \Rightarrow \frac{x^3 + y^3}{(xy)^3} = \frac{p}{q} \Rightarrow \frac{(x + y)^3 - 3xy(x + y)}{512} = \frac{p}{q} \Rightarrow \frac{2024^3 - 24 \cdot 2024}{512} = \frac{p}{q} \Rightarrow \frac{2024(2024^2 - 24)}{2^9} = \frac{p}{q} \Rightarrow \frac{253(2^6 \cdot 253^2 - 8 \cdot 3)}{2^6} = \frac{p}{q} \Rightarrow \frac{253 \cdot 2^3(2^3 \cdot 253^2 - 3)}{2^6} = \frac{p}{q} \Rightarrow \frac{253(2^3 \cdot 253^2 - 3)}{2^3} = \frac{p}{q}$

[gilded halo]

samuel

[hushed charm]

$p = 253(2^3 \cdot 253 - 3) \quad \text{and} \quad q = 8 \Rightarrow 253(2^3 \cdot 253 - 3) \bmod 5 = 3(3 \cdot 4 - 3) = 2 \quad \text{and} \quad 8 \bmod 5 = 3 \Rightarrow 2 - 3 = -1 \Rightarrow -1 \bmod 5 = 4$

[gilded halo]

samuel

[wise cape]

$p = 253(2^3 \cdot 253 - 3) \quad \text{and} \quad q = 8 \Rightarrow 253(2^3 \cdot 253 - 3) \bmod 5 = 3(3 \cdot 4 - 3) = 2 \quad \text{and} \quad 8 \bmod 5 = 3 \Rightarrow 2 - 3 = -1 \Rightarrow -1 \bmod 5 = 4$

[gilded halo]

yangbegeflex

$p = 253(2^3 \cdot 253 - 3) \quad \text{and} \quad q = 8 \Rightarrow 253(2^3 \cdot 253 - 3) \bmod 5 = 3(3 \cdot 4 - 3) = 2 \quad \text{and} \quad 8 \bmod 5 = 3 \Rightarrow 2 - 3 = -1 \Rightarrow -1 \bmod 5 = 4$

[wise cape]

$x^3 + y^3 = (x + y)^3 - 3xy^2 - 3x^2y = (x + y)^3 - 3xy(x + y)$

[gilded halo]

yangbegeflex

[wise cape]

dude this cool

[gilded halo]

loremz

loremz

loremz

[keen ingot]

nice

[obsidian herald]

this is a channel for discussing contest math

• for linear programming problems, consider #linear-algebra or #optimization
• for testing LaTeX code, consider #bots or #latex-testing

[obsidian herald]

if you need help in proving the inequality that you've proved, then you need to show your work

otherwise, you may try approach0 to search that

https://approach0.xyz/

[tulip coral]

I want to prepare for Math competitions

But I don’t have a routine

Any recommendations?

For Summer.

[soft vigil]

aops books to beginwith

[tulip coral]

Aops books?

Any online resource?

For free?

[soft vigil]

idk

aops books r always good

[tulip coral]

I’ll just do it through YouTube

I have a study strategy that might help

Active recall strategy

[soft vigil]

i know aops has those free like

mathcounts minis

[tulip coral]

Amc?

Online practice or YouTube

[soft vigil]

idk

theres lots

theres mathdas

amctrivial

[tulip coral]

Ok

[tulip coral]

Im a Sophomore

[soft vigil]

juust start one and stick to it

[tulip coral]

In algebra 2

Im taking summer classes for calc

[soft vigil]

shouldnt you learn precalc first 💀

[tulip coral]

Yeah both

[soft vigil]

alsocalc does not matter for comp math

[tulip coral]

That’s what I mean

[tulip coral]

What matters?

[soft vigil]

comp math matters ig

geo is important

[tulip coral]

Its just to build foundational knowledge to recap on basic skills

[soft vigil]

its rly different than school math

[tulip coral]

I think it’s helpful

[soft vigil]

pickup like aops volume 1

[safe oriole]

are amc questions good for other olympiads

[soft vigil]

im not condoning this but you may or may not be able to find the books free online

[soft vigil]

ehh

[safe oriole]

like which amc questions should i start with

[soft vigil]

amc is good if you are in the USA

[tulip coral]

Use YouTube

See how others practice

[safe oriole]

ok

[tulip coral]

Understand the topics

Than solve

It on your own

[soft vigil]

yes thats the strat

for basically everything in math

[tulip coral]

Yeah

[safe oriole]

thanks

[warm mantle]

Yo can anyone help with this problem? Let ABC be a triangle in which we have drawn the median BE from vertex B, and a segment that goes from C to a point D on side AB. This point D can slide along side AB.

