inequalities calculus algebra polynomial function derivative quadratics

[quote="iura"]I don't know but it seems that multiplying every fraction by $a+1 etc.$ we can apply the Andreescu Lemma
$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}$.
Am I right?[/quote]

Cool, this Andreescu Lemma is a special case of what has been known to me as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url]...

Actually your method doesn't generally work. Strangely enough, it works (i. e. the term you get on the right hand side when you apply your Andreescu Lemma is still $\geq \frac{36}{5}$) if and only if $\left( b+c-2a\right) \left( c+a-2b\right) \left( a+b-2c\right) \leq 0$, but whether this condition is true depends on the numbers a, b, c. And I have no idea for the general case. Well, Harazi's inequalities are usually not so simple to kill...  :D

  Darij

[quote="harazi"]Here is one of the craziest inequalities I have ever found:
    If $ a,b,c>0 $ add up to 1, then $\frac{a+1}{c(2-b)}+\frac{b+1}{a(2-c)}+\frac{c+1}{b(2-a)}\geq\frac{36}{5} $. Isnt it nice?[/quote]

The LHS can be written as $\left(\frac{1}{c} - \frac{1}{2-b}\right) + \left(\frac{1}{a} - \frac{1}{2-c}\right) +\left(\frac{1}{b} - \frac{1}{2-a}\right)  = \frac{2(1-a)}{a(2-a)}+ \frac{2(1-b)}{b(2-b)} + \frac{2(1-c)}{c(2-c)}$.

[b]Lemma[/b]: If $0<x< 1$, then $\frac{2(1-x)}{x(2-x)} \geq \frac{1}{25}(138-234x)$.

It's easy to verify this is equivalent to $(3x-1)^2(25-13x) \geq 0$.

The result immediately follows from the Lemma.

You're an expert in breaking inequalities! Good job, splendid solution!

In fact one doesn't need calculus here.
Since $\frac{1}{2-a}=\frac{1}{2b+2c+a}$
$ \leq \frac{1}{5} (\frac{1}{a}+2\frac{1}{b}+2\frac{1}{c})$
And now we are left with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9$ which is obvious.

Actually there is a difference between using calculus to find the proof and using calculus in the proof itself. I actually find both Fuzzylogic's and Iura's proof elementary and very nice.

Iura's proof is correct but it has a little typo: The 5 should be a 25. (Actually it took me 10 minutes to get it.) Here is the corrected version:

$\frac{1}{2-a}=\frac{1}{2\left( a+b+c\right) -a}=\frac{1}{2b+2c+a}\leq \frac{1}{5^{2}}\left( \frac{2}{b}+\frac{2}{c}+\frac{1}{a}\right) $,

where the last inequality is because $\left( \frac{2}{b}+\frac{2}{c}+\frac{1}{a}\right) \left( 2b+2c+a\right) \geq 5^{2}$ (Cauchy-Schwarz). Similarly,

$\frac{1}{2-b}\leq \frac{1}{5^{2}}\left( \frac{2}{c}+\frac{2}{a}+\frac{1}{b}\right) $;
$\frac{1}{2-c}\leq \frac{1}{5^{2}}\left( \frac{2}{a}+\frac{2}{b}+\frac{1}{c}\right) $.

Summing up results in

$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}$
$\leq \frac{1}{5^{2}}\left( \frac{2}{b}+\frac{2}{c}+\frac{1}{a}\right) +\frac{1}{5^{2}}\left( \frac{2}{c}+\frac{2}{a}+\frac{1}{b}\right) +\frac{1}{5^{2}}\left( \frac{2}{a}+\frac{2}{b}+\frac{1}{c}\right) $
$=\frac{1}{5^{2}}\left( \frac{5}{a}+\frac{5}{b}+\frac{5}{c}\right) =\frac{1}{5}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $.

Thus, using a + 1 = 2 - (1 - a) = 2 - (b + c) = (2 - b) - c and similarly b + 1 = (2 - c) - a and c + 1 = (2 - a) - b, we get

$\frac{a+1}{c\left( 2-b\right) }+\frac{b+1}{a\left( 2-c\right) }+\frac{c+1}{b\left( 2-a\right) }$
$=\frac{\left( 2-b\right) -c}{c\left( 2-b\right) }+\frac{\left( 2-c\right) -a}{a\left( 2-c\right) }+\frac{\left( 2-a\right) -b}{b\left( 2-a\right) }$
$=\left( \frac{1}{c}-\frac{1}{2-b}\right) +\left( \frac{1}{a}-\frac{1}{2-c}\right) +\left( \frac{1}{b}-\frac{1}{2-a}\right) $
$=\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) -\left( \frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}\right) $
$\geq \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) -\frac{1}{5}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $
$=\frac{4}{5}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $.

So it remains to show that $\frac{4}{5}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq \frac{36}{5}$, i. e. that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9$. But this is clear, since Cauchy-Schwarz gives $\left( a+b+c\right) \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9$, and a + b + c = 1. This completes the proof.

Thanks again, Fuzzylogic and Iura!

  Darij

[quote="manlio"]Fuzzylogic method of breaking inequality is very powerful and very nice but

it includes calculus to find the bounding or interpolating function.

In fact you must put:

1) f(x) >= g(x) =ax+b  or f(x) >= g(x)=ax^2+bx+c (it depends on the ineq)

2) f(p)=g(p)  where p is the point that makes equal the original inequality

3) f'(p)=g'(p) and also f''(p)=g''(p) if g(x) is of second degree

4) check that f(x)-g(x) >=0 using algebra

So I am also very interested in seeing your solution, Harazi that I suspect

doesn't include calculus.[/quote]

Manlio,

1. Calculus is not necessary. You can use polynomial long division to find the bounding function as you called.

2. I never used 2nd derivative in my solutions. The only use is first derivative for convenience.

3. Derivative of polynomials can be defined formally in algebra which doesn't involve calculus at all (See P.8 of [url=http://www.amazon.com/exec/obidos/tg/detail/-/0821819151/qid=1093873344/sr=8-2/ref=sr_8_xs_ap_i2_xgl14/002-7979132-1252055?v=glance&s=books&n=507846]Kvant Selecta: Algebra and Analysis II (Mathematical World, Vol 15)
[/url]).

Dear Fuzzylogic, when I studied your very nice solution of Harazi problem

[quote="fuzzylogic"][quote="harazi"]For any positive real numbers $ a,b,c$ we have the inequality
\[\frac{(b+c-a)^2}{3a^2+(b+c)^2}+\frac{(c+a-b)^2}{3b^2+(c+a)^2}+\frac{(a+b-c)^2}{3c^2+(a+b)^2}\geq\frac{9(a^2+b^2+c^2)}{7(a+b+c)^2}.\] Hope someone will solve them[/quote]

Here is the proof of the second one:

WLOG, we may assume $a+b+c=3$. Then the inequality becomes:

\[\frac{(3-2a)^2}{3a^2+(3-a)^2}+\frac{(3-2b)^2}{3b^2+(3-b)^2}+
\frac{(3-2c)^2}{3c^2+(3-c)^2} \geq \frac{a^2+b^2+c^2}{7}.\]

[b]Lemma[/b] For any $0\le x\le 3$, the following inequality holds:

$\displaystyle f(x) = \frac{(3-2x)^2}{3x^2+(3-x)^2} \geq \frac{1}{49}(7x^2-44x+44)$.

The result clearly follows from the lemma since $\displaystyle f(a)+f(b)+f(c) \geq \frac{1}{49}(7(a^2+b^2+c^2) - 44(a+b+c) + 44\cdot3)=\frac{a^2+b^2+c^2}{7}$.

Proof of the Lemma: Let $g(x) = (7x^2 - 44x + 44)(3x^2 +  (3 - x)^2) -  49 (3 - 2x)^2$. Simplifying and factoring, we get $g(x) = (x-1)^2 (28x^2-162x-45)$. It remains to show: $h(x) = 28x^2-162x-45 \leq 0$ when $0\le x\le 3$. But this is almost trivial. Since $h(x)$ is convex, $h(x) \leq \max\{h(0),h(3)\} < 0$.[/quote]

to find the bounding function

[quote]$ f(x) = \frac{(3-2x)^2}{3x^2+(3-x)^2} \geq \frac{1}{49}(7x^2-44x+44)$ [/quote]

I have had to use the first and second derivatives, so I would like to know your method without the use of second derivatives.

And, please, can you explain your very nice general method without derivatives as for me it is very hard to think of it.

Thank you very much.

[quote="manlio"][...]to find the bounding function

[quote]$ f(x) = \frac{(3-2x)^2}{3x^2+(3-x)^2} \geq \frac{1}{49}(7x^2-44x+44)$ [/quote]

I have had to use the first and second derivatives, so I would like to know your method without the use of second derivatives.

And, please, can you explain your very nice general method without derivatives as for me it is very hard to think of it.

Thank you very much.[/quote]

I made a "educated" guess first. If the method works, the lower bounding function has to be a quadratic polynomial with leading coefficient $\frac{1}{7}$.

Then I anticipated the resulting polynomial by clearing the denominator has a double root at the point at which the "equality" holds, in this case $x=1$.

Divided by $(x-1)^2$ and setting the remainder to be 0 will give you a system of linear equations for the unknowns in the bounding function. (This is the part I find more convenient to use the first derivative, $p(1)=p'(1)=0$ gives me two equations for two unknowns).

Thank you very much. Your method is very interesting!!!

When I solved this question I assumed it was an application of the Jensen inequality.However nobody has mentioned this in their posts

f(x) = 2(1-x)/x(2-x)  has a positive second derivative for 0<x<1.The function f is therefore convex in this interval and noting that f(1/3) =12/5 the result follows from{ f(a) +f(b) +f(c)}/3 >= f((a+b+c)/3)

Hopefully this is correct


calculus
integration
calculus computations

[img]http://img207.imageshack.us/img207/483/integralzd1.jpg[/img]

Please help explain to me on how to solve this integral by trig. substitution? Because the denominator is not in a square root, I am confused on how to solve this.

Thanks. :)

Substitute x = 4secu/3

The integrand reduces to cotu*cosecu du which you should be knowing.

The denominator can be rewritten as $\left(\sqrt{9x^{2}-16}\right)^{3}$. Do you see the method now?

[color=darkblue][b]Definitions.[/b] $\sinh x=\frac{e^{x}-e^{-x}}{2}\ ;\ \cosh x=\frac{e^{x}+e^{-x}}{2}\ ;\ \tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\ .$
[b]Remarks.[/b] $\cosh^{2}x-\sinh^{2}x=1\ ;\ (\sinh x)'=\cosh x\ ;\ (\cosh x)'=$ $\sinh x\ ;\ (\tanh x)'=\frac{1}{\cosh^{2}x}\ .$

You can apply the substitution $x=\phi (t)=\frac{4}{3}\cdot\cosh t$ and $dx=\phi '(t)\ dt=\frac{4}{3}\sinh t\ dt\ .$[/color]


:maybe:  What's the problem in the drawing ?

A bad drawing is not a counterexample.

It is indeed wrong because you can take a square as counterexample (= hexagon with 2 opposing sides "very very small"), as the $4$ points on the longer side will have an symmetry group of order $8$, but no four points in a regular hexagon have a symmetry group of order $8$ (that's always in $\{1,2,4\}$.

The correct is that their centers form an also centrally symmetric hexagon at the same center as the given (easy to prove).

Kostas Vittas.

Then there is nothing to prove, as none of the steps in the construction are asymmetric...

[quote="silouan"]Consteuct regular hexagons on the sides of a centrally symmetric hexagon .Prove that their centers  form the vertices of a regular hexagon .[/quote]

What is the Barlotti Theorem???


Given the graph of $ y \equal{} \frac12 x$ as pictured, vertical line segments are drawn perpendicular to the $ x$-axis and rectangles are formed, as shown. How many square units are in the area of the shaded region?
[asy]size(150); defaultpen(linewidth(0.8));
yaxis("$y$",-1,3);
xaxis("$x$",-1,6);
draw(Label("$y = \frac{1}{2}x$",align=2*N),(-1,-0.5)--(6,3));
void drawBar(int x){
 filldraw((x,0.5*x)--(x+1,0.5*x)--(x+1,0)--(x,0)--cycle,gray(0.5));
}
drawBar(1); drawBar(2); drawBar(3);[/asy]

Yes; there are an infinite number of possible answers...

