competition-math

[sturdy panther]

so our odd series' denominator repeats every 50 numbers

[sturdy panther]

but i need to double check this statement

oh no

its 24!

so our odd AND even summations repeat themselves every 24 counts.

but i really don't understand man

if they do repeat themselves how is this series convergent?

and how can it be expressed as p/q

[sick tapir]

is the answer 4 ?

man i m gettin 97 and 4 both

by 2 different approaches

its been a while since a math question got me this good

[sick tapir]

i got this

it has to be one of these 2

and dont tell me you dk the soln

method 2

[fallen magnet]

interesting way to write sigma

[fallen palm]

thats what im thinkin

[sick tapir]

well when you'r confident then you don't c how much space you occupy

so thats now my habit

also my hands were just flowing while attempting this question for the 3rd time

i m surprised that its been almost 4 months since a question took me more than 2 attempts to do

[sick tapir]

i rem u

[hexed oak]

i just had a class on zsigmondy, and i blanked out (figuratively) when they started explaining polynomial LTE (they were doing something invloving primitive cyclotomic polynomials) cans someone help me

[sick tapir]

so just watch the recording again and again and again ....................

[hexed oak]

its a long lecture

[sick tapir]

then give up crying ........

[hexed oak]

i am not giving up

[sick tapir]

brahh my brain aint braining that why u even wrote this shit

how can any1 even ????

[hexed oak]

even what

[sick tapir]

just study man u r here crying you sins

[sturdy panther]

wait a second

i have a question and it might sound stupid

but hear me out

when we have the summation from 1 to inf

there are infinite amt of times where cos pi*n/24 is equal to cos pi m/24

where n and m are two distinct integers

won't the series be divergent??

i found that after the even and odd simplification, in both cases, the cos term repeats at an interval of 24

so we are basically summing up the output from 1 to 24 over and over again endlessly.

the result would be undefined, right?

im not sure

oh wait?

the 1/2^k prevents divergence i guess

so i have an idea but i have no guess abt how it would be p/q form

No no!

the series IS divergent!

[sleek slate]

It should be likr 12 diff geom sequences so it should converge

[sturdy panther]

question is unsolvable im pretty sure

[sturdy panther]

yeah that was what i was thinking earlier, but when i plugged the question into a cas system it said it was divergent

[sleek slate]

Oh what

[sturdy panther]

and how would we evaluate cos ( pi/12) , cos(pi/6), cos (pi/4), cos (5pi/12), etc 12 times (with some difficult trig values)

this surely wouldn't the type of problem from a math competition

5pi/12 would be a little long

7pi/12

[sleek slate]

im sure theres a simplification thats why its in a math comp

[sturdy panther]

hmm

[sleek slate]

U put 2 insteaf of n at the end of the numerator

[sturdy panther]

where?

ohh ok

oooh

well

it doesn't say its divergent anymore

it approaches a number close to 1/100

[sturdy panther]

i think 97 is correct

but ill check the geo series

[sullen crypt]

how u got selected in mit

[sturdy panther]

what you did wrong in your solution was take out the geo series without handling the trig. you need to go 12 or 24 times depending on whether you simplify the odd/even classes

[sturdy panther]

what do you mean?

[sullen crypt]

he is in mit

[sturdy panther]

so?

[sullen crypt]

so what exam did he do

[sturdy panther]

have you researched the mit admit process?

its not just one exam like our jee system

[sullen crypt]

nah

[sturdy panther]

the us handles college apps differently

[sullen crypt]

i am not going there

i will serve my country till death

[sturdy panther]

so you"ll make assumptions of people going there?

[sullen crypt]

for knowledge

[sturdy panther]

anyway, i apologize, let's continue with the problem.

[sullen crypt]

i to

apologize to disturb u

[sturdy panther]

nah no worries

its good to be patriotic

but sometimes we need save arguments for later

[sullen crypt]

ur idol

??

[sullen crypt]

i didn;t understand

[sturdy panther]

like the discussion was about a problem right?

if you want to know abt how he got admitted,

dm him instead 👍

[sullen crypt]

my bad i am sorry

[sturdy panther]

no worries man

dm him if you want 👍

[sturdy panther]

but...

in our odd summation, we would be forced to deal with an extra pi/24 because of the 1 in the form 2n-1

so we would ditch the effort of getting into the odd form form for odd summations

and instead evaluate by hand?

that sounds sad :(

[sand hull]

Blud jee is nothing

[sand hull]

How's your geometry

[sand hull]

OMC?

[hexed oak]

from different discord

but yeh

[sand hull]

Mmmm

[sand hull]

Up to inmo or smth?

[hexed oak]

wdym?

[sand hull]

Are you preparing for INMO?

[hexed oak]

yeh

u?

[sand hull]

Yeah for next year

[stable path]

are you in iit

?

[sand hull]

Apparently my geometry proof of p5 in RMO wasn't up to the mark and only got 10 marks in it , missed the inmo cut off by 2 marks

[hexed oak]

did you do the rotation qn?

[sand hull]

No but doing things better and way tougher than it

[hexed oak]

i got it

no

[sand hull]

?

[hexed oak]

inmo is easier the jee trust me

[sand hull]

Not really

[hexed oak]

yes it is

[stable path]

like what

[hexed oak]

jee has chemistry

[stable path]

class

[sand hull]

Inmo is easy though , I plan to appear for IMO in 2028

[sand hull]

8

[stable path]

wtf

[hexed oak]

i plan to appear in 2025

[stable path]

where are you from

?

[sand hull]

Woah great

[sand hull]

India

[hexed oak]

i got into camp last year

so have high hopes

[sand hull]

Woah!!!

[stable path]

ik state

[sand hull]

I wish you good luck dude

[sand hull]

I'm not saying in a public platform dude

[stable path]

kk no prblm

check dm'

[sand hull]

Chemistry is tough ...?

[hexed oak]

it has memorisation

[sand hull]

Chem never catched my eyes

[sand hull]

Ah so you mean it like that

I understand

No like if we just talk about maths and phy

[hexed oak]

oh yeh those are easy

[sand hull]

Phy is still sometimes good

[sand hull]

Btw which grade ?

[hexed oak]

12th rn

[sand hull]

Ok sure

[sand hull]

Ohh

[hexed oak]

U?

[radiant jasper]

nope

it's a pumac question

[radiant jasper]

What in the cursed

oh princeton maths

[woven ivy]

hey, does anyone have any tips on how to prep for a maths olympiad?

[spiral tapir]

Has anyone ever done MATH OLYMPIAD?

lol

[spiral tapir]

same here

[deft marsh]

f(xf(y))+xy=f(x+y) -2y

[deft marsh]

past papers help a lot but firstly u should get maybe a book or smth

for number theory , geometry , algebra , combinatorics

or just search on the internet for free ones

[deft marsh]

i found that f(x) = 2x + 1

[jade widget]

I haven’t get my amc score yet…

Is this normal?

