competition-math

[gilded halo]

Ruben

[waxen vapor]

ok probably it wont

time to take the pen

[waxen vapor]

wait a second, since i know that $(a^n - b^n)$ divides $a - b$, i can use the extended induction principle

[gilded halo]

Ruben

[waxen vapor]

for "extended induction principle" i mean that i suppose that P(0), ..., P(n-1), P(n) are true

[near hill]

Yes.

[waxen vapor]

oooh thank you!

i didn't think about it

[gusty knoll]

I dont understand what this question is asking

[near hill]

Do you know what a "balance scale" is?

[radiant jasper]

A balance scale is something that is level to the horizontal plane when equal weights are placed on it's two sides

It's basically asking how many combinations of 3, 5 and 7 can you make

[gusty knoll]

Ohh

Alr thank you

[near hill]

"weights can be placed on either side" makes it a bit more complicated than that.

E.g. we can weigh out 2 grams of something by putting our 3-gram weight on the left and 5-gram weight on the right, and then adding "something" to the left until the scale balances.

[gusty knoll]

I see

That seems hard

[near hill]

I'm not sure there's a simple clever way to do it, but each of the three fixed weights can be either on the left or on the right or not on the scale at all; that gives only 3·3·3=27 combinations to try, but some of them actually weigh the same difference.
We can cut down the range of possibilities a bit by deciding that the heaviest weight must always be on the left place; then there are only 14 possibilites (of which one is "all three weighs are off the scale" which measures 0 grams and may or may not count ...)

[gusty knoll]

I counted all the possibilities and removed the ones that weigh the same as another possibility

And got 11

Someone else might figure out a clever way

But I guess this works too

[timber comet]

chat is this easy ?

[gusty knoll]

I got -15*(sqrt(3)/27) is that correct?

I can’t seem to get any other answer

[ivory ember]

,w find (x^3-y^3)/(x^3+y^3) if x+y=sqrt(6), xy=1

[gilded halo]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[ivory ember]

Yeah

[gusty knoll]

Nice

So the paper’s wrong

[barren iron]

No

Author can be wrong paper can not be

[silver juniper]

hello

can somebody help me pls

[scarlet root]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[silver juniper]

help me with this

find x so that 1 / sqrt(x + 1) + 1 / sqrt(x + 4) = 1 / sqrt(x + 2) + 1 / sqrt(x + 3)

[ornate blade]

,w 1 / sqrt(x + 1) + 1 / sqrt(x + 4) = 1 / sqrt(x + 2) + 1 / sqrt(x + 3)

[gilded halo]

Results provided by WolframAlpha

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

Requested by sooouth.

[ornate blade]

Wolfram says no solutions

but the way to typically approach this sort of question is to set x = u - 2.5

so you'd have u - 1.5, u + 1.5, u - 0.5, and u + 0.5 which is symmetric

[silver juniper]

ok

lemme try

[ornate blade]

try another question btw

hopefully it's just a typo in your book

[silver juniper]

no

it's rly my HW

lemme see if my friends online to ask

[ornate blade]

ok

[gilded halo]

Cube

[ornate blade]

not quite
if you follow this process and see what you get, you have 7 + 4 + 1 + 8 + 5 + 2 + 9 + 6 + 3 + 0 + ... with 66 digits
then you have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + ... with 22 digits
then you have 3 + 6 + 9 + 3 + 6 + 9 + .. with 7 digits

so all the 3rd digits have become 9s

I don't get the hint

the key is that you can loop back around, 0, 7, 4, given that the pattern repeats in the 1st sequence

[ornate blade]

ah I guess it's similar, like the digits that remain after each pass are first 1 mod 3, then anything mod 3, then 0 mod 3

[ornate blade]

wow that's smart

[acoustic nova]

source?

I’m very bad with these but f(0) = 0 or 1

wait oops

one sec

[ornate blade]

no thanks

[slim ermine]

My humble opinion, competition math is nothing but grindMAXXing until you can find out "that" pattern from reading the question. I swear geometry solutions are insane.
"Uh draw a parallel line to bisector blablablabla- oh this is just 2013 "insert country here" selection P2!"
Insane.

[willow anvil]

not really true. recognizing a config in a geo problem and citing it isn’t that crazy, and in contest you’re probably not going to remember the name of whatever problem so you’re just going to prove that problem anyways

[gusty verge]

$f(x)=c$, for $c=0\cup 1\le c<2$

[gilded halo]

Max

[slim ermine]

I wanted to point out you have to grind the questions. The geometry one was one of the examples. You can't ignore the advantage of knowing obscure questions and theorems. Take some of the algebra questions for example, I have seen plenty of solutions consisting of obscure theorems, just they happened to know it because of grinding

[willow anvil]

this is not from imo or shortlist, psure this is someone’s OC

Well ofc you have to grind for all kinds of math

[slim ermine]

yes but not in this degree

[willow anvil]

If you take math seriously you probably will

[slim ermine]

Usually you are in a comfy position to understand the topic you are reading by doing the exercises and questions the book gives you

it is not the case in competition math

[willow anvil]

The reason why people find geometry theorems obscure is because it’s barely taught in school

[willow anvil]

Why?

[slim ermine]

I would disagree, they just feel " random"

[pallid tundra]

for oly geo you definitely have to know a lot more "nonstandard" results

[willow anvil]

I feel like geometry is the most intuitive subject

[pallid tundra]

standard high school geo is just so dogshit that there's a significant content gap between the two

[slim ermine]

Combinatorics is the really the only one where you feel it is the mathematics

[gusty verge]

Proof is just by considering cases. Set $x=0$, you get two cases.

One of the cases being $\floor{f(y)}=1$, inputting that into the original equation gets you $f(x)=c$ for $1\le c < 2$. Then consider the other case which is $f(0)=0$. If that is the case then choose $x\in[0,1)$. You then find that $f(x)=0$ for $x\in[0,1)$. Take $y\in[0,1)$, then for $z\in\mathbb{R}$ you can choose a $y\in[0,1)$ and $x\in\mathbb{R}$, such that $\floor{x}y=z$. Using this you find $f(z)=0$.

[gilded halo]

Max

[willow anvil]

Well ik this isn’t from imo or shortlist but this was posted on aops iirc

And it wasn’t from a contest

bro what

😭

[slim ermine]

Again, it is because you have to recognize the patterns and configurations.
I hate the geometry most, since practically, all the grinding is useless because computers immediatly output the answer

[willow anvil]

ok yeah idk how I didnt recognize that

[radiant jasper]

lol

[willow anvil]

There was a similar problem posted recently

[slim ermine]

dunno maybe I am delusional

[willow anvil]

No it was someone creation

[willow anvil]

Computers can only do so much geometry, same with other branches of math

[radiant jasper]

don’t you just binge competition math problems till you are good at competition math

[slim ermine]

Computers can d any IMO geometry questions with no sweat. I agree on your second point

[radiant jasper]

but being good at competition math doesn’t translate to anything meaningful

[slim ermine]

No need to generalize

[willow anvil]

I don’t think computes can do any imo problem

[radiant jasper]

alphageometry exists

[slim ermine]

I would disagree

[willow anvil]

Many imo problems are just angle chase (which computes can do), but harder ones don’t use angle chases

[slim ermine]

they can do, especially the geo ones.

[willow anvil]

I think I could make a problem alpha geometry could not solve

[slim ermine]

you can just rewrite the question in algebraic, bayesian or whatever numerical method and bash through it really

[radiant jasper]

if it’s about being creative, why train for it? Training makes you more inherently creative? Sounds tautological

[willow anvil]

Well this is part of the problem, how do you know the coordinates of a certain point for example?

[slim ermine]

Again, I disagree. For P6 questions, maybe. But for other questiosns you know what you will do after reading

[radiant jasper]

Sounds like you’re just stealing someone else’s creativity and then making your own “creativity” based off that

[willow anvil]

Well ig every geo can be bashed theoretically

Other than special antibash problems

[slim ermine]

Like, my point is grinding and such. Putting them to numerical problems takes skill, and is actually practial AND you dont have to grind

[willow anvil]

Like weird angle conditions

Wdym putting them to numerical problems

[slim ermine]

it is antibash for the competitiors in that competition, it is still very bashable

[slim ermine]

I meant conveying the given conditions to algebraic expressions

[willow anvil]

Hmm not sure if this is true

An example I’ve seen is angles being in an arithmetic progression
How do you encode this algebraically?

