competition-math

[ripe iron]

Thanks guys now I  got it

[ornate blade]

no worries!

[radiant jasper]

hi everyone

[modern prawn]

wrong server

💀

[winter sable]

What

[vital glen]

yo

anybody preparing for imo?

[high goblet]

as in the international maths olympiad?

[vital glen]

yahh

[high goblet]

i almost made it to the team for the internationals but i didn't quite make it

anyway i'm at uni now so i can't do the IMO anymore

[high goblet]

r u preparing for IMO?

wait r u talking about that one random indian thing called the IMO even though it is not international nor an olympiad?

[vital glen]

haven't started studying for it yet

[vital glen]

i dont really know much about it tbh

[high goblet]

send me what ur taking about cus the actual international maths olympiad happened a few months ago in the UK already

[vital glen]

ahh okay

[high goblet]

the next one is in australia and won't be held until next summer

[vital glen]

Yes, that's correct! The International Mathematical Olympiad (IMO) is held annually, but the dates can vary from country to country. In the UK, the selection rounds and national competitions leading up to the IMO are often completed before the international event takes place, allowing the selected team to prepare for the IMO.

In India, the selection process typically involves multiple rounds, including regional and national competitions. The INMO (Indian National Mathematical Olympiad) is usually held a bit later in the year, which means that students in India may still be in the process of preparing for their selection to the IMO team when the event has already occurred in other countries.

[vital glen]

what was your score?

[high goblet]

i didn't do the IMO

[vital glen]

oh

[vital glen]

damnn mann you must be smart

got any tips?

[high goblet]

ngl i'm still slightly confused by what ur asking abt

are u saying ur preparing for the regional olympiads in u area?

[vital glen]

yup

[high goblet]

ic, well gl for those!

there's a discord server called maths olympiad discord server which is better suited for maths olympiad questions

[vital glen]

ohh thanks for letting me know , i'll join it 😄

[radiant jasper]

problems, problems, and problems

problems like these:

but easier

try looking for math olympiad problems more your style

like AMC or something easier like comps from big companies that steal your money

[vital glen]

ohh thanks dude : ) , but when i'm trying to do problem and i am stuck at them i lose all the motivation to do so

[radiant jasper]

just do problems from AoPS, since they give out solutions, dont do IMO problems from AoPS tho since most of them are above your level

just do other problems since they have a whole library of past olympiads

[vital glen]

man thank you very muchh!

[vital glen]

hey do you got any book suggestions?

[radiant jasper]

i only practice online

if i were to recommend a book for you maybe some basic fundamentals in olympiad math

[radiant jasper]

oh also forgot to mention you should prioritize proofs

most problems are asking you to give proof

ok bye

[vital glen]

like proving theorems etc?

[winter sable]

Hi guys

[winter sable]

International maths Olympiad?

[vital glen]

yess

i was talking about the regional level

[winter sable]

Yea me

India

[modern prawn]

BRUH

how smart are you

wth

[high goblet]

it may have been a 'legendary' problem back in 1988

but it's really not that hard in the modern olympiad landscape

[modern prawn]

dude if thats not a legendary problem then i’m a kindergartener

[high goblet]

sorry i didn't mean legendary sarcastically lol

[high goblet]

basically this question was hard bcus u had to come up with the idea of vieta jumping to solve it

that was a very novel idea that no one had done before

but then basically "oh look that's a cool olympiad trick" and now suddenly you have a billion questions that all use vieta jumping

it's not hard in the modern landscape cus basically ppl overused the idea

to the point where it's more "oh look another vieta jumping question" rather than "damn how do u do this"

[modern prawn]

wot is vieta jumping

I only know vieta’s formula or whatever and i know that probably is not related at all to “vieta jumping”

[high goblet]

questions like IMO 2007 Q3, IMO2007 Q6, IMO 2020 Q6 are considered the hardest IMO qs in recent history

[high goblet]

basically it's the trick to do IMO 1988 Q6

[modern prawn]

do u compete in imo

[high goblet]

the rough tldr is that this question was really hard but only had 1 trick

then ppl came up with a load of other questions that all use the same trick

[modern prawn]

So its hella easy if you know the trick but impossible if you dont

[high goblet]

but if u have a bunch of questions that have the same trick, it eventually becomes 'learn the trick' rather than 'come up with the trick'

hope that explanation makes sense

[high goblet]

yh

the main thing is ppl overused it so it got really boring

[modern prawn]

I feel dumber every day I am in this server

[high goblet]

almost

[modern prawn]

I should leave so my ego regains

[high goblet]

didn't quite make the IMO team for my country but i got to the final stage

[modern prawn]

dude how are u so smart wwttthhh

[high goblet]

ahaha yeah it can feel like it sometimes

[high goblet]

lol ty <3

[high goblet]

i'm sure u're smart as well

[modern prawn]

are u in college yet

[high goblet]

given that u recognised the problem source

[modern prawn]

i am dumb as hell

[high goblet]

yh i'm at uni now so i can't compete in these things anymore

[modern prawn]

back to get my ego destroyed more, what is 1 + 1

[hot shore]

2

[normal granite]

what's that? is that the national olympiad for your country

[normal granite]

could you dm me the link 🙏

[winter sable]

We are skibidi indians

Maths lives in us🤣

[high niche]

what country would u have represented if you were selected?

[dry forum]

How do I get faster with calculation?

Should I focus on techniques, training, or working memory?

[radiant jasper]

any hints plzz

[radiant jasper]

tf

[ornate blade]

,w sum n/(n^2 + 2kn + k^2) from k = 1 to n

[gilded halo]

Results provided by WolframAlpha

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

Requested by sooouth.

[ornate blade]

what to compare to hmmm

[radiant jasper]

+ig s infinity

[ocean narwhal]

Thats just a riemann sum no?

[ornate blade]

oh RIGHT

[ocean narwhal]

Ye so its just integral of 1/(t²+t+1) from 0 to 1

,w integrate 1/(t²+t+1) from 0 to 1

[gilded halo]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[ocean narwhal]

So that

[ornate blade]

wait

[ocean narwhal]

The answer is b,c 💀

Its a jee pyq

I dont know why

[ornate blade]

yeah it has to do with the summation bounds

[ocean narwhal]

Yea that

But well n is not approaching infinity so i got no clue

[ornate blade]

1/(x^2 + x + 1) is a decreasing function in [0, 1] so the left Riemann sum will be larger than the integral
but the right Riemann sum will be smaller than the integral

the left Riemann sum is T_n and the right Riemann sum is S_n

that makes perfect sense now that I know the answer lol

also yeah we are not taking the limit, so the sum will be a finite number of rectangles

[ocean narwhal]

Um what 💀 like since Sn = Tn + something shouldnt Sn be greater

[ornate blade]

left Riemann sum

[ocean narwhal]

The Tn is not a proper riemann sum tho i think?

