competition-math

[dusty fable]

ur not wrong however if i generalised the question to can u think of an interval of N where none are prime, u cant assume N! +1 is nonprime as u cannot factor out the 1 to show that it is a factor

[torpid cairn]

u can also prove that it's composite with wilson's theorme

[limpid folio]

can someone suggest me some free courses (articles and videos) on olympiad polynomial (with graph ) and inequalities

[turbid lodge]

guys

[severe eagle]

Welcome to mathcord!

[turbid lodge]

i said help not welcome me

[severe eagle]

so i cannot welcome people?

[turbid lodge]

so can u help me in dis question

[severe eagle]

send lah

[severe eagle]

Really now?

[turbid lodge]

yea

[sweet pewter]

the most 0/10 troll ever

[severe eagle]

lmao

this aint comp math

[sweet pewter]

look at their pronouns tho

[severe eagle]

What have you tried?

[sweet pewter]

bruh they left

[severe eagle]

smh

[candid fern]

Check little fermat yt channel inequality playlist
It has the main olympiad ineq (am-gm, Muirhead, cauchy, holder, rearrangement,... )

[limpid folio]

k thanks

[dusty fable]

try this i haven’t don’t it yet (i cba to work with such a big number lol)

i actually think it’s not too hard at all tho

[ornate blade]

,w prime factorise 446617991732222310

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[ornate blade]

,w divisors of 420

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[reef condor]

Bruh

Well ig yk they’re all 1 mod a divisor of 420

[dusty fable]

lol i did that it just looks so long

[radiant jasper]

ew

[dusty fable]

fr it was the oxford maths puzzle of the day

[radiant jasper]

The whole problem is just factoring that number

Then you're done

So cringe

[dusty fable]

proving 43,61,71,211,421 are factors isn’t easy

u can’t do that by factoring

u have to do induction i think

[radiant jasper]

I'm not saying the factoring part isn't easy

[dusty fable]

but what part of a factored form will show that 421 is a factor

obviously things with 6 etc are easy

coz it will give consec integers

but i dont see where u can pull out 421 or any of the large prime factors

hence i sent it here

[radiant jasper]

ok you have to do skibidi reasoning to factor it

[dusty fable]

just realised how long this would be

[torpid cairn]

send the solution here I'm curious now

[dusty fable]

i don’t have it

[torpid cairn]

I'm not even gonna attempt this tho

[torpid cairn]

once u get it I mean

[dusty fable]

haven’t started it just looked at it

[torpid cairn]

fair

[dusty fable]

i will give it a go

[torpid cairn]

I feel like it's ||Euler totient|| but I'm not sure how to apply ot

[radiant jasper]

Want to prove that if p|N then p|mn(m⁴²⁰-n⁴²⁰)
Rule out p|n,m and say we want p|N->p|n⁴²⁰-m⁴²⁰.
n⁴²⁰=m⁴²⁰ (p)
(n/m)⁴²⁰ for all n,m coprime with p.
Choose n/m=g primitive root mod p.

so ord(n/m)=p-1 | 420

[dusty fable]

i haven’t learnt this stuff yet lol

maybe i don’t know enough maths to solve it

i just saw prove x^7-x is a multiple of 42 and thought it would be similar

[torpid cairn]

oh yeah I've done that one

[dusty fable]

yeah it’s pretty straight forward i feel

ooo i saw a nice question yesterday which i enjoyed

not very difficult but nice

why the reaction?

[radiant jasper]

Because 119 and 539

[dusty fable]

it’s from BMO

[sweet pewter]

nah just use Fermat's little theorem once you've factored it

[radiant jasper]

The problem was factoring it not that

[dusty fable]

u should try it the numbers don’t get in the way

[radiant jasper]

Do the numbers simplify nicely when you ||multiply these together||

If they're small enough then it's easy to finish

[dusty fable]

i didn’t do that

i wouldn’t recommend that either the thing i did seemed to be exactly what it wanted coz everything became very simple

if u want i can post the initial step and im sure u will be able to do it easily

[radiant jasper]

dw i can do it without

Lol I just multiplied wrong in myhead

Like completely wrong

[dusty fable]

how r u doing it?

[sweet pewter]

had a nice idea

multiply the first one by 25 and the second one by 4

[torpid cairn]

ik how you could ||reduce it to 1 linear equation|| but I'm not even sure how to solve that lmao so this is probably above my level

[dusty fable]

very smart

exactly what i did (well i multiplies by 75,12 and divided by 3)

[sweet pewter]

ye it's a really nice problem

[dusty fable]

who can be bothered to solve SOEs tho lol

[sweet pewter]

well I do

[dusty fable]

anyone here applying to UK uni math courses for 2026

[dusty fable]

can u check if someone is valid from a problem i did earlier

[sweet pewter]

valid?

[dusty fable]

like it’s true

and correct

[sweet pewter]

the divisible by 4466... one?

[dusty fable]

no i will send it in 2 mins im just writing it out

[sweet pewter]

this has 2 solutions surprisingly

[dusty fable]

@sweet pewter

[sweet pewter]

mm I have literally never solved a function equation like this

[dusty fable]

it’s also from BMO

[sweet pewter]

I never got into a maths olympiad (aside from a few very small-scale ones)

[dusty fable]

and they haven’t posted solutions

[dusty fable]

same i’m hoping to get into BMOthis year as i have only started rlly enjoyed competition style maths this year

[sweet pewter]

good luck

[radiant jasper]

when you say BMO you mean british or Balkan?

[sweet pewter]

I did almost get into the national chemistry olympiad btw

[radiant jasper]

btw multiplying in that problem works

[dusty fable]

british

[dusty fable]

ellaborate

[dusty fable]

can u check this after pls

[radiant jasper]

,w (12n-119)(75n-539) expand

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[radiant jasper]

Too lazy to do it

[dusty fable]

what does this show tho

@radiant jasper ?