Even if you are able to, it’ll be extremely messy and I doubt even the computer will be able to bash through it all

[slim ermine]

Maybe, but this is really the extreme case isn't it? My point still stands

[willow anvil]

There are plenty of problems with weird angle conditions

[slim ermine]

If you can construct the problem in geogebra (or in paper really it doesnt matter) , and result is immediately shown, it is very bashable

[willow anvil]

And if you can’t?

[slim ermine]

even maybe it is not, grindMAXXIng is still the case

[willow anvil]

Wdym grindmaxxing

[slim ermine]

sorry it is an internet slang, just grinding questions like a maniac

is the canadian IMO guy still here?

[willow anvil]

What Canadian imo guy

[slim ermine]

the gold winner that was active here

[willow anvil]

Yeah he’s in this server

canbankan

He is also notoriously a geo antimain

[slim ermine]

I don't think it is him tho

well it doesnt translate to anything practical

other than "art"

[willow anvil]

How do you know?

I was saying how do you know that it doesn’t translate to anything

[slim ermine]

Like, in the practical world, applied mathematics. The grinding stuff is really not, practical but more like art

That is why juries also like traditional solutions for geo problems

and hate bashings

[willow anvil]

It’s not about the discipline of grinding, it’s about what you learned

[slim ermine]

it is not about if I would apply it. If I WOULD it would be practical

[rich cradle]

I think jury prefers non bash solutions because they are more “in the spirit” of Olympiad geometry

[slim ermine]

indeed

[slim ermine]

From personal experience, some of them just think that bashing is ugly, meanwhile the traditional solutions are "beatiful"

[rich cradle]

Of course

[radiant jasper]

Why would it be transferable when comp math is timed and requires no advanced mathematics

[slim ermine]

which is in essence should be not allowed in a competition, especially in a rigirous math competition

[slim ermine]

well majority of algebra questions can be solved with advanced math

[radiant jasper]

The opportunity cost of devoting your time to comp math is time you could’ve spent learning more pure math

[radiant jasper]

By advanced math, i’m talking about math you’d see at research level

[willow anvil]

You learn a lot of pure math learning comp math

[slim ermine]

yeah of course

but the grinding is still there

[willow anvil]

Yes

[slim ermine]

The Chinese bootcamps where they raise comp children starting from elementary grade

just honestly, it is just grinding

[willow anvil]

It is grinding, and I agree that you grind more in comp math in other types of math

But I wouldn’t disregard comp math as useless

[slim ermine]

it is not useless

specific parts of it is

[willow anvil]

Like what

[slim ermine]

like geometry stuff

and some of the algebra questions

that can be easily solved with uni level math

[willow anvil]

what kinda algebra questions

[slim ermine]

maximizing minimizing questions for example

you could bash throught it with lagrange or something I really dont remember

even if it was more than 4 variables

[willow anvil]

Euclidean Geometry is pretty much useless in higher level math (for now), so maybe I can see an argument for that

[willow anvil]

Is it the easiest way though?

Lagrange multipliers are a powerful technique but they often take too long and someitmes don’t work on problems

[slim ermine]

By easy if you mean fast then no

and if you will make this argument, then my point still stands

[willow anvil]

Yes I see

[slim ermine]

Like I wish they would reconstruct the competition to one or two question a day with long timers. Maybe it would halt the grinding a bit or something. The Soviets really did it badly

[willow anvil]

Many olympiads are like that

[slim ermine]

Well many olympiads are after IMO

I was talking about IMO

[willow anvil]

I think mot putting such “useless” questions in competitions is better

I haven’t really struggled with time that much in Olympiad so idk

[slim ermine]

I meant change the structure of the questions ( I really don't know how) so it would require lots of time

Maybe I am just mad I could not get selected for IMO idk

[willow anvil]

Hmm, research questions are expected to take decades to solve, and Olympiad problems are expected to take ~an hour to solve, I’m not sure if there’s a reasonable in between

[gusty verge]

The proof is always the hardest part.

Many times in competitions I've reached the answer but not figured the proof fully.

In one competition I got exactly the answer, missed one small part of rigour and lost 5 out of the 7 marks

[willow anvil]

Other than to just make the problems harder

[slim ermine]

Making new questions is also a problem. Majority of questions are really just redesign of older questions. Not even talking about generalisations

[willow anvil]

I think new ideas will always pop up, I wouldn’t say a majority of questions are

[slim ermine]

well a bit exagiration, but a good sum of them are

Like combining two different questions, ( I definetly saw this in IMO, I don't remember), using multiple different question's parts on just one question, generalising a question for n, changing the topic of the question (for example from algebra to geometry)

[acoustic nova]

I had to go to bed

[brave heron]

Congrats, that is science, making your own based on ones output etc

[sacred stone]

starting to hate calc

[barren iron]

[midnight frigate]

8

[junior talon]

wait this chat exists

orz

[mellow thorn]

bro this youtube channel should help for competion math

https://www.youtube.com/channel/UCQuPPuiyPNM-PlV7aFMzKWg

[pallid tundra]

and the channel itself appears to be nothing special

[finite wolf]

Good video but it would be great if you add videos about combinatory, chances, and building

[mellow thorn]

thanks and will do

thanks

[radiant jasper]

how do I help people with math?

[mellow thorn]

mb

[mellow thorn]

sorry i couldn't meet your expectations

[radiant jasper]

Ok thanks

[reef condor]

Be at a much higher level than then

Certainly helps a lot, although not necessary

It can be made up for with teaching experience

[pallid tundra]

the “contest math channel” space is way too saturated as is

as participation numbers additionally continue to plummet

[reef condor]

I mean handouts >>> videos anyway

[radiant jasper]

Can someone ask me a question and see if I am good

[pallid tundra]

compute TREE(3)

😁

(no don’t actually)

[reef condor]

It just depends on your audience

[radiant jasper]

do I go to the help section

[reef condor]

I mean

What are you teaching? What is your experience in what you are teaching?

Oh wait you mean in this discord

💀

[radiant jasper]

yeah

[reef condor]

Yeah just use the help forum

Or channel

[thorn rune]

You want?

[acoustic nova]

uh no

sorry

yeah but he look it literally i think

its ok though

some of the content is decent though and i commend the effort still!

[slate saffron]

Anybody from MIT ??

[finite wolf]

how to get good at math-competition. Yea, can anyone recommend me any channel?

[hushed charm]

do you mean youtube channels?

[hushed charm]

you can check this server, isn't very large but maybe you find something cool there
https://discord.gg/mods

idk any youtube channels but you will only get better solving a lot of problems i think, is generic but what i have already heard is that you know, maybe you can try to avoid seeing the solutions and if you are struggling with a problem you can ask for help and learn more about the subject, that will probably help you to think better and have more creativity.

I don't have experience, that is all things that i have already heard

[finite wolf]

joined right way thanks

[finite wolf]

oh um sometimes i check the answer immediately if I can't do it hehe

[hushed charm]

me either, it is hard to stop

[finite wolf]

Are you on college or high school?

[hushed charm]

high school

there's some time to go to imo yet

[finite wolf]

1 year to go from college for me :v

[hushed charm]

hmm, bad

i think

do you think you can go to imo before that?

[finite wolf]

where is it

i dont rlly know about imo cuz im asian

[hushed charm]

international math onlympiad

[finite wolf]

OH hold on

[hushed charm]

dont you know?

[finite wolf]

you mean the students that was chosen from school

after being executed 1 by 1, some last man standing will be send there?

is it like that

[hushed charm]

well, it depends. in my country you have to take a obrigatory math exam, depending of your score you can take the national math olympiad, depending of your score you can take a exam to go to the imo team

where are you from

[finite wolf]

Indonesia

[hushed charm]

https://en.wikipedia.org/wiki/International_Mathematical_Olympiad_selection_process

In Indonesia, National Mathematical Olympiad is held as a part of National Science Olympiad (Olimpiade Sains Nasional), and has been held annually since 2002. About 100-120 students who pass the province-level test will be eligible to participate in the National Mathematical Olympiad, which is held in August or September. About thirty students are chosen to get into the first training camp, which is held at October through November. About half of them will go to second training camp and participate in the Asian Pacific Mathematics Olympiad. At the end, six students are selected to represent the country. The selection depends on the results of regular tests held every week in every training camp, IMO simulation test and APMO.