I just looked at the picture and guessed 3...which was correct.

Well that is correct if the x-value of the rectangles are 1,2,3.

[quote="athunder"]i do not understand how youdo this with such less information, they should give you coordinates of the x axis right?[/quote]

That's why I asked this. ;)


geometry
3D geometry
rotation
calculus
integration
analytic geometry
vector

Moments of inertia can be computed through integral formulas.  See [url=http://en.wikipedia.org/wiki/Moment_of_inertia]here[/url] for some details.  Your first step would be to parametrize the cone.

i was thinking about using cylindrical coordinates because the side part would just be $ z$ but i'm not sure what to do...

i'm not really sure how to use the link...

That just explains more on moments of inertia and some of the formula.  I don't have my vector calculus book on me but inertia is given by an integral of a density function.  You need to parametrize the cone in order to set up your limits of integration.

"Slicing" the cone in "coins" of infinitesimal thickness an applying Steiner's parallel axis theorem, it's not hard to get \[ {3\over 5} m ({{r^2}\over 4} \plus{} h^2)\]

why do you need the parallel axis theorem? doesn't the axis of rotation go through the center of mass?

[quote="jli"]what is the inertia of a cone when rotated about an axis perpendicular to the height of the cone and passing through the tip of it[/quote]
First compute the M.I about the height (which is easy using a little bit of calculus), then using the Perpendicular axes' Theorem, you may find the required M.I.

[quote="Immanuel Bonfils"]Doesn't the Perpendicular axes theorem applies only to plane "objects"?[/quote]
Oops! Sorry! Yes the theorem is applicable only to plane objects. :oops:

[quote="Immanuel Bonfils"]"Slicing" the cone in "coins" of infinitesimal thickness an applying Steiner's parallel axis theorem, it's not hard to get
\[ {3\over 5} m ({{r^2}\over 4} \plus{} h^2)
\]
[/quote]
Isn't the $ AB$ axis about which we have to find the M.I? You have computed the M.I about $ OO'$, haven't you?

[quote="Immanuel Bonfils"]The M.I. about $ OO'$ would be $ {3\over{10}} mr^2$.[/quote]
Yes, I too get M.I about $ OO'$ as $ \frac {3}{10} MR^2$.

[quote="JRav"]You may want to fix your syntax.  First of all, there is no 3 in the numerator and second of all, if the $ mr^{2}$ is in the denominator, your answer is nonsense since the units don't match.[/quote]
Who is that idiot with such a non-sense answer?  :wink:

Huh? :huh:   I hope you're referring to yourself. :wink:


geometry
rectangle
perimeter
congruent triangles

The diagonals of a rectangle intersect at point $ P$. Point $ P$ is $ 5$ centimeters further from the shorter side than the longer side. The perimeter of the rectangle is $ 44$ centimeters. What is the number of square centimeters in the area of the rectangle?

Draw these diagonals on the rectangle ABCD. Now, draw perpendiculars to two adjacent sides, say AB and BC, creating PE and PF. Within this smaller rectangle which is 1/4 of the whole, call one length of perpendicular x and the other x=5. These are actually congruent triangles PEB and BFP you have drawn, so EB is x+5 and BF is x. Doubling these gives you side lengths 2x+10 and 2x.

$ 2x\plus{}10\plus{}2x\equal{}4x\plus{}10\equal{}22$, by half of perimeter
$ x\equal{}3$

Sides are 16 and 6, area is $ 96$ sq. cm.


I didn't turn my entry form yet and I was thinking of turning them with my solutions.

But since I'll be emailing my PDF solutions, it doesn't really matter when I send the solutions (because email is fast). What about entry form? At least what date should I mail so that I can make sure the form goes within the deadline?

Thank you.

Try to send the form in by the first round deadline (postmark).

I didn't get any time to get my parent signed paper till tonight. I just put the mail in mailbox (currently, it's 11:32 PM Friday) and I don't know if this mail will get to Alpine, California by Monday.

I live in East USA so this is why I'm asking. Will this be okay?

I'll send my solutions in PDF so it won't matter..

If you're mailing, we don't need to receive it by Monday.  It must be postmarked by Monday.


To me logic is defined as something higher in existens than what just seems right for the moment. Being logical is to make logical correct conclusions from given data.

When I am talking about logical, I mean that there is a reason for every single thing that ever happens. Every even that happens, for example, you choosing what to write in the forum, the particular words you choose, the way you move, the sounds you speak, the way the clouds move, the results of the lottery, everything is logical which means that all these things can be explained and have reason in them.

I have invented (discovered) a theory of universes. I know that everyone who reads this thinks I am crazy but still. Anways, you can suppose our universe in this way.  Everything is logical in our universe and the only reason that people dont act perfectly is because they dont know that logic. This can obviously be more rigioruously argued. Anways, people keep posting here, we must develop this theory here, the most logical theory in the world, on which even mathematics is based.

[quote="hello"]First of all I noticed that I am like the only one posting today, where is everybody? I think that all the world can be deducted into pure logical axioms? Do you people agree? Also, I think that everythink illogical is really logical, what I mean is if a person does something wrong, say solves a math problem incorrectly, he is still being logical from the aspect that he is doing something wrong for a reason, not just out of the blue. So all actions that happend in this world are very logical, do you agree? The only problem is to understand that logic, and to see where the person needs to correct his logical process.

Post your ideas here. This is a philosophical discussion. :rotfl:[/quote]

Well.. IMHO everything a person does can be defined as 'logical' to him but it doesnt make it logic. Real logic is whats correct, logically. When a person does something and he thinks its right and its logical - its like a guitar thats out of tune, but is in tune to itself. Its not on the standard, correct way of sounding, but it's 'right' from the standpoint of working in the way it knows how to. Logical things can be reasons, but not all reasons are logical.

ToBe, what you are saying is true but that is not the point. I am postulating the simple idea that something happens for a reason, and because it happens for a reason, it is logical that it happens and not something else, I understand that it doesnt mean that it is correct, I am just saying that it is logical because it happens, if it would not be logical for it to happen, it would not happen. QED.

hello:  Are you postulating that the universe as it exists is ontologically necessary, and could not have been otherwise?  That is, are you answering in the affirmative to the question, "Is all truth tautological?"?

I'm just a little unclear on what precisely your position is.

i dont know ur terminology? explain tautology and that other stuff.

Well, any oterh people wil lhelp me develop the theory?

A tautology is a statement in mathematical logic (propositional calculus), such that the final column of it's truth table contains only T's. This doesn't help if you never studied introductory logic. Here's an example which might make more sense: ~(A & B) $\rightarrow$ ~A or ~B. In real life, this might be equivalent to "If it is not the case that both teams win, then at least one of the teams must lose (or tie)". Such a statement is always true.

Ok, so what is that meant in the context of what I wrote?

Many physicist will probably agree with your 'reduction' of the world. In physics, it's called scientific determinism or reductionism. They think that once you know the initial condition of everything in the universe (for example, the position and velocity of every particle), the laws of physics will dictate how the universe evolves, so you can work out the state of the universe at any given time after that.
The consequence is that everything is predetermined, even the function of your brains. So free will does not actually exist!
However, uncertainty principle is introduced in modern physics and determinism has lost some popularity.

i think world can't be reduced to purely logical axioms or assertions. The illogical can be  obviously explained with logic, but irrationality can't be explained this way
The posture that the world can be reduce to loigc it is a old philosophical question not interested to modern philosophers . Semantics, gramatology, ambiguity are some of the many new themes of the post modern philosophy.

An example of something that cannot be exaplined with logic? I can't really give one because my human language works by logical assertions; which limits greatly the human mind.
but i can try.. i found this really far from logic:
there is something that exist but i can explain logically that it exists. I can't give a definition or a word to this thing. this thing is something found between the concept or idea of an obect and the word. in the word-object relation there is something that it is not part of the definition of the word but neither it explains the obejct. it is like a predetermined attraction between the word and the object that lies in an uknown space that human language can't explaine correctly.
sorry, i know that i sound crazy and maybe nonsense but it is the language's fault!  :D
for more interesting and 'real'  postmodern philophy there are good writers like: Derrida and Focault.
also a lot of european films express some of the postmodern ideas, some of them are: Persona (by Ignman Bergman), all of the movies by Tarkovsky, Songs from the second floor (surrealist film from Denmark i think, not sure) and many more films that have as theme silence, language, etc...
but i really don't know much about philosphy, but i am 100% sure that the modern concern of philosphy is does not has to do with 'world logic' or with any kind of positivsm.
That is why i see math as a complete art, and I think mathematicians should stop looking for "logic" in the world and should stop looking for applications of math to other sciences to 'understand' the world, and should stay studying the purity of mathematics in its own internal logic (logic is one of my favorite math areas!). i think mathematicians should evolve to the postmodern world...

Come down. U the physics guy, I think u r almost right. With all known, we can derive everything that happens. But that is not 100% not free will. Free will still exists when a human makes a decision, it is just that he/she will make it for a certain reason, not irrationaly.


[b][u]Problem[/u][/b] A closed conical container of height 12 cm is partially filled with liquid and the level of the liquid is marked when the vertex is pointed down. When the container is inverted, it is noticed that the liquid reaches exactly the same mark. What is the depth of the liquid in the first position?

---
if anyone has the solution to this problem, please post it here. Thx.

Is that even possible considering a cone has less of an area at the top than the bottom?  :|

The mark is not the same distance from the base and the top of the cone.  It is possible.

Took a bit too long but I got
[hide]$6 \sqrt [3]{4}$.

Set the radius of the entire cone equal to $R$. Set the radius of the liquid when the cone is upside down to $r$. Since the height of the cone is $12$, we can find the height of the liquid at that position by forming the proportion $\frac {R}{12} = \frac {r}{x}$ and solving we get $x = \frac {12r}{R}$. When the water is in the other position (touching the circular base), the height of the frustrum that is formed would then be $12 - \left(\frac {12r}{R}\right) = \frac {12(R - r)}{R}$. To find the value of $r$, since the volume of the water is the same in both positions, we set the volume of the cone-shape of the water to the volume of the frustrum-shape of the water which is

$\frac {12r^3 \pi}{3R} = \left(\frac {12(R - r)}{R}\right)(R^2 + r^2 + Rr)$.
$\frac {12(R^3 - r^3)}{R} = \frac {12r^3}{R}$

Then solving for $r$ we get $r = \frac {R}{\sqrt [3]{2}}$. Plugging that into $\frac {12r}{R}$ for the height we get $\frac {12R}{R\sqrt [3]{2}} = \frac {12}{\sqrt [3]{2}}= 6 \sqrt [3]{4}$[/hide]

[hide]Let the depth of the water in its original position be $x$, the height be $h$, the radius of the whole cone be $R$ and the radius of the cone at the original water be $r$.

Using similar triangles, $\frac h R=\frac{h-x}{r}\to r=R\frac{h-x}{h}$
The volume of the water is one-half the volume of the whole cone because when you flip the cone, the water covers exactly what was previously empty. Therefore, half the volume of the whole cone is equal to the volume of the empty cone above the water in its original position.
$\frac12(\frac13 \pi R^2h)=\frac13\pi(R\frac{h-x}{h})^2(h-x)$
$\frac12 R^2h=R^2\frac{(h-x)^3}{h^2}$
$\frac12 h^3=(h-x)^3$
$\sqrt[3]{\frac12}=\frac{12-x}{12}$
$\boxed{x=12-6\sqrt[3]{4}}$[/hide]


Compute the value of $ f(\minus{}4)$ given $ f(x) \equal{} x \minus{} |x|$.


We can choose n points on a circle and draw all chords joining these n points. Find the number of parts into which the circular disk is cut.


What is the smallest 3-digit number that can be obtained from 4921508 by crossing out 4 digits? (Note: the order cannot be changed)

[hide=" click to reveal hidden content"]

$ 150$

[/hide]

[hide]
$ \large \boxed{108}$
You cross out $ 4, 9, 2,$ and $ 5$.[/hide]

[hide]Your answer would be 108, because you would cross out the 4,9,2, and the 5..[hide][/hide][/hide]


Evaluate:
$ \sum_{r \equal{} 1}^{n}( \minus{} 1)^r(1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ...... \plus{} \frac {1}{r})C_{r}$
where $ C_{r}$ is the binomial coefficient

I assume your $ C_r$ means $ C_n^r$ (also known as $ \binom{n}{r}$), and your sum can be simplified using the formula [color=green][b](5)[/b][/color] http://www.mathlinks.ro/viewtopic.php?t=255493 for $ m\equal{}1$.

