Let $ G=(V,E)$ be a graph on $ n$ vertices. Prove that there exists a subset $ F \subseteq E$ such that:
1) $ |F|\le n-1$
2) In the graph $ G-F$ (meaning, $ G$ with the edges in $ F$ removed) all of the vertices' degrees are even.

Color all the edges of the graph blue.  Choose a path joining two vertices of odd blue-degree and color all the edges of the path red.  Repeat the previous step until there are no vertices of odd blue-degree.  Now, among the red edges, if any cycle exists, color the edges of the cycle blue.  Repeat the previous step until there are no red cycles.  At the end of this process, the red edges form a forest, so there are at most $ n-1$ of them, while every vertex has even blue-degree.  Thus, the red edges are exactly our desired set $ F$.

$ n-1$ is achieved in the case that $ G$ is a tree.


1.Find max $y=4sin3x-4cos2x-5sinx+5$
2. For any x , we have $f(x) = 1+a_{1}cosx+a_{2}cos2x+a_{3}cos3x \geq 0$.
Find $P_{max}=a_{1}+a_{2}+a_{3}$

I will post solution of problem 1.
Because $\sin{3x}=3\sin{x}-4\sin^{3}{x}, \cos{2x}=1-2\sin^{2}{x}$, we have $y=-16\sin^{3}{x}+8\sin^{2}{x}+7\sin{x}+1$, Let $\sin{x}=t,-1\leq t \leq 1$. Now, we have to find max of $f(t)=-16t^{3}+8t^{2}+7t+1,t\in [-1,1]$, but this is easy.  :wink:


inequalities AMC USA(J)MO USAMO function vector trigonometry

Are there going to be inequalities(pure inequalities) on the USAMO because for some reason I heard a rumor that they took it off the USAMO, but then there was one on the IMO?

I heard they took them off because they were too easy to dumbass. Of course, that's been said about the IMO for several years now, so who knows.

I don't know for certain, but I seriously doubt inequalities are gone.
That said, ever since a certain IMO question I think people are less willing to put a clever variation of Cauchy/Holder on a test than they are something which requires some sort of novel step. Accordingly, 3 variable cyclic / symmetric inequalities are probably going to be a lot less common than they once were (with respect to their n-variable and acyclic counterparts). Its possible that there won't ever be a 3 variable inequality on the USAMO again.
Of course, inequalities in some form will always be around, like the easy part of #4 last year.

Don't blame me if I'm making stuff up though.

Inequalities are a fundamental to the algebra component of an olympiad.  Although some are quite easy to dumbass (IMO 2005 #3),there are certainly very beautiful and hard inequalities (which are difficult or impossible to just expand+am-gm+schur).

Even some 3 variable cyclic inequalities may be difficult to dumbass, although there are certain clever ways to expand products.

Therefore, I think beautiful inequalities will always feature onto the USAMO and the IMO.

[quote="fourierseries"]there are certainly very beautiful and hard inequalities (which are difficult or impossible to just expand+am-gm+schur).[/quote]

What is Schur?

The Vornicu-Schur Inequality states that for $f$ a increasing function on $[,\infty)$ or $f$ a positive convex function on the interval containing $a,b,c$, \[f\left(a\right)\left(a-b\right)\left(a-c\right)+f\left(b\right)\left(b-c\right)\left(b-a\right)+f\left(c\right)\left(c-a\right)\left(c-b\right)\geq 0\]

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=381983#381983]here.[/url][/url]

[quote="solafidefarms"]The Vornicu-Schur Inequality states that for $f$ a increasing function on $[,\infty)$ [/quote]

What is that interval supposed to be? It looks incomplete.

[quote="13375P34K43V312"]Can't one dumbass any inequality by multiplying out and doing tedious manipulations to reduce an inequality to like, $\sum a_{k}^{2}\ge0$?[/quote]
Not any inequality, however a lot can be dealt in that way.
Try proving AM-GM for n=3 that way :P

[quote="solafidefarms"]The Vornicu-Schur Inequality states that for $f$ a increasing function on $[,\infty)$ or $f$ a positive convex function on the interval containing $a,b,c$,
\[f\left(a\right)\left(a-b\right)\left(a-c\right)+f\left(b\right)\left(b-c\right)\left(b-a\right)+f\left(c\right)\left(c-a\right)\left(c-b\right)\geq 0 \]
See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=381983#381983]here.[/url][/url][/quote]

This is actual stronger than the usual version of Schur, which is just
\[x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geq 0 \]

I personally love the good ol' 3 variable cyclic inequalities, as there is a small, defined toolbox which you can always pull out. Of course, a thorough understanding of each tool is vital; but I simply liked being able to see what is availability for me to use, and what I need to try to do.

But maybe that's just me. And, of course, unfortunately, there are ways to just dumbass almost every pretty inequality ever dreamt up on a shortlist... So, in that respect, I guess if the USAMO coordinators did want to prevent dumbassing and/or easy points, the inequalities can be fairly removed. Oh well.

But I don't know for sure if these "rumors" are true. I sure hope they aren't  :( .

[quote="solafidefarms"]The Vornicu-Schur Inequality states that for $f$ a increasing function on $[,\infty)$ or $f$ a positive convex function on the interval containing $a,b,c$,
\[f\left(a\right)\left(a-b\right)\left(a-c\right)+f\left(b\right)\left(b-c\right)\left(b-a\right)+f\left(c\right)\left(c-a\right)\left(c-b\right)\geq 0 \]
See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=381983#381983]here.[/url][/url][/quote]

Actually, the conditions are that $f$ is a non-negative function that is monotonic or convex.  And it can be defined for any subset of the reals.  Orthodox Schur is this result for $f= x^{r}$, where $x,r$ are nonnegative.  It is important that $f$ be non-negative, e.g., consider the increasing function $f(x) =-\frac{1}{x}$.

And I heard a rumor that there [i]would[/i] be an inequality on this USAMO.  By the way, has anybody read the dumb solution for USAMO 2004/5?  It was [i]dumb[/i].  And then there were about six to eight variations on the elegant solution you could have used.

I don't see why inequalities should be taken off the USAMO. If you just put radicals into a 3-variable ineq, then you cannot bash it.
I like inequalities as some of them have very elegant solutions and require quite a lot of creativity. I would be happy if there was an ineq on USAMO this year.

Can somebody explain the Cauchy-Schrwz inequality to me. I looked it up but I didn't get it.

Basically,
$(a_{1}+\dots+a_{n})^{2}(b_{1}+\dots+b_{n})^{2}\geq a_{1}^{2}b_{1}^{2}+\dots+a_{n}^{2}b_{b}^{2}$

where $a_{1}\dots a_{n}$ and $b_{1}\dotsb_{n}$ are a sequence of nonnegative real numbers.

Note:Cauchy-Schwartz is just a special case of the Holder Inequality.

One can also derive variations by utilizing the cross product in this formulation as well...

$\left(\vec{a}\circ \vec{b}\right)^{2}+\left(\vec{a}\times \vec{b}\right) \circ \left(\vec{a}\times \vec{b}\right) = \left|\vec{a}\right|^{2}|\vec{b}|^{2}$

Expanding the above and removing some terms gives some powerful inequalities.  (and equalities :lol:)

whoops switched them  :blush: ...for some reason i cant edit?

Its that new 24-hour rule. if you want to edit your post, you have to edit it in 24 hours, I believe.

I hope there is an inequality on the usamo. While yes, they're generally a subset of the easier usamo questions, it provides easy points for people who don't know a ton (i.e. me). The #3s and 6s that are inequalities are generally very hard to dumbass, anyway. I'd be okay if they had them on there. I don't see why they should be taken out? They're definitely a part of olympiad math. They're like the definition of olympiad.

[quote="t0rajir0u"]Er... not quite.  The actual statement of Cauchy-Schwarz is given by

$(a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2}) \ge (a_{1}b_{1}+...+a_{n}b_{n})^{2}$

Where $a_{1}, ... a_{n}, b_{1}, ... b_{n}$ are sequences of real (not necessarily non-negative) numbers.  In vector notation, given $\vec{A}= \left< a_{1}, ... a_{n}\right> \vec{B}= \left< b_{1}, ... b_{n}\right>$ this simply becomes

$| \vec{A}| | \vec{B}| \ge \vec{A}\cdot \vec{B}= | \vec{A}\parallel \vec{B}| \cos \theta$

Where $\theta$ is the angle between the vectors.[/quote]

actually cos is defined analytically to be the ratio $\frac{\vec{A}\cdot \vec{B}}{ | \vec{A}\parallel \vec{B}|}$, and you can do this, because cauchy guarantees it to be be <=1


calculus
integration
logarithms
real analysis
real analysis unsolved


Let O(n) be the number of ways of expressing the integer n as a sum of odd positive integers.
Show that O(8) = 6.
Let D(n) be the number of ways of expressing the integer n as a sum of distinct positive integers.
Show that D(9) = 8.
Evaluate O(n) and D(n) for n = 1, 2, ..., 10.

Consider the infinite polynomial Q(x) = (1+x)(1+x^2)(1+x^3)(1+x^4)...
Write down the different ways of getting x^7 from the polynomial Q(x).
Explain why it is no coincidence that the number of ways it can be done is the same as the value of D(7)
and show that Q(x) = 1 + D(1)x + D(2)x^2 + D(3) x^3 + ...
Prove that P(x) = Q(x)  and so O(n) = D(n) for all n.

Surely someone can do this before it falls off the bottom of the page. Is it just too easy to be interesting?


[hide="Voila"]Using the restrictions on $x_{2}$, $x_{3}$, and $x_{4}$:

$60 < x_{2}+x_{3}< 69$

Breaking this down by cases and using the restrictions on each variable's value:

Case 1:
$x_{2}+x_{3}= 60$

This can happen in only one way ($x_{2}= 4$ and $x_{3}= 56$)

Thus, $x_{1}+x_{4}= 9$, which can happen in 9 ways.

Case 1 can happen in $1*9 = 9$ ways.

Case 2:
$x_{2}+x_{3}= 61$

This can happen in two ways.

Thus, $x_{1}+x_{4}= 8$, which can happen in 8 ways.

Case 2 can happen in $2*8 = 16$ ways.

It is easy to verify that this pattern continues all the way through $x_{2}+x_{3}= 68$, so there are 9 cases. So, the number of solutions is the sum of the number of solutions for each case, or:

$\sum_{i=1}^{9}i(10-i) = 165$ solutions.[/hide]

[hide="my solution"]Since they are integers, $x_{1}\ge0$, $x_{2}\ge4$, $x_{3}\ge56$, $x_{4}\ge1$. Then the minimum of $x_{1}+x_{2}+x_{3}+x_{4}$ is clearly $4+56+1=61$. Thus, the equation with restrictions is equivalent to $a+b+c+d=8$. By balls and urns, this is just ${11\choose3}=165$.[/hide]

well the answers are the same..so i dont know what you mean by different..

Oh, I thought he got something different, never mind.

[quote="Altheman"]if we write it like (x_1)+(x_2-4)+(x_3-56)+(x_4-1)=69-4-56-1

then we can just use the ball and urn thing because each parenthesis is a nonegative integer[/quote]
That's what I did.


probability
invariant
real analysis
integration
limit
real analysis unsolved

Let $(X, \mathcal B)$ be a measurable space, $f: X \to X$ a measurable function. Suppose that $\mu_{1}$ and $\mu_{2}$ are $2$ distinct probability measures, $f$-invariant, $f$-ergodic. Show that there are mutually singular.

Write $\mu_{2}=\mu+\nu$, with $\mu\ll \mu_{1}$ ($\mu$ is absolutely continuous wrt $\mu_{1}$), and $\nu\perp\mu_{1}$ ($\nu$ is singular wrt $\mu_{1}$). This is possible by the Lebesgue Decomposition Theorem.

We can find sets $U,V\in\mathcal B$ such that $U\cap V=\emptyset,\ U\cup V=X$, and $\mu_{1},\nu$ are concentrated on $U,V$ respectively. Because $\mu_{1}(V)=0$, the $\mu_{1}$ measure of $\widetilde V=\bigcup_{n\ge 0}f^{-n}V$ is also zero. This, in turn, implies that $\mu(\widetilde V)=0$, so $\mu_{2}(\widetilde V)=\nu(\widetilde V)\ (*)$.

If $\nu\ne 0$, then $\nu(V)>0\Rightarrow\mu_{2}(V)>0$. By the $f$-ergodicity of $\mu_{2}$, we get $\mu_{2}(\widetilde V)=1$, so $\nu(\widetilde V)=1$. This means that the whole weight of $\mu_{2}$ is carried by $\nu$, i.e. $\mu=0$, and $\mu_{2}\perp\mu_{1}$, as desired.

The other case is $\nu=0$, i.e. $\mu_{2}\ll\mu_{1}$. By Radon-Nikodym, we can find $h\in\mathcal L^{1}(\mu_{1})$ such that $\mbox d\mu_{2}=h~\mbox d\mu_{1}$. Because $\mu_{2}$ is $f$-invariant, for all $E\in\mathcal B$ we have $\int h\chi_{f^{-1}E}~\mbox d\mu_{1}=\int h\chi_{E}~\mbox d\mu_{1}$. By Birkhoff's Ergodic Theorem, $\frac{1}{n}\sum_{i=0}^{n-1}\chi_{f^{-i}E}$ converges $\mu_{1}$-a.e. to $\mu_{1}(E)$, so $\mu_{2}(E)=\lim_{n}\frac{1}{n}\sum_{i=0}^{n-1}\int h\chi_{f^{-i}E}~\mbox d\mu_{1}$ (because all the terms of the average are equal to $\mu_{2}(E)$), which, in turn, is equal to $\mu_{1}(E)\int h~\mbox d\mu_{1}=\mu_{1}(E)$. This proves that $\mu_{2}=\mu_{1}$.

i don't understand something : $\widetilde V$ is not $f$-invariant. So how can one conclude that $\mu_{2}( \widetilde V) = 1$ if $\mu_{2}(\widetilde V) >0$ ?

