competition-math

[mental rose]

cause i don't know that much

everything's always been "derive geometry from its roots"

[zinc tundra]

gng wha

js know ur special right triangles

and get a job too

[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

[ivory ember]

Use the discriminant to get a condition relating a and b. The rest is standard geometric probability.

[granite stump]

Guys if u like MATHCOUNTS respond to the message

😢

😢

https://tenor.com/view/dog-crying-meme-doggo-crys-megan-soo-crying-dog-gif-5276199764143986284

[torpid cairn]

just doing practice tests were in fact enough for me

special right triangles are the main one, geometry has a lot of steps so I find solving past problems particularly useful in this subject - you'll see a lot of the same steps (forming special right triangles, finding area of circles based on inscribed triangles)

the problems won't map 1 to 1 but often the individual steps will, just in a different order

here's a list of formulas I've learned that help on AIME if you feel like you want to learn them, but I think the shoelace theorem is the only one that could even possibly help here

some of these aren't formulas just things I'm not extremely familiar with

[granite stump]

I like how no one likes MATHCOUNTS here

😢

[granite stump]

Whyyy

[torpid cairn]

I never did math counts and I think I'm the only one active lol

[granite stump]

Oh

Ok

[glacial berry]

never did mathcounts either

[drifting shell]

||no way i got gold prize (aimo)||
-# ||even with just maximum 71 marks||
-# ||out of 150||
|| ||

[novel juniper]

i just finished amc and i did so bad😢

[pallid ginkgo]

aimo aint really that bad

iirc

wait final?

[drifting shell]

yep

[pallid ginkgo]

oh i have no say in that then

[hollow flax]

set the two equations equal to each other, get everything to one side, use the quadratic formula, and look at the discriminant

this ends up turning into a geometry problem

[high goblet]

i mean 1) AOPS has such a large database of problems that there's bound to be enough AIME difficulty problems and

2. if you need more, then ur probably revising wrong

ur probably spending too much time on problems which are too easy for you

when ur nearing the end of ur AIME problem bank, you should start moving to easy olympiads

[graceful palm]

What is the answer?

[sly token]

surely it wasnt that bad

[hollow flax]

it’s more fun if you find it yourself

[graceful palm]

Like where is it from because I wanna check my answer that's why

[hollow flax]

the answer is ||E. 61||

[sly token]

did u do amc?

[novel juniper]

i guessed 50%

for the questions

[sly token]

oh... lol damn

[novel juniper]

jnr

[subtle sundial]

dont worry man

if its hard for you it should be just as difficult for the others

considering you put in the effort

[novel juniper]

[earnest hull]

I need some exercises of algebra or pdf for preparation olympiad maths

[molten lotus]

ty

[solid cloud]

anyone have the source of the problem?

my answer is (1,1) (1,5) (5,1) for all pair (m,n)

[solid cloud]

yeah i know bro

i mean i have found the answer + proof

i want to check my answer

i want to check from the bulgarian math olympiad

since mont doesnt give solution

[weary vortex]

the furthest progress I've made is finding that:

$$
If we assign f(x) = 2^{2^x} + 1, then:

f(x) = ((f(x - 2))^2 + 1)((f(x - 2) - 2)^2 + 1) - 3
$$

[solid cloud]

not only that

x is co prime with 2^2^x +1

so m | 2^2^n + 1

[weary vortex]

yeah i gotta learn NT more deeply to have a chance against these types of problems

this is pure algebra, the field im somewhat good in

[solid cloud]

im learning this kind of problem too

[solid cloud]

yeah you are right

[gilded halo]

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

[weary vortex]

Totally useless info but this expression can be simplified to $f(x) = (f(x - 2) - 1)^4 + 1$

[gilded halo]

azart

[calm wagon]

[sour cobalt]

angelkiller411 on wordpress has a pdf of most bulgarian olympiad problems - regional and national. However many from Titu Andreescu’s NT book which are referenced don’t appear

[novel karma]

what

[edgy vapor]

that's actually all the answer exist

the source is || BMO 2016 P1 of 6 ||

If I remembered correctly

[novel juniper]

?

[shut lark]

I’m doing the Australian Math competition tomorrow and the shit inside that is hard as hell. So yeah wish me good luck lol

[soft vigil]

good luck

[weary vortex]

good luck mate

[half arrow]

Does anyone know how difficult the regional stage of the All-Russian Math Olympiad is? Like how does it compare with other contests like USAJMO or BMO2

[edgy vapor]

good luck, I like the number theory thing on AMO

[novel karma]

amc isn't till november

[novel juniper]

its different

for everyone😢

[edgy vapor]

Ah

I'll send it again

like I don't know what you meant but here's what I like

[edgy vapor]

emmmmmmm maybe diophantine , sequence , construction (number only)

It’s AMO P5 of 8

[frail geyser]

I am trying this BMO problem but I'm stuck. I have tried breaking the numbers down into digits and see what combinations can be done but it looks like this way may end in me having the brute force each way which im almost certain is the most efficient way for this. I have attached some of what i have done so far(the rest is just me trying cases with different numbers). Could someone give me a hint or suggest an alternative way of approaching this? Thanks!(The handwriting is a bit messy so if you want clarification on what i have written pls ask). Edit: I have also figured out that the squares which give 4 digits are between 32 and 99 if that helps

[dusty fable]

i’ve got a solution for a question very similar

basically, consider calling the digits abcd = N in decimal form, then u know the other integer M=. some permutation of M= ab(c-1)(d-1) moving the minus 1 digits around

and u know they are both squares so u could say for the case i showed N= M-11, therefore as N,M are squares. p^2=q^2 - 11 and this is just a basic diophantine Eq

[dusty fable]

this is for ur case

[kindred wraith]

omg gorgeous handwriting

[dusty fable]

r u being sarcastic

coz idk ive always thought it was mid on the bad side

[kindred wraith]

no not sarcastic at all, that is very pretty

well organized, clean, comfy to look at

somehow the letters can be absolutely tiny and yet i can tell exactly what the word is from just a quick eyeball

a tiny tiny tiny tiny gripe is i wish the a's were closed on the top, they look a bit like u's

but id kill for handwriting like this

[dusty fable]

oh yeh i always do that

but tysm that’s so nice

[misty raptor]

hii! i was wondering if any of you guys have any methods that have worked for you when studying for the AMC10? I'm currently trying to lock in and pass (for i am not incredibly smart or anything, but need a tried & tested method that I can stick to until competition day)

[shut lark]

Lmao I know nothing about number theory

[edgy vapor]

doing their past exam , gathering content focus on your subject which you good at it..(in another word just doing exam non stop) …?

[misty raptor]

okok thank you!

[frail geyser]

oh yeah that is very similar

[frail geyser]

ah that makes sense thank you, so after this you just need to solve the 6 dipohatine eqs then?

[dusty fable]

yeah but some of them clearly can’t work so it reduces it

[gray pollen]

Stop procrastinating math or you will be unemployed

[gray pollen]

https://youtu.be/NRy6FQeKmVg?si=vKxlMvD_U6pu_mgQ

Tf is wrong with this guy?

What is a 174M as the answer

This is why we don't use self imposed brackets

8⁸ . 8 - 1 cannot be 8⁸. (8 - 1)

[pallid tundra]

this is what passes as “olympiad” these days? 😭

[warped folio]

Hello

does anyone know about ikmc here?

i dont think ikmc's in north america

[sly pawn]

How do you evaluate markov chains?