[finite wolf]

yep yep

[slender saddle]

Sounds accurate

[finite wolf]

well im cooked i failed :v

[hushed charm]

i think it is late for you to take the imo, you have to have less than 20 years, be enrolled in high school or under and dont be enrolled in college

[finite wolf]

ye my school not allowing third year students joining province or the imo things anymore, only first or second

[hushed charm]

sad

[finite wolf]

well imma just tryna win other olympiad outside there

[hushed charm]

i don't know any olympiad for college, maybe the majority of them are for high school students but maybe you find something

[finite wolf]

I'm high school

im 16

[hushed charm]

the problem is that the most prestigious exams take some time to get in you know

[finite wolf]

it is :v

[hushed charm]

in my coutry i will be able to take the imo in 2 years only if i make everything perfect

anyways, good luck for you

[finite wolf]

thanks u too

[finite wolf]

[hushed charm]

in october i will make an exam, the 300 best scores can take the national olympiad and those who get a gold medal can take some exams, the best scores go to the imo team

the national olympiad is in december of 2025

so i will be able to go to the imo in 2026

i think it is even harder and take more time in countries like usa and china

yes

oh

have you already took obm?

[finite wolf]

idk rules

maybe who knows but look at this

olympiad for each subjects but they only give chances for X and XI students

[hushed charm]

yeah, this year is OBMEP and i will be able to go to OBM only in 2025

i could take another exam and go this year but i was not prepared and it was 2 days before the first phase of obmep so i preferred to study more for obmep

and about imo?

👀

do you have more chances?

good to know, at least i have more than 1 year between obmep and obm

idk, i think i go well studying by myself

of course

[rancid kettle]

Evaluate

[limpid shore]

no thanks

[nocturne valley]

Understandable

[remote island]

Hi, I'm currently studying for the AMC 8 and 10. I have been considering about taking John Hopkins CTY course for High School Competitive Mathematics Prep. https://cty.jhu.edu/programs/online/courses/high-school-competitive-mathematics-prep-cpm I am already a CTY student but I have not taken any courses yet. Has anyone taken this course? If so, how good is it?

[dapper magnet]

probably 0

[radiant jasper]

yeah

Cause the sums can simplify or even result in zero due to symmetry

[south ridge]

oh

[pallid tundra]

it’s probably roughly isomorphic to a lot of the other popular contest prep classes out there - maybe it’ll be helpful to you, maybe it won’t

[pallid tundra]

vol 1 is a bit outdated content wise

but it’s an ok starting point

[pallid tundra]

try a past test first

identify areas you’re weak in

and study accordingly

[radiant jasper]

Is it good now?

Lol, alright

[radiant jasper]

What are you’ll doing?

[umbral mica]

I'm going to 8th grade, and I'm trying to study for the AMC 8. I'm new to math competitions in general and took a few classes over the summer. Do you guys have any tips on how to study and what topics I should focus on?

[pallid tundra]

try a few past tests first to get an idea of what the problems are like

[umbral mica]

I did the test last year ( didn't study) so I have a rough guess of what the problems are

[pallid tundra]

try another one, just to jog your memory

[umbral mica]

ok

apart from that what else can I do

[pallid tundra]

people usually recommend books like AoPS volume 1 or some of their intro books for studying

the AoPS forums have tons of resources to study from as well

though your primary goal should be to

do lots of problems just above your level

[umbral mica]

what are some good sources for these problems

[pallid tundra]

AoPS forums have thousands of problems from a variety of sources

if you want AMC 8 problems specifically

you can look through their contest collections

[umbral mica]

ok

thanks

[pallid tundra]

np

[umbral mica]

also

do you have some study tips or strategies I can use

[hushed charm]

love math

[pallid tundra]

if you get a problem wrong be sure to read through the solution carefully

studying for contests is not a passive activity

you constantly update your own intuitions/processes as you work through problems

[hushed charm]

And never go directly for the soluction, the most important is thinking about the problems, that is the unique way you can improve or problem solving and creativity

[umbral mica]

thanks guys

appreciate the help

[hushed charm]

np, good luck for u

[umbral mica]

thanks

[pallid tundra]

perseverance is also a big thing

don’t give up on a problem that you can’t solve until you’ve made a good honest effort at it

(in a contest, you’d probably want to more quickly skip to a problem that you can do, but when studying generally the above works best)

[hushed charm]

i saw a guy saying that if you are struggling with a problem you can study more the theory, make some easier problems and then come back

[umbral mica]

The main roadblock I hit whenever I try practice problems is that I make silly mistakes or I forget the principles needed to solve it

[hushed charm]

maybe that can help you

[pallid tundra]

practice practice practice!

[hushed charm]

practice more i think

[reef condor]

Yes.

[umbral mica]

ok

will try

thank you

[hushed charm]

👍

[umbral mica]

Are there any other math competitions in the fall

I can use to prepare

[pallid tundra]

aops wiki has a big list id imagine

and lots of universities and other organizations have competitions in a variety of settings

[pallid tundra]

are you doing MATHCOUNTS as well? ik those don’t start until winter but just curious

[umbral mica]

I'm considering it

[pallid tundra]

doesn’t hurt to try

[hushed charm]

taking exams for practice i think is only useful for practicing mind things and avoid anxiety i think, so i think you can focus on making previous exams

[pallid tundra]

I don’t have a good pulse on how modern AMC 8s compare in difficulty to the ones I took in middle school

start with some of the older ones?

[umbral mica]

ok

much appreciated

[pallid tundra]

np

[limpid tide]

modern ones are getting harder

[soft vigil]

Do Mathcounts

If you do good you are either gonna get a free trip to dc or Orlando

[ornate blade]

sus, considering that you need to be in middle school to do that

and the minimum age for Discord is 13

[limpid tide]

???

[ornate blade]

2024 AMC 10A problems and solutions. The test will be administered in November 2024.

[limpid tide]

im a time traveller

dw abt it

[limpid tide]

i didnt spell it wrong, its jus a diff spelling from the future.

[ornate blade]

or you're Bri'ish

[limpid tide]

ew

[soft vigil]

8th grade exists

[ornate blade]

yeah they'd have to be in a narrow age range is what I'm saying

[unborn quartz]

Hey guys do u know any good guides and tips if I am trying to qualify for IMO

I am from India

so my qualification lvl may be different

[supple flame]

bro im doing amc in a few days 💀💀💀

[remote island]

@supple flame huh???? amc 10 and 12 is in november

and amc 8 in on jan 22-28

[ornate blade]

wrong AMC, the Australian one should be in a few days

this is the American one

[supple flame]

amazing math competition 💀💀💀

[ornate blade]

cringe

[supple flame]

how is that cringe

are you like born in 2000

[ornate blade]

how is that relevant

[ornate blade]

I mean that acronym for AMC is cringe

[brave heron]

what even is this?

[hollow bear]

heyy can someone please help with this ^^ (thanks for posting max)

[slim ermine]

I am not from USA, on my country yes

upsie sorry for the ping

[radiant jasper]

What????

[radiant jasper]

what if you were held back?

[tawny hull]

Find a recursive expression for a_n and use it to get the generating function as a rational function.

[gusty verge]

Yeah we said recurrence relation earlier but so many difficulties with it

[tawny hull]

Make sure that the recursive formula is correct by computing some small values. Changing the order of the sums can be helpful for evaluating double sums if you have those

[gusty verge]

Have you actually found a recurrence relation?

Or just suggesting the use of one

[tawny hull]

I have a recurrence relation and I managed to use it to get the generating function as a rational function

(and then rest is easy)

The number of terms depends on n though, it is not constant. But you can still solve for the generating function

[ivory ember]

isn't this legit just the gen func at x = 1/11

[reef condor]

Ok but what is the gen func?

It seems nontrivial to get

[tawny hull]

ping me if you want hints or spoilers

[ivory ember]

@reef condor ^

[granite stump]

Yo, I love AMC8, AMC10, AIMEs and MATHCOUNTS!

Those are my favorite things to do in the world!

I'm so glad I found this channel.

[reef condor]

But there’s other serves that are solely dedicated to comp math

[umbral mica]

can I post questions I need help with in this channel

[hushed charm]

Yes but usually people only post questions here when that is about math competitions and maybe you will have to wait more time to someone answer you so i recommend you to post your questions in math help like #help-5 or maybe post in #prealg-and-algebra #geometry-and-trigonometry #precalculus

at math-help you can ask about every math problem, but in the other channels a good practice is to ask only about the specific subject like pre calculus in #precalculus

[tawny hull]

IMO if they are about competition math they are fine. I feel like help is mostly saturated with pre-university stuff. While competition math is pre-university, you need to know stuff that even most people doing pure math degree won't learn until late into their degree or at all

Like competition math number theory knowledge can be made into an undergraduate number theory course, and then you still can't solve more difficult number theory problems without knowing how to apply it

[hushed charm]

I have two years for study with 6 hours per day, do you guys think is it possible to study for IMO and IOI with that time? XD

[reef condor]

it is possible.

but you might not be successful...