[radiant jasper]

Actually it has a bit too many calculations I don't like it

But basically notice this is a square less than (30n-256)² or something like that, and prove that for n>like 5 or 6 this is more than (30n-265)² then bash

Don't do this at home

[sweet pewter]

might work

crazy idea tho

[radiant jasper]

no I just wanted to complete the square

[dusty fable]

oh cool

it’s cool finding multiples ways to do something

[radiant jasper]

But I end up having to check n≤70 so I had to fix it and do worse bounds

[dusty fable]

can u check if this is right @radiant jasper

[radiant jasper]

R to R?

[dusty fable]

i believe so

yeah it would be

[radiant jasper]

But f(x)≠0 for all x?

[dusty fable]

that wasn’t part of the question i just added that

[radiant jasper]

Ok

[dusty fable]

coz the question was find functions

[radiant jasper]

What's your answer

[dusty fable]

i cba to find multiple so just found that one

[radiant jasper]

bruh what

[dusty fable]

plusminus root (1-x^2)

[dusty fable]

it said find all f

i found f = 0 or plusminus root (1-x^2)

but maybe there are more

[radiant jasper]

are you saying that for any x, f(x) is either + or - square root of that?

That seems unreasonable as a solution

[dusty fable]

just pls check my workings

[radiant jasper]

Then I'll just say it's wrong

[radiant jasper]

Or you're crazy lucky and this works

But it clearly doesn't

[dusty fable]

why is it wrong mate

i don’t see what’s wrong with my workings

i got f(x)^2 = 1-x^2

so i just rooted for f(x) and accounted for both the plus minus

[radiant jasper]

f(x)²=1-x² implies f(x) \in {+√1-x², -√1-x²}

You got point wise trapped

[sweet pewter]

it is wrong if you consider x on reals

[dusty fable]

that’s the Q

[sweet pewter]

1-x^2 can't always be non-negavtive

[dusty fable]

i guess my solution isn’t defined for all reals tho

[sweet pewter]

it is if only -1 <= x <= 1

[dusty fable]

but i wanna know what is wrong with it

[sweet pewter]

this

[dusty fable]

i asked gauth

[sweet pewter]

and when the left hand side is [f(x)]^2 there's no function that satisfies the equation for all real number x

[radiant jasper]

How did you get f(x)=f(-x)

[dusty fable]

it said x+1

[radiant jasper]

this is wrong

[dusty fable]

rearrange x^2 = h(f(x)) = h(f(-x)) thus f(x)= f(-x) a

[sweet pewter]

that also

[dusty fable]

so how would u approach the problem

[radiant jasper]

you can't write that expression as a function of f(x), and you can't conclude there's any particularly nice relation between f(x) and f(-x) anyway

[sweet pewter]

let y=0

[dusty fable]

i’m going out now i’ll still read this

[sweet pewter]

oh nvm you already got f(0)=1

[dusty fable]

f(0) can = 0 if x,y=0

[sweet pewter]

but it has to be true for all x and y right?

[dusty fable]

ye

[worldly swift]

ooh this problem is fun!

immediately my gut tells me either a linear polynomial or constant polynomial would work

I would firdt confirm whether f(0) is 1 or 0

then try and find which one is consistent

then find some values for f(1) and f(2)

i feel like that should give you enough base info to work on

[sweet pewter]

yeah I solved it and supposing f(1) or f(-1) is enough

[worldly swift]

i see i see, yeah seems so

the tricky part will be arguing what the values for ax+b will be

but any polynomial of degree 2 or higher is likely not going to work

[sweet pewter]

yep

[worldly swift]

these problems are always familiar and fun

especially the case work loll

[sweet pewter]

I never solved these kinds of problems lol

[worldly swift]

I find them interesting! Especially when you change your domain/range to be things other than the reals

[sweet pewter]

that sounds rather scary

[worldly swift]

like only positive integers or prime

[worldly swift]

teah but usually the changed domains/ranges help give you other pieces of info you know the function must have

[sweet pewter]

if the answer includes some modular function I would rather quit

[worldly swift]

like a saw a problem where they give you f(3) =5, f(2)=2 and you can just claim f(1)=1 because the function has to be monotomically increasing

[worldly swift]

mods are confusing but pretty fun

especially in abstract algebra

[subtle sundial]

considering these are possibly not integer values this would be a pain to work with

[reef condor]

Ye
Short FE => probably not too hard fr

[dusty fable]

can people keep sending questions to try it’s rlly fun to collaborate

[acoustic nova]

what was that thing u sent, why did u delete it

[dusty fable]

it was some questions i made however i made a tiny mistake which i wanted to correct- i can send to u in DMs now and just show u the mistake

if u want

[acoustic nova]

its alright

but ok

[dusty fable]

i sent it

try this but on Q2 : in this the n and m are constant coefficients non the same as the n in U_n. it should be U_n = λU_n-1 + μU_n-2 + … ( the mistake is shown in the SS below)

Q2 is the best question tho- last part of a and b are good Qs tbf

[torpid cairn]

I solved the other 3 besides the calc I'll do 2 tmrw probably

[severe eagle]

your typical geom

[subtle sundial]

average olympiad geometry problem

[polar patio]

"trivially..." ahh

[subtle sundial]

the values can probably be solved for using some ancient babylonian technique

[potent patrol]

wassup

anyone here thats still participating in junior high school olympiads?