[trail berry]

I suck at this already lol

[torpid cairn]

they're saying this was the hidden question 7 on the imo

[torpid cairn]

YouTube commenters are brilliant man

bro he just keeps playing with the exponents he's not actually simplifying any of the calculations lmfao

[hybrid inlet]

hi could anyone help me w this

what i tried:
since im only looking for the first non zero digit, i'm ignoring any 0s after the first non-zero digit and ignoring any digits before the first non-zero digit

then i got
= (2 x 3 x 4 x 5 x 6 x 7 x 8 x 9)^5 x 20 x 30 x 40 x 50

then i got rid of all the 10s
= (2 x 5)^5 x (3 x 4 x 6 x 7 x 8 x 9)^5 x 20 x 30 x 40 x 50
= (3 x 4 x 6 x 7 x 8 x 9)^5 x 2 x 3 x 4 x 5
= (3 x 4 x 6 x 7 x 8 x 9)^5 x 3 x 4

since x^5 has x as it's ones digit, i just got rid of the power since i only care about the ones digit
= 3 x 4 x 6 x 7 x 8 x 9 x 3 x 4
the last digit of the above number is 6 and that's what i got as my answer

but thats not the answer so ive been stuck

[ivory ember]

(11)(12)(13)(14)…(19) doesn’t have the same last digit as (1)(2)(3)…(9) (and etc) for one

,w 19!/10!

,w 9!

[gilded halo]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

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

Requested by civil_service_pigeon

[hybrid inlet]

oh

oh i get it now

the multiples of 5 cause the change

5 15 25 35 45

[verbal jetty]

Problem #2 (vertices)

this is what i have (updated):

Notice that all the points are part of one big web
There are three subsets of the vertices: The first one has vertice 1 (call this S1), the second has vertice 2 to 511 (call this S2), the third one has vertices 512 to 1023 (call this S3).
We can break up the second subset into more subsets: The first has vertices 2-3 (call this SS1), the second has vertices 4-255 (call this SS2), the third has vertices 256-511 (call this SS3).
In SS1, there is one endpoint in S1 and two in S2. In SS2, there are three endpoints in SS2. In SS3, there is one endpoint in S2 and two endpoints in S3.

There are six methods of deletion:

Delete 1 edge that is S1-SS1 and 1 that is SS1-SS2. (+1 of degree 1)
Delete 1 SS2-SS1 and 1 SS2-SS2 (+1 of degree 1)
Delete 2 SS2-SS2 (+1 of degree 1)
Delete 1 SS2-SS2 and 1 SS2-SS3 (+1 of degree 1)
Delete 1 SS2-SS3 and 1 SS3-S3 (+1 of degree 1)
Delete two SS3-S3 (-1 of degree 1)

The current place I am stuck on is calculating the probabilites of each amount of vertices of degree 1. Also, I think it would be helpful to establish an upper and lower bound, but I'm also not sure how to do that. All help is appreciated! Thank you

[hollow flax]

I’m so smart I can solve an olympiad problem in under a minute 😎

[soft vigil]

it seems to work on youtube

gets more views and stuff

[weary vortex]

i couldnt come up with a really "mathematical" way to solve it but, just in case, you're interested, here is a recursive programm solution:

def last_nonzero_digit(n):
    if n <= 1: return 1
    return 6 * [1, 1, 2, 6, 4, 4, 4, 8, 4, 6][n%10] * 3**(n//5%4) * last_nonzero_digit(n//5) % 10

with an input 50, it returns 2

[hybrid inlet]

thanks

[scarlet thorn]

You drop all 10s
And then find it module 10

[dusty fable]

is that not quite tedious tho

but that’s what i would do

[scarlet thorn]

Well, I just tried, and got 2, it didn't take me more than 5 mins
It simplifies fast
you can easily cross all but 1 6 immediately (as 6 x 6 = 6 mod 10) after removing 10s
all pairs of 2 and 3 too, if there's at least one 6 (as, well, 2 x 3 = 6, and we can drop it)
by the same logic all pairs of 4 and 9s
and speaking of 9s, all pairs of 9 can be dropped too, as 9 x 9 = 1 mod 10
and so on

[torpid cairn]

is there really not a faster way to solve this lol

[dusty mauve]

Fr just thinking the same thing 😭

[scarlet thorn]

Well

If you actually think about it a bit more
The product
x1 x2 x3 x4 x5 x6 x7 x8 x9 (x+1)0

Can be turned into:
1 2 3 4 5 6 7 8 9 x+1 for odd x and X=0
1 2 3 4 5 5 6 7 8 9 x+1 for even

And these are equivalent to 8 (x+1) and 4 (x+1) respectively

So
For 50!
It’s 8 1 8 2 4 3 8 4 6
Or
2

[sour cobalt]

Can someone confirm if the answer is ln(2)?

No answer key is available, work below

[scarlet thorn]

Yeah, you can add and subtract ln(3^(1/n) -1)/(2^(1/n) -1))

And (x-1)(x+1) = x^2 - 1
So, the infinite product collapses into 2

[vernal axle]

no, there is a mistake in the last limit

[sour cobalt]

Where exactly?

[vernal axle]

it's not 0

[sour cobalt]

Oh, i see

[frail geyser]

I cant find the answer to this question so was hoping if someone could see if they get the same answer as me. I got 1996/1997.

[vernal axle]

write the same thing for f(n-1) and subtract, to eliminate that sum f(1)+...+f(n-1). Then you get a very easy thing, which is ||f(n)=2f(1)/(n(n+1))||

[verbal jetty]

does anyone know how to solve this problem using graph theory?

currently i have that we start with a complete graph from 5 vertices. an edge represents a friendship between two people and the absence of an edge means that two people are not freinds.

i've been told there are three "isomorphic states", but idk what to do with this information let alone find the number of vertices/edges in the "isomorphic state"

[pallid void]

Yo

How do I win UIL math competitions

If anyone here has experience with them

[slow basin]

i got 2/1997?

i have working out

not sure if this is the quickest way to arrive at “the answer” (probably wrong anyways) but the q looked too interesting, had to give it a go

could someone look over it please 🥹 😁

[gray pollen]

What is the 4th root of -1?

[pallid tundra]

the principal 4th root is probably just $e^{\pi i/4}?$

[gilded halo]

elrichardo1337

[pallid tundra]

google roots of unity

over the complex numbers there are 4 4th roots

[vestal pond]

any AMC10 tips? also can someone explain mass points, I get the concept but I still need some further research

[gray pollen]

Let's try and solve this

First we rewrite 4thsqroot of -1 as (-1)^ 1/4

We know that i² = -1

So, (i²)^1/4

I.e. i^2 × 1/4 = i^2/4 = i^1/2

[pallid tundra]

think seesaws

[gray pollen]

Rewrite i^1/2 as sqroot i

[pallid tundra]

sqrt is not defined for imaginaries

try again

[gray pollen]

...

[pallid tundra]

are you familiar with polar/exponential form for complex numbers

[gray pollen]

We know that sqroot i = a + bi

One of the many forms of sqroot i

Squaring both sides we get i = (a + bi)²

[pallid tundra]

what if it were -i?

