inequalities calculus derivative logarithms function limit integration

Prove or disprove:

$ \forall x\ge 1, $

$ ln( 1 + \frac 1x ) - \frac 1 {1 + x} \ge 0 $

I'm fairly sure this is true, but I ran into several brick walls when I tried to prove it.

I was trying to prove that the derivative of $ y = ( 1 + \frac 1x )^x, $ is positive, which it is, i think.

I would have posted it in the calculus forum but i don't think calculus can be used to prove whats left of the problem.

I might be wrong...

You mean there's probably a calculus solution, seamusoboyle?

You could just observe that the derivative of

$ ln( 1 + \frac 1x ) - \frac 1 {1 + x} $

is always negative, and that as x goes to infinity,

$ ln( 1 + \frac 1x ) - \frac 1 {1 + x} = 0 $

which would be a proof, but not a very elegant one.

I was sort of hoping there would be a non-calculus proof.

EDIT: Amirhtlusa,

$ y = (1 + \frac 1x)^x $

$ ln(y) = (x)ln(1 + \frac 1x) $

$ \frac {dy}{dx} \frac 1y = ln( 1 + \frac 1x ) - \frac 1 {1 + x} $

as $ x \ge 1, y \ge 0 $

Therefore, $ \frac {dy}{dx} \ge 0 $ iff $ ln( 1 + \frac 1x ) - \frac 1 {1 + x} \ge 0 $

Which is the problem I posed...

i probably should have explained...

This is what you might want to try.

We need to show that

$1+1/x\geq 1+1/(1+x)+1/[2(1+x)^2]+\ldots$

Take a few terms out of the RHS and sum the rest by comparison to a smaller geometric series.  I don't have time now; I might try it later.

Ive already tried that way, not so sure it work too well though... By the way, i was trying to rile you stephen! :)

[quote="seamusoboyle"]Actually, that function that you started off with, that you took the deritive of, thats well know to be monotonically increasing sequence with limit e.[/quote]

I know that, thats what inspired it... I wasn't just messing around with random equations. I've seen the graph of it too, which is why i was so sure its derivative was positive. Like you say, its monotonically increasing. I tried to prove it was increasing, which led to an interesting inequality, and here we are...

I like trying to prove things you've already proved another way... In this case, because we know its increasing, and there's that calculus proof. Its still interesting to try find another way... What blahblahblah said is probably a good possibility...

I tried it but to no aviail! See can you do better!

Does this prove the interesting result that

$(1+\frac{1}{x})^x \leq e \leq (1+\frac{1}{x})^{x+1}$

???

The 'calculus' proof is probably the nicest way though.  Lets make it rigorous.

We know that $f(x)=\ln(1+1/x)-1/(1+x)\to 0$ as $x\to\infty$ with $f'(x)<0$ for all $x\geq 1$.

Say that there exists $a$ such that $f(a)<0$.  Then because $\lim_{x\to\infty} f(x)=0$, for every $\epsilon>0$, there exists $n$ such that $x>n$ implies $| f(x) |<\epsilon$.

Set $\epsilon=|f(a)|/2$.

Then for $x=n+1$, we have $f(n+1)-f(a)=f'(c)(n+1-a)>|f(a)|/2$ for some $c\in (a,n+1)$.  But since $f'(c)$ is strictly negative, this is impossible.

This inequality is trivial. it follows from the well known ln(1+y)>=y/1+y for y=1/x.

By the covexity of $y=\frac{1}{x}\ (x>0)\ $,

$\int_x^{x+1} \frac{1}{x}dx>\frac{1}{x+1}\cdot (x+1-x)$, we are done.

Ah! The old confexity thingy! I always have trouble with that. How do you prove a function is confex on a certain interval? Is it that the second derivitive is positive on that interval? And what inequality did you use there :blush: ?

He is using the fact that if $f$ is continuous on $[a,b]$ and achieves a minimum at $b$, then

$f(b)(b-a)\leq \int^b_a f(x)dx$ with equality iff $f$ is constant.

[quote="seamusoboyle"]

$(1+\frac{1}{x})^x \leq e \leq (1+\frac{1}{x})^{x+1}$

???[/quote]
I can prove that :
$(1+\frac{1}{x})^x \leq e \leq (1+\frac{1}{x})^{x+1/2}$
but for
$(1+\frac{1}{x})^x \leq e \leq (1+\frac{1}{x})^{x+λ}$
[0<λ<1/2]
which λ can satisfy this inequality?
Can somebody slove this problem?
Thanks...

Just find the minimum of that function, and set it equal to $e$ ?

I think 1/2 is the lower limit -- no proof, just empirical evidence.

For $\frac{1}{2}$ mathematica gives us the solution:

{x -> 168689.}

Can you show the proof for this?

Really?  Then perhaps 1/2 doesn't work after all.  I suppose it is possible that it might fail.

Mathematica solves it through re-iteration, so it might no be exact. But it cant find solutions to anything lower than a half. Whats the proof for that?

I am sorry that my English is not very good,and I can't understand all your reply clearly.
But I just find that:
$(1+\frac{1}{x})^{x+2/5} \leq e \leq (1+\frac{1}{x})^{x+1/2}$ [x≥1]
Can somebody give more advanced solution?

Are both these functions monotinacilly increasing? I think they might be...


algorithm
algebra
polynomial
linear algebra

Is there any way (some algorithm) to compute the matrices $P,D$ s.t. $A=PDP^{-1}$ without knowing before hand the eigenvalues/eigenvectors of $A$?

The question is a little bit silly. If you do find $A=PDP^{-1},$ then the diagonal elements of $D$ are the eigenvalues and the columns of $P$ are the eigenvectors. Thus any algorithm for finding $P$ and $D$ would be an algorithm for finding the eigenvalues and eigenvectors.

Are there practical algorithms for finding or approximating $P$ and $D$ that do not involve the characteristic polynomial? That's a different question, best addressed in a numerical anyalysis textbook.

sorry if the question was silly. i'm not very familiar with these notions  :blush:


Let $ V$ and $ W$ be vector spaces over the field $ F$ and let $ U$ be an isomorphism of $ V$ onto $ W$. Prove that $ T\rightarrow UTU^{\minus{}1}$ is an isomorphism of $ L(V, V)$ onto $ L(W, W)$.

Let $ \varphi_U(T) \equal{} UTU^{\minus{}1}$. Then certainly $ \varphi_U$ is a homomorphism from $ L(V,V)$ to $ L(W,W)$. Furthermore, it is easy to see that $ \varphi_U\varphi_{U'} \equal{} \varphi_{UU'}$ for any other map of spaces $ U'$. In particular, $ \varphi_U\varphi_{U^{\minus{}1}} \equal{} id_{L(W,W)}$, and $ \varphi_{U^{\minus{}1}}\varphi_U \equal{} id_{L(V,V)}$, so $ \varphi_U$ is an isomorphism.


if we have a spherical hollow shell of radius R and a mass M and another point mass m at a distance 2R from the center what would be the gravitional field at the centre of the shell?
If we were deallng with electrostatics, such a situation would give a zero electric field at the center. But would it be same with a gravitional field or would the answer be based on superposition principle?

Due to the shell theorem, a mass inside the body experiences no gravitational force due to the body it is inside.  However, since it is at a distance from the center...  I think your instinct is right and there is no force experienced.


Which do you use for FTW?

I use a slide rule for fun, but otherwise I'll do it in my head or occasionly write some stuff down

[quote="r15s11z55y89w21"]I always have scratch paper in front of me just in case; most problems can be calculated mentally, though. I strongly advise against using a calculator, as that eliminates the bulk of the joy of doing math.[/quote]

With FTW-ers, calcs are "cheating".

[quote="isabella2296"]Sometimes it's annoying though, because some problems clearly require calculators. Unless you know your powers of 27, of course.  :)[/quote]
bigger version of powers of 3 :P

[quote="xpmath"]For even more fun, try to do the problems having no paper at all :P . That's what I do most of the time and it can actually help.[/quote]

Yeah, I can't find paper, ever.  And it's great practice for the CD round.

How do you do the target questions on FTW with no paper???

Well, you can either learn something and use a calculator, or you can just not do the problem. :wink:

Or you can learn ''memming'' ability, as called.

Seriously, don't mem intentionally. It is a plain downright incorrect practice. Accidental memming is okay, though, since it's hard to deviate from that.

yeah, using calculators takes all the joy of multiplying five-digit numbers, dividing into a decimal with a 7-digit repeating block, adding together a non-telescoping series of fifty fractions, etc. out of math. endless computation is my only pleasure left in math.

Uh, no, if you see a problem you know the answer to, what's the point of not answering? There's no difference either way.

[quote="Temperal"]yeah, using calculators takes all the joy of multiplying five-digit numbers, [b]dividing into a decimal with a 7-digit repeating block[/b], adding together a non-telescoping series of fifty fractions, etc. out of math. endless computation is my only pleasure left in math.[/quote]

unlikely unless you have to deal with fractions with a denominator divisible by 239 or 4649...  :lol:

if you mean 6-digits, memorize 1/7=0.142857 repeating, and 1/13=0.76923


ARML
geometry
3D geometry
rotation
trigonometry
analytic geometry
geometric transformation

The mathclub at my school would like to participate in the Santa Clara University High school mathematics competition (it's not SCVMA) that's coming up this year.

Can anyone give us some registration etc. information?

thx in advance

Yeah, i'd also like to know when Santa Clara is; don't they usually run one in November?  i'm planning to organize a contest/ARML thing sometime in November, and i'd like to avoid conflicting with their contest.  In other news, i'll try to get some information to you all about ARML within the next 2 or 3 weeks..

I'm guessing it's going to be on November 8--they like to hold it on a holiday weekend so they don't have more participation than they can handle. They should send out invitations to the schools sometime before then. I'll try to post more info when we get an invitation.

anyone know who runs the event?  i need to contact them to make sure i'm not double scheduling on that date, and no one i've contacted yet has bothered to answer my emails..

Looks like i found the information we're looking for; the contest is on November 15 apparently:

http://cms.scu.edu/cas/math/mathcontest.cfm

That said, those of you who are planning to try out for ARML this year can expect our first event to take place on the afternoon of November 8.  More details will be forthcoming as soon as Silas and i get the details worked out..

So how did it go?
I thought the tie breaker was unusual.  Did anyone get more than 9 bricks in?
For those who weren't there, here was the problem:  How many 3*7*9 bricks can fit in a 13*13*13 cube?

I think it's weird how they added an extra problem this year. Plus the problems seemed harder than last year's.

6 was crazy. I wasted so much time on it and didn't get anywhere! I was so mad about that.

The tie breaker was just to explain 8b right? But since you have to show your solutions on all the problems, the tie breaker seemed redundant. But it didn't matter to me. I didn't get 8b. -_-

By the way, my teacher got the answers:

1. 12:04:57
2. 2
3. 99/20 = 4.95
4. 23958 x 4 = 95832
5. 90
6. 150 degrees, 5/6 pi radians
7. yes, no, yes, no
8a. 3 x 7 x 9
8b. 9

Did anyone else get 5 for number 2?

Hmm my answers match except for #7... are you sure it isn't Y/N/N/Y?  There was a problem with the answer choices, so maybe there was a problem with the correct answer as well.
I hope I don't get points taken off for 8a tiebreaker--I forgot to show that it couldn't be 1 x 7 x 27, 3 x 3 x 21, etc...

