In a country everyone knows the area of the country, but none knows the total population of the country. But everyone assumes that the population density is uniform everywhere and same as the density in their neighborhood. The government conducts a census. The authority decides to ask every person his opinion about the total population and calculates the arithmetic mean of the figures and publish the result. Suppose you don't know the actual population, the result and how the population is distributed over the country.

[b]State true or false:[/b]

$ \boxed{1}$ The result depends on the distribution of population, and thus can't be predicted with the given data.
$ \boxed{2}$ With the given data nothing can be predicted regarding whether the result can be less than, more than or equal to the actual population.
$ \boxed{3}$ With the given data the result is less likely to be more than the actual population than it to be less, but possible.
$ \boxed{4}$ With the given data the result is less likely to be less than the actual population than it to be more, but possible.
$ \boxed{5}$ With the given data it is sure that the result can never be more than the actual population.
$ \boxed{6}$ With the given data it is sure that the result can never be less than the actual population.

I think I might be wrong but:

$ \boxed{1}$ True
$ \boxed{2}$ True
$ \boxed{3}$ False
$ \boxed{4}$ False
$ \boxed{5}$ False
$ \boxed{6}$ False

Nice problem, by the way!  :)

[quote="isabella2296"]$ \boxed{1}$ True
$ \boxed{2}$ True
$ \boxed{3}$ False
$ \boxed{4}$ False
$ \boxed{5}$ False
$ \boxed{6}$ False[/quote]
Doesn't match entirely with my answers. :wink:
I could myself be wrong of course :P ...
I'll post my answers later if none finds my answers. :wink:

[quote="isabella2296"]Can you please tell me which ones(if any) were correct?[/quote]
Some are correct. But I won't disclose entirely bcoz that will be equivalent to disclosing my answers. And then ppl will be less likely to try on their own which I don't want. If you want, I will pm you my answers.

[quote=".Ana."]1.True
2. True
3. False
4.False

and hat about  5 and 6. it is hard to say if they are correct or not. if 5 is correct, then 6 must be false  and if 6 is correct, 5 must be false.
 but it is impossible I think one both of them to be true or false. So if With the given data it is sure that the result can never be more than the actual population  is true it can't be also true that
 With the given data it is sure that the result can never be less than the actual population.  and so is if they both are false.

I don't think I said clearly what  I wanted but  one of them "5" or "6" is true and another also or "5" or "6"  is false.

I don't know if this is right but I think it is :)[/quote]

First of your answers doesn't match entirely with mine again.

I think you misunderstood the statement of $ \boxed{5}$ and $ \boxed{6}$. Note that I said "never".

I meant that if we consider all possible circumstances of the distribution of population, under [b]any[/b] circumstances the result can [b]never[/b] be more than the original population.
In other words, suppose there are several such countries where they conduct census the same way. Those countries have different distributions of population etc. In $ \boxed{5}$ it says that there will be [b]no[/b] country where the result could be more than the real population.
Similarly $ \boxed {6}$

So both $ \boxed {5}$ and $ \boxed {6}$ could be false. This implies $ \boxed {2}$ is true.
Also $ \boxed {5}$ and $ \boxed {6}$ both could be true. Meaning that the result is [b]always [/b] same as the actual result. This implies $ \boxed {1}$ is false. This you may have ruled out. I am not still saying your particular answer matches with mine though.

Also if my reasoning is true, $ \boxed{5}$ and $ \boxed{6}$ can be surely answered.

Warning: If my answer is true, the problem is [b]*extremely*[/b] tricky. However it will require only insights. That's why I posted the problem in classroom math. However, now I think I should have used another forum.

I still hope that someone will find my answers.

If anyone is interested, see my answers and the intuitive reason behind them in [url=http://www.mathlinks.ro/viewtopic.php?p=1189482#1189482]this thread[/url] I started.

I think I am wrong

$ \boxed{1}$ True
$ \boxed{2}$ True
$ \boxed{3}$ False
$ \boxed{4}$ False
$ \boxed{5}$ False
$ \boxed{6}$ True

[quote="myyellowducky82"]I think I am wrong[/quote]
Then why did you post? :P  But your answer is nearest to my answers, Which is $ \boxed{TFFFFT}$
You are saying that $ \boxed{6}$ is true(I've proved it [url=http://www.mathlinks.ro/viewtopic.php?t=214847]here[/url]). Then why is $ \boxed{2}$ true? [b]Something[/b] can be predicted, that the result can never be less than the actual population. But it says that [b]nothing[/b] can be predicted. Then $ \boxed{2}$ should be false.
And please tell us how you got that $ \boxed{6}$ is true. I too got at that, but I want to hear your reasoning.

I put True for $ \boxed{2}$ accidentally... oops. I'll post my explanation in a second.

1. True
2. False
3. False
4. True
5. False
6. False

That would've been my guess. :)


induction
number theory
least common multiple
arithmetic sequence
number theory proposed

A positive integer $n$ is said to be a [i]perfect power[/i] if $n=a^b$ for some integers $a,b$ with $b>1$.
$(\text{a})$ Find $2004$ perfect powers in arithmetic progression.
$(\text{b})$ Prove that perfect powers cannot form an infinite arithmetic progression.

a) Let's go for an induction procedure :
First note that 1 and 4 are the first two terms of the arithmetic progression $1+3n$ where $n=0,1,2,...$.
Now suppose that for some $k$, you have an arithmetic progression $x+nr$ and that for $n=0,1,...,k-1$ we have $x+nr = a_n^{b_n}$ for some positive integers $a_n,b_n$ and $b_n >1$.
Let $q= lcm \{b_1,...,b_n \}$.
Let $t$ be any positive integer. Then, using that $t^q(x+nr) = t^qx + n(t^qr)$, we see that $ \{t^q(x+nr) \}$ is an arithmetic progression.
Moreover for $n=0,1,...,k-1,$ we have let $q=b_nc_n$, we then have $t^q(x+nr) = t^{b_nc_n}a_n^{b_n} = (t^{c_n}a_n)^{b_n}$ is a perfect power.
Now choose $t=x+kr$, clearly $t^q(x+kr)$ is a perfect power too, and so we have an arithmetic progression formed with perfect power of kength $k+1$.

It follows that we may find arithmetic progressions of perfect powers with arbitrary long, but finite, length.

b) Suppose that $x+nr$ is an infinite arithmetic progresion form with perfect powers.
Let $d = gcd(x,r)$, $x=dy$ and $r=ds$, where $y,s$ are coprimes.
Then for each $n$, we have $x+nr = d(y+ns)$.
Now, from the well-known theorem of Dirichlet about primes contained in the arithmetic progressions, we know that there are an infinite number of primes of the form $y+ns$. Thus, we may find $n$ such that $y+ns$ is a prime $p$ which does not divide $d$.
In that case $d(y+ns)$ is a term of the sequence, it is divisible by $p$ but not by $p^2$. Therefore it cannot be a perfect power. A contradiction.

Pierre.

It was in our exams last year (Iran 2003)

for a) you can find  number $d$ such that all numbers $d, 2d... nd$ are all perfect powers. Solution is boring, based on CRT. First you have to choose all primes less or equal than $n$ and carefully choose their exponents in factorization of $d$. Choosing is similar like  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=117503&sid=c792bd67b26d3da8c862eb112a28e1fb#p117503 -  here of course you need to use $n^k$ primes- there  only $ n^2$ were enough.

[hide="a"]We inductively construct a sequence $x_1,x_2,\cdots ,x_n$ of natural numbers in arithmetic progression such that $x_1=y_1^{t_1},\cdots,x_n=y_n^{t_n}$, with the $y_i,t_i$ natural numbers greater that one.

[b]Base case[/b] - $n=2$:
We put $x_1=4=2^2$, $x_2=9=3^2$.

[b]Induction step[/b]:
Let $x_{n+1}=2x_n-x_{n-1}$ (so that $x_1,x_2,\cdots,x_{n+1}$ are in arithmetic progression), $k=x_{n+1}^{\prod_{j=1}^nt_j}$.
Then we have that $x'_i=kx_i$ are $n+1$ naturals in arithmetic progression (we multiplied an arithmetic progression by a constant), and each is a perfect power (because if $i\leq n$, $x'_i$ is the product of two $t_i$-th perfect powers, and if $i=n+1$ we have that $x'_{n+1}=(x_{n+1})^{1+{\prod_{j=1}^nt_j}}$)

This concludes the proof that for any natural number $n$ there is such a sequence, as desired.[/hide]
[hide="b"]
Suppose that there is such an arithmetic progression $x_n$. Then we have that $\sum_{n=1}^{\infty}\frac{1}{x_n}=\infty$.

But on the other hand $\sum_{n=1}^{\infty}\frac{1}{x_n}\leq 1+\sum_{a=2}^{\infty}\sum_{b=2}^{\infty}\frac{1}{a^b}=1+\sum_{a=2}^{\infty}\frac{1}{a(a-1)}=2$, contradiction, as desired.
[/hide]


algebra
functional equation
algebra solved

Determine all pairs (a,b) of nonnegative real numbers such that the functional equation f(f(x)) + f(x) = ax + b has a unique continuous solution f : R -> R.

Well, it is easy to see that if we look for solutions of the form f(x) = rx + t, we have r = (-1  \pm  \sqrt (1 + 4a) )/2 and t(r+2) = b.
Thus, if a =/= 2, then r =/= - 2 and there are at least two continuous solutions. Thus, we must have a = 2 and then r = -2).
Moreover, in that case, if b = 0 there still are two continuous solutions(the same as above).
Thus, we must have b > 0 (since b is nonegative).
In that case, there is at least one continuous solution, which is f(x) = x + b/3.

It remains to prove that it is the only one....if true.

Pierre.

I think I have a solution to this one. Let g(x)=f(x)-x-b/3. One can see that f doesn't have fixed points , so g is bounded (superior or inferior, it depends on how f(x)>x for all x or f(x)<x for all x). But also g(f(x))=-2g(x) and so g(f^n(x))=(-2)^ng(x). Using the bounded property, we find that g is zero.


Is there a 2000-element subset $A$ of $\{1,2,...,3000\}$ such that $2x$ is not in $A$ whenever $x$ is in $A$?

No ... you would need at least $4000$ elements in your other set ...

By the principle of inclusion-exclusion, we have that such a set selected from $\{1,...,3000\}$ can have a maximum of $\sum_{k=0}^{\infty}(-1)^{k}\lfloor \frac{3000}{2^{k}}\rfloor=1999\neq 2000$.

Masoud Zargar

[quote="neelnal"]Answer is no.

2x=3000
x=1500 is the highest we can go. Why do we need to use all of those complicated sums?[/quote]

That's not true. Not all of the numbers have their corresponding double in the set. We can choose all of the numbers $1501, 1502, \ldots, 3000$ and still be ok. Clearly the answer is more than $1500$. We can then proceed to pick the ranges

$[376,750]$
$[94,187]$
$[24,46]$
$[6,11]$
$[2,2]$

to get a total of $1999$ numbers. So the problem is NOT trivial.

Dang, I got misleaded and everybody else followed  :rotfl:  Sorry for that guys ... but a hint for next time: usually I answer olympiad problems only if I think they're trivial, cuz they're big time suckers !!!


Can anyone help me get this question?

What is the integer $x$ which satisfies
$10^{x}<\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times\frac{7}{8}\times...\times\frac{99}{100}<10^{x+1}$

Oh this is a fun problem.  The main idea is to multiply the expression by another expression x to get a simple result.  Now compare the original expression to the original expression times x.

[hide]That can be re-written as

It's clear that  is less than , so the expression will be between  and . Thus, .

Somehow I think what I did on line 4 is wrong (Making the product distributive).[/hide]

Um, that's impossible. You start out with $\frac{1}{2}$, and it becomes less and less. How can the answer be close to 1?

wait nvm...
no i think he's right...
cuz it says its less than .9
not that it is..
so its no where close to 1
but how do we kno its not between.... $10^{-2}$ and $10^{-1}$?

I haven't worked with product of series long enough to know what manipulations are legal, but somewhere there must be an error, but I can't see it.  :(

[quote="Elemennop"]$\prod_{n=1}^{50}\frac{2n-1}{2n}$

$\frac{1}{2}\prod_{n=1}^{50}\frac{2n-1}{n}$
[/quote]I think you messed up on the second step.  Wouldn't it be $(\frac{1}{2})^{50}\prod_{n=1}^{50}\frac{2n-1}{n}$?  But I'm not sure about this, I'm not very familiar with the product notation.

[hide]Well I found that the bottom is just $50! * 2^{50}$ and the top is $\frac {100!}{50!*2^{50}}$ so we get $\frac {100!}{50! 50! * 2^{100}}$. After a lot of killing stuff off and a filled up piece of scratch paper, I have determined that this falls between .01 and .1. therefore $x = -2$.[/hide]

tarquin's answer is correct.  Can you explain your solution in a bit greater detail?

I would have to have factored out $\frac{1}{2^{50}}$, but that just makes the answer even larger.

Probability is right, the way I distributed was incorrect (what I was guessing to begin with).

I need to work with products more.

I might as well give the hint that's about 95% of the solution.  Let the expression be x.  I'll let you prove that x < 1/(100x).  Then x^2 < x/(100x) = 1/100.

