competition-math

[crystal rock]

I'll just note a few things that can help in case the idea isn't clear enough:
Note 1: ||Notice where the circumcenter of triangle BIC is.||
Note 2: ||Shooting Lemma||

[radiant jasper]

Oh, I'm doing it for year 10

[deep field]

wow really!!!!!!

how are you preparing?

[radiant jasper]

I have past papers of it from previous years

Have you ever done it before?

[hallow dome]

Hello

[deep field]

nope

im in year 7 so im just starting junior

[deep field]

oh wow

[hallow dome]

are you living in the US @deep field

[radiant jasper]

In year 7, I found it quite easy. Many questions are in multi choice, and the test progressively gets more challenging.

[deep field]

nope

im talking about australian math competition

same acronym thouygh

[deep field]

i see

i struggled a bit with the practice questions tho lol

[hallow dome]

so youre australian?

[deep field]

considering it was 2014 i can only assume it'll be harder this year

[deep field]

yeah ig

[radiant jasper]

Would you like me to send you more recent tests?

[deep field]

yes pls

[hallow dome]

so wait, im having some problems in the competition now

[deep field]

that would be super duper kind

[hallow dome]

i need some helps

[deep field]

u also doing amc?

american or aus?

[hallow dome]

nope

[hallow dome]

im trying tranlaste the ques into eng now

[hallow dome]

is it 7th grade math?

[deep field]

ah i did these already

didnt get a single one right💔

[radiant jasper]

This is a past test for juniours (so year 7 and 8)

[deep field]

i guess they give everyone the same practice questions

oh

[hallow dome]

oh thanks u, its easier than my country math^

oops sr, i mean its a reply for RRemember..

not salami

i think im going to show some exercises in my country, is it annoying?

[radiant jasper]

Have you done this?

[deep field]

no i havent!!

thank you so muchh

[radiant jasper]

It's okki

[hallow dome]

... i wanna know how old are u @deep field

[deep field]

are these actually the questions

wow they're kind of easy

i assume they get harder later on though

[deep field]

ummm why though😔😔

[deep field]

i'll tell u i js wanna know why

[deep field]

omg thank you so much actually ur so nice like i was stressing because i didnt know how to prepare😭😭😭

[hallow dome]

because i see these excersies in amc is quite as same as our 3rd-4th grade math .

[deep field]

ooh im 13 though lol

[deep field]

australia is pretty behind curriculum wise i guess

[hallow dome]

im aslo 13=)))

[deep field]

woah

you must be pretty smart

[hallow dome]

but ourr math test..

[deep field]

the tests here are fairly easy

school is super laidback so not a lot of high achievers

nothing compared to school in asia for example

[hallow dome]

our math test=)))

[radiant jasper]

It's ok, the AMC is still quite a while away : ), make sure to relax whilst you study hard

[deep field]

thanak you!!

i wish you the best of luck on it

is it at a venue or do we do them at school?

i think it would be fun to go to a place to do it

[radiant jasper]

I did mine at school because I'm at a private institution, however some schools do it at a specific place

[deep field]

ah i see i see

ooo yeah i think its held at school

i hope i can at least get credit

high distinction is obviously just a pipe dream but it's worth a shot !

[radiant jasper]

I'm sure you'll do fine, most of the multi choice can be found within the year 7 curriculum!!

[deep field]

i see i see

have you done it before?

[radiant jasper]

Do you particularly enjoy mathematics?

[deep field]

i do

i love it

[radiant jasper]

Yep, since 2022. This year is my last

[deep field]

wow!!

did u get any awards?

[radiant jasper]

I'm happy : ), I love mathematics too

[radiant jasper]

I always got distinction, but not high distinction: ((, but I wanna aim for it this year

[deep field]

wow!!!!!!

i really hope i can get an awward

anything is possible i believe you could definitely do it

[radiant jasper]

Here is the scoring system

[radiant jasper]

Thank youu : ), you will absolutely obliterate the AMC with your capabilities

All students receive an award according to the criteria below.

Participation
Awarded to a student who has participated in the competition and received no higher award.

Proficiency
Awarded to a student who has received a pre-set score and received no higher award.

Students who achieve proficiency have demonstrated competency at mathematical problem solving to Australian standards.

Credit
Awarded to a student who is in the top 55% of their year and region (60% for Senior division) and has received no higher award.

Distinction
Awarded to a student who is in the top 20% of their year and region (25% for Senior division) and has received no higher award.

High Distinction
Awarded to a student who is in the top 3% of their year and region (top 5% for Senior division) and has received no higher award.

Prize
Generally awarded to no more than 1 student for every 300 students within a region and year group. Prize winners are given a lapel pin and a voucher for our online shop. The value of the voucher may vary year by year.

Best in School
Awarded to the student with the best overall result in a school, after adjustment for year level. It is only awarded if the achievement is at least a Distinction, and there are at least 50 entries (secondary divisions) or 30 entries (primary divisions) in the school. There may be more than one winner if adjusted scores are equal.

Cheryl Praeger Medal (Secondary divisions only)
Awarded to the female student with the highest score in Australia, in each year level, where preset criteria are met.

AMC Medals (Secondary divisions only)
Awarded to a student with outstanding results within their year group for their state, territory or country. Generally, no more 1 in 5000 students receive a medal.. Medals are awarded at the discretion of the Australian Maths Trust.

[radiant jasper]

I copied this info from the official website, but the award system is here if you're interested

[deep field]

thank you!

[novel karma]

is this wrong

i got s = 99*2^101 +2

and can someone recheck if the sum is correct

[scenic sun]

I’m pretty sure what your circle is wrong (the bottom line of what you circled)

[scenic sun]

this is correct

[novel karma]

oh

i got it now

cuz the 2 gets subtraced after

[scenic sun]

Oh okay, in that sense they should done S+2 in the solution because the solution stated is incorrect at the bottom stage. Still arrives at the same answer for accounting the -2

[novel karma]

thank you

[candid harness]

@pallid tundra I got a 12/20

Thanks bro

[scenic sun]

np

[glossy berry]

u knoe rd sharma

RD sharma

[hollow otter]

hello

[turbid cipher]

I have trouble with Trigonometry

Nice

Geometry as well.

Can't waste any more time.

Then I will finish the calc book

first volume

[crystal rock]

I'm not trying to know the solution, but how would you guys solve this equation? (x is a real number)

[vernal axle]

Hint: ||note that sums of those expressions under roots on both sides are equal.||
Solution: ||#1382370897511448716 message||

[crystal rock]

Oh, I guess someone else took that test too

I managed to solve it during the test but thanks anyways

[crystal rock]

This one is easier and I already know the solution, but I'll still send it for some people to try
-# someone might have sent it before like the other equation but I'm not sure

[vernal axle]

Replace 3-xy in the second equation with x+y and extract the full cube on the right hand side.
Then you get (2y+x)^3=(x+y+1)^3
So, y=1, x=1.

[crystal rock]

Thanks, well I guess that's the best way to do this anyways :D

[nocturne narwhal]

HI ALL, how to get into AIME and jmo?? 2 yrs ago i got a 48.5 on the AMC10 (its very bad ik). But I wanna qualify this year. im pretty sure test is in November, so I'm a bit cooked for timing idk u tell me, but yeah, what did u use, (if ur comfortable, u don't have to tell, but your score) and yeah ig that's it. and study tips thanks!! sorry for bad grammar and punctuation.