Please tell me if I have misinterpreted the problem.

  darij

Well nice post there Darij. Thanks for enlightening us yet another time.


integration
real analysis
function
limit
real analysis solved

Let f:[a,b]->R such that f+sinf is integrable.Prove that f is integrable.

I don't wich definition are taken for integrable, here a answer
 with this one

Definition: f is interable on [a,b] if  \int _[a,b]|f| is finite number
(f is Lebesgue integrable)

In this case  f = (f+sinf) - sinf =>

|f| \leq | f+sinf | + |sinf|

|f| \leq | f+sinf | + 1

\int _[a,b]|f|  \leq \int _[a,b]|f+sinf| + b-a < +oo

If g = f + sinf is Riemann integrable then g is bounded

There exist A>0 such that |g| \leq A

|f| \leq A+1 => f is bounded

This does not prove that f is integrable.

I don't think youre right,Moubi.Are you saying that a bounded function is always integrable?

Indeed there exist functions bounded wich is
not Riemann integrable.

 moubi
" Cantor it's me moubi ?
Would you mind if I use your example for answer xxxxtt=x^4t^2 ? "

Cantor
"No problem, it is a real pleasure that you remind my caracteristic
 \varepsilon - Cantor set"

Let 0< \varepsilon <1 consider  \varepsilon -Cantor set C_ \varepsilon
 of [0;1]
K(C_ \varepsilon ) is caracteristic function on [0,1]
This is bounded and not Riemann integrable

So your solution is not correct,Moubi.
Another example of bounded not integrable function is Dirichlet function

Hi xxxxtt,

What do you think about the following ?

Let g(x) = k(f(x))

where k(y) = y +siny
k'(y) = 1 + cosy = 2cos^(y/2) \geq 0

Consider the case that the interval [a,b] such that f([a,b]) is included
in a compact J ( it is possible since f is bounded see above post)
where the restriction of k to J is srictely increasing

k is bijective its inverse k^(-1) is continous

Now there is a theorem of the great Bernhard Riemann

Theorem: w:[a,b]-->R Riemann integrable, h a continous function
from a compact K wich contains w([a,b]).
Then h(w(x)) is Riemann integrable.

Apply this theorem

h is k^(-1)  inverse of k, wich is continuous
w is g = f + sinf = k(f) Riemann integrable

f(x) = k^(-1)(g(x)) is Riemann integrable.

x^4t^2 can you remind me the definition
of the Dirichlet function please ?

I think it's the characteristic function of Q.

I wonder if Lebesgue's criterion works:

Let D(f) be the set of points of discontinuity of f on [a,b]. We must show that given that D(f+sin(f)) has Lebesgue measure 0, then so does D(f). We can see that D(f) includes in D(f+sin(f)) (*).

Now we have to look at those points in which f isn't continuous, but f+sin(f) is. Let a be such a point. Then there is a sequence (an)n>=1 which tends to a s.t. bn=f(an) tends to c=/=f(a). f+sin(f) is continuous in a so bn+sin(bn)->f(a)+sin(f(a)). From the fact that x+sinx is a bijective continuous function and the last relation we easily get bn->f(a), so we have a contradiction. This means that there are no points in which f is not continuous but f+sin(f) is, so D(f) is included in D(f+sin(f)) (**).

(*) & (**) => D(f) has Lebesgue measure 0 iff f is Riemann integrable.

Is it Ok?

[quote="grobber"]
Now we have to look at those points in which f isn't continuous, but f+sin(f) is. Let a be such a point. Then there is a sequence (an)n>=1 which tends to a s.t. bn=f(an) tends to c=/=f(a)....[/quote]

Nice idea to try Lebesgue criterion nearly 01h am  :)
A bounded function f:[a,b]-->R is Riemann integrable iff
it is continuous almost everywhere.

But grobber I don't see why f(a_n) should have a limit c
maybe \lim f(a_n)  does not exist, no ?

We can't not have a limit for ALL sequences an. I didn't say that f(an) has a limit for every sequence an, but I can find ONE such sequence for which f(an)has a limit different from f(a). Try to show this, it's really easy.

Yes,grobber, that was also my idea, with the Lebesque criterion.


There are some communication problems in a meeting of four mathematicians A,  B , C and D:
Each of the mathematicians can only speak two languages among Chinese, English, French and Japanese, and they don't share a common language. But there is a common launguage for three of the mathematicians.
Besides,
(1) B can't speak English, but he translates for A and C;
(2) A knows Japanese and D doesn't know Japanese, but they can talk with each other;
(3) B, C and D have no common languages;
(4) No one knows both Japanese and French.

So, what languages does each of the mathematicians know?

[hide]
Ok i basically mae a table with A,B,C,D on the top and the languages and fellwo mathmaticians on the side..

Its obvious that A,B,D have the common language. It can't be English (B doesnt know it), it can't be French (A already knows Japanese), and it cant be French b/c A knows Japanese but can still talk with D. Thus the common language between A,B, and D is Chinese.

A is then full (Jap, Chi)
B has Chinese and one more spot to fill.
C cannot know Chinese, so he must either know french or english (becuase if he knows Jap or Chi, then A and C can talk to each other). B must also then know Fra or Eng.
D has Chinese as well, and will have the other language that B and C do not know.

I'm going to propse that B (Fra, Chi) $\rightarrow$ C(Fra), so C cannot know Jap or Chi so C must know (Fra,Eng). D cannot know Fra or Jap so D knows (Eng, Chi). A has already been determined to know (Chi, Jap).
So we have a solution with: $\boxed{ \text{A (Chi, Jap), B(Fra, Chi), C(Eng, Fra), and D(Eng, Chi)}}$.

Dunno what happens when B knows enlgish, it gets a little messy... anyone want to pick up where i leave off?[/hide]

By conditions (1), (2) and (4), we immediately have the following table.
We may start from there.

$\begin{array}[b]{ccccc} \hspace&A&B&C&D\\ {Chinese}&\hspace&\hspace&\hspace&\hspace\\ {English}&\hspace&\times&\hspace&\hspace\\ {French}&\times&\hspace&\hspace&\hspace\\ {Japanese}&\surd&\hspace&\times&\times \end{array}$

[quote="cincodemayo5590"]
Ok i basically mae a table with A,B,C,D on the top and the languages and fellwo mathmaticians on the side..
[b]
Its obvious that A,B,D have the common language[/b]. It can't be English (B doesnt know it), it can't be French (A already knows Japanese), and it cant be French b/c A knows Japanese but can still talk with D. Thus the common language between A,B, and D is Chinese.

A is then full (Jap, Chi)
B has Chinese and one more spot to fill.
C cannot know Chinese, so he must either know french or english (becuase if he knows Jap or Chi, then A and C can talk to each other). B must also then know Fra or Eng.
D has Chinese as well, and will have the other language that B and C do not know.

I'm going to propse that B (Fra, Chi) $\rightarrow$ C(Fra), so C cannot know Jap or Chi so C must know (Fra,Eng). D cannot know Fra or Jap so D knows (Eng, Chi). A has already been determined to know (Chi, Jap).
So we have a solution with: $\boxed{ \text{A (Chi, Jap), B(Fra, Chi), C(Eng, Fra), and D(Eng, Chi)}}$.

Dunno what happens when [b]B knows enlgish[/b], it gets a little messy... anyone want to pick up where i leave off?[/quote]

Hi! Cinco,
I think the assumption in the line with bold face is too early.
Besides, B does not know English by (1).
However, your final answer is correct.  :)

[quote="demji"][hide]
A knows English and Japanese
B knows Japanese and Chinese
C knows Chinese and French
D knows English and French
[/hide][/quote]

Sorry, you've got the wrong answer.


calculus
integration
inequalities
function
derivative
real analysis
real analysis unsolved

Suppose that f is a differentiable function with f(0) = 0 and 0 < f' <= 1.
Prove that for all x >= 0, we have:

  $ \int$0 to x  f^3   <= ( $ \int$0 to x  f )^2

I tried using the cauchy-schwarz inequality, but didn't get much. I've also tried to do something with f(x) <= x and that it's increasing and always positive. Didn't get much either.

 I'd appreciate the help.

Thanks

[quote="leshik"]Just consider the function $ g(x) \equal{} (\int\limits_{0}^x f)^2 \minus{} \int\limits_{0}^x f^3$. Then  you can easily check that $ g''(x)\ge 0$ which implies $ g'(x)\ge 0$ (because $ g'(0) \equal{} 0$) for $ x\ge 0$.The rest should be clear. :)[/quote]

thanks! That seems to be a common strategy. I will keep that in mind next time. Anyway, I tried showing that g''(x) >= 0, but I couldn't really.. I think my algebra just stinks. Could you show me please?

Thanks

Ok.
We have $ g'(x)\equal{}2f(x)\int\limits_{0}^{x} f(t)dt \minus{}f^3 (x)\equal{}f(x)(2\int\limits_{0}^{x} f(t)dt \minus{}f^2 (x) )$. We now want to prove that the function $ g'$ is increasing. From the condition of the problem ($ f'\ge 0$) follows that $ f(x)$ increases therefore it is enough to show that $ h(x)\equal{}2\int\limits_{0}^{x} f(t)dt \minus{}f^2 (x)$ is an increasing function. But $ h'(x)\equal{} 2f(x)\minus{} 2f'(x)f(x)\equal{}2f(x)(1\minus{}f'(x))\ge 0$ because of the conditions of our problem. That' all. :)

[quote="leshik"]Ok.
We have $ g'(x) \equal{} 2f(x)\int\limits_{0}^{x} f(t)dt \minus{} f^3 (x) \equal{} f(x)(2\int\limits_{0}^{x} f(t)dt \minus{} f^2 (x) )$. We now want to prove that the function $ g'$ is increasing. From the condition of the problem ($ f'\ge 0$) follows that $ f(x)$ increases therefore it is enough to show that $ h(x) \equal{} 2\int\limits_{0}^{x} f(t)dt \minus{} f^2 (x)$ is an increasing function. But $ h'(x) \equal{} 2f(x) \minus{} 2f'(x)f(x) \equal{} 2f(x)(1 \minus{} f'(x))\ge 0$ because of the conditions of our problem. That' all. :)[/quote]

 alright, thanks! so we don't have to look at the second derivative of g?

[quote="Aile"][quote="leshik"]Formally, we do not, but instead of this we used some  trick (about monotonicity of $ f$).But we could just calculate second derivative and get the same result.[/quote]
ya maybe you are talking about this ? :roll:  $ 2\int_0^xf(t)\,\mathrm{d}t \minus{} f(x)^2 \equal{} 2\int_0^xf(t)\big(1 \minus{} f'(t)\big)\,\mathrm{d}t$,
since we have $ f'(x)\ge 0 \implies f(x)\ge f(0) \equal{} 0$, we conclude the result.[/quote]

I was presented with this problem, and I understand that g starts at 0 and increases, and g' increases and is monotonic.

I set $ g(x) \equal{} (\int\limits_{0}^{x}f)^{2}\minus{}\int\limits_{0}^{x}f^{3}$

Our explanation(s) about g"(x) and g'(x) were nearly the same, but my teacher wants a better explanation than just g"(x) >= 0, which implies g'(x) >= 0, etc.
Any help here?

I did the first derivative of the problem.
I understand that g'(x) was named h(x) and the derivative of it was found, which is the second derivative, right?
My problem/my teacher's problem is the part with
$ 2\int_{0}^{x}f(t)\,\mathrm{d}t\minus{}f(x)^{2}\equal{} 2\int_{0}^{x}f(t)\big(1\minus{}f'(t)\big)\,\mathrm{d}t$

I need to show that this is positive, just like how the other parts are.
Basically, can I see how this proof is fully written and/or explained, because my teacher lost me on some of this?!

Also, can this problem be proven another way, like MVT for integrals, etc?

Thanks for any response :roll:


The Happy Animals Kennel has 18 cages in a row. It will allocate 6 to the dogs and 12 to the cats. Let $ A$ be the number of times in the row of cages that a dog cage and a cat cage are adjacent.  For example, in the arrangement cdcdddcdccccccdccc, we have A=8.  Find the expected value of A.

is the only way to do this is to list each one out?  it seems too tedious to be the best solution.