Everywhere it says n

It should be n-1

I think

[ornate blade]

nah, the variable in both summations is the k, so it's fine

basically you know Tn is the left Riemann sum cause it's from k = 0 to k = n - 1
the right Riemann sum would range from k = 1 to k = n

both have n rectangles

[ocean narwhal]

I

Didnt notice the lower bound

Fuck

Sorry

Nvm

[ornate blade]

okok

[radiant jasper]

whats left ansd right reimann sum

[solid cloud]

Helppp plss

I don't quite understand how I could turn gcd or lcm into a variable in this

[ornate blade]

you probably need the fact that ab - a - b + 1 = (a - 1)(b - 1)

so add 1 to both sides

I'm not sure how to proceed

[high goblet]

hint: write a = dx, b = dy where d=gcd(a,b)

[high goblet]

the final stage of TSTs for my country, doesn't really have a name

like only ~10ppl take it, it's not really a stage of our national olympiads

[high goblet]

UK

[high goblet]

https://discord.gg/mods

[solid cloud]

Can you please explain it to me

[high goblet]

it's quite a common trick to rewrite our variables a,b

in terms of a/gcd(a,b), b/gcd(a,b) cus then they're coprime

[high goblet]

anyway so let's call gcd(a,b) = d

then write a = dx, b = dy

then we know gcd(a,b) = d and lcm(a,b) = dxy

etc.

[solid cloud]

Oh shi that is rightt

Let me do it now

Tyy

[high goblet]

nw!

[solid cloud]

I'm sorry for asking again, for my question I change all of a into dx and b into dy

So I write it again as

D^2xy -dx-dy = d+dxy

After that I make

D^2xy = d(x+1)(y+1)

Then I cross the d

Dxy=(x+1)(y+1)

Then I got stucked

Did I do something wrong here

@high goblet tyy

[high goblet]

so there's possible another way to finish from here

but my thought process was to write it as

d = (x+1)(y+1)/xy

i.e. (x+1)(y+1)/xy is an integer

this is a useful condition - why?

[solid cloud]

Cuz we may find d from there

I'm lost bro

[high goblet]

well intuitively

if x=100 and y=100

then (x+1)(y+1)/xy is just very slightly larger than 1

and definitely smaller than 2

and certainly not an integer

[high goblet]

in fact for a lot of values of x,y, this is between 1 and 2

so let's wlog x <= y

can u find a restriction on x?

if x>=3, what can u say about (x+1)(y+1)/xy

[solid cloud]

4(y+1)/3y at max

I'm truly lost here

[high goblet]

so if x>=3, y>=3 which means

$(\frac{x+1}{x})(\frac{y+1}{y}) \leq \frac{16}{9} < 2$

[gilded halo]

LY

[high goblet]

can u use this to finish off the problem?

[solid cloud]

Alr let me tryy

By the way @high goblet why did you use x>= 3 and y>=3 why Did you use 3 in here

[high goblet]

guess

(maybe if x>=3 then we can force it to be less than 2)

(oh it works)

[solid cloud]

Huh how??

This is new math knowledge I must learn bro

[high goblet]

as in this is quite a common trick that appears in quite a few questions

[high goblet]

you get an expression that has to be an integer

but isn't likely to be an integer for very many values of x,y

because you can bound it between i.e. 1 and 2 for a lot of values of x,y

then wlog x <= y

and take a guess for x (i.e. x>=3)

then if that inequality works, you know x=1 or x=2

[high goblet]

otherwise if you couldn't establish a useful ineq.

try x>=4 etc.

[solid cloud]

Ohh I see can you continue solving it so I can review it and potentially understand it

[high goblet]

i mean u should be able to finish it off right?

we have x=1 or x=2 now

try to finish from there

[solid cloud]

Alr

[high goblet]

yh so i believe it's the same trick for this q

and this q

[solid cloud]

@high goblet is there a name for this trick so I can learn it

[high goblet]

not really?

bounding?

finding inequalities to bound a quantity is very common and comes up like everywhere

[solid cloud]

For x = 1 there is no suitable y because 2(y+1)/y <= 16/9

Dividing both by 2

(Y+1)y <= 8/9

Not possible for positive integer

Is this explanation right??

[high goblet]

well no that's not what's going on

[solid cloud]

💀

[high goblet]

^

so we've said that (x+1)(y+1)/xy is an integer

[solid cloud]

Alr ty let me try to do it

[high goblet]

if x>=3 then y>=3 so that is between 1 and 2, a contradiction

so we need to test x=2 and x=1

[solid cloud]

And after I check it

What should I done

This material is beyond my level rn

I join some math competition but never see this material come out

[high goblet]

i think it might be helpful to try to understand the solution

Write $a=dx, b=dy$ where $d=gcd(a,b)$. Then our equation becomes:
$$d^2xy - dx - dy = dxy + d$$
i.e.
$$d = (1+\frac{1}{x})(1+\frac{1}{y}) \in \mathbb{N}$$
Clearly $(1+\frac{1}{x})(1+\frac{1}{y}) > 1$. Now wlog $x\leq y$. Then if $3 \leq x \leq y$ then
$$(1+\frac{1}{x})(1+\frac{1}{y}) \leq (1 + \frac{1}{3})(1 + \frac{1}{3}) = \frac{16}{9} < 2$$
and so is clearly not an integer. Thus, $x=2$ or $x=1$. \
If $x=2$ then $\frac{3}{2}(1+\frac{1}{y}) \in \mathbb{N}$. If $y > 3$ then we get another contradiction and thus $y=3$, giving us $d=2, x=2, y=3$ as a possible solution. \
If $x=1$ and $y \geq 3$ then
$$2 < 2(1+\frac{1}{y}) \leq \frac{8}{3} < 3$$
and so not an integer. Thus, $y=1$ or $y=2$. It is trivial to see that these work and thus $x=1, y=1, d=4$ and $x=1, y=2, d=3$ are 2 further possible solutions. \
Hence, the solutions to our original quesiton are $(a,b) = (4, 6), (6, 4), (4, 4), (3, 6), (6,3)$

[gilded halo]

LY

[solid cloud]

I'll try to understand this

Thank you

[high goblet]

nw

[solid cloud]

Bruhhhh

That is whyy

I understand

I make the x a internet but I didn't make sure the original "a" variable also an interger

Oooh I forgot

Can anyone explain me how should I try to approach the number 2 question about 100 digit real number

[high goblet]

you wanna make the sum of the digits nice

fix the sum of the digits then try to find a number that satisfies the properties you want

remember that we don't really have any good ways to test if i.e. n is divisible by 19 etc.

so u want to make the sum of the digits nice

think about how u can do that

[solid cloud]

Alr

[solid cloud]

I make number where the sum of the digits will be 125

1......125

I though it would work

@high goblet

Where there is 96 spot left for the rest of the number

[solid cloud]

This is the 100 digit number

[high goblet]

yh that works

u just have to fudge ur numbers a bit

so that the sum of the digits is 125

[solid cloud]

Yeah

There is many variation using 1 and 2

[solid cloud]

How should I factor a+b=a^2-ab+b^2

And how should I prove number 9

[gusty verge]

Sum of squares is usually destroyed by Cauchy Schwarz

[novel forge]

this when it's on the larger side of the inequality though (in this case it is so no problem)

[gusty verge]

Cauchy Schwarz:

[\left(\sum_{i=1} u_i v_i\right)^2\le \sum_{i=1}u_i^2\sum_{i=1}v_i^2]

Then take $v_i=1$ and $u_i$ as in the sum they showed, then rearrange.