[vestal pond]

seesaws?

[pallid tundra]

imagine you're on a playground with a weird seesaw whose fulcrum is not right at the center

[vestal pond]

guys predict of amc10a or amc10b will be harder this year -

[pallid tundra]

let's say it's like uh 2/3 of the way from one end to the other

you sit on the end farther away from the fulcrum

what weight do you have to put on the other end to balance it?

twice your own weight

[vestal pond]

ok but how do you use it in problems 😭

[gray pollen]

That's something like what we are solving here right now

[vestal pond]

and how do you assign lengths weights? is it corresponding to adjacent points??

[pallid tundra]

ratios of lengths when you otherwise can't apply the other stuff like similarity/etc easily

[gray pollen]

Now expand the bracket (a + bi)²

[vestal pond]

oh

[gray pollen]

We get: a² + (bi)² + 2abi

[pallid tundra]

you assign points weights, not lines

that's why it's called mass points lol

[vestal pond]

hm

but what does it do

[gray pollen]

So, a² + b²i² + 2abi = i

[pallid tundra]

https://www.mathcounts.org/sites/default/files/Mass Points (M) - DSEC.pdf

here's a handout MATHCOUNTS put out a couple years ago

these are the kinds of problems (among others) that you can solve with mass points that would otherwise be pretty gnarly to solve

[gray pollen]

We know that i² = -1

[vestal pond]

thanks

[gray pollen]

So a² - b² + 2abi = i

Can we write this as

a² - b² + 2abi = 0 + i?

This doesn't change the equation

Now we can solve by parts

a² - b² = 0, 2abi = i

Solving for 2abi = i, we get 2ab = 1

To find b, we get b = 1/(2a)

Substituting 1/(2a) into equation a² - b² = 0,

a² - {1/(2a)}² = 0

We get that, a² - 1/ (4a²) = 0

a² = 1/(4a²)

Therefore, 4a⁴ = 1 => a⁴ = 1/4

Taking square root of both sides

a² = 1/2

Again, a = 1/ sqroot 2

Rationalizing the denominator, we get sqroot 2/ 2

Or better still let's leave it at 1/ sqroot 2

So b = 1 ÷ 2a right?

Substituting the value of a we have , b = 1 ÷ 2 (1 / sqroot 2)

Can we write 2 as the product of sqroot 2 and sqroot 2?

A sqroot 2 cancels out the sqroot 2 in the denominator. We get that b = 1 ÷ sqroot 2 × 1/1

This means that b = 1/ (sqroot 2)

Now we know that both a and b is 1/(sqroot 2)

Substituting these values into the equation a + bi = i, we have
1/ (sqroot 2) + 1/(sqroot 2) × i = i

This means
1/ (sqroot 2) + i/ (sqroot 2) = i

[gilded halo]

Results provided by WolframAlpha

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Requested by normie0988

[gray pollen]

Not equals to i sorry, equals to the fourth power of -i

And we are done! Simplify it further if you want

We can say that sqroot 2 + i sqroot 2 = sqroot -i

,rotate

[gray pollen]

@pallid tundra so this is the answer to your question

[pallid tundra]

magnitudes dont match up

[gray pollen]

Oh I see

It's because we used 2 real numbers that weren't part of the question

And multiplying a negative number 4 times will never result in a negative number

So technically, the fourth root of -i does not exist!

[pallid tundra]

https://www.mas.ncl.ac.uk/ask/numeracy-maths-statistics/images/Expo_form_complx_num.pdf

[gray pollen]

I did more research and found that we can actually find the fourth root of -I using Polar forms

[pallid tundra]

there are 4 fourth roots but yes that’s the method

[gray pollen]

A complex number z = a + bi can be written in polar form as z = r(cos(theta) + i sin (theta)) where r = sqrt{a²+ b²} is the magnitude and theta is the angle.
For z = -i, we have a=0 and b=-1.
The magnitude is r = sqrt{0² + (-1)²} = sqrt{1} = 1.
The angle is theta = arctan (-1 /0), which lies on the negative imaginary axis, so theta = - pi/ 2 radians or 270°
Therefore, -i = 1 (cos (- pi/2) + i sin (- pi/2 ) ).

[icy drift]

bruh are you using AI

[gray pollen]

For research purposes so yes

[icy drift]

quicker imo to write as -i = e^(-iπ/2 + i2πn), n ∈ {0, 1, 2, 3} then find the 4th root that way

and wtf are you talking about...

[gray pollen]

Hey AI, wtf are you talking about?

[acoustic nova]

??

[gray pollen]

Yes?

I am not helping anyone so I can use AI

[turbid cipher]

Competition math is scary fiys

guys

@half mulch, are you in competition math?

I want to get a sense of what to learn for this

like tricks and all that

[acoustic nova]

That’s not what it is

[turbid cipher]

Oh, okay

I don't know what they are

[pallid tundra]

!nogpt

[scarlet root]

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

[turbid cipher]

Can you explain to me what they are?

[pallid tundra]

low level problems might seem like they’re all gimmicks

[turbid cipher]

but?

Any spicy twist to it?

[pallid tundra]

but they’re really not for the most part once you’ve really solidly learned your basics

[acoustic nova]

I think the same could be true for all levels but there are good problems in any case

[turbid cipher]

Wait, so it's just low-level math?

So it just focuses on fast computation?

[pallid tundra]

no.

basically all high school contests are no calculus for example

but that doesn’t mean you can’t still write some really crazy problems

the point is the problems are actually nontrivial unlike what youd see in a typical math class

[acoustic nova]

u can also learn similar types of math without as much of the competition aspect

Or in something like USAMTS

[pallid tundra]

while it is kinda true that the low level contests (in the US, this includes MATHCOUNTS, AMCs, etc) are more focused on computational speed

that becomes less true the higher you go

all Olympiads at the national level are proof based

[turbid cipher]

heh, imagine the folks that already know how to write somewhat decent proofs

also, what's the age limit to enter a national math olympiad?

Is it actually worth it?

[pallid tundra]

all the major Olympiads are for high schoolers

to actually qualify for one is usually pretty hard tho

[turbid cipher]

What if I dropped out of high school

[pallid tundra]

you’d have to go look at each individual contest’s eligibility rules

[turbid cipher]

well, I guess I'll be preparing till then

[half mulch]

yah

wait what kind

[turbid cipher]

idk

[half mulch]

my fav one does calc!

I do one called mu alpha theta

but the common route you'll see

is amc, aime, usamo, imo

u should defo take the amc

[pallid tundra]

^in the US

[turbid cipher]

I thought that it would include analysis or something

I think that they do

[half mulch]

this one has crazy calculus and analysis

and it's for highschoolers

[turbid cipher]

nice

😼

[pallid tundra]

MA\theta has analysis??

[pallid tundra]

oh huh

[half mulch]

well kinda

[half mulch]

this question is from this year

[turbid cipher]

I'll be learning and learning and learning and learning and learning...

[half mulch]

but I've seen questions using stolz cesaro, Abels theorem

even residue theorem

those are crazy

[turbid cipher]

Any graduate level math?