The intersection point of both segments is point F.

We call p the ratio between the segments into which D divides side AB, q the ratio between the segments into which F divides the median BE, and r the ratio between the segments into which F divides CD.

What will be the ratio between triangle CEF and triangle FDB for any value of p?

I'm pretty sure it's pr/2 but I don't know how to prove it?

[tulip coral]

Picture?

[tulip coral]

….

I need the question

[tulip coral]

There

Ok

I honestly don’t know where to start.

Wait

So

[warm mantle]

The only thing I know from previous parts of the exercise is that r = p+1

[tulip coral]

I know we’re told that

CF/FD IS 7

So

If the triangles

Share the same height, the ratios of their areas is the ratio of their bases.

That being said

Yeah

I might need more info

Are you sure this is the question?

[warm mantle]

Yes, it is. Maybe you have to do it through Menelau's theorem, the other parts of the exercise were done with that

[tulip coral]

Ok

[warm mantle]

Also, BF × AD = 2 × FE × BD if that helps

[tulip coral]

Yeah thanks

Im reading the theorem to refresh memory

I just need heights and

Other ratios

Knowing AD/DC doesn’t help me determine the other segments

@warm mantle

[warm mantle]

So the problem doesn't say anything about the heights or the other ratios, but maybe you can find an auxiliary point parallel to AE in the FDB triangle and then say CEF is congruent to FDX and maybe something with that afterwards idk

[tulip coral]

Never

Mind

I see

What I need to do

[warm mantle]

Ok

[tulip coral]

@warm mantle

Is it 126

Tell me where you found the problem

[warm mantle]

It's a website in Spanish

It doesn't say the answer

I dont have the link but I can send screenshots

It's the 9th question

[tulip coral]

Oh

[pallid tundra]

mass points

[obsidian herald]

for this type of geometry problems involving ratio of segments and areas, you may want to use the area method to make your solution organized

.

[ornate blade]

<@&268886789983436800> here too

[gilded halo]

hx62d

[hushed charm]

<@&268886789983436800>

[radiant jasper]

who does math and code at the same time in lua

hopefully, sending files like these isnt a violation. its like vector mathematics in lua, its a simple module and i plan on integrating quaternions

if its a violation then the mods can delete this

[sullen tapir]

using roblox as a refence lol

[hushed charm]

It is not

But here isn't really the perfect channel to send that, focus on olympic maths here

#1213610885277421588 Would be a better channel

[radiant jasper]

lollll

[radiant jasper]

I made it for a competition lollz

And I lost 💔

[obsidian herald]

I revisited an old post: #1190382943663816756 message
Actually, L3435 showed a simple proof by rearrangement inequality in AoPS: https://artofproblemsolving.com/community/c6h1712301p11047177
The strategy is to start by homogenization of the given inequality using the constraint.
Inequality's LHS degree = -1
Inequality's RHS degree = 1
so use the "degree 2 = degree 0" equality constraint to raise inequality's LHS degree by 2 (by multiplying by "1")

[balmy spoke]

theta real -2pi < theta <2pi

(please tag me)

[scarlet root]

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

[balmy spoke]

1

[ivory ember]

ok idk what the point of this question is

but as a hint, consider $(\cos \theta+\sin \theta)^2$

[gilded halo]

Civil Service Pigeon

[balmy spoke]

$(\sin\theta+3\cos\theta)^2=0=\sin^2\theta+6\sin\theta\cos\theta+9\cos^2\theta$

[gilded halo]

lumi

[balmy spoke]

,tex \sin^2\theta+9\cos^2\theta&=-3\sin2\theta \ (\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta+8\cos^2\theta&=-3\sin2\theta \ (\sin\theta+\cos\theta)^2&=-3\sin2\theta+\sin2\theta-8\cos^2\theta \ \therefore\sin\theta+\cos\theta&=\sqrt{-2(\sin2\theta+4\cos^2\theta)}

[gilded halo]

lumi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[balmy spoke]

can i say this?