[crystal rock]

I'm not really sure, but last year I just prepared for AMC10 by going over a bunch of AMC10 questions (sometimes AMC12) and ended up with 117 (could've been better oof)
May I know what grade you're in? That would make it a bit easier for me :D

[vernal axle]

There are no positive integers satisfying the second condition

[raven coyote]

whats this three lines???

n three lins 5 mod 7 whats the meaning of this

[raven coyote]

how do you get it to write like this

[crystal rock]

$n \equiv 5 \pmod 7$

[gilded halo]

Rando.

[frank valve]

plug it into wolframalpha

[raven coyote]

,help

[gilded halo]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[raven coyote]

,list

[crystal rock]

I'm sending a proof related problem next time :3

[crystal rock]

Also this was given to 9th grade students in a test, so things were kinda limited

[raven coyote]

i cant solve it

[crystal rock]

oof.

[raven coyote]

helpp

[crystal rock]

you can refer to this hint, the solution isn't that friendly for grade 9 tho

[raven coyote]

i refered to the hint and still i cant i refered to sol. and i got scareed

[crystal rock]

Try replace the roots with variables a,b,c,d, you'll have:
a+b=c+d
a⁴+b⁴=c⁴+d⁴

[raven coyote]

ohh

[raven coyote]

how did you getthe second expression

i get it

[crystal rock]

It's just true, look at the stuff in the roots

[raven coyote]

i solving

[raven coyote]

yess

i cant

i have reACHED THIS EXPRESION

ab + cd = 2(c+d)^2

[crystal rock]

I'm surprised you got that far :3

Anyways

[raven coyote]

🥲

[crystal rock]

Split the 2(c+d)² into (a+b)²+(c+d)²

[crystal rock]

After all you can kinda tell that it can't happen

[raven coyote]

what cant happen??

[crystal rock]

That equation

[raven coyote]

how

[crystal rock]

First of all, a,b,c,d>=0

[raven coyote]

yea

[crystal rock]

(a+b)²>=4ab

(c+d)²>=4cd

[crystal rock]

So this only happens when a=b=c=d=0 which can't happen

[raven coyote]

what to do then

[crystal rock]

Wait

Let me see

[raven coyote]

what was your sol.??

[crystal rock]

I did some stuff and get 2 cases: either ab=cd or your equation is true

[raven coyote]

the eqn is tru only if a,b,c,d=0

since on solving like you said i got a3+b3+c3+d3 =0

[crystal rock]

all of them can't be 0 because then x would be -1... but then the other ones wouldn't be all 0

btw how did you get that equation

[raven coyote]

yes

i took a+b=c+d to the fourth power

and then a4+b4 canceled out with c4+d4

then i took ab and cd common

both sided

and then ab/cd = a term which i cant type out

[crystal rock]

After this and simplifying I got
2ab(a+b)²-a²b²=2cd(c+d)²-c²d²

[raven coyote]

yes and since a+b = c+d therefore (a+b)^2 = (c+d)^2

[crystal rock]

yeah

[raven coyote]

substitute a+b for c+d

take ab and cd on one side in a fraction and apply componendo dividendo

[crystal rock]

huh

[raven coyote]

lemme show

wait

[crystal rock]

I'm pretty sure you can factor out (ab-cd) and some other stuff that can't be 0

[raven coyote]

yes

this is componendo and dividendo

[ornate blade]

meanwhile if you do ext 2 or the equivalent you've got to be gunning

of course there are still some people who will find ext 2 straightforward

Australia's system isn't that bad

a lot of Asian countries have a memorisation-based curriculum for humanities and it absolutely sucks

remember that the Australian Curriculum has eight equally important curriculum areas after all

[raven coyote]

@crystal rock

[raven coyote]

yea they do

[ornate blade]

but yes Australia traditionally values sports (the footy) much more than academics

[deep field]

it definitely does

[ornate blade]

so it's become much more normalised to not try at school

[deep field]

people genuinely just dont care

kids in my class are proud of getting ungraded which is basically a fail

its below 40%

it's honestly harder to get a ug than an A

and an A+ is 90% and above so it's not difficult

[ornate blade]

oh no...

[crystal rock]

huh, sure

Although I don't like it when the denominator can be 0

[raven coyote]

can it be in this case??

[crystal rock]

I think so, but we can just check before doing it

cd=0 means c=0 or d=0, but c and d are positive

We only need to check ab=cd then

If that's the case then:
(a-c)(a-d)=a²-a(c+d)+cd
=a²-a(a+b)+ab=0
So a=c or a=d, this is easy

Overall we can easily solve each case to get x=0 or x=1.

[raven coyote]

im burnt out

byby

[rain ravine]

On this

You are on the money with that

So I take O level geography

And its very simple, they in fact give you the answer key(the textbook), its just you have to memorise it

[ornate blade]

yeah and Cambridge is supposed to be the more open and less traditional board...

really my experience growing up in HK, and my friends' experiences in Singapore: those are more of the exception

Japan and Taiwan are also trying out curriculum reforms

they've been trying for the last 30 years in fact

[rain ravine]

Singapore’s curriculum as of current is terrible

I asked one of my classmates have you started calculus yet (We take Additional Mathematics), bro said he started differentiation but not calculus

[ornate blade]

I mean if the government rots from the top like a fish from the head up, whatever they teach for the humanities will be terrible

[rain ravine]

Fax

[ornate blade]

there have been a million papers about teaching English to Japanese students, but Japanese students' conversational abilities remain as bad as ever

[rain ravine]

You should see a Singaporean candidate take an oral exam

You an Aussie?

[ornate blade]

well, failing at oral exams is a universal thing

[ornate blade]

yeah

[rain ravine]

Brisbane is a W

[ornate blade]

I mean he probably meant he didn't start A level calculus
but if you look at the actual Cambridge A level syllabus, it's a joke: Add Maths students are learning some pure 3 content already

cause it's been pushed back so late to pure 3

like logs and exponentials

[rain ravine]

Dude, they just add maclaurin and integration techs

Ok fine volumes of revolution

But that’s it

[ornate blade]

a few extensions of course

mhm

[rain ravine]

Yeah

Functions are completely compromised 💀

[ornate blade]

how so? I'm curious

[rain ravine]

Well over here in Singapore, basically how an A level student answers a question to find an inverse function is just switch x with y, write $x\in \mathbb{R}$ and hope for the best!

[ornate blade]

fair

[gilded halo]

Lost fruit

[ornate blade]

well I mean high school students have so many demands on their time

[rain ravine]

That’s true

[ornate blade]

if they don't enjoy maths already they're going to take the easiest shortcut

[rain ravine]

Yeah

[ornate blade]

that's just the nature of our education system

[rain ravine]

Sadly

[ornate blade]

write down a bunch of shit for induction they don't understand themselves

[rain ravine]

A-level maths here doesn’t even cover proving

Unless you take H3 maths

But barely anyone does it (because they all take H1 and H2)

The culture arises from how they teach at O’s

In an O level math exam I find myself typing shit on the calculator more than I derive stuff during the exam

Like most kids here memorise the formula for the angle in the law of cosines, and have no clue how to get it from the actual law

[mighty pine]

no one here chooses stem-related classes because stem jobs in general be underpaid asf

[ornate blade]

that is interesting as fuck

[mighty pine]

how it be feeling being paid less than global minimum wage for engineering job

[ornate blade]

oh true

[mighty pine]

immaculate

[ornate blade]

engineering has a huge oversupply in the subcontinent

according to basic laws of economics, there's no way each engineer can be paid much

is it similar where you're from?

