AMC
AIME
trigonometry
trig identities
Law of Cosines
geometry
power of a point

I already posted this elsewhere on this board, but I'm going to recap in this thread.

Basically, 1-15 were easier than usual, 16-20 were harder, and 21-25 were about the same.  I think there will be more people qualifying for AIME but fewer with scores above 130.

JB

I think it might've been easier. Based on my range from various practice tests - my score on the actual test was toward the upper end.

I agree with you jackb...harder questions in the middle while the ends were easier than usual.

Well yeah I agree, #1-15 was easier, #16-20 were harder, #21-25 were about the same (#24 was a bit disappointing).

Now I really need some redemption because I was so much slower than expected, off by 12 minutes!!! from answer every problem. Not to mention I got #4 wrong.

I'd say 1-16 were easier than usual, 17-22 are harder than usual, 23-25 are about right.

1-15 were pretty easy yeah. And #16 hard? I thought that was fairly trivial. Anyway 17 and 19 were indeed hard. And #21-25 were hard enough, but they seemed slightly easier than usual (unless I was taking harder previous exams or something)

[quote="Mildorf"]My sources have indicated that it was much easier to qualify for AIME with a low 100+ than usual. (TJ will have about twice as many as normal I am told, with lots of people scoring just over 100.)[/quote]
yeah...
my school had a record of highest qualifiers as well...most of which (except 3) were lower than 110

What kinda ballpark figures are we talking for qualifiers?

While taking the AMC, I breezed through 1-16 without much difficulty. Then I looked at 17.... ooooooommmmmmmmmmmgggggggggggggggggggggggggg!!!!!
The first six out of the last nine problems were generally much tougher than normal. Of course, that excludes exceptions such as 18 and 21. Fortunately, that's exactly what I did: pick off the easier ones and then go back and do the others (including 17).

um...
i don't see why people are complaining so much about 17....wasn't just a simple Power of a Point ?

the test itself was easier than usual, except for the geometry...especially that forsaken 4th page.

the later questions were easier than usual, but 17 was tricky...serves people right for assuming things, how do you do it with power of a point, i used pythaogrean on AFE and law of cosines on ABE

[quote="amirhtlusa"]solution with POP for 17:
extend AD to hit the circle and G
using POP on A we have:
$AF^2=AD\cdot AG=s(s+r\sqrt{2})\Rightarrow 9+5\sqrt{2}=s^2+rs\sqrt{2}\Rightarrow s=3,r=\frac{5}{3}$ so $\boxed {B}$[/quote]

That is pretty slick. There's a quick Law of Cosines solution too (since we have a handy 135 degree angle). I thought it was medium difficulty overall, and that's all I have to say about it...

I did 17 with twice pythagorean theorem with hypotenuse $AE$.
$\sqrt{9+5\sqrt2}^2 + r^2 = AE^2 = (s+\frac{r}{\sqrt2})^2 + (\frac{r}{\sqrt2})^2$

The only tricky part I thought was setting the parts equal since $r$ and $s$ are rational (that simplifies to $9\sqrt2 + 10 = s^2\sqrt2 + 2rs$ so yeah.

I thought it was significantly easier than any past ones I've taken.  Maybe difficult to find [i]nice[/i] solutions for some, but even the last few could be bruted quite painlessly.  25 tho I bruted for some time and came up with $exactly$ 277, so I figured it had to be right.  But you get that if it had been "S contains no number less [u]or equal to[/u] k  :(

The AMC exam is a little different each year.  I'm not sure if this year's was easier or harder than usual.  However, I do feel certain that more students than ever before are very well prepared for the exam.  I'm seeing dozens of students who have taken the AMC or AIME Problem Series at AoPS reporting scores of 125 to 150, with several 150s.  Perhaps this is just an indication that there was always a certain amount of untapped potential within the American student population.  I certainly believe that.

Hopefully, instead of the top students at 50 or so schools around the nation getting good preparation, that total is expanding quickly.  This means that we should expect more students to find the problems easier -- and we should expect more high scores -- even if the test remains the same in difficulty.

While it's clear that I have vested interest in suggesting that providing resources for top students achieves results, I think the evidence is beginning to speak for itself.  The number of prepared students will probably increase at a greater absolute rate moving forward.


How many questions one has to ans to crack ISI BStat or BMath entrance test?
What types of questioins are asked in the interveiw????  :|

[quote="Aravind Srinivas L"]The questions are very easy. I dunno how many will be there but they are very simple.[/quote]
I Know that questions are very easy as u say....
But the interveiw is not as easy as the written test.
there are usually 8 to 10 questions to be answered in 2 hours in the short answer type test which is actually the 2nd phase v. :)
In the 1st phase there are 30 MCQ questions to be answered in 2 hours........  :lol:
and I ve no idea about the interveiw....  :(


Find all positive integers $n$ such that $3^{n}\mid 5^{n}-2$

No, this is obviously not easy, it is really very difficult. It was a conjecture of Nathanson and the proof given in AMM for this is more than complicated. Anyway, as I said in the topic started by Vess (and to which I still wait for some remarks, comments, hints, etc from his part...) this follows immediately from "Vess's estimations", but unfortunately I have no idea how to prove those estimations.

This make me relax... it wasn't my fault if i hadn't be able to solve it...

thx a lot Harazi... p.s. can you give me some links related to the methods you mentioned? thanks in advance   :)

Ooops, sorry, my reference was wrong. I attach the article with the proof. I just hope one day (in the next 200 years, of course), Vess will enlighten us with his magical and marvelous proof of his $p$-adic estimations. But this will probably never happen...


Using the map to the right, John is creating a
new map with a scale of 1 cm = 5 km. The
representation of the portion of road between
the two points on the map shown here is
3.75 cm. In centimeters, how long will this
same stretch of road be on John's new map?
[asy]unitsize(0.5inch);
draw((0,0)--(20,0)--(20,50)--(0,50)--cycle);
label("$1 cm = 8 km$",(10,0),N);
draw((5,10)--(18,43));
dot((18,43));
dot((5,10));[/asy]

[quote="GameBot"]Using the map to the right, John is creating a
new map with a scale of 1 cm = 5 km. The
representation of the portion of road between
the two points on the map shown here is
3.75 cm. In centimeters, how long will this
same stretch of road be on John's new map?
[asy]size(250);
draw((0,0)--(20,0)--(20,50)--(0,50)--cycle);
label("$1 cm = 8 km$",(10,0),N);
draw((5,10)--(18,43));
dot((18,43));
dot((5,10));[/asy][/quote]
I just fixed the asy. Im not sure whether the scale is 1:5 or 1:8, but it was 1:8, the real length would be $ 3.75\cdot 8 \equal{} 30$ kilometers. If it was 1:5, then the answer would be $ 3.75\cdot 5\equal{}18.75$ km.

im  not sure if its just me but ftw says its 6

i also got 18.75 originally though

please pm me if im [size=85] i dont mean to revive this old thread [/size]wrong

Yeah, this problem needs to be fixed. Also note that there is a typo at the end, where it asks for the length of the road in [b]centimeters[/b].

I don't think so... the 1cm=8km map means that the length of the road is 30km, and John wants to create a map that has a scale of 1cm=5km so he needs a 6cm line for the 30km road

This problem probably isn't wrong. It says that John is creating a NEW map with a scale of 1 cm to 5 km. On the old map, it was 1 cm to 8 km.
8 * 3.75 = 30 km.
The 30 km means that the road is really 30 km.
The new map scale says that 1 cm is equal to 5 km. 30/5 = 6 cm, which is indeed the answer. I don't understand how you guys got 18.75 or how it shouldn't be centimeters. If it was km, the real road would be HUGE.
Read carefully.


3 squared ends in 9, 3 cubed ends in 7. What digit does 3 to the thirty-fifth power end in?

!Too big for me calculatora!

First, you have to find a pattern in the last numbers.
[hide]
The pattern is 3,9,7,1 thus if you keep on going, you'll keep on getting 3,9,7,13,9,7,13,9,7,13,9,7,13,9,7,13,9,7,13,9,7,1 etc. Thus, The last digit is your 1st number.[/hide]

 :)

[hide] the pattern is 3 9 7 1
and the remainder when you divide 53 by 4 is 1
so, 3[/hide]


[quote="stergiu"][quote="Nick Rapanos"][quote="stergiu"]Για να μη μου ξεφύγει καμιά ειδική περίπτωση σε αυτό που θα γράψω, θέλω και τη γνώμη σας :

 Αν τέσσερα σημεία έχουν (το καθένα)  ίσες  δυνάμεις ως προς τρεις  κύκλους , τι συμπεραίνουμε  για τα σημεία και για τους κύκλους;

 Ευχαριστώ - Μπάμπης[/quote]

To kathe simeio exei ises dunameis os pros tous treis kuklous, ara einai to riziko kentro ton trion kuklon.ara ola ta simeia subiptoun me to riziko kentro ton trion.

auto tha elega ego an den me "ponireuate" ...[/quote]

Νίκο, και ο πιο δύσκολος χρησμός της Πυθείας , είναι παιγνιδάκι μπροστά σε αυτό το ωραίο σου μήνυμα  :) . Και γω δυστυχώς αυτό σκέφτηκα, αλλά το ζητούμενο της άσκησης με γλύτωσε(πρόκειται για μια απόδειξη στην ευθεία Gauss - Newton στο πλήρες τετράπλευρο).
 Περιμένω λοιπον τη γνώμη και των άλλων φίλων και αν θέλεις επανέρχεσαι στο τέλος !

 Μπάμπης[/quote]

okay...vsk den itan "xrismos" to parapano apla itan statement! tora an enoeitai tin periptosi pou oi kukloi sxhmatizoun desmh, einai sxetika gnosto (asxeta dld me to an se ponirepsei kapoios i oxi...uparxei sta palia vivlia geometrias pou den exo mazi mou)-alla kai pali den allazei tpt me tin ennoia oti tote exoun koino riziko aksona kai ola ta simeia tou boroun na theorithoun rizika kentra...(xoraei kai ligh parapano eksigisi...alla prota prepei na mou peite an theoreite tis dunameis me prosimo i xoris.)

Νίκο, το ερώτημα προέκυψε από μια απόδειξη στην πρόταση ότι τα μέσα των διαγωνίων πλήρους τετραπλεύρου είναι συνευθεικά σημεία και πως τα ορθόκεντρα των τριγώνων του βρισκονται σε ευθεία κάθετη με την ευθεία των μέσων. Σε αυτή την απόδειξη θεωρούμε τρεις κύκλους με διακέντρους κατάλληλα τμήματα, οπότε τα ορθόκεντρα έχουν ίσες δυνάμεις ως τους τρεις αυτούς κύκλους. Από κει προκύπτει ότι και τα μέσα είναι συνευθειακά και τα ορθόκεντρα. Δεν είχα ξαναδεί αυτή την απόδειξη και προβληματίστηκα για τη σχέση των κύκλων, των δυνάμεων κλπ.Ανέτρεξα μετά στh γεωεμετρία του Johnson για τις δέσμες κύκλων κλπ. Ζήτησα όμως τις απόψεις σας μήπως και κάποιος έχει ασχοληθεί πιο επισταμένα με αυτό το θέμα και μας γνωστοποιήσει κάποια πρόταση κλπ που απλοποιεί τα πράγματα και απότελεί και μια μέθοδο συγγραμμικότητας , πέραν αυτής του ριζικού άξονα.

   Καλό απόγευμα - Μπάμπης


[hide]She will have $\left(\frac12\right)\left(\frac23\right)\left(\frac34\right)=\frac14$ of her money left.

If $\frac14$ is $\$7777$, then she earned $4\times\$7777$ dollars over 12 months.  Thus her monthly allowance is $\$\frac{7777}3$.[/hide]

Umm, that is a typo.  It is really $\frac{7777}3$  :blush:

[hide][quote="john lai"]Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had 7777 dollars left. Assuming she only gets money by allowance, how much money does she earn every month?
 :D[/quote]

7777*4/3*3/2*2 /12 (for months)
7777/3
dollars[/hide]


Find all functions F : R --> R satisfying :
$f(xf(y)) = f(xy) + x$
for all real x,y .

[quote="Myth"][quote="kvmedya"][quote="kueh"]y=0 shows surjectivity (everything has a preimage)
x=1 shows that f(k) = k + 1.[/quote]

Your $solution$ is completely wrong.  ;)[/quote]
Your $remark$ is completely wrong. ;)[/quote]
Could you please $explain$ it for me ? ;)

What should I explain? kueh's solution is correct.