[quote="probability1.01"]You can't distribute products like that!  That's like saying (a+b)^2 = a^2 + b^2.  Hint: first prove that the expression is less than 2/3*4/5*6/7*...*100/101[/quote]Well, we know that $\frac{2}{3}>\frac{1}{2}$ and $\frac{4}{5}>\frac{3}{4}$, and this works all the way up to $\frac{100}{101}>\frac{99}{100}$, I guess this isn't really a formal proof...
Judging from your first hint, am I supposed to multiply these two expressions together?  The cancel out quite nicely.

Edit: oops, I didn't see those two posts above while making this post.

Ah, good catch! That's the problem there.

What would this look like using products?

probability, can you tell me your thought process in coming with the solution?

If you multiply the fraction by the the fraction 2*4*6*...*100/2*4*6*...100, cant you simplify it greatly.

Sorry, I didn't make my post very clear.  I understand how to do the problem now, what I was asking was what you thought when you read the question and how you came up with the idea of using the second sequence.

First of all, I've seen this problem before.  When I first solved it, it was with some guidance.  But it should pop into your mind pretty early to try somehow multiplying by that second sequence to simplify things.  Then if you're lucky/experienced you'll think of the rest.

telescoping comes to mind...so you think of those missing intermediate terms

Actually The answer  can be easily shown  between $1/12$ and $1/13$

In fact:

  $\frac 1 2 \times \frac 3 4 \times \dots   \times  \frac {2n-1}{2n} $  can easily be shown between

$\sqrt {\frac {1}{n \pi}} $  and $\sqrt {\frac{2}{(2n+1)\pi}}$

For more detail see the post  [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=%281%2F2%29%2A%283%2F4%29&t=15144]here[/url]


Prove that :
$\frac{a^{4}b}{2a+b}+\frac{b^{4}c}{2b+c}+\frac{c^{4}}{2c+a}\geq 1$
where a;b;c are positive numbers satisfying the condition $ab+bc+ca \leq 3abc$.
When does equality occur?

[quote="arqady"][quote="cefer"]
I will use well known inequality $\frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z}\geq \frac{(a+b+c)^{3}}{3(x+y+z)}$
[/quote]
It's the Holder's inequality! :P
Nice proof, cefer. :)[/quote]

sorry, but is isn't that power mean inequality(by some transformation), i don't know the holder's inequality you were referring to  :blush:

[quote="andy_tok"] i don't know the holder's inequality you were referring to  :blush:[/quote]
see http://en.wikipedia.org/wiki/Holder%27s_inequality :D


f:[0,oo]->[0,1) continuous and (x_n), x_0=1, x_(n+1)=x_nf(x_n)
a)Prove that x_n is convergent and find it's limit
b)Compute: lim{n->oo} \sum {k=1,n}{x_k(1-f(x_k))}

???? If x_0= 0 then x_n = 0 for each n.

Pierre.

a) All x_n  \geq 0
x_(n+1)-x_n = x_n(f(x_n)-1)  \leq 0
x_n  decreasing positive => lim x_n = L  \geq 0 exist

Suppose that L>0, f is continous => L=Lf(L) => f(L)=1 contradiction since the value of f is in [0;1[
So lim x_n = 0

b) \sum {k=1,n}x_k(1-f(x_k)) =

\sum {k=1,n}x_k - \sum {k=1,n} x_k.f(x_k) =

\sum {k=1,n}x_k - \sum {k=1,n}x_(k+1) =

x_1 - x_(n+1) ---> x_1=f(1)


AMC
AIME
calculus
real analysis
real analysis unsolved

Find $ x,y,u,vt\in R$ such that they are satisfying simultan the conditions:
$ u\plus{}v\equal{}2$
$ ux\plus{}vy\equal{}1$
$ ux^2\plus{}vy^2\equal{}\minus{}1$
$ ux^3\plus{}vy^3\equal{}\minus{}5$                                                                  (Exam 2006,UPB)

1990 AIME number 15 is a very similar question, and the solution method that works there works here, too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80286
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68753
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6820

This also has nothing to do with calculus ;)

I was too lazy to do a single non-trivial computation (not even to solve a 3 by 3 linear system), so I came up with the following "solution".

The curve $ t\mapsto(1,t,t^2,t^3)$ has the nice property that every 4 distinct points on it are linearly independent in $ \mathbb R^4$ (the corresponding determinant is a Vandermonde determinant). The equations tell us that a certain non-zero point belongs to the plane generated by 2 points on that curve. Note that the point itself doesn't lie on the line passing through any point on the curve ($ 1\cdot( \minus{} 1)\ne 2\cdot( \minus{} 5)$). Thus, the 2 points in question are unique (otherwise the corresponding planes would either have the trivial intersection or intersect by a line containing the common point on the curve). Thus, the solution is essentially unique (up to the order of points). Looking into the crystal ball attentively, we see that $ (2,1, \minus{} 1, \minus{} 5) \equal{} 3\cdot(1,1,1,1) \plus{} ( \minus{} 1)\cdot(1,2,4,8)$ (I understand that this point in my argument is somewhat weak, but it is still formally correct :P). The end.

Of course, the recursion trick is better in the sense that the crystal ball is not needed for solving the problem, but I decided to post this anyway.


algebra
polynomial
linear algebra
matrix

can anybody tell me what's the minimal polynomial for $ T: \mathbb R_2[x]\to\mathbb R_2[x]$ where $ T(a\plus{}bx\plus{}cx^2)\equal{}b\plus{}2cx\plus{}2(a\plus{}bx\plus{}cx^2)$ ?

i already have the answer, i wanted to know if it's the correct one or not.

$ T$ has the following matrix: $ A\equal{}\begin{bmatrix}0&1&0\\0&0&2\\2&2&2\end{bmatrix}.$

(Some details depend on what order you choose to write the basis in.)

Now, $ A^2\equal{}\begin{bmatrix}0&0&2\\ *&*&*\\ *&*&*\end{bmatrix}.$

Just from looking at the top rows, we see that $ \{I,A,A^2\}$ is a linearly independent set. Hence, $ A$ satisfies no degree two (or lower) polynomial. That means that the minimal polynomial must be the characteristic polynomial, which we just compute directly:

\[ x^3\minus{}2x^2\minus{}4x\minus{}4\]

Note that this has no rational roots.

Right. But in any other basis, the matrix would be similar and the minimal polynomial would be the same.

ohhh... can you please explain how you got that matrix?


We usually measure angles in degrees but we can use any other unit we choose. For example, if we choose $ 30^{\circ}$ as a new unit, then the angles of a $ 30^{\circ}, 60^{\circ},90^{\circ}$ triangle would be equal to $ 1, 2, 3$ new units respectively.

A Triangle $ ABC$ has another triangle $ DEF$ inscribed in it with $ D$ on $ BC$, $ E$ on $ CA$ and $ F$ on $ AB$.

$ \angle BAC \equal{} a, \angle ABC \equal{} b, \angle ACB \equal{} c, \angle FDE \equal{} d, \angle DEF \equal{} e, \angle EFD \equal{} f, \angle AFE \equal{} g, \angle AEF \equal{} h, \angle BFD \equal{} i, \angle FDB \equal{} j, \angle EDA \equal{} k, \angle DEA \equal{} l$.

All the angles are whole number multiples of some new (unknown) unit; their sizes $ a, b, c, d, e, f, g, h, i, j, k, l$ with respect to this new angle unit are all distinct.

Find the smallest possible value of $ a \plus{} b \plus{} c$ for which such an angle unit can be chosen, and find the corresponding values of $ a$ to $ l$.


Out of my own curiosity, what kinds of levels of expertise do you guys have in computer science coming out of high school?  (or still in it, if you're there)

What kinds of capabilities do kids have who, say, get a 4 or 5 on Computer Science AP?  Can you program a game or other small piece of software on your own?  Or do you just know theory?

I'm asking because I've had some discussions with some corporate folks about launching a new competition involving high schoolers and computer programming...the idea we're kicking around is giving teams of high schoolers - say, teams of 5, like the AHSIMC, or perhaps as many as 10 - a month to program a game.

The best game, as voted on by a panel of selected judges from within the industry, would win.

Each member of the winning team would get a top-of-the-line PC from Alienware or another high-end computer firm.

Comments?  Is this type of programming knowhow required for this type of competition within current school curriculum, or only something students would know how to do based on outside personal experience?

Any insight you can provide would be great.

Three members of my AHSIMC team, including myself, got 5s on computer science AB.  Another got a 4 on the A.  I haven't done any programming lately, though.  One guy made a sort of Monopoly game for his computer science final project.  It was pretty cool.  It even had a cheating feature.  He's probably good enough to make the IOI (I think that's what it's called) team, but he's too lazy.  He did all of our team's brute-force stuff.

I my school, the only computer classes are for Word and Powerpoint, not really any programming.  I know some, from what I've learned, but I can't program a game, the graphice and the user interface part would be really hard for me.

Questions:

Does the game have to work on a certain OS?
Does it have to be a certain programming language?


If $ x$, $ y$, and $ z$ are positive numbers satisfying \[x \plus{} 1/y \equal{} 4,\quad  y \plus{} 1/z \equal{} 1,\quad\text{and}\quad z \plus{} 1/x \equal{} 7/3,\] then $ xyz \equal{}$

$ \textbf{(A)}\ 2/3 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 4/3 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 7/3$

Ok, I stink at tex, so it'll be a little messy, but here goes nothing...

[hide="solution"]
First you can multiply the original equations together:
4*1*7/3=(x+1/y)(y+1/z)(z+1/x)

After multiplying this out, we get:
28/3 = xyz+x+y+z+1/x+1/y+1/z+1/xyz

Then, we can alter the original equations slightly:

x+1/y=4
x=4-1/y

Doing the same to the other equations, we get:

x=4-1/y
y=1-1/z
z=7/3-1/x

Then, we can substitute these into the product of these equations:

28/3 = xyz+4+1+7/3+1/x+1/y+1/z-1/x-1/y-1/x+1/xyz
28/3 = xyz+1/xyz+22/3
2 = xyz+1/xyz
Then, we just substitute u in for xyz
2 = u+1/u
2u = u^2+1
u^2-2u+1=0

This is easily factorable:
(u-1)^2=0
Then take the square root:
u-1=0
u=1
xyz=1
So the answer is B.[/hide]

[hide="Quicker?"]
Adding the equations, we see that

$ x\plus{}y\plus{}z\plus{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{22}{3}$.

Multiplying the equations, we find

$ x\plus{}y\plus{}z\plus{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}xyz\plus{}\frac{1}{xyz}\equal{}\frac{28}{3}$.

Subtracting the first equation from the second,

$ xyz\plus{}\frac{1}{xyz}\equal{}2$.

AM-GM implies that $ xyz\equal{}\frac{1}{xyz}$, so $ xyz\equal{}1$.  Answer is $ B$.
[/hide]

It would be nice to demonstrate that $ x,y,z$ actually exist.

Since $ xyz\equal{}1$, we can eliminate $ z$, and get

\[ x\plus{}\frac{1}{y}\equal{}4\qquad y\plus{}xy\equal{}1\qquad \frac{1}{xy}\plus{}\frac{1}{x}\equal{}\frac{7}{3}\]

The third equation can be rewritten as $ 1\plus{}y\equal{}\frac{7}{3}xy$. Using that with equation $ 2$ gives us a linear system in $ (y,xy)$. It follows that $ (y,xy)\equal{}\left(\frac{2}{5},\frac{3}{5}\right)$. So $ x\equal{}\frac{xy}{y}\equal{}\frac{3}{2}$. Then $ z\equal{}\frac{1}{xy}\equal{}\frac{5}{3}$. That means that a satisfactory triple is \[ (x,y,z)\equal{}\left(\frac{3}{2},\frac{2}{5},\frac{5}{3}\right)\] we can see that these values satisfy the original equation.

Altheman's solution is also far more obvious on a 75-minute test.

[quote="E^(pi*i)=-1"]
[hide]
AM-GM implies that $ xyz \equal{} \frac {1}{xyz}$, so $ xyz \equal{} 1$.
[/hide]
[/quote]

Why is this true? Also, is it true for all xyz, or just in this situation?

Since $ x$, $ y$, and $ z$ are all positive, we know that their product will be positive.  So we can apply AM-GM:

$ \frac {xyz \plus{} \frac {1}{xyz}}{2}\geq \sqrt {xyz\left(\frac {1}{xyz}\right)} \equal{} 1$.

We already know that $ xyz \plus{} \frac {1} {xyz} \equal{} 2$, so equality holds.  AM-GM only has equality when everything is equal, so $ xyz \equal{} \frac {1}{xyz}$.

AM and GM stand for arithmetic mean and geometric mean, btw.

[quote="E^(pi*i)=-1"] AM-GM only has equality when everything is equal, so $ xyz \equal{} \frac {1}{xyz}$.[/quote]
Oh, I never knew that. Thanks.

The third equation gives z = (7/3) - (1/x). Then substituting this into the second equation, y = 1 - (1/z) = (4x-3)/(7x-3). Finally substituting in the first equation, x + (7x-3)/(4x-3) = 4 which gives 4x^2 -12x +9 = 0, that is, (2x-3)^2 = 0, so that x = 3/2. Then y=2/5 and z = 5/3.