[gilded halo]

Max

[solid cloud]

Alr let me try it

[cunning bloom]

Hello guys
I need references for which book I can use for Maths for class 12th boards?

[tribal notch]

How do I actually learn from my mistakes? I understand how to solve that particular problem now, but couldn’t apply the exact same thing in a different context

[modern prawn]

wat

can you elaborate further

your description is kind of weird

sorry, just cant understand

[solid cloud]

Sorry I'm kinda lost here

[jolly solstice]

I'd multiply both sides by (a+b) since that seems potentially interesting

[warped wedge]

has anyone here done the official aops amc12 mocks

how does it compare with actual amc12 in terms of difficulty

[jolly solstice]

well I guess we know it's some conic section, and it's symmetrical across the line a=b. Solving gets us 2 points, so that means it's either an ellipse (good finitely many things to check) or a hyperbola (boo infinitely many things to check). We know it's not a hyperbola because... well I just checked the points at infinity in the projective plane but probably less fancy way to do this by trying to find asymptotes or something.

of course this is all ad hoc rationalization, I just graphed it in desmos lol

[solid cloud]

Then I used quadratic formula

Then just found x and y

X = a, b=y

[jolly solstice]

can you elaborate on this step

[solid cloud]

There

So there is 6 pair

How should I try to approach this question

[radiant jasper]

can some solve htese tqo

two

jsut give me hints

[ivory ember]

1. Note that $$(ad-bc)^2=(a^2+b^2)(c^2+d^2)$$

[gilded halo]

Civil Service Pigeon

[ivory ember]

2. Direct Cauchy Schwarz question

(Question 1 isn’t well posed imo tho)

[radiant jasper]

nah the questiobn all clear to me

can some one just solve thje first question

q1

[solid cloud]

Does my question here could use Cauchy schwarz too

[solid cloud]

.

[ivory ember]

As in there’s no answer b/c ac never achieves a maximum

[scarlet root]

Please do not ping individual helpers unprompted.

[radiant jasper]

what abotut thw seconfd question the question in which ac + bd is min

[gusty verge]

Nah I tried it yesterday, Cauchy Schwarz reduces it too much

(once you apply CS and remembering you can choose the different letters arbitrarily, you can essentially show the minimum of the expression is 0, which is too low)

[high goblet]

yh

[high goblet]

first expand out our entire thing

often in inequalities, you have something that looks nice, but often the first step to solving it is just to expand out

anyway u end up with this

this looks horrific

partially because we have ws multiplied with xs and ys but not zs etc. etc.

can u rewrite this in a nicer way?

then u can apply C-S and finish

[solid cloud]

Ly saving me once again

[solid cloud]

What do you mean nicer way

Like give me a hint how to use Cauchy schwarz in here

[high goblet]

well this was actually kinda a hint

notice that w^2 is multiplied by x^2 and y^2

and z^2 is multiplied by x^2 and y^2

anyway stare at it and realise that we have

(w^2 + z^2)(x^2 + y^2) + 2wx + 2yz + 2 >= 1

now u can CS that

[tall quartz]

Find all natural numbers (a,b) such that ab| a^2024 + b. Can anyone help me with this??

[high goblet]

funnily enough someone posted this problem earlier today lol

[shadow dirge]

(a,b) = (1,1)

[high goblet]

anyway the idea here is clearly:

a | a^2024 + b

so a | b

[shadow dirge]

Maybe thats the ans

[high goblet]

then if a | b write b=ab'

then a^2 | a^2024 + ab'

so a | b'

repeat that basically until u get a conclusion

[shadow dirge]

I also got that but not in this way. Can you explain this step?

[high goblet]

a | ab and ab | a^2024 + b so by transitivity

a | a^2024 + b => a | b

[shadow dirge]

Oh i see

[wheat kettle]

does anyone have any tips on studying for short-answer contests? aka for getting faster and doing less mistakes

[dusk tendon]

short answer as in a short statement or just a number?

[wheat kettle]

I mean those boring ass contests with like 20 questions for 3 hours yk

[dusk tendon]

ohh

i cant help you i have no experience w that

sorry

[wheat kettle]

its ok

[dusk tendon]

i completed the square on the RHS, then multiplied everything by negative 1 to get ab-a-b=-(a-b)^2, then i added 1 to factor the LHS by sfft (simons favorite factoring trick). when we factor, we get (a-1)(b-1)=1-(a-b)^2. we factor RHS by difference of squares for (a-1)(b-1)=(1-a+b)(1+b-a). now we see that 1-a+b=a-1 and 1+b-a = b-1. solving the system of equations gives a = -2 and b = -2 which are the only ones i think (i could be wrong)

[elfin crown]

idrk if I should be asking this here, but is this channel used for discussion about how to solve comp questions (ie, the last 5 questions of AMCs) or just kinda talking about the comps in general

I joined like, five minutes ago so I know nothing 😭

[dusk tendon]

both ig

[elfin crown]

alrightyy

[dusk tendon]

starting with amcs is easy

just make an aops account and do the mocks and use their books

[elfin crown]

I took an aops class and I lowkey hated it

[dusk tendon]

dang which one

[elfin crown]

i dont really remember, apart from the fact that it was on the amc10 and discussion strategies

I also did the seminar

[dusk tendon]

oh amc 10 problem series probably

[elfin crown]

yeah probably that

I prefer face to face teaching, which was probably why it was so tedious

[dusk tendon]

yeah

i havent done any classes yet but if you just use the books from aops (intro series and volume 1) and do questions it will boost your score

and also do mocks

[elfin crown]

are the books free??

[dusk tendon]

no

[high goblet]

you claim that 1-a+b = a-1 but why?

that doesn't follow from (a-1)(b-1)=(1-a+b)(1+b-a)

[high goblet]

what contest r u exactly referring to?

[dusk tendon]

i tried setting 1-a+b to b-1 but i got infinite solutions

idk i may be wrong

[high goblet]

i think in general, u've probs just gotta do more qs to build up speed etc.

but yeah generally multiple choice qs don't really matter

idk how it works in the US system

[wheat kettle]

me neither

[high goblet]

ah k

[wheat kettle]

two brazilian contests, jacob palis and obmep, but if you're not brazilian you prob have no idea what these are

[high goblet]

ah kk

what r the questions like?

i think in general, just do qs

multiple choice qs are usually followed by like the harder more prestigious competitions anyway

[wheat kettle]

do you think I should do like, them timed or just do one after the other

[high goblet]

probably do them timed

[wheat kettle]

yeah I really regret only practicing for the harder and cooler ones so that's why I'm kinda desperate for something rn

I thought if I just got really smart I would automatically think these contests are easy but they are not

I mean, each question is easy

[wheat kettle]

ty

[dusk tendon]

can you give an example of a question

just want to see the difficulty

[wheat kettle]

hum I'll find one with pretty median difficult of each contest

[dusk tendon]

ok

[wheat kettle]

how many ways can we color each labeled point with red, blue and green such that no 3 colinear points have the same color

[high goblet]

eww trial and error

yeah probably just do questions to eventually build up speed

[wheat kettle]

yeah it's honestly abhorrent

well from obmep every question is straight up extremely easy so it's REALLY just not getting anything too wrong or thinking too much

I'll send one that I could not solve fast because my brain hurt a little

Benício put together, twice, three dice in such a way that faces in contact have the same number, each time he summed the numbers in all faces that are not in contact and the difference between the sums was 16. which were the numbers in the faces that were never in contact?