SALAM
ie nafar ie source khub vase ajsame aflatuni moarrefi mikone?

khob chi mikhai rajebeshoon bedooni...!? :maybe:
midoonam too ketabe esbat ye chizai neveshte rajebeshoon,ya masalan mitooni ye negahi be in bendazi:
[url]http://mathworld.wolfram.com/PlatonicSolid.html[/url]

[quote="parinaz.sa"]yadame tuye yeki az shomarehaye majaleye riaziate fatemi ye maghaleye khub  darbareye ajsame aflatuni bud vali nemidoonam daghighan kodum shomarash bud ,dar zemn mathworld aslan ettelaate kameli nadare har matlabo faghat dar hadde yek ta'arife mokhtasare gofte va kamel hich matlabiyo tozih nadade albate shayad baraye in mozu khub bashe vali dar kol man har mozuii niga kardam az tush aslan kamel nabud va faghat baraye ashnayi moghadamati bud!!![/quote]

salam
age lotf koni bebini kodum shomareie majale on maghalaro dashte mamnun misham :oops:

man majalehaye riaziatamo dada be kasi va nadarameshunke niga konam va begam ke che shomareiish hast sharmande!


I have two questions,

1)  In a styrofoam cup calorimeter, if only one cup is used (instead of two) will the heat capacity of the calorimeter increase, decrease or stay the same?
I said decrease since it has less insulation, but I am not sure.

2)  I have a 3 part question where I find the first two parts and express my answer using the correct sig figs, but in a third part I have to use the first two values to calculate another value.  Do I use the unrounded numbers or the rounded ones?

Thanks

1. Right

2. I believe you are technically supposed to use only the sig figs of the previous answer, but being a math person, I usually don't round 4, 5, and 6's cuz it seems like too much.. but you can't get points off for using sig figs properly, so...


I have heard and read that UChicago arguably has one of the top five math departments in the country, and that their Honors Analysis course is superb. But does the hefty core curriculum at UChicago distract undergrad math majors from their math studies? Does it restrict them to less math classes than a math major at a different college would be able to take?

The department is top 5.... but how would the [b]undergraduate[/b] mathematical experience compare? Would it also be top 5 or top 10? Does the core take away from the math experience?

I'd appreciate any opinions.

I applied to Chicago (I'm not there though).  While they have a great, great department, the core is pretty rigorous, and their classes are very hard.  While I doubt that the core "distracts" math majors, it is something you will have to work around.

I am quite biased, but sonsider Johns Hopkins.  The department is very strong, there is no core curriculum, and it just as rigorous as Chicago.

[quote="math312"]I have heard and read that UChicago arguably has one of the top five math departments in the country, and that their Honors Analysis course is superb. But does the hefty core curriculum at UChicago distract undergrad math majors from their math studies? Does it restrict them to less math classes than a math major at a different college would be able to take?[/quote]

If you have a strong foundation (read, write, and think well), the core is more than manageable, provided that you go to class, do most of the readings, and turn in all your papers. The typical core class requires no more than 5-10 hours a week outside of class.

I personally don't think that the core detracts from the math experience. I think it provides a good balance (and a healthy break every now and then). I know people who take grad classes their 3rd year and resent the core that they still need to complete. I would simply advise that you choose your core wisely (talk to upperclassmen), choose your profs wisely (very important), and get as much core done ASAP so that you can focus on your major later on.

[quote="math312"] The department is top 5.... but how would the [b]undergraduate[/b] mathematical experience compare? Would it also be top 5 or top 10? Does the core take away from the math experience?[/quote]

I wouldn't have it any other way. I've had professors e-mail me back at 4am, in response to my questions, instructors sit w/me for 3+ hours outside of class on a Friday afternoon to go over homework/class. Students tend to work in groups -- so if there is suffering, at least you get to suffer with others and chuckle about it.

I kind of have a similar concern.

At a state school, I'd be taking 5 math classes in my first semester, and be finished with my degree in about 2 years because of all the AP credit they give. (granted, yeah, I know how bad an idea that would be).

On a typical Chicago math major's schedule, how many math classes does that person take?

If it is $ /ge3$, that'd be pretty decent. It is not that I'm against the idea of the core, I actually really like it as an idea. However, you have to consider how much you're paying to take Core courses.

I mean with a state school, I'm getting paid to go and live there, I'm doing exactly what I want. I'm getting done ridiculously fast. I mean sure, it isn't a UChicago education, but does that mean alot if I get a 4.0 at a state school, versus a 2.8 at Chicago because of the Core?

I dunno, these concerns are really bothering me. It kind of applies to a bunch of other situations as well, I'm sure.

Is a mediocre GPA at a good school really that much better than a great GPA at a mediocre school? These concerns are really going to affect my decisions as I look at college.

I think you should to take into account not only the number of math classes a math major takes, but also factors like the quality of the instruction, how demanding the math class is, the math opportunities outside of class, etc. Just to give you an example, you can get a dose of topology, algebra, and measure theory in a calculus class here.  But I think there are a lot of both public and private schools with good programs -- much more than the names that pop up on this site -- so you should choose the one that's right for you. Nine quarter-courses of hum/civ/sosc/arts may not be what fits you.

To answer your question though, I'll give you a sample, approximate math major's schedule, which maximizes the number of math classes. Note that we're on the quarter system, that there are a maximum of 4 classes per quarter, and that and students usually take 3 quarters a year (the 4th often being devoted to being at home, a math REU, or some other alternative).

1st year: 3 quarter-classes in math.
2nd year: 3-6 quarter-classes in math.
3rd year: 8-12 quarter-classes in math.
4th year: 12 quarter-classes in math.

This would be if you're going for a BA. Remember that this is just a sample, and there's a chance I made a mistake regarding some requirements. You can look into the core at http://collegeadmissions.uchicago.edu/level3.asp?id=396 and the math degrees at http://math.uchicago.edu/undergraduate/concentration.shtml#math.

Other things you may want to be aware of:
- a joint BA or BS / MS degree, which you can do if you place high enough as an incoming first year
- directed reading program (DRP), a year-round or at least 3-quarter-round program where you study a topic with a grad student 1on1, or something like that, during the quarter.

/rant


Guys, it's Feb. now and you just have a couple of months to go before JEE.
I would like all of you, esp. newcomers, to post your name, hometown, school, favourite subjects.
Most importantly I'd like you to post which college you wanna go to, which course, and why you want to do that.

[hide]My name: Aditya G Date :D
Hometown: Chennai, Velachery
College: Indian Institute of Technology, Kharagpur
Department:Physics and Meteorology (5 Yr. Integrated Msc.)
Areas of interest: Physics, Programming in C/C++ and AVR, Image Processing using MATLAB & OpenCV.
To put it in simple words, I love physics and robotix.

What I wanna do in life: Do research in Physics from India or abroad(preferably :P ).

It's nice to see so many guys with a high tempo for learning above JEE level stuff. Hope each of you ends up in your dream college or course!

[/hide]

Name- Gaurav Shukla
Hometown- Lucknow
College- Massachusetts Institute of Technology
UC Berkley
IIT
IISc Bangalore
Dept.- Depending on wether in India(Only B.Tech/B.Sc) or US(Integrated M.Sc) I'd like to pursue a course in nuclear physics. If not then whatever closes possible in the field of physics. If not even that, then computer sciences.
Interests- Physics, specially Modern Physics, Astronomy(Origin & fate of the Universe), Origin of life, genetics, solving good probs and algorithms, writing poetry, stories, philosophy, psychology, music( I love performing with my band!), rebellions! and last but certainy not the least... GIRLS!  :P
Aim of life- Understanding who I am, developing nuclear fusion as an economical, efficient and unlimited energy source(hence my name), fulfilling my interests and  answering the questions that eat me away and changing the socio-political condition of India plus a few other things  :lol:

I really apreciate what you wrote in your aim :omighty:

u didnt leave ne stone unturned,man....go on!!! :lol:  :D

What about the rest of the junta?
Guys, come on. This is the time that members are most active( atleast in my year).
You can have a look at what happened last year.

[hide]And also where they finally ended up in. :P [/hide]

[quote="adityagdate"]Nice to see that you're passionate about physics. You should go to IIT-Kanpur, it's really good for physics.
OR, if you get to go abroad, then it's even better.

[hide] :D IIT's are planning to start courses on nuclear science from this year :lol: [/hide][/quote]
WHAT?HOW?WHEN?  :o
How did you come to know! This is ASTONISHING! AMAZING!

It was there in the news sometime ago. Something about IIT's being given the permission to start their own courses without approval of the HRD ministry (thank god :roll:  :!: ).
Anyway, you'll find out more about it in your jee application brochure or your counselling brochure.
I'm not sure that it will happen [b]this[/b] year though :| . It may take some time.
Since your aim is to do nuclear physics, I'd advise you to take either Msc. Physics at IIT-K/KGP or Engineering Physics at IIT-B/M/D/G.

Anyone else interested in basic science??

[hide]I am probably the only one from last years batch to pursue it in college :( [/hide]

What exactly are you doing bhaiya?
P.S. You are quite a mysterious person.....hidden content in every post!  :rotfl:

[quote="adityagdate"][hide]I am probably the only one from last years batch to pursue it in college :( [/hide][/quote]

Soumya too..as long as u include maths in science :P


limit
function
algebra
floor function
calculus
calculus computations

Evaluate

$\lim_{x\to 0}\left( \left[\frac{n\sin(x )}{x}\right] + \left[\frac{n\tan(x )}{x}\right] \right).$

where [ ] is the greatest integer function and n is an integer

[Moderator edit: I gave you your bigger brackets.]

$sinx\sim x$ when $x\to 0$ same as $tanx\sim x$ when $x\to 0$,
so our limit is equal to $2n$

That would be true if there were no integer part. ;)

since $\frac{sin(x)}{x} \leq 1$ and $\frac{tan(x)}{x} \geq 1$ you get $\lim_{x \to 0} \left( \left[ \frac{n\sin(x )}{x}\right] + \left[\frac{n\tan(x )}{x}\right ] \right) = (n-1)+n=2n-1$


how many $(x1,x2,x3)$ is a positive number so that $x1+2x2+3x3=n$ (n>=4)

Sorry Thanhliem but your post is not an easy read. Below is the rephrase:
How many triples $(x_1,x_2,x_3)$ of positive numbers such that $x_1+2x_2+3x_3=n$, $n\geq4$?

And the subject title is nice too  :D

Pierre.

This one is relatively easy and not so nice but anyway: (i think that positive integers means not including 0)

Denote by f(n) the number of solutions to the equation,
Denote by g(m) the number of solutions to x+2y=m, x,y>0.

g(2n+1)=n, since we can have x=2n-1,2n-3...1.
g(2n)=n-1, since we can have x=2n-2,2n-4...2.

Obviusly, the number of solutions x1+2x2+3x3=n=g(n-3)+g(n-6) ... + g(t), where t=1,2 or 3.

if n=3k,
f(6k)=g(6k-3)+g(6k-6)...+g(3)=(g(3)+g(9)+g(15)...+g(6k-3))+(g(6)+g(12)...+g(6k-6))=(1+4+7...+(3k-2))+(2+5..+(3k-4))=$\frac{3\cdot k^2-k}2+\frac{3k^2-5k+6}2=3k^2-3k+3$.
Similarly consider f(6k+1), etc.
Sorry for any computational errors i may have made :P  :blush:


The inequality is false for say $(a,b,c) = (1000,1,1)$.

Maybe there is another condition that pvthuan forgot to mention like: $abc=1$, $a+b+c=1$,...etc

[quote="Maverick"]Maybe there is another condition that pvthuan forgot to mention.[/quote]
Indeed, you are right.

If $x,y,z> 0$ and $yz+zx+xy=7$, then

$\frac1{(1+y)(1+z)}+\frac1{(1+z)(1+x)}+\frac1{(1+x)(1+y)}\leq \frac3{(1+\sqrt[3]{xyz})^{2}}$.

Thanks, Thuan.

If $x,y,z$ be positive real numbers such that $x+y+z=5$, then

$\frac1{(1+y)(1+z)}+\frac1{(1+z)(1+x)}+\frac1{(1+x)(1+y)}\leq \frac3{(1+\sqrt[3]{xyz})^{2}}$.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
AMC 8
geometry

5 Round Tournament:

The exact format has not yet been decided, but the tournament will have 5 rounds, a AMC 8, Mathcounts, AMC 10, AMC 12, AIME round. Each round has 10-15 problems (10,12, or 15)

I welcome your participation. More details as scheduled.

I will join!

Will this be by advancement (i.e people with a certain score will move to the next round)?
Also, will we have to give solutions or only answers?

\join

hopefully this will be as successful as AIMEs 10 round tourney

You need solutions.

The scoring and exact format will be announced later.

