.1 + .2 + .25 = .45
the median question...
as a result i thought 90 was the median so i got a difference of 4 instead of 1

the probability question. oops. cost me a 130+ on AMC12B

Zero, if I did not make any mistakes in bubbling.

But I didn't answer much...

misread (0,k) and (k,0)... misread conjugate as norm (which made the problem much easier heh) >.<

on the abcdefg phone number problem i had all the possible solutions but i accidentally filled in 7 instead of 8

miss counting combinations with the OTHER $20 bill on #11.


What is Polar Inertial Momentum Inequality?
Does anyone know? Please post a reply.

You should post in "Inequalities Open Questions".

Ok, I did, so please see the same post at

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151397[/url]


How will you prove by [b]mathematical induction[/b] that:
$p+1$ divides $C(2p,p)$?

Do you know how to do the calculation, showing the steps, when P = 1?

of course i know the first step and can do it. But i am stuck in the 2nd step i.e to assume the statement true for $p=k $and then prove it true for $p=k+1$

By induction?  Are you sure?  That doesn't seem at all like the nicest way to prove that theorem.

Here's an inductive proof: Let $a_n=\frac{1}{n+1}\binom{2n}{n}$. Assume that $a_0,\ldots,a_{n-1}$ are integers. Then since

\[
a_n = a_0a_{n-1}+a_1a_{n-2}+\cdots+a_{n-1}a_0,
\]
$a_n$ is also an integer. (the numbers $a_n$ are known as the catalan numbers... that's how I know the recurrence)

OR, if you're familiar with the catalan numbers, you can just say that $\frac{1}{n+1}\binom{2n}{n}$ is equal to a catalan number and the catalan numbers are known to be integers!

Sorry zabelman I can't get u.I asked to prove it by mathematical induction.I can not do the second step i.e tp prove the statement true for $p=k+1 $by assumin the satatement to be true for $p=K$
                                  :( [b]Please help me.[/b] :?

What zabelman has done is slightly stronger: that if its true for 0, 1, 2, 3, ..., k then its true for k+1. That works just as well. You can't really do it by straight induction assuming its true for k - you can't really use the fact that something is divisible by k to prove something else is divisible by k+1..


Prove that the sum of 5 odd squares cannot be an odd square.

[hide]
Try different mods.
[/hide]

Yay, I can get this one! ^_^

Odd numbers are congruent to $1,3,5,7\pmod{8}$.
Thus, odd perfect squares are congruent to $1\pmod{8}$, so the sum of five odd perfect squares is congruent to $5\pmod{8}$. Again, since all odd perfect squares are congruent to $1\pmod{8}$, this sum cannot be a perfect square.


How many prime numbers are divisible by $ 39$ ?

[quote="vallon22"]1

the only prime number divisible by 39 is 39, because any other number will have 39 and another number as factors, so it's not prime[/quote]

$ 39 \equal{} 3 \times 13$.
Thus, $ \boxed{2}$.

oops
wow that was stupid

since 39 is not prime, and no other number divisible by 39 is prime, our answer is 0

@Kouichi Nakagawa
you found the numbers that 39 are divisible by, not the numbers that are divisible by 39


Let $p$ be an odd prime such that $p-2$ and $p+2$ also are primes. Show that $p=5$.

$2$ and $3$ isn't solution.
for $p> 5$, $p\equiv 1\ or\-1\ [6]$
if $p+2$ is prim then $p\equiv-1[6]$
but we will have $p-2\equiv 3[6]$ => $3|p-2$ and $p-2$ prim $> 5-2=3$
so, $p=5$


You may need to post these problems in the GETTING STARTED or INTERMEDIATE forums.

[hide="solution"]By the first two statements, one is negative and two are positive. That means that x=1+1-1=1. So the answer to the problem is the sum of the coeficciants of the polynomial:

16+1-7=10[/hide]


i know this is not a math topic, but i thought it fits here because it only applies to students in canada:

just curious who applied the young physicists camp organized by the primere institute?

post "yes" if you did.

Did you know that Edward Witten (M-theory genius) attended the institude not to long ago!  I was on Coop so I didn't had a chance to attend it!

I also applied.

Today was supposed to be the day which they release the invitees list (as said on the application form). But they instead put up the message: "2005 ISSYP candidates will be notified over the first week of June and a full list will be posted on this webpage." Oh well, guess we'll have to wait another week or so.