[wheat kettle]

quite saddening

maybe I'll do like a daily amount of questions or something as the actual shore it is

maybe weekly for flexibility idk

[high goblet]

jesus christ these questions look like pain

[radiant jasper]

can you translate to english

[wheat kettle]

I did

[radiant jasper]

oh

mb

ew counting

[wheat kettle]

to be fair obmep is meant for people with no experience in math to be incentivized to think about it in a more fun way
but for someone like me it's just painful I already know all the tricks

[high goblet]

ic ic

[wheat kettle]

+the standard is each school forcing all students to take it but my school doesn't do that so only 2 people per level can pass to the second phase which made me unable to do it this year

[wheat kettle]

jacob palis is similar but in a higher level, it incentivizes people to start learning the first little olympiad basic tricks like cyclic quadrilaterals and stuff

[warped wedge]

`has anyone here done the official aops amc12 mocks
how does it compare with actual amc12 in terms of difficulty

[radiant jasper]

they are very orz

[ripe iron]

x²+4y²+9z²-2x-12y+6z+10=0
Show that 1/3≤x+y+z≤4

[ripe iron]

x²+4y²+9z²-2x-12y+6z+10=0
⇔x²-2x+1+(2y)²-12y+9+9z²+6z=0
⇔(x-1)²+(2y-3)²+9z²+6z=0
⇔(x-1)²+(2y-3)²+9z²+6z+1-1=0
⇔(x-1)²+(2y-3)²+(3z+1)²=1
0≤(x-1)²≤1
⇔0≤|x-1|≤1
⇔|x-1|≤1
⇔-1≤x-1≤1 |+1
⇔0≤x≤2
⇔x∈[0;2]

0≤(2y-3)²≤1

⇔0≤|2y-3|≤1
⇔|2y-3|≤1
⇔-1≤2y-3≤1 |+3
⇔2≤2y≤4 |:2
⇔1≤y≤2
⇔y∈[1;2]

0≤(3z+1)²≤1

⇔0≤|3z+1|≤1
⇔|3z+1|≤1
⇔-1≤3z+1≤1 |-1
⇔-2≤3z≤0 |:3
⇔-2/3≤z≤0
⇔z∈[-2/3;0]

min (x+y+z)
=min x+min y+min z
=0+1-2/3
=1/3

max (x+y+z)
=max x+max y+ max z
=2+2+0
=4

⇒ 1/3≤x+y+z≤4

[gusty verge]

Nice

But you can use latex on discord

A lot easier than having to copy $\in$ and $^2$

[gilded halo]

Max

[dusk tendon]

slightly easier imo

[royal estuary]

$$ 2^p \neq 3m + 1 $$ prove for positive odd integer values of p and m

[gilded halo]

Mr.Bombastic

[royal estuary]

Need help

[ornate blade]

without loss of generality, note that the inequality is homogenous, as in $(a,b,c) \mapsto (ka,kb,kc)$ for $k \in \mathbb R \setminus {0}$ doesn't change the inequality

[gilded halo]

higher's secret twin brother

[ornate blade]

so we can let $abc = 1$

[gilded halo]

higher's secret twin brother

[ornate blade]

AMGM gives $\ge 3 \left( \left| \frac{abc}{(b + c)(a + c)(a + b)} \right| \right)^{1/3}$

hmmmm

oh if you can show that (a+b)(b+c)(c+a) >= 8abc you're done

wait that's the wrong direction sorry

[gilded halo]

higher's secret twin brother

[ocean narwhal]

i dont know the original que

but this part is easy

just apply am gm in a, b then in b,c then in a,c

and multiply

[ornate blade]

I think we need another approach since the inequality is in the wrong direction

[ocean narwhal]

what was the original que

[ornate blade]

.

[ocean narwhal]

this?

lemme think

nvm i give up 💀 how would you even

that looks an awful lot like it need the GM-HM

but then there's inequality sign

nd HM is smallest of them all

you cant really compare

[ornate blade]

https://approach0.xyz/search/?q=OR content%3A%24\left|\frac{a}{b%2Bc}\right|%2B\left|\frac{b}{c%2Ba}\right|%2B\left|\frac{c}{a%2Bb}\right|%24&p=1

oh the inequality isn't sharp

if in doubt just Approach0

[ocean narwhal]

thats cheating

[ornate blade]

no it's not

if you've made an honest attempt

[ocean narwhal]

true that

[ornate blade]

also comp maths people rarely give detailed solutions

you always have to fill in the gaps yourself

[south birch]

I agree

[ripe iron]

wow. Thanks a lot

https://math.stackexchange.com/questions/4980637/a-b-c-are-non-zero-reals-such-that-abacbc-neq0-prove-left-frac

[jolly solstice]

I suppose (1+sqrt(5))(1-sqrt(5))=-4 is not a coincidence that 4 is in the denominator, idk

[untold thunder]

It is 🥶

[jolly solstice]

you have the answer?

[untold thunder]

Yes I have a full solution

It's very long so there may be some error but I think I have the general idea

[jolly solstice]

gotcha, yeah I guess there may be more than one way to do it, but if you didn't use that fact that's good to know

oh I think I see a way to do it

,texsp ||think we can replace the $a_k$ and $b_k$ with these then work out the geometric series we get, assuming they converge $$\sum_{k=1}^\infty \frac{\frac{(1+\sqrt{5})^k+(1-\sqrt{5})^k}{2} + (1+\sqrt{5})^k-(1-\sqrt{5})^k}{4^k}$$||

no idea if that actually spoilers the texit, I know there is/was a way for texit to spoiler and I have spoilers to auto-show

[cerulean ibex]

,texsp

[gilded halo]

Please give me something to compile, for example latex ,tex The solutions to \(x^2 = 1\) are \(x = \pm 1\).See ,help and ,help tex for detailed usage and further examples!

Oct 5, 2024, 7:55 PM

[cerulean ibex]

this is not spoilered

[gilded halo]

Merosity

[jolly solstice]

that do it?

[cerulean ibex]

yep!

[jolly solstice]

ty ty

[cerulean ibex]

nw i just learned abt it a couple days ago

[untold thunder]

Good luck computing that

[jolly solstice]

it's really just 2 geometric series

[untold thunder]

Oh wait

[cerulean ibex]

it's not as bad as it looks

[jolly solstice]

you can kinda work it through with +- on the sqrt too to get both in one go if you're really wanting to be cute

[untold thunder]

Yeah I think that works, but the idea that I thought was cool with that problem is you can actually find a closed form for a_i and b_i (Kinda)

If you want to try and solve it that way you can

By all means it's harder though

[jolly solstice]

interesting, would be curious to know

well do we get the same answer I just worked it out and got -2/sqrt(5)

I wouldn't be surprised if I made a mistake though lol

[untold thunder]

I checked my longer solution through demos and stuff (because you can) and honestly, it would probably never have a radical form like that in the end I probably made a mistake too

[cerulean ibex]

binomial thm i think

[jolly solstice]

I don't see how that gets us a closed form though

[cerulean ibex]

maybe the sums end up nice

i dont see any other way to elementarily get a closed form

[jolly solstice]

that was the first thing I did tbh but didn't see anything

then just noticed I could write a_k, b_k in terms of the exponentials so it'd all thread through the geometric series

so that's where I gave up, maybe some combinatorics trickery can be done though on those binomial coefficients

maybe we want to expand 5=(1+4) and do another binomial expansion there, cause, idk 4

swap the sums

[cerulean ibex]

i'm not sure i believe in a 'true' closed form

i think you can show some things

[untold thunder]

It's kinda one

Like not completely but better than a bunch of terms

[cerulean ibex]

i think they both have to be divisible by some power of 2

[untold thunder]

I have a hint to start

[jolly solstice]

I feel like I've already finished lol

[untold thunder]

You don't have to explain your process what was your "closed" form for a_i and b_i?

