AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 12
AMC 10 A

Is it true that traditionally B test was a little harder than A test? If that is the case for this year, do A and B takers have the same cut off score for AIME? Also, will 1% of AMC 10B takers and 5% of 12B takers get to qualify for AIME test?

There is no substance to that rumor. The two tests are interchangeable, and the differences in difficulty are simply random.
AIME qualification is determined independently. For each test, at least $ 1\%/5\%$ qualify, and everyone with $ 120/100$ qualifies. Most of the time, the second standard is the one that matter.s

Actually many schools (like mine) only offer the AMC B, and some people have to go to a university since their school doesn't offer it at all.

[quote="the future"]I dont think the test in content is harder, but I think its harder to get top 1% or top 5% on the B because the people that take the B are better than the people that take the A test.  Since the A test is the default test so the people that have "no chance" at passing will most likely only take the A, whereas mostly everyone that takes the B is really serious about math and also took the A.[/quote]

but don't the "really serious" people take the AMC 12? idk if people usually choose to take the 12A or 12B.. or both?

[quote="the future"]I dont think the test in content is harder, but I think its harder to get top 1% or top 5% on the B because the people that take the B are better than the people that take the A test.  Since the A test is the default test so the people that have "no chance" at passing will most likely only take the A, whereas mostly everyone that takes the B is really serious about math and also took the A.[/quote]
Serious? not really, I know some people and have heard of a good deal of people who take the B because they totally failed the A and want to see if they can pass on the B. Besides 120/100 is the default passing score pretty much, and the cutoffs are always very close to them, so I think in terms of difficulty is the same.

uhh okay but not everyone that fails the A takes the B.  However, most people that pass the A take the B (well i guess not most in your terms, but that percentage should be higher than the people who fail the A and take the B).

I dont know im just guessing thats the case becuase most schools only order As...like our school, normal students cant take Bs, they have to be in math club or something.

If your school does not offer A test, does it mean your only choice is to take the B test through the university?

It is unfair that some students can take two while some can only take one and the test will not even be in their own school.

If you take a look at the perfect scorer list, it seems it is very rare for those who take the B test  get perfect score. The overall score seems to be a little lower than A as well. Just interesting observation, there may not be a lot to it.

Most schools offer one date or the other, and most students simply take that one. A few schools offer both dates, and a few students do everything in their power to take both dates. These are disproportionately represented among the top scores.
The college sites are primarily meant to be an opportunity for interested students to take the AMC even if their school doesn't offer it. The AMC has chosen to restrict this to the B date, and I think they're happy with the additional students it brings in. Getting all those other schools would be better, but it's hard- in the district I came from, serving half a million residents, only [i]one[/i] of the high schools offers the AMC at all (That happens to be on the B date, by the way). Of course it's unfair- but it's better than shutting them out completely.

Perfect scores are [i]extremely[/i] dependent on test difficulty, and it's not even the same difficulty that matters for marginal qualifiers. Just take a look it the historical list of perfect scores, and how much the number of names fluctuates year to year. I don't think you can read much into it, except that perfect scores are a lot more common since the move from 30-question tests to 25-question tests.

For the AMC index, they use the higher of the A/B scores right?

Or do they do some percentage thing with the A/B.

[quote="Zmastr"]They use the higher of your A and B score to calculate the index.[/quote]

Alright, so qualifiers arent based on what test they took?  Like a porportional: (amount of qualifers/total number that took the test), for the A/B should be about even?

If by some chance the B test was a lot easier than the A test, and more percentage of the B takers make USAMO then A takers (by more than like 5%), they wont care?

I don't know if that has ever occurred. The AMC is a very good test maker, though not perfect.

[quote="daman71692"]Does anyone know a sight where I can find out where the AMC 10B is being given near me?

If anyone happens to live near New Rochelle and knows off of the top of their head where it is being given, don't hesitate to inform me  :lol:[/quote]

http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml

If your school doesn't offer AMC, just ask your school's head math teacher if they can add it in. All you really need to register is 10 students and a proctor (and small amount like 60 dollars)

That's what I did and now my school offers AMC. (We're taking it on the B date)

[quote="ZzZzZzZzZzZz"][quote="supercircle"]I wonder how the 1% for AMC10 (or 5% for AMC12) actually work in reality.

Suppose 100,000 students take AMC10A and 1% of them, 1000 students will qualify for AIME.

Suppose 10,000 students take AMC10B, and 1% of them, 100 students will qualify for AIME. However, among the 100 students, say 40 of them already took AMC10A and qualified for AIME.

QUESTION: will only 60 students from AMC10B qualify for AIME? Or, additional 40 students will be added to make up the 100 students from AMC10B to qualify for AIME?[/quote]

I don't think they care about A or B.  They take 1% of all students who took the AMC 10.[/quote]

They do, there were different cutoffs last year.

"I don't think they care about A or B. They take 1% of all students who took the AMC 10."

This is not true. AMC10A/12A and AMC10B/12B had different cutoff scores for some years, this means the 1% and 5% rule applied to each individual contest every year.

Yuck, snow forecast for Tuesday night here.

Just hope it's not much.

[quote="undefined117"]but don't the "really serious" people take the AMC 12? idk if people usually choose to take the 12A or 12B.. or both?[/quote]

I have taken both the 12A and the 12B since I was 10, and the only year I didn't qualify for the AIME so far was when I was 10.

I am 12 now.

If it snows as much as for the As...I don't think I can take the 12B. Oh well.

this is the first year our school is offering any of the B tests. i'm excited.

hmm it looks like theres 2-5 inches of snow by wednesday morning and it will keep on snowing until like 12:00 wednesday...

if its two hour delay, then its okay

if its a snow day, ill just force the school to still allow us to take the AMC...just like last year when we had that dumb fog day.

good thing our school board doesnt usually have a lot of snow days...which on wednesday will be good for me.

Our school only offers the A - wanted to take the B, but can't find a school that will let me.  Last year there were some colleges not too far away that offered the B.  But this year in all of New York state, there are only 2 colleges - and they're both upstate - over 400 miels from where I live. :(

I'm taking the 10B, and Daivdyko, an AoPSer is coming to our school! I'm so excited to learn from a master of the arts... or problem solving.

I had to take the B because we got hit with a snowstorm on the day of the A.  :lol:


Let $a, b, c$ be side lengths of a triangle. Prove that at least one of the
equations:

$x^{2}-2bx+2ac=0, x^{2}-2cx+2ab=0, x^{2}-2ax+2bc=0$

has no real solutions.

I've solved it by contradiction, but i didn`t use the fact that a,b,c are sides of a triangle. I wonder if the condition is really neccesary. :huh:

Yeah, WLOG suppose $a\geq b\geq c$. The discriminant of the second quadratic is $\Delta_{2}= 4c^{2}-8ab = 4(c^{2}-2ab)$. But $a\geq c$ and $b\geq c$ and all three variables are positive, so $ab\geq c^{2}$, so $-2c^{2}\geq-2ab$. Therefore, $\Delta_{2}= 4(c^{2}-2ab) \leq-4c^{2}\leq 0$. $c$ is a side of a triangle, so it can't equal zero. Thus $\Delta_{2}< 0$, which gives the result.

I think they just said they were sides of a triangle to suggest that $a, b, c$ are positive reals?  :huh:

[color=darkblue][b]You are not right, M4ario ![/b]

Given are $a,b,c>0$ or $a,b,c<0$. Then $a^{2}\ge 2bc\ ,\ b^{2}\ge 2ca\ ,\ c^{2}\ge 2ab$ $\Longrightarrow$ $a^{2}b^{2}c^{2}\ge 8a^{2}b^{2}c^{2}$ $\Longrightarrow$ $1\ge 8$, what if falsely.

[b]Remark.[/b] $a=3\ ,\ b=-1\ ,\ c=-2$ or $a=-3\ ,\ b=1\ ,\ c=2$ then $a^{2}\ge 2bc$, $b^{2}\ge 2ca$, $c^{2}\ge 2ab\ .$

[b]Lemma.[/b] Let $ABC$ be a triangle ($AB=c$, $BC=a$, $CA=b$) with the area $S$. Then :

$\blacktriangleright 1.\ \ 16S^{2}=2\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)$ and $2\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)>a^{4}+b^{4}+c^{4}\ .$
$\blacktriangleright\ 2.$ The values $\sqrt a$, $\sqrt b$, $\sqrt c$ are the side-lengths of a triangle and $\boxed{2(ab+bc+ca)>a^{2}+b^{2}+c^{2}}\ .$

[b] Proof of the proposed problem.[/b] $a^{2}\ge 2bc\ ,\ b^{2}\ge 2ca\ ,\ c^{2}\ge 2ab\Longrightarrow 2\sum ab\le \sum a^{2}$, what is falsely.[/color]

[hide]Here is the proof as suggested by [b]Virgil Nicula[/b].

Let's assume, for the sake of contradiction, the three equations all have real roots. Then, we must have $b^{2}\geq 2ac, c^{2}\geq 2ab, a^{2}\geq 2bc.$ Adding all three, we get $a^{2}+b^{2}+c^{2}\geq 2(ab+bc+ca)$...(I)

But, for any three sides $a,b,c$ of a triangle, we always have $a^{2}< a(b+c), b^{2}< b(c+a), c^{2}< c(a+b).$ Adding all three again, we obtain $a^{2}+b^{2}+c^{2}< 2(ab+bc+ca),$ which clearly contradicts (I). Hence, our initial assumption is false.[/hide]

[quote="M4RI0"]I think $a,b,c \in \mathbb{R}^{+}$ would have been enough.[/quote]

So do I.

[quote="FiervHvdra"] For any three sides $a,b,c$ of a triangle, we always have $a^{2}< a(b+c), b^{2}< b(c+a), c^{2}< c(a+b).$ Adding all three again, we obtain $\boxed{a^{2}+b^{2}+c^{2}< 2(ab+bc+ca)}\ .$[/quote][color=darkblue][b]Very nice, FiervHvdra ! This is even the classical proof.[/b][/color]


On a large, flat field, $n$ people ($n > 1$) are positioned so that for each person, the distances to all other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $n$ is odd, show that there is at least one person left dry. Is this always true when $n$ is even?

[hide]We will proceed by induction.  For $n=1$, the person is left dry.  Suppose at least one person is left dry for $n=2k-1$.  Consider $n=2k+1$.

Consider the pair of people with the least distance between them.  These two people shoot each other, and both become wet.  If anyone else shoots one of these two people, then there are $2k-1$ dry people and only $2k-3$ people to shoot them, leaving at least one dry person.  Otherwise, we have $2k-1$ dry people who haven't shot yet, which is just like the case for $n=2k-1$.  The induction is complete.

If $n$ is even, there is not always a person left dry.  A simple counterexample is $n=2$.  It is also possible that no one is left dry for any even $n$.  One way this can happen is if the people are grouped in pairs so that each person shoots the other person in their pair.[/hide]

Base case is for $n=3,$ not $n=1$ though this doesn't change things much.

Nice proof!  :)


geometry
similar triangles
geometry proposed

on sides of $AC$ and $AB$ draw 2 similar  triangles $ACE,ABF$ such that$\angle E=\angle F=90$prove that $BE,CF$ and altitude from $A$ are cuncurrent.

Equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=39101 .

  darij


summer program
MathPath
algebra
system of equations

Show that there is no solution to the following system of equations for the real numbers.

                              x^4 + 6x :^2: y :^2: +y^4
                                 xy(x :^2: +y :^2: )

You have forgotten to include the right-hand sides of the equations.  In any case, I ask folks to please not post a solution.  This problem appears on the MathPath qualifying test.

asdfd, where did you get this problem from?


Let a,b,c positive reals and $abc=1$ what is the maximal value of $(a+b+c)-(ab+ac+bc)$?

:rotfl: There isn't. The expression is equal to $a+b+c-(\frac1a+\frac1b+\frac1c)$.
Take $a=n^2, b=c=\frac1n$ => no maximum as n grows
$a=\frac1{n^2},b=c=n$ => no minimum as n grows

oops I just saw that you talk about integers. Then there is not that much choice, a=b=c=1 :rotfl:  :rotfl:

so there isn't the maximal value.ok
and if $abc=1$ what is the maximal value of $ab+ac+bc$?

i find just a minimal value of $a+b+c$ :blush:
and i want the minimal value of $ab+ac+bc$

That is a theorem which is well known AM:Atithmetic Mean and GM:Geometric Mean and AM>=GM

and min value of that inequality is 1.

Davron

can you write the proof for $ab+ac+bc<1$?

[quote="akhrif"] $ab+ac+bc<1$?[/quote]

I do not think about this......
We have $ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}=3$   but i don't think if this would help to solve your problem...

Yes it won't help in solving, but if we're talking about...=)

if $abc=1$, then $ab+ac+bc$ doesn't have any minimum, Take just $a$ and $b$ nearly infinity and $c$ nearly 0. $ab$ is nearly infinity (and it hasn't any minimum), rest is nearly 0


calculus
integration
logarithms
function
calculus computations

I tried much but i couldn't resolve..
Integral of  (:sqrt: 1+ln(x))/xln(x) dx
thanks =)

$
\int \frac{\sqrt{1+\ln x}}{x\ln x}dx = \int \frac{\sqrt{1+\ln x}}{\ln x}d(\ln x)
$
We are going to compute $\int \frac{\sqrt{1+u}}{u}du$

Let $\sqrt{1+u}=t$, $u=t^2-1$.
Thus
$\int \frac{\sqrt{1+u}}{u}du = \int \frac{t}{t^2-1}\cdot 2t\ dt = 2\int \left(1+\frac{1}{t^2-1}\right)dt = 2t+\ln \frac{t-1}{t+1}+c$

Hence
$
\int \frac{\sqrt{1+\ln x}}{x\ln x}dx = 2\sqrt{1+\ln x} + \ln\frac{\sqrt{1+\ln x}-1}{\sqrt{1+\ln x}+1}+c$

well i got an easier solution
the above function is the derivate of the bottom so the integral is ln|xln(x)|+C
right?