[jolly solstice]

letting x=0 in the functional equation gets us f(0)=f(y)-2y so f(x)=2x+C for some constant C would have to be the form at least. When I plug that in to solve for C I get,

2x(2y+C)+C+xy = 2(x+y)+C-2y
simplifies to
5y+2C=2

But this can't be right because y is a free variable, so there's not a solution unless there are some more constraints in the problem you might have left out

[deft marsh]

yh mb i set y=0 and i got f(xf(0))=f(x) but it was meant to be f(y) not f(x)

😭

[swift imp]

No

[blissful tulip]

i got mine like over a month ago

maybe more

ask ur teacher/proctor

[swift imp]

I got mine late November so yea

[blissful tulip]

amc 10 or 12

[woven ivy]

i will do this, thank you
last year I tried just doing past papers and never end up doing the olympiad because I knew I was gonna get cooked

[jade widget]

10a

[deft marsh]

U got it this yearr

[woven ivy]

it's in March and I still don't know what I'm doing 😭

[deft marsh]

U got time dww

[sick tapir]

well i might say but geometry is the second toughest after calculus with trigo

i m good in it but not like i discovered my own formulas in it till now

but i did for trigonometry and integration

[sick tapir]

just don't die

[sick tapir]

ik it very well

[vernal axle]

In fact it doesn't matter if it is Pi/24 or any other angle inside the cos. You can take any. If you start summation from n=0 then this sum is always 0. So, p/q=-a_0=1/96. In order to see that, you may notice that this long expression is just a real part of a fairly simple expression in roots of unity, which in turn is a difference of two equal series, so it is 0.

[sick tapir]

my answer was right

[ornate coyote]

its joever

[radiant jasper]

the answer is 97

you can find the official solution on pumac archives

it's pumac 2018 algebra division a

[radiant jasper]

fr lmao

[alpine delta]

@sand geode

[plain jacinth]

How do I study for aime

I plan on doing aops precalc + vol 2

How bout after that tho

[lament frigate]

hi guys

today MAA updated the AMC 12a cutoff for AIME and now I'm able to go. The cutoff is 76.5 and I got 76.5. I'm just wanted to know how they send out the invitations to participate in AIME

[sick tapir]

whats pumac arvhives ?

[vernal axle]

Here is my solution.

[kind spear]

is it the case that one has to get better at amc

then aime

then move to regional mos?

because idk i've heard some people say that amc and mo's are vastly different and like not related kinda?

[sick tapir]

you dint solve till end

it goes further too after coming 1/96

[sick tapir]

and btw this method is long thats why i dropped in middle

[sick tapir]

hats off for thinking like doin like a professr

[faint mist]

is this australian or american maths competition?

[shadow spruce]

However if time is sufficient you can redo any question to ensure that answer is right

🤔

[jade widget]

Nah American one

[small rock]

Pardon me if this doesn't belong here. When you guys look at the solution after failing to solve a problem, do you guys also ponder on what lacked you to find the solution, what you could have thought that would have lead to the solution? I think I should be asking myself these questions while self studying. What other questions can I ask myself?

[wet grove]

I find it astonishing beautiful and elegant how there are connections between the simpliest of mathematics to the most hardest of concepts. If you had failed to find a solution, perhaps you may need to widen your scope of understanding of the problem and see how you can connect it to something else, which can then help you find the solution to the problem you're trying to solve. You're bounded by your imagination, is what is ultimately is

[faint mist]

I also ponder on why i couldn’t find the solution, especially if it was simple to understand.

[latent bough]

wsg guys

[latent bough]

I usually ask myself "how could i have seen this" or "what hints to this solution were there"

be careful not to get overly focused on deriving like specfic algorithms and just try to find connections between problems

problem solving requires creativity

[plain jacinth]

How to study for aime

[glass pelican]

Mock past aimes

[sick tapir]

Study 10 hours each day with 3 hrs 3 slots and 1 hr for practicing question based on the topics studied on the day. But only 7 question for that 1 hr and the questions should be of high level which should take only 7 min/ question and the rest approx 10 min to check the answers/steps/marking on the solution given.

so that you analyze your stuff..

[grand heron]

Where do you usually find online math exercise?

[sick tapir]

i got books

like euclidian geometry , Number theory , Real and Complex problems , etc

Online man u need ai to make you a question but then you wont have its solution or final answer

and the number of questions you get online are very few

which aint enough to qualify for competitive examinations

[coarse shuttle]

How do you actually swap order of summation?

Usually I just write out a grid and figure it out, but it takes some time

[sick tapir]

1. Understand the region of summation
2. Visualize the region
3. Determine new bounds
4. Write the swapped summation
5. Tips for Speed
   If you’re comfortable with inequalities, algebraically express the region directly.
   Practice with common summation regions like rectangles and triangles to build intuition.
   Use symmetry: Some problems naturally have symmetric limits that make swapping simpler.

[surreal knoll]

hi

i want to ask you about hte imc

and how can i revise to this competition

and if you can give some recources

[gusty verge]

IMC past papers and similar competition past papers, everytime a solution uses a certain trick study in depth that trick until you understand it to a high lvl

It's probably the hardest one to prepare for as the range of topics is insanely wide

Do Putnam papers too and Putnam and beyond is a good resource for that

[surreal knoll]

what does it mean Putnam

and i understand you but which recourse can i use ?

and this compitetion is for the student of the first year of universty or what

can you explain more about this competition

[gusty verge]

Putnam is the American equivalent of IMC

You can find resources online that tell you about it

[acoustic nova]

What is imc

[gusty verge]

Search up IMC Maths comp

[surreal knoll]

so just solving this problem

this problems

[prime raft]

disboard

[surreal knoll]

what do you mean

[prime raft]

nothin

absolutely nothin

[surreal knoll]

ok

[ornate coyote]

dawg aime isnt that serious 😭

[plain jacinth]

Any materials for AIME

[plain jacinth]

I plan on doing AoPS Precalc + vol2

after that are there any other Aops like books?

[sick tapir]

na

[sick tapir]

ik

but the way he's asking looks he is serious abt it

[sick tapir]

Try this and send answer pics pls

[burnt trail]

Alice chose five positive integers and found that their product was even. What is the maximum number of odd integers she could have chosen?

[sand hull]

4

[gusty verge]

Surely no competition maths? 😭

To be even you need a factor of 2, a single factor of 2.

So you only need one of the numbers to be even.

[gusty verge]

Spam identities until it's simpler, might require converting to cosx and sinx everywhere to make spotting the identities simpler.

Personally I multiplied top and bottom by $\cos^4 x$

Which leaves you with:

$$\int \frac{\sin^3x\cos^2x+1}{\cos^4x(1+\cos^2x)}dx$$

This integral is easily done by splitting the numerator and doing both integrals separately (the left fraction will require a $u=\cos x$ sub after using the Pythagorean identity on $\sin^3=\sin x \sin^2 x $, the right fraction the $1+\cos^2x$ hints at using the tan-sec Pythagorean identity, so divide top and bottom by $\cos^6x$ and use $u=\tan x$.)

[gilded halo]

Max

[dapper blade]

"easily"

[sick tapir]

well this is certainly right but it aint easy to think if u dint use this tool or bot or whatever it is

[sick tapir]

this expression comes in the solving

[dapper magnet]

what are u even talking about

[gusty verge]

Wdym by bot or tool? As in the latex bot? That's just to make the maths easy to read

[toxic imp]

S is the sum of two distinct positive 3-digit integers, both of which factor as the product of exactly three different primes. Compute the minimum possible value of S.

Best way to go about this? I could only come up with prime factoring brute force starting from 100 or just testing combinations

[gusty verge]

S is the sum of two numbers. To minimize S you need to hence minimize them numbers.