Yeah, I got y/n/n/y too, and so did most of the people I talked to.

But the answers I posted above came from the "official" list of answers my teachers obtained. Maybe there's an error.

#6 is a fairly well known problem  (I've seen it copied on at least 2 or 3 different exams including this one)
The basic idea is to [hide]rotate the triangle about C 60 degrees and 60 degrees in the other direction[/hide] although I'm sure it can be trig bashed.

Number 2 seems sketchy. It should not be x2008 at the end, but -2008, if the pattern follows through, with all multiples of 3 getting +'s. (the official answer seems to go with x2008...)

archimedes1 can you elaborate on how to solve #6?

At first I tried bashing it with Heron's formula, quickly realized that would be way too crazy, and ended up trying to trig-bash it. I didn't get anywhere though... but the problem can be coordinate bashed. I realized that after the competition.  :|

[hide="coordinate bashing"]Let $ A, B, C$ be $ (0, \frac{\sqrt{3}}{2} s), (\minus{}\frac{1}{2}s, 0), (\frac{1}{2}s,0)$ and let $ P$ be $ (x,y)$. You can write a system of 3 (crazy) equations to get $ s\equal{}\sqrt{25\plus{}12\sqrt{3}}$ and use law of cosines to get the answer. -_-[/hide]

I'm assuming the x 2008 was a typo. I didn't even realize that during the competition.

I wasn't there, but...

[hide]
Rotate ABC clockwise 60 degrees around B, let C map to C' and P map to P'. Rotate ABC by 60 degrees clockwise around C, let A map to A' and P map to P''. We have PBP' = 60 because of the rotation, so PBP' is equilateral. Similar, PCP'' is also equilateral. We have PC = 5, PP' = 4, P'C = 3, and P''A = 4, P''P = 5, PA = 3. These two triangles are congruent. Then BPP' + P'PC + CPP'' + P''PA = 60 + 60 + P'PC + P''PA = 120 + 90 = 210, so APB = 150.
[/hide]

Number 7 was definitely Y/N/N/Y.

Number 6 can be Law of Cosines bashed. I did that, and messed up :(

I was in a hurry to get to piano class, so I kind of rushed the last few after wasting too much time checking the early problems, since I haven't really done much math lately.

Here are the problems for those who didn't go:
1. Stacy takes 9 minutes to walk from A to B and Tracy takes 11 minutes to walk from B to A.  If Stacy starts at A and Tracy starts at B at 12:00 and it both walk at a steady pace, at what time will Stacy and Tracy meet?
2. In calculating the following, rather than use the standard order of operations, we perform any operation on the left before the operations to its right.
$ 1 \cdot 2 \plus{} 3 \minus{} 4 \cdot 5 \plus{} 6 \minus{} 7 \cdot 8 \plus{} 9 \minus{} 10 \cdot 11 \plus{} \dots \cdot 2008$

So after the first operation we get 2, then adding 3 we have 5, subtracting 4 from that we get 1, multiplying that by 5 gives 5, etc.  Find the last digit of the above expression.

3. On a rectangle with height 11, a straight line is drawn from the upper left corner to the lower right corner.  Then a second line is drawn from the lower left corner to a point on the right edge 2 inches below the top right corner. How high is the intersection point from the bottom of the page?

4. In the following multiplication, each letter represents a different digit.  Find the value of each letter:
$ RECAP \cdot 4 \equal{} CAPER$

5.  Find the coefficient of $ x^{25}$ in the expansion of $ (x^3 \plus{} x \plus{} 1)^{10}$.

6.If $ A$,$ B$,and $ C$ are the vertices of an equilateral triangle and $ P$ is a point in the interior such that $ P! \equal{} 3$, $ PB \equal{} 4$, $ PC \equal{} 5$, then find $ \angle APB$.

7. Do there exist solutions of $ 3^{\cos \theta} \equal{} 2.008 \plus{} \sin \theta$ in the interval
a) $ (0, \pi/2)$?
b) $ (\pi/2, \pi)$?
c) $ (\pi, 3\pi/2)$?
d) $ (3\pi/2, 2\pi)$?

8.  A solid rectangular brick whose dimensions are $ r \cdot s \cdot t$, where $ r,s,t$ are integers, is placed entirely inside an empty $ 13 \cdot 13 \cdot 13$ container.
a) If the remaining space in the container is 2008 cubic units, then what are the dimensions of the brick?
b) How many bricks of that size can be placed completely within the container without breaking them?

9. tiebreak -- explain your answers to problem 8 as clearly as possible.

EDIT: Nevermind!I realize where my mistake is. :(

I FACTORED $ 189$ INTO $ 3\cdot 3\cdot 31$!!!  :cursing:

[quote="chemequilibrium"]So the answer for #2 really should be 5, not 2?[/quote]
I believe it said ...X2008, so no?

[quote="chemequilibrium"]So the answer for #2 really should be 5, not 2?[/quote]
#2 contradicts itself, so I don't see how there is a valid answer.  If you looked at the expression, you'd get 2, whereas if you followed the pattern up to 2008, you'd get 5.

[quote="serialk11r"]Basically, the test was pretty bad as usual.[/quote]
Basically, I didn't attend to this contest but I had a chance to see the problems in this year. I feel that it's not as bad as you say it was. There are some nice problems like: #2, #8b and #6. Although the solution to #2 is probably wrong(yeah I think we have to look at the pattern of the alternative signs, so it must end up with+2007 -2008). Anyway, it's just my opinion and I don't meant to offense yours :wink:

[quote="archimedes1"]
Does anyone know the original source of #6?  I've definitely seen it before somewhere.[/quote]
its generally pretty well known but i guess most people know it from math olympiad challenges

[quote="archimedes1"][quote="bookaholic"]For 8a, can anyone explain how to rigorously eliminate 1x9x21 and 3x3x21? 13sqrt(3) is about 22.1, which eliminates all but those two and 3x7x9. It's not clear to me how to make sure either of them couldn't fit in diagonally.[/quote]
Well, 1x9x21 can be eliminated by examining its space diagonal, which is larger than the space diagonal of 13x13x13.  I'm not sure about 3 x 3 x 21.

Does anyone know the original source of #6?  I've definitely seen it before somewhere.[/quote]
Maybe take sin45 times the diagonal and compare to 13? (either dimension is minimized)
I don't want to work through it so yea.
EDIT: nevermind it's more complex than that, I better finish up math homework :(

Please let me know as soon as any of you hear anything about results from this contest.  The contest coordinator told me the results would be available in January, but i want to get them as soon as i can so i can do more ARML recruiting..


A spherical, 3-dimensional planet has center at $(0,0,0)$ and radius $20$. At any point $(x,y,z)$ on the surface, the temperature is $T(x,y,z): =(x+y)^{2}+(y-z)^{2}$. What is the average temperature of the surface?

[hide="I got"] EDITED to 1600/3, thanks Ravi B[/hide]

I think you are close, but don't forget that $x^{2}+y^{2}+z^{2}$ is $20^{2}$, not $20$.

What's the general idea to solution with these problems. I want to do it myself, but I think I need a hint.

A hint would be to [hide]impose symmetry![/hide]

[hide="Hint"]What are $E(x^{2})$, $E(y^{2})$, and $E(z^{2})$?  (Here, $E$ means expected value.) What are $E(xy)$ and $E(yz)$?[/hide]

Alternately, you could use area of the sphere formula


How many non-empty subsets of $ \{ 1 , 2, 3, 4, 5, 6, 7, 8 \}$ consist entirely of odd numbers?

[hide="Solution"]For each odd element, there are two choices: whether it is in or out of the subset.  For the even elements, they have to be out of the subset, so there are $ 2\cdot 1\cdot 2\cdot 1\cdot 2\cdot 1\cdot 2\cdot 1\equal{}16$ subsets.  However, one of these is empty, so the answer is $ 16\minus{}1\equal{}\boxed{15}$[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide your solutions[/url].[/color]

[hide="or:"]
to choose no odds there is 1 choice
to choose 1 odd there are 4 choices
to choose 2 odds there are 6 choices
to choose 3 odds there are 4 choices
to choose 4 odds there is 1 choice
1+4+6+4+1=16
-1 because to choose no odds isn't allowed

p.s. 1 4 6 4 1 is pascal triangle!!
why??[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide your solutions[/url].[/color]

[hide]The subset we need to pull out is $\{1,3,5,7\}$  There are 16 subsets of this.  However, we counted the empty set, which is not allowed.  The total is 15[/hide]

[quote=Bestestsocks]Is there a way to find the number of subsets without bashing?[/quote]

There are $2^n$ subsets where $n$ is the number of elements in the set total.

[quote=abcb1030]Why’s it $2^4 - 1 = \boxed{15}$ not 4!?[/quote]

It's probably too late now but the reason is since we don't care about the order. Now if we didn't care about order, then that would mean that we would be choosing ... and that's exactly what we're doing. Another interesting way to view this problem is that there are $\dbinom{4}{1}$ 1-element subsets meeting the requirements, $\dbinom{4}{2}$ 2-element subsets meeting the requirements, $\dbinom{4}{3}$ 3-element subsets meeting the requirements, $\dbinom{4}{4}$ 4-element subsets meeting the requirements. Adding all of these up gives $4 + 6 + 4 + 1 = 15$, which is the way I originally did it. It's a more casework-oriented way of doing it (this also kind of relates mathematically powers of two and combinations since $\sum_{k=0}^{n}\dbinom{n}{k} = 2^n$).

oops I didn't read non-empty and got it wrong first try

anyways,

[hide=sol]To get a subset with only odd integers, we can only choose elements which are odd (i.e 1,3,5,7) because the evens each have one choice: They can't go into the subset.

There are $2^4 = 16$ ways to do this, and subtract one to get $\boxed{15}$, empty set doesn't count.  (here's where I missed the empty set).[/hide]


function
real analysis
real analysis unsolved

if $ T: X \rightarrow Y$ is a linear operator , where $ X, Y$ are Banach spaces , then if $ e$ is an extreme point of $ B[X]$ then $ T(e)$ is also an extreme point .

I assumed $ T$ is onto operator  to prove last statment.

Is surjective (onto) codition necssary for $ T$ ??? if yes , give me a counter example when $ T$ is not  onto operator.

Yes, it's necessary. Here is a finite-dimensional example: let $ T$ be the embedding of $ \ell_{\infty}^n$ into $ \ell_{\infty}^{n\plus{}1}$ given by $ T(x_1,\dots,x_n)\equal{}(x_1,\dots,x_n,0)$. Then $ (1,1,\dots,1)\in \ell_{\infty}^n$ is an extreme point on the unit ball, but its image under $ T$ is not.

You can take $ n\equal{}1$ to make this example even simpler.

What is $ \ell_{\infty}^n$ ? What is the norm on it ? Is it supremum norm?

Thx

$ \mathbb{R}^n$ with $ \|\cdot\|_{\infty}$ norm. But I'm just guesing... :wink:

What if we assumed $ T: X\rightarrow Y$ is an [b]isometry[/b] , Is the onto condition still necessary ???

If yes , then we need a counterexample in infinite dimensional , since if $ T$ is an isometry , then $ T$ is (1-1) ,and if $ X, Y$ are Banach finite dimensional spaces , then $ T$ is onto iff $ T$ is (1-1)

May be it's not necessary , What do u see guys ??

 :)

All right, is mlok's example an isometry?