[mighty pine]

i chose to because i liked it and now i'm locked in a vicious cycle of having to study calculus, la, ra, ca, etc, etc, etc over holidays and vacation

otherwise i will end up dead broke on the streets

[mighty pine]

kind of

engineers are purposely underpaid here

• the fact there aren't that many universities that connect it to technical, hands-on activity

the curriculum is disjoint

[mighty pine]

i just want to be 13 again man

ts shit is going to drive me insane

[crystal rock]

I swear the shit I went through in grade 9 is crazy, like wdym I have to be able to solve stuff like this 😭

[crystal rock]

,w x = \sqrt{(3 - x)(4 - x)} + \sqrt{(4 - x)(5 - x)} + \sqrt{(5 - x)(3 - x)}

[gilded halo]

Results provided by WolframAlpha

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

Requested by randomintruder.

[crystal rock]

What is that x≈6.1828 doing lmao 😭

Well it's valid :(

Nevermind I have to split the square roots, or else that 6.1828 will be a solution and I don't think I can solve for it

So it should be sqrt(3-x)sqrt(4-x)
not sqrt((3-x)(4-x))

$x = \sqrt{3 - x} \sqrt{4 - x} + \sqrt{4 - x} \sqrt{5 - x} +\sqrt{5 - x} \sqrt{3 - x}$

[gilded halo]

Rando.

[crystal rock]

Now this works better.

[vernal axle]

it's not hard just by moving 3,5 term to the left and squaring. x=671/240

[glossy falcon]

I want to get better at math oly but I have no resources 😭 anyone mind giving a few playlists or book recommendations (that include solutions) for geom?

[raven coyote]

which country are you in??

[crystal rock]

vietnam :3

[raven coyote]

ohk

[crystal rock]

scary 😭

[raven coyote]

what grade are u in rn??

[crystal rock]

I just finished grade 9
-# help

[raven coyote]

wow

[crystal rock]

somehow that works, thanks for the easy solution

[mystic shale]

Why do you have the undergraduate role in grade 9 lmao

[crystal rock]

:) I'm kinda stupid

-# sorry for trying to learn more than I should ig 😔

[mystic shale]

?

Learning things doesn't depend on what level of education you are at lol

[crystal rock]

Oof, then I guess I'm stupid.

[mystic shale]

What.

Are you trying to incite something on purpose?

[crystal rock]

I just don't understand what stuff means, that's all lol

[mystic shale]

Yeah the undergraduate role is for when you enter college, until then people usually have the pre-high school role, and having the undergraduate role is fine and all, but it won't really help you learn any more than what you are right now, it'll just expose you to math that you might not understand in places like the undergraduate channel, which might lead you to feel demotivated(or motivated, depending on you)- but keeping it is your choice i was just asking lol

[crystal rock]

Hmm, I see.

I'll still keep the role since I've been learning things about that level for 2-3 years, but thanks for the suggestion :)

[pure yarrow]

You are lucky bro I am 17 and I started in competition maths just two months ago, I somehow feel it's late and I should go into something else but I still have two years in high school so I think It's worth a try

Anyways I have this problem does any one know how to solve it

Consider a square with a side length of 2. Prove that, for any choice of 5 points within this square, there are at least two points whose distance from each other is less than or equal to .

[crystal rock]

hm

Well if you divide it into 4 squares of side length 1, there will always be 2 points in the same square, which have a maximum distance of sqrt2.

[quartz sleet]

This. In German we call it the "Schubfachprinzip", not sure whether in English you actually call it "drawer principle" or something... However, it describes that if you distribute these 5 objects (your points) among the 4 "drawers" (the subsquares of side length 1), there will always be a "drawer" containing at least two objects. The rest is the argument you brought up

[crystal rock]

Yeah, I think it's called the "pigeonhole principle" in English

[quartz sleet]

Right, I remember that name 👍

[torpid igloo]

Hey everyone!

Wanted to ask if you all have heard of something called 'MathDash' and have any raw thoughts about it 🙂

[soft vigil]

isnt MathDash your website 💀

[novel karma]

that;s cool

[untold thunder]

THis bad bad boy was on hmmt 2024 november team round

[sweet pewter]

god I always forget this (a+b)(a+c) grouping trick in these problems

[crystal rock]

Yeah I thought of something like this too, although I didn't expect the "squaring both sides" thing the other guy mentioned to also work lmao

[pallid ginkgo]

euclidian geometry in math olympiads, its the otis excerpts of geometry (hell im p sure its mentioned there?)

wait although it doesent have complete solutions, only some problems jave sokhtuons

[crystal rock]

I kinda had a stroke reading that

[glossy falcon]

Nah there isn't geom in Otis excerpts

[pallid ginkgo]

? what does that have to do with what i said

[glossy falcon]

Misunderstanding sorry

[austere wedge]

wanna start comp math, heard aops is a good place to start, but what aops book should I start ith

[digital mason]

i loive gd cologne+

[pallid ginkgo]

im not sure where you got it from, but aops could also refer to the aops site, where it has aton of past olympiad papers and alcumus which basically gives you questions and answers

[sweet pewter]

imagine playing gd

[subtle sundial]

is po shen loh's compilation of solving quadratic equation worth it?

i feel like brahmaguptas way is more practical

[tame acorn]

If A and B are positive integers, and when divided by 5 they leave remainders 2 and 3 respectively, what is the remainder when A(A+1)+5B is divided by 25?

my answer is 21 but idk if it's correct or nah

[summer roost]

I think it is correct

Although quite easy to fakesolve

[near hill]

Metagaming it a bit: if there's any single answer it must be what you get when A and B are 2 and 3 themselves.

[tame acorn]

okay"

[tame acorn]

so just substitute 2 and 3 to A(A+1)+5B?

[sweet pewter]

yep

[near hill]

Yeah.

[tame acorn]

hmmm

oh same

21

[sweet pewter]

fakesolve trick used in comp math lol

[tame acorn]

okay okay

thanks

[gritty notch]

yup

[near hill]

It's not obvious that there's only one answer, though.

[sweet pewter]

you have to write in terms of 5m+2 and 5n+3 and expand that

[near hill]

Yeah, it's not as simple as saying "we know the factors modulo 5, so therefore we know the product modulo 25" -- which isn't true in general.

[sweet pewter]

mod rules can't apply here ye

[winter sable]

only the prodigies remember me

[pallid tundra]

it’s just the quadratic formula window dressed slightly differently

I don’t see any meaningful distinction between that and the standard quadratic formula

[slate meadow]

it’s right and I think you can just fake solve this and it will be quite easy

[pure yarrow]

Yeah I solved it using this principle

[eager palm]

hello people

how can one start studying for competitive math

[ornate blade]

Alcumus on AoPS

[untold thunder]

If you're going to use AoPS start with the books because it's not the best idea to just use alcumus

The books are under bookstore and alcumus is under reasources on the home page.

[violet elbow]

What topics are consistently tested on amc 10?

[plain jacinth]

All of the common topics that are tested on the amc 10

[ornate blade]

you'll know if you practice

[stark swallow]

@round gazelle

Are you online

[round gazelle]

Uh yes but I'm going to sleep

[ripe snow]

hello im sbhs student

i need help with yr 12 mathematics

cuz we're doing yr 12 maths in yr 7

some calculua

[mystic shale]

How old are you?

[blazing leaf]

hey

anyone who can help me out with some questions?

related to IMO you can say

yeah, IMO.

[untold thunder]

Join the mathematical Olympiads discord server

[sullen dock]

Hey everyone who is the youngest in this chat?

