algebra
polynomial
induction
modular arithmetic
binomial coefficients
algebra proposed

What a cool problem!

Suppose $ f(x) = a_n x^n + a_{n - 1} x^{n - 1} + \ldots + a_0$. Note that if we divide $ f$ by $ x - 1$, we get

$ \frac {f(x)}{x - 1} = a_n x^{n - 1} + (a_n + a_{n - 1}) x^{n - 2} + (a_n + a_{n - 1} + a_{n - 2}) x^{n - 3} + \ldots + (a_n + a_{n - 1} + \ldots + a_1) + \frac {a_n + a_{n - 1} + \ldots + a_0}{x - 1}.$

That is, the coefficients become partial sums. Then by induction, we can show that for any positive integer $ p$,

$ \frac {f(x)}{(x - 1)^p} = \sum_{i = 0}^{n - p} \left( \sum_{j = i + p}^n \left( {j \choose p - 1} a_j \right) x^i \right) + \frac {R(x)}{(x - 1)^p},$

for some remainder polynomial $ R(x)$ of degree less than $ p$. If $ f(x)$ has $ 1$ as root with at least $ k$ multiplicity, $ R(x) = 0$ and $ \frac {f(x)}{(x - 1)^p}$ is a polynomial. Let $ g_p(x) = \frac {f(x)}{(x - 1)^p}$. Then as long as $ 1$ is a root of $ f$ with multiplicity at least $ p$, Hockeystick gives that

$ g_p(1) = {n \choose p} a_n + {n - 1 \choose p} a_{n - 1} + \ldots + {0 \choose p} a_0 = 0. \quad (\ast )$

This equation is true for all $ 0 \leq p \leq m - 1$, since $ g_p(1) = 0$ for those values.

[b]Lemma[/b]: For positive integers $ n$, $ {n \choose 2^k - 1}$ is odd if and only if $ 2^k \mid n + 1$.

[b]Proof[/b]: The base 2 representation of $ 2^k - 1$ is $ 111\ldots1$ with $ k$ $ 1$s. Suppose $ n$ has binary representation $ d_m d_{m - 1} \ldots d_1$. By Lucas's Theorem,
$ {n \choose 2^k - 1} \equiv {d_m \choose 0} {d_{m - 1} \choose 0} \cdots {d_k \choose 1} {d_{k - 1} \choose 1} \cdots {d_1 \choose 1} \pmod{2}.$
For this to be odd all the binomial coefficient factors on the right side must be $ 1$. Thus it is required $ d_k = d_{k - 1} = \ldots = d_1 = 1$, equivalent to $ n = 2^k p + 2^k - 1$ for some nonnegative integer $ p$. That's one direction. Also note $ d_m, d_{m - 1} \ldots d_{k + 1}$ can be anything for the remaining binomial coefficients to evaluate to $ 1$, so $ p$ can be arbitrary and $ n + 1$ can be any multiple of $ 2^k$. This shows the other direction.

Suppose $ m \geq 2^k$. Also suppose $ n < 2^{k + 1} - 1$, so that $ 2^a \leq n < 2^{a + 1} - 1$ for some $ 0 \leq a \leq k$. Take $ p = 2^a - 1$ (which is always at most $ m$) in $ (\ast )$ and consider it mod $ 2$. Since $ a_i \equiv 1 \pmod{2}$ for all $ i$, this becomes

$ g_{2^a - 1}(1) \equiv {n \choose 2^a - 1} + {n - 1 \choose 2^a - 1} + \ldots + {0 \choose 2^a - 1} \equiv 0. \pmod{2} \quad (\ast \ast ).$

By the lemma, the only odd terms in the sum are of the form $ {2^a p - 1 \choose 2^a - 1}$ for some positive integer $ p$. But since $ 2^a \leq n < 2^{a + 1} - 1$, the only term of this form is $ {2^a - 1 \choose 2^a - 1}$, since the next one would have $ 2^{a + 1} - 1$ on top which is greater than $ n$. Thus there is exactly one odd term in that summation, and so the whole thing must be odd. But we have it is equivalent to $ 0 \pmod{2}$. Contradiction! Thus $ n \geq 2^{k + 1} - 1$.

MellowMellon, your formula for $ \frac {f\left(x\right)}{\left(x \minus{} 1\right)^p}$ alone is way cooler than this trivial problem. The problem has nothing to do with $ \pm 1$'s, it is enough that all coefficients of $ f$ are odd. This yields that the reduction $ \overline{f}$ of $ f$ modulo $ 2$ equals to the polynomial $ x^n \plus{} x^{n \minus{} 1} \plus{} ... \plus{} x \plus{} 1\in\mathbb{F}_2\left[x\right]$. On the other hand, $ 1$ as an $ m$-fold root of $ f$ yields $ \left(x\minus{}1\right)^m\mid f\left(x\right)$ (as polynomials in $ \mathbb{Q}\left[x\right]$, and thus also in $ \mathbb{Z}\left[x\right]$, because $ \left(x \minus{} 1\right)^m$ is a monic polynomial), so that $ \left(x\minus{}1\right)^{2^k}\mid f\left(x\right)$ (since $ m\geq 2^k$ leads to $ \left(x\minus{}1\right)^{2^k}\mid \left(x\minus{}1\right)^m$). Thus, so that $ f\left(x\right) \equal{} \left(x \minus{} 1\right)^{2^k}g\left(x\right)$ for another polynomial $ g\left(x\right)\in\mathbb{Z}\left[x\right]$. Reducing modulo $ 2$, this yields $ x^n \plus{} x^{n \minus{} 1} \plus{} ... \plus{} x \plus{} 1 \equal{} \left(x \minus{} 1\right)^{2^k}\overline{g}\left(x\right)$ in $ \mathbb{F}_2\left[x\right]$. But $ \left(x \minus{} 1\right)^{2^k} \equal{} x^{2^k} \minus{} 1$ in $ \mathbb{F}_2\left[x\right]$, so this becomes $ x^n \plus{} x^{n \minus{} 1} \plus{} ... \plus{} x \plus{} 1 \equal{} \left(x^{2^k} \minus{} 1\right)\overline{g}\left(x\right)$. Hence, $ \deg\overline{g}\geq 2^k \minus{} 1$ by comparison of coefficients (otherwise, the left hand side would be a polynomial with coefficient $ 1$ before $ x^{2^k \minus{} 1}$, but the right hand side would be a polynomial with coefficient $ 0$ before $ x^{2^k \minus{} 1}$). Thus, $ \deg g\geq 2^k \minus{} 1$, and the rest is clear. I have no idea what is the point of such a problem as a [b]third[/b] problem of a Chinese TST (even a quiz).

  darij

[quote="darij grinberg"]I have no idea what is the point of such a problem as a [b]third[/b] problem of a Chinese TST (even a quiz).[/quote]
Dear darij, you will see the very point after seeing the previous two questions of the quiz.
In fact in day 3 the students did far worse than expected (the couches' words).

In fact, I think in general you must have $ n \ge 2m\minus{}1$, so the powers of 2 are irrelevant.

Hm... Why don't you post a proof?

Addendum (more than two years later!): MellowMelon's proof may be shortened by noting that 1) the hypothesis that $x-1$ divides $f^{(p)}$ is enough to derive $(*)$ and 2) the lemma may be proved using the fact that $\binom{n+1}{2^k}=\frac{n+1}{2^k}\binom{n}{2^k-1}$ (or at least the lemma's relevant implication).

Second addendum: I have partially answered my own question, use darij's approach. If $(x-1)^m$ is a factor of $1+x+\ldots+x^n$, then it is also a factor of $x^{n+1}-1$. It is then easy to see, considering the algebraic closure of $\mathbb F_2$, that each root of $x^{n+1}-1$ has multiplicity at least $m$. If there are two such roots, then $n+1\geq 2m$, as we required. Else $x^{n+1}-1=(x-1)^{n+1}$, which implies $n+1$ is a power of $2$. However, this is all we can get working in $\mathbb F_2$, since $(x-1)^m$ divides $(x-1)^{n+1}$ whenever $m\leq n+1$.

[i]The solution of this problem and some others in China TST can be found here :lol: [/i]

Consider the algebraic closure $\bar{F_2}$ of the field $F_2$. We work in $F_2[X]$ and see that $f=x^n+\dots +x+1=\frac{X^{n+1}-1}{X-1}$. Moreover, by Frobenius we have $(X-1)^{2^k}=X^{2^k}-1$. Let $\alpha$ be a root of the cyclotomic polynomial $\phi_{2^k}(X)$ in $\bar{F_2}$. We see that the cyclotomic polynomial is a factor of $X^{2^k}-1$ and so $(X-1)^{2^k}$ vanishes at $\alpha$. Therefore, we have $\alpha^{n+1}=1$ and since the order of $\alpha=2^{k}$, we deduce that $2^k \mid n+1$. Now, clearly, comparing degree gives that $n \ge m \ge 2^k$. Therefore, $n+1\ge 2\cdot 2^k$ and the conclusion holds.

Comment: much more is true.
https://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A532&l=en

[quote=feliz]Hm... Why don't you post a proof?

Second addendum: I have partially answered my own question, use darij's approach. If $(x-1)^m$ is a factor of $1+x+\ldots+x^n$, then it is also a factor of $x^{n+1}-1$. It is then easy to see, considering the algebraic closure of $\mathbb F_2$, that each root of $x^{n+1}-1$ has multiplicity at least $m$. If there are two such roots, then $n+1\geq 2m$, as we required. Else $x^{n+1}-1=(x-1)^{n+1}$, which implies $n+1$ is a power of $2$. However, this is all we can get working in $\mathbb F_2$, since $(x-1)^m$ divides $(x-1)^{n+1}$ whenever $m\leq n+1$.[/quote]
Let me explain probability1.01's claim-for this you really need the coefficients are $\pm 1$ rather than just being odd. (With only odd coefficients, as people explained above by mod $2$ one can show $v_2(n+1)\geq\log_2{(m+1)}$ and thus $n\geq 2^{\lceil\log_2{(m+1)}\rceil}-1$. This estimate is sharp and $m=2^k\Rightarrow n\geq 2^{k+1}-1$ happens to be its most powerful case.)

By MellowMelon's method or simply taking formal derivatives we can get his equations (*):
\[\sum_{k=0}^n f(k)a_k=0\]
for $f(x)=\binom{x}{0},\binom{x}{1},...,\binom{x}{m-1}$ and hence also for any polynomial of degree $\leq m-1$. In particular we have
\[\sum_{k=0}^n k^la_k=0,0\leq l\leq m-1.\]
Let $S=\{k|a_k=1\},T=\{k|a_k=-1\}$, then once $m\geq 1$ we have $|S|=|T|=\frac{n+1}{2}$ and
\[\sum_{i\in S}i^l=\sum_{j\in T}j^l,1\leq l\leq m-1.\]
As $S\neq T$ we must have $m\leq\frac{n+1}{2},n\geq 2m-1$.

Remark: by this method we get if $f(x)\in\mathbb{Z}[x]$ is a nonzero multiple of $(x-1)^m$, then the sum of its positive coefficients $\geq m$.

[quote=Fang-jh]Let $ f(x)$ be a $ n \minus{}$degree polynomial all of whose coefficients are equal to $ \pm 1$, and having $ x \equal{} 1$ as its $ m$ multiple root. If $ m\ge 2^k (k\ge 2,k\in N)$, then $ n\ge 2^{k \plus{} 1} \minus{} 1.$[/quote]

The problem basicaly says that $(x-1)^{2^k} \mid f(x)$ and now we work on $\mathbb F_2$.
On $\mathbb F_2$ we have that $f(x)=\frac{x^{n+1}-1}{x-1}$ and $(x-1)^{2^k}=x^{2^k}-1$ hence the divisibility condition can be re-writen as $x^{2^k}-1 \mid x^{n+1}-1 \implies 2^k \mid n+1$ but since by the first divisibility condition get $n \ge 2^k$ we have that $n \ge 2^{k+1}-1$ thus we are done  :blush:

Considering the problem in $\mathbb{F}_2[x],$ we see $f(x)=x^n+x^{n-1}+\dots+1.$ Notice $(x+1)^{2^k}=x^{2^k}+1$ (induction) divides $f,$ so $$(x^{2^k}+1)g(x)=x^n+x^{n-1}+\dots+1.$$ If $n<2^{k+1}-1,$ then $\deg g\le 2^k-2.$ Hence, there is no $x^{2^k-1}$ term on the LHS while there is one on the RHS since $n\ge 2^k,$ so we have a contradiction. $\square$


Let $2abc$ ($a$,$b$, and $c$ are digits) be the next year that is equal to $300$ times the sum of its digits. Find $a+b+c$

[hide] well since its 300 times we know 2abc must be a multiple of 300 so b and c are 0 and we test the values for a - 1 4 and 7 and we find that 7 works.  so 7+0+0 = 7?  did I misinterpret the question? that seems too simple.[/hide]

Yeah its supposed to be simple. That's why its in this forum. I'll be posting some more, generally a bit more difficult, in a little.

So as per the condition -

$2000 + 100a + 10b + c = 300(2 + a + b + c)$

$2000 + 100a + 10b + c = 600 + 300a + 300b + 300c$

$1400 = 300a + 290b + 299c$

Now last digit of RHS is 0, and terms involving a and b have 0 as the last digit. Hence c has to be 0. So the eqn now becomes,

$1400 = 300a + 290b$
$140 = 30a + 29b$
Again applying the same logic,
b has to be 0.
$140 = 30a$
Therefore
$a = 7, b=c=0$
Hence the year is $2700$

[hide]There are only $3$ muliples of $300$ between $2000$ and $3000$: $2100$, $2400$, and $2700$. Testing these, we find that $2700$ works and the answer is $7 + 0 + 0 = \boxed{7}$[/hide]


function limit integration logarithms calculus algebra polynomial

$f$ bounded measurable function period 1. Is it true
$\lim_{n\rightarrow + \infty}\frac{1}{n}\sum_{k=1}^{n}f(ka)=\int_{0}^{1}f(x)dx$
for almost all $a \in R$

The $\frac1k$ in this statement is wrong; that shouldn't be there. If $f$ is any bounded function, $\left|\frac1k f(ka)\right| \le \frac Mk$ where $M$ is a bound for the function. Then

$\left|\frac1n\sum_{k=1}^n\frac1kf(ka)\right|\le \frac Mn\sum_{k=1}^n\frac1k$.