[quote="kennis"]the above function is the derivate of the bottom so the integral is ln|xln(x)|+C
right?[/quote]
That would be true if there were no square root. But with the square root it is wrong and liyi's solution is right.


geometry circumcircle AMC AIME ratio trigonometry

This is a question from the 2005 Mock AIME series of Mildorf (#5/it's posted as #7 on AOPS wiki). I couldn't find the solution anywhere.

#12: ABC is a scalene triangle. The circle with diameter AB intersects BC at D, and E is the foot of the altitude from C. P is the intersection of AD and CE. Given that AP = 136, BP = 80, and CP = 26, determine the circumradius of ABC.

thanks a lot

The triangle with it's vertices as the midpoints of the sides of $ \triangle ABC$ is called the medial triangle of $ \triangle ABC$. It is similar to the original triangle with similarity ratio $ \frac {1}{2}$, and its sides are parallel to the those of $ \triangle ABC$.  Let O be the circumcenter of $ \triangle ABC$, H be the orthocenter.
Lemma 1: O is the orthocenter of the medial triangle.
Lemma 2: AH is twice the distance from O to BC.
Try to prove these on your own, they're not that difficult.
So let M,N,P be the midpoints of sides AB,BC,CA, respectively, in the given problem. Since AB is a diameter, $ \angle ADB\equal{}90$, and since also $ \angle AEC\equal{}90$, P is the orthocenter of \triangle ABC. Let O be the circumcenter, from our lemmas, ON=13, ON=68, OP=40. Let R be the circumradius of \triangle ABC. We have $ \angle AON \plus{} \angle BON \plus{} \angle COP \equal{} 180, so cos^{ \minus{} 1}(\frac {13}{r}) \plus{} cos^{ \minus{} 1}(\frac {40}{r}) \plus{} cos^{ \minus{} 1}(\frac {68}{r}) \equal{} 180$. Solving this gives x=85, I don't really feel like writing it out.

can you post the method of solving that trigonometry equation (I'm not too strong in trig). all I can think of is taking the cos of both sides and using sum of cosines and that becomes really ugly.

The notation in this problem is bad but anyway this is pretty simple if you know the right stuff...

$ AP\equal{}2R\cos A$, etc. Also $ \cos^2A\plus{}\cos^2 B\plus{}\cos^2 C\plus{}2\cos A\cos B\cos C\equal{}1$, plugging in your can easily solve for $ R$.


in how many ways can we choose 3 numbers x,y,z such that x+y+z=n for some natural number n?

what is the general formula for $i$ terms?

You forgot to mention that they are natural or nonnegative(tell me which one)

Imagine that we have $n$ apples in a row. And we can put a wall beetween two nearest apples. So... we should put 2 walls.
If u mean that x,y,z are nonnegative integers than the answer will be ${(n+1)^2}/ {2}$
Am i true??

If x,y,z are natural than ... i think the answer will be for $i=3$: $\frac{(n-1)^2}{2}-(n-1)$
now we can put the wall only in one of (n-1) places as $x,y,z\geq1$ and $\"-(n-1)\"$ as we should exclude that walls which goes into one.

sorry
[b]its for distinct positive numbers x,y,z[/b]

shouldnt it be like: $\binom{n+i-1}{i}$ or something similar?

[quote="BOMBOM"]Continuing i thing that general formula will be like that:
\[ \frac{(n+1)^{i-1}}{(n-1)!}  \][/quote]

This is not generally an integer for $i > 2$, and even for $i = 2$ it is not an integer for $n, n+1 \in \mathbb{P}$.   :huh:

(Edit:  The likely flaw in your counting is that cases where one or more of the numbers are equal are counted differently.  For example, for $n = 7$ the $2, 2, 3$ case is only counted three times, where as the $1, 2, 4$ case is counted six times.)

[quote="nateharman1234"]n-1 holes between apples, i walls to place so isn't it just: (n-1) choose i ?[/quote]

To split up $n$ objects into $i$ different groups, you need $i-1$ walls. These walls have "slots" of their own, so the number of ways to partition them is: $\displaystyle \binom{n+i-1}{i-1}$. This allows you to put more than one wall in the $n-1$ gaps between the objects, and at the very outside edges of the string of objects.

EDIT: This way of counting includes empty subgroups, so if the problem is asking for positive integers, then it's slightly different.

[quote="amirhtlusa"]also, its asking for distinct numbers...which is more diffrent ;)[/quote]

Right, :blush:


Cool problem and solution. What is the source?

[quote="jeez123"]Find the real value of $k$ for which $\log(\frac{3x+2y}{k})=\frac{\log(2xy)}{3}+\frac{\log(3x+2y)}{3}$
for all positive values of $x$ and $y$ such that $27x^{3}+8y^{3}=3018x^{2}y+2012xy^{2}$.

Solution:
[b]1[/b]
Rewrite as $\log(3x+2y)-\log k = \frac{1}{3}\log(3x+27)$..
[/quote]

Can someone please help me that first line above?
Thanks!

[quote="jeez123"][b]1[/b]
Rewrite as $\log(3x+2y)-\log k = \frac{1}{3}\log(3x+27)$..

[b]2[/b]
This gives $\frac{2}{3}\log(3x+2y)-\frac{1}{3}\log(2xy)=\log k$, or $\log\frac{(3x+2y)^{\frac{2}{3}}}{(2xy)^{\frac{1}{3}}}=\log k$[/quote]

I think that the $27$ is a typo.  Looking at Line [b]2[/b], it appears that the $27$ should be $2y$.

oh duh!!  :rotfl: I was banging my head last night trying to figure out where the 27 comes from
thanks! should not be a problem now...


agha inam mese baghie marathonast yeni khaheshan ta vaghti ye masale hal nashode masale badi ro nazarid
age ham ye masale modate ziadi hal nashode moondo hosele hame sar raft oon vaght daste jami idehamoono midimo ba ham hallesh mikonim :!:

[quote="BaBaK Ghalebi"]agha sharmande vali man dige hoselam sar rafte va bakhshe tarkibiato shoroo mikonam:
[b]noghate $P,A_1,...,A{}_2{}_n$ noghati dar safhe hastand ke hich $3$ta hamkhat nistand
neshan dahid noghte $P$ daroone tedade zoji az majmooeye mosalashaye ba rooose az beine noghate $A_1,...,A{}_2{}_n$ zaher mishavad[/b] :?:
mohamad hosein sharmandeha vali oon masalehato hanooz natoonestam hal konam tarkibiate in siteham kheili vaghte khabide pas man marathon ro be ham mirizamo tarkibiato shoroo mikonam[/quote]
oonahash dige:
[b]noghate $P,A_1,...,A{}_2{}_n$ noghati dar safhe hastand ke hich $3$ta hamkhat nistand
neshan dahid noghte $P$ daroone tedade zoji az majmooeye mosalashaye ba rooose az beine noghate $A_1,...,A{}_2{}_n$ zaher mishavad[/b] :?:

age 3ta noghte mesle $A_1,A_2,A_3$ dar nazar begirim va yek noghte daroone  mosalasi ke ba$A_1,A_2,A_3$ dorost mishe yetorayi dorost dar nemiyad

na dige
tavajoh nakardi noghteha ro ba $A_1,...,A{}_2{}_n$ neshoon midim
yani tedade $A_i$ ha zoje
 ;)

[quote="mohamadhosein"]khob oon noghte vasati ham hesab mikonim[/quote]
agha khaheshan soalo dorost bekhoon
ma dar majmoo $2n+1$ noghte darim $A_1,...,A{}_2{}_n$ in $2n$ ta HAMRAHE $P$ mishan $2n+1$ noghte
hamchenin hich se tai az $A_i$ ha hamkhat nistand
hala bia hame mosalasai ke roooseshoon in $A_i$ han ro dar nazar begir
sabet kon $P$ tooye tedade zoji azina gharar dare

[quote="mohamadhosein"]agha sharmande
vaghaen hagh ba toe ;)[/quote]
khahesh mikonam
 :D

[quote="pirhoosh"]bale in doostihas ke mimoone
 hala agha babak age azat beporsam in soal ro too kodoom bakhsh mizari chi migi
be khatere in migam ke beram injoor bakhsh hai ro bekhoonam[/quote]
manzooret ine ke injoor soala marboot be che mabhasi too tarkibiate??
valla manam nemidoonam vali mese inke be ina migan [b]hendese tarkibiati[/b]
albate fek konama
chon mehdiam ye soal too in mayeha porsido aghaye hatami behesh goft ke hendese tarkibiatie
ye ketabam goftana!!vali esmesho yadam nemiad
mikhai ye dor dige bepors shayad javabeto begiri
 ;)

ketaabe:"mabahes va masaaele olampiad riyazi"...
va hamin tor ketaabe"a course in combinatorics,van lint"(albatte dovvomi kaamel tare...kheli kaamel tar
bache hya man emrooz har chi khastam beram dars bekhoonam ya khabam gereft ya delam be dars khoondan naraft...chikar konam?daram mimiram :(  :(  :stretcher:  :wallbash_red:

salam ye dooste khoobi biyado lotf kone bege "mabaahes va masaaele olampiaade riazi" male kiye va che entesharaati . oon yeki ketaab zabaan asliye dorost fahmidam?
bedrood

bale ketabe (a course in combinatorics) engilisiye. :)

salam bar hame bebakhshid khodam in ketaabe "mabaahaes." ro daashtam  :D
vali oon yeki chi ? :(

mitooni rahat giresh biyari(az daneshgah ya ye jaye dige...)

inam az javam
badihi ast ke agar noghteye $A_i$ bar rooye nimkhate $PA_i$ jabaja shavad dar tedade mosalashaye matloob taghiri ijad nemishavad pas baraye rahatiye kar farz mikonim noghat bar rooye yek dayere be markaze $P$ vagheand.
hal bar rooye $n$ esteghra mizanim:]
halate paye ke badihi ast:
hal farz mikonim ke baraye $n=k$ bargharar ast yani $2k$ noghte bar rooye dayere ba sharayete masale vojood darand ke tede zoji mosalase matloob ba raas haye $A_i$
vojood darad.
hal do noghteye $X,Y$ra ezafe mikhonim be tori ke sharte hamkhat naboodan bargharar bemanad.khotoote $XP,YP,A_i$ dayere ra baraye bar dovom be tartib dar $X`,Y`,A_i`$ghat mikonand badihi ast ke agar yek noghte mesle $A_i$ dar kamane $X`Y`$ bashad(kamane $AB$ manzoor kamane koochektar ast)(tedade $A_i$ ha ra dakhele  kamane $AB$ ra ba$Q_{AB}$ neshan midahim)
mitavan ba$A_i,X`,Y`$ yek mosalase matloob sakht.
tedade mosalas haye matloob dar halate $n=k+1$ barabar ast ba :
tadade mosalashaye matloob dar halate $n=k$($T_1$) $+$ tedad mosalasyaye matloob ke do raseshan$X,Y$ ast($T_2$) $+$ tedad mosalas haye matloob ke yeki az ras hayeshan $X$ ya$Y$($T_3$)
$T_2$ barabar ast ba $Q_{X`Y`}$
$T_3$ barabar ast ba $\sum {Q_{X`A_i`}}+\sum{Q_{Y`A_i`}}$
ke azlahaze zojiat barabare $((2n-Q_{X`Y`})* Q_{X`Y`})+(Q_{X`Y`}*(Q_{X`Y`}-1))$ ke az lahaze zojiat ba
$Q_{X`Y`}$ barabar ast
pas $T_2+T_3$ adadi zoj ast va shon $T_1$ niz zoj ast pas tadade mosalas haye matloob zoj miba shad.
vaghean cheghadr bad neveshtam :oops:

age fekr mikonin rehal doroste ta masaleye jadido post konam ;)

salam
in web log kia ro hatman sar bezanin va hatman maghalasho raje be Godfather bekhoonin:
http://www.kiakarbasi.blogsky.com