It should have Fibonacci numbers in there

[jolly solstice]

I could see that

[jolly solstice]

sure, I can explain that in a couple ways

you can see it from the binomial theorem when you combine them, the sqrt(5) and -sqrt(5) terms will either cancel or double

$$a_k = \frac{(1+\sqrt{5})^k+(1-\sqrt{5})^k}{2}$$

[gilded halo]

Merosity

[untold thunder]

basically my "answer" is just the exponential of the radicals replaced by Fibonacci numbers because they're easier to work with

[jolly solstice]

this is a general trick, for instance you might have seen people do this to get the real part of an imaginary number

here they're replacing sqrt(5) with sqrt(-1), but you're still adding/subtracting conjugates

[untold thunder]

yeah I've seen that

[jolly solstice]

actually my b_k should have a sqrt(5) in the denominator

to divide that out

so my answer in the end was wrong, but I'm not gonna go through and fix it

[untold thunder]

This is what I got in terms of Fibonacci I did this really weird method that might not be necessary though $a_n=2^{n-1}(F_n+F_{n-2})$, $b_n=2^{n-1}F_{n-1}$

wait I think tyop

[gilded halo]

Cube

[jolly solstice]

I see what you mean about fibonacci I've seen something similar before so I can see that working out that's a cool idea

ok worked out some recurrence relations, by taking the kth term and multiplying by (1+sqrt5) to get the k+1 term and collectin gcoefficients, but doesn' tquite line up how I was expecting. But reading your message a bit closer now I'll see if that works
$$a_{k+1}=a_k+5b_k$$
$$b_{k+1}=a_k+b_k$$

[gilded halo]

Merosity

[jolly solstice]

or rather, I can just diagonalize the matrix now to get the answer

[untold thunder]

I used vietas formulas to get that

[jolly solstice]

oh?

I collected terms the were on sqrt5 or not on this, if that's what you mean $$(a_k +b_k\sqrt{5})(1+\sqrt{5})=a_{k+1}+b_{k+1}\sqrt{5}$$

[gilded halo]

Merosity

[untold thunder]

You can find the quadratic with the root (1+sqrt5)/2 and then do some induction to find a_k and b_k

[untold thunder]

But I see what you did

[ebon sigil]

Who else doin math counts

[soft vigil]

sadly i took it for the last time last year

good memories

[modern prawn]

in mathcounts do you compete infront of a lot of people or is it just a boring classroom environment

like if ur just doing a local school mathcounts

[elfin crown]

You do it in a school environment

[soft vigil]

Mathcoutns is fire at the national levels

224 kids

4 kids per table who are on the same team

Environment is great

[shadow spruce]

😭

[lyric parrot]

solve in 2 mins

hint: find tan (a+b)/2

[ornate blade]

oh so you already know the answer

smh

[lyric parrot]

its a really good question for olympiads so i wanted to share it

[ornate blade]

probably square both equations and add them to get $2 + 2 \cos(\alpha - \beta) = \frac{6}{9} + \frac{3}{9} = 1 \implies \cos(\alpha - \beta) = -\frac{1}{2}$

[gilded halo]

higher's secret twin brother

[serene pewter]

i geneuinly dk

[ocean narwhal]

how'd that help you to find cos(a - b/2)

[ornate blade]

idk

[rose mortar]

Anyone good at complex numbers dm me I have a problem

Thx

[fast coral]

send it here

[rose mortar]

It’s quite simple 😅😅

[fast coral]

dont dm me bruh ask it here

[rose mortar]

Uhhh ok

So my teacher had shown me a method to generate Pythagorean triplets using 2 other triplets
Using complex numbers

So I thought that I should generalise it

And I have an equation
It would be brilliant if someone could check it

[slim oriole]

will you provide the solution?

[fast coral]

go on

[lyric parrot]

Alright so
Find [2sin(a+b)/ 2]cos (a-b)/2 which is basically sin a+ sin b( its a basic identity you can look up the proof on yt)
And similarly [2cos (a-b)/2]cos (a+b)/2 = cos a+ cos b (eq1)
Divide both the equations then u get tan (a+b)/2
Then just apply tan^2(x)+1=sec^2(x)
With this you find cos (a+b)/2
Substitute in eq1 to find cos (a-b)/2
Simply square it and multiply by 100, boom you solved it

The question is find 100cos^2([a-b]/2)

It's a fun question to solve

Sure shot olympiad question

[slim oriole]

????? i was exploding my brains out trying to find cos(a-b/2) WHEN IT WAS COS((A-B)/2)???

the request is fucking wrong

[modern prawn]

how are people so smart lol

[ornate blade]

jesus

yeah I was scratching my brain and all that time

[winter sable]

hi guys i m back

[long wharf]

im the smartest

[ebon sigil]

lol math counts is light work

it's 6-8

but it really shouldn't matter

cuz math counts should rep the smartest schools

[ornate blade]

I mean the time pressure is the real deal

I bet you couldn't do the questions as fast as the finalists

if you could work on each question for as long as you like then it'd be easy

also yeah they are challenging cause they are extensions of middle school material without going to more difficult stuff

[novel karma]

why won't my school let me take both amc 10a and 10b

i thought you could

am i wrong

[ornate blade]

cause it's your school

[novel karma]

but most schools do right

[ornate blade]

the AMC doesn't care but the AMC isn't the one organising the details of the contest

[novel karma]

huh

[ornate blade]

I'm not sure but at my international school you could only sit it once

they'd choose either the A date or the B date

[novel karma]

oh

[modern prawn]

the mathcounts has really easy questions

a standard 8th grader doing algebra 2 could easily answer every question they give even in national mathcounts

its just you gotta do it fast as hell

[fossil loom]

fun fact, ... standard 8th graders don't do algebra 2

[soft vigil]

yeah no

this is just false

[modern prawn]

its true

[soft vigil]

its not true

[modern prawn]

so what level would you AHVE to be as a student to be able to solve these problems without a time limit in your perspective?

