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MATHCOUNTS
number theory

I am in 8th grade and this is my 1st year participating in Math Counts (long story - forget about it..)

I've always been interested and pretty good at Math (aced my classes, blah blah)

However, Math Counts is at a whole new level... and I want to do good this year since it's my last chance. (I got selected in my school team)

I ordered AoPS volumes 1 and 2.
I read most of one (spent a long time especially on number theory, how to count, and probability)

Two seems really tough- kinda beyond me, and my Chapter competition is on February 24th. (I am in Cincinnati)
I have a 4 day weekend now- I can spend 6-7 hours each day studying and also spend 1-2 hrs on weekdays until next Saturday.

My team is ok.. not amazing or anything, so I don;t think I can get into State by team, I'll need to do it individually.

Question 1: How many people go to the State in my Chapter? (Cincinnati)

I've been doing Sprints and targets at the chapter level, from 2001-2006, besides the ones that my teacher at school gives me... I seem to only get between 22-25 on sprints and 6-7 on targets.

Question 2: I know this is kind of hard to answer but what would be the average score I'd need to get on my sprint and target to go to state?

Question 3: Also, what can I do to prepare for Math Counts. As said earlier, my chapter competition is next weekend (not this one) and I have a 4 day weekend in which I can do a solid 6-7 hrs of study everyday.

Please help me out. My main goal is to first get to state, and get a decent score at the state level, (not necessarily go to Nationals)

1. it depends on the size of the chapter, from 4 individuals to top 12. i'm guessing its top 12 because it's cincinatti.

2.probably about a total of 35 at least

3. problems usually get harder and harder as years pass, so it's a good idea to do state problems to practice for chapter

1. it really depends. you might want to search up on the internet for past results of cincy chapter. my chapter is top 4 teams and 5 ind. while a chapter as big as jhredsox's allow 7 teams and 12 ind., i think.

2. this also depends on how tough your chapter is. for example, in my chapter last year, a 44 was needed to advance individually.  :!: other chapters, a 31 could let you advance...

3. figure out you weakness area(s) and do more problems in that specific area. don't cram the last few days before chapters, instead let you mind cool off...

[quote=".:Logic:."]I am in 8th grade and this is my 1st year participating in Math Counts (long story - forget about it..)
...
Please help me out. My main goal is to first get to state, and get a decent score at the state level, (not necessarily go to Nationals)[/quote]

One Word: Relax. ( I've been told that for another test coming up)

and by do good, don't you mean you want to do well?

I'm sure doing tests would help; and rest? bleh this doesn't sound like me.

and at least top 4 get in, maybe more through district.

Ask Gyan for questions regarding Cincinatti. He's like the only active poster I know from there. Otherwise, post this in the Ohio local forum and see what they have to say.

I'm from Cleveland, but I can probably say that a [b]40[/b] would get you to States.

For the chapter competition, I would prepare with past chapter competitions under a 30 minute time limit for Sprint (not 40) and a 10 minute time limit for all 8 target round questions (pretty fast).

Good Luck and Hopefully I'll see you at states. :)

remember it depneds on the year, but just try your best

my chapter's big. we advance 8 teams and 20-30 individuals

making for 50-60 that go to states

jorian

Okay.. I can confidently get atleast a 25 on old chapter sprints. (taking full 40 mins)

However on state sprints, I get 16-20 :( (taking full 40 mins)

:(

I don't have much of a chance of making the state do I??  :|

Hmm... Will the chapter sprint this year, be as hard as state sprints from previous years?

I can guarantee myself ATLEAST 5 right on the target. That means I'll need a 27-28 to get a good chance at the state... :maybe:

 :stretcher:

[quote="mathmom"][quote=".:Logic:."]
Hmm... Will the chapter sprint this year, be as hard as state sprints from previous years?[/quote]

There's really no way to predict that.  But remember, if it's hard for you, it's hard for other people too, and conversely, if it's easy for you, it's easy for other people too.  What matters is how good you are relative to the other people in your chapter, not the number you get for a score.[/quote]
Also, the more prone you are to careless errors the harder you want the test to be.
I'm less prone than most people, so a nationals similar to last years, this year would benefit me.

[quote="vishalarul"][quote=".:Logic:."]Hmm... Will the chapter sprint this year, be as hard as state sprints from previous years? [/quote]

Do not discuss the difficulty of the 2007 Chapter Competition yet.[/quote]
He is asking, will it be?

[quote="#H34N1"]Can we say the cutoff this year for our chapter?[/quote]

No, that would give quite a lot of information about how hard the Chapter contest is.  Not giving that kind of information is the point of not talking about scores before everyone is done.

What I reccommend in the way of studying is to do 5 problems, take a 30 minute break, and repeat. If you cram too hard, it won't help. You'll get sick and tired of it and lose your motivation. I also reccomend you get as much sleep as possible.

It worked for me.  :D

Hmm I don't mind doing 6-7 hours a day....  :)
I still have a bit of today and tomorrow left for quality studying without school stuff..

[b]Are there any specific chapters in AoPS: Volume One that I should focus on?? I found Learning to Count and Probability since there's always 2-3 of those problems in sprint, any other chapters like that?[/b]

Sorry to double post.. but could someone please just answer the above question :?:

[quote=".:Logic:."]Ok last time I'm asking.. I'm desperate :P

Someone suggest chapters from AoPS Volume 1 that will greatly help for the Chapter Round- its this Saturday :([/quote]

[hide="Here you go"]Sych! Keep going
You that desperate?
Go Go Go
You really want to know?
Fine

I read chapters XX, XX, XX, XX, XX, XX, XX, XX, XX, XX, XX, XX and XX. These chapters are usually what many tests in mathcounts have. I'm sure there are more but these are what I read before my competition.

[color=red][size=75]Specific chapter number edited by moderator rcv[/size][/color][/hide]

[quote=".:Logic:."]Are there any specific chapters in AoPS: Volume One that I should focus on?? I found Learning to Count and Probability since there's always 2-3 of those problems in sprint, any other chapters like that?[/quote]
Logic:  You should not ask such a specific question while the Chapter or State rounds are in progress.  While your question might be innocent, it might also result in discovery of some of the content areas of this year's Chapter test.  That is absolutely forbidden.

Study all of AoPS Volume I.  Try all old contests you can get your hands on.  Try all of this year's and previous year's handbook problems you can get your hands on.

You certainly should practice.  Practice will increase your speed, your confidence, and your score.  But don't expect to become a MATHCOUNTS Champion with just one week of cramming.

I am locking this thread, because of the high risk that further responses will reveal areas of emphasis on the current Chapter competition.


Please give me a proof of Muirhead! Or tell me where I can find it in this forum. Thanks! :wink:

Already posted. Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15896]here[/url] and scroll down.

Except of it, a proof of very similar theorem which can be transformed into proof of the exact Muirhead can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17288]here[/url]. I heared that it also can be proved by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]Karamata inequality[/url]. A very short proof is in [url=http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf]this pdf by Dr. Kiran Kedlaya[/url] at the beginning of the 13-th page (12-th numbered page).
I hope it will help.


geometry
3D geometry
sphere
circumcircle
tetrahedron
geometric transformation
reflection

Given any five distinct points on the surface of a sphere, show that we can find a closed hemisphere which contains at least 4 of them.

Not sure about this, but here goes nothing.

[hide]
[b]Case 1:[/b] Let us first take any three points. If the circumcircle of these points passes through the center, then adding the fourth one defines a hemisphere, and we are done.

[b]Case 2:[/b] Otherwise, we add a fourth point. If the tetrahedron formed by the four points does not pass through the center of the sphere, then we clearly have such a hemisphere (cut the sphere with a plane passing through the center that is parallel to the face closest to the center).

[b]Case 3:[/b] Otherwise, we add a fifth point and repeat the above procedure with each of the four faces of the tetrahedron we have.

[b]Aside:[/b] [i]For points $ A$, $ B$, and $ C$ on a sphere, their respective reflections across the center $ A'$, $ B'$ and $ C'$, and a point $ D$, the tetrahedron $ ABCD$ contains the center iff. $ D \in$ spherical triangle $ A'B'C'$.[/i]

The four spherical triangles formed by the points are non-overlapping, thus the four spherical triangles formed by the projections of the points are also non-overlapping. Wherever we place the fifth point on the sphere, there is at least one such region that it is not in (it is either in none of them, or in only one of them).

Hence, upon adding the fifth point (and creating 4 new tetrahedrons), there is at least one tetrahedron that does not pass through the center, which implies a group of four points with a common hemisphere.
[/hide]


trigonometry
algebra
polynomial
sum of roots

Consider $ \sin 7x\equal{}0$, so that $ 7x\equal{}n\pi$ where $ n$ is an integer.

$ \sin 7x \equal{} Im (c\plus{}is)^{7}$, where $ c\equal{}\cos x, s\equal{} \sin x$.

Expand by the binomial and write $ c^{2}\equal{}1\minus{}s^{2}$, to get $ s(64s^{6}\minus{}112s^{4}\plus{}56s^{2} \minus{}7)\equal{}0$

So the sextic has roots $ \pm sin \frac{n\pi}{7}, n\equal{} \pm1, \pm2, \pm3$

Let $ s^{2} \equal{} t$, so $ \sin^{2}(\frac{n\pi}{7}), n\equal{}1, 2, 3$ are the roots of $ 64t^{3} \minus{}112t^{2} \plus{}56t \minus{}7\equal{}0$.

Then sum of roots $ \equal{}\frac{112}{64} \equal{} \frac{7}{4}$.

Then square of the sum of the roots of the sextic = sum of roots squared + twice cross products gives required sum $ \equal{}\sqrt{\frac{7}{4}} \equal{} \frac{\sqrt{7}}{2}$.

Sorry, a bit rushed at the end.

[quote="kushal.sharma09"]Prove that sin(2pi/7) + sin(4pi/7) + sin(8pi/7)= (root of 7)/2
Please help me out.
Thanks.[/quote]

Is this $ \sin \frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}=\frac{\sqrt_7}{2}$ what you mean ?


[quote="stancioiu sorin"]Pro[color=red]u[/color]vez que $8|13^{23} \cdot 27^{41} + 1$. :lol:[/quote]

:)

EDIT: Et une réponse rapide :
$13^{23} \equiv 5 \pmod 8$ (puisque congru à 5 mod 8 si puissance impaire et à 1 mod 8 si puissance paire)
Ensuite : $27^{41}=3^{123} \equiv 3 \pmod 8$ (car congru à 3 mod 8 si puissance impaire et à 1 mod 8 si puissance paire)
Donc, le produit est congru à 15 mod 8, d'où, en ajoutant 1, à 16 mod 8, soit 0 mod 8, ce qui conclut. :)


the sequence (x_n) is defined by x_0 =4, x_1 = 0, x_2 = 2c, x_3= 3b, and
 x_{n+4} = a*x_{n} + b*x_{n+1} + c*x_{n+2} ; where a , b and c are integers with b odd. prove that p^m|x_{p^m} for all primes p and positive integer m.


I noticed one day that

$\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot..\cdot\frac{k^{2}-1}{k^{2}}\cdot....=\frac{1}{2}$.

I can't find a very elegant proof for this. Does anyone know a good proof?


Hi everyone,
Sorry for the delay. Technical difficulties have forced us to temporarily use an alternate server for hosting JHMT materials. The online registration form, old problems/solutions, results from previous years, and other general information about the contest can now be found online at:

http://webhost5.nts.jhu.edu/~liondance/jhmt/index.php

There are still glitches with the website. Please do not register until the newest registration form ("eApplication for JHMT 2007") is posted online (probably tonight).

Once again, the tournament will be held on November 11, 2006. Thanks for your patience.

Regards,
JHU Math Club

And it has just come to my attention that the original server is now back in place.

http://www.tripleintegral.org/jhmt

The finalized registration form will be up shortly. Go register  :).


Prove or disprove that
\[ \frac1{3a^3\plus{}4}\plus{}\frac1{3b^3\plus{}4}\plus{}\frac1{3c^3\plus{}4}\le\frac37,
\]
where $ a,b,$ and $ c$ are positive integers whose product is 1.

positive integers!!!
 $ R^ \plus{} \rightarrow$  wrong for $ a,b \rightarrow 0$

[quote="Johan Gunardi"]Prove or disprove that
\[ \frac1{3a^3 \plus{} 4} \plus{} \frac1{3b^3 \plus{} 4} \plus{} \frac1{3c^3 \plus{} 4}\le\frac37,
\]
where $ a,b,$ and $ c$ are positive integers whose product is 1.[/quote]

If $ (a,b,c)\in\mathbb{N}^3$ (positive integers) such that $ abc\equal{}1$ (whose product is 1) then $ a\equal{}b\equal{}c\equal{}1$ and
$ \frac{1}{3a^3\plus{}4}\plus{}\frac{1}{3b^3\plus{}4}\plus{}\frac{1}{3c^3\plus{}4}\equal{}\frac{3}{7}\le\frac{3}{7}.$


calculus integration trigonometry geometry

A triangle has sides of 13, 14, and 15. The altitude to one of these sides is integral. What is that altitude?

[hide]Let A be opposite the 13 side. By the cosine rule, cos A = 0.6 so sin A = 0.8. BE = c sinA = 15x0.8=12[/hide]

[hide="or"]By Herons, the triangle's area is $\sqrt{(21)(6)(7)(8)}=84$. The area is also $\frac12bh$. The only value of $b$ that is a side length that makes $\frac{168}{b}$ an integer is $14$, so the altitude is of length $12$.[/hide]


vector
geometry
3D geometry
combinatorics proposed
combinatorics

Given 30 non-zero vectors in space, show that there must be two which have the angle between them less than $ 45^\circ$.

Kalva provides a kind of heavy-machinery solution. Does anyone have an elementary one?

Oh, sorry,

Each vector ~ u, join a cone with a circular base with axis of revolution is extended
by ~ u, whose length of a generator is 1 which is the angle between the axis and the generator is
equal to = 22, 5. The design was cut following the situation:
   ~ u
1
On this figure, the cone is the triangle shaded and obtained by gure tridimentionnelle
by turning the design around the axis ~ u.
Let the absurd assumption that every angle between vectors considered
exceed 2. This meant that all e interiors cones associated with vectors are separated.
The volume of a cone is the third of the proceeds from the surface of the disk base with the length
of the arrivals. It is therefore
3 sin2 cos. At their thirties, cones are therefore an equal volume
to:
10 sin2 cos> 4, 25.
But this is impossible because the volume of the ball radius 1, which contains all that is
3
4 6 4 19.