This is approximately $\frac{M\ln n}n$ and always tends to zero, no matter what the integral of $f$ is.

The challenging question is whether $\lim_{n\to\infty}\frac1n\sum_{k=1}^nf(ka) = \int_0^1f(x)dx.$

It can be shown that this is true whenever $a$ is irrational for any trigonometric polynomial $f$. Making reasonable estimates in the infinity norm gets that it is true for any periodic continuous function, and I think we may be able to extend that to any Riemann integrable function (again, for irrational $a$.) But none of those is exactly the question asked.


complaint for #3
[hide]i felt that 3 was very vague.  i noticed that 512 + 103 = 522 but because of lack of restrictions, -51 also worked.   thats why i put 0.  also, without restrictions, p could be say + or - 3 :sqrt: (2)[/hide]

You're right.

It should be 0 then. No wonder you two have 0 as that answer.

Thanks for pointing out.

#5 u can brute force.  after listing out the last digit of 60 terms, u get 1 + 0 = 1 for the 61st term, so it repeats every 60 terms.  then the 96th term = the 36th term, which ends in 2

Here is the final result:

alan with 12 right

RC-7th with 9 right

As I said, blank answer is no point and guessing gains no penalty. So, maximum score you could have was 15.

If anyone has question on how to do some of the questions, please post on this thread.

Congratulation to the alan for champion of COP The Second and also to RC-7th to two-time second placer.

Congratulations!

just out of curiosity, how did you do #11
[hide]why isnt it 12C4 + 8C4 + 4C4? :? [/hide]

Firstly, I think you meant to multiply those, not to add them..

But the other mistake with that is..
[hide]
12C4 * 8C4 * 4C4 counts everything 6 times. Why? Because the order of the three teams doesn't matter: we are counting ABCD EFGH IJKL, EFGH ABCD IJKL, etc etc once each. So we have to divide by 3! to get the final answer.

Well, thats one way of doing it anyway.
[/hide]

TripleM's right.

We don't care about what kind of team we want to have so we have to divide by 6!.


The numbers $ 1,2,\ldots,n$ are catenated in this order to get a number $ x \equal{} 1234567891011...(n \minus{} 1)n$. What is the least positive $ n$ such that $ 777777777\mid x$?

It's more than 68900 :P. Anyway, this sounds more like a mathematical problem than a computer science problem.

Really?  :| I don't see how you could possibly solve this mathematically.

Its just a simple brute force with a computer though - the answer turns out to be 8 digits long, so as long as you use 64 bit integers and carefully make sure you do all calculations mod 777777777 without any chance of overflow, a program will solve it in seconds.

[quote="bus"]I thought the answer would be that such n doesn't exist. Oh well I guess I used a little too much brute force:
[code]
a = 0;
q = 10;
for (i = 1; true; i ++) {
	if (i == q) q *= 10;
	a = (a * q) + i;
	if (a % 777777777 == 0) return a;
}
[/code]
Yes, it's slow :blush:. (I run that in bc.)[/quote]
Does this works! I don't know how bc works by when I copy-pasted the code it said

(standard_in) 6: Return outside of a function.

and this is all output I got. Do I have to wait more?

As TripleM suggested, do your calculations mod 777777777. Here's working C++ code:
[code]
#include <iostream>
using namespace std;

int main(void)
{
	long long i, a = 1, q = 10;
	for (i = 2; a != 0; i++)
	{
		if (i == q) q *= 10;
		a = (a * q + i) % 777777777;
	}
	cout << i - 1 << endl;
	return 0;
}
[/code]
This gives $ n\equal{}42702705$. It takes 3 or 4 seconds.

[quote="MeKnowsNothing"]
Does this works! I don't know how bc works by when I copy-pasted the code it said

(standard_in) 6: Return outside of a function.

and this is all output I got. Do I have to wait more?[/quote]

No this is not the actual bc code. I've rewritten it into C to make it more readable.

durt: Duh. I so stupuid :). When TripleM said you could use brute force, I immediately realized that there's a much faster way than what I tried - but it [i]wasn't[/i] doing the calculations mod 777777777. What you've written here is so much simpler... I think I'm getting too old for this :lol:.

Anyway this is a working bc code:
[code]
a = 0;
q = 10;
for (i = 1; ; i ++) {
	if (i == q) q *= 10;
	a = ((a * q) + i) % 777777777;
	if (a == 0) break;
}
i;

[/code]
Note that bc is slower than a compiled C code, especially if you only need 64 bit integers.


function
logarithms
algebra
functional equation
algebra proposed

Find at least one function $f: \mathbb R \rightarrow \mathbb R$ such that $f(0)=0$ and $f(2x+1) = 3f(x) + 5$ for any real $x$.

All odd number x can be written $x=m*2^k-1$, m is odd. We have
$f(x)=f(m2^k-1)=5+3f(m2^{k-1}-1)=5+3*5+9f(m2^{k-2}-1)=\dots=\frac{5(3^k-1)}{2}+3^kf(m-1)$
This formula give means f(x) when x is odd if f(x) is certain for even x.  One of f(x) is
$f(2m)=0;f(m2^k-1)=\frac{5(3^k-1)}{2}$.

[quote="Rust"]All odd number x can be written $ x \equal{} m*2^k \minus{} 1$, m is odd. We have
$ f(x) \equal{} f(m2^k \minus{} 1) \equal{} 5 \plus{} 3f(m2^{k \minus{} 1} \minus{} 1) \equal{} 5 \plus{} 3*5 \plus{} 9f(m2^{k \minus{} 2} \minus{} 1) \equal{} \dots \equal{} \frac {5(3^k \minus{} 1)}{2} \plus{} 3^kf(m \minus{} 1)$
This formula give means f(x) when x is odd if f(x) is certain for even x.  One of f(x) is
$ f(2m) \equal{} 0;f(m2^k \minus{} 1) \equal{} \frac {5(3^k \minus{} 1)}{2}$.[/quote]

sorry for reviving this topic, but i don't understand why Rust just discuss when $ x$ is integer and he does not construct it for all reals $ x$.. it's just for odd and even $ x$ of integer.. :(

Indeed, we need $ f: \mathbb{R}\to\mathbb{R}$. Consider the sequences given by $ a_{n \plus{} 1} \equal{} 2a_n \plus{} 1$ and $ b_{n \plus{} 1} \equal{} 3b_n \plus{} 5$ where $ (a_0,b_0) \equal{} (0,0)$. From the functional equation, it is clear that $ f(a_n) \equal{} b_n$ for $ n\in \mathbb{Z}^*$ satisfies the F.E. Furthermore, we can let $ n\in\mathbb{R}\Leftrightarrow a_n\in( \minus{} 1,\infty)$. We can find $ a_n \equal{} 2^n \minus{} 1$ and $ b_n \equal{} \frac {5}{2}\left(3^n \minus{} 1\right)$ so solving yields $ f(x) \equal{} \frac {5}{2}\left((x \plus{} 1)^C \minus{} 1\right)$ where $ C \equal{} \log_2 3$ which satisfies the F.E. for $ x\in( \minus{} 1,\infty)$. We WLOG set $ f( \minus{} 1) \equal{} \minus{} 1$ and for $ x < \minus{} 1$ we set $ f( \minus{} 2) \equal{} 0$ and let $ f(x) \equal{} f( \minus{} 2 \minus{} x)$ where $ \minus{} 2 \minus{} x > \minus{} 1$. This satisfies the F.E. as well: $ 3f(x) \plus{} 5 \equal{} 3f( \minus{} 2 \minus{} x) \plus{} 5 \equal{} f(2( \minus{} 2 \minus{} x) \plus{} 1) \equal{} f( \minus{} 3 \minus{} 2x) \equal{} f(2x \plus{} 1)$ for $ x < \minus{} 1$. Thus, we have:
\[ f(x) \equal{} \begin{cases} \frac {5}{2}\left((x \plus{} 1)^C \minus{} 1\right) & \mbox{if }x\geq \minus{} 1 \\
\frac {5}{2}\left(( \minus{} x \minus{} 1)^C \minus{} 1\right) & \mbox{if }x < \minus{} 1 \end{cases}\mbox{ where }C \equal{} \log_2 3\]
And we can check that this works.


quadratics
LaTeX
function
algebra
quadratic formula
special factorizations

[quote="junggi"]solve:
1)$x^{2}+(1-i)x+i=0$

2)$3x^{2}+(1+2i)x+1=0$

is there any general method to solve quadratics in complex numbers?[/quote]

[hide]Quadratic formula still works  :| $\frac{i-1\pm\sqrt{-6i}}{2}=\frac{i-1\pm i\sqrt{6}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})}{2}=\frac{i-1\pm i(\sqrt{3}+i\sqrt{3})}{2}$ and i have to go. I think you can do #2 the same way[/hide]

hmmm...ok thanks!
somehow I thought i inside a root was a bad thing

Actually, the quadratic forumula states "for all quadratics with real coefficients", so I would use completing the square:

[hide="Here is what I got..."](1)
$x^{2}+(1-i)x+\frac{(1-i)^{2}}{4}-\frac{(1-i)^{2}}{4}+i = 0$
$(x-\frac{1-i}{2})^{2}= \frac{1-2i+i^{2}}{4}-i =-\frac{3i}{2}$

$x-\frac{1-i}{2}=\sqrt{-\frac{3i}{2}}$
$x-\frac{1-i}{2}=\sqrt{-1}\sqrt{\frac{3i}{2}}=i\sqrt{\frac{3i}{2}}$

$x=\frac{1-i}{2}+i\sqrt{\frac{3i}{2}}$

$x=\frac{1-i+i\sqrt{3}-\sqrt{3}}{2}$

(2)
For the other one, I get (by completing the square):

$x=\frac{\sqrt{4i-15}-2-4i}{6}$

Too hard writing things in $\LaTeX$.[/hide]

Only real coefficients? Isn't the quadratic formula derived from completing the square though...?

[quote="vishalarul"]
$(x-\frac{1-i}{2})^{2}= \frac{1-2i+i^{2}}{4}-i =-\frac{3i}{2}$

$x-\frac{1-i}{2}=\sqrt{-\frac{3i}{2}}$
[/quote]

Shouldn't you add a modulus function here so you had obtain 2 solutions?

The quadratic formula can be used for quadratics with real or complex coefficients. :wink:


Hey i got few questions please if i can get their solutions ...

Q1- The value of "k" for which the function of itself g(x)=kx+1 , (for k is not equal to zero) is the inverse of itself.

Q2- Let R --> R be any funtion and g: R-->R is defined as g(x)= |f(x)| for all x. Then g is ______ funtion?

Q3- Let E= {1,2,3,4}, F= {1.2} then number of on to functions from E to F is ?

[hide="3"]
Well, we have four inputs, and two outputs. We want to choose only one input for each output, so the answer is $\binom{4}{2}$ which is 6.
[/hide]

the number of functions from E-->F is 4^2
there is no injective ones cause Card(E)>Card(F)
there is two function wich are noninjective and nonsurjective
then the number of onto functions is 6


geometry
inequalities
complex numbers
triangle inequality

Here's a problem I probably shouldn't have such difficulty figuring out (it is supposed to be the easiest problem on the exam :().

Draw the set $M$ of complex numbers $z$ such that
\[|z - 2 + i| \le 1\]
Give the range of $|z|$ for $z \in M$.

Drawing the set is easy. It's just everything inside the circle with radii 1 and midpoint (2,-1) in the complex plane. The equation of the circle can also be given by (with x = Re z, y = Im z) $(x - 2)^2 + (y+1)^2 = 1$. From that I realize that we want to decide the two values of $z$ for which the distance from the origin to the circle will be smallest and largest, respectively. Then the range of |z| will be between those two values. However, I simply cannoy figure out how to determine those two values. I'd guess some geometric reasoning is in order. Thank you for all your help and patience :)

[hide="Hint"]Draw the line passing through the origin and the center $(2, -1)$.  That line will pass through two points on the circle.  One of those points will be the closest to the origin, and the other point will be the farthest.  To prove those extremal claims, you can use the triangle inequality $|a + b| \le |a| + |b|$.[/hide]

Hi kalle!

This is actually an easy and maybe routine problem.however I would like to help you.

Firtstly note that $\forall z_1,z_2 \in C$ we have
$\mid\mid z_1\mid-\mid z_2\mid \mid\leq\mid z_1+z_2 \mid \leq\mid z_1 \mid+\mid z_2 \mid$
which is easy to prove.In fact This is the triangle inequality....

Now see that $\mid z \mid=\mid (z-2+i)+(2-i) \mid\leq \mid z-2+i\mid+\mid 2-i\mid \leq 1+\sqrt{5}$

Besides $\mid z \mid=\mid (z-2+i)+(2-i) \mid\geq \mid\ \mid z-2+i \mid-\mid 2-i\mid\ \mid\geq \mid\ \mid z-2+i \mid-\sqrt{5} \mid=\sqrt{5}- \mid z-2+i \mid \geq\sqrt{5}-1$
......

For a geometrical approach let $k$ the center of the circle. Then the point of the circle that is nearest to the origin $O$ is the intersection of the circle circumference and the line $OK$. There are two two intersection points, the one that has the largest and the one that has the smallest distance from $O$...

ok??