[quote="8th Army"]Do you guys know approximately how many people apply for PI each year?[/quote]

I think they had nearly 400 for the first year (2~3 years ago). I would expect something like 500~800 this year.

Given that, picking 10 guys and 10 girls solely from an essay and a teacher evaluation can be a pretty formidable task.

[quote="billzhao"]

Given that, picking 10 guys and 10 girls solely from an essay and a teacher evaluation can be a pretty formidable task.[/quote]

At the time I applied, I remember they select one guy and one girl from each province, which makes your task even tougher  :(  and I remember Amirali got selected that year, so others in Ontario simply didn't have a chance

The [url=http://www.perimeterinstitute.ca/activities/community/highschool/summerinstitute.php]list[/url] is out.


inequalities
function
calculus
derivative
inequalities unsolved

Assume that n is a positive integer and a is real with a > 0.
Proof the following inequality:

[tex]\frac{1+a^2+a^4+....+a^{2n}}{a+a^3+a^5+...+a^{2n-1}} \geq \frac {n+1}{n}[/tex]

We need only consider the case $a \ge 1$, since
$\frac{1+a^2+\cdots+a^{2n}}{a+a^3+\cdots+a^{2n-1}} = \frac{1+a^{-2}+\cdots+a^{-2n}}{a^{-1}+a^{-3}+\cdots+a^{1-2n}}$
but, also,
$\frac{1+a^2+\cdots+a^{2n}}{a+a^3+\cdots+a^{2n-1}} = \frac{a^{2n+2}-1}{a^{2n+1}-a} \ge \frac{n+1}{n}$
if and only if (*)
$na^{2n+2} - (n+1)a^{2n+1}+(n+1)a-n = (a-1)\left(na^{2n+1} - a^{2n} - a^{2n-1} - \cdots - a^1 + n\right) \ge 0$
but we have
$a^{2n+1} - a^{2n+1-r}-a^{r}+1 = \left(a^{2n+1-r}-1\right)\left(a^r-1\right) \ge 0$
for $r = 1, 2, \dots, n$. Adding these $n$ inequalities proves that the second factor in (*) is also non-negative, completing the proof.

once you computed the geometric sums, and are left to show that (a 2n+2 - 1) / (a 2n+1 - a) - (n+1)/n  >= 0. you can also treat it as a function of a for every given n, and taking the derivative wrt a will look sometimes like (a 2n+1 + 1)(a - 1) >= 0 since a >= 1. But LHS, or f(a)=0 when lim(a)->1 for all n. Hence f(a) >= 0 for all a and n.  :D

I thin Ulysses should have read http://www.mathlinks.ro/Forum/topic-21455.html before posting.... you shouldn't use "very difficult" as the name of the topic.


inequalities
superior algebra
superior algebra unsolved

Assume that $\left(P,\leq\right)$ is a partial order on a set with $n$ elements. For any $i\in P$, let $U_i=\left\{j:j\in P,j>i\right\} $  and $L_i=\left\{j:j\in P,j<i\right\} $. Let us label each $i\in P$ with a real number $X_i$ such that

$X_i=\frac {1-\sum\limits_{j\in L_i}X_j}{n-|U_i|}$     if $L_i\neq \emptyset$.
 and $X_i=\frac{1}{n-|U_i|}$    if $ L_i=\emptyset$ .

Now prove that for any $i\in P$, we have $ 0\leq X_i\leq 1$.

this one deserve to take your time try it :D

I really like it, but, unfortunately, I have brute-force-like solution only.

[b]1.[/b] We can write $n\cdot X_v-|U_v|\cdot X_v+\sum_{u\in L_v}X_u=1$. Summing it up over $P$ we obtain:
\[n\sum_{v\in P} X_v - \left(\sum_{v\in P}|U_v|\cdot X_v-\sum_{v\in P}\sum_{u\in L_v}X_u \right)=n.\]
Now you just need to use your imagination to realize that
\[\sum_{v\in P}|U_v|\cdot X_v=\sum_{v\in P}\sum_{u\in L_v} X_u.\]
It follows that $\sum_{v\in P} X_v=1$. Thus we are only to show that $X_v\geq 0$ for all $v\in P$.

