competition-math

[high goblet]

what the fuck why is there a y9 kid here and why is he 100 times better at maths than me

[pallid tundra]

need some help on this ineq - I’ve tried writing everything in terms of 2^x and 3^x but am unable to do anything useful with that

[ivory ember]

Consider multiplying the left hand side by 2. Then ||complete some squares||.

[glad zinc]

He asked that because it was a question in a Olympiad we both did a couple weeks ago, he’s not asking for help

South’s secret twin brother replied with the answer but I recommend looking at the solutions and investigations of the SMC 2024 to see similar problems to this and other notes they have added to the question. Good way to improve. The question we re talking about is Q24 if I’m not mistaken

[ornate blade]

it's q25 SMC 2024

[glad zinc]

Oh yeah mb

[fallen magnet]

tbf theres a yr8 at my school who did his further maths a level already with an a* 💀

[glad zinc]

Eh thats not that good, there's a kid i know who doubled s ed on STEP in kindergarten

[fallen magnet]

wait just realised he goes to tonbridge school

very close to mine lmao

[gilded halo]

NicknamedTwice

[fallen magnet]

this one hurt my brain

[jade widget]

WTH

Yr 8 is grade 7 right?

Wth

[high goblet]

as in alex did the IMO in Y7

[high goblet]

for a level FM, the main barrier is schools

most schools will not let you do FM that early even if you are good enough

because it hurts your chances of getting into good unis

[fallen magnet]

nah my school let us, technically i couldve done it but whats the point of doing it early

[ornate blade]

lucky

[glad zinc]

can we petition mods to make a uk maths channel or something like that

[pallid tundra]

oh aight ty

[high goblet]

there's already a UK maths olympiad discord server

it's much smaller and a lot quieter tho

[vocal geyser]

yo anyone here preparing for Sharygin Geometry Olympiad??

[celest pollen]

YOOOOOOOO IMO IS PEAK

Im going to HKIMO TIMO WIMO from 7yo

WIMO got glass medals 2019

[fallen magnet]

glass medals look insane

[thorny latch]

Does anyone know how to do this

[vague temple]

Do you know about power sets?

The collection of all subsets of a set

This definition is equivalent to S being a subset of its own power set.

Analyze a generic set of 4 elements and its power set

I’m realizing that a lot of competition maths is pretty much covered in real analysis, maybe I should do myself a favor and read through a textbook properly

[warped wedge]

Hey guys Im doing AMC 12 next week and I'm wondering whether its worth grinding alcumus rn if i can usually solve 12 ish problems

Im trying to qualify for AIME

[thorny latch]

For our nsml prep

We never even covered anything like this

Even our leaders had no clue what was going on

But I’m worried that on the competition it’s going to be like this

[swift imp]

math is a year round grind so dont do too much since amc is next week

as in like youve done all the prep now

dont get too tired

[warped wedge]

Certainly, its just if i dont keep practicing my brain gets rusty u know

[swift imp]

Yea

But no long hours grinding

Just enough to stay where u r

[celest pollen]

and also you go to different countries for each one of thenm

[radiant jasper]

tecxhnically an upsetting set is a power set

and the formula for number of subsets or n(Power set) is technically formula of combinations (nCr). the formula of number of subsets for a set having n elements is $$2^n$$

[gilded halo]

Professor Uchiha

[radiant jasper]

so u can use that

since 4 element

$$2^4 = 16$$

[gilded halo]

Professor Uchiha

[timber wraith]

what could I be missing?

the construction makes me think that EY = EF

but the given figure says otherwise

[radiant jasper]

i would suggest the use of trigonometry to determine whether BEF is a right triangle

after you’ve done that, accept that diagrams will often be drawn inaccurately to avoid solvers guessing properties off the diagram without proof

[radiant jasper]

hint on how to do this: ||sine rule on EF and <EBF vs. EB and <EFB||

[timber wraith]

oh right

well, it seems that EY = EF indeed

[radiant jasper]

it would in fact seem that F=Y

[timber wraith]

yeah

if it wasnt, my idea was to get value of EY, then subtract EF from it

[timber wraith]

i feel like im very close😭

[radiant jasper]

what happened here

[radiant jasper]

you can use cos law on BOF, i think

you have OF and BO in terms of x, BF as a value, and you should be able to find one of the angles (i shall not divulge which)

[timber wraith]

i bruteforced the similar triangles

[thorny latch]

The answer is 9

[alpine bobcat]

$$2^4 = 16$$

[gilded halo]

warbled

[mortal tinsel]

I was trying to go to sleep when I got an idea of how to solve for average distance of two points in [0,2] using recursion

Let x be equal to this average distance.
Let's bisect [0,2]. There is a 50% chance the points are in the same section (in which case the average distance is 0.5x), and there is a 50% chance the points are in a different section (in which case average distance is 1).
So this gives us x=50%(1)+50%(0.5x)
or x=0.5+0.25x
or x=2/3

This seems really interesting, does anyone have resources on a more general way to use this technique? Thanks

[fast coral]

like

there are similar problems i can recommend

[mortal tinsel]

sure, if they have similar solutions

[timber wraith]

i dont have OF yet

[radiant jasper]

in terms of x you should

[timber wraith]

i dont have BO either

or am i not getting what youre tryna imply

@radiant jasper

[barren sparrow]

how do i solve this? i feel they missed something in this question

[shy meteor]

Hello

Can someone help me out on

34A3 = 1(mod33)

When I solve it I get A = 24 which is incorrect bc how can it be two digit

It’s 1 but idk why it’s 1

I guessed 1 because 34A2 = 0(mod33) and 12 is divisible by 3 so I guessed 1, but that’s js a coincidence yes?

Nvm A = 3 apparently, but I don’t still don’t know why

[gusty verge]

Use: $34A3=3403+10A\equiv 4+10A\mod 33$

[gilded halo]

Max

[gusty verge]

Not really the topic of this channel.

Essentially draw it, then draw the perp distance

To the line

[vague temple]

Yes, it’s a very common strategy in Markov chains and probability theory

Recursive decomposition or recursive computation of expectation

[barren sparrow]

mb but like there can be infinite points right?