 :)

Thank you!

edit: is there a better way to do it geometrically than finding that the line has the equation y = -x/2, substituting it into the equation for the circle and solving for x and y, to then find |x + iy| ? This process is quite time-consuming for a ten-minute problem.

edit: obviously there is a much better solution. by pythagoras theorem the distance to the circle's midpoint is $\sqrt{5}$. And then just add or subtract one for the circle radii. I can't believe I overcomplicated it that much.

[quote="socrates"]Ravi B you are faster and more brief........

 :)[/quote]

As the admins of this site, we feel Ravi B is one of the best, most experienced math teachers we know about anywhere.  Many students in our forums and elsewhere benefit from his crisp understanding of mathematics.


Given triangle $\triangle ABC$ that $B(2,-7)$. The perpendicular from $A$ is named $(d): 3x+y+11=0$ and the median from top $C$ is $(d)^*: x+2y+7=0$. Find the equations of lines $AB,\ BC,\ CA$.

I will try to find equation of BC
$m_(BC)*m_(d_1)=-1$
$m_(d_1)=3$
so
$m_(BC)=\frac{1}{3}$
so EQUATION of BC is
$y=\frac{1}{3}x-\frac{23}{3}$
$x-3y-23=0$

Hint
[hide]Use that A and B are at the same distance from $x+2y+7=0$[/hide]


Let $ m$ and $ n$ be integers greater than $ 1$ Prove that $ m^n$ is the sum of $ m$ odd consecutive integers.

I think induction must help since for
$ n\equal{}2$  we have  $ 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2m \minus{} 1) \equal{} m^2$.

I also think you can build from the $ n \equal{} 2$ case.

[hide]

Call the average of the arithmetic series $ y$.  There are $ m$ terms in the series.  So the problem could be thought of as $ (m)(y) \equal{} m^{n} \rightarrow y \equal{} m^{n\minus{}1}$.  We already know that the series $ 1, 3, 5.... 2m \minus{} 1$ satisfies this for $ n \equal{} 2$ because the value of $ y$ is $ m$ for this series.  If you slide all the terms one number to the right you get $ 3, 5, 7.... 2m \plus{} 1$.  The difference between the value of $ y$ for these two series is $ 2$.  As a result, we can increase the value of $ y$ by increments of $ 2$ which means we can make $ y$ any  number with the same parity as $ m^{1}$.... including all powers of $ m$ :lol: [/hide]

Suppose that we have the odd numbers $ 1$ to $ 2k\minus{}1$. $ 1\plus{}...\plus{}2k\minus{}1\equal{}k^2$,

Likewise, the sum of $ 1,...,2(k\minus{}m)\minus{}1$ is $ (k\minus{}m)^2$
So the sum of $ m$ consecutive odd numbers can be written as:

$ k^2\minus{}(k\minus{}m)^2\equal{}(2k\minus{}m)(m)\equal{}m^n$

$ 2k\equal{}m\plus{}m^{n\minus{}1}$

Now $ k\equal{}\frac{m\plus{}m^{n\minus{}1}}{2}$ which is even (as long as $ n\ge 2$) so we're done.


In the 360 Problems for Mathematical Contests book, the notations "tg" and "ctg" are used for trig. Are they equivalent to tan and cot?

Yes. I've seen that frequently from the Europeans on the college forum.


For each of the next four days, there is an 80% chance that the weatherman will say that it will snow on that day. Each day, there is a 70% chance that the weatherman will be wrong. What is the probability that it will snow on exactly two of the next four days?

[hide]There is a $30\%$ chance that the weatherman is right and thus there is a $0.3\cdot 0.8=24\%$ chance that it will snow on a given day.

Then, the probability that it will snow on exactly 2 of 4 days is:
$\binom42(0.24)^{2}(0.76)^{2}\approx 0.2=\frac15$.[/hide]

Don't forget when the weatherman says it's not going to snow, but he's wrong.

[hide]There are two cases:  The weatherman says it will snow and is correct; and the weatherman says it won't snow and is incorrect.  Thus, the probability of it snowing on any given night is

$20\%\times70\%+80\%\times30\%=14\%+24\%=38\%$.

There are $\binom42=6$ ways that it could snow on 2 of 4 nights, and the probability is $6\left[(38\%)^{2}\cdot(62\%)^{2}\right]\approx33.3\%$[/hide]

Oh yeah :blush:

That was pretty tricky.

You sort of have to make the assumption that the only things the weatherman says are "It will snow" and "It will not snow". What if he says "It will rain"?

I've learned that it's best not to think about these things during competitions.


Check out this cool website about Klein bottles. The are quite a few humorous bits in there. Also make sure to check out all the links.
[url]http://www.kleinbottle.com/whats_a_klein_bottle.htm[/url]
Oh, and moderators--don't wory, I am not affiliated with that company in any way.

Cool!!  I was wondering what objects were in 4 dimensions.

The Klien bottle can be inefficient when storing and pouring water, i think.
[size=10]Nice Revive.[/size]


There is a "phone tree" that is rather ineffective because of the poor memory of the people. SInce they can't remember who they are supposed to call, they just call people completely at random. Once they are called, they immediatly start calling other and all the calls take the same length of time. Also if they call 3 people in a row who are already informed, then they get bored and forget what they were doing. They jsut ist there untill someone calls them to remind them what they were doing. Also there is an infinte number of people.

If the amount of people who are sitting there bing bored, never changes, what amount of the people are making calls?

If all calls are synced, then no one can get bored as no one can call an already informed person who is not bored as they are calling someone else, and the call gets a busy signal.  So an infinite number of people are making calls ;)

A busy signal counts as an unsucessful call so if a person gets 3 busy signals in a row they get bored.

Yeah a busy signal would mean the person they were trying to call was calling other people.... meaning they were informed.

And i would think that if there were $\infty$ people then there'd be a $\frac{n}{\infty}$ chance that they call one of the $n$ people already informed..... and since
$\frac{n}{\infty}=0$
Then noone would ever call a previously-informed person... meaning an infinitely expanding tree of callers with no bored people.  So i'd say the answer is either inconsistent, undetermined, infinite, or a combination of the above.

The amount of people that are informed is a certain fraction of the total people. It is not a finite number.

I was asking for the fraction.


x=1/1+1/2+1/n,let 2^k divide n,but 2^(k+1) can't divide n, if n=2m+1,M=lcd(1,3,5n),if n=2m,M=lcd(1,3,5,n-1),
if x is an integer,(2^(k-1))*x*M is an integer ,too.but when i=1,2,3
(1/i)*(2^(k-1))*M is an integer

but I lose my way,
soory
7615

correct
let 2^a<=n<2^(a+1),if n=2m+1,M=lcd(1,3,5…n),if n=2m,M=lcd(1,3,5…,n-1),
if x is an integer,(2^(a-1))*x*M is an integer
 but when i=1,2,3… (1/i)*(2^(k-1))*M is an integer except i=2^a,they
are contradictory

[quote="mrblueskies"]show that

$1+\frac12+\frac13+...+\frac{1}{n}$

can never be an integer if n>1[/quote]

Chebyshev said it, i'll say it again.
There's always a prime between n and 2n.

That's a hand-waving solution, but I remember someone posting that before and found it funny.

Otherwise, you should try to show that there is a power of two in the bottom which is greater than the greatest power of two on the top.

1/k+ 1/(k+1)+..+1/(k+n) is never an integer i think ...

It was extended to AP of any starting natural number, and d of any natural number. i think it was Erdos.


Since when it is uncommon that sometimes a bird tries to fly through glas¿

It is actually well-documented, and a particular case led to the 2003 Ig Nobel Prize in biology. Note: the following is scientific, but mildly graphic. Readers under 18 are cautioned.
http://www.nmr.nl/deins815.htm

[quote="thatin"]Returning from a 4 day trip I found one dead bird in my balcony, blood around it... :o
After a closer examination I found that it hitted to the window, broke it's head and then died...
poor bird but... do you think that it might had a virus or something like that or it is common? (I have cleaned the windows reaaaaaaaaaaally good :lol: )[/quote]

? i've seenv that happen too. the4y can't tell

[quote="K81o7"]It once happened during my english class in 6th grade.  But the bird was not killed...it was just knocked out for a little while, stunned, and it got up and flew away later.[/quote]
To my knowladge, chances are high that they die the next days on such things.
Birds have, due to flying (=mass) reasons, not so (but still) strong bones, thus it's not so good for them. And they fly head-ahead.

If you do keep your windows really clean, the bird was probably flying (maybe trying to escape some preadator) and didn't see the window there and flew right into it.  Think of how people (well, some people  :blush:  walk through windows because they don't see them).

I volunteer at a bird refuge of sorts and we get a good number that hit windows, many die of head trauma, some recover, but if there was blood I'd bet it died on or close to impact.  I'd guess the bird wasn't sick or anything, just a mistake on its part.  You can hang things in your window if you have a chronic problem of it happening, or even just putting a screen over the window helps.  You don't need to worry though, it happens a lot...

Poor bird.
Such an event is dramatized in :
[hide][img]http://img148.imageshack.us/img148/4176/birdyff5.jpg[/img][/hide]

[quote="byrdgyrl"]If you do keep your windows really clean, the bird was probably flying (maybe trying to escape some preadator) and didn't see the window there and flew right into it.  Think of how people (well, some people  :blush:  walk through windows because they don't see them).[/quote]

I once tried to walk through a glass door. *nods* Our family had one of those sets of sliding glass doors to the back yard in our old house. Educational experience, that was.

But as for your bird, I think it's more likely that it saw something potentially good to eat or nice and shiny.

nice and shiny... hm... TV?
I think that it might tried to escape from a predator

[quote="thatin"]nice and shiny... hm... TV?[/quote]

I was thinking more along the lines of jewelry.

Somehow the predator scenario...I mean it's not like American residential zones are similar to the African Safari or anything like that. Where would you find a predator?

[quote="MithsApprentice"]Somehow the predator scenario...I mean it's not like American residential zones are similar to the African Safari or anything like that. Where would you find a predator?[/quote]
You know, according to his location, he doesn't live in an American residential zone. :P You are succumbing to the fallacy that fredbel6 always points out! I don't know anything about the avian population of Greece, but it may be that your assumptions about American residential zones also apply to Greek ones. :P

[quote="MithsApprentice"]I somehow conveniently missed the part about him being located near Belgium. In that case, I can definitely see where the predator might come in.[/quote]
I live in Flanders (Belgium) :ninja:, Thatin lives in Athens, Greece?

But hey, watch Birdy, it's so sad. :(

[quote]Somehow the predator scenario...I mean it's not like American residential zones are similar to the African Safari or anything like that. Where would you find a predator?[/quote]
Leaving aside the part where you shouldn't have used the word "American," the answer is that there are pleny of predators. The ones that would require the most dire of escape attempts would be other birds - especially hawks, falcons, and related birds.

Those can easily be found in urban areas. In several U.S. cities, peregrine falcons nest on tall buildings in downtown areas - and perigrine falcons primarily attack small birds, striking them in flight.

Other, ground-bound predators who would kill small birds also abound in urban areas, notably domestic cats.

Actually I don't know if there is a predator for that bird in the area.
ground-bound predators... hm I think that this is impossyble because my home is at 5th floor

And fred is right!

It's pretty normal especially when dealing with drunk birds. But then they fly into all sorts of things :P


function
integration
real analysis
real analysis unsolved

Find an example of a sequence of measureable functions $ (f_n)$ on the unit interval (0,1) such that
$ \displaystyle \int (\liminf f_n) dx < \liminf(\int f_n dx) < \limsup (\int f_n dx) < \int (\limsup f_n ) dx$

sorry to top my own thread, but could anyone help?

thanks you so much!

happy holidays!

For even $ n,$ let $ f_n(x)\equal{}\begin{cases}0&0<x\le\frac12\\ 2&\frac12<x<1\end{cases}$ and

for odd $ n,$ let $ f_n(x)\equal{}\begin{cases}3&0<x\le\frac12\\ 1&\frac12<x<1\end{cases}.$

I'll let you work out the details.


This is already beeing discussed in http://www.mathlinks.ro/Forum/viewtopic.php?t=4561 throughly. Topic closed.


abstract algebra
function
superior algebra
superior algebra solved

Show that any finitely generated torsion abelian group is finite.

It follows directly from the structure theorem of finetly generated abelian groups...

Here is an easier way: Let $G$ be the finitely generated torsion group.
Let $x_1,\ldots,x_n$ be a system of generators of $G$ of orders $d_1,\ldots,d_n$.
The map $\varphi: \mathbb{Z}^{d_1}\times\ldots\times\mathbb{Z}^{d_n}\to G$ defined by $\varphi(e_i)=x_i\ \forall i=\overline{1,n}$ and extended by linearity is obviously a surjective morphism. $e_1,\ldots,e_n$ deonte the obvious generatros of $\prod_{i=1}^n\mathbb{Z}^{d_i}$. Now the image of a finite group through any function is finite, so $G$ is finite.


Don't post the same thing in two different forum sections: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318185


geometry
3D geometry
probability
ratio
trigonometry
logarithms
algebra

[b]1.[/b] A [i]space diagonal[/i] of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0  (B) 2  (C) 4  (D) 8  (E) 28

[b]2.[/b] Find the units digit of the product of the first 2004 odd primes.

(A) 1  (B) 3  (C) 5  (D) 7  (E) 9

[b]3.[/b] The area of the triangle bounded by the lines [tex]y = x[/tex], [tex]y = -x[/tex], and [tex]y = 6[/tex] is

(A) [tex]12[/tex]  (B) [tex]12\sqrt{2}[/tex]  (C) [tex]24[/tex]  (D) [tex]24\sqrt{2}[/tex]  (E) [tex]36[/tex]

[b]4.[/b] How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224  (B) 280  (C) 504  (D) 512  (E) 729

[b]5.[/b] For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18  (B) 19  (C) 20  (D) 21  (E) 22

[b]6.[/b] How many right triangles with integer side lengths are such that their side lengths are in a nonconstant geometric sequence?