I decided it would be more convenient and better organized for 3 rounds; 1 MC/AMC 8, 1 AMC 10/12, 1 AIME Be aware: I may add a proof round
[hide="Format"][hide="Round 1"]This is Mathcounts/AMC 8 level.
1. Easy AMC 8
2. Medium AMC 8
3. Hard AMC 8
4-5. School MC
6-7. Chapter MC
8-11. State MC
12-15. National MC[/hide]
[hide="Round 2"]This is AMC 10/12 level
1. Easy AMC 10
2. Easy AMC 12
3. Medium Easy AMC 10
4. Medium Easy AMC 12
5. Medium AMC 10
6. Medium AMC 12
7. Medium Hard AMC 10
8. Medium Hard AMC 12
9. Hard AMC 10
10. Hard AMC 12[/hide]
[hide="Round 3"]This is AIME level
1-3. Easy AIME
4-8. Medium AIME
9-10. Hard AIME[/hide][/hide]
[hide="Rules"]You must submit a full solution for full credit on a problem. You are not able to look up answers/problem, as these are made up.[/hide]
[hide="Submitting solutions"]\pm me the solutions. Please do not put them in a \pdf.[/hide]

Scoring/deadlines will be posted in another post. I'm having a problem with space here.

[hide="Signups"]Canton, PowerOfPi, Thunder365, joshisunni16[/hide]

Any questions, \pm me.

I join, but I would fail on the last two(particularly AIME)

in[hide=" "]Please pm me to remind me about this:P [/hide]

You'll get a few days to work on each problem. EDIT: :mad: ERRR I mean each set of problems

Also I am a little lenient on the solutions on the harder problems.

I am not a naturally harsh person :D

[hide="Scoring"]A single answer gets no more than 1 point. The max. points varies for some problems. Each max. per problem is listed in ()[/hide]
[hide="Format"][hide="Round 1"]1-2. Easy AMC 8 (2)
3-4. Medium Easy AMC 8 (2)
5-6. Medium AMC 8 (3)
7-8. Medium Hard AMC 8 (4)
9-10. Hard AMC 8 (4)[/hide]
[hide="Round 2"]1-2. School MC (2)
3-4. Chapter MC (3)
5-7. State MC (4)
8-10. National MC (6)[/hide]
[hide="Round 3/4"]Round 3 is AMC 10 and Round 4 AMC 12. So 1. AMC 10/12 means Round 3: 1. AMC 10 and Round 4: 1. AMC 12
1-2. Easy AMC 10/12 (3)
3-4. Medium Easy AMC 10/12 (4)
5-6. Medium AMC 10/12 (5)
7-8. Medium Hard AMC 10/12 (6)
9-10. Hard AMC 10/12 (8)[/hide]
[hide="Round 5"]1-2. Easy AIME (3)
3-4. Medium Easy AIME (4)
5-7. Medium AIME (6)
8-9. Medium Hard AIME (8)
10. Hard AIME! (10)[/hide][/hide]
[hide="Advancement"]Qualification for:
[hide="Round 1"]You sign up[/hide]
[hide="Round 2"]Score 20+ on Round 1[/hide]
[hide="Round 3/4"]Score 25+ on Round 2 (subject to change)[/hide]
[hide="Round 5"]Score 40%R3+60%R4$ \ge$35[/hide][/hide]
[hide="Disclaimer"]AwesomeToad is not liable for any physical or emotional damage to yourself as a result of entering this competition[/hide]

[hide="Signups"]Canton, PowerOfPi, Thunder365, joshisunni16, cgyao15, Maybach, King![/hide]
After you sign up, please copy the signup list with your username in there, thanks.

I think each because otherwise it would be impossible to qualify for Round 2.

[quote="PowerOfPi"]For example, in round 1, where you put "1-2. Easy AMC 8 (2) ", does that mean 2 points for each question or two points total for both questions?[/quote]

2 points per question sorry if I didn't make that clear

[quote="Maybach"]I think he's going to grade it over the weekend.......[/quote]

Yes, possibly today.

Sorry I know you guys are really impatient...

I seriously don't want to make things worse, but I thought more people would submit solutions round 1...

Some of them haven't even responded...

I [b]may[/b] cancel tourney, considering the mock AMC 10/AIME and Mock Mathcounts.

I know you guys probably don't trust me much to run something anymore, but I learned to make it REALLY clear so that people are COMITTED before signup unless otherwise specified.

Thanks, and I'm really sorry :(

[quote="SonyWii"]Don't stop it!

You gotta be stern! You gotta be like, YO, SIGNUPS AND THE DEADLINE HAS PASSED SO TOO BAD.

That's the way life is man - life isn't all great.

But whatever you, don't bother stopping it now. You just gotta be strong, and not be a softie. That's the way people survive.
 :weightlift:  :weightlift:  :weightlift:[/quote]

OK, OK, OK.

But I do have Mock AMC 10/12/AIME also. Not alone... but...

Ill Join even though it probably means that i am going to get embarrassed.

It already started, so I think you are too late. Try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311120]this[/url].

so... when is the next round going to b posted??

CANCELLED?!?!?!?!? WHY?  :mad:  :(  :spider:

[quote="Maybach"][/quote][quote="Maybach"][/quote][quote="Maybach"][quote="AwesomeToad"][quote="SonyWii"]Seeing as the deadline passed, can you announce how many people passed?[/quote]

Sorry, for now, this is canceled :(   I would like to give credits to the people who quit :mad: or didn't submit solutions :mad:[/quote]
That's why.....[/quote][quote="Maybach"][/quote][quote="Maybach"][/quote]
That isn't a reason... :|

[quote="SonyWii"]OH man.

that really stinks. my first real tourney and it stops. bleh.

Oh well, it was fun... i suppose.[/quote]

As I said, I was sorry to disappoint those who DID sign up and DID the problems. (includes you)

EDIT: @AIME15, I [b]am[/b] putting effort into Mock AMC, as you may see if you \pm "summai" and ask.

Y'know, advertising that on a thread that shows you're not going to put effort into your mock competitions might not be a good idea.


category theory
abstract algebra
superior algebra
superior algebra unsolved

If $ R, S$ are rings with unity, and $ {_R}M, {_R}N_{S}, P_S$ are modules, show that

$ Hom_{S}(N, P) \otimes_{R} M \cong Hom_{S}(Hom_{R}(M, N), P)$.

It's supposed to be a group isomorphism.

It's not true.  Take $ P\equal{}0$, for example, then it says $ M\cong 0$ for all $ M$... :maybe:

My mistake.. sorry. It's really a tensor product.

Presumably, then the map going right takes $ f \otimes m$ to $ g$, where $ g(h) \equal{} (f \circ h)(m)$. Not sure on the inverse map...

I tried this one, but then I can't figure out how to construct an inverse,
because it seems I need to arbitrarily choose an element of $ M$.

Another possibility would be to show that the right side satisfies the
universal property of the tensor product. Couldn't do that either.

Still false?  $ M\equal{}\mathbb{Z}/2$, $ N\equal{}P\equal{}\mathbb{Z}$.

Was this what you were thinking of?

$ Hom_{S}(N\otimes_R M, P) \cong Hom_{S}(N, Hom_{R}(M, P))$

The statement I've got is really that one, so I think there's
a problem with my source.

Anyway, thanks for the replies.  :)

I just found the problem.

It's missing an hypothesis: M is a finitely generated free module.

That's a rather enormous hypothesis... please take more care in posting your problem next time.

In that case, the (finite) direct sum distributes over everything and it follows from the case $ M\equal{}R$. (it's false for an infinitely generated free module because distributivity doesn't work on the right side)


geometry
ratio
trigonometry
angle bisector
trig identities
Law of Cosines

All right guys, here's a pearl:
You have a triangle ABC with <B=120. You draw the angle bisectors from angles A,B, and C. You connect all three points where these angle bisectors hit the opposite side to make a triangle. Prove that the angle created in this triangle, opposite <ABC, is 90 degrees.

Let AD,BE,CF are the three bisectors of ABC.
I is a point on AC such that AD=DI=AI
Let show that DE is the bisector of angle ADC
                    DF is the bisector of angle ADB :D
This is my fovorite problem when I'm in grade 7. :)

Another hint: using the ratio with the bisector. :lol:

Extend BC to G so ABG is equilateral. Then $ ACG\sim DCB$, so $ \frac {AG}{GB \plus{} BC} \equal{} \frac {BD}{BC} \implies \frac {AB}{AB \plus{} BC} \equal{} \frac {BD}{BC} \implies BD \equal{} \frac {AB \cdot BC}{AB \plus{} BC}$
Also, by the angle bisector theorem, $ \frac {BE}{AB} \equal{} \frac {BC \minus{} BE}{AC} \implies BE \equal{} \frac {AB \cdot BC}{AB \plus{} AC}$
Similarly, $ BF \equal{} \frac {AB \cdot BC}{BC \plus{} AC}$
By the Law of cosines,
$ FE^2 \equal{} FB^2 \plus{} BE^2 \plus{} FB \cdot FE$
$ FD^2 \equal{} FB^2 \plus{} BD^2 \minus{} FB \cdot BD$
$ DE^2 \equal{} BE^2 \plus{} BD^2 \minus{} BE \cdot BD$
FDE is right only if $ FD^2 \plus{} DE^2 \equal{} EF^2$, only if $ 2BD^2 \equal{} FB \cdot BE \plus{} BE \cdot FE \plus{} FB \cdot EF$, which can be proven by the previous formulas and one more law of cosines on the whole triangle.

[quote="dgreenb801"]Can someone post a full solution?[/quote]$ \frac{{AD}}{{DC}} \equal{} \frac{{DI}}{{DC}} \equal{} \frac{{BA}}{{BC}} \equal{} \frac{{AE}}{{EC}}$
 $ \Rightarrow$DE is the bisector of angle ADC :D

Oh I see, you made A the 120 degree angle. Nice proof, much better than mine!

Yeah, I've taken A=120.That is my mistake.Sorry :blush: .But the solution is still right. :) .


A set of three nonnegative integers $\{x, y, z \}$ with $x<y<z$ is called historic if $\{z-y, y-x\}=\{1776,2001\}$. Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.

If a number is $ 0,1,$ or $ 2\mod 9$ then say it is the "x" part of a historic set. If it is $ 3,4,$ or $ 5\mod 9$ then say it is the "y" part of a historic set. Finally, if it is $ 6,7,$ or $ 8\mod 9$ then say it is the "z" part of a historic set. Then every nonnegative integer is part of a historic set, because if $ x\equiv 0,1,2\mod 9$ then $ y\equal{}2001\plus{}x\equiv 3,4,5\mod 9$, and $ z\equal{}1776\plus{}y\equiv 6,7,8\mod 9$.

tjhance, there is an error in your solution. Your statements imply that numbers like 3,4, and 5 are in y, but for these, there exist no nonnegative x such that x+1776=3,4, or 5. The same applies for z, and for a couple hundred other numbers (all those y under 2001 or those z under 2001+1776).

btw...the owner of this username is not the one who is posting this (aka it's not the guy you were punching at ARML).

Drat. I thought it said all integers.

I thought that seemed suspiciously easy to not be solved already...

EDIT- On second thought, is this even possible? 0 and 1776 must both be the "x" part of a historic set in the partition, because if they were y or z, then the x part would be less than 0.

But then 2001+1776 would be the "z" part of the historic set with x=0, but also the "y" part of the historic set with x=1776, a contradiction.

Maybe I am misunderstanding again?

EDIT EDIT- Oh. You could either have z-y = 1776 and y-x = 2001 OR z-y=2001 and y-x=1776. I can you can just ignore all of my posts.

Set $ a\equal{}1776$ and $ b\equal{}2001$.

We will partition the nonnegative integers into sets by assigning each number as an x, y, or z using the following algorithm: Choose the smallest unassigned integer (call it $ q$) and assign it as an x, and assign $ q\plus{}a\plus{}b$ as a z. If $ q\plus{}a$ is unassigned, then assign it as a $ y$, otherwise assign $ q\plus{}b$ as a y. Either way, we will have made a historic set $ \{q,q\plus{}a,q\plus{}a\plus{}b\}$ or $ \{q,q\plus{}b,q\plus{}a\plus{}b\}$. I will show that this algorithm is always possible to complete.

All numbers that are currently assigned to $ x$ must be less than $ q$. Therefore, $ q\plus{}a\plus{}b$ will be unassigned, so we can let it be $ z$. To show that it is possible to assign y to some value, I will show that $ q\plus{}b$ must currently be unassigned. It cannot be an x, because it is not less than $ q$. It cannot be a y, because then its associated x value would not be less than $ q$. If it were a z, then $ q\plus{}b\minus{}(a\plus{}b)\equal{}q\minus{}a$ must be its associated x value. However, when we assigned $ q\minus{}a$ to x, we should have assigned $ (q\minus{}a)\plus{}a\equal{}q$ as its y, a contradiction because we are assuming $ q$ is unassigned.


For a positive integer $ n$:
Let $ C_n\equal{}\frac{1}{n\plus{}1}\binom{2n}{n}$

And $ S_n\equal{}C_1\plus{}C_2\plus{}...\plus{}C_n$

Prove that $ S_n\equiv1(mod3)$ if and only if there exists a $ 2$ in the base $ 3$ expansion of $ n\plus{}1$.