Now unless I'm reading this wrong it appears the only link between these two numbers is that they are not the same.

So essentially we need to find the smallest 3 digit number that factors with exactly 3 different primes and then the next smallest.

Yes it'll require a bit of brute force:

1. Factoring each number from 100
2. Creating the numbers, ie. 2×3×5=30, 2×3×7=42, 2×3×11=66, 2×5×11=110 ext.

[pallid dragon]

it's likely 2×3 something

not necessarily but likely

so we get 102

so you only need to factor up to 114

basically yeah there's no good way

it's only solveable because you would likely find both 102 and 105

early

[toxic imp]

Yeah that was my conclusion as well

Thanks 🙂

[pallid dragon]

you can mergesort the sequences, it's solveable without factoring just you need the computer

[gusty verge]

Now try make the most asymptotically efficient algorithm for two n digit numbers (assuming you have a perfect prime finder)

[pallid dragon]

if it was semiprimes you probably like go towards middle

i don't want to figure out how this works with 3 factors

[pallid dragon]

same way recursively

but what's the first line

2 3 something

[dry quartz]

Let $a_1, a_2, a_3, a_4, a_5 \in [-2,2] \subseteq \mathbb{R}$ such that $\sum a_i = 0$, $\sum a_i^3 = 0$, and $\sum a_i^5 = 10$. Find $\sum a_i^2$

[gilded halo]

fluX

[pallid dragon]

i thought you mean there's something better but maybe merging is already that?

actually there's no way it wouldn't reuse multiplications, i'm not thinking in the right direction

nvm, that's not slower asymptotically

[vernal axle]

You can use Newton's identities and get the polynomial P(x) with roots a1,...,a5 of the type x^5+ax^3+bx-2. Then the sum of squares equals -2a. Then you use the bounds for the roots and the fact that it has exactly 5 real roots. It looks like this polynomial is unique (and has multiple roots). And it equals x^5-5x^3+5x-2.

[pallid dragon]

@gusty verge it was easier than i thought but unfortunately unreadable

no reason to think it's optimal, but maybe
1309, 1310, 1311 are 3 consecutive, but i can't find 4

[pallid dragon]

it's ultra slow and can't do 10 digits

and factoring obviously can

oh but i can't tell how much of it is the prime finder

[ornate coyote]

how am i supposed to draw a diagram that looks like tihs

[gusty verge]

With a pen 🌚

[gusty verge]

Yeah I imagine that's the case

It relies on having a lot of knowledge about the distribution of primes tbh and obviously that is super uncharted territory

[pallid dragon]

i'll try the thing where you walk towrds the middle now really not viable

is it surprising that 3 in a row is abundant and 4 isn't?

can't find 4 in a row anywhere

[gusty verge]

You mean 4 numbers in a row that are a product of 3 distinct primes?

It cannot be possible tbh, at least two have to be even

[vernal axle]

You can use Geogebra

[pallid dragon]

oh it's cuz there's 2 and 2

in the same number

it's impossible

[gusty verge]

Yep

Kinda sad it's not possible tbh haha would've been a fun challenge

[pallid dragon]

i know right

[vernal axle]

You may allow equal primes among those three prime factors

[pallid dragon]

10003892,10003893,10003894,10003895

10092601,10092602,10092603,10092604,10092605,10092606

cool

[vernal axle]

find the longest possible sequence.

[pallid dragon]

it stops growing?

why

[vernal axle]

because every p numbers in a row have one divisible by p

[pallid dragon]

oh duh

16 would have 2 2 2 2

[vernal axle]

it has <=7 elements.

6 is already there

[pallid dragon]

why not 8?

oh

[vernal axle]

then it must contain 8 itself

[pallid dragon]

yes

[vernal axle]

do 7 exist at all?

[pallid dragon]

yeah

it dfinds the first one instantly

[vernal axle]

so what is the first one?

[pallid dragon]

further out 10092601,10092602,10092603,10092604,10092605,10092606,10092607 (same as 6!)

[pallid dragon]

211673,211674,211675,211676,211677,211678,211679

[vernal axle]

ok we have a function f(k) which is the starting number of the first longest sequence of consecutive integers having exactly k prime divisors. Thus f(3)=211673. Graph logarithm of this functon and find asymptotics ^))

[pallid dragon]

xd

[vernal axle]

and maybe g(k) is the length of that longest sequence, i.e. g(3)=7.

This one looks simpler

looks like g(k)=2^k-1 ?

[radiant jasper]

[dry quartz]

Ah, thanks

[pallid dragon]

32535999,32536000
experiment yourself https://onlinegdb.com/fork/uG8mBT03M
factors go to ~10 and length to 9, though not at the same time

[ornate coyote]

chat am i cooked

[hexed oak]

whats the qn

[latent bough]

hey guys

[keen nest]

hi

[balmy hound]

anyone want to find the determinant of a 10x10 matrix (random)

[gusty verge]

Yeah, what a fun activity

[gusty verge]

Let's make it more interesting, you have to do it only using the matrix minors method for find a determinant

So there will be 10!/2 2by2 determinants to work out

[ornate coyote]

chat is it hard to get into awesomemath summer camp

[reef condor]

not hard, if you know the math

[reef condor]

its mostly pay to learn and less prestigious

[ornate coyote]

K cool

I got 9 questions on the admission test

[warped anvil]

it is so expensive

i did all the problems and was gonna apply then saw 1.2k and i was out bruh

[radiant jasper]

hi

[radiant jasper]

wha

u do competition math too?

[hard crag]

Yo can anyone help verify it's 5

The answer key is non-existent

[ornate blade]

okay assuming there are 7 five-cent coins and 5 ten-cent coins

5, 10, 15, 20, 25, 30, 35 can be made from the 5-cent coins
so can 40, 50 from the 10-cent coins, and so can 40 + 5 = 45

hence 50 + 5, 10, 15, 20, 25, 30, 35 can be made
yes and no higher can be made, since 85 is the sum of all the coins

so yeah this is correct

[balmy hound]

there's another method?

[hard crag]

Thanks

[ornate blade]

no wait that's not 17 different values, I misread the q

there are only 10

[hard crag]

Wym

The answer is 10?

[hard crag]

It's 17 diff values wym?

[ornate blade]

sorry give me time

[hard crag]

K

[ornate blade]

you're correct I checked

[hard crag]

Alr thanks

[radiant jasper]

how ru so good

at competition math

and how can i get good fast

by the UKMT

date

[ornate blade]

I literally checked their solution and made a mistake before I edited

[radiant jasper]

doesnt matter

ur still damn good

i want to get good in UKMT

which is mcq btw

but i am struglling

any recommendations

[ornate blade]

great, UKMT is one of the more achievable competitions

which one are you doing? intermediate or senior?

[radiant jasper]

intermediate

[ornate blade]

ah okay

[ornate blade]

which topics are you struggling in?