Recall that $\theta$ is a real number, then $\sin\theta + \cos\theta \in \mathbb{R}$. By contradiction, if $\sin2\theta+4\cos^2\theta$ is non-negative, then $\sin\theta + \cos\theta$ is part of the complex plane, which is not possible. Hence, \underline{\underline{$\sin2\theta+4\cos^2\theta\leq0$}}.

[gilded halo]

lumi

[ivory ember]

I was intending for you to consider $(\sin \theta +\cos \theta)^2$ in general

[gilded halo]

Civil Service Pigeon

[ivory ember]

This is fine though

[ivory ember]

this isn't my favorite wording

[balmy spoke]

):

how would you word it

[ivory ember]

Since $\theta$ is real, $\sin \theta+\cos \theta$ must also be real. Hence, $\sqrt{-2(\sin 2\theta+4\cos^2 \theta)}$ must also be real, which implies that $$-2(\sin 2\theta+4 \cos^2 \theta) \geq 0$$ $$\sin 2 \theta+4\cos^2 \theta \leq 0$$

[gilded halo]

Civil Service Pigeon

[ivory ember]

cause the part about "is part of the complex plane" is a bit iffy

just say it's non real if you're gonna say that

[ivory ember]

the other issue is that you never excluded the possibility of the $\sin 2 \theta+4 \cos^2 \theta$ being zero

[gilded halo]

Civil Service Pigeon

[balmy spoke]

hahaha

[maiden panther]

maybe i'm missing something but what is the point of the expression at the top

and what is part ii even saying

[ivory ember]

,w sin 2x + 4cos^2 x, x = pi

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

,w sin x + 3 cos x, x = pi

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

take that as you will

[ivory ember]

There is some $\alpha \geq 0$ such that $$\cos^2 \theta-\sin^2 \theta=4 \alpha \cos \theta$$

[gilded halo]

Civil Service Pigeon

[ivory ember]

Anyway r u good now

[balmy spoke]

yep thank you pigeon

[ivory ember]

[balmy spoke]

i think i remember you

from like 2 years ago

got to national and scored a zero hahaha

im proud of it tho

[ivory ember]

Although $\frac{\cos^2 \theta-\sin^2 \theta}{4 \cos \theta}$ isn't always non-negative

[gilded halo]

Civil Service Pigeon

[ivory ember]

so maybe they just want you to find the value of this in the case where it's non negative

and they're calling that alpha

not my favorite wording

in which case it would be ||sqrt(2/5)||

[maiden panther]

i may be missing something, but doesn't $\sin\theta + 3\cos\theta = 0$ already imply $\cos^2\theta = \frac{1}{10}$?

[gilded halo]

flying_fly

[ivory ember]

well this is awkward...

[ivory ember]

Yeah you could say that

there's a few ways that you can do this

but I doubt that they intended for you to just bash out the value of the expression explicitly

[ivory ember]

which is why I kinda like this

[balmy spoke]

👉 👈

[maiden panther]

it's kind of silly imo but 🤷

[ivory ember]

I did say this lol

[balmy spoke]

you got plenty of time ❤️

[maiden panther]

like, why is there a 4 in part ii if it's just saying the ratio is nonnegative

[balmy spoke]

what are you aiming to join next

[ivory ember]

yeah that too lol

it's giving the sat/jee just adding extra stuff for no reason

[maiden panther]

fr

[sand river]

guys I solved an olympiad number theory problem

🤯

[tulip coral]

😱

[soft vigil]

this be looking familiar

[balmy spoke]

slayyy

[glossy pawn]

How do you solve it

Dm the solution

[keen pelican]

dm solution too it looks intresting

[radiant jasper]

💀

[sand river]

ok

[hushed charm]

Guys

What are your opinions on khan academy vs aops books? I am currently reading introduction to algebra by aops, I really like it but some people prefer khan academy, what do you think?

[pallid tundra]

apples to oranges

my opinion of KA has considerably degraded over the years

[hushed charm]

why?