Anyone can do this? I have a solution, but I want to see another, please, try it.


Find the remainder when $ 4^{101}$ is divided by $ 101$

hello, note that $ 4^{25}\equiv\minus{}1 \mod 101$, hence $ 4^{101}\equiv 4 \mod 101$.
Sonnhard.

Or we could use the Fermat-Euler theorem.

$ a^{p \minus{} 1} \equiv 1 \bmod{p}$ where p is prime and a is coprime to p.

Since 101 is prime:

$ 4^{100} \equiv 1 \bmod{100}$

$ 4^{101} \equiv 4 \bmod{101}$

Can it be done using binomial theorem.??

I dunno, but since $ 101$ is prime you can directly use [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url]:
$ 4^{101}\equiv 4 \pmod {101}$...


MATHCOUNTS
number theory
Diophantine equation

Target round problem 6

What is the highest schore that you cannot get on the dart board shown if you may use as many darts as you wish?  There is an inner circle with a score of 9 and a concentric outer circle with a score of 4.

how would you do it?

theres a formula mn-(m+n). so it would be 9*4 - (9+4) giving you 23.

The main point of the problem is as follows:
If you start checking early cases, you will eventually find that you can make 24 (4+4+4+4+4+4), 25 (9+4+4+4+4), 26 (9+9+4+4), and 27(9+9+9). From then on, you can make every value from adding a 4 to any of these. Then a simple check shows that 23 doesn't work.

And after you've done enough of these problems, instead of needing to guess that starting number, you can use the wellknown result (not surely exactly what its called) that the number is ab-a-b. But the main point is the above.]

Oh, by the way - that formula only works when the two numbers are coprime. If they aren't, then obviously you can only make numbers that are divisible by their greatest common divisor.

[quote="cobra"]what is the formula called?

It would be helpful for me to get to the root of the solution.

Thanks.[/quote]

I'm not sure if there is a serious name for this theorem, but I often see it called the Chicken McNugget Theorem which is how I teach it.

Proving the theorem is a great exercise for top Mathcounts students.  To get started, think of the problem for m and n (instead of the specific case 4 and 9).

We write the Diophantine equation
mx + ny = c.

The idea is to find all the positive integers, c, for which the equation has at least one solution (x, y) in nonnegative integers.

[quote="MCrawford"]
We write the Diophantine equation
mx + ny = c.

The idea is to find all the positive integers, c, for which the equation has at least one solution (x, y) in nonnegative integers.[/quote]

You attempt to make a chart:

 1  I [u]0     1     2     3     4     5[/u]
 2  I 0     m     2m  3m  4m   5m
 3  I n     n+m n+2m  n+3m  n+4m
 4  I 2n   2n+m  2n+2m  2n+3m  2n+4m  2n+5m
 5  I 3n   3n+m ...

That's the best I can do.

You then examine each row's (horizontal) modulo residue.  Obviously the first row is c :equiv: 0 (mod m).
The second row would be c :equiv:  n (mod m)
The thrid row would be c :equiv:  2n (mod m)
4th- c :equiv:  3n (mod m)

Obviously after a certain amount of rows, it will just start to repeat becuase of the modular congruences.  So, we want to go until c :equiv: (m-1)n (mod m) becuase if it were c :equiv: (m)n (mod m), then it would be the same as the first row - it repeats.

Hence, all values c :ge: n(m-1) are attainable.  Now for the final part.  The largest value of c unattainable is the largest modulo residue of c :equiv: (m-1)n (mod m) that is less than n(m-1).  That is because n(m-1) is the lower bound of all attainable values.

The largest c unattainable is then n(m-1) [the bound] - m [mod operation thing, cycles basically]=nm-n-m.

Same thing.

Note that I did not think of this by myself.  I learned it in aops v2 about a month ago, but I'm at least proud of myself for remembering it all after redoing the proof over and over for practice.


number theory
greatest common divisor
number theory unsolved

Show that the equation

$ y^2\equal{}x^3\plus{}xz^4$

has no solutions in nonzero integers.

$ y^2 \equal{} x (x^2 \plus{} z^4)$.  From here, it's easy to see that $ x$ must be a perfect square.  Otherwise, some prime $ p$ divides it a total of odd number of times.  But $ x^2 \plus{} z^4$ only contributes even number of factors of $ p$.  So $ x \equal{} x'^2$ for some integer $ x'$.  Then this reduces to solving $ y'^2 \equal{} x'^4 \plus{} z^4$ where $ y' \equal{} \frac{y}{x}$.  It's well known that this has no nonzero solutions in integers.

[quote="jinjohn"]But $ x^2 \plus{} z^4$ only contributes even number of factors of $ p$.[/quote]

Please clarify a bit more on this statement.

[quote="jinjohn"]Then this reduces to solving $ y'^2 \equal{} x'^4 \plus{} z^4$[/quote]

Please clarify a bit more on this statement.

[quote="jinjohn"]It's well known that this has no nonzero solutions in integers.[/quote]

Please consider ignorant people like me. Maybe a link perhaps...

[quote="Akashnil"][quote="jinjohn"]It's well known that this has no nonzero solutions in integers.[/quote]

Please consider ignorant people like me. Maybe a link perhaps...[/quote]

It's the statement that proves the $ n \equal{} 4$ case of Fermat's Last Theorem and was known to Fermat.
The proof is by infinite descent.

Oh.. Thank you t0rajir0u. Is it complicated?

[quote="jinjohn"]But $ x^2 \plus{} z^4$ only contributes even number of factors of $ p$.[/quote]

I understand it now. Consider $ (x,z) \equal{} 1$.if $ p|x$ then $ p\not |x^2 \plus{} z^4$. And $ 0$ is an even number.
EDIT1: Sorry. I am wrong.
EDIT2: suppose $ j,k$ be the respective highest powers of $ p$ dividing $ x,z$. Let $ j > 0$. then highest power of $ p$ in $ (x^2 \plus{} z^4)$ is $ \min(2j,4k)$ which is even.

[quote="jinjohn"]Then this reduces to solving $ y'^2 \equal{} x'^4 \plus{} z^4$[/quote]

I realized it will be $ y' \equal{} \frac {y}{x'}$ instead of $ \frac {y}{x}$.

[quote="t0rajir0u"][quote="jinjohn"]$ x^2 \plus{} z^4$ only contributes even number of factors of $ p$.  [/quote]

$ x \equal{} z \equal{} 1, p \equal{} 2$?
$ x \equal{} z \equal{} 2, p \equal{} 5$?
$ x \equal{} z \equal{} 4, p \equal{} 17$?[/quote]

My $ p$ divides $ x$.

[quote="jinjohn"][quote="t0rajir0u"][quote="jinjohn"]$ x^2 \plus{} z^4$ only contributes even number of factors of $ p$.  [/quote][/quote][/quote]

To make things clearer, my $ p$ divides $ x$ total of odd number of times.  This means $ x \equal{} p^{2r \plus{} 1}m$ for some nonnegative integer $ r$ and gcd($ m$,$ p$) $ \equal{} 1$.

Suppose $ z \equal{} p^{t}v$ where gcd($ v$,$ p$) $ \equal{} 1$.  Then $ x^2 \plus{} z^4 \equal{} p^{4r \plus{} 2}m^2 \plus{} p^{4t}v^4$.  Now the point is that $ 4r \plus{} 2$ and $ 4t$ are never equal.  So $ x^2 \plus{} z^4$ contributes min($ 4r \plus{} 2$, $ 4t$) number of factors of $ p$, which is even.


summer program
PROMYS
Mathcamp
Ross Mathematics Program

List all the numbers between 0 to 600 that can't be expressed as the sum of two or more consecutive integers.
i.e.   10=1+2+3+4

[quote="paladin8"]Just FYI, a problem like this is on the PROMYS application...[/quote]
And on the Mathcamp application.  And on the Ross application.  I wonder who came up with it first.   :D

AND my [url=http://www.artofproblemsolving.com/Resources/Papers/LaurendiPartitions.pdf]paper on partitions[/url] deals with a similar, more general problem (example 3).  I would say that this is quite a famous problem.  When I put it in my paper I hadn't seen the Mathcamp or PROMYS applications.  Rather, I thought it would be interesting to generalize a problem that I saw on a local competition.


calculus
derivative
calculus computations

How many tangent lines to the curve y = x\(x+1) pass throught the point (1,2)? At which point do these tangent lines touch the curve?

Take the derivative and plug it into the formula for your specific point.

No, don't. Sketch a picture; there are two points on the line that matter, and only one is on the curve. Call that point $ Q$; you want to find all $ Q$ such that the tangent line at $ Q$ passes through $ (1,2)$.


Consider the diofantine equation $ 1 + x + x^2 + \ldots + x^a = y^b$ where $ (x,y,a,b)\in \mathbb b{Z}^4$ are integers with $ \min\{a,b,|x|\} > 1$.

1) Find all solution in the case $ 2 \mid b$.

2) Find all solution in the case $ 4 \mid a + 1$.

(A corollary [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=286622]here[/url])

(a) We consider two cases.
Case 1: $ a$ is even.
If $ x>1$ then $ x^a  < \sum\limits_{i \equal{} 0}^a {x^i }  < (x \plus{} 1)^a .$
If $ x<\minus{}1$, set $ t\equal{}\minus{}x$. Then $ (t \minus{} 1)^a  < \frac{{t^{a \plus{} 1}  \plus{} 1}}{{t \plus{} 1}} < t^a$
Thus, this case doesn't work.
Case 2: $ a$ is odd.
If $ x$ is even then $ x^{\frac{{a \plus{} 1}}{2}}  \plus{} 1 \equal{} u^2 ;\frac{{x^{\frac{{a \plus{} 1}}{2}}  \minus{} 1}}{{x \minus{} 1}} \equal{} v^2.$
If $ (a\plus{}1)/2$ is odd then $ 2|a\minus{}1$. Hence
$ x^{\frac{{a \minus{} 1}}{2}}  < \frac{{x^{\frac{{a \plus{} 1}}{2}}  \minus{} 1}}{{x \minus{} 1}} \equal{} v^2  < (x \plus{} 1)^{\frac{{a \minus{} 1}}{2}} .$
Thus $ (a\plus{}1)/2$ is even. That gives $ x^{\frac{{a \plus{} 1}}{4}}  \plus{} 1 \equal{} t^2.$ Thus $ u^2\minus{}1\equal{}(t^2\minus{}1)^2\equal{}>u\equal{}t\equal{}\minus{}\plus{}1.$
IF $ x$ is odd then $ x^{\frac{{a \plus{} 1}}{2}}  \plus{} 1 \equal{} 2u^2 ;x^{\frac{{a \plus{} 1}}{4}}  \plus{} 1 \equal{} 2t^2.$
Thus $ 2u^2\minus{}1\equal{}(2t^2\minus{}1)^2\equal{}>u^2\minus{}(t^2\minus{}1)^2\equal{}t^4$=>exist $ a;b\in Z$ such that $ ab\equal{}t^2;a^2\equal{}u^2\minus{}t^2\plus{}1;b^2\equal{}u^2\plus{}t^2\minus{}1.$
So $ b^2\minus{}a^2\equal{}2(a^2b^2\minus{}1)\equal{}>a\equal{}b\equal{}1.$
Anyone has solution for part b?
Another problem: Find $ P(x);Q(x)\in Z[x]$ and $ a;b\in N$ such that: $ 1\plus{}P(x)\plus{}(P(x))^2\plus{}...\plus{}(P(x))^a\equal{}(Q(x))^b$ .