[gray pollen]

5^x × 5^x = 60. Solve for x

[ornate blade]

bruh

[summer roost]

Is it

,tex 5^{x5^{x}}

[gilded halo]

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

[gray pollen]

No

It is 5^x . 5^x

[summer roost]

This isn't competition maths though

[gray pollen]

Cool

I am sorry

[wide sinew]

This should be it

[gray pollen]

Let's see

[sweet pewter]

too difficult

😭

[wide sinew]

Ah sarcasm

[gray pollen]

Idk if I am correct here

It's actually Log 5/(Log 5) + Log 12/Log 5

[ornate blade]

careful

$\frac{\log 60}{\log 5} = \frac{\log 5 + \log 12}{\log 5}$

[gilded halo]

south

[ornate blade]

which can't be cancelled like that

[gray pollen]

This should do it

[ornate blade]

sure, now that's correct

,w log(60)/log(25)

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[ornate blade]

,w (1 + 2 * log_5 2 + log_5 3)/2

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[gray pollen]

,w 5^(1 + 2log_5 2 + log_5 3)/2 × 5^(1 + 2log_5 2 + log_5 3)/2

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[gray pollen]

,w (5^(1 + 2log_5 2 + log_5 3))\2 × (5^(1 + 2log_5 2 + log_5 3)\2

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[gray pollen]

,w 5^((1 + 2log_5 2 + log_5 3)/2)× 5^((1 + 2log_5 2 + log_5 3)/2)

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[summer roost]

How would we solve this

Or is this just Lambert w

[gray pollen]

That isn't an equation to solve for x

[mystic shale]

it's not that tho

it's 5^(2x)= 60

[summer roost]

I meant how would we solve rhis

[subtle sundial]

this is impossible

[wide sinew]

No wait

[mystic shale]

take log base 5 both sides

youll get some ugly number on rhs

,w log_5 (60)

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[mystic shale]

this

and then divide by 2

[wide sinew]

Yeah

The initial method

[summer roost]

Hmmmm

Why divide by 2

[mystic shale]

it's 2x

[wide sinew]

2x

[summer roost]

Wait I was talking about the equation i sent

[mystic shale]

o

5^x^(5^x)?

[summer roost]

Yes

[mystic shale]

idk, i'd still try log and see if something comes of it

5^x \cdot x = log_5 (60)
x+ log_5 (x) = log_5 {log_5_(60)}

but nah

gives nothing

[summer roost]

These types of questions often involve Lambert w but I don't know how to apply it here

,w x(5^x)= log_5 60

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[prisma python]

i don't know how to write it using latex but i think you rewrite 5^x with base e

[radiant jasper]

Geo is life

[potent patrol]

anyone here doing junior high school olympiads

i be practicing alot but idk if im practicing the wrong problems cause i dont have any strong intuition when it comes to actually doing the test

constantly*

when im doing the test, theres always a couple of problems where i know most of the way, but just miss one simple step

and i waste a lot of time on problems that i eventually wont be answering

i dont really know which ones to really try on

got discouraged alot recently bcuz the competition ive been having high hopes for a year+ (yes im a bit new) has just been held and i did significantly worse than the other participants from the selected in my school

[frank valve]

Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.

[gilded halo]

Robert

[frank valve]

where are you getting your practice problems

[potent patrol]

honestly social media

some AoPS, youtube, and occasionally seeing past jhs olympiad problems from tiktok or insta

ive done problems in the past years of the olympiad but never really the full thing

watched the full solution for the olympiad the year before but yeah i didnt really redo the whole thing because its known that the contest makes its questions completely different than the problems the year before

[frank valve]

you can find practice tests for AMCs and AIMEs and such on google

like official tests from previous years

[potent patrol]

hm okayokay

do i directly do the whole test?

which amc should i do

[potent patrol]

is it better to do amc-8 or amc-10

btw this olympiad has city, province, and national levell

[frank valve]

what grade are you in

[potent patrol]

7 transitioning to 8

dw im 13 alr

[frank valve]

probably amc 8 then

[potent patrol]

oh alr

[frank valve]

makes more sense

it might be too easy for you if so that's great

[potent patrol]

the competition junior level is only for grade 7 and 8

[potent patrol]

okay let me see

thanks for helping btw

[potent patrol]

ok i just did it for the intended time

and i really fall short on the time

but the test questions are way harder than this, the questions would be freebies if they appeared in the competition

[orchid linden]

https://mathdash.com/home I think this is a decent resource for practice

https://amctrivial.com/ or this but idk how well the website is working

[potent patrol]

oh wow i really like this

i never seen a website like that

[orchid linden]

ya

it got posted on aops community a while ago

[potent patrol]

thank you for sharing

[orchid linden]

👍 np

[potent patrol]

oo cool

[pallid tundra]

AMC 10/12 are a pretty big leap up from 8

try working through some past tests and see what you need to study

[subtle sundial]

mhm..

😭

[pallid tundra]

that’s like AIME/USAMO wtf

😭

[subtle sundial]

had a stroke trying to visualize this

i gave up halfway and went to solutions

😗

[subtle sundial]

https://tenor.com/view/zidane-confused-reaction-champions-league-gif-11552616

the games gone

[orchid linden]

💀

[hexed lion]

do u guys have advice to limit sillying

;-;

i feel i know how to do a lot, but i keep messing up

[willow anvil]

whats the original problem?

[torpid cairn]

whoaa this is cool

ty for this

[torpid cairn]

is this actually what Olympiad geometry looks like

I'm convinced you guys can't actually do this

[misty torrent]

No, that image looks like AI. I don’t mean AI generated, I mean it looks like a proof written by an Olympiad AI, which tend to be technically correct but terrible in all other aspects

[torpid cairn]

well the problem is real right? I fundamentally do not understand how people can visualize and work with so many shapes lmfao

dawg they put like 3 shapes on top of each other on an aime #5 and I spend 30 minutes looking for a connection I hate geometry

[torpid cairn]

there might not be a single problem I've ever wanted to do less

[sweet pewter]

Rough translation: Let P and Q be 2 random points inside ∆ABC. Let X be the intersection of the pedal circles formed by point A, ∆BPQ and ∆CPQ. Let Y be the intersection of the pedal circles formed by point B, ∆APQ and ∆CPQ. Let Z be the intersection of the pedal circles formed by point C, ∆APQ and ∆BPQ (X, Y and Z are different from the projection of A, B, C on PQ). Prove that ∆ABC ~ ∆XYZ.

[near hill]

This is another integer-recurrence problem in disguise.
Set A = 2008+sqrt(4032000), B = 2008-sqrt(4032000).
Then A and B are the roots of x² - 4016x + 64, which by standard tricks means that a(n) = A^n + B^n is an integer sequence defined by the recurrence a(n) = 4016a(n-1) - 64a(n-2), with (by direct computation) a(0) = 2 and a(1) = 4016.
So if you add B^2000+B^2001+....+B^2008 to your long sum, you get a(2000)+...+a(2008), an integer you should be able to find the last digit of using the recurrence mod 10. (The period of which turns out to be very short).
And B is positive and close to 0, so B^2000+....B^2008 is a very small positive number, and you can correct for that afterwards by subtracting 1.