P.S.

I can't link to kalva. What should I do? :(

[quote="kunny"]
I can't link to kalva. What should I do? :([/quote]Go to [url=http://web.archive.org/web/20040211000554/www.kalva.demon.co.uk/index.html]web archive[/url].


algorithm
number theory
Diophantine equation
algebra unsolved
algebra

give [a^2/b]+[b^2/a]=[(a^2+b^2)/ab]+ab
    [x] is a integer y so that y  \leq x
   find integer a,b :D  :?

The only solutions of the Diophantine equation

$(1) \;\; {\textstyle [\frac{a^2}{b}] + [\frac{b^2}{a}] = [\frac{a^2+b^2}{ab}] + ab}$,

are $(a,b)=(n^2+1,n)$ and $(a,b)=(n,n^2+1)$, where $n \in \mathbb{Z} \setminus \{-1,0\}$.

[u][b]Proof:[/b][/u]
First of all, we may WLOG assume $a \geq b$ since $a$ and $b$ are symmetric in (1).  Secondly $a,b \neq 0$ by (1).
Let us consider the following three cases depending on the signs of $a$ and $b$:

[b]Case 1: [/b] $a,b<0$. Then in (1), $LHS < 0$ and $RHS>0$. Hence there is no solution of (1) in this case.

[b]Case 2: [/b] $a>0>b$.  Set $c=-b>0$. According to (1)

$(2) \;\; {\textstyle ac + [-\frac{a^2}{c}] = -[\frac{c^2}{a}] + [-\frac{a^2+c^2}{ac}]}$.

In (2), $LHS<0$ since $RHS<0$.  Therefore

$(3) \;\; {\textstyle ac +1 \leq -[-\frac{a^2}{c}]}$.

The fact $-[-x] < x + 1$  for all $x$, meaning

$(4) \;\; {\textstyle -[-\frac{a^2}{c}] < \frac{a^2}{c} + 1}$.

Combining (3) and (4) yields ${\textstyle ac < \frac{a^2}{c}}$, i.e. $a>c^2$.  Consequently $0 < {\textstyle \frac{c^2}{a} < 1}$, implying ${\textstyle [\frac{c^2}{a}]=0}$, which inserted in (2) gives

${\textstyle ac = - [-\frac{a^2}{c}] + [-\frac{a^2+c^2}{ac}]}$.

Now $-[-x] \geq x$ and $[x] > x-1$ for all $x$, which means

${\textstyle ac > \frac{a^2}{c} - \frac{a^2+c^2}{ac} - 1}$,

i.e.

$(5) \;\; a^2(a-c^2) < a^2 + ac + c^2$.

Dividing both side of (5) by $a^2$ we obtain

${\textstyle (a-c^2) < 1 + \frac{c}{a} + (\frac{c}{a})^2 < 3}$

since ${\textstyle 1 > \frac{c^2}{a} \geq \frac{c}{a}}$. Hence $0 < a-c^2 < 3$, i.e. $a - c^2 \in \{1,2\}$.

Assume $a - c^2 = 2$. Then by (5)

$0 > a^2 - ac - c^2 > (c^2)^2 - c^2 \cdot c - c^2$

since $a = c^2 + 2 > c^2$, i.e.

$c^2(c^2 - c - 1) \leq -2$,

yielding $c<0$.  This contradiction implies $a \neq c^2 + 2$.

Consequently $a = c^2 + 1$. Then in (2)

$ {\textstyle LHS = ac + [-\frac{a^2}{c}] = c(c^2+1) + [-\frac{(c^2+1)^2}{c}] = c^3 + c - (c^3 + 2c) + [-\frac{1}{c}] = -c-1}$

and

$ {\textstyle RHS = -[\frac{c^2}{a}] + [-\frac{a^2+c^2}{ac}] = [\frac{c^2}{c^2+1}] + [-\frac{c^2+1}{c} - \frac{c}{c^2+1})] = -c + [-\frac{1}{c} - \frac{c}{c^2+1}}]$.

If $c=1$, then $LHS = -2 \neq -3=RHS$. Thus $c > 1$, which means

${\textstyle 0 > -\frac{1}{c} - \frac{c}{c^2+1} > -\frac{1}{c} - \frac{c}{c^2} = -\frac{1}{c} - \frac{1}{c} = -\frac{2}{c} \geq -1}$.

Therefore ${\textstyle [-\frac{1}{c} - \frac{c}{c^2+1}] = -1}$, giving $RHS = -c-1 = LHS$.

So in this case the solutions of (1) are $(a,b)=(b^2+1,b)$, where $b \geq -2$.

[b]Case 3: [/b] $a,b>0$.  Combining (1) and the facts $a^2+b^2 \geq 2ab$ for all $a,b$ and $x-1 < [x] \leq x$ for all $x$, yield

${\textstyle \frac{a^2}{b} + \frac{b^2}{a} \geq [\frac{a^2}{b}] + [\frac{b^2}{a}] = [\frac{a^2+b^2}{ab}] + ab > \frac{a^2+b^2}{ab} + ab - 1 \geq 2 + ab - 1 = ab + 1}$,

hence

$(6) \;\; {\textstyle \frac{a^2}{b} + \frac{b^2}{a} > ab}$.

Multiplying (6) by ${\textstyle \frac{b}{a^2}}$ we obtain

${\textstyle \frac{b^2}{a} < 1 + (\frac{b}{a})^3}$,

implying ${\textstyle \frac{b^2}{a}<2}$ since $a \geq b > 0$.

First assume ${\textstyle 0 < \frac{b^2}{a} <1}$.  Then ${\textstyle [\frac{b^2}{a}]=0}$, which inserted in (1) gives

$(7) \;\; {\textstyle [\frac{a^2}{b}] = [\frac{a^2+b^2}{ab}] + ab}$.

According to the Division algorithm there exist integers $q>0$ and $0 \leq r < b$ s.t. $a = qb + r$.
Now $b^2 < a=qb+r < b(q+1)$ since $0 \leq r < b$, implying $q \geq b$ and

${\textstyle \frac{a^2+b^2}{ab} = \frac{a}{b} + \frac{b}{a} =\frac{qb+r}{b} + \frac{b}{qb+r} = q + \frac{r}{b} + \frac{b}{qb+r}\leq q + \frac{r}{b} + \frac{1}{q} \leq q + \frac{r}{b} + \frac{1}{b} \leq q + \frac{r+1}{b} \leq q+1}$,

so ${\textstyle \frac{a^2+b^2}{ab} \leq q+1}$ with equality when $r=0, q=b, r=b-1$, i.e. $q=b=1$ giving $a=1=b^2$.  This contradiction of the fact $a>b^2$ means ${\textstyle q \leq \frac{a^2+b^2}{ab} < q+1}$.  Hence ${\textstyle [\frac{a^2+b^2}{ab}] = q}$, which inserted in (7) with $a=qb+r$ gives

${\textstyle [\frac{(qb+r)^2}{b}] = q + b(qb + r)}$,

i.e.

${\textstyle q^2b + 2qr + [\frac{r^2}{b}] = q + qb^2 + br}$,

or alternatively

$(8) \;\; {\textstyle [\frac{r^2}{b}] = qb(b-q) + r(b-2q) + q}$.

Assume $q \neq b$, i.e. $q>b$.
Let $r=0$.  Then by (8)

$0 = qb(b-q) + q = q(b(b-q)+1)$,

implying

$b(q-b) = 1,$

since $q>b>0$, which also implies $b=1$ and $q=2$, yielding $a = qb+r = 2 \cdot 1 + 0 = 2$. So the only solution when $r=0$ is $(a,b) = (2,1)$.

Next assume $r>0$. Now $q>b$, so $b-q \leq -1$ and $b-2q < -q$, which according to (8) means

$(9) \;\; {\textstyle [\frac{r^2}{b}] < -qb - qr + q = q(1-b-r)}$.

Evidently ${\textstyle [\frac{r^2}{b}] \geq 0}$, so $1-b-r>0$ by (9). Thus $b+r<1$, which is impossible since $b > r > 0$.  Therefore the only solution of (1) when $q \neq b$ is $(a,b)=(2,1)$.

Finally let $q=b$. Inserted in (8) we obtain

$(10) \;\; {\textstyle [\frac{r^2}{b}] = b(1-r)}$.

Now ${\textstyle 0 \leq [\frac{r^2}{b}] < [\frac{b^2}{b}] = b}$ since $0 \leq r<b$, so $0 \leq b(1-r) < b$ according to (10).  Hence $0 \leq 1 - r < 1$, i.e. $r=1$.  Thus $a=qb+r = b \cdot b + 1 = b^2+1$, where $b > 1=r$.  Observe that $b=1$ gives the solution $(a,b)=(2,1)$.  In other words, the only solutions of (1) in this case are $(a,b)=(n^2+1,n)$, where $n>0$.

Summa summarum, the only solutions of (1) with $a \geq b$ are $(a,b)=(n^2+1,n)$, where $n \neq -1,0$. $\;\;$ [b]q.e.d.[/b]

I did that...but my solution is almost same as http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1218612&sid=030abdcaaf3c9ef2979fb264df0eba70#p1218612


Given a natural number, one may repeatedly

a) multiply it by any other natural number, and
b) delete zeros in its decimal representations.

Prove that, starting with any natural number, one can perform a sequence of these operations that transforms this number into a one-digit number.


inequalities
Cauchy Inequality
inequalities unsolved

[quote="Vasc"]For $a,b,c$ positive real numbers and  $p \ge 7/8$ prove that

$\displaystyle \frac{1}{a^2+b^2+pab}+\frac{1}{a^2+c^2+pac}+\frac{1}{b^2+c^2+pac}\ge \frac{27}{(2+p)(a+b+c)^2}$[/quote]

hmmm....  :lol:
that is a generalisation of the inequality :
$\frac{1}{a^{2}+b^{2}+ab}+\frac{1}{b^{2}+c^{2}+bc}+\frac{1}{c^{2}+a^{2}+ca}\geq \frac{9}{\left( a+b+c\right) ^{2}}$
but both of them are hard !!!  :lol:

i just found a solution in case $p\geq2$

and here my solution:

the ineq is equivalent to:
$\sum\frac{(a+b+c)^2}{(a+b)^2+(p-2)ab}\geq\frac{27}{2+p}$
applying CAUCHY we get:
$\sum\frac{(a+b+c)^2}{(a+b)^2(1+\frac{p-2}{4})}\geq\frac{27}{2+p}$

it reduces to:

$\sum(1+\frac{c}{a+b})^2\geq\frac{27}{4}$

let $\frac{c}{a+b}=x$

$f(c)=(1+c)^2$  is concave

appying jensen and nesbit ;)

this tell that  they are very weak :D

only in case p>=2

to erdos:

your ineq can prove by brute force and :

$a^6+b^6+c^6+3b^5c+3ab^5+3ac^5+3bc^5+3a^5b+3a^5c\geq3a^4c^2+3a^4b^2+3b^4a^2+3b^4c^2+3c^4a^2+3c^4b^2+a^3b^3+b^3c^3+c^3a^3$

which is true by CAUCHY :D

something tell me that : brute force for this one may get the result to the generalized ineq ;)

There is a mistake where you write

[quote="siuhochung"]
 $(4 - p)(ab + bc + ca)( {\sum {\frac{1}{{a^2  + b^2  + pab}}} }
  \ge (4 - p)(ab + bc + ca)( {\sum {\frac{1}{{\frac{{p + 2}}{4}(a + b)^2 }}} }$
 [/quote]

as $a^2+b^2 \geq \frac{(a+b)^2}{2}$

Can you please give a hint for a proof, Vasc?

Thank you.

This ineq can be obtained adding up the inequalities
$\displaystyle \sum \frac{2a^2+(1+2p)bc}{b^2+pbc +c^2} \ge \frac{3(3+2p)}{2+p}$,
$\displaystyle 4\sum \frac{a(b+c)+(p-1)bc}{b^2+pbc+c^2} \ge \frac{12(1+p)}{2+p}$.

Thanks, Vasc.

But these two inequalities seem harder than original inequality. :?

However this is really a nice inequality to solve. ;)

[quote="Vasc"]For $ a,b,c$ positive real numbers prove that

$ \frac {23}{8b^{2} \plus{} 7bc \plus{} 8c^{2}} \plus{} \frac {23}{8c^{2} \plus{} 7ca \plus{} 8a^{2}} \plus{} \frac {23}{8a^{2} \plus{} 7ab \plus{} 8b^{2}}\ge\frac {27}{(a \plus{} b \plus{} c)^2}.$[/quote]Using Cauchy inequality,

$ \left(\sum{\frac {1}{8b^2 \plus{} 7bc \plus{} 8c^2}}\right)\left[\sum{(19a \plus{} 4b \plus{} 4c)^2(8b^2 \plus{} 7bc \plus{} 8c^2)}\right]$

$ \geq\left[\sum{(19a \plus{} 4b \plus{} 4c)}\right]^2 \equal{} 729\left(\sum{a}\right)^2,$

we must prove that

$ 621\left(\sum{a}\right)^4\geq \sum{(19a \plus{} 4b \plus{} 4c)^2(8b^2 \plus{} 7bc \plus{} 8c^2)}.$

This comes down to

$ \frac {621\left(\sum{a}\right)^4 \minus{} \sum{(19a \plus{} 4b \plus{} 4c)^2(8b^2 \plus{} 7bc \plus{} 8c^2)}}{5}$

$ \equal{} 73\sum{a^4} \plus{} 180\sum{a^3(b \plus{} c)} \minus{} 506\sum{b^2c^2} \plus{} 73\sum{a^2bc}$

$ \equal{} 73\sum{a^2(a \minus{} b)(a \minus{} c)} \plus{} 253\sum{bc(b \minus{} c)^2}\geq 0,$

which is true by [url=http://www.mathlinks.ro/viewtopic.php?t=1166]Schur inequality[/url].