[quote="tronghieu"](a) We consider two cases. Case 1: $ a$ is even.
If $ x > 1$ then $ x^a < \sum\limits_{i \equal{} 0}^a {x^i } < (x \plus{} 1)^a .$[/quote]

And so? lhs must be a b-power. Idem for the case x<-1.
I'm have not already checked your second part, but a olympic solution for part (b) exists  :wink:

Well...for the 2nd part we have(we suppose that $ b$ is odd since the first part is solved).
$ \frac {x^{a \plus{} 1} \minus{} 1}{x \minus{} 1} \equal{} y^{b},4|a \plus{} 1 \equal{} 4k \implies \frac {x^{k} \minus{} 1}{x \minus{} 1}.(x^{k} \plus{} 1).(x^{2k} \plus{} 1) \equal{} y^{b}$,
there r 2 cases:
1-$ x$ is even: then $ x^{k} \plus{} 1,x^{2k} \plus{} 1,\frac {x^{k} \minus{} 1}{x \minus{} 1}$ are mutually coprimes so it implies that all of these numbers are the $ b$-th power of some integers:$ x^{k} \plus{} 1 \equal{} s^{b},x^{2k} \plus{} 1 \equal{} q^{b},\frac {x^{k} \minus{} 1}{x \minus{} 1} \equal{} t^{b}$.We get
$ (s^{b} \minus{} 1)^2 \plus{} 1 \equal{} q^{b}$. but u can easily see that this expression can be put between two consecutive $ b$-th power integers(namely $ (s^2)^{b},(s^2 \minus{} 1)^{b})$.Note that i didn't consider trivial cases.
2-$ x$ is odd: then the gcd of any two of these numbers($ x^{k} \plus{} 1,x^{2k} \plus{} 1,\frac {x^{k} \minus{} 1}{x \minus{} 1}$) will be a power of $ 2$.
Let $ v_2(k) \equal{} \alpha,2^{\alpha}k' \equal{} k, \alpha \geq 0$.then we have the following equalities:
$ x^{2^{\alpha \plus{} 1}k'} \plus{} 1 \equal{} 2s_{\alpha \plus{} 1}^{b},x^{2^{\alpha}k'} \plus{} 1 \equal{} 2s_{\alpha}^{b},...,x^{2k'} \plus{} 1 \equal{} 2s_1^{b}$,
and $ x^{k'} \plus{} 1 \equal{} 2s_0^{b},\frac {x^{k'} \minus{} 1}{x \minus{} 1} \equal{} t^{b}$ where $ t$ is odd.
we see that $ (s_{i}^{b} \minus{} 1)^2 \plus{} (s_{i}^{b})^2 \equal{} s_{i\plus{}1}^{b},0 \leq i \leq \alpha$
the solution of the remained case will be posted (if i'll find it :D ).


hi, i just want to know, has anyone ever seen the harry potter puppet pals? i know it sounds lame....and it kind of is....but it's really funny!

there are like 5 videos:
-bothering snape
-trouble at hogwarts
-ppp potions class
-the mysterious ticking noise
-wizard angst

my fav is the mysterious ticking noise

if you want to check it out, go to http://www.potterpuppetpals.com. it's hilarious! :rotfl:

I've seen this. Although I thought that Neil will have thought of a new one by now, since the 7th book came out. Who here thinks that Neil looks like the actor for James Potter in the movie for Harry Potter 5?

i've seen them too.

but i think the mysterious ticking noise is the only one worth watching.

that was stuck in my head for ages...

theres a secret vid too that ill tell u guys how to get to after someone asks?

go to harry potter south park, my friend told it was good...

I already know how to get to the secret video, and it's sorta lame. It's funny, though.

i know about that one. my brother told me but i still havent found it. and theres more than 5. go one youtube and search. theres like bananafone, harry potter idol-lame, and others

There's also
-Ginny
-Mustache Buddies
-the Christmas card
-Ron's Parents
-Snape's Diary

wizard swears is weird.  i don't think anyone mentioned that one yet...

also i think there was one called "albus dumbledore lists your good qualities"
the curtain opened and closed and he didn't say anything.  so i was like :rotfl:


geometry
geometric transformation
geometry proposed
U sviti matematyky

In convex pentagon $ ABCDE$, $ \angle ABC\equal{}\angle AED\equal{}90^{\circ}$ and $ AB\cdot DE\equal{}BC\cdot AE$. Let $ F$ be the intersection of $ CE$ and $ BD$.

Prove that $ AF$ is perpendicular to $ BE$

[hide]Let $ \frac{AB}{BC}\equal{}\frac{AE}{ED}\equal{}k>0$, the point K lies on the angle EAB such that $ AK\perp BE$ and $ AK\equal{}k.BE$.
$ \triangle KAB\sim\triangle EBC\Rightarrow EC\perp BK$
$ \triangle KAE\sim\triangle BED\Rightarrow BD\perp EK$
The altitudes AK,BD,CE of triangle KBE are concurrent, so AK passes through F, hence result.[/hide]

[quote="discredit"]In convex pentagon $ ABCDE$, $ \angle ABC \equal{} \angle AED \equal{} 90^{\circ}$ and $ AB\cdot DE \equal{} BC\cdot AE$. Let $ F$ be the intersection of $ CE$ and $ BD$.

Prove that $ AF$ is perpendicular to $ BE$[/quote]
[hide="Solution"]
Let $ X$ be on $ AC$ and $ Y$ be on $ AD$ so that $ BX\perp AC$ and $ EY\perp AD$.  Let $ BY$ meet $ CE$ at $ N$ and let $ EX$ meet $ BD$ at $ M$.  Also, let $ BE$ meet $ AF$ at $ Z$.  Since $ AB\cdot DE \equal{} BC\cdot AE$, we see that $ \frac {AB}{BC} \equal{} \frac {AE}{ED}$.  Also, $ \angle ABC \equal{} \angle AED\equal{}90$, so $ \triangle ABC\sim \triangle AED$.  Since $ EY$ is an altitude of right $ \triangle AED$, we have that $ \triangle AEY\sim \triangle ADE\sim \triangle ACB$, so a spiral similarity maps $ \triangle ABY\sim \triangle ACE$.  This means that $ \angle ABN \equal{} \angle ACN$, so $ ABCN$ is cyclic.  since $ \angle ABC \equal{} 90$, we have that $ \angle ANC \equal{} 90$.  Similarly, we can find that $ \angle AMD \equal{} 90$.  This means that $ \angle AYE \equal{} 90 \equal{} \angle ANE$, so $ AYNE$ is cyclic.  This implies that $ \angle DAE \equal{} \angle YNF$.  Similarly, $ \angle FMX \equal{} \angle BAC$.  Yet $ \triangle ABC\sim \triangle AED$, so $ \angle BAC \equal{} \angle DAE$, so\[ 180 \minus{} \angle FME \equal{} 180 \minus{} \angle BAC \equal{} 180 \minus{} \angle DAE \equal{} 180 \minus{} \angle FNB \] Yet, since \[ \angle BME\equal{}180\minus{}\angle FME\equal{}180\minus{}\angle FNB\equal{}\angle BNE\] we see that $ BMNE$ is cyclic, so $ \angle FNM \equal{} \angle FBE$.  Also, $ \angle AMF \plus{} \angle ANF \equal{} 90 \plus{} 90 \equal{} 180$, so $ AMFN$ is cyclic, so $ \angle MNF \equal{} \angle MAF$, so $ \angle MAZ \equal{} \angle MBZ$, which means that $ ABMZ$ is cyclic.  We conclude that $ \angle AZB \equal{} \angle AMB \equal{} 90$, so $ AF\perp BE$, as desired. [/hide]

[quote="discredit"]In convex pentagon $ ABCDE$, $ \angle ABC \equal{} \angle AED \equal{} 90^{\circ}$ and $ AB\cdot DE \equal{} BC\cdot AE$. Let $ F$ be the intersection of $ CE$ and $ BD$.

Prove that $ AF$ is perpendicular to $ BE$[/quote]

[color=blue][i]Vittasko had sent in this forum a solution to this difficult problem , based on Jacobi theorem, if I remember well. After some time Huen Vu ( the name is not written in this way- it is in east languages) sent a nice solution  (in three lnes!)with cyclic quadrilaterals. I hope that my friend Vittasko will send us these links. The problem comes from a Russian competition.It was also proposed in the greek subforum , months ago.I'm glad  :) that now we have two more nice solutions - the above ![/i][/color]

[b]NOTE[/b]

[b][color=red]Mmm! I tried a little and I think that the link is this :[/color][/b]

 http://www.mathlinks.ro/viewtopic.php?t=79051

 [u]BAbis[/u]


geometry
perimeter
inequalities
geometry proposed

Given nonisosceles triangle $ ABC$. Consider three segments passing through different vertices of this triangle and bisecting its perimeter. Are the lengths of these segments certainly different?

Call $ E,F$ the points respectively on $ CA,AB$ such that $ BE,CF$ bisect the perimeter, and call $ s$ the semiperimeter.  Clearly $ CE \equal{} s \minus{} a$, $ AE \equal{} s \minus{} c$, and by Stewart,
\[ BE^2 \equal{} \frac {(s \minus{} a)c^2 \plus{} (s \minus{} c)a^2}{b} \minus{} (s \minus{} a)(s \minus{} c),
\]
and similarly for $ CF$.  Therefore,
\[ BE \equal{} CF\leftrightarrow\frac {(s \minus{} a)c^2 \plus{} (s \minus{} c)a^2}{b} \minus{} (s \minus{} a)(s \minus{} c) \equal{} \frac {(s \minus{} a)b^2 \plus{} (s \minus{} b)a^2}{c} \minus{} (s \minus{} a)(s \minus{} b)\leftrightarrow(b \minus{} c)(s \minus{} a)^2(a \plus{} b \plus{} c) \equal{} 0.
\]
Clearly $ a \plus{} b \plus{} c > 0$ and by the triangular inequality $ s \minus{} a > 0$.  Hence two such cevians as described in the problem are equal iff the corresponding sides are equal.  It follows that the answer is yes, the lengths are necesarily different.

No computations! In the notations of the previous post, $ BC \plus{} CE \equal{} CB \plus{} BF \equal{} \frac {1} {2} (AB \plus{} BC \plus{} CA)$, hence $ BF \equal{} CE$. Then, if $ BE \equal{} CF$, triangles $ \Delta BCF$ and $ \Delta CBE$ are congruent, hence $ \angle BCE \equal{} \angle CBF$, so $ \Delta ABC$ is isosceles.


[quote="PepeCamello"]When I solved a problem, I would just write up the answer in Word. I did about a problem or two every weekend when I remembered and finally have typed out all my solutions.
Now, normally, Word saves stuff as .doc. Can I just save it as .rtf and send it in like that? Or do I have to hand-write it?
Also, if I mail the solutions, can I just mail my printed out Word copy or do I have to hand-write it?
If that didn't make sense, ask and I'll try and explain better. Thanks for the help.[/quote]
You do not need to handwrite it.  We can accept RTF files, and we also accept mailed copies of your Word printouts.

One more question:

Am I supposed to receive a confirmation e-mail?
I sent the problems yesterday with gmail and I haven't received anything yet.
I had the subject as Name, ID, round, so I did all that. Was there anything that I missed?
Thanks.

If you have the correct email in your profile on the Forum, you will get a confirmation email within a few days of submitting your solutions.  We have received several hundred papers in the last couple days, so it will take us a while to sift through them all.

Ok, cool.
Thanks for the quick replys on both my questions.


[url=http://tmp.mathideas.org/public/math/mock_aime-2007.4.pdf]Here you go.[/url]

Thanks to [b]paladin8[/b] for editing the problems.

Just a note: it's probably too easy...

PM me your answers sometime before mid-january.

Wow! That was unexpected! I'm going to spend my entire day at school tomorrow doing this :rotfl: . Thanks for taking the time to write an AIME.

i will probably do it, school ends this week for me so yea...

[size=200]Instructions for submitting solutions[/size]

1. PM me and I will PM you back with a special number.

2. Go to http://tmp.mathideas.org/public/math/submit.

3. Login and enter your answers.

The advantage of doing it this is (1) I don't have to do much work :D and (2) all your scores will be entered into a database, so we can calculate cool statistics and stuff. :)

If I submit my answers, can I go back and edit them?  Or do I need to submit them all at once??

You have to submit them all at once... just do it when your 3 hours are finished.

When are the answers due? Because I'm leaving tomorrow morning and I'm going to be gone until next Sunday...

[quote="PenguinIntegral"]For 9 do you want such k so it is a positive integer or so that both sides of the fraction have a common factor?[/quote]

So that both sides of the fraction have a common factor.

the form for # 15 makes no sense... find the least integer less than or equal to ...either you mean greatest integer less than or equal to, or you mean least integer greater than or equal to...

EDIT: yea...i sent you a pm with my answers...i can resubmit them on your site if you want...(pending my questions resolved and i get the password thing)

Yes, sorry that should be greatest integer less than or equal to :blush:

Yeah you don't need to submit them on the site I put it in myself.

Uh mid-january like I said in the first post.

[quote="chess64"]Bump... 7 people that said they were going to do it never submitted their answers.[/quote]
How many have submitted their answers?

holy **** I got the answer for #1 but I typed it wrong...

Submit your answers before Sunday morning. That's when I plan on posting results.

[quote="SplashD"][quote="Altheman"][hide="umm"]i don't think the form for the answer in #11 works[/hide][/quote]

yeah, i got a different answer form as well[/quote]

yes.. i am currently investigating this =D

is #15 also correct in this sentence: Find the [b]least[/b] integer less than or equal...?  :maybe:

least integer greater than

greatest integer less than

It should be greatest integer less than or equal to.

Are you sure the answer to #5 has 3 digits or less? Or am I not understanding the wording?

Yes, I am 100% sure. (They must have _at least_ two consecutive 1's.)

[quote="purple plume"]i hate the aime dude[/quote]
Yo, you otta using others avatars.

How would you compare the difficulty of your contest to that of the yearly AIME?

I personally thought it was much harder than the real deal, but I made a multitude of careless mistakes.