Hmm, it seems to me that we did something like this last year.

However, I forget what we did.

i think i saw something like this in the AoPS vol 1 articles and excerpts

could we take the max and min values of A and somehow use harmonic mean or something?

I'll take a shot at it...
[hide]There are 17 ways to have consecutive pairs.The probability of dog-cat= (1/3)(12/17)=4/9 and of cat-dog is (2/3)(6/17), meaning the sum is 24/51. Multiply by 17 since there are 17 ways, and get 8.[/hide].

Woah, where did that come from?

What is the expected value?
Is it like a mathematical term or something?

[quote="ZhangPeijin"]Wait, I read the definition and I still don't understand...[/quote]The expectation of a random variable can be thought of as the mean value or statistical average of the random variable.

(Alternatively, (if you have done mechanics): say you have a bar with a certain number of masses on it. Then the expectation is just the centre of gravity, which is computed by finding the moments if I recall correctly.)

This is a problem from an old AHSME, and is also included in AoPS 2 and Intermediate Counting and Probability. It's not appropriate for this forum; I'm pretty sure you don't even have to know what Expected Value means on MathCounts.

If you're having difficulty with the definition, think of it intuitively; if you were to take every possibility, what would the average of what you're trying to count be?

Uh, I gave an answer, but I don't know if it's right? Someone want to help?

[hide="Just a guess"]
$ \frac{2^6}{6}\equal{}10.\overline{6}$
[/hide]

There's a problem with the same basic idea on page 4 of [url]http://www.usamts.org/Forms/ExpectedValue.pdf[/url]

Okay, so I looked in Volume 2 (specifially ch. 17, problem like 297), and if there's m of something and n of something else like this, it's $ 2mn/(m\plus{}n)$. Like harmonic mean. The proof is kind of like my solution I guess.


[color=darkblue]Daca $x, y, z$ sunt numere reale pozitive astfel incat $x+y+z=1$ atunci are loc inegalitatea : $\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\le 2$ ([b]Autori[/b]: [u]TituZvonaru, Bogdan Ionita[/u], Bucuresti).[/color]

Se scrie $x+yz= x*1+yz=x*(x+y+z) +yz=(x+y)(x+z)$  Analog celelalte numere si dupa se aplica inegalitatea Mg-Ma pt perechile (x+y;x+z) si permutarile circulare,rezulta concluzia..eu am facut altfel la concurs..prin ridicare la patrat.Dar rezolvarea este  mai lunga

Sau putem aplica Cauchy si inegalitatea rezulta imediat, pentru ca asa cum spuneam din C.B.S, avem

          \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\leq\sqrt{3(x+y+z+xy+yz+zx)}.  \]

  Insa cum, $x+y+z=1$ , iar $xy+yz+zx\leq\frac{(x+y+z)^{2}}{3}=\frac{1}{3}$, concluzia decurge imediat. :)


$ \{a_n\}_{n\equal{}1}^\infty$,a sequnce of real numbers,$ n\in \mathbb {N}$

Suppose $ a_1\plus{}2a_2\plus{}3a_3\plus{}\cdots\plus{}na_n\equal{} n^2\plus{}3n\plus{}1$

Find $ (1) a_n ,(2) a_{20}$

Let $ a_n \equal{} a_1 \plus{} 2a_2 \plus{} \cdots \plus{} (n \minus{} 1)a_{n \minus{} 1} \plus{} a_n$,
For $ n \equal{} 1$, $ S_1 \equal{} 1\cdot a_1 \equal{} 1^2 \plus{} 3\cdot 1 \plus{} 1\ \therefore a_1 \equal{} 5$.

For $ n\geq 2$, $ S_n \minus{} S_{n \minus{} 1} \equal{} (n^2 \plus{} 3n \plus{} 1) \minus{} \{(n \minus{} 1)^2 \plus{} 3(n \minus{} 1) \plus{} 1\}$

$ \therefore na_n \equal{} 2n \minus{} 1 \plus{} 3 \equal{} 2(n \plus{} 1)$.

$ \therefore a_n \equal{} \frac {2(n \plus{} 1)}{n}\ (n\geq 2)$.


Suppose x, y, z are positive reals such that x + y + z = 1.
Determine the maximum value of xy + yz + zx.

This is certainly not easy ! :o

CMO 98 or something.  max occurs at 2/3,1/3,0

CMO 1999 :) (Canadian Mathematical Olympiad)
Can you prove it bugzpodder ? (Without looking at the official soln)

A serious hint. Let f(x, y, z) = xy + yz + zx.
Compare f(x, y, z) and f(x + z, y, 0). Which one is larger ?
:mrgreen:

We'll ... it's not too complicated to prove that WLOG x :ge: y :ge: z.
Now using x :ge: y :ge: z you can prove that f(x + z, y, 0) :ge: f(x, y, z).
So we can assume that z = 0. Then the problem becomes :
Given x, y > 0 such that x + y = 1, maximise xy.
This is easy using derivatives or the more elegant way :
xy = 1/2 (2xy) :le: 1/2 ((2x + y + y)/3) = 1/2 8/27 = 4/27.

[f(x) = xy + yz + zx, see the hint above.]

I like this one but it's difficult.

Hehe, all 30 inequalities are done now :mrgreen::mrgreen::mrgreen:


vector
geometry
parameterization
Asymptote
geometry unsolved

Dear All,

I have the following problem.

Let C(x,r) be the notation for a circle centered in x (vector in R^2)
and having radius r.

Let us define the following four circles:

C(0,gamma)
C(a,alpha-gamma)
C(b,beta-gamma)
C(a+b,2+gamma-alpha-beta)

Now, suppose we are given a quadruple of the form <a,b,alpha,beta>.

Is there a necessary and sufficient condition in a,b,alpha,beta such
that there exists gamma for which the four circles have a nonzero
overlapping area?

A necessary condition is that ||a+b||+||a-b|| is smaller or equal than
2. However, this is not sufficient.

Thanks a lot for your help,
Simone
-
Simone Severini
http://www.iqc.ca/~sseverin

I just want to note that, if you have any family of circles in which every triple has a non-trivial intersection, then all of them together have a non-trivial intersection. Thus, there is no need to talk of more than 3 circles. The condition for three circles $C(a,\alpha)$, $C(b,\beta)$ and $C(c,\gamma)$ to overlap exists and I have even seen it somewhere but, if I remember it right, it is not what you'd call beautiful. If you desperately need it for something important, we'll try to derive it but if it is just pure curiousity, I'd rather spend my time on something else :).

It turned out that condition is not as bad as I thought when written in trigonometric terms (I saw it as an algebraic relation between the parameters, which was bad, indeed). The idea is that the following two pictures exist or don't exist simultaneously:
[asy]import cse5; size(300); void routine(pair A,pair B,pair C,string La, string a, string Lb, string b, string Lc, string c) { pair P=(A+B+C)/3; D(D(A)--D(B)--D(C)--A--D(P)--B); D(P--C); void M(string Lx, pair X) { MP(Lx,X,9,(0,0)); } M(La,(B+C)/2); M(a,(A+P)/2); M(Lb,(A+C)/2); M(b,(B+P)/2); M(Lc,(B+A)/2); M(c,(C+P)/2); } routine((0,0),(3,0),(2,2),"a","<\alpha","b","<\beta","c","<\gamma"); routine((5,0),(8,0),(7,3),">a","\alpha",">b","\beta",">c","\gamma");[/asy]
The first one is what we need and the condition for the existence of the second one is just
\[\arccos \frac{\alpha^{2}+\beta^{2}-c^{2}}{2\alpha\beta}+\arccos \frac{\alpha^{2}+\gamma^{2}-b^{2}}{2\alpha\gamma}+\arccos \frac{\gamma^{2}+\beta^{2}-a^{2}}{2\gamma\beta}<2\pi\]
(the sum of angles at the center point shouldn't be too large). That's it (if I didn't mistake anywhere).


MATHCOUNTS
ARML
Gauss
floor function
geometry
3D geometry
number theory

Teams:
1. Kennedy Junior High School
2. Avery Coonley School
3/4. University Laboratory High School or UOC Lab Schools (can't remember which was which)
5. Plum Grove Junior High School
6. (don't remember)
7. Lincoln Junior High School

Individuals:

Nats Team:
1. Patrick Liscio, Lincoln Junior High School
2. Megan Chen, Wood Oaks Junior High School
3. James Tao, Plum Grove Junior High School
4. David Wang, Hinsdale Middle School

Also noted: stargirl got 15th place, ravi jagadeesan 5th place, suma_milli 60something place, helagha was somewhere in top 16

Countdown:

1. Eric Lullo, Kennedy Junior High School
2. Ravi Jagadeesan, Avery Coonley School
3. Megan Chen, Wood Oaks Junior High School
4. Thomas Wu, Kennedy Junior High School

Congrats everyone! :D  :D  :D

i think? im not sure exactly tho

but he isnt active

kevin chen should have beasted arml last year. i mean, that first problem was just a drawing problem. it doesnt take that long to draw like 69 straight lines  :mad:  :mad:  :mad:
EDIT: someone sitting near me got it right (by drawing of course). i didn't have anything to draw on

Please do not spam this, or any other topics in the IL forum.

Please keep all posts relevant to the correct topics. *cough cough IL outsiders this is about mathcounts not arml* :wink:

To say something about nats teams from IL, I notice how funny it is when every year, the people on the team are always different from the people who were on the team in the preceding year. :rotfl: Lol, that must tell something about how IL is a black horse state where kids beast on surprise. (and yes, A LOT of unusual trends happen)

I can't believe you guys get to go to Orlando
They should've done that last year

Wasn't last year TX too? That was rather boring. And there was a :censored: -load of mosquitoes everywhere.

For the teams wasn't it:

3> Plum Grove/UOC lab school
4> UOC lab school/Plum Grove
5> University Labratory High School

Remember how the lady messed up and had to swap their trophies?

lol Uni had to give their 3rd place trophies back and then get 5th place trophies.

[quote="dragon96"]"At Nats, the judge calls out, 'Kevin Chen!'
From the audience arise 12 different people, looking around.
'I'm Kevin Chen! Why are you standing up?'
'I'm Kevin Chen too!'"

Is Patrick Liscio an AoPSer?[/quote]

:rotfl: :rotfl: :rotfl:

I beat David Wang in CD and he's going to nat's  :mad:
ah, meh :dry:

[quote="cheesymath"]The fourth person on the team was David Wang, and he went to Hinsdale Middle School, not Kennedy.[/quote]
Oops. Fixed. :blush:

Don't forget me!!! I got 65th! haha
Though, if you know Russell Lee, he got 15th last year i think, he really should've done better this time around...  im surprised!

Um, no offense, but I don't know you...sorry I can't put you on the list.

Most common name in mathcounts history: David Wang?

(There were at least two in 06).

[quote="guineapiggies"]OH MY GAWD JAMES TAO IS GOING TO NATS?!?!??!?! I KNOW THAT KID!!! and i went to PGJH. arg. we sucked. dang. my team uber failed last year. but that doesn't count cuz our team was thrown together 3 days before chapter and we got to practice exactly once. during lunch. w/e[/quote]
OH DANG WAIT A SEC?

Uh...guinea I GUESS I SAW U AT CHAPTER LAST YEAR!!!! Yes, I was the nervous Asian kid who phailed countdown but took 4th.
Maybe I saw you but you didn't see me...maybe you said GJ to me...lawlz who knows

OMG GAUSSY i think i remember you!! you were the uber happy hyper asian girl right??? lol. i just remember this really super hyper asian girl that was i think either 4 or 6th on the lineup for countdown!!!!! and i think you wore a red shirt......... lol

yes that was me. Actually I was the only asian girl who sucked at countdown.
And I still suck at countdown.
And I'm really slow.