See also here : http://www.mathlinks.ro/viewtopic.php?t=123636


geometry
trigonometry
trig identities
Law of Sines
Law of Cosines
perpendicular bisector
Pythagorean Theorem

Someone help me on this problem:

Consider two concentric circles. The circumference of the larger circle is one foot longer than the circumference of the smaller circle. Find the [positive] difference between the two radii of the circles.

[hide="Long Way"]
Let the circumference of the smaller circle be $ x$ and the larger circle's circumference is then $ x \plus{} 1$. The radii of the circles are then $ \frac x{2\pi}$ and $ \frac {x \plus{} 1}{2\pi} \equal{} \frac x{2\pi} \plus{} \frac 1{2\pi}$, respectively. Thus the difference between the radii is simply $ \frac 1{2\pi}$.[/hide]

[hide="The Short Way"]
We notice that the initial radius doesn't matter. Just divide 1 by $ 2\pi$ to get $ \frac 1{2\pi}$ to get the answer.[/hide]

[hide]
Let the radius of the smaller circle by a and the larger one b. So, 2b\pi - 1 = 2a\pi. Divide by 2\pi to get b-1/(2\pi) = a. So, b is 1/(2\pi) larger than a.
[\hide]

Edit: Woops. Yongyi beat me.

Ok,

C circle I= $ 2\pi$
C circle II: $ 2\pi\plus{}1$

Because C=$ \pi$d, we divide each by $ \pi$ to get the diameter, and divide that by 2 to get the radius.
$ \frac {2\pi} {2\pi}$ $ \equal{}$ $ 1$
$ \frac {2\pi\plus{}1} {2\pi}$ $ \equal{}$ ... no simplifying to do so its just $ \frac {2\pi\plus{}1} {2\pi}$.
The positive difference is $ \frac {2\pi\plus{}1} {2\pi} \minus{}1$. We want to make a common denominator so....
$ \frac {2\pi\plus{}1} {2\pi} \minus{}\frac {2\pi} {2\pi}$ $ \equal{}$ $ \frac {1} {2\pi}$

Dang lots of people beat me...

Um...What are concentric circles? Are they circles who share the same center?

Yes - con = same, centric = well, obviously, center

I get it! Thanks!
NOTE: I will probably be asking any other Geometry questions I have here.

Note that proofs do not belong here though... :P

Are they really proofs, though, not just simply answering questions?

I know that this was just answering a question, I just wanted to remind everyone that no proofs should really be posted here.

i.e. For this question, if the problem had been: "Prove that the difference between the lengths of the radii is...", then it would probably not belong here. (As simple as it would be.)

Um...I just completed Alg II Honors, so I used Law of Sines/Cosines...
 ----------------------------------------------------------------------------------------------------------------
Relabel triangle CDE as follows:

Side CD is now a.
Side DE is now b.
Side CE is now c.

$ \gamma\equal{}120$ (obviously)

[u][b]Law of Sines Method:[/b][/u]

$ \alpha\equal{}\beta\equal{}30$ (Because this is isosceles.)

$ \frac{\sin30}{2}\equal{}\frac{\sin120}{c}$

$ c\equal{}\frac{2\sin120}{\sin30}$

$ \equal{}\frac{2\left(\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}}$

$ \equal{}2\sqrt{3}$

[u][b]Law of Cosines Method:[/b][/u]

$ c^2\equal{}2^2\plus{}2^2\minus{}2(2)(2)\cos120$

$ c^2\equal{}4\plus{}4\minus{}8\left(\minus{}\frac{1}{2}\right)$

$ c^2\equal{}8\plus{}4$

$ c^2\equal{}12$

$ c\equal{}\sqrt{12}$

$ c\equal{}2\sqrt{3}$

Yes, but you're assuming it's a regular hexagon! Maybe you're supposed to... :D

Non-Law of Sines or Cosines Method:

Notice that $ CDE$ is a 120-30-30 triangle, and if you know enough about 120-30-30 triangles, the base is the leg times $ \sqrt 3$. Therefore, the answer is $ 2\sqrt 3$.

Proof (not rigurous, trying to be understood clearly):

Drawing the perpendicular bisector of $ CE$ creates two 30-60-90 triangles, both with hypotenuse 2. Therefore, the desired length is twice the longer leg of one of the triangles, so it follows that the length of $ CE$ is $ 2\sqrt 3$.

Darn it! I knew there was some way to do this using Pythagorean Theorem, except it just didn't occur to me at the time. (At least I have 3 months of summer vacation to get Alg II out of my head... :lol: )

Yeah, I guess it could be a non-regular hexagon, but that wouldn't be solvable. Maybe the problem should have been clearer.

Oh, I get it! But I don't get why this was moved to High School math. Probably because it is.

Sorry. That was probably my fault. I used Law of Sines/Cosines which is Alg II/trig.

[quote="qwertythecucumber"]Let hexagon $ ABCDEF$ have a side length of 2. What is the distance between $ C$ and $ E$?[/quote]

Please create new a thread when asking a seperate question. We're not trying to start a marathon here.  :wink:

Yeah, marathons do seem to end up locked... :D

[quote="Yongyi781"]Yes, but you're assuming it's a regular hexagon! Maybe you're supposed to... :D [/quote]

Hes saying it has [b]side[/b] length of 2. That means all sides are equal. Doesn't that mean that it is a regular hexagon?

No, because the angles could still be different.

OK, I forgot to say it was a regular hexagon.


Natural numbers x,y,k,n where n>1 is odd. Prime number p>2 satisfied:

x^n + y^n = p^k

prove that n=p^m for some natural m.

 :help:


If you want to practice in posting replies than you should do it in the Test Section of this forum. ;)

[quote="bellahsayn"]FIND ALL FONCTIONS SATIS[/quote]

are u mad?


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}\frac{1}{2}$
prove that
$ \frac{1}{(a\plus{}b\plus{}1)^4}\plus{}\frac{1}{(b\plus{}c\plus{}1)^4}\plus{}\frac{1}{(c\plus{}a\plus{}1)^4} \le \frac{9}{2^{11}abc}$
 :!:

[quote="devil_maths"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} \frac {1}{2}$
prove that
$ \frac {1}{(a \plus{} b \plus{} 1)^4} \plus{} \frac {1}{(b \plus{} c \plus{} 1)^4} \plus{} \frac {1}{(c \plus{} a \plus{} 1)^4} \le \frac {9}{2^{11}abc}$
 :!:[/quote]

$ \frac {1}{(a \plus{} b \plus{} 1)^4} \plus{} \frac {1}{(b \plus{} c \plus{} 1)^4} \plus{} \frac {1}{(c \plus{} a \plus{} 1)^4} \equal{} \sum \frac {1}{(a \plus{} b \plus{} 6* \frac {1}{6})^4}$ $ \le \sum \frac {1}{(8\sqrt [8]{(\frac {1}{6})^6ab})^4} \equal{} \sum \frac {27}{512\sqrt {ab}}$

we only prove

$ \sum \frac {27}{512\sqrt {ab}}\le \frac {9}{2048 abc}................(*)$

substitution $ a \equal{} \frac {x^2}{2(x^2 \plus{} y^2 \plus{} z^2)},a \equal{} \frac {y^2}{2(x^2 \plus{} y^2 \plus{} z^2)},a \equal{} \frac {z^2}{2(x^2 \plus{} y^2 \plus{} z^2)}$

$ (*) \Longleftrightarrow \frac {9(x^2 \plus{} y^2 \plus{} z^2)^3}{256x^2y^2z^2} \minus{} \frac {27(x^2 \plus{} y^2 \plus{} z^2)}{256}$ $ (\frac {1}{xy} \plus{} \frac {1}{yz} \plus{}$ $ \frac {1}{zx})\geq 0$

$ \Longleftrightarrow 9(x^2 \plus{} y^2 \plus{} z^2)(x^4 \plus{} 2y^2x^2 \plus{} 2x^2z^2 \plus{} y^4 \plus{} 2y^2z^2$ $ \plus{} z^4 \minus{} 3yz^2x \minus{} 3yzx^2 \minus{} 3y^2zx)\geq 0$

$ \Longleftrightarrow 9(x^2 \plus{} y^2 \plus{} z^2)(\sum x^2(x \minus{} y)(x \minus{} z) \plus{} \sum x(y \plus{} z)(x \minus{} y)(x \minus{} z) \plus{}$ $ 4\sum yz(x \minus{} y)(x \minus{} z))\geq 0$

 :lol:


Again, this isn't all that difficult, but it has the word "converge" in it so I'll put it in advanced.

Prove that for non-negative [tex]a_n[/tex], the infinite product [tex]\displaystyle \prod_{n=1}^\infty(1+a_n)[/tex] converges if and only if the series [tex]\displaystyle \sum_{n=1}^\infty a_n[/tex] converges.

(Edited to say "non-negative" instead of "real," although I don't actually know if that change is necessary or not.)

This is a standard theorem in mathematical analysis, is it?

Probably, but it's nice to prove it anyway.

[quote="polymorphic"]how is this true? let ai = -1 for instance. Product is 0.  sum is clearly -infinity[/quote]

Check out the definition of convergence/divergence of infinite product carefully, you might find a surprising answer.

In the original statement, we probably need to assume absolutely convergence on the :Sigma: a_n part. [b]zableman[/b]? (Again, I am not sure about this, I don't have an analysis reference book here to check upon this).

By the way, this is what I found (of course, on the web again):

http://www.math.ucsb.edu/~mckernan/Teaching/02-03/Autumn/202A/l_26.pdf

interesting.  I think that the other way is true though without any other restrictions.   Namely, for any reals an, convergence of the sum implies convergence of the product.  The product clearly cannot diverge since log(1+x) < x.  THen we just have to show that it can't oscillate, but i don't have a very rigorous way for showing that.

Let [tex]\displaystyle a_{2k-1}= -\frac{1}{\sqrt{k+1}}[/tex] and [tex]\displaystyle a_{2k}= \frac{1}{\sqrt{k+1}}[/tex] for [tex]k\ge 1[/tex].

Define [tex]\displaystyle S_n = \sum_{k=1}^n a_k[/tex] and [tex]\displaystyle  P_n = \prod_{k=1}^n (1+a_k)[/tex].

Then [tex]S_{2n} = 0[/tex] and [tex]S_{2n-1} = a_{2n-1}[/tex], So [tex]\displaystyle \lim_{n\rightarrow\infty}S_n = 0[/tex]. That is, [tex]\displaystyle  \sum_{k=1}^{\infty} a_k[/tex] converges (but not absolutely converges).

On the other hand [tex]\displaystyle 0 < P_{2n-1}\le P_{2n} = \frac{1}{n+1}[/tex], So [tex]\displaystyle \lim_{n\rightarrow\infty}P_n = 0[/tex]. That is, [tex]\displaystyle  \prod_{k=1}^{\infty} a_k[/tex] diverges.

There is a very interesting fact (I think it's due to Riemann if I remember it correctly):

If the infinite series  :Sigma: a_n converges conditionally (which means  :Sigma: |a_n| diverges, but :Sigma: a_n converges), then for any given real x, we can rearrange the terms so that the series converges to x.

I took a second look at the place where I got this fact, and their proof is unsatisfactory (and incomplete) the way it was originally given. But I'm pretty sure that replacing the 'real' with 'non-negative' makes it work, (hint in spoiler) [hide]since then both the sum and the product are monotinically increasing and we just have to show that if one has a finite value than the other does as well.[/hide]

[quote="zabelman"]... But I'm pretty sure that replacing the 'real' with 'non-negative' makes it work, ...[/quote]

Sure it works if all a_n  :ge: 0, since absolutely convergence is the same as convergence in this case.

However, if you limit a_n  :ge: 0, then your criterion can not be applied to infinite product such as [tex]\displaystyle \prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{n^{1+\alpha}}\right)[/tex], where [tex]\alpha > 0[/tex].

A tweak on fuzzylogic's example, put to a slightly different (but related) purpose:

Let a_(2k-1) = -1/sqrt(k)
Let a_(2k) = (1-1/sqrt(k))^(-1) - 1

Then (1+a_(2k-1))*(1+a_(2k)) = 1 and the infinite product of (1+a_n) converges to 1.

But if you expant out a_(2k) by the binomial series, you'll see that a_(2k-1) + a_(2k) is comparable to 1/k for large k and hence the infinite sum of a_n diverges to infinity.

But these are the only two directions you can go. You can have the sum converge and the product diverge to zero (as in the first example) and you can have the product converge and the sum diverge to + :inf: . That's it for bad examples.

[quote="Kent Merryfield"]But these are the only two directions you can go. You can have the sum converge and the product diverge to zero (as in the first example) and you can have the product converge and the sum diverge to + :inf: . That's it for bad examples.[/quote]
So the Pro painted the whole picture. :-) That's nice to know.

[quote="polymorphic"]hm. diverge to zero?  that seems like a strange concept.[/quote]

Loosely speaking, Prod (1+a_k) diverges to 0 is same as saying :Sigma: log(1+a_k) diverges  to  -:inf: since Prod_(k: 1 to n) (1+a_k) = exp [:Sigma:_(k: 1 to n) log(1+a_k)].

This is some of my thoughts when a_i :ge: 0

[hide]Case I: Does pro(a_i) converge if sum(a_i) converges?

Since a_n are nonnegative, then we can apply AM-GM on pro(1+a_i). We have

pro(1+a_i) = ((pro(1+a_i))^(1/n))^n :le: (1+sum(a_i)/n)^((n/sum(a_i))*sum(a_i)) :le: e^(sum(a_i))

So if sum(a_i) converges, e^(sum(a_i)) converges, and pro(a_i+1) converges.

Case II: Does sum(a_i) converges if pro(1+a_i) converges?

Well, since a_i  :ge: 0. It is obvious that pro(1+a_i) :ge: sum(a_i). Thus, sum(a_i) does converge if pro(a_i +1) converges.[/hide]

Thus, if one of pro(a_i +1) and sum(a_i) converges, so does the other.


topology
advanced fields
advanced fields unsolved

The problem is as follows:

Let P be the set of all lines in $ R^{2}$. We establish the following notion of convergence in P: a sequence $ L_{n}$ converges to the line $ L_{0}$ if and only if $ \forall x\in{L_{0}}$  $ d(x, L_{0})$ converges to 0, where d is the standard Euclidean distance between a point and a line. Is there a metric D on P such that convergence in terms of metric D is exactly the convergence in the sense explained above?