[hide]Yes, it is.[/hide]

OK, here's an infinite-dimensional example, based on the same idea:

Let $ X\equal{}Y\equal{}C[0,1]$, the space of continuous functions on the unit interval with the $ \infty$-norm. Define $ T$ as follows: $ Tf(x)\equal{}\begin{cases}f(2x)&0\le x\le \frac12\\ (3\minus{}4x)f(1)&\frac12\le x\le\frac34\\ 0&\frac34\le x\le 1\end{cases}$

That's also an isometry.

mlok's example is similar to shift operator , which it's not onto and it's an isometry .

Thank u Jmerry .


geometry
trigonometry
angle bisector
algebra proposed
algebra

Given $x, y, z$ real numbers for which:
$x^{2}+xy+y^{2}=2$,
$y^{2}+yz+z^{2}=5$,
$z^{2}+zx+x^{2}=3$.
Find the value of: $xy+yz+zx$

Consider a geometric interpretation.  Given a point $P$, we construct three lines at $120^{\circ}$ angles from each other to points $A, B, C$ such that $AP = x, BP = y, CP = z$.  Then $x^{2}+xy+y^{2}= AB$ and so forth.

But notice that $AB+CA = BC$; in other words, the triangle is degenerate.  This can only occur either when at least two of $x, y, z$ are zero (not possible here) or at least one of $x, y, z$ is negative.

The negative side becomes an angle bisector to the longest side $BC$, so it must be $x$.  Stewart's Theorem should finish the calculation...

At least one of x,y,z negative, because it had 2 solutions
\[x=-\theta \sqrt{\frac{10-6\sqrt 2}{7}},y=\theta\sqrt{\frac{3-2\sqrt 2}{7}},z=\theta\sqrt{\frac{19-3\sqrt 3}{7}},\theta=\pm 1. \]

Well I got a differant answer : $xy+yz+zx=\sqrt{8}$

[quote="t0rajir0u"]  Then $x^{2}+xy+y^{2}= AB$ and so forth.[/quote]

 :blush:

Of course this should be $x^{2}+xy+y^{2}= AB^{2}$, etc. which gives a [b]right[/b] triangle with side lengths $\sqrt{2}, \sqrt{3}, \sqrt{5}$.

Then it's a simple area calculation; we find that

$\frac{\sin 120^{\circ}}{2}\left( xy+yz+zx \right) = \frac{ \sqrt{6}}{2}\implies xy+yz+zx = 2 \sqrt{2}$.

Curiously, this solution implies $x, y, z$ are all positive.  Did I do something wrong?  And if so, how did I still get the right answer?

$sign(y)=sign(z)=-sign(x)$, but your solution is true for negative sides too.


AMC
USA(J)MO
USAMO
induction
MATHCOUNTS
AoPS Books

Well, it is on Tuesday.  Any last words?  Advice? Random Remarks? etc...  The forum will be locked soon so hurry up and post!

I wish everyone good luck and hope that I can at least get a few points  :D .

Good luck!!!

Just to say something before the AMC forum locks :D .

I would like to have a simple funeral, and I would like to donate the remains of my brain to science to further the study of brain fry.  :)  Oh yeah and my math books and other possessions go to the highest bidder, all proceeds go to fund the study of my fried brain.

If I get too suicidal, I will consider the various methods to die and see which one of those is the most painless one :lol:

No, just claim anything you don't know how to do is trivial.

The solution is left as an excercise to the reader.

Quick Question --

If there is a "find with proof" question where we must find a value for n and then prove it, do they give any points if you correctly find n but your proof is not good ...

For example, lets say the problem is:

Find, with proof, a value in terms of n for

(Sum from k=1 to n) 1/[(n)(n+1)]

And you write "the sum comes to the quantity n/(n+1)" along with a poor proof, will they give you, say, a point for FINDING the value?

JB

Good luck everyone! Especially me :D! But you guys too!

[quote="yif man12"]If all else fails, the solutions follows easily by induction.[/quote]

The proof is trivial. For further details, see USAMO participant ___'s paper.

Please note that you will need ___ such that ___ is the USAMO number of a very mathematically successful individual for this strategy to work.

Good luck to everyone.

[hide]Not that I need any. Heh.

[hide]Yeah, that's because I'm not taking it!   [/hide]

[/hide]

Just say in your paper:

please refer to _______'s paper (______ being a really good participant). I concur with all his explanations and proofs.

Why not replace ______ with your own number?  Or, alternatively, say "I'll prove this claim at the end of my paper".  Then, at the end of your paper, say "I proved earlier that..."  Or, begin the proof with "Proof without words:" and then draw extensive and confusing diagrams that could believably follow from each other, the last one of which shows the thing you want to prove.  :P

Good luck!

Good luck to everyone tomorrow! I hope everybody comes out alive!

get a hatchet, write something that looks like a proof, i.e. "we prove by induction that .... QED" and smear blood in the middle. that won't do anything. but it'd be pretty weird

I have a wonderful proof of this proposition,but this margin is too small to contain it.
-Fermat

good point: or if the question asks: Find with proof, a number n satisfying _______(conditions). The answer to that is very simple:

If they specify that n is a real positive number, your answer is always: n is in the set (0, pos. inf)
If  n must be any real number: n is in (-inf, +inf.)

Proof of my claim: If n satisfies the above conditions, then it is tangible, that is, it must have a value. Since all values of n must be be real and between two numbers, a and b, then n is in the set: (a,b).
 I know, they will not give it anything for providing an answer to a trivial interpretation

[quote="probability1.01"]I still think a touching story about growing up on the streets and literally working to get into the USAMO would be better :P.[/quote]Let's propose a sequel "Good Will Hunting, The Legacy" to the studio. The story has the junior with not quite the math talent of his dad, worked extremely hard against all odds (his dad wouldn't even buy him the AoPS books, he had to work 2 shifts just to afford a set), barely squeezed out an USAMO qualification, and in the process regained his father's love in grand tear jerker style!

just please don't put your teacher through what I had to deal with when one of my more dramatic students could not answer any of the questions at a math tournament and decided to while away the time filling the answer sheet with suicidal poetry. The grader freaked out and took it to the tournament director, who brought it to me, then I had to go to the student's parents....

So, keep a positive mental outlook (especially on paper), and good luck everyone!

[quote="texas137"]...at a math tournament and decided to while away the time filling the answer sheet with suicidal poetry...[/quote] Sounds like this year's MathCounts State  :D

USAMO in about 2 hours for me, the final good luck everyone!!!

I'm surprised this forum is not locked yet!!!

in an hour for me.... I am stilll  home..!!! GOOD LUCK EVERYONE!


calculus
inequalities proposed
inequalities

Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}2$. Find the minimum and maximum value of
\[ f(a,b,c,d) \equal{} \sqrt[a]{a}\plus{}\sqrt[b]{b}\plus{}\sqrt[c]{c}\plus{}\sqrt[d]{d}.\]

Is no one interested in posting a solution?  :)

I cannot think of a solution involving elementary method and how it gets terrible to apply Calculus here so you could post your solution as a reference.

For the maximum, we see that $ x^{\frac{1}{x}} \leq x$ with equality iff $ x\equal{}1$ or $ x\equal{}0$.

Therefore
\[ f(a,b,c,d) \equal{}\sqrt[a]{a}\plus{}\sqrt[b]{b}\plus{}\sqrt[c]{c}\plus{}\sqrt[d]{d} \leq a\plus{}b\plus{}c\plus{}d \equal{} 2\]
with equality if $ \{a,b,c,d\}\equal{}\{0,0,1,1\}$.

But, we still have to find the minimum... :P


Show that for all positive real numbers $a$ there exsist infinitely many positive integers $n$  , such that the number $[na]!$  is divisible by$n^{2}+1$ .

It's enought to prove that for $a=1/k$. We noticed that equation $n^{2}+1=(t^{2}+1)s^{2}$ (where $t$ is establish $t^{2}+1> 9k^{2}$ and $n$ and $s$ are variables) have infinity many solution (we have 1 solution when $n=t$ and $s=1$). Next we make using Pell theory. Divisors of $n^{2}+1$ are $t^{2}+1$, $s^{2}$. If we want $n^{2}+1|[an]!$. It's enought $t^{2}+1 < n/k$,
$2s < n/k$.  First condition is comply because $t$ is establish and $n$ is enough big. $s<\frac{n+1}{3k}<\frac{n}{2k}$.


algebra
polynomial
number theory proposed
number theory

Let p is a prime of the form $ 3k\plus{}2$
$ P(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ and $ a^2\equiv 3b (\mod p)$
Prove that exist an integer k such that $ P_k(x)\equiv x (\mod p),\forall x\in N$

What is $ P_k(x)$?  Does it denote $ P(P(...(x))$?

Yes it means that $ P(...P(x)...)$ (take k time )


A convex quadrilateral $ABCD$ with $AB=1,\ BC=1,\ CD=2,\ DA=3$ is inscribed in the circle $O.$ Find the area of the circle which touches to three sides of $BC,\ CD,\ DA.$

[quote="Altheman"]blah...extend the sidesBC, DA, to intersect at X, use similar triangles to find the lengths, then you want the radius of the ex circle XCD, which is trivial (tanX/2)*s[/quote]

Of the ex-circle or the incircle?

And also, which similar triangles?

oh i guess it would be incircle...i envisioned wrong...anyway

let $AD \cap BC =X$, trivially, $\triangle ABX\sim\triangle ACD$, so if $XA=a$, and $XB=b$, then we have

$\frac{a}{b+1}=\frac{b}{a+3}=\frac{1}{2}$, so we get $(a,b)=(\frac{5}{3},\frac{7}{3})$, and then our circle is the incircle of XCD, which is easily calculated since we have the sides...i am getting $\frac{\sqrt{3}}{3}$

Your answer is incorrect, Altheman.
Are you sure that $\triangle ABX\sim \triangle ACD?$

[quote="delta"]I cheked [b]Altemans [/b]solution and got the same answer $r=\frac{1}{\sqrt 3}$. So area is $S=\frac{\pi}{3}$.
Obviously he was meaning $\triangle ABX \sim \triangle CDX$. :D
By the way [b]kunny[/b] are you sure of the source? Never heard of such exam  :o .[/quote]

Your answer is correct. :)

Sophia University in [b]Japan[/b] is as follows.

[url=http://www.sophia.ac.jp/E/E_toppage.nsf/]Take a look[/url]


put $ a,b,c,d$ respectively the $ ?$'s then,
$ 8\plus{}3\plus{}1 \equal{} 12$
$ 4\plus{}a\plus{}b \equal{} 12 \rightarrow a\plus{}b\equal{}8$
$ 6\plus{}c\plus{}d \equal{} 12 \rightarrow c\plus{}d\equal{}6$
and $ 3\plus{}a\plus{}c\equal{}16 \rightarrow a\plus{}c\equal{}13$
and $ 1\plus{}b\plus{}d\equal{}2 \rightarrow b\plus{}d\equal{}1$
then we get the system
$ a\plus{}b\equal{}8$
$ b\plus{}d\equal{}1$
$ c\plus{}d\equal{}6$
$ a\plus{}c\equal{}13$
whish is easy to solve

Oh ... that's ok. I understand now. Thanks for your help ;)


prove that if 30|a+b+c

then 30|a^5+b^5+c^5

[hide]
\begin{align*}a^5 + b^5 + (- a - b)^5 &\equiv -5ab(a^3 + 2a^2b + 2ab^2 + b^3) \equiv -5ab[(a+b)^3 - (a+b)ab]\\ &\equiv -5ab(a+b)[(a+b)(a+b) - ab] \equiv 5abc[-ca - cb - ab]\\ &\equiv -5abc[ab + bc + ca] \pmod{30} \end{align*}Thus we need to show that $ 6|abc(ab+bc+ca)$. Clearly at least one of $ a,b,c$ is even, otherwise their sum is odd, but then $ 30 \nmid a+b+c$. If none of $ a,b,c$ are divisible by $ 3$, then we see from $ 3|a+b+c$ that $ a \equiv b \equiv c \pmod{3}$, so $ ab+bc+ca \equiv 0 \pmod{3}$, as desired.