[severe eagle]

me

[severe eagle]

I am a geo main thus i cannot disagree

[severe eagle]

if you dont mind which country are you from

[torpid cairn]

lol genuinely how

I quite literally hate everything about it

like there's no way u enjoy solving these problems

[severe eagle]

i do

[prisma python]

i once made my own method to solve a geometry problem using algebraic equations lol

[torpid cairn]

like coord bashing?

[prisma python]

yeah

sometimes it uses gradients too

[torpid cairn]

idk what those are

oh it's just slope

[potent patrol]

Indonesia

[prisma python]

yo i'm from indonesia too

[severe eagle]

ah

hi neighbor

[potent patrol]

you particiapte in OSN?

participate*

[prisma python]

yeah

[potent patrol]

oo

which grades did u participate

they made it online for my grade this year

[prisma python]

i just participated this year as a g10, and i used to do it once at g6 too

[prisma python]

me too, i think it's been online for a few years now

2020 was offline though

[potent patrol]

ooo

how was it?

[prisma python]

i think i did well, but i messed up some questions and i counted that i probably got like 30-40 out of 70 points

[potent patrol]

wow thats still nice

im just worried about people cheating

many instances of people taking pictures of thw questions in the actual test

[prisma python]

from your school?

[potent patrol]

nah

social media

[prisma python]

oh

[potent patrol]

if they take photos of the question, that itself is cheating

top 5 may be really high this year, dont rlly know

[prisma python]

yeah i hope something will be done about it

[potent patrol]

especially in competitive places like jakarta

[prisma python]

what city are you from? jakarta too?

[potent patrol]

nah

medan

i dont think the cutoff score is really high here

for SMP, 15 correct should be safe

[prisma python]

i think in jakarta most spots are taken by penabur, i hope i continue to the next stage though

[potent patrol]

yes lol theyre never absent in these competitions

goodluck to you

where do you usually practice?

KTOM should be great practice for senior levels, but i think its a bit too hard for junior

[prisma python]

i don't practice like with a timer, i usually just study at home

[prisma python]

oh ok thanks, i'll keep that in mind for next year

[potent patrol]

oh so you dont do online tests?

[prisma python]

no, i just study from past papers

[potent patrol]

yes, they do monthly tests to prepare for osk osp osn sma

also for smp too but only simulation

the monthly test itself is usually sma level

completely free, just make an account

if not wrong theyre about to have a test this month

[prisma python]

it follows the format of osk?

[severe eagle]

why does halo, walnut sound so funny

[potent patrol]

for all the simulations, yes the exact format is followed

before each phase in osn, they will release a simulation but this is different from their monthly tests

[potent patrol]

𝐡𝐞𝐥𝐥𝐨 𝐰𝐚𝐥𝐧𝐮𝐭

[prisma python]

oh nice

[potent patrol]

for the monthly tests i dont relly know the format but i can send u the test for last month

[severe eagle]

i think i am cooked

[potent patrol]

i swear there was a fella named something along the lines of carbonite with your pfp and tag

hmmm..

[severe eagle]

hmm...

i wonder who could it be

[potent patrol]

So basically

[severe eagle]

btw i did it

too lazy to translate

[potent patrol]

my basic level chinese should do

[severe eagle]

good luck

Welcome to mathcord btw

apologies if i spoiled the question

[potent patrol]

at the center of ∆abc, ab is greater than ac, ∆abc 的内something 圆 I BC, Ca, AN, something D, E, F, EF, BC something k.
DG is the height of DEF, IG is ∆abc outside smth, etc

i got butchered the translation

idk the math terms in chinese

[potent patrol]

thanks

[prisma python]

i just use gpt to translate it

[severe eagle]

indeed

dont feel like translating but i prob will

for you

ABC is a triangle with AB > AC

[potent patrol]

i probably would still be confused on how to solve but okay

[severe eagle]

If the incircle with center I is tangent to AB at D, AC at E, AB at F

And EF and BC intersect at K

DG is the height from G in triangle DEF

IG intersects circumcircle of ABC at H

Prove HGDK concyclic

[potent patrol]

oohh

ive never done an olympiad level geo proof question

are those considered common questions in chinese olympiads?

[prisma python]

yeah i haven't had an olympiad requiring a proof either

[radiant jasper]

Wow this looks extremely cooked

I've never seen a geo with two random points P, Q inside ABC

I would try it if i didn't have 19383792934 other problems to do

8 days remaining before death

[subtle sundial]

fuck geometry

[severe eagle]

get out

[subtle sundial]

yeah, the chinese olympiads are way harder compared to other countries

[potent patrol]

reasons why china is the lebron james of math

[subtle sundial]

truth

[potent patrol]

china imo team gives usa olympics avengers team

[severe eagle]

lmao fr

i love china tst

did some questions

the dificulty is goated

[severe eagle]

yes

but this is considered easy

[potent patrol]

[potent patrol]

wow

[severe eagle]

as a geo main i need to get better

i couldnt do an IMO G1 for some weird reason

[sweet pewter]

Arent US geometry problems just chugging regular polygons together?

[severe eagle]

im not accepting that

[potent patrol]

is this just common

[sweet pewter]

it is

[potent patrol]

how many structures do you need to sketch 💔😔

im too new for this

[sweet pewter]

they arent too difficult

[potent patrol]

geo mains have too much aura

they invented aurafarming

[sweet pewter]

geometry is like a dictionary

[potent patrol]

why

[sweet pewter]

pretty sure this is just analyzing properties of the orthocenter and incenter

[potent patrol]

let me see how many orthocenters i can spot

[severe eagle]

normal shit

[severe eagle]

cool analogy

[potent patrol]

wait what

u guys are two diff ppl

[sweet pewter]

true...