[b]2.[/b] Suppose we have already proved that for all $v\in L_{v_0}$ inequality $X_v\geq 0$ holds. We will show that $X_{v_0}\geq 0$ too.
Consider
\[S=\sum_{v\in L_{v_0}}\left(n\cdot X_v-|U_v|\cdot X_v+\sum_{u\in L_v}X_u\right).\]
We know that $S=|L_{v_0}|$. On the other hand (use your imagination again!)
\[S=\sum_{v\in L_{v_0}}X_v\left(n-|U_v|+|U_v\cap L_{v_0}|\right).\]
Note that $n-|U_v|+|U_v\cap L_{v_0}|=n-|U_v\setminus L_{v_0}|\geq |L_{v_0}|$ !!!! Since all $X_v\geq 0$ ($v\in L_{v_0}$) we conclude
\[|L_{v_0}|=S \geq \sum_{v\in L_{v_0}}X_v|L_{v_0}|,\]
i.e. $\sum_{v\in L_{v_0}}X_v\leq 1$, so $X_{v_0}=\frac{1-\sum_{v\in L_{v_0}}X_v}{n-|U_{v_0}|}\geq 0$.

[b]3.[/b] Final step. It is clear that we can enumerate all elements $P$ into sequence $v_1$, $v_2$, ..., $v_n$ in such way that $L_{v_n}\subset\{v_1,...,v_{n-1}\}$. Now using [b]1[/b] and [b]2[/b] we immediately obtain desired result.

P.S. I believe that there is easy combinatoric or probabilistic interpretation.


Solution:
[hide]Remember, any perfect square is in the form $(a+b)^2=a^2+2ab+b^2$. We see that there are radicals involved, and any square-root-of-a-rational radical squared is rational, so we know that, for the first radical,

$11=a^2+b^2$, $2ab=-2\sqrt{30}$, then $ab=-\sqrt{30}$. Note that 5+6=11 and 5x6=30-- aha!

So the first radical is $\sqrt{(\sqrt{6}-\sqrt{5})^2}=\sqrt{6}-\sqrt{5}$.

By the same logic, the second denominator is $\sqrt{5}-\sqrt{2}$.

The third is a little harder, but factoring a 2 gives $\sqrt{2(4+2\sqrt{3})}=\sqrt{2{(1+\sqrt{3})^2}}=\sqrt{2}+\sqrt{6}$

So we have $\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}-\frac{4}{\sqrt{6}+\sqrt{2}}.$

Then you just rationalize denominators (multiply by conjugates; the first two cancel quite nicely. I actually suspect the + sign in the denominator of the 3rd should be a - sign, then the simplified expression would be 0; instead it's a bit less nice).[/hide]

Actually, I think it is 0 with the plus sign, because the last part becomes $-\sqrt{6}+\sqrt{2}$ and it all cancels out. :huh:

[quote="jensen"][quote="anirudh"]$\text{Simplify: }$

$\displaystyle \frac{1}{\sqrt{(11-2\sqrt{30})}}-\frac{3}{\sqrt{(7-2\sqrt{10})}}-\frac{4}{\sqrt{(8+4\sqrt{3})}}$[/quote]
i think this problem is good but it is not useful. :D[/quote]Yeah I think this comment is good but it is not useful. :D

[quote="jensen"][quote="anirudh"]$\text{Simplify: }$

$\displaystyle \frac{1}{\sqrt{(11-2\sqrt{30})}}-\frac{3}{\sqrt{(7-2\sqrt{10})}}-\frac{4}{\sqrt{(8+4\sqrt{3})}}$[/quote]
i think this problem is good but it is not useful. :D[/quote]

It doesn't have to be useful jensen, if you love solving it, it doesn't matter really.


superior algebra
superior algebra unsolved

Let R be any ring, let s be an element in R, and suppose we have some f(X)∈R[X] with f(s)=0, and D[f(X)] also giving value 0 at s. Prove that f(X)=g(X)(X-s)^2 for some g(X)∈R[X].
(hint. divide f(X) by (X-s)^2 with remainder, and compute D[f(X)] from that expression.)

the hint is the solution ...

we may write $ f \equal{} g * (X\minus{}s)^2 \plus{} h$ for some $ h \in R[X]$ with $ deg(h)<2$. using the leibniz-formula and the assumptons yields $ h(s)\equal{}0$ and $ D(h)(s)\equal{}0$. since we may write $ h \equal{} aX \plus{} b$, this means $ a\equal{}b\equal{}0$ and thus $ h\equal{}0$.