[half mulch]

they try to teach it to you before the question

[half mulch]

nah not really

[turbid cipher]

mhm

[turbid cipher]

Well, it's still a challenge fs

[half mulch]

here's a crazy looking one

I'm also studying this year to make usamo

lowkey gonna be a stretch tho

[turbid cipher]

I am not sure if I want to enter those

[half mulch]

does ur school have a mu alpha theta team

[turbid cipher]

no, I don't go to school

[half mulch]

huh

what

[turbid cipher]

I dropped out of hs

[half mulch]

oh

interesting!

[turbid cipher]

not in the way you might be thinking...

[half mulch]

I'm curious

[turbid cipher]

cool

there's a group for young mathematicians

[fickle hill]

damn. I dont start see integrals, i´m in derivates

[turbid cipher]

I can send you an invite

[fickle hill]

it´s looks hard[

good luck for u guys

[half mulch]

am I a young mathemmatician

[turbid cipher]

yes

[half mulch]

:0

yipee

[turbid cipher]

there aren't many people in there, but there are 11- 15-year-olds with upper undergraduate to graduate level math

[half mulch]

11 year olds

[turbid cipher]

and olympiad medalists

[turbid cipher]

yes

[half mulch]

that know graduate level math

the fuck

[turbid cipher]

yes

[half mulch]

uh excuse my language

[turbid cipher]

they just started young

[half mulch]

like

[turbid cipher]

there are only two of those

[half mulch]

abstract algebra??

topology?

[turbid cipher]

yeah

[half mulch]

br

[turbid cipher]

a 12-year-old learning arithmetic geometry lol

[half mulch]

I'm 17 and I'm still at mid undergraduate

[turbid cipher]

That's really nice

[half mulch]

u mean algebraic?

[turbid cipher]

nope

[half mulch]

what is arithmetic geometry

[turbid cipher]

uhh

[turbid cipher]

this but applied

pretty much

[half mulch]

oh gosh

wait like

u learn all ts stuff

but can u solve problems with them?

or they I guess

[turbid cipher]

Yeah, sure

[half mulch]

like are they actually good at arithmetic geometry

[turbid cipher]

we are still organizing the group

[turbid cipher]

no, only one of them just got started on it

[half mulch]

okay that makes me feel a bit better about myself

are any of them good at integration bee

[turbid cipher]

yeah

[half mulch]

heh

lemme 1v1 them

[turbid cipher]

yeah fs

[half mulch]

I can teach them a thing or 2

I sent u a friend req

I gotta sleep now

night night

[turbid cipher]

Cya

I will send you the link to the group in a bit

[frail geyser]

That looks correct i think, I got the same function f(n)/f(n-1)=n-1/n+1 but i didnt do the cancelling out which you did after, idk why i just thought that it would boil down to just f(1)/n+1 cause i kept subbing in f(n-1) over and over like a recurrence equation and it seemed it would all just cancel out to f(1) but yh your method seems better and easier to understant asw

[timber galleon]

Can anyone help me study for a trig test dm me I need help lol

[slow basin]

Oh I see

How difficult would you say that question is as a competition problem?

If you know

[slow basin]

Apparently you can just solve it by listing a few terms..

[foggy nest]

Hi

[torpid cairn]

ur 17 in undergrad?

[half mulch]

uhh not literally

just math wise

[torpid cairn]

when did u start lol

[feral drift]

Guys how do i prove this

[half mulch]

hear me out

nvm

I'm not cooking

[lavish dagger]

Hi

[half mulch]

try considering the harmonic sum

I'm tempted to say just approximate it as 1-ln(2)

but I don't think that's rigorous

and that's probably outside the scope of the solution

[half mulch]

hi

[steel solar]

did you really have to use 1-ln(x) then x=2?
yes. yes i did

also guess what?
its less than 2/5

[half mulch]

lmfao

can u actually use stuff like that in proof based math like usamo

I assume this is like a usamo question

[turbid cipher]

No

just brains 😼

That's for usamo

[torpid cairn]

I'm tempted to say u can integrate this

and then prove that the integral is greater than the sum

but there would be a problem if the integral is >2/5

nvm idk what to do with the neg terms

[vernal axle]

you can write it as
1/(2 * 3)+1/(4 * 5)+1/(6 * 7)+... which is less than
1/6+[1/(3 * 5)+1/(5 * 7) +...]
The sum in brackets telescopes, and you get
1/6+ 1/2 * [1/3-1/5+1/5-1/7+...]<1/6+1/6=1/3<2/5

[acoustic nova]

This definitely isnt a usamo problem

[hot moss]

too easy to be a USAMO question lmao

[acoustic nova]

yeah

[turbid cipher]

I need to master algebra, number theory, combinatorics, geometry and trigonometry.

I will study and do MO problems

[swift imp]

Why do just that

Just master math

[subtle sundial]

that is math

ohj

im dyslexi

[vestal pond]

I need to do well on AMC10. does anyone have tips

[frank hornet]

Well, I would suggest doing more Mock Tests (in the given time), Practice Exams..... Also, mastering with the AoPS Volume 1 and Intro Books (Alg, Combo, NT, Geo), and maybe doing some Alcumus Problems?

[short rune]

Guys does anyone know a good resource and yt channel for Number theory for IOQM

I would be really grateful for any help

[scenic kestrel]

is this time of year when i should start getting ready for mathcounts/amc8 or can i wait a few weeks until the club starts

[worldly swift]

Never hurts to start early

Especially because stronger competitors with a deeper math background who are experienced with these kind of contests still practice a lot despite already being "good"

But don't practice for the sake of competing. You should do it mainly because you just genuinely enjoy doing math in your free time 😁

[vestal pond]

Ok so I have done all the mock tests up to 2014 and I am planning to do all of them before november. I have done the AoPS volume 1 and all intro books plus inter. algebra and inter. count/prob. (I am working on precalc now) However, I still have a lil trouble doing some of the solid geometry problems, and some combo problems.

also I am going into 7th grade is that late, or... ?

[vestal pond]

honestly mathcounts is not too bad if your goal is states and beyond. I'd just say do a lot of practice tests, to get into states you maybe have to score 40+, so a split of 27-28 on sprint and 12+ on target should be fine 🙂

[scenic kestrel]

got it

[ornate blade]

I think we have a future Mathcounts finalist on our hands /jk

[vestal pond]

I need to make AIME this year 🙂

[scenic kestrel]

u got it

[frank hornet]

Oh wow, you must be prepared then! I guess maybe recap the important concepts?

[frank hornet]

Same!

[turbid cipher]

No, I won't do annoying high school problems just to prove that I like math

not doing them in your free time has nothing to do with not liking math.

[frail geyser]

it was a british math olympiad round 1 problem so it is difficult but it seems much on the easier side, Im actually quite new to math olympiads and so far this is the one i have actually been able to make progress relatively quickly, most times i spend an hour just stuck on 1 part so yh

[worldly swift]

Oh I mean yeah, you don't have to do anything to "prove" you like math.

[worldly swift]

Yeah, I agree with you.

If you don't like competition math, no need to do it. However, the question was asking about specifically preparing for a math contest to begin with... So the assumption is that they are interested in doing math contests in conjunction with liking math. I just was stating that you should of course be doing these contests/training for them because you simply love doing the math (i.e. this kind of math puzzling/flavor of math contests) over the glamour of "winning."

[slow basin]

Ah, I see. I don’t really do Olympiad problems either, I just watch them get solved, but this one actually looked doable. Are you from the UK too?