[quote="alekk"]$\widetilde V$ is not $f$-invariant.[/quote]

Yes it is, not in the strict sense that $f^{-1}(\widetilde V)=\widetilde V$, but in the sense that $\mu_{2}(f^{-1}(\widetilde V)\Delta\widetilde V)=0$ ($\Delta$ is the symmetric difference, but we could just as well have written $\widetilde V\setminus f^{-1}(\widetilde V)$ instead, since $f^{-1}(\widetilde V)\subseteq\widetilde V$). I'll let you figure out why this is so.


There was a site that used to be posted on the forum. It had a lot of old state/chapter tests. Anyone have a link?

goodlymaths.0catch.com ? But be warned it has on-0page unblockable popups...and tons of advertising...spyware...things like that. It's a high price to pay for them.

I have all of the stuff from that site archived.

Try using HTTrack Website copier.

Really useful.

It has spyware? :( mathfanatic should get a new server!

[quote="math92"]this is the only one i know:
http://www.yashar.biz/mathcounts/[/quote]

tHis one has back to 2001. Is that what you mean by old?

There is also the mathcounts saab page, of equivilant MC level math, at mathcounts.saab.org/mc.cgi.

[quote="math92"]this is the only one i know:
http://www.yashar.biz/mathcounts/[/quote]

I bookmarked that site, and later on I tried to access it. It prompted for a password, and when I didn't specify one, it gave me an error message in Spanish. What went wrong?

Goodlymaths has spyware?! Why didn't anyone tell me?! Noooo... But thankfully I have an antispyware program :)

Farazy's site's been down for quite some time now... unfortunately, yashar.biz doesn't work anymore. Talk to Farazy about getting it back up... :)

Hmm...
Yashar.biz is still working for me
I clicked the link and the site came up and I was able to download the tests and stuff

I normally just try to google them and normally i think someone has put them in a pdf file. If not, I think your coach can go to the coach's forum and ask someone else for it. I beleive they have it.

Oh, I was wondering why I had some spyware on my computer...
Anyway, after goodlymaths, I have all that stuff.  If you need it, PM me.


Let $ a,b,c$ be real numbers. Prove the inequality

$ \frac {(a^2 \plus{} 1)(b^2 \plus{} 1)}{(c^2\plus{}1)(a\minus{}b)^2}\plus{} \frac {(b^2 \plus{} 1)(c^2 \plus{} 1)}{(a^2 \plus{} 1)(b \minus{} c)^2} \plus{} \frac {(c^2 \plus{} 1)(a^2 \plus{} 1)}{(b^2 \plus{} 1)(c \minus{} a)^2}\geq 2$

[hide="approach"]    :idea:

$ \sum \frac{(a^2\plus{}1)(b^2\plus{}1)}{(c^2\plus{}1)(a\minus{}b)^2}\geq 2$

$ \iff$  $ \sum (\frac{(a^2\plus{}1)(b^2\plus{}1)}{a\minus{}b})^2\geq 2\prod (a^2\plus{}1)$

$ \iff$  $ \sum (\frac{(a\minus{}b)^2\plus{}(ab\plus{}1)^2}{a\minus{}b})^2 \geq 2\prod (a^2\plus{}1)$

$ \sum (\frac{(a\minus{}b)^2\plus{}(ab\plus{}1)^2}{a\minus{}b})^2 \geq \sum \frac{2(a\minus{}b)(ab\plus{}1)}{a\minus{}b})^2\equal{}4\sum (ab\plus{}1)^2$

and i think it is easy to prove that : $ 4\sum (ab\plus{}1)^2 \geq \prod (a^2\plus{}1)$  :ninja:   [/hide]

[quote="mehdi cherif"]

and i think it is easy to prove that : $ 4\sum (ab \plus{} 1)^2 \geq \prod (a^2 \plus{} 1)$  :ninja:   [/quote]

Really?  :maybe:

After some work, I don't think that inequality is true (or if it is, it's hard)

Can anybody come up with a solution?

I can solve it!!! :D
Well, this is a very interesting problem and it is solved by using Cauchy- Schwartz ( believe or not  :) !)
Frankly, I assure that it is a true problem and actually, I wonder where you took this problem from. Anyway, here is the full proof of the problem! Hopefully, everyone will enjoy it! :)
[color=olive][u]Proof:[/u][/color]
By using Cauchy- Schwartz, we have
$ \frac {(a^2 \plus{} 1)(b^2 \plus{} 1)}{(c^2\plus{}1)(a\minus{}b)^2}\plus{} \frac {(b^2 \plus{} 1)(c^2 \plus{} 1)}{(a^2 \plus{} 1)(b \minus{} c)^2} \geq 2\frac{b^2\plus{}1}{|(b\minus{}a)(b\minus{}c)|}$
Likewise, we have the similar inequalities like above.
It suffices to prove that
$ \frac{b^2\plus{}1}{|(b\minus{}a)(b\minus{}c)|}\plus{} \frac{a^2\plus{}1}{|(a\minus{}b)(a\minus{}c)|}\plus{} \frac{c^2\plus{}1}{|(c\minus{}a)(c\minus{}b)|} \geq 1$
However, we have the identity and an absolute inequality:
1/ $ \left|\frac{b^2\plus{}1}{(b\minus{}a)(b\minus{}c)}\plus{} \frac{a^2\plus{}1}{(a\minus{}b)(a\minus{}c)}\plus{} \frac{c^2\plus{}1}{(c\minus{}a)(c\minus{}b)}\right|\equal{}1$

2/ $ |a|\plus{}|b|\geq|a\plus{}b|$

That is all clear! :)

What can I say, great math?

This is wonderful!  :)

Problem is from "Old and new inequalities" and the solution given there is same as yours.


Let $n$ be a natural number greater or equal to $3$, $a_1,a_2,\ldots,a_n$ natural numbers different from $0$, $a_k<n$ for all $k \in \{1,2,\ldots,n\}$ and $\gcd(a_k,n)=1$, for all $k \in \{1,2,\ldots,n\}$.

Prove that $\gcd(a_1+a_2+ \cdots +a_n, n )\neq 1$.

On the contrary, then $\gcd(a_1+a_2+\cdots +a_n,n) = \gcd(n,n) = n \neq 1$.
A counterexample is e.g. $n=3$, $a_1=a_2=1$, $a_3=2$, when $\gcd(a_1+a_2+a_3,3) = \gcd(4,3) = 1$.


A six-question test has three true/false questions and three multiple choice questions with four choices each. If Ojeni guesses at all the answers, what is the probability that she has all answers correct? Express your answer as a common fraction.

[hide] Just multiply probablitity of each-

1/2*1/2*1/2*1/4*1/4*1/4= 1/512[/hide]


I saw these problems in my book today and they blew my mind see how u do with them

prove that sin squared x (1+ cot squaredx)=1
prove that tan x (cotx+tanx)=sec squared x

just distribute and use trigidentities.
note that $tanx=\frac{sinx}{cosx}$ and $cotx=\frac{cosx}{sinx}$


For  a,b,c >0 .
        Prove that
                               $ \sum_{cyc} \frac {a^4}{a^3\plus{}b^3} \geq \sum_{cyc} \frac {a}{ 2}$

hint : if it's right
$ \sum a^3b \geq \sum a^2b^2$

Your ineq is wrong ,Hong Quy ^^.Try :$ a\equal{}0;b\equal{}1;c\equal{}2$
The above ineq(of thanhak5) was old :D

a, b, c > 0.

[hide="Hint"]Hong Quy's inequality is true by Muirhead or weighted AM-GM.[/hide]

[quote="thanhak5"]For  a,b,c >0 .
        Prove that
                               $ \sum_{cyc} \frac {a^4}{a^3 \plus{} b^3} \geq \sum_{cyc} \frac {a}{ 2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113

[quote="thanhak5"]1. For $ a,b,c > 0$ . Prove that
        $ \frac {a^2 \plus{} 1}{\sqrt {2a^{2} \plus{} 3ab \minus{} c^2}} \plus{} \frac {b^2 \plus{} 1}{\sqrt {2b^{2} \plus{} 3ab \minus{} a^2}} \plus{} \frac {c^2 \plus{} 1}{\sqrt {2c^{2} \plus{} 3ac \minus{} b^2}}\geq 2(a \plus{} b \plus{} c)$[/quote]
Will the expression in the root always exist? what happens if we let $ a\equal{}2,b\equal{}1,c\equal{}1$?

[quote="can_hang2007"][quote="thanhak5"]1. For $ a,b,c > \frac {1}{\sqrt {6}}$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
. Prove that
        $ \frac {a^2 \plus{} 1}{\sqrt {2a^{2} \plus{} 3ab \minus{} c^2}} \plus{} \frac {b^2 \plus{} 1}{\sqrt {2b^{2} \plus{} 3ab \minus{} a^2}} \plus{} \frac {c^2 \plus{} 1}{\sqrt {2c^{2} \plus{} 3ac \minus{} b^2}}\geq 2(a \plus{} b \plus{} c)$[/quote]
Will the expression in the root always exist? what happens if we let $ a \equal{} 2,b \equal{} 1,c \equal{} 1$?[/quote]


geometry
circumcircle
radical axis
geometry unsolved

Let $\Omega$ be the circumcircle of a triangle $A_1A_2A_3$, and $\omega_1$, $\omega_2$, $\omega_3$ be the corresponding excircles. Let $B_1$ be a radical center of the circles $\Omega$, $\omega_2$ and $\omega_3$; $B_2$ be a radical center of the circles $\Omega$, $\omega_3$ and $\omega_1$; $B_3$ be a radical center of the circles $\Omega$, $\omega_1$ and $\omega_2$. Prove that the lines $A_1B_1$, $A_2B_2$ and $A_3B_3$ are concurrent.

Please see: http://www.acm.org.ve/
My paper posted 2004 in section Entrenamientos:
Concurrencias y colinealidades
http://www.acm.org.ve/files/imovzlajcsfinal.pdf
Sorry for the spanish.


a and b are  positive integers and  a<b.
İf x,y element of [a,b] closed interval, 1/x+1/y element of [a,b].
Find a and b integers.

Nice problem! You set x=y=a and get a<=2/a<=b. If you set x=y=b you get a<=2/b<=b. From these inequalities you get a^2<2 and 2<=ab<=2. So you have a=1, b=2. The function 1/x+1/y increases if x or y decreases, so it has its maximum for x=y=1 and the minimum for x=y=2. So the interval [1,2] fits all the conditions of the problem and is the only such interal.

Misha


[quote="MCrawford"]What is the range of the following set of scores?
\[ \{43, 7, 65, 5, 19, 71, 3, 59, 79, 29\} \][/quote]

[hide]
Range=highest value-lowest value
highest value=79
lowest value=3
79-3=76
The range of the set of scores is 76. :)
[/hide]

[hide]highest: 79
lowest: 3
Range: 79-3 = 76[/hide]

I like those answers. ;)

and DarkKnight, please show your works

[hide]The range is:
h-l=r
where h = highest and l = lowest.
79-3=76[/hide]


Acá adjunto algunos problemas que no me quedan claros para resolver.

Determinar dominio y recorrido (codominio) de la función $f(x)=(x-1)^{2}$.

Hallar dominio y codominio de la función $g(x) = \left\{ \begin{gathered}-4{\text{ si }}x <-2 \hfill \\-1{\text{ si }}-2 \leqslant x \leqslant 2 \hfill \\ 3{\text{ si }}2 < x \hfill \\ \end{gathered}\right.$

Hallar dominio y recorrido de la función $h(x)=[[x+2]]$

Sea la función $f(x)=|4-x^{2}|$. Defina la función a trozos sin emplear las barras de valor absoluto.

Gracias.

Creo que esto no corresponde a esta sección... que dirán los moderadores?

$Tipe$

Verdad...

Debiese estar en el sector [url=http://www.mathlinks.ro/Forum/index.php?f=288]Matemáticas de preparatoria[/url]

Aún así espero lo trasladen.

A todo esto, aprovecha de ayudarme con los problemas por favor.

Saludos.

[color=blue]Estamos moviendo este topic, de manera temporal a esta seccion,
puesto que no se ajusta al contenido de la Seccion Olimpiadas de Mate,

Saludos,

Carlos Bravo :)
Lima - peru[/color]

Para $f$ es claro que su dominio es $\mathbb{R}$, por otro lado, sea $y$ un elemento del rango tenemos que $y=(x-1)^{2}\geq 0$, recíprocamente, si $y\geq 0$ vamos a demostrar que está en el rango, para esto basta tomar   $x=1+\sqrt{y}$. Luego el rango de $f$ es $[0,+\infty>$.

Por la definición de $g$ su dominio es $\mathbb{R}$, su rango es $\{-4,-1, 3\}$

Observación: Rango = Codominio

$Tipe$


superior algebra
superior algebra theorems

At most $3$ files on any post?  :( This is file p_4.

[quote="N.T.TUAN"]Hi all,
I have some questions on this example, please see on file p_4

1st red: What is $A_{4}$? is it group of all even permutations on $\{1,2,3,4\}$?
2nd red : Prove this assertion :)
3th red: I don't understand? What do you mean?  :maybe:
Thanks for all help of you.[/quote]

1st red: Yes

1nd red: Note that if G be a nonabelian group of order 12 and G has not element of order 6 then G isomorphic to A_4.

3th red: $U(R_{3}) \cong Q \rtimes \mathbb{Z}_{3}$ (Q is the quaternion group)


calculus
integration
logarithms
symmetry
calculus computations

Well, just do it in polar coordinates. (And it's a double integral, not a triple integral.)

$ \int_0^1\int_0^{2\pi}r^2\sin(r^2)r\,d\theta\,dr\equal{}4\pi\int_0^1r^3\ln r\,dr\equal{}\minus{}\frac{\pi}4.$

[quote="bookworm_vn"]Calculate
\[ \iint\limits_{x^2 \plus{} y^2 \leqslant 1} {\left( {x^2 \plus{} y^2 } \right)\ln \left( {x^2 \plus{} y^2 } \right) dx dy} .
\]
[/quote]
since it is a simple double integral, what i will do is this:
let $ x^2\equal{}u,\;\;\;y^2\equal{}v$

then, by symmetry and dirichlet,
$ I\equal{}4\;\frac{1}{4}\iint\limits_{0<u\plus{}v \leqslant 1} \left( {u\plus{}v } \right)\;\ln \left( {u\plus{}v } \right) \;u^{\minus{}1/2}\;v^{\minus{}1/2}\;du\;dv$

$ \equal{}\dfrac{\Gamma^2\left(\frac{1}{2}\right)}{\Gamma(1)}\int_0^1\;w\;\ln\,w\;dw$

$ \equal{}\pi\left(\minus{}\frac{1}{4}\right)\;\equal{}\;\minus{}\frac{\pi}{4}$


May someone please tell me what I need to get in order to use the \LaTeX presentations system? I just heard about it. Also, can you explain how to use it and how to install it? Thanks a lot for your help. :?