[gusty verge]

Just draw it, but if we are talking about the shortest distance then it's unique

[radiant jasper]

“in terms of x” remember what x is?

[radiant jasper]

i’m not entirely sure how to solve it but nothing seems to be missing

[barren sparrow]

well ok thanks

[vague temple]

Hi everyone, I proved that every odd square is 1 mod 4 using the sum of first 2k + 1 odd numbers, but is there another solution?

[reef condor]

Do casework on the value of the odd number mod 4

Or expand (2x+1)^2

[vague temple]

Oh yeah that’s smarter nvm

4k^2 + 4k + 1

I didn’t see that

Thanks

[radiant jasper]

take 1^2 and 3^2 mod 4; note that they are both equal to 1

[vague temple]

Yeah I thought of that too but not the factorization for some reason

I practiced too much casework now all I see everywhere is casework

[shy meteor]

I am confused on why though

[ivory ember]

you're expanding it in base 10

In general, $$\left( \overline{d_n d_{n-1} d_{n-2} \cdots d_2 d_1 d_0} \right)n=\sum^{n}{k=0} d_k n^k$$

[gilded halo]

Civil Service Pigeon

[ivory ember]

You probably don't need the full on definition for this since you use base ten 95%+ of the time

but the general definition is def good to know for anything other than base 10

[shy meteor]

I’m confused

My teacher just said to find a number that made it divisible by both 11 and 3

[shy meteor]

Okay thank u

[ivory ember]

I mean yeah that's the same thing

[shy meteor]

As that definition?

[ivory ember]

expanding it like that is how you derive the divisibility tests for 11 and 3

so

[shy meteor]

I am so confused

So 3 + 4 + A + 3 = number that is divisible by 11 and 3

?

No it is not this..

nvm

[ivory ember]

divisibility test for 11 tells you that 3 - 4 + A - 3 is a multiple of 11
divisibility test for 3 tells you that 3 + 4 + A + 3 is a multiple of 3

Taking 4 + 10A to be a multiple of 3 and 11 gives you the exact same thing

[shy meteor]

Where did the 10 come from?

Omg

[shy meteor]

What is divisibility test

[radiant jasper]

have you learnt about place values

[shy meteor]

What is place values

Maybe I did but don’t know that name

[radiant jasper]

12 = 10+2

[shy meteor]

Yes

[radiant jasper]

123 = 100+20+3

[shy meteor]

Yes

[radiant jasper]

1234 = ?

[shy meteor]

1000 + 200 + 30 + 4

[timber wraith]

wow

[shy meteor]

But my teacher writes as 10^3

[radiant jasper]

34A3 = ?

[radiant jasper]

that works too

1000 = 10^3

[shy meteor]

ohhhhhhhh

oh😭

[timber wraith]

PF?

[shy meteor]

I see

[radiant jasper]

yes, and also the radius of the circle

[timber wraith]

yeah

[radiant jasper]

what relation does the radius have to OB, and what about OF?

[timber wraith]

one sec, ill repost my soln (too tired to backread)

[timber wraith]

radius = OB=OF

[radiant jasper]

radius = OB, yes

OF is a bit different

[timber wraith]

oh wait yeah

OF is sqrt 2 times the radius

[radiant jasper]

and BF is known to be 4rt(3)

you can find one of the angles in BOF

[timber wraith]

BFO = 15deg ?

since PFB = 30, and PFO = 45, then BFO = 45-30=15

[timber wraith]

,rotate

[timber wraith]

@radiant jasper

[radiant jasper]

why ping

[timber wraith]

mb

[radiant jasper]

yes to BFO

[timber wraith]

can i find equivalent ratio if cos 15 w/out some table

[radiant jasper]

4rt3 should be squared but yes

[timber wraith]

woopsies

[radiant jasper]

i suppose you could use the double angle formula for cosine to determine cos30 in terms of cos15

i have cos15 memorized so it’s not usually a problem

[timber wraith]

this one?

[radiant jasper]

yep

[timber wraith]

do i just plug this in the quad formula

[radiant jasper]

wait

i believe the expansion is incorrect

[timber wraith]

ooh

[radiant jasper]

from line 4 to line 5 (the first line without brackets), you missed a minus sign

[timber wraith]

oh right

dang my arithmetics

i now got 0 = x² - x(12+4sqrt.3) + 48

[radiant jasper]

i think you can factorize that

[timber wraith]

i just plugged it in😭

[radiant jasper]

i suppose that works?

did you get an answer

[timber wraith]

one min, still simplifying jt

dang image wont send

i got x = 6 + 2sqrt3 +- 2(sqrt6)(4throot3)

[radiant jasper]

,w x^2 - (12+4sqrt(3))x + 48 roots

[timber wraith]

what

[timber wraith]

,rotate

[gilded halo]

Results provided by WolframAlpha

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

Requested by micabo_l8r

[radiant jasper]

wrong prefix

that is a strange number

i don’t like it

[timber wraith]

which

[radiant jasper]

all

i don’t trust it

[timber wraith]

lol what

[radiant jasper]

it’s the answer to a geometry question

it shouldn’t be like this

[timber wraith]

yeah

tbh what the fck is this problem honestly

it even came from the ministry of defense

[radiant jasper]

well, i see nothing wrong that i could’ve done

but still it is strange to me that it is so ugly

[timber wraith]

ikr

the soln is seem so wacked

[radiant jasper]

i suppose i’m trying not to “cope”, as they say

do you have an answer to this question?

as in, from a book?

[timber wraith]

ill try to see one

dang, cant find it anywhere

[modern prawn]

I’m not well versed in modular arithmetic. Can someone explain the modular arithmetic used here? I dont understand. How did they get 2^606 - 1 congruent to (-1)^606 - 1?

[radiant jasper]

2 is equivalent to -1 in mod 3

[quick frost]

To add to that, if a ≡ b (mod n) and c ≡ d (mod n), then ac ≡ bd (mod n). You can apply this with n = 3, c = 2 and d = -1 many times (605 times to be precise) to get 2^606 ≡ (-1)^606 (mod 3).