[quote="mohamadhosein"][size=150][color=red]Problem 2 [/color][/size]
dar khanehaye sefide yek jadvale shatranjiye $1983*1984$ khanei adad haye $1,-1$
ra be goone yi neveshte im ke baraye har khaneye siah hasele zarbe khane haye sefide mojavere an brabare $1$ bashad .
sabet konid in vaz vaghti pish miayad  ke hameye adad haye nevesh te shode barabare $1$ bashad. :?:[/quote]
farz konid $m>0$ ta khane sefid havie adade $-1$ bashand
hal midanim ke haslele zarbe aadade jadval barabare $(-1)^m$ ast amma az tarafi hasele zarbe in aadad barabare haselezarbe aadade khanehaye siah ast pas darim:
$(-1)^m=\frac{1}{(-1)^r(-1)^n}$ ke dar an $r,n$ be tartib barabare tedad khanehaye sefide gooshe va vasate jadval ke meghdare $-1$ ra darand mibashad (manzoor az khanehaye vasat,khanehaye yek jadvale $1981*1982$ mibashad ke ba jadvale avalie ham markaz ast)
tasavie fogh az anja natije mishavad ke har khane gooshei $2$ khane siahe mojaver darad pas dar hasele zarbe fogh $2$ bar shemorde mishavad hamchenin har khane vasat niz daraye $4$ khane siahe mojaver ast pas $4$ bar shemorde mishavad yani adade $-1$ dar khanehaye fogh dar tasavi zekr shode barabare meghdare $1$ khahad bood pas niaz ast ke adade hasel bar $(-1)^n,(-1)^r$ taghsim shavad ta har khane ba adade haghighie khodash zaher shavad
pas darim $m \equiv r+n (\bmod 2)$
amma agar $r',n'$ ra be tartib barabare tedad khanehaye gooshe va vasate jadval ke havie adade $1$ hastand va $q,q'$ ra be tartib barabare tedad khanehaye navarhaye bala,pain,chap,raste jadval ke havie adade $-1,1$ hastand dar nazar begirim angah be vozooh darim $i) q+q' \equiv e 0 (\bmod 2),r+r' \equiv 0 (\bmod 2)$
amma az tarafi [b]tedad khanehaye sefide jadvale $(1983*1984) \equiv 0 (\bmod 2)$[/b]
va az tarafi [b]tedad khanehaye sefide jadvale $(1983*1984)=r+r'+q+q'+n+n'$[/b]
pas $ii)r+r'+q+q'+n+n' \equiv 0 (\bmod 2)$ pas tebghe $i),ii)$ darim:$n+n' \equiv 0 (\bmod 2)$
amma $n+n'$ barabre tedad khanehaye sefide jadvale $1981*1982$ ke tedade fardi ast mibashad
pas be tanaghoz resideim
pas $m=0$
doroste??  :?

baba yeki javab bede ta man soale badio post konam dige
age fek mikonid ghalate begid ghalate
age tikeisho nemifahmid begid bishtar tozih bedam
yechi begid dige
 :(

khob dige ye zare satho mibarim balatar:
[b]dar jadvali $n*n$ aadade $1$ ta $n^2$ neveshte shodeand be tori ke majmooe aadade har satr va majmooe aadade har sotoon barabare meghdare sabeti shode.baraye har $2$ khane az in jadval,markaze khane koochektar ra be markaze khane bozorgtar be vasile bordari vasl mikonim sabet konid jame in bordarha barabare $0$ mibashad[/b]
bacheha age in hal nashod asan negaran nashid chon sathesh momkene az marhale dovom ye zare balatar bashe
vali oon ghadr ham sakht nist rahat hal mishe age natoonestid begid ta man hallesho bezaram
 ;)

az do rah inja hastesh
khastid begid ta vasatoon tarjomashoon konam ya bishtar tozih bedam ;)
http://www.mathlinks.ro/Forum/topic-79208.html

[quote="BaBaK Ghalebi"]$2000$ noghte dar safhe darim.hich $8$ noghtei ham rasta nistand sabet konid az beine parekhathaye vasele in noghat haddeaghal $250$ parekhat ba toolhaye motamayez vojood darand :?:[/quote]
farz konid $k$ no toole mkhtalef $(x_1,x_2,...,x_k)$ dashte bashim hal parekhate beine do ras ra ba range $i(1\leq i\leq k)$ rang mikonim agaro faghat agar toole parekhate beine in do noghte $x_i$ bashad
hal rooye amoode monasefe do noghte $W,Y$ tebghe farze masale haddeaksar $7$ noghte motamayez vojood darad pas baraye har noghte $X$ rooye amoode monasefe parekhate $WY$ midanim ke range do parekhate $WX,XY$ yeksan ast

hal biaid tedade masirhaye be toole $2$ takrang ra beshomarim:
ebteda in tedad ra az tarighe rase vasati mishomarim baraye har $2$ noghte az beine in $2000$ noghte chos haddeaksar $7$ noghte rooye amoode monasefe anhast pas darim:
[b]tedade masirhaye be toole $2$ takrang$\leq 7{2000 \choose 2}$[/b]
hal yekbar niz az tarighe do rase  entehai mishomarim:
farz konid az rase $i$,$x{}_i{}_j$ yal be range $j$ kharej shavad amma midanim ke baraye har rase $i$ jame $x{}_i{}_j$ha meghdare sabeti ast pas darim:
$2000k{\frac{1999}{k}\choose 2}\leq \sum{x{}_i{}_j\choose 2}$

tarafe raste namosavie fogh az anja natije mishavad ke agar baraye rase $i$ hame $x{}_i{}_j$ ha barabar bashand ya be ebarati dar rase $i$ az har rangfe be tedade barabari amade bashad yani har rangi $\frac{1999}{k}$ bar amade bashad

ba sade kardane namosavie fogh be ebarate $k\geq 250$ mitavan resid

baba cheghad shomaha be tarkibiat ahamiat midid
mashalah hezar mashalah masaleha be zamin nareside javabesh roo site  :(
baba ye zare faal bashid dige
yedoone sakhte dige:
[b]$k$ no moozaik az har no namotanahi ta mojood ast ke har kodam be shekle morabai $1*1$ ast ke ghotrhaye an rasm shodeand va dar har yek az chahar bakhshe be vojood amade adadi tabii neveshte shode
do moozaik mitavanand kenare ham gharar girand agaro faghat agar adadhaye bakhshhaye mojavereshan yeksan bashad
$i)$sabet konid ke agar robe avale safhe mokhtasate $\mathbb{R}^2$ ra betavan be tore kamel pooshand angah kolle safhe $\mathbb{R}^2$ ra mitavan pooshand
tavajoh dashte bashid ke har moraba azlaash ba mehvarhaye mokhtasat movazist
hamchenin davaran niz mojaz nemibashad
$ii)$masale ra baraye $\mathbb{R}^3$ va taghsime har mokaab be $8$ ghesmat niz tamim dahid[/b]


If anyone has any spare time, he/she may attempt to solve the following problem:

For which natural numbers n, is the sum of any n subsequent primes the product of n and some prime number? Examples:
for n=1, 2=1 (=n)*2 (= some prime);
for n=2 (there aren't any);
for n=3, 3+5+7=3 (=n) * 5 (= some prime), and so on!

Have fun!  :lol:

Oh and yeah, have you any ideas, please do post!

The sum of n primes!?! That'll be one tough motherf***** to solve!


1.what is the gen. form of the parabola is it y=4ax or y=(x-h)^2_g
2.provide a prove forthe lenght of the latus rectum in a parabola
3.is algebra2 by meril a good textbook?

Are these homework problems?
Posting your homework is often frowned upon on this forum.

I'll provide you with a hint as to the first question.
If you have y = (4a)x (with a as a constant) then what you are describing is not a parabola, but something else.


Help!! Right after I watched one of the mini videos and submitted the comment, the computer said I had no access to the video.  Then, I tried to watch another one,  but that didn't work either.  Also that already happened twice. :(  :(  :|

It happened again today. Does anyone know why?

Hi everyone! You should not be getting any more of those access denied messages errors. Sorry for the inconvenience. :)

Thank you very much. I'm not having anymore trouble now!!!! :huh:


trigonometry
function
algebra unsolved
algebra

[quote="conan_naruto236"]Compute the sum :
$ S \equal{} \sum_{k \equal{} 1}^{n} sin{\frac {x}{2^k}}$
by $ x$ and $ n$[/quote]
use the complexe formulas of $ sin(x)$

$ sin(x)\equal{}1/2*(e^(x)\minus{}e^(\minus{}x))$

[quote="Ibn Rochde"][quote="conan_naruto236"]Compute the sum :
$ S \equal{} \sum_{k \equal{} 1}^{n} sin{\frac {x}{2^k}}$
by $ x$ and $ n$[/quote]
use the complexe formulas of $ sin(x)$

$ sin(x) \equal{} 1/2*(e^(x) \minus{} e^( \minus{} x))$[/quote]
That is not true. $ \sin x \equal{} \frac{e^{ix} \minus{} e^{\minus{}ix}}{2i}$.

I don't think we can represent it throw well-known function, only as series:

$ \sin\left(\frac{x}{2^k}\right)\equal{}\sum_{n\equal{}0}^{\infty}{\frac{(\minus{}1)^{n}x^{2n\plus{}1}}{(2n\plus{}1)!2^{k(2n\plus{}1)}}}$

$ \sum_{k\equal{}0}^{\infty}\sin\left(\frac{x}{2^k}\right)\equal{}\sum_{k\equal{}0}^{\infty}\sum_{n\equal{}0}^{\infty}{\frac{(\minus{}1)^{n}x^{2n\plus{}1}}{(2n\plus{}1)!2^{k(2n\plus{}1)}}}\equal{}$

$ \equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^{n}x^{2n\plus{}1}}{(2n\plus{}1)!}\sum_{k\equal{}0}^{\infty}\frac{1}{2^{k(2n\plus{}1)}}\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^{n}2^{2n\plus{}1}}{(2n\plus{}1)!(2^{2n\plus{}1}\minus{}1)}x^{2n\plus{}1}$


inequalities
algebra
system of equations

I have a question about the minimum about this problem.

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} + \frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c}$

where $a,b,c$ are positive reals.

In my opinion, since $a,b,c$ are positive, I assumed that I could use AM-GM on six variables which gave me 6. But I don't think this is possible for any values of $a,b,c$.. Or does it?

Can anyone help with this?

[quote="Silverfalcon"][quote="frt"][hide="My approach"]
Consider the minimum value of $\frac{a}{b+c}+\frac{b+c}{a}$.

Using AM-GM, we get $\frac{a}{b+c}+\frac{b+c}{a}\ge 2\sqrt{\frac{a}{b+c}\cdot\frac{b+c}{a}}=2$

Repeating this two more times, we see that the minimum value of the original expression is $6$.[/hide]

EDIT- I don't think I answered your question, sorry. In my opinion, the minimum is $6$ only if $a$, $b$, and $c$ are positive reals, as the problem states. (I don't know how to find the minimum if one or more variables are nonnegative numbers.)[/quote]

I agree. I stated that $a,b,c$ are positive and all signs in there were positive so I clearly could use AM-GM. I just don't know when the equality states.

According to usual AM-GM, it should be when $a=b=c$ but then it doesn't give 6...[/quote]
actually, AM-GM states that the items being applied to the inequalityare equal which means a/(b+c)=b/(a+c)=c/(a+b)=...=(a+b)/c, and NOT a=b=c. However, this system of equations is impossible to solve (it gives that a, b, and c are all 0 and thus the expression is undefined) and thus in positive reals, the minimum will be larger than 6

[quote="frt"]Maybe something is wrong...

$\frac{a}{b+c}=\frac{b+c}{a}$ gives $a^2=(b+c)^2\Longleftrightarrow a=b+c$ ($\because a,b,c>0$)

Repeating this three times, we get $a=b+c$, $b=c+a$, and $c=a+b$. Adding these equations, we get $a+b+c=2(a+b+c)$, which means $a+b+c=0$...[/quote]
using substitution on your equations, we also receive a=b=c=0

We cant have equality but we can have something extremely close to equality.  I could be wrong with this but I believe the expression simply must be greater than 6.

The *actual* minimum is $3/2+6=15/2$.  Let's prove that.

We'll do it in parts:

1)

$\sum \frac{a}{b+c}\geq \frac{3}{2}$

Proof:  I know about nine million of them (this is Nesbitt's inequality), so lets do one that only involves AM-GM.  Put $x=a+b$ and so on, so that we transform the inequality into

$\sum \frac{a}{b+c}=\sum \frac{1}{2}\left(\frac{y+z-x}{x}\right)=\frac{1}{2}\left(\sum \frac{y}{x}+\frac{z}{x}\right)-\frac{3}{2}\geq \frac{6-3}{2}=\frac{3}{2}$

as desired.

2)

$\sum \frac{b+c}{a}\geq 6$.

Proof:  We already handled this in the above.  AM-GM everything at once to get the result.

Adding 1) and 2), we get a maximum of $\frac{15}{2}$ as desired.

Of course, the problem was exactly what you guys figured out - the equality was unattainable because we would have had to have $a=b=c$ for the first three terms to be equal, but also $a^2=(b+c)^2$ for the first and fourth term to be equal.

I think I understood the reason now.

The definition of AM-GM stated that it should be written as $\frac{x_1+x_2+x_3+\cdots x_n}{n}$ and I just applied AM-GM without having common denominator.

I think that would've been a problem.  :D I think..

BTW, thanks for your proof blahblahblah. I know, there are MILLIONS of proofs for Nesbitt's and I think I saw it appeared about MILLION/2 times appearing on math contests.

No, the common denominator was not a problem, it was that the terms that you took the arithmetic mean of could not simultaneously be equal.


Basically its a trivia game. Depending on the number of applicants, the number of competitors will be determined and will be expressible in form of $ 2^{n}$.

For example, if there are 30 people who apply, then 16 will get in. The largest number of competitors will be 16. If there are less than 16 applicants, the number of competitors will be the largest power of two under the number of applicants.

The trivia game will be multiple choice and possibly fill in the blank.

Each round will have exactly ten questions. Tie breakers will be determined by the amount of time it took for someone to answer the questions.

Please apply by posting here.

I'll apply.

(This message is...... stupid)

meh, w/e

The message is too dumb. Please make the message more pointless before submitting.

I will apply.

Signup list (4):
kimmystar94
BOGTRO
mz94
Brut3Forc3

i guess ill do it.  it wont hurt me, right?

(me join)

/in
[hide]The message is too small. Please make the message longer before submitting.
 [/hide]

Everyone may apply [url=http://mersenne.zapto.org/Code/Trivia/index.php]here[/url]. Tiebreaker question will be used only in case of at tie. Acceptance will be based on the accuracy of the answers.