[severe eagle]

yes

[subtle sundial]

is this some type of cult

[severe eagle]

this was a china tst btw @sweet pewter

one of my favourite geoms

P is a point on 9 point circle of triangle ABC, connect AP and draw the perpendicular line to AP at P which meets BC extended at Q. Let X be on PQ such that XA is perpendicular to AQ. If H is the orthocenter of ABC, and D and M are the midpoints of BC and AQ respectively provs that's HX perpendicular to DM.

for any of yall who want to do

[sweet pewter]

problem no.2?

[severe eagle]

ye

2 out of 24

[sweet pewter]

how many problems are there in a test

[sweet pewter]

💀

[severe eagle]

3 a day

imo format

[sweet pewter]

time restriction?

[severe eagle]

imo format

4.5h for 3 qn

[sweet pewter]

tbh when I first solved this I was tricked into trying to analyze the lines that have 0 properties

until I reread the question I realized it is not difficult at all

[severe eagle]

not difficult at all 💀

[sweet pewter]

I suspected DM and HX had some special property at first because I haven't read about 9-point circles yet

[sweet pewter]

to me at least

[severe eagle]

nah bro comes here and dosent know 9 point circle

and bro does china tst

orz

[sweet pewter]

no like I haven't done anything related to special properties of it

like these

[sweet pewter]

in this problem it just uses cyclic quads

[severe eagle]

ok

[sweet pewter]

and ratio spam

[severe eagle]

tbh i didnt know that

[severe eagle]

the most important thing was (to me)

seeing the similarity that fnished the question

[sweet pewter]

lol

[sweet pewter]

When I first see that configuration with points A, X, P and Q I knew there are a lot of ratio properties I could use

[severe eagle]

pro

[sweet pewter]

but then I got into this problem of trying to analyze these segments

[prisma python]

how hard are number theory problems in the chinese olympiad?

[severe eagle]

hard to say

china tst is the hardest olympiad known to man

so yeah

[prisma python]

can you send an example question? just curious

[white blaze]

does anyone know how to start getting into comp math

[gritty notch]

umm weell u can ask ur school to see if they offer anything

i think to start out the amcs are nice

and then you could try and find smaller competitions online (like more local ones)

[gritty notch]

https://www.scribd.com/document/151836078/China-Tst-2011

if u get any of these bro...

[prisma python]

what does the {} notation mean in question 3?

[gritty notch]

lowk no idea lemme check

yeah idk

[prisma python]

it usually means decimal but i'm not sure

[gritty notch]

hmmm...

[prisma python]

wait it makes sense if it's decimal

[gritty notch]

yeah yr right cz

yeah no that makes sense

why do i have to look at this

[prisma python]

i think my solution would involve modulo in some way, i'll try solving it tomorrow

[summer roost]

What is this 😭 😭 😭 😭

[subtle sundial]

china imo team would look at that and say its normal

[sly pawn]

hi

can someone help with this problem

[potent patrol]

hm

[sly pawn]

I was thinking these are usually contradiction but maybe induction might work too

[potent patrol]

what if p = 2

oh i read that wrong

how did u solve it?

[sly pawn]

then it equals 19

ok I found the solution

for real ts time

[sly pawn]

lets say k+1=p, so that k is even

then the expression subsituting k in becomes 3*3^k+7k+3

k is even so 3^k is a perfect square

the difference between perfect square n^2 and (n+1)^2 is 2n+1

[potent patrol]

k=1 can work too

[sly pawn]

so 2*3^k +7k+3=2n+1

meaning n=1+3^k+7k/2

but then n^2 would also have to equal 3^k because 3^k was our base square

but (1+3^k+7k/2)^2 does not equal 3^k

contradiction

therefore it does not work for any p prime numbers

and also for any odd numbers

[potent patrol]

ohh

[sly pawn]

yeah but we already checked that 2 doesn't work so we doin't need to check it again

[potent patrol]

nice solution

[sly pawn]

thx

[radiant jasper]

whats the problem statement

[torpid cairn]

I am genuinely so stupid

[subtle sundial]

we are all stupid

[potent patrol]

nah no one is stupid

[gritty notch]

my death

[gritty notch]

idk my friend sent it to me

[torpid cairn]

how would you induct across the primes

genuine q

[orchid linden]

all primes are odd except 2

So I think he just considered all odd p and 2 separately ?

[torpid cairn]

hmm right

[sly pawn]

yeah you could do that

my solution ended up just subsituting and using contradiction

[ornate blade]

alternatively, this expression -1 (mod p) using Fermat's little theorem
so if this is a perfect square, n^2 + 1 = kp

[vernal axle]

wait, we have 3*3^k+7k+3 is a square and 3^k is a square. Why do you think they should be consecutive squares?

[vernal axle]

I would do it this way:
Suppose it is a square and p is odd. Reduce it modulo p, then it equals -1 mod p. So, -1 is a quadratic residue mod p which means that p=1 (mod 4).
Then reduce the initial expression mod 4. You get 7*1-1-0=2 (mod 4) which means it is not a square.

[radiant jasper]

mod 3 then mod 7 also works

[sly pawn]

because if 3^k is a square then 2*3^k+7k+3 must be equal to the difference between a square

so that 3*3^k+7k+3 becomes a square too

[prisma python]

but what if it's like (n+2)^2-n^2? then it becomes 4n+4 instead of 2n+1

[dusty fable]

anyone here sitting the MAT in october?

[sly pawn]

hmm

then I would have to prove that for all of them

[vernal axle]

yes, they are both squares. If the difference between them were small (kind of less then 2n+1 between n^2 and (n+1)^2) we could conclude that there are no other squares between them — which would imply that the two squares are equal, potentially leading to a contradiction. But in this case, the difference is large, so when k grows, there are many other squares between 3^k and 3*3^k+7k+3.

[sly pawn]

using induction ig

[prisma python]

how is it possible to prove for primes using induction?