$x+\frac{1}{y}=4\ \cdots (1),\ y+\frac{1}{z}=1\ \cdots (2),\ z+\frac{1}{x}=\frac 73\ \cdots (3)$

From $(1),\ (2)$, $x=\frac{4y-1}{y},\ z=\frac{1}{1-y}\ \cdots (4)$, thus from $(3)$, we have $\frac{1}{1-y}+\frac{y}{4y-1}=\frac 73\ \therefore 25y^2-20y+4=0\Longleftrightarrow (5y-2)^2=0$, yielding $y=\frac 25.$
Then from $(4)$, we get $x=\frac 32,\ z=\frac 53\Longrightarrow xyz=1.$


calculus
integration
function
real analysis
real analysis unsolved

Discuss the convergence of the squence $ \int_0^1 x^{\minus{}1/n}|f(x)|dx$ for $ f \in L^1[0,1]$.

Anyone having any idea? Thanks.

It is a non-increasing sequence of non-negative functions ;)

That is right. But how can you use the Lebesgue dominated convergence theorem if you have no dominator? Or am I missing something?

There are two cases: either one of the integrals is finite (then it provides the majorant), or all of them are infinite (then there isn't much more to say).


Add parenthesis to make this equation true.

$\frac{9 + 10 - 17 + 3}{4 - 5 + 3}= 5$

[hide]

( 9+10-17+3) / (4-5+3) = 5

is there really a way to do this?  my thinking was that since its all addition, the paranthesis dont matter except for 2 spots: around the 17 and 3 at the top or around the 5 and 3 at the bottom.  none of those work! i hope this isn't oneof those riddle things :)

Oops...there isn't supposed to be a 9 in the numerator.
Here is the new problem:
$\displaystyle \frac{10 - 17 + 3}{4 - 5 + 3}= 5$

(10-(17+3))/(4-5+3) = -5

hmm are you sure there's an answer even now?  :lol:

the best you can get is -5 I think

[hide]yes i agree with Barnacle, i got $\frac{[10-(17+3)]}{(4-5+3)}=-5$[/hide]


algebra polynomial geometry limit trigonometry function

can someone tell me what pi equals as a fraction? Or am I just asking for too much? PLEASE REPLY!

$\pi$ is irrational as well as transcendental, which means that it cannot be expressed as a fraction or as a root of a polynomial with integer coefficients.
However, there are two good fraction approximations of $\pi$: $22/7$, and $355/113$.
And I'm assuming you know the first few digits, at least.

There's the Gregory-Leibniz Series:
$\pi$ $=$ $\frac{4}{1}-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}...$
Not a fraction but it has fractions  :oops: .
Source: http://en.wikipedia.org/wiki/Pi

Pi is cirumference divided by diameter and the area of a circle divided by its radius squared. Also, pi is about 3.141592653589793238462643383279502884197169399375105820974944.
And I think pi can be a continued fraction.

[quote="7h3.D3m0n.117"]Well isn't there a limit definition? for example e can be expressed by

[img]http://www.artofproblemsolving.com/Wiki/images/math/2/1/5/2158d71ea12e95f58290eb18897a0c3d.png[/img]

but that is beside the point

$\pi$ is irrational but you can (to a certain nubmer of digits) make a fraction[/quote]
There is a limit defition.
$\pi=\lim_{n\rightarrow\infty}{\left(\frac12n\sin\frac{360^\circ}n\right)}$

There is also the Wallis product

$\pi=2\cdot\frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdots}$

Or $\pi=2\prod_{n=1}^\infty\left(\frac{4n^{2}}{4n^{2}-1}\right)$

Also, applying the Zeta function, which is $\zeta(x)=\sum_{i=1}^\infty \frac{1}{i^{x}}$, results in some patterns involving pi when $x$ is even.

$\zeta(2)=\sum_{i=1}^\infty \frac{1}{i^{2}}=\frac{\pi^{2}}{6}$
$\zeta(4)=\sum_{i=1}^\infty \frac{1}{i^{4}}=\frac{\pi^{4}}{90}$

I have no idea why.

there is no definite fraction, but most people use 22/7 or the easy to remember 113355 fraction (335/113).
like davidyko said :rotfl:

I use $\frac{1087}{346}$ because I like to be original!

I changed it!
it can't be better than the "official" good ones
because I just found two random digits

[quote="t0rajir0u"][quote="davidyko"]However, there are two good fraction approximations of $\pi$: $22/7$, and $355/113$. [/quote]

Only two?

There are in fact an infinite series of such [b]convergents[/b] $\frac{p_{n}}{q_{n}}$, which have some fascinating properties.

http://mathworld.wolfram.com/PiContinuedFraction.html

[quote="8parks11"]I use $\frac{1178}{375}$ because I like to be original![/quote]

Unfortunately, this is not a convergent, and is in fact much worse than $\frac{355}{113}$.[/quote]

Hmm, now that is interesting.
I guess I meant "there are only two reasonably accurate fraction approximations of $\pi$ that I can think of right now".  :lol:

Or, you could check my signature  :D

(only true for the 83+ line of calculators)

see

http://en.wikipedia.org/wiki/Continued_fraction

in Continued fraction expansions of π

I usually approximate with $\frac{104348}{33215}$

OK, I think the question has been answered and this doesn't need to become a "Who can name the most accurate representation of Pi Contest."

Locked.


geometry
trigonometry
3D geometry
perimeter
circumcircle
inequalities
ratio

1.[EASY] THe MIDpoints P, Q, R, S of the sides of quadrilateral ABCD are connected to form quadrilateral PQRS. Also, THe MIDpoints W, X, Y, Z of the sides of quadrilateral PQRS are connected to form quadrilateral WXYZ. If the area of quad ABCD is one-half the product of the diagonals. What kind of quad is WXYZ?

2. [EASY] What is the area of the largest semincircle that can be drawn in an isoscles right triangle with hypotenuse 4?

3. [EASY] Lines SP and UE are secants of a circle that intersect at R, an interior point of the circle. IF measure of angle SPE is 26 deg, and the measure of minor arc UP is 46 deg., find the measure of angle SRU.

4. [EASY] the sine value of  each internal angle of a heptagon is greater than 0.5. WHat is the maximum number of acute angles that the heptagon can have?

5. [Moderate] The two tangent lines to a circle from a point make an angle of 45 deg., Of the dianmeter of the circle is 8, find the area of the region bounded by the two tangent lines and the circle.

6. [Moderate] 2009 unit cubes are glued together to form a solid cube. What is the maximum surface area of the figure.

7. [Moderate] B and X are points on angle CAX (w;c measure 45 deg), such the B is between A and C. If AB=8, BC=10, and D lies on the other ray, how far shud B be from A so that angle CDB is a maximum.

8. [EASY] A circle of radius 1 is inscribed in a 3-60-90 triangle. Find the perimiter of the triangle.

9. [EASY] Sqaure ADGJ is inscribed in a unit circle. Points B, C, E, F, H, I, K, and L are selcted on the circumference of the unit circle such that a regular dodecagon ABCDEFGHIJKL is formed. Find the area of the region outside the sqaure but inside the dodecagon.

10. [MOderate] Find all values of x if 10, 11, and !x+12! (abs value i mean)
 are the lengths of the sides of the triangle.

11. [HARD] IN triangle ABC, AB=6, BC=8, and AC=7. D is on side AC such that AD =3, and E is on side AB while F is on side BC. IF the areas of triangle ADE, CDF, AND quadrilateral BEDF are equal, find BE+DF.

#11
first use heron's foruma find the area of ABC
sqr root [(21/2)(9/2)(7/2)(5/2)] = 21*sqr rt(15)/4

the triangle ABC is divided in to 3 parts ADE = CDF = BEDF
so divide the area by 3

ADE = CDF = 7*sqr rt(15)/4   (these are the areas of the 2 little triangles)

Now use Law of Cosine to find angle A and angle C

A = Arc cos (1/4),     C = Arc cos (11/16)

Sin(A)= sqr rt(15)/4,   Sin(B) = 3*sqr rt(15)/16

Use the foruma:

Area of a triangle = 1/2*a*b*sin(C)

ADE = 1/2 AE*AD*sin(A)
AD = 3 (Given),   Sin(A)= sqr rt(15)/4,  area of ADE = 7*sqr rt(15)/4

AE = 14/3
AB = 6,

BE = AB - AE =4/3
(we just found the first part of the problem)

do the same thing with the other little triangle CDF

CDF= 1/2 CF*CD*sin(C)
CD = 7-3=4,   Sin(C)= 3*sqr rt(15)/16,  area of CDF = 7*sqr rt(15)/4

CF = 14/3

Then use law of cosine
DF^2 = CD^2 + CF^2 - 2(CD)(CF)*cos(C)
DF = sqr rt(109)/3

DF+BE = (4+sqr rt(109))/3

(i am not sure if the numbers are right...)

Key lemma for #7 (well-known): The point D on AX which will maximize $ \angle BDC$ is the point of tangency of the circle which passes through B and C and is tangent to AX.
Proof:For any other point K on AX, the circumcircle of KBC will be larger because the circle will cut through the line, not just touch it, and thus $ \angle BKC$ will be smaller since it is a bigger circle subtending the same size chord.
I actually forgot the lemma but rediscovered it once I wanted to find the locus of points such that BDC is constant, which is a circle.

[hide="#8"]

I take it you mean 30-60-90 right triangle. Let the shorter leg be $ s$, the longer leg be $ s\sqrt3$, and the hypotenuse $ 2s$. We can express the area in two ways:

$ \frac {s^2\sqrt3}{2}$ by $ bh/2$ and $ \frac {s \plus{} s\sqrt3 \plus{} 2s}{2} \equal{} \frac {s(3 \plus{} \sqrt3)}{2}$ by $ rs$

If we set these equal and solve for s, we have $ s \equal{} 1 \plus{} \sqrt3$. So, the perimeter is $ 6 \plus{} 4\sqrt3$

[/hide]

[hide="#10"]

By the triangle inequality, we have \[ 10\plus{}11>|x\plus{}12|\] \[ 10\plus{}|x\plus{}12|>11\] \[ 11\plus{}|x\plus{}12|>10\]

Simplifying the first gives $ \minus{}33<x<9$. The second yields $ x>\minus{}11$ and $ x<\minus{}13$. Since the third works for all real x, we ignore this one. Combining the inequalities gives $ \minus{}33<x<\minus{}13$ and $ \minus{}11<x<9$

[/hide]

COrrect answers were given to 8 and 10.

#11 14/3-> correct answer

#2
[hide]You can position the semicircle such that it touches the legs of the triangle and its center lies on the midpoint of the hypotenuse.  Drawing the line from this midpoint to the tangent point on one of the legs, you get similar triangles in ratio $ 1: 2$.  The leg of the triangle has length $ 2\sqrt {2}$, so the leg of our smaller, similar triangle is $ \sqrt {2}$.  The area of this semicircle is $ \boxed{\pi}$.[/hide]
#5
[hide]Let the tangent points to the circle be $ A$ and $ B$ and the center of the circle be $ O$.  Also, let the tangents meet at $ E$. Draw $ AB$. We now have an isosceles triangle with $ < E \equal{} 45^{\circ}$.  Drawing the altitude from $ A$ to $ BE$ to $ F$, and letting it have length $ \beta$, we get that $ AE \equal{} \sqrt {2}\beta$.  We then find that $ BF \equal{} \beta\sqrt {2} \minus{} \beta$. Using right triangle $ \triangle {AFB}$, we get that $ AB \equal{} \beta\sqrt {4 \minus{} 2\sqrt {2}}$.  Now, let's look at triangle $ AOB$.We have that $ < AOB \equal{} 135^{\circ}$.  Draw the altitude from $ A$ to the extension of $ OB$ to the point $ G$.  $ < GOA \equal{} 45^{\circ}$!  So, since $ AO \equal{} 4, AG \equal{} OG \equal{} 2\sqrt {2}$.  By the Pythagorean Theorem, and then setting it equal to our other expression for $ AB$ we then find that $ /beta \equal{} \frac {4\sqrt {2 \plus{} \sqrt {2}}}{\sqrt{4 \minus{} 2\sqrt {2}}} \equal{}$.  Finally, we can multiply this by $ \sqrt {2}$ to get our tangents.  After some simplifying and some algebra, we get that this equals $ 1 \plus{} \sqrt {2}$.Now, we just have to find the areas of the two triangles, which is just $ 4 \plus{} 4\sqrt {2}$.  The area of the circle is just $ 16\pi$.  The area of their overlap is $ \frac {135}{360} \bullet 16\pi \equal{} 6\pi$. Adding and subtracting the necessary results, we get that the area of the total figure is $ \boxed {10\pi \plus{} 4 \plus{} 4\sqrt {2}}$.[/hide]
And all of that without any trig =]

EDIT: fixed latex

Correct for number 2. Thanks!

#5 should be 16(sqrt2)+16-6pi


[quote="ProblemSets"]If we were to simplify the following...

(1+3+...+(2n-1)) / (2+4+...+2n)

would this simplify to

n/(n+1) or (n-1)/n

One strategy leads me to the first answer and the other to the second answer.