LaTeX
inequalities
algebra proposed
algebra

solve the following equation:
27^x +5^(5x)+196^(4x)=sqrt[(3^(5x)) * (5^(3x))] +sqrt[20^(7x)] * 7^3x+sqrt[(12^x) *(49^x))]*7^(4x)
what do u think about the level of this problem??is it at least nice?please try it and tell me your opinion.

nobody wants to try?is it to easy? :blush:

Firstly: wait some more time - don't be impatient - only two hours gone after you posted your problem and there are posts which haven't ever received a solution

Secondly: put your problem into LaTeX - it will make it more readable for others.

is this to ugly? :? i think the idea is not very easy to see(neither hard) and it is nice.

hint:try to use some inequalities

Are we solving $\in \mathbb{Q}$ or $\in \mathbb{N}$?  I have a strategy for the second, but the first...  :|

bodom: please stop spamming - if somebody get interested then he surely post a solution.


limit
absolute value
calculus
calculus computations

Is it possible to find sequences of real numbers $(a_{n})$ and $(b_{n})$ such that (satisfying the below 2 conditions)

[b]1[/b] $lim_{n \to \infty}(a_{n})=+\infty$; and $lim_{n \to \infty}(b_{n})=0$
[b]2[/b] $(a_{n}b_{n})$ is unbounded but neither $lim_{n \to \infty}(a_{n}b_{n})=+\infty$ nor $lim_{n \to \infty}(a_{n}b_{n})=-\infty$

It is.
Let a_n diverge to +infinity, and b_n converge to 0, but with alternating signs, s.th. the absolute value of a_n b_n grows unbounded.
[hide="Example"]
$a_{n}= n^{2}, b_{n}= (-1)^{n}/ n$
[/hide]

Another possibility is that the product a_n b_n is unbounded, but always comes back to a bounded value, e.g. by setting b_2n = 0.
LiMa

[quote="LiMa"]
$a_{n}= n^{2}, b_{n}= (-1)^{n}/ n$

LiMa[/quote]

but in this example: $lim(a_{n}b_{n})_{ n \to \infty}={+\infty,or-\infty}$

could you explain this please (I am new to these topics) :lol:

For a sequence $a_{n}$,
divergence to $+\infty$ ($\lim_{n\to\infty}a_{n}=+\infty$) means that for every bound $R > 0$, there exists an index bound $N \in \mathbb N$, s.th. for all $n > N$, we have $a_{n}> R$,
divergence to $-\infty$ ($\lim_{n\to\infty}a_{n}=-\infty$) means that for every bound $R < 0$, there exists an index bound $N \in \mathbb N$, s.th. for all $n > N$, we have $a_{n}< R$.

Our sequence is $a_{n}b_{n}= n^{2}(-1)^{n}/ n = n (-1)^{n}$. This sequence is unbounded to both sides. But it does not diverge to +infinity or to -infinity, as you can see by checking the definition.

LiMa

Did spider_boy mean for the second condition to read "neither $\lim_{n\to\infty}(a_{n}b_{n})=+\infty$ nor $\lim_{n\to\infty}(a_{n}b_{n})=0$"?

If so, here's an example:

For $n=2k,$ even, let $a_{n}=2^{k^{2}}$ and $b_{n}=2^{-(k^{2}-k)}.$

For $n=2k+1,$ odd, let $a_{n}=2^{k^{2}}$ and $b_{n}=2^{-(k^{2}+k)}.$

These are sequences of positive numbers. Both $(a_{n})$ and $(b_{n})$ are monotone: the former nondecreasing, the latter nonincreasing.

$\lim_{n\to\infty}a_{n}=\infty;\ \lim_{n\to\infty}b_{n}=0;$

but $\limsup_{n\to\infty}a_{n}b_{n}=\infty$ and $\liminf_{n\to\infty}a_{n}b_{n}=0.$


linear algebra
matrix
algebra
polynomial
modular arithmetic
linear algebra unsolved

1) Find the $ X\in\mathcal{M}_5(\mathbb{Q})$ such that $ X^3\plus{}X\equal{}\minus{}I_5$.
2) Find the smallest prime $ n>3$ such that there exists $ X\in{SL_5(\mathbb{Z})}$ such that $ X^3\plus{}X\equal{}\minus{}I_5$  $ mod(n)$.