[hushed charm]

that is tough

[hollow bear]

Spoilers please

[snow dirge]

Hi everyone, this problem might be old, but can someone explain the simple solution to me? Twenty random cards are placed in a row all face down. A turn consists of taking two adjacent cards where the left one is face up and the right one can be face up or face down and flipping them both. Show that this process must terminate. (with all the cards facing up).

From the movie the brilliant young mind(2014).

[tawny hull]

Your problem statement is flawed. You need to be able to turn a face down card face up and then turn the card to the right of it (if it exists).

Interpret the row of cards a number written in binary where face up corresponds to zero and face down corresponds to one. Now notice that any move you make will decrease the number because any digit contributes more to the number than all digits that come after it combined. Since the smallest number that can be represented like this exists (zero) the process must terminate after finitely many steps. Also if we have any face down cards i.e. our number isn't zero, we can still make at least one move so the process hasn't terminated. Therefore the process terminates precisely when all cards are face up.

Also, the number of cards doesn't matter and the initial state of the cards (face up or down) doesn't matter.

[tawny hull]

Okay first we want to find a recurrence relation.

||Consider a string of length n+1. It begins with some number of ones.

If the number is 0 i.e. it begins with a star then it has the same value as the length n string without the star and all length n strings appear exactly once by truncating a length n+1 string starting with a star. This gives us the term a_n into our recurrence relation.

If the number is n+1 then the number consists only of ones and has the value (10^{n+1} - 1)/9 and we include this into our recurrence relation.

If the number is k such that k is between 1 and n (inclusive) then it will be followed by a star and some length n+1-k-1 = n-k string and in fact each length n-k string appears exactly once. We get n-1 terms sum k=1 to n (10^k - 1)/9 a_{n-k} into our recurrence relation. Note that a_0 = 1 because because the empty string corresponds to empty product which is equal to one.

Our recurrence relation is a_{n+1} = a_n + (10^{n+1} - 1)/9 + sum k=1 to n (10^k-1)/9 a_{n-k}. It will be more convenient to work with a_k instead of a_{n-k} so we will reindex the sum and the final form of the recurrence relation is a_{n+1} = a_n + (10^{n+1} - 1)/9 + sum k=0 to n-1 (10^{n-k}-1)/9 a_k.||

Now that we have a recurrence relation we will find the generating function.

|| We take the recurrence relation for each n, multiply both sides by x^n and sum from n = 0 to inf to get

sum n = 0 to inf a_{n+1} x^n = (sum n = 0 to inf a_n x^n) + (sum n = 0 to inf (10^{n+1} - 1)/9 x^n) + (sum n = 0 to inf (sum k=0 to n-1 (10^{n-k}-1)/9 a_k x^n)).

We have four sums, let's evaluate them independently. Denote the generating function of a_n by f(x).

First term is just the generating function shifted and the second term is the generating function i.e.
sum n = 0 to inf a_{n+1} x^n = (f(x) - 1)/x
sum n = 0 to inf a_n x^n = f(x).

The third term is difference of two geometric series i.e.
sum n = 0 to inf (10^{n+1} - 1)/9 x^n = 10/9 sum n = 0 to inf (10x)^n - 1/9 sum n = 0 to inf x^n = 10/(9(1-10x)) - 1/(9(1-x)) = 1/((1-10x)(1-x))

The fourth term is the difficult one. We have a double sum. We notice that iterating from n = 0 to inf and from k = 0 to n-1 iterates through precisely the same numbers as iterating from k = 0 to inf and from n = k+1 to inf i.e.

sum n = 0 to inf (sum k=0 to n-1 (10^{n-k}-1)/9 a_k x^n) = sum k = 0 to inf (sum n=k+1 to inf (10^{n-k}-1)/9 a_k x^n)

Now we perform an index shift on the inner sum because it is nicer to start from n = 0. We also take a_k and out of the inner sum because it doesn't depend on n.

= sum k = 0 to inf a_k (sum n=0 to inf (10^{n+1}-1)/9 x^{n+k+1})

We can now take the x^{k+1} part of the power of x out of the inner sum and we notice that we are left with the difference of geometric series we already evaluated as the third term. We will manipulate it further and notice that the sum we are left to evaluate can be represented in terms of the generating function

sum k = 0 to inf a_k (sum n=0 to inf (10^{n+1}-1)/9 x^{n+k+1})
= sum k = 0 to inf a_k x^{k+1} (sum n=0 to inf (10^{n+1}-1)/9 x^n)
= sum k = 0 to inf a_k x^{k+1}/((1-10x)(1-x))
= x/((1-10x)(1-x)) sum k = 0 to inf a_k x^k
= x/((1-10x)(1-x)) f(x)

Combining the terms gives us

(f(x)-1)/x = f(x) + 1/((1-10x)(1-x)) + x/((1-10x)(1-x)) f(x)

We can now solve for f(x)

(1 - x - x^2/((1-10x)(1-x))) f(x) = 1 + x/((1-10x)(1-x))
=> f(x) = (1 + x/((1-10x)(1-x)))/(1 - x - x^2/((1-10x)(1-x))) = ((1-10x)(1-x)+x)/((1-x)(1-10x)(1-x) - x^2) = (1-10x+10x^2)/(1-12x+20x^2-10x^3)
||

Now that we have the generating function we can find the answer

|| Notice that the sum is just our generating function in the power series form evaluated at 1/11 with the zeroth term missing. Thus we get f(1/11) - 1 = 142/89 so p+q=231.
||

[near hill]

It would be a bit more interesting if we insist that there must be a "card to the right of it" so we can never flip the rightmost card alone.
Then we would need a parity argument that we can never end with just the rightmost card needing to be flipped.

[jaunty nova]

hi guys, just wanted to ask a question about something i am unsure about and i am asking here because idk where else i can ask, i am currently studying in vancouver and i was wondering if i can take the amc 10 there or if i need to go to somewhere else i tried searching online but i didnt really find anything that tells me

maybe i am bad at research but i will greatly appreciate it if somebody helps me out <3

[sterile helm]

Mario, Yoshi, and Toadette play a game of "nonconformity": They each choose rock, paper, or scissors. If two of the three people choose the same symbol, and the third person chooses a different symbol, then the one who chose the different symbol wins. Otherwise, no one wins.

If they play 4 rounds of this game, all choosing their symbols at random, what's the probability that nobody wins any of the 4 games? Express your answer as a common fraction.

[gusty verge]

Not really the right channel #probability-statistics

(unless this is much harder than I think it is? Seems like a common probability problem)

[sterile helm]

it is easy for many

but I'm just bad at probability

[gusty verge]

Like I said you'll get more luck in #probability-statistics

Hint:
What's the probability on no one winning on a particular round?

[radiant jasper]

<@&286206848099549185>

[shy sorrel]

Yes

[formal plover]

Which one do you think it is?

[radiant jasper]

I know √7 is bigger than √45

But i can't compare the third one

Anyone wanna help 🙏

Like i know we can take lcm of all these powers which is 12

[shy sorrel]

It is III

[ \sqrt[3]{11} \approx 2.223 < \sqrt[4]{45} \approx 2.545 < \sqrt{7} \approx 2.646 ]

[formal plover]

Got 3 from using the calculator. But is this a calculator paper ?

[pallid tundra]

this seems like it would be no calculator

[shy sorrel]

I said the answer is the third option

[pallid tundra]

raise each of them to the 12th power

then compare

[radiant jasper]

Ok i solved 🙂

But i need to do it fastly

I only have 36 sec to solve this

@shy sorrel @pallid tundra @formal plover Thanks for the response 😊

[shy sorrel]

Your welcome, so was it the third option

[formal plover]

No probs.

[radiant jasper]

Yes

[shy sorrel]

OK, well it honour to help such a brilliant individual. I'm out.

[brave heron]

this is 1/81 right

probability that everyone chose the same symbol = 3/27, probability that everyone chose differently = 6/27, that means that for each round, there is 1/3, so 1/3^4 = 1/81

[sterile helm]

Yes

[brave heron]

yup thats right

[violet idol]

One plus one equals window accept it

[pallid tundra]

if you're going to troll at least try harder

<@&268886789983436800>

[spark garden]

Referring to the tig780 post?

[spark garden]

Don't troll here.

[pallid tundra]

ye

this is their only message so far

and they just joined today

judging by that i doubt they joined to partake in any actually productive discussion

[spark garden]

I see a couple less trollish posts in their history

I'll just note it in the log

[pallid tundra]

this is all i see 🤔 are their other messages in the social channels?