I've tried a metric described as follows:

Let r(L) be the distance between a line L and the point (0, 0) taken with sign, i.e. it's positive if the orthogonal projection of (0, 0) onto the line lies below y=x and it's negative otherwise, and let a(L) be the angle between L and the line y=0 taken modulo $ \pi$. The distance D between L and K is

 d[(r(L), a(L)), (r(K), a(K))],

where d is the standard Euclidean metric on $ R^{2}$. Does this metric satisfy conditions of the problem?

If my metric is incorrect, please don't spoil the correct answer to problem - just show me why it's wrong.

That should be $ d(x,L_n)$ in the definition- it doesn't make sense as stated.

It's almost right, but you have to be more careful about "taken with sign". Look at the family of lines $ L_k\equal{}y\plus{}kx\equal{}1$, as $ k\to 1$. If $ k<1$, we get $ r(L_k)\equal{}\frac1{\sqrt{k^2\plus{}1}}$ by your definition, and if $ k>1$, we get $ r(L_k)\equal{}\minus{}\frac1{\sqrt{k^2\plus{}1}}$.

This space of lines is homeomorphic to a Mobius strip (with open edges).


floor function
number theory proposed
number theory

Let $f(x)=\lfloor x^2 \rfloor + \{x\}$ for all positive reals $x$.
Prove: There exists an arithmetical progression $<a_n> = <a_0 + nd>$ of positive rational numbers for which all  (canceled) denominators are $3$ and $a_n$ is not in $f(x)$ for all $n$.


function
calculus
integration
algebra unsolved
algebra

Suppose function $f: N\times Z \to N$ satisfies the conditions:
 (i) $f(0,0) =5^{2003}$ , $f(0,n)=0$, for all number integer  $n \neq 0$
 (ii) $f(m,n) = f(m-1,n)-2\left[\frac{1}{2}f(m-1,n)\right]+\left[\frac{1}{2}f(m-1,n+1)\right]$ for all natural  $m > 0$ and for all integer $n$.

Prove that exist $M \in N$ such that :
1) $f(M,n)=1$ for all $n \in Z$ that $|n| \leq \frac{5^{2003}-1}{2}$
2) $f(M,n)=0$ for all $n \in Z$ that $|n| > \frac{5^{2003}-1}{2}$.

P.S. $[x]$ denotes the integral part of $x$.

[i]Edited by Myth[/i]

Hm... It seems the statement is not correct, since $f(m,n)=0$ for all $m$ and $n>0$.


I'm trying to put pictures in my pdf but whenever I do, it makes the text after it go down farther than it should

so I try to use vskip with the negative component, and the only lengths that will bring the text back up is \vskip -4 in+ and when I do this it brings the text back up too far

the picture is made by geogebra
help?

There could be blank space in the image. Could you post your code plus the image here (or a different one with the same problems if you don't wish to make it public)?

Thanks, but could you post the 3 pngs as I suspect it's a problem of extra space in those that is causing the problem? I tried a different image and it was fine.

the second image was causing the problem

here they are

I think you can see from your post the problem. Each png consists of a diagram with loads of blank space below them and this is being faithfully reproduced in your pdf.

If you can tell GeoGebra not to do this that will be best (GeoGebra has a [url=http://www.geogebra.org/forum/]forum[/url] if you need help) but meanwhile you could open the images in an image program and cut out and resave the part of the image you need.

yes but it's only the second picture that's giving the major problem with the other pictures \vskip takes care of that space

The problem is that the (A) (B) (C) etc line has gone on to the next page so overriding the negative vertical space. You can use \newpage after the (A) (B) (C) etc line which will work, though I still think it best to use the right size image.

what image program could I use to wipe out that extra space?

Which operating system are you using? In Windows, Paint will do the job (cut and paste into a small canvas). You may also find you have an image or photo editing software on your computer eg Microsoft Photo Editor that you can use.

If not then [url=http://www.openoffice.org/]OpenOffice[/url] Draw program has a crop command which will work. [url=http://www.gimp.org/]GIMP[/url] is a powerful image program. Both are free and work on most operating systems.

ok it works now. thank you for your help


induction
number theory solved
number theory

Let p_n denote the nth prime number and let pi(m) denote the number of primes \leq m, e.g. pi(12)=5. Prove for n \geq 6 that pi((p_1p_2*...*p_n)^0.5) > 2n.

Use induction on n.

It is not difficult to verify that it holds for n = 6.

Suppose it holds for n :
Since n  \geq 6, we have p_(n+1)  \geq 17.
It follows that  \sqrt (p_1....p_(n+1))  \geq 4 \sqrt (p_1....p_n) = 4x.
Now, from Bertrand's postulate :
(pi(4x) - pi(2x)) + (pi(2x) - pi(x))  \geq 2
Then pi(4x)  \geq 2 + pi(x). And we are done.

Pierre.


In a sequence, each term is obtained by calculating the sum of the preceding two terms. the eighth term is 81, and the sixth term is 31. What is the fourth term?

It's a Fibonacci-like sequence. To calculate the seventh term, we have $ 31 \plus{} x \equal{} 81 \implies x \equal{} 50$. Working backwards, we find that the fifth term is $ 19$, and the fourth term is $ 31 \minus{} 19 \equal{} \boxed{12}$.


I heard there is a math jam after every round to dicuss how to solve those problems. Do you have to do USAMT to have access to those math jams.

In general Math Jams are free to anyone with an AoPS account.

Wait Im looking at the upcoming math jams and I dont see a USAMTS math jam anywhere . Where did you hear that? It would be a cool idea though.

Well, the deadline for Round 1 submissions is still more than a month away, so I assume they'll post the date for the Math Jam later.

Because (starting last year) we now post full solutions to every problem shortly after the conclusion of each round, we have decided to not do the USAMTS Math Jams anymore.

On a slightly related question, are commendations to solutions still begin given out? I was under the impression that commendations were abolished last year, but on my grading page it says "Note: If your number is [b]bold[/b], then your solution to that problem has been commended as particuarly excellent"

no, they said that last year on my profile too, and again this year.  They will most likely not commend anymore, or at least this year.


inequalities
limit
real analysis
real analysis unsolved

Show that f is Lipschitz if one of its Dini derivates is bounded.

I tried this with $ D^{\plus{}}f$ bounded and I got something like for each $ x$, $ \frac{f(y)\minus{}f(x)}{y\minus{}x} \leq M\plus{}1$ for $ 0<y\minus{}x<\delta_{x}$ and I didn't know how to continue. Maybe not a good approach?

Bounded means $ |D^\plus{} f|\le M$, so you actually have $ \minus{}M\minus{}1\le \frac{f(y)\minus{}f(x)}{y\minus{}x}\le M\plus{}1$ for $ 0<y\minus{}x<\delta_x$. Try to show that this inequality is true for all $ y>x$.

I would have thought that from the definition of $ D^{\plus{}}$ as a $ \limsup$, the lower bound for the expression isn't for all $ y \in (x,x\plus{}\delta_{x})$ but rather for each $ \delta$, there is a $ y \in (x,x\plus{}\delta)$ such that it holds.

[quote] you actually have $ \minus{}M\minus{}1 \leq \frac{f(y)\minus{}f(x)}{y\minus{}x} \leq M\plus{}1$ for $ 0<y\minus{}x<\delta_{x}$ [/quote]

This would be true if $ D^{\plus{}}f(x)$ is defined as $ \lim_{y \rightarrow x^{\plus{}}} \frac{f(y)\minus{}f(x)}{y\minus{}x}$. If I follow the definition given as $ \limsup_{y \rightarrow x^{\plus{}}} \frac{f(y)\minus{}f(x)}{y\minus{}x}$ instead, does it still hold?

Then to show that the inequality is true for all $ y>x$, is some compactness argument required?[/quote]

I think I may need more help with this problem. I haven't been able to solve it.


I don't know if anyone has seen the playing cards with politicians or terrorists on them. However, I am wondering if similar playing cards exist except with mathematical formulas or theorems on them?

I don't think I've ever seen those kinds of cards, but I think they'd be cool!
[url=http://imageshack.us][img]http://img146.imageshack.us/img146/2949/6d2nm.png[/img][/url]

Well, one of my friends has a deck of CERN cards; that's almost as cool.


number theory
greatest common divisor
relatively prime

Show that if $ c|a$ and $ c|b$, then $ c|\gcd (a,b)$.

let gcd(a,b)=d
dx=a, dy=b where x and y are relatively prime
(x,y)=1: so there is always mx+ny=1
c|dx, c|dy,
c|dmx, c|dny
thus  c|dmx+dny
c|d(mx+ny)
c|d

I'm Pretty sure I overcompicated it...

$ c|a\implies a\equal{}ck$

$ c|b\implies b\equal{}cj$

$ \text{gcd}(ck,cj)\equal{}c\cdot\text{gcd}(k,j)$, which is clearly divisible by $ c$.


Let $ (O)$ be a circle.Let $ A$ be a fixed point outside the circle .$ BC$ be a changleful diameter of $ (O)$.$ AB;AC$ cut $ (O)$ at $ D;E$,respectively.Prove that the line $ DE$ has a fixed point.

the problem is equivalent to the following one:
[b]show that the circumcircle of $ \triangle ABC$ passes thruogh a fixed point (except $ A$ of course...)[/b]  :D


[quote="ckck"]On a recent trip, Ellie drove 273 km in the same length of time Carol took to drive 198 km. Ellie's speed was 17 km/h greater than Carol's speed. How many km/h was Ellie driving?[/quote]
[hide]
$\frac{273}{x}=\frac{198}{x-17}$[/hide]


symmetry
geometry
inequalities
number theory proposed
number theory

we have $ (y^2 \plus{} z^2 \minus{} 2) \minus{} (z^2 \plus{} x^2 \minus{} 2) \equal{} z(x \minus{} y)$ so $ (x \minus{} y)(x \plus{} y \plus{} z) \equal{} 0$ and with the same way $ (y \minus{} z)(x \plus{} y \plus{} z) \equal{} 0$ and $ (z \minus{} x)(x \plus{} y \plus{} z) \equal{} 0$.
if $ x \plus{} y \plus{} z \equal{} 0$ then $ y \plus{} z \equal{} \minus{} x$ so the first equation gives $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$ so solution are $ (\plus{}\minus{}1,\plus{}\minus{}1,0) (\plus{}\minus{}1,0,\plus{}\minus{}1)$ and $ (0,\plus{}\minus{}1,\plus{}\minus{}1)$.
if $ x \plus{} y \plus{} z$ is non zero then two of these integers are equal if $ x \equal{} y$ then the first and the last equation give $ z^2 \equal{} xz \plus{} 2$ and $ xz \equal{} x^2 \minus{} 1$ so $ x^2 \minus{} 1 \equal{} z^2 \minus{} 2$ so $ (z \minus{} x)(z \plus{} x) \equal{} 1$ so $ (0,,0,1)$ and $ (0,0, \minus{} 1)$ are solution in this case.

[quote="Johan Gunardi"]Find all triples $ (x,y,z)$ of integers which satisfy

$ x(y \plus{} z) \equal{} y^2 \plus{} z^2 \minus{} 2$

$ y(z \plus{} x) \equal{} z^2 \plus{} x^2 \minus{} 2$

$ z(x \plus{} y) \equal{} x^2 \plus{} y^2 \minus{} 2$.[/quote]
Add them all together, subtract $ 2(xy \plus{} yz \plus{} zx)$ and add $ 6$. Then we have $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 \equal{} 6$. With the only solution $ {|x \minus{} y|,|y \minus{} z|,|z \minus{} x|} \equal{} {1,1,2}$ or a permutation.
Because of symmetry assume $ x \ge y \ge z$. Then we must have $ x \minus{} z \equal{} 2$ and $ x \minus{} y \equal{} y \minus{} z \equal{} 1$. So we have all possible solutions $ (x,y,z) \equal{} (k \plus{} 2,k \plus{} 1,k)$ or a permutation.

Inserting gives: $ (k \plus{} 2)(2k \plus{} 1) \equal{} (k \plus{} 1)^2 \plus{} k^2 \minus{} 2$ $ \iff$ $ 2k^2 \plus{} 5k \plus{} 2 \equal{} 2k^2 \plus{} 2k \minus{} 1$ $ \iff$ $ k \equal{} 1$. So we just have to check the solutopn $ (x,y,z) \equal{} (1,2,3)$. Which gives:
$ 1(2 \plus{} 3) \equal{} 2^2 \plus{} 3^2 \minus{} 2$ But this is false. Hence: No solutions!

Edit: Wooops! Of course $ (k \plus{} 2)(2k \plus{} 1) \equal{} (k \plus{} 1)^2 \plus{} k^2 \minus{} 2$ $ \iff$ $ k \equal{} \minus{}1$ :) This gives the solution $ (\minus{}1,0,1)$ or any permutation, which all works.