In fact, Mathematica reduces $ a^5 + b^5 + c^5 = e_1^5-5e_1^3e_2 + 5e_1e_2^2 + 5e_1^2e_3 - 5e_2e_3 = 0$, where $ e_1 = a+b+c, e_2 = ab + bc + ca, e_3 = abc$.
[/hide]

[hide="Casework?"] Fermat's Little Theorem gives $ a^5 \equiv a \bmod 5, a^5 \equiv a^3 \equiv a \bmod 3, a^5 \equiv a^4 \equiv a^3 \equiv a^2 \equiv a \bmod 2$. [/hide]


trigonometry
complex numbers
trig identities
Law of Cosines
geometry
Law of Sines

[quote="Altheman"]you could just drop a bunch of perpendiculars, but i think the best way is to use

[hide="hint"]complex numbers[/hide][/quote]

[hide]
I got $\boxed{9.5}$. But how would you use complex numbers?  :huh: [/hide]

EDIT: Ignore this. It's wrong.

[hide] Both 135 and 120 degree angles are easy to work without a calculator. Draw AC, and use the law of cosines to get AC. The law of since reveals angle BCA=30 and then ACD is a right triangle. Then AD comes out to about 8.5, I'll work the precise fraction/radical later... but that's my idea for the trig approach. [/hide]

OK. Ignore mine. It's wrong. I assumed it was a trapezoid. This sounds like a textbook trigonometry problem with less friendly numbers. :D  And I got $\angle{BCA}=34.66$ but I'm probably wrong...again.  :( For a final answer, i got $AD=\boxed{9.5}$.

[quote="Silverfalcon"]In a quadrilateral $ABCD, AB = 2 \sqrt{6}, BC = 7 - 2 \sqrt{3} , CD = 5, \measuredangle B = 135^\circ,$ and $\measuredangle C = 120^\circ$. Compute $AD$.[/quote]

(using the complex plane)
$A=0$, and $B={2\sqrt{6}}$, $C=(7-2\sqrt{3})*cis 45+B$, $D=C+5*cis 105$

then compute $|D|$

I like the trig approach best.  Law of Sines and Law of Cosines are easier than the complex plane for something like this...

The main thing I like the complex plane for is DeMoivre's Theorem.


100 positive numbers 0<x1<x2<...<x100 satisfy
x1^2+x2^2+...+x100^2>10000
x1+x2+x3+..+x100<300

prove x100+x99+x98>100

(xi is not x times i, i should appear smaller.)

Hi, let $ z_i\equal{}x_i/100$, $ a\equal{}z_{100}$, $ b\equal{} z_{99}$, $ c\equal{} z_{98}$. If $ a\plus{}b\plus{}c\le1$ then $ a\plus{}b\le1\minus{}c$ and $ a\le 1\minus{}2c$ thus we have $ a^2\plus{}b^2\le (1\minus{}c)(1\minus{}2c) \;(*)$. Moreover, $ \sum_{i\le98} z_i<3\minus{}a\minus{}b\le 3\minus{}2c$ implies that $ \sum_{i\le98} z_i^2\le c \sum_{i\le98} z_i<c(3\minus{}2c)  \; (**)$.  Summing up $ (*)$ and $ (**)$ we get  $ \sum_{i\le100} z_i^2 < 1$, which is contradiction.


Solve root integer equation:
$(x^2+y)(x+y^2)=(x-y)^3$

but Rust about 2. when is that an integer ?
and how did you manage to solve it ?

I find all solutions (forget about integer in condition).

1)please show me your solution for real values.
2)how does one restrict it to integers?


quadratics
modular arithmetic
number theory proposed
number theory

Prove that primes of the form 3k+1 is expressible as a^2+b^2-ab

The same method, used to prove $ p \equal{} 4k\plus{}1$ can be expressed as $ a^2\plus{}b^2$ works here.
Letting $ a \equal{} k\plus{}l, b\equal{}2l$ we find $ a^2\plus{}b^2\minus{}ab \equal{} k^2\plus{}3l^2$, so it suffices to prove that $ p \equal{} k^2\plus{}3l^2$ for some $ (k,l)$. If $ p \equal{} 3k\plus{}1$, quadratic reciprocity gives us $ \minus{}3$ is a quadratic residue in $ Z_{p}$. Choose $ t$ so that $ p | t^2 \plus{} 3$. Using Thue's Theorem (which states for any value $ t$, there exist $ m,n$ with $ m, n < \sqrt p$ and $ m \equiv tn (mod$ $ p)$), choose $ m,n < \sqrt p$ such that $ m \equiv tn (mod$ $ p)$, we have $ p | m^2 \plus{} 3n^2$. But since $ m, n < \sqrt p$, then $ m^2 \plus{} 3n^2 < 4p$, so $ m^2 \plus{} 3n^2 \equal{} p, 2p$ or $ 3p$. If $ m^2 \plus{} 3n^2 \equal{} 2p$, since $ p$ is an odd prime, $ m^2 \equiv n^2 \plus{} 2 (mod$ $ 4)$, which is a contradiction. If $ m^2 \plus{} 3n^2 \equal{} 3p$, let $ m \equal{} 3q$ and we have $ p \equal{} n^2 \plus{} 3q^2$. If $ m^2\plus{}3n^2 \equal{} p$, we are done - so in either case, $ p \equal{} 3a^2 \plus{} b^2$ for some integers $ a,b$ as required.

http://www.mathlinks.ro/Forum/viewtopic.php?t=55714 (not exactly the same, but with a small bit of algebra, it gets the same, see the post above)

can you please apply the same for prime 31 and illustrate ?

Above mentioned explanation, and Thue's theorem. I am not able to find the solution a and b with the above mentioned explanation.
I have got t as 11.

For a particular value of $ p$, finding $ m,n$ that satisfy the condition is simple, since you know they must satisfy $ m^2\plus{}3n^2\equal{}p$. In this case, $ 31\equal{}2^2\plus{}3\cdot 3^2$ and we get $ 2\equiv 11\cdot 3\pmod{31}$.

(Now if $ t$ were -11, we would have to negate one of 2,3, but it still works. In fact, finding a representation of $ p\equal{}m^2\plus{}3n^2$ is an easy way to find the square roots of -3 for relatively small $ p\equiv 1\pmod{6}$. Here, our square roots are $ \pm 3\cdot 2^{\minus{}1}$. As $ 2^{\minus{}1}\equiv 17$, the square roots are $ \pm 51$, which reduces to $ \pm 20$, or $ \mp 11$.)


inequalities
trigonometry
inequalities unsolved

[color=blue][b]Let any $ \Delta ABC$, Prove that:

$ \sin \frac{A}{2} \plus{} \sin \frac{B}{2} \plus{} \sin \frac{C}{2} \plus{} tan\frac{A}{2} \plus{} tan\frac{B}{2} \plus{} tan\frac{C}{2} \ge \frac{3}{2} \plus{} \sqrt 3$[/b][/color]

This should be in the inequalities forum.

BTW this is a direct consequence of Jensen's inequality

Let $ f(x) \equal{} \sin(x) \plus{} \tan(x)$
$ f''(x) \equal{} \dfrac{2\sin(x) \minus{} \sin(x)\cos^3(x)}{cos^3(x)}$
Since $ 0\le \cos(x)\le 1 \rightarrow sin(x) > \sin(x)\cos^3(x) \Rightarrow f''(x)>0, x \in \left(0,\frac{\pi}{2}\right)$

$ \sum \sin\left(\frac{A}{2}\right)\plus{}\tan\left(\frac{A}{2}\right) \ge 3\left[sin\left(\frac{A\plus{}B\plus{}C}{6}\right)\plus{}\tan\left(\frac{A\plus{}B\plus{}C}{6}\right)\right] \equal{} \frac{3}{2} \plus{} \sqrt{3}$


Picture:
[img]5813[/img]

You are looking for the darkest shaded region's area.

From this you can see that we are (roughly) looking for half the area of the circle with center $(-3,-3)$ and radius 4 (the top of the circle is one unit above the X axis).

The area of this circle is $16\pi$, and we want (roughly) half the area, or $8\pi$, which is roughly equal to $25.132$, or choice $\boxed{E}$

any non computer-aided way to do it?

this was on an AMC  :D

... Graphing calculators are allowed.. anything I did on my comp you could easily do (and visualize) on a TI-83.

[quote="iamagenius"]Let $f(x)=x^{2}+6x+1$, and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that $f(x)+f(y) \leq 0$ and $f(x)-f(y) \leq 0$. The area of $R$ is closest to

$\left( \bf{A}\right) 21 \,\,\,\,\, \left( \bf{B}\right) 22 \,\,\,\,\, \left( \bf{C}\right) 23 \,\,\,\,\, \left( \bf{D}\right) 24 \,\,\,\,\, \left( \bf{E}\right) 25$[/quote]

$f(x)=(x+3)^{2}-8$, i will apply a shift, so $=x^{2}-8$

then we have a circle $x^{2}+y^{2}=16$, and two lines $(x-y)(x+y)=0$, the area is two quarter circles $2*\frac{1}{4}*4^{2}\pi=8\pi\implies\boxed{E}$


We're trying to make the 1st stage of the Canadian Computing Competition online sometime in the next few years. Right now, the grader is almost done, so we're going to run a series of onilne programming contests to test it. The first Canadian Test the Server Contest is scheduled to run sometimes in late Novemember/early December.
As this is basically an online warmup for the CCC, the format is basically the same (5 problems, 3 hours). However, the differences are that each problem is out of 100 points, stricter time/memory limits and languages are restricted to C/C++/Java as of now. As this is a trial, don't count on the grader to work 100% of the time during the contest.
If you're interested in doing this contest, please contact me (peng_richard@hotmail.com) or Simon (fat_simon@hotmail.com)

I am currently doing [u]DWITE Online Computing Competition[/u] to practise for the CCC. Does Your school have this?


Given the real numbers $x$, $y$ and $z$ s.t.: $x+y+z=1$ .

Show That we have:
     $x^2+y^2+z^2 = x^3+y^3+z^3+x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)$

Conclusions?  ;)  or  :mad:

We're dreamed: Dream is creation & God is dreamer
Are you dreaming : :P

When factored, the polynomial is:
$(x^2+y^2+z^2)(x+y+z-1)=0$
So, the last factor forces the equation to hold. How did you arrive at your equation?

(Oh, and I know that I am much more than a "dream"...unless you want to redefine "dream.")