[spark garden]

Mine has more as far as server wide stuff.

[pallid tundra]

icic

[snow dirge]

Prove that 1^∞ is not equal to 1

[gusty verge]

Surely $\lim_{x\rightarrow \infty}1^x=\lim_{x\rightarrow\infty}1=1$

[gilded halo]

Max

[hushed charm]

i think you could use log too

[snow dirge]

What if I transform 1 to 2/2, 3/3 and so on

So 1^∞ becomes (2/2)^∞ = 2^∞/2^∞ which gives the undefined answer

[hushed charm]

or maybe you can think that log 1^∞ = ∞ * log 1

you can't multiple numbers by infinity but you know that log 1 is 0

so it would be ∞ * 0

[snow dirge]

From your solution it gives 0=0 am I right?

[hushed charm]

no

you can't multiple things by infinity

but

it is not 1

[snow dirge]

Affirmative

Much obliged

[gusty verge]

This is not competition maths tho, #calculus

[gusty verge]

They are essentially "equal infinities"

[snow dirge]

So next time I should ask this kind of problem in other categories?

[gusty verge]

Ye

[ocean narwhal]

But log1 is not 0 as the 1 is not exact

Its ever so slightly bigger than 0

So 0*infinity indeterminate form

Nvm sorry i didnt see that you had answered it

[hushed charm]

https://www.youtube.com/watch?v=-UvTWUEGDO8&themeRefresh=1 you can check it later if you want

[slender saddle]

1. No you can’t
2. Infinity is not a real number
3. Google indeterminate form’

[gusty verge]

I asked Google Gemini to come up with a probability problem at the level of IMO, this is what it came up with: A $n \times n$ grid is painted with $n$ distinct colors, such that each color appears exactly once in each row and each column. Let $S$ be the set of all possible colorings of the grid. For two colorings $C_1, C_2 \in S$, we say that $C_1$ and $C_2$ are related if there exists a sequence of colorings $C_1 = D_1, D_2, \ldots, D_k = C_2$ in $S$ such that for each $i$, $D_i$ can be obtained from $D_{i-1}$ by swapping the colors of two adjacent rows or two adjacent columns.
Determine the number of equivalence classes of $S$ under this relation.

[gilded halo]

Max

[near hill]

Well, it's not a "probability problem", but otherwise looks interesting.

[gusty verge]

yeah true

i asked for another and it gave the classic recurrence relation probability problem

Alice and Bob play a game with a standard six-sided die. Alice rolls the die first. If she rolls a 6, she wins. Otherwise, Bob rolls the die. If Bob rolls a 6, he wins. Otherwise, they continue taking turns rolling the die until one of them rolls a 6.
What is the probability that Alice wins the game?

[ashen jacinth]

is it 6/11

i tried solving it a diff way than i usually do

[gusty verge]

yeah i believe so

you just solve it iteratively, so if no one wins on the first go, it's like you've started again then u got an equation for p

not sure how u did it but thats the go to way on this

[ashen jacinth]

yeah typically i use infinite geometric series but im trying to use that way more

so this time i did

[gusty verge]

yeah that would also work

nice

[ashen jacinth]

now i’m ready for putnam trust guys

[near hill]

You can also say: Alice and Bob keep rolling dice simultaneously until at least one of them get a 6. Alice wins if she has a 6 in the final roll.
The we're just looking for a conditional probability.

[ashen jacinth]

interesting perspective

[gusty verge]

Good luck sir

Did this problem earlier, important trick to get the result

[ashen jacinth]

wow

actually not seem too bad but i can’t do in my head

i try when i go home fr

[fleet lynx]

How would you even start this

[ashen jacinth]

tbh i wasn’t sure what the notation on the top meant but im assuming it’s saying the recursive sequence is formed by repeatedly applying the function f

essentially the recursion is a recursion

it is recursioning

so i’m assuming it’s irrelevant for this lol

[fleet lynx]

Yeah but when they say S_n are the referring to a set

[ashen jacinth]

no

S_n refers to the sum of the first n terms

[fleet lynx]

Oh that makes sense, the {a_n} threw me off

[ashen jacinth]

does it make more sense where to start now

[fleet lynx]

Hmmm

Maybe

Could we compare that part to the function, like the sum of the first n terms and the function using the first n+m-1 terms to find the first n+m terms, so we could adjust the S_n equation to include m?

Actually

That might not be useful

[ashen jacinth]

idk i’m ignoring the function part lol

||its interesting if you consider n=0||

[fleet lynx]

Okay

[gusty verge]

It's natural numbers tho, pretty sure it starts at n=1

[gusty verge]

I'll give a hint and this is the important trick you should at least try on questions like this:

||consider $S_{n+1}-S_n$||

[sly veldt]

ive been reading about characteristic equations of recurrence relations in this competition prep book but it doesn't really go in depth about why they work

can someone here provide a proof for that?

[jolly solstice]

forgive the poor answer, but I believe it's cause they end up being the eigenvalues of translation operator

[pallid tundra]

you see smth similar come up in constant coefficient homogenous linear differential eqns

[shadow niche]

can i put competition math for any grade here?

[pallid tundra]

sure lol

(generally we see middle and high school contest stuff here)

[shadow niche]

oh okay

[distant barn]

How do i start learning competition math?

[gusty verge]

Search up art of problem solving and do problems from their website

[hushed charm]

guys, what aops books do you recommend me? Should i start with the first one?

[tawny hull]

That's because you pretty much need a course in basic linear algebra which is usually taught in the 1st year of college to have the prerequisites

[pallid tundra]

if you’re starting out with no contest background I’d recommend the intro series (intro algebra, number theory, counting, geometry)

intro algebra might be the gentlest introduction (and closest to what you’d be doing in a standard math class)

[spring spade]

yo is anyone giving amc

[distant barn]

Isnt aops paid?

[sharp lake]

Not a proof but more like intuition.

when you have fib sequence you know its related to :
$$x^2 - x - 1 = 0$$
But that basically just means
$$x^2 = x + 1$$
Now if you multiply by $x$ you will get
Or just
$$x^3 = x^2 + x = 2x + 1$$
Multiply by $x$ again.
$$x^4 = 2x^2 + x = 2(x + 1) + x = 3x + 2$$
$$x^5 = 5x + 3$$
So basically you have
$$x^n = F_n x + F_{n - 1}$$
Now denote the solutions of $x^2 - x - 1 = 0$ with $u , v$
Then you have
$$F_n (u - v) = u^n - v^n$$
Or just $$F_n = \frac{u^n - v^n}{u - v}$$

[sharp lake]

Ofc not

[gilded halo]

Pluton

[distant barn]

sooo where do i start

[gusty verge]

Forums

[hushed charm]

i have no contest background but i am already studying algebra, combinatories, geometry... for 2/3 months, do you think that the intro esries keep being a nice series for me?

And about the contest preparation, do i need a contest background to undertand better the book?

[pallid tundra]

the intro books (for algebra and geometry) are written to be somewhat more difficult/rigorous versions of the standard textbooks

and the difficulty curve is a lot gentler than some of the higher level aops books

[umbral mica]

Do you guys have any suggestions for books for the AMC 8

More specifically on number theory, combinatorics, and probability

[patent thistle]

the aops intro books

[thorn verge]

What book/s should I read for BMO1?

[tropic grove]

help?

[pallid tundra]

!status

[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

[prime tulip]

1status

!status

[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

[prime tulip]

1

!1

!status 1

[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

[prime tulip]

1

[near hill]

Could you stop spamming that?

[gusty verge]

Give that for all $n$: [0<a_n\le a_{2n}+a_{2n+1}] Prove that $\sum_{n=1}^\infty a_n$, diverges

[gilded halo]

Max

[gusty verge]

Nice problem, me and a friend found a method for proving it but not sure where our proof goes wrong (the solution from the competition is way more complicated), I'll post later in case people wanted to try it first

[gusty verge]

(from Putnam)

[near hill]

||a1 <= a2+a3 <= a4+a5+a6+a7 <= a8+a9+...+a15 <= ...||

This works in particular when a_n = 1/(n+1), so the harmonic series diverges.

[gusty verge]

||We did S (a_n)<=S(a_2n+a_2n+1), subtracting both sides you get a_1<=0, which contradicts the original statement, but feels wrong, perhaps the Riemman rearrangement theorem has been broken here? we werent sure||

[near hill]

The assumption forces all terms to be positive, so Riemann rearrangement doesn't apply.