[pallid tundra]

its coverage of the material is far too surface level

it's like if you wanted to sparknotes middle and high school math without any of the underlying rigor

• not enough exercises for a lot of the topics (especially for AP calc)

and their whole chatgpt enshittification

[hushed charm]

alright, that makes sense

[sand river]

aops books are great, but solving previous years competition problems and then learning what you dont know through other online resources is great too

but aops books are good for introduction

[hushed charm]

Yeah, I agree, that is what I am most focused on doing btw, but between these resources that you mentioned, I always liked aops books, there are tons of problems and some really nice explanations

What other online resources do you like?

[sand river]

Previous tests like amc, aime, usamo, bmo, tournament of towns, and others are good

youtube has many good courses that are useful and free

some previous olympiad medalists or coaches have websites

[hushed charm]

Indeed, organic chemistry channel has saved me a lot

yeah

the only one I know is evan chen's

[sand river]

https://youtube.com/@shefsofproblemsolving?si=tsQHxjBwjla2ygxw

this guy has a really great channel

[hushed charm]

Interesting, going to check his videos later

Are you studying for math olympiads either?

[pallid tundra]

organic chemistry tutor is also far too surface level to actually help at all for contests

[tulip coral]

Anyone knows a good site to review limits?

[sand river]

yes

[sand river]

probably if you can find the stewart single variable calculus or precalc book online somewhere

[tulip coral]

You’re studying for Olympiad math?

[sand river]

yes

[tulip coral]

Nice, how’s it going?

[sand river]

good

[tulip coral]

Elaborate.

I want information

On another person’s experience.

With Olympiad

[sand river]

I have been practicing previous years

and so far it's going well

Im able to solve much more problems than I used to be able to

especially with things like geometry which used to be by far my worst subject

[tulip coral]

Like how bad.

Hopefully you suceed.

[sleek ivy]

ong fuck geometry

[hushed charm]

Indeed depending only on his channel might not help at all

[pallid tundra]

his channel is like

remedial math

not the kind you need for contests

[pallid tundra]

the math you’d need to pass an intro level undergrad class or a typical high school class

neither are particularly deep or difficult

[hushed charm]

But like, his short videos can help a little with some introductions to new subjects and remember some other subjects

Usually when I forger something kinda easy, his videos helped me

but indeed, he doesn't show a lot of problems

[hushed charm]

Do you like any other sources besides these we already talked about?

[pallid tundra]

the problem is not the quantity

it's the "quality"

he only covers the most basic trivial plug and chug examples

if you need it to pass your standard algebra 2 math test it's probably fine

if you need to do anything slightly more substantial

hell no

[hushed charm]

Indeed studying only with his channel isn't a good idea

[hushed charm]

What about u

[sleek ivy]

is it just me or is aops website the slowest thing ever

[sand river]

I used to not be able to solve any competition geometry problems, not even amc level

[tulip coral]

😬

I was in a similar situation.

[spark gorge]

is anyone here participating in ARML local this year? i have some clarification questions about the event

[tulip coral]

Hm

[hazy reef]

im in 8th grade and doing some comp math and soon one of my math classes will end

any suggestions for another math curriculum or class u would recco

i alr do aops

and some other classes

[wooden ice]

Guys i need help understanding this

i already know the answer

it's kangaroo math 2024

[hazy reef]

can u translate to english

[ionic smelt]

It's D i think

if I understood the question, i'm honestly bad at french

[wooden ice]

yup that is correct but how did you get it knowing that they both have opposite directions and we dont even know the pace in which they make a turn, since its freely

[ionic smelt]

I think it can be inferred that the hexagons are rigid and they don't swap positions randomly, so the distance from C to A is 2 counterclockwise, so you move F counterclockwise twice, and you get from F - E - D

[wooden ice]

yes sure
Two hexagons with the same center, one with letters and the other with numbers, can rotate freely around their center. In the figure, the letter F is next to 6. Which letter will be next to 1 when the F is next to 3?

[ionic smelt]

Thus:
3 - 2 - 1
F - E - D

1 corresponds to D

[hazy reef]

you dont need to account for pace bcz the angles r same and 1 is 2 counterclockwise from 3

so just find what is 2 counterclockwise to F

[wooden ice]

1 is 2 counterclockwise?

[hazy reef]

directions of turns dont matter

[hazy reef]

mb 2 spots away from 3

[wooden ice]

yeah

ah

but how did u think that it should be counterclockwise

or either

[ionic smelt]

because to move from 3 to 1 you need to rotate counterclockwise

[wooden ice]

yeah

[ionic smelt]

Imagine a clock with only 6 hours

[wooden ice]

yes i get it

[ionic smelt]

alr

[wooden ice]

but if they turn at the same time its not possible right

[wooden ice]

i see i see

saf i got it thx

thanks!