[mystic shale]

Random

[blazing leaf]

can u pls dm me the server link

[summer roost]

thet probably wont reveal that

[mystic shale]

And shouldn't

[blazing leaf]

tru

[sullen dock]

I am 16

[mystic shale]

Unless you are under 13, in which case you should

[wanton ridge]

lol

[sullen dock]

What about you guys

[summer roost]

it is in the description

[blazing leaf]

same

[blazing leaf]

thnx

[summer roost]

of this channel

[sullen dock]

Which grade?

[summer roost]

have you qualified for imo ???

[blazing leaf]

11

[blazing leaf]

nop

u think if someone has qualified for it, will seek help instead of doing what they have done over the time?

[sullen dock]

I think I am the dumbest here

[blazing leaf]

nahh, talking 'bout me, i dont judge no one

ur pretty fresh and alive

lol

[sullen dock]

I am all fresh

[acoustic nova]

that’s a very interesting question to ask

[sullen dock]

Then answer it buddy

[zealous steeple]

Hello

[novel karma]

you called?

[formal ice]

my 2 year old cousin who has scored a 15 on the aime first try while he was sleeping

anyways I derived triple cos angle identity on the fly during a mock yesterday I feel like a god

[unique cedar]

hello yall

do uk some free competitions i can participate in

i am a year 10 in india and ik most competition math

[summer roost]

Free ???

[sweet pewter]

free tibet

[vocal hedge]

easy

[ripe snow]

13

[ornate blade]

really??

honestly just send the image

[reef trail]

Im 16

And I have started to think that I have the lowest level of mathematics here.

[reef trail]

combined operations?

my brother is watching that in 9th grade

well he watched

[charred kettle]

anyone wants to enter a stem online olympiad?

[sleek ivy]

aint it just compound angle

[formal ice]

tes

[sacred perch]

Find the least n such that gcd of (n^3, n!) ≥ 100.
this is just bashing after ruling out n>4 and n can't be a prime, yeah?
i mean i got the answer as 8 but i wanted to know the most optimal solution

[near hill]

n can be prime, e.g. n=101 works :-)

[simple gulch]

I don't understand what does the "," mean

[near hill]

Would it be clearer for you to say "the gcd of n³ and n! is at least 100"?

[simple gulch]

Yeah sorry, I'm not a native so it's often a bit confusing 😭

[simple gulch]

How can you have clue with this one?

[rain ravine]

15

[sullen dock]

I am also with you!

[edgy vapor]

maybe this?
if n≥100 ,this always work

now consider n<100
note that if n is prime
this lead to failed

consider composite number n
4,6,8,9,10,...99 ?

don't have a full sol yet,I'm bashing too bruh

maybe try case $$ n=\prod_{i=1}^{m} p_i^{k_i} $$ where pi prime, ki non-negative and $$ n=p^k $$ where $3 \leq k \leq 6$

[gilded halo]

S

[edgy vapor]

I think I just done full sol(?)

if p is odd (case n = p^k), the only
solution for gcd(n³,n!) = n gcd(n²,(n-1)!) = n which fail

and what about p=2 k=3,4,5,6...

and for case 1
4,6 lead to fail so 8 is the least number n

[severe eagle]

does anyone have guides on how to improve combinatorics

(combi is my weakest topic)

so for example im taking about imo shortlist C1-2 difficulty

[edgy vapor]

weakest but still doing imosl 1,2...

orz

[severe eagle]

ty ig

my strongest is geom

solved some years g6

tbf i got lucky and saw the shit

[edgy vapor]

My strongest is N but the best I can do is N3-4...

[severe eagle]

das good

i cant go g2-3 consistently

only if its easier

[edgy vapor]

have u ever read Evan chen syllabus for olympiad?

[severe eagle]

no

dont think will help me

i got other material

i learnt about hensel lemma in NT and my brain died

btw i heard some IMO year has a N10

so i will try my very best to do it and then i can say my highest solve is a 10 💀

theres 0.01% chance i do it tho

i like algebra but can't do shit about sequences

only thing i can do is muirhead

[edgy vapor]

try C10..

[severe eagle]

which one

send pic pls ty

[edgy vapor]

jk I don't have one

[severe eagle]

okay

[edgy vapor]

but they're C8,9

in sl2001

[severe eagle]

okay

[sacred perch]

yup but we need to find the least n ! so i thought it would be pretty good to rule that out first because all primes > 100 like 101,103 etc satisfy the condition but they're not the least

[sacred perch]

i see, a pretty good intuition which i thought of would be to choose a composite number with the greatest number of prime factors (preferably not distinct).

[sacred perch]

it's just bashing and smart guessing i think
i can explain the conditions tho if you want
n>4 because 4^3 = 64 which can't have a gcd exceeding 64 but we want gcd greater than 100 so that's ruled out
the least number's cube greater than 100 is 5 so we start from that

[near hill]

Also, since 8 obviously works, all that remains is to show that 5, 6, 7 don't. There’s no requirement of being all smart and generic here, when a more general method wpuld almost certainly be more complicated than just checking those three cases already. If the problem setter wanted that, they'd have chosen a higher threshold than 100.

[sacred perch]

agreed was just curious for more approaches, it was a very easy problem after all

[near hill]

For larger thresholds, I think the best way is still trial-and-error starting from the cube root of the treshold. Primes near the cube root cannot work; nor can two times a prime (there wouldn’t be enough copies of the prime in the factorial), but the first number not of that form I'd expect to work in the sense that the gcd is the cube.

Any number with at least 6 divisors clearly satisfies gcd(n³,n!)=n³, so the exceptions can only have the form pq or p^k with k<5.

[frank valve]

n! requires n to be an integer and you’re looking for the least n, so you can just count up from 0 until you find a number that works and that’s your answer

[near hill]

For pq, unless p or q is 2, the factorial contains p, 2p, q, 2q, pq, so there gcd(n³,n!)=n³ too.

[frank valve]

I don’t really get the appeal of this question

it’s obviously 8

[near hill]

I'm just considering whether it would be more interesting if it had said eg gcd(n³,n!)>1234567890 instead.

[hidden ruin]

For anyone who solves this combinatorics question is a genius: How many 7 digit integers can be formed using exactly 3 distinct digits?

hint: Zero cant be first digit

[crystal rock]

Sounds doable

[sweet pewter]

that sounds very doable lol

[crystal rock]

Just work out 2 cases ig

[sweet pewter]

no need to

[crystal rock]

Yeah ok I was probably stupid

[sweet pewter]

nvm when you consider the first digit as 0 you need 2 cases

[crystal rock]

👍

-# 🧿

[crystal rock]

I worked it out but it doesn't feel right lmao

[soft olive]

has anyone here done the IOQM maths olympiad, my school is proviing free tutoring this year for this exam and i enrolled myself not expecting much

[hidden ruin]

What is the final answer guys

If it was that easy

[queen moss]

is anyone here familiar with the contest collection of Art of Problem Solving website? i was wondering what contest to study if i'm an amateur proofwriter

[reef trail]

same...

[hushed charm]

it's impossible to say what would be nice to start for you with so little information, so the only way you can discover it is by doing problems. Start with the easy ones, if it is too easy, try to solve harder problems, if it is hard, keep with it, if it is tooo hard, try something easier...
Focus on doing hard problems, always problems that you have a certain difficulty to do, but never something easy, or too hard to be done by you.

[reef trail]

Oh, younger than me

[crystal rock]

I don't even know if I'm allowed to send it

[simple gulch]

And I'm 16 me too so we have a lot in common😔

[crystal rock]

If you only want the final answer then maybe just write a code to count it.

[reef trail]

Its cause we are in differents countries

[crystal rock]

oh hi
-# 🧿

[near hill]

Just count solutions without the no-leading-zeros condition at first. By symmetry, 1/10 of them will start with a 0, so you can subtract those afterwards.