[radiant jasper]

cant do the questions from 20-25

[ornate blade]

https://ukmt.org.uk/wp-content/uploads/2024/02/IMC_2024-Paper.pdf - last years' paper for example

[ornate blade]

ah okay that helps clarify a lot

those are supposed to be challenging

I'm guessing that you can't do the geometry questions which are unfamiliar at first

cause you haven't been taught, hey

[radiant jasper]

i did exam last year

got gold

but didnt qualify

[ornate blade]

okay keep at it then

[radiant jasper]

for the next olympiad thingy

[radiant jasper]

but i want to qualify though that needs to be my improvement

[ornate blade]

@high goblet you can talk to this guy in this channel

he's done BMO 1 and 2

[radiant jasper]

oh ok ty

[ornate blade]

there's really nothing I can say other than keep practicing

https://ukmt.org.uk/wp-content/uploads/2024/02/IMC-2024-Extended-Solutions.pdf

check q23 for example

[radiant jasper]

are there any books though?

[ornate blade]

they've used Pythagoras and tangent perp to radius in a smart way to show OTP is collinear

[ornate blade]

yes, the AoPS ones for the AMC (US)

[radiant jasper]

whats that

whats name of the book

i need to know

[ornate blade]

American maths competitions

[ornate blade]

https://artofproblemsolving.com/store

[radiant jasper]

which particular book though

should i use

[ornate blade]

https://artofproblemsolving.com/store/book/aops-vol1

[radiant jasper]

for UKMT

[ornate blade]

this one

[radiant jasper]

ty

would this have all i need to know?

[radiant jasper]

also are there no reps?

in this thing

[ornate blade]

nah

[hard crag]

Unfortunately I have skill deficit syndrome. thereforth I am not able to complete this wretched riddle. So I must say, help me.

Help

[high goblet]

for the maths challenges, you just have to practise a lot

they don't really require much theory so i don't think books etc. are necessary

make sure you do every problem when ur practising

you can obviously time yourself when u do mock papers but you should spend time after ur time is up to do all the problems

there is a big difficulty spike for the last 5 problems, but that's normal & it's often what separates the ppl who qualify for the olys and the ppl who don't

make sure you are spending a lot of time practising the last 5 qs

you don't have to like do them separately but like basically don't finish ur 1hr, then not really know how to solve the last 5 qs, give up and read the answer

like when i first started practising for SMC, like the first 20qs took like 30m and the last 5 took like 1hr (i think, it's been a long time lol)

anyway make sure you do spend time on the hard problems, that'll be how you improve quickly

if you think you've spent sufficient time, then you can have a look at the solutions

but you should always be thinking & trying to answer "why didn't i think of that" and like "why was that a good idea to consider" when u read the sols

sometimes you make simple slip-ups on earlier problems, i wouldn't focus too much on those

being more accurate will come with practise

• once u've gotten to the point where ur not worrying about "can i answer all 25 problems" but rather "can i not make any small slip-ups", then you can practise checking lol

[high goblet]

whilst it is true that smth like answering like 1-21 all correctly will get u to the olympiad or smth like that, it's generally better to focus on being able to do maths rather than accuracy

ppl who think "i'll answer 1-21 all correctly" usually never make it cus they slip up still or they can't solve one of the later problems etc.

anyway yeah just some advice

[radiant jasper]

are there any books in particular though that i can use

to get fundamentals for UKMT specifically

[high goblet]

^

a lot of ppl i know didn't use books for IMC or SMC so i can't really help you if you really do want books

but like there isn't any theory for IMC

so you just need to apply the theory you know to problems

[ornate blade]

by dividing the hexagon into congruent right triangles, 12 of those triangles make up the entire hexagon -> 1 triangle = 1/12 of the area

[hard crag]

Can you uh

Draw the congruent triangles

If you can

[ornate blade]

divide this equilateral triangle into 2 halves

6 equilateral triangles * 2

[hard crag]

Ok thanks

Man the explanations for 50% of the mock papers are just gone

[ornate blade]

then you need the fact that the centroid splits the median in the ratio 2:1

as in

[hard crag]

Area smilarity?

Similarity*

[ornate blade]

these two are similar

wait let me fix the ratio

also you know this is a perpendicular bisector, cause it passes through two equilateral triangles

with the same side length of course

[ornate blade]

then you just need to figure out the side length ratio of the blue and red triangels and square it, to get the ratio of the areas

this is trickier cause you need to compare hypotenuse to hypotenuse, or longer leg to longer leg

but then you can just use the side length ratios in these 30-60-90 degree triangles

1 : sqrt(3) : 2

or actually I'm overthinking this, cause you just have

these 3 areas are congruent, by rotational symmetry

[hard crag]

Thanks

[radiant jasper]

They really just want you to build what you know from regular GCSE maths and think a bit outside the box

No really complicated theorems appearing until BMO 2

and even then, its more about problem solving than any really complicated maths that wouldnt be taught at IGCSE and beyond so I agree that theory is mostly useless when it comes to UKMT competitions

[ornate blade]

yeah, @radiant jasper if you look at AMC10 which corresponds to your grade level, you'll be like wtf

you'll seriously be challenged trust me

[gusty verge]

Yes ofc haha, transform the matrix into a upper / lower triangular matrix and just multiply the diagonal entries

It'll take a lot of time still

But more efficient than matrix minors for sure

[balmy hound]

why is this not taught in schools?

[gusty verge]

It usually is

[ornate blade]

competition questions rely on this

[gusty verge]

Both are taught

[ornate blade]

yes India teaches this quite thoroughly I'd say in class

so does China, senior high mathematics

it's really not that special, it's just a consequence of taking minors and then all the terms except by 1 will be multiplied by 0

it is pretty cool though

[gusty verge]

If you want to do it super efficiently then calculate the eigenvalues first then multiply the eigenvalues, there are better tricks for finding eigenvalues

But tbh you'd need a computer

[balmy hound]

what are eigenvalues

[gusty verge]

Search them up if interested, come back if ur unsure on anything 🙂

[radiant jasper]

Hehe have fun with eigen-anything

[balmy hound]

k

[ornate blade]

also highly recommend 3b1b's essence of linear algebra series

[balmy hound]

ok

will probably see that one today (its holiday)

[stone cipher]

Sending exorcices

[reef condor]

Wdym “too”

Yes I do it, I did AMSP last year

[open kite]

hey guys does anyone know any prepartion tips for my foundation year at university

planning on doing accounting and finance after my foundation

so the subjects ill do will be associated with the course , which probably includes additional math which will be a bit new to me cs i only did extended math in igcse

[hollow aurora]

maybe just review some math on khanacademy if you feel rusty

I don't have any more specific advice, do you have a particular math topic you feel like you want to review?

[open kite]

yeah there is some topics i failed to fully get the concept off which will actually be useful when i get there

for example probability and complex trig

they r usually like 5 marks or more on the past papers

and i did a past paper today and i was so rusty almost felt like i forgot almost every trig formula lol

turning points and quadratics

[hollow aurora]

yeah khanacademy is a good resource for reviewing then

it has a list of a bunch of topics, with videos and exercises for each one

[open kite]

alr thanks

[deft marsh]

Let ( n \geq 2 ) be an integer. Prove that:
[
\sum_{\substack{1 \leq k \leq n \ \gcd(k, n) = 1}} \frac{1}{k} > \frac{\phi(n)}{n} \cdot \ln n,
]
where ( \phi(n) ) is Euler's totient function.