 :wallbash_red:  :wallbash_red:  :wallbash_red: <----that's me phailing at state countdown

And yes, guinea, I took 2nd written and 3rd countdown at state. (Though It's also a relief that I'm the 1st placed person of all the 4 pplz on IL nats team 2009 in countdown)
BTW, last year did I remember seeing you, or did I have bad memory and not notice you? hmm Maybe you said hi/congrats to me but I think I ignored you?

aww come on gaussy you can be awesome!!

nice. gj gaussy!!! idk... i didn't know this site so i hadn't met you yet XD. but idk... i said congrats to a bunch of people that day XD hahaha  :D

i thought u had a weird reaction when u saw me specifically? :maybe:
oh and if you wanted to know:

AT STATE COUNTDOWN:

First match: Beat Xiaoyu Li, an 8th grade girl from Dunlap Valley MS
Second match: Beat Emily Mu, a 6th grade girl from Kennedy (YES KJHS BEASTS)
Third match: Phailed against Ravi Jagadeesan, a seventh grade boy from Avery Coonley School :wallbash_red:  :wallbash_red:  :wallbash_red:
(DANG I was so close to beating him after tying after two questions but he slammed the buzzer like a fraction of a nanosecond before I tried to reach my buzzer and I collapsed like dead on the table)
Fourth match: Beat Thomas Wu, an 8th grade boy from Kennedy, in order to win the semifinals tiebreaker, resulting in me getting 3rd.

Oh by the way I was wrong.  Uni got 4th place and Edo Roth came in 9th place.

[quote="guineapiggies"][quote="gauss1181"]i thought u had a weird reaction when u saw me specifically? :maybe:[/quote]

me did? idk... lol. if you met a super short asian girl. then that's probably me XD but i was really out of it that day XD cuz i saw my worst nightmare (5/6th grade teacher)[/quote]
i saw you though....
i thought u were shocked by me??!??!?! :o
And it seemed like when I was looking out in the audience I was so nervous because I saw a super short asian girl looking at me like "yay gaussy"? :maybe:

Top 10 Things to know for Nationals:

:O DID YOU NOTICE THE LIST IS IN BASE 12 ahahaha... ... ...

[hide="Important Question 1"]

1. What is the last digit of $ 2^{2009}$ ? Of $ 3^{2009}$? etc.

[/hide]

[hide="Important Question 2"]

2. What are the last two digits of $ 2^{2009}$ ?

[/hide]

[hide="Important Question 3"]

3. What are the last three digits of $ 2^{2009}$ ?

[/hide]

[hide="Important Question 4"]

4. What are the last four digits of $ 2^{2009}$ ?

[/hide]

[hide="Important Question 5"]

5. What is the prime factorization of 2009?

[/hide]

[hide="Important Question 6"]

6. QUICK WHAT IS THE SUM OF THE FACTORS OF 2009?

[/hide]

[hide="Important Question 7"]

7. What is lcm(a, b) * gcd(a, b)? Also, generalize to three variables.

[/hide]

[hide="Important Question 8"]

8. How many ways are there to misspell MATHCOUNTS?

[/hide]

[hide="Important Question 9"]

9. What is $ \lfloor \sqrt{2009} \rfloor$ ? and ceilings. cube root may also be good.

[/hide]

[hide="Important Questions 10, 11, and 12"]

10. Where is the bathroom?

11. Where can you get (nontoxic) food? ("nontoxic" optional)

12. Where you left your room key.

[/hide]

lol maybe i did but as i said... sadly i didn't find out about this website until like a month after chapter. so i don't think i knew you as gaussy XD but yes it is confirmed i did see you. lol i talked to this other kid on the team and he's like 'yeah! (insert gaussy's name)" so i did in fact see you =] idk if i said anything XD hahaha i was kinda cheering for the whole CD thing cuz the highest place our school got was 11th XD hahaha

uhh idk. lol
uhh i super failed didn't make top 40 XD haha go me!!
lol. well i'm too old to compete this year so i really didn't try or need to XD... right now i'm working on my chinese skills... i have a competition on saturday...lol

2) 12
3) 512
4) 0512
second half of 7) abc/(gcd(a,b,c))^2 unless 2 o the #s share a factor not common to the third.  then you have to divide by that factor also.
Ex: a=45 b=30 c=35
45*30*35=47250
divide by gcd^2 (25) to get 1890
However, 45 and 30 share another factor (3), so you divide 1890/3=630

[quote="gauss1181"]What are your current thoughts about MC now?
Also what was your place at chapter 2008?
I'm pretty sure that if you stepped up your game with AoPS and not waste too much time on G&FF then you might have made state if you were to compete this year...lol[/quote]

I waste all my time in G&FF!

(Except that I usually also look at the problems in olympiad section)

Another mathcounts level problem:

Prove that for every prime p and integers n and d, n is congruent to a sum of d dth powers modulo p. If you can pass chapter you can solve this.

[size=9]I mean if you can make MOP you MIGHT be able to solve this. Source: Problem Gabriel Carroll proposed for MOP.[/size]


induction
graph theory
combinatorics unsolved
combinatorics

Given an integer $n\geq6$.Find the least integer $m(n)$s.t.for any graph$G$ with $n$vertexes and $m(n)$edges,there must exist two circles in $G$ without common vertexes.

I think the answer is $m(n)=3n-6$. Let my explain why I think that:

First of all, $m(6)=12$, the exemple is: draw a complete graph with $6$ vertexes, then if it is labeled ${v_{1},v_{2},...,v_{6}}$ on clockwise direction, erase the edges $(v_{1},v_{3}),(v_{3},v_{5})$ and $(v_{5},v_{1})$.

It is the only way we have to draw such a graph:
[i]
There are ${\left(\begin{array}{l}6\\3\\ \end{array}\right) = 20}$ triangles in a complete $6$-graph. Consider a triangle, so the vertexes left are also a triangle, but we must "kill" at least one of these triangles to get to one graph that satisfies the statement. But one edge erased "kills" at most four triangles, then we must erase at least three edges. Now there are only five possible ways to erase that three edges;
$(i)$they have no vertex in common
$(ii)$two of them have one vertex in common
$(iii)$they are like a "Y"
$(iv)$they are like a "N"
$(v)$they are a triangle

It is very easy to see that the only erasing way that works out is $(v)$, and we're done. [/i]

Now, by induction, we'll prove that:
$(i) m(n+1)=m(n)+3=3(n+1)-6$
$(ii)$ there are only one graph that satisfies it. The exemple with $n$ vertexes is:[i] three vertexes have degree $n-1$ and the others have degree $3$.[/i]
$(iii)$ in any $n$-graph that satisfies the statement, there exists a vertex with degree at most $3$.

First, to prove the initial case of $(iii)$, note that if a $6$-graph satisfies the statement and it has less than $12$ edges, otherwise is trivial, then the minimum degree $d$ is such that $6d < 24$, so $d \leq 3$.

OBS:[i]
If we prove that there exists a vertex with degree at most $4$ we prove $(i)$, because if we erase that vertex and it's edges, we'll have a $n$-graph with  $m(n+1)-4 \leq m(n)$ edges, but the equality can't hold, because when we redraw the vertex we will add more $4$ edges on that unique $n$-graph, which is impossible, so $m(n+1) \leq m(n)+3$ and follows the equality by the given exemple.[/i]

Consider a $n+1$-graph. Fix a vertex. So if we erase that vertex, we'll have a graph with $n$ vertexes and at most $m(n)$ edges, but following $(iii)$, it has a vertex with degree at most $3$, then when we redraw the first vertex, it will have degree at most $4$, and we are done. Then $m(n+1)=m(n)+3=3(n+1)-6$.

Lets prove the uniqueness. Again, consider a $n+1$-graph fix a vertex. Erase that vertex, and one may find a vertex with degree at most $3$, let look for the worst case, so it have degree $3$, and when we redraw the first vertex we'll find out that it actually has degree $4$. Consider the graph erasing that vertex, in the worst case we'll find another vertex with degree $4$. Let's do it untill we can. If that worst case doesn't hold everytime, we'll find a vertex with degree exactly $3$, and the uniqueness we be proved. If the worst case occurs, then each vertex has degree $4$. So $2m(n+1)=4(n+1)$ and $3(n+1)-6=2(n+1)$ which gives $n=5$, contradiction.

The proof of $(iii)$ follows from the proof above. So we finish the induction.

Great job.But a little mistake:m(n) should be 3n-5.Maybe my statement is not clear since my English is not good.

Sorry.... you are right... the statement is cleary... i didn't pay attention to this... I calculated the greatest integer such that it could be possible to find an $n$-graph with $m(n)$ edges satisfying the statement. So $M(n)$ you were looking for is $M(n) = m(n)+1 = 3n-5$.

Thank you!


Prove: A number is divisible by 3 iff the sum of its digits is divisible by 3.

[quote="SplashD"][quote="Astronum"][hide]$\text{Let}$ $abcd$ $\text{be a 4 digit number.}$

$abcd = 1000a+100b+10c+d = 999a+99b+9c+a+b+c+d$

$\text{Since}$ $999a+99b+9c$ $\text{is always divisible by 3,}$ $a+b+c+d$ $\text{must be divisible by 3 for}$ $abcd$ $\text{be divisible by 3.}$

$\text{QED}$[/hide][/quote]

That only proves it for a 4 digit number.[/quote]
But if we apply a very similar process to a $n$ digit number it can be proved.

[quote="fireemblem13"]that's right. try considering a general form for a number expressed in terms of the digits and powers of 10.[/quote]
I will try.  :lol:

[hide]$\text{Let}$ $abc...uv$ $\text{be a n digit number, where u and v are the last digits}$

$abc...uv =$
$=10^{n-1}a+10^{n-2}b... 10^{1}u+10^{0}v = (10^{n-1}-1)a+a+(10^{n-2}-1)b+b+...+(10-1)u+u+v$

$\text{Since the integers of the form}$ $10^{k}-1$ $\text{are always divisible by 3,}$ $a+b+...+u+v$ $\text{must be divisible by 3 for}$ $abc...uv$ $\text{be divisible by 3.}$

$\text{QED}$[/hide]

[hide]$d=\sum_{k=0}^{n}a_{k}10^{k}\implies s\left(d\right)=\sum_{k=0}^{n}a_{k}$

$d-s\left(d\right)=\sum_{k=0}^{n}a_{k}\left(10^{k}-1\right)\equiv 0\bmod 3\implies d\equiv s\left(d\right)\bmod 3$[/hide]


superior algebra
superior algebra unsolved

Suppose that in a ring (no unit) one has $x^3=x$ for all $x$. Find such an infinite ring and prove that all such rings are commutative.

For an example, take any infinite Boolean ring. If you really really want it not to have a unit, take the direct sum of infinitely many copies of $\mathbb F_2$ (the field with $2$ elements) with the obivous, pointwise operations.

First let's prove it for a ring $A$ with unit $1$ (and the property that $x^3=x,\ \forall x$).

In such a ring, $x^2=0\Rightarrow x=x^3=0$. Now, for any idempotent $t$ and any element $x$, we have $(xt-txt)^2=0\Rightarrow xt=txt$. Also, $(tx-txt)^2=0\Rightarrow tx=txt=xt$. This means that the set of idempotent elements is contained in the center $Z(A)$ (old trick :)). We have $(x^2)^2=x^2\Rightarrow x^2\in Z(A),\ \forall x\in A$. Also, from $(x+1)^3=x+1,x^3=3$ we get $3(x^2+x)=0,\ \forall x$. Since $(-x-x^2)^2=2(x+x^2)=-x-x^2,\ -x-x^2$ is idempotent and hence contained in $Z(A)$. Since $x^2\in Z(A)$ as well, we get $x\in Z(A),\ \forall x$, and we're done.