[quote="loup blanc"]1) Find the $ X\in\mathcal{M}_5(\mathbb{Q})$ such that $ X^3 \plus{} X \equal{} \minus{} I_5$.[/quote]There is none: Since $ f(Y) \equal{} Y^3 \plus{} Y \plus{} 1$ is irreducible in $ \mathbb Q[Y]$, it is the minimal polynomial of $ X$. Thus the characteristic polynomial having degree 5 is a power of $ f$ - which is absurd.
[quote]2) Find the smallest prime $ n > 3$ such that there exists $ X\in{SL_5(\mathbb{Z})}$ such that $ X^3 \plus{} X \equal{} \minus{} I_5$  $ mod(n)$.[/quote]

[color=red]EDIT: I erased the previously released section, since it was nonsense.[/color]

Here some facts on situations which cannot happen to the polynomial $ f(Y)$ from above over $ \mathbb Z/n\mathbb Z$:

[b]$ f$ cannot be irreducible:[/b] See 1.)

[b]$ f$ cannot have a triple zero:[/b] If $ a$ is this zero, then expanding $ (Y\minus{}a)^3$ yields $ 3a\equal{}0$, hence $ a\equal{}0$ - which is impossible.

[b]$ f$ cannot have a double zero:[/b] In this case $ n\equal{}31$ and $ f(Y)\equal{}(Y\minus{}3)(Y\minus{}14)^2$. But no matrix with eigen values 3 and 14 (with any multiplicities between 0 and 5 summing up to 5) has determinant 1.

[b]$ f$ cannot have a single zero and an irreducible factor of degree 2:[/b] This case yields $ a^2\equal{}1$ - which is absurd, too.

That means that $ f$ splits over $ \mathbb Z/n\mathbb Z$ with three different zeros. As olorin already stated, the minimal n is 47 with the factorization $ f(Y)\equal{}(Y\minus{}25)*(Y\minus{}34)*(Y\minus{}35)$ and $ 25\cdot 34^4\equiv 1 \mod 47$.

Hi third edition,
for 1) you are right.
for 2) you are wrong.:wink:

For b) we need $ n \equal{} 47$ and find $ X\equiv \begin{pmatrix} \minus{} 13 & & & & \\
& \minus{} 13 & & & \\
& & \minus{} 13 & & \\
& & & \minus{} 13 & \\
& & & & \minus{} 22\end{pmatrix}\pmod{47}$.
See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=225595]here[/url].

EDIT
[quote="Third Edition"]As olorin already stated, the minimal n is 47 [/quote]
Actually it's the only possible prime $ n>3$.

Take $ \alpha,\beta,\gamma$ the three roots of $ f(Y)$ over the field $ F\equal{}\mathbb{Z}/n\mathbb{Z}$.
Then $ \alpha\plus{}\beta\plus{}\gamma\equal{}0$ and $ \alpha\beta\gamma\equal{}\minus{}1$.
Let $ n_\alpha,n_\beta,n_\gamma$ be their multiplicities as eigenvalues of $ X$ over $ F$.
Then $ n_\alpha\plus{}n_\beta\plus{}n_\gamma\equal{}5$ and assume $ n_\alpha\ge n_\beta\ge n_\gamma\ge 0$.
There are $ 5$ possible cases $ (n_\alpha,n_\beta,n_\gamma)\equal{}(5,0,0),(4,1,0),(3,2,0),(3,1,1),(2,2,1)$.

Assume $ (n_\alpha,n_\beta,n_\gamma)\equal{}(4,1,0)$. (the other cases are analog and give no solutions)
Then $ \det X\equal{}1$ gives $ 1\equal{}\alpha^4\beta$ and
$ 0\equal{}\alpha^8\beta(\alpha\plus{}\beta\plus{}\gamma)\equal{}\alpha^9\beta\plus{}\alpha^8\beta^2\minus{}\alpha^7\equal{}\alpha^5\plus{}1\minus{}\alpha^7$.
And calculating the $ \gcd$ gives
$ 0\equal{}(\alpha^3\plus{}\alpha\plus{}1)(171\alpha^6\minus{}108\alpha^5\minus{}36\alpha^4\plus{}45\alpha^3\minus{}162\alpha^2\minus{}9\alpha\plus{}117)\\
\hspace*{0.2in}\plus{}(\alpha^5\plus{}1\minus{}\alpha^7)(171\alpha^2\minus{}108\alpha\plus{}306)\\
\hspace*{0.06in}\equal{}3^2\cdot 47$
in $ F$, so $ n\equal{}47$.


number theory
relatively prime
combinatorics proposed
combinatorics

Let $ S\equal{}\{1,2,\ldots,2005\}$. If there is at least one prime number in any subset of $ S$  consisting of $ n$ pairwise coprime numbers, find the minimum value of $ n$.