Are we allowed to use the internet for our app?

You may use all sources at your disposal.

ALL APPLICATIONS FOR ENTERING THIS CONTEST ARE DUE BY TUESDAY, MAY 6, 2008. LATE APPLICATIONS WILL NOT BE ACCEPTED.

BRIBES WILL BE ACCEPTED xD

If the tiebreaker does not break the tie, the odd man out on the tiebreaker will be eliminated.

I have received applications from:
[list]
[*]Brut3Forc3
[*]BOGTRO
[*]DonkeyKingKong
[*]ajain
[*]distracted523
[*]kimmystar94
[*]mz94
[*]andersonw[/list]

Meh, can you PM the cut people their scores?

[quote="#H34N1"]cf2489 - 5[/quote]
Ahem?

Don't worry about it, I just like giving people a hard time about mistyping my username.   :D

erm whatever

you put cf2489 in the application :D

Problems for Round 2 will be (hopefully) posted sometime Monday

Trivia Round II has been posted!

http://mersenne.zapto.org/Code/Trivia/?q=node/6

Please use your same passkeys. ONLY QUALIFIERS MAY SUBMIT.

Thank you.

Answers due May 26, 2008 at 6:00 P.M. E.S.T.

Those qualifying for the final round will be posted around 9 the same day.

Dude that's not fair. This round is a lot easier than the first round.

That is only because it is multiple choice.

I [i]highly[/i] doubt anyone will be able to get more than a 7 on that without choices.

If the other two people could also submit, I'd like to see the finals squeezed in before next Friday, as I am leaving for Intl FPS

We will be administering a tie breaker between distracted523 and mz94, as they both tied.

I think that timing is not a cool way since it depends on when they read this post

The qualifiers for Round III - the final round - are:

[hide="Qualifier I -  :winner_first: "]DonkeyKingKong[/hide]

and

[hide="Qualifier II -  :winner_second: "]distracted523[/hide]

Congratulations to all who participated in Round II.

 :clap2:

The third place winner (:winner_third:) is [b]mz94[/b]! Congratulations!

And, we are posting the correct answers.

[hide="Answers"]1	C
2	D
3	B
4	B
5	C
6	B
7	D
8	C
9	A
10	D
11	B
12	C
13	D
14	B
15	C
16	A
17	B
18	D
19	C
20	C
21	A
22	B
23	D
24	D
25	C
[/hide]

The highest score was 15. The lowest score was 11. The tiebreaker was between mz94 and distracted523. Scores for both were 14.

There were 25 questions.

In round I, both mz94 and distracted523 got 2/3 questions correct. In round ii of tiebreakers, distracted523 had 3/3 and mz94 2/3. The final round will be composed of

[b]distracted523 [/b]and [b]DonkeyKingKong[/b]

There will be 30 questions. The test will not be posted until Saturday at minimum. The test will have no choices. Resources will not be permitted, as usual.

The person with the higher score of the 30 will be declared the winner of Trivia Game I.

Wow...so close. And I guessed #1 to the second tiebreaker...

It seems DonkeyKingKong has not responded to his PMs, and I will therefore declare distracted523 as the champion.


I am wondering what competition the question with M&IQ are from?

Mathematics & Informatics Quarterly - a journal for problem solving mathematics and informatics.


A science test has 10 questions worth 5 points each, 7 questions worth 6 points each, and 4 questions worth 2 points each.  None of these questions will be given partial credit.  How many scores between 0 and 100 are impossible to score?

[hide]4[/hide]

--Is there a quicker way to solve this besides trying every integer from 0 to 100?--

--Thanks.--

[quote="kimmystar94"]i'm not sure but the Chicken McNuggets Theorem?[/quote]

THis is really dumb but...whats the Chicken McNuggets Theorem?

the chicken mcnugget theorem is that if u have 2 nonnegative relatively-prime integers, X and Y,
then the largest integer that u can't make in the form aX+bY where a and b are nonnegative integers is xy-x-y

[quote="alanchou"][hide="hint"]
If a score x is impossible, 100-x is impossible
[/hide][/quote]

Why does that work?  Thanks.

[hide]
1 and 3 are obviously impossible, and its clear that any other value less than or equal to 50 works, so only 1,3,99,97 are impossible, for a total of 4.[/hide]

Is there any possible way you could use mods, like any number that is not equal to 5 mod 6 or mod 2, any number that is not equal to 6 mod 5 or mod 2, and any number that is not equal to 2 mod 5, and then cross out any repeats?

Well the problem with that is that you can only use a certain number of each.


Problem. Give $ A \in M_n(k)$ such as $ A^2\equal{}I_n$. Prove that:
$ rank(A\plus{}I_n)\plus{}rank(A\minus{}I_n)\equal{}n$

This was solved quite recently. Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308915]here[/url].


Let a,b,c are non_negative such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \sum_{cyc}{\sqrt {a \plus{} b^2}}\ge 2$
Can anyone helpme?

[quote="thegod277"]Let a,b,c are non_negative such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that
$ \sum_{cyc}{\sqrt {a \plus{} b^2}}\ge 2$
Can anyone helpme?[/quote]
Maybe $ a \plus{} b \plus{} c \equal{} 1$ $ ?$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782


A and B independently do a multiple choice quiz with 10 questions, each having 5 options.
A knows the answers to 5 questions and B knows the answers to 6 questions.
Both answer the remaining questions by pure guesswork.
It is already know that A answers at least one of the remaining questions correctly.
Now, what are the chances of A beating B, B beating A, and drawing?

The probability of A beating B is more than the probability of B beating A or both of them 'drawing'.

The probability of both of them drawing is more than the probability of B beating A.

Sorry, I did this pretty intuitively, my knowledge of probability needs to be refreshed.

Are my answers right?

P.S:The appropriate place for a problem like this is- "Probabilities and Statistics" in the "College Playground section".

Well, there is technical language to describe what's going on here (shifted binomial distribution, shifted truncated binomial distribution), but actually computing the various probabilities is just some straightforward combinatorics (with a bit of conditional probability for A).

For example, if $ p \equal{} \frac {1}{6}$ then the probability that A gets exactly 6 questions correct is
\[ \frac {\binom{5}{1} p^1\left(1 \minus{} p\right)^4}{\sum_{i \equal{} 1}^5 \binom{5}{i} p^i (1 \minus{} p)^{5 \minus{} i}}.
\]
Make the similar computation for the other scores.  (The analogous computation for B is easier -- no truncation, so no denominator.)  Then it's easy enough to read off from the resulting collection of numbers (some arithmetic required) what the probability of any given outcome is.


calculus
integration
analytic geometry
geometry unsolved
geometry

Find the maximum real number $ k$ such that: For any three integral points $ A,B,C$ on a coordinate plane,if
$ (|AB|\plus{}|BC|)^2<8S(ABC)\plus{}k$  , then $ A,B,C$ must be three vertices of a isosceles triangle. (Hint: On a coordinate plane, a point with its $ x$ and $ y$ coordinate both integer, is called a integral point).

Why do anyone not answer?I think this problem not dificult


algebra
function
domain
superior algebra
superior algebra unsolved

Prove that
$ T_5\equal{}\lbrace \frac{m}{n} : m \in \mathbb{Z}, n \in \mathbb{N}, hcf(5,n)\equal{}1 \rbrace$
is a Euclidean Domain.

HINT $ |k \; 5^i/n| : \equal{} 5^i$ for $ (kn,5) \equal{} 1$ is trivially a Euclidean valuation since  $ b\not|\;a \Rightarrow |a| < |b|$


geometry
3D geometry
search
algebra unsolved
algebra

An arithemetic progression whose terms are positive integers contains the square of an integer and a cube of another integer.Prove that the progression contains the sixth power of some integer.

ISL 1997,in PEN,
http://www.mathlinks.ro/viewtopic.php?search_id=1959085089&t=150830


trigonometry
inequalities unsolved
inequalities

$x\in[0,\pi]$. Prove that $sinx (1+ cosx)\leq \sin\frac{x+\pi}{4}(1+\cos\frac{x+\pi}{4})$

let $\frac{\pi+x}{4}=t.$ Thence, $\frac{\pi}{4}\leq t\leq\frac{\pi}{2}.$
Hence, $sinx(1+cosx)\leq sin\frac{x+\pi}{4}(1+cos\frac{x+\pi}{4})\Leftrightarrow$
$\Leftrightarrow-sin4t(1-cos4t)\leq sint(1+cost)\Leftrightarrow$
$\Leftrightarrow-32cos^3t(2cos^2t-1)(1-cost)\leq1\Leftrightarrow$
$\Leftrightarrow64a^6-64a^5-32a^4+32a^3-1\leq0\Leftrightarrow$
$\Leftrightarrow(2a-1)^2(16a^2(a+\frac{1}{\sqrt2})(a-\frac{1}{\sqrt2})-(2a+1)^2)\leq0,$
where $a=cost,0\leq a\leq\frac{1}{\sqrt2}.$
Equality, when $a=\frac{1}{2},$ id est $x=\frac{\pi}{3}.$ :)


calculus
integration
logarithms
trigonometry
calculus computations

$ \minus{} \ln\left(1 \minus{} b\, \cos x\right) \equal{} \sum_{n \equal{} 1}^\infty \frac {b^n}{n} \, \cos^n x$

$ \Rightarrow \minus{} I(b) \equal{} \sum_{n \equal{} 1}^\infty \frac {b^n}{n} \underbrace{\int_0^{\frac {\pi}{2}} \cos^{n \minus{} 1} x \, dx}_{ \equal{} A_{n \minus{} 1}} \equal{} \underbrace{\sum_{n \equal{} 0}^\infty \frac {b^{2n \plus{} 1}}{2n \plus{} 1} A_{2n}}_{ \equal{} S} \plus{} \underbrace{\sum_{n \equal{} 0}^\infty \frac {b^{2n \plus{} 2}}{2n \plus{} 2} A_{2n \plus{} 1}}_{ \equal{} T}$

Because of $ A_{2n} \equal{} \frac {\pi}{2} \, \frac {1}{2^{2n}} \, {2n\choose n}$ and $ A_{2n \plus{} 1} \equal{} \frac {1}{2n \plus{} 1}\left[\frac {1}{2^{2n}} \, {2n\choose n}\right]^{ \minus{} 1}$

$ S \equal{} \frac {\pi}{2} \arcsin b$ and $ T \equal{} \frac12 \arcsin^2 b$

Hence $ \minus{} I\left(\minus{}\frac12\right) \equal{} \frac {\pi}{2} \left( \minus{} \frac {\pi}{6}\right) \plus{} \frac {1}{2} \left( \minus{} \frac {\pi}{6}\right)^2 \equal{} \minus{} \frac {5\pi^2}{72}$

I agree with your answer , but I don't agree with this
$ \minus{} \ln\left(1 \minus{} b\, \cos x\right) \equal{} \sum_{n \equal{} 1}^\infty \frac {b^n}{n} \, \cos^n x$

 :(
also with integration step by step
 because if $ b\geq 1$ then this serie
$ \sum_{n \equal{} 1}^\infty \frac {b^n}{n} \, \cos^n x$ will diverge and you can't integrate ...  :(
and if $ b\equal{}1$ and when $ x$ is near zero then you also can't integrate this sum...
but in this case you can do this  :) because

  $ \minus{} 1 \leq b\cos x < 1$   for $ b\equal{}\frac{1}{2}$

[quote="Extremal"]I agree with your answer , but I don't agree with this
$ - \ln\left(1 - b\, \cos x\right) = \sum_{n = 1}^\infty \frac {b^n}{n} \, \cos^n x$[/quote]
This is easy to justify:

\begin{eqnarray*}- \ln (1 - b\cos x) &=& \int_0^{b\cos x} {\frac{1}{{1 - u}}\,du}\\&=& \sum\limits_{n = 1}^\infty  {\int_0^{b\cos x} {u^{n - 1} \,du} }\\&=& \sum\limits_{n = 1}^\infty  {\frac{{b^n \cos ^n x}}{n}} \quad\blacksquare\end{eqnarray*}

I know all of them ..
I noted that it's not true for : $ b\cos\alpha >1$

(I also note that it's not important and my comment is not to Dr.Doe because I believe that he know all of them, this my comment is for young students, for freshman). :)


We already know the mean value theorem :"Let I=[a,b] ,K=(a,b) , f:I-->I f is continuous on I and differentiable on K . Then exists c in K s.t. f(b)-f(a)=f'(c)(b-a) ". If we replace I by open intevals such as (a,b) , [a,b),(a,b] , is the conclusion still true now ?

It can't be I an open interval!
 :P Because then it will not exist c so that f(b)-f(a)=f'(c)(b-a), because the function is not deffinate on a close interval I, so will not takes the a and b value!
It has to be an closed interval! :D

Specific counterexample:
[itex]f(x)=0[/itex] if [itex]x=a[/itex] or [itex]b[/itex], and [itex]f(x)=x[/itex] if not.

Specific counterexample:
$f(x)=0$ if $x=a$ or $b$, and $f(x)=x$ if not.

[size=75]There was a change to the latex code since the above message was posted, making it incomprehensible so I have redone it[/size]

it can be pushed a little further

if f is continuous on [a,b] and differentiable on ]a,b[ it has a point f'(c)=(f(b)-f(a)/(b-a)

example : sqrt(1-x*x) is continuous on [-1,1] and differentiable on ]-1,1[ and
f'(0)=0

so this theorem has an obvious analogue for open intervals :

suppose f is continuous and differentiable on ]a,b[ (the latter implies the first) and has a one sided limit near both border points, there exists such point
(it is obvious because you can extend it to a function on [a,b])


Unfortunately, my school only offered the A testdate for the amc's, and even more unfortunate is how I am really sick right now and I had to drag myself to school at 12pm just because I did not have a backup plan.  Now, my question is how I can take the B test.  I do not know which schools in my area offer it, or how I can just sign up for it individually and take it at a library or something.  I don't really know what to do.