Now for the general case:

Let $a\in A$ be non-zero. Then $a^2$ is a non-zero idempotent element (which we’ll denote by $e$). It’s also central, since just as above, we can prove that $x^2\in Z(A),\ \forall x$. The ideal $(e)$ can be organized as a ring $\widetilde A$ with unit $e$, satisfying the requirement that $x^3=x,\ \forall x\in\widetilde A$. By the above, it’s commutative, i.e. $exy=eyx,\ \forall x,y\in A$. In particular, $eax=exa=xae,\ \forall x\in A$, but $e=a^2$, so $ax=xa,\ \forall x\in A$. Since $a\in A\setminus\{0\}$ was arbitrary, this finishes it.


function
AMC
AIME
algebra
domain
functional equation

I am not sure if this has been posted before,but it's not up on the contests page. Anyways, I'd like to see some solutions to this:
A certain function has the properties that $f(3x) = 3f(x)$ for all positive real values of x, and that $f(x) = 1 - |x - 2|$ for $1 \le x \le 3$. Find the smallest x for which $f(x) = f(2001)$

[hide]
from the functional equation, we see that $f(x) = 3^nf(\frac{x}{3^n})$ for positive integer $n$.\  we split the second property into two, basically $f(x) = 3-x$ for $2 \le x \le 3$, and $f(x) = x-1$ for $1 \le x < 2$. It is easy to see that $3^7 = 729$ and $\frac{2001}{729}$ will make $x$ fall in this domain.\  Therefore, to find the smallest $x$, we need $f(x) = 729f(\frac{2001}{729})$ for some $x$.\  We can see that $f(x) = 186$, and we see that we can obtain this x from some multiplying from the domain $2 \ge x \ge 1$.\  so $x-1 = \frac{186}{3^n}$, and $x = \frac{186+3^n}{3^n}$, this will fall in the interval for the least $n$ is $n = 5$, and therefore $3x = 243 + 186 = 429$.
[/hide]

Sorry if I didn't explain it well

:lol: didn't want to lose any readers, you see :D


[color=blue] [size=150]$2^{nd}$  Iberoamerican Olympiad[/size]
Salto y Paysandú, URUGUAY. [1987]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1.[/b] Hallar las funciones $f(x)$ tales que $f(x)^2 f \left ( \frac{1-x}{1+x} \right)=64x \ , x \neq 0, \pm 1.$[/quote]
Sea $x=\frac{1-y}{1+y},$ luego $\frac{1-x}{1+x}=y$ $\Longrightarrow$ $f \left (\frac{1-y}{1+y} \right)^2 \cdot  f(y)=64 \cdot \frac{1-y}{1+y}$ $\Longrightarrow$

$f \left (\frac{1-x}{1+x} \right)^2 \cdot f(x)=64\cdot \frac{1-x}{1+x}.$ Pero $f(x)^4f \left (\frac{1-x}{1+x} \right)^2=64^2 \cdot x^2$ $\Longrightarrow$

$f(x)^3=\frac{64(1+x)x^2}{1-x} \Longrightarrow f(x)= 4\sqrt[3]{\frac{(1+x)x^2}{1-x}}.$

[quote][b]Problema 2.[/b] En un triángulo $ \triangle ABC,$ $ M$ y $ N$ son los punos medios de $ AC,AB,$ respectivamente y $ P$ es el punto de intersección de $ BM$ y $ CN.$ Probar que si es posible inscribir una circunferencia en el cuadrilátero $ ANPM,$ entonces $ \triangle ABC$ es isósceles.[/quote]
$ ANPM$ es circunscriptible en una circunferencia $ \Longleftrightarrow$ $ AM + PN = AN + PM$

$ \Longrightarrow  AB + \frac {_2}{^3}BM = AC + \frac {_2}{^3}CN$

Si $ L$ es el punto medio de $ BC$ y $ Q$ el simétrico de $ P$ con respecto a $ L,$ se tiene

$ AB - AC = \frac {_2}{^3}(CN - BM) = QB - QC^$

Es decir que $ Q$ yace en el mismo brazo de la hiperbola $ \mathcal{H}$ con focos $ B,C$ que pasa por $ A.$ Por la simetría de la hiperbola con respecto a su centro $ L,$ toda recta que pase por $ L$ y la corte en $ A$ ha de cortarla por segunda vez en  su otro brazo. Entonces es contradictorio que $ LA$ corte dos veces el mismo brazo de  $ \mathcal{H},$ a menos que esta degenere en la mediatriz de $ BC$ $ \Longrightarrow$ $ \triangle ABC$ debe ser isósceles con ápice $ A.$

[quote][b]Problema 3.[/b] Si $ x,y,z$ son entero positivos y $1 \plus{} x \plus{} \sqrt {3}y \equal{} (2 \plus{} \sqrt {3})^{2z \minus{} 1},$ pruebe que $x$ es cuadrado perfecto.[/quote]
Si $(2\plus{}\sqrt3)^{2z\minus{}1}\equal{}(x\plus{}1)\plus{}y\sqrt 3,$ entonces $(2\minus{}\sqrt3)^{2z\minus{}1}\equal{}(x\plus{}1)\minus{}y\sqrt 3$ $\Longrightarrow$

$x\plus{}1\equal{}\frac 12(2\plus{}\sqrt3)^{2z\minus{}1}\plus{}(2\minus{}\sqrt3)^{2z\minus{}1})\equal{}\cosh((2z\minus{}1)2a).$

Donde $a>0$ y $\cosh(2a)\equal{}2.$ Luego $x\equal{}2\sinh((2z\minus{}1)a)^2,$ así hay que probar que $\sqrt 2\sinh((2p\minus{}1)a)$ es entero.

Se tiene $\cosh(2a)\equal{}2 \ , \ \sinh(2a)\equal{}\sqrt 3 \ , \ \sinh(a)\equal{}\frac 1{\sqrt 2} \ , \ \cosh(a)\equal{}\sqrt{\frac 32}$

Sean $a_p,b_p$ números reales reales, tales que $\sinh((2p\minus{}1)a)\equal{}\frac {a_p}{\sqrt 2}$ y $\cosh((2p\minus{}1)a)\equal{}b_p\sqrt{\frac 32}.$ Luego

$\sinh(2p\plus{}1)a)\equal{}\sinh((2p\minus{}1)a)\cosh(2a)\plus{}\cosh((2p\minus{}1)a)\sinh(2a)$  $\Longrightarrow$ $a_{p\plus{}1}\equal{}2a_p\plus{}3b_p$

$\cosh(2p\plus{}1)a)\equal{}\cosh((2p\minus{}1)a)\cosh(2a)\plus{}\sinh((2p\minus{}1)a)\sinh(2a)$ $\Longrightarrow$ $b_{p\plus{}1}\equal{}a_p\plus{}2b_p$

Como se tiene $a_1\equal{}1$ y $b_1\equal{}1,$ obtenemos que $a_p,b_p\in\mathbb N.$ Por consiguiente, $x\equal{}2\sinh((2z\minus{}1)a)^2\equal{}a_z^2$ es cuadrado perfecto.

[quote][b]Problema 4.[/b] Se define una sucesión $p_n$ de la siguiente forma: $p_1=2$ y para todo $n$ mayor o igual que $2,$ $p_n$ es el mayor divisor primo de la expresión $p_1 p_2 p_3 \ldots p_{n-1}+1.$ Pruebe que $p_n$ es diferente de $5.$[/quote]
$p_1=2$ y $p_2=3.$ Para todo $n>2,$ $p_1p_2 p_3 \ldots p_{n-1}+1$ no es múltiplo de $2$ ni es múltiplo de $3.$ Si el mayor divisor primo de $p_1 p_2p_3 \ldots p_{n-1}+1$ es $5,$ la expresión anterior es una potencia de $5.$ Por ende $p_1 p_2 p_3 \ldots p_{n-1}+1=5^k-1.$ Como $5^k-1$ es multiplo de $4$ para todo k entero, resulta $p_1p_2p_3 \ldots p_{n-1}+1$ multiplo de $4.$ Esto es absurdo, ya que $p_1=2$ y $p_1p_2 p_3 \ldots p_{n-1}$ es impar.

[quote][b]Problema 5.[/b] Sean $r,s,t$ raices de la ecuación $x(x-2)(3x-7)=2.$ Mostrar que $r,s,t$ son reales positivas y determine el valor de $\arctan r+\arctan s +\arctan t.$[/quote]
Escribimos $f(x) = x(x - 2)(3x - 7) - 2 = 3x^3 - 13x^2 + 14x - 2.$ Entonces $f (0) = -2, f (1) = 2,$ ésto indica que hay una raíz entre $0$ y $1.$ $f (2) = - 2,$ entonces hay otra raíz entre $1$ y $2.$ $f (3) = 4,$ así que la tercera raíz está entre $2$ y $3.$ $f (x) = 0$ tiene tres raíces, así que son todas soluciones reales y positivas. Por otro lado, se tiene la identidad

$\tan (x+y+z)=\frac{\tan x+ \tan y+ \tan z - \tan x \tan y \tan z}{1-( \tan x \tan y+ \tan y \tan z+ \tan z \tan x)}$

Luego sustitutendo $x=\arctan r , \ y=\arctan s , \ z=\arctan t,$ se tiene

$\tan (x+y+z)=\frac{r+s+t-rst}{1-(rs+st+tr)}=\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}=-1 \Longrightarrow x+y+z=-\frac{\pi}{4}+k\pi$

$\Longrightarrow x+y+z=\arctan r+\arctan s +\arctan t=\frac{3\pi}{4}.$

[quote] [b]Problema 6.[/b] $ABCD$ es un cuadrilátero convexo y $ P,Q$ son dos puntos en $ AD, BC$  tales que $ \frac {_{AP}}{^{PD}} \equal{} \frac {_{AB}}{^{CD}} \equal{} \frac {_{BQ}}{^{QC}}.$ Probar  que $ PQ$ forma igual ángulo con $ AB$ y $ CD.$ [/quote]
Las paralelas por $P$ a $DC,AB$ cortan a $AC,BD,$ respectivamente en $X,Y.$ Entonces $\tfrac{BQ}{QC}=\tfrac{AP}{PD}=\tfrac{AX}{XC}$ $\Longrightarrow$ $QX \parallel PY \parallel AB$ y analogamente $QY \parallel PX \parallel DC.$ Además $\tfrac{PX}{PY} \cdot \tfrac{AB}{DC}=\tfrac{BQ}{CQ}$ $\Longrightarrow$ $PX=PY,$ así el paralelogramo $PXQY$ es un rombo $\Longrightarrow$ $PQ$ es bisectriz de $\angle XPY,$ paralela pues a la bisectriz del ángulo $\angle (AB,CD)$ y por ende forma ángulos iguales con $AB$ y $CD.$


[color=darkblue]Let number sequence $ (u_n)$ so that $ u_1\equal{}u_2\equal{}1$ and $ u_{n\plus{}1}\equal{}u_n\plus{}u_{n\minus{}1}$, $ n\ge2$. Prove that $ \frac{ku_{n\plus{}2}\plus{}u_n}{ku_{n\plus{}3}\plus{}u_{n\plus{}1}}$ is a fraction in its lowest terms for $ \forall k\in N$.[/color]

It is fairly easy by induction on $ n$.  The base case implies $ \frac{2k\plus{}1}{3k\plus{}1}$ is in lowest terms which is true since $ (2k\plus{}1,3k\plus{}1)\equal{}(2k\plus{}1,k)\equal{}1$ since $ 2k\plus{}1$ leaves remainder $ 1$ when divided by $ k$.

Suppose it was true that $ (ku_{n\plus{}2}\plus{}u_n,ku_{n\plus{}3}\plus{}u_{n\plus{}1})\equal{}1$.  Then $ (k(u_{n\plus{}2}\plus{}u_{n\plus{}3})\plus{}(u_n\plus{}u_{n\plus{}1}),ku_{n\plus{}3}\plus{}u_{n\plus{}1})\equal{}1$ which from $ u_{n\plus{}1}\equal{}u_n\plus{}u_{n\minus{}1}$ implies that $ (ku_{n\plus{}4}\plus{}u_{n\plus{}2},ku_{n\plus{}3}\plus{}u_{n\plus{}1})\equal{}1$ which is the claim for the next value of $ n$.