Let $ a, b, c$ be positive real numbers satisfying $ a^2b\plus{}b^2c\plus{}c^2a\equal{}3$ and $ k \ge 7$. Prove that:
\[ \frac{1}{a^3\plus{}k}\plus{}\frac{1}{b^3\plus{}k}\plus{}\frac{1}{c^3\plus{}k} \le \frac{3}{k\plus{}1}\]
:)

Will your way works for the large condition: $ a^2b\plus{}b^2c\plus{}c^2a \ge 3?$  :maybe:

[quote="nguoivn"]Will your way works for the large condition: $ a^2b \plus{} b^2c \plus{} c^2a \ge 3?$  :maybe:[/quote]
Yes, it works. Because for $ a^2b \plus{} b^2c \plus{} c^2a \ge 3,$ we have
$ \frac {1}{a^3 \plus{} k} \le \frac {a^2b \plus{} b^2c \plus{} c^2a}{3a^3 \plus{} k(a^2b \plus{} b^2c \plus{} c^2a)}.$

[quote="nguoivn"]Yes, you are right   :)
Besides, I had nice proof for the following problem (with the condition $ a^2b \plus{} b^2c \plus{} c^2a \equal{} 3$):
\[ \frac {1}{ka^3 \plus{} 1} \plus{} \frac {1}{kb^3 \plus{} 1} \plus{} \frac {1}{kc^3 \plus{} 1} \ge \frac {3}{k \plus{} 1}\]
With $ k \ge \frac {3}{2}$

@VIMF: With $ k \equal{} 1$, this ineq is not trues  :wink:[/quote]

I think $ k \equal{} 1$

$ \frac {1}{ka^3 \plus{} 1} \plus{} \frac {1}{kb^3 \plus{} 1} \plus{} \frac {1}{kc^3 \plus{} 1} \ge \frac {3}{k \plus{} 1}$

no ture

try $ a \equal{} \frac {1}{\sqrt [3]{19210}},b \equal{} \frac {3\sqrt [3]{19210^2}}{1921},c \equal{} \frac {6\sqrt [3]{19210^2}}{1921}$

[quote="nguoivn"]Thank you very much, hedeng123 for your check  :) . Indeed, I spent many times to solve the case $ k \equal{} 1$  :([/quote]

You are welcome :lol:

Another result:
Let $ a, b, c$ be positive real numbers satisfying $ a^3b\plus{}b^3c\plus{}c^3a\equal{}3$.
Prove that: $ \frac{1}{5a^4\plus{}3}\plus{}\frac{1}{5b^4\plus{}3}\plus{}\frac{1}{5c^4\plus{}3} \ge \frac{3}{8}$  :)

[quote="nguoivn"]Let $ a, b, c$ be positive real numbers satisfying $ a^2b \plus{} b^2c \plus{} c^2a \equal{} 3$ and $ k \ge 7$. Prove that:
\[ \frac {1}{a^3 \plus{} k} \plus{} \frac {1}{b^3 \plus{} k} \plus{} \frac {1}{c^3 \plus{} k} \le \frac {3}{k \plus{} 1}\]
:)[/quote]

$ k_{min} \equal{} 2.3547...$

[quote="nguoivn"]Yes, you are right   :)
Besides, I had nice proof for the following problem (with the condition $ a^2b \plus{} b^2c \plus{} c^2a \equal{} 3$):
\[ \frac {1}{ka^3 \plus{} 1} \plus{} \frac {1}{kb^3 \plus{} 1} \plus{} \frac {1}{kc^3 \plus{} 1} \ge \frac {3}{k \plus{} 1}\]
With $ k \ge \frac {3}{2}$

@VIMF: With $ k \equal{} 1$, this ineq is not trues  :wink:[/quote]

$ k_{min}\equal{}1.0933...$

[quote="nguoivn"]Another result:
Let $ a, b, c$ be positive real numbers satisfying $ a^3b \plus{} b^3c \plus{} c^3a \equal{} 3$.
Prove that: $ \frac {1}{5a^4 \plus{} 3} \plus{} \frac {1}{5b^4 \plus{} 3} \plus{} \frac {1}{5c^4 \plus{} 3} \ge \frac {3}{8}$  :)[/quote]

I can use computers to solve.

 :D

[quote="nguoivn"][quote="nguoivn"]Another result:
Let $ a, b, c$ be positive real numbers satisfying $ a^3b \plus{} b^3c \plus{} c^3a \equal{} 3$.
Prove that: $ \frac {1}{5a^4 \plus{} 3} \plus{} \frac {1}{5b^4 \plus{} 3} \plus{} \frac {1}{5c^4 \plus{} 3} \ge \frac {3}{8}$  [/quote]

I solved it by Cauchy Schwarts and AM-GM  :oops:
How about with the following, my friend (it also exits a long time)  :) :
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \sqrt [9]{\frac {a^3}{b \plus{} c}} \plus{} \sqrt [9]{\frac {b^3}{c \plus{} a}} \plus{} \sqrt [9]{\frac {c^3}{a \plus{} b}} \geq\ \frac {1}{\sqrt [9]{2}}(\sqrt [9]{ab} \plus{} \sqrt [9]{bc} \plus{} \sqrt [9]{ca})$
[/quote][/quote]

sorry. :blush:

[quote=can_hang2007][quote="nguoivn"]Let $ a, b, c$ be positive real numbers satisfying $ a^2b \plus{} b^2c \plus{} c^2a \equal{} 3$ and $ k \ge 7$. Prove that:
\[ \frac {1}{a^3 \plus{} k} \plus{} \frac {1}{b^3 \plus{} k} \plus{} \frac {1}{c^3 \plus{} k} \le \frac {3}{k \plus{} 1}\]
:)[/quote]
I have proved that this inequality holds for all $ k \ge \frac{12}{5}.$[/quote]
There is a nice  proof of this inequality for $k=3$. ;)


Let p be a prime and a(1),a(2),...,a(p+1) be (p+1) distinct positive integers. Prove that there exist i and j such that:
a(i)/gcd(a(i);a(j))  \geq  p+1

Leningrad Mathematical Olympiad (considered as being very unconventional and very difficult).

[quote="lhvietbao"]Let $p$ be a prime and $a_1, a_2, \ldots, a_{p+1}$ be $p+1$ distinct positive integers. Prove that there exist $i$ and $j$ such that:
$\frac{a_i}{\gcd{(a_i,a_j)}} \geq p+1$[/quote]

 :D

EDIT : Nevermind, see : http://www.mathlinks.ro/Forum/viewtopic.php?t=6334 .


geometry
inequalities unsolved
inequalities

x,y,z>0 , ABC  triangle then [itex][/itex] \frac {xa^4}{y+z} + \frac {yb^4}{z+x} + \frac {zc^4}{x+y}  \geq  8\sigma [ABC]^2 [itex][/itex]

Sourch : D. Kontogiannis - ''inequalities  in  triangle''  Athens  1996 , Private  edition , page 140 .  This  problem  is  unsolved....


Hi, this problem has been bugging me like no other, because I don't think it should work. Any help would be greatly appreciated.

The polynomial of degree $4$, $P(x)$ has a root of multiplicity $2$ at $x=1$ and roots of multiplicity $1$ at $x=0$ and $x=-2$. It goes through the point $(5,168)$.
Find a formula for $P(x)$.

So this is what I did:

$x(x-1)^{2}(x+2)$
$x^{4}-3x^{2}+2x$
and then to go through the point:
$x^{4}-3x^{2}+2x-392$

but that makes it wrong. I'm so confused.

You forgot that the polynomials which have all the same roots STILL can differ in a constant factor. Hence your polynomial is $P(x)=Cx(x+2)(x-1)^{2}$ - take it from there.

Oh snap totally forgot about that. Thanks a ton!!!


A circular cylindrical post with a circumference of 4 feet has a string wrapped around it, spiraling from the bottom of the post to the top of the post. The string evenly loops around the post exactly four full times, starting at the bottom edge and finishing at the top edge. The height of the post is 12 feet. What is the length, in feet, of the string?

[asy]draw((0,0)--(0,20)..(1,19.5)..(2,20)--(2,0)..(1,-.5)..(0,0),linewidth(1));
draw((0,20)..(1,20.5)..(2,20),linewidth(1));
draw((1,19.5)--(0,18.5),linewidth(1));
draw((2,.5)--(1,-.5),linewidth(1));
draw((2,16)--(0,14),linewidth(1));
draw((2,11)--(0,9),linewidth(1));

draw((2,6)--(0,4),linewidth(1));[/asy]

[quote="JSGandora"][quote="aleph0"]"Unwrap" the cylinder and you find that the string forms the hypotenuse of a right triangle with legs 12 and 16. Therefore the string is 20 feet long.[/quote]
Wait, can you please elaborate on that?  I don't understand how it the other leg is $16$ feet.  Thanks.  :lol:[/quote]

It's 16 because the circumference is 4 and the string wraps around it 4 times
(i.e. 4*4=16).

Yea, interesting problem. When I first saw it, I didn't notice that you can "unwrap" it to make a right triangle :P


Given $ a, b, c$ are the [b]reals[/b] satisfy $ a\plus{}b\plus{}c\equal{}0$. Prove that:
\[ \left(ab\plus{}bc\plus{}ca\right)^2\left(\frac{1}{(a\minus{}b)^4}\plus{}\frac{1}{(b\minus{}c)^4}\plus{}\frac{1}{(c\minus{}a)^4}\right) \ge \frac{33}{16}\]
 :)

[quote="nguoivn"]Given $ a, b, c$ are the [b]reals[/b] satisfy $ a \plus{} b \plus{} c \equal{} 0$. Prove that:
\[ \left(ab \plus{} bc \plus{} ca\right)^2\left(\frac {1}{(a \minus{} b)^4} \plus{} \frac {1}{(b \minus{} c)^4} \plus{} \frac {1}{(c \minus{} a)^4}\right) \ge \frac {33}{16}\]
:)[/quote]

Let : $ a<0\le b<c$

Because the inequality is homogeneous , we can assume that : $ c\minus{}b\equal{}1$

So : $ a \equal{} \minus{}1\minus{}2b$ , $ b$ , $ c\equal{} 1\plus{}b$; ($ b \ge 0$)

$ \Leftrightarrow f(b)\equal{} (3b^2\plus{}3b\plus{}1)^2(1\plus{}\frac{1}{(3b\plus{}2)^4}\plus{}\frac{1}{(3b\plus{}1)^4})\ge \frac{33}{16 }$

$ min(f(b))\equal{} \frac{33}{16}$ - holdes for $ b \equal{} 0$.


A uniform rectangular wood block of mass M, with length b and height a, rests on an incline as shown. The incline and the wood block have a coefficient of static friction, μs. The incline is moved upwards from an angle of zero through an angle θ. At some critical angle the block will either tip over or slip down the plane. Determine the relationship between a, b, and μs such that the block will tip over (and not slip) at the critical angle. The box is rectangular, and a ≠ b.
[img]http://img193.imageshack.us/img193/4254/81244.jpg[/img]
A. μ > a/b
B. μ > 1 - a/b
C. μ > b/a
D. μ < a/b
E. μ < b/a - 1

[img]http://i48.tinypic.com/13zbpdj.jpg[/img]

First, for the block to not slip, $ mg\sin\theta \equal{} \mu mg\cos\theta\implies \mu \equal{} \tan\theta$. Now look at the picture to the right. In order for the block to not topple, the CM has to be above a base of support. Consider an imaginary right triangle normal to the inclined plane as drawn. One leg, $ a/2$ is held constant. The other leg would depend on the value of theta. This leg has to be less than $ b/2$ in order for the block to not topple. This means that $ \tan\theta < (b/2)/(a/2)\implies \tan\theta < b/a$ for the block to not topple. Since $ \mu\equal{}\tan\theta,$ $ \mu<b/a$ for the block to not topple and the converse is also true so $ \mu>b/a$ so the answer is C.