What I did:

hstran@amc.unl.edu   I emailed that person, and they gave me a list of schools in my state registered for the B date.  Then i asked the school if i could take the test there, and when they said yes, i emailed

Ms. Bowers at dbowers@unlnotes.unl.edu   and asked for a Visiting Student Registration form.  Then just fill it out and send it in.  I have no clue if there is a deadline, though, as i did this about a month ago.

good luck.

LOL the one i took AMC 12 with isn't churchilljrhigh (the school i go to) and the one i plan to tkae the AMC 10 with isn't churchiljrhigh. LOL


Two players by turn paint the vertices of triangles on the given picture each with his colour. At the end, each of small triangles is painted by the colour of the majority of its vertices. The winner is one who gets at least 6 triangles of his colour. If both players get at most 5, then it is a draw. Does any of them have winning strategy? If yes, then who wins?
\[ \begin{picture}(40,50) \put(2,2){\put(0,0){\line(6,0){42}} \put(7,14){\line(6,0){28}} \put(14,28){\line(6,0){14}} \put(0,0){\line(1,2){21}} \put(14,0){\line(1,2){14}} \put(28,0){\line(1,2){7}} \put(14,28){\line(1,2){7}} \put(14,0){\line( \minus{} 1,2){7}} \put(28,0){\line( \minus{} 1,2){14}} \put(42,0){\line( \minus{} 1,2){21}} \put(0,0){\circle*{3}} \put(14,0){\circle*{3}} \put(28,0){\circle*{3}} \put(42,0){\circle*{3}} \put(7,14){\circle*{3}} \put(21,14){\circle*{3}} \put(35,14){\circle*{3}} \put(14,28){\circle*{3}} \put(28,28){\circle*{3}} \put(21,42){\circle*{3}}} \end{picture}\]

the first player has winning strategy by changing the color of the middle point.  (I did all the cases hahaha)

A first occupy the central one.then he can paint three of the six surrounding vertices which are not all adjacent.


1. If $L$ is a context free language prove that there exists $n\in N:$ for every $\alpha \in L$ $|\alpha|\geq n$ and $\alpha = xyuvw$ :
1) $|yuv|\leq n$
2)$xy^{i}uv^{i}\in L$ for every $i\geq 0$
3)$y\neq \varepsilon$, $v\neq \varepsilon$

The second problem is this
2.$L=\{ xy|$ $|x|=|y|,$  $x\neq y$ $, x,y\in\{0,1\}^{*}\}$
Prove that $L$ is context free language

First problem looks exactly like pumping lemma for CF languages except for point 2) which reads (in the pumping lemma) $xy^{i}uv^{i}w \in L$ for every $i \geq 0$. Are we supposed to prove your modified version? (Are you sure it holds?)

In the second problem it's pretty easy to find the grammar, I advise everyone to try it :).

[quote="bus"]Are we supposed to prove your modified version? (Are you sure it holds?)[/quote]
If you can prove it i'll be more than thankful (it holds for sure)
[quote="bus"]In the second problem it's pretty easy to find the grammar, I advise everyone to try it :).[/quote]
I've been trying to find this grammar for a week and i couldnt. Will you please tell me which grammar is this(if you want the other people watching this post to try and think about it you can send me a private message :D )

10x

Wow, I'll have to be more careful next time I post, I misread the second problem :blush:. (I thought it was $x \neq y^{R}$.) I'll think about both, hopefully I'll be able to come up with at least some useful idea...

All right so here goes the answer to question #1:

$\sigma \rightarrow \gamma B$
$\gamma \rightarrow \gamma A$
$\gamma \rightarrow \varepsilon$

You start from $\sigma$ and the language you get is $A^{n}B$. Note that this is a CF language but the lemma doesn't hold... or at least that's what I think... I may be wrong again... :).

[quote="bus"]My third post in a row, sorry for such bad manners. Anyway, I didn't find a CF grammar for your language but I found a (non-deterministic) push down automaton for the following laguage:

$\{x\#y: |x|=|y|\wedge x\neq y\wedge x,y\in\{0,1\}^{*}\}$

And since CF languages are closed under homomorphism, your language is context free as well.

EDIT: You could probably find a similar automaton for exactly your language, it would be just a little more complicated.[/quote]

Can you please write the solution or tell whats the idea for the push down automata you have found.
10x a lot by the way :)

Well, seems like I got it wrong again :(. I didn't really believe your language was CF anyway... does anybody else have ideas?

All right, I've spent quite a while trying to prove this and I'm very close to being sure that this time I got it right - the language $\{x\#y: |x|=|y|\wedge x\neq y\wedge x,y\in\{0,1\}^{*}\}$ I mentioned above is not context free (by pumping lemma). However, I did this only to find out that... it's not going to help up with the original problem at all! What a bummer... Well, at least we know that this is not the way to go.


Yes, I think so too, \[(ab + b^2 + a^2)(a^2 + c^2 + ac)(b^2 + bc + c^2) \geq (ab + bc + ca)^3\] by the Hölder inequality.

I have a question, does holder hold for negative numbers? I don't really know about this.

I am not sure, but I don't think so.

Oops - I misread the question, I thought the question was about positive reals...

Hölder surely doesn't hold for negative numbers: Take a standard Hölder application like $\left(a^3+b^3\right)\left(c^3+d^3\right)\left(e^3+f^3\right)\geq\left(ace+bdf\right)^3$. This inequality clearly can't hold for negative numbers, since its degree is odd.

As for the original problem, the solution posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=27480 post #7 works for arbitrary reals a, b, c, provided that you use directed segments.

  Darij


algorithm
geometry
geometric transformation

Try to present a [b] definition of an [/b]     [color=green][b]algorithm[/b] [/color] !

A possible answer to my question is in:

[b]A. A. Markov (1954) [i]Theory of algorithms,[/i] [Translated by Jacques J. Schorr-Kon and PST staff]
 Imprint Moscow, Academy of Sciences of the USSR, 1954 [i.e. Jerusalem, Israel Program for Scientific Translations, 1961;
 available from the Office of Technical Services, U.S. Dept. of Commerce, Washington] Description 444 p. 28 cm. Added t.p. in Russian Translation of Works of the Mathematical Institute, Academy of Sciences of the USSR, v. 42. Original title: Teoriya algerifmov. [QA248.M2943 Dartmouth College library. U.S. Dept. of Commerce, Office of Technical Services, number OTS 60-51085.]
[/b]


On my car, a particular brand of tyre lasts 40000 km on a front wheel and 60000 km on a rear wheel. By interchanging the front and rear tyres, what is the greatest distance in kilometres my car can travel?

wait... does interchanging the wheels have anything to do with how far the car goes?

Even if the wheels are interchanged, the new back wheel will still wear out before the front wheel. hm can the car drag itself along with one set of wheels? cuz then we need to know whether the car has front wheel drive or back wheel drive.

[hide]Let a tire's "health" represent how much wear it went through.  A health of $ 1$ represents a brand new tire, and a health of $ 0$ represents the tire is completely worn.  Any value in between shows the distance.  For example, if a front tire went 20,000 km, its health would be $ 0.5$.

Let the total health of a set of four tires be the sum of the healths of each of the tires.  So it starts at $ 4$.

Every kilometer, the total health decreases by $ \frac{1}{40000}+\frac{1}{40000}+\frac{1}{60000}+\frac{1}{60000}$, or $ \frac{1}{12000}$.  So the answer has an upper bound of $ \frac{4}{\frac{1}{12000}}$ or 48,000 km.

A 48,000 km trip is possible.  Let $ T_{1}$, $ T_{2}$, $ T_{3}$, and $ T_{4}$ be the tires, and let $ H_{n}$ be the health of $ T_{n}$.
Drive 24,000 km with $ T_{1}$ and $ T_{2}$ in the front and $ T_{3}$ and $ T_{4}$ in the back.  After 24,000 km, $ H_{1}=H_{2}=0.4$ and $ H_{3}=H_{4}=0.6$.  Then switch the tires so that $ T_{3}$ and $ T_{4}$ are in the front and $ T_{1}$ and $ T_{2}$ in the back.  After another 24,000 km, $ H_{1}=H_{2}=H_{3}=H_{4}=0$.  So the answer is $ \boxed{\text{48,000 km}}$.[/hide]

[quote="lingomaniac88"]After another 24,000 km, $ H_{1}=H_{2}=H_{3}=H_{4}=0$.[/hide][/quote]

Wouldn't $ T_{1}$ and $ T_{2}$ be "dead" (I'm assuming that these are the $ 40000$ tires) only after $ 16000$ kilometers? Since they are already at $ 0.4$ health, they can only go for $ 16000$ more kilometers.

I remember a Mathcounts problem similar to this, but the vehicle only had three tires.

I said that you switch the tires around.  Move the front ones to the back and the back ones to the front.


function
inequalities
inequalities unsolved

[quote="blahblahblah"][quote="FOURRIER"][size=150]Its very Bizarre because yo didnt use f(0)=0 or f(1)=0
U just used f(0)=f(1).
There Must be an  ERROR[/size][/quote]

no, there isn't.  Nothing is changed if we replace $f$ with $g(x) : = f(x)+c$ for some constant $c$.[/quote]

your right


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
geometry

I got a 138. :( Made two stupid errors but what's done is done. Preparing for the AIME now!

141. could've done better, oh well. considering that this was a pretty easy AMC 10/12 year, assuming AIME is also easy, (like in 2008) do people think a 7 will cut it for USAMO (im pretty sure an 8 would get me in)

I think a 127.5, made very dumb mistakes. I just can't control them. I thought 24 and 25 were surprisingly easy though, didn't you?

overall, i thought it was much easier than the 10A. does anyone agree with me?

Yes, I totally blanked in the middle of it, but still did pretty well. Got #20 last minute, and spent too much time on #23. It was fun :]

[quote="DarkKnight"]I actually did better on the AMC10A with a 144, but if it hadn't been for stupid mistakes on both I think I would've been able to get perfect scores. But stupid mistakes are part of the test after all, so I'm not going to complain about any of my scores. :) Planning on making USAMO this year, I'm thinking a 7-9 will be the cutoff. I want to get at least 10.[/quote]

you think the cutoff will be 7-9?!?! that high even with the extensions? i thought the floor cutoff only depends on the difficulty of the AIME, but the cutoff index for <=10 graders depends on the difficultly of the AMC 10/12 and AIME combined. due to the trend of math competitions this year being really easy, i think the cutoff will be around 6-7 MAYBE (unlikely) 8. but i've been wrong (many times) before.

incidentally, DarkKnight, you got the same thing on 10A and 10B as my sister, 144 on the A and 138 on the B.

138 on A, 132 or 138 or B.  Multiplied wrong and added wrong.

For the clock one, just count directly: x:yz. 8 possibilities for x, 5 possibilities for y, 9 possibilities for z, and 60*12 total possibilities.

My score: look at sig.

[quote="dgreenb801"]Talk about stupid mistakes, here's what I did:
For the one with the clock, at first I only thought 1:00-1:59 would work with 1 in the hours place, it took me a while to realize that 12 was too and even more to realize 10 and 11 were also possibilities (I kept getting the wrong answer). Then, it took me even more time to realize that a clock only goes up to 59, and not 99. Then, I made two subtraction errors, realized both eventually, but I forgot that :01 was a possibility in addition to :11 (which was already counted), :21, :31, :41, and :51. So I guessed 5/8 and got it wrong.
I have huge trouble with these problems with all of these details (like 23 also). Right when I see them I am disappointed, which kind of makes it a self-fulfilling prophecy to get it wrong, but still how can I improve on this stuff?[/quote]

I think the best way to prevent such errors is to think of a more systematic approach. Here's how I did it (this was on 12B too): the clock is correct iff the hour has no digit 1 and the minute has no digit 1. There are $ 8$ correct values for hours (2,...,9) out of $ 12$ and $ 5\cdot 9 \equal{} 45$ for minutes out of $ 60$. The answer is then $ 8/12\cdot 45/60 \equal{} 1/2$.

Also, solving it in two ways can reduce chance of errors on these types of counting problems as well.

EDIT: 1=2 beat me.

138.
sad thing is, i missed #7 and #12.
lawl.

I love how everyone's complaining about their near-perfect scores  :roll: I got 130.5. You think your stupid mistakes were bad? I got #4 wrong, marked the diagram incorrectly. I would've gotten 18 and 22 (left them blank, along with 24) if I hadn't forgotten that similar triangles even existed  :roll: Ah well, what's done is done. Time to go through AoPS volume 2 in 19 days...

Oh dang i epic phailed. This was my first AMC10, and i am only in 5th grade, but i still phailed. I got 105.
Question 1: I thought that bagels were 50 cents and muffins were 75 cents. Muffins taste better, so I assumed that muffins cost more.

Question 15: Somehow I messed up on multiplication when i did trial and error.
Question 16: NO CLUE HOW TO DO THAT.
Question 17: Why did I get this wrong? How was the answer C? Can someone please tell me?
Question 19: I forgot how to read numbers and forgot that 10 had a 1 in it.
Question 20: After the test, I looked over this. HOW COULD I HAVE FORGOTTEN AA SIMILARITY?!?!
Question 21: Didn't even try, thought you needed the first to digits of every power of three.
22: Didn't even try.
23: Didn't try.
25: Somehow i thought that 2^6 was 256... don't know how that happened...