We have a whole tutorial:
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

Try this link : http://www.miwie.org/presentations/
SSN.


4. One day I found a receipt from my grandfather's store. The bill was faded so that I couldn't read the price of the gadgetrs at all, and I could read only 3 of the 5 digits in the total. How many dollars did one gadget cost? Express your answer as a decimal to two decimal places.
The bill showed the follwing:

72 gadgets at $ \$$_._ _ each =$ \$$_74.5_.

[hide="Answer"]7.98 dollars[/hide]

Since there's a 72 on the left, the right must be divisible by 8 and 9

[hide]Well, _74.5_ must be divisible by 8. Since 456 is divisible by 8, that means that the last digit is a 6. The sum of all the digits must be a multiple of 9. 7+4+5+6+X=22+X.  So X=5. 574.56/72=7.98[/hide]

Just remember to divide that by 72 :wink:


[i]Παιδιά , ας μην ανοιγόμαστε σε πολλά μέτωπα (topics), διότι δεν θα βγαίνει άκρη σε λίγο. Να γεμίσουμε με ανισότητες  αυτό το χώρο , ώστε να ανατρέχουμε εύκολα και να τις βρίσκουμε[/i].

[b]ΑΣΚΗΣΗ  - 11[/b]

Αν $ x , y , z > 0$  και $ x \plus{} y \plus{} z \equal{} 1$, να βρείτε τη μέγιστη τιμή της παράστασης :

  $ A \equal{} xy \plus{} yz \plus{} zx \minus{} 3 xyz$

  Μπάμπης

Ta $x,y,z$ den mporoun na paroun tin timi $0$? Rwtaw dioti simfwna me mia akompsi methodo vrika to elaxisto na einai to $\frac{1}{4}$ p.x. gia $x=0$, $y=z=\frac{1}{2}$.

Alexandros

μαλλον μπορει.δοκιμασες  μηπως με συναρτησεις πολλων μεταβλητων ?

Oxi Kwsta... Dokimasa me poli apla mesa (triwnimo ktl). Den empleksa to thewrima tou Lagrange gia euresi akrotatwn. Apla einai ligo akompso ti stigmi kata tin opoia tha iparxei sigoura enas wraios tropos me xrisi gnwstwn anisotitwn pou tha vriskei to akrotato.

An pantws mporoun na paroun tin timi $0$ kai i elaxisti timi einai auti pou egrapsa, mporw na ekthesw ton tropo pou skeftika.

Alexandros

Ωραία άσκηση .... :)

ΑΣΚΗΣΗ 11-ΛΥΣΗ
[hide]Νάτη για σένα [/hide]

Έστω $f(x,y,z)=xy+yz+zx-3xyz$ . Θα χρησιμοποιήσω τη μέθοδο πολλαπλασιατών Lagrange .
Έστω $g(x,y,z)=x+y+z=1$
Θα λύσουμε το σύστημα  $\nabla f=m\nabla g$  .
και τότε παραγωγίζοντας μερικά έχουμε
$\nabla f=\ <y+z-3yz,\ z+x-3zx,\ x+y-3xy>$   και
$\nabla g=\ <1,\ 1,\ 1>$

Άρα $m=y+z-3yz= z+x-3zx= x+y-3xy$  από όπου βρίσκουμε ότι
η $f$ παρουσιάζει ακρότατο (μέγιστο ή ελάχιστο ) για $x=y$ ή $z=\frac{1}{3}$

Η $f$ πράγματι παρουσιάζει τέτοιο ακρότατο αφού είναι συνεχής και σε κλειστό διάστημα αφού $x,y,z\in\left[0,1\right]$

Αν $x=y$ τότε $2x+z=1$  και άρα $f(x,y,z)=f(x)=6x^{3}-6x^{2}+2x$ ,  $x\in\left[0,\frac{1}{2}\right]$  αφού x=y το χ δεν μπορεί να είναι μεγαλύτερο του $\frac{1}{2}$

Τότε $f'(x)=18x^{2}-12x+2=(x-\frac{1}{3})^{2}\geq 0$
άρα $f$ αύξουσα άρα
$f_{max}(x,y,z)=f_{max}(x)=f(\frac{1}{2})=f(\frac{1}{2},\frac{1}{2},0)=\frac{1}{4}$

Τώρα αν $z=\frac{1}{3}$ τότε $x+y=\frac{2}{3}$
και άρα $f(x,y,z)=f(x)=\frac{2}{9}$

Τελικά αφού $\frac{2}{9}<\frac{1}{4}$
το μέγιστο της παράστασης είναι $\frac{1}{4}$
και η ισότητα επιτυγχάνεται όταν $x=y=\frac{1}{2}$ και $z=0$

Ap'oti fainetai mporei na parei tin timi $0$ opote ekthetw ki egw ton tropo mou. Mou fainetai omws ligo akompsos. An o kos Stergiou exei diaforetiki lisi me xrisi gnwstwn anisotitwn as mas ksestravwsei meta tis 2 diaforetikes liseis pou exoume dwsei.

Afou $z=1-x-y$, h parastasi $A$ me aples prakseis ginetai

$A=(3x-1)y^{2}+(3x^{2}-4x+1)y-x(x-1)$ i opoia gia na exei lisi ws pros $y$, tha prepei

$\Delta\geq 0 \ x\leq \frac{1}{3}$ (afou etsi ki alliws $x\leq 1$)

Etsi loipon i parastasi $A$ einai ena triwnimo ws pros $y$ tou opoiou o sintelestis tou $y^{2}$ einai arnitikos ara parousiazei megisto

gia $y_{0}=-\frac{b}{2a}=\frac{1-x}{2}$ to $A(y_{0})=-\frac{\Delta}{4a}=-\frac{(x-1)(3x^{2}+1)}{4}$. Afou etsi ki alliws i megisti timi eksartatai apo to $x$, tha dialeksoume mia timi gia to $x$, gia tin opoia i megisti timi $A(y_{0})$ tha megalwsei akomi perissotero. Omws i $A(y_{0})$ einai einai mia gnisiws fthinousa sinartisi ws pros $x$ gia $0\leq x\leq\frac{1}{3}$ i opoia profanws parousiazei tin megisti timi tis gia $x_{0}=0$. Opote apo ekei $y_{0}=\frac{1-x}{2}=\frac{1}{2}=z_{0}$ ki etsi i megisti timi einai to $\frac{1}{4}$.

Alexandros

Ευχαριστώ Κώστα . να μια elementary λύση
ΑΣΚΗΣΗ 11-2ος τροπος

Λόγω συμμετρίας υποθέτουμε $x\geq y\geq z$ .
Τότε $z\leq \frac{1}{3}$ άρα $1-3z\geq 0$ .
Είναι $xy\leq\frac{(x+y)^{2}}{4}=\frac{(1-z)^{2}}{4}$
Άρα αν $A=xy(1-3z)+z(x+y)$ είναι η παράστα'σή μας τότε
$A\leq (1-3z)\frac{(1-z)^{2}}{4}+z(1-z)$

Τώρα πλεόν με τριώνυμο ή $f(z)$ εύκολη υπόθεση .... :wink:

και εγώ έτσι την έλυσα αλλά δεν ήξερα ότι το [b]x [/b]μπορεί να πάρει τιμή μηδέν....

[quote="stergiu"][i]Παιδιά , ας μην ανοιγόμαστε σε πολλά μέτωπα (topics), διότι δεν θα βγαίνει άκρη σε λίγο. Να γεμίσουμε με ανισότητες  αυτό το χώρο , ώστε να ανατρέχουμε εύκολα και να τις βρίσκουμε[/i].

[b]ΑΣΚΗΣΗ  - 11[/b]

Αν $x , y , z \geq 0$  και $x+y+z = 1$, να βρείτε τη μέγιστη τιμή της παράστασης :

  $A= xy+yz+zx-3 xyz$

  Μπάμπης[/quote]

 [color=red]Αγαπητοί φίλοι , SORRY ! [/color]

Να θεωρήσετε ότι  $x ,y , z \geq 0$ και όχι
     $x ,y , z > 0$ , όπως αρχικά έχει γραφεί. Δυστυχώς , έτσι ήταν γραμμένο και στην πηγή της άσκησης , όπου ήταν άλυτη. Όταν την πρότεινα εδώ , δεν την είχα λύση ο ίδιος για να διαπιστώσω την παράλειψη. Μου φάνηκε ότι δε θα είχε δυσκολία. Δυστυχώς ξεγελάστηκα , αφού η υπόθεση δεν ήταν σωστή, με την έννοια ότι με θετικά τα $x ,y , z$ δεν υπάρχει μέγιστο αλλά supremum.

 Μπάμπης

[quote="fotis"]Σε αυτό το topic βάζουμε μόνο αλγεβρικές ανισότητες;[/quote]

Νομίζω όχι βάλε ότιδήποτε ανισότητα θες .Εκτός αν έχει σχέση με ολοκληρώματα που υπάρχει ειδικό τόπικ για αυτά....

[quote="mostel"][b]ΑΣΚΗΣΗ 12[/b]

Για $xyz=1$ και $x,y,z\geq0$, να βρεθεί η μικρότερη τιμή της παράστασης:

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$[/quote]
Stelio ise siguros oti iaskisi ine etsi akrivos..?
p.x to $x,y,z\geq 0$dn m aresi,gt an itan kapios $0$ tote pos 8a ixan ginomeno $1$?
episis i sin8iki $xyz=1$,xriazete?
gt me BCS  vgeni eykola oti i parastasi ine $\geq \frac{3}{2}$ aneksartita apo to $xyz=1$...
den ksero....mpori na kano+la8os...ala nomizo oti i moni sin8iki pu xriazete ine ta $x,y,z$na ine 8etiki pragmatiki..

 :D

Kalhmerizw kai euxomai kales Apokries,anoiksi tou triwdiou gar ki edw stin Patra ginetai xamos.Brika kati kales anisotites-katalipseis exoumew opote apofasisa na asxolithw ligo kai me kati allo...

Check...

Let x, y, z be non-negative real numbers such that x + y + z = 3,
prove that

x2+y2+z2+xyz>=4

blog[1+(a/b)]>=dlog[1+(c/d)] δεν συνεπάγεται πάντα log[1+(a/b)]>=log[1+(c/d)] (με b>=d) εδώ είναι το λάθος μου,sorry :blush:

Fis/man δεν ισχυει η ανισωση  :maybe:

Όταν προσπαθούμε να αποδείξουμε κάτι φτάνοντας σε κάτι που ισχύει χρησιμοποιηούμε συνεπαγωγή που μας δίνει την δυνατότητα να πάμε και προς τα πίσω..
Βέβαια πρέπει να προσέχουμε(και εδώ είναι το λάθος σου..)ότι δεν μπορούμε να την χρησιμοποιήσουμε πάντοτε...

Το γεγονός ότι καταλήγεις να αποδείξεις ότι (a/c)>=1 δεν σημαίνει πως άμα αρχίσεις από αυτή την παραδοχή μπορείς να αποδείξεις ότι (a/c)>=(b/d)>=1 που το χρειάζεσαι για να αποδείξεις την ανίσωση.Δηλαδή δεν ισχύει ότι αν (a/c)>=1 τότε αναγκαστικά έχουμε ότι (a/c)>=(b/d)>=1....

Bασικά το λάθος δεν ήταν ακριβώς εκεί.Το λάθος εντοπίζεται σε προηγούμενο στάδιο αν και είναι το είδος του λάθους που επισημαίνεται στο προηγούμενο post.Ξεκίνησα από την προς απόδειξη (και με δοσμένο το b>=d) έφτασα (a/c)>=(b/d).Όμως (b/d)>=1 (από υπόθεση) άρα (a/c)>=1 ==> a>=c ,που ισχύει.Το λάθος ήταν ότι απ' το blog[1+(a/b)]>=dlog[1+(c/d)] συμπέρανα βιαστικά ότι log[1+(a/b)]>=log[1+(c/d)] αφού b>=d,κάτι που δεν ισχύει πάντα,φυσικά  :blush:
Αυτό που θέλω να ρωτήσω (αν και σε κάποιους θα φανεί αστείο),το έχω ,όμως, απορία πολύ καιρό:μπορώ να χρησιμοποιήσω την προς απόδειξη και μέρος των δοσμένων για να φτάσω σε κάτι που ισχύει.Μόνο όταν διατηρείται η συνεπαγωγή,έ;
Ευχαριστώ πάντως για τις απαντήσεις/υποδείξεις/διορθώσεις (γι αυτό σας υποσχομαι μέχρι το καλοκαίρι να καθίσω να μάθω Latex-τώρα βαριέμαι :rotfl: ) :)

Proti skeyi pu mu r8e ine oti
[hide]i akolu8ia $x_{b}=(1+\frac{a}{b})^{b}$ me $a$ sta8ero ..ine gnisios auksusa (+btw tini sto $e^{a}$ otan to $b$ tini sto apiro...)[/hide]
e...+ara[hide]
$(1+\frac{a}{b})^{b}\geq (1+\frac{a}{d})^{d}\geq (1+\frac{c}{d})^{d}$[/hide]

 :D

γιατι να μπορουμε να υποθεσουμε το $a$ σταθερο?

[color=red]ASKHSH 26[/color]

Ean oi $x,y,z$ einai thetikoi pragmatikoi na apodeiksete oti:

$\frac{x^{3}}{yz}+\frac{y^{3}}{xz}+\frac{z^{3}}{xy}\geq x+y+z$.

Pote isxiei i isotita?