[jade widget]

What is the Rem thing

[radiant jasper]

Rem(x, y) = remainder of x upon division by y

i must admit that this is my first time seeing such notation as well

[jade widget]

Oki thx

[pallid ginkgo]

which comp is this?

[round jackal]

hello. i am currently creating a math formula that doesn’t do much just for fun but

if someone here is a genius.
knows tetration pls help me out

this is with exponentiation

you have c to the power of m you need to divide by c to the power of m-1 to get c

example 10^10/10^9=10
10 = c (constant)
put c to the power of m which is 10 in this case
divide by m-1 which is ^9

10^10//10^9 = 10

it always works

i want to expand this to tetration

if you can help me pls dm or add me

basically the point is

you need to divide c by c and it has to equal c
but it has to be to the power of something and then to the power of whatever that is -1

to stay at c

i’m stuck on that
3^3 = 3 • 3 • 3 (c^m/c^m-1=c)
and
3^3 = 3^3^3 (?)

this is what i need help with

[ornate blade]

hold on

cause for example, if you take the base to be 3

[gilded halo]

south's secret twin brother

[ornate blade]

but you could do $\sqrt[3^3] {3^{3^3}} = 3$ for example

[gilded halo]

south's secret twin brother

[ornate blade]

it gets trippy when you try to generalise this

[karmic rune]

Addition is extended to Multiplication. Multiplication to Exponentiation, and Exponentiation to Tetration. So logically, the expression $\frac{c^n}{c^{n-1}}$ involves the exponentiation of $c^n$ and $c^{n-1}$ with a connecting relation of the inverse of its primitive. So you divide it. So, logically extending this to tetration means you have to relate the tetration $^{n}c$ and $^{n-1}c $ with the inverse of exponentiation.

so $\log_{^{n-1}c} \left( ^{n}c \right) = c$ is my logical extension

[gilded halo]

Bacter10Fr4g is not fr0g

Bacter10Fr4g is not fr0g

[karmic rune]

but is that true?
I do not want to calculate that

[round jackal]

wiat what

$\sqrt

{3+3=6}

@karmic rune how do you use bot

@gilded halo

[karmic rune]

You learn Latex and format your content according to its rules. If stuff is proper, bot detects it and converts it automatically

[round jackal]

what

type in

[karmic rune]

what do you wanna type?

[round jackal]

“3^3^3” square root “3^3^3^3”

[karmic rune]

ok

[round jackal]

how do you make the bot work

[karmic rune]

$\sqrt[3^3^3]{3^3^3^3}$ is correct I hope

knew it

dang

[round jackal]

$\sqrt[3^3^3]$

[gilded halo]

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

[karmic rune]

Just gonna do
${3^{3^{3^{3}}}}^{\frac{1}{3^{3^{3}}}$

[gilded halo]

Bacter10Fr4g is not fr0g
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[karmic rune]

btw its not a calculator

[round jackal]

${c^m/c^m-1}=c$

[karmic rune]

$\frac{c^m}{c^{m-1}} = c$

[gilded halo]

max

Bacter10Fr4g is not fr0g

[round jackal]

but that’s for exponentiation

i need to create the formula for tetration

[round jackal]

HOLD UP

what if we’re calculating c?

no that’s not what i’m doing

so i need to find the value of C i guess??

idk

(c^m/c^m-1=c) exponentiation
(? = c) Tetration

so i just want to find the formula.

[round jackal]

and it has to be consistent

who here has a phd

[hallow finch]

Are any of you guys doing the IMO

[radiant jasper]

hopefully next year

[vocal tundra]

Hopefully in 2027

[high goblet]

unfortunately not 😢

narrowly missed out on selection, and now i'm at uni so i can't do it anymore

[autumn stirrup]

Hello im going to be doing the WMTC or World Mathematics Team Championships. Are there any resources, tips, or advices for me to help review for it?

[novel karma]

guys when are yall taking amc 10

mines this wednesday 😭

[reef condor]

Yeah everybody has it this Wednesday

[stoic hatch]

guys i need help on this quiestiion
question

[vague temple]

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24

[vague temple]

Hi, can someone explain an example on how to use mass point geometry?

[modern prawn]

How would you solve this? I know how to solve the intersection if i know how far their centers are from each other, but 25% of the circumference seems weird. How to solve?

I think what i tried was spliting the intersected area into two equal parts, and then forming a triangle

Yea i tried doing this

[vague temple]

That should work

1/2 * pi * 2^2 (2 quarter disks = half disk) - 2^2 (2 45-45-90 triangles = square) = 2pi - 4

[modern prawn]

How do we know that angle is a right angle though

[radiant jasper]

25% of circumference

[modern prawn]

We would get sector of circle radius two for that would be pi, right? Then, we have to subtract that from triangle which we know is 2,2,(sqrt8), so we get 4(sqrt8)

[radiant jasper]

that means 25% of the angle at the origin or 25% of 360

[modern prawn]

Pi - sqrt(128) which means the total area of the intersection is 2pi - 32

[radiant jasper]

the area of the triangle would not be 4rt(8)

[modern prawn]

where did i go wrong aaaaaaaaa

Oh wait… its HALF

im dumb

ok so we get sqrt 8

[radiant jasper]

the area of the triangle would not, in fact, contain any sort of square root

base times height over 2

which one is your base and your height?

[modern prawn]

I’m so dumb

I multiplied the sides not the altitude by the base

[radiant jasper]

all good

what’s your area?

[modern prawn]

More problem presented, base is clearly sqrt 4 but idk the height

I’m too dumb for this crap

How did i get above 90 on the aops practice test omg i cant even solve this

[ornate blade]

I don't think you can use mass point geometry cause you have a rectangle instead of a triangle

furthermore you have two transversals AG and EF

so even if you could, you would need to find split masses

[vague temple]

Oh ok, so is there no shortcut compared to Cartesian coordinate bashing or drawing similar triangles?

[ornate blade]

I can't say in general but there's not enough information when you split ABCD into two triangles
You need to compare areas or use coord geo

[radiant jasper]

isn’t your triangle 2-2-rt(8)?