[gilded halo]

saintyzy

[ornate coyote]

chat competition math is east

easy

you just gotta like

spent a couple thousand hours

maybe ten thousand hours

[jolly solstice]

you mind demonstrating that by giving the solution to the previous problem? 😎

[barren sparrow]

what even is a euler totient?😭

[jolly solstice]

it's function that gives the number of numbers relatively prime to and less than a number. So for instance phi(4)=2 because both 1 and 3 are relatively prime to 4.

[ornate coyote]

so like you take the first number and put in the gcd and take the value into account and plug it in and divide it by the first part

and then you put the thing the variable thing and substitute the value of the other onee

[latent bough]

Bruh I just got cooked by imosl a1

2022

2021

Even the solution is cooking me

“By trivial induction”

[dry forum]

If the book says trivial, just try o1 or Sonnet chat

the chat bot will be probably wrong but you at least have a clue where to start

[radiant jasper]

damn

[latent bough]

It’s imosl

Imo shortlist

I finally solved with help tho

[vale osprey]

It counts all the numbers which are relatively prime to n.... Phi(n)

[vale osprey]

I think I will try this in my free time..

[balmy hound]

what?

[jade widget]

I think someone in my school cheated in amc12a…

she scored 108, she doesn’t even know binomial coefficient, struggle to understand Bayes’ Theorem, and asked me about school level trigonometry problems

Idk what to do

[ornate blade]

Sadge

[gusty verge]

Did the questions require that stuff

Either they cheated, they'll get found out for being a fraud in the future

Or they didn't cheat and calling them a cheat and spreading it is a really bad thing to do

Either way I wouldn't get involved, it would be very impressive to figure a way out of cheating in an exam like that anyways lmao

[karmic rune]

do you need to submit the whole solution in amc or just the correct option in the mcq?

[sleek ivy]

do you rlly need those to answer those questions?

i wouldnt get involved

[latent bough]

nah you dont even neeed allat for amc

its just computation man

the probability (bayesian) of probability coming up on amc more than 3 times

oh well ig there is a trig problem on 12a

[ornate blade]

you do actually

AMC 12 is much harder than the UK counterpart, UKMT senior

especially if she scored 108 so that means she scored on some of q21 to 25

108 is 18 questions correct ah wait, well

but with 108 you can qualify for AIME

not even can, will

[ornate blade]

have you not heard about mainland Chinese students leaking the answers?

bribing proctors and so on

[ornate blade]

but yeah this isn't your business, it's up to your school to handle this

after you tell the teacher who organised the comp it's out of your hands

[ornate blade]

correct option

and same with the AIME, the answer is an integer from 0 to 999

[lilac niche]

They can do that wow

[ornate blade]

it's insane

[jade widget]

oki

[lilac niche]

Exactly

[deft marsh]

considering the polynomials P(x) = x^2 + x + b , Q(x)= x^2 +cx + d where b,c,d are real numbers and P(x)Q(x)=Q(P(x)) find the roots of P(Q(x))

someone help ples

[plain jacinth]

I have a question for the sumac, ross, and promys summer camps, which one is the easiest in terms on entrance exam?

[blazing dagger]

Hint: ||P•Q (-1) = Q(b) = P•Q(0)||

[ornate coyote]

Brain not braining

Circumcircle isnt circumcircling

Its so over

[high goblet]

don't be put off when ur trying IMO/SL problems for the first time

remember that the solutions are written for i.e. ppl selecting the problems etc.

of course, every step you do might feel hard

but with practice, you'll get better at them and u might actually agree with the label of trivial induction

anyway if ur looking for easy ISL problems

ISL 2022 N1 is fairly easy (but it's also a really boring problem)

ISL 2022 N2 is alright

and also 2006 P1 and P4 are both very approachable

[radiant jasper]

How the hell is the N constructed here? 🥲 I'm dying

[radiant jasper]

Ah, makes sense. The solutions making my feel so dumb

[high goblet]

i think they've just slapped on a bunch of modulo conditions that are useful and then used CRT to construct N

if ur asking for the motivation behind such a construction, it's hard to say without actually doing the problem

[radiant jasper]

Whatever they did- how did they assign q(i,j) to each N + i or M + i.

See they said 2012 × 2013 primes aaarghhhh

[high goblet]

they've picked 2012*2013 primes bigger than n+p+2013

[radiant jasper]

Nuh uh, I get that for N. But it's properties- haha, this is insane

[high goblet]

and then they've just put them together in a 2012*2013 table

then label the columns N+1, N+2, ..., N+2012 and the rows M, M+1,...,M+2012

then they want basically everything in the row to divide M or M+1 or etc.

and everything in the column to divide N+1 etc.

you know you can always find such an M,N by CRT

[radiant jasper]

Feel totally arbitrary to me rn 🥲

[radiant jasper]

Wait- yeah!

Ohhhh

Let me brain storm again

Brain-kill.

[hard crag]

Bro my brain is fried

And this

[radiant jasper]

Is this for real numbers or complex as well?
(Also is the answer '6'?)

[hard crag]

Gimme a sec

Should be real tho

I don't think imaginary numbers are given at 8th grade olynpiads

Yet

[hard crag]

Should be 6

Dk the explanation

[radiant jasper]

@hard crag in terms of real numbers, notice at least one of x²-2/5x-4 and x²-2/4-5x must be negative, or both must be zero.
If we stay in real numbers, they must be zero as we dont allow negatices in radicals.
x, thus, is root 2.
y subsequently is 2.
x²+y² = 2 + 4 = 6

[hard crag]

Wait

Lemme process

[hard crag]

Ah I see it now

Thanks

[radiant jasper]

A lot of times bigger expressions are used on purpose to confuse.
Substitute the stuff under radical as a

[hard crag]

Thanks bro

[pallid knoll]

anybody wants a challange?

[fresh hawk]

I was searching for one

[pallid knoll]

aight

u sure?

[fresh hawk]

100%

[pallid knoll]

alright

here ya go

[deft wraith]

competition math

[fresh hawk]

I am pretty certain this ahs nothing to do with competition math

[pallid knoll]

oh sry then

cya

[deft wraith]

lmao

[fresh hawk]

If you actually need those proofs you can direct message me.

[reef hollow]

Let A, B, and N be random positive integers =<500. Two numbers A and B are called co-combinable for N if NcA=NcB. What is the probability A and B are co-combinable for N?

I wrote this question for a non-profit I’m a member of that hosts competitions.

[fresh hawk]

That's interesting

[reef hollow]

I have no idea how to solve it💀

[fresh hawk]

Intuition tells me that because A,B,N are uniformly chosen from {1,2,...,500} the number of possible (A,B,N) tripples are 500500500=125000000

Then we know that binomial coefficient satisfies the symmetry property: $\binom{N}{k}=\binom{N}{N-k}$ and this suggests that $$A=B \text{ or} A+B=N\iff B=N-A$$

[gilded halo]

trigonometria

[fresh hawk]

For the trivial case where A=B there are exactly 500 choices and for the other case , since N is fixed we can just solve B=N-A

(this was my first attempt at the problem so maybe there are some mistakes but I am pretty certain this way of approaching this can theoretically yield a resault)

[reef hollow]

You are honestly closer than me i have no idea what I’m doing

[pallid dragon]

2/1000 that N=A
gives (2/1000)(1/500)
499/1000 that N > A
gives (499/1000)(2/500)

[blazing dagger]

N is fixed?