[vernal axle]

I don't see how mod 3 and mod 7 work. Say if p=6k-1, then it equals 1 both mod 3 and mod 7. A square may well be 1 mod 3 and 1 mod 7. Say, 13^2= 169 mod 3=1 and 169 mod 7 = 1.

[radiant jasper]

Then i might have done some mistake in the calculations in my head lol

Bruh how did I get 3^5=2 mod 7

I'm a fakesolving machine in this server

Good that I keep fakesolving away from harder problems

[subtle sundial]

you can use properties of primes

[foggy jewel]

I’m preparing for it, but I’m taking it next year

[prisma python]

how?

[untold thunder]

I got $p \mid m + n$, but I am not sure how to continue without LTE lemma. Can I get a hint?

[gilded halo]

ProjectTime

[gritty notch]

could someone plz explain problem 9 aime 1, 2025

cz algebra ens me

ends

The parabola with equation $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$.

[gilded halo]

nightshade🥀

[misty torrent]

(x+y)^p == x^p + y^p (mod p)
this is known as the freshman's dream

[sweet pewter]

the real freshman's dream

[untold thunder]

But this just implies $ p \mid x^p+y^p$ which is obvious, but I need $p^2 \mid x^p+y^p$

[gilded halo]

ProjectTime

[misty torrent]

oh, I missed that you mentioned that already

in this case, it suffices to prove that if p divides x^p +1, for some integer x, then p^2 divides x^p + 1, can you tell me why
(actually, you don't need to reduce it into this step to prove the question, but it makes seeing the proof a lot easier)

[untold thunder]

No I'm not sure how to do that

Can you give me a nudge

[misty torrent]

If n is an integer that is not equal to 0 mod p, then there exists another integer m such that nm is equal to 1 mod p

in other words, when you are doing arithmetic mod p, you are actually allowed to divide by numbers as long as you are not dividing by 0

[misty torrent]

as an extra hint, to prove that it suffices to show that if n and m are not equal to 0 mod p, then nm is also never equal to 0 mod p

[untold thunder]

Okay I think I got it doing it a different way from you but thanks for the help

[vernal axle]

You can write m=-n+pk and then expand binomial m^p=(-n+pk)^p=-n^p+p^2(...).

[untold thunder]

Yeah that's exactly what I did

[tribal magnet]

Is Amie 10 goofy or na

[pallid tundra]

what the fuck is an amie 10

[torpid cairn]

I've done this one lol pretty fun problem

I can re-solve it rq so I can explain

[orchid linden]

wahh aime p9

havent even qualified aime

[torpid cairn]

this one is probably on the easy side for a p9 so if you know how to interpret the rotation I'm sure u could solve it

[prisma python]

do you use trig for it or what?

[blazing pilot]

just the fact that a rotation around the origin by an angle a maps (x,y) to (xcos(a)-ysin(a) , xsin(a)+ycos(a))

[surreal knoll]

hi , i am new here and i need help

can some one help

me

[tribal magnet]

aime

[blazing pilot]

yes, what's your question ?

[orchid linden]

y’all have tips for getting to usa(j)mo🤔

[surreal knoll]

i want to prepare for IMC , and i need some resources or the topic in this competition

and thanks

[blazing pilot]

just learn enough undergrad well and do a lot of Putnam, IMC, SEEMOUS and other similar competitions problems
work on these : https://artofproblemsolving.com/community/c13_contests

Miklós Schweitzer is an exception tho, just work on the simple ones, good enough to be good at solving IMC problems

[surreal knoll]

just study the undergrad then solving problems

thanks a lotbro

[surreal knoll]

but do you have some resources to learn undergrad??

and another quetion i have no idea on imo it will not be an excuse to preparing for IMC

@blazing pilot

[blazing pilot]

take any linear algebra/real analysis/number theory/combinatorics/geometry books and learn from them
you can also watch videos
there are many ressources on the internet

[blazing pilot]

are you trying to prepare for the IMO or for the IMC ?

[surreal knoll]

IMC

[potent patrol]

is IMC just IMO but the olympiad is replaced with competition

[blazing pilot]

IMC is undergrad but if we compare difficulty i'd say imo is slightly harder

[potent patrol]

ohh.

[warm wing]

is there a putnam server around here or something

just asking as I start attempting to prep

[worthy steppe]

how does one start studying for comps

just wanna know

[potent patrol]

maybe read a beginner friendly olympiad book

you can see the techniques in the solution of the problem

[severe eagle]

i failed imc selection test

Gives me trauma fr

[blazing pilot]

you don't become good if you just read the solutions
meditating and focusing on one hard problem trying to solve it with different methods is way better than doing 100 problems in that way

that's how the russians, chinese and koreans work

[potent patrol]

yea i completely agree with u

its just a small tip hehe

[potent patrol]

is it just researching more about the problem?

going deeper into it rather than just finishing it and moving on

if that makes sense

[blazing pilot]

yes
you read the theory needed for that problem and try again
then you keep researching and master things very deeply until you solve it

that's why you gotta master the proofs and not just try to apply theorems blindly

[potent patrol]

oooh okay

[blazing pilot]

yeah exactly

it's even better to work in teams and solve the same problem with at least 3 different ways

[ocean gale]

Hi guys, i need some help from you with a last year South African University Maths competition paper

Can someone please help with question 4 and 7

[ivory ember]

4. Work inside out. $$||||t|-4|-3|-2|=1$$ $$|||t|-4|-3|-2=-1, 1$$ $$|||t|-4|-3|=1, 3$$ $$||t|-4|-3=-3, -1, 1, 3$$ etc. \ \ 7. Recall that $\lim_{x \to \infty} \arctan(x)=\frac{\pi}{2}$.