Add them, divide by two, subtract each of them, you get
$ yz \equal{} x^2 \minus{} 1$
$ xz \equal{} y^2 \minus{} 1$
$ xy \equal{} z^2 \minus{} 1$
Multiply them:
$ x^2y^2z^2 \equal{} (x^2 \minus{} 1)(y^2 \minus{} 1)(z^2 \minus{} 1)$
But $ x^2 > x^2 \minus{} 1$, etc. since x,y,z are integers.
Thus, one of them must be 0, another 1.
This gives $ x \equal{} 0$ $ y \equal{} 1$ $ z \equal{} \minus{} 1$ and permutations, which does indeed work. This is the only solution, some given by tchebytchev don't work.

adding $ x^2$ to the first equation,
$ x(x\plus{}y\plus{}z)\equal{}x^2\plus{}y^2\plus{}z^2\minus{}2$
the same can be done to the others showing that $ x\equal{}y\equal{}z$ or $ x\plus{}y\plus{}z\equal{}x^2\plus{}y^2\plus{}z^2\minus{}2\equal{}0$
If $ x\equal{}y\equal{}z$ then $ 2n^2\equal{}2n^2\minus{}2$ so no solution.
In the other case, $ x^2\plus{}y^2\plus{}z^2\equal{}2$ means one of the variables is zero, and the other two have norm 1. Thus all solutions are permutations of $ (0,1,\minus{}1)$

let us write the equation $ x(y \plus{} z) \equal{} y^2 \plus{} z^2 \minus{} 2$ on the form $ y^2 \plus{} z^2 \minus{} \frac {1}{2}\cos(120) yz \equal{} xy \plus{} yz \plus{} zx \plus{} 2$. The $ LHS$ of the given equation can be seen as the side of a triangle. Let us put:

$ x^2 \plus{} y^2 \minus{} \frac {1}{2}\cos(120)xy \equal{} a^2$
$ y^2 \plus{} z^2 \minus{} \frac {1}{2}\cos(120)yz \equal{} b^2$
$ z^2 \plus{} x^2 \minus{} \frac {1}{2}\cos(120)zx \equal{} c^2$

so we can organize a, b, c, x, y, z as in the figure.
Adding the equations
$ a \equal{} xy \plus{} yz \plus{} zx \plus{} 2$
$ b \equal{} xy \plus{} yz \plus{} zx \plus{} 2$
$ c \equal{} xy \plus{} yz \plus{} zx \plus{} 2$

we get: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3(xy \plus{} yz \plus{} zx) \plus{} 6$  (1)
But the area of the triangle in the picture is: $ \frac {1}{2}xy\sin(120) \plus{} \frac {1}{2}yz\sin(120) \plus{} \frac {1}{2}zx\sin(120) \equal{} \frac {\sqrt {3}}{4}(xy \plus{} yz \plus{} zx)$
And so if we call $ A$ this area we have $ xy \plus{} yz \plus{} zx \equal{} \frac {4}{\sqrt {3}}A$
And so (1) becomes:  $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4\sqrt {3}A \plus{} 6$
but a well known inequality states that $ a^2 \plus{} b^2 \plus{} c^2 \leq 4\sqrt {3}A$, so we conclude that at least one of a, b, c must be zero, and so at least one of x, y, z must be zero. From here the solution is easy


[b]1.[/b] Find the area of a quadrilateral ABCD inscribed in a circle with sides AB, BC, CD, and DA of 15, 20, 24, and 7 respectively.

[b]2.[/b] In the harmonic sequence 6, 3, 2, 3/2, 6/5, . . . , what will the eighth term be?

[b]3.[/b] Find the real values of x if x :^4: + 8x :^3: + 24x :^2: + 32x - 713 = 0

[b]4.[/b] Let x =  :sqrt:(3 + :sqrt:3) - :sqrt:(3 - :sqrt:3).  Find 192x :^2: - x :^4:.

[b]5.[/b] For x, y, z, and w:ge:0, compute the smallest value of x satisfying the following system:

y = x - 2002
z = 2y - 2002
w = 3z - 2002

2. challenge. finding the nth term of a harmonic sequence (or indeed of any sequence where each term is the jth mean* of the two surrounding terms) is not very difficult, once this hint is used: [hide]considering taking all terms of that sequence to the jth power and seeing what you get[/hide]. However, try to find a formula for the sum of a jth mean series. I am not sure one exists, and it in fact seems rather unlikely for |j|:ge:2 (:sqrt:2+:sqrt:5+:sqrt:8+:sqrt:11+:sqrt:14, for example, doesn't seem likely to come out of a nice simple formula), but see if you can find one for j=-1 (the harmonic sequence).

*the jth mean of two numbers a and b is defined as ((a^n+b^n)/2)^(1/n)

[i][b]Why isn't anybody trying these problems?????[/b][/i]

Now try number 1 with only theorems you know how to prove ;)

And in so doing, see if you can use your approach to prove the cyclic version of Brahmagupta that you cited above.

On number 1:

[hide]

I drew a picture and recognized the adjacent sides as pythagorean triples 7, 24, 25 and 15, 20, 25, so they are 2 right triangles that share a hypotenuse. The area is then (15)(10)+7(12)=234. A bit easier than Brahmagupta

[/hide]

1) Cheap solution:

[hide]Notice that 72 + 242 = 252, and 152 + 202 = 252. Thus ADC and ABC could be two right triangles, with <D = <B = 90o. If AC is any longer than 25, both <D and <B are more than 90o, thus ABCD is not cyclic; similarly, if AC is any shorter than 25, both <D and <B are less than 90o, and ABCD won't be cyclic. Therefore <D and <B are right angles. Then triangle ADC has area 7 * 24 / 2 = 84, and triangle ABC has area 15 * 20 / 2 = 150. The quadrilateral then has area 150 + 84 = 234.[/hide] <-- argh too slow...

3)

[hide]Rearrange to yield (x + 2)4 = 729 = 36. Thus x + 2 = :pm: 33/2, and x = 33/2 - 2 or x = -33/2 - 2.[/hide]

Yes, you said what I wanted to say on number one.


Eleven Teachers run a conference. Every hour, one or more teachers give a one-hour presentation, while all of the other teachers observe the presentations. Find the least amount of time during which it is possible for each teacher to observe all otherpresentations at least once.

What is this problem asking?? I think you need to polish the problem statement up a bit.


superior algebra
superior algebra unsolved

Find infinitely many pairwise distinct extensions $M$ such that $F_{p}(x^{p},y^{p}) \subset M \subset F_{p}(x,y)$.

Consider $M_{c}= \mathbb{F}_{p}(x^{p},y^{p},x+cy)$, for $c\in\mathbb{F}_{p}(x^{p},y^{p})$.


limit
geometry
3D geometry
combinatorics solved
combinatorics

Some of the vertices of the unit squares of a n x n chessboard are colored such that any k x k square formed by these unit squares has a colored point on at least one of its sides. If l(n) denotes the minimum number of colored points require to ensure the above condition, prove that  \lim n -> 00 l(n)/n 2  = 2/7.

This one has been proposed earlier in Math Magazine (pb. 1296, proposed in 1988. Solved in 1989).

Let's say that we color the points in red. And suppose that we have colored l(n) points to have the desired property.

Let P be a red point. Let S be a unit square with P as one of its vertices. Let n(P,S) be the number of red points which are on the boundary of S (in fact its vertices), including P, so that n(P,S) > 0.
Let N(P) =  \sum 1/n(P,S) , where the sum is over all the unit squares with P as vertex.
We have :
 \sum {P red} N(P) =  \sum {P red} ( \sum {S such that P vertex of S} 1/n(P,S) )
=  \sum {S unit square} ( \sum {P red and vertex of S} 1/n(P,S) )
= \sum {S unit square} 1
= n 2 .  (1)

In another hand :
- If P is a corner of the chessboard, then clearly N(P)  \leq 1 (there is only one unit square having P as vertex)
- If P is on the boundary of the chessboard, but not a vertex, then there are two unit squares having P as vertex, so that N(P)  \leq 2.
- If P is an interior point of the chessboard, then there are 4 unit squares having P as vertex. But the 2x2 square centered at P must have at least one red point on its boundary, say Q. If P and Q are adjacent vertices of a unit square, then they are for two, so that N(P)  \leq  3. If P and Q are opposite vertices of an unit square, then N(P)  \leq 7/2.
Thus, for each red point P, we have N(P)  \leq 7/2.

It follows that \sum {P red} N(P)  \leq  7*l(n)/2, and using (1), we deduce that l(n)  \geq  2n 2 /7.  (2)

Conversely, let's consider the following 7x7 square, where x stands for a non colored point, and r for a red point :

xxxrxrx
xrxxxxr
xxrxrxx
rxxxxrx
xrxrxxx
xxxxrxr
rxrxxxx

Note that is has the desired property with 14 red points.

Any nxn chessboard can be considered as a corner of a mxm chessboard with m = 0 mod[7] and n  \leq  m+6. Let m = 7p.
We use p 2  copies of the 7x7 chessboard above to tile the mxm chessboard (by translated copies...), so we have a total of 14p 2  = 2m 2/7 red points, and it is easy to verify that any kxk square on the chessboard has at least a red point on its boundary. Thus the mxm chessboard has the desired property, so that the nxn chessboard has too even if we delete m-n rows or columns from the top and from the right, respectively.
Thus l(n)  \leq 2m 2 /7  \leq  2(n+6) 2 /7.  (3)

From (2) and (3), we then deduce that lim l(n)/n 2  = 2/7.

Note that it can be proved that in higher dimensions (square boundaries are replaced by surfaces of cubes), the general limit is 2/(2 (d+1)  - 1), where d is the dimension of the space.

Pierre.


Our office will be closed on Thursday and Friday this week due to Thanksgiving.

We will try to process as many of the Round 2 submissions as we can by Wednesday.  But please do not panic if you do not see the checkmark when you log in at http://www.usamts.org; we may not be able to process them all on Wednesday.  And, needless to say, if you don't have the checkmark on Wednesday, then it will not show up until next Monday (at the earliest).

Update: all of the Round 2 papers have been checked in.

If you submitted Round 2, but do not have a green checkmark next to your name, then there was some sort of problem with your submission.  Please email usamts@usamts.org (or in many cases we have already emailed you) to find out what the status is.  When doing so, please include your USAMTS ID#, name, and the date, time, and method that you submitted your solutions.


group theory
abstract algebra
modular arithmetic
linear algebra
matrix
algebra
polynomial

What is struture of Sylow 2-group of simple group L_2(q) with q odd?

Set $ Z: \equal{} \{\pm I_2\}$ the centre of $ SL_2(q)$.

Assume $ q\equiv 1\pmod{4}$ and take $ \langle \zeta\rangle$ the Sylow 2-subgroup of the cyclic group $ \mathbb{F}_q^\times$.
Then $ |\langle \zeta\rangle | \equal{} 2^n$ with $ n\ge 2$ and $ i: \equal{} \zeta^{2^{n \minus{} 2}}$ satisfies $ i^2 \equal{} \minus{} 1$.
Then $ r: \equal{} {\zeta \plus{} \zeta^{ \minus{} 1}\over 2},s: \equal{} {\zeta \minus{} \zeta^{ \minus{} 1}\over 2i}\in\mathbb{F}_q$ satisfy $ r \plus{} is \equal{} \zeta,r \minus{} is \equal{} \zeta^{ \minus{} 1}$ and $ r^2 \plus{} s^2 \equal{} (r \plus{} is)(r \minus{} is) \equal{} 1$.
So $ A: \equal{} \begin{pmatrix}r & s \\
\minus{} s & r\end{pmatrix}\in SL_2(q)$ has characteristic polynomial $ X^2 \minus{} 2rX \plus{} 1 \equal{} (X \minus{} \zeta)(X \minus{} \zeta^{ \minus{} 1})$ and order $ 2^n$.
This gives $ a: \equal{} AZ\in PSL_2(q)$ of order $ 2^{n \minus{} 1}$ and $ b: \equal{} \begin{pmatrix}i & \\
& \minus{} i\end{pmatrix}Z\in PSL_2(q)$ of order $ 2$ with $ b^{ \minus{} 1}ab \equal{} a^{ \minus{} 1}$.
So we find the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] $ \langle a\rangle\rtimes \langle b\rangle$ of order $ 2^n$ as Sylow 2-subgroup of $ PSL_2(q)$ with central involution $ \begin{pmatrix} & 1 \\
\minus{} 1 & \end{pmatrix}Z$.

Analog for $ q\equiv \minus{} 1\pmod{4}$ we take $ \langle \zeta\rangle$ the Sylow 2-subgroup of $ \mathbb{F}_{q^2}^\times$ of order $ 2^n,n\ge 3$
and $ i: \equal{} \zeta^{2^{n \minus{} 2}}$ with $ i^2 \equal{} \minus{} 1$.
But $ \mathbb{F}_{q^2}/\mathbb{F}_q$ is galois with the only nontrivial automorphism $ x\mapsto x^q$,
which fixes $ r: \equal{} {\zeta^2 \plus{} \zeta^{ \minus{} 2}\over 2},s: \equal{} {\zeta^2 \minus{} \zeta^{ \minus{} 2}\over 2i},t: \equal{} {\zeta \minus{} \zeta^{ \minus{} 1}\over 2},u: \equal{} {\zeta \plus{} \zeta^{ \minus{} 1}\over 2i}\in\mathbb{F}_{q^2}$.
So $ r,s,t,u\in\mathbb{F}_q$ with $ r \plus{} is \equal{} \zeta^2,r \minus{} is \equal{} \zeta^{ \minus{} 2},t \plus{} iu \equal{} \zeta,t \minus{} iu \equal{} \minus{} \zeta^{ \minus{} 1}$
and $ r^2 \plus{} s^2 \equal{} (r \plus{} is)(r \minus{} is) \equal{} 1$ and $ t^2 \plus{} u^2 \equal{} (t \plus{} iu)(t \minus{} iu) \equal{} \minus{} 1$.
So $ A: \equal{} \begin{pmatrix}r & s \\
\minus{} s & r\end{pmatrix},B: \equal{} \begin{pmatrix}t & u \\
u & \minus{} t\end{pmatrix}\in SL_2(q)$ have characteristic polynomials
$ X^2 \minus{} 2rX \plus{} 1 \equal{} (X \minus{} \zeta^2)(X \minus{} \zeta^{ \minus{} 2})$ and $ X^2 \plus{} 1 \equal{} (X \minus{} i)(X \plus{} i)$.
This gives $ a: \equal{} AZ,b: \equal{} BZ\in PSL_2(q)$ of orders $ 2^{n \minus{} 2},2$
and $ AB \equal{} BA^{ \minus{} 1}\equal{}\begin{pmatrix}rt\plus{}su & ru\minus{}st \\
ru\minus{} st & \minus{}rt\minus{}su\end{pmatrix}$ gives $ b^{ \minus{} 1}ab \equal{} a^{ \minus{} 1}$.
So again we get the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] $ \langle a\rangle\rtimes \langle b\rangle$ of order $ 2^{n \minus{} 1}$ as Sylow 2-subgroup of $ PSL_2(q)$
with central involution $ \begin{pmatrix} & 1 \\
\minus{} 1 & \end{pmatrix}Z$.


logarithms
complex numbers
imaginary numbers

[b]How can I arrive from the proper way of solving this kind of logarithmic equation: $log_2(4-x)-log_4(x-1)=1$. What is x in this case? I use trial and error and I get x=2 but how can I solve it in proper way because the case is they have different base?[/b]

by the properties of logs, $log_ab=\frac{log_cb}{log_ca}$. So $\displaystyle log_4(x-1)=\frac{log_2(x-1)}{log_24}$

Hydron, I get x=10 and x=2....What is the valid solution?