[ornate blade]

you should only use #geometry-and-trigonometry next time, instead of posting your question here as well

I'll leave it to you to show the green angles are in fact equal

[gusty verge]

True, so the argument is valid?

[near hill]

Looks valid to me.

[gusty verge]

So weird, the answer on the Putnam site is so long, was convinced it couldn't be correct

[spiral granite]

Someone explain the solution

5 (ii)

[radiant jasper]

in number theory, do you have some videos you can recommend? bcuz I can't seem to find anything on YT. mod and everything is not explained well in all videos, please rec some resources for it

[silver vault]

not really competition math but how do u get good at tmua papers?

[untold thunder]

I am studying algebra with a book right now and I have three more for geometry, number theory, and counting. I want to get them done as soon as possible, but I don't want to make it that hard, so should I do them all simultaneously?

[gusty verge]

GL to those in the IMC this week

8 new goated uni level comp problems to be released this week 🙏

[soft vigil]

@eternal acorn hi

[eternal acorn]

WHY HERE OF ALL PLACES

[gilded halo]

Vanellope von Schmugz

[south wharf]

2 variable and 2 equations, you can get the value of x and y frm there and then compute it in the below eqn

[untold thunder]

I don't know how much this will help but try multiplying the top equation by x and the bottom by y

[ivory ember]

Subtract the two equations: $$4(y-x)-4(x^2-y^2)=0$$ $$-(x-y)-(x-y)(x+y)=0$$ $$(x-y)(x+y+1)=0$$

See what you can do from here

It’s pretty straightforward

[gilded halo]

Civil Service Pigeon

[ivory ember]

[ivory ember]

Uh it should

You just get two quadratics

[scarlet root]

Show your work, and if possible, explain where you are stuck.

[ivory ember]

It’s easy to confirm the second case fails by considering the discriminant

[gilded halo]

Vanellope von Schmugz

[ivory ember]

the easiest way is to chest the unique answer and see x=y by inspection lol

[jolly solstice]

I feel like that's valid since it's literally the reflection across the line y=x so that's necessarily the points of real intersection

[ivory ember]

Touché

[jolly solstice]

like I'm saying, assuming what the question assumes, that there is only one real solution and praying it works out

where does that come from, the partial derivatives or something

[ivory ember]

#competition-math message

Fewer calculations -> good

[jolly solstice]

well, that misses the point of this way of thinking, since you're going through a different path

you're talking about a different solution than @ivory ember

that is a true statement..

oh, your original message to me refers to your solution when I'm talking about the other solution

[jolly solstice]

the point is to side step all that work like he's saying here

[granite silo]

How do I solve any world problems. I am bad at solving them

[lilac citrus]

Read the sentence and write the infos that are given

Use ur logics to see what the question asks and what is given

[soft vigil]

you can probably troll this with x=y

[stoic hemlock]

well, the top part defines two curves. So intuitively there should exist finitely many points which satisfy the top

[acoustic nova]

exactly

[soft vigil]

x=y=1/2 i think

[wanton bolt]

Is this a valid proof?

[reef condor]

Yeah

[wanton bolt]

Felt kinda lack luster... Thank you!

[cloud ermine]

Hello, sorry if this is not my channel, I'm new to this but I need help please 😦 (I'm not good at English and I don't have time to put everything in the translator)

Hello, I introduce myself, my name is Axel and I need help with these exercises (ignore the Spanish words), please, if you could help me, I would appreciate it.

please 😦

[pallid tundra]

#precalculus

for number 1, add the coordinates of A to the coordinates of P, to get the coordinates of Q

for part a, (7,-3)+(-3,7)=(4,4) so Q=(4,4)

and similarly for the other ones

for number 5, you’re looking for the vector you have to travel to get from P to Q

subtract the coordinates of P from the coordinates of Q: the first coordinate of the result will be your coefficient for i, and the second will be your coefficient for j

[brave heron]

I heard AMC got ridiculously hard today, can anybody confirm?

the australian one btw

[fleet fractal]

What year?

[brave heron]

the senior one

[fleet fractal]

I didnt take the senior one but I took the intermediate one

[brave heron]

was it harder than usual(assuming you joined back then)

[fleet fractal]

Well I didnt join back then but its harder than what I usually take

For me based on past papers it seemed easier but maybe it was because of heat of the moment

[brave heron]

I've had seniors telling me that

[fleet fractal]

Albeit most of the answers I just guessed in the final 3 minutes so

[brave heron]

easy back then

[fleet fractal]

I mean that past papers were harder and this was easier

[brave heron]

ohhh

well its the opposite for them

[fleet fractal]

But maybe because I just guessed some of the questions and I didnt bother doing 3 of the 0-999 questions

Oh well

[brave heron]

Tim uses plug-in timers to automate his indoor plant lighting. Each timer has a plug on the back, a socket on the front, and a motorised dial that rotates once for every 24 hours that the plug has power. When the time on the dial is between 9:00 and 17:00, the timer switch is on, so that any power on the plug is switched through to the socket. For fun, he plugs together three of his timers, all set to time 0:000:00, with a lamp plugged into the front socket. How many hours until the lamp turns on?

familiar(from amc allegedly)

?

the wording is so confusing though I think its just 24*9

basically the timer is plugged 24/7 but the dial on the timer only rotates once every day, and if the dial is on the 9:00 part then the power is switched on through the socket

[near hill]

The dial on one of the timers only moves when it has power.

[reef condor]

Yeah the timers are plugged in series

ie. The socket of one is plugged into the output of the previous

[brave heron]

is it or its a separate for each?

[near hill]

So for the first 9 hours, only the first timer rotates.
Then the first timer switches on, and the second timer starts rotating.
At 17:00, the first timer switches off, and the second timer stops rotating. By this time, the second timer has only reached 08:00 on its dial, so it hasn't ever switched on yet.
The next day at 09:00, the second timer gets power again, and at 10:00 the second timer has finally reached 09:00 and starts powering the third timer.

[brave heron]

only the third timer is required to power up the lamp?

i'm confused, theres a lot of variables

[reef condor]

It’s just a series electrical circuit

[reef condor]

But the third timer is powered by the output from the second

Which is powered by the output from the first

Which is (finally) powered by the socket

[brave heron]

its so mindboggling, how does the dial helps here?

[reef condor]

The dial determines when power will leave the socket of a clock

It’s value must be between 900 and 1700

And when the dial is within that range, the next device will power - if it is the lamp, we are done, but if it is another clock, then that clock will start ticking

[brave heron]

I feel like the solution is simple but you need a lot of comprehension

probably didn't help the fact that I don't use plug-intimers

[brave heron]

how does the first timer rotates on the first 9 hrs?

[near hill]

The first timer always has power.

[brave heron]

i mean the dial rotates every 24 hours?

or I misunderstood smth?

[near hill]

Behold my mad diagramming skills!

Each clock turns at a rate of one revolution per 24 hours but only when it has power.

So the second clock only ticks when the first switch is on, and the third clock only ticks when the first and second switches are both on.

[brave heron]

hmmmm

[brave heron]

OH I COMPLETELY misunderstood

dial = revolution, now I understand

[brave heron]

thanks for this

would that mean that after 17:00 on the first day, the dial on the 1st timer would go to 10:00 until it go again to 9:00?

[near hill]

The dial on the first timer always turns. So it always shows the actual current time.

[reef condor]

Yeah that part (and the whole problem tbh) is very poorly written

[brave heron]

ohhh so it only took 33 to get it work again

the 2nd

[near hill]

Yes, 33 hours after starting, the first switch will go on for the second time.

Thus, 34 hours after starting, the second clock will have collected 9 hours of power, so then the third clock gets power for the first time.

[brave heron]

yup then it ends at 67 hours right?

that part make the 3rd timer at 8

though the 2nd timer will end at 17

or it demands another return?

[near hill]

I haven't simulated that far out.

[brave heron]

its easy to look after 33*n hours, thats the time where 2nd timer opens

in hours

[near hill]

Uh, no, there's no special significance to 33·n hours.

[brave heron]

I mean at 66 hours, 2nd timer reopens again and after an hour, it reaches 17:00 and 3rd reaches 8:00

I'm using it as reference for the timer of the 1st

[near hill]

Definitely not. 66 hours after starting, the first timer will be at 66 mod 24 = 18 hours, so by that time it won't even be sending power to the second timer.