[reef trail]

where are u from @simple gulch ?

[reef trail]

Somone knows "El traductor de ingenería"?

[crystal rock]

[X] Doubt

[crystal rock]

Wait can you explain the symmetry thing wha

... oh nevermind

That's a bit quicker thanks

[simple gulch]

I'm from France, croissant cochon baguette béret

💥💥

[reef trail]

REALLY?

[sweet pewter]

oh yeah

I think I should utilize it more in combinatorics

I've never used that before

[jagged current]

"engineering translator"...?

[reef trail]

YES

Its a Channel in youtube

some videos are translated to english

but the native idiom of Damian is in spanish

[novel karma]

https://programs.mcs.cmu.edu/arml/wp-content/uploads/sites/3/2023/09/algebra-jv-10-06-19.pdf
anyone know where to find the answers to these

pls ping

[ivory ember]

AMCs and AIMEs can be found on the aops pages for the respective past contests

Here’s NYCIML: https://www.nyciml.org/free-past-contests

This is the same as 3.10 with different numbers, or just use wolfram: https://artofproblemsolving.com/community/c4h565120p3305179

4.1 is 2014 AMC 12A p18

4.2 is this: https://artofproblemsolving.com/community/c4h61810p370838

4.3 is 2008 amc 12b p23

4.4 is 2013 amc 12b p22

4.5 is p2 here: [https://www.dropbox.com/scl/fi/4t45xopz5eganlxo4f5wr/MockAIMEI2015Solutions.pdf?rlkey=6o58y4epm641oxhwr4atfjokn&e=1&dl=0]

[novel karma]

thank you 👍

[high goblet]

have u seen evan chen's OTIS excerpts?

i think that can be a useful starting point for improving

idk what country ur trying to represent but start off with like the hardest level of C u can do and just attempt problems slightly above that

tbh i was discussing with some of the ppl who run the UK olympiad training program and like we think combi is just the one area of maths u just get better at over time

exposure to more maths etc. helps a surprising amount

anyway otherwise just grind the C shortlist ig

[high goblet]

can't go geo to save my life

(don't have to do it now though!)

like the only IMO geo i did by myself was IMO 2006 P1 or P4 (can't remember which)

this lack of geo skill is a bit of an issue if i am trying to cook up olympiad problems involving L and Y

[dusky isle]

yoo what is that name 😂

[severe eagle]

yes i have

[severe eagle]

fair yes

[severe eagle]

[severe eagle]

am from singapore btw, so i will do smo open round 2 combi questions ty

but the thing is that i cant do 90% of combi in smo open round 2 but i can do 90% of geom there 😭

btw smo open r2 is essentially singapore TSTST its not the prestigious because singapore is smol nation (less people)

got quite long time ahead of me so can grind

[rain ravine]

$(\cos^2\theta-\sin^2\theta)\sqrt{\sin^2\theta+\frac{2gh}{v^2}}=\sin\theta[\frac{2gh}{v^2}-(\cos^2\theta-\sin^2\theta)]$

[gilded halo]

Lost fruit

[rain ravine]

How does one get $\sin \theta =\frac{1}{\sqrt{2+\frac{2gh}{v^2}}$ from that

[gilded halo]

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

[misty torrent]

you expand everything out and solve for sin(theta) lol

all the terms cancel out

[rain ravine]

I hate physics

Wait did you actually work through it 😭

[misty torrent]

yes

[haughty portal]

Hey!!
Anyone from India here? Targetting ISI?

[edgy vapor]

I like N in SMO, like 2023 senior p2 p4 is really fun

[severe eagle]

really

lemme see

Which country are you from (if you dont mind)

[edgy vapor]

here

If I’m not wrong it’s from Singapore right?..

[verbal breach]

hi recently joined the server here for a competetive exam n yaa i am new to discord so sry if i make a mistake

[ornate blade]

all good, you're in the exact right place

but to let you know, we also have #prealg-and-algebra , #geometry-and-trigonometry , #precalculus for regular school maths

[verbal breach]

hahah glad i found the right place

[ornate blade]

depends how much like competition maths your exam is

some of those can be quite like olympiads, and others not so much

I just had a feeling that you'd be okay here

[verbal breach]

ohh okok actually i am preparing for jee 2026

[ornate blade]

ahhhh

yeah then this is the right channel for JEE honestly

[verbal breach]

ok coolll

[light cape]

Hello , solve this .

[vernal axle]

Make a sub t=e^x, then it becomes 2arcsinh(t)/(t+1)^2, then do it by parts. You get the integral of 1/((t+1)sqrt(t^2+1)). The rest is a standard exercise from textbooks with sub z=1/(t+1)

[calm wagon]

hey guys im tryna study for a competition for year 13 how can i be fully prepared?

[rare bluff]

I just found this complicated formula in Luschny's old works.

man so beautiful

[proud gust]

4th question is really nice

[proud gust]

For the second question:
||If 3 doesn’t divide k, RHS is divisible by k therefore LHS is divisible by k therefore a^2=2(mod 3) contradiction
So 3|k. Will continue in a bit||

[viscid atlas]

<@&268886789983436800>

[proud gust]

Thanks

[versed dagger]

Man, if I want to continue being a mod here I gotta work on my APM

[stark swallow]

guys how and why cosnπ = +-1

[severe eagle]

I have an olympiad tomorrow and my combi sucks do yall have any advice (olympiad is 3 hour 5 question, i can complete 3.5/5 usually but to get a good placing i need better marks)

[stark swallow]

where are u from

[severe eagle]

why do you ask

[soft vigil]

Arent you from singapore or smth

[severe eagle]

mhm

[soft vigil]

Ok

Good luck

[severe eagle]

thanks

[soft vigil]

What oart of combo sucks

Just make sire to like

[severe eagle]

everything

[soft vigil]

Remember php and invariant

[severe eagle]

except graph theory

[severe eagle]

thats the thing

how tf do i find invariance

[soft vigil]

Wdym

You just play around

[severe eagle]

this is easy af

but couldnt find the solution after a while

ended up using Mantel's theorem on graphs

@soft vigil

like what are common things to consider

[soft vigil]

Invariant

Pigeonhole

[severe eagle]

Okay

Thanks ig

[soft vigil]

Also like

Some problems can be turned inti graphs

[severe eagle]

Yes

[pallid tundra]

this is not competition math.

#geometry-and-trigonometry

[edgy vapor]

Any hint for this?

my mentor said make it in pairs but I didn't try

[versed dagger]

maybe try a pair of small numbers explicitly to see if a pattern becomes obvious?

like perhaps 11 and 7?

[fair parcel]

I don't know where to put this so it might as well go here.

I guess there should be countably infinite x>0 and y>0 rational solutions to x^3+y^3=9, by drawing a line or a tangent line that intersects the curve at a rational solution.

How can I somehow "minimize" the numerator and denominator? Is there some constraint that bounds the numerator(denominator) that is easy to work with in Q?

https://math.stackexchange.com/questions/2473850/prove-that-x3-y3-9-has-infinitely-many-rational-solutions
https://math.stackexchange.com/questions/1794726/rational-solutions-of-x3y3-9

[vernal axle]

The first sum is just a number of integer points inside a triangle under the line y=ax/b on [0,b-1] (including the line itself) and the second sum is the same thing for the line y=bx/a on [0,a-1]. Since these two triangles are equal (one is a rotated by 90 degrees copy of another) so the number of points with integer coordinates in them are equal.
You can even count the value for each one in terms of a and b as these two triangles make together a rectangle of integer points. Each sum equals (ab-a-b+gcd(a, b))/2.