[quote="E^(pi*i)=-1"]10 is an extraneous solution because the first logarithm wil then be $\log_2 -6$, which is an imaginary number that does not satisfy the requirements.  2 is the only valid solution.[/quote]

Yea, you can't have negatives there


$ x,y,z$ are positive reals and $ \sum x=1$

Proof: $ \sum \frac{xy}{\sqrt {xy+yz}} \le \frac{3 \sqrt{3}}{4} \sqrt{(x+y)(y+z)(z+x)}$


modular arithmetic
number theory proposed
number theory

Does there exist a multiple of $ 103$, say $ n$, such that $ 2^{2n \plus{} 1} \equiv 2 \pmod{n}$?

If $ n$ is even, then $ n\equal{}2m$, were m odd. Let $ m\equal{}n$ if n is odd.
Then condition equavalent to $ T_m|2n, \ 103|m$, were $ T_k$ mean minimal period for 2 mod k. Becouse $ 51\equal{}T_{103}|T_m|2n$, we get $ 51|m$. It give $ 8\equal{}T_{17}|T_m$ or $ 8|4m$ contradition with odd m.


I am planning to take the October SAT II in Mathematics 2.

I want to know how much detail and rigor is need for the exam. I've seen the questions - they are easy. I'm asking about the topics - how many different, which ones and in what detail should they be done. Which topics are covered (the official syllabus is a bit vague)Is the Barron's book perfect. Where can I get more practice??

Please help! I can clarify my question if you want.

I don't think you'll have much of a problem with the test.  Most people on this forum would consider it easy, I believe.  Although I haven't taken it yet and i haven't prepared very much for it, from what I've seen it doesn't look too bad.

As for the test, it's all multiple choice and probably grid-ins.  Therefore, since there aren't any proofs, short answers, or whatever, you don't have to be rigorous at all.  The official syllabus should be enough to tell you the topics.  And Barron's should be adequate.

I was very surprised to find a matrix question on my test. It is not covered anywhere in my high school curriculum, and it's not explicitly listed on the syllabus that I found online. I made a lucky guess and got it right. It was only one question, though, and considering how the raw scores are curved, I think I would have gotten 800 even if I got it wrong.

The Math 2C has had one matrix question on about half of the most recent administrations.  If you know how to compute the determinant of a 2x2 matrix, and know that multiplying a 2x3 matrix with a 3x4 matrix gives a 2x4 matrix, then you should be okay.

Another fairly recent topic is standard deviations.  You don't need to know any quantitative formulas for it.  Just know that standard deviation measures how far apart numbers are.  In particular, numbers bunched together have small standard deviation, and numbers spread apart have large standard deviation.

http://augustusmath.hypermart.net/tofchtml.html

Do these topics cover EVERYTHING tested on the SATII Math 2c exam?

Any one precalculus book may or may not cover all the topics that are fair game on the SAT II Math Level 2 test. But don't worry about that too much. It is possible to get the highest scaled score, 800, even if you miss a few of the items on the test. That's how the scaled scores work on that test.

I'm sure that you want to get a good review of precalculus math to learn that math, and you have found some good sources. They will help you learn more math, and that is important. Don't worry too much about the test--that will just spoil the fun of learning more math. You should do well on the test if you focus on the math ideas and how interesting they are while you are studying.

Best wishes for your math studies. Good luck on the test.

I wrote the SAT II Math Test today..
overall, it wasn't very hard
I was glad that I read this post.  :D
There was a question on STANDARD DEVIATION..

so, when i took the Math SAT IIs i first thought i shoudl take the IC exam, thinking it would be easy and i would do well- but no... i got a 750... (that was because i did something like 6+6=18, but that's not the point- i think i missed like 3 questions) then i decided that being on math team with a 750 on math SAT II is a bad thing, so i took the IIC- i finished pretty fast, took a 10 minute bathroom break (because they wouldn't let me go!) and then a short nap- that was good relaxation after writing and physics  ;)


inequalities
algebra
binomial theorem
inequalities open

I need a solution to this problem as soon as possible. I think there's a trivial proof, but I can't find it. Can you?

Prove or disprove that ${a \choose b} \leq (a/b)^{2b}$, when $b \leq a/2$.

Here is one approach.  By the binomial theorem, we know that ${a \choose b} p^b (1-p)^{a-b} \leq 1$ for every p between 0 and 1.  Choose $p = b/a$ to optimize.  We get the inequality
\[{a \choose b} \leq \frac{a^a}{b^b (a-b)^{a-b}} .\]
That bound turns out to be better than yours.  Do you see how to compare the two bounds?

The inequality between the two bounds is equivalent to $(x+1)^{x-1} \leq x^x$, for $x \geq 1$. I suppose this is trivial...
Thanks a lot!

It could also be proved by a two-fold induction...


Balance the following unbalanced equations: $ \ce{S2Cl2} \plus{} \ce{NH3} \rightarrow \ce{S4N4} \plus{} \ce{S8} \plus{} \ce{NH4Cl}$, and $ \ce{KCl} \plus{} \ce{HNO3} \plus{} \ce{O2} \rightarrow \ce{KNO3} \plus{} \ce{Cl2} \plus{} \ce{H2O}$ (state symbols ommitted).

Also, I don't understand this:

[hide="question"]Balance the unbalanced chemical equation: $ \ce{NH3}\plus{}\ce{O2} \rightarrow \ce{NO}\plus{}\ce{H2O}$.[hide="solution"]Firstly, I balanced the hydrogen and nitrogen to get $ \ce{2NH3}\plus{}\ce{O2} \rightarrow \ce{2NO}\plus{}\ce{3H2O}$. I then added a stoichiometric coefficient of $ \frac{1}{2}$ to the $ O2$ on the LHS, and then I multiplied the whole equation by 2 to get integral stoichiometric coefficients. Thus the answer is $ \ce{4NH3}\plus{}\ce{O2} \rightarrow \ce{2NO}\plus{}\ce{3H2O}$. Why can we simply add $ \frac{1}{2}$ and multiply the equation. Also, in what circumstances do I need to add non-integer real values of $ x$ as stoichiometric coefficients? Please give some practice problems. [/hide][/hide]

Does anyone know where to get the answers for the problems in the textbook Chemistry: Principles, Patterns and Applications (volume 1) by Bruce Averill and Patricia Eldredge?

for second
$ 4\ce{KCl} \plus{} 4\ce{HNO3} \plus{} \ce{O2} \rightarrow 4\ce{KNO3} \plus{} 2\ce{Cl2} \plus{} 2\ce{H2O}$
for first
$ 6\ce{S2Cl2} \plus{} 16\ce{NH3} \rightarrow \ce{S4N4} \plus{} \ce{S8} \plus{} 12\ce{NH4Cl}$

for doing these, [u]STEP I[/u] first see which element is having most atoms unbalanced and balance it.
[u]STEP II [/u] repeat step I.

Which element did you first balance? Any advice on which elements to balance first (generally) in a given unbalanced chemical equation?

theta pi, u can READ THE EDITED POST.for wut u wanted 2 know.

i have first heard the word stoichiometric coefficients :oops: . i thought u had great knowledge as it is a powerful word.but when i googled it i came 2 know that it was nothing.  :rotfl: i jus got decieved by a powerful sounding word. not criticizing y're knowledge though :|

[quote="ThetaPi"]
Also, I don't understand this:

Balance the unbalanced chemical equation: $ \ce{NH3} \plus{} \ce{O2} \rightarrow \ce{NO} \plus{} \ce{H2O}$.[hide="solution"]Firstly, I balanced the hydrogen and nitrogen to get $ \ce{2NH3} \plus{} \ce{O2} \rightarrow \ce{2NO} \plus{} \ce{3H2O}$. I then added a stoichiometric coefficient of $ \frac {1}{2}$ to the $ O2$ on the LHS, and then I multiplied the whole equation by 2 to get integral stoichiometric coefficients. Thus the answer is $ \ce{4NH3} \plus{} \ce{O2} \rightarrow \ce{2NO} \plus{} \ce{3H2O}$. Why can we simply add $ \frac {1}{2}$ and multiply the equation. Also, in what circumstances do I need to add non-integer real values of $ x$ as stoichiometric coefficients? Please give some practice problems. [/hide]

[/quote]

err after you balanced H and N there, you can see that there are 5 Os in the right hand side and 2 on the left. So now you want to make it a 5,so multiply it by $ 5\frac{1}{2}$ and then to get integers multiply the whole thing by 2.

theta pi, that happens with everyone. u have learn those confusing things. but they are basics. and wud be applied afterwards. however , i enjoy chemistry now, learning about carbon and all,

shadysaysurspammed pointed a good thing there that  whenever u have a element unbalanced (isolated on one side i.e. either only element or homoatomic compound) and on one side the element is a fraction of that of other side, multiply by whole equation by the reciprocal of that fraction.

[quote="ThetaPi"]My chemistry teacher taught us that. It's no use really. That's why I don't really like Chemistry: too many unnecessary definitions, like Avogadro's number equalling a mole etc., formula mass, molecular mass, atomic mass, blah. Oh please! They are all masses![/quote]

Let's be fair if they are all exactly the same thing then people won't bother with different names  :D .

Anyway, Chemistry is my favourite (science wise) and it becomes really interesting when you delve into deep theories of bonding and organic chemistry.


a,b,c are real number, they are not equal and a,b,c>o
prove:
(m^2)/(mp+mq-pq)+(p^2)/(mp+pq-mq)+(q^2)/(qm+pq-mp)<3
 :maybe:

I will assume m=a, b=p, c=q.  Then we want to prove

\[ \sum_{\text{cyc}}\frac{a^{2}}{ab+ac-bc}<3\]

Okay, try $ a=b=1, c=\frac{1}{2}+\epsilon$.  The expression goes to infinity.
Unless m, p, and q are some totally different numbers, it appears the inequailty doesn't hold (even if "not equal" meant "no two are equal", just let $ b$ be really close to $ a$).


Here is another one:
 The sides of AB and BC of \[ \triangle\]ABC

Please tell me how I can draw an image of a geometric figure and insert it here.Then I can post the sum and also what i have done .Sorry for the incomplete sum.

see the "Geogebra" forum--and u'll understand better :D

Here is the correct sum: The sides AB and Ac of $ \triangle$ ABC are respectively equal and parallel to the sides XY and ZX of the $ \triangle$ XYZ.Show that AC and XY are equal and parallel.

This is what I have done. I have joined Z and B,Z and A,A and X,X and C,C and Y and B and B and Y.
XZ $ \|$ BC and XZ = BC.
So, XZBC is a parallelogram.
ZB=XC

AB $ \|$ XY and AB=XY
So, ABYX is a parallelogram.
AX=BY
What next?
Am I going in the right direction?
If yes, how should this be solved?In fact, I am trying to prove that $ \triangle$ ZBY $ \cong$ $ \triangle$ XYC and that $ \triangle$ AZB $ \cong$ $ \triangle$ XCY.
Thank you.

 [geogebra]562df5d3508734afa260e4d19a39ebcc95bbd847[/geogebra]

The problem isn't true. What they required to be proven should be $ BC$ and $ YZ$ are equal and parallel.

As a general note on geometry, never draw a special case. It is sometimes misleading as it shows special properties which is not true in general, but provokes you to try to prove them. Your figure has the triangles equilateral, which is kind of a special case. And also the other triangle is kind of symmetric to the main triangle. That is why it seemed $ \Delta ZBY\cong \Delta XYC$. But unfortunately this doesn't hold in general. So you can't prove it.

Try this picture:

 [geogebra]f18ee2ab6d2671a38c77fd7a524046f1ede33ca4[/geogebra]

Note that you can move around the blue points and still the conditions of the problem hold (that is because I defined the points precisely in geogebra) . Though this luxury is not available with pencil-paper, it's best if you draw a figure that is not a special case. For example draw triangles with no 2 sides equal, no right angles etc. like my figure.

You were in the right direction. Try to prove the two parallelograms again. But there was some mistake in there before. You stated at first that $ XZ\parallel{}BC$ but that isn't given.

I've joined the points which should be joined (red lines),  that is what you too wanted to, but messed up a bit because of the misleading diagram.

another advice.. muks..
i hope that u have actually tried the problems for some time b4 asking for solutions...
some time = quite some time...

I am sorry.I unnecessarily made this complex while the sum was quite easy.In fact, having made the sum complex, I worked on it for quite some time.
Thank you.(I will be very careful next time before posting a sum)

[quote="muks"]I will be very careful next time before posting a sum[/quote]
No really. You need not be all that careful. Though Keshav is right you should spend some time yourself, but that was meant for your own improvement. Don't take this as wasting someone's time. You are always welcome to post any problem in here. :)

muks make good use of Akashnil's experience here. Frankly to say, I have become better at math than what i was in the beginning of 9th std only after being inspired by Akashnil.

It seems people are here to help! Thank you. If I am not wrong, I will have to complete the class 12 syllabus before May-June while I have completed $ \frac {2}{3}$ of class 9 syllabus and $ \frac {3}{11}$ of Class 10 syllabus.

Thank you again.I will make it a point to really elevate my standard,which is very poor at the moment.

[quote="muks"]I will have to complete the class 12 syllabus before May-June [/quote]
For what? If you think that you'll have to complete the class 12 syllabus before May-June for Math Olympiads, I can assure you that you are wrong :o  :P  :wink: .

Well muks, u have a completely wrong understanding of olympiad maths......olympiad maths is not about theory at all.....though u need to know some basics in all fields...its about problem solving...

Then I will give myself another year after May-June. In fact, I do not have any idea of this RMO and such olympiads.

[quote="muks"]Then I will give myself another year after May-June.[/quote]
No. What I meant to say is that you need not even complete upto class 12 syllabus at all for math olympiad.

Akashnil, why do u go offline in just 1 min or something and again come online and go offline?
BTW, ur soln for my $ \pm1\pm2....$ problem can be made more rigorous by a simple statement which is very very very very obvious but still to be written for rigority. Guess urself......
There is only 1 thing tht prevents u from the IMO team : 100% rigority in proof writing. Please take it in the right sense. I am just saying as a friend, though I am not even fit to comment about someone who know 1000...0000{100 zeros} times the theory I know and is the same no. of times better than me.