Here is my updated list

Chapter Round

2000 Sprint Round Only Answers
2001 Sprint Round  Only Answers
2001 Target Round No Answers or Solutions
2001 Team Round No Answers or solutions
2002 Sprint Round With Solutions and answers
2002 Target Round With Solutions and answers
2002 Team Round With Solutions and answers
2003 Sprint Round With Solutions and answers
2003 Target Round With Solutions and answers
2003 Team Round With Solutions and asnwers
2004 Sprint Round With Solutions and answers
2004 Target Round With Solutions and answers
2004 Team Round With Solutions and answers
2005 Sprint Round With Solutions and answers
2005 Target Round With Solutions and answers
2006 Sprint Round With Solutions and answers
2006 Target Round With Solutions and answers
2006 Team Round With Solutions and answers
2007 Sprint Round With Solutions and answers
2007 Target Round With Solutions and answers
2007 Team Round With Solutions and answers
2008 Sprint Round With Solutions and answers
2008 Target Round With Solutions and answers
2008 Team Round With Solutions and answers
2009 Sprint Round With Solutions and answers
2009 Target Round With Solutions and answers
2009 Team Round With Solutions and answers
2009 Countdown Round Only Answers
2008 Countdown Round Only Answers
2007 Countdown Round Only Answers
2006 Countdown Round Only Answers
2005 Countdown Round Only Answers
2004 Countdown Round Only Answers
2003 Countdown Round Only Answers
2002 Countdown Round Only Answers
2001 Countdown Round No Answers

State Round

2001 Sprint Round Only Answers
2001 Target Round Only Answers
2001 Team Round Only Answers
2002 Sprint Round With Solutions and answers
2002 Target Round With Solutions and answers
2002 Team Round With Solutions and answers
2003 Sprint Round With Solutions and answers
2003 Target Round With Solutions and answers
2003 Team Round With Solutions and answers
2004 Sprint Round With Solutions and answers
2004 Target Round With Solutions and answers
2004 Team Round With Solutions and answers
2005 Sprint Round With Solutions and answers
2005 Target Round With Solutions and answers
2006 Sprint Round With Solutions and answers
2006 Target Round With Solutions and answers
2006 Team Round With Solutions and answers
2007 Sprint Round With Solutions and answers
2007 Target Round With Solutions and answers
2007 Team Round With Solutions and answers
2008 Sprint Round With Solutions and answers
2008 Target Round With Solutions and answers
2008 Team Round With Solutions and answers
2009 Sprint Round With Solutions and answers
2009 Target Round With Solutions and answers
2009 Team Round With Solutions and answers
2009 Countdown Round Only Answers
2008 Countdown Round Only Answers
2007 Countdown Round Only Answers
2006 Countdown Round Only Answers
2005 Countdown Round Only Answers
2004 Countdown Round Only Answers
2003 Countdown Round Only Answers
2002 Countdown Round Only Answers
2001 Countdown Round Only Answers Maybe

School

2006 Sprint Round With Solutions and answers
2006 Target Round With Solutions and answers
2006 Team Round With Solutions and answers
2006 Countdown Round Only answers
2007 Sprint Round With Solutions and answers
2007 Target Round With Solutions and answers
2007 Team Round With Solutions and answers
2007 Countdown Round Only answers
2008 Sprint Round With Solutions and answers
2008 Target Round With Solutions and answers
2008 Team Round With Solutions and answers
2008 Countdown Round Only Answers
2009 Sprint Round With Solutions and answers
2009 Target Round With Solutions and answers
2009 Team Round With Solutions and answers
2009 Countdown With Solutions and Answers
1999 Sprint Round No Answers
1999 Target Round No Answers
1999 Team Round No Answers
1999 Countdown Round No Answers

Please email me @ dwade884@gmail.com if you are intersted in trading with me.....

Thanks.

Mods, please move this to MATHCOUNTS RESOURCES HERE!! thread. Thanks.


Let's make a spin-off of Jeff Foxworthy's famous routine.  This time, it's about video game addicts.

If you go to the zoo and start covering the hedgehogs with blue paint...

you might be a video game addict.

If you're playing golf, you grab the flag, jump into the cart, and scream "Floor it!"

you might be a Halo addict

If you try to buy a Wii with Rupees

You might be a Video Game Addict

if you starve to death because you though you already fed your character on WOW...then you might be a world of warfract addict

If you take a stick and try to make a tree explode,

you might be a star wars addict.

If you name your son Flavius The Handsome, you might be a Rome: Total War addict.

if you throw your backpack at someone because they are hurt, you might be a bf2 addict

If you walk into people's houses, break their pots, and steal whatever is inside, you might be a Zelda addict.

If you have ever seen a row of 10 boxes and wondered why they were still there, you might be a Tetris addict.

if you play video games a lot you might be a video game addict

If you jump on all of the turtles in the zoo, you might be a Mario addict.

If you try to run a loop-the-loop, you might be a Sonic the Hedgehog addict.

If you see a robot, and you shoot it, you might be a Terminator addict.

If you try to swing from house to house on a string, you might be a Spider Man addict.

If you do a bunch of housework for quarters, you might be an arcade game addict.(or more specifically, a pinball addict)

If you go to college, and you are sitting in front of a computer, and you say "How do I attack and defend?", you might be a video game addict.

If the only two Italian names you know are Mario and Luigi....
you might be a video game addict.  :rotfl:

if you have the Halo release date tatooed on your arm
u might be a halo addict

if you buy a PS3 jsut to turn it into a grill u might be a george foreman addict

if u buy Viva Pinata for Xbox 360 just for acheivement points y ou might be an Xbox live addict

haha if you buy Crackdown just to get into Halo 3 Beta you might be a halo addict

If you jack your dad's car and keys, get in the car, and race laps, all the while accidently crashing along the edges, you might be a race car game addict

[quote="jli"]if you spend 400 dollars on windows vista to get halo 2 on PC, you might be a halo addict (btw, halo wars looks awesome)[/quote]

agreed haha halo wars looks awesome except that it's a strategy game on the 360 wtf? oh well...

hey runescape...yeah you're right runescape does kinda suck (although i luv pwning noobs in the wildy =D)

If your grenade button twitches when seeing large crowds you might be a first person shooter addict.

[quote="ra5249"]If a guy with a gun comes up to you and you shout: "Stop! You can't hurt me, you aren't in the wilderness!" you might be a runescape addict.  :rotfl:[/quote]

haha thats great

If you see a mushroom, and you jump on it, you might be a Mario addict.

If you put a frog on the road, and run to the other side to see if it could cross safely, you might be a video game addict.

If you spell Dear d-e-e-r, you might be a hunting game addict.

If someone sticks a gun into your ribs, and you laugh, you might be dreaming.(sorry, the pun was irresistible)

if you push your sister onto the floor and stomp on her head, you might be a gears of war addict

if you throw your toy cars and action figures at street gangs you might be a crackdown addict

if u try too hard to b funny by posting stuff about video game addicts, u might b in denial   (It aint just a river in Egypt u no)

If you remember complex cheat codes, but not your own phone number...

you might be a video game addict.

[quote="mathnerd314"][quote="7h3.D3m0n.117"][quote="mathnerd314"]If you use Age of Empires screenshots for your World History class...[/quote]

actually they work extremely well for history projects...i got an A on mine  :D [/quote]

Seriously?! :rotfl:[/quote]

of course...we had a history day project and i think my partner or something used a very nice image taken and modded from Age of Empires to creat some pictures for our slideshow

if you're rank in the top five of PKers an an mmorpg

when you play paintball and hit someone in the back with your weapon (halo)

when you belive you can breath underwater when you wear a blue polo shirt, and walk through fire with a red one (zelda)

when you belive you can cast fire bolt while holding white round rocks with symbols on them (runescape)

[quote="jli"][quote="7h3.D3m0n.117"][quote="mathnerd314"]If you use Age of Empires screenshots for your World History class...[/quote]

actually they work extremely well for history projects...i got an A on mine  :D

if you camp out at best buy 3 weeks before the release of Halo 3 you mite be a halo addict[/quote]

the halo 3 beta came out on xbox 360 btw[/quote]

i was talking about the ACTUAL game...not the beta...

and btw...if you are planning and trying to save up to $\$$130 to buy the legendary edition, you mite be a halo addict

if u try to drift in a supermarket cart u might be a fast and furious tokyo drift addict

if you manage to respawn you might be addicted to video games

if you knock out your friend and help him back up again, you might be a bf2 addict


inequalities
logarithms
inequalities open

I hope everyone remembers the inequality proposed by Vasile Cartoajein GMA: if sum x^3=3, then sum (xy)^4<=3. So, it follows that the maximum of the sum (xy)^4/3 is attained when x,y,z are equal, supposing that x+y+z=1. Now, you all remember a problem from Mathlinks contest: if a,b,c>0 add up to 1, then sum (ab)^5/4<1/4. Deduce that 1/4 is not the best bound and find that bound ( I know the solution to this one).
   SEcond question (I don't have a solution):
  Find the greatest p such that the maximum of sum (xy)^p (considered when x+y+z=1) is attained when x=y=z. All I know is that p>=4/3 and that p<=2. Can you solve it? I really believe this is a very difficult problem.

It is also my opinion, but I do not think that is easy to prove that.

[quote="Myth"]I think, p \leq 0.5*ln(3)/ln(1.5) = 1,3547.[/quote]
I really have no idea for this. Can you tell me the reason that you think this number the result :?  :?  ?

It is condition that set (1/2,1/2,0) is not optimal, i.e. (1/4)p \leq (1/9)p+(1/9)p+(1/9)p.

[quote="Ji Chen"][quote="harazi"]Find the greatest $p$ such that the maximum of $\sum y^{p}z^{p}$ (considered when $x+y+z=1$) is attained when $x=y=z$. I really believe this is a very difficult problem.[/quote]
[quote="Myth"]I think, $p \leq\frac{\ln{3}}{\ln{9}-\ln{4}}= 1.3547\cdots$.[/quote]
It's true.

[b]Theorem[/b] If $x,y,z$ be nonnegative real numbers, then the following inequality
\[\sqrt{\frac{yz+zx+xy}{3}}\leq\sqrt[p]{\frac{x^{p}+y^{p}+z^{p}}{3}}\]
holds if and only if $p\geq \frac{\ln{9}-\ln{4}}{\ln{3}}=0.73814049285708512580094\cdots$.

This theorem was established in our paper:

Shichang Shi, Ji Chen, Separation of the power mean by the elementary symmetric mean and its applications (Chinese), Journal of [url=http://www.cdu.edu.cn/]Chengdu University[/url] (Natural Science), Vol. 14 (1995), No. 2, pp. 2-8.

( ISSN 1004 - 5422 )[/quote]

Do you have solution in english?

I also posted before in Mathlinks the similar one for $4$ number
\[(abc)^{k}+(bcd)^{k}+(cda)^{k}+(dab)^{k}\le \max(4,(\frac{4}{3})^{3k}).\]
This can be done by SMV.

The initial of harazi and Jichen can be done easily by classical mixing variable.

Alternatively, the general for $n$ numbers is also proposed and solve in my book by UMV (AC) theorem. It was published in Vietnam only.

[quote="hungkhtn"]I also posted before in Mathlinks the similar one for $4$ number
\[(abc)^{k}+(bcd)^{k}+(cda)^{k}+(dab)^{k}\le \max(4,(\frac{4}{3})^{3k}). \]
This can be done by SMV.

The initial of harazi and Jichen can be done easily by classical mixing variable.

Alternatively, the general for $n$ numbers is also proposed and solve in my book by UMV (AC) theorem. It was published in Vietnam only.[/quote]

I want buy your book. What muss i do? I will pay by viza.

[quote="Ji Chen"][quote="harazi"]Find the greatest $p$ such that the maximum of $\sum y^{p}z^{p}$ (considered when $x+y+z=1$) is attained when $x=y=z$. I really believe this is a very difficult problem.[/quote]
[quote="Myth"]I think, $p \leq\frac{\ln{3}}{\ln{9}-\ln{4}}= 1.3547\cdots$.[/quote]
It's true.

[b]Theorem[/b] If $x,y,z$ be nonnegative real numbers, then the following inequality
\[\sqrt{\frac{yz+zx+xy}{3}}\leq\sqrt[p]{\frac{x^{p}+y^{p}+z^{p}}{3}}\]
holds if and only if $p\geq \frac{\ln{9}-\ln{4}}{\ln{3}}=0.73814049285708512580094\cdots$.

This theorem was established in our paper:

Shichang Shi, Ji Chen, Separation of the power mean by the elementary symmetric mean and its applications (Chinese), Journal of [url=http://www.cdu.edu.cn/]Chengdu University[/url] (Natural Science), Vol. 14 (1995), No. 2, pp. 2-8.

( ISSN 1004 - 5422 )[/quote]

And here is my weighted problem :

[color=blue][b]Theorem. [/b]If $a,b,c$ are non-negative real numbers such that $a+b+c=3$ and $\lambda$ is a real number in $[0;1]$, then the following inequality holds
\[(ab)^{k}+(bc)^{k}+(ca)^{k}+\lambda (abc)^{k}\le \max \left(3+\lambda; \frac{3^{2k}}{2^{2k}}\right) \]
for all positive real number $k$.[/color]

I think it's very hard to solve.