[soft vigil]

there are a lot of problems that require higher level maths to be solved quickly

without a time limit is much different

with time limit you need a lot of knowledge on tricks and stuff

[modern prawn]

What is the highest “higher level math” that you’re talking about here

[soft vigil]

i know for sure a standard 8th grader would be scared shitless of this

and this isnt even a nats problem

[soft vigil]

often times calc is used for some problems

to speed up the process a lot

[modern prawn]

I mean i would be scared because i AHVE to do it fast but its like a final end of term exam for geometry kind of question

[soft vigil]

then solve it right now

[modern prawn]

im eating

[soft vigil]

ok

go eat instead of typing

[modern prawn]

ok

cya

ima just watch youtube then

[radiant jasper]

mathcounts?

this problem is sus

@soft vigil was this a sprint problem

[soft vigil]

yeah

p27 of states in 2023-2024 year

[radiant jasper]

wtf

ngl i would ksip this

perfectly normal quadrilateral

[soft vigil]

oh this

just extend AD

to intersect CB

[radiant jasper]

yeah

[soft vigil]

then CD=3/2

wait no

its $\sqrt{3}$

[gilded halo]

buboblakistoni

[fossil loom]

how would you do this lol

[cerulean ibex]

coordinate bash to find B, three points uniqely define a circle, find the center with said three points, use it to figure out radius, pir^2

is the obvious way

there's probably a nice trick

[fossil loom]

oh yea, just make everything into coordinate geo

valid

[cerulean ibex]

not in the spirit of the original quote but yes, coordinate bashing, the devil's marvelous machine

i would just use linalg or something here, i'm sure it works

but that's the obvious soltuon

[fossil loom]

not even sure I can solve it, granted I don't do comp math, but I'm still a grade 11 honors math student, no way in hell an average 8th grader who hasn't taken geometry is getting that

[fossil loom]

i have now filled a page with diagrams and am not anywhere nearer solving it

[cerulean ibex]

do u know how to get to the center from the three points

[fossil loom]

i can do it with a lot of trig, but mathcounts is no calculator right

[compact lion]

hi please someone can tell me how to prepare for olympiad geometry from basic/intermediate to olympiad level/

[idle fjord]

Any math helpers dm me

[cerulean ibex]

sps a,b,c define a circle, then consider the lines ab and bc. then draw the lines perpendicular to them. the point where those lines intersect is the center of the circle

[fossil loom]

yea but the obvious solution to that system uses trig

[cerulean ibex]

mm ig but can't you just solve for the center directly

like

take the three points you have

then consider the equation for a circle

plug them in to get three equations

with a little bit of manipulation this is a linear system so you can just solve for the center

then boom

[high goblet]

you don't need olympiad geometry skills, i almost made it to my national team whilst not solving a single geometry problem along the way!

that's also probably why i almost made the team

[compact lion]

wow where are you from

[high goblet]

anyway what sort of olympiad level r u talking about? USAMO?

[high goblet]

UK

[compact lion]

international mathematical olympiad

[high goblet]

oh as in r trying to improve ur geometry skills to be good enough at IMO geo?

or like trying to get to IMO?

anyway i can offer some basic advice though:

[compact lion]

i like to get in imo team

[compact lion]

sir can you please dm me

[high goblet]

ah k

[winter sable]

hi guysy i m back i hope you all recognise me

i am sig,a

m

[reef condor]

Oinks

[reef condor]

Ok

[south socket]

Visiting the farm where several dwarves live, Bob organizes 4 forest trips, each of which is attended by all the dwarves on the farm. Each gnome has a red, a white and a black cone. The dwarves use one of their cones on each trip. After examining the colors of the cones used by the dwarves on the trips, Bob notices that no two dwarves use the same color cone on more than one trip. According to this, what is the maximum total number of dwarfs on the farm?

[pallid dragon]

9?

[radiant jasper]

xooks

[reef condor]

ook

[radiant jasper]

skibidi

[reef condor]

trivel

[radiant jasper]

trivial?

[idle fjord]

any math helpers?

[south socket]

idk

what is your solution

[pallid ginkgo]

what does 3-6 and 2-4 mean? is there like a consistent way of ranking olympiad difficulties?

[ornate blade]

yes, I'll link to the AoPS page

https://artofproblemsolving.com/wiki/index.php/AoPS_Wiki:Competition_ratings

wait they're not using this difficulty scale

on Google

[pallid ginkgo]

whats the diffrence between competition and olympiads

[ornate blade]

I don't think there is a difference

[pallid ginkgo]

why is the intermediate competition split from the others and why is it not called intermediate olympiads?

[ornate blade]

oh wait on that page

yeah so basically they mean competitions are anything not at Olympiad level

i.e, in the IMO sequence

olympiads test the foundations of IMO level material or are IMO level themselves

[high goblet]

yeah there's several difficulty rankings

i suspect that's probably the AOPS ranking

MODS 1-10 is also sometimes used

Evan Chen has also rated like all the olympiad qs in MOHS

[high goblet]

competitions are generally easier, u don't need to write proofs etc.

[ornate blade]

it's complicated by the fact a lot of competitions style themselves as olympiads

[pallid dragon]

https://en.wikipedia.org/wiki/Kirkman's_schoolgirl_problem

specifically this

so you can do 9 dwarves i don't know if it's the largest

[south socket]

Thanks

[winter sable]

do you remeber me

i was here 5 texts ago

[broken halo]

please do not

[cerulean ibex]

<@&268886789983436800>

sounds scammy

[brave ocean]

please don't advertise unsolicited here.

[cerulean ibex]

thanks smay!

[solid cloud]

100a + 64 and 201a+64 is both squared number Less than 10000 what is a

I try to make it to

A^2-b^2

(A-b)(a+b)

So I could get 101a = (x-b)(x+b)

After that I'm confused

Or did I do something wrong

[high goblet]

yeah that's the right idea!

[high goblet]

unfortunately the problem ehre is that we still have a

[high goblet]

can u do something so that we get rid of the a?

[solid cloud]

How bout I make x-b is equal to a and make x+b equal to 100

[solid cloud]

Then I make it into linear equation

[high goblet]

why would x-b = a etc. give an answer?

why is the optimal not x-b = 2a or whatever?

[brave heron]

Actually this can be solved even with a on the equation, but i'm curious on the way you did it

[brave heron]

Try continuing on this and see what you can do

[brave heron]

It's just a matter of substitution and sheer will, and you can easily form a quadratic

[loud gull]

To quantify diffiiculty level,
https://jaymaron.com/fermi.html

[loud gull]

The MIT high school math contest has an algorithm to dynamically decide problem weights based on how many people solved the problem.

[loud gull]

Putnam publishes data for problem score distributions so you can judge the difficulty of a given problem. It also publishes the function of total score vs. rank. The above plot is an averaging of this function over many years. Do other contests publish such data?

[grizzled kestrel]

Hello

May i know whats this channel for

[high goblet]

oh interesting

also i realised my thing isn't as straightforward as i assumed it would be

u can eliminate the a and end up with a pell's equation

pell's equations are quite well-studied so if u end up with that u can just use theory to finish the problem

[boreal sky]

Is $f:[0,\infty) \to [0,\infty)$ such that
[f(x+y)=f(x)+f(y)]
For all $x,y\in [0,\infty)$ only have $f(x)=cx$ as the solution? If so, what's the proof on it?