[pallid dragon]

but we overcounte cases when N−A = A

[blazing dagger]

Are we looking for P(NCA = NCB | N) ?

[blazing dagger]

.

I thought it'd be $(500)^{-3} \cdot \left(\frac{500 \cdot 501}{2} + \frac{498 \cdot 499}{2}\right)$

[gilded halo]

Arya

[pallid dragon]

so it's like, minus (1/62500) minus (1/500000)

250 outcomes where 2A=2B=N

or how am I supposed to count N < A

would it count when 0 = 0 or not

[blazing dagger]

Which case you counting? A = B? Or A + B = N?

[pallid dragon]

it's not a case it's just a question

is 3c5 = 3c11

[blazing dagger]

=_= you don't want any such choices for A,B,N that leads to this

[pallid dragon]

makes sense either way tbh

yeah it's clearly easier to count outcomes and not multiply fractions

include blue optionally

[blazing dagger]

why not simply count A = B ≤ N and A + B = N

[blazing dagger]

why not simply count A = B < N and A + B = N

[pallid dragon]

it's wrong

i mean mine

[pallid dragon]

that's what it's doing idk it feels like it's not easier

yeah i agree still wrong

[pallid dragon]

this one was right

the red part at least

[pallid dragon]

they intersect, you have to count intersection

if that's what you meant

[blazing dagger]

Intersection is A = B = N/2 => 250 outcomes

[pallid dragon]

yes

[blazing dagger]

So. (500)^{-3} [ 501C2 + 499C2 - 250 ]

[pallid dragon]

i don't get it

neither part

what's 499C2

computer says mine is right

[gilded halo]

Arya

[stable shore]

Only m,n: (1, 1), (2, 4)

[tiny fern]

i have the same conclusion

i tried cases for m =1 and m = 2

and m = 3

m = 1 and 2 gave me solutions

and if observe it for m > 3 7ᵐ would be even larger, making the gap between 7ᵐ and the next lower multiple of 3·2ⁿ always greater than 1

[tiny fern]

i think this statement can be prooved more rigrously but this is how we will solve this problem i think

[stable shore]

m = 0 mod 2 (m > 1) and n = 1 (mod 3)

But we need to achieve a limit on m or n from above, mod x won't help

[blissful hemlock]

any and all help is appreciated

[slow brook]

is there somone who understand frensh

[deft wraith]

$\exists$

[gilded halo]

user95040

[vernal axle]

Let G be the intersection of AD and EP.
AXDY is cyclic -> AFDP is cyclic -> ∠APY=∠DEY -> APY and DEY are similar -> EY/YP=DY/AY -> AGYP is cyclic -> ∠AGP=∠AYP=90°.

[burnt trail]

George is planning a dinner party for three other couples, his wife, and himself. He plans to seat the four couples around a circular table for $8$, and wants each husband to be seated opposite his wife. How many seating arrangements can he make, if rotations and reflections of each seating arrangement are not considered different?

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[burnt trail]

sorry if its not olympiad per se, is prolly amc 8 level

can I get some hints tho?

[frank birch]

Hi, any and all help will be appreciated

[radiant jasper]

First off thats a lazy ass way to give a real world example problem.

[blazing dagger]

You keep stating "mod x" won't help, but the way to solve this is using mod only ^^". For m > 2, n > 4, ||you can rewrite the equation as 49(49^p - 1) = 48(8^q - 1), p ≥ 1, q ≥ 1, and that is how you arrive at the conclusion "there's no solutions to this"||

[wide mulch]

You just integrate and write the coefficients as( x-(a))(x-(b))(x-(c))(x-(d)) and just multiply it out

And get the coefficient of x^2

[stable shore]

I solve it, ||m = 1 n = 1, now m > 2 => 7^m = 1 (mod 8) => m = 2k => (7^k + 1)(7^k - 1) = 2^n * 3; gcd(7^k - 1, 7^k + 1) = 2 => 2 cases:

1. 7^k - 1 ≠ 0 mod 4 => 7^k - 1 = 2a (a odd) => 2a(7^k + 1) = 3*2^n => a = 3 or a = 1 only solution a = 3 => n = 2 and m = 4;
   Same for 2) (7^k + 1) ≠ 0 mod 4 but here no any solution||

[ornate blade]

do you know how to integrate polynomials in general?

power rule for integrals, search it up if you don't know

[radiant jasper]

my city is pretty impoverished and the nearest professional coaching Institute is like 90 km away

i can ask for help at some colleges near my hs but i feel like they are too busy

[dusky oyster]

Zsigmondy’s trivializes the problem

[blazing dagger]

Can you show how to use Zsigmondy's here?

[radiant jasper]

hello! i'm preparing for mathcounts school and chapter level. though embarassing to admit, i have not been preparing much and as wonderig if just the aops "competition math for middle school" would be enough for me to qualify for states or perhaps perform well at states, too

i alr have competition math experience and have qualified for states twice, but last year i just barely made it due to the change in chapters

[high goblet]

rewrite as 7^m - 1^m = 3*2^n

[reef condor]

Yeah you should avoid admitting

(Joke)

[radiant jasper]

yeah like i legit did not do ANYTHING 💀

[reef condor]

Probably aops vol 1 is good too
Imo if you lock in on the MS book though, you can do well

[radiant jasper]

ooh so vol1 + MS book = success...? 😭

[reef condor]

The word “admitting” is an in joke in the comp math community

[reef condor]

Vol 1 is more than enough

[radiant jasper]

sigh am not familiar 😔

[radiant jasper]

ohh okiee

[reef condor]

States is a big jump from chapter but chapter is not that hard

What kind of score do you need to make states?

[radiant jasper]

i think id need like... 40 maybe?

i did rly well on the sprint round but was iffy on the target ngl

[deft wraith]

Yeah i forward this

Good luck!

[violet dagger]

what is the nature of improvement in competition math?

as in, how does one go from being able to do AMC problems to being able to do AIME problems?

what is the mental process, I suppose

[jagged hearth]

at what level?

[buoyant gust]

I found some of my old math books we used in my class from 2 years ago and 3 years ago. Guess the grade by some of the problems:

1. Find the integers x and y so that: x² = 120-7x³y

2. Let a and b be 2 prime numbers. Let x and y be 2 natural numbers
   What are all the divisors of a^x? How many of them are there?
   What are all the divisors of (a^x)•b? How many of them are there?
   What are all the divisors of (a^x)•b²? How many of them are there?

[jagged hearth]

hmm then you're probably asking for things like putnam and stuff for which i don't really have much to say since im just a high school student.. but idk if it helps but..you may try Modern olympiad number theory (MONT) [its written for high school Olympics so idk if itll help but some stuff of putnam is similer to that of IMO so it might work]

hmm👍

[digital hedge]

Let $a, b, c$ be positive reals such that $(a+b)(b+c)(c+a) \geq 1$. Prove that $$\sqrt\frac{a}{b+c} + \sqrt\frac{b}{c+a} + \sqrt\frac{c}{a+b} \leq \frac{b^2+c^2}{\sqrt{a}} + \frac{c^2+a^2}{\sqrt{b}} + \frac{a^2+b^2}{\sqrt{c}}$$

[gilded halo]

1Rebensol

[digital hedge]

Turned out the main claim of the solution is to prove that for all positive reals $a, b, c$, $\sum_{\text{cyc}} \sqrt{a(a+b)(a+c)} \leq 2\sqrt\frac{(a+b+c)^3}{3} \leq \sum_{\text{cyc}} \frac{b^2+c^2}{\sqrt{a}}$

[gilded halo]

1Rebensol

[digital hedge]

But anyone has any other idea than this?