What about this one guys

[ionic smelt]

ok now i don't understand lmao

please translate it

[wooden ice]

lol

[ionic smelt]

my french is horrible

[wooden ice]

gpt 4 doesnt work for me idk where i should translate it

cuz i already used it

[ionic smelt]

use google translate

[wooden ice]

can u use it? like send it the pic and tell it to translate it for u

how? with pic?

i cant copy paste

its kangaroo

olympiad

[ionic smelt]

oh it's pic

rip

[wooden ice]

lol

u have gpt

[ionic smelt]

i don't pay

[wooden ice]

me neither

but its free

[ionic smelt]

oh then i'll try

[wooden ice]

for limited time

ya

[ionic smelt]

ok

Six packages are placed on the platform of a truck. Manu unloads them to place them on the ground. He takes one package at a time and only if there isn’t another package on top of it. He places it either on the ground or stacks it on top of those already placed, and he won’t move it again.

Which of these stackings could not be obtained by Manu?

hmm... lemme think

hell nah D is sus as hell

[wooden ice]

yay

[ionic smelt]

because 2 is stacked below 4

which is impossible unless you are a magician

[wooden ice]

whaa

really

how

[ionic smelt]

yea it's D

as you see, 4 can't be stacked on top of 2

[wooden ice]

wait only now i got it

i didnt know that the pic at the right was in the camion

lol

[ionic smelt]

yea

[wooden ice]

he took em from here

[ionic smelt]

wait you arent french right?

[wooden ice]

ok yea makes sense

[wooden ice]

im not

[ionic smelt]

lmao

[wooden ice]

why

[ionic smelt]

because neither of us were understanding the question at first instance

lol

[wooden ice]

no but i understand french really well

[ionic smelt]

ohh

[wooden ice]

im on this question now

how do we calculate the perimiter

or idk how to spell it

peremiter

?

of the gray part

[ionic smelt]

well first of all i think it says that the white parts are square

if i understood it correctly

[wooden ice]

yes

benjamin cut those white squares

2 3 6 and 1 cm

ok i see that its 36 cm perhaps

[ionic smelt]

wait lemme see

[wooden ice]

cuz 40 is of the whole square

and smaller than 40 is only 36

the only option

[ionic smelt]

yea but it's easily calculable

[wooden ice]

which doesnt make sense?

yea go on

[ionic smelt]

in the perimeter you have at least 3+3 (2 sides of the 3x3 square), 6+6 (2 sides of the 6x6 square), 1+1 (yea u understand it) and 2+2

[wooden ice]

yes

[ionic smelt]

and then you have 10-3-2=5 on top, 10-3-6=1 on right, 10-6-1=3 bottom and 10-2-1=7 left

[wooden ice]

not times* 4?

[ionic smelt]

no because 2 of the sides of the squares aren't in the perimeter

[wooden ice]

oh?

[ionic smelt]

so you have 6+12+2+4+5+1+3+7 which gives you 36

[wooden ice]

niice

[soft vigil]

you do know that these are available en anglais

[sand river]

here I have another problem today:

[soft vigil]

why is this so familiar

[spark gorge]

for the ARML Local, are teams supposed to find the nearest school that is competing in or is every school team supposed to host the competition at their school?

[soft vigil]

if there is no nearby site then just get like a parent to host it

thats what happened in chicago at least\

[wooden ice]

yes but i have a practice olympiad today at school and its in french not english
Like im good in french dw, i speak and write french better than some of my teachers, and we study math, physics, biology in french
nothing to worry about

Ok guys they said 3 identical cubes are on a table, what is the sum of the 3 numbers written on the faces that we put on the table.

is it a b c d or e

[summer roost]

A general method might be to find the opposite faces

That might be slow but works all the time

Which competition is it ??

[ionic smelt]

13+22+8

by the second cube you see that beneath 8 is 13, by the third cube you see that beneath 13 is 22 and by the third cube you see that beneath 22 is 8

[ionic smelt]

it's C

[karmic rune]

unfold the cubes, somewhat like this and check which numbers are adjacent and which are opposite. If you do it properly, you can solve the problem easily