[quote="Aravind Srinivas L"]Akashnil, why do u go offline in just 1 min or something and again come online and go offline? [/quote]
Dunno. I think it is some bug on the site. I never do that. It just displays wrong things to you. I've got the web-browser to remember my passwords and stuff. So it auto logs in when I open the site. o that can't happen right?

[quote="Aravind Srinivas L"]BTW, ur soln for my $ \pm1\pm2....$ problem can be made more rigorous by a simple statement which is very very very very obvious but still to be written for rigority. Guess urself......[/quote]
Can't. :(


LaTeX
irrational number
number theory unsolved
number theory

This guy insists that he found much simpler way to prove Fermat's Last Theorem, not only one method, but two, and he says he also proved 4 Color Theorem without using computer. Please find some errors so that this guy can live usual life!! This guy is demonstrating alone for 18 weeks for these proof to be accepted!!

[b]4 Color Theorem proof of the regions on global surface [/b]
3 inter relationship of the regions on global surface
sharing a common boundary line. sharing a common point. not adjacent regions.
[1] 4 colors suffice for the distinguishing all regions about one region and the regions those share a common one region`s boundary line.
[Proof] The reason is this. 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[Example1] 3 colors suffice for the distinguishing all hexagonal shape regions. The reason is this. 2 colors suffice for the distinguishing 6 hexagonal shape regions those share a common one hexagonal shape region`s boundary line.
[Example2] 2 colors suffice for the distinguishing all square shape regions. The reason is this. One color suffice for the distinguishing 4 square shape regions those share a common one square shape region`s boundary line.
[2] 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[Proof] The reason is this. When there are lined from a region`s inner one point to the points on one region`s boundary line those are met the adjacent regions boundary lines, all extension regions are the regions those share one common point. 3 colors suffice for the distinguishing all regions those share one common point.
[3] 3 colors suffice for the distinguishing all regions those share one common point.
[Proof] The reason is this. When one region is selected, 2 colors suffice for the distinguishing all regions those share a common one selected region`s boundary line.

[b]2 methods of FLT proof and Pythagorean triples [/b]
$X^{n}+Y^{n}=Z^{n}$
Fermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n.
$Y+A=X+B=Z, A=Z-Y, B=Z-X X-A=Y-B=Z-A-B=X+Y-Z G=(X-A)/(AB)^{(}1/n)=(Y-B)/(AB)^{(}1/n)=(Z-A-B)/(AB)^{(}1/n)=(X+Y-Z)/(AB)^{(}1/n) X=G(AB)^{(}1/n)+A, Y=G(AB)^{(}1/n)+B, Z=G(AB)^{(}1/n)+A+B{G(AB)^{(}1/n)+A}^{n}+{G(AB)^{(}1/n)+B}^{n}={G(AB)^{(}1/n)+A+B}^{n}$
When n=1, G=0. When n=2, G=2^(1/2). When n>2, G=function(A,B) is the positive real number.
$X=(2AB)^{(}1/2)+A, Y=(2AB)^{(}1/2)+B, Z=(2AB)^{(}1/2)+A+B$
(X,Y,Z) are the irrational numbers or all Pythagorean triples in all natural number (A,B).
We can translate the upper form into this.
$AB=2k^{2}, B=2k^{2}/A X=2k+A, Y=2k(k+A)/A, Z=2k+A+2k^{2}/A XY=2k(2k+A)(k+A)/A$
When A is the odd number, $k=hA, XY=2A^{2}h(2h+1)(h+1)$and $hk=A, XY=2k^{2}(2+h)(1+h)/h$
When A is the even number, $2k=hA, XY=A^{2}h(h+1)(h+2)/2$and $2hk=A, XY=2k^{2}(1+h)(1+2h)/h$
Therefore XY cannot be the power numbers in all Pythagorean triples.
[b]* * * * * 1st method of FLT proof * * * * * [/b]
$G(AB)^{(}1/n)$ is the irrational number in all natural number (A,B), so (X,Y,Z) are the irrational numbers.
${G(AB)^{(}1/n)+A}^{n}+{G(AB)^{(}1/n)+B}^{n}={G(AB)^{(}1/n)+A+B}^{n}$
When $A=B, 2{GA^{(}2/n)+A}^{n}={GA^{(}2/n)+2A}^{n}{2^{(}1/n)-1}GA^{(}2/n)={2-2^{(}1/n)}A G=[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]A^{(n-2)/n}$
We make new form with $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]. [2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2$
This form is the irrational number in all natural number (A,B).
$G(AB)^{(}1/n)$ divide and multiply by $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2$, and now we can get two forms. And when $A=B, q=1. G(AB)^{(}1/n)=q[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]/2 q=2G(AB)^{(}1/n)/[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)][{A^{(}n-1)B}^{(}1/n)+{AB^{(}n-1)}^{(}1/n)]$
If G(AB)^(1/n) is the natural number (N) in some (a,b), $G(ab)^{(}1/n)=N$ can not have $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]$, and $G(AB)^{(}1/n)$can not have $[2^{(n-1)/n}+...+2^{(}2/n)+2^{(}1/n)]$.
So when A=B, q cannot be 1. That is an apparent contradiction.
Therefore $G(AB)^{(}1/n)$ is the irrational number in all natural number (A,B). 1st method. end.
[b]* * * * * 2nd method of FLT proof * * * * * [/b]
$X^{n}+Y^{n}=Z^{n}{X^{(}n/2)}^{2}+{Y^{(}n/2)}^{2}={Z^{(}n/2)}^{2}$
When n=2, we can display ${X^{(}n/2),Y^{(}n/2),Z^{(}n/2)}$ with (a,b).
$a=Z^{(}n/2)-Y^{(}n/2), b=Z^{(}n/2)-X^{(}n/2) X^{(}n/2)=(2ab)^{(}1/2)+a, Y^{(}n/2)=(2ab)^{(}1/2)+b, Z^{(}n/2)=(2ab)^{(}1/2)+a+b$
When n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number.
$ab=Z^{n}-(YZ)^{(}n/2)-(XZ)^{(}n/2)+(XY)^{(}n/2)$
We multiply $X^{(}n/2)$ and $Y^{(}n/2)$.
$(XY)^{n}=2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$
If (X,Y,Z) is the natural number, $(XY)^{n}$ is the natural number, but $2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$ is the irrational number.
That is an apparent contradiction.
Therefore (X,Y,Z) must be the irrational number. 2nd method. end.

There is some problem with latex..So..

Original text:

4 Color Theorem proof of the regions on global surface
3 inter relationship of the regions on global surface
sharing a common boundary line. sharing a common point. not adjacent regions.
[1] 4 colors suffice for the distinguishing all regions about one region and the regions those share a common one region`s boundary line.
[Proof] The reason is this. 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[Example1] 3 colors suffice for the distinguishing all hexagonal shape regions. The reason is this. 2 colors suffice for the distinguishing 6 hexagonal shape regions those share a common one hexagonal shape region`s boundary line.
[Example2] 2 colors suffice for the distinguishing all square shape regions. The reason is this. One color suffice for the distinguishing 4 square shape regions those share a common one square shape region`s boundary line.
[2] 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[Proof] The reason is this. When there are lined from a region`s inner one point to the points on one region`s boundary line those are met the adjacent regions boundary lines, all extension regions are the regions those share one common point. 3 colors suffice for the distinguishing all regions those share one common point.
[3] 3 colors suffice for the distinguishing all regions those share one common point.
[Proof] The reason is this. When one region is selected, 2 colors suffice for the distinguishing all regions those share a common one selected region`s boundary line.

2 methods of FLT proof and Pythagorean triples
X^n+Y^n=Z^n
Fermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n.
Y+A=X+B=Z, A=Z-Y, B=Z-X
X-A=Y-B=Z-A-B=X+Y-Z
G=(X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n)=(X+Y-Z)/(AB)^(1/n)
X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B
{G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n
When n=1, G=0. When n=2, G=2^(1/2). When n>2, G=function(A,B) is the positive real number.
X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B
(X,Y,Z) are the irrational numbers or all Pythagorean triples in all natural number (A,B).
We can translate the upper form into this.
AB=2k^2, B=2k^2/A
X=2k+A, Y=2k(k+A)/A, Z=2k+A+2k^2/A
XY=2k(2k+A)(k+A)/A
When A is the odd number, k=hA, XY=2A^2h(2h+1)(h+1) and hk=A, XY=2k^2(2+h)(1+h)/h
When A is the even number, 2k=hA, XY=A^2h(h+1)(h+2)/2 and 2hk=A, XY=2k^2(1+h)(1+2h)/h
Therefore XY cannot be the power numbers in all Pythagorean triples.
* * * * * 1st method of FLT proof * * * * *
G(AB)^(1/n) is the irrational number in all natural number (A,B), so (X,Y,Z) are the irrational numbers.
{G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n
When A=B,
2{GA^(2/n)+A}^n={GA^(2/n)+2A}^n
{2^(1/n)-1}GA^(2/n)={2-2^(1/n)}A
G=[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]A^{(n-2)/n}
We make new form with [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)].
[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2
This form is the irrational number in all natural number (A,B).
G(AB)^(1/n) divide and multiply by [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2, and now we can get two forms. And when A=B, q=1.
G(AB)^(1/n)=q[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2
q=2G(AB)^(1/n)/[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]
If G(AB)^(1/n) is the natural number (N) in some (a,b), G(ab)^(1/n)=N can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)], and G(AB)^(1/n) can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)].
So when A=B, q cannot be 1. That is an apparent contradiction.
Therefore G(AB)^(1/n) is the irrational number in all natural number (A,B). 1st method. end.
* * * * * 2nd method of FLT proof * * * * *
X^n+Y^n=Z^n
{X^(n/2)}^2+{Y^(n/2)}^2={Z^(n/2)}^2
When n=2, we can display {X^(n/2),Y^(n/2),Z^(n/2)} with (a,b).
a=Z^(n/2)-Y^(n/2), b=Z^(n/2)-X^(n/2)
X^(n/2)=(2ab)^(1/2)+a, Y^(n/2)=(2ab)^(1/2)+b, Z^(n/2)=(2ab)^(1/2)+a+b
When n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number.
ab=Z^n-(YZ)^(n/2)-(XZ)^(n/2)+(XY)^(n/2)
We multiply X^(n/2) and Y^(n/2).
(XY)^n=2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2)
If (X,Y,Z) is the natural number, (XY)^n is the natural number, but 2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) is the irrational number.
That is an apparent contradiction.
Therefore (X,Y,Z) must be the irrational number. 2nd method. end.

I have totally no idea about what the "4 Color Theorem proof" are trying to express. Seems that it only considers parts of the whole graph separately, but when the regions are put together, the claim may not be true.

You should be very doubtful if someone not belonging to the world's best mathemticians claims to have proven some old open problem (or one that was some years ago).
You should be even more doubtful if the proof is short, simple, and elementary (by what the author says).
The doubts should grow to $\infty$ if the author claims to have more than one proof or to have proven more than one of these hard problems.

[quote]This guy is demonstrating alone for 18 weeks for these proof to be accepted!! [/quote]
So what¿

[quote]

[quote]
This guy is demonstrating alone for 18 weeks for these proof to be accepted!!  [/quote]

So what¿
[/quote]

I think that guy is so pitiful, believing that he's right, but he's not and is fighting for nothing.

Well... if you say "who cares?" or something..I don't know.

Anyway Thanks.

All that he's asking for is to find a fault in the proofs; we know there is one, of course, ZetaX.

View it as the opposite of a proof problem :P

Its kind of necessary to fix the latex, so:
[quote="car202"]
[b]2 methods of FLT proof and Pythagorean triples [/b]
[hide="Preliminary Part"]
$x^{n}+y^{n}=z^{n}$
Paraphrased: Since FLT proof for $n=4$ is well known, it remains to prove it for any odd prime exponent.
Define $a,b$ as follows: \[y+a=x+b=z\implies a=z-y, \ b=z-x, \ x-a=y-b=z-a-b=x+y-z\]