[modern prawn]

yes

WAHT

I EMANT SQRT8

also if its a triangle like that

the area is 4

I just realized lol

oh wait

No

Its 2 x 2/2 = 2

P - 2 = 2pi - 4

[radiant jasper]

yes

[radiant jasper]

so half of your shaded region will be the sector (which you earlier found to be π), minus the triangle (which is 2)

double that for the whole shaded area

[radiant jasper]

so the answer is 2π-4

[winter sable]

i am back

only sigmas remember me

[jade widget]

4 points fall randomly into a circle, what is the probability that 4 point fall into a same semicircle

I get that for 3 points it is 3/4, but I don’t get the forth one

[stoic hatch]

ya

ya

aya

ya

[wanton bolt]

I am very lost in this proof, what is happening here?
(Example 8)

[modern prawn]

what textbook is that it looks cool

tell me the name

[novel karma]

stfu

when should i start learning proofs? i'm doing amc and aime only rn

[modern prawn]

Nvm.

[wanton bolt]

Why?

[modern prawn]

I thought it was some cool algebra proofs book but its pre-college i wont understand crap 💀

[vale lynx]

real bro

[fallen magnet]

how come u got the pre uni role without knowing college stuff?

isnt university after college?

[hearty tendon]

multiply by cube root of p

[wanton bolt]

buyt like what does the b*(1)-c (2) mean?

[hearty tendon]

(1) and (2) are equations

so take b*(1) (the equation) and subtract by c*(2)

[wanton bolt]

that is so obscure

how was I ever subbosed to find that

[hearty tendon]

its not too bad cuz theres not much you can do in general
if you multiply stuff by cube root of p or cube root of p^2 you get something "cyclic" with structure so its worth a shot

with that being said by first approach would be to say that the quadratic cx^2 + bx + a = 0 has cube root of p as one of its roots and try to push a contradiction from there

which im 90% sure should work- you should give it a shot (good exercise)

[wanton bolt]

My first approach was to isolate a single cube root then cube, but I kept running into some form as the original

[wanton bolt]

I'll give this a go, thank you!

[hearty tendon]

yeah a common theme in these sorts of problems (that isnt applicable in this case) is if you have something of the form then the following holds:

$(a^{\frac{2}{3}} + a^{\frac{1}{3}}+1)(a^{\frac{1}{3}}-1) = a-1$

[gilded halo]

cow

[hearty tendon]

a lot of comp algebra is just learning lots of tricks and summoning them whenever needed

[wanton bolt]

notes

this textbook assumes you don't know any comp tricks, I haven't even gotten to the algebra section yet

all the problems have been proofs of irrationality (except a few)

[hearty tendon]

well now you know a trick ig LOL

[wanton bolt]

couldn't hurt

¯_(ツ)_/¯

[hearty tendon]

consider this: you rearrange 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 in some random order (the no. of ways of doing this is 16C2 (for the 1's spots) * 14C2 (for the 2's), etc.) = 16c2 x 14c2 x 12c2 ... 2c2.

Now in this ordering, since the sock must be placed before the shoe, the first 1 MUST be a sock, and the second 1 MUST be a shoe. so any rearrangement of 11...88 gives a unique way of dressing, and each way of dressing has a unique rearrangement- thus a bijection.

for example, 1234567812345678 means you put on sock 1 sock 2 ... sock 8 shoe 1 ... shoe 8

i mean for your original thing you dont really put on things "randomly"

if you say the spider has 3 legs for simplicity

sock 1 sock 2 shoe 2 sock 3 shoe 3 shoe 1

yeah

do you know about bijections

basically what we're doing here is showing that the ways to dress up is a 1-1 matching with the ways to rearrange 11...88

yeah

are u tryna aime qual lol

yeah

icic

gl

combo is super hard

i cant do it for my life

10 or 12

thats pretty solid

which years though

before 2017 is easier

still worth to do the problems

just that its not an accurate indicator

??

amc12 is on the 6th

lol

lol yeah

i heard this years gonna be hard

💀

fr good luck

im cooked

[kindred wraith]

hi, im compiling a list of competition math problems that involve obscure concepts and ideas, anyone able to share any examples?

by "obscure" i mean using ideas that are as far away from the standard algebra/geometry/combinatorics/etc as possible, which is not only a bit vague but also subjective, so anything remotely fitting is fine

ideally, i would like difficulty around amc10/12 but ill take any difficulty

[modern prawn]

what the hell i thought it meant anything before uni

[verbal jetty]

do logic problems count?

i know they are also pretty hard, even though it's not within the standard topics

[verbal jetty]

one problem is one from the 1968 AHMSE (problem 10): Assume that, for a certain school, it is true that

I: Some students are not honest. II: All fraternity members are honest.

A necessary conclusion is:

(A) Some students are fraternity members.
(B) Some fraternity members are not students.
C) Some students are not fraternity members.
(D) No fraternity member is a student.
(E) No student is a fraternity member.}

if you would like further information, you can check https://artofproblemsolving.com/wiki/index.php/1968_AHSME_Problems/Problem_10 out (solution) or https://artofproblemsolving.com/wiki/index.php/Category:Introductory_Logic_Problems?srsltid=AfmBOop_-IH-4h4jIYchUmc8ZT4vWf3VM4jzc8NjatcHAavLDpznnhWZ (page including seven logic problems from AHSME and iTEST)

[kindred wraith]

yeah thats an ok example

thanks

[sleek ivy]

Yup

Anyone as cooked as me

I’ve barely had time to prepare

[radiant jasper]

how do i approach??