I think I got a 126. I answered 21 questions, but I made a stupid mistake on the one about the remainder when you divide the sum of 3^0+3^1....+3^2009 by 8. I saw it went 1,3,1,3 so I multiplied 4 by 2010, not realizing that every 2 add up to 4. So I came up with 8040 mod 8=0 instead of 4020 mod 8=4.

Edit: Make that a 120. On the one about the birds in seeds, I thought the bird feeder was just being refilled, I didn't realize a whole quart was being added every time.

I got Number 1 wrong, number 4 wrong and divided wrong on 25... I think I got <120.... Please dont make fun of me.

I got a 145.5, left 23 blank because I misread and got a fraction that was ugly. I agree with the people who said number 25 and 24 were really really easy.

Man, I totally felt that the AMC 10B was way easier than the A form. Sadly I got a 138, I made two really stupid errors. #3(I swear the 25 in the corner of the page said 30) and #22 the cake problem where I thought it said cubic inches of icing for some reason.

I qualified for AIME with a 138! This was my first AMC (other than the A, I guess :P) on which I got a 114 (haha. 24 point improvement.) so I'm super excited. I don't even know where I'm going to take the AIME, and I have seen maybe one AIME test...  :wow: I'm completely not thinking about my dumb dumb dumb error on number 16 (I also missed the cake problem which I know understand, but I was wondering: who eats 2 inch cubical cake and slices it that way?)

Last year I thought I got a 144 on the B, but I actually got a 138.  This year I circled my answers on the test booklet and checked them, and I actually did get a 144.

Yeah I'd agree that this years AMC 10B was fairly easy (Unofficial 150), but I havent seen the AMC 10A yet
My school registered late and missed the A date, so I didn't get to take both (for better or for worse, b/c I might've taken the 12B)
Quite a bit of luck I'd say, no crazy geometry problems (Luckily I reviewed trig identities the morning of the test [#20]) this time
Since my USAMO fail last year I hadn't even touched math problem solving ^^
So I suppose now i have to start reviewing for AIME/USAMO >:O

How did you still get that 1.5 on the question if you got it wrong?

121.5, hopefully just qualifying for AIME :heli: I answered 20 questions, so I could get one wrong and make it. I was pretty sure of most of my answers (except 21, powers of three). I got 14 (millet seeds) wrong, but the rest of my answers were correct! I jumped up and down like crazy. So much euphoria... :D

Time to worry about the answer form  :maybe:

Oh, and he could've left it blank...

Sadly, the only one I got wrong was #1, but then again, there's no room for excuses.

145.5, had 2 minutes to do #25, didn't understand the question the right way, so I left blank.

oh well at least I did better than on 10A

w00t I tied shentang

I answered 21 questions for a potential of 132. 1,3,1,3...etc. for 2010 terms does not equal 4*2010. 126. Adding a quart of seeds, is not the same as making the total one quart. 120. I knew I got 2 wrong, so I was really nervous when I checked my answers off my scratch paper at school today. I got all the other ones right! I'm so happy, I qualified. I got a 91.5 last year. Yeah, I could've gotten a 132, but I qualified while sick. The AIME is going to be intense, but I'm glad to have a shot at it.

I got a 132 (21 right, 4 blank)
I made a really stupid mistake on the clock problem when I thought seconds counted. :(

But at least I made AIME(and 27 point improvement over 10A :oops: )


[quote="DCTPKTCSPHN"]Let $ a,b,c\in Z^ \plus{} ,a\neq c,b\neq c$.Prove that $ \exists$ infinite $ p\in {P}$ such that $ \exists n\in Z^ \plus{} : a^n \plus{} b^n \minus{} c^n\vdots p$[/quote]

What is the set P?

Primes.
Problem was posted before by the way.


MATHCOUNTS
Princeton
college
calculus
AMC
AMC 10
ARML

Any memorable moments of 2007?

The "DONT TAZE ME BRO" quote was very funny and it was Yahoo!'s favorite all year quote.

I graduated 8th grade in June  :) .

hushed the audience at state CD
saw a former MathCounts teammate get top 10 princeton
got to AIME.  :blush:

I think we should all thank the AoPS Staff for a wonderful year on this brilliant forum.

Thank you AoPS.

Oh yeah and memorable things........

-AoPS helped me greatly improve my Mathematical skills (though it's still not that good)
-Got into the Chemistry Olympiad Scholar School
-Many exciting opportunities to improve myself
-Developed a love for literature (no, not Harry Potter)
-Harry Potter came to and end (how sad)
-A girl said 'I love you' to me but I think she was drunk (or a hangover)
-Bush said OPEC instead of APEC, and Austria instead of Australia
-The Chasers dressed up as terrorists and managed to get pass the guards during APEC
-A great IMO performance for everyone
-A great year in general

[quote="BanishedTraitor"]-A girl said 'I love you' to me but I think she was drunk (or a hangover)
[/quote]

dude that is pretty awesome, was she hot?

Very, but as I said she was probably drunk, hung over or just simply high.

P.S. A Potasium Iododisulphide would have been nice  :D  lol get it?

Uhh....

"He has an ax!" "But he has Bud Light!" commercial premiered.

USA commercial for you who don't get it...

Lol I remember that commercial. This couple was driving in the dead of the night, when out appears some random dude on the side of the street. He is carrying an axe and a case of Bud Light. He is looking for a ride. The girl says "Oh no! He has an ax! Let's move on." However, the guy says "But he has Bud Light! He's coming in."  :rotfl:

[quote="Nerd_of_the_Ages"]Best event in my life: AoPS brought me in to the realm of math. in 2006, i had no love for math whatsoever[/quote]

Ditto, but I was only good at math, not a lover of it. :)

[quote="LevatorNasolabialis"]Bush still did not learn to say nuclear correctly...[/quote]

 :rotfl:  I remember that. Classic.

2007 was just an all-around good year for me.  I can't really think of anything bad that's happened in the last year.

(Btw, what's the 2008th post of GFF?)

2007..
6th grade.
Nice year.
Better than 7th grade anyways.

[quote="miyomiyo"](Btw, what's the 2008th post of GFF?)[/quote]

There are a total of 2002 curernt GFF topics. I'm not sure what you mean by the "post" though, that might be harder to track down.

[quote="Temperal"][quote="Nerd_of_the_Ages"]Best event in my life: AoPS brought me in to the realm of math. in 2006, i had no love for math whatsoever[/quote]

Ditto, but the only math I knew back in 2006 was how to solve one-var linear equations, and I only learned that in November or something... before that I didn't even know what algebra was.[/quote]
thats same with me, only i didnt go from pre alg to calculus is less than 2 years :P

i lost the game too many times to count.
I joined AoPS :)
i got smarter(by a lot)
i made states MC.
we exploded 3 coke and mentos at once and covered a good portion of a parking lot(sci oly team)
we had a spitball fight in a mexican restaruant(sci oly team)

Dope I just lost the game. Bad start to the year.

[quote="abacadaea"]
I joined AoPS :)
[/quote]

That was probably the best thing that happened to me this year too. :lol:

Oh Oh Oh I know!
I got into my school math team for the first time.
 :lol:

Actually, that's not determined yet :wink:

let's see, I joined AoPS, and fell down the stairs.  Ok, not the one with the stairs.

I finally found out that when you type A S S you get ass.

I joined.

I explained the process of milk making.

The terror began, the muffin-man was born.

lol

Let's see...I...well...uh...read the 7th harry potter? I did like nothing last year. I honestly can think of nothing of particular merit that I did in 2007.  :oops_sign:

:o So you are to blame for this absurd madness of the Muffin Man.

Personally, I felt 2006 was a better year for me.  I was closer to getting a girlfriend in 2006 than in 2007.  I let a perfect score on the AMC10 slip through my fingers because I filled in the wrong bubble on the answer sheet.  I managed to correctly guess (due completely to luck) how to factor the ugly quartic I got from ARML tiebreaker #1, but thought I was wrong, and let a thousand dollars slip through my fingers.  I'm hoping that 2008 will be better for me.

[quote="7h3.D3m0n.117"][quote="lingomaniac88"]Personally, I felt 2006 was a better year for me.  I was closer to getting a girlfriend in 2006 than in 2007.[/quote]

Based on your definition of a good year then methinks 2008 will be a great one for me  :D .[/quote]

Lol, good luck (no sarcasm intended).

[quote="kimmystar94"]i know!
i went to CTY Lancaster 2007 session 2!
i love cty..[/quote]

i went to CTY Saratoga Springs session 1.
i hated it.

Copper, not wood!

(Sorry, Siddhant, I just had to say that.)

...that post was not posted.
Uh...we sweeped the quiz bowl tourneys in MI?

[quote="erabel"]I finally got SSBM? I need to unlock Ganondorf, any suggestions?[/quote]

Beat the event match called "Triforce Gathering." If you don't have it yet, just play 1P until you get it. The event match is pathetically easy; it should be no problem.


How hard would it be to get a perfect score on the amc 10 if you are getting 120's on the amc 12.  I'm in tenth grade although i'm a bit rusty with mental math.  So do you think its possible for me to get a perfect?  Last year i got a 121.5 on amc 10 and i could do all of the problems but the time limitation is just extreme.  I can't seem to get that insane speed down.

I got 120 on the AMC 12 last year and 144 on AMC 10   ;)  If you're having problems with the time limit do practice tests and time yourself.


ABCD is a rectangle.  P is the point on CD such that CP=CB.  R is the point on AD such that AR=AB.  If PR=4 and BR=3*BP, find the area of the rectangle ABCD.

Edit: The letters are right now.

[quote="E^(pi*i)=-1"]ABCD is a rectangle.  P is the point on CD such that DP=DB.  R is the point on AC such that AR=AB.  If PR=4 and BR=3*BP, find the area of the rectangle ABCD.[/quote]

how come it says "R is the point on AC" ? isn't AC a diagonal? or are we supposed to draw a diagonal in...  :?

I can't seem to draw a rectangle using the conditions given, even if I switch the letters of the rectangle.

[quote="catcurio"]I can't draw the rectangle either... How is BR longer than BP? :?[/quote]

That's exactly what I am thinking. Since $P$ is not inside the rectangle ($P$ is on the lasting part of $CD$) and $R$ is inside the rectangle, how come $BR$ is longer than $BP$? :!:


I can't find it in the forum...
Find all primes $ p,q$ such that $ pq|2^p\plus{}2^q$.

Let $ p\le q$. If $ p\equal{}2$, then $ q\equal{}2$ or $ q\equal{}3$.
If $ p\ge 3$, then p,q - odd and $ 2^{p\minus{}1}\equal{}\minus{}1\mod q, 2^{q\minus{}1}\equal{}\minus{}1\mod p$ it give $ ord_2(p\minus{}1)<ord_2(q\minus{}1)<ord_2(p\minus{}1)$ - contradition.

I don't think this is correct... Can someone write a solution (or link to the solution)?

You could see a proof in here: http://www.mathlinks.ro/viewtopic.php?t=57586

In fact, the Rust's solution is right and the same.

By the way, this problem looks similar to 2. question of 2009 Chinese Mathematical Olympiad.

Thanks pluricomplex :)  I didn't know that in Rusts solution $ ord_2(p\minus{}1)\equal{}v_2(p\minus{}1)$.

http://matematyka.pl/17679.htm
http://matematyka.pl/57146.htm
[b]limes123[/b] I suppose you speak Polish well  :D


A function $ f: \mathbb{N}\mapsto \mathbb{N}$ fulfilles the following

($ i$) $ f(a+b)=f(f(a)+b)$
($ ii$) $ f(a+b)=f(a)+f(b)$ if $ 0\leq a+b < 10$
($ iii$) $ f(10)=1$

How many three-digit numbers $ n$ are there such that $ f(n)=f(2^{3^{4^{5}}})$


What it is IMO? I have seen it here many times.. Is it some abbreviation? Or some famous strategies? I am interested in it.


Prove that with $n \ge 2$  there are a set $S$ has $n$ elements such that $a.b$ divisible by $(a-b)^2$ with any $a,b$ distinc in $S$

A set which has this property is $S = \{n!\,(n! + k) \,|\, 0 \leq k \leq n-1\}$. Assume $a$ and $b$ are two different integers in $S$. Hence $a=n!(n! + k)$ and $b = n!(n! + m)$ where $0 \leq m \neq k \leq n-1$. Consequently

$\frac{ab}{a \,-\, b} = \frac{(n!)^2(n! + k)(n! + m)}{n!(k - m)} = \frac{n!(n! + k)(n! + m)}{k - m}$

is an integer since $0 < |k - m| < n$ implies that $(k - m) \,|\, n!$.

Let $S_{2}=\left\{1,2\right\}, S_{n+1}=\left\{x+A|x \in S_{n}\right\}\cup \left\{A\right\}, A=\prod_{x\in S_{n}}x$

Then you can prove $S_{n}$ is the answer.

[quote="leecom"]I think you should to read the problem again.[/quote]
This problem has many solutions
It is easy problem

It is an easy problem??????
 tanlsth, I don't think so.