THANKS![/quote]

[hide]$\frac{1+3+...+(2n-1)}{2+4+...+2n}$

$\frac{n^{2}}{2(1+2+...+n)}$

$\frac{n^{2}}{n(n+1)}$

$\frac{n}{n+1}$[/hide]

yeah
its clearly
[hide]
(n^2)/(n^2+n)=n/(n+1)

[/hide]
I SO WISHED I could have the time to learn Latex

[quote="ProblemSets"]If we were to simplify the following...

(1+3+...+(2n-1)) / (2+4+...+2n)

would this simplify to

n/(n+1) or (n-1)/n

One strategy leads me to the first answer and the other to the second answer.[/quote]

A very simple way to check your answer:  Plug in $n = 1$.  The expression is $\frac{1}{2}$, which invalidates the second answer.


Find the sum of the first $ 30$ odd numbers

The formula for the sum of the first $ n$ odd numbers is $ n^2$.
Thus, we have $ 30^2$ or $ 900$.


I am using texniccenter and miktex on windows. IS there any way to get the font that it uses to use on mS word. The TTF file is fine.

yes, but don't expect it to be easy. See [url=http://www.radamir.com/tex/ttf-tex.htm]Using TrueType fonts with TeX (LaTeX) and pdfTeX (pdfLaTeX)[/url]

No. I mean like the default font in latex. I want the TTF for that or the font name at least so I can use the latex font in MS Word.

The default font is probably Computer Modern. What do you mean by [quote]The TTF file is fine[/quote] - what TTF file? The cm fonts come in lots of different sizes eg ftp://ftp.dante.de/tex-archive/fonts/cm/ps-type1/bakoma/ttf/ and the UM fonts in um.zip at http://www.truetex.com/ Notice that there are also separate versions of the fonts for bold and italics.

Nono. i mean if I type a report in latex (no math though) and don't use math mode. Then I get a font not Times New Roman. I want that

Yes, the default text fonts are Computer Modern. It's not a single font but a family of fonts. You'll see the fonts I mentioned above in the LaTeX log for your document.

For example, cmr10 is for upright text 10 point and cmmib10 is bold face math font as you'll see on page 5 of [url=http://www.latex-project.org/guides/fntguide.pdf]LaTeX2e Font Selection[/url]

What do you want to do with Word that you can't with LaTeX? Why would you want to use Word anyway?


well , linear algebra will be the course taught for us fromm the upcoming semester starting from january ,

i want to know which is the most widely used book for this branch of mathematics


blogs
geometry
rectangle
combinatorics proposed
combinatorics

2.A room in the shape of a rectangular has been covered by some rectangular-shaped carpets such that each point in the floor is covered by exactly one carpet. Prove that the sum of the width of the carpets is not smaller than the width of the room.

(Note. The width of a rectangular is it's smallest side's lenght)

here's it's link in my blog:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=749422#749422

Let p1,...,pn and q1,...,qn be lines which contain horizontal and vertical
sides of rectangles (p1 and pn as well as q1 and qn contain sides of big rectangle).
Then one of 2 statements must be true:
(1)	in sequence p1,...pn, distances between two consecutive lines
      are parts of smaller sides of some little rectangles
(2)	the same thing about q1,...,qn.
In any case the conclusion follows immediately.
I havent explained it in details, but do you think its correct?


Bun si care-i scopul? Ele pot fi gasite in cartea mea "Inegalitatii alese in matematica", Editura Niculescu.
O mica observatie. Inegalitatea lui Ballieu ( data in 1937) are loc si intr-un context mai general si anume:

  Daca $t\in (0,1]$ sau $t=2$, atunci in orice triunghi are loc inegalitatea \[ 2^{t-1}\cdot\sin^{t}\frac{A}{2}\leq\frac{a^{t}}{b^{t}+b^{t}}. \]

Uite o aplicatie asa de inceput.... :)

Sa se demonstreze ca in orice triunghi $ABC$ are loc inegalitatea \[ \frac{R}{r}\geq\frac{(a+b)(b+c)(c+a)}{4abc}. \]

Caesar, scopul este de a informa utilizatorii acestui site cu privire la anumite inegalitati trigonometrice, pe care eu le-am considerat a fi utile in concursurile de matematica din Romania si nu numai.Da, e adevarat ca acestea se gasesc si in cartea, impropriu zis <a ta>(mai sunt si altii:prof.I.V.Maftei, George Popescu...), dar poate nu toti au cartea <ta>.Am aflat(din G.M.) ca tu esti de origine din Constanta(tatal tau cred ca e profesor).Ai putea sa-mi dai si mie niste numere de telefon(sau adrese de email) ale Revistei de matematica si informatica din Constanta?As dori sa trimit si eu probleme propuse de mine(evident corecte-sper).Poti sa ma ajuti?
P.S.:vezi si sectiunea <Reviste de matematica din Romanai> de pe forumul romanesc. :)


Trying many different ways, there doesn't seem to be a way to involve a dollar sign in a hide tag without arousing $ \text{\LaTeX}$ coding.

Backslashes are banned in the code; "hide=string" requires that string to avoid several special characters, and for good reason. The dollar sign works just fine in hidden text.


Hi all,

This is exercise 6 in section 1.3 of Allen Hatcher's "Algebraic Topology."

Let $ X$ be a shrinking wedge of circles (that is, a union of circles $ C_n$ of radius $ 1/n$ with center $ (1/n,0)$ for $ n\equal{}1,2, \ldots$), and let $ \tilde{X}$ be the covering space of $ X$ that is a linear chain of $ X$s with a line connecting them, meeting each copy of $ X$ at the wedge point.  The problem is to construct a $ 2$-sheeted cover $ Y \rightarrow \tilde{X}$ such that the composite $ Y \rightarrow \tilde{X} \rightarrow X$ is not a covering space.

First, I am having trouble seeing that $ \tilde{X}$ is a covering space of $ X$.  Where do the line segments connecting each copy of $ X$ map to under the covering map?

My intuition is that the space $ Y$ that we're looking for looks like $ \tilde{X}$ but with another copy of $ X$ for each copy of $ X$ in $ \tilde{X}$, glued to the wedge point.  This would seem to be $ 2$-sheeted, but I can't visualize the composition.

If anyone has trouble seeing the spaces $ X, \tilde{X}$, and $ Y$, I will try to clarify further.

Any help would be greatly appreciated.  Thanks.

Okay, I don't think that $ Y$ works because the fiber above an $ x \in \tilde{X}$ on a line segment connecting two copies of $ X$ has cardinality $ 1$.  Should I somehow introduce two line segments, like two copies of $ \tilde{X}$ glued together in some way?  Every time I try this though, there is still a point whose fiber has $ 1$ element.


probability AMC AIME induction limit ratio

Suppose you flip a coin $n$ times such that two consecutive heads never appear. Let $p_n$ be the probability that, given such a sequence of $n$ flips, it ends in a tail. Find $p_n$ as $n$ approaches infinity.

[hide]
AIME 1990 Problem 9[/hide]

Ummm, its just the last two numbers that matter? Oh hold on thats wrong...

Well the maximum number of heads is n/2, so you could set up a summation.  I think that it would be easier to find the probability that it ends in a head, and then subtract that from 1.

[quote="nr1337"][hide]
Let F_1 = 1, F_2 = 2, F_n = F_(n-2) + F_(n-1).  We have

p_1 = F_1/F_2
p_2 = F_2/F_3
p_3 = F_3/F_4
p_4 = F_4/F_5

(easily verified)  So I would conjecture that p_n = F_n/F_(n+1) and lim(n to infinity) p_n = 1/phi = 2/(1 + sqrt(5)).

Now I just have to prove that conjecture :D. Induction probably works...
[/hide][/quote]

Yes. That is correct. But I don't think anybody can really tell what you are doing or how you did it :P

[quote="nr1337"]Heh, okay I'll try again :)

[hide]
So first I try to calculate some small values of $p_n$.  We can easily see that  $p_1 =\frac{1}{2}$, $p_2 = \frac{2}{3}$, $p_3 = \frac{3}{5}$, $p_4 = \frac{5}{8}$ just by listing out possibilities.  We define a sort of shifted Fibonacci Sequence by $F_1 = 1$, $F_2 = 2$, and $F_n = F_{n-1} + F_{n-2}$.  We can see that, for all $1\leq n \leq 4$, $p_n = \frac{F_n}{F_{n+1}}$.  So we conjecture that this is true for all $n$.  Ummmm..... "we leave it to the reader to prove this." :D  Now we have $\lim_{n \to \infty} p_n = \lim_{n \to \infty} \frac{F_{n}}{F_{n+1}} = \frac{1}{\phi}$ where $\phi$ is the golden ratio $\frac{1+\sqrt{5}}{2}$.[/hide][/quote]

Can you think of a proof for it?

[hide="hint"]
For a sequence of length $n$, look at the last coin of a sequence of length $n-1$.[/hide]

I got a proof now; I just needed a piece of cake to get my  brain working again :D :D

[hide]
Suppose p_n = F_n/F_(n+1) for some n.  Let p_(n+1) = a/b.  We will prove that a = F_(n+1) and b=F_(n+2).  Let's start with b.

Let R be a sequence of coin flips satisfying the condition of no consecutive heads.  Suppose R has n flips and is of the form ...T (T tails, H heads).  From that, we can construct 2 more sequences of coin flips of the forms ...TH and ...TT.  If R is of the form ...H, we can only construct 1 more sequence, ...HT.  There are F_n sequences R of the form ...T and F_(n+1) - F_n = F_(n-1) sequences of the form ...H, so there are a total of b= 2F_n + F_(n-1) = F_n + F_n + F_(n-1) = F_n + F_(n+1) = F_(n+2) sequences with no consecutive heads.  Going back to our sequence R with n flips, we knew that there are F_n sequences R of the form ...T and F_(n-1) of the form ...H.  So we should have F_n sequences of length n+1 of the form ...TT and F_(n-1) of the form ...HT, for a total of a = F_n + F_(n-1) = F_(n+1).  p_(n+1) = a/b = F_(n+1)/F_(n+2).  By induction, we are done.
[/hide]

That was kind of hard to explain....

Yeah that's basically it. Simply put...

Let $h_n$ be the number of working sequences of length $n$ that end in a head and $t_n$ be the number of working sequences of length $n$ that end in a tail.

Notice that $t_{n+1} = h_n+t_n$ because you can add a tail to either previous sequence and you won't violate the condition. Also notice that $h_{n+1}=t_n$ because you can add a head to a previous sequence iff the previous sequence ended in a tail. The base case is easy to find ($h_1=1, t_1=1$), so we see that $t_n = F_{n+1}$ and $h_n = F_n$. So $p_n = \frac{t_n}{t_n+h_n} = \frac{F_{n+1}}{F_n+F_{n+1}} = \frac{F_{n+1}}{F_{n+2}}$ and $\lim_{p \rightarrow \infty}{p_n} = \frac{1}{\phi}$.

[quote="probability1.01"]This is a classic Fibonacci problem.  I don't see how they would have converted the answer into an integer though. :?[/quote]

Go look at the problem - this is a variation/extension.


Show that any number greater than n^4/16 (n is positive integer) can be written in at most one way as the product of two of its divisors having difference not exceeding n.

But a question says:"can be written in at most one way"
You have found one way but you should prove that you can not find two or more ways to decompose that number in a such way.

I will show that if $m$ can be written in two ways as the product of two of its divisors having difference not exceeding $n$ then $m\leq n^{4}/16$

Let $m=ab=cd$ and $a>c\geq d>b$

As $ab=cd$, we can let $a=ef, b=gh, c=eg, d=fh$ with $e>h$ and $f>g$

Denote $u=a-b$ and $v=c-d$ thus $0\leq v\leq u\leq n$

then $16efgh\leq (e+h)^{2}(f+g)^{2}\leq (e^{2}-h^{2})^{2}(f^{2}-g^{2})^{2}=(u^{2}-v^{2})^{2}\leq n^{2}$


A triangle  and  a  square are  drawn on  a  piece  of  paper.  What  are  the  least  and  greatest number  of  regions  into  which  they  divide  the  paper?

the least is 3 since each one has to make a new one. the most intersections between theirs lines is 3,  The most times their line could intersect is 7, creating 8 regions total.


Nα βρεθουν οι ακεραιοι που ικανοποιουν την σχεση
$x^{2}+y^{2}=(x-y)^{3}$

 :D  :D

Τη γράφουμε στη μορφή
$(x-y)^{2}(x-y-1)=2xy$  και παρατηρούμε ότι x-y kai x-y-1 πρώτοι μεταξύ τους .
Και επίσης για x>4y  έχουμε $(x-y)^{2}>2xy$ άρα
4y>x>y
H ιδέα είναι αυτή..... :wink:

Εγραψα μια ολοκληρη λυση μιας και δεν εστειλε κανεις.Η εξισωση μετασχηματιζεται στην εξισωση α^2+β^2=2β^3(εθεσα χ+ψ=α, χ-ψ=β)   αρα  ο β διαιρει τον α αρα εστω οτι βτ=α τοτε τ^2+1=2β εστω οτι τ=2κ+1 τοτε β=κ^2+(κ+1)^2 και α=(2κ+1)[κ^2+(κ+1)^2] οποτε οι λυσεις της εξισωσης ειναι οι χ=[κ^2+(κ+1)^2](κ+1) και                          ψ=[κ^2+(κ+1)^2]κ.