Alexandros

[quote="mostel"][b]ΑΣΚΗΣΗ 19[/b]

Να αποδειχθεί ότι για τους $x,y,z>0$, ισχύει:

$x+y+z\leq\frac{x^{3}}{yz}+\frac{y^{3}}{xz}+\frac{z^{3}}{xy}$[/quote]

H Άσκηση ξαναπροτάθηκε  :D ......

Kosta ..dn 8eorisa stin lisi mu to $a$ os sta8ero..
pi8anon na mnperdeytikes apo ton tropo pu ta egraya..gia ayto ksanagrafo oli tin lisi ..

i sinartisi (+oxi i akolu8ia opos ixa grayi..auto isos na se mperdeye..)  $f(x)=(1+\frac{t}{x})^{x}$ me $t$  pragmatiko..ine gnisios auksusa (apodikniete poli eukola pernontas paragogo...) opote afu $b\geq d$ pernume $(1+\frac{a}{b})^{b}\geq (1+\frac{a}{d})^{d}$  $(1)$
kai episis epidi $a\geq c$ exume $(1+\frac{a}{d})^{d}\geq (1+\frac{c}{d})^{d}$  $(2)$
apo tis $1,2$ prokipti to zitumeno....

 :D

[color=red][b]Άσκηση 27[/b] [/color]

[i]Αν για τους πραγματικούς θετικούς a,b,c ισχύει [/i]

$a^{3}+b^{3}+c^{3}\leq 3$ και $abc=1$

[i]Nα βρεθεί η τιμή της παράστασης [/i]

$A= a^{5}+2b^{5}+3c^{5}$

[color=darkblue]Έχει κανείς λύσει για την άσκηση 26?[/color]

$3\geq a^{3}+b^{3}+c^{3}\geq 3abc=3$
opote $a=b=c=1$ kai $A=6$...
XANO MIPOS TIPOTA??...

@giannis
i lisi tu derksaga stin askisi 19 dn se kani? :P

 :D

askisi 26

kanontas apaloifi paronomastwn stin zitoumeni prokyptei oti arkei
$ x^4 \plus{} y^4 \plus{} z^4 >\equal{}x^2yz \plus{} xy^2z \plus{} xyz^2$
twra simfwna me thn anisotita twn barwn exw
$ 1/2x^4 \plus{} 1/4y^4 \plus{} 1/4z^4 >\equal{} x^2yz$
$ 1/4x^4 \plus{} 1/2y^4 \plus{} 1/4z^4>\equal{} xy^2z$
$ 1/4x^4 \plus{} 1/4y^4 \plus{} 1/2z^4>\equal{}xyz^2$
oi opoies treis sxeseis me prosthesi katameli dinoun tin zitoumeni.

Είναι από το βιβλίο του Andreescu "360 problems" κεφάλαιο "Comprehensive problems", άσκηση 38
[hide]Θέτεις $ y_{n} \equal{} 2^{2^{n}} \minus{} 1$ και αποδεικνύεις ότι $ \frac {1}{y_n} \minus{} \frac {2}{y_{n \plus{} 1}} \equal{} \frac {1}{x_n}$ :wink: .[/hide]

Για 26: αναδιάταξη


The $ 2^N$ vertices of the $ N$-dimensional hypercube $ \{0,1\}^N$ are labelled with integers from $ 0$ to $ 2^N \minus{} 1$, by, for $ x \equal{} (x_1,x_2,\ldots ,x_N)\in \{0,1\}^N$,
\[v(x) \equal{} \sum_{k \equal{} 1}^{N}x_k2^{k \minus{} 1}.\]
For which values $ n$, $ 2\leq n \leq 2^n$ can the vertices with labels in the set $ \{v|0\leq v \leq n \minus{} 1\}$ be connected through a Hamiltonian circuit, using edges of the hypercube only?

[i]E. Bazavan & C. Talau[/i]

I suppose the constraint of $n$ is actually $2\leqslant n\leqslant 2^N$. The answer is $n=4,6,8,...,2^N$.

For the prove, it's enough to show that there not exists odd-length cycle. This is obvious by noting that each edge connect two vertices whose coordinate differ by exactly one point. If there's odd cycle, then there's a coordinate that was changed by odd number of times through the cycle, which means the coordinate can't be back to the initial one.

For the construction, simply use induction on $N$. Also use the fact that when $n=2^N-1$, we can start the cycle by any edge that is adjacent to vertex label $0$ (by virtue of symmetry of the hypercube.)


Find all rael-valued functions f on the reals which have at most finitely many zeros and satisfy $ f(x^4 \plus{} y)$ $ \equal{}$ $ x^3 f(x)$ $ \plus{}$ $ f(f(y))$ for all $ x,y$

This problem is solved somewhere in the forum... But here is a solution (I hope I have not miscalculated):

$ x \equal{} 0$ gives $ f(f(y)) \equal{} f(y)$. $ y \equal{} 0$ gives $ f(x^4) \equal{} x^3f(x) \plus{} c$, where $ c \equal{} f(0)$. Substituting both in the original equation, we get

$ f(x^4 \plus{} y) \equal{} f(x^4) \plus{} f(y) \minus{} c$.

Then $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} c$ whenever $ x \ge 0$. In particular, $ f(2x) \equal{} 2f(x) \minus{} c$ and $ f(16x) \equal{} 16f(x) \minus{} 15c$ for $ x \ge 0$. We know that

$ f(x^4) \equal{} x^3f(x) \plus{} c \Rightarrow f(16x^4) \equal{} 8x^3f(2x) \plus{} c \equal{} 16x^3f(x) \minus{} 8x^3c \plus{} c$ if $ x \ge 0$.

But $ f(16x^4) \equal{} 16f(x^4) \minus{} 15c \equal{} 16x^3f(x) \plus{} 16c \minus{} 15c \equal{} 16x^3f(x) \plus{} c$. Then $ 8x^3c \equal{} 0 \Rightarrow c \equal{} 0$. Now we know that:

$ f(f(x)) \equal{} f(x)
f(x \plus{} y) \equal{} f(x) \plus{} f(y)$

(it is easy to see that the second relation generalizes for $ x \le 0$)

If $ f(x) \equal{} 0$ for some $ x \not \equal{} 0$, we have $ f(nx) \equal{} 0 \forall n \in \mathbb{Z}$, contradicting the initial assumption (the function has finitely many zeros). Then we have $ f(f(x) \minus{} x) \equal{} f(f(x)) \minus{} f(x) \equal{} 0 \Rightarrow f(x) \minus{} x \equal{} 0$.


Solve in whole positive numbers

1)  x^3 + y^3 = 111*10^4 + 111*10^2 + 111*\frac{{10}}{2}*10 + 111 + 89 - 2*3
         ( '  :lol: There is a hint in writeln form :lol: ')


:maybe: Just wondering how everyone did this year

:( I could've done better...
But, I didn't remember Stewart's Theorem for number 4

And, when I decided to use the cosine law, it was too late... :(
I rushed my answers, so i'm sure I got at least a marked docked off from 4 a...
I didn't even get to attempt 4 b and c...

I got all the other questions right tho ... :lol:

got all of part I.
got all of part II, except for 4b and 4c.

lets just hope i didnt make any silly mistakes, which i have been known to do so often...

BLah.. Im just guessing my scores...

But I think the unofficial results might be out, and
the teachers would be able to access the scores...

Unofficial... pfff
They don't really change anyways :P

I don't think so, because that is pretty much Stewart's Theorem.

So, did you prove the median length formula by cosine law

I just introduced a lemma determining the length generally, using Stewart's Theorem.

wow, I see lots of people getting a 75+ here. Congrats! Good luck with CMO to these people.  :)

Yeah, I got an 80/80.  Improved by 1 from last year.

[quote="carpo"][quote="gf4848"]I have a question
If we cite the appolonius's theorem without the proof, will they take marks off?[/quote]

No, as long as it was spelled correctly.[/quote]

lol..I said "a well-known theorem" and directly give the median length formula since I thought it was acceptable and the proof was indeed trivial. I believe this is why they took marks off...


Find x,y,z>0 naturals with
(xy-1)/z ,(yz-1)/x , (zx-1)/y integers.

There are clearly no solutions to your question because:

you have 3 inequalities. $xy-1\leq z$  $xz-1\leq y$ $yz-1\leq x$

where the three cannot all be true unless $x=y=z=1$

do you mean that: $x|yz-1$ $y|zx-1$ $z|xy-1$?

[quote="sen"]
do you mean that: $x|yz-1$ $y|zx-1$ $z|xy-1$?[/quote]
From his post, that's indeed what he meant.

The only solutions are : $(2,3,5)$ and $(1,1,z)$ ($z \ge 1$), and all the permutations.

I'll soon post my proof.

yep ive got a proof for that as well  :D

$x\le y \le z$
well if $x=1$ then  $z|y-1$ and $y\le z$ so $y=1$ and we can chceck $(1,1,z)$ is solution

now $x>1$
$xyz|(xy-1)(yz-1)(zx-1)$
$xyz| xy+yz+zx-1$
$xyz\le xy+yz+zx -1<3yz$
$x<3$

so $x=2$
$2|yz-1$ and $z|2y-1$ and $y|2z-1$
$yz|(2y-1)(2z-1)$
$yz|2y+2z-1$
$yz\le 2y+2z-1<4z$
$y<4$
now if $x=y=2$ we have $2|2z-1$ contardiction
so $y=3$
$2|3z-1$ and $z|5$ and $3|2z-1$
so $z=5$

solutions $(1,1,z)$ $(2,3,5)$ and permutations

[quote="mathmanman"]Ok. :)

Well, thanks [b]_el_doopa[/b] for having done the job for me! :P[/quote]
That is also what I want to say to you!!! :rotfl:


A river flows between two houses $ A$ and $ B$, the houses standing some distances away from the banks. Where should a bridge be built on the river so that a person going from $ A$ to $ B$, using the bridge to cross the river may do so by the shortest path? Assume that the banks of the river are straight and parallel, and the bridge must be perpendicular to the banks.

Translate A into C by vector $ \vec{PQ}$, where PQ is the bridge, with Q and B on one side of the river

Then the path is $ AP\plus{}PQ\plus{}QB\equal{}AC\plus{}CQ\plus{}QB\ge AQ\plus{}QB$, with equality when A,C,Q are collinear.

Construction: Construct C as described, one end of the bridge is intersection of straight path between C and B, other is easily derived.

let the river flows along y axis.

see where ever we create the bridge the person must go through a path (a polygonal path) which consists of three line segments of which the middle one has a constant length(wrt to the position of the bridge).
so enough to minimize the total length of the rest two segments.
it is then equivalent to the following question
 we can squeeze the river so that two ends of the bridge coincides.then what is the shortest path from A to B ?
it is the line joining A and B.
let the perpendicular distance from river bank to A is $\ h_{A} $
let the perpendicular distance from river bank to A is $\ h_{B} $
let the horizontal distance between A and B is $\ h $
then consider a straight line starting from A which is inclined towards B (wrt x-axis) at an angle $\theta $
where $\theta= arctan(\frac{h}{h_{A}+h_{B}})$
let that line intersect the river bank at the point C.
create the bridge at the point C.

See the drawing to understand my notations:

$l, a, b, X$ and $Y$ are constants. The distance changes as soon as m and n changes.
Assume that the mininal path is such that this relation is valid: $\frac{m}{a}=\frac{Y}{a+b}=\frac{n}{b}$

To prove it's mininal let's assume the oposite, there's a path shorter than the one I showed:

$\sqrt{a^2+(\frac{aY}{a+b})^2}+l+\sqrt{b^2+(\frac{bY}{a+b})^2}>\sqrt{a^2+m^2}+l+\sqrt{b^2+n^2} \Rightarrow \sqrt{(a+b)^2+Y^2}>\sqrt{a^2+m^2}+\sqrt{b^2+n^2} \\ \Rightarrow a^2+b^2+m^2+n^2+2ab+2mn>a^2+b^2+m^2+n^2+2\sqrt{a^2+m^2}\sqrt{b^2+n^2} \Rightarrow ab+mn> \sqrt{a^2+m^2}\sqrt{b^2+n^2} \\ \Rightarrow a^2 b^2+m^2 n^2+2abmn>a^2 b^2+m^2 n^2 + a^2 n^2+b^2 m^2 \Rightarrow (an-bm)^2<0$


Let $ a,b,c,x,y,z$ be positive numbers such that: $ ax\plus{}by\plus{}cz\equal{}xyz$.Prove that:
$ x\plus{}y\plus{}z >\sqrt{4(a\plus{}b\plus{}c)\plus{}2\sqrt{3(ab\plus{}bc\plus{}ca)}}$

[quote="quykhtn-qa1"]Let $ a,b,c,x,y,z$ be positive numbers such that: $ ax \plus{} by \plus{} cz \equal{} xyz$.Prove that:
$ x \plus{} y \plus{} z > \sqrt {4(a \plus{} b \plus{} c) \plus{} 2\sqrt {3(ab \plus{} bc \plus{} ca)}}$[/quote]
It follows from
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1899


This comp is open to undergraduates and Honours students at any Australian university or tertiary institution.

http://www.maths.usyd.edu.au/u/SUMS/


linear algebra
matrix
induction
algorithm

Given an n x n matrix A let A*
denote the principal submatrix made up of the first к rows and columns. Show
(by induction) that if the first n - 1 principal submatrices are nonsingular, then
there is a unique lower triangular matrix L with unit diagonal and a unique upper
triangular matrix U such that A = LU

Run the LU factorization algorithm (Gaussian elimination, tracking the row operations, without normalizing the diagonal). What could go wrong? You might need a row interchange, because the next entry you would divide by is zero. Since you've only looked at the entries in the principal submatrix so far, this means that that whole row in the row-reduced principal submatrix is zero, and the original principal submatrix didn't have full rank.
That's existence.

For uniqueness, I'll give the invertible case: if $ LU\equal{}L'U'\equal{}A$ and $ A$ is invertible, let $ U'\equal{}BU$; we must have $ L'\equal{}LB^{\minus{}1}$ for equality of products. Now since $ U'$ and $ U$ are upper triangular, $ B\equal{}U'U^{\minus{}1}$ is as well. Since $ L$ and $ L'$ are lower triangular with unit diagonal, $ B\equal{}(L')^{\minus{}1}L$ is as well. A matrix which is both upper triangular and lower triangular with ones on the diagonal is the identity, so $ L'\equal{}L$ and $ U'\equal{}U$.


modular arithmetic
quadratics
geometry
3D geometry
calculus
integration
ARML

Prove that if 2n + 1 and 3n + 1 are both perfect squares, then n is divisible by 40.