[edgy vapor]

I never dream bout this sol even once ,thanks

[sacred perch]

is there any good resource like video or text/book on Chinese remainder theorem? I understand mod and stuff perfectly but for some reason i'm not able to make much sense of CRT especially when the mod equations are 2>

[subtle sundial]

right, i’ve had this problem where if the variables arent in the order x_3>x_2>x_1 the variable values end up in the negatives or in fractions. Prime Newtons has a beautiful solution though which for me has worked 100% of the time.

i’d also search up Errichto Algorithms explanation on the crt. since his method also works on non-coprime modulos albeit more tedious.

[sacred perch]

Thanks, i'll check it out!

[subtle sundial]

happy to help man! goodluck

[slate hatch]

Solve question number 12

[dry quartz]

no

[slate hatch]

This question contains multiple correct .One or more than one are correct

[sweet pewter]

try drawing a rough graph of it

both f(x) and y=x are odd functions

[slate hatch]

I solved
I just find inverse of the given curve
Then find out derivative and it’s always positive . So no maxima or minima

[slate hatch]

Thanks for the help
I am going on by your method

[sweet pewter]

that's a nice method as well tho for some reason we are never taught about inverse functions here

after 12 years

[slate hatch]

Got this

[sweet pewter]

so my immediate thought is just to draw a graph

[slate hatch]

Yeah

[serene perch]

I love catss

[tiny grove]

I wanna do amc 10

Im struggling bro😭

Someone told me to study hard and try to do the olympiad but i cant even do amc 10's first problems

[trim bramble]

do u got a problem ur struggling on?

[tiny grove]

Eh idek

Just overall

Maybe its number theory

Ive never learnwd number theory ive just covered the 4 core hs classes

[trim bramble]

yea that’s the same problem i ran into

[tiny grove]

Oh you took the amc 10

[trim bramble]

yea

[tiny grove]

Wooo

[trim bramble]

did bad last year, my school for some reason put a mandatory 1 hour assembly right before

[tiny grove]

Was it that bad in actuality

[trim bramble]

and i was exhausted starting the test

[trim bramble]

i didn’t think it was horrible

did some dumb things cause was tired

[tiny grove]

Not being like rude but are your smart and i dont mean like wtv i mean like are you an avg guy or someone who like has credit if you get what i mean

Like id like someone to talk to a little deeper

[trim bramble]

nah i get what u mean

[tiny grove]

Not smart just know how to get further than the amc 10 and the amc

Did you stop at the amc

Some hs senior apperwntly got qualified for the usamo at my hs so yep

[trim bramble]

damn

they’re crazy

[tiny grove]

Self taught multi var calc year 12😭

I was gonna do multi var calc like in 2 or so months but ive been moping

You there

[tiny grove]

Mate come back

[trim bramble]

sorry eating lunch

[tiny grove]

Your the only socials ive had all day

Oh it might bw my wifi

[trim bramble]

😭

there’s a climbing competition on rn that my friends are competing in

[tiny grove]

Oh sorry

My wifi is atrocios

0.52 mbps a second upload😭

Did you get past the amc 10 btw. The aime i mean

[trim bramble]

i did not last year

i honestly learned about all this super late

and played catch up for like 3 years

if i took it now i could probably make aime

[tiny grove]

Dawg my internet crashed fkr 26 mins after this

Did you graduate

Also whats thw diff between amc and aime

Com3 back

[trim bramble]

nope

going into junior

[trim bramble]

tougher problems, not multiple choice

[dapper path]

.

wait are you guys from?

[mystic shale]

Yes, my name is indeed from, wbu?

[dapper path]

like what continent are you from?

your name is from?

[mystic shale]

My name is from my perpetual tree cutting habits

[mystic shale]

And my continent is atlantis

[urban parcel]

are there any good remote online math competitions with certificates/awards that I can join for free? or almost free?

[reef condor]

International math bowl

Is free

[urban parcel]

thank you

just signed up

anything else?

[wanton ridge]

whoa :o what's it like down there?

[mystic shale]

Cold.

[tiny grove]

Bro

How much tougher btw i hearf like thw hardest amc problems are the first aime problems

Also fo you have to get elected into aime like usamo or is it only usamo

And to get into usamo do they only check your score or your score and time on amc and aime too

[pallid tundra]

time?

what does time have to do with any of this?

where did you get this info???

late (final 5) AMC problems are roughly comparable to mid AIME

for AIME qualification your AMC score is what they use to determine cutoffs

then for USAMO they use both your AMC and AIME scores to calculate an "index" which is your AMC score plus 10x your AIME score (max 300)

and use that to determine qualification

[stark swallow]

I am unable to understand why the area of triangle CFE is CX^2

[sweet pewter]

∆CFE is right isoceles so CE=CF=√2 CX

[ornate blade]

You can rearrange the two smaller isosceles triangles into a square

[tiny grove]

So the cutoff is only score

[glossy nacelle]

im too old for competitions now rip

[tiny grove]

Im too stupid for ts compition bro🤞 😭

[glossy nacelle]

"ts competition" 🥀

[tiny grove]

🥀

I heard aops book, volume I is good for amc 10 and aops volum 2 is good for aime but gow much would it help

Also whats covered in volume I besides from standard us ciriculum

Number theory?

[tiny grove]

Wait question

If i buy volume I of aops and just go through that and then do problems on alcumus am i good enough for amc10

Also what does volume I vover and samr for Volume II and how far would both grt me (compitition wise)

[summer roost]

Volume 1 seems outdated for amc 10

The difficulty is easier than amc 10 problems

[tiny grove]

Which volume to get

Volume 2?

[tiny grove]

What does volume I cover btw and what foes volume II vover also is II good enough for aime too

[pallid tundra]

look up its table of contents

the problems are approaching too easy even for AMC 8

[coarse shuttle]

Does anyone know where I can find solution manual for everaise math II

$Let a,b,c,x,y,z$ be real numbers such that $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 52$$ $$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 147$$ $$\frac{xyz}{abc} = 9$$ Find $\frac{x^2}{a^2} +\frac{y^2}{b^2} + \frac{z^2}{c^2}$

[gilded halo]

donut123

[frank valve]

I found it, it's right there

[ivory ember]

,w 52^2-21479

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

I’m getting this if you want to cross check

[coarse shuttle]

alr well for some reason im getting divided by 9 instead of multiplied

but i think i had the right ideas

prob some calculation mistake

[coarse shuttle]

wdym

[ivory ember]

[ivory ember]

Probably trolling (similar logic to the meme where it says “find x” and the kid literally circled the x)

[acoustic nova]

if u let a = b = c = 1, it shouldn’t be too bad. I don’t know if the algebra as is, is better though

[ivory ember]

I mean it’s basically a system in x/a, y/b, z/c so whatevs

[amber nexus]

Can anyone help me with this question?

Let A be a real root of x³=x+1:

1 - Prove that there exists an integer M such that |M-A⁹⁰⁰|<4⁻¹⁰⁰
2 - Find a pair of integers p and q such that |A- p/q|<1/100
3 - Find the unity digit of ⌊A²⁰²⁵ + A⌋

[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

[amber nexus]

I don't know where to begin

[acoustic nova]

do u have a source for this question?