[gilded halo]

Its_TheMathCraft

[brave heron]

basically 101a = (x-y)(x+y) where 100a + 64 = x² and 201a + 64 = y²
You only get (101,a) or (101a, 1)
But we use (101,a) cause why not
y+x = 101
y-x = a
y = (101+a)/2

Then x = (101-a)/2
Substituting on one

100a + 64 = ((101-a)/2)^2
400a + 256 = a² - 202a + 10201
a² - 602a + 9945 = 0
(a-17)(a-585)

Obviously a = 17

[high goblet]

why are those the only possibilities?

what if you can split a=12 say and it's actually split 101*2 and 6

[high goblet]

i agree in that gets a potential solution

but you haven't found all a

[high goblet]

actually i suspect this is true

you can probably extend ur original f by setting

g(x) = -f(-x) for x negative

probably can then show g satisfies the F.E.

so f is the +ve side of a cauchy FE solution

and bcus pathological cauchy FE solutions are dense in R^2, we must have f(x) = cx

[high goblet]

i haven't properly checked this though

it might not be true

[deft wraith]

Is it worth doing later AIME problems as practice just for mathematical/ problem solving skills? My country uses BMO1/ 2 which are easier than JMO but are proof based. I think my proof writing skills are fine but is it worth doing those problems just for the sake of improving maths for olympiads?

[glad zinc]

Learn the A level math and further math pure content, then use AoPS and past papers to prepare. Personally thats what I would do, unfortunately I didn't do it last year out of pure laziness and I missed out on alot of potential I had

[ornate blade]

not everyone does A levels

[glad zinc]

He said his country uses BMO1/2 so I assumed he would be following the british spec

[ornate blade]

oh shit I didn't read, right

[deft wraith]

further math isn't really related to BMO1/2 though

[ornate blade]

agreed there's even a huge difference between further and STEP

in terms of format and everything

[glad zinc]

Some parts of the core pure and further pure spec are useful, also studying pure maths in general just helps problem solving

[ornate blade]

like even if STEP isn't about proofs, the standard of working is pretty high

they prompt you very little

[glad zinc]

Are you doing this for STEP prep?

[ornate blade]

you're supposed to just write a page of working naturally without needing hints

[deft wraith]

no i'm doing this for bmo1/2 prep

[glad zinc]

If so then you should definitely do most if not all of normal and further a level maths, then do the STEP prep modules

[deft wraith]

step prep is adjacent too as well

also trying to understand a level fm concepts a bit more ref:

[glad zinc]

Then as I said, do pure maths, not all of it is needed obviously, then AoPS is really really helpful (at least it was for me) and just spam past papers

[glad zinc]

What year are you in?

[deft wraith]

i just started year 13

ive finished maths and further maths

[glad zinc]

I see

[deft wraith]

do you mean the forum or the books?

the books are a bit easy

[glad zinc]

Which book did you use

[deft wraith]

ive had a look at all of the official aops ones

[glad zinc]

Weird, did you do the BMO last year?

[deft wraith]

yeah

[glad zinc]

Result/

?

[ornate blade]

you applying to Oxbridge then lol

[deft wraith]

yes cambridge

[deft wraith]

low distinction

i got lucky -- last years paper was very easy compared to other years

and i wasnt very good last year

[glad zinc]

Quite strange because I found their books quite useful and I also got low distinction on BMO1 and iirc merit on bmo2

[deft wraith]

vol2?

[glad zinc]

vol1 was almost fully easy, just that odd thing here and there, vol2 was decent

[deft wraith]

i'm using mont and art and craft of problem solving by paul zeitz

i need to learn how to do functional equations properly

[glad zinc]

I was in y11 when I sat the BMO btw so maybe thats the reason

[deft wraith]

maybe yeah

what year are you?

[glad zinc]

Y12 now

SMC was so free this year

got full marks

[ornate blade]

bro

jacked

[glad zinc]

ill get cooked in BMO dont worry i have done 0 prep cause im a lazy mf

[deft wraith]

i got 120 haha i just made a stupid mistake on the dice question

but bmo qualification is all that matters

anyway for me i just want to get into cambridge over all of that

[glad zinc]

Bro I actually feel so bad

Getting the dice one wrong is just

wild

Did you not know that the opposite sides add up to 7?

[deft wraith]

i did the algebra and i went to plug in the solutions so i could see straight away 4 would work so i thought thats the answer

if i had tried 2 i wouldve realised oh right they sum to 7

[glad zinc]

I guessed the circle area question

[deft wraith]

a lot of my friends got question two wrong so hahaha

[glad zinc]

Time to make new friends

I want to sit STEP in y12, I need to do 2.5 chapters a week then go through all the step modules and past papers 😭

[deft wraith]

both step papers?

[glad zinc]

Yeah

[deft wraith]

nice

i think just doing questions is the main thing

[glad zinc]

My goal is s,s but its probably not gonna happen

not probably

like definitely

[deft wraith]

then just sit it in year 13

[glad zinc]

Yeah but I want to increase chances and also want to use it to show other UK and US unis

[deft wraith]

could you help me with my question that i linked?

[glad zinc]

Let me have a look

[glad zinc]

I recently started this so I may not be the best person to ask but cant you just plug in x+iy, multiply the numerator and denom by conjugate of the denom then set the real part equal to zero?

[deft wraith]

Yeah that's what I mean by I can do the algebra but I don't get how they affect it geometrically

[glad zinc]

Honestly mate, I don't know, my class is starting this content in a couple lessons and my teacher gave me a question so I could get started early and my research from thats all I know

[deft wraith]

yeah no worries

what a levels are you doing?

[glad zinc]

Math, FM, CS, Econ

This is the question $Let ( z_1 = a + bi ) and ( z_2 = c + di ). Determine the locus of ( z_3 = \frac{z_1}{z_2} ) as ( c ) varies.$

[gilded halo]

Bitwise
Compile Error! Click the reaction for more information.
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[glad zinc]

Oops

[ornate blade]

you both British btw?

[glad zinc]

Not British nor live in Britain, just go to a british curriculum school

[high goblet]

i'd probably just recommend going

BMO1 -> BMO2/IMO easy -> IMO/ISL

do u have the BMO1 book?

forgot the name of it

[ornate blade]

cause LY is British so I was just asking, cheers m8

[glad zinc]

IMO easy is on the level of BMO2??

[high goblet]

basically i just worked through all the BMO1 problems from 1996 - current and that worked for me

and then i just started doing problems i enjoyed

[high goblet]

the harder end of BMO2 is comparable to IMO medium

[deft wraith]

The primer? yes i've been working through it

[high goblet]

i.e. something like BMO2 2022 Q3 is probably ~IMO medium

(so p2/p5 standard)

[deft wraith]

i've been doing around 4-6 questions a day

[high goblet]

yeah u'd be surprised, IMO easy questions are a lot easier than u mightthink

[high goblet]

yh yh fairs

[high goblet]

yeah i'd suggest working through the entire book

once ur done with that work through the entire BMO2 book

and that should certainly be enough for camps

[deft wraith]

what about theory specific i.e mont

[glad zinc]

So if I got merit in BMO2 what does that mean

[high goblet]

mont?

[deft wraith]

modern olympiad number theory

it's quite good

[high goblet]

oh right

with theory i think it's just something u'll pick up over time

[deft wraith]

i'm trying to brush up on number theory for itnerviews too

[high goblet]

as in like u'll do problems

often with BMO1 questions u can learn the theory bit by bit

[glad zinc]

@high goblet What was your best placement in BMO2?