[quote="Αρχιμήδης 6"] οποτε οι λυσεις της εξισωσης ειναι οι χ=[κ^2+(κ+1)^2](κ+1) και                          ψ=[κ^2+(κ+1)^2]κ.[/quote]

Αρχιμηδη 6 εισαι σιγουρος?εβαλα οπυ χ και y  αυτεσ τισ τιμεσ στην αρχικη εξισωση ..αλλα δεν μου βγηκε ταυτοτητα.... Θα το ξαναδω,μηπως εχω κανενα λαθος ,αλλα δες το και εσυ....
Silouan,αυτη την φορα δεν πολυκαταλαβα την κεντρικη ιδεα..Μπορεις να την εξηγησεις λιγο πιο αναλυτικα?
Με την ασκηση δεν εχω ασχοληθει περετερω απλα μονο μια φορα γι αυτο +λεω οτι μπορει να εχω κανει και εγω λαθος στις λυσεις μου...
Παντως ειχα βρει[hide]$(x,y)=(k^{2}+1)(k+1)/4,(k^{2}+1)(k-1)/4$kai $-(k^{2}+1)(k+1)/4,-(k^{2}+1)(k-1)/4$
Θα το ξαναδω...[/hide]
 :D  :D

Edit:sorry,παραλειψη μου...κ=περιττος....

Ραλλη τις ιδιες λυσεις βρηκαμε (x,y)=......  Aπλα εγω παρεληψα να γραψω και τις λυσεις (-x,-y) που ειναι προφανες.Εξαλλου και η ταυτοτητα επαληθευεται. :wink:

Ναι,εχεις δικιο οι ιδιες λυσεις ειναι,απλα εγω εκανα ενα λαθος στις πραξεις οταν το ελεγχα... :o
Εγω ειπα οτι το α μελος ειναι θετικο και αρα $x>y$,εθεσα $x=y+a$ με $a>0$ kαι μετα πηρα τριωνυμο ως προς χ.Η διακρινουσα δινει $a=(k^{2}+1)/2$ με κ περιττο και τελικα προκυπτουν οι λυσεις που βγαινουν παραπανω...

 :D  :D

Θα μου επιτρεψεται να προτεινω και γω μια παραλλαγη της ασκησης του Ραλλη.(δεν εχω βρει τις λυσεις απλα προεκυψε). Να λυθει η εξισωση χ^2+y^2+z^2=(x-y+z)^3 στους ακεραιους.Αμβροσιε,Σιλουαν,Ραλλη......και ολοι σας περιμενω τις λυσεις  η τις ιδεες σας. :wink:

Φαινεται οτι η εξισωση που προτεινα βαζοντας μια ακομη μεταβλητη (την μετεβλητη z) παρουσιαζει πολλες δυσκολιες σε σχεση με αυτην του Ραλλη.Kαμια ιδεα?

Αρχιμηδη 6,καλη σαν σκεψη,αλλα εχουμε τρεις μετεβλητες σε μια μονο σχεση....!!!!! :maybe:
Μηπως αν χρεισημοποιησουμε την σχεση που ειχες πει σε ενα αλλο τοπικ...$3(x^{3}+y^{3}+z^{3})\geq (x+y+z)(x^{2}+y^{2}+z^{2})$?????οποτε αντικαθιστωντας το $x^{2}+y^{2}+z^{2}=(x-y+z)^{3}$ παιρνουμε οτι <κατι> τριτου βαθμου ειναι μεγαλυτερο απο <κατι> μεγαλυτερουυ βαθμου(4ου),οποτε τα x,y,z φρασσονται.Αυτο μονο σαν ιδεα....Δεν το εχω δουλεψει ακομη...

 :D  :D

Καλη η ιδεα σου Ραλλη.Ψαξτο ομως. :) Περιμένω την επιλυση της εξισωσης.Αν καποιος την ελυσε παρακαλω να μας το πει. :wink:


Two congruent squares are placed so as to form an 8-point star with a regular octagon as their common area. If a side of one of the squares is of length 10 + (5 sqrt 2), what is the perimeter of the octagon in units?

I don't understand the question, but the answer is
[hide]
40 sqrt 2
[/hide]

Imagine an {8/4} star.
[url=http://en.wikipedia.org/wiki/File:Regular_Star_Polygons.jpg]Look at the second figure in row 8[/url]

The isosceles leg portion of the square (jutting out of the octagon) is rt(2) smaller, but there are 2 of them, so the total length is rt(2) larger.

Thus we are looking at x(1+rt(2))=10+5rt(2), for which x=5rt(2).

There are eight sides so 8x=40rt(2), our answer.


inequalities
function
inequalities proposed

Let be $ a,b,c \in \mathbb{R}_+^*$ Prove that:

1)$ a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\le \sqrt{2(a+b+c)(ab+bc+ca)}$ ;

2)$ \sum_{cyc}\frac{a}{\sqrt{b+c}}\ge (a+b+c)\sqrt{\frac{a+b+c}{2(ab+bc+ca)}}$

consider the concave function $ f$ , such that $ f(x) \equal{} \sqrt {x}$

$ af(b \plus{} c) \plus{} bf(c \plus{} a) \plus{} cf(a \plus{} b) \le (a \plus{} b \plus{} c)f \bigg(\frac {2(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c} \bigg)$

$ a \sqrt {b \plus{} c} \plus{} b \sqrt {c \plus{} a} \plus{} c \sqrt {a \plus{} b} \le \sqrt {2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$

[quote="alex2008"]Let be $ a,b,c \in \mathbb{R}_ + ^*$ Prove that:

1)$ a\sqrt {b + c} + b\sqrt {c + a} + c\sqrt {a + b}\le \sqrt {2(a + b + c)(ab + bc + ca)}$ ;

2)$ \sum_{cyc}\frac {a}{\sqrt {b + c}}\ge (a + b + c)\sqrt {\frac {a + b + c}{2(ab + bc + ca)}}$[/quote]
By Cauchy-Schwarz inequality ,we have:
1)$ LHS \le \sqrt{(a+b+c)(a(b+c)+b(c+a)+c(a+b))} =RHS$
2)$ LHS \ge \frac{(a+b+c)^2}{a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}} \ge RHS$
Then,we have Q.E.D

we can use jensen inequality , the convex function $ f(x) = \frac {1}{\sqrt x}$

$ \sum_{cyc} af(b + c) \geq (a + b + c)\sqrt {\frac {a + b + c}{2(ab + bc + ca)}}$


Alcumus
videos
algebra
binomial theorem
Support

Hey guys,
  I'm just wondering what do I do!
I have been using Alcumus for quite sometime.
I didn't "speed" through the first sixty questions or anything, but what happened was:
I was going on with questions until it came back to one category that it said I had "mastered" just because I got all of those questions right during the first 60 questions, fine. Those were easy. But now I am getting towards the stuff that I do not have access to, and naturally, I am getting them wrong. So my question is:
How do I move forward and pass my current lessons if I don't get anymore questions from them.
AND, won't I always be in a rut? If I am missing those new questions now, won't they eventually count towards my score (if I somehow get this fixed) and mess me up then?

Thanks in advanced.

you're confusing me...
i think i know what you mean
you should maybe review the videos if you still have them
or study the types of problems/learn them if you're getting them wrong???
don't know what you mean so i hope this helps???
don't really know what you're asking.... :arrow:   :?:

Basically, I've gotten to the point where it is asking questions that have not been taught to me yet.
For example, a bunch of Binomial Theorem questions have been thrown my way, but none of those videos are available to me. And it's gotten to the point where I am no longer being asked the questions for things I've learned. For example, If it only asks me about Binomial Theorem, but I haven't learned it, how do I progress (answer them correctly) if I never learn how to!

hmmm....
it seems to me (through what i have read on this forum) that alcumus only gives you videos once you have answered a few of those questions correctly
this doesn't seem to work out that well.....

i think alcumus should provide video lessons as soon as the first problem of that category is given
ex: as soon as a binomial theorem question is asked, the video lesson should be availiable

is it possible for someone to program alcumus so that this can be done?

You are given access to videos just as soon as you are ready to work on problems from the associated topics. You do not have to answer any particular number of those problems before being granted access to videos.

I tink taht we shood be given the video once we solve a problem of taht lesson. we can understand things better. also, we learn, then we do the problems, isnt that wat the book is doing, FIRST explaining, then giving u problems? I believe that we shood get the videos as soon as we get the problem corresponding to the video.
At least, i learn better this way......

I like the videos too. :)

You can certainly choose not to watch the videos until you get a problem in the topic correct. That's the benefits of technology, you get to choose!

but, dont we haveta pass the topic BEFORE we get the videos???
I haveta first solve some prblms correct, THEN i can see the videos.....
Now im confused...... :huh:


Find an equation for the plane through the point (4,1,3) and perpendicular to the line x=2+5t, y=1-3t, z=2t.

The answer is 5x-3y+2z=23 but i'm not sure how to get this... I keep coming up with -11x-11y+6z=-29.

Rather than me write out a solution, which may or may not help you, I think it would be more beneficial if you put up your solution and then we try to find your mistake.

hello, make the ansatz $ (\vec{x}\minus{}\vec{x_P})\circ \vec{n}\equal{}0$ where $ \vec{n}$ is the normal vector and $ \circ$ stands for the dot product.
Sonnhard.


How do you draw geometric figures using LaTeX? For example, if I'm doing the USAMTS, how would I send in a copy of my figure (or can I just describe what lines, triangles, etc. I drew on the original figure provided for the problem)? What program could you use to draw it if not LaTeX?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217[/url]

I recommend the geometry program Cinderella ([url]http://cinderella.de/tiki-index.php[/url]), which feels a lot like Geometer's Sketchpad, has a free educational version and can export directly to PNG or PDF.

Is the educational version the demo version?  If it is, how will they make it inconvient for you unless you but the full version?  If I can make do with that program I would not need to buy one.

This might be a little off topic, but what is a good Drawing program that is easy-to-use and will last me a long time?

I am thinking about getting a program for drawing on usamts, but I want some extra reach with my dollar. Is Cinderella the best choice?

I use geogebra.  It might take a little bit of getting used to, though (but that applies to all programs like this).


Prove this inequality with $5n \geq m \geq 0$
$\sum \frac{a}{\sqrt{mb^{2}+mc^{2}-na^{2}}}\geq \frac{3}{\sqrt{2m-n}}$

no one?????????
is there any problem in my inequality?


search
real analysis
real analysis unsolved

f(x) is continuous on [a,b], f(a)<f(b). for all x in (a,b); the limit
\[ lim_{t\minus{}>0}\frac{f(x\plus{}t)\minus{}f(x\minus{}t)}{t}\equal{}g(x)\]
exist and is not infinity.
Prove that there exists c in [a,b] s.t. g(c)>=0.

$ lim_{t \minus{} > 0}\frac {f(x \plus{} t) \minus{} f(x \minus{} t)}{t} \equal{} lim_{t \minus{} > 0}(\frac {f(x \plus{} t) \minus{} f(x)}{t} \minus{} (\frac {f(x \minus{} t) \minus{} f(x)}{t}) \equal{} lim_{t \minus{} > 0}\frac {f(x \plus{} t) \minus{} f(x)}{t} \plus{} lim_{t \minus{} > 0}\frac {f(x \minus{} t) \minus{} f(x)}{ \minus{} t} \equal{} 2 f'(x)$

so $ 2 f'(x) \equal{} g(x)$

1* if exist x,y in [a,b]  s.t. f(x)=f(y) using Rolle's theorem we have that exist c in (x,y) such that f'(c)=0
so exist c in [a,b] such that g(c)=2f'(c)>=0

2* if don't exist x,y in [a,b]  s.t. f(x)=f(y) that mean ('cause f is continuous) that f is monotonious and because f(a)<f(b)  ,  f is growing/increasing (don't know how to call it in english:)) in [a,b]
so always in [a,b] f'(x) >=0 so exist c in [a,b] such that g(c)=2f'(x)>=0

1*,2* ->   exist c in [a,b] s.t. g(c)>=0    q.e.d

sorry for my bad english  :roll:

The problem with the above solution is that we don't know if $ f'$ exists.


Find all continuous functions $\large f\colon\mathbb{R}\to\mathbb{R}$, such that
                     $f (x+y+f(xy))=xy+f(x+y)$
for all $\large x,y \in R.$

[hide="Check it!"]Given $y=0$ then $f(x+f(0))=f(x)$
So $f(x)$ is a periodic funtion.
$y=\frac{x}{x-1}$
$t=\frac{x^{2}}{x-1}$
$x+y=xy=t$
$f(t+f(t))=t+f(t)$
Then if $f(t)\neq-t$ for $x$ is sufficiently great, $f(x)=x$ holds onwards. Then $f(x)=x$ for all $x \in\mathbb{R}$
So the equation has two roots $f(x)=x$ and $f(x)=-x$[/hide]

It give only one solution. $f(x)=-x$ is solution too.