(A) 0  (B) 1  (C) 2  (D) 4  (E) Infinitely many

[b]7.[/b] A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is [tex]\frac{D}{Q}[/tex], with D and Q relatively prime positive integers, find D + Q.

(A) 3  (B) 17  (C) 331  (D) 346  (E) 349

[b]8.[/b] Find the number of integers [tex]k[/tex] such that [tex]k^3 + k^2 + k + 7[/tex] is divisible by [tex]k+1[/tex].

(A) 2  (B) 4  (C) 6  (D) 8  (E) Infinitely many

[b]9.[/b] Circle A is inscribed in an isosceles right triangle and circle B is circumscribed about the same triangle. The ratio of the area of circle A to the area of circle B is

(A) [tex]\frac{1}{2}[/tex] (B) [tex]3 - \sqrt{2}[/tex]  (C) [tex]3 + \sqrt{2}[/tex]  (D) [tex]3 - 2\sqrt{2}[/tex]  (E) [tex]3 + 2\sqrt{2}[/tex]

[b]10.[/b] A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn randomly and without replacement. Given that the probability that the value of the coins drawn is at least 50 cents is [tex]\frac{m}{n}[/tex], with m and n relatively prime positive integers, find m + n.

(A) 961  (B) 1015  (C) 1051  (D) 1056  (E) None of these

[b]11.[/b] Given that [tex]i^2 = -1[/tex], for how many integers [tex]n[/tex] is [tex](n + i)^4[/tex] an integer?

(A) 0  (B) 1  (C) 2  (D) 3  (E) 4

[b]12.[/b] A quadrilateral circumscribed about a circle has three of its side lengths 132, 162, and 201. What is the positive difference between the minimum and maximum lengths for the fourth side?

(A) 60  (B) 78  (C) 93  (D) 138  (E) 231

[b]13.[/b] For how many primes [tex]p[/tex] is [tex]p^4 + p^5[/tex] the square of an integer?

(A) 0  (B) 1  (C) 2  (D) 3  (E) Infinitely many

[b]14.[/b] If [tex]x[/tex], [tex]y[/tex], and [tex]z[/tex] are positive numbers satisfying [tex]x + \frac{1}{y} = 4[/tex], [tex]y + \frac{1}{z} = 1[/tex], and [tex]z + \frac{1}{x} = \frac{7}{3}[/tex], then find [tex]xyz[/tex].

(A) [tex]\frac{2}{3}[/tex]  (B) [tex]1[/tex]  (C) [tex]\frac{4}{3}[/tex]  (D) [tex]2[/tex]  (E) [tex]\frac{7}{3}[/tex]

[b]15.[/b] If [tex]\frac{\pi}{2} < \alpha < \frac{3\pi}{4}[/tex], [tex]\cos{\alpha} + \sin{\alpha} = \frac{1}{3}[/tex], and [tex]\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}[/tex], where [tex]n[/tex], [tex]p[/tex], and [tex]q[/tex] are relatively prime positive integers and [tex]p[/tex] is not divisible by the square of any prime, find [tex]n+p+q[/tex].

(A) 34  (B) 35  (C) 36  (D) 37  (E) 38

[b]16.[/b] If [tex]r_1[/tex], [tex]r_2[/tex], and [tex]r_3[/tex] are the solutions to the equation [tex]x^3 - 4x^2 - 4x + 17 = 0[/tex], find [tex]r_1^3 + r_2^3 + r_3^3[/tex].

(A) -68  (B) -4  (C) 0  (D) 61  (E) 68

[b]17.[/b] Find [tex]\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}[/tex].

(A) [tex]0[/tex]  (B) [tex]-2^{49}[/tex]  (C) [tex]2^{49}[/tex]  (D) [tex]-2^{50}[/tex]  (E) [tex]2^{50}[/tex]

[b]18.[/b] A [i]rising[/i] number is a positive integer, each digit of which is larger than each of the digits to its left, and no digit of which is 0. For example, 12689 is a rising number. When all five-digit rising numbers are listed from smallest to largest, the 97th number in the list does not contain the digit

(A) 4  (B) 5  (C) 6  (D) 7  (E) 8

[b]19.[/b] [tex]\bigtriangleup ABC[/tex] has all side lengths distinct. Two altitudes of this triangle have lengths 4 and 12. If the length of the third altitude is also an integer, find its maximum value.

(A) 4  (B) 5  (C) 6  (D) 7  (E) None of these

[b]20.[/b] It is given that [tex](\log_z{3})(\log_x{3})+ (\log_y{3})(\log_x{3}) + (\log_y{3})(\log_z{3}) = 6(\log_z{3})(\log_x{3})(\log_y{3})[/tex] and [tex]x<y<z[/tex]. If [tex]x[/tex], [tex]y[/tex], and [tex]z[/tex] are in a geometric sequence with common ratio 2, and [tex]x + y + z = \frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m + n[/tex].

(A) 61  (B) 62  (C) 63  (D) 64  (E) 65

[b]21.[/b] Consider the following polynomial equation with [tex]q_4, q_3, q_2, q_1[/tex], and [tex]q_0[/tex] real constants:

[tex]q_4z^4 + iq_3z^3 + q_2z^2 + iq_1z + q_0 = 0[/tex]

If [tex]z = a + bi[/tex] is a solution to this equation, where [tex]a[/tex] and [tex]b[/tex] are real constants and [tex]i^2 = -1[/tex], which of the following must also be a solution?

(A) [tex]-a-bi[/tex]  (B) [tex]a-bi[/tex]  (C) [tex]-a+bi[/tex]  (D) [tex]b+ai[/tex]  (E) None of these

[b]22.[/b] Find the number of distinct pairs of integers (x, y) such that [tex]0<x<y[/tex] and [tex]\sqrt{1984} = \sqrt{x} + \sqrt{y}[/tex].

(A) 0  (B) 1  (C) 3  (D) 4  (E) 7

[b]23.[/b] Four balls of radius 1 are mutually tangent, three resting on a plane and the fourth resting on the others. A tetrahedron, each of whose edges has length q, is circumscribed around the balls. Find q.

(A) [tex]4\sqrt{2}[/tex]  (B) [tex]4\sqrt{3}[/tex]  (C) [tex]2\sqrt{6}[/tex]  (D) [tex]1 + 2\sqrt{6}[/tex]  (E) [tex]2 + 2\sqrt{6}[/tex]

[b]24.[/b] Find the smallest integer n such that for real x, y, and z, the following inequality holds:

[tex](x^2 + y^2 + z^2)^2 \leq n(x^4 + y^4 + z^4)[/tex]

(A) 1  (B) 2  (C) 3  (D) 4  (E) 8

[b]25.[/b] In [tex]\bigtriangleup JSR[/tex], [tex]\angle R[/tex] is a right angle and the altitude from [tex]R[/tex] meets [tex]JS[/tex] at [tex]Z[/tex]. The lengths of [tex]\bigtriangleup JSR[/tex] are integers, [tex]SZ=1331[/tex], and [tex]\cos{S} = \frac{p}{q}[/tex], where [tex]p[/tex] and [tex]q[/tex] are relatively prime positive integers. Find [tex]p+q[/tex].

(A) 71  (B) 72  (C) 142  (D) 143  (E) 144

I have received a few complaints about the difficulty of this test...I apologize for this. The AIME qualifying score for this test will be lowered as a result.

[quote="JSRosen3"]PLEASE RELOAD THE PAGE WHEN YOU HAVE A MOMENT. There was a slight error in #22. It should read:

[b]22.[/b] Find the number of distinct pairs of integers (x, y) such that [tex]0<x<y[/tex] and [tex]\sqrt{1984} = \sqrt{x} + \sqrt{y}[/tex].

(A) 0  (B) 1  (C) 3  (D) 4  (E) 7[/quote]

I was just about to PM you about that :)

[quote="nr1337"][quote="JSRosen3"]TIME IS UP. Please PM me your answers.

I have received a few complaints about the difficulty of this test...I apologize for this. The AIME qualifying score for this test will be lowered as a result. I will elaborate on this later.[/quote]

It seemed much harder than this years AMC 12.  It was fun anyway.[/quote]

AIME qualifying score!?!?!? Please say you're kidding... :)

I recognized some of the problems from old AIMEs, etc.

I answered 17 questions :( , pretty sure i will make some stupid mistakes.

[quote="Rep123max"]Geometry killed me. I spent too much time on the geometries and didn't answer them! I wont be surprised if I didn't break 100.[/quote]

Exactly same here.  I burned 30 minutes on a geometry problem, and I called myself !@$!@\#$ retard.

I won't be surprised if I didn't make 100.

I want to live up to my name. (UltimateMathGeek)

[quote]I really shouldn't have listened to music while taking the test.[/quote]

I really shouldn't have been sitting on the floor, and putting my work on a chair.  It was quite uncomfortable.

I answered 19, but I know for certain that I got one wrong.

That was umm... unexpectedly difficult, but I appreciate the challenge, though I only answered 16. Breaking 100 seems to be out of the question. This seemed much more like an AMC 12 than an AMC 10.

Thanks anyways JSRosen!

Hey, I'd be surprised if I broke  70, and I got 105 on the AMC 12 last year!

I answered 15... didn't even bother to finish reading the geometry problems.

I might just break 100, but I doubt it. Think I'll come close, at the very least.

Heh, someone should make a mock USAMO...it'd be interesting to see some of the solutions.

The questions are posted on the first page of this topic...

Has everyone else gotten their score yet?

I answered 15 and missed 2(#5 and #21).  I believe that is a 103... that is close to my score from last year.  At least I would have made AIME!!! That wasn't as bad as I thought it was going to be (the score not the test).

Please discuss solutions in the "Discussion of the Mock AMC Contest" thread. Also, if you believe that I have made a mistake in either a question or grading your test, post in that thread and we can work out the situation.

Also, if you found this difficult, don't worry...it was indeed very difficult. The AIME qualifying score has been lowered by 11 points as a result. And if you were disappointed with your score...go over the questions without a time limit! And if you have any questions, ask and they will be answered.

I got a 80.5.  I missed 4 problems all because I can't read.  For example, I missed the word "odd" in number 4, thought number 8 said "positive integers" and not "integers", etc.  I would've gotten an AIME qualifying score. :( :( :( :( :(

When do we get our scores?  I have haven't reciewved them, and well... it's been a day, so I'm kind of wondering.

I think you just score yourself.  He posted the answers in spoiler in this post.

Individual scores have been sent. If I missed you, let me know.

[quote="Peter VDD"]for what age group is this AIME? I guess it's for younger students? Or is there a very hard time limit, if yes how much time should we take for this?[/quote]

The AIME is the second round in quallifying for the USA's team for the IMO.  Since you have to take it to be on the USA's math team, there are people from 12 to 17 who take it.

And there is a 3-hour time-limit.

De-stickified to save space at the top of the forum.

[b]1.[/b] A [i]space diagonal[/i] of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0  (B) 2  (C) 4  (D) 8  (E) 28

[b]2.[/b] Find the units digit of the product of the first 2004 odd primes.

(A) 1  (B) 3  (C) 5  (D) 7  (E) 9

[b]3.[/b] The area of the triangle bounded by the lines $y = x$, $y = -x$, and $y = 6$ is

(A) $12$  (B) $12\sqrt{2}$  (C) $24$  (D) $24\sqrt{2}$  (E) $36$

[b]4.[/b] How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224  (B) 280  (C) 504  (D) 512  (E) 729

[b]5.[/b] For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18  (B) 19  (C) 20  (D) 21  (E) 22

[b]6.[/b] How many right triangles with integer side lengths are such that their side lengths are in a nonconstant geometric sequence?

(A) 0  (B) 1  (C) 2  (D) 4  (E) Infinitely many

[b]7.[/b] A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is $\frac{D}{Q}$, with D and Q relatively prime positive integers, find D + Q.

(A) 3  (B) 17  (C) 331  (D) 346  (E) 349

[b]8.[/b] Find the number of integers $k$ such that $k^3 + k^2 + k + 7$ is divisible by $k+1$.

(A) 2  (B) 4  (C) 6  (D) 8  (E) Infinitely many

[b]9.[/b] Circle A is inscribed in an isosceles right triangle and circle B is circumscribed about the same triangle. The ratio of the area of circle A to the area of circle B is

(A) $\frac{1}{2}$ (B) $3 - \sqrt{2}$  (C) $3 + \sqrt{2}$  (D) $3 - 2\sqrt{2}$  (E) $3 + 2\sqrt{2}$

[b]10.[/b] A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn randomly and without replacement. Given that the probability that the value of the coins drawn is at least 50 cents is $\frac{m}{n}$, with m and n relatively prime positive integers, find m + n.

(A) 961  (B) 1015  (C) 1051  (D) 1056  (E) None of these

[b]11.[/b] Given that $i^2 = -1$, for how many integers $n$ is $(n + i)^4$ an integer?

(A) 0  (B) 1  (C) 2  (D) 3  (E) 4

[b]12.[/b] A quadrilateral circumscribed about a circle has three of its side lengths 132, 162, and 201. What is the positive difference between the minimum and maximum lengths for the fourth side?

(A) 60  (B) 78  (C) 93  (D) 138  (E) 231

[b]13.[/b] For how many primes $p$ is $p^4 + p^5$ the square of an integer?