It's pretty hard to come up with the form 2√((a+b+c)³/3) as the stepping stone

[high goblet]

where's the problem from?

[high goblet]

some vague motivation for the LHS is if we multiply the LHS by (a+b)(b+c)(c+a), we homogenise the inequality so we can probably ignore the condition in the q

hard to say what to do for the middle without actually writing down inequalities and seeing what happens

[digital hedge]

Yeah I think that's the hardest part, the rest is standard ineq problem

[digital hedge]

Idk my teacher found it somewhere or he created it

What level do you think this problem is

[reef condor]

That is kind of standard though - homogenizing inequalities

It may not be completely obvious how, but that is a natural way to do it

[high goblet]

i think they meant for the middle part

[high goblet]

u can try posting this in MODS, it's been a while since i had to do any oly ineqs

[digital hedge]

Ohh ok thanks

[real scaffold]

in a 3x3 grid of integers, the sum of each row is 0, the sum of each column is 0, and the sum of each diagonal is 0. The top row middle square is -4, the far right column middle square is 2, and the bottom left square is x. What is the value of x? (x has to be either -2, -1, 0, 1, or 2). Here is a photo of the question (its question 9)

[jade widget]

Assign an unknown value to one grid, express the remaining grids in terms of that unknown value, solve for the unknown, and substitute it back into the grid

This is what I get

[dry moat]

geniuses

can one of you help

😭

nvm

[hard crag]

Someone help verify my answer

I got 1039

My friend got 337

So uh

Yes

[reef condor]

Ur friend

2025/(8-2) floor

[hard crag]

Yeah but it's decimal

The result

[vague temple]

Is this a valid proof that the radical axis is perpendicular to the line connecting the centers? I’m trying to avoid coordbash/vectors

[pallid ginkgo]

whoa you participated?

[hard crag]

Yuh

[pallid ginkgo]

neat

[hard crag]

Hoping I get top 20% atleast

[pallid ginkgo]

best of luck

[hard crag]

Last time I got 0% correct

[hard crag]

Thanks

[pallid ginkgo]

? 💀 😭

[hard crag]

Nah, last final round I got 0/15

Hoping I don't be dumb and get the same score

[jovial snow]

Can someone please solve this

[ornate blade]

find the midpoint = centre and hence you can find the radius by distance formula

[sleek ivy]

You could find the distance and half it, that’d give you the radius, then you can solve a system to find the equation

[ornate blade]

yes note that the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$

[gilded halo]

south

[ornate blade]

(h, k) is the centre, r is the raidus

[sleek ivy]

Yeah you got the radius u got the centre

You’re done

[jovial snow]

Aight thanks a lot.
I got more questions. Should I bring them?

[sleek ivy]

Sure I can try to help

[stuck umbra]

theres a formula that can directly give you equation with diameter points only

i'd recommend to use this only if your course has it though, if not just use the general equation of circle.

find midpoint using the two points, that will be center, then calculate the distance from the center to one point which will be radius, then just plug those into the equation

[frozen glade]

So I’m 4th year in highschool (next year I go to university). What are some books for competitive math? Or in general resources.

[azure bloom]

What is an Ai project atm IMO?

Does this website have anything to do wiht competition math?

[warped anvil]

usamts r2 is very sus

[vague temple]

When you take the Fibonacci sequence you have F(n) = F(n-1) + F(n-2). However, in the generating function form, you have to add a correction term for the initial conditions F(x) = xF(x) + x^2F(x) + x. Is there a general procedure to find this correction term?

[tepid surge]

when you say generating function

do you mean $\sum_{n=1}^{\infty} F_n z^n$?

[gilded halo]

hiidostuff

[tepid surge]

nvm im being silly

i suppose lets just look at how to develop this function for F(x)

so let $F(x) = \sum_{n=1}^{\infty} F_n z^n$

[gilded halo]

hiidostuff

[tepid surge]

where F_0 = 0 and F_1 = 1

we then get $F(x) = x + \sum_{n=2}^{\infty} F_n z^n$

[gilded halo]

hiidostuff

[tepid surge]

also sorry for the mistake but i meant to let the inner variable be x and not z

ill just fix it from here on out

because of the definiton of the fibonnaci sequence

we get $F(x) = x + \sum_{n=2}^{\infty} (F_{n-1} + F_{n-2})x^n = \sum_{n=2}^{\infty} F_{n-1}x^n + \sum_{n=2}^{\infty} F_{n-2} x^n$

[gilded halo]

hiidostuff

[tepid surge]

we can factor out x from the first sum and x^2 from the second

and we get n-1 and n-2 in the sums respectively

which can be shifted back to gives us how F(x) was initially defined

so $F(x) = x + xF(x) + x^2 F(x)$

[gilded halo]

hiidostuff

[tepid surge]

so it seems that first x came from starting the sum at n=2

[vague temple]

Thanks

I saw this one

[radiant jasper]

ANY HELP IN stem international maths olympiad

class 12th

where to start

[real scaffold]

Appreciate it!

[real scaffold]

many thanks to your way!

[radiant jasper]

hello! um on the aops competition math for middle school book it says in the introduction/etc. part to go to j.batterson's website for the solutions in order (i wanted to see them easily) but when i went to the website it just redirects me to aops, which just shows me options for purchasing the book itself

does anyone know by chance if the solutions are available and where to find them TT

[pallid tundra]

it's possible that that website no longer exists, try looking it up on the wayback machine?

[hexed oak]

did anyone here solve inmo 2025 p6, dont give the pw walah fake solve, afaik no one from mumbai solved it(in contest)

[pallid ginkgo]

#1330509538658353212 message

is this not it?

[gusty verge]

Maybe send the question?

[hexed oak]

why is n^2+1 less than (b-1)^2?

[hexed oak]

basically the qn is (not these words, but equivalent)
consider the function f(x) = x - s_b(x) where s_b(x) is the sum of digits in some constant base B, show that if f(x) = n^2+1 for some natural n and x, there are infinite such pairs

[radiant jasper]

Literally no one here solved p6 or p3

[hexed oak]

someone in mumbai solved p3

i wont tell name for privacy

but the method works

[spiral thicket]

There is a question in my head

Currently, with the development in Deep Learning, do traditional ML algorithms such as SVM, Decision Trees, K-Means, etc. need to be known, or is there no need for one to know them and focus only on Deep Learning, For someone who wants to specialize in ML Research ?