[tiny grove]

I see

Uh

Should i just get a number theory book as ive never covered it

Or volume 2

[acoustic nova]

Nah

ok I don’t love the vol 1 vol 2 books but content from the rest of the books is enough to get 25/25 on this https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems for instance

[sacred perch]

weird question

this is A^900

idk

[tiny grove]

So then

Like

Could i get Volume I to cover amc 10 and amc 12,m

[radiant jasper]

Bro this is horrible

[radiant jasper]

Consider the sequence defined by a_(n+3) = a_(n+1) + a_n for appropriate integer values of a_0, a_1, a_2.

you should know that if B,C are the complex roots of x³-x-1 then you have the closed formula
a_n = xAⁿ+yBⁿ+zCⁿ
for some reals x,y,z

If you never did problems like this one maybe it can feel a bit weird, but the power of this approach is that if a_n=xAⁿ+yBⁿ+zCⁿ and all the roots r except one have |r|<1 then a_n for n big is very very close to kRⁿ where R is the only root with |R|>1, and in particular a_n is an integer (because the sequence is an integer sequence), so you get kR^n is super close to an integer and this allows you to solve this problem

then I think it's useful to
||notice that here |A|>1>|B|=|C| ||
||choose appropriate values of a_0, a_1, a_2 so that x=y=z=1 so a_n ≈ Aⁿ|| (you can do this with ||vieta's formulae||)

Then the skull part in the part 1 of your problem is to prove that |B⁹⁰⁰+C⁹⁰⁰|<(1/4)¹⁰⁰ 💀

so I'm a bit sussed, I'm not sure if this is the best way to do it

But then parts 2 and 3 are much easier. For part 2 just do approximations and for part 3 use what I was telling you about earlier: ||A²⁰²⁵≈a_2025 and you can just look at the sequence mod 10. Then the +A will just add 1 because A≈1.3||

[radiant jasper]

Bro this approximation is so fucking sharp

Fuck

Ok I'm missing something, you can probably do something very similar to what I said

how is the first part so much harder than the second and third bruh

Probably just a skill issue

I think I got it

nvm probably not

[near hill]

This may be easier said than done -- the sequence mod 10 seems to be quite long too.

[radiant jasper]

Well you do mod 2 and mod 5 separately and surely mod 5 is not too long

[near hill]

Ah!

[radiant jasper]

Anyway mod 5 it's quite fast to do so even if it's a bit long it's not too bad

[near hill]

Okay, mod 5 gives length 24.

[radiant jasper]

XD

probably not the best but ehhhhhh idk what else to do

[near hill]

Unless I did Viète wrong and used an incorrect starting state ...

[radiant jasper]

💀

I think it was 3,0,6?

[near hill]

Ah, no, the single shorter cycle mod 5 is 1, 2, 4, 3, 1, 2, 4, 3, ....
and a1 definitely is 0.

[radiant jasper]

The cycle should be at most 5³ which is very small :)

[near hill]

Yeah, I computed the 24-cycle starting from 1,0,0 and it doesn't have n,0,0 for any other n, so there are 4 different cycles that are multiples of that, totaling 120 triples. Then there is the trivial 0,0,0-cycle, and 1-2-4-3, with length 4, and then all 125 triples are accounted for.
Edit: Whoops, I cannot arithmetic -- there was a 4×24 is only 96, not 120. I had missed a final 24-cycle starting with 1,1,0,...

[near hill]

3,0,2: we have A²+B²+C² = (A+B+C)² - 2(AB+BC+AC) = 0² - 2×(-1) = 2.

[radiant jasper]

Oh yeah i accidentally multiplied by 3 again

[near hill]

Hmm, the magnitude of each of B and C is sqrt(1/A), so |B^900| = |C^900| = A^-450.
So we want to prove that 2 A^-450 < 4^-100, or A^450 > 2^201.

Wolfram plus my calculator says that A < 4/3 < 2^(201/450), but establishing the second inequality with pencil and paper still looks a bit forbidding.

But wait! that's an inequality in the wrong direction. 💀

[radiant jasper]

yeah it looks like you can't just use 2|B⁹⁰⁰| instead of |B⁹⁰⁰+C⁹⁰⁰|

Maybe you can do something with induction

But idk and I don't have time now

[near hill]

Wolfram also gives the argument of B (or C) with many digits, and 900 times that is about 65° (mod 360°) . So |B^900+C^900| is close to |B^900|, but the last factor of 2 won't make a large enough difference.

But the difference between A^900 and the nearest integer is definitely exactly B^900+C^900, so I think part 1 of @amber nexus's problem is simply wrong.

[amber nexus]

Thx for the help guys

A friend sent me this problem, he said the problem is from a class given by an IMO medalist

But he doesn't have the solution

[calm tangle]

Hi, I was wondering if anyone have some interesting math problem that they can share , thank you!

[wanton ridge]

for me, not atm, however you can look at help channels; you don't need to reply to them, but you can try out the problems :>

welcome to the mathcord btw

[calm tangle]

Alright, tysm

[coarse shuttle]

Let $F_i$ be the $i$th term of the Fibonacci sequence, defined by $F_0 = 1$, $F_1= 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n > 1$. Show that the sum $$\frac{1}{F_0F_2} + \frac{1}{F_1F_3} + \cdots $$ never exceeds 1

[gilded halo]

donut123

[near hill]

Experimenting with a calculator, it gets pretty close, so there's probably not any room for approximation.
This smells of telescoping, somehow.

[blazing pilot]

it's exactly 1, just use partial fractions and telescope

[novel karma]

Did anyone here to to mao nats

[coarse shuttle]

how do u partial fractions it?

[near hill]

The question was how do you pick numerators such that the difference works out to exactly 1/F(i)F(i+2).

[scenic sun]

Ohhh okay. I’m sorry, I misinterpreted the question

[near hill]

Presumably we'll need to find A and B such that AF(n+2) - BF(n) = 1.
It's true that F(n) and F(n+2) must be coprime, so we can Bezout them somehow, but the precise form of the coefficients is not ready knowledge for me.

Ah, googling for "fibonacci" "bezout" led to http://www.ryanhmckenna.com/2015/03/investigating-bezouts-identity-for.html which derives some coefficients.
However, it won't be a simple telescope with those coefficients; the tail terms don't go to 0. But it looks like one can it such that the tails for odd and even n go towards phi and -phi, such that they cancel out in the limit.

[coarse shuttle]

Yeah I doubt that is the intended solution

It’s cool

But given the context from where I got the problem lkkelh not

[vernal axle]

I don't see a lot of difficulties. Just write 1/(F(n)F(n+2))=1/(F(n)F(n+1))-1/(F(n+1)F(n+2)) and it telescopes nicely. According to Fibonacci's definition F(n+2)-F(n)=F(n+1).

[near hill]

Oh, clever. But I'd dispute the "just" -- you need to pull 1/(F(n)F(n+1)) out of thin air to get that argument going.

[vernal axle]

no, it is quite natural. You have 1/F(n)F(n+2) and you definitely want 1/F(n)-1/F(n+2) which equals (F(n+2)-F(n))/(F(n)F(n+2))=F(n+1)/(F(n)F(n+2)). So, the obvious thing is to divide 1/F(n)-1/F(n+2) by F(n+1) to get our n-th term. And that's it

[near hill]

you definitely want 1/F(n)-1/F(n+2)
But at the end of the day that's not what you're getting.

[vernal axle]

sure. But that's a standard thing to start with when you want to sum something like 1/n(n+1) or 1/n(n+2) aiming at some later telescoping. As someone mentioned above just do partial fractions. And that's usual I think, when you have something like 1/(AB) to try to represent it as 1/A-1/B.

[near hill]

Yes, trying to represent it as 1/A - 1/B is natural, but when that doesn't get the right numerator after you multiply out, I definitely think it's stretching it to describe the inspired save here as a "just" thing.