[frail geyser]

yh same, thought I would start to get into them tho so i could see if i can qualify for BMO1 this year

got a bronze in the senior maths challenge last year so it is gonna be quite a turn around😭

[slow basin]

Same 😂

It was at the start of the year tho, I was much much slower than I am now

I’m sure if I retook it I would do a lot better

[turbid cipher]

I am still at my humble beginnings

[slow basin]

Can’t you take BMO1 even if you didn’t qualify? I heard you just have to pay an extra fee or something

[frail geyser]

yh you can but it would probs look much better to universities if you got a certificate of qualification for BMO instead of participation

[slow basin]

Ye true

you in year 13?

[frail geyser]

going into yr 13 rn

you?

[subtle sundial]

you're too nice

🥹

[slow basin]

Ye same

[soft vigil]

Mathcounts is goated

Not for the competition but for the experience

If you put in the work and make it to nationals

Its very great experience

[pallid tundra]

missed nats by a single point on my last chance bc I sillied literally the first question on state sprint 😭

fun times

[half mulch]

what are math counts questions like

send some

[half mulch]

nvm

it's for middleschool

[summer lynx]

is it still being worked on? i did well on a mock test and it lowered my rating 😭

[green lotus]

I've been working on a problem where I was supposed to find the exact value of this sum when x = 6 without a calculator, and I found that it was equal to 221/8. After I was able to find solutions for x = 2 and x = 4, being 89/2 and 133/4 respectively. Are all solutions of this sum when x is even rational? After plugging x = 8 through 20, I found that it seems to be the case, but does it generalize to all even natural numbers?

[vernal axle]

Yes, sure. sin^2(t)=(1-cos(2t))/2. So raising this to x/2 power and using binomial we get sums over k=1..89 of the type cos(pi * x * k/90), where x is an integer. There is a well known formula for this (x is not a multiple of 180). By putting N=89 we see that the thing is always rational.

[weary vortex]

Definition: Δn = 1 + 2 + 3 + ... + n (n ∈ ℕ). Δ0 = 0.

Question: Prove that, ∀N ∈ ℕ, there exists a way to represent N as, N = Δi + Δj + Δk. (i, j, k are all non-negative integers)

[vernal axle]

This is a well known result https://math.stackexchange.com/questions/1028404/why-is-every-positive-integer-the-sum-of-3-triangular-numbers
It is reduced to Legendre's three square theorem https://en.wikipedia.org/wiki/Legendre's_three-square_theorem

[little lintel]

im NOT making a 240 cutoff jmo 😭

[calm wagon]

this uni olympiads?

[vernal axle]

idk

[torpid cairn]

240 is insane lmfao

[calm wagon]

what are you in at the moment?

[vernal axle]

I've graduated from all I could 🙂

[calm wagon]

wow good for you bro 🙏

[radiant jasper]

does anyone have advice for USAMO? I average 130s for the AMC 12 and a 11-12 for the AIME, but I get 0s on the USAMO bc i cant do proofs

like cant do proofs for the life of me

and my parents keep telling me its sus i cant do usamo proofs 💀

[vestal pond]

I need to make

[little lintel]

or else what

[half mulch]

then he doesn't make jmo 🙁

[vestal pond]

I cry

[green lotus]

Thanks!

[green lotus]

It's a high school level bonus problem.

[hot moss]

orz orz orz

[radiant jasper]

bruh 😭

i wish I was orz

[hot moss]

130 on AMC 12 and 11-12 on AIME is really good, in fact if u have $900 to spare try AOPS WOOT program.

you’ll learn proofs from there

you can probably sign up for level 2

[radiant jasper]

i dont have the money to spare sadly

[hot moss]

a lot of IMO medalist and USA(J)MO perfect scorers used WOOT

[hot moss]

aww 😭

[radiant jasper]

😭

well rip

i rlly dont think i can even partial score

and the average score is like 1 and 1/2 score

bruh

[hot moss]

you could probably try to do a very old USAMO test and like try 3 problems at a time for 3 hours. then if you give up, read the solution

[radiant jasper]

Ahh alright

thanks for the advice

[hot moss]

Hopefully it works if u still have a month till school u should probably make a schedule

[radiant jasper]

Yeah, I started studying for like 3 hours everyday since june

😭

[hot moss]

u got this!!

[calm wagon]

well im done for then xDD

[round crow]

hwy guys what are the best mathematics competetions that are currently opened for high schoolers

[frank hornet]

Hi, I think you can check out these websites from AoPS for more information about open contests: https://artofproblemsolving.com/wiki/index.php/List_of_United_States_high_school_mathematics_competitions https://artofproblemsolving.com/wiki/index.php/List_of_Canada_mathematics_competitions

[round crow]

ty man i wish i can improve my cv with those

[frank hornet]

yes ofc

[flat jewel]

how old are you?

just curious

is there anyone here?

emm

[half mulch]

https://cdn.discordapp.com/attachments/1286891015646482483/1287938893425021041/togif-5-1.gif

[fluid tapir]

what are some maths competitions i can join (i live in scotland, am 15) & how best can i study for them
already do the SMC & UKMT

[pure mauve]

You can probably see if there's enough interest to make a team for the Senior Team Maths Challenge

[fluid tapir]

I'm a year too young for that (S4), but thanks

[pure mauve]

ah shet you right

Wait, to clarify, you've started (or about to start) S4, right?

(also technically, the SMC is the UKMT)

[fluid tapir]

about to start (in two days help)

[fluid tapir]

ohh sorry
i mean the scottish mathematical challenge
didnt realize it has the same initials as senior mathematical challenge

[pure mauve]

ohhhh

you mean this? https://www.wpr3.co.uk/MC/

(whose webpage looks like it hasn't changed in 20 years wtf)

[half mulch]

well top 6 out of 60 people i think

i know someone who made mop in both freshman and junior year (gonna be a senior now) but no imo

i think its a very big jump

as are all stages of the amc route

[golden oak]

I think the gap between MOP and IMO is going from scoring around 15–25 on the USAMO to scoring 35–42, plus ranking in the top 6 after mastering the TSTST and TSTs. It’s still a big gap in between in think. However in general the jump from MOP to IMO is generally bigger than from USAMO to MOP, as it shifts from national to world-class competition

[fluid tapir]

yea lol

[cedar perch]

Now what's this

[radiant jasper]

wtf is that in your pronouns?

i don't think that's allowed?

<@&268886789983436800>

moderators???

[radiant jasper]

supermute lmao

[radiant jasper]

bro got banned ☠️

[frail geyser]

How do you do this Q?I got tan36 but the answer was || tan54||. I tried using the fact you can make an isosceles triangle with sides length 2 and r but that got me a sin value of r instead of tan. I have tried using basic sohcahtoa but that didn't work

[near hill]

I tried using the fact you can make an isosceles triangle with sides length 2 and r
That would be true if the cords RS, ST, etc. had length 2, but they don't -- they're not diameters in the circles. Instead I would consider this triangle:

[agile gazelle]

Did anyone here had the chance to compete in the IMC

International math competition

[blazing comet]

U mean the imo?

Probably not

[subtle sundial]

the imc is different

[blazing comet]

The one for undergrads?