(A) 0  (B) 1  (C) 2  (D) 3  (E) Infinitely many

[b]14.[/b] If $x$, $y$, and $z$ are positive numbers satisfying $x + \frac{1}{y} = 4$, $y + \frac{1}{z} = 1$, and $z + \frac{1}{x} = \frac{7}{3}$, then find $xyz$.

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

[b]15.[/b] If $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$, $\cos{\alpha} + \sin{\alpha} = \frac{1}{3}$, and $\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}$, where $n$, $p$, and $q$ are relatively prime positive integers and $p$ is not divisible by the square of any prime, find $n+p+q$.

(A) 34  (B) 35  (C) 36  (D) 37  (E) 38

[b]16.[/b] If $r_1$, $r_2$, and $r_3$ are the solutions to the equation $x^3 - 4x^2 - 4x + 17 = 0$, find $r_1^3 + r_2^3 + r_3^3$.

(A) -68  (B) -4  (C) 0  (D) 61  (E) 68

[b]17.[/b] Find $\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}$.

(A) $0$  (B) $-2^{49}$  (C) $2^{49}$  (D) $-2^{50}$  (E) $2^{50}$

[b]18.[/b] A [i]rising[/i] number is a positive integer, each digit of which is larger than each of the digits to its left, and no digit of which is 0. For example, 12689 is a rising number. When all five-digit rising numbers are listed from smallest to largest, the 97th number in the list does not contain the digit

(A) 4  (B) 5  (C) 6  (D) 7  (E) 8

[b]19.[/b] $\bigtriangleup ABC$ has all side lengths distinct. Two altitudes of this triangle have lengths 4 and 12. If the length of the third altitude is also an integer, find its maximum value.

(A) 4  (B) 5  (C) 6  (D) 7  (E) None of these

[b]20.[/b] It is given that $(\log_z{3})(\log_x{3})+ (\log_y{3})(\log_x{3}) + (\log_y{3})(\log_z{3}) = 6(\log_z{3})(\log_x{3})(\log_y{3})$ and $x<y<z$. If $x$, $y$, and $z$ are in a geometric sequence with common ratio 2, and $x + y + z = \frac{m}{n}$, with m and n relatively prime positive integers, find $m + n$.

(A) 61  (B) 62  (C) 63  (D) 64  (E) 65

[b]21.[/b] Consider the following polynomial equation with $q_4, q_3, q_2, q_1$, and $q_0$ real constants:

$q_4z^4 + iq_3z^3 + q_2z^2 + iq_1z + q_0 = 0$

If $z = a + bi$ is a solution to this equation, where $a$ and $b$ are real constants and $i^2 = -1$, which of the following must also be a solution?

(A) $-a-bi$  (B) $a-bi$  (C) $-a+bi$  (D) $b+ai$  (E) None of these

[b]22.[/b] Find the number of distinct pairs of integers (x, y) such that $0<x<y$ and $\sqrt{1984} = \sqrt{x} + \sqrt{y}$.

(A) 0  (B) 1  (C) 3  (D) 4  (E) 7

[b]23.[/b] Four balls of radius 1 are mutually tangent, three resting on a plane and the fourth resting on the others. A tetrahedron, each of whose edges has length q, is circumscribed around the balls. Find q.

(A) $4\sqrt{2}$  (B) $4\sqrt{3}$  (C) $2\sqrt{6}$  (D) $1 + 2\sqrt{6}$  (E) $2 + 2\sqrt{6}$

[b]24.[/b] Find the smallest integer n such that for real x, y, and z, the following inequality holds:

$(x^2 + y^2 + z^2)^2 \leq n(x^4 + y^4 + z^4)$

(A) 1  (B) 2  (C) 3  (D) 4  (E) 8

[b]25.[/b] In $\bigtriangleup JSR$, $\angle R$ is a right angle and the altitude from $R$ meets $JS$ at $Z$. The lengths of $\bigtriangleup JSR$ are integers, $SZ=1331$, and $\cos{S} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

(A) 71  (B) 72  (C) 142  (D) 143  (E) 144

non-'texified' version yay


In acute triangle $ ABC$, let $ AD,BE,CF$ be the altitudes, and let $ H$ be the orthocenter. Let $ EF$ intersect $ BC$ at $ X$, and let $ XH$ intersect $ AB,AC$ at $ P,Q$ respectively. Prove that $ PE,QF$ intersect on $ AH$.

Menelaus's and Ceva's Theorem should work

We can also playing with three arbitrary concurrent cevians, not necessary the altitudes of a given triangle.

It is a simple application of harmonic conjugation theory, proved by [b][size=100]Menelaus[/size][/b] and [b][size=100]Ceva[/size][/b] theorems, as [b][size=100]Mathangel[/size][/b] mentioned.

Kostas Vittas.


[b]Theorem 1.[/b] Let $ 0 < x_{1}\leq x_{2}\leq\cdots\leq x_{n}$ and $ 0 < y_{1}\leq y_{2}\leq\cdots \leq y_{n}$. Now, if we have all the following conditions fulfilled:

  $ x_{1}\geq y_{1}$;
  $ x_{1} \plus{} x_{2}\geq y_{1} \plus{} y_{2}$;
  $ \cdots$
  $ x_{1} \plus{} x_{2} \plus{} \cdots \plus{} x_{n}\geq y_{1} \plus{} y_{2} \plus{} \cdots \plus{} y_{n}$,
  (generally $ \sum_{i \equal{} 1}^{k}x_{i}\geq \sum_{i \equal{} 1}^{k}y_{i}$ for any natural number $ k$ with $ 1 \leq k \leq n$)

  then $ M_{p}(x) \geq M_{p}(y)$ for $ p < 1$,

where the power mean of order $ p$, of $ (a) \equal{} (a_{1}, a_{2},\cdots,a_{n})$ be defined by

$ M_{p}(a) \equal{} \left(\frac {1}{n}\sum_{i \equal{} 1}^{n}a_{i}^{p}\right)^{\frac {1}{p}}(p\neq0)$, and $ M_{0}(a) \equal{} \left(\prod_{i \equal{} 1}^{n}a_{i}\right)^{\frac {1}{n}}$.

[b]Theorem 2.[/b] Let $ x_{1}\geq x_{2}\geq\cdots\geq x_{n} > 0$ and $ y_{1}\geq y_{2}\geq\cdots \geq y_{n} > 0$. If all the following conditions be fulfilled:

  $ x_{1}\leq y_{1}$;
  $ x_{1}x_{2}\leq y_{1}y_{2}$;
  $ \cdots$
  $ x_{1}x_{2}\cdots x_{n}\leq y_{1}y_{2}\cdots y_{n}$,
  (generally $ \prod_{i \equal{} 1}^{k}x_{i}\leq \prod_{i \equal{} 1}^{k}y_{i}$ for any natural number $ k$ with $ 1 \leq k \leq n$)
then $ \sum_{i \equal{} 1}^{n}x_{i}\leq \sum_{i \equal{} 1}^{n}y_{i}$.

The weak majorization inequalities were proved in :

Ji Chen, Congjie Chen, Linear inequalities in a triangle (Chinese), Geometric Inequalities in China (Edit by Zun Shan, Ji Chen, Yao Zhang, Shiguo Yang), Jiangsu Education Publishing House, 1996, pp. 87-110.

( ISBN 7 - 5343 - 2810 -1 )

[quote=Ji Chen][b]Theorem 3.[/b] Let $ x_{1}\geq x_{2}\geq\cdots\geq x_{n} \geq0$ and $ y_{1}\geq y_{2}\geq\cdots \geq y_{n} \geq 0$. If all the following conditions be fulfilled:

   $ x_{1}\geq y_{1}$;
  $ x_{1} \plus{} x_{2}\geq y_{1} \plus{} y_{2}$;
  $ \cdots$
  $ x_{1} \plus{} x_{2} \plus{} \cdots \plus{} x_{n}\geq y_{1} \plus{} y_{2} \plus{} \cdots \plus{} y_{n}$,
  (generally $ \sum_{i \equal{} 1}^{k}x_{i}\geq \sum_{i \equal{} 1}^{k}y_{i}$ for any natural number $ k$ with $ 1 \leq k \leq n$)

  then $ \sum_{i \equal{} 1}^{n}x_{i}^p\geq \sum_{i \equal{} 1}^{n}y_{i}^p$ for $ p > 1$.[/quote]
It's follows immediately from Karamata because $f(x)=x^p$ is an increasing convex function.


inequalities
function
inequalities unsolved

let $ a$ and $ b$ positives numbers $ \in$ $ R$  prove that:

$ \forall n$ $ \in N$  $ (a\plus{}b)^{n} ( a^{n}\plus{}b^{n}) \equal{}< 2^{n} ( a^{2n} \plus{} b ^{2n})$

By Holder we have:
$ (a\plus{}b)^n(a^n\plus{}b^n)\le 2^{n\minus{}1}(a^n\plus{}b^n)^2\le 2^n(a^{2n}\plus{}b^{2n})$ :P

could you do this inequality by recurrence please?
i think it's diffucult by recurrence :o

If one of the variables is zero, the inequality is trivial. Hence we may normalize to $ a\plus{}b\equal{}2$. Hence it suffices to prove

$ \frac{(a^{2n}\plus{}b^{2n})}{a^n\plus{}b^n}\ge 1$.

But $ \frac{1}{a^n\plus{}b^n}*\sum a^{2n}\ge \frac{1}{2}\sum a^n\ge \frac{1}{2}*\sum n(a\minus{}1)\plus{}1\equal{}\frac{1}{2}*2\equal{}1$

Since the function $ a^n$ is increasing in $ a$ so it is greater than or equal to it's tangent line with equality when $ a\equal{}b\equal{}1$. Unnormalizing, that gives equality iff $ a\equal{}b$.


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Hello, it's my first thread in this forum. All the users seem to be very good math student. Thus you can help the bad whom I am :D  :blush:
I search an "élementary" proof of this proposition: The family $( \sqrt{p} )$, where the $p \in \mathbb{P}$, is free on $\mathbb{Q}$
i.e if $(p_1, p_2, ..., p_n) \in \mathbb{P}^{n}$ and $( \alpha_1, \alpha_2, ..., \alpha_n ) \in \mathbb{Q}^n$ then: $ \alpha_1 \sqrt{p_1} + \alpha_2 \sqrt{p_2} + ... + \alpha_n \sqrt{p_n} = 0 \ \ \Rightarrow \ \ \alpha_1 = \alpha_2 = ... = \alpha_n = 0.$

search on the forum. Pierre has already posted a solution for this one ..

Since I just answered this question in a PM, I'll dump my argument here. For $n>4$, you really need some advanced tools- basic field extension and automorphism ideas work best.

We work by induction and start by assuming that $\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_{n-1}}$ and their products are independent, so the field $K$ generated by them has degree $2^{n-1}$ over $\mathbb{Q}.$ Those square roots and 1 are a basis for this field over $\mathbb{Q}.$
The degree of the extension of $K$ generated by $\sqrt{p_n}$ must be either one or two, so either $\sqrt{p_n}$ is in the field or it and the associated square-free numbers are linearly independent of what we already have.
The automorphisms of $K$ include $n-1$ automorphisms $\sigma_k$ which fix $\sqrt{p_i}$ for $i\neq k$ and negate $\sqrt{p_k}.$ If $\sqrt{p_n}$ is in the field, it must be either fixed or negated by this automorphism. The fixed field of this automorphism is the field $F_k$ generated by $p_i: i\neq k$ and the set of elements negated by this automorphism is $\sqrt{p_k}F_k.$ Using all $\sigma_k,$ we find that for each $k$ either all terms or no terms of the expansion of $\sqrt{p_n}$ are products involving $p_k,$ and therefore the expansion of $\sqrt{p_n}$ reduces to a single term $c\sqrt{a},$ where $a$ is a product of the primes $p_1,p_2,\dots,p_{n-1}.$ This is obviously not the case, so $\sqrt{p_n}$ is not in the field and the square roots are independent.


[hide]This problem was posted by me a long time ago.
Here it is http://www.mathlinks.ro/viewtopic.php?t=277315[/hide]


We have an infinite sequence of positive integers. The sequence is strictly increasing. Prove that there are infinitely many terms of the sequence a_m s.t. a_m=xa_i+ya_j for some i and j and positive integers x and y.

Problem 2 IMO 1975 Bulgaria
Here the solution from book IMO 1959-1977 by Greitzer
-------------------------------------------------------------------------------
For 0 \leq r<a_p, denote by Br the subsquence of a_1,a_2,a_3,...
consisting of those members wich are congruent to r modulo a_p
The remainder are finite upon division by a_p (namely 0,1,...,a_p-1)
the sequence  a_1,a_2,...is infinite, there is at least one value of r for Br
has infinitely many members.Let r such value.
Suppose a_q is the smallest member of Br wich exceeds a_p.
Then for all a_m in Br with m<q, we have

a_m - a_q congru 0 (mod a_p)

a_m - a_q = xa_p with x integer

a_m = xa_p+a_q

take x= (a_m-a_q)/a_p , y =1

Thx. I found a soln too meanwhile (after posting the prb): it used the fact that if (a, b)=1 then any number > ab+a+b can be written as xa+yb with positive integers x, y. But the idea is KIND OF the same, so there's no point in posting my soln..

Bye! :D

grobber!
Your idea needs more argumentation. Are you sure your solution is correct?

This is a theorem actually:

If (a, b)=1 then ab+a+b is the greatest natural number that cannot be represented as xa+yb with positive x and y. I know a soln (I found it in a book) but I'll post it a bit later, Ok?

I know this theorem!
I don't see easy way for applying this theorem to the problem. That is why I asked you about your solution but not about this theorem!

Sorry abt misunderstanding, Myth! I hope that, deep in your heart, you can find the power to forgive me!  :D  :D  :D (just kidding-I really am sorry, though)

Anyway, here's my proof:

If wehave 2 coprime terms of the sequence, a and b, it's finished since we have infinitely many terms > ab+a+b, so let's assume that for any 2 terms of the seq, a and b, we have (a, b)>1.