In so many places, I had the thought: "If I remembered AoPS 2, this would be trivial".

I don't think the answer to #11 is in the right format, and I'm suspicious that #15 is not in the right format also.

[quote="beta"]and I'm suspicious that #15 is not in the right format also.[/quote]

It is, you're probably making a false assumption like I was.  :wink:

Okay, what are we supposed to do about #11?  I have the answer in a different form...

Right... results are going to be posted soon and yes #11 was in the wrong form. K81o7, you can just PM me your answers. Like, now =D

Solution document please...

Or someone AoPSWiki up the answers, because I didn't get enough sleep last night...

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3751276_2007_mock_aime_4]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


By considering the Euler Mascheroni constant I was able to 'show' that,
$ \sum_{n = 1}^{\infty}\frac{1}{n}=\sum_{n = 1}^{\infty}\frac{{\zeta}(2n)}{n}$

Now from the analytic continuation of the zeta function ${ {\zeta}(1)}$ in not analytic and hence not zeta regularisable.
So I assume the result I derived is a consequence of Riemann's theorem that a divergence or conditional convergence can be 'summed' to any required value. Is this the case?

Perhaps I ought to elaborate,

$ {{\gamma}_{n}}\equal{}\sum_{r \equal{} 1}^{n}\frac{1}{r}\minus{}\log{n}$
This is the same as,
$ {{\gamma}_{n}}\equal{} 1\plus{}\sum_{r \equal{} 2}^{n}\frac{1}{r}\plus{}\log{(1\minus{}\frac{1}{r})}$

thanks to log rules, now with the MacLaurin expansion of the logarithm(this is absolutely convergent here since r ranges in integers) we have,
$ {{\gamma}_{n}}\equal{} 1\minus{}\sum_{i \equal{} 2}^{{\infty}}\frac{1}{i}\sum_{r \equal{} 2}^{n}\frac{1}{r^{i}}$

Limiting n we get,
$ {\gamma}\equal{} 1\minus{}\sum_{i \equal{} 2}^{{\infty}}\frac{{\zeta}(i)\minus{}1}{i}$

Now consider,
$ log (1\plus{}\frac{1}{r}) \equal{}\sum_{i \equal{} 1}^{{\infty}}\frac{(\minus{}1)^{i\minus{}1}}{ir^{i}}$
summing 1 to n we have,
$ \sum_{r \equal{} 1}^{n}\log{(r\plus{}1)}\minus{}\log{r}\equal{}\log{(n\plus{}1)}\equal{}\sum_{r \equal{} 1}^{n}\frac{1}{r}\plus{}\sum_{i \equal{} 2}^{{\infty}}\sum_{r \equal{} 1}^{n}\frac{(\minus{}1)^{i\minus{}1}}{ir^{i}}$

Now limiting n and using the same technique as in the first case we have,
$ {\gamma}\equal{}\sum_{i \equal{} 2}^{{\infty}}\frac{(\minus{}1)^{i}{\zeta}(i)}{i}$
Equating the two results we are left with the rather baffling,
$ \sum_{i \equal{} 1}^{{\infty}}\frac{1}{i}\equal{}\sum_{i \equal{} 1}^{{\infty}}\frac{{\zeta}(2i)}{i}$

The step where I ignore the 1 when limiting n in the second result may be questionable but we do limit to infinity after all!

They're both divergent. The partial sums of the first are strictly greater than the partial sums of the second. I see no meaningful way that they're equal.

The steps you showed are all fine- it's the rearrangement step after that that doesn't work.

I thought we'd gotten over the divergence bit, equality in this case doesn't mean much in the classical sense but I was looking for a 'sum' of the harmonic series. Which becomes all the more difficult since $ {\zeta}(s)$ has a simple pole at $ s\equal{}1$ and the neat tricks with analytic continuation and hence zeta regularization fail.
But this new $ {\zeta}$ series, which I 'equated' to the harmonic may be more easily 'summable' (though admittedly I doubt it). I was looking for someone else's take on this.


quadratics
number theory solved
number theory

The quadratic $ax^2+bx+c$ has integer coefficients. For some prime $p$ its values are squares for $2p-1$ consecutive integers $x$.
Show that $p$ divides $b^2-4ac$.

I solved this problem. But I want to see your soln :D

I've only thought about the case when $b^2-4ac$ is a quadratic residue of $p$. In this case it reduces to $(x-t)(x+t)$ being a quadratic residue of $p$, where $t$ is an element of $\mathbb{Z}_p$ s.t. $t^2=b^2-4ac$, for all $x\in \mathbb {Z}_p$. This easily reduces to $t=0$. In this case we don't even need $2p-1$ consecutive integers, just $p$. I guess that other thing with $2p-1$ is required for the assumption $b^2-4ac\ne 0,\ b^2-4ac$ isn't a quadratic residue (here I mean $0$ in $\mathbb{Z}_p$).

If $p\nmid a$:

Let $\Delta=a^{-2}\cdot (b^2-4ac)$. We actually get $x^2-\Delta$ is always a quadratic residue or always not a quadratic residue, depending on a.

Let P be the set of quadratic residues and NP be the set of nonquadratic residues. Then $|P|=\frac{p+1}2$ and $|NP|=\frac{p-1}2$.
We have $P+\Delta \subseteq P$[b][1][/b] or $P+\Delta \subseteq NP$[b][1'][/b]. We have $|P+\Delta|=|P|$, because the transaltion is bijective, so (1') is impossible because |NP|<|P|.

So $P=P+\Delta=P+2\Delta=...$. Since $0\in P$ we get $k\cdot\Delta\in P\ \forall k\in \mathbb{Z}_p$. This means $\Delta=0$ in $\mathbb{Z}_p$, so $p|\Delta$.
-----------------------------------------------------------------------------------------
If p|a:
We must prove p|b.
Assume $p\nmid b$. Then f(x)=bx+c is a bijective map of $\mathbb{Z}_p$ onto itself. So it can't take values only in P.

^^Where do you guys learn that advanced math?


geometry
3D geometry
tetrahedron
geometry proposed

$In$ $tetrahedron$ $ABCD$ $angles$ $between$ $edges$ $at$ $apex$ $D$ $have$ $90^{o}$.
$Prove$ $that$ $square$ $of$ $ABC$ $area$ $is$ $equal$ $sum$ $anothers$ $squares$ $of$ $walls$. $Sorry$ $about$ $my$ $english$. $I$ $hope$ $content$ $is$ $comprehensible$

Hmm...this is a well known result. as a adition, you can try to show that:
$AH \perp (BAC)$.


What is the sum of the first 100 positive odd integers?

[hide]100^2=10000[/hide]

the formula for it is x^2 where x is the number of odd integers etc

[hide]10000

Since the sum of the first nth odd numbers equal n^2[/hide]

[hide]N :^2:  is the sum of N amount of odd integers, so 100*100 which is 10000. If you want the sum of N even integers, the formula is N :^2: +N, so the sum of the first 100 even integers in 100*100 + 100 which is 10100.[/hide]

Correcto, for all of you. Nice explanation there MathFiend.

Here's my favorite proof for the sum of odd numbers (looks better if you get the horizontal and vertical scales the same)...

o e o e o e o e o
e e o e o e o e o
o o o e o e o e o
e e e e o e o e o
o o o o o e o e o
e e e e e e o e o
o o o o o o o e o
e e e e e e e e o
o o o o o o o o o

Try to build pictures for the sum of the first n positive integers and for the first n cubes - there are a few cool ones.

I get that the 'o' is odd and the 'e' ios even, but how do you know where to arrange them?

I think it looks better now, si?
Although I have no clue what this is suppose to mean... :)

me either, all I see are a lot of ooooooeeeeeeeeeoooooooooeeeeeeeeoooooooooo's like from the
W of O

...okay...kind of looks like spam, but I'll let it fly...
Mr. Rusczyk's thought are way beyond our comprehension..."All hail the great ones"

Basically what he's saying (I think) is that you should replace the es and os with 1s, and in each little backward L, there will be an odd number of 1s. In the es you will get an even squrae, and in the os you will get an odd square.

agreed, still not getting it at all...where's the backward Ls?

o's and e's not meant to be odd and even - they were just the two characters I knew would be roughly the same size - when I built it with x's and o's, it didn't space right (though that it matched the odd/even squares is a bit of a bonus).

Look at the diagram - how many o's in the upper left corner ([hide]1[/hide]).

It's surrounded by how many e's? ([hide]3[/hide])

Those are surrounded by how many o's ([hide]5[/hide])

What shapes are made when we add each successive set of e's or o's?  (how many total letters are there after we add each successive set)

:o Oooooooh, so each "ripple" is made up of odd components. Each odd number is represented by either an "e" or an "o", and they all form a "backward L." Each ripple with all the previous ripples make a square, thus proving the sum of odd numbers is a square numbers...it's all coming together now...


geometry
geometric transformation
reflection

Hey can anyone tell me the cutoff for NSEA(Sr).I m expecting 91..
Also there r mistakes in Q.5..There r 2 answers as per IAPT..Also in Q.31..Total internal reflection should also take place..Wht will they do in such cases??

i hope u get selected coz i am getting 93

@"VP_91"Any idea about last year's cutoff??

please read post no 99 by advaitvedant (nsep nsec ......discussions)
vedant has already cleared nsea twice(even got a silver in 9th)
he is expecting 85 to be the cutoff
lets hope he is right.........

I cleared it in 12th and even calculated my score. I think the cutoff would be around 100.

@aditya the cutoff u r talking about is out of 180/150
please reply soon

Well... I would like to tell that I talked about the cutoffs.... with Dr. Aniket.... who conducts the inao and camp etc... He told me that they generally hover around 50%.... It was 50% last to last year.....
I am getting 85/150 = 56%.... So quite hopeful..... And since it seems IU am the lowest scorer in mathlnks.... So i dont know why everyone else should worry......

Qualified NSEA :)  :) Congrats 2 all who qualified...

who all are appearing for inao senior.........

reply with you name and class.......

Anurag Sharma X

Can anyone tell me where i can get past INAO papers???I hav only 2008 paper..

cool asish........
u the same one who got 100/100 in both science and maths from bhubhnes..........??????? :P  :D

Which board? ICSE?CBSE? Or the famous KVPY interview one?  :rotfl:

[quote="rituraj007"]perhaps boards... :(  :P[/quote]

crrect :rotfl:  :rotfl:

i am frm ddun.......dehradun..........uk...aka uttarakhand :rotfl:
btw do tel ur intrvew expo........ :P


geometry
perimeter
greatest common divisor
inequalities
number theory
relatively prime
triangle inequality

What is the fewest possible number of units in the perimeter of a triangle with side lengths that are relatively prime integers?

Well, relatively prime means that the Greatest Common Factor of the integers is 1.

So, we could then have an equilateral triangle with side lengths 1, 1, and 1.

After all, GCF (1,1) =1

So the smallest possible perimeter of the triangle is 3.

I don't think that really counts because when considering prime/composite (even though they're relatively prime, same thing) 1 is neither.
In the game, the solution is 12. I sort of experimented with it, not sure if anyone has a better way:
We need to find the smallest group of 3 integers side lengths that satisfy both the Triangle Inequality (sum of two smallest sides is greater than longest side) and are relatively prime. Three consecutive integers would probably make it as small as possible, and would satisfy the Triangle Inequality. However, they need to be relatively prime.
Try: 2,3,4. 2 and 4 are not relatively prime.
       3,4,5. Relatively prime and as small as possible. Sum: 3+4+5=12.

1,1,1 looks fine to me. Not sure why the given answer would be 12.

[quote=jeff10]*AWKWARD REVIVE* I also got 3. I am not sure why the don't allow that.[/quote]
Same Here

[quote=Guam21]How come 2,3,5 doesn't work????
And jeff10 - 1 is not a prime nor composite number[/quote]

2+ 3 = 5, which means the triangles is degenerate.

I am pretty sure its $3-4-5$. Hmm.. now to think about it, I am also not sure of why its not $1-1-1$ :(

No they mean regular triangles, its just that 1 and 1 share no other common factors, but they aren't relatively prime either because they aren't prime or composite, so they don't behave like regular numbers in number theory. I searched it up  :P .

BUgMaster told me they mean right triangles.

Oh ok they should report that problem then.

The "coprimeness" of numbers do not depend on whether a number is prime or composite. Two numbers are defined to be coprime if their GCD is 1.

In this case, $\text{gcd}(1,1)=1$. So 1 and 1 are coprime.

$\boxed3$ should be the correct answer.

https://www.quora.com/Are-1-and-1-coprime

I won't quote the private message, but I believe (will edit in confirmation) BugMaster has added the fact that it is a right triangle to the problem.


Asymptote
USAMTS
analytic geometry
LaTeX

I'm not too good at asymptote, and I don't want to asymptote that picture on USAMTS 2 anyway.
So I'm trying to use includegraphics. But I'm still not sure in what ways it works.