Can you give me another solution,tanlsth.


linear algebra
matrix
search
linear algebra unsolved

Let $ PAQ\equal{}LR$ be the triangular descomposition with Pivot's search (?) in the remaining matrix of a matrix $ A\in\mathbb R^{n\times n}.$ Prove that for $ A^{\minus{}1}\equal{}(a_{ij}^{*}),$ \[ \big|a_{ij}^{*}\big|\le n\cdot2^{2n\minus{}4}\max\Bigg\{\frac1{|r_{11}|},\ldots,\frac1{|r_{nn}|}\Bigg\}.\]

Spanish:

Hola ..¿eres dela universidad de Concepcion y estudias ingenieria civil matematica y estas tomando Analisis Numerico 2 por si acaso?

Eso es porque yo postie la misma pregunta en este link:

http://www.mathlinks.ro/viewtopic.php?t=195918

Esta pregunta me salio en una tarea, y no pude obtener su respuesta

 saludos

PD: ahh ya sé quien eres (krizalid de fmat xD)..se agradece tu ayuda :)


ARML algebra polynomial function induction

Hi. I am from the United States of America. i tried asking this question to a couple guys in the SoCal ARML forum but they're a bit too advanced for me(freshman). I was hoping you'd understand and answer it without using any freaky theorems or postulates no one's ever heard of. just use simple polynomial factoring.

Prove:
if $n\in\&\&Z$
then prove $n^{7}-n$ is divisible by $7$.

I will give the reverse of the answer I posted in the other forum.

An example of a polynomial that is divisible by $7$ for any integer is

$(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)$

Because each factor is distinct $\bmod 7$.  We would like to compute the difference between this polynomial and the given one to prove that the given is also divisible by $7$.  Let's expand.

$n(n^{2}-1)(n^{2}-4)(n^{2}-9)$
$n(n^{4}-5n^{2}+4)(n^{2}-9)$
$n(n^{6}-14n^{4}+49n^{2}-36)$
$n^{7}-14n^{5}+49n^{3}-36n$

The difference between the polynomials is

$-14n^{5}+49n^{3}-35n$

Which is always divisible by $7$.

factorize n
n(n^6-1)
and by FLT
if n is coprime to 7
7|(n^6-1)
else n is divisible by 7
so for all integer n, n^7-n is divisible b 7, QED

or if your feeling bored check n=-3 to 3 mod 7 (the fact it is an odd function makes it a bit faster for you) and conclude by the naturally inductive property of modulus that for all integer n, n^7-n is divisible b 7.

This problem [i]is[/i] (the full statement of) Fermat's Little Theorem for $ p \equal{} 7$, so citing it does nothing.

[quote="bigboy82892"]Hi. I am from the United States of America. i tried asking this question to a couple guys in the SoCal ARML forum but they're a bit too advanced for me(freshman). I was hoping you'd understand and answer it without using any freaky theorems or postulates no one's ever heard of. just use simple polynomial factoring.

Prove:
if $ n\in \& \& Z$
then prove $ n^{7} - n$ is divisible by $ 7$.[/quote]

To be divisible by $ m$, the expression must evaluate to $ 0\mod m$ for each of $ n = 0,1,\dots,m-1$. Let's factor:

$ n^7 - n = n(n^6 - 1) = n(n^3 - 1)(n^3 + 1) = n(n - 1)(n^2 + n + 1)(n^2 - n + 1)(n + 1)$

Plugging in $ n = 0,1,2,3,4,5,6$ you can see that the expression evaluates to $ 0\mod m$.

[quote="t0rajir0u"][quote="hello.world"]To be divisible by $ m$, the expression must evaluate to $ 0\mod m$ for each of $ n \equal{} 0,1,\dots,m \minus{} 1$. [/quote]

You need to show that the value of the expression $ \bmod m$ is defined up to the value of $ n \bmod m$, which you haven't.

[quote="hello.world"]Plugging in $ n \equal{} 0,1,2,3,4,5,6$ you can see that the expression evaluates to $ 0\mod m$.[/quote]

You can do that to the original expression without any factoring; it requires more computation, but it's essentially the same proof.  Dadeyemi's suggestion makes this a little easier.

Consider, in any case, [url=http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem]proofs of FLT in full[/url].[/quote]

 :surrender: Yeah...I wasn't trying to be completely rigorous. Your proof does seem to be the best one here.

This can be done by induction.

It is clear that

 $ 7|1^7 \minus{} 1 \equal{} 0$

Now in inductive step v assume

$ 7|n^7 \minus{} n$

For $ n \plus{} 1$ we have

$ (n \plus{} 1)^7 \minus{} n \minus{} 1 \equal{} n^7 \plus{} 7n^6 \plus{} 21n^5 \plus{} 35n^4 \plus{} 35n^3 \plus{} 21n^2 \plus{} 7n \plus{} 1 \minus{} n \minus{} 1 \equal{} (n^7 \minus{} n) \plus{} 7n^6 \plus{} 21n^5 \plus{} 35n^4 \plus{} 35n^3 \plus{} 21n^2 \plus{} 7n$

The term is divisible by $ 7$ wich concludes the induction step.(we can do the same thing for negative numbers)
	$ \blacksquare$


How can I solve the equation:

$x^{3}-6x^{2}+9x-3=0$ :?:

Con la solucion general de una cubica, haciendo $x=y+k$ de manera que la ecuacion se transforme en una en y sin termino cuadratico (calculas el valor de $k$), luego sustituyendo $y=u+v$ y asumiendo que $u+v$ y $uv$ son linealmente independientes.

Hallas $u$ y $v$ y luego hallas $y$ y luego $x$.

OK that looked very ugly....But as I remember, the formula for solving cubics is pretty ugly.


how do i generate the table of contents without the whole file being generated?
i just want to send my toc to my supervisors but the actual content is still crappy so i don't want that attached to the document.

thanks

I don't know about just generating the table of contents, but, if the toc is the only thing on its page, and you have imagemagick, you can just extract that page (there are probably other programs that do similar tasks).  For information on downloading imagemagick see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148457]this thread[/url].

What I will do is to comment all the content and leave it with just the sections and subsections headers.

sounds a little bit too much like hard work.

Most pdf programs can take out pages from a pdf, i think there are some free online services that do it as well.  You can compile your paper and then just take the table of contents.


calculus
integration
algebra
binomial theorem
calculus computations

Can someone please calculate this integral?
$ \int\frac{x^{n}+x^{n-1}+\dots+x+1}{(x-1)^{n}}dx,\hspace{0.2cm}x>1,$ and $ n\in\mathbb{N^{*}}.$

MOD: Moved to the calculus forum.

$ \textrm{Integral calculus}\subset \textrm{Calculus}$, and as such it belongs in the calculus forums.  (You may notice the announcement at the top of this forum titled, "PLEASE do not post calculus problems in this forum!")

[hide="Solution"]$ \int\frac{x^{n}+x^{n-1}+\dots+x+1}{(x-1)^{n}}dx = \int\frac{x^{n+1}-1}{(x-1)^{n+1}}dx.$  Make the substitution $ y = x-1$ to express this as $ \int\frac{(y+1)^{n+1}}{y^{n+1}}dy$.  From here, it is easy: expand out by the Binomial Theorem, integrate term-by-term, and back-substitute to get the final expression.[/hide]


conics ellipse inequalities vector quadratics

The focuses of an ellipse are: $F_1: (-1,0)$ and $F_2: (1,0)$. The line $y=x+5$ and the ellipse have intersection(s).
Find the least value of $a$.

I also got $\sqrt{13}$. I used $\Delta$ to get it and I consider it not beautiful, so I am waiting for another solution.

Well, I would have gotten that if I said the focal length was 1.  But... as far as I can remember the focal length is the distance between the foci, not from the center to a focus, so I used c=2 instead.

The equation of the given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{a^2-1}=1\ (a>1),$ thus use Cauchy-Schwalz's inequality, we have

$(y-x)^2=\left(\sqrt{a^2-1}\cdot \frac{y}{\sqrt{a^2-1}}-a\cdot \frac{x}{a}\right)^2$

$\leq \left\{(\sqrt{a^2-1})^2+(-a)^2\right\}\left(\frac{y^2}{a^2-1}+\frac{x^2}{a^2}\right)$

$\Longleftrightarrow 5^2\leq (2a^2-1)\cdot 1$

$\Longleftrightarrow a^2\geq 13\ (a>1)\Longleftrightarrow a\geq \sqrt{13}.$

The equality holds when $\left(\frac{x}{a},\ \frac{y}{\sqrt{a^2-1}}\right)$ is pararell to $(-a,\ \sqrt{a^2-1})$ in the same direction, and since the latter vector is unit vector, we have

$\left(\frac{x}{a},\frac{y}{\sqrt{a^2-1}}\right)=\frac{1}{\sqrt{2a^2-1}}(-a,\ \sqrt{a^2-1})$

$\Longleftrightarrow (x,y)=\left(-\frac{a^2}{\sqrt{2a^2-1}},\frac{a^2-1}{\sqrt{2a^2-1}}\right).$

Therefore desired minimum value of $a$ is$\sqrt{13}.$

Yeah, so you're using c=1.  I would have gotten that too, but I used c=2.  So, the focal length is half the distance between the foci?
I didn't use cauchy, I set up the ellipse equation and plugged in the $y=x+5$ and solved, setting the resulting quadratic equal to 0 so the lines are tangent.  So you get $a^2 + b^2 = 25$ and $a^2 - b^2 = c^2$, and the two answers result from the two values of c.


I wonder if the old problems had an official solution?Where can I find it?

You will have to wait the section is in reconstruction.


Mamy nieujemne $a,b,c$ z suma rowna jeden. Udowodnij:

$a\sqrt{b}+b\sqrt{c}+c\sqrt{a} \leq \frac{1}{\sqrt{3}}$

Bardzo proste, ale calkiem fajne. Zamienmy $a: =a^2$, $b: =b^2$, $c: =c^2$ no i mamy dowiesc, ze:
dla $a^2+b^2+c^2=1$ zachodzi $a^2b + b^2c + c^2a \leq 1/\sqrt(3)$
Podnoszac do kwadratu, ujednoradniajac i redukujac dochodzimy do rownowaznej postaci:
$a^6 + b^6 + c^6 + 3(a^2b^4 + b^2c^4 + c^2a^4) + 6a^2b^2c^2 \geq 6(a^3bc^2 + a^2b^3c + ab^2c^3)$

Zauwazmy teraz, ze na mocy AM-GM zachodzi:
$a^2b^4 + a^2b^2c^2 \geq 2a^2b^3c$

Ukladajac podobne nierownosci, sumujac stronami i mnozac razy 2 dostajemy:
$2(a^2b^4 + b^2c^4 + c^2a^4) + 6a^2b^2c^2 \geq 4(a^3bc^2 + a^2b^3c + ab^2c^3)$

Wystarczy, ze dowiedziemy teraz:
$a^6 + b^6 + c^6 + a^2b^4 + b^2c^4 + c^2a^4 \geq 2(a^3bc^2 + a^2b^3c + ab^2c^3)$

W celu dowiedzenia tej nierownosci zauwazmy, ze jak poprzednio, na mocy AM-GM mamy:
$a^6 + b^2c^4 \geq 2 a^3bc^2$
ukladamy jeszcze dwie podobne nierownosci, sumujemy stronami i dostajemy teze, QED.

Można też skorzystać z Jensena dla $f(x) = \sqrt{x}$ i faktu, że $ab+bc+ca \leq a^2+b^2+c^2$.

$af(b)+bf(c)+cf(a) \leq f(ab+bc+ca) \leq \sqrt{\frac{3(ab+bc+ca)}{3}} \leq \sqrt{\frac{2ab+2bc+2ca+a^2+b^2+c^2}{3}} = \sqrt{\frac{(a+b+c)^2}{3}} = \sqrt{\frac{1}{3}}$

Druga nierownosc to by chyba nawet rownoscia mozna zastapic :)?

Mozna tez z Czebyszewa i QM-AM:

$(a+b+c)(\sqrt{\frac{a+b+c}{3}}) \geq \frac{1}{3}(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c}) \geq (a\sqrt{b}+b\sqrt{c}+c\sqrt{a})$

 :)

Na [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110370[/url] jest cały temat o tej nierówności. Moje rozwiązanie jest bardzo podobne do rozwiązania Michała Marcinkowskiego:
Z ważnonej $\mathrm{QM-AM}$:
$a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\leqslant \sqrt{ab+bc+ca}.$
Wystarczy więc udowodnić, że
$\sqrt{ab+bc+ca}\leqslant \frac{1}{\sqrt{3}}.$
Po podniesieniu do kwadratu i ujednorodnieniu:
$ab+bc+ca\leqslant \frac{(a+b+c)^{2}}{3}.$
Mnożymy stronami przez $3$ i skracamy $2(ab+bc+ca)$:
$ab+bc+ca\leqslant a^{2}+b^{2}+c^{2}.$
A to jest już oczywiste.


Let $ a$, $ b$, $ c$ be nonnegative reals such that $ a\plus{}b\plus{}c\equal{}1$. Let $ m\geq 2$ and $ n\geq 2$ be two integers. Prove that

$ a^mb^n\plus{}b^mc^n\plus{}c^ma^n \leq \frac{m^mn^n}{\left(m\plus{}n\right)^{m\plus{}n}}$.

When does the equality hold?

I have replaced the original problem (which was old and discussed several times) by the problem posted by Sbos in http://www.mathlinks.ro/Forum/viewtopic.php?t=2549 post #4.

  darij


Find the locus of $z\in\mathbb{C}$ which satisfy the following

$arg(z-1)-arg(z+1)=\frac{\pi}{4}$

where arg = argument

[hide]The triangle defined by 0, z-1 and z+1 has a right angle at 0 and a hypothenuse parallel to the real axis, of length 2. We want to find locus of the hypothenuse midpoints of all such triangles.