Now let's take the first term a_1. All the other terms have a common divisor with a_1, so there is a divisor d_1 of a_1 s.t. there are infinitely many terms in the sequence which are divisible by d_1. Now take the sequence formed by all these terms which are divisible by d_1. The first term is still a_1. If a_m=xa_p+ya_q then for a common divisor d of a_m, a_p, a_q we have a_m/d=xa_p/d+ya_q/d, so it's ok to divide all the terms of the newly formed sequence by d_1 (it's a common divisor of all the terms of our second sequence). The first term of this new sequence is now a_1/d_1 with d_1>1. We repeat this process until we eventually get an infinite strictly increasing sequence which contains 2 coprime terms, so we are finished.

It looks correct to me. What do you think Myth?

I think you are right!
Now I realized for myself rigorous proof. Thank you!

How about generalization of Ptolemei's theorem?

i haven't checked your proof completely but i think there is a sharper result

(for me , strictly positive is more than zero, positive is zero included)

if a and b are strictly positive and coprime then a*b-a-b is the highest number without positive representation

proof : a and b are coprime so there is exactly one solution modulo b for x

let us take x0, the solution for x   so that 0<=x0<b-1

if then y=(n-a*x)/b

some easy swapping and multiplying in the equations give us that , thanks to n>a*b-a-b , y is >=0

it is obvious that a*b-a-b is not representable in that way

the equation then becomes a*(x+1)+b*(y+1)=a*b

a and b are coprime , so that means b has to divide x+1
x+1 can't be 0 or less, because then x is too low, but it can't be 2*b or higher because that would render y+1 negative

so x+1= b---->b*(y+1)=0--->y=-1 --->contradiction

however! : i heard that exactly half the integers

0,1,....a*b-a-b-1   can be represented in this way

i suspect even more :  k can be represented <---> a*b-a-b-k cannot be represented

i have no idea how to prove this
anyone?

And what is the country where positive numbers include zero, fredbel6?

Belgium of course ! (Fred[u]bel[/u]6) ...

As regards natural numbers, it is very simple to understand
what a man means looking on his flag: if he belongs to the German branch
of mathematics (and our country, dear Myth, imported mathematics from Germany), then his natural numbers start from 1 ("God created 1", as Kronecker says). If he belongs to French branch (as Belgium or, say,
Poland does), he knows that before adding two numbers or even one number he should study at school adding zero numbers...  :)
   Cautious Anglo-Saxons, afraid of this mess, say "positive integers".

in romania postive integer means any integer \geq 0. A positive real is a real x\geq 0.

a negative number though, cannot be 0. I guess this comes from the fact that we  are relative with the french. :D


Hi,
I need some help in solving the following problem:

Calculate the value of the function (1 - x^2)/(1 - x) where x go toward 1.

Thanks for your help,
Marin

Factor $\displaystyle \frac{1-x^2}{1-x}$ into $\displaystyle \frac{(1-x)(1+x)}{1-x}$ and simplify.  After canceling, you get $1+x$.  Plug in and you get 2.  That's the limit, there's a hole at x = 1.

yeah, that's correct. by the way if you know any calculus, then there's a methode called the L'Hopitals Rule, for limits that get in a form of $\frac{0}{0}$, you can take the derivative of top and bottom, and plug in the value for $x$,
this is method is useful, for the limits that are not factorable.but in this case, factoring is easier

is that $\lim_{x\rightarrow 1}{\frac{x^2-1}{x-1}}$??

I just wondering if this function too can be solved in the same way ?
 (e^h - 1)/h and h goes toward 0

The problem I see is that "h" is not rational, how to solve that?

the answer to $\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1$
for this one you have to know a little bit of Calculus, you have to use the LaHopital's rule,
all you have to do is to take the derivative of Top and bottom and find the value:
$\lim_{x\rightarrow 0} \frac{e^x-1}{x}=\lim_{x\rightarrow 0} \frac{e^x}{1}=1$

Thank you Amirhtlusa,

I see now your point.

BRG
Marin

[quote="4everwise"]Cool, this has gotten me interested in higher mathematics.[/quote]If the idea of limits (or L'Hopital's rule, which really wasn't his  :D ) interests you, you might want to find a book on its historic development. There are tons of them out there so look for a good one (I would suggest one but I never got around to reading any). You won't really learn much about calculus and what it is in AP Calculus for example.. mostly just applying theorems that are given without proof to computational problems.

Actually I have a book in mind. I've had my eye on [i]The Calculus Gallery[/i] for a while now..

[quote="amirhtlusa"]yes, "AntonioMainenti" is right, the Lahopital's rule actually is not from Lahopital. as far as i know. it belongs to John Bernoulli. the reason it's called Lahopital's rule is that, Lahopitar was a stupid rich person who was John Bernoullies studnet, since John Bernoulli and his brother were both very smart, they always made different theorems, then when John figured out this rule. he told this to his student Lahopital and since he was rich and stuff like that, he published it as his own...(rich basterd... :lol: )

this is what i've heard... :D[/quote]

It can be bastard to people who stand in point of John Bernoulli but if you're rich and want to have your name on mathematics, I don't think Lahopital isn't really a bastard.

Maybe a big math loser? :D

isnt limits too hard for Getting Started??

[quote="joml88"]Ok, please try to spell his name correctly...it is spelled either L'Hopital or L'Hospital.[/quote]

Really? L'Hospital is right? I thought I was just imagining things when I remembered seeing that ;)

And yes, limits is too hard or, rather, not topical for Getting Started... but the first intentions of this thread were at the level of Getting Started, so I don't know if we should move it. Not that I have any say in this in the first place.

[quote="paladin8"]
Really? L'Hospital is right? I thought I was just imagining things when I remembered seeing that ;) [/quote]

http://scienceworld.wolfram.com/biography/LHospital.html


quick question:

How come Duke has a approx 20% of acceptance while colleges such as columbia univ have much lower rate at 10%. But if you look at the SAT scores, duke is definitely higher in all sections.  and according to collegeboard, both school list test scores as an very important factor.

why?

Have you compared the number of applicants to each school? Self-selection of applicants is a part of admission process that is not discussed much.

Ivy League colleges in general look for student characteristics besides test scores (notably, extracurricular activities) more than some non-Ivy League colleges.


algorithm
algebra
polynomial
inequalities
inequalities proposed

For a positive constant $ p$, let $ x,\ y,\ z\neq p$ be real numbers such that $ xyz \equal{} p^3$.
Find the minimum value of  the expression:
\[ \left(\frac {px \plus{} 1 \minus{} p^2}{x \minus{} p}\right)^2 \plus{} \left(\frac {py \plus{} 1 \minus{} p^2}{y \minus{} p}\right)^2 \plus{} \left(\frac {pz \plus{} 1 \minus{} p^2}{z \minus{} p}\right)^2\]
Last Edited

[quote="kunny"]For a positive constant $ p$, let $ x,\ y,\ z\neq p$ be real numbers such that $ xyz \equal{} p^3$.
Find the minimum value of  the expression:
\[ \left(\frac {px \plus{} 1 \minus{} p^2}{x \minus{} p}\right)^2 \plus{} \left(\frac {py \plus{} 1 \minus{} p^2}{y \minus{} p}\right)^2 \plus{} \left(\frac {py \plus{} 1 \minus{} p^2}{y \minus{} p}\right)^2\]
[/quote]

$ [\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2}]{min}\equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(x,y,z\in{R},p>0).$

But if $ {x,y,z}\in{R^{\plus{}}}$,then we have
$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} 2p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(0 < p\le{\frac {\sqrt {2}}{2}});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} 2\left(p \minus{} \frac {1}p\right)^{2}(\frac {\sqrt {2}}{2} < p\le{1});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(p > {1});$

[quote="kunny"]How did you solve it?[/quote]

Let  $ x \minus{} > \frac {px}{y},y \minus{} > \frac {py}{z},z \minus{} > \frac {pz}{x},$then it is equivalent to
\[ (yp^2x^2 \plus{} xy^2 \minus{} y^2p^2x \plus{} x^2z \minus{} x^2p^2z \minus{} 3xyz \plus{} y^2p^2z \plus{} p^2xz^2 \plus{} yz^2 \minus{} z^2p^2y)^2\ge{0}(p\geq{1}),......\]

Nice, but how can we find the substitution?

[quote="kunny"]Nice, but how could find the substitution?[/quote]

We have established a new mechanical proving algorithm  for a kind of polynomial inequality ...

[quote="fjwxcsl"][quote="kunny"]For a positive constant $ p$, let $ x,\ y,\ z\neq p$ be real numbers such that $ xyz = p^3$.
Find the minimum value of  the expression:
\[ \left(\frac {px + 1 - p^2}{x - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2\]
[/quote]

The minimum value of  the expression: $ \frac {2p^4 - 2p^2 + 1}{p^2}.$[/quote]
This is the infimum if $ p\le\frac{\sqrt{2}}{2}$, see $ (x,y,z)\to (0,\infty,\infty)$.

But for $ p\ge\frac{\sqrt{2}}{2}$, we have $ \inf\sum \left(p+\frac1{x-p}\right)^2= p^2+2\left(p-\frac1p\right)^2^$, see $ (x,y,z)\to (0,0,\infty)$.

Don't trust computers too much, they are rather stupid after all :lol:

[quote="spanferkel"][quote="fjwxcsl"][quote="kunny"]For a positive constant $ p$, let $ x,\ y,\ z\neq p$ be real numbers such that $ xyz = p^3$.
Find the minimum value of  the expression:
\[ \left(\frac {px + 1 - p^2}{x - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2\]
[/quote]

The minimum value of  the expression: $ \frac {2p^4 - 2p^2 + 1}{p^2}.$[/quote]
This is the infimum if $ p\le\frac {\sqrt {2}}{2}$, see $ (x,y,z)\to (0,\infty,\infty)$.

But for $ p\ge\frac {\sqrt {2}}{2}$, we have $ \inf\sum \left(p + \frac1{x - p}\right)^2 = p^2 + 2\left(p - \frac1p\right)^2^$, see $ (x,y,z)\to (0,0,\infty)$.

Don't trust computers too much, they are rather stupid after all :lol:[/quote]

dear spanferkel:
\[ \left(\frac {px + 1 - p^2}{x - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2 + \left(\frac {py + 1 - p^2}{y - p}\right)^2\ge{p^2 + 2\left(p - \frac1p\right)^2}\]

holds $ \forall{p\ge{\frac{\sqrt{2}}{2}}???}$

oops I did a silly mistake. Maybe I should have used a computer  :lol:  :maybe:

If $ {x,y,z}\in{R^{\plus{}}},$ then

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} 2p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(0 < p\le{\frac {\sqrt {2}}{2}});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} 2\left(p \minus{} \frac {1}p\right)^{2}(\frac {\sqrt {2}}{2} < p\le{1});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(p > {1}).$

[quote="spanferkel"][quote="fjwxcsl"]We have
$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} 2p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(0 < p\le{\frac {\sqrt {2}}{2}});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} 2\left(p \minus{} \frac {1}p\right)^{2}(\frac {\sqrt {2}}{2} < p\le{1});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(p > {1}).$[/quote]
 :?: I must have troubled you  :blush:

I think now your first solution is correct...
The minimum $ \frac {2p^{4} \minus{} 2p^{2} \plus{} 1}{p^{2}} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}$ is attained for $ x \equal{} y \equal{} \minus{} \frac1p \plus{} \sqrt {\left(p \minus{} \frac {1}p\right)^{2} \plus{} 1}$. Note that [quote="kunny"]$ x,\ y,\ z\neq p$ be [color=red]real [/color]numbers [/quote]  :)[/quote]

Thank you,spanferkel! I forgot $ {x,y,z}$ is real  numbers , edited.

The correct conclusion should be

$ [\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2}]{min} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}{p}\right)^{2}(x,y,z\in{R},p > 0).$

But if $ {x,y,z}\in{R^{ \plus{} }}$,then we have

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} 2p^{2} \plus{} \left(p \minus{} \frac {1}{p}\right)^{2}(0 < p\le{\frac {\sqrt {2}}{2}});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} 2\left(p \minus{} \frac {1}{p}\right)^{2}(\frac {\sqrt {2}}{2} < p\le{1});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}{p}\right)^{2}(p > {1}).$

[quote="fjwxcsl"][quote="kunny"]For a positive constant $ p$, let $ x,\ y,\ z\neq p$ be real numbers such that $ xyz \equal{} p^3$.
Find the minimum value of  the expression:
\[ \left(\frac {px \plus{} 1 \minus{} p^2}{x \minus{} p}\right)^2 \plus{} \left(\frac {py \plus{} 1 \minus{} p^2}{y \minus{} p}\right)^2 \plus{} \left(\frac {py \plus{} 1 \minus{} p^2}{y \minus{} p}\right)^2\]
[/quote]

$ [\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2}]{min} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(x,y,z\in{R},p > 0).$

But if $ {x,y,z}\in{R^{ \plus{} }}$,then we have
$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} 2p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(0 < p\le{\frac {\sqrt {2}}{2}});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} 2\left(p \minus{} \frac {1}p\right)^{2}(\frac {\sqrt {2}}{2} < p\le{1});$

$ \inf\sum\left(p \plus{} \frac {1}{x \minus{} p}\right)^{2} \equal{} p^{2} \plus{} \left(p \minus{} \frac {1}p\right)^{2}(p > {1});$[/quote]

That's correct answer. :lol:

Could you show your full solution?


Hello,
I have got miktex and texnic centre installed on my computer. I want to type in Georgian. How can I do it? I hava heard that there is a packege babel that allows to type in other langueges but it doesn't include Georgian. What can I do?