I am kind of just posting ideas randomly that might be useful; If you wrote the equation 2n + 1 = x^2, then their might be some way to simplify it. Same with the 3n + 1, but I don't have any idea about how to solve that problem, but you can't blame me, because I'm just 7th grade.

ok, it's not the prettiest solution, but it's still the first solution
[hide]
The problem is equivalent to proving that n is divisible by 8 and 5.
We can see that perfect squares are either 0, 1, or 4 for both modulo 8 and 5.
We are given that 3n + 1 is a perfect square.
Let us assume that 3n + 1 = 0 or 4 (mod 5)
then n is 3 or 1 (mod 5) respectively.
2n + 1 must also be a perfect square. we see that when n = 3 or 1 (mod 5), 2n + 1 is 2, and 3 respectively, which is a contradiction, since 2 or 3 (mod 5) cannot be a perfect square.
thus 3n + 1 = 1 (mod 5)
and 3n = 0 (mod 5)
I'm too lazy to type it in, but the same reasoning can be applied to (mod 8) to show that
3n + 1 = 1 (mod 8).

Since n is divisible by both 5 and 8, n must also be divisible by 40.
QED[/hide]

We have
$ 2n \plus{} 1 \equal{} a^2$
$ 3n \plus{} 1 \equal{} b^2$
Adding,
$ a^2 \plus{} b^2 \equal{} 2$ (mod 5)
Squares are 0,1,or 4 mod 5, so both a and b must be either 5k+1 or 5k+4
It is not too hard to show 8 divides n (I don't feel like typing it). Then we have two cases:
$ 16q \plus{} 1 \equal{} (5k \plus{} 1)^2$ or
$ 16q \plus{} 1 \equal{} (5k \plus{} 4)^2$
In both cases, it is easy to show 5 divides q.

Actually, I'm almost positive I've seen this as an application to a camp. Are there any possibilities other than n=0?

Let $ 2n\plus{}1\equal{}a^2$ and $ 3n\plus{}1\equal{}b^2$.

$ 3a^2 \minus{} 2b^2 \equal{} 1$

mod 5, we get $ 3a^2\equiv0,2,3$ and $ 2b^2\equiv0,2,3$, the only 2 that subtract to 1 mod 5 are $ 3a^2\equiv3, 2b^2\equiv2$ therefore, $ a^2 \equiv 1$ and $ b^2\equiv 1$ and therefore:

$ 2n \plus{} 1 \equiv 1$ mod 5 and $ 3n \plus{} 1 \equiv 1$ mod 5, subtracting we get $ n \equiv 0$ mod 5.

Now, mod 4, we get $ a^2$ is odd, $ b^2$ is odd. Mod 16, we have $ 2b^2\equiv 2$ so therefore $ 3a^2 \equiv 3$ which gives $ a^2\equiv 1$ so $ 2n \equiv 0$ mod 16, so n is a multiple of 8.

Q.E.D.

heres what I have :
$ 2n \plus{} 1 \equal{} x^2$
$ 3n \plus{} 1 \equal{} b^2$
$ \therefore x^2 \plus{} b^2 \equal{} 2\pmod 5$
Clearly $ x \equal{} 1\pmod 5$ or $ 4\pmod 5$
$ 2n \plus{} 1 \equal{} (5k \plus{} 1)^2 \implies 2n \equal{} 10k \plus{} 25k^2 \implies n$ divisible by $ 5$.
$ 6n \plus{} 2 \equal{} 2b^2 \implies a^2 \plus{}2b^2 \equal{} 8n \plus{} 3 \equal{} 3\pmod 8$
The quadratic residues in $ \pmod 8$ are $ 0,1,4, \implies a^2 , b^2 \equal{} 1\pmod 8$
$ 3n \plus{} 1 \equal{} (8k \plus{}1)^2 \implies 3n \plus{} 1 \equal{} 64k^2 \plus{} 16k \plus{} 1 \implies n$ is divisible by $ 8$
Hence $ 40$ divides $ n$
QED

I'm not an expert at modular arithmetic, but I do know a little bit about it and how it is used, but I don't know when to use it at certain times. These are some really good solutions.

Im no expert either; Ive only read aops posts, and the chapter in aops v1 about modular arithmetic. i should be better after reading the chapter in aops v2.

I've known about modular arithmetic for about a year now, but I forgot to use it.

cool...
heres a similar problem where you dont need mod:
The product of Jack ages 7 years from now, and 7 years ago is the cube of a prime. Find Jack's age now.

Well, x(x-14) = x^2 - 14x, which has to equal the cube of a prime number. Let's say that the prime number is 'a'. Then we have the equation; x^2 - 14x - a^3 = 0. If we factor we know that it will be in the following form; (x - m)(x + n) = 0. We know that the product mn has to equal a^3, and n - m = 14. We could try out different values of a, like 2, 3, 5, 7, or 11, but I don't know what to do from there. I'm sure I got pretty close to the answer.

read the problem more carefully. The product of Jack's age 7years ago and 7 years from now is $ (j\minus{}7)(j\plus{}7) \equal{} J^2 \minus{}49$ not $ j^2 \minus{}14J.$

He is coming well, so you knew that $ j(j \plus{} 14) \equal{} a^3$ anda $ a$ is a prime, so $ j \equal{} a$ and $ j \plus{} 14 \equal{} a^2$, its the same but he is consedering $ j$ as the age of Jack 7 years ago, btw, how you finish your solution, mathmon.?

Is J a multiple of 6?  If so, I will post how I got that.

I actually did get to a point where i had j^2 - j = 14, but I couldn't figure it out. There is no solution to that. Or at least there is, but just no solution that is an integer. Maby using mathmonsters equation J^2 - 49, then that would equal a^3, so the equation would be; J^2 - 49 = A^3. I don't know how to solve it from there on. You could do guess and check which looks to be the easiest solution.

since $ a$ is prime, you could have by de problem that $ (j-7)(j+7)=a^3\Rightarrow j-7=a$ and $ j+7=a^2$ corect?
Then you have $ (j-7)^2=j+7\Leftrightarrow j^2=15j-42$ and use modular aritmetic to solve that, I will see'a

I don't think there is a solution. I've tried looking at the problem with so many different ideas, but they all end up with radicals. What is the solution mathemonster?

Sorry all, replace 7 with 21 and you should get an integral answer for J. Raul's method is correct.
here is the solution:
$ (j\minus{}21)(j\plus{}21) \equal{} p (p^2) \implies J\minus{}21 \equal{} p , J \plus{}21 \equal{} p^2 \implies J \plus{} 21 \equal{} (j\minus{}21)^2 \implies J \equal{} 28$
(ARML)

Oh, I probably could've gotten it if it were 21, but that's okay.

So we could also do it a different way;

We know that J^2 - 42J - 441 = J+ 21
So J^2 -43J - 462 = 0.
Which simplifies to 52.
Is that the answer?

Oh, well now I get it. I don't know what I did wrong though.


1.  Why don't \fbox and \mbox work in math mode?  Is there a replacement for them?

2.  Also, I have found that \mbox generally doesn't render in the output file.  Why?

3.  Is there some way to suppress the page number that appears in the center of the footer of LaTeX documents?

4.  Is there some way to increase the height of each row in a tabular environment?  As it is, fractions within a table do not render well.

[quote="roadnottaken"]1.  Why don't \fbox and \mbox work in math mode?  Is there a replacement for them?[/quote]They do work, for example,
$x^{2}=1\mbox{ if and only if }x=\pm 1$
$x^{2}=1\ \fbox{if and only if}\ x=\pm 1$
Can you give me a mnimal example where you are having problems?

[quote="roadnottaken"]2.  Also, I have found that \mbox generally doesn't render in the output file.  Why??[/quote]This could be linked to 1. Are you getting any error messages? Give a minimal example of a complete document where \mbox doesn't work.

[quote="roadnottaken"]3.  Is there some way to suppress the page number that appears in the center of the footer of LaTeX documents??[/quote]\thispagestyle{empty} for just one page or \pagestyle{empty} for all pages until overridden

[quote="roadnottaken"]4.  Is there some way to increase the height of each row in a tabular environment?  As it is, fractions within a table do not render well.[/quote]\renewcommand{\arraystretch}{[i]factor[/i]}The standard value of the distance between rows is 1 so a factor of 1.5 increases it by 50%.
I can't show it here but
[code]\begin{tabular}{|c|}
\hline
$\frac{1}{2}$\\
\hline
\end{tabular}[/code]looks a lot better if you add \renewcommand{\arraystretch}{1.5} before the table.
You can also have finer control if you use \vspace as in this example:[code]M=\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \vspace{-2mm} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{pmatrix}[/code]which looks like this
$M=\begin{pmatrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \vspace{-2mm}\\-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{pmatrix}$

Boxes in math mode seem to be working now (earlier I got an error message when I tried to place a box within math mode), and I've attached an example of the difference between fbox and mbox.

The pdf looks perfect.

Line 1: In maths mode \mbox is for text (though it has now been superceded by amsmath \text). Thus $\mbox{word}$ instead of $word$. So the line $\$$1+1=\mbox{2}$\$$ treats the 2 as text though you can't tell as it's a number.

Line 3: The line \mbox{This should be in a box.} is in an invisible box which is how it should be ie all the text is being kept together in one unit even if something else happens to the page - it won't go onto two pages for example.

Lines 2 & 4: \fbox puts a frame round the text correctly whether or not it's in maths mode.

[quote="stevem"][quote="roadnottaken"]3.  Is there some way to suppress the page number that appears in the center of the footer of LaTeX documents??[/quote]\thispagestyle{empty} for just one page or \thispagestyle{empty} for all pages until overridden[/quote]
In the second case, you mean \pagestyle{empty}, right?

I am using the fancyhdr package.  will \pagestyle{empty} interfere with the header and footer commands?

Yes, because \pagestyle{empty} means no header or footer and anyway you would need \pagestyle{fancy} for fancy headers and footers.
Use something like \renewcommand{\thepage}{} to get no page numbers but keep headers and footers.

When I use the following line in a \LaTeX document, it pops up a bunch of errors, but when I remove the \fbox command, it works fine.  Apparently it is boxes around fractions that it can't handle, rather than boxes in math mode.  Are there any workarounds for this?

[code]\[\frac{y(y+3)+x(12-x)}{2}=\fbox{\frac{y^2+3y-x^2+12x}{2}}\][/code]

Edit: it can't be that it doesn't handle fractions, because I used \fbox for a previous fraction in the same document which turned out fine.

\fbox puts things back into text mode so you need dollar signs to put the boxed formula back into maths mode, plus a way of putting back display mode using \displaystyle or \dfrac. [code]\[\frac{y(y+3)+x(12-x)}{2}=\fbox{$\dfrac{y^2+3y-x^2+12x}{2}$}\][/code]Of course Ravi B's suggestion of \boxed is much simpler and, I suspect, is why amsmath introduced it.


factorial
AMC
AIME
geometry
3D geometry
sphere
function

Rate the avatar of the person above you!
When you post, say if you made the avatar yourself. (I did.) This should add 1 too your score.
Score rates from 0 to 10.

I rate yours a 7, plus 1, 8.

I did not make mine.

YOU have to make it.
I'll give it a 6. (reason- crappy 2-frame animation doesn't help your score. No originality points either.)

edit:
Oh, sorry. 3-frame animation. :roll:

9. Cute little eggs and it fits your username.

Um.....8 i guess because I love ice cream. I'm not sure if it's original, though.

i doubt yours is original; but either way, since im nice unlike some people COUGH GAG WHEEZE i say 8. skateboarding is cool

i obviously made mine. hence my name in the corner

i give you 7. mine is not original, by the way.

8 - I love eclipses

Mine is original - my guitar.

@eggylv999
It's not mean, it's constructive criticism.
@1=2
sorry.
@jain_harshi
Rock out, dude! 7+1=8!

its original (you made it) but its still based on something else (iPod ads) so 6+1=7.

turtle looks dumb, but 6.5+1=7.5

not mine

Nice... but it's not original, so I'll give it an 8.

An abbreviation, [b]A[/b]mir [b]H[/b]ossein [b]P[/b]arvardi.

9, cool as heck, but what is it?

9.. cool..

It's Noble 6 from Halo Reach.

0... no offense, but
[i][size=200][color=#FF0000]BOOOOOORRRRRIIIINNNNNNGGGGG!!!!!!!!!!!!!![/color][/size][/i]


Find the minimum K such that:

Sum of 1/An <= K

Where An are all natural numbers without the digits 9.

Clearly K<= 10 (1+ 1/2 + ... + 1/8)
Why is it the minimum?

See the following thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32625[/url].

With computer work, someone showed that the minimum $K$ is approximately 22.9.


Hi!

$ a,b,c,x,y$ and $ z$ are positive real numbers such that $ y^2\leq xz$ and $ z^2\leq xy$. Prove that

\[ \frac{a}{ax\plus{}by\plus{}cz}\plus{}\frac{b}{bx\plus{}cy\plus{}az}\plus{}\frac{c}{cx\plus{}ay\plus{}bz}\leq\frac{3}{x\plus{}y\plus{}z}.\]

Thanks.

[quote="yuyang"]Hi!