Since the median to the hypothenuse has half the length of the hypothenuse, this locus is the unit circle centered at the origin.

However, z-1 has a greater argument than z+1 only for z with positive imaginary parts, so we only want the upper semicircle.[/hide]

$arg(z-1)-arg(z+1)=\frac{\pi}{4}$      ($1$)

$z=x+yi$ where $\{x,y\}\subset\mathbb{R}$

note that $arg(a+bi)=\tan^{-1}\frac{b}{a}$, take tangents of both sides,

$\tan\left(\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}\right)=1$

Then using the expansion formulas

$\frac{ \frac{y}{x-1}- \frac{y}{x+1} }{ 1+\frac{y}{x-1}*\frac{y}{x+1} }=1$

$y[(x+1)-(x-1)]=x^2-1+y^2$
$2=x^2+(y-1)^2$

but i am pretty sure that i changed the solution set when i took tangents of both sides, because at ($1$), the LHS is only equal to $\frac{\pi}{4}$, but now it is also equal to $\frac{5\pi}{4}$, i am not sure how to resolve that right now...

Bictor717 , I am pretty much sure the circle is center at $(0,1)$ with radius $\sqrt{2}$ (look at Altheman's post) . Im also not very sure how to proceed after that , but by checking through mathematica , the locus is the upper part of the circle which is above the real-axis .

Oops, I thought pi/4 is a right angle. :blush:


Here is a problem:

There is a truck driver with a truck that can transport 1000 apples at a time. This driver has the habit of eating 1 apple per each km while he is driving. His/her task is to transport 10000 (10^4) apples from one city to another. The distance between the cities is 500km.

He/she may use any strategy of transporting apples. He/she may have less than full loads, may unload/reload at any place along the way from the first city to the second.

(So he/she may take a full load for 20km and then unload the remaining 980 apples and then go back for another 40, which will be 20 by the time the driver gets back to the unload point; then he may load the 1000 apples back to the truck and continue.)

The question is: what is the greatest number of apples that will survive the trip?

One trivial solution is to transport full loads from one city to the other, in which case 5000 apples will survive.

My task for you is to prove that the greatest possible number of apples that survive is 5868.
(What strategy produces this?)

Evidently you may ask yourselves why this problem appears in the college forum.

1. First, the way one gets 5868 leads to something that looks like $\frac{1}{n}+\cdots+\frac{1}{2n}\sim \log 2$, which is not the standard math olympiad type of solution.

2. Secondly, the problem appeared on an interview for a job at one of the most respected Wall Street hedge funds.

My challenge is that this may start a little discussion about why on earth we care about olympiads. Are they useful in terms of college mathematics? How about real-life mathematics?

Sincerely,
Andrei

(http://www.ajorza.org/math/)

I don't think anyone should pretend to understand all the mysteries of the human mind, especially not the minds of college admission officers, but it does seem that "play" with math problems, which is mostly what Olympiad competition is about, has a proven ability for [b]many[/b] (perhaps not all) participants to develop personal knowledge and problem-solving characteristics that make former Olympiad participants productive research mathematicians. Colleges, noting this, pay considerable attention to achievement in Olympiads in deciding whom to admit. Other people get in for other reasons, but Olympiad participants are, as Miths has already said, at least not disadvantaged when applying to college.

[quote="MithsApprentice"]I don't know about studying math in college or whatnot, but as far as getting into a college, I'm sure a noting on your app that you happen to be a 2-time USAMO qualifier or something like that can give you a nice edge.[/quote]

Considering that he's an IMO medalist and and finished in the top 20 on the Putnam one year, I suspect that he knows that.

His question was more of an academic (e.g, math for the sake of math) than a pragmatic (i will do math to get me into school A) nature.

[quote="blahblahblah"]Considering that he's an IMO medalist and and finished in the top 20 on the Putnam one year, I suspect that he knows that.

His question was more of an academic (e.g, math for the sake of math) than a pragmatic (i will do math to get me into school A) nature.[/quote]

That's quite interesting. I was just trying to think of something to say...lol

But anyways, yes I'm sure Olympiad math will help you in more ways than you can probably think of at the moment. Sometimes training like this helps indirectly, even subconsciously, in situations that have absolutely nothing to do with math.

Indeed, my question was less "why do high school students do olympiad math" as "why do grown-ups do olympiad math?".

Besides the evident fun-factor in high-school olympiads, their advantages are evident: it's the one place where you prove you can do math.

However, are olympiad problems used only for testing intellectual abilities, or is there something to the ideas in the olympiad world that are actually useful for real-life applications?

Andrei

This has not much to do with solving the problem, but I heard it discussed on a radio talk show a while ago.  Twas rather painful :(


A white crystalline ionic solid $ A$, because of its thermal instability, is used extensively as a direct explosive in mining and quarrying operations. At high temperatures ($ >300^{\circ}C$) it can explode violently giving off two colourless and odourless gases $ B$ and $ C$ and $ \ce{H2O}$. Under controlled thermolysis at about $ 250^{\circ}C$, a gas $ D$ is liberated with $ \ce{H2O}$ as a by-product. $ D$ is a moderately unreactive gas comprised of linear unsymmetrical molecules, as expected for a $ 16$-electron triatomic species. Like $ A$, $ D$ is also thermodynamically unstable and decomposes exothermically upon heating to give $ B$ and $ C$. With atomic oxygen, $ D$ gives $ E$ which exists in equilibrium with $ F$, which is paramagnetic. At high pressures and moderate temperatures $ F$ rapidly disproportionates to $ D$ and a brown gas $ G$. In the presence of $ C$, $ F$ is also quickly converted to $ G$, which exists in equilibrium with colourless $ H$. In a medium of high dielectric constant such as nitromethane, $ H$ undergoes a self-ionization equilibrium with $ I$ and $ J$. $ I$ is isoelectronic with $ \ce{CO}$ and $ J$ is the anionic constituent of $ A$. $ H$ can be oxidised by ozone to give a highly deliquescent light sensitive solid $ K$. X-ray diffraction studies show that solid $ K$ consists of an ionic array of linear $ L$ and planar $ J$. In the gas and solution state, $ K$ is probably molecular, $ L$ is isoelectronic with $ \ce{CO2}$.

Identify all the lettered compounds and write down all the equations as described.

[hide="First thought"]From the description given, it seems likely that A is ammonium nitrate and between the other species there are several nitrogen oxides.[/hide]

[hide]Maybe L is $ \ce{N2O}$ and  E is $ \ce{N2O4}$ and F is $ \ce{NO2}$ and I is $ \ce{N2} , J is \ce{NO3^\minus{}}$
[/hide]

Carcul is on the right track.

Madness, I can't really answer you unless you show some reasoning.

valeriummaximum, that is wrong.

[hide="Second thought"]D is $ N_2O$, since this molecule is triatomic, linear, and has 16 valence electrons. This gas decomposes on heating to $ N_2$ and $ O_2$, which are B and C.[/hide]

reasonings
[hide]
L is planar and isoelectronic with CO2 so N2O
E is a guess and so is F (maybe you could say that they were rather NO and NO2)
I is clearly N2 bcos its isoelectronic with CO
J is clearly NO3 - bcos A is NH4NO3 as its explosive terribly  :)

[/hide]

Planar does not always mean it's linear. If you read carefully then you should notice that $ \ce{N2O}$ is another species on the list.

$ I$ is not $ \ce{N2}$.

No one has given a satisfactory answer yet (only good suggestions).

Here are a few hints:

[hide="Hint 1"]

You have all identified that these are a class of reactions involving nitrogen.

[/hide]

[hide="Hint 2"]

What is the brown gas?[/hide]

[hide="Hint 3"]

What is the shape of $ \ce{N2O}$?[/hide]

[hide="Hint 4"]

You probably know by now what $ A$ is hence what is $ J$?[/hide]

I have lost the answers so I have to figure them out again  :P .

[/hide]


function
limit
real analysis
real analysis unsolved

let $ f: R\rightarrow R$ a twice-derivable function (if that's the name), such that $ f(x)>0$ and $ f''(x)>0$ for all reals $ x$. Suppose that $ lim_{x\rightarrow \plus{}\infty}f(x)\equal{}0$ (this hypothesis may be unnecessary, it was a problem with several parts). Prove that $ lim_{x\rightarrow \minus{}\infty}f(x)\equal{}\plus{}\infty$.

the stipulation $ \lim_{x\to\infty} f(x) \equal{} 0$ is necessary: take $ f(x) \equal{} e^x$.

You don't really need $ f$ to be twice differentiable (that's the name). As long as $ f'$ is strictly increasing, you know that the limits $ a\equal{}\lim_{x\to \minus{}\infty}f'(x)$ and $ b\equal{}\lim_{x\to \plus{}\infty}f'(x)$ exist in the extended sense (as elements of $ [\minus{}\infty,\plus{}\infty]$), and $ a<b$. If $ b\ne 0$, then $ \lim_{x\to \plus{}\infty}f(x)$ is infinite. Hence $ b\equal{}0$, but then $ a<0$, which implies the claim.


inequalities
Cauchy Inequality
inequalities unsolved

For positive reals x,y,z such that xyz=1 prove that

$(1+x)(1+y)(1+z)\geq 2(1+\sqrt[3]{\frac yx}+\sqrt[3]{\frac zy}+\sqrt[3]{\frac xz})$

This one can be proved by Cauchy inequality:
[tex] (1+x)(1+y)(1+z) \geq 2(1+\sqrt[3]{y/x}+\sqrt[3]{z/y}+\sqrt[3]{x/z}) [/tex]
[tex]\Leftrightarrow x+y+z+xy+xz+yz \geq 2(\sqrt[3] {y/x}+\sqrt[3] {z/y}+\sqrt[3]{x/z} [/tex]
By Cauchy inequaltity we have:
 [tex]y+zy \geq \sqrt{y^2z} =\sqrt {y/x} [/tex]
 [tex]z+xz \geq \sqrt{xz^2} =\sqrt {z/y} [/tex]
 [tex]x+yx \geq \sqrt{yx^2} =\sqrt {x/z} [/tex]
 [tex]2\sqrt {y/x} +1 \geq 3\sqrt[3] {y/x} [/tex]
 [tex]2\sqrt {z/y} +1 \geq 3\sqrt[3] {z/y} [/tex]
 [tex]2\sqrt {x/z} +1 \geq 3\sqrt[3] {x/z} [/tex]
 [tex] \sqrt {y/x}+ \sqrt{z/y}+ \sqrt{z/x} \geq 3[/tex]
Add them up and we have what we need. :)

Setting $X^3 = x$, $Y^3 = y$, and $Z^3 = z$, homogenizing, and expanding the given inequality yields the equivalent one $\frac{X^2}{YZ} + \frac{Y^2}{ZX} + \frac{Z^2}{XY} + \frac{XY}{Z^2} +\frac{YZ}{X^2} + \frac{ZX}{Y^2} \geq 2(\frac YX + \frac ZY + \frac XZ)$.
However, $\frac{X^2}{YZ} + \frac{XY}{Z^2} + \frac13(\frac YZ + \frac ZX + \frac XY) \geq  \frac{X^2}{YZ} + \frac{XY}{Z^2} +1 \geq 3\frac XZ$ where both inequalities follow by AM-GM.
Adding this inequality to its cyclic counterparts gives the required result.  Equality holds iff $X=Y=Z$ or $x=y=z$.

This is basically the same as APMO 1998 #3, which asks to prove
\[
\left(1+\frac ab\right) \left(1+\frac bc\right) \left(1+\frac ca\right) \geq 2 \left( 1 + \frac{a+b+c}{\sqrt[3]{abc}}\right)
\]

Where by basically, you mean exactly, right?

Setting $\frac ab = x$, $\frac bc = y$, and $\frac ca = z$, we get EXACTLY the given inequality.


inequalities calculus integration logarithms

(i) Solve the simultaneous inequalities, $x<\frac{1}{4x}$ and $x<0$; i.e. find a single inequality equivalent to the two simultaneous inequalities.

(ii) What is the greatest integer that satisfies both inequalities $4x+13 < 0$ and $x^{2}+3x > 16$.

(iii) Give a rational number between $11/24$ and $6/13$.

(iv) Express 100000 as a product of two integers neither of which is an integral multiple of 10.

(v) Without the use of logarithm tables evaluate \[\frac{1}{\log_{2}36}+\frac{1}{\log_{3}36}.\]

I'm bored, so I'll just post these solutions. :P

i) $4x^{2}>1\implies x>\frac{1}{2}, x<-\frac{1}{2}$
ii) Too boring.
iii)$\frac{\frac{11}{24}+\frac{6}{13}}{2}$
iv)$100000=2^{5}\cdot 5^{5}$
v)$\log_{36}2+\log_{36}3=\frac{1}{2}$

EDIT: Typo in i). I had forgotten a negative in the result.

Ok so boxedexe got i,iii,iv,v...

That leaves problem [hide="ii"]
$x^{2}+3x>16\implies x^{2}+3x+9/4 > 73/4$
$\implies x> \frac{\sqrt{73}-3}{2}\approx 2.77\;\text{or}\;x<\frac{-\sqrt{73}-3}{2}\approx-5.77$

$4x+13 < 0 \implies x <-3.25$

Largest number is $x=-6$
[/hide]

why are there 5 parts? also i don't see how this can possibly get on an olympiad...

its from 1973..plus its question 1..so free marks i guess

Yeah, I forgot to write down that the case $x>\frac{1}{2}$ does not count, but you forgot to flip the inequality, thus you obtained a wrong answer.

In V, I wrote a * instead of a + in the last sentence.