I have found this: http://www.tug.org/tex-archive/fonts/georgian/
Can this help me? Can install new by fontins package?


calculus
integration
logarithms
real analysis
real analysis unsolved

Hey :)

i'm a univeristy student from new zealand, and im confident in all sorts of integration substitutions, but today during a project i came across one which i couldn't get my head around

In a simplified form, i wish to integrate this: dx/(K^2 - X^2)
looking up a book of tables, i saw that the integral should fit (1/2K)*ln((K + X)/(K - X)) but im not sure if this is correct

can anyone confirm this integral? or have another solution?

thanks
-brad

Strictly speaking your answer is incorrect.

The right answer is as follows.

When $k=0,\ \int \frac{1}{k^2-x^2} dx=-\int \frac{1}{x^2}dx=\frac{1}{x}+Const.$

When $k\neq 0,$

$\int \frac{1}{k^2-x^2} dx=\frac{1}{2k}\int \left(\frac{1}{k-x}+\frac{1}{k+x}\right) dx=\frac{1}{2k}\ln \left|\frac{x+k}{x-k}\right|+Const.$

[quote="kunny"]Strictly speaking your answer is incorrect.[/quote]

Yeah the absolute value is absolutely needed.

[quote="kunny"]
When $k\neq 0,$

$\int \frac{1}{k^2-x^2} dx=\frac{1}{2k}\int \left(\frac{1}{k-x}+\frac{1}{k+x}\right) dx=\frac{1}{2k}\ln \left|\frac{x+k}{x-k}\right|+Const.$[/quote]

Or $\int \frac{1}{k^2-x^2} dx=\frac{1}{k}arctanh \left( \frac{x}{k} \right )$

Sorry for that Japanese high school students don't study arc tangent x and hyperbolic function.  :)

[quote="braddles"]thanks guys, thats exactly what i got

now i have (1/2K)*ln((K + X)/(K - X)) = -zt

if i wish to rearrange this in terms of X, is the following correct?

X = (exp^(-2*(z)(k)t))/2[/quote]

Your answer is incorrect.

The answer is $x=\frac{e^{-2kzt1}-1}{e^{-2kzt}+1}.$

Maybe Ramanujan will simplify this solution as certain function.

By the way is this problem is related to physics problem? I guess the movement for falling body with $kv^2.$

ah cool, that solution works

yes it is a physics related problem. the project im doing is for a math paper as part of my engineering degree. we are using matlab to model second order ode's. this year the project is on a person parachuting from a moving plane, including the effects of air resistance. the question im working on at the moment, is a model of the person jumping from a stationary helicopter, as it is not possible to obtain an analytic solution for the person jumping from the moving plane, once i obtain this solution i am to compare it with the numerical solution of vertical velocity from the moving plane. the X term i have used is the velocity in the vertical direction :)

if someone could derive the above expression of X for me i would be grateful, as i don't have much experience with natural logs

thanks again

[quote="kunny"][quote="braddles"]thanks guys, thats exactly what i got

now i have (1/2K)*ln((K + X)/(K - X)) = -zt

if i wish to rearrange this in terms of X, is the following correct?

X = (exp^(-2*(z)(k)t))/2[/quote]

Your answer is incorrect.

The answer is $x=\frac{e^{-2kzt1}-1}{e^{-2kzt}+1}.$[/quote]

Oops! My answer is also incorrect.

On the condition that $\frac{k+x}{k-x}>0,$

$\frac{1}{2k}\ln \frac{k+x}{k-x}=-zt\Longleftrightarrow \ln \frac{k+x}{k-x}=-2kzt\Longleftrightarrow \frac{k+x}{k-x}=e^{-2kzt}$

$\Longleftrightarrow k+x=(k-x)e^{-2kzt}\Longleftrightarrow (1+e^{-2kzt})x=k(e^{-2kzt}-1),$

yielding $\boxed{x=\frac{k(e^{-2kzt}-1)}{e^{-2kzt}+1}}.$

cool, thanks a lot :)

my graphs are looking better now!

thanks for your time guys

If you like, please take a look 8th post of this.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=57578]www.mathlinks.ro/Forum/viewtopic.php?t=57578[/url]


inequalities
linear algebra
matrix
linear algebra unsolved

Let $A=(a_{ij})_{1 \leq i,j \leq n} \in {\mathbb R}^{n \times n}$ such that $\sum_{j=1}^n {a_{ij}}^2 = i$, for $i=1,2,...,n$. Show that $|det(A)| \leq \sqrt {n!}$.  :ninja:

It is exactly Hadamard inequality: for every matrix $X$ the following inequality holds $|detX|^2 \leq \sum_{i}|x_{1i}|^2...\sum_{i}|x_{ni}|^2$

yap. that's it  :D . But can u give a proof of hadamard ineq. ? pls   :roll:


Call a "bisector" a line that divides a finite region into 2 parts of equal area.  Prove that there will always exist at least one point through which 2 bisectors of the same triangular region can be drawn.

EXTENSION 1: Prove that this works for any finite region.
EXTENSION 2: Prove that this works with a line, a plane and a 3D region.

[quote="Altheman"]take the plane or w/e and move it through...take a function that is the area to the left of the plane, it is a continous function, that starts at 0, and ends at the whole area, by the intermediate value theorem, there is some place where this divides the area in half, take planes coming in mulitple directions to finish this[/quote]

That's the basic idea, but be more specific.


If you divide both sides with $xyz$ you get the inequality from BMO 2006 :)

it's easy.
put
x=a+1, y=b+1, z=c+1 with a,b,c>0
multiply by abc(1+abc) and simple use of A-G in the end :)

Thank you very much for your idea and your interest !  :)

So you're saying:
$\sum_{sym}\frac{(a+1)(b+1)}{a}\geq \frac{3(a+1)(b+1)(c+1)}{1+abc}$
Now multiply through:
$(1+abc)\frac{\sum_{sym}(a+1)(b+1)(bc)}{3}\geq abc(a+1)(b+1)(c+1)$
I don't see how there is an easy AM-GM here...

Let $x-1=a,y-1=b$ and $z-1=c$, the inequality become a wellknown one in forum. Sorry, I can't post solution because there was a key died.

me@home, it's little bit harder  :lol:
you multiply it out, some terms cancel, and then simple A-G
try, it's trivial   :P


There are 8 red, 8 blue and 8 yellow marbles in a jar. What is the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7?

Let the number of red, blue, and yellow marbles be $ r,b,y$ so that our goal is to have

$ \frac r{b \plus{} y} \equal{} \frac37\implies7r \equal{} 3b \plus{} 3y$ and $ \frac y{b \plus{} r} \equal{} \frac37\implies7y \equal{} 3b \plus{} 3r$, so subtracting the first from the second,

$ 7y \minus{} 7r \equal{} 3r \minus{} 3y\implies10y \equal{} 10r\implies y \equal{} r$ (this conclusion could have been reached through common sense, but it doesn't hurt to be rigorous). Now, let the number of blue marbles removed be $ x$ such that $ 0\le x\le8$ and the number of red marbles removed be $ y$ such that

$ 0\le y\le8$ as well (so that the number of yellow marbles removed is also $ y$). Then we now have the requirement

$ \frac {8 \minus{} y}{16 \minus{} x \minus{} y} \equal{} \frac37\implies56 \minus{} 7y \equal{} 48 \minus{} 3x \minus{} 3y\implies8 \plus{} 3x \equal{} 4y$; taking both sides $ \pmod4$ gives

$ 3x\equiv0\pmod4\implies x\equiv0\pmod4\implies x \equal{} 4m$ for some integer $ m$, so either

$ x \equal{} 0\implies y \equal{} 2\implies x \plus{} 2y \equal{} \boxed4$, or $ x\ge4$, but since $ x \plus{} 2y\ge4$ for the latter cases, we don't need to consider them.

Notice that $ 3$ to $ 7$ is the same ratio as $ 6$ to $ 14$. If you remove 2 red marbles and 2 yellow marbles, you get exactly that ratio. Thus the answer is $ 2 \plus{} 2 \equal{} \fbox{4}$.

EDIT: whoops got beaten to writing answer

Well using intution 3+7=10 20 is a multiple of 10 so 4.


geometry
perimeter
inequalities
ratio
rectangle
geometry proposed

Suppose $F$ is a polygon with lattice vertices and sides parralell to x-axis and y-axis.Suppose $S(F),P(F)$ are area and perimeter of $F$.
Find the smallest k that:
$S(F) \leq k.P(F)^2$

It appears to be $\frac 1{16}$. We have an equality for a square and the inequality holds for rectangles.

When the polygon is concave, we can easily extend its border so that it encompasses more squares but the perimeter stays the same or decreases, so for a certain positive integer $n$, the polygon with perimeter $\le n$ and maximal area must be convex, and since the polygons we're working with have right angles, it must be a rectangle. This means that the ratio $\frac{S(F}{P^2(F)}$ is smaller for a concave figure than for a certain rectangle, and that of the rectangle is $\le\frac 1{16}$, which means that $\frac 1{16}$ is Ok.

Only simpler solution.Consider it's circumscribed rectangle .It's Area is bigger and it's perimeter is smaller.Inequality is obvious for rectangles.

That's exactly what I thought, but I thought we needed some justification for the fact that the perimeter of the circumscribed rectangle is smaller (you can't just say it's smaller :)).


geometry
circumcircle
incenter
vector
angle bisector
geometry proposed

Given a triangle $ ABC$. Denote $ p \equal{} \frac {AB \plus{} AC \plus{} BC}2$.
Let $ L$ and $ N$ be points on the rays $ [AB$ and $ [CB$, respectively, such that $ AL \equal{} CN \equal{} p$.
Let $ K$ be a point on the circumcircle of triangle $ ABC$ such that $ BK$ is a diameter of this circumcircle.
Let $ I$ be the incenter of triangle $ ABC$.
Prove that $ KI$ is perpendicular to $ NL$.

[i]Alternative formulation:[/i] Let $ p \equal{} \frac {BC \plus{} CA \plus{} AB}{2}$ be the semiperimeter of a triangle $ ABC$.
Let $ L$ be the point on the ray $ AB$ such that $ AL \equal{} p$. Let $ N$ be the point on the ray $ CB$ such that $ CN \equal{} p$. Let $ K$ be the point diametrically opposite to $ B$ on the circumcircle of triangle $ ABC$. Prove that the perpendicular to the line $ NL$ through the point $ K$ passes through the incenter of triangle $ ABC$.

Are you sure $K$ is the antipode of $B$ instead of $A$?

[color=red][Moderator edit: Yes, Grobber, it should be B. I have corrected the typo and will post my proof later. Darij][/color]

a small vector calculation and the fact that the vector $\frac{AB}{c}+\frac{BC}{a}+\frac{CA}{b}$ is orthogonal to $OI$ should suffice i guess, but that's only what the problem sounds like to me(i looked at it for 10 seconds). it could be complete nonsense.

it must work (i didn't do all the algebra, but if the result is true, one can prove it this way): the vector OI is has equal length and is perpendicular to the vector $\frac{R}{a}BC+\frac{R}{b}CA+\frac{R}{c}AB$, where R is the radius of the circumcircle (ask darij for a proof, i only his beautiful one for that result, but is too lengthy to state her). now we want to prove that (KO+OI)(BL-BN)=0. but it is easy to verify that $BL-BN=\frac{a+b+c}{2}(\frac{BC}{a}+\frac{CA}{b}+\frac{AB}{c})-\frac{a-b+c}{2b}CA$
or something like that (maybe i did an error in the calculation). but since the scalar products of sides and the scalar product of a side with KO=OB are easy to compute, we must get a solution :D
but darij sure has a nicer solution as i now him  :maybe:

Peter

Thanks, Peter, for the nice compliments... actually you don't need the length of the vector $\frac{R}{a}BC+\frac{R}{b}CA+\frac{R}{c}AB$, you just need it's direction, and it is not difficult to show that the vector $\frac{\overrightarrow{BC}}{a}+\frac{\overrightarrow{CA}}{b}+\frac{\overrightarrow{AB}}{c}$ is perpendicular to OI. [In brief: $\frac{\overrightarrow{BC}}{a}\cdot \overrightarrow{OI}$ is the projection of OI on BC, i. e. the distance from the midpoint of BC to the point of tangency of the incircle with BC, and this distance is (c-b)/2; now sum up, and get the scalar product 0, so that the vectors are perpendicular.]

Indeed, if our calculations are not wrong, then we can nuke the problem with your method.

My proof of Omid's problem is synthetic but quite "heavy" (i. e. some facts from the triangle geometry arsenal are applied):

[quote="Omid Hatami"]Given a triangle $ABC$. Denote $p=\frac{AB+AC+BC}2$.
Let $L$ and $N$ be points on the rays $[AB$ and $[CB$ such that $AL=CN=p$.
Let $K$ be a point on the circumcircle of triangle $ABC$ such that $BK$ is a diameter of this circumcircle.
Let $I$ be the incenter of triangle $ABC$.
Prove that $KI$ is perpendicular to $NL$.[/quote]

I will use the [b]Orthologic Triangles Theorem[/b]:

[i]If $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2}$ are two non-degenerate triangles such that the perpendiculars from the vertices $A_{1}$, $B_{1}$, $C_{1}$ of the former to the respective sidelines $B_{2}C_{2}$, $C_{2}A_{2}$, $A_{2}B_{2}$ of the latter concur, then the perpendiculars from the vertices $A_{2}$, $B_{2}$, $C_{2}$ of the latter to the respective sidelines $B_{1}C_{1}$, $C_{1}A_{1}$, $A_{1}B_{1}$ of the former also concur.[/i]

Now to the problem. At first, since p is the semiperimeter of triangle ABC, from AL = CN = p, it follows that N is the point where the C-excircle of triangle ABC touches the line BC, and L is the point where the A-excircle touches the line AB. Let the C-excircle touch CA at N', and let the A-excircle touch CA at L'.