$ a,b,c,x,y$ and $ z$ are positive real numbers such that $ y^2\leq xz$ and $ z^2\leq xy$. Prove that
\[ \frac {a}{ax \plus{} by \plus{} cz} \plus{} \frac {b}{bx \plus{} cy \plus{} az} \plus{} \frac {c}{cx \plus{} ay \plus{} bz}\leq\frac {3}{x \plus{} y \plus{} z}.\]
Thanks.[/quote]
See here: http://canhang2007.wordpress.com/2009/11/03/inequality-40-unknown-source/

LOCK THIS TOPIC. IT'S PROBLEM 3459 FROM CRUX MATHEMATICORUM


as my brother shouted out 4 consecutive integers, i divided each by my age (an integer), then added all four remainder to get 40. as he shouted out 4 other consecutive integers, i divided each by his age (a different integer), added all 4 remainders and again got 40.  What's the sum of our ages?

thanks

A pretty simple question. :)
All four reminders may include a zero, or no zero at all

For non-zero situation, no answer since 40 cannot be expressed like that.
When the zero is at the first place or the fourth place, no answer, since 40 is not a multiple of 3
When is at the third place, $19+20+0+1 = 40 \Rightarrow \text{one guy is } 21$
When the zero is located at the second place, $37+0+1+2 = 40$ Another guy's age is $38$
$21+38=59$

This was also a CA math league question. Is it a national thing because from my knowledge it's just regional. Funny coincidence maybe.

It's national.

Why is this in the olympiad section? It should be moved.


In how many ways can 1009 distinct integers be chosen from the set {1,2,3, ,......, 2009, 2010} so that no two chosen integers differ by 10?

[hide]
The task is equivalent to:
Pick 1009 distinct numbers from the sequences
{1, 11, 21, ..., 2001}, {2, 12, 22, ..., 2002}, ..., {10, 20, 30, ..., 2010}
such that no two numbers in the same sequence are consecutive.

The maximum number of numbers we can pick from any one sequence is 101, obtained by taking the first number and every other number afterwards. Thus the 1009 integers must be obtained by taking 101 numbers from each sequence except for 100 from one of them.

Obviously there is only one way, as we explained above, to take 101 numbers from a sequence. So now the question is, how many ways are there to take 100? Let's consider this in the context of the sequence, {1, 2, 3, ..., 200, 201}, since clearly the values of the numbers are no longer relevant, merely whether they are consecutive, and the number of ways to take 100 numbers from any other sequence with 201 values, under the same conditions, will be the same.

Now, assume we've picked the numbers $ a_1, a_2, \ldots, a_{100}$. Let $ a_0 \equal{} \minus{}1, a_{101} \equal{} 203$. Let $ d_i \equal{} a_{i\plus{}1} \minus{} a_i \minus{} 2$. All $ d_i$ is nonnegative iff no two selected numbers are consecutive. Furthermore, $ \sum_{i\equal{}0}^{100} d_i \equal{} a_{101} \minus{} a_{100} \plus{} a_{100} \minus{} a_{99} \plus{} \ldots \plus{} a_{2} \minus{} a_{1} \minus{} 201(2) \equal{} 2$. So either we have two different $ d_i \equal{} 1$ and the rest 0, or one single $ d_i \equal{} 2$ and the rest 0. This means $ \binom{201}{2} \plus{} 201 \equal{} 20301$ ways. Or, using balls and bins, $ \binom{202}{2} \equal{} 20301$ ways.

Finally, since we could have picked any of the 10 sequences to be the one with 100 numbers instead of 101, the answer is $ \boxed{203010}$.

[/hide]


limit
logarithms
integration
real analysis
real analysis unsolved

Let there be a sequence $ (x_{n})_{n\geq 1}$ such that $ x_{1}>0$ and $ n(x_{n\plus{}1}\minus{}x_{n})\equal{}1$ for every $ n\geq 1$.Calculate:
\[ \lim_{n\to\infty} \frac{n\ln{n}}{x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}}\]

$ x_n\equal{}x_1\plus{}\sum_{k\equal{}1}^{n\minus{}1}\frac1k\equal{}\ln n\plus{}O(1).$

Hence, $ \sum_{k\equal{}1}^nx_k\equal{}\sum_{k\equal{}1}^n\ln k\plus{}O(n)$

$ \equal{}\int_1^n\ln x\,dx\plus{}O(n)\equal{}n\ln n\plus{}O(n).$

The limit in question is $ 1.$

$ \lim_{n\to\infty}n(x_{n \plus{} 1} \minus{} x_n) \equal{} \lim_{n\to\infty}\frac {n(x_{n \plus{} 1} \minus{} x_n)}{\ln\left(1 \plus{} \frac {1}{n}\right)^n} \equal{} \lim_{n\to\infty}\frac {x_{n \plus{} 1} \minus{} x_n}{\ln(n \plus{} 1) \minus{} \ln n} \equal{} 1$.Hence from Stolz's lemma follows that $ \lim_{n\to\infty}\frac {x_n}{\ln n } \equal{} 1$.

We also have from Stolz's lemma that
$ \lim_{n\to\infty}\frac {n\ln n}{x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n} \equal{} \lim_{n\to\infty}\frac {(n \plus{} 1)\ln(n \plus{} 1) \minus{} n\ln n}{x_{n \plus{} 1}} \equal{}$

$ \lim_{n\to\infty}\frac {\ln \left(1 \plus{} \frac {1}{n}\right)^n \plus{} \ln(n \plus{} 1)}{x_{n \plus{} 1}} \equal{} 1$.

Note that in this way the conclusion would have been also true if the stronger condition, $ \lim_{n\to\infty}n(x_{n \plus{} 1} \minus{} x_n) \equal{} 1$, had been given.


10 nonnegative integers have an average of 6. What is the greatest possible median?

Because the answer by itself doesn't mean much, especially from someone who knows exactly what they're doing, I think I'll expand on probability's answer.

Let's imagine the ten integers in non-decreasing order. We want the term in the middle to be as large as possible. However, the average cannot be more than 6. Let's call our median $'x.'$ So there will be 4 terms less than or equal to $x$ and four terms greater than or equal to $x.$ Therefore, every term other than $x$ should be as small as it can be. The smallest the first four terms can be is $0$ (the question says non-negative), and the smallest the rest of the terms can be is $x$. So now we have a sequence of terms that looks like: $\displaystyle 0,0,0,0,x,x,x,x,x,x$

We have one more condition, however. The average is $6,$ so: \[
\displaystyle \frac {6x}{10} = 6
\displaystyle 6x = 60
\displaystyle x = 10\]

So, the median (as well as every turn after it) is 10. The end :)

according to Hikaru ...
If we just use  6x/10= 6 => 6x=60=>x=10 why would we  use those steps above?


induction
number theory
relatively prime

Find an explicit form for the recursion $a_{0}=a_{1}=1$ and $a_{n+1}= \frac{a_{n}a_{n-1}}{a_{n-1}+a_{n}}$

[hide]
Experimenting, we see that the first few terms are $1,1,\frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{8}\ldots$.  We conjecture that $a_{n}=\frac{1}{F_{n-1}}$ for all positive integers $n$, where $F_{n}$ is the $n$th Fibonacci number.

We will prove this by induction.  Clearly the base case is true.  Say that $a_{k-1}=\frac{1}{F_{k-2}}$ and $a_{k}=\frac{1}{F_{k-1}}$.  Then

$a_{k+1}=\frac{\frac{1}{F_{k-2}F_{k-1}}}{\frac{1}{F_{k-2}}+\frac{1}{F_{k-1}}}$.

Using the fact that any two consecutive Fibonacci numbers are relatively prime,

$a_{k+1}=\frac{\frac{1}{F_{k-2}F_{k-1}}}{\frac{F_{k-1}+F_{k-2}}{F_{k-2}F_{k-1}}}=\frac{1}{F_{k-1}+F_{k-2}}=\frac{1}{F_{k}}$.

This completes the induction and the proof.
[/hide]

[quote="JBL"][quote="E^(pi*i)=-1"]Using the fact that any two consecutive Fibonacci numbers are relatively prime, [/quote]

I'm curious in what way you believe you used this fact.[/quote]

You're right, I didn't.  I guess what I was thinking was that the fraction $\frac{F_{k-1}+F_{k-2}}{F_{k-1}F_{k-2}}$ wouldn't be in simplest form unless the consecutive Fibonacci numbers were relatively prime, but of course that's irrelevant.  I guess I'm in AMC/AIME mode, thinking all fractions must be in simplest form.  :roll:


[quote="unimpossible"]$ z_1$, $ z_2$, and $ z_3$ are complex numbers.

$ z_1 \plus{} z_2 \plus{} z_3 \equal{} 1$
$ z_1^2 \plus{} z_2^2 \plus{} z_3^2 \equal{} 3$
$ z_1^3 \plus{} z_2^3 \plus{} z_3^3 \equal{} 7$

find $ z_1^7 \plus{} z_2^7 \plus{} z_3^7$[/quote]

[hide]Let $ s_i \equal{} z_1^i\plus{}z_2^i\plus{}z_3^i$

Then we can establish a relationship between $ s_i$ and $ s_{i\minus{}1}, s_{i\minus{}2}$[/hide]


inequalities
trigonometry
inequalities proposed

[quote="Virgil Nicula"]Prove that in an acute triangle $ ABC$ exists the inequality $ \boxed {\ \ \frac {a^2\cos A \plus{} b^2\cos B \plus{} c^2\cos C}{a^2 \plus{} b^2 \plus{} c^2}\le \frac rR\ }\ \le\ \frac 12$ .[/quote]
I think, it's true for all triangle.

I think I have solved out the right part. Here's my soln.

r/R=(A/S)/(abc/4A)=4A^2/Sabc=(b+c-a)(c+a-b)(a+b-c)/4abc
or, r/R=(1/4)[(2abc+...)/abc] {I am too lazy and too sleepy to calculate the dotted portion, so far I guess it would be=>0. I'll see it tomorrow morning again}
or,r/R<=(1/4)*2=1/2...Done!

If there is any mistake in the proof, pls. inform! Good night! :sleeping:


linear algebra
matrix
linear algebra unsolved

[color=blue]
Let $A$ be a matrix of order $2$ and $d=det(A)$ and $a\in\mathbb{R}$ such that $ad>0$. Prove that if $A^{*}$ is
the adjoint of $A$ and $det(A+aA^{*})=0$, then $A=-A^{*}$. ;)
[/color]

Let´s see!

Let $B$ such that $A+aA^{*}=B$. Note that $A^{*}=dA^{-1}$ and hence $A+adA^{-1}=B$, and so we have $A^2+adI_2=BA$, so we have that $det(A^2+adI_2) = det(B)det(A)$. But $det(B)=0$, by hipothesis. So we conclude that $-ad$ is an eingenvalue of $A^2$. So the eingenvalues of $A$ are $\sqrt{ad}i$ and $-\sqrt{ad}i$.

This last is true because I´m considering a real matrix, and must be this way, since we have $ad>0$, with real a.

But $A$ has order $2$. So we have $z^2-tr(A)z+d=0$. And substituting $z$ by $\sqrt{ad}i$, we have $-ad-tr(A)\sqrt{ad}i+d=0$. So, since $d>0$, we conclude that $a=1$ and $tr(A)=0$. So by the Cailey-Hamilton Theorem, we have $A^2+dI_2=0$. So $A=-dA^{-1}$, and hence $A=-A^{*}$.


the difference in area between a circle's circumscribed and inscribed triangles is 25 cm^2. what is the radius of the circle?

this problem makes me feel stupid.

oh whoops i forgot to say the triangle are equilateral

This probably should go in the mathcounts forum

[hide="solution"]
Diagram would be appreciated
Let $ O$ be the center of the circle, let $ ABC$ be the inscribed triangle, and let $ DEF$ be the circumscribed triangle.  Additionally, let $ X$ be the midpoint of $ AB$ and let $ Y$ be the midpoint of $ DE$.
Draw $ OA$ and $ OX$.  Then, $ OAX$ is a 30-60-90 triangle, so $ AX\equal{}\frac{r}{\sqrt{3}}$ and $ AB\equal{}\frac{2r}{\sqrt{3}}$.  The area of $ ABC$ is \[ \frac{\left(\frac{2r}{\sqrt{3}}\right)^2\sqrt{3}}{4}\equal{}\frac{r^2\sqrt{3}}{3}\]
Draw $ OD$ and $ OY$.  $ ODY$ is a 30-60-triangle, so $ DY\equal{}r\sqrt{3}$ and $ DE\equal{}2r\sqrt{3}$.  The area of $ DEF$ is \[ \frac{\left(2r\sqrt{3}\right)^2\sqrt{3}}{4}\equal{}3r^2\sqrt{3}\]
It is given that the difference in the areas is $ 25$, so we have the equation \[ 3r^2\sqrt{3}\minus{}\frac{r^2\sqrt{3}}{3}\equal{}25\]
You can solve this yourself
[/hide]

The most important part is noticing the $ 30\minus{}60\minus{}90$ triangles. (Refer to [url=http://mathworld.wolfram.com/30-60-90Triangle.html]this.[/url])

The area of an equilateral triangle is $ \frac{s^2\sqrt3}4$.

The side of the larger triangle is $ 2r\sqrt3$, and $ r\sqrt3$ for the smaller triangle.

Thus, we have $ \frac{(2r\sqrt3)^2\sqrt3}4\minus{}\frac{(r\sqrt3)^2\sqrt3}4\equal{}25$

$ \implies12r^2\sqrt3\minus{}3r^2\sqrt3\equal{}100\implies9r^2\sqrt3\equal{}100\implies r^2\equal{}\frac{100}{9\sqrt3}\equal{}\frac{100\sqrt3}{27}\implies r\equal{}\boxed{\sqrt{\frac{100\sqrt3}{27}}}$.

Of course, we simplify our radical to $ \frac{10}{3}\sqrt{\frac{\sqrt{3}}{3}}$.


quadratics
function
algebra
number theory unsolved
number theory

It's equivalent to the quadratic equation
$x^{2}+x(3y+4006)+(4006y+(2003)^{2})=0$

The determinant must be a square, so for some $A \in \mathbb{Z}$, we have
$A^{2}= (3y+4006)^{2}-4(4006y+(2003)^{2})$
$\Leftrightarrow (9y+4006+3A)(9y+4006-3A) = (4006)^{2}$
and $x = \frac{-3y-4006+A}{2}$ or $x = \frac{-3y-4006-A}{2}$

Notice that $S-T \equiv S+T \equiv 1 (mod 3)$.
Let $9y+4006 = S$ and $3A = T$. Since $(4006)^{2}$ is even, both $S-T$ and $S+T$ must be even.
Thus, $S-T = S+T = 4006 \Rightarrow S = 4006, T = 0 \Rightarrow y = 0, A = 0, x =-2003$.