[quote=PiAreSquare][quote="#H34N1"]
In order for the block to not topple, the CM has to be above a base of support..[/quote]

I don't understand the reason for this[/quote]

If there is no base of support, then the normal force cannot counteract the torque

[quote=Quarky]I think that the phrase "such that the block will tip over" isn't important, because tipping has no effect on the relation between $ \mu$ and $ \theta$. If we only prevent the block from slidding, then $ \mu mgcos\theta > mgsin\theta \implies \mu > tan\theta$. By simple geometry, $ tan\theta  \equal{} \frac{b}{a}$, so $ \mu > \frac{b}{a}$

Is this an exact solution, or am I making some unjustified assumptions?[/quote]

I don't think that you can assume that $ tan\theta  \equal{} \frac{b}{a}$ because the dimensions of the box are not necessarily related to the angle itself. You need to actually find the relationship.


Source: 2005 AMC-10B #19

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ poitns, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?

$\text{(A)}\ 0 \qquad\text{(B)}\ 1 \qquad\text{(C)}\ 2 \qquad\text{(D)}\ 4 \qquad\text{(E)}\ 5$

[hide]To get the average, multiply each score by it's occurence, $.1\times70+.25\times80+.2\times85+.15\times90+.3\times95=86$ (30% of the scores must be 95) and the median is 85 (assume a 100 test sample and count 50 tests from one direction) so the answer is 1.[/hide]

[hide]
Assume that there are 100 students.

Listing the # of scores:
70$\times$10
80$\times$25
85$\times$20
90$\times$15
95$\times$30

From that, you can see that the median is 85.

Using a calculator, the mean is 86.  So the difference is 1.
[/hide]

[hide]30% got 95.

Assume that there were 100 students. The average would be $\frac{10(70)+25(80)+20(85)+15(90)+30(95)}{100}=86$. The median would be 85 (since 85 contains the 36th-55th students which contains the median). The difference is 1. [/hide]


modular arithmetic
number theory unsolved
number theory

Prove that if $\left[\sqrt{n}\right]\equiv a_{n}\pmod{10}$ and $-1<a_{n}<10,$ then the sequence $a_{n}$ is not periodic.

As $n$ gets very large, there will be arbitarily long intervals between squares, and $[\sqrt{n}]$ will take the same value arbitarily many times in a row- hence the sequence cannot be periodic.

$k\in N^{*}$
if $\forall n,\ \ [\sqrt{n+k}]-[\sqrt{n}]\equiv 0[10]$
$\forall (n,s)\in N^{2}\ [\sqrt{n+sk}]-[\sqrt{n}]\equiv 0[10]$
we take $n=i^{2},s=j+i,k=j-i$ and ($i<j$)
then, $10|j-i$ ; $\forall (i,j)\in N^{2}$
impossible


$\all a,b$;
if $a=b$ $\rightarrow$ $a^2=b^2$ $\rightarrow a^2-b^2=a-b \rightarrow (a-b)(a+b)=a-b \rightarrow a+b=1$ ??? :?:

BUT a-b=0, so in the last step you're dividing by zero. Nice try.

[hide]you can't divide by 0. I remember a similar "proof" that stated that 1=2 :lol: [/hide]

I rather prefer this proof.  The problem is much harder to find :)

$\sqrt{1} = 1$
$\sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{(-1)} \times \sqrt{(-1)} = i \times i = -1$
$1 = -1$

The version I had seen of that is:

$i=\sqrt{-1}$
$i^4=1\Rightarrow i^2=1$
$i^2=-1\Rightarrow -1=1$

The problem with that is that $1$ has four distinct fourth roots. I believe it's a similar problem with the "proof" which you gave. ;)


Please only respond if you're eligible to participate in the USAMTS, also...

also what?

I was "eligible" but I was too busy to do it...

Gr...I sent my solutions (unfortunately only to 3 of them :-x ).  I don't have time...which explains why I post about 1 post/week nowadays....  The ones that I answered were 2, 3, and 5.  How about you guys?

-interesting_move

[quote="Tare"]also what?[/quote]

I meant "also, please only respond if you're eligible to participate in the USAMTS".  That being so that we have a more accurate figure of the number of AoPSers who read Other Contests & Programs who submitted USAMTS solutions.  For instance, people like Eve and Simon shouldn't respond to this poll, since they couldn't have submitted solutions anyway.

I don't see any reason why I couldn't have submitted solutions. They might not have graded them, but I certainly could have submitted them.

...I guess so, but that's just wasting the grader's time because they have to separate out the non-eligible solutions (and trash it in some cases; if they have enough time they might have glanced over it)

(h)Oy, too busy... :)

I got answers to all five, but I forgot to send them in :oops:


[quote="can_hang2007"]We have discussed this problem once time in the past. I try to search it but I cannot find the link. So I repost it here, please forgive me:

(Belarus 1998) Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} 1.$

I wait for some nice solutions using the classical inequalities.[/quote]

Well, I found a link http://www.mathlinks.ro/viewtopic.php?search_id=1187594133&t=107534

multiplying by $ b\plus{}c$ the ineq is reduced to :

$ \frac{b^2}{c} \plus{} \frac{bc}{a} \plus{} \frac{ac}{b} \plus{} \frac{c^2}{a} \geq \frac{ (b\plus{}c)^2}{a\plus{}b} \plus{} b \plus{}c$ , which is true by summing :

$ \frac{c^2}{a} \geq \frac{ (b\plus{}c)^2}{a\plus{}b} \minus{} b$   ( cauchy shwarz)
$ \frac{bc}{a} \plus{} \frac{ac}{b} \geq  2c$   ( am-gm)
$ \frac{b^2}{c} \geq  2b\minus{}c$  ( am-gm)

regards  :)

[quote="anas"]multiplying by $ b \plus{} c$ the ineq is reduced to :

$ \frac {b^2}{c} \plus{} \frac {bc}{a} \plus{} \frac {ac}{b} \plus{} \frac {c^2}{a} \geq \frac { (b \plus{} c)^2}{a \plus{} b} \plus{} b \plus{} c$ , which is true by summing :

$ \frac {c^2}{a} \geq \frac { (b \plus{} c)^2}{a \plus{} b} \minus{} b$   ( cauchy shwarz)
$ \frac {bc}{a} \plus{} \frac {ac}{b} \geq 2c$   ( am-gm)
$ \frac {b^2}{c} \geq 2b \minus{} c$  ( am-gm)

regards  :)[/quote]
Very nice your solution, dear anas. The following is my solution:

We write the inequality as
$ \left( \frac{a}{b}\minus{}\frac{a}{b\plus{}c}\right) \plus{}\left(\frac{b}{c}\minus{}\frac{b}{b\plus{}c} \right)  \plus{}\left( \frac{c}{a} \minus{}\frac{c}{a\plus{}b}\right) \ge \frac{b}{a\plus{}b}\plus{}1,$
or
$ \frac{ca}{b(b\plus{}c)}\plus{}\frac{b^2}{c(b\plus{}c)}\plus{}\frac{bc}{a(a\plus{}b)} \ge \frac{a\plus{}2b}{a\plus{}b}.$
Now, using the Cauchy Schwarz Inequality, we get
$ \left[ \frac{ca}{b(b\plus{}c)}\plus{}\frac{b^2}{c(b\plus{}c)}\plus{}\frac{bc}{a(a\plus{}b)} \right]\left[ (b\plus{}c)\plus{}\frac{a(b\plus{}c)}{b}\plus{}(a\plus{}b)\right] \ge \left( \sqrt{\frac{ca}{b}}\plus{}\sqrt{\frac{ab}{c}}\plus{}\sqrt{\frac{bc}{a}}\right)^2 \equal{}\frac{(ab\plus{}bc\plus{}ca)^2}{abc}.$
Thus, it suffices to prove that
$ \frac{(ab\plus{}bc\plus{}ca)^2}{abc} \ge \frac{a\plus{}2b}{a\plus{}b} \left[ (b\plus{}c)\plus{}\frac{a(b\plus{}c)}{b}\plus{}(a\plus{}b)\right].$
Because $ (b\plus{}c)\plus{}\frac{a(b\plus{}c)}{b}\plus{}(a\plus{}b)\equal{}\frac{(a\plus{}b)(2b\plus{}c)}{b},$ this can be written as
$ (ab\plus{}bc\plus{}ca)^2 \ge ac(a\plus{}2b)(2b\plus{}c),$
which is true since by the AM-GM Inequality, we have
$ ac(a\plus{}2b)(b\plus{}2c) \le \left[ \frac{a(2b\plus{}c)\plus{}c(a\plus{}2b)}{2}\right]^2 \equal{}(ab\plus{}bc\plus{}ca)^2.$
This ends our proof.

[quote="szl6208"]
[3].
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\le{3+\frac{(a-b)^2}{b^2}+\frac{(b-c)^2}{c^2}+\frac{(c-a)^2}{a^2}}\][/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[(7-3\sqrt3)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)\le\frac{(a-b)^2}{b^2}+\frac{(b-c)^2}{c^2}+\frac{(c-a)^2}{a^2}\]

[quote=luofangxiang][quote=sqing](Belarus 1998 ?) Let $ a,b,c$ be positive real numbers. Prove that\[ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 3\plus{}\frac {(a\minus{}c)^2}{b^2\plus{}ab\plus{}bc\plus{}ca}.\][/quote]

The following inequality is also true
[img]http://s13.sinaimg.cn/middle/006ptkjAzy75EWqFFy47c&690[/img][/quote]

$\Leftrightarrow c{{b}^{2}}{{\left( a-c \right)}^{2}}+a{{\left( {{b}^{2}}-ac \right)}^{2}}\ge 0$

[quote=sqing](Belarus 1998 ?) Let $ a,b,c$ be positive real numbers. Prove that\[ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 3\plus{}\frac {(a\minus{}c)^2}{b^2\plus{}ab\plus{}bc\plus{}ca}.\][/quote]
 Let $ a,b,c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h1669861p10617572]Prove that[/url]
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant 3 + \frac{(c-a)^2}{a(b+c)}.\]


Here we go: some true (in my opinion) intermediate challenge problems. If these problems seem easy or trivial, don't solve them! Also, please spoiler your solutions as much as possible. - Thanks.

1. Let S$ = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots$

a) Find, in terms of S, the value of $\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \cdots$

b) Find, in terms of S, the value of $\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots$

c) Find, in terms of S, the value of $\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots$

2. Let N be a positive integer. Prove the following:

a) If N is divisible by 4. then N can be expressed as the sum of two or more consecutive odd integers.

b) If N is a prime number, then N cannot be expressed as the sum of two or more consecutive odd integers.

c)  If N is twice some odd integer, then N cannot be expressed as the sum of two or more consecutive odd integers.

3. For any real number x, [x] denotes the greatest integer that does not exceed x. Given natural number N, donote as f(n) the follwing sum of N integers:

$f(N) = [N/1] + [N/2] + [N/3] + \cdots + [N/N]$.

a) Evaluate $f(7) - f(6)$.

b) Evaluate $f(35) - f(34)$.

c) Evaluate (with explanation) $f(1996) - f(1995)$.