Now, we will use a nice theorem (see below a link to the proof) stating that the lines NN' and LL' meet at a point B' which lies on the B-altitude of triangle ABC. In other words, this point B' lies on the perpendicular from the point B to the line CA. Now, if I is the incenter of triangle ABC, the line CI is the angle bisector of the angle BCA; but the points N and N' are symmetric to each other with respect to this angle bisector. Hence, $NN^{\prime }\perp CI$, i. e. the line NN' is the perpendicular from the point N to the line CI. Hence, we can say that the perpendicular from the point N to the line CI passes through B'. Similarly, the perpendicular from the point L to the line AI passes through B'. Finally, remembering that the perpendicular from the point B to the line CA also passes through B', we see that the perpendiculars from the points N, L, B to the lines CI, AI, CA concur at one point (namely at B'); hence, after the Orthologic Triangles Theorem applied to the triangles NLB and ACI, it follows that the perpendiculars from the points A, C, I to the lines LB, BN, NL also concur. But the perpendicular from the point A to the line LB passes through K (since < KAB = 90�, for BK is a diameter of the circumcircle of triangle ABC). Similarly, the perpendicular from the point C to the line BN also passes through K. This forces the perpendicular from the point I to the line NL to pass through K, too. In other words, $KI\perp NL$, what was to be proven.

I remember having discovered the problem above for myself more than a year ago. See http://mathforum.org/epigone/geometry-college/whudytax

Finally, I have used the lemma that the lines NN' and LL' meet on the B-altitude of triangle ABC. In order to formulate this lemma more symmetrically, we change the notations:

Let the A-excircle of triangle ABC touch the lines CA and AB at $B_{a}$ and $C_{a}$. Similarly define the points $C_{b}$, $A_{b}$, $A_{c}$, $B_{c}$. [Of course, we then have $B_{a}$ = L', $C_{a}$ = L, $A_{c}$ = N, $B_{c}$ = N'.] Let the lines $A_{c}B_{c}$ and $C_{b}A_{b}$ meet at A', and similarly define the points B' and C'. Then, these points A', B', C' lie on the altitudes of triangle ABC. Or, in other words, the lines AA', BB', CC' are the altitudes of triangle ABC and concur at its orthocenter. In this form, the lemma was proved in the note "Synthetic proof of Paul Yiu's excircles theorem" on [url=http://de.geocities.com/darij_grinberg/]my geometry website[/url]. (In fact, I wrote this note in 2002 because I liked the theorem very much, and its only proof I had seen to that time was a brute force proof using barycentric coordinates. Now I know another synthetic proof, it doesn't use any calculation, but it is rather long. It is, by the way, not really "Paul Yiu's" excircle theorem, since it was one of the excercises in Jacques Hadamard's geometry book, but I have not seen his solution.)

I can understand you very well if you are not disposed now to read the references...  :D

  Darij

Well there is another beautiful solution Suppose R and T are on CK and AK and IR is parallel to BC and IT is parrrel to AB.  Easily Triangles ITR and LBN are congruent.
Suppose S is intersection of KI and LN.
So $\angle AKI=TRI=BNL=ANS$ so KASN is cyclic so $\angle NSK=NAK=90$
So KI is pependicular to LN

Oh yes, indeed simpler than all what I wrote... thanks  :blush:

  dg

[quote="Omid Hatami"]Easily Triangles ITR and LBN are congruent.

[/quote]

Why is that?

radio:
If $ s \equal{} \frac {a\plus{}b\plus{}c}{2}$, we have $ BN \equal{} s\minus{}c$, $ BL \equal{} s\minus{}a$. Since $ IT, IR$ are parallel to $ AB, BC$, we have $ \angle TIR \equal{} \angle ABC$. To prove $ ITR$ is congruent to $ BLN$, suffices to prove that $ IT \equal{} s\minus{}a \equal{} BL$, and $ IR \equal{} s\minus{}c \equal{} BN$.
Let the parallel to $ AK$ through $ I$ meet $ AB$ at $ F$ - we then have $ IF$ is perpendicular to $ AB$, so $ F$ is the point where the incircle of $ ABC$ meets $ AB$, and it's well known that $ IT \equal{} AF \equal{} s\minus{}a$. Analogously, $ IR \equal{} s\minus{}c$.

[color=darkblue][size=117][b]Yes, [u]Omid Hatami[/u], the your proposed problem is very nicely ![/b][/size][/color]

[quote="Omid Hatami"] [color=darkred]Given is a triangle $ ABC$ . Denote the its incircle $ w = \mathcal C(I,r)$ , the points $ L\in [AB$ and $ N\in [CB$ for which $ AL = CN = p$ and the point $ K$ such that $ [BK]$ is a diameter in the its circumcircle . Prove that $ KI\perp NL$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the points $ \|\begin{array}{c} D\in (BC)\cap w \\
 \\
F\in (BA)\cap w\end{array}$ . Observe that $ B\in (AL)\cap (CN)$ and

$ \{\begin{array}{c} BD = BF = p - b \\
 \\
BN = p - a\ ,\ BL = p - c\end{array}\|$ $ \implies$ $ \{\begin{array}{c} ND = NB + BD = c\ . \\
 \\
LF = LB + BF = a\ .\end{array}$

$ 1\blacktriangleright$ $ \{\begin{array}{c} IN^2 = DN^2 + DI^2 = c^2 + r^2 \\
 \\
IL^2 = FL^2 + FI^2 = a^2 + r^2\end{array}\|$ $ \implies$ $ \boxed {IN^2 - IL^2 = c^2 - a^2}\ .$

$ 2\blacktriangleright$ $ \{\begin{array}{c} KB^2 = KC^2 + a^2 = KA^2 + c^2 \\
 \\
KN^2 = CN^2 + CK^2 = p^2 + CK^2 \\
 \\
KL^2 = AL^2 + AK^2 = p^2 + AK^2\end{array}\|$ $ \implies$ $ \boxed {KN^2 - KL^2 = c^2 - a^2}\ .$

From the above relations obtain $ IN^2 - IL^2 = KN^2 - KL^2$ , i.e. $ KI\perp NL\ .$[/color]


Show that any positive integer has at least as many positive divisors of the form $ 3k\plus{}1$ as of the form $ 3k\minus{}1$.


What's the funniest aops topic you've read?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82754

^that's mine. :D

"... but some moron deleted my thread..."

"that moron was me..."

lol.

sorry, i'm bored x___x

haha that was hilarious

[url=http://ihavealife.justgotowned.com/]that guy got OWNED!![/url]

This one -

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82927[/url]

The second topic of The Never Ending Story was quite good!


There is a planet of mass $M$ and radius $R$.It has a hole along its diameter.A small object with mass $m$ and initial velocity $v_{0}$ is set in motion from infinity.It moves along the straight line connecting it with the hole.Find the relative speed of mass-planet system when the mass $m$ passes through the center of planet?

If $m \ll M$, I get $v_{r}\approx 3v$, is that ok?

Djole,please write the full solution.I also got approximately same answer.But I didn't use m<<M.

Well, I just used conservation of energy and momentum: \[mv_{0}^{2}= mv^{2}+MV^{2},\] \[m\vec v_{0}= m\vec v+M\vec V.\] Solving for the relative speed $v_{r}= v+V$ I got \[v_{r}= \frac{3M+m}{M-m}v_{0}\approx 3v_{0}.\] Is that ok?

No,answer isn't like that.I also don't know the original answer.But I know it must contain also gravitational constant $\gamma$. :blush:


Prove that the equation x^3+y^3+1=z^3 has an infinite number of solutions.
I know that in such cases you should find a solution (x,y,z) and then generalise it in some way, for example (x*a^2,y*a,z*a^3) (which obviously doesnt work in this case).
all i could do was finding a solution which is (6,8,9).

perhaps this is silly, but if x = -1, we have y=z...do we need positive integer solutions?

Yes we do, in fact we are looking for solns in the set of positive integers.
I see that i have forgot to write it. My appologies

The most productive thing I've done on this problem is to write a program that trolls for other possible solutions (which are reasonably plentiful), but I was curious if this could be linked to also solving the equation $x^3 + y^3 = z^3 + 1$ in positive integers somehow.

Assuming
x = 6n - 144 n^4
y = 144 n^4
Z = 72 n^3 -1

we see that x^3+y^3 = z^3 + 1

(If you like only positive values, just use negative values of n and move the two neagtive cubes around)

This will not give  *all* possible integer solutions but will give infinite solutions   by taking integer values of n.

(There are infinite solutions for the Ramanujan's x^3+y^3= a^3+b^3 - This is just a special case when b=1)

[quote="Magnara"]So...how did you come up with those equations?  Just out of curiosity.[/quote]

By putting m=3n  ( and  r=1) in:

x = r(1 - (m - 3n)(m^2 + 3n^2))
y = r((m + 3n)(m^2 + 3n^2) - 1)
u = r((m + 3n) - (m^2 + 3n^2)^2)
v = r((m^2 + 3n^2)^2 - (m - 3n))

(for x^3+y^3=u^3+v^3)

By the way  if you put r = -361/42, m = 10/19, n = -7/19  in above equation you get Ramanujan's  Taxi Cab story: ( Ramanujan's comment on the  interesting taxicab number 1729=10^3+9^3=12^3+1)


linear algebra
matrix
superior algebra
superior algebra unsolved

Pg 265 of Burton's Abstract Algebra.
Suppose that R consists of all matrices in $ M_2\{Z\}$ of the form $ \left [\begin{matrix} n & 0 \\
2n & 0 \end{matrix} \right ]$.
Prove that R is a ring isomorphic to the integers.

Edit : Forgot to type my stuff. I was just wondering if its as easy as this :
Define a map $ \begin{matrix} \phi : \mathbb{Z} \to M_2\{\mathbb{Z}\} \\
n \to \left[ \begin{matrix} n & 0 \\
2n & 0 \end{matrix} \right ] \end{matrix}$

Then $ \phi(a \plus{} b) \equal{} \left [\begin{matrix} a \plus{} b & 0 \\
2(a \plus{} b) & 0 \end{matrix} \right ] \equal{} \left [\begin{matrix} a & 0 \\
2a & 0 \end{matrix} \right ] \plus{} \left [\begin{matrix} b & 0 \\
2b & 0 \end{matrix} \right ] \equal{} \phi(a)\plus{}\phi(b)$
And $ \phi(ab) \equal{} \left [\begin{matrix} ab & 0 \\
2(ab) & 0 \end{matrix} \right ] \equal{} \left [\begin{matrix} a & 0 \\
2a & 0 \end{matrix} \right ] \left [\begin{matrix} b & 0 \\
2b & 0 \end{matrix} \right ] \equal{} \phi(a)\phi(b)$ which is verified just by multiplying out the right side.

Then for injectivity, if $ \phi(a) \equal{} \phi(b)$ then $ \left [\begin{matrix} a & 0 \\
2a & 0 \end{matrix} \right ] \equal{} \left [\begin{matrix} b & 0 \\
2b & 0 \end{matrix} \right ] \implies \left [\begin{matrix} a \minus{} b & 0 \\
2(a \minus{} b) & 0 \end{matrix} \right ] \equal{} \left [\begin{matrix} 0 & 0 \\
0 & 0 \end{matrix} \right ]$ which is true only if $ a \minus{} b \equal{} 0$.

And surjectivity is just as obvious. For some matrix $ \left [\begin{matrix} a & 0 \\
2a & 0 \end{matrix} \right ]$ take $ a\in \mathbb{Z}$. THen $ \phi(a) \equal{} \left [\begin{matrix} a & 0 \\
2a & 0 \end{matrix} \right ]$

I was just trying to see if the $ 2$ has any real significance in this. Because it seems like if each entry was zero except the top left, it would still all be true. so also that is isomorphic to this, and $ 2$ can be changed to any integer. :\

This is true if $ \left[ \begin{array}{cc} 1 & 0 \\
2 & 0 \end{array} \right]$ is replaced by any matrix $ \mathbf{M}$ such that $ \mathbf{M}^2 \equal{} \mathbf{M}$.  In particular, yes, you can change the $ 2$ to any integer, and in fact any element of a commutative ring.


My English is not good.
Give x,y,z >0 ;   xy+yz+zx=1.
Prove that
 \sum {x,y,z}(1-x  2 ) 2 /1+x 2
 \geq 1

I remember this problem was discussed a long time ago in $Advanced$ $section$ by the topic similar to "too hard for beginner".Does anyone know the links?

I will tex it.

Given $x,y,z >0$, $xy+yz+zx=1$

Prove that
$\sum_{cyc} \frac{(1-x^2)^2}{1+x^2} \ge 1$

Finally I found a link!!!
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard+for+beginners&t=4575


limit
algebra
function
domain
calculus
calculus computations

oh , i think when $x\rightarrow 1^{-}$then the limit $=+\infty$ , and when $x\rightarrow 1^{+}$ then it hasn't an answer.  :)

The definition domain of the function $f(x)=\frac{\sqrt{x^2-3x+2}}{x^2-5x+4}$ is

$D_f=(-\infty ,1)\cup[2,4)\cup(4,\infty )\ .$ Therefore, $1\in D'_f$ and

$\lim_{x\to 1} f(x)=\lim_{x\nearrow 1} f(x)=\lim_{x\nearrow 1}\frac{\sqrt{(2-x)(1-x)}}{(4-x)(1-x)}=\lim_{x\nearrow 1} \frac{1}{4-x}\sqrt{\frac{2-x}{1-x}}=\infty\ .$

[b]Remark.[/b] $f(1+)\equiv \lim_{x\searrow 1} f(x)$ hasn't sense !