Define $g$ also: \[g=\frac{x-a}{\sqrt[n]{ab}}=\frac{y-b}{\sqrt[n]{ab}}=\frac{z-a-b}{\sqrt[n]{ab}}=\frac{x+y-z}{\sqrt[n]{ab}}\] Thus,
\[x=g\sqrt[n]{ab}+a, \ y=g\sqrt[n]{ab}+b, \ z=g\sqrt[n]{ab}+a+b\]
so that \[\left(g\sqrt[n]{ab}+a\right)^{n}+\left(g\sqrt[n]{ab}+b\right)^{n}=\left(g\sqrt[n]{ab}+a+b\right)^{n}\]
Notice that $n=1\implies g=0; \text{ and }n=2\implies g=\sqrt{2}$. When $n>2, \ g=f(a,b)$ is a positive real number.
\[x=\sqrt{2ab}+a, \ y=\sqrt{2ab}+b, \ z=\sqrt{2ab}+a+b\]
Where $(x,y,z)$ are irrational numbers [forming?] all Pythagorean triples in all natural number [pairs] $(a,b)$.
Let $k=\sqrt{\frac{ab}{2}}$; now solve for $x,y,z$ in terms of $a, \ k$ (not sure why this is done)
\[x=2k+a, \ y=\frac{2k(k+a)}{a}, \ z=2k+a+\frac{2k^{2}}{a}\implies xy=\frac{2k(2k+a)(k+a)}{a}\]
When $a$ is odd, $k=ha, \ xy=2a^{2}h(2h+1)(h+1)$ and $hk=a, \ xy=\frac{2k^{2}(2+h)(1+h)}{h}$
When $a$ is even, $2k=ha, \ xy=\frac{a^{2}h(h+1)(h+2)}{2}$ and $2hk=a, \ xy=\frac{2k^{2}(1+h)(1+2h)}{h}$
Therefore $xy$ cannot be the power numbers in all Pythagorean triples.
(I'm not sure what that last sentence actually means)
[/hide]
[hide=" $1^{st}$ method of FLT proof"]
$g\sqrt[n]{ab}$ is irrational in all naturals number $(a,b)$ (?), so $(x,y,z)$ are irrational.
\[\left(g\sqrt[n]{ab}+a\right)^{n}+\left(g\sqrt[n]{ab}+b\right)^{n}=\left(g\sqrt[n]{ab}+a+b\right)^{n}\]
When $a=b$, \[2\left(g \cdot \sqrt[n]{a^{2}}+a\right)^{n}= \left(g\cdot \sqrt[n]{a^{2}}+2a\right)^{n}\]
[Added these steps, using fact that each side is in the reals and that $n$ is odd] \[\iff \sqrt[n]{2}\left(g \cdot \sqrt[n]{a^{2}}+a\right)=g\cdot \sqrt[n]{a^{2}}+2a \iff \]
\[\left(\sqrt[n]{2}-1\right) g\cdot \sqrt[n]{a^{2}}= \left(2-\sqrt[n]{2}\right) a \implies g = \frac{\left(2-\sqrt[n]{2}\right) a}{\left(\sqrt[n]{2}-1\right) \cdot \sqrt[n]{a^{2}}}\] \[\iff g = \left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right) \sqrt[n]{a^{n-2}}\]
Consider the expression \[\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)\]
And use it in: \[\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)\left(a^{n-1}b^{1/n}+(ab)^{\frac{n-1}{n}}\right)/2\]
This expression is irrational for all naturals $(a,b)$.
Consider the quantity $g\sqrt[n]{ab}$ divided and multiplied by \[\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)\left(a^{n-1}b^{1/n}+(ab)^{\frac{n-1}{n}}\right)/2\]
, which yields two expressions. When $a=b, q=1$ we have that: \[g\sqrt[n]{ab}= q\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)\left(a^{n-1}b^{1/n}+(ab)^{\frac{n-1}{n}}\right)/2 \] \[q=\frac{2g\sqrt[n]{ab}}{\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)\left(a^{n-1}b^{1/n}+(ab)^{\frac{n-1}{n}}\right)\]
If $g\sqrt[n]{ab}=N$ is natural for some $(a,b)$, then $g\sqrt[n]{ab}=N$ can not have $\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)$ (what does that mean), and $g\sqrt[n]{ab}$ can not have $\left(\sum_{i=1}^{n-1}\sqrt[n]{2^{i}}\right)$ (oh I think it means divides.
So when $A=B, q \not = 1$; contradiction.
Therefore $g\sqrt[n]{ab}$ is an irrational number for all naturals $(A,B)$.
[/hide]
[hide=" $2^{nd}$ method of FLT proof"]
$X^{n}+Y^{n}=Z^{n}{X^{(}n/2)}^{2}+{Y^{(}n/2)}^{2}={Z^{(}n/2)}^{2}$
When n=2, we can display ${X^{(}n/2),Y^{(}n/2),Z^{(}n/2)}$ with (a,b).
$a=Z^{(}n/2)-Y^{(}n/2), b=Z^{(}n/2)-X^{(}n/2) X^{(}n/2)=(2ab)^{(}1/2)+a, Y^{(}n/2)=(2ab)^{(}1/2)+b, Z^{(}n/2)=(2ab)^{(}1/2)+a+b$
When n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number.
$ab=Z^{n}-(YZ)^{(}n/2)-(XZ)^{(}n/2)+(XY)^{(}n/2)$
We multiply $X^{(}n/2)$ and $Y^{(}n/2)$.
$(XY)^{n}=2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$
If (X,Y,Z) is the natural number, $(XY)^{n}$ is the natural number, but $2a^{3}b+2ab^{3}+13(ab)^{2}+6ab(a+b)(2ab)^{(}1/2)$ is the irrational number.
That is an apparent contradiction.
Therefore (X,Y,Z) must be the irrational number. 2nd method. end.
[/hide][/quote]

I'm not really understanding what the original proof was, because in key places the messy writing is leading to misunderstandings etc.

Anybody else want to try to decipher the "proofs"... ?
(It will be a good lesson in writing convincing fallacious proofs)

[quote="ZetaX"]My point is:
- proofreading is mostly boring
- this  stuff is very badly written
- I doubt that the author could be called mathematician, looks like a swank

If he really want's to know if his reasoning is true and not to show off, he could always try to read is stuff himself, write it more clearly and especially pay someone to proofread.

I will never read that kind of stuff and hope others don't so, either. That's like feeding trolls.[/quote]
I agree with ZetaX.
I seen SO many of these stupid claims.

You know who that guy is. Some person why knows barely any math, and understood the Last theorem and got interested in it, because it is so elementary and then made up some argument that he found it.

I think this guy's name is Lee Jae Yuel (South Korea)
He's the most stubborn(and the most famous) of people who claim to have  :) proven Fermat's last theorem and the four color theorem.
 He enjoys terrorizing the Korean Mathematic Society Homepage,
along with another of his friends who claims to have proven the Goldbach conjecture.
 like Zetax said, your time is better spent not reading this proof.

At a glance, you can already tell that he doesn't have a definite concept with numbers (like introducing the quantity g, which is a real number in the first place) and working with integers at the same time with rationals and irrationals??? His variables are badly defined. Aww, he just wasted 2 minutes of my time.  :furious:


$a$ $b$ $c$ $d$ are reals
find
$\lim_{x \to 0}\frac{1-cos(ax)\time cos(bx)\time cos(cx) \time cos(dx)}{x^{2}}$

Maybe L'Hospital's might help, even if it is Calculus.

Limits = calculus = belongs in the calculus forum.

limits=in my precalculus book and in AoPS vol 2. :wink:

So I think it's okay here.


If $a , b$ are positive integers, prove that

   $\sqrt[3]{( \frac{a^{4}+b^{4}}{a+b})^{a+b}}\geq{a^{a}b^{b}}$

 Note that $a+b$ is an exponent on  the fraction.

[quote="Jan"]I think this problem is true for positive reals, and not only for integers.

Weighted AM-GM:
\[\sqrt[3]{\left(\frac{a.a^{3}+b.b^{3}}{a+b}\right)^{a+b}}\geq \sqrt[3]{\left(\sqrt[a+b]{a^{3a}b^{3b}}\right)^{a+b}}=a^{a}b^{b}\]
[/quote]

YES !!! Your solution and generalization for positive reals  are good . Just a friendly note : usually for positive reals we do not use radicals but rational fractions. So , the above solution [color=red]is perfect [/color]for positive integers ! A little modification is necessary for the other case(positive reals).

 Babis

Thanks for the comment.
I'm not sure I understand, though...

Do you mean I should have written:
\[\sqrt[3]{\left(\frac{a.a^{3}+b.b^{3}}{a+b}\right)^{a+b}}\geq \sqrt[3]{\left((a^{3a}b^{3b})^{\frac{1}{a+b}}\right)^{a+b}}=a^{a}b^{b}\]

Exactly !  Now your solution is OK for all positive reals $a , b$ .

  This is the power of   '' weighted means '' !!!

 babis


Na lythei i parakato eksisosi stous thetikous akeraious:

                               $ x^2\plus{}y^2\equal{}1997(x\minus{}y)$

Αυτο που θα ρωτησω ισως ειναι βλακεια :blush:
Μπορουμε να ξεχωρισουμε,ποτε οι λυσεις που θα δοσουμε σε μια διαφαντικη θα ειναι συγκεκριμενοι αριθμοι η θα εχουν γενικοτερη μορφη :?: (π.χ $ x \equal{} t^2 \plus{} 1$ και $ y \equal{} 3t \plus{} 5$)

[quote="Dimitris X"]Αυτο που θα ρωτησω ισως ειναι βλακεια :blush: [/quote]

Καθόλου βλακεία δεν είναι. (Αλλά και να ήταν, εδώ επιτρέπεται να ρωτάμε βλακείες.)

Λοιπόν, γενικά δεν μπορούμε να ξέρουμε. Βλέπε για παράδειγμα το τελευταίο θεώρημα του Fermat που δεν γνωρίζαμε καν αν υπάρχουν λύσεις ή όχι. (Μέχρι το 1994 που ο Wiles απέδειξε πως δεν υπάρχουν.) Ακόμη χειρότερα, έχει αποδειχθεί πως δεν μπορεί να υπάρξει αλγόριθμος που να του δίνουμε μια διοφαντική εξίσωση και να μας απαντάει αν έχει λύσεις ή όχι. (Κάνοντας πεπερασμένο αριθμό βημάτων.)

Υπάρχουν κάποιες μορφές διοφαντικών εξισώσεων που ξέρουμε πως να τις λύνουμε και από εκεί και πέρα αν μοιάζει κάπως με κάποια διοφαντική που έχουμε δει, ίσως μπορέσουμε να μαντέψουμε τι μορφή θα έχουν οι λύσεις.

Σε αυτήν την περίπτωση η πρώτη μου σκέψη ήταν να συμπληρώσω το τετράγωνο. Έχουμε $ (x \minus{} 1997/2)^2 \plus{} (y \plus{} 1997/2)^2  \equal{} 1997^2/2$.  Πολλαπλασιάζωντας με το 4 παίρνουμε $ (2x \minus{} 1997)^2 \plus{} (2y \plus{} 1997)^2  \equal{} 2 \cdot 1997^2$. Από εδώ βλέπουμε ότι πρέπει η εξίσωση να έχει πεπερασμένο αριθμό λύσεων.

Από εδώ και πέρα παίζει ρόλο και η εμπειρία και τι παρόμοιες διοφαντικές έχουμε δει. Ένας προγραμματιστής θα έφτιαχνε ένα πρόγραμμα που θα έβρισκε όλους τους τρόπους για να γραφτεί το $ 2 \cdot 1997^2$ σαν άθροισμα δύο τετραγώνων και για κάθε τέτοιο άθροισμα θα υπολόγιζε γρήγορα τα χ και ψ.

Κάποιος πιο έμπειρος θε γνώριζε ότι επειδή ο 1997 είναι πρώτος της μορφής 1mod4 το $ 2 \cdot 1997^2$ μπορεί να γραφτεί σαν άθροισμα δύο τετραγώνων με ακριβώς 3 τρόπους (και πως να το γράψει) και έτσι θα γλύτωνε αρκετό χρόνο.

Ίσως εξηγήσω αργότερα πως γράφουμε κάποιον αριθμό σαν άθροισμα δύο τετραγώνων. (Αν έχει ξανασυζητηθεί στο παρελθόν ας βάλει κάποιος κανένα link.)

Μια πιο βολική μορφή είναι η παρακάτω . Η εξίσωση πολλαπλασιάζοντας με το 2 και προσθέτοντας το $ 1997^2$ στα δύο μέλη , γίνεται :

$ (x\plus{}y)^2\plus{}(1997\minus{}x\plus{}y)^2\equal{}1997^2$ και αφού το 1997 είναι πρώτος μπορούμε να δουλέψουμε με τις primitive solutions των πυθαγόρειων τριάδων .Δηλαδή πρέπει  $ 1997\equal{}m^2\plus{}n^2$ , $ x\plus{}y\equal{}2mn$ και $ 1997\minus{}x\plus{}y\equal{}m^2\minus{}n^2$ . Επίσης στη συnέχεια παίρνοντας mod 3 kai mod 5 μειώνουμε κατά πολύ το πλήθος των αριθμών ου πρέπει να τσεκάρουμε . Τελικά μένει να τσεκάρουμε μόνο τα $ m\equal{}34,41,44$ από τα οποία
$ (m,n)\equal{}(34,29)$ η οποία τελικά μας οδηγεί σε δύο λύσεις . $ (x,y)\equal{}(170,145)$ και
$ (x,y)\equal{}(1827,145)$

 :)

ωραια ασκηση. ελπιζω να μην εκανα καπου λαθος.
αρχικα γραφουμε την εξισωση μας ισοδυναμα,
$ (x \plus{} y)^2 \plus{} (x \minus{} y)^2 \minus{} 2*1997(x \minus{} y) \plus{} 1997^2 \equal{} 1997^2$
$ \leftrightarrow (x \plus{} y)^2 \plus{} ( x \minus{} y \minus{} 1997)^2 \equal{} 1997^2$
αν θεσουμε,
$ k \equal{} x \plus{} y , l \equal{} 1997 \minus{} x \plus{} y$ ή αντιστροφα $ k \equal{} 1997 \minus{} x \plus{} y , l \equal{} x \plus{} y$
τοτε εχουμε την εξισωση,δεδομενου οτι $ 0 < k,l < 1997$
$ k^2 \plus{} l^2 \equal{} 1997^2$ ,
παρατηρουμε οτι $ gcd(k,l) \equal{} 1$ αφου ο 1997 ειναι πρωτος.
επομενως απο τις πρωταρχικες πυθαγορειες τριαδες  πρεπει να υπαρχουν $ a,b$ με $ a > b$ που να ικανοποιουν τις σχεσεις,

$ a^2 \plus{} b^2 \equal{} 1997$(3)
$ 2ab \equal{} l$
$ a^2 \minus{} b^2 \equal{} k$

ομως ισχυει,

$ 1997 \equiv 2 (mod7) , 1997 \equiv 2 (mod5)$
αρα
$ a,b \equiv \plus{} \minus{} 1 (mod7)$ και $ a,b \equiv \plus{} \minus{} 1 (mod5)$
δλδ

$ a,b \equiv 6,24,31,34 (mod35)$

ομως προφανως $ 1997 > a > \frac {1997}{2}$

αρα ελεγχοντας με τις σχετικες πραξεις ποια τετραγωνα αριθμων ικανοποιουν τις (1),(2) παιρνουμε,
$ a \equal{} 41$ ή $ a \equal{} 34$

ομως ικανοποιει την (3) μονο τ ζευγος $ (a,b) \equal{} (34,29)$
αρα εχουμε
$ k \equal{} 315 , l \equal{} 1972$
Αρκει λοιπον να λυσουμε τα συστηματα,

$ x \plus{} y \equal{} 315$
$ y \minus{} x \plus{} 1997 \equal{} 1972$

kai

$ x \plus{} y \equal{} 1972$
$ y \minus{} x \plus{} 1997 \equal{} 315$

λυσεις των οποιων ειναι οι,
$ (x,y) \equal{} ( 170,145), ( 1827,145)$
QED
EDIT 2 λεπτα πιο πριν ρε silouan! !!  που θα μου παει ομως ! θα ειμαι σε ετοιμοτητα next time  :P

Σιλουανε και Ηλία ωραίες οι λύσεις σας.

Βρήκα σήμερα λίγο χρόνο και έγραψα [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1443992#1443992]εδώ[/url] κάποια θεωρία και πως αυτή βοηθαει στον τρόπο που πρότεινα στο αρχικό μου post.