[gilded halo]

Civil Service Pigeon

[ocean gale]

What do you mean by inside out

[ornate blade]

yeah, 'inside out' isn't that clear to me too

so from the 1st line to the 2nd line, you shed a pair of absolute values

and then the right hand side becomes -1 or 1 to compensate

I find the working to be very clear

[ocean gale]

Ohh yeah i think i see it too it kind of clear for me too, but "inside out" was vague i didnt get it😅

[stark swallow]

what is the definition of excentre

[radiant jasper]

ok so first extend two sides of the triangle

on the same side

then create the angle bisectors of the exterior angles

[stark swallow]

as shown

[radiant jasper]

The point where they meet is the excentre

[stark swallow]

oh

[radiant jasper]

u can put it in a better way but that's all for excentre

[stark swallow]

alr

[radiant jasper]

then theres excentre incentre lemma

stuff

[stark swallow]

yes thats next topic after that

[radiant jasper]

yeah

are u doing egmo?

[stark swallow]

IOQM

[radiant jasper]

oh

nice which lectures? or book?

[stark swallow]

i am enrolled in an online crash course, plus i have pathfinder and the crash course provided free module book that is very nice book alot of pyqs and theres imo's 1975 questions as well

there are so many examples that i find difficult to understand altho

understanding those exmaples self is tricky part

[radiant jasper]

oh nice

[stark swallow]

what about you what are you preparing for

[radiant jasper]

math olympiads yeah

ioqm rmo

[stark swallow]

oh

have you qualified ioqm last year?

[radiant jasper]

nop

delhi ncr was too high + i studied for like two weeks 💀

[stark swallow]

oh

[radiant jasper]

i wasnt expecting to quali

[stark swallow]

here cutoffs are also high

rajasthan

[radiant jasper]

hyd?

oh

yeah kota stuff

[stark swallow]

exactly

[radiant jasper]

are u in kota?

[stark swallow]

yes

[radiant jasper]

oh man

which grade?

[stark swallow]

11th

[radiant jasper]

same

[stark swallow]

nicee

[radiant jasper]

@jagged current are we allowed to go sidetracked here

💀

and yes i got reminded of this server by your bio

[jagged current]

What

[jagged current]

Uhh, no, this is a subject-specific channel

[radiant jasper]

alr damn sorry

[stark swallow]

and if i draw a circle tangent to ab past b any point and ac past c point and and it touches bc and those 2 points then it will be called excircle right?

[radiant jasper]

yeah

[stark swallow]

and if i join radii from that point that is ex-radii

[jagged current]

I am not a head helper here, somewhere else

[radiant jasper]

yep

[radiant jasper]

oh? 💀

[stark swallow]

how are you preparing btw like any book ?

[radiant jasper]

pathfinder sr + pw

i did a bit of titu's

for number theory

really a bit of it

a bit of egmo by evan chen as well

which edition of pathfinder are u using? jr or sr

[stark swallow]

i heard this alot but never understood this, what is this

[stark swallow]

2024-25

[radiant jasper]

😭 it is a book by evan chen

[radiant jasper]

uhhh pj sir author?

[stark swallow]

exactly

[radiant jasper]

okay it is the jr version

pw course? 🔥

ok lets take this to dms because this is going irrelevant

[stark swallow]

sure

[ocean gale]

Anyone assist with number3

[rain edge]

sure what u need help for it?

i can help

[ocean gale]

Please help bro

[rain edge]

its a type of number theory prob

u gotta think abt wat all three terms have in common

not in terms

of the numbers themselves

but whats around them

[ocean gale]

Ohh wait i think i did manage to find some sols

I got (3, 2, 12)

[rain edge]

yea

tats wat i got

my method was that all 3 are non-negative

square, abs, and root

so a+b+c = 0

a, b, c >= 0

[ocean gale]

I just equated each term to zero so that when you add you get zero so that how i got those values

[rain edge]

ye, thats right

[rain edge]

tis is hte formal way to prove that a = b = c = 0

where a, b, c are the three terms

[ocean gale]

GGs bro. number 9 has to be a joke or something 😂

[rain edge]

lemme try

[ocean gale]

Have a look

[rain edge]

oh idk integrals and derivatives, im in 9th grade

i havent started learning precalc and calc yet

[ocean gale]

Ohh not a problem bro, fact that you solved that at 9th grade is impressive keep it up

[rain edge]

thx

what grade r u/

?

[ocean gale]

Am a uni student bru

[rain edge]

oh damn!! 💀 💀

[ocean gale]

Ye, but schooling systems are kind of different a bit. Where yoy from .... am from South Africa

[rain edge]

usa

[ocean gale]

GGs, all best mate

[ocean gale]

Keep that passion for maths

[rain edge]

ty

[torpid cairn]

I just did another AIME mock and got a 5 bro I can't stop wasting 3 hours

I hate knowing exactly how to solve the problem and then getting it wrong regardless

[rain edge]

YE LOL

u jst have like a claculation error once u find the method

[rain edge]

damn wat grade r u?

[torpid cairn]

incoming sophmore

my goal is USAMO idk if I can pull it off

[torpid cairn]

right lmao these problems have so many steps

especially the combinatorics ones where you can miss one edge case and you'll have no way of knowing whether your answer is off or not

[rain edge]

same

im an incoming freshman tis yr

prob gonna get into aime, but the prob is passing aime

cos ur aime score * 10 + amc score gotta be >200 i think

approx 200

[rain edge]

ye

[torpid cairn]

yep

if ur already prepping as an incoming freshman ur doing better than I was tbf

gl bro

[rain edge]

im not lol

im lagging

cos of robotic

s

I missed the easiest question on amc 8 last yr, and got a 22 😭

i made a counting mistake

tis one

lmao

[rain edge]

ty, same to u

[torpid cairn]

lowkey I think my highest was 17 lol I was ass for some reason

the other way of looking at it is I'm getting better faster than everyone else and thus I'll make usamo (I'm coping)

[acoustic nova]

What would be more interesting is if you got more points and weren’t given a fixed order

[worthy steppe]

ok thank you!

[orchid linden]

how do yal like effectively train for olympiads

[worthy steppe]

I would like to know as well

[rain edge]

im not exactly fast lmao, i had to guess last one on amc 8

i got it right!!

[rain edge]

as in??

all paths?