For example, the command I'm using is \includegraphics[viewport=0 0 500 150]{PICTURE.png}
But what do those 4 numbers mean? And is there a better way of using includegraphics than using the viewport thing?

The viewport numbers are llx lly urx ury where ll=left lower, ur=upper right, x,y are coordinates relative to the lower left corner of the bounding box.

If the picture is the right size you don't need viewport at all. For more on includegaphics see [url=ftp://cam.ctan.org/tex-archive/info/epslatex/english/epslatex.pdf]Using Imported Graphics in LaTeX2e[/url] (viewport is in Table 2 on page 15) and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures#Picture_Manipulation]LaTeX:Pictures[/url].


algebra
polynomial
rotation
geometry
geometric transformation
reflection

Heres a nice problem that you can work on..

A polyomino (no I didn't misspell polynomial :-P) is a shape made up of n squares of the same size. I'm having trouble thinking of how to explain it the best, squares can only be connected side to side and they can't be rotated or anything.. hmm.. and they all have to be connected in some way.. argh.

Well you all know what a domino is, theres only one of them. There are two trionimoes, one with three squares in a row and one in an L shape. When we get up to 4 squares, there are 5 possible arrangements (rotations and reflections are counted as the same) - the 'square', 's shape', 'L shape', 4 in a row, and the 'T shape'. Then we get pentominoes, which consist of 5 squares. Etc.

Anyway, hopefully you have an idea of what these polyominoes are. Can anyone explain this better?

The question! Finally! How many polyominoes with n squares are there?

Try it for as many small cases as you want. After you have worked on it for, say, a few hours, you can read the spoiler below. No sooner

[hide]

This problem is unsolved!! Sorry  but it is so amazing that such a very mathematical way of putting squares together has no known formula. And also, just because it is unsolved, don't give up.. it is real fun to work with. Maybe someone here could come up with a formula.. you never know.

[/hide]

Think of them like Tetris pieces, but there aren't necessarily 4 squares connected.

OKay...Tetris pieces with and real number of squares connected to each other, and we're looking for all the possible ways to arrange them??

No wonder theres no formula...

[quote="MithsApprentice"]
No wonder theres no formula...[/quote]

Well there may be, but just no one has found it yet. Similar to the Hardy-Ramanujan formula for partitions.

I worked on this problem quite a bit in high school, but got nowhere on it.

nice,but it's a heavy one. :P
in some papers i have read some reccurence method for enumerating them wich seems the only method for this job,there is a nice job that remove some extras,but the problem is that there is still some repeated polyominoes wich we made from one less order and our duty is to compare them toghether in order to remove the extras.and with big computers we can enumerate 28-ordered polyominoes along 2weeks working.


greatest common divisor
number theory
least common multiple

What is the least common multiple of 6, 8, and 10?

6, 8, 10 gcf is 2
$ \frac62\cdot\frac82\cdot10\equal{}120$

You should explain your answer more clearly, in terms of your process, for example, explain how you divide by two because of the relation between GCF and LCM and involved numbers.

ok. So to find the LCM, you must multiply the most amount of factors for each number. This gives 2*2*2*3*5. This equals [b]120[/b]


function
LaTeX
calculus
calculus computations

[quote="techno-freak"]suppose f(x) is a differentiable function,  g(x) is a doubly diff function   such that f'(x)=g(x)      and  mod(f(x)) <(or equal to)1   for all x E(-3,3)

if further [g(0)]^2 +[f(0)]^2 = 9

prove there exists some c E (-3,3) such that g(c).g''(c)<0[/quote]
I was going through some old posts and found this unsolved problem. I am putting it into LATEX just so that everyone can read it properly.

Suppose $f(x)$ is a differentiable function,  $g(x)$ is a doubly differentiable function   such that $f'(x)=g(x)$ and  $|f(x)| \le 1$ for all $x \in (-3,3).$ If further ${[g(0)]}^{2}+{[f(0)]}^{2}= 9,$ prove there exists some $c \in (-3,3)$ such that $g(c) \cdot g''(c) < 0.$


Is it always possible to find a monochromatic isosceles triangle inscribed in a given circumference coloured with k > 2 colors? (... i assume that k is finite! )

This problem is similar (but I don't think identital) to finding two points (on a line) whose midpoint is of the same color.

I can prove it for $k=2$. (I know that randomized asked about $k>2$, but this is what I have so far.)

Choose 5 evenly spaced points on the circle and lable them (in order) $1$, $2$, $3$, $4$, and $5$.

Let the colors be red and blue.

Assume no two neighboring points are the same color.  Let $1$ be red.  Then, $2$ is blue, $3$ is red, $4$ is blue, and $5$ is red.  But $1$ and $5$ are neighbors and both red, contradicting our assumption.

Let $1$ and $2$ be the neighboring points that are the same color and let that color be red.  In order for no such isoscles triangle to exist, $3$, $4$, and $5$ must be blue.  But then $3$, $4$, and $5$ are form the monochromatic isosceles triangle.

PS: The line problem that I mentioned is not analogous to this circle problem.

Clearly if it is always possible to find a,b,(a+b)/2 of the same color in a coloured interval of the reals then the answer to my question is positive.
I don't know a solution when there are more than two colors.

I was thinking at the following. Suppose that A is a subset of the interval [0,1] such that there are no distinct x,y such that x,y,and (x+y)/2 are all in A (let's call this the midpoint condition). Is it possibile to proof that A has null "measure" or that it is contained in a null measure set, in some measure theoretic sense? If the answer is ''yes'' we can easily derive that there are always a,b,(a+b)/2 of the same color.
For example, if we take the Cantor set it satisfies the midpoint condition (i think!) and has null measure in standard Lebesgue measure. However this seems to be more difficult than the original problem ...!


continued fraction
number theory unsolved
number theory

Looking at the simple continued fraction convergents of $\sqrt{k}$, we can find solutions to the equation $x^{2}-yk^{2}=\pm1$.  For $x^{2}-3y^{2}=1$, we have $x=\frac{(1+\sqrt{2})^{2n}-(1-\sqrt{2})^{2n}}{2\sqrt{2}}$, $y=\frac{(1+\sqrt{2})^{2n-1}-(1-\sqrt{2})^{2n-1}}{2\sqrt{2}}$


Solve the system of equations  :

$ y^2 \plus{} (3x \minus{} 1)^2 \equal{} \sqrt [3]{4x(8x \plus{} 1)} (1) \\
87x^2 \plus{} 2 \equal{} 2y \sqrt {14x \plus{} 1} (2)$


number theory
greatest common divisor
modular arithmetic
relatively prime
number theory proposed

Express in a (very) closed form the sum of the $ m$-th powers of the primitive roots mod $ p$.

$ n$ is a $ m$-th power of a primitive root if and only if n is a k-th power of a primitive root, where $ k = \mbox{gcd}(m,p-1)$, if and only if n has order $ \frac{p-1}k$.
Moreover, if g is a primitive root and $ n = g^{k}$, then $ g^{a}$ is a primitive root iff $ (a,p-1) = 1$, and $ (g^{a})^{k}= g^{k}$ if and only if $ p-1 \mid ak-k$, and since $ k \mid p-1$, if and only if $ a \equiv 1 \pmod{\frac{p-1}k}$. There are $ k$ solutions (mod p-1) of this conguence, if we want a to be relatively prime to k (and therefore to p-1), we have $ \phi(k)$ solutions.

So, the answer is $ \phi(k)$ times the sum of all the elements of order $ \frac{p-1}k$. If $ f(n)$ is the sum of the elements of order n, then we have $ f(1) = 1$, and $ \sum_{d \mid n}f(n) = 0$ for n>1, since the elements whose order divides n are the (distinct) roots of $ x^{n}-1$.
But then, $ f(n) = \mu(n)$, where $ \mu$ is Moebius' function.

So, the final answer is:
The sum of the m-th powers of the primitive roots mod p is $ \phi((m,p-1)) \cdot \mu\left(\frac{p-1}{(m,p-1)}\right)$.
In particular, if m divides p-1, it reduces to $ \phi \left(m \right) \mu \left (\frac{p-1}m\right)$, which is a nice closed form.  :)

Very beautiful! My solution was much uglier, using the inclusion-exclusion principle, but this one is so nice.


Hello all!  I'm looking for people to put problems up so that I can solve them and see if I need more work in Algebra, combo., NT, trig., geometry, or whatever else.  Ranging difficulty is especially nice too.  I've made AIME two years in a row, so I'm about that level, but I haven't made the USAMO cut and I'm from Virginia.  I hope you can help me out on this thread or at least point me in the direction of some formal diagnostic that will help me see and remove my kryptonite.

you might want to do a search of the site. there are a great number of problems posted that you can go back and look at.

In each forum, I believe that there's a 'sticky' about the types of problems that should be posted in that forum with many examples. Look through those and see ones that suit your level--not too easy, but not too too challenging. They should be ones that require you to work. Since you made AIME twice, You should probably go to Intermediate and Pre-olympiad forums.

Good luck with USAMO qualifying.

An easy way to tell is just to examine AIMEs you have taken and what problems you missed. You should have a feel for what problems you're "afraid of" though.

Of course you can always check out the contests section: http://www.mathlinks.ro/Forum/resources.php (click on USA for American competitions). Click on the problem number to get you to the topic of discussion of each problem ;)

Thanks, I was just feeling bad about getting a 3 on a Nyciml.  Thanks for your help.

[quote="yif man12"]If you don't mind some sarcastic and abusive comments, hire Dr. Gregory House.[/quote]

Who's that?

He is a character from House M.D., which is a pretty good television show airing on Tuesday nights in the US.  Unfortunately, I have no idea why someone would write something like that on a thread.


I'm only posting partial results, and only for large schools:

[u]Overall[/u]
[b]1st Place Team:[/b] Spring Valley - Score: 102
[b]2nd Place Team:[/b] Summerville - Score: 101
[b]3rd Place Team:[/b] A.C. Flora - Score: 98

[u]Division 1[/u]
[b]1st Place Team:[/b] Dobbyns Bennett
[b]2nd Place Team:[/b] Wando
[b]3rd Place Team:[/b] A.C. Flora
[b]1st Place Individual:[/b] Me - Score: 20
[b]2nd Place Individual:[/b] ? - Score: 19
[b]3rd Place Individual:[/b] ? - Score: 18

[u]Division 2[/u]
[b]1st Place Team:[/b] J. T. Hoggard - Score: 67
[b]2nd Place Team:[/b] Summerville
[b]3rd Place Team:[/b] ?
[b]1st Place Individual:[/b] John Berman - Score: 25
[b]2nd Place Individual:[/b] Joe Laurendi - Score: 24
[b]3rd Place Individual:[/b] Sunni Xi - Score: 23

I edited some of the stuff you had ? marks by.

I don't see any edits.  Get on AIM and we'll talk.  Also, add team scores if you know them.

Coastal was a very dirty competition.  I won't point fingers, but much cheating occurred, including use of calculators on the test!  What's more, proctors didn't care!

EDIT: No Shobhit, I was talking about a senior from some out of state school who used a calculator on the LEVEL 1 TEST.  If I had a calculator I could have gotten a perfect score...

[quote="arkmastermind"]div 2, 3rd place: spring valley.

and sean i think your team's overall scorew was 98? not sure.[/quote]
hey Asif, I've heard some things about your team, not you obviously, but about other people on your team ;)  ;)

Hoggard got 67. I don't know anything else you don't already have.

I will get the results from Mr R. as soon as he receives them, and then I will post them here.  Until then, does anyone know who got 3rd place in division 2?

EDIT: Congrats to jb05!  I just learned he is none other than John Berman, the only perfect scorer at CCU this year!


logarithms
induction
integration
geometric series
calculus
calculus computations
Summation

Prove that : \[\sum_{k=0}^{\infty} \binom{2k+p}{k} x^k = \frac{2^p}{\left(1+\sqrt{1-4x}\right)^p\sqrt{1-4x}}.\]

OK, let us start with the easy part: case $ p \equal{} 0$.

Calling $ f_{0}(x)$ the given sum when $ p \equal{} 0$, we may write
\[ \frac{df_{0}}{dx}\equal{}\sum_{k\equal{}1}^{\infty}2\left(2k\minus{}1\right)\binom{2k\minus{}2}{k\minus{}1}x^{k\minus{}1}\equal{}2f_{0}(x)\plus{}4x\sum_{k'\equal{}0}^{\infty}k'\binom{2k'}{k'}x^{k'\minus{}1}\equal{}2f_{0}(x)\plus{}4x\frac{df_{0}}{dx},\]
where we have performed the variable change $ k' \equal{} k\minus{}1$.  Now, it is obvious that $ f_{0}(0) \equal{}\binom{0}{0}\equal{} 1$, or the solution to the resulting separable differential equation $ \frac{1}{f_{0}(x)}\frac{df_{0}(x)}{dx}\equal{}\frac{2}{1\minus{}4x}$ is $ f_{0}(x) \equal{}f_{0}(0)\cdot\exp\left(\minus{}\frac{1}{2}\ln\frac{1\minus{}4x}{1}\right) \equal{}\frac{1}{\sqrt{1\minus{}4x}}$, qed.  I guess the best form of solving the general case for arbitrary $ p$ will require possibly induction, and most probably a geometric series, although I have not worked out the details yet.  More later... unless somebody's faster than me!