[vernal axle]

In most cases yes, you have to divde by B-A. Which may be constant or just suitable for your purpose of telescoping. Just like with these Fibonacci. Anyway it is a popular trick. So, for me it falls into a category of 'just' 🙂

[near hill]

Generally if B-A is not constant, dividing each of 1/A and 1/B by it would destroy the telescoping property. It seems extremely specific to this problem that it just happens to work out such that a different telescoping becomes possible. Dismissing that as "just partial fractions" still sounds arrogant to me.

[vernal axle]

idk, it is the first thing that came to me, and it worked

[near hill]

I'll retract my "out of thin air" comment, though.

[inner oar]

who want to play roblox with me @everyone

[severe eagle]

Wtf

[sweet pewter]

game rách dell ai hỏi

[mystic shale]

Me, send me friend request: imdefinitelynottrollingyou@123

[sweet pewter]

given acute triangle ABC (AB<AC) inscribed in (O), altitude AD, H is the orthocenter. BD = DE. EH intersects AC at F. Prove ODF = 90°

[severe eagle]

nice diag

lemme try

[sweet pewter]

(it's very difficult)

[severe eagle]

we'll see about that

[sweet pewter]

(or very easy if you remember a particular theorem)

[severe eagle]

[severe eagle]

||ok so first i reflected H and BHEH' is a rhombus and ABH'C concyclic||

dont tell me if im right

dont give me a hint pls

is this complex bashable

where are similar triangle

gonna eat

[stark swallow]

How can i apply mn cot theorem that is (m+n)cotx = ncotB - mcotC in this trianlge along the lenght AD. I am not able to identify which one is m and which is n

[vernal axle]

I don't know what the original problem was (I hope BAC=90), but if you are given only these angles, then you can denote EC=t, then using the low of sines you find AE in terms of t, then similarly find DE, and finally BE from the triangle ABE. So you will find BD:DE:EC is (supposedly) 1:2:sqrt(3).

[severe eagle]

blyat

@sweet pewter can i have a hint

cant find any similarity

NO

I FOUND IT

for one angle haiya

[sweet pewter]

did you solve it?

[severe eagle]

not yet

[sweet pewter]

if I do give a hint it'll prob spoil the whole problem

[severe eagle]

i need to prove spiral sim

and i got an angle

[severe eagle]

okay then i will wait

ty

[sweet pewter]

actually I'll give a hint

since this is not meant to be easy

[severe eagle]

is this imo p1/4 level

[sweet pewter]

I've seen solutions involving Brokard's theorem

[severe eagle]

[severe eagle]

what

heard of that

[sweet pewter]

idk I found this on Facebook and solved it

[severe eagle]

okay

[sweet pewter]

took me a long time to remember this theorem

[severe eagle]

you did it first try?

[severe eagle]

but you remembered it

pro

[sweet pewter]

not Brokard

[severe eagle]

ah

[sweet pewter]

https://en.m.wikipedia.org/wiki/Butterfly_theorem

this

[severe eagle]

i know that theorem

but didnt think of using

hmm

[sweet pewter]

I breifly remember having to prove this in one of the first lessons on cyclic quads

[severe eagle]

but how is that applicable

[sweet pewter]

you first step is correct

[severe eagle]

yay

[sweet pewter]

hint: use the inverse of the butterfly theorem

[severe eagle]

okay

i do tmr

got homework i didnt finish

https://tenor.com/view/water-spray-stop-sprinkle-water-gif-11895255

[crystal rock]

hmm.. that will definitely require something else on DF

yeah that works because of the rhombus lol

[mystic shale]

<@&268886789983436800>

[torn solar]

bro the diagram is soo neat XD

[sharp sandal]

I thought it was on notepad but it's hand drawn, damn

[mossy viper]

id say its a proof level version of an aime p7 so wayyy lower

[soft vigil]

Aime p7 💀

[rare bluff]

is this integration solvable??

[severe eagle]

Aime p7

[vernal axle]

yes, it equals ||arcosh²(a)/2||
which is ||ln²(a+√(a²-1))/2, if a>1, ||
||and -arccos²(a)/2, if -1≤a≤1.||

[rare bluff]

thx for help me solve it!!

[vernal axle]

This can be bashed with Ceva's (2 times) and some trigonometry

[mystic shale]

<@&268886789983436800>

[formal ice]

yo so uh.........

what do yal do when u get stuck ona prpblem you dont know how to solve

lwk sunds dumb

BUT

sometimes i encounter probelms with skills i havent practiced in a while

even though theyre easy i cant figure it out sometimes

in general i get stuck on "easy" problems because i forgot something

does naybody know a remedy to this?

[wanton ridge]

review textbooks & lecture notes / open help channel here

or just keep doing problems that way you'll never forget

[formal ice]

oh true

[torpid cairn]

has anyone here done usamo?

[torpid cairn]

if so, what'd you guys get in your HS years on the 3 MAA comps? rn I think I'd get ~120 AMC 10 and ~5 aime as an incoming sophomore and I'm curious how that compares to others

[dusty fable]

anyone know any good courses to get good at BMO

also i have made a sort of problem set with olympiad style questions if anyone wants to try them i rlly want someone to give it a go

[daring totem]

if its basic math for 9th grade then give it to me

[pallid tundra]

just making AIME already places you well above average

https://artofproblemsolving.com/wiki/index.php/AMC_historical_results?srsltid=AfmBOopvk5mVkgqyEE2MoOeVTwsEtbXZG9KSWiP-qJkZLoxRpeIpmLGS

see here for past score summaries

[dusty fable]

it’s not that basic but i have one u can think about “ give me an interval of 999 consecutive integers where none of them are prime”

[torpid cairn]

I'm sure, but my goal is USAMO and I'm more interested in how other usamo qualis improved over time

ty tho

[torpid cairn]

||[lcm(2,3...1000)+2, lcm(2,3,...1000)+1000]|| right

i could probably optimize

a little bit

[acoustic nova]

Yeah

Or if u want ||1000! + 2||

[soft vigil]

Ive done jmo once at got hm but thats it

[soft vigil]

Hmm

I know people who have done like

Sub 120 amc and made jmo through like 10+ aime

[torpid cairn]

deadass?

that's insane to me wtf

bc I'm genuinely horrible at aime problems but 120 is light on amc

[torpid cairn]

still impressive but I'm afraid it's probably too late for me in that regard

[ornate blade]

120 is way above the AIME cutoff though

it would only make sense to practice AIME difficulty questions after qualifying

[torpid cairn]

AMC score is factored into USAMO qualification, too

I have done quite a significant amount of aime problems but I have not improved at them to the degree I did with AMC - I went from like 80 to 120 on the previous tests

[ornate blade]

oh I mean you guys are American/Canadian? ah

[torpid cairn]

yep

lol is there another aime?

[ornate blade]

no it's just that you can't enter for USAMO otherwise

there's residency/citizenship requirements

there is another AMC though, the Australian one

it's pretty damn popular

[torpid cairn]

ohhh ic

[dusty fable]

this is the one i thought of

[dusty fable]

but yes this suffices

[plucky shard]

When Houston university contest

[modern torrent]

<@&268886789983436800> sorry for ping but spammers above

[daring totem]

500

[zealous quest]

can somebody that trains for a NMO tell me how how he learns to solve problems

[reef condor]

1000! To + 1000
O edit it a bit but I doubt 1000!+1 is prime probabilistically speaking

[honest barn]

But it could be

[near hill]

Factordb.com says 6563 divides it.

[honest barn]

Aww…

[acoustic nova]

https://oeis.org/A002981

yeah it’s not prime