[ornate blade]

$(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)$

[gilded halo]

south's secret twin brother

[radiant jasper]

what's this formula called? i never saw it in my entire life

ive never seen it before

[ornate blade]

it just follows by expanding (a+b+c+d)^2
that this is equal to a^2 + b^2 + c^2 + d^2

this problem is creative cause equations 1, 2 are not cyclic (in 4 variables)

[radiant jasper]

ah so u just subtract the remaining terms

why would you set it as d though?

oh because the 3rd equation

ah

[ornate blade]

but this is the way to go

[teal basin]

I have so much to learn lol

[ornate blade]

and then yes sub into the 3rd equation

[radiant jasper]

when i see the solution i understand

but i just can't think of the solution by myself

[ornate blade]

yeah I mean

me too

[teal basin]

is it bad that I'm struggling with completing the square?-

[ornate blade]

but I definitely think there are other solutions

(ab + bc + ca)(a + b + c + d) - (abc + bcd + cda + dab)
= (a + b)(a + c)(b + c)

[ornate blade]

no

are you interested in competition maths btw, or no?

[teal basin]

what is it exactly

[ornate blade]

it's in the name

[radiant jasper]

what's the point of getting that value though

[teal basin]

compete against what tho?

or who?

[ornate blade]

well just playing around with expressions

[ornate blade]

ah okay

[radiant jasper]

competiting with yourself

[radiant jasper]

ah

[ornate blade]

so competition maths is about creatively applying your knowledge

it's not about how much you know exactly

[teal basin]

oh WOW!

[ornate blade]

it's about if you can look at a question you've never seen before

and dig your way into it

[teal basin]

yes!

I'd LOVE to do that!

dude that's what I LIVE for

[ornate blade]

I swear the number of geometry questions which are just applications of Pythagoras in AMC 8 and 10

or that tangents to the circle from the same point are equal

[radiant jasper]

also did you multiply everything to get that final value or

[ornate blade]

you'd beat yourself up once you realised you only need to do that

[teal basin]

I have a pen and paper >:)

(I'm in Pre-Calc btw)

[ornate blade]

(a + b + c)(a + b + c) = a^2 + b^2 + c^2 - 2(ab + bc + ca)
(-3 + c)(-3 + c) = (a^2 + b^2 + c^2) - 2 * -4

yeah it seems like everything is pointing to reducing the problem into a, b, c using abcd = 30

like if you play around you can see why it's motivated like that

[radiant jasper]

hmm

[teal basin]

are there problems I have to find or what's the source here?

[radiant jasper]

im solving amc questions

it's on the website

[teal basin]

amc?

okie!

[ornate blade]

https://artofproblemsolving.com/wiki/index.php/Main_Page

lifesaver

[teal basin]

which one we doing?

should I start off with a certain one?

[radiant jasper]

help me with what im doing

[teal basin]

okay!

[hearty tendon]

funnily enough, you can solve for d:

$abc + d(ab+bc+ca) = abc - 4d = \frac{30}{d} - 4d = 14$ gives a quadratic; solve for $d$, plug in, iterate (note that some values of d lead to complex roots, you need to plug and remove those

[gilded halo]

cow

[ornate blade]

yeah that was what the solution suggested on AoPS

[hearty tendon]

oh

[ornate blade]

which SYA and me looked at

[hearty tendon]

didnt read oop

[radiant jasper]

what's oop?

[hearty tendon]

oops - s

but this feels like a solid aime 6?

[radiant jasper]

im definitely not aime level

[hearty tendon]

anyone have the source

[hearty tendon]

have you qualed for aime

[radiant jasper]

no

i wish to

i have amc 10 on wednesday

[hearty tendon]

😢

i wish it was just aime

amc is so bad

[radiant jasper]

have u tried aime before??

[hearty tendon]

yeah

[radiant jasper]

really??

[hearty tendon]

but do you know which year aime this is from

[radiant jasper]

no

wait do you have any tips for the amc

[ornate blade]

https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems

[hearty tendon]

make sure to get the early ones right ig?

[ornate blade]

problem 7

[hearty tendon]

no sillies

[hearty tendon]

gotcha thanks

pretty accurate prediction lol

[radiant jasper]

that's actually impressive

[hearty tendon]

ive done a lot of aime mocks 💀

[ornate blade]

yeah no shit

[radiant jasper]

how

[hearty tendon]

LMAO

[radiant jasper]

like i can't get above 90 on amc

😭

[hearty tendon]

grind since birth

[radiant jasper]

what grade are you in??

[hearty tendon]

probably

11

[radiant jasper]

wow

it's not like ur a lot older than me

which math course are you taking??

[hearty tendon]

college stuff

[radiant jasper]

ah

too many smart people in the world

[hearty tendon]

im lwk bad u should see the moppers

[radiant jasper]

who r they

If i qual for aime

i would be extremely happy with that

[hearty tendon]

u got this

im cooked bro

😢

[radiant jasper]

y?

[hearty tendon]

12 is hard

[radiant jasper]

oh amc 12?

wait do you live in america?

[hearty tendon]

yes

[radiant jasper]

ah

private school?

[hearty tendon]

wait are u intl

no

[radiant jasper]

public?

[hearty tendon]

yeah

[radiant jasper]

damn

[hearty tendon]

can intl take amcs this year??

[radiant jasper]

international?

[hearty tendon]

yeah

[radiant jasper]

i live in canada

[hearty tendon]

i thought they blocked that off cuz cheaters

oh

[radiant jasper]

im an international student

[radiant jasper]

cheaters?

[hearty tendon]

okok

[hearty tendon]

yeah 💀

[radiant jasper]

how can there be cheaters

that's crazy

ok so

d = 3/2 or -5

abc = -6, 20

[radiant jasper]

leaking amc papers beforehand and having other people buy them

it was a thing last year, i believe

[jade widget]

geometric questions are so hard!

I’m not even gonna get 100

[shadow spruce]

anyone knows how to make geometry sketchs in latex?

[fallen magnet]

kinda, why?