[u]Solution.[/u] $ n\geq 16$; otherwise, we can take $ \{1,2^2,3^2,5^2,\hdots,43^2\}\subset S$ as $ 14$ numbers, any two of which are coprime, that does not contain a prime number.  I claim that $ n \equal{} 16$ is sufficient.

Let $ a_1,\hdots, a_{15}$ be $ 15$ coprime numbers from $ S\setminus \{1\}$, and let $ p_1,\hdots, p_{15}$ be the smallest prime divisors of $ a_1,\hdots,a_{15}$, respectively.  Suppose that none of $ a_i$ are prime numbers.  Denote by $ p$ as the largest of $ p_1,\hdots,p_{15}$.  Because $ a_1,\hdots, a_{15}$ are coprime, $ p_i$ are distinct.  Suppose $ p \equal{} p_j$.  Becuase $ p_i$ are distinct, $ p_j\geq 47$ (the fifteenth prime number).  Therefore, $ a_j\geq p_j^2\geq 47^2 > 2005$, which is a contradiction, as desired. Therefore, the smallest such $ n$ is $ 16$.  $ \Box$

[quote="Johan Gunardi"]You (like many people) forgot the number $ 1$.[/quote]
1 doesn't work. For instance, the set {4} is a 1 element subset of S, every pair of distinct elements from the set {4} are relatively prime (vacuously true), but {4} doesn't contain any primes.

Edit: whoops :blush:

[quote="Johan Gunardi"][quote="boxedexe"][u]Solution.[/u] $ n\geq 15$; otherwise, we can take $ \{2^2,3^2,5^2,\hdots,43^2\}\subset S$ as $ 14$ numbers, any two of which are coprime, that does not contain a prime number.  I claim that $ n \equal{} 15$ is sufficient.

Let $ a_1,\hdots, a_{15}$ be $ 15$ coprime numbers from $ S$, and let $ p_1,\hdots, p_{15}$ be the smallest prime divisors of $ a_1,\hdots,a_{15}$, respectively.  Suppose that none of $ a_i$ are prime numbers.  Denote by $ p$ as the largest of $ p_1,\hdots,p_{15}$.  Because $ a_1,\hdots, a_{15}$ are coprime, $ p_i$ are distinct.  Suppose $ p \equal{} p_j$.  Becuase $ p_i$ are distinct, $ p_j\geq 47$ (the fifteenth prime number).  Therefore, $ a_j\geq p_j^2\geq 47^2 > 2005$, which is a contradiction, as desired. Therefore, the smallest such $ n$ is $ 15$.  $ \Box$[/quote]You (like many people) forgot the number $ 1$.[/quote]

Well, then make it $ 16$ (I have edited my post)


[hide="solution"]
Let $ \Lambda \equal{} \cos\left(\frac {\pi}7\right) \cdot\cos\left(\frac {2\pi}7\right)\cdot\cos\left(\frac {4\pi}7\right)$. Use the identity $ \cos{(a \plus{} b)} \plus{} \cos{(a \minus{} b)} \equal{} 2\cos a\cos b$. $ \Lambda \equal{} \frac 12(\cos{\frac {\pi}7} \plus{} \cos{\frac {3\pi}7})\cos{\frac {4\pi}7} \equal{} \frac 14(\cos{\frac {\pi}7} \plus{} \cos{\frac {3\pi}7} \plus{} \cos{\frac {5\pi}7} \plus{} \cos{\frac {7\pi}7}) \equal{} \frac 14\sum\limits_{i \equal{} 1}^7{(\cos{\frac {i\pi}7})} \minus{} \frac 14(\cos{\frac {2\pi}7} \plus{} \cos{\frac {4\pi}7} \plus{} \cos{\frac {6\pi}7})$
From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197735]yesterday's trig problem[/url], we know that $ \cos{\frac {2\pi}7} \plus{} \cos{\frac {4\pi}7} \plus{} \cos{\frac {6\pi}7} \equal{} \minus{} \frac 12$ Also, since $ \cos{\frac {\pi}7} \equal{} \minus{} \cos{\frac {6\pi}7}$, $ \cos{\frac {2\pi}7} \equal{} \minus{} \cos{\frac {5\pi}7}$, and $ \cos{\frac {3\pi}7} \equal{} \minus{} \cos{\frac {4\pi}7}$, we have $ \sum\limits_{i \equal{} 1}^7{\cos{\frac {i\pi}7}} \equal{} \cos{\frac {7\pi}7} \equal{} \minus{} 1$. Hence, $ \Lambda \equal{} \frac 14[ \minus{} 1 \minus{} ( \minus{} \frac 12)] \equal{} \boxed{ \minus{} \frac 18}$[/hide]