The solution is only $(x,y) = (-2003,0)$  :)

I'm sorry, kunny. But I don't understand what's the function of $y(3x+4006) \leq 0$. Can you explain more clearly ? :maybe:


The Erdos Selfridge theorem states that the product of consecutive numbers cannot become powers.
many special cases can be seen as olympiad problems.
I have searched a lot for the proof of it, but have been unable to obtain one.
Can anybody show me an elemantary proof?

Also, a elemantary proof of the prime number theorem would be appreciated.

The paper of Erdös and Selfridge is 10 pages long, so I don't think anybody will copy it here...if someone wants to scan it...

P. Erdös, J.L. Selfridge, "the product of consecutive integers is never a power", Illinois J. Math. 19 (1975), 292-301.

Pierre.

For "the product of consecutive integers is never a power", see [url]http://www.amazon.com/Proofs-BOOK-Martin-Aigner/dp/3540404600/ref=sr_1_1/002-3688268-8179211?ie=UTF8&s=books&qid=1194593941&sr=8-1[/url].
For an elementary proof of the prime number theorem, see [url]http://www.amazon.com/Elementary-Methods-Number-Theory-Nathanson/dp/0387989129/ref=sr_1_2/002-3688268-8179211?ie=UTF8&s=books&qid=1194594042&sr=1-2[/url].

Also, you can see "the product of consecutive integers is never a power" in the SNU library. See [url]http://library.snu.ac.kr/DetailView.jsp?uid=1&cid=14720[/url].

[quote="ywl"]Well How could I obtain the article? It sure isnt in JSTOR or libraries I know.[/quote]

Many papers of Paul Erdos are available at the site:  http://www2.renyi.hu/~p_erdos/
A listing with explicit links to the papers is at: http://www2.renyi.hu/~p_erdos/Erdos4ht.html
The paper with Selfridge is available at: http://www2.renyi.hu/~p_erdos/1975-46.pdf


Hola,

Por que no hacemos algo de olimpiadas. Vamos, resolvamos algo, hagamos algo, para algo esta la comunidad en espaniol.

De momento recomiendo que calculen
\[\sum_{k=1}^n\left\lfloor\frac{n+2^{k-1}}{2^k}\right\rfloor\]
donde $\lfloor x\rfloor$ es la parte entera de $x$.

Ciao,

El problema es de una IMO vieja, no se la fecha exacta, pero es de antes de los 80's (creo).
Basta con conocer la identidad:
\[\lfloor x\rfloor+\left\lfloor{x+\frac{1}{2}\right\rfloor}=[2x]\]
Esta identidad es sencilla de demostrar, ya que se reduce a ver que si se tiene un real $1>\alpha\geq 0$, entonces
\[\left\lfloor{\alpha+\frac{1}{2}\right\rfloor}=\lfloor{2\alpha\rfloor}\]
Una vez demostrada esta identidad vemos que la suma se convierte en
\[\sum_{k=1}^{\infty}{\left\lfloor\frac{n}{2^{k-1}\right\rfloor-\left\lfloor\frac{n}{2^k}\right\rfloor}}\]
Es una suma telescopica y de aqui resulta evidente que la suma es igual al primer termino menos el utlimo de donde queda que
\[\sum_{k=1}^\infty\left\lfloor\frac{n+2^{k-1}}{2^k}\right\rfloor=n\]
De aqui se sigue lo pedido

Nota: El problema que se dio en el foro, el indice de la suma termina en n, pero creo creo que es un typo. En el problema original lo que te piden es hallar la suma hasta cualquier punto, de todos modos se puede hacer exactamente como se hace aqui pero con un termino restandole a n.

[quote="Leonardo"] El problema que se dio en el foro, el indice de la suma termina en n, pero creo creo que es un typo. En el problema original lo que te piden es hallar la suma hasta cualquier punto, de todos modos se puede hacer exactamente como se hace aqui pero con un termino restandole a n.[/quote]

No, no es un typo, aunque no sea identico al planteamiento original. A partir de un punto (cuyo termino es por su puesto menor que $n$) todos los terminos son cero, entonces, terminar en $n$ o en $\infty$ es basicamente la misma cosa.

Alguien propone algun otro problema?? HAGAMOS OLIMPIADAS

Saludos,


Some questions to consider:
1. If a tree falls in the forest and no one is around, does it make a noise?
2. If a tree falls in the forest and no one is around, did it actually ever fall?
3. If a tree falls in the forest and no one is around, does the tree exist?

By the way, I don't think that anything behind me exists.  I don't see it, so I don't believe it.  How about you?

Have you been reading Berkeley?  He talks a lot about this.  (His answers:  No, no, no.  Esse est percipi.)

The relationship of quantum mechanics to these questions is highly tenuous at best.  Questions 2 and 3 are logical tautologies: $\textrm{If} \exists \textrm{a tree, then the tree exists}$.  The answer to question 1 depends at least partially on the definition of "noise."

I don't think this has anything at all to do with quantum physics. Okay, I actually know nothing about quantum physics, but I vaguely remember from somewhere that when it talks about observers and so on, it means photons. Or if not photons, something else physical. It's got nothing to do with "conscious" beings like we pride ourselves on being.

[u]HERE MY SUPER DUPER AWESOME POOPY HEAD THEORY[/u]
Nothing in the the world is there. Our mind draws images and thus brings things like trees into life. The whole universe is vast emptiness. Our minds creates our own universe. Maybe our lives are false in the sense that everyone and everything around us isn't really there but just a pigment created by our minds. As I am typing this, the whole world, outside my room, is unknown to my brain. If i am speaking to someone on the telephone, my brain can only hear but not see, therefore it is unecessary for an image. The reason why we are here and observe these pigment, i don't know. So my answer is no no no, the tree doesn't exist. SNAP!

You are forgetting about superpositions and Schrodinger's Cat! If you cannot know something for sure, that does not mean you do not believe it. The tree and its noise are in a superposition at the time it is not observed. As soon as it is observed, it is immediately certain.


A carton of juice with a 50-ounce capacity contains 49 ounces of juice. If Pete drinks 38 ounces of the juice, what percent of the carton will be empty?

First we must subtract $ 38$ from $ 49$.

$ 49 \minus{} 38 \equal{} 11$

Now we just find the percent.

$ \frac{11}{50} \equal{} 0.22$

So, we have $ \boxed{22 \text{ percent}}$

we have 22 percent, so 78 percent of the carton is empty.

answer : 78

:oops_sign:

Aha! I have figured out why I never get that one right.....so much for good reading skills.

[quote="isabella2296"]:oops_sign:

Aha! [b]I have figured out why I never get that one right[/b].....so much for good reading skills.[/quote]

 :rotfl:

[quote="mewto55555"]Is the carton 78% empty or 22% full? :rotfl:[/quote]

 :rotfl: again

ill answer this in a series of steps   [b][i][u]*i think i have a much easier way*

[/u][/i][/b]

[hide="THIS IS MY EXPLANATION"]1.we know that the juice box has a 50 ounce capacity but it can only hold 49 ounce

2.then the remaining 1 ounce is empty space

3.pete drank 38 ounces and those 38 ounces go away and that 1 ounce is not there so all together 39 out of 50 ounces are empty so you can get the equation  39/50

4.[hide="ANSWER"]78%[/hide][/hide]

[hide="."][size=200][color=#0080FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH [/u][/i][/b][/color][/size][/hide]

[size=50]please be fair! and PM if i was wrong thank you so much!![/size]


trigonometry
calculus
calculus computations

[quote="Dr Sonnhard Graubner"]hello,it is
${ \text{(i)}\qquad \prod_{k = 3}^\infty \; (1 - ( \frac {2}k})^3 ) = \frac {1}{126}\frac {\sin(\pi\sqrt {3})\sqrt {3}}{\pi}$
and
${ \text{(ii)}\qquad \prod_{k = 4}^\infty \; (1 - ( \frac {3}k})^3 ) = \frac {8}{15561}\frac {\cos(\frac {3}{2}\pi\sqrt {3})}{\pi}$
Sonnhard.[/quote]

Can you please explain what method you used while solving those problems?  :huh:  :oops:


\[\displaystyle Daca \ \ a+2b+3c+4d\geq30 \ \ atunci \ \ \sum a^2\geq30\]

Rockymariancosmin, te rog nu mai pune text in codul $\LaTeX$! Multumesc.

Am inteles

my name is Rockymarcianocosmin99 (marciano nu marian) :P

:oops_sign: eu sunt cunoscut ca mai busesc numele :D

[quote="Cristi Sav"]$(1+4+9+16) \sum a^{2} \geq (a+2b+3c+4d)^{2} \geq 30^{2}$ de unde rezulta concluzia[/quote]
     Problema putea fi rezolvata in diverse moduri, dar mie acum imi vine in cap Inegalitatea lui Cebisev.Cred ca merge su cu asta, dar si cu Cauchy. :D

cred ca mai repede Inegalitatea lui Cebisev

Generalization:

For $ a_k>0\ (k\equal{}1,\ 2,\ \cdots n)$,

$ \sum_{k\equal{}1}^n a_k^2\plus{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}\geq 2\sum_{k\equal{}1}^n ka_k$


MATHCOUNTS probability geometry rectangle ratio inequalities quadratics

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

i know this topic is over 2 and a half years old...who cares.

[hide="1"]1. the area inbetween is $ 3^{2}\pi\minus{}2^{2}\pi$ which is $ 5\pi$. the area of the entire dartboard is $ 16\pi$. divide them. [b]5/16[/b][/hide]

[hide="2"]2. the height is 4*6=24 the area of the base is $ 16\pi$. the volume of the cylinder is $ 384\pi$. the volume taken up by the tennis balls is $ 4/3\pi r^{3}*3 \equal{} 256\pi$ subtract the 2 volumes. $ 128\pi$[/hide]

Is it okay if I answer some of these too, even if I'm not from Pennsylvania?

i guess so

3 and 5
[hide="3"]
we have the 2 long sides and the short side of the rectangle added together. thats 11.5 then the arc of the semicircle which is 3.5/2pi which rounds to 6 [b]17.5[/b]
[/hide]
[hide="5"]
we have 3/4 of a dozen roses, 1/2 a dozen daises, and 3/2 dozen chrysanthemums. so thats 15 dollars for roses, 2.5 dollars for daises, and 15 dollars for chrysanthemums. thats 32.5 dollars
[/hide]

theres a formula for number 6. i just acant think of it

[hide="9"]Quadratic formula, then ... just working it out: $ \frac{8\pm{2\sqrt{5}}}{3}$[/hide]

[hide="hint to 4"]The total product is 8!=40320. The largest is the one closest to the square root. Then all I can think of is guess and check.[/hide]

well, atleast that limits some work. thanks


geometry
3D geometry
sphere
combinatorial geometry
geometry solved

(1)
If a circular planet lies completely within the convex hull H of a similarly defined system of circular planets, its entire circumference is visible from the other planets.

(2)
Consider a cluster of n spherical planets, all the same size, which drift rigidly together through outer space, that is, each with no motion relative to the others in the cluster. In general, there is some region R on the surface of a planet which cannot be seen from any point on any of the other planets (if a planet is in the midst of the others, then its surface is completely visible and this region is empty). Prove that the sum of the areas of these visible regions R is exactly equal to the area of the surface of one planet.

Wasn't it proposed for some old IMO? (my teacher showed it to me a long time ago).

The version (2) is an ISL problem from 1981. The version (1) is the easier 2-dim. version of the 3-dim. version (2).

I think (1) isn't that hard (not even the 3D version).

Let's assume a certain point from the circumference of the planet isn't visible from any point from any of the other planets. We draw the tangent to the circle at that particular point. The semiplane defined by this tangent, which doesn't contain the circle, can't contain any points from any circles, because otherwise the point would be visible, but this means that the point is on the convex hull, and not inside it, which ends the proof.

We do the same in the case of spheres, by drawing the plane which is tangent to the sphere at the point which we want to show is invisible.

This only defines the set of points which are invisible from any other points: those points which are ON the convex hull, and not INSIDE it. Proving that this region has an area equal to that of each of the spheres is another problem... :D

An observation: for each sphere which has invisible points, the region which is invisible is a polygon on the sphere; by this I mean a polygon on the sphere, which has arcs of large circles of the sphere as its "sides".

Let a_1, a_2, ... , a_n be a "corresponding" points of planets A_1,A_2, ... A_n. Then it is easy to see that exactly one of them holds the "invisible" property. So if we paint the invisible points on each planet in black and then translate all planets into one each point on the surface will be black exactly one time. So the sum of the areas of all "invisible points" is the area of the sphere of the same radii.


An ordered n-tuple (x_1 x_2,...x_n)
in which each term is either 0 or 1 is called a binary sequence of length n. Let a_n be the
number of binary sequences of length n containing no three consecutive terms equal
to 0, 1, 0 in that order. Let b_n be the number of binary sequences of length n that
contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove
that b_n+1 = 2a_n for all positive integers n.

I solved his awhile ago once (it's an old USAMO), though I don't remember there being a $ \plus{}1$ in the statement you have to prove. Probably me forgetting.

[hide="Hint"]Come up with a recurrence relation for both $ a_{i}$ and $ b_{i}$. You'll end up getting the same one with different initial conditions, and the conclusion will follow immediately.[/hide]


Mr. Reader has six different Spiderman comic books, five
different Archie comic books and four different Garfield comic
books. When stacked, all of the Spiderman comic books are
grouped together, all of the Archie comic books are grouped
together and all of the Garfield comic books are grouped
together. In how many different orders can these 15 comic
books be stacked in a pile with the covers facing up and all of
them facing the same direction? Express your answer as a
whole number.

OK, there are $ 3!$ ways to arrange the groups.  Within them, there are $ 6!$, $ 5!$, and $ 4!$ ways to arrange the comics in the groups, respectively.  So we have $ 3!6!5!4!$ or $ \boxed{12441600}$.

[quote=cw357]Is it just me or is the formatting messed up on the GameBot post?[/quote]

IDK. Why is the text squeezed into 1 side of the page?