Thanks Rust, I edited my previous post. Can you check it again?

Your solution is not correct and complete.

[b](*)[/b] $f(x+y+f(xy))=xy+f(x+y)$
$a=x+y,b=xy$ gives
[b](**)[/b] $f(a+f(b))=b+f(a)$ for all $a,b\in\mathbb{R}$ with $a^{2}\geq 4b$.

[b]1)[/b] $f$ is injective:
If $f(b)=f(c)$ for some $b,c\in\mathbb{R}$,
then take $a\in\mathbb{R}$ with $a^{2}\geq 4b,4c$,
and then we get $b+f(a)=f(a+f(b))=f(a+f(c))=c+f(a)$ and $b=c$ using (**).

[b]2)[/b] $f(0)=0$ and $f(f(b))=b$ for all $b\leq 0$:
$a=b=0$ in (**) gives $f(f(0))=f(0)$ and $f(0)=0$ using 1).
Then for $b\leq 0$ taking $a=0$ in (**) gives $f(f(b))=b$.

[b]3)[/b]  $f(a+b)=f(a)+f(b)$ for all $a,b\in\mathbb{R}$ with $b\leq 0$ and $a^{2}\geq 4f(b)$:
Replacing $b$ with $f(b)$ in (**) gives the result using 2).

[b]4)[/b] $f(b+c)=f(b)+f(c)$ for all $b,c\in\mathbb{R},\ b,c\leq 0$:
Take $a\in\mathbb{R}$ large enough with $a^{2}\geq 4f(b+c)$, $(a+b)^{2}\geq 4f(c)$ and $a^{2}\geq 4f(b)$.
Then 3) gives $f(a)+f(b+c)=f(a+b+c)=f(a+b)+f(c)=f(a)+f(b)+f(c)$
and $f(b+c)=f(b)+f(c)$.

Now $\alpha: =-f(-1)\not =0$ by 1) and 2). Then
[b]5)[/b] $f(x)=\alpha x$ for all $x\leq 0$:
For all $m,n\in\mathbb{N}$ we have
$f(-{m\over n})={1\over n}(\underbrace{f(-{m\over n})+\cdots+f(-{m\over n})}_{n\text{ times}})={1\over n}f(\underbrace{-{m\over n}-\cdots-{m\over n}}_{n\text{ times}})={1\over n}f(-m)\\\hspace*{0.65in}={1\over n}f(\underbrace{-1-\cdots-1}_{m\text{ times}})={1\over n}(\underbrace{f(-1)+\cdots+f(-1)}_{m\text{ times}})={m\over n}f(-1)=\alpha (-{m\over n})$
which gives the result, as $f$ is continuous.

And now
[b]6)[/b] $f(x)={1\over \alpha}x$ for all $x\in\mathbb{R}$:
For $z\in\mathbb{R}$ we have $\lim_{x\to-\infty}(x+{z\over x})=-\infty$.
So take $x\in\mathbb{R}_{-}$ small enough with $x+{z\over x}\leq 0$ and $x+{z\over x}+{1\over \alpha}z\leq 0$.
Then taking $y={z\over x}$ in (*) we get using 5)
$f(x+{z\over x}+f(z))=z+f(x+{z\over x})=z+\alpha(x+{z\over x})=\alpha(x+{z\over x}+{1\over \alpha}z)=f(x+{z\over x}+{1\over \alpha}z)$
and 1) gives $f(z)={1\over \alpha}z$ for all $z\in\mathbb{R}$.

Then 5) and 6) give $f(-1)=-\alpha=-{1\over \alpha}$ and $\alpha=\pm 1$.
So the continuous solutions of (*) are $f(x)=x$ and $f(x)=-x$.
(ugly proof, I know [img]http://www.emoticons.online.fr/smileys/Default/Various_Artists-doubt.gif[/img])

There are uncorrect moments.
[quote="olorin"][b](*)[/b] $f(x+y+f(xy))=xy+f(x+y)$
$a=x+y,b=xy$ gives
[b](**)[/b] $f(a+f(b))=b+f(a)$ for all $a,b\in\mathbb{R}$ with $a^{2}\geq 4b$.[/quote]
Correct $a^{2}\ge 4|b|$ or $-a^{2}\le 4b\le 4b$.
Therefore uncorrect
[quote]2)[/b] $f(0)=0$ and $f(f(b))=b$ for all $b\leq 0$:
$a=b=0$ in (**) gives $f(f(0))=f(0)$ and $f(0)=0$ using 1).
Then for $b\leq 0$ taking $a=0$ in (**) gives $f(f(b))=b$.[/quote]
and etc.

[quote="Rust"]Correct $a^{2}\ge 4|b|$ [/quote]
No, if $a^{2}\geq 4b$, then $x={a+\sqrt{a^{2}-4b}\over 2}$ and $y={a-\sqrt{a^{2}-4b}\over 2}$ exist in $\mathbb{R}$ with $x+y=a$ and $xy=b$.


LaTeX
geometry
geometric transformation
reflection
trigonometry
trapezoid
trig identities

Given ABCD is a regulated quadrilateral inscribed in a circle (O) (which has $AB\cdot CD=BC\cdot DA$).
Let K, H be the midpoints of AC and BD respectedly. Prove that AK+CK=BH+DH.

[color=red][Moderator edit: LaTeXed the equation. Please use the standard notation $x\cdot y$ for the product of two numbers x and y, rather than the obsolete notation x.y (note that LaTeX has the command \cdot for a dot like $\cdot$).][/color]

That is equivalent to prove that $AC=BD$, which is not true (at least it doesn't seem to be )... or perhaps you mean that we have to prove $AH+CH=BK+DK$.

He's referring to a harmonic cyclic quadrilateral. It has lots of nice properties. I'm sure he meant $AH+CH=BK+DK$, as RaMlaF said.

I think similar problems have been discussed before. For instance, I'll use the fact that $AC,BD$ bisect the angles $\angle BKD,\angle AHC$ without proof (take it as a nice problem :)). From this we can deduce that it suffices to show that the chords obtained by producing $AH,DK$ are equal, which, in turn, would be derived from $\angle DAH=\angle ADK\ (*)$, so let's focus on proving $(*)$.

Now I'll use another property of the harmonic quadrilaterals: the tangents through $A,C$ to the circle concur in $P\in BD$, and the symmetric situation also holds. From here we find $K$ to be the image of $O$, the center of the circle, through the inversion of pole $P$ and power $PB^2=PD^2$, so $BDKO$ is cyclic, meaning that (look at the second equality) $\angle DKA=\frac{\angle DKB}2=\frac{\angle DOB}2=\angle DCB$, and since $\angle DAK=\angle DBC$, we find $DKA,DCB$ to be similar. In the same way we show that $AHD,ABC$ are similar, so $\angle DAH=\angle CAB=\angle BDC=\angle ADK$, which is precisely $(*)$.

Sorry for my stupid mistake. I really mean AH+CH=BK+DK :D

But Grober, can you explain to me what we should do after we have $\angle DAH=\angle ADK$? :?

I don't see so clearly what to do with $\angle DAH = \angle ADK$ ... but i have another solution (without magic properties ...  :lol: just kiddin'):
Let $D'$ be the reflection of $D$ with respect to $A$. Then $\angle BAD = \angle BCD$ and $\frac {BA}{AD'} = \frac{BA}{AD}=\frac{BC}{CD}$ since $ABCD$ is an harmonic quadrilateral. Therefor $BAD'\simeq BCD$, with $BD'=\frac{BD.AB}{BC}$ And, since $DH=BH$ and $DA=DA'$ we have $BD'=2AH$ so $AH=\frac{BD.AB}{2BC}$. Similarly, $CH=\frac{BD.BC}{2AB}$. And doing the same with $BK$ and $DK$ the problem reduces to show that $BD(\frac{AB}{BC}+\frac{BC}{AB})=AC(\frac{BC}{CD}+\frac{CD}{BC}) \iff BD(\frac{AB}{BC}+\frac{CD}{AD})=AC(\frac{AB}{AD}+\frac{CD}{BC})$ (because $ABCD$ is harmonic) $\iff BD(AB.AD+BC.CD)=AC(AB.BC+CD.AD) \iff \frac{AC}{BD}=\frac{AB.AD+BC.CD}{AB.BC+CD.AD}$. And we see that $(AB.AD+BC.CD).\sin BCD = (AB.BC+CD.AD).\sin CDA = 2.Area(ABCD)$. Using that, and the Extended Law of Sines with $\sin BCD=\frac{BD}{2R}$ and so on, we obtain the desired result.

Since $\angle DAH=\angle DAK$, let $P,Q$ be the intersections of $DK,AH$ with the circle. Then $DPQA$ is an isosceles trapezoid, meaning that $DK,AH$ produced form equal chords of the circle. This is what we wanted.

YEAH, I finally find another simply solutions  :D
Use the condition $AB\cdot CD=AD\cdot BC$ and the Ptoleme theorem we have the triangle DAH and BAC are similar.
$DH=\frac{2R\cdot S(ABD)}{AC\cdot BD}$
And qith the same way we will get the result
Sorry if I did make mistakes because I must be very hurry now  :D

this problem is not difficult
to prove the equality you can use this
the tangent to the circle at B,D and the line AC are concurrent after you use ptoleme theorem; :D  :D


I found this in a lateral thinking puzzle(I currently can't find it so these aren't exact replicates, but here are the problems.  It's kinda like a dummy test
1. What is 14 times 2 divided by a half?
2. How many seconds are in a year?
3. What is the largest island in the world before Australia was discovered?
4. If it takes 20 men 5 days to dig 1 hole, how many days does it take 10 men to dig 1/2 of a hole?
5. Who starred with Jack Nicolson in the movie "Who flew over the Robin's Nest"?
Can't remember any more............... :(  :blush:
Here are the answers:(post results here)
[hide]
1. 56
2. 12.......there's January the second, Febuary the second and so on....
3. Greenland...Australia is a continental landmass.
4. You can't dig half of a hole.
5. Such a movie was never created(It was "Who flew over the cukoos nest)
[/hide]
Here's the scoring ranges
5 right: Genius
4 right: Smart Alec
3 right: Average
2 right: Wally
1 right: Super Wally
0 right: Ultra Wally

Note:  You DON't want a "wally" ranking.

I got average yay.

And btw #2 is just stupid  :wink:

I have some other WALLY test questions, also from a lateral thinking puzzles book. :)

1. How can you stand behind someone while he or she stands behind you?
2. What looks like a horse, moves like a horse, and is as big as a horse but weighs nothing?
3. Who is bigger: Mr. Bigger or Mr. Bigger's son?
4. What has four fingers and a thumb but isn't a hand?
5. What multiplies by division?
6. What's white when it's dirty, and black when it's clean?
7. What gets higher as it falls?
8. Why did the overweight actor fall through the theater floor?
9. How did an actor get his name up in lights in every theater in the country?
10. Where would you find a square ring?
11. How do you make a slow horse fast?
12. Why did Sam wear a pair of pants with three large holes?
13. Where do you find a no-legged dog?
14. Approximately how many house bricks does it take to complete a brick house in England?
15. What cheese is made backward?
16. Take away my first letter; I remain the same.
Now, take away my fourth letter; I remain the same.
Now, take away my last letter; I remain the same.
What am I?
17. If a white man threw a black stone into the Red Sea, what would it become?
18. If an atheist died in church, what would they put on his coffin?
19. Who was it that went into the lion's den unarmed and came out alive?
20. A man rode down the street on a horse, yet walked. How come?
21. How can you eat an egg without breaking the shell?
22. In China they hang many criminals, but they will not hang a man with a wooden leg. Why?
23. A circular field is covered in thick snow. A black cow with white spots is in the middle. Two white cows with black spots are on the edge of the field. What time is it?
24. What was the problem with the wooden car with wooden wheels and a wooden engine?

Okay, that was a lot. I won't post the answers yet, but I know them if you want to ask me. :)

[hide="answers?"]1. You stand back to back
2. The horse's shadow
3. Mr. Bigger's son
4. A glove
5. A single-celled organism
6. A blackboard
7. Snow
8. It was a stage he was going through
9. He changed his name to Exit
10. At a boxing arena
11. Give it no food
12. To get his feet in it
13. Where you left it
14. One
15. Edam
16. A mailman
17. Wet
18. A lid
19. The lion
20. The horse was named "yet"
21. Get someone else to break the shell
22. They hang people with a rope and a noose
23. Wintertime
24. It wooden go[/hide]

Okay, okay, I have the book.  :D

I don't have the book.

[hide="my answers"]1. Face him.
2. Horse's shadow.
3. Both of them.
4. Blender that hasn't been cleaned out yet.
5. Black adder.
6. Black cat with a barrel of flour.
7. Pothead fatally falling off a cliff.
8. Trap door.
9. He was overweight and he kept falling through the trap door.
10. On a square wife. Or husband. It doesn't matter.
11. Get off of it.
12. All pants have three large holes.
13. Chiropractor.
14. The same number of bricks it takes to complete it in the USA.
15. eseehc
16. Pen pal that lost all his letters.
17. Wet.
18. Dirt.
19. A lion.
20. He walked to the stable, he walked into the house, etc.
21. Have somebody else cook it, he breaks the shell not you.
22. ???
23. Time to bring the cows in.
24. Answer 1: It needed a wooden road. Good luck finding that. Answer 2: It had a wooden windshield.[/hide]


topology
advanced fields
advanced fields unsolved

Let X,p be a metric space where for each uncountable subset A of X there is an accumulation point in X.  Show that X,p is separable.