I can only prove for $\lambda \in [0;1]$. How about for $\lambda \in R$, or $\lambda \ge 0$ ?
Can you help me, Ji Chen ?  :P

[color=blue][b]Open question.[/b] Find all real values of $\lambda$ such that the following inequality holds for all positive real number $k$ and for all non-negative real numbers $a,b,c$ satisfying $a+b+c=3$ :
\[(ab)^{k}+(bc)^{k}+(ca)^{k}+\lambda (abc)^{k}\le \max \left(3+\lambda; \frac{3^{2k}}{2^{2k}}\right) \][/color]


$ y\equal{}4\minus{}x^2$

How would you go about sketching $ y\equal{}f(|(x)|)$

What does this actually mean? $ y\equal{}4\minus{}|x^2|$??

I assume your first statement should read $ f(x)\equal{}4\minus{}x^2$.
In that case, $ y\equal{}f(|x|)$ means $ y\equal{}4\minus{}|x|^2$.
Which obviously doesn't make too much of a difference.


A fair, twenty-faced die has 19 of its faces numbered from 1 through 19 and has one blank face. Another fair, twenty-faced die has 19 of its faces numbered from 1 through 8 and 10 through 20 and has one blank face. When the two dice are rolled, what is the probability that the sum of the two numbers facing up will be 24? Express your answer as a common fraction.

I'm probably wrong but is it $ \frac{2}{25}$ ?

No it isn't $ \frac{2}{25}$

Assuming you roll the second dice first(it doesn't make a difference), you can roll any face other than the ones numbered $ 1, 2, 3, 4,$ and [b]blank[/b]. So you have $ 15$ out of $ 20$ faces you can roll and still get a sum of $ 24$. Then, there is exactly $ 1$ corresponding face on the other dice that when added to the number you just rolled sums to $ 24$.
Probability = $ \frac{15}{20}\times \frac{1}{20} \equal{} \boxed{\frac{3}{80}}$


Hey can anyone give me some usefull Geometry formulas because i am practicing for mu alpha theta

Tanks

Just use the search function, that will really help you....

[quote="shybox"]I sorry im new so can you tell me what that is and if anyone could tell me good ggeometry website that would be great

tanks[/quote]

Just look to the top of your screen where the various buttons of private messages,logout,etc are there.There is a search button there as well.
Besides you got look up formulas on [url=http://mathworld.wolfram.com/]mathworld[/url]

Well, shadysaysurspammed has already answered.......


abstract algebra
integration
trigonometry
calculus
Euler
function
real analysis

Let $L_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}|D_{n}(t)|dt \qquad (n=1,2,3\ldots)$, where $D_{n}(t)=\sum_{k=-n}^{n}e^{ikx}$. Prove that the sequence $\{ L_{n}-\frac{4}{\pi^2}log\,n\}$ is bounded.

I can't get a hold of the right technique. Can anyone please help me?

-Laura

Find a closed form of $D_n(t)$. If you graph it, it goes up and down between 1/x. So we should expect it to do what you're supposed to prove. I don't have time to do more right now (and probably shouldn't if this is your homework) but most likely you can finish it off with some clever approximation of the kernel that you can integrate.

For the record it has nothing to do with school. I've been trying to do it for quite a long time now, but just can't seem to find the right way.  The "clever approximation" is the thing I can't come up with. :?

Okay, it isn't as obvious as I thought, I must admit. But here's what you need to do:

Put the Dirichlet kernel in closed form, with sines (as I suggested before). Then split it into a sum of integrals over small intervals on which you know the sign of $D_N$. Then assume that $N$ is large enough that $\sin(t)$ is constant on that interval, and pull that bit out of the integral. Then you can integrate the bit that's left, and you should come out with a summation of a harmonic series (including some constant like $\frac{4}{\pi^2}$). This is well known to behave like log (modulo the Euler-Mascheroni constant, $\gamma = 0.577...$ (that's all I remember off the top of my head)). So it does what you want.

If you really want it to be rigorous you also have to show that the error you introduce by approximating sine is negligible, which I was too lazy to check but it is certainly true. You'd just subtract your approximation from the actual thing and do some other clever approximation and you should be able to make it look like $\frac{1}{N}$ times something bounded. That's what I'd expect at least.

My first analysis professor always told us that we needed to be crude (with approximations) in order to write elegant analysis proofs.

$D_n(x)=\frac{sin(n+\frac{1}{2})x}{sin(\frac{1}{2}x)}$.
Notice that it is easier to study $E_n(x)=\frac{sin(n+\frac{1}{2})x}{\frac{1}{2}x}$ and that
$D_n=E_n+f(x)$ where $\int_0^{\pi} |f_n(x)|dx < C$ where $C$ is a constant that does not depends of $n$.
Hence if one proves that $\int_{-\pi}^{+\pi} |E_n(x)|dx - \frac{4}{{\pi}^2}Ln(n)$ is bounded, conclusion will follows.

Because $|E_n|$ is an even function one can restrict the study of the integral to $[0; \pi]$.
There is a little problem in zero, so one cuts the integral near zeo:
$\int_0^{\pi} |E_n(x)|dx=\int_0^{\frac{\pi}{n+\frac{1}{2}}} |E_n(x)|dx+\int_{\frac{\pi}{n+\frac{1}{2}}}^{\pi} |E_n(x)|dx=A_n+B_n$.

1/Prove that $|A_n|<A$ for a certain constant $A$ independant of $n$

2/$B_n=\int_{\pi}^{(n+\frac{1}{2}) \pi} \frac{|sin(t)|}{t}dt$, and the mean value of $|sin(t)|$ is $\frac{2}{\pi}$.
So you can try to study $\int_{\pi}^{(n+\frac{1}{2}) \pi} \frac{|sin(t)|-\frac{2}{\pi}}{t}=\int_{\pi}^{(n+\frac{1}{2}) \pi} \frac{g(t)}{t}$
where $G(x)=\int_{\pi}^{x} g(t)dt$ is a bounded function of $x$. Integrate by part and the result will follow (I think ).

Oh yeah -- I forgot to mention, with my method you also have to do the interval around 0 separately. But it should work out okay. Thanks alekk for a different estimation method -- seems to work just as well, and it is always good to see multiple ways of doing something.

Phew I finally got it done. I did it more or less xevarion style. In case someones interested it turned out  that the total cumulative error introduced by approximating  $sin(t/2)$ on the integration intervals $[\frac{(k-1)\pi}{n+1/2},\frac{k\pi}{n+1/2}]$  stays bounded but doesn't become negligible as $n$ tends to infinity (basically what alekk pointed out). But considering the objective of the problem that is just as good.

Thank you very much Xevarion and Alekk for your help!!! Especially without xevarion's quick response I wouldn't have had the courage to keep trying.

-Laura


A number is randomly selected from the one-digit prime numbers. What is the probability that its square has a units digit of 9? Express your answer as a common fraction.

One digit primes are 2, 3, 5, 7.
$ 3^2\equal{}9$ and $ 7^2\equal{}49$

Probability is $ \frac{2}{4}\equal{}\boxed{\frac{1}{2}}$


[quote="Slizzel"]Eliminate $\theta$ from the following relations: $\sin\theta+\cos\theta=a$ and $\sin^5{\theta}+\cos^5{\theta}=b$.[/quote]

$a^2  = 1 + 2\sin \theta \cos \theta  \Rightarrow \sin \theta \cos \theta  = \frac{{a^2  - 1}}{2}$

Now, $\sin^5{\theta}+\cos^5{\theta}=b$

$\Rightarrow a(\sin ^4 \theta  - \sin ^3 \theta \cos \theta  + \sin ^2 \theta \cos ^2 \theta  - \sin \theta \cos ^3 \theta  + \cos ^4 \theta ) = b$

Combining first, third and fifth terms inside the bracket, we can write

$a\left[(\sin ^2 \theta  + \cos ^2 \theta )^2  - \sin ^2 \theta \cos ^2 \theta  - \sin \theta \cos \theta (\sin ^2 \theta  + \cos ^2 \theta )\right] = b$

$\Rightarrow a\left[ {1 - (\frac{{a^2  - 1}}{2})^2  - \frac{{a^2  - 1}}{2}} \right] = b$

Hence $\theta$ is eliminated.


geometry
3D geometry
dodecahedron
geometric transformation
rotation
combinatorics unsolved
combinatorics

How many way to use the least number of color to color the Dodecahedral such that adjacent faces have different colors.
[img]http://www.liv.ac.uk/~spmr02/calendar/dod.gif[/img]

I think four is the number. Obviously we cannot colour it with less than four, since any face and the five faces around it require four colours.

Suppose that we use the colours A, B, C and D. Colour one face A, and then colour the five faces around it B, C, B, C, D (in clockwise order, say).

Now consider the five faces touching these five. If we number the face touching the D and the B 1, and then proceed in clockwise order, we get face 2 touching a B and a C, face 3 touching a C and a C, face 4 touching a B and a C, and face 5 touching a C and a D. We assign these the colours 1-A, 2-D, 3-A, 4-D and 5-B. Finally, we colour the last face C.

Sorry if this is a bit hard to read, it's tricky to give a clear description, but I'm pretty sure this solution works.

I think the OP was looking for the number of ways to color the dodecahedron with four colors.

Label a face $A$, and let's say we color it with color $a$.  Let $B,C,D,E,F$ be the faces adjacent to $A$, labelled in the clockwise direction.  Let's say they are colored, respectively, $b,c,d,c,d$.  Any other coloring is equivalent to this coloring by rotation about $A$ and permutation of the colors.  Now, label $A',B',C',D',E',F'$ to be the faces opposite $A,B,C,D,E,F$ respectively.  We see that $F',B',C'$ must be colored $b,a,b$;  for $D'$ we have a choice between $a$ and $c$, and for $E'$ we have a choice between $a$ and $d$.  We can't choose $c$ and $d$, for that would leave no color left for $A'$.  So we must choose $a$ for one, and the remaining two faces are automatically colored.

Now we can count the number of possibilities:  there are 4 choices for coloring $A$, 5 choices for which face adjacent to $A$ gets the unique color, 3 choices to pick that color, 2 ways to color the remaining four faces, and 2 ways to finish up the coloring for the rest of the dodecahedron.  This gives 240 colorings.  If we only want colorings distinct under rotations of the dodecahedron, we use the Cauchy-Frobenius theorem.  The rotational group of the dodecahedron is $A_{5}$;  every member of that group besides the identity switches two adjacent faces, so there is no coloring fixed by any of those elements. So the number of colorings is $240/60 = 4$.

I think there should be 2 colorings under arbitrary symmetries of the dodecahedron, but I'm not sure exactly what the orientation-reversing symmetries are.


how do u access the solution on the resource page?

Clicking on the number next to the problem should bring you to a topic in which solution(s) are included.


limit
algebra
polynomial
logarithms
real analysis
real analysis unsolved

You can study the more general problem:

Let $ P(x)\equal{}a_{n}x^{n}\plus{}\cdots\plus{}a_{1}x\plus{}a_{0}$,    ($ a_{n}>0$),  be a polynomial of degree $ n\geq 1$.

Find the limit:

$ \lim\limits_{x\rightarrow \infty}P(x)\left(\left(1\plus{}\frac{1}{P(x)}\right)^{P(x)}\minus{}e\right)$

That's not much of a generalization. As long as $ P(x)\to\infty,$ the substitution $ u\equal{}P(x)$ will simply return this to being the original problem.

$ \ln\left(1\plus{}\frac1{x}\right)\equal{}x\minus{}\frac1{2x^2}\plus{}O(x^{\minus{}3})$

$ x\ln\left(1\plus{}\frac1{x}\right)\equal{}1\minus{}\frac1{2x}\plus{}O(x^{\minus{}2}).$

Exponentiate to get that

$ \left(1\plus{}\frac1{x}\right)^{x}\equal{}
e\cdot\left(1\minus{}\frac1{2x}\plus{}O(x^{\minus{}2})\right)$

So $ \left(1\plus{}\frac1{x}\right)^{x}\minus{}e\equal{}\minus{}\frac{e}{2x}\plus{}O(x^{\minus{}2})$

Finally, $ x\left(\left(1\plus{}\frac1{x}\right)^{x}\minus{}e\right)\equal{}
\minus{}\frac{e}2\plus{}O(x^{\minus{}1})$

The limit is $ \minus{}\frac{e}2.$