[subtle sundial]

im sure some uni students here have competed in the imc before

[subtle sundial]

yep

[blazing comet]

I see

[weak swan]

💔💔

first week back

[acoustic nova]

Limits in only a week?

wow that’s fast

[hollow flax]

really? I find a week to be plenty of time to learn limits at a beginner level

they’re pretty intuitive

[weak swan]

its because its ap calc BC 2 semestars of college calculus and we are supposed to move hella fast

[weak swan]

i am just struggling with like finding the limit using graphs

its throwing me off a lil

but like algebrically i can

[agile gazelle]

Nah not the imo

[frail geyser]

thanks, i just found out i misread the question, looks a lot easier now

Is my solution along the lines of what the question is asking? I cant find any published solutions so not really sure if this is correct or where i went wrong

[near hill]

I cannot follow everything you're doing, but it definitely looks wrong that you're suddenly assuming n²(n²+16) = 720I out of the blue (which was what you were supposed to prove) and deriving consequences from that. Doing that can be a useful way of exploring the question for yourself, but it doesn't belong in a written-down proof.

[frail geyser]

cool, could you give me a hint on what to do instead?

only thing i am sure of is that n must be a multiple of 6 right?

[near hill]

I would structure it something like:

Suppose n-1 and n+1 are both prime.
Then n must be divisible by 6 because (bla bla bla). So n² is a multiple of 36, and n²+16 is a multiple of 4. These two facts make n²(n²+16) a multiple of 36·4 = 144 = 720/5. In order for it to be a multiple of 720, either n² or (n²+16) needs to be a multiple of 5 too.
Now split into cases according to the remainder of n modulo 5:
n == 1 (mod 5) or n == 4 (mod 5) are impossible (bla bla bla).
If n == 0 (mod 5) then (bla bla bla)
If n == 2 (mod 5) or n == 3 (mod 5), then (bla bla bla)

Your answer to "is the converse true?" seems sound.
(The math-hipster approach here would be: Because of Dirichlet's theorem on primes in arithmetical progressions, there is a prime of the form 720k+1. But of course n=720k satisfies 720 | n²(n²+16).)

[frail geyser]

i understand up to the modulo part. why is it n instead of n^2 for modulo 5?

[frail geyser]

interesting!

[near hill]

You need to be looking at n itself in order to use the assumption "n-1 and n+1 are prime" to exclude the cases n == 1 and n == 4.

(Each case for n mod 5 lets you compute n² mod 5 next, of course).

[frail geyser]

so after finding the cases of n mod 5 which are valid would you try to find what multiples n must be to give those remainders from modulo 5?

then would that give you a formula for all n terms with n-1 and n+1 being prime?

[near hill]

No, in each case I would simply look for proof it n is such-and-such modulo 5, then 5 | n²(n²+16).

You don't need to find n. Just to prove that whatever n is, then as long as the other assumptions hold, 720 | n²(n²+16) will be true.

[near hill]

Whoops my hipster solution was horribly backwards and doesn't actually work.
Better hipster solution: Because of the Chinese Remainder Theorem there exists an n such that n == 0 (mod 720) and n == 1 (mod 7). Then n-1 is divisible by 7 and so cannot be prime, but n²(...) is still divisible by 720.

[frail geyser]

ok, why can we not just find what multiples of n allow you to factorise 720 out of the equation such that whatever integer n is, n-1 and n+1 being prime along with 720 | n^2(n^2+16) are satisfied. Though, i feel like this sounds a little similar to what you said about just prove whatever n is it is divisible.

[near hill]

Because you do not know when you start the proof that 720 divides n²(n²+16).
That is the thing you need to prove.

[frail geyser]

so how would you prove 720| n^2(n^2+16) for each of the valid cases of n modulo 5 whatever n is, I see how it works with n modulo 5 being 0 cause then you have (56n)^2((56n)^2 + 16) then you can factor out the prime factorisation of 720 but correct me if i am wrong, i dont think you can do that with the cases of n modulo 5 being 2 and 3

[near hill]

Hint: In those two cases, compute n²+16 modulo 5.

[frail geyser]

ah i see

it becomes 0

[near hill]

Yes.

[frail geyser]

ok that makes sense, then you can just show that it is a multiple of 720, thanks!

[acoustic nova]

Yeah but still

[acoustic nova]

Ok sure. If for example you take continuity for granted a lot of the time

[sullen blaze]

just write p1 = 6k + 1 and p2 = 6k - 1

solution is then straightforward.

ok I see you have used 6k, you can prove 9k² + 4 is always divisible by 5

let k be 5s + 1, but it back into the p1 and p2, you will see one of them is divisble by 5, making k = 5s + 1 invalid, same for 5s + 4.

the other three are perfectly valid, i.e k = 5s, 5s + 2, 5s + 3

so any prime will be of form of k = 5s, 5s + 2, 5s + 3

which makes 9k² + 4 divisble by 5, or k² divisble by 5

[radiant jasper]

yo anyone got any good lectures for number theory? need it, preparing for olympiad

[frail geyser]

thank you!

[frail geyser]

i tried recalling what we discussed, i am trying to learn the chinese remainder theorem but i havent added it here for the converse but i think my contradiction by using 48 was good. Does this look along the lines of what the solution should be?

[near hill]

In case 2 and 3, it looks you're substituting 2 or 3 mod 5 into 3²·2^4·R²·(9R²+4) -- but they should just be the n in n²(n²+16).

E.g. for case 2, even if you want to be very specific, you can just say

If n = 5p+2, then n² = 25p² + 20p + 4, and then n² + 16 = 25p² + 20p + 20, which is clearly divisible by 5. So n²(n²+16) will be divisible by 5 too.

[high goblet]

looks okay but your write up could be better

[high goblet]

also as tropo points out here, you actually only need to analyse n^2+16 for the other cases

[high goblet]

i'd phrase it like this: "to prove n^2(n^2+16) is a multiple of 720, it suffices to prove it is divisible by 2^4, 3^2 and 5"

(do your stuff with n must be a multiple of 6), so n^2(n^2+16) is always divisible by 2^4 and 3^2

now we show it is always divisible by 5

if n is divisible by 5 then we are done, otherwise (do your stuff with n is 2 more/less than a multiple of 5)

(btw you can often combine cases like these together by doing like 5k+-2)

[peak kestrel]

"Is the converse true" is actually a wild question to even be asking

Like you must have learned about primes yesterday if that's a question you can even imagine to ask expect to not have a negative answer

In particular, it means that every number of the form 60m±1 is prime, which means pi(n) = O(n), which contadicts the prime number theorem Q.E.D.