For every $k$, let $N_{k}\subset X$ be a maximal set s.t. the distance between every two points in $N_{k}$ is at least $1/k$ ($N_{k}$ exists by Zorn's Lemma). The set $N_{k}$ has no accumulation points. Therefore, it is at most countable. Now consider the set
\[D = \bigcup_{k}N_{k}.\]
The set $D$ is a countable dense set in $X$. Thus $X$ is separable.


[quote="#20002 pwns"]If $w^9=1$ and $w\not=1$, then find the sum: $1+w+w^2+...+w^8$
[hide="hint"]You can do this in your head... also, for more information $w$ is a complex number[/hide]

[hide="-"]
[hide="Warning... here is the solution... but still attempt the problem, I am looking to see what are the reactions by people who see this problem for the first time...DO NOT CLICK, attempt first"]
If $S$ is the sum we are trying to find, $S*w=w+w^2+...w^9=w+w^2+...1=S$. Since $S=Sw, w\not=1, S=\boxed{0}.$[/hide][/hide][/quote]
[hide]$1+w+w^2+...w^8=\frac{1-w^9}{1-w}=0$[/hide]


geometry
3D geometry
tetrahedron
pyramid

On the Scholastic Aptitude Test (an American college entrance exam), students were once asked to determine the number of faces of the polyhedron obtained by gluing a regular tetrahedron to a square pyramid along one of the triangular faces. The answer expected by the test authors was 7, since the two polyhedra have 9 faces, 2 of which are moved by gluing, but a student taking the exam pointed out that this is incorrect.
What is the correct answer?

[hide="This is a classic"]  When you glue the two figures together, 2 pairs of the triangular faces ("pair" meaning one face from the tetrahedron and one lateral face from the square pyramid) align in such a way as to create 2 uniform face, due to the angles involved , i.e., the internal angles of the two figures add to 180 degrees.  I figured this out by actually using paper to build the figures, I think. [/hide]


[quote="banned UBER"]If 2 standard 6-sided dice are rolled, then what is the probability that at least 1 of them shows a prime number and that their sum is also prime?[/quote]
Make a chart. Cross out all the numbers that aren't prime.You're left with
6,5,11,
5,6,11,
6,1,7,
5,2,7,
4,3,7,
3,4,7,
2,5,7,
1,6,7,
4,1,5,
3,2,5,
2,3,5,
1,4,5,
2,1,3,
1,2,3,
1,1,2,
Crossing out (6,1,7), (1,6,7), (4,1,5), (1,4,5), (1,1,2), you're left with 10/36, 0r 5/18 chance of doing what banned UBER said


modular arithmetic
geometry
inequalities
algebra
polynomial
power of a point
AMC

Discuss the solutions here and request anything that you would like to be explained.

I didn't do this one but as a moderator, it's my job to set these stuff up to prevent multiple threads.

That's cool but I still can't figure out how to do 25.

I'm having a boredom attack during my MOP free time right now, so I'll just post bunches of solutions...  I'll have to get back to a couple of these (for diagrams, and I can't remember how I did the uber problem! I can brute force it, but I remember having a nicer way...)

[hide="1"]2007+xy = 2007 + (4)(5) = 2027.  (C)
[/hide]
[hide="2"]-13-12-11-...-0+1+2+...+13+14.  Everything cancels but 14.  (B)
[/hide]
[hide="3"]  The minute hand is 1/3 of the way around, or 120 degrees.  Between the 3 and the 4, there are 30 degrees, and the hour hand is 1/3 of the way between.  This is 90+30/3 = 100 degrees past the hour: 120-100=20.  (C)
[/hide]
[hide="4"]$3+3^2+3^3+3^4+3^5 = 363$.  (E)
[/hide]
[hide="5"]2007 = 2 mod 5.  By Euler's theorem, $2^4 \equiv 1 \pmod 5,$ so $2^{2007} \equiv 2^3 \equiv 3 \pmod 5$. (D)
[/hide]
[hide="6"]Simply count the ones that don't have that letter and subtract from the total: $6^3-5^3 = 91$. (B)
[/hide]
[hide="7"]If you put everything over a common denominator of $abcdef$, you get the sum of six odd number in the numerator (which is even) over the product of them (which is odd) and thus cannot be 1.  (A)
[/hide]
[hide="8"]Just do synthetic division and you'll notice the pattern... 37*73 - 37*73 = 0. (B)
[/hide]
[hide="9"]Just using the 3-4-5 triangle, it can be at most 12.  $ab$ must be divisible by four since squares are $0$ or $1 \pmod 4$.  Also, $ab$ must be divisible by 3 since squares are $0$ or $1 \pmod 3$. So the answer is 3*4=12.  (B)
[/hide]
[hide="10"]
Haha, this was funny.  Our number has to be more than one hundred digits (obviously).  And it must be divisible by 3, so the number of digits must be divisible by 3 (since it's all 1s).  300 is the only answer that fits. (C).
[/hide]
[hide="11"]
Do this a different way: assume you have 7 students and you want to pick where to put five such that none are next to each other.  There are ${8 \choose 5} = 56$ ways to do this.
[/hide]
[hide="12"]
Plug in $x_1=0$, and use the ball-and-urn formula to get $165$ such solutions.  Plug in $x_1=1$ and use ball-and-urn to get $9$ solutions; add to get 174.  (D)
[/hide]
[hide="13"]I just listed out the possibilities... I'll get back to this one.
[/hide]
[hide="14"]Suppose the diagonal of length 6 divides the other into lengths $x$ and $y$.  So the area is $\frac{6x + 6y}{2} = 3(x+y) = 21$.  (A)
[/hide]
[hide="15"]$f(x) = ax+10$ from the first equation.  From the second, we can solve for $a$ to get $a=2$.  So we end up with a triangle with vertices $(-2,0)$, $(0,0)$, and $(0,10)$.  10*2/2 = 10.  (C)
[/hide]
[hide="16"]Note to self: don't make this harder than it needs to be!  There are 18! ways to arrange them, and pretend there is a divider every 2.  The sum of the sum of the digits is 18! mod 9, which is 0.  Since the sum obviously can't be 0, it must be 9. (E)
[/hide]
[hide="17"]
Plug in x=1 and get 5f(1) = 4, so f(1) = 4/5.  (A)
[/hide]
[hide="18"]
I can't explain this well without a diagram, but I'll just say you use power of a point to find the length of the tangents from A to the incircle and use the Pythagorean Theorem.  (B)
[/hide]
Ok, I'm going to stop trying to explain these geometry problems without diagrams.  Someone else can do that, or I'll make diagrams when I get home...
[hide="20"]Crap, I forgot how I did this one... getting back to it...
[/hide]
[hide="21"]AM-GM says that it will have it's minimum when $a^4$, $b^4$, and $c^4$ are equal.  And lo-and-behold, (2,2,2) satisfies the conditions!  3*2^4 = 48. (B)
[/hide]
Another geometry problem with no diagram... skip!
23's already been posted...
along with 24....
[hide="25"]
$\frac{a}{b+c} = \frac{(a+b+c)-(b+c)}{b+c} = \frac{7}{b+c} - 1.$  Add them cyclicly to get $7 \left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3 = \frac{19}{10}.$  (A).
[/hide]
done!

[quote="Silverfalcon"]By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]

:D Yeah, I couldn't think of any more hard problems.

[quote="chess64"][hide="23"]
$f(x)=\frac{x+1}{x}$

$\iff x\cdot f(x)-x-1=0$

$g(x): =x\cdot f(x)-x-1$

The roots of $g(x)$ are $1,2,\ldots,2006$. We can write

$C(x-1)(x-2)\cdots(x-2006)=x\cdot f(x)-x-1$

Substituting $x=0$, we get

$C(2006!)=-1$

$\therefore C=\frac{-1}{2006!}$

$\Rightarrow x\cdot f(x)-x-1=-\frac{1}{2006!}(x-1)(x-2)\cdots(x-2006)$

At $x=2007$, $2007f(2007)-2008=-\frac{1}{2006!}(2006)(2005)\cdots(1)$

$f(2007)=1$

Oh dang I screwed this one up too...

The right answer is not even on there :(
[/hide][/quote]

Haha, I was wondering about that... :)

[quote="chess64"]
$C(x-1)(x-2)\cdots(x-2006)=x\cdot f(x)-x-1$

Substituting $x=0$, we get

$C(2006!)=-1$
[/quote]
How did you get this? $f(0)$ is undefined.

$f(k)=1+1/k$ only for $k=1,2,\ldots,2006$. It's a polynomial that just happens to satisfy that equation for those specific values. But since it's a polynomial it's defined on all reals.

EDIT: And if you're asking how I got that $0\cdot f(0)-0-1=-1$, it's because of the 0 multiplied by $f(0)$.

Did you guys like my $\textit{uber}$ problem? :D

[quote="chess64"][quote="SplashD"][quote="Silverfalcon"][hide="Solution to 21"]

This is an excellent problem. I use Cauchy's Inequality here but I don't know if this was necessary because normally, AMC problems do not use Cauchy.

By Cauchy:

$(1^2+1^2+1^2)(a^2 + b^2 + c^2) \geq (a + b + c)^2$

So, the minimum value of $a^2 + b^2 + c^2 = 12$ after doing some algebra and using $a+b+c = 6$.

Now, note that:

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$

Since we minimized $a^2+b^2+c^2,$ we can now see that $ab+ac+bc = 12$.

Now, observe these two squares:

$(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$

$(ab+ac+bc)^2 = a^2 b^2 + a^2 c^2 + b^2 c^2 + 2abc(a+b+c)$

Using what we know, we can get $a^2 b^2 + a^2 c^2 + b^2 c^2 = 48$. So, $a^4 + b^4 + c^4 = 12^2 - 2(48) = \fbox{48}$. [/hide]

As I said, nice job on making this problem. By the way, I'm pretty sure that #24 is not original. I think this is in Mathematical Circles (Russain Experience).[/quote]

I solved that faster than I solved #3 just by plugging in 2 for all the variables :rotfl:[/quote]

Oops, I made that one too brute-forcable :([/quote]
Yeah, I did the same thing.  Problems like that (and I've done a couple of similar ones recently) almost always seem to work out that way.

I would like to see a solution to 22.  It was very confusing, you didn't specify which point had the right angle!  Although, I assumed it was C since that made it have nice integer side lengths.  However, I solved it that way twice and didn't get integer answers.  It was very simple, just using coordinates...

[quote="Silverfalcon"][quote="mathgeniuse^ln(x)"]That was the problem I was unable to solve as well as 23.  I don't know what other question I got wrong.

SilverFalcon, for #21, couldn't u just use AM-GM, since a^4+b^4+c^4 is always positive as well as abc?[/quote]

You can't and in this case, it was lucky.

Note that the problem stated that $a,b,c$ are REAL not necessarily positive. In order for AM-GM to work, all three variables must be positive.

That's why I disagree with your idea and nr1337's solution about using AM-GM.[/quote]

$a^4, b^4, c^4$ are always positive
By am-gm
$a^4 + b^4 + c^4 \ge 3 (abc)^{ \frac 4 3}$
$a^4 + b^4 + c^4 \ge 48$

[hide="another sol for 25"]Let what we have to find =x

Add 3 to both sides(on l.h.s,add them as +1,+1,+1 to each of the fractions and u'll obtain):-

$\frac{7}{b+c}+\frac{7}{c+a}+\frac{7}{a+b}=x+3$

$\implies 49/10=x+3$[/hide]

nice questions chess!!! It wasn't easy to complete it in the time limit,had to think for some  :D

[quote="matt276eagles"][quote="futuremontrealian"]About no 16,

I informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16.

My experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for:

$\left(\begin{array}{c}18 \\ 2 \end{array} \right) \left(\begin{array}{c}16 \\ 2 \end{array} \right) \left(\begin{array}{c}14 \\ 2 \end{array} \right) \left(\begin{array}{c}12 \\ 2 \end{array} \right) ...$

Or

$\frac{18!}{2^9}$

Why is this different?  Do I have the wrong idea?  Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same?  I have no idea... I'd just like a bit of help).[/quote]

I used the same method, and you are basically right, but you still overcounted by $9!$ for the different orders to choose the pairs.  Still, the key to this problem is that the sum of digits is equivalent to the number $\pmod{9}$, which is clearly $0$ because the number is divisible by 9.  Then the sum of digits is 9, because it obviously can't be 0.  This isn't such a great problem because you can use $18!$ and still get the right answer, but oh well...  :)[/quote]

Oh!  I forgot about that mod 9 thing.  I will have to keep that in mind for future sum of digits problems.  Good solution.

[quote="Xantos C. Guin"][quote="nr1337"]
[hide="25"]
$\frac{a}{b+c} = \frac{(a+b+c)-(b+c)}{b+c} = \frac{7}{b+c} - 1.$  Add them cyclicly to get $7 \left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3 = \frac{19}{10}.$  (A).
[/hide]
[/quote]

Wow, that pwns the way I got the answer.