[shadow spruce]

Did geometric sketch pad and import by using /includegraphics

Terns out that it is bit difficult to do so with latex

Geometric sketch pad output as .svg, not normal .png

Auto-correction works just fine as usual 😦

[winter sable]

Learn smtg from him guys

[soft vigil]

then dont do them

skip em

[soft vigil]

THIS PROBLEM

[gilded halo]

< ? >

[jade widget]

That is about 25% of scores

[soft vigil]

well

for the most part

first 15 shouldnt have hard geo

just solve forst 15 all correctly and ignore the other 10

easy aime

[jolly solstice]

I'm guessing LTE

[willow anvil]

n=ab where a,b are integers greater than 1. then prove 3^a-2^a|3^(ab)-2^(ab) and get a contradiction using zsigmondys

[reef condor]

Orz

[willow anvil]

nounks

[sleek ivy]

easier said than done

[trim bramble]

real

[soft vigil]

i mean

look at 2023 amc10

it was easy compared to previous years

first 15 is manageable

[sleek ivy]

well theres no indication this year will be easy

and you stilll need preparation and shit

and i thought u meant amc 12

[trim bramble]

yeah but like don’t stress

[sleek ivy]

i mean personally i havent prepared at all so ik im cooked so my attitude rn is js fuck it

[trim bramble]

😭

[soft vigil]

yeah thats the problem

you need to do mocks

and develop a strat

sometimes theres hard problems in like 16-20 range and an easy one in the 21-25 range

[sleek ivy]

yea ik ive done it in the past

this year i js found out abt the competition too late and didnt have time due to other schoolwork and shit

but theres always next year

[vague temple]

Sometimes they troll and put an impossible question at like question 13

That you have to cheese unless you plan on spending 15 minutes on 1 problem

[soft vigil]

this problem was funny

[radiant jasper]

that doesn’t seem too bad

let x be the left handers, y = 2x be the right handers
x(x-1)/2 LvL
xy = 2x^2 RvL
y(y-1)/2 = x(2x-1) RvR
we know that the right handers won at least x(2x-1) games
2x^2 - x < (2.5x^2 - 0.5x)/1.4

nope, i’m stuck

x=3 seems to be the only solution

gives B)

[soft vigil]

insane variable names

the total number of games must be 3x*(3x-1)/2 due if we let there be x left-handed people

[radiant jasper]

it’s 3x(3x-1)/2, isn’t it?

[soft vigil]

only answer that works here of that form is 36

oh yeah

right

[radiant jasper]

15 is the 5th triangle number, 36 is the 8th and 66 is the 11th

unless i am very much mistaken

[soft vigil]

yeah

36=9*8/2

so its right

[radiant jasper]

yes, but we cannot discard the other options just yet

you may however note that it must be divisible by 2.4

66 does not satisfy that and 15 certainly won’t

[soft vigil]

yeah that too

no need for silly variables

[radiant jasper]

x and y are the most common variable names

and i wouldn’t bother with that in contest, i just enjoy solving questions as if they weren’t multiple choice

[fair vale]

Hey guys, quick question here:

Let a, b, c be positive real numbers such that abc = 1. Show that if
a + b + c > 1/a + 1/b +1/c,
then exactly one of the three numbers is greater than 1.

If anyone can give me a good hint as to how to solve the problem, that would be appreciated!
Ive already worked on it and there seems to be 4 cases for how many numbers are above one (all 3, 2, 1, or none) and i have disproved the all 3 and none cases
Thank you!

[radiant jasper]

have you tried setting a, b > 1 and c = 1/ab?

then move all the reciprocal terms to one side and the non-reciprocal to the other?

(i haven’t solved this, just throwing out ideas)

[fallen magnet]

how do i do this?

Let $a, b, c$ be positive real numbers with $abc = 1$.\[0.3cm]
Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.$

[gilded halo]

NicknamedTwice

[fallen magnet]

i thought about maybe using the power mean inequality to help but i'm not quite sure

[radiant jasper]

have you tried dividing each denominator by abc to get a^2(1/b + 1/c)? (i haven’t solved this, just throwing out ideas)

[fallen magnet]

sorry i dont see how that works out

[soft vigil]

isnt this like the most famous example of titus lemma

[fallen magnet]

idk

[soft vigil]

yeah it is

[radiant jasper]

oh, i really ought to learn that

what’a the lemma?

[soft vigil]

google it

i cant explain it well

[radiant jasper]

oh

[radiant jasper]

am i on the right track? @bubo

[soft vigil]

hmm

try to get a perfect square on the top of each fraction

[radiant jasper]

1/a^2 / (1/b+1/c)

[soft vigil]

this includes squares of ractions too

[soft vigil]

no

thats not right

[radiant jasper]

it’s a perfect square

[soft vigil]

yeah but does 1/a^2 /(1/b+1/c)=1/a^3(b+c)

[radiant jasper]

yes

[soft vigil]

no

it doesnt

[radiant jasper]

abc=1

[soft vigil]

oh

ig

[radiant jasper]

that’s the same as 1/a^2 / (ab+ac)

[soft vigil]

yeah that works ig

then you apply titus and amgm and it works

the other way to do it is to just do:

$\frac{\frac{1}{a^2}}{a(b+c)}$

[gilded halo]

buboblakistoni

[soft vigil]

then titus give
$\geq \frac{(1/a+1/b+1/c)^2}{2(ab+ac+bc)}$

[gilded halo]

buboblakistoni

[soft vigil]

this is equal to $\frac{ab+ac+bc}{2}$ and then apply amgm to get the $3/2$

[gilded halo]

buboblakistoni

[radiant jasper]

ok!

[teal basin]

i’ve been trying this for over an hour now

[fair vale]

lol me tooo!

[teal basin]

all i have gotten is that b>-1/2

but even then

[fair vale]

?

[teal basin]

idrk

[fair vale]

ab and c are all positive lolol

[soft vigil]

Try what Micabo said

[teal basin]

ik

[fair vale]

mentioned in the question

[teal basin]

here’s the thing tho

it says GREATER than

[soft vigil]

Let a>=b>1>=c

[fair vale]

yeah?

[soft vigil]

And show that in that case it isn’t possible

[fair vale]

yeah thats what IM TRYING to do

bruh

[soft vigil]

Just try to make it homogenous

[teal basin]

is there like a calling thing to discuss math on this server?

[soft vigil]

There is no public vc

[teal basin]

darn

[fair vale]

btw if i prove all other cases wrong am i done or do i have to also prove the original case right?