[brave heron]

oh i'm dumb, it should be 24n+9

[near hill]

(FWIW, I found a video showing a representative example of these timers in use. Beware weird kitschy soundtrack: https://www.youtube.com/watch?v=By8J4LreT9E)

[brave heron]

at n = 1, 2nd timer starts at 8:00 and ends with 16:00 after 8 hrs

then n = 2, 2nd timer starts at 16:00 and ends with 00:00 after 8 hrs

then n = 3, 0:00 starts then 8:00 with 16:00 at ends, but we only need one more hour for all 3 to actually work together, so its 24(3)+9+1 = 82

[near hill]

So, now that we've hopefully agreed what the task is about, I think it's easiest to solve it by working backwards:
The third timer needs 9 hours of power before it turns on.
How many hours of power does the second timer need to receive in order to pass on 9 hours? That must be 9 + 24 + 1 = 34 hours.
How long does it take for the first timer to have delivered 34 hours of power? 34 is 4×8 + 2, so it takes 9+4×24+2 = 107 hours from start.

[brave heron]

wait huh?

its not 82?

no its still 82 nvm

basically 5th day at 10:00

[near hill]

4 days is more than 82 hours.

[brave heron]

it should be 106 then

where did the extra 1 cme from 107?

[near hill]

Consider the situation after 4 days:

• The first timer ticks correct time, as it always does.
• The second timer has seen 4×8=32 hours of power, so it is one hour from turning on.
• The third timer has seen 8 hours of power.
  At 09:00 on the fifth day, the second timer starts getting power.
  At 10:00 on the fifth day, the second timer starts sending power to the third one.
  At 11:00 on the fifth day, the third timer has finally seen 9 hours of power, and switches on.

[brave heron]

first day 17:00 -> 8:00 - > 0:00 -> lamp(off)
2nd day 17:00 -> 16:00 -> 7:00 -> lamp(off)
intermission(3rd day) 10:00 -> 17:00 -> 8:00 -> lamp off
3rd day 17:00 -> 0:00 -> 8:00 -> lamp(off)
4th day 17:00 -> 8:00 -> 8:00 -> lamp(off)
5th day 11:00 -> 10:00 -> 9:00 ->lamp(on)

OKAY I GET IT

I fr though 8 it opens

[gilded halo]

Vanellope von Schmugz

[tawny hull]

https://discord.com/channels/268882317391429632/326138737606262786

Also, why do you ask about basic quadratic equation as a Pending Postgraduate?

[tawny hull]

Well, have you tried just plugging this directly into the quadratic formula or completing the square? Your substitution does make the coefficients into rationals, but it isn't necessary to do that and feels like extra work for nothing.

[gusty verge]

The solutions are imaginary

But yeah even so just use quadratic formula

[tawny hull]

You are going to use the same tool to solve the equation after doing your substitution as without doing it in the first place.

[tawny hull]

Doesn't matter how you factorize it, using quadratic formula is perfectly valid way to factorize a quadratic polynomial. The other method I suggested, i.e. completing the square, is the standard way of factoring quadratic polynomial and that is how you prove the quadratic formula

Your substitution doesn't achieve anything useful in terms of factoring that polynomial. Same method works before and after doing that and the substitution is just extra work when you do it and when you reverse it

[soft vigil]

Quadratic formula?

[jolly solstice]

I don't think it's worth calling it factoring once you stop using integer coefficients

7sqrt(3) is not an integer

oh ok I see

maybe substitute x by x*3sqrt3 to cancel out the 9 no idea if that makes it more convenient or not

looks like it does actually

it becomes 6x^2+7x+1=0 so you have (6x+1)(x+1) now you're done basically

[fleet fractal]

This seems familar, yes

Albeit im not sure if I would be able to talk about it because the paper says you cant distribute it until a certain date

I wasnt able to solve it because the wording is weird and I couldnt comprehend it

[brave heron]

and there was two easy algebra questions

aabb*c = cbbc and that easy sfft about xy +x + y = 2024 etc

[brave heron]

I agree

[fleet fractal]

i didnt solve the aabb thing because i was more focused on the question after it

the xy + x + y question was so easy when you realize the trick

no need for sfft

[brave heron]

its 341 assuming a b c are distinct

[brave heron]

add 1

[fleet fractal]

nope

[brave heron]

?

[fleet fractal]

xy + x + y = 2024
zy + z + y = 2024

[brave heron]

then you get x = z < y

where 729 = (x+1)(z+1)

[fleet fractal]

then you get x=z then you can solve the third equation as a quadratic

[brave heron]

of course x = z = 26

[brave heron]

that is definitely harder

[fleet fractal]

idk but it was easier for me

im not that good at sfft

[brave heron]

then you get y = 74 easily by dividing 2025 by 27

[fleet fractal]

the problem after that was the dial problem

[brave heron]

that is definitely the intended trick

[fleet fractal]

way too hard to comprehend

then after that was a triangle problem which im not sure where to even start

[brave heron]

it was really easy to see after you realize that the dial that rotates means that it makes a full revolution after 24 hrs

[brave heron]

is that the 18 24 30?

[fleet fractal]

yea thats what i think it meant but hindsight is 2020

[fleet fractal]

its the right triangle one yes

with the line passing through P or something

Im sure theres a really smart way to do that problem but im lost on how to find it

[brave heron]

go look at help 22 and scroll back(I accidentally ghosted the two helpers 😭 😭

or search incenter at the search bar and you would see the thread

they got 630, and I can see it already

[fleet fractal]

nah im not going to be able to make that up within 75 minutes

[brave heron]

its a 30 item qs, I assume that the first 25 was piss easy and can be solved atleast 30 mins

[fleet fractal]

i really need to think smarter

[fleet fractal]

nah the first 10 are baby

11-20 is slightly harder but if you think harder it will be fine

21-25 is harder

by the time i reached 21-25 i had like 5 minutes left because i focused on 26-30 too much lo

[brave heron]

joining a f2f contest, you should have priorities like solving problems you can easily see a solution

and not waste time to experiment(unless you are in a 3 hour contest then you are good)

to maximize the points and get a lot of time toyour advantage

[fleet fractal]

im guessing that the first 10 problems are just there to make you comfortable because damn

ill just think harder next time

[brave heron]

basically the easy trick is

(x+1)(y+1) = 2025
(y+1)(z+1) = 2025
(z+1)(x+1) = 729

thus it suffices that x = z < y

thus sqrt(729) = z+1 = x+1 = 27 thus x = z = 26

2025/(z+1) or (x+1) = 75 as they are both equal to 27, thus y+1 = 75, y = 74, so 74+26+26 = 126

1 minute solve

idk where the quadratic come from

[fleet fractal]

bec x=z

xz + x + z = 728
x^2 + 2x - 728 = 0
(x-26)(x+28)=0

and since the problem states x,y,z is positive you can discard the negative solution

[brave heron]

thats one way but i'm too lazy to see 26*28 but it works

[jolly solstice]

the result is easier for me to factor

[hushed charm]

does anyone know what subjects aops algebra book covers? I am thinking about start to read but idk if it is below my level

[acoustic nova]

algebra?

[hushed charm]

yeah

[acoustic nova]

the subject is algebra idk what u mean then?

[glass kite]

It's a wide spectrum tho

[acoustic nova]

introductory algebra or intermediate algebra? what level do u want?

[hushed charm]

Well, I would like to know about introductory algebra. Is this a high school book?

Actually, it isn't I saw here in the site

Don't worry, I'll search about the topics of the book, thx

[random nimbus]

https://images-ext-1.discordapp.net/external/AoxzAIWcYNXvfZDpU6cjn4Tmri2llS8_LBtKCBEXGfM/https/images-ext-2.discordapp.net/external/Qt5VF-OjeYN9HRzBWLA7ND6-MSb_qEqTsYtKivOGpqI/https/i.imgur.com/mQU5Nj0.gif

[acoustic nova]

Intermediate algebra is good

a lot of stuff in introductory can be pretty easy

[pallid tundra]

intro covers more or less the content of a standard algebra 1 course

intermediate covers algebra 2 + a number of contest topics

FEs, harder sequence/series problems, inequalities

[acoustic nova]

intro covers a bit more and some alg 2 stuff

Inter covers way more

[soft vigil]

I don’t remember it covering FE’s

But it’s been a few years so that might be the case

[pallid tundra]

chapter 19

[soft vigil]

Is the FE chapter in depth

[pallid tundra]

it goes over some of the common techniques but is probably not enough on its own for olympiad FEs

[radiant jasper]

I would like to get into olympiads

How do I start

[hushed charm]

i am starting too but i can help you with something i think. what have you been doing and where are you from?

[radiant jasper]

I’m from the us

I haven’t prepared yet I’m just

Not rlly sure where to start

[hushed charm]

alright, what grade are you currently in?