I decided to let a=4 b=2 and c=1

That worked surprisingly well.[/quote]

I magically found a very simple answer...

$(a+b+c)(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=1+\frac{a}{b+c}+1+\frac{c}{a+b}+1+\frac{b}{a+c}=7(\frac{7}{10})$
Answer = $\frac{49}{10}-3\rightarrow \frac{19}{10}$

Oh yeah. You're right. I forgot about raising to 4th power. Haha.. I did way too much work.

[quote="Adunakhor"]Hmm, turns out I wasn't using the latest version of the test ... none of the choices for Question 3 on the version I downloaded was 20. No wonder I couldn't answer that one  :P[/quote]

Same happened to me, I asked on the forum what were the answer choices, and some guy just said "its just changed 940 to 320." -,- he didnt give me the answer choices so I got a retarded one wrong

I also wrote down the wrong answer thinking the right answer for another problem.

Also when i saw #21 i skipped it because it looked like a big expansion problem but then when i went back after i finished the test and turned in my answers i was like OMG. 2 2 2 because i didnt read the actual sum and product thing, i just didn't think it would be so nice.

[quote="pianoforte"][hide="13"]$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{1}{2007}$
$\implies xy-2007x-2007y=0$
$\implies (x-2007)(y-2007)=2007^2$ from Simon's Favorite Factoring Trick.
The prime factoriztation of $2007^2=3^4\times223^2$. $2007^2$ has $5\times3=15$ divisors.
There are 7 ordered pairs. C.[/hide][/quote]

Heh, I made that way harder than it needed to be :D

[quote="futuremontrealian"]futuremontrealian wrote:
About no 16,

I informally took the test (I only discovered this mock exam an hour ago), and I am wondering about question 16.

My experience from reading Discrete and Combinatorial Mathematics says that the way to solve the problem would be to solve for:

Or

Why is this different? Do I have the wrong idea? Obviously the answer is different on the test, because that one is based off of just plain 18! (or is it the same? I have no idea... I'd just like a bit of help).

I used the same method, and you are basically right, but you still overcounted by 9! for the different orders to choose the pairs. Still, the key to this problem is that the sum of digits is equivalent to the number , which is clearly  because the number is divisible by 9. Then the sum of digits is 9, because it obviously can't be 0. This isn't such a great problem because you can use 18! and still get the right answer, but oh well... (sorry, too lazy to do quote)

--------------------------------------

Are you sure I overcounted?  I understand now why it doesn't matter, I'm just curious about the 9! thing.  I will gladly admit a flaw in my logic, just tell me what's going wrong so I can fix it.

18C2 is $\frac{18!}{16!2!}$  Order doesn't matter here because person a paired with person b is the same as the opposite.

This simplifies to $\frac{18 x 17}{2!}$  The same thing is going to happen with 16c2 and 14c2 and the rest, so the end result will be

$18 x 17 x 16 x15...$ or $18!$ over $2!2!2!2!2!2!2!2!2!$ I hope that's nine twos.
That's what gave me $\frac{18!}{2^{9}}$

Where'd I go wrong (if I did)?

Note: those x's should be read "times" :)  I'm too lazy to type \ast six million times.[/quote]

There are 9 pairs of people. The order of the pairs does not matter. Call the 9 pairs of one possible system of pairing A, B, C, ..., J

ABCDEFGHIJ has the same people paired with each other as JABCDEFGHI. There are 9! ways to order the 9 pairs, so we must divide by 9! to correct the overcounting.

But why is that so?  ABCDEFGHIJ has a with b and c with d, etc, whereas a is with j on the second list.  I'm going to look this up again in Grimaldi's book.

---edit:
wait I get it now!  the 9! is because it doesn't matter what order you choose the pairs themselves in, so long as the people in them are the same!  hazzah!

Yeah, those letters represented pairs of people, not individuals. That's why there were only 9 of them  :P

So, how did people like the uber problem?  I didn't get that question.


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

To all sophomores, freshmen, and middle schoolers:

What was your USAMO index (AMC 10(A/B) + 10*AIME)?

$ 135 (\text{AMC 10B}) \plus{} 10\cdot8 (\text{AIME}) \equal{} 215$
So close...

$ 132 \left(\text{AMC 10A}\right) \plus{} 10\cdot7 \left(\text {AIME I}\right) \equal{} \boxed{202}$
Stupid mistakes...

[quote="pythag011"]144+10(9) = 234. 2 careless mistakes on AIME, 1 on AMC 10. Maybe I'll stiill make USAMO though.[/quote]You definitely will make USAMO...

[quote="vishalarul"][quote="davidyko"]$ 135 \text{ AMC 10B} \plus{} 10(9 \text { AIME I}) \equal{} 225$.  :P[/quote]
$ 127.5 \text{ AMC 10A} \plus{} 10(10 \text { AIME I}) \equal{} 227.5$.  :P

Hmm... AIME does make a big difference. (BTW, I could've got a 145.5 on the 10A if I hadn't made any stupid mistakes...)[/quote]

Lawl, go and die, Vishal. Just kidding.  :wink:
I feel like my 10B score was vaguely unsatisfying. Maybe because I got a higher score two years ago.

$ 145.5 \text{ AMC 10A} \plus{} 10(9 \text { AIME I}) \equal{} 235.5.$

Stupid mistakes on AIME, changed correct answers to incorrect answers, due to doubts... :(
Hopefully, I'll make USAMO...

ehh i completely and utterly failed the 10, (better on the 12, but still failed  :D ) so
$ 118.5\plus{}60\equal{}178.5$...i doubt i'll make USAMO

$ 144 (\text{AMC 10A}) \plus{} 10\cdot8 (\text{AIME}) \equal{} 224$

Hope I didn't add wrong :D

$ 144 \plus{} 80 \equal{} 224$ right?

I failed the 10.

$ 120 \plus{} 7 \cdot 10 \equal{} \boxed{190}$.

126 (AMC 12)/11 AIME -> 236.. I failed miserably on the AMC..

[quote="FloodFilter:"][quote="miller4math"]ANYONE here in 7th grade and has decent scores?
oh well, i got a 138/9... so 228
I [i]PROBABLY[/i] made USAMO
im the only optimistic one around here these days...

EDIT: heh spelled optimistic wrong[/quote]

Johnny (his name hehe i spoiled it) is uber pro!  :mad:  He can totally destroy all the middle schoolers in california except Allissa Zhang (did i spell it right). As for pythag, I bet he's pro too. Anyways, Johnny knows a bunch of crazy trig formulas so you can't blame him (even though he got the trig problem on AIME wrong  :huh: )[/quote]

You're in California???(So am I).

144/8 = 224 :(, but the AIME I i took was unofficial. gonna take the AIME II this wednesday.

you people are making me feel stupid...
123/between 6 and =between 183 and 213
but im an 8th grader. :D

139.5+8*10=219.5

I really am nervous about this with all the "easy AIME" rumors that have been going around... I dunno if i'll make it...

EDIT: The AMC looks at both AMC scores right? I put my slightly worse 12A score because I couldn't confirm my 10B score.

[quote="Zmastr"]Stop making us all feel bad!  You're in 8th grade and probably qualified for the USAMO by like 20 points.[/quote]

 :) Thanks. Now I have a little more hope  :ninja:

i would any day trade my 126/60 for your 139.5/80.  :roll:

[quote="alkjash"]Wow, I was looking at the poll and I was thinking it was a poll for the index cutoff for about 10 seconds.
Which kind of scared me considering the number of people who put super-high cutoffs.

But anyways, 138 (10A) + 9*10 = 228.[/quote]

AHAHA same. I was like "Holy S*** 250???" before actually figuring out what the poll was for.
Anyways,
AMC10 A - 144(pretty good, I think) and
AIME I - 7(phail XD)
= 214 index

same here too.

AMC 10a -- 120
 :mad:

 :mad:

  :mad:
AIME I -- 70
for a total of 190.

[quote="Moldytape"]

I probably won't make the =<10th grade index though, since the floor will probably be a 7, and with my bad AMC10 score,  :([/quote]

you think you have it bad, think of the =<10th graders who are forced to take the amc12.

[url=http://www.imo-official.org/participant_r.aspx?id=21095]david yang[/url] versus [url=http://carp.di.unipi.it/]johnny ho[/url]???


Let $ x_1,x_2,...,x_n$ be positive numbers. Prove that

[list]$ \frac{x_1^2}{x_2}\plus{}\frac{x_2^2}{x_3}\plus{}...\plus{}\frac{x_{n\minus{}1}^2}{x_n}\plus{}\frac{x_n^2}{x_1} \geq x_1\plus{}x_2\plus{}...\plus{}x_n$[/list]

[quote="4865550150"]Let $ x_1,x_2,...,x_n$ be positive numbers. Prove that

[list]$ \frac {x_1^2}{x_2} \plus{} \frac {x_2^2}{x_3} \plus{} ... \plus{} \frac {x_{n \minus{} 1}^2}{x_n} \plus{} \frac {x_n^2}{x_1} \geq x_1 \plus{} x_2 \plus{} ... \plus{} x_n$[/list][/quote]
It is straight B-C-S(known as Andreescu inequality)....
Just multiply both sides with:
$ (x_2\plus{}x_3\plus{}...\plus{}x_n\plus{}x_1)$

you can also directly use CBS.Titu Andreescu is another way to express CBS...but a very useful one


trigonometry
algebra
polynomial
algebra open

Let $P,Q$ be nonconstant real polynomials. Prove that every $x\in\Bbb R$ is an
accumulation point of the sequence $a_n=P(n)\sin Q(n)$. At least prove that the
sequence $(a_n)$ always has a finite accumulation point, or prove the general statement for some special polynomials $P$ and $Q$ (the case $P(n)=Q(n)=n$ is already hard enough).

??? Why do you say that the case $P(n)=Q(n) = n$ is hard enough? As far as I know it is an open question to know if the sequence $U_n = n \cdot sin(n)$ is dense in $R$...
Have you some results on it ?

Pierre.

That problem is open? I think it was posted on ML, and not as an open problem..

I answered here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=108295#108295

Pierre.

Is it my misunderstanding or did you say that it's unknown whether 0 is an accumulation
point of $n\sin n$?

I can't prove it, but it was my exam problem, that's why I was confused.

I've heard of teachers which give some unsolved problems in exams, just in case a brilliant student has a good idea without being discouraged by the fact that they're unsolved :). Maybe you have such a teacher.

Oh, I think all 7 Clay's Millenium Problems will be in ML Open Questions soon :)


If the 25th day of the year 2003 falls on a Saturday, on what day of the week did the 284th day of the year 2003 fall?

$ 25 \equiv 284 \mod 7$ so it falls on a [b]Saturday[/b].


How would you proof the theorem: If a quadrilateral has both pairs of opposite angles congruent, then it's a parallelogram.  I can't seem to make any progress

If a quarilateral ABCD with $\angle A \cong \angle C $ and $\angle B \cong \angle D $ then $ m\angle A+m\angle B+m\angle C+m\angle D=360^\circ$, so $ 2m\angle A + 2m\angle C=360^\circ $ which means that $m\angle A + m\angle B=180^\circ$.  By one of the basic axioms of geometry if a line interesects two other lines and two interior, same side angles that add up to 180 are created, then the two intersected lines are parallel.  Thus AD||BC.  The same argument works for AB and DC, so the figure is a parallelogram.


What do you think about that story ?

Should he have suffered more? Or was that what he deserved?

Please post your thoughts and opinions.  :lol:

there's a topic about this in round table


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Find explicit form for $x$ using $y$ such that $x^y=y^x$ nad $x,y \in \mathbb{C}$.

Pose y/x=k and modify the equation to keep only k.

The detailed process is on http://www.les-mathematiques.net , search y^x=x^y in archives.

The link doesn't work because you have a comma at the end.

can you give more specifik link, i can't get it

[url]http://les-mathematiques.u-strasbg.fr/phorum/read.php?f=2&i=97844&t=96315#reply_97844[/url]

If you have any problem with the french just ask me.

if I unedrstood it, it's only for $(x,y)\in R^2$. I need for complex....


calculus
derivative
function
real analysis
real analysis unsolved

Let $f:R->R$ be a $n$-times bounded derivable function such that $ f^{(n)}$ is Lipschitz. Then $f,f',...,f^{(n)}$ are bounded.

Since the boundedness of $f$ and $f^{(n)}$ implies the boundedness of all the other derivatives, as was discussed before, it suffices to prove the boundedness of $f^{(n)}$, given that it's Lipschitz and $f$ is bounded.

The boundedness of $f^{(n)}$ follows from this (which is pretty obvious and easy to prove):

If $g$ is a differentiable function s.t. for all $M>0$ and all $t$ (no matter how large) we can find an interval of length $t$ on which $|g'|\ge M$, then $g$ has the same property.

If $f^{(n)}$ were unbounded then we could apply the sentence to $g=f^{(n-1)},g=f^{(n-2)},\ldots,g=f$, until we reach the conclusion that $f$ must be unbounded, a contradiction.

I hope I didn't say anything stupid.