4. (A bit more challenging) A,B,C are three equally spaced points on a circular hoop. Prove that as the hoop rolls along a horizontal line, the sum of the distances of the points A,B, C above the line is constant.

Also prove that S (from Q1 above) is well defined,   that is:   1<S<2 .

[hide](It is, of course, (pi^2/6)[/hide]

As far as I can tell I can only do 2b:

[hide]If N could be expressed as the sum of two or more consecutive odd integers, it could be written like this:

(2k+1)+(2k+3)+...+(2k+1+2x)

Where k is some whole number, and x is one less than the number of consecutive odd integers. Simplifying the expression gives

2k(x+1)+(1+3+5+...+(2x+1))

Which then simplifies to

2k(x+1)+(x+1)^2

(2k+x+1)(x+1)

Since there are two or more consecutive odd integers, x+1 is at least 2, so N has two factors other than 1 and itself, contradiction.[/hide]

Hi - can we edit the original question  [quote]b) If N is a prime number, then N cannot be expressed as the sum of two or more consecutive odd integers. [/quote]  so as not to leave  a little    loop-hole lilke:
(-1) + (1) + 3 =  3 and  3 is prime.

may  be  change "odd integers" to " positive odd integers" ?

Thanks.
(Nice problems  by the way)

[hide]1a. = 1/4 (1/1^2 + 1/2^2 + ... ) = S/4

1b. = (1/1^2 + 1/2^2 + ... ) - (1/2^2 + 1/4^2 + ...) = S - S/4 = 3S/4

1c. = (1/1^2 + 1/2^2 + ... ) - 2(1/2^2 + 1/4^2 + ...) = S - 2S/4 = S/2

2a. Let the number be 4x for some integer x.  Then it can be expressed as (2x-1) + (2x+1), which represent the sum of any two consecutive odd integers.

2b. Any number that is the sum of an even number of odd integers is even, and 2 can't be the sum of any number of consecutive integers, a sum of an even number of consecutive odd integers cannot be prime.  Any odd number > 1 of consecutive integers can be expressed as m - 2x, m - 2x + 2 , ... m - 2, m, m+2, ... , m + 2x, so their sum will be (2x+1)m which is clearly not prime.  This eliminates every possibility.

2c. A sum of two consecutive odd integers is a multiple of 4 (see 2a).  A number that is twice some odd number is not (it's :equiv: 2 mod 4).  Therefore no.[/hide]


Every point in a plane is colored either blue or red. Prove that
there exist two blue points $B_{1}$ and $B_{2}$ such that $B_{1}B_{2}=1$ or
four collinear red points $R_{1},R_{2},R_{3},R_{4}$ such that
$R_{1}R_{2}=R_{2}R_{3}=R_{3}R_{4}=1$.

Nice link, but I was looking for some more competition-like
solution, never mind...


We have one square of each of the following side lengths:1,3,5,7,10,11,13,14,15,18 and 21.Can we form a rectangle using these eleven squares without overlap?

$1^2+3^2+5^2+7^2+10^2+11^2+13^2+14^2+15^2+18^2+21^2 = 1660$. The length and width must both be greater than or equal to 21 (and integers) because of the square of sidelength 21. $1660 = 2^2*5*83$. The closest the length and width can get is 20 and 83. So, there is no way to get dimensions both greater than 21 unless at least one is non-integer. Is there a flaw in my argument, or am I misinterpreting this?

Ahh, I should have seen this, I attempted to construct it.  Didn't work very well.  I'm glad I didn't try to make it into casework.


three positive number $x,y,z$ such that $x+y+z=3$ and $p \ge 1$
can we prove:$\left(1+x^p-x \right)\left(1+y^p-y \right)\left(1+z^p-z \right) \ge 1$

I'm not sure about it.
(btw it is easy to see if change $x+y+z=3$ to   $xyz=1$)

I've proved that for p=>3, the inequality is right.
And I wonder if p is a interger number or real

in my problem p is real.]
but if you can prove when p is an interger number,please post it.thanks ;)

when p=>4  and is a integer
(1+x^p-x)=>((p-1)x^p+1)/p
then we can use Calson(I dont sure about this name ,but in Chinese it is “卡尔松”) Ineq , or use Holder Ineq can solve the problem.

but when p=2,3  I have no idea .(I found I haven't prove correctly when p is 3.sorry!)

[quote="zhaobin"]thanks for your nice idea,
but I don't know how to use calson ineq or holder ineq?
can you tell me :)[/quote]
I felt really sorry that I was wrong .
It is the load noise of the history teacher why I made the mistake so stupidly.
But this time ,however ,I proved that when p=2 the Ineq is right :
x^2-x+1=((2x-1)^2+3)/4
and we can prove that (a^2+3)(b^2+3)(c^2+3)=>64
when a+b+c=3 by some method(I don't know how to say it which is usually used in the university)
I'm trying to generalise it .

[quote="zhaobin"]do you mean largrange multiply?
yes I think it is useful. :) thanks
but I don't like to use it...[/quote]

Yes ,I do.

Neither I do.

[quote="perfect_radio"]how do you solve the inequality when $xyz=1$?[/quote]
ok,hint$x^p-x \ge x^{p-1}-1$ :)

Thanks, very nice solution.

Good luck in solving it for $x+y+z=3$.

[quote=zhaobin]three positive number $x,y,z$ such that $x+y+z=3$ and $p \ge 1$
can we prove:$\left(1+x^p-x \right)\left(1+y^p-y \right)\left(1+z^p-z \right) \ge 1$
[/quote]

Any progress on this?


I have taken the intro to number theory course and know basic modular arithmatic, but can't seem to figure this one out.

Not in mathcounts please.  Also, look it up in the wiki.


4,16,37,58,89,145,42,20.

I can't find the pattern or rule that's being used here. I'm pretty sure it has something to do with adding the digits of each number together and doing something else.

Doing a search on [url=http://www.research.att.com/~njas/sequences/]OEIS[/url]: [url=http://www.research.att.com/~njas/sequences/A080709]Sequence #A080709[/url]

Thanks for the. I appreciate it. I knew it was something simple, and spent far too much time on this one problem. I'm still frustrated that I needed help. I wanted to get it on my own, but have other work to do. Thanks again though. :)

Don't beat yourself up. Look-and-see sequences are often complicated because you know what you're looking for but can't find it. Like lost keys, they're always in the worst place :D.

Here is one of my favorites for you to try. $ 1,11,21,1211,111221,312211,....$ Strictly think of "look and say" sequence.


can anyone explain me in detail the third,sixth and the seventh problem's solution in detail of RMO 1990 paper.
the problems are
1. A square sheet of paper ABCD  is so folded that B falls on the mid point of M  of CD . Prove that the crease will divide BC  in the ration 5:3

2. N is a 50-digit number (in decimal representation). All digits except the 26th digit (from the left) are 1. If N is divisible by 13, find its -26th digit.

3. A census man on duty visited a house in which the lady inmates declined to reveal their individual ages, but said "We do not mind giving you the sum of the ages of any two ladies you may choose". Thereupon, the census man said, "In that case, please give me the sum of the ages of every possible pair of you". They gave the sums as: 30, 33, 41, 58, 66, 69. The census man took these figures and happily went away.

How did he calculate the individual ages?

Hello.

suppose BC ,is devided to BN=x , NC=y

then we have:

NM=NB , MC=BC/2

So:  tan(CBM)=1/2

and  tan(CNM)=tan(2*CBM)=4/3

CM/CN=4/3         BC/y=8/3      x/y=5/3

Question 2 is in this thread here: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56053]Find the digit[/url] (see the other post)
For Question three, obviousl  all ages are different, let us assume  the women are  ordered from youngest to oldest as $A,B,C,D$

Let the age of B be $x$
So age of       A is  $30-x$    (AB do form the youngest pair)
And age of      C is $x+3$     (AC does form the second youngest pair with sum = 33)
And age of     D is   $66-x$    (CD is the oldest pair)

Now the pair  BC  which should give you $2x+3$ and odd number so, $2x+3=41$   which gives $x=19$

So you have $11,19,22,47$


It seems this is your homework. Donot ask your homework problems to be solved here. If its not, then see below.

[hide]$(a+2b)/5= (3b-c)/3 = (2c+a)/7 =k$
Get a, b, c in terms of k.
Thus obtain a:b:c.[/hide]

I got .....

[hide]a:b=1:2[/hide]

If this much is right,i'll post my solution,all i did was cross multiply the first 2 fractions and the last and the first fraction,then solve the equatons...though i'm not too sure.


[quote="rnwang2"]The AM-GM inequality

$\left(\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}\right)\geq\sqrt[n]{a_{1}a_{2}\cdots a_{n}}$

always holds for $a_{1},a_{2},\ldots,a_{n}\geq 0$.

But how do you know there isn't a better bound? Is it true this always provides the best bound?[/quote]

n=1, a=1
n=2, a=2
n=3, a=3
n=4, a=4
n=5, a=5
etc.

[quote="rnwang2"][quote="chickendude"]how does AM-GM present a bound in that sense?

a bound is a value that something doesn't exceed

as long as the Arithmetic mean and the Geometric mean aren't equal, there will always be a better bound (in between the two values).

there is no "best bound" unless I'm misinterpreting your question[/quote]
Oh wow, ok. I have no idea what I was thinking.  :maybe:

EDIT: Wait I do know what I was thinking. I was solving the problem:

Find the minimum value of $\frac{9x^{2}\sin^{2}x+4}{x\sin x}$ for $0 < x < \pi$.

By AM-GM, the expression is $\geq 12$, so the minimum value is 12. But what if there was a stronger inequality (or is there never)?[/quote]

Let me see if I can understand you.

$\frac{\frac{9x^{2}\sin^{2}x}{x\sin x}+\frac{4}{x\sin x}}{2}\geq \sqrt{\frac{36x^{2}\sin^{2}x}{x^{2}\sin^{2}x}}= 6$
$\frac{9x^{2}\sin^{2}x+4}{x\sin x}\geq 12$, with equality iff $\frac{9x^{2}\sin^{2}x}{x\sin x}= \frac{4}{x\sin x}$ or
$9x^{2}\sin^{2}x = 4$
$x\sin x = \frac{2}{3}$
This seems like it should have a solution on the given interval (although I'm not sure), so there exists an equality case, and there is not a better bound. If this does not have a solution, then we don't have an equality case by AM-GM and there should be a better value.

[quote="Hamster1800"]This seems like it should have a solution on the given interval (although I'm not sure), so there exists an equality case, and there is not a better bound. If this does not have a solution, then we don't have an equality case by AM-GM and there should be a better value.[/quote]
Right, but how do we know AM-GM provides the best bound? What if, by some other inequality, we got $\geq 10$ (random number)? Or is that not possible?

Am I making any sense?  :(

AMGM doesn't always provide the best bound...that's why its neccesary to check for equality cases...if there is one, then you have the best bound...

AM-GM provides the best bound in the equality case. If the equality case has no solution, then it won't provide the answer.

oh and
$x sinx = \frac23$ has a solution:

let $f(x) = x sinx$

$f(0) = 0$
$f(\frac{\pi}{2}) = \frac{\pi}{2}$

$0 < \frac23 < \frac{\pi}{2}$
so by the intermediate value theorem, there is an $x \in [0,2\pi]$ such that $f(x) = \frac23$

[quote="rnwang2"]The AM-GM inequality

$\left(\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}\right)\geq\sqrt[n]{a_{1}a_{2}\cdots a_{n}}$

always holds for $a_{1},a_{2},\ldots,a_{n}\geq 0$.