[twilit zodiac]

what should i do to prepare for amc 10 (i dont know much number theory and im not the best at combinatorics, but i think i should be fine on most of the alg and geom in it)

[half mulch]

have you looked at volume 1?

also intermediate cp is a good book

[pure mauve]

True, but this is the BMO - at this point, answering "no" is insufficient; you have to prove this, either by how you've done it or by CE (as hawklo did)

[remote light]

amc 10 is kinda ez, you must study nt and combi ig

a little gisp

js use aops for problems

[tribal berry]

hey anyone giving IOQM

[deft nexus]

Yes, probably

[neon moss]

I’m trying to self learn AOPS but I find it kind of hard to understand some of their solutions or strategies

I have volume 1

Any tips to help, I am also studying for AMC 10

[kindred wraith]

volume 1 is sort of like a quick summary

its not as deep as if you grabbed like intro to algebra, intro to geometry, etc

its for sure not enough practice problems if youre not already super comfortable

i definitely recommend grabbing the entire introduction set of books if that is an option

if not, youll have to find other sources, perhaps their website, and work your way through a checklist of things

in the meantime, if there is anything in particular you are stuck on, you can use help like this discord and ask in the proper channels, free and limitless

[remote light]

true

[remote light]

me

[remote light]

do take help from chatgpt too but chatgpt is unreliable and is not a math tool, and will frequently give you results that look right but are not

if you dont understand once counter ask them at least 3-4 times , if you still dont understand you need a pro teacher

might take help from your maths teacher in school

sometimes guys on yt have posted solutions , you can check them out (that's all I can tell you from my side as I use these techniques )

or people are here too

[kindred wraith]

do NOT take math help from chatgpt

[remote light]

I do confirm the answers

[kindred wraith]

chatgpt is unreliable and is not a math tool, and will frequently give you results that look right but are not

[remote light]

I know, I match them with solutions

[brave ocean]

was this from a spammer or from this conversation

[kindred wraith]

spammer

[brave ocean]

oh ok

[ping dealt with] <@&268886789983436800>

you don't have to delete the ping

we will just give it a thumbsup

[kindred wraith]

yeah you check your answers, but then 1. why aren't you just using a math engine in the first place and 2. that doesn't mean you should recommend it to someone who may not have literacy in either AI usage or math

if someone is struggling with concepts pre-AMC 10 then they are likely not mathematically mature enough to verify solutions on their own

you dont need it, dont touch chatgpt

[reef condor]

💀

[bold tiger]

huh

[near hill]

Not really a math question.

But it's F to F, which is ...

[acoustic nova]

loll

[bold tiger]

?

[rocky haven]

XD

[bold tiger]

?

[twilit zodiac]

but the problem is i dont rly know number theory, where can i learn it, also its been so long since i did combinatorics so i dont remember much and my teacher had a thick russian accent

[subtle sundial]

youtube has alot of basic number theory principles

prime newtons, michael penn dive in depth into it

[twilit zodiac]

ok ty

[tribal berry]

Hey tommorow , i asked for IOQM , guy replied , is he here ?

[remote light]

sowy

[tribal berry]

hey @remote light

[remote light]

yo

[remote light]

ah hi

[tribal berry]

Hey when you started your Ioqm prep

[remote light]

1 month ago (the real one) before that I was just focusing on the basics

to be exact 29 days ago

[tribal berry]

Yo i started on june 😅
but i am more focused on other things

[remote light]

nsep?

or academics

[tribal berry]

Coding

[remote light]

oh great

which language?

you can learn C++ and give ZIO

[tribal berry]

python , basic js

[remote light]

ohk

[tribal berry]

Bro which class

[remote light]

10

[tribal berry]

Same here

[remote light]

if ur learning js then u must have learnt html and css

[remote light]

oh gr8

[tribal berry]

Yes , basic stuff of html and css

[remote light]

ok

[tribal berry]

Bro IOQM topic , is more pain in the neck ( but i enjoy everything)

[remote light]

hmm

have you completed your syllabus?

I guess, I have but need to work upon problem solving

[tribal berry]

IOQM don't have a syllabus , logic is everything

[remote light]

it definitely has and logic is an integral part of it

except calculus everything is there of high school maths

[tribal berry]

but yes , you should now the basic stuff

[remote light]

Algebra, combi.nt and geometry

[tribal berry]

bro you studies calculas ?

[remote light]

nope

it aint a part

ill do it in the next grade

obv I have done basics for 11th physics

[tribal berry]

ok great , it's great , its your choice

[remote light]

have u done?

[tribal berry]

Yes 😅

[remote light]

DAMN

[tribal berry]

Bro it's actually very easy

[remote light]

hmm

[tribal berry]

More than Geometry ☠️

[remote light]

(of ioqm)

not jee

[tribal berry]

Yes\

[remote light]

Ioqm doesn't even has calculus tho

[tribal berry]

Jee , no
it doesn't have synthetic geometry

[remote light]

ye

[tribal berry]

No not directly
bro it helps a lot

Still i can't solve problems

[remote light]

alr cya

gotta do nt

[severe eagle]

good luck

i am addicted to geo but i am degenerating

into a degenerate triangle

[remote light]

thank you

[severe eagle]

i cant even see some trivial G2 shit

[remote light]

lol

[tribal berry]

👍🏼

Hey studied transformative geometry

[remote light]

yeah

inmo imo level

[tribal berry]

But yeah , i tell's about different ideas

[remote light]

which olympiad will you give this year?

[severe eagle]

give?

[remote light]

the name

[severe eagle]

the major one finished and i kinda died

[severe eagle]

smo

[remote light]

oh

[severe eagle]

singapore math oly

[remote light]

sasmo

[tribal berry]

senior ?

[severe eagle]

no

[remote light]

is lower

[severe eagle]

thats small olympiad

[remote light]

ye

[severe eagle]

smo is the main

[tribal berry]

coaching ?

[severe eagle]

anyway gtg go geom bye

[remote light]

yeah

cya

[remote light]

It exists in India only

coz the schools dont care about olympiads

[tribal berry]

I am asking to you😅

[remote light]

most of them*

oh

PW

[tribal berry]

oh , I am self study guy

[remote light]

ohk

I also mostly do self study

coz they aint doing nothing

[tribal berry]

you flex in front of someone ?

[remote light]

like they cant even complete the syllabus

[remote light]

in what terms?

[tribal berry]

☠️

[remote light]

about coaching?

I dont like to flex

[tribal berry]

No , about being different

[remote light]

Actions speak louder than words

[tribal berry]

I also don't , but sometimes , when i diidn't complete my school notes school teacher , and she ask's about " what you were doing ?"
i don't even flex but still ☠️

[remote light]

I ignore them, they just keep following us like simps

but they also want the development of us

in academics

I gotta handle that calmly

[tribal berry]

yeah you are right
i regret a lot for my words 😭

[remote light]

nah its fine

alr cya in the evening

gotta complete some part of my syllabi

[tribal berry]

any prblem to discuss

of maths

[remote light]

I'll come back to you

maybe I got some

related to combi

[tribal berry]

oh yeah always on fire 🔥

[remote light]

come on dm, Ill send

[neon moss]

That is possible, but I’m studying for AMC 10, so how could I study all 4 intro books before the competition on November 12th??

[frank hornet]

I guess maybe review the most important concepts from each book (you don't have to do all of the problems, just the ones you think that are important/could appear on AMC 10).

[kindred wraith]

oh i didnt have this context

you dont, you just grind and do your best

its not realistic to learn all of that by the time the test rolls around

just do what you can

[zinc tundra]

how much was aime qual for amc 10 last year

[golden oak]

The AIME floor cutoffs for AMC10 were 94.5 for AMC10a and 105 for AMC10b

[novel karma]

Any predictions about amc 10? I feel like there’s gonna be more geometry

There were very few geometry problems last year

[radiant jasper]

fr

[radiant jasper]

there might be a tonna aritmetic

[bronze smelt]

Could someone explain permutations and combinations for me and give me some problems?

[half mulch]

do you know stars and bars

its pretty intuitive

[bronze smelt]

No

[half mulch]

want an explanation?