[teal basin]

now all i have is b > 3/2

i know i have to do more but idek if this is right 😂

[radiant jasper]

admits

[novel karma]

guys can yall share your fav amc 10 mocks

like not past tests but mocks

[modern prawn]

ask amc trivial and it will summon a mock for you bro

[zealous violet]

how should i study for the amc 10 in this final homestretch

[swift imp]

Reinforce key ideas

Basically

And get a good night sleep

[zealous violet]

okay thanks

[high goblet]

Muirhead's also works here

[radiant oyster]

anyone here writing the amc 10a on wednesday

[ornate blade]

just use long division, so write out 3.000000

30 goes into 7 four times, so it starts with 0.4

[radiant jasper]

since 3 doesnt divide 7

keep a point after 0

so like

and then add 0 to the back of 3

so u divide

30/7

essentially what ur doing is

$$\frac{30}{7 *10} $$

[gilded halo]

Professor Uchiha

[radiant jasper]

and u divide 30/7

and get the decimal

and u repeat this process

untill u are satisfied

and then finally move the decimal places accordingly

[novel karma]

thanks

[soft vigil]

memorize it fr

142857 but shifted i think

its like 428571

,w 3/7

[gilded halo]

Results provided by WolframAlpha

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

Requested by buboblakistoni

[soft vigil]

yeah

[hollow orchid]

Ah, okay! I get it. Thank you!

[sleek ivy]

long division by hand until u get to a specific digit u need

[sleek ivy]

im doin 12 tmr

[jade widget]

Stop memorizing everything bubloblakistoni

[jade widget]

I’m on Thursday, could be ur Wednesday though

[soft vigil]

no

it has actually shhown up on a lot of problems for me

just memorized iti accidentally

[jade widget]

Ok

👍

[winter sable]

m for mat

[deft wraith]

Find all integers (n, m) such that (10n + 4)^m + 4 is a perfect square

Can someone help me with this problem?

[ornate blade]

wait

mb

[radiant oyster]

isnt there a chance it could get leaked by then tho

[surreal willow]

Good luck to anyone who’s taking the AMC 10/12 a tomorrow !!!

[foggy obsidian]

anyone here doing CSMC next week?

[surreal willow]

R u in California or some western state?

Cause that’s the only way it can work

[tame burrow]

Doesn't everybody take it on the 6th?

[surreal willow]

Yes but some people are in later time zones so they may take it later/earlier

It also depends on the schools decision on when to take the test

[tame burrow]

OK

[fair vale]

We draw a 2024 × 2024 grid of unit squares. We call the vertices of the unit squares in the
grid lattice points; there are 2025^2 of these in our grid. Someone chose 10 of these lattice
points, and drew all the line segments that connect any two of them. Show that at least
one of these drawn line segments will contain at least two more lattice points besides its end
points.

anybody pls help

[warped wedge]

Should i even bother studying today or should i just try to rest and clear my mind

[strong kestrel]

how hard was the amc 12 today? were the questions alright?

[radiant jasper]

Isnt amc 10 tomorrow??

[radiant jasper]

Me

[swift imp]

I’m doing 12 a

Hopefully it shard

[tame burrow]

Depends I think

[tame burrow]

Is that a test question?

[swift walrus]

rest

unless u are unprepared then cram

[torn gate]

Gn everyone taking AMC 10/12 tomorrow!

[lime karma]

🤗 good luck!!

[vague temple]

So, I way overcomplicated this problem. Other than looking at the graph, why do we only consider integer values of x, not rational factors that cancel to an integer?

https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25

[plain jacinth]

Im taking the amc 10 A tmrw. Any last minute tips?

Who else is taking it im kinda nervous

[plain moon]

i'm taking it

i think at this point just get a good sleep

[soft vigil]

good luck people

[radiant jasper]

im so scared 😭

[foggy solstice]

Good luck everyone

[jade widget]

<@&268886789983436800>

[radiant jasper]

LMFAO

[swift imp]

Did someone leak it

[radiant jasper]

no

[swift imp]

Last year he like

💀

[torn gate]

Gl everyone

[plain moon]

GL!!

[thick hinge]

Can anyone share the questions after they attempt it (AMC 10 A)?

[radiant jasper]

For what?

Wait isnt amc 10 a tomorrow??

I thought someone said they took it today

Odd

[thick hinge]

Diff time zones

[radiant jasper]

So some people took it today?

[thick hinge]

Yes

[radiant jasper]

But what if the questions leak

[thick hinge]

They will

From 6:30pm today people can attempt

[radiant jasper]

Actualy?

Attempt what?

[thick hinge]

The exam

So, the time period is of one day

[radiant jasper]

Im taking it tomorrow

[thick hinge]

Me too

[radiant jasper]

But then some people said they took it today so i was wondering about the leak

[thick hinge]

I think they can leak it

[radiant jasper]

Thats crazy lol

[thick hinge]

But who will

If someone does it’ll be extremely helpful

[radiant jasper]

Lmao

[radiant jasper]

Yeah

[jade widget]

I’m taking it 25 hours later

[radiant jasper]

Tell me how it goes

[torn gate]

Not cool bro

[undone spindle]

I am pretty lost

Nvm i got it

[lone wasp]

Is x and y are real numbers ?

Interesting

[thick hinge]

My bad

[dreamy saddle]

HIIII

[thick hinge]

Can anyone tell when do we get the marks for AMC

[sleek slate]

Wtf

[surreal willow]

Good luck to anyone who’s taking the AMC 10/12 a today !!!

[plain jacinth]

Bro what yall talking bout share the questions after they attempt💀

[winter sable]

@everyone amogus

[jade widget]

Amoooguss

[ionic pivot]

amogus

[soft vigil]

Came back to say good luck again

[trim bramble]

good luck yall

[radiant jasper]

good luck!

[sleek slate]

No

[radiant jasper]

people take it at different times

please do not share the problems

[jade widget]

when will the latest test finish

[radiant oyster]

745

[jade widget]

wt time zone

[radiant oyster]

IST

in 3 mins basically

from right now

itll be over

[radiant oyster]

lol

[warped wedge]

Did anyone else think it was much harder than previous years tests

[plain moon]

you took it already??