From the [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x)#2.15]Mellin transform[/url] $ \int_0^\infty \frac{1}{p} \left(\frac{2}{1+\sqrt{1+4x}}\right)^p \, x^{s-1}\, dx=\Gamma(s)\cdot\frac{\Gamma(p-2s)}{\Gamma(p-s+1)} $

it follows $ \sum_{k=0}^\infty \frac{\Gamma(p+2k)}{\Gamma(p+k+1)}\, \frac{(-x)^k}{k!}=\frac{1}{p} \left(\frac{2}{1+\sqrt{1+4x}}\right)^p $ by [url=http://mathworld.wolfram.com/RamanujansMasterTheorem.html]Ramanujan's Master Theorem[/url].

Replace $ x $ by $ -x $ , then differentiate with respect to $ x $ :

$ \sum_{k=1}^\infty \frac{\Gamma(p+2k)}{\Gamma(p+k+1)}\cdot\frac{x^{k-1}}{(k-1)!}=\frac{2^{p+1}}{\sqrt{1-4x}\cdot\left(1+\sqrt{1-4x}\right)^{p+1}} $

After that replace $ p $ by $ p-1 $ and move the index $ k $ to $ k+1 $ :

$ \sum_{k=0}^\infty \frac{\Gamma(p+2k+1)}{\Gamma(p+k+1)}\cdot\frac{x^k}{k!}=\frac{2^p}{\sqrt{1-4x} \cdot \left(1+\sqrt{1-4x}\right)^p} $


let <n> denotes the integer closest to sqrt(n). Show that sum( n = 1 to infinity) [2^(<n>) + 2^(-<n>)] / [2^n] = 3


Let $ n$ be a fixed natural number $ n \geq 1$ . Find $ a_0 , a_{kl} \in \mathbb{R} \text{ with } k,l \equal{} \overline{1,n} ,  k>l$ with the propriety that the equality is true :

\[ \frac{\sin^2{nx}}{\sin^2{x}} \equal{} a_0 \plus{} \sum_{1 \leq l < k \leq n}{a_{kl} \cdot \cos{2(k\minus{}l)x}}\] ,  $ \forall x \in \mathbb{R}\minus{}\{m \pi | m \in \mathbb{Z}\} .$

[quote="turcas_c"]Let $ n$ be a fixed natural number $ n \geq 1$ . Find $ a_0 , a_{kl} \in \mathbb{R} \text{ with } k,l \equal{} \overline{1,n} , k > l$ with the propriety that the equality is true :
\[ \frac {\sin^2{nx}}{\sin^2{x}} \equal{} a_0 \plus{} \sum_{1 \leq l < k \leq n}{a_{kl} \cdot \cos{2(k \minus{} l)x}}
\]
,  $ \forall x \in \mathbb{R} \minus{} \{m \pi | m \in \mathbb{Z}\} .$[/quote]

Ok ,  [hide="here  is a proof"] :
[b] It is easy to compute the sums : [/b]
$ 1\ \blacktriangleright\ \boxed{ S_1\equal{}\sum_{j\equal{}1}^n{\cos{2jx}} \equal{} \frac{\sin{nx}\cdot \cos{(n\plus{}1)x} }{\sin{x}} }$
$ 2\ \blacktriangleright\ \boxed{S_2\equal{}\sum_{j\equal{}1}^n{\sin{2jx}} \equal{} \frac{ \sin{nx}\cdot \sin{(n\plus{}1)x} }{\sin{x} }}$ .
[color=darkred] It is obvious that $ S_1^2\plus{}S_2^2 \equal{} \left( \frac{\sin{nx}}{\sin{x}} \right)^2$ .

On the other hand , we have
 $ S_1^2\plus{}S_2^2 \equal{} (\cos{2x} \plus{} \dots \plus{} \cos{2nx})^2 \plus{} (\sin{2x}\plus{} \dots \plus{} \sin{2nx})^2 \equal{}  n \plus{} \sum_{1 \leq l < k \leq n}{(\cos{2kx}\cos{2lx}\plus{} \sin{2kx}\sin{2lx})} \equal{} n \plus{} \sum_{1\leq l < k \leq n}{\cos{2(k\minus{}l)x}}$ .

 Therefore we have the identity :

\[ \boxed{\left( \frac{\sin{nx}}{\sin{x}} \right)^2 \equal{}  n \plus{} \sum_{1\leq l < k \leq n}{\cos{2(k\minus{}l)x}} }\]  So ,  $ a_0 \equal{} n$ and $ a_{kl}\equal{}2$ , where $ 1 \leq l <k \leq n$ . [/color]

  [/hide]


inequalities
algebra
polynomial
inequalities proposed

Let a,b,c be positive reals such that $a+b+c=1$.
Find min of
 \[ \sum_{cyc}\frac{a}{\sqrt{a+b}} \]

I think there 's no min value of this expression. But the [b]inf[/b] value of it is 1.
Because  $S= \sum_{cyc}\frac{a}{\sqrt{a+b}} \geq \sum_{cyc}\frac{a}{\sqrt{a+b+c}}=1$
First, let b tend to 0, then $S \to \sqrt{a} + \frac{c}{\sqrt{c+a}}$ , then let a tend to 0 , c tend to 1 then $S \to 1$
By the way, can you find max of $S$,  I found it difficult!

[quote]By the way, can you find max of [S] , I found it difficult![/quote]
No, I cannot. Does anyone have any ideas?

I found that it was posted before in Mathlinks:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lovely+inequality&t=1459

Thank you. Let me give you one more problem as a gift.
Let a,b,c be positive reals such that $a^2+b^2+c^2=1$ .
Find min of

$\sum_{cyc}\frac{a}{\sqrt{a+b}}$

$\sum_{cyc}\frac{x}{\sqrt{x+y}}\geq\sqrt{x+y+z}\geq\sqrt[4]{x^{2}+y^{2}+z^{2}}$,

with equality if $x=1,y=z\rightarrow0$.

The infimum is still $1$.

By the way, the following inequality is true and not trivial:

If $x,y,z$ be positive reals, then

$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\geq\sqrt[4]{\frac{27(yz+zx+xy)}{4}}$,

with equality if and only if $x=y=z$.

The following is even better

\[\frac{x}{\sqrt{5x+4y}}+\frac{y}{\sqrt{5y+4z}}+\frac{z}{\sqrt{5z+4x}}\ge 1\]

if $x,y,z\ge 0$ and $xy+yz+zx=3$.

[quote="Ji Chen"]

If $x,y,z$ be positive reals, then

$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\geq\sqrt[4]{\frac{27(yz+zx+xy)}{4}}$,

with equality if and only if $x=y=z$.[/quote]
[quote="GacKiem"]
\[\frac{x}{\sqrt{5x+4y}}+\frac{y}{\sqrt{5y+4z}}+\frac{z}{\sqrt{5z+4x}}\ge 1 \]
if $x,y,z\ge 0$ and $xy+yz+zx=3$.[/quote]
Wow! :ninja:  :maybe:


Merhaba Arkadaşlar,

Bence Türkiyenin Forum sayfasının da bir moderatöre ihtiyacı var.Gerçekten diğer ülkelerin moderatörleri var. Türkiye'mizin mod. un olmaması ayıp doğrusu.Lütfen birisi moderatörlük için başvuruda bulunsun...
Saygılarımal,

ihsan YÜCEL

 :?:  :?:  :?:

moderatörlük için kime başvuru yapılacak acaba ben yapabilirim moderatörlük...
valentin e mi ?

ben de yaparım

ve moderator ben seçildim


Ich finde das Article sehr "accurate" und also es ist lustig wie ein Pferd speaking German.

(Yes, I butchered that sentence on purpose)

German is an .. interesting language. My friend speaks it to me like it's english, so I have to put up with his butchered sentences as he tries to demonstrate the improper use of the German language.

[quote="PiDeltaPhi"]Hey liyi  :D
Lets consider german vs chinese
Let A be the average time which is required by some average foreign student to learn german
Let B  be the average time which is required by some average foreign student to learn chinese
A/B=1/20  :D
This post is directed to those who think german is difficult to learn  :cool:  ;)[/quote]

Depends... For english people or other europeans that may be true

But for other asian people? I'm sot quite sure on that...

German isn't that hard to learn if you know english.  Its hard to learn to write it and speak it but I have been studying the laguage off and on for the past year and can read it fairly well.

[quote="confuted"][quote="BHorseMath"]ihr sprechen German Confuted?

(gah my spelling is worse in German than in English please excuse any spelling mistakes)[/quote]
Ich hab' drei Jahr Deutsch, aber ich kein mehr Deutsch lernen. Ich geh' jetzt zum Universitt fr Physik und Elektrotechnik.[/quote]

Well I appreciate people who are English/American native speaker and try to learn foreign languages as they know all other people learn their language. Congrats to your silver medal at IMO and efforts to learn German. Just some minor grammar errors but I understand you pretty well. I don't think German is too hard. Latin is much more heavy on grammar. I think English can be learnt easily too. But I feel Russian is easier than English but unfortunately I have just little practise right now. So you tend to forget things.  I think Chinese is hard. Grammar is easy. But for Europeans it is hard to pronounce the words. For me several words just sound too similar or the same for me. But pronouncing it wrongly you say something completely different which turn out to be very embarrassing. Keep learning German, confuted.  :)

He isnt Matt Ince. Altough that would be cool. :)

I still think I should have won a gold medal at IMO, but no, I'm not Matt Ince, and I actually didn't even make it to AIME. ;) I've considered continuing in my German studies, but it seems rather unlikely that I will; while the language is somewhat neat, the grammatical constructs are nightmarish. I'm not great at memorizing vocabulary words either, but that's something I would consider working around if not for the grammar. Also, it appears that I'll have little time for extra studies at Michigan Tech, since I'm trying to cram a dual major into three years (24 credits/sem, youch). However, if the opportunity to learn a language comes up conveniently for me at some point in my life, I'd likely resume German (although I might instead opt for Spanish, which is a good deal easier).

Yeah nice! German is gramtically much harder than English I think so, too. And in Germany we say: Deutsche Sprache schwere Sprache.
But it's quite hard for me to learn English because the English language has more words then our language I think.

[quote="confuted"]
Ich hab' drei Jahr Deutsch, aber ich kein mehr Deutsch lernen. Ich geh' jetzt zum Universitt fr Physik und Elektrotechnik.[/quote]

I think your German is quite well but this two sentences are not 100% correct. I think you wanted to say:
Ich hatte drei Jahre Deutsch, aber ich lerne kein Deutsch mehr. Ich geh' jetzt zur Universitt fr Physik und Elektrotechnik.

But you've got my respect I would be happy if my English would be as good as your German.

[quote="Idril"]
But it's quite hard for me to learn English because the English language has more words then our language I think.
[/quote]
German has long words compounded by basic words.
Take an example, "hippopotamus" in German is "Flusspferd" (river-horse)...  :D
"rhinoceros" in German is "Nashorn" (Nase, Horn)
"pirate" is "Seeraeuber"... which is clearer  :D

ya, ganz genau. Deutsch ist eine sehr schwierige Sprache besonders fr Auslnder.
German grammar is full of exceptions and things you really just need to memorize and know.

Idril: "Deutsche Sprache schwere Sprache."  yep. lol

German is also an agglutinative language, which means that words chance a lot with tense and articles and so on. But it's still a beautiful language !  :) not to mention a beautiful country ^_~

[quote="joyofpi"]German is also an agglutinative language, which means that words chance a lot with tense and articles and so on. But it's still a beautiful language !  :) not to mention a beautiful country ^_~[/quote]I'm glad you like it ... so why don't you practice in the [url=http://www.mathlinks.ro/Forum/index.php?f=269]German Forum[/url] ? :)

Wow, you made languages subsection.Great Valentin, just great! :D
Happy new year to you all! ;)

[quote="Peter Scholze"]? where do you live that at the time it's 10pm the 30th decembre where i live you already celebrate the new year  :?[/quote]I think he lives closer to you than me :)

[quote="Peter Scholze"]? where do you live that at the time it's 10pm the 30th decembre where i live you already celebrate the new year  :?[/quote]

December has 31 days mate  :D . Happy New Year!

[quote=derpy_chimp]Error 404 - Page Not Found.[/quote]