But how do you know there isn't a better bound? Is it true this always provides the best bound?[/quote]

Do you mean a "sharper" inequality, where the LHS and RHS are always [u]closer[/u] to each other, as opposed to an inequality like $x^{2}+1 > 2x$ (which is AM-GM)?

From what I know (not much  :P ), AM-GM is not one of the "sharpest" inequalities. I think there are quite a few that are much "sharper" than it.

But ask the others on this forum for those inequalities. I do not know them off the top of my head.  :)

You can't have better bounds, because there is an equality case. If there were a constant that could be multiplied on the right hand side, we would have a contradiction of the inequality.

If you can prove an inequality in general and you can find an equality case (or a LIMITING equality case, i.e. as $x,y\to \infty, z\to 0$ or something crazy) then you clearly have the best [b]constant[/b] bound, just be the definition of inequality with an equality case. However, there are always infinite (but increasingly too messy to even compute) "sharper" or "stronger" inequalities, for instance if we have $A(m,x_{i})\ge G(m,x_{i})$ then we can find some messy $F(m,x_{i})$ with fixed $m$ such that $A(m,x_{i})\ge F(m,x_{i}) \ge G(m,x_{i})$


geometry calculus trigonometry derivative

[quote="ch1n353ch3s54a1l"]In quadrilateral $ABCD$, $AB=BC=3$ and $CD=7$. What is the greatest area that $ABCD$ could have?[/quote]

for any set of 4 fixed sides, the max area happens when the quadrilateral is cyclic, now, we can just use bragmupta's formula from here, and maximize that with calculus

Is there a way without calculus.

I think the answer is 21 for some reason but I don't know why.

It's $16\sqrt{2}$

Let $x=m(ABC)$

$[ABCD]=[ACB]+[ACD]$

$[ACB]=\frac{1}{2}(3)(3)\sin (\pi-x)=\frac{9}{2}\sin x$

$AC^{2}=3^{2}+3^{2}-2(3)(3)\cos x \implies AC=3\sqrt{2}\sqrt{1-\cos x}$

$[ACD]$ is maximized when $AC$ is perpendicular to $CD$, giving $[ACD]=\frac{1}{2}(7)(3\sqrt{2}\sqrt{1-\cos x})=\frac{21}{\sqrt{2}}\sqrt{1-\cos x}$

If you maximize $[ABCD]=\frac{21}{\sqrt{2}}\sqrt{1-\cos x}+\frac{9}{2}\sin x$, you get $\cos x =-\frac{7}{9}\implies [ABCD]=16\sqrt{2}$

But that requires calculus too (thank goodness for TI-89's).

[b]Edit:[/b] Or you could do what Altheman said and maximize Brahmagupta's formula, I donno if that would be cleaner or uglier. But you'd get that it's maximized when the fourth side is $9$...

[b]Hmmm...[/b]
If this was from a lecture, how was it solved there?

[quote="ch1n353ch3s54a1l"]In quadrilateral $ABCD$, $AB=BC=3$ and $CD=7$. What is the greatest area that $ABCD$ could have?[/quote]
[hide]Let's say the third side is x. We know that x<17.
$\sqrt{(3.5+.5x)(3.5+.5x)(6.5-.5x)(.5x-.5)}$
$\frac{\sqrt{(7+x)(7+x)(13-x)(x-1)}}{4}$
$\frac{\sqrt{(x^{2}+14x+49)(14x-x^{2}-13)}}{4}$
How the heck could you maximize this!?!?!?!?[/hide]

(as suggested by ALTHEMAN) you use calculus..so i think you find the derivative of that..and then the maximum value.. :maybe:

$\frac{d}{dx}\left[\frac{(x+7)\sqrt{(13-x)(x-1)}}{4}\right]=0$
$\frac{18+7x-x^{2}}{2\sqrt{-13+14x-x^{2}}}=0$
Solutions of $18+7x-x^{2}=0$ but not of $x^{2}-14x+13=0$
$x=-2,x=9$

Clearl $x>0$, so $x=9$ for an area of $16\sqrt{2}$ as AMY said.
It's a bit nicer to solve than the trig derivative she used though... I'm guessing anyway.

Basically when you minimize or maximize something, take the derivative and set it equal to zero, solve, and plug in the values to find the minimum. (or maximum)

[quote="bpms"]$\frac{\sqrt{(x^{2}+14x+49)(14x-x^{2}-13)}}{4}$[/quote]

[hide]
By AM-GM, the maximum value of $\frac{1}{4}\sqrt{(x^{2}+14x+49)(-x^{2}+14x-13)}$ is attained when the two factors are equal. However... setting the two factors equal gives $x=i\sqrt{31}$, and then the area is $\frac{1}{4}\sqrt{128+504i}=\frac{1}{2}(9+7i)$.

Some of those numbers do appear in the posts above... :D
[/hide]

I'm sorry, my inferior mind cannot visualize a quadrilateral with complex area  :roll:


Rep123max wrote:By any chance, does somebody have the answers to the 3rd contest from last year?

I have a LaTeX file and the PDF file created from it that cover the individual-round problems from that meet. They are left over from when I coached the local Math League team last year. If you want them, I can try to post them here. I don't have the team-round problems written up.

Or if you'll settle for just the answers, they are (in spoiler mode):

3-1.  [hide]3[/hide]

3-2.  [hide]7[/hide]

3-3.  [hide]n = 169[/hide]

3-4.  [hide]16/3[/hide]

3-5.  [hide]20[/hide]

3-6.  [hide]1/3[/hide]

Erin Schram

[quote="Rep123max"]I think we have two different tests. The one that I am talking about has its first question "At most how many sundays can occur in a single calendar year?"[/quote]

Sorry about that. Here in Howard County, Maryland, teams from the different high schools compete in monthly meets to see who will team up to represent Howard County in the American Regions Mathematics League. We call this the Math League. I sent you the answers to the third meet of last year. I always assumed that the questions were provided by the ARML, but I suppose they could be locally written. In which case, no-one outside Howard County would care about them.

I did see once that another competition was really named the Math League, but I still think of the local ARML contests as the Math League.

Erin Schram


What is the largest number that leaves remainder of 1 when divided by 2,3,4,5,6,7,8,9,10?

Any number of the form 10!*k+1,did you mean the smallest?

Suppose that a+1 is the smallest  such number.Then however it gives remainder 1 when divided by 9 it is of the form 9k+1,hence 9|a,by the same way 8|a,7|a,5|a.Hence the answer is LCM(1,2,3,4,5,6,7,8,9,10)+1=9*8*7*5+1=2521.
Am I wrong?

[quote="Silverfalcon"]Yeah, smallest. :D[/quote]

Really? Then isn't [b][size=150]1[/size] [/b]the samllest postive answer. (I  can give even smaller numbers  if you need, but they would not be positive...) :D

P.S. One has to be careful around here to use the trems like "smallest  number greater than 1 or such thing.. :)

[quote="Silverfalcon"]I assumed that 1/10 doesn't necessarily mean that the remainder is 1. I mean, it's true but on the math competitions, those aren't usually the case....[/quote]

What do you mean?

1 divided by 2,3,4,5,6,7,8,9,and 10 leaves a remainder of 1.

[quote="Silverfalcon"]Hmm..

I assumed that 1/10 doesn't necessarily mean that the remainder is 1. I mean, it's true but on the math competitions, those aren't usually the case....

Hmm...

I'm not sure for this since I don't have the answer keys...

If someone can tell it, I'll be grateful.[/quote]

In a math competition, they would (should) say something like
[quote]
What is the smallest positive integer greater than 1 that leaves a remainder ...
[/quote]

They don't assume you will ignore the easiest answer :)

Silverfalcon- [quote]Hmm..

I assumed that 1/10 doesn't necessarily mean that the remainder is 1. I mean, it's true but on the math competitions, those aren't usually the case....

[/quote]

Make NO mistake about it. Yes,   1 is the  remainder when you divide  1 by 10.    Just  like  when   you divide 21, or 31  etc by 10... There isn't an iota of doubt there.

Far  from “aren’t usually the case?,  if, for example,  the question in [i]any [/i]reputable (eg Mathcounts) math test   was [quote]What is the smallest positive number that leaves remainder of 1 when divided by 2,3,4,5,6,7,8,9,10?   [/quote]

 and if some one  did not answer 1 ,  (s)he is not going to get any  points.  Of course,  most likely the question will ask for a number greater than 1 etc - [i]and it will be expicitly stated as such [/i]-  (and in that case, your answer would   be other number greater than 1 etc..)
Hope this is helpful...

Edited later: Sorry din't know others would post while i am posting .. can delete this, but I will keep it.. hope it is useful.


I have found some problems in an Olympiad book of grade 8. Enjoy it.

(1) $2x^2+xy-y^2-4x+5y-6$

(2) $(a-b)^3+(b-c)^3+(c-a)^3$

(3) $x^4+\frac56 x^3 + \frac12 x^2 - \frac12 x - \frac13$

for the others, you could just use the fact for a quadratic $ax^2 + bx +c$ can be written a$(x-r_1)(x-r_2)$ where $r_1$ and $r_2$ are roots of the quadratic, and this works for any polynomial generally

[quote="Melissa"][hide="Nr.2"]$(a-b)^3+(b-c)^3+(c-a)^3$
$=3(-a^{2}b+ab^2-b^{2}c+bc^2-c^{2}a+ca^2)$
$=3[a^2(c-b)+bc(c-b)-a(c+b)(c-b)]$
$=3(c-b)(a^2+bc-ac-ab)$
$=3(c-b)(a-b)(a-c)$[/hide][/quote]

That, of course, is correct.  Also One useful trick to know  is:

[hide]You can see right  away that if a=b, the expression is zero (and that is true if b=c or c=a) hence
$(a-b)(b-c)(c-a) $  are factors of above.
Now since this is of the third degree, so only possible remainig factor is just a constant, hence:
The answer is $k(a-b)(b-c)(c-a)$  if  you put  a=1,b=-1 and c=0 we see 8 - 1 - 1 =2k which gives k=3.
[/hide]

Not to mention that [hide]if $x+y+z=0 $ then $x^3+y^3+z^3 = 3 x y z $ if pretty well known
 .[/hide]
Similarly a possible hint for 3:
[hide]Easy to see the expression is  negative when x=0 and positive when x=1 and also x=-1, so there must be a root each  between  -1 and 0, and 0 and 1 , one is lucky enough to find that   these are -1/2 and 2/3 so (x+1/2)(x-2/3) are factors, rest is easy... (Of course there are other , equally easy way to do these .. but this trick is sometimes very helpful. [/hide]

frt, how did you get to the 4th line of your latex???

Just by factoring.  There's a lot more factoring practice you can get than the regular high school curriculum.

Let $2x^2+(y-4)x-(y-2)(y-3)=(2x+a)(x+b)=2x^2+(a+2b)+ab$.

Then we want to find $a$ and $b$ such that $a+2b=y-4$ and $ab=-(y-2)(y-3)$.

That gives $a=-(y-2)$ and $b=y-3$.

Hope this makes it clear.