[torn gate]

Let’s not discuss problems on AMC 10A until tomorrow

There are still people writing it

[brave hare]

why r the timezones schedules not matched up for amc

some today some tmr is crazy

[thorn rune]

@brave hare can you help me solve a problem in ahsme

[radiant oyster]

I SWEAR

[warped wedge]

Have u taken it

Dm me

I had it this morning

[radiant oyster]

yeah

[radiant oyster]

i have, check

[radiant oyster]

u have to accept my friend request for me to dm u 😢

[radiant jasper]

How difficult was the tree st??

Test*

[vague temple]

Is there a name for the kind of polynomials (x-a)^n + (x+a)^n

[noble dune]

thats just an ordinary polynomial tho

replace the + with • and itll be a difference of squares (i guess powers if exponent is >2?)

[restive pier]

I dm'd you.

[plain jacinth]

I havent taken 10 a yet

Was it more difficult than previous years

[tame burrow]

I hope that the cutoff is low

[plain jacinth]

Does that mean it is difficult?

What are your cutoff predicions

[tame burrow]

Maybe 90-95, based on the recent trends (2023 doesn't count)

[plain jacinth]

How about honorable mention (top 5%)

[tame burrow]

100? maybe a little higher?

[swift imp]

imo yes

but it depends what ur good at

ofc

[radiant jasper]

How hard was amc 10

Im taking it in 20 mins💀

[tame burrow]

I mean it always gets more difficult every year

[tame burrow]

I'm taking it in 3 hours so last minute cram time

[radiant jasper]

I honestly dont think u can prepare for it

Unless you lack basic math formulas or theories

[tame burrow]

I know just like last minute review

[radiant jasper]

What should I review

[tame burrow]

Whatever you're weak at

[radiant jasper]

Im good as long as there arent many probability or sequence questions

[radiant jasper]

Ah

[tame burrow]

SAME

[radiant jasper]

😭

[fallen magnet]

im not even american but lowk the questions for amc dont seem that hard

aime 2 is hard af tho

[radiant jasper]

Smart

[tame burrow]

I just need to not screw up the last 2-3 and I'm good

[radiant jasper]

U guys all pros bro

[tame burrow]

I have to do good this year or I'm quitting comp math

[radiant jasper]

Do u live in america?

[tame burrow]

Yes

[radiant jasper]

Like in canada we have canadian math contests which are extremely easy

Compared to amc

[fallen magnet]

honestly, the second last question for 2023 amc 10 kinda looks like a gcse question for higher maths 😭 🙏

[tame burrow]

The second to last one was free

[tame burrow]

ngl they kind of are

[radiant jasper]

2023 seems so easy now lol

Prob cuz i looked at it several times

[tame burrow]

Nah 2023's cutoff was way higher than it should have been

[tame burrow]

It was a year ago

[radiant jasper]

Like

I see the solutions i fully understand

But coming up with that solution..

I guess thats where the skill comes in

Being creatuve

[tame burrow]

Whenever you see the solutions, you think "oh well why didn't I think of that?" The truth is, they're disguising the problems so that finding the solution isn't easy. The content is not hard, so they have to find other ways.

[radiant jasper]

Right

[tame burrow]

Being completely real, if you're like a genius, you can get a very good score on the AMC 10 with just regular school knowledge and critical thinking skills

[fallen magnet]

u dont even gotta be a genius

just a top scoring student

[radiant jasper]

All my friends qual for aime

So idk

Maybe just experience issue

[fallen magnet]

hopefully u will too

[tame burrow]

AIME qual isn't that hard anymore; the cutoffs are going lower much faster than the difficulty is higher.

[tame burrow]

But good luck nevertheless

[radiant jasper]

Thank you

4 mins left🥶

Wait do grade 11s take amc 10 or 12??

[tame burrow]

12

[radiant jasper]

Ah

Last chance then lol

[fallen magnet]

whats the difference?

[radiant jasper]

12 is more hard

[tame burrow]

Only cuz it has like logs and complex and extra stuff; it actually is easier to AIME qual through 12.

[radiant jasper]

Ah cuz the cutoff is lower?

[tame burrow]

Yes, and also cuz the 10 and 12 have overlapped questions

A LOT of overlappig questiong

[radiant jasper]

Right

Well anyways i hope i get the result i want

[fallen magnet]

i dont get the point of test a and test b

[radiant jasper]

Ill be back after a hour and half

[radiant jasper]

Different time ig

But yeah its weird lol

[fallen magnet]

good luck

[tame burrow]

Yep

You'll do fine

[sleek slate]

So if u sell one u can do better on the othrr 🙏

[fallen magnet]

oh i see

[plain jacinth]

Is there any point in getting a higher score than aime qual (if your not trying to go to jmo or amo)

like if you have a 110 is there any point in aiming for a higher score

[trim bramble]

sold a gonna cook on b fr

[autumn stirrup]

Hello im going to be doing the WMTC or World Mathematics Team Championships. Are there any resources, tips, or advices for me to help review for it?

[drifting nacelle]

did you took the amc today?

[drifting nacelle]

what amc did you took

[drifting nacelle]

can i take the a and then the b?

im from latin america and i took the amc today, the amc12, is the amc already post like the questions

[sleek slate]

No discussion aint allowed so dont talk abt jt

Tmrw at 8 its open

[drifting nacelle]

what is jt?

yet?

[reef condor]

The amc

Do not discuss the amc until 8am eastern

MAA will not post anything until that time either

[drifting nacelle]

im latin american so my school isnt that good, is my grade posted tomorrow?

[drifting nacelle]

ok, srry

im lk for friends interested in this world

[tame burrow]

10

[tame burrow]

JMO quals require a generally higher score unless you like get a perfect AIME

[fallen magnet]

i live in europe

i dont think i can do it

[ruby egret]

got the amc in 2 hours

am i cooked?

[tame burrow]

N

i do too

[latent bough]

Yay I just got cooked on the AMC

[tame burrow]

wdym

[ruby egret]

how was it

please

[latent bough]

I got destroyed but that’s bc I just didn’t know how to dove some problems

[vestal yacht]

avg difficulty

[ruby egret]

cutoff predic?

[vestal yacht]

105