[high goblet]

but like yeah ofc u'd also want to develop theory alongside

EGMO is a very good book

[deft wraith]

egmo is a bit hard i've been avoiding geo

[high goblet]

(not european girl's maths olympiad lol, euclidean geo book)

[deft wraith]

i've heard no euclid geo in interviews so like

[high goblet]

basically everyone in like trincamp will have a copy

[high goblet]

yeah i really dislike geometry too

[glad zinc]

?

[deft wraith]

i can sort of bash around/ just do bmo1 geo anyway without much but for bmo2 i mght use it like post interview

to prep for bmo2

[high goblet]

sorry i'll reply to u in a bit lol too many msgs

[high goblet]

but like the tactic for if u dislike geometry is basically to learn a bit of synthetic geo

[glad zinc]

no worries just thought you didnt see

[high goblet]

and then once u've learnt enough u can then often bash the end of the problem

[high goblet]

NT doesn't really come up on interviews

where r u applying out of interest?

[deft wraith]

trinity, cambridge

i heard they give oly style so yeah

[glad zinc]

Why trinity?

[ornate blade]

this chat is so British my water turned into tea

[high goblet]

basically NT is quite oly-specific and spec-specific (i.e. we did 0 NT in our school) so any NT you get in interviews will basically be approachable without having to know i.e. FLT etc.

[high goblet]

ahaha

[high goblet]

hopefully i shall be seeing u next year then!

[high goblet]

yeah the interview qs they do are quite olympiad-y

[deft wraith]

i was going to ask where to find mechanics questions of suitable difficulty?

[high goblet]

the question u find online are probably slightly easier than the ones u'll get

[ornate blade]

Armand from KoreanEnglishman goes to Cambridge
apparently this other guy from Cambridge saw him around campus a few times

[deft wraith]

yeah i'm not too comfortable with orders/ flt but things like being really good with mod arithmatic is something i'm focusing on

[ornate blade]

sorry for derailing chat

[high goblet]

https://www.trin.cam.ac.uk/subjects/mathematics/

[high goblet]

what do they study?

[high goblet]

just don't do mechanics lol

on my test, there were 9 non-mech questions and 1 mechanics question

looked at the mechanics question and went 'nope'

[ornate blade]

history and politics

[high goblet]

so i did all 9 other qs

[ornate blade]

oh damn they're doing Tripos then

right

[high goblet]

unfortunately i don't now the history and politics ppl too well

[high goblet]

but as a serious answer, probably just STEP and trinity's sample tests?

[ornate blade]

also Cambridge is insane for not allowing resits
I heard some really bad things about students who have a medical emergency and the administration has no mercy

[deft wraith]

to solve diophantine equations as well mainly

easy ones

[deft wraith]

okay

i've done the sample ones

it's only like 1 or 2

[high goblet]

we're not meant to share our interview questions lol

[deft wraith]

there's this really weird sample question as well let me show you

[deft wraith]

yep i know

[high goblet]

but i can DM u like my rough memory of my interview questions if u want them

[deft wraith]

sure that'd be nice

[high goblet]

i personally don't have any medical issues but i've heard that u should be able to get compromises for medical emergencies?

[deft wraith]

this second part was weird in explaining it

[high goblet]

but yeah unfortunately it is just 1 exam at the end of the year

[deft wraith]

i don't know if it's really a good question either it's just algebra

[ornate blade]

https://www.varsity.co.uk/features/24163

[high goblet]

(you also get ur ranking in the year so ig only 1 exam means u can have an official ranking)

[ornate blade]

yeah it's this elite uni culture

[high goblet]

ooof didn't know abt that

[ornate blade]

impressive

I found something you don't know about your uni

[high goblet]

yeah the rankings are a bit outdated

but if u do well it's kinda nice knowing ur ranking lol

10th/259 :)

[ornate blade]

jesus

man like people are cracked

[glad zinc]

for what exam?

[high goblet]

remind me roughly what merit is?

[high goblet]

cambridge 1st year exams

[glad zinc]

6+

[high goblet]

ah right

the easier questions for BMO2 are comparable to the harder questions for BMO1

but yeah if u want easy IMO easy problems

[glad zinc]

Yeah, geometry always fucks me over no matter how easy though

[high goblet]

oh ahaha it's fine

i suck at geo too

[high goblet]

i am doxing myself lol but 20/40 in 2022, 35/40 in 2023

[glad zinc]

A month until BMO1, I want to qualify for BMO2 this time, should I spam BMO1 papers? Or do something else to prepare

[deft wraith]

did you alpha farm

is there alpha and beta system in ia

[glad zinc]

"i suck at geo" gets 35/40

[high goblet]

yeah i tried to do as many alphas as possible

[high goblet]

i improved at geo in Y13 but in Y12 i really sucked

i did 0 geometry questions in BMO1 and BMO2 in Y12

[glad zinc]

But y13 results dont even matter no?

Its after uni apps

[ornate blade]

https://www.reddit.com/r/cambridge_uni/comments/1fq8i9j/hot_water_and_house_keeping_at_downing/

[high goblet]

and then i proceeded to do a further 0 geometry questions in TST1,2,3 and 4

[ornate blade]

this is interesting, talking about accom

[deft wraith]

for fun / if you want to make a team / camp

[high goblet]

yh true but it's for team

[glad zinc]

Not british : (

[high goblet]

huh interesting

[glad zinc]

And my country has a different weird system

[high goblet]

will u have spent 3 years in the UK educational system?

[glad zinc]

Well yes but its a school in the UK

not

[high goblet]

oh wait no they don't allow non-UK schools

do u hold a british passport?

[glad zinc]

Nope

[high goblet]

oh unfortunate

is there any country u can represent?

[glad zinc]

The country Im originally from, which I dont happen to live in and they have a weird system with little information online for qualifying for olympiads

sooo yeah

tough

I guess Ill just do some undergrad olympiads

[high goblet]

oh yeah it's a bit mean lol but yeah the d/dx are operators

[high goblet]

oof unfortunate

[deft wraith]

it feels like just weird notation over anything else i don;t think it's hard per se

what even is an operator

[glad zinc]

And olympiads would be very helpful for getting into the job I want to do so it hurts even more : (

[ornate blade]

which job

[high goblet]

have u done this problem?

it surprisingly doesn't require much theory

[glad zinc]

quant trader or more generally a quant job

[high goblet]

and if u do it ur smarter than 1/2 the chinese team and most of the south korean team

[ornate blade]

interesting

I was at a Jane Street event just today

[ornate blade]

like you clearly have the passion and engagement for maths

[deft wraith]

i have heard many annoyances about this problem haha

[ornate blade]

it could be something small such as tutoring secondary school students

[high goblet]

yeah it's like when you expand, you apply the d/dx

[glad zinc]

Am I missing something because part i and ii seem easy?

[ornate blade]

basically they said the interview questions will be about getting you to explain your thinking, like Google or big tech if you know about those

[deft wraith]

it's easy it's just weird

[glad zinc]

I see

[deft wraith]

like explaining bit