[gusty verge]

1. Wrong channel, 2) Yes they are important to know, Deep learning often uses the ML algorithms within itself, ie. for a simple neural network you may have each neuron uses logistic regression (very common)

[gusty verge]

cheers

[deft wraith]

put the link into wayback engine

if u get stuck send me the exact book title or the url to solutions and ill try and find it

[oblique fossil]

Does anyone have any tips for getting better at AMC

I did rly bad this year and I wanna make Amie next year

[acoustic nova]

is there any pattern with these numbers? i came up with a problem im just curious if there could be anything nice about it (the problem isnt just the numbers im just curious if there is anything to these numbers)

[wet grove]

I want to say that if you attribute the even numbers to dots and the odd numbers to dashes (vice versa), and if you evenly split the 22 dots and dashes into 11 sections, if you translate the morse code into letters and coalesce the letters into a possible word, you get the word "MainMan" with an extra 2 Ms and 2Ns left over. But you said that you came up with the problem, so I'm wondering if there's a context to the numbers that you chose to select

[high goblet]

context?

oeis doesn't find anything

[fallen magnet]

i found this online

[fallen magnet]

for this

[acoustic nova]

it was this question i came up with

[fallen magnet]

and this website also comes up with similar results

[acoustic nova]

For positive integers n, let f(n) be the (nonnegative) distance from n and the nearest square. Find the number of positive integers n less than 2025 such that f(2n) = 2 * f(n).

its probably a bad question

[high goblet]

nah i think it's fine

yeah i mean it's not in the oeis so probably isn't that interesting

[fallen magnet]

is it 22 perchance?

im probs wrong tho

[acoustic nova]

Yeah that’s the number of integers in the set above

How did u get it though? or did u just use that

[fallen magnet]

oh i didnt even notice lmao

[acoustic nova]

I mean it could be it’s just probably something that wasn’t thought of. Could there be a good way to do it though?

[acoustic nova]

How did u get it then

[fallen magnet]

i coded it lol

[acoustic nova]

yeah i did as well

[high goblet]

nah it's unlikely that there'll be a non-bashy soln

[fallen magnet]

u do it like this?

f = lambda x: min(x - (k := int(x**0.5))**2, (k+1)**2 - x)
print(sum(f(2*n) == 2*f(n) for n in range(1, 2025)))

[vague temple]

When I do geometry problems, I often hit nested radicals when solving for values. Are there ways to avoid them, like are there patterns I should recognize to realize I’m 100% headed toward nested radicals without doing the full computation? For AMC 12 problems there is always a clever solution around it

[vague temple]

They are both just using Newton forward difference interpolation, if there are not enough terms or the pattern can’t be fit by a finite polynomial then it might not work

[reef hollow]

Where have you guys found quality questions and instruction on geometry and trig? I feel like they are my biggest weaknesses for math, that or creativity.

[ornate blade]

try Cut the Knot, covers a wide range of topics and levels including *plenty of Olympiad geometry

the only disadvantage is the info can be quite disorganised

[pallid tundra]

do you know the method to simplify radicals of the form sqrt(a+sqrt(b))

[hexed oak]

for geometry problems, just never do coordinates. (at oly level)

[sleek ivy]

What are nested radicals?

[blazing dagger]

$\sqrt{1 + 2\sqrt{1 + 2\sqrt{1 + 2\sqrt{1 + \cdots}}}}$

[gilded halo]

Arya

[vague temple]

I presume you let the inside be a perfect square so that let a = p^2 + r^2 and sqrt b = 2pq

[whole fern]

Just a general question, I got a 60 on the last AMC 12 and haven't studied since. I'm trying to quality for aime this year, which should be in 9 months. Is this achievable, and more importantly, what resources do I need?

[sleek ivy]

If you’re taking a rigorous math course and are planning on practicing yes

[whole fern]

I do plan on practicing, but the thing is I don't know where else other than practice tests to practice

[jaunty oriole]

hi

[jaunty oriole]

hi

wanna solve this?

[shadow spruce]

😭

[gusty verge]

$x=\sqrt{1+2x}\Rightarrow x^2=1+2x\Rightarrow x=1+\sqrt{2}\text{ or }1-\sqrt{2},$ clearly $x>0$ in fact we can see from the definition $x>1$ so $x=1+\sqrt{2}$

[gilded halo]

Max

[hexed gazelle]

sols pls hv tried a ton no luck

[vague temple]

Well it’s a pretty basic recursion setup, consider the possible ways you can reach $J_n$. The action space is placing a 1x2 block, 2 2x1 blocks or a 2x2 square. Hence $J_n = J_{n-1} + 2J_{n-2}$.

[gilded halo]

victor

[tepid surge]

Not quite sure myself

[neat blaze]

I have no idea maybe a hexagon is involved

[tepid surge]

I tried counting the total amount of sides

It's close to an arithmetic progression but not quite

[neat blaze]

Yeah i can't see a real correlation

[tepid surge]

Ah

If you don't count the black shape, it's a progression

The first part has 9 total sides on each white shape

Then 13

Then 17

Then 21

So you need to choose a picture with 25 total sides on its white shapes

Seems to be the bottom left @neat blaze

[neat blaze]

You're the best thanks bro

[tepid surge]

No problem gang

[buoyant pilot]

guys

is

48pi the right answewr for 2

also

pls dont give the right answewr

ust whether 48pi is right or wrong

[safe yarrow]

do any of you know IMO tsts and national mos which are INMO level?Especially in Number thekry

[ornate blade]

correct

[lapis cloak]

What is wroskian of a differential equation?

[ornate blade]

https://tutorial.math.lamar.edu/Classes/DE/FundamentalSetsofSolutions.aspx#Wronskian

[acoustic nova]

where is this from? NYML? It kinda looks like that

[hexed oak]

this goes to root(2)-1

[buoyant pilot]

thx

[buoyant pilot]

CAML

california math league

[primal cloud]

Is anyone given the amc 8 early

[timber yacht]

Lim x --> 1
f(x) = (√((x³ - x² + x⁰)*(x² - x³ - x⁰)⁷))/((e^(x) - e) ( sin²(x) - cos²(x))

Anyone can solve it?

Function made by @timber yacht

[proper estuary]

x^0 is so personal 💀

[timber yacht]

[timber yacht]

U have undergraduate math's role.
Dont tell me u cant solve it

[proper estuary]

im on my bed rn so i cant do complex math

Although i can probably tell its inf

Since e^x -> e as x -> 1

Everything else is just a joke

[tepid surge]

Could you potentially write this in latex

[timber yacht]

e - e

[tepid surge]

Or send a picture or smth like that

[timber yacht]

The whole numbers under squar are ^⁷

^7

[tepid surge]

Is this not nonexistent

[timber yacht]

Solve V2 of this

[timber yacht]

Idk, i just create it

[tepid surge]

Pretty sure it doesn't exist

The numerator approaches a nonzero value while the denominator approaches 0

[timber yacht]

Try imaginary world, VR world

[tepid surge]

Regardless of this, the limit doesn't exist

[timber yacht]

Hmm

[tepid surge]

And even if the intent was to use lhopitals, I'm pretty sure it fails over the complex numbers

As in lhopitals doesn't work for complex limits

[timber yacht]

Nah

(e^(x-5) - e)

Fix it

[tepid surge]

Ok so

[timber yacht]

Is it exist now?

[tepid surge]

Something like i/(e - e^(-4))

Oh nvm it doesn't go to 0

The trig values just make this whole thing awkward

[timber yacht]

I have an additional process to warm up your brain before start,

(a - x ) × (b - x) × (c - x) × ... (z - x)

Solve it

@tepid surge

[tepid surge]

0

[timber yacht]

Owh

U have good brain

Congratulations

[tepid surge]

Thanks man

[timber yacht]

Solve V2 of this

[hallow seal]

does anyone know of the ukmt