[quote=goober123]$                                       $[/quote]

don't post unnecessary stuff

[quote=GameBot]Mr. Reader has six different Spiderman comic books, five different Archie comic books and four different Garfield comic books. When stacked, all of the Spiderman comic books are grouped together, all of the Archie comic books are grouped together and all of the Garfield comic books are grouped together. In how many different orders can these $15$ comic books be stacked in a pile with the covers facing up and all of them facing the same direction? Express your answer as a whole number.[/quote]

ftfy


function
real analysis
real analysis unsolved

Let $X$ and $Y$ be metric spaces, $f: X\rightarrow Y$ a continuous function, $W$ and infinite subset of $X$ such that $f\,|\,W$ is injective, and $x\in X$ a limit point of $W$.  Prove that $f\left(x\right)$ is a limit point of $f\left(W\right)$.

From hypothesis there exist a sequence $\{x_{n}\}\subset W$ tend to $x$ such that $x_{m}\neq x_{n}$ for all $m\neq n$. But then $\{f(x_{n})\}\subset f(W)$ is the sequence tend to $f(x)$ and $f(x_{m})\neq f(x_{n})$ for all $m\neq n$, and we're done!.


Let $ a,b,c$ be non-negative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that
\[ 4\le\left(\frac{1}{\sqrt{1\plus{}a^2}}\plus{}\frac{1}{\sqrt{1\plus{}b^2}}\plus{}\frac{1}{\sqrt{1\plus{}c^2}}\right)(a\plus{}b\plus{}c\minus{}abc)\]

[quote="nayel"]Let $ a,b,c$ be non-negative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ 4\le\left(\frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}} \plus{} \frac {1}{\sqrt {1 \plus{} c^2}}\right)(a \plus{} b \plus{} c \minus{} abc)
\]
[/quote]
$ 4\le\left(\frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}} \plus{} \frac {1}{\sqrt {1 \plus{} c^2}}\right)(a \plus{} b \plus{} c \minus{} abc)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\plus{}b)\sqrt{(a\plus{}c)(b\plus{}c)}\geq4(ab\plus{}ac\plus{}bc).$
But $ \sqrt{(a\plus{}c)(b\plus{}c)}\geq c\plus{}\sqrt{ab}.$
Thus, it remains to prove that $ \sum_{cyc}((a\plus{}b)(c\plus{}\sqrt{ab})\minus{}4ab)\geq0,$
which is equivalent to $ \sum_{cyc}(\sqrt a\minus{}\sqrt b)^2\sqrt{ab}\geq0.$

What about the following?
Let $ a,$ $ b$ and $ c$ be non-negative real numbers such that $ ab \plus{} ac \plus{} bc \equal{} 1.$ Prove that
\[ \frac {19}{4}\le\left(\frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}} \plus{} \frac {1}{\sqrt {1 \plus{} c^2}}\right)\left(a \plus{} b \plus{} c \plus{} \frac {1}{2} abc\right)
\]


Putnam
irrational number
Irrational numbers

Prove that for any irrational number $\xi$, there are infinitely many rational numbers $\frac{m}{n}$ $\left( (m,n) \in \mathbb{Z}\times \mathbb{N}\right)$ such that \[\left\vert \xi-\frac{n}{m}\right\vert < \frac{1}{\sqrt{5}m^{2}}.\]

[quote="Peter"]Could it be this is not from Putnam 1980?  :maybe:[/quote]

Yeah, it is not from Putnam 1980. It seems that some problems in G section has wrong references.
I'm gonna check it.

This is known as Hurwitz's Theorem. Note that the following beautiful proof is not mine.

It's enough to consider only $\xi \in (0,1)$.
For each positive integer $n$, consider the [url=http://mathworld.wolfram.com/FareySequence.html]Farey sequence[/url] $F_n$.
Suppose $\frac{a}{b},\frac{c}{d}$ are two consecutive terms in $F_n$ that $\frac{a}{b}<\xi <\frac{c}{d}$.
We'll show that at least one of $\frac{a}{b},\frac{c}{d},\frac{a+c}{b+d}$ work. Note that this complete the problem by inductively choosing large $n$.

Suppose to the contrary that none of three fractions work. We get that $\xi -\frac{a}{b}\geqslant \frac{1}{\sqrt{5}b^2}$ and $\frac{c}{d} -\xi \geqslant \frac{1}{\sqrt{5}d^2}$.
Adding two inequalities and using $bc-ad=1$ gives us $\frac{1}{bd}\geqslant \frac{1}{\sqrt{5}b^2}+\frac{1}{\sqrt{5}d^2}\implies bd\geqslant \frac{b^2+d^2}{\sqrt{5}}$.
Now assume $\xi >\frac{a+c}{b+d}$ (the other case can be done in similar way.) We get $\xi -\frac{a+c}{b+d} \geqslant \frac{1}{\sqrt{5}(b+d)^2}$.
So, $\frac{1}{d(b+d)}=\frac{c}{d}-\frac{a+c}{b+d}\geqslant \frac{1}{\sqrt{5}d^2}+\frac{1}{\sqrt{5}(b+d)^2}\implies d(b+d)\geqslant \frac{d^2+(b+d)^2}{\sqrt{5}}$.
Hence, $d^2+2bd\geq \frac{b^2+2d^2+(b+d)^2}{\sqrt{5}}\implies -\frac{((\sqrt{5}-1)d-2b)^2}{2\sqrt{5}} \geqslant 0$, which is impossible, done.

Actually it's sufficient to consider $n$ only as a prime number, because then none of fractions $\frac{a}{b},\frac{c}{d}$ repeat and we know there are infinitely many prime numbers.


calculus
integration
logarithms
trigonometry
absolute value
calculus computations

Calculate the following indefinite integrals.

[1] $\int (x^2+4x-3)^2(x+2)dx$

[2] $\int \frac{\ln x}{x(\ln x+1)}dx$

[3] $\int \frac{\sin \ (\pi \log _2 x)}{x}dx$

[4] $\int \frac{dx}{\sin x\cos ^ 2 x}$

[5] $\int \sqrt{1-3x}\ dx$

[quote="djimenez"][hide="Number 2"]This one is nice. Use $u=\ln x$ and then

$\int \frac{\ln x}{x(\ln x+1)}dx=\int \frac{u}{(u+1)}du=u-\ln(u+1)=\ln x-\ln(\ln x+1)$[/hide][/quote]

You are wrong. :(

[quote="amirhtlusa"]#5 seems very simple:
$\frac{-2}{9}\sqrt{(1-3x)^3}$[/quote]

That's right. :)

[quote="amirhtlusa"][quote="kunny"][quote="amirhtlusa"]yeah, for #2 you should let $u=\ln{x}+1$
the final answer:
$\ln{x}-\ln{(\ln{x}+1)}+1$[/quote]

You forgrot something. :([/quote]
hmm i got the same thing..
if you let $u=\ln{x}+1$
you get some thing like this to integrate:
$\int\frac{u-1}{u}$ which is obviously: $u-\ln{u}$[/quote]

What's the sign of $x$ and $\ln x +1$?

[quote="amirhtlusa"]3
$\frac{-\ln{2}}{\pi}\cos({\pi\log_{2}{x}})$[/quote]

That's wrong. :(

come on you mean the absolute value?? :lol:
$\ln{|x|}-\ln|{(\ln{|x|}+1)}|+1$

[quote="amirhtlusa"]come on you mean the absolute value?? :lol:
$\ln{|x|}-\ln{(\ln{|x|}+1)}+1$[/quote]

Yes!

[quote="djimenez"][quote="kunny"][quote="amirhtlusa"]come on you mean the absolute value?? :lol:
$\ln{|x|}-\ln{(\ln{|x|}+1)}+1$[/quote]Yes![/quote]The 1 is an integration constant, therefore it is not that important. The absolute value, well, you are right, but ... just assume x>0 and my solution was right anyway ;)[/quote]
yeah i kind'a agree...


trigonometry
number theory proposed
number theory

How many real solutions does the equation \[x=1964\sin x-189\] have ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Radii $\overline{OA}$ and $\overline{OB}$ of circle $O$ meet at an angle of $120^{o}$.  A square has one vertex on $\overline{OA}$, another on $\overline{OB}$, and two more on minor arc $AB$.  If $OA=13$, the area of the square can be written as $p+q\sqrt{3}$, where $p$ and $q$ are rational.  Find the ordered pair $(p,q)$.

draw a perpendicular from the center to the closest side of the square, if s is the side, then this is s/(2root 3), also we have the perpendicular line bisects the side of the square, s/2, then using the distance formula, 13^2=(s^2)(1+1/2root3)^2+(s^2)(1/2)^2, result follows


geometry
function
real analysis
real analysis unsolved

x(t)=t^3+2t, y(t)=t^4+1, find the area bounded by the curve and x-axis from 0.785398163397 and 1.0471975512.
Plz show me the work.

Bring y as a function of x and then integrate between the specified limits..


probability
combinatorics proposed
combinatorics

8 identical dice are arranged in a 2x2x2 cube. Two touching dice are said to "match" if the side they share has the same number on both dice. If these 8 dice are put together randomly, what is the probability that all 12 pairs of touching dice will match?

Here's a hint:
The answer can be expressed in the form (1/2^a)(1/3^b). Solve for a and b.


Consider the function $f: \mathbb{N} \longrightarrow \mathbb{Z},f(x)=ax+b,a,b \in \mathbb{Z}$.If m,n are positive integers such that $m\neq n$ and
$f(1)+f(2)+...+f(m)=-n$ and
$f(1)+f(2)+...+f(n)=-m$,find the value of $\frac{f(1)+f(2)+..+f(m+n)}{m+n}$

My result to this problem was m+n / n-m ...so it`s " blat " :|

Funny! :D My proof is very similar to hydro's, and I get $1$!

$s(k)=f(1)+f(2)+...+f(k)=\frac{k(k+1)}{2}a+kb$
or, simplified,
$\frac{s(k)}{k}=\frac{k+1}{2}a+b$

We have
$k=m$, $-\frac{n}{m}=\frac{m+1}{2}a+b$.
$k=n$, $-\frac{m}{n}=\frac{n+1}{2}a+b$.
Substracting, we immediately eliminate $b$,
$\frac{-n^2+m^2}{nm}=\frac{m-n}{2}a$, that can be simplified if $m-n\ne 0$, and then $a=2\frac{m+n}{mn}$.
We can also immediately isolate $b$, for instance, $b=-(\frac{n}{m}+\frac{m+1}{2}a)$.

We have
$k=m+n$, $\frac{s(m+n)}{m+n}=\frac{m+n+1}{2}a+b$.
We substitute $b$,
$\frac{s(m+n)}{m+n}=\frac{m+n+1}{2}a-(\frac{n}{m}+\frac{m+1}{2}a)=\frac{n}{2}a-\frac{n}{m}$
And then we substitute $a$,
$\frac{s(m+n)}{m+n}=\frac{m+n}{m}-\frac{n}{m}=1$.

yes, you are right, I just got wrong in the factorization and I wrote -1 instead of 1... ;)

:idea: There's a nice generalisation:

Consider the function $f: \mathbb{N} \longrightarrow \mathbb{Z}$, $f(x)=ax+b$, $a,b \in \mathbb{Z}$. If $m$, $n$ are positive integers such that $m\neq n$, and
$f(1)+f(2)+...+f(m)=g(n)$ and
$f(1)+f(2)+...+f(n)=g(m)$,
with a given function $g: \mathbb{N} \longrightarrow \mathbb{Z}$,
find the value of $h(m,n)=\frac{f(1)+f(2)+..+f(m+n)}{m+n}$ (not a function of $a$ and $b$),
1) with a general $g(z)$, and
2) with the particular case $g(z)=kz^d$, with $k,d\in\mathbb{Z}$.
In the case 2, for every $d>0$ show that $h(m,n)\in\mathbb{Z}$, so $m+n$ divides $f(1)+f(2)+..+f(m+n)$.

[hide="Partial solutions here"]
$h(m,n)=-\frac{g(m)-g(n)}{m-n}$
$h(m,n)=-k\sum_{i=0}^d{m^{d-i}n^i}$, with $d\ge 0$.
And just a little more difficult with $d<0$.
[/hide]
[Edit: I added a simpler relation]

Hey all we need to do is express$\frac a2(1+m+n)+b$ in terms of m,n,g(m) and g(n).And by some boring computations I get that $\frac a2(1+m^2+n^2)+b=\frac{g(m)+g(n)}{m+n}$ from which we get that $\frac a2(1+m+n)+b=\frac{g(m)+g(n)}{m+n}-\frac a2(m^2+n^2-m-n)$ and in order to express $\frac{f(1)+f(2)+f(3)+\cdots f(m+n)}{m+n}$ in terms of m,n,g(m),g(n),all we have to do is express a in terms of m,n,g(m),g(n),which I think is not that hard(it requires merely solving the sistem
$\frac{m^2+m}2a+bm=g(n)$
$\frac{n^2+n}2a+bn=g(m)$,which sets $a=\frac{g(n)\cdot n-g(m)\cdot m}{(m^2+m)n-(n^2+n)m}2$ therefore our desired value is (get ready for this)
$\frac{g(m)-g(n)}{m+n}-\frac{ng(n)-mg(m)}{m(m+1)n-n(n+1)m}\cdot(m^2+n^2-m-n)$.Now I really hope there is no mistake,but I thinks the general case requires just a bunch of computations..anyway if there is a mistake please express it(I`m too tired to go over boring computations and verify them) :D
ok bye

the blat is not over...and yes the result was 1 :| :|

(what's a "blat", and why isn't it over?)

[quote="andyciup"]In the contest I got the result as $\frac{amn}{m+n}-1$ and did not finish the solution because I thought it was enough..looking on the solution I understant that if I had showed that $a=2\frac{m+n}{mn}$,like lordWings did,then I would have gotten a full mark..[/quote]
It would be interesting to know how did you get $\frac{amn}{m+n}-1$. Was it a whole different approach to the problem?
I understand what you said, "I thought it was enough". Until I reached the simple $1$, I didn't exactly know when it was enough. I think the problem should have said they wanted a solution independent of $a$ and $b$.

"blat" is a romanian word and it`s used in many expresions.I got to that form by allmost the same aproach as you..I just added two relations.