[bronze smelt]

Ye

[half mulch]

okay

so

the original question

is how many distinct ways we can put 8 balls into 4 baskets

I'll just start there

so we can draw this out

we're gonna use lines to divide the balls and represent the baskets

so one such case would be

oo |oo |oo | oo

this would be like 2 balls in every basket

another could be

[bronze smelt]

Ik

[half mulch]

oo |ooo |ooo|

which leaves one basket empty

but anyay

anyway

the trick comes in

when u realize

you always have the same number of o's and |'s

you're just rearranging them

so in total

there are 8 balls and 3 lines

so 11 total spaces

and we pick 3 of them to be lines

the ways to do this

is 11 choose 3

which is like

165 I think?

idk

wait preface

you know how to use the choose function and what it represents right?

[bronze smelt]

No

[half mulch]

oh

uhh

maybe I should start there

[bronze smelt]

Are you sure this is combination and permutations?

[half mulch]

ya

[bronze smelt]

Isn’t it like factorials?

[half mulch]

factorials are a super important part to combinations and permutations

n! represents the number of ways u can arrange n distinct objects

okay okay

back to the choose function

let's say we had

uhh

7 distinct rocks

and we wanted to know how many ways we can pick 3 of them

7 options for the first rock

then we pick that up

then 6 options for the second rock

then 5 options for the last rock

but right

the order we pick them up doesn't matter

so we gotta account for thag

so we have to divide off what fraction of the ways would be over counting

because when we pick up those 3

no matter what order we picked them up in

we still got the same 3 rocks

so essentially

we wanna know how many ways we could order 3 rocks

which is 3!

cuz 3 options for the first rock

2 options for the 2nd

and one for the last

combining this all

the ways to pick up 3 rocks from 7

is 7(6)(5)/3(2)(1)

notation wise

usually you see the choose function in a different form

this can be found by multiplying the top and bottom by 4! (in general its (n-k)!)

which gets you n! / k!(n-k!)

[bronze smelt]

i know

but i had a question

its on mhy aops homework

u see, i took an aops mathcounts/amc 8 basics

[half mulch]

do u understand that

[bronze smelt]

and it was going fine until we hit permutations/combinations

[half mulch]

ya

did my explanations help at all

[bronze smelt]

yes i know that formula

no

[half mulch]

💀

[bronze smelt]

but its okahy

but i already kniw

[half mulch]

I'm a bad teacher mb

[bronze smelt]

here is a problem

[half mulch]

u already knew the intuition for choose function

[bronze smelt]

How many equilateral triangles in the plane have two vertices in the set ${(0, 0), (0, 1), (1, 0), (1, 1)}$

[gilded halo]

Critical Peak

[neon moss]

Ohhh that’s a really smart idea, thanks a lot 🙏

[bronze smelt]

can u just guide me

i was thinking mahybe arrange cases

[half mulch]

okay so first off

I'd graph this

[neon moss]

lol ok

[half mulch]

draw it out

you have a square of vertices in the first quadrant

[bronze smelt]

but is that necessary

[half mulch]

drawing it is a good first step for combinatorics

[bronze smelt]

i think we can just use cases

[half mulch]

I mean yeah

[bronze smelt]

so triangles with 4 vertices

that'd be 1

[half mulch]

what

[bronze smelt]

sorrhy two

wait i got it

correct

also this off topic

[half mulch]

is the answer 12?

[bronze smelt]

preciselhy

[half mulch]

do u want an explanation

[bronze smelt]

but this is off topic

[near hill]

For each choice of two points in the set, there's exacly two places we can put a third point to make an equilateral triangle.

[half mulch]

yaa

[bronze smelt]

whenever i type "y" it makes it"hy" on my keyboard

[half mulch]

true

[half mulch]

lmao what

[bronze smelt]

what do i do to fix it

ik its random

[half mulch]

hm

[bronze smelt]

but i need help so bad

[half mulch]

is it that keyboard

or the computer

could u try switching keyboards

[bronze smelt]

how do i do that

im on mac btw

[near hill]

Is it a laptop with integrated keyboard?

[bronze smelt]

idk

[half mulch]

are u on a laptop

[bronze smelt]

correct

[half mulch]

okay try resetting your keyboard driver

can u even do that on mac

[bronze smelt]

what is that

[half mulch]

like the program that let's ur keyboard work

[bronze smelt]

where is it

[half mulch]

I'm not sure abt mac

[bronze smelt]

whhy is trophospher tyhping forever

[near hill]

My guess would be there's an electrical problem in the keyboard itself (such as corroded or shorted traces). I've had several (external) keyboards fail on me with similar symptoms over the years. It's not really feasible to repair other than replacing the keyboard. Some laptop designs are modular enough to swap out the keyboard with a spare part (if you can get one or it's under warranty) -- in other cases it's the end-of-life for the entire laptop unless you take to always using it with an external keyboard.

[bronze smelt]

but im pretthy sure this new

u see how it switches to hy

[near hill]

Sorry.

[half mulch]

Mac probably cant

[novel karma]

Examples?

[radiant jasper]

idk

[half mulch]

I think he meant algebra

[elfin falcon]

What are prerequisites to START this type of stuff?

[humble charm]

I do this math competition every year, do pretty well without studying. I was thinking of actually trying to study so I can steal first place. This competition seemed to have advanced high school math like proofs, trig, probability, statistics, geometry, and algebra. But it doesn’t have calculus from my research.
Are there any resources to practice and better myself in these math areas? Any tips and tricks?

If you for some reason want to research this competition its the SCSU Math Contest and i will competing against 11th graders. They have previous years tests on their website which I will be using as a resource but would appreciate better resources.
Thank you.

[fossil vessel]

There’s no better prep than doing every old test you come across. Additionally, set a timer, if applicable. Get as close as you can to simulating testing environment

Load past exams into an AI and have it generate more.

As you do them, you’ll know which areas you need help in. Then I’d double back and ask for resources on those subjects.

Also this is a wonderful resource for general formulas that you should have memorized.

https://cdn.artofproblemsolving.com/attachments/9/0/0e7904bb539cd6056457058f60d7a1c7864bb8.pdf

[kindred wraith]

probably going to find limited use throwing problems at an AI, because the AI solutions might not be correct and you might not know

i wouldnt waste time here getting stuck into dead ends, there are enough human-curated problems on the internet

LLMs are really bad at math, dont bother

[polar patio]

aops 🤔

its got tons of problems

[neon moss]

I asked ChatGPT for a list of the most important things to study in all four AOPS intro books, is it a reliable source?

I don’t have enough time since studying for AMC 10 in November

[half mulch]

what did it tell you

[half mulch]

3 months is plenty imo

if u work consistently and learn everyday

[neon moss]

For all 4 books 😭

Plus practice tests V

[keen lake]

how do I get good at geometry

[magic skiff]

do egmo

[keen lake]

done most of that

[magic skiff]

do plenty of problems

[magic skiff]

thats good do more

[keen lake]

I can never remember Miquel point stuff

[keen lake]

I’m not doing the bash chapters

[magic skiff]

i think if you do a good number of problems using miquel properties then it will become more ingrained in your brain

[magic skiff]

the bash chapters are sometimes pretty useful

good tool to have in your pocket

in case you cant think of a synth solution