💀

without visiting same point twice

too much counting for da 5th prob of da amc8

[ocean gale]

you not alone with that bro i also have the same issue with tackling problems

[potent patrol]

mine is like

i know the steps to solve it

but im just missing one or two steps

[distant night]

Hi there! I'm new here, and I have a equation system that I'm sure can be resolved with inequalities (ik median inequality, CS, Titu's Lemma, etc). I've been stumped on it for a while. x,y,z are positive reals, such that the aforementioned equalities hold true. We have to solve for x,y,z

[hexed oak]

I don't see any a,b,c in this

[distant night]

my bad, I meant x,y,z are pozitive reals 🫡

[sweet pewter]

hint: $\frac {2x^3}{(2+y)(2+z)} + \frac {2+y}{4} + \frac {2+z}{4} \geq \frac {3x}{2}$

[gilded halo]

DeC∆rbonizeD

[sweet pewter]

combine 3 equations together then use this

[distant night]

Thanks so much for the help!

[sweet pewter]

np

[gritty notch]

is guessing numbr 3 on 2001 aime correctly a reasonable crashout

why can thjis never happen on actual testtsss

[gritty notch]

bru for that one i spent 5 minutes thinking you could go between the streets

[gritty notch]

BRO HOWW

[dusty fable]

can someone explain to me how using inqualities by saying things like WLOG x>y>z => f(x,y,z) < g(x) is actually helpful- how to build up a solution from it and when to use it

as i rlly struggle w those problems

[near hill]

That's very problem-specific.

The typical case where one would say spmething like that is when the property you want to prove is symmetric in x, y and z, and it helps (or at least helps your imagination along) to pretend you know which of them is larger/smallet than the others.

(Saying "wlog a<b<c" of course requires either that you know none of a,b,c can be equal, or that it is so obvious that your goal is immediately true in that case that you can get away with not explaining why.)

[rain edge]

lol, i got one number with calculations i think it was like 125 or 25 or smth

it was the only number divisible by tat

and it looked like a neat num lol

so i picked it

[rain edge]

wats a wlog

wat tat equation mean

??

very weird equation

wats a wlof

*wlog

[subtle sundial]

without loss of generality

[rain edge]

oh ok

ty

[subtle sundial]

ofc

[gritty notch]

great ida

[gritty notch]

wait can someone explain this cz i have no idea how to do withough guessing

[rain edge]

2001 aime??

lemme see

aime 1 or aime 2

tis one??

im pretty sure theres no roots?

[ornate blade]

they're considering complex roots so there must be some by FTA

well ||this is literally Vieta||

[rain edge]

idk wat ||vieta|| is

havent learnt that formula

[ornate blade]

doomed

[rain edge]

oh tat thing

[ornate blade]

oh there's a clever way without Vieta actually

[rain edge]

its called ||VIeta's||?????

i didnt know

i use it like evry time for quadratic equations

[ornate blade]

indeed and it generalises to polynomials

[rain edge]

ive used it for cubics too

[gritty notch]

if u didn't know vieta's u mightve been cooked

[rain edge]

never known it was called tat

[gritty notch]

ye

[rain edge]

so the answer is ||0||??

[gritty notch]

nope

[rain edge]

nvmd

hold up

wait

wait, lemme cook

[gritty notch]

bet

[ornate blade]

it's mentioned in the solutions but just to restate, if ||r is a root then 1/2 - r must be a root, and you know that r must never equal 1/2 - r so...||

[gritty notch]

bru im slow

[rain edge]

||1/4|| is my answer

hopefully

tats right

[gritty notch]

nope

[rain edge]

||-1/4||

i forgot its -b/a

[gritty notch]

nope

[rain edge]

fudge

wat i do wrong bruh

[gritty notch]

idk bro

thats why im askinggggg

[ornate blade]

there isn't an x^2001 term

[rain edge]

oh!!

i forgot is -x^2001

not x^2001

2

*2001

sht

ok wait

[gritty notch]

oop

[rain edge]

a = 2001/2 then

for coefficient of 2000th term

sry to anyone who im spoling solution to

[gritty notch]

aime is mean

[ornate blade]

(you need the binomial theorem)

[rain edge]

yeah ik

[gritty notch]

i js guessed the most random s and it worked

[rain edge]

im tryna use it

so coefficient of x^1999 = ||1/4 * 2001000||

so ||2001 * 1000 * 2/2001||

= ||500||

is it correct?

[gritty notch]

ye

congrats

[rain edge]

finally!!!!!!!!!!

[gritty notch]

oop

[rain edge]

oh wat?

so the exclamation activates texit?

or watever it calleed

[gritty notch]

i guess

[rain edge]

ok

[gritty notch]

orz

alr bye

thanks

[rain edge]

can someone help me prove tat for all prime x, the complex roots of zeta(x) all lie on the line a = 1/2 on the complex plane??

[storm mortar]

angle chase

[rain edge]

?

wats an angle chase

[ornate blade]

no one's giving you USD 1 million for free bruv

[rain edge]

lmao

lmao

[storm mortar]

its a joke

[rain edge]

also did yall know zeta(3) is a definite irrational constant

its caleld da apery's constant

my fav nu

irrational num

1.202...

[storm mortar]

@rain edge you should discuss this in #math-discussion instead of #competition-math

[rain edge]

ok

[torpid cairn]

ngl I didn't know vietas until like a week ago

I think I'm just way behind on all theory but I love solving problems more than learning so that's never gonna change

[torpid cairn]

AIME answers are always positive integers <1000

maybe it can be 0 too idk

[rain edge]

oh lol

did not know

[gritty notch]

wait rlly.

[torpid cairn]

yess lol

[gritty notch]

thats amazing

1/1000 chance

[torpid cairn]

it doesn't really change much

[gritty notch]

eh

ig

[torpid cairn]

there are questions where the answer is a fraction and it'll say "express it as m/n, find m+n for coprime m and n" or something similar

[gritty notch]

oh ye

those

[rain edge]

oh ya lol

[storm mortar]

you have heard of vieta but have you heard of vieta jumping?

[torpid cairn]

what the fuck is that

I'm looking at the Wikipedia this looks like unnecessary information for me