[hushed charm]

do you know aops? there are really great problems there to you practice. That is probably one of the best ways to study, practicing. People really like their books too

The mainly subjects at imo for example are algebra, geometry, counting, number theory and precalculus

there are aops books for each subject as you can see here : https://artofproblemsolving.com/store

you should know about the selection process to get into usa team, i guess you are pretending to take imo right?

try always to think about thee problems, that is important. Reading answer without thinking about the problem arent usefull at all

[remote loom]

im grade 9

[remote loom]

does this work for me as well?

[burnt trail]

try to do alcumus, its free

I think that helps a bit

[wary magnet]

aops also has classes for high school comp math

try those

but unfortunately its not free

also practice past competitions

[hushed charm]

of course

aops books have a lot of exercises that is the most important thing in my opnion. You can check aops exercises too. There are a lot of resorces

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

start by looking at previous tests of easier tests and as you improve you can try harder problems

amc, aime, usamo and imo i think

if you are from us

[remote loom]

thanks

[radiant jasper]

what is competition math?

[distant barn]

is this the correct way to study for math comp?

i practice -> if cant solve look at the solution -> understand the solution -> answer the question myself

[ornate blade]

it's in the name

most high school competitions have a decent sampling of algebra, trigonometry, functions, Euclidean and coordinate geometry, maybe classical inequalities as well

it's an entirely different mindset and focus to the maths you do at school

[gusty verge]

Yea

You should also study random courses

But depends on what lvl of study you are at

For university level, I've seen problems where you needed a good understanding of lie groups, I've seen others that required p-adic numbers 🤷

[distant barn]

thanks

[finite wolf]

Why olympiad questions is so hard?

[vivid ingot]

Because it is olympiad?

[hushed charm]

no, i don't agree at all. I think that a batter way to study would be

Thinking about the problem for some time seems to be really helpful, looking at the solution if you don't understand the problem won't help you, try to understand the problem before looking the solution, you answer the problem and then see the solution, if your answer is wrong, that is not a problem at all, you read the solution, understand and answer the question with your own words ( if you understood the solution you will be able to do that ). There are some problems that a hint won't help or will almost answer the question when the question is solved with a single idea, in this cases you can just try more and study more about. If you have already tried for a lot of time, looking at the solution isn't a big problem but that means you have to practice even more

[hushed charm]

btw, i saw that here
https://github.com/bqi343/cp-notebook/blob/master/Resources/Competition Math.md
https://www.youtube.com/watch?v=bSdp2WeyuJY

it is not specifically about math the second one but a lot of things makes sense for math competitins too

[pallid tundra]

because it actually involves nontrivial applications of a much wider range of mathematical tools?

[radiant jasper]

A bit late but thanks for the help guys

To my previous question

[radiant jasper]

is competition and olympiad the same thing

[weary light]

wat

[novel karma]

how to get better at amc 10 problems 22 - 25

[swift imp]

Do aime

[reef condor]

Depends on how far you are

Cuz if the problem is solved with a single idea then asking for hints is hardly useful

(Which is most amc) vs. oly where it’s many steps

[radiant jasper]

i dont do competition math but usually solutions should only be viewed after seriously trying the question

you shouldn't be running to solutions for every question you see and can't solve

[olive saffron]

Hi I’m gold medalist on math Olympiad 3x

My parents tortured me 10 hours a day as a little kid until I got all questions right on every AMC 12 in one sitting

Then they did the same for the USAMO

I poisoned my teammates at the MOP and secured my spot in the IMO

[radiant jasper]

wow! did you have to poison the competitors at the IMO to get first place?

[olive saffron]

No I actually had answer key

[radiant jasper]

was it hard to memorize?

[olive saffron]

Well my life was on the line so no

Parents said they would kill me

Chinese water torture

[radiant jasper]

do you have any tips for how to discreetly administer poison?

[fossil loom]

what's the best type of poison to use?

[thin quail]

What are your future goals ?

[rich cradle]

noob i

What the fuck are you saying

[nocturne valley]

Might consider this for a coding olympiad

[hexed vapor]

who does the australian AMC

[pallid tundra]

how does one approach this besides bashing small cases

[radiant jasper]

australians

[reef condor]

How so Xor ???

[radiant jasper]

exclusive or?

[reef condor]

You probably can’t

Actually

n must be even since otherwise contradiction (four odd^4 has even sum)

Then it’s just 1,2,p,2p (and this one has bounds on p), or 1,2,4,p (also bounds on p) or 1,2,3,4 and that’s probably it

[radiant jasper]

1 2 3 5 may also be an option

[reef condor]

Yeah

[pallid terrace]

no

[reef condor]

The answer is probably in 1,2,4,p
Nvm

[reef condor]

Well it doesn’t actually work, but you still have to check

[radiant jasper]

1 2 p 2p ends up being 17(p^4 + 1) which forces p=17

[pallid tundra]

34^4 by hand hooray

[radiant jasper]

i would advise against it unless strictly necessary

it may be easier to check that 17^4+1 has no smaller prime factors in other ways

[novel karma]

what the fuck is this kid saying

[reef condor]

(Orz NT ways)

Actually it’s not that bad with QR bashing or maybe not idk

[radiant jasper]

what are you on about

[reef condor]

Sus

[radiant jasper]

it’s not divisible by 3, 5, 7, 11 or 13

should work

[reef condor]

Idts

[radiant jasper]

(unless i messed up) that’s the 12p2p case complete with 17(17^4+1) as the answer, forcing floor(n^(1/5)) = 17

1234 is not an option as 1^4 + 2^4 + 3^4 + 4^4 not divisible by 4

1235 not divisible by 2

[reef condor]

O wait nvm

[radiant jasper]

124p results in p|1^4 + 2^4 + 4^4

[reef condor]

I see what u meant

[radiant jasper]

which is around 273

[reef condor]

Well QR eliminates 3 mod 4 primes so it helps ig

[radiant jasper]

QR?

[radiant jasper]

this is divisible by 3, 7 and 13

oh but 4 won’t divide

this case is also impossible

[radiant jasper]

answer is 17 working out shown above

@pallid tundra

[reef condor]

We have -1 as a quadratic residue

Because 17^4 = -1 mod p

It’s also a quartic residue but whatever

https://en.m.wikipedia.org/wiki/Quartic_reciprocity#:~:text=A quartic or biquadratic residue,biquadratic nonresidue (mod p).

Reduces our bashing to 1 mod 8 primes so gg

[reef condor]

Quadratic residue

[radiant jasper]

that looks complicated

besides i didn’t use it and still solved the question

[reef condor]

Oars

[radiant jasper]

oars?

[reef condor]

Well it’s pretty simple in the quadratic case - x^2 = -1 mod p has solutions only when p is 1 mod 4

[reef condor]

= orz

[radiant jasper]

orz?

[lavish lily]

if a,b and c are in gp, then x cannot be-

a+b+c=xb

[radiant jasper]

3?

assuming a, b, c are not equal

also likely cannot be 0 but don’t quote me on that

b/r + b + rb = xb
r + 1 + 1/r = x

[lavish lily]

please give a logical solution

[radiant jasper]

i think |r+1/r|>=2 so x is either >=3 or =<-1

this should be the solution

[lavish lily]

but the answer is -3

explain me how

[radiant jasper]

what if i don’t want to

[lavish lily]

??

[ivory ember]

Is this is a multiple choice question that you forgot to include the options to

[lavish lily]

yes @ivory ember

[lavish lily]

why the heck did you start to answer it then?????

[radiant jasper]

i enjoy solving math questions

[ivory ember]

Then just pick the option that’s not in the interval

[radiant jasper]

not answering to the demands of “explain me how”

have a good day

[ivory ember]

Tbf, they provided a full solution. Your poor communication isn’t their fault.

[lavish lily]

@ivory ember fine. next time i would see the internet rather than seeing your dumb discord channel!

[lavish lily]

ohk

[lavish lily]

are u a robot or a person?

[soft vigil]

Bruh

Keep the Comp math brainrot in the comp math servers

[ornate blade]

a text emoji for kneeling on the ground

[near hill]

"orz" is a stylized rendering of a person genuflecting towards the left.

[ornate blade]

o is the head, z is the legs

[near hill]

The joke is it would be more or less homophone with "orz" if you pronounce the latter as a word.

[soft vigil]

Oars has the same pronounciation

I would suggest not caring too much about comp math slang

[willow anvil]

comp math brainrot is bad

[near apex]

this is true

[gilded halo]

Vanellope von Schmugz

[peak kestrel]

An easier way is to just know that this is true (for example, see unique lagrange interpolation polynomial).