Let
\[ P = \cos\left(\frac {\pi}7\right)\cdot\cos\left(\frac {2\pi}7\right)\cdot\cos\left(\frac {4\pi}7\right).
\]
Then
\begin{eqnarray*}8\sin\left(\frac {\pi}{7}\right)P & = & 4\cdot2\sin\left(\frac {\pi}{7}\right) \cos\left(\frac {\pi}7\right)\cdot\cos\left(\frac {2\pi}7\right)\cdot\cos\left(\frac {4\pi}7\right) \\
& = & 4\cdot\sin\left(\frac {2\pi}{7}\right)\cdot\cos\left(\frac {2\pi}7\right)\cdot\cos\left(\frac {4\pi}7\right) \\
& = & 2\cdot2\sin\left(\frac {2\pi}{7}\right)\cos\left(\frac {2\pi}7\right)\cdot\cos\left(\frac {4\pi}7\right) \\
& = & 2\cdot\sin\left(\frac {4\pi}{7}\right)\cdot\cos\left(\frac {4\pi}7\right) \\
& = & \sin\left(\frac {8\pi}{7}\right) \\
& = & - \sin\left(\frac {\pi}{7}\right). \end{eqnarray*}
Therefore, $ P = -\frac {1}{8}$.


Find the great value of the expression:
$N= \frac{a}{4+bc} + \frac{b}{4+ca} + \frac{c}{4+ab}$
Where a,b,c are real number in segment [0; 2]

this one has been posted by tranthanhnam twice

okie, where is solution ? :D
If using Mixing Variable ........ :D

This is easy problem.
We suffice to prove that:
               a/(4+bc) <= a\(a+b+c)
similar      b/(4+ca) <= b\(a+b+c)
               c/(4+ab) <= c\(a+b+c)
Add them we have max N =1 when (a,b,c)=(0,2,2)

[b]q: find maximum of f(a,b,c) = a/(4+bc) + b/(4+ac) + c/(4+ab) with a,b,c are in [0 , 2].
a: f is convex in each of the variable, for example let b, c be fixed, then
f''(a) = 2.b.c^2/(4+ac)^2 + 2.c.b^2/(4+ab)^2 > 0. Similary, f''(b) > 0, and f''(c) > 0 when the other two variables held fixed. So f assumes maximum value at one of the 8 end points in the set {(0,0,0), (0,0,2), (0,2,0), (2,0,0), (2,2,0), (2,0,2), (0,2,2), (2,2,2)}. We can verify that the max value of f is: 1 and it occurs at (2,2,0) or (2,0,2) or (0,2,2).
____________________________________________________________[/b]


This should be easy.

The sides of a rhombus are two radii and two chords of a circle. If the radius of the circle is 10cm, then the area of the rhombus is ...?

DO you mean the diagonals are two chords in a circle? Because 2 does not equal 4.

One of the diagonals is also a radium of the circle. Then the rhombus is formed by two equilateral triangles.

The area is $ 2\cdot\frac{R^2\sqrt{3}}{4}\equal{}\frac{R^2\sqrt{3}}{2}$

Just wondering, what does AMTI stand for?

[b]@Silverfalcon :[/b] AMTI = Association of Mathematics Teachers of India.
See [url]http://amtionline.com/[/url] for details.

Oh lol, I thought u meant that each side equalled 2r and that two of the sides were chords. Lol.

Just wondering red_dog,

What is a radium?
I thought it was a white highly radioactive metallic element. Symbol Ra.

And Silverfalcon, the AMTI is an organisation in India which conducts math workshops and olympiads. Those who qualify in these olympiads are eligible to write the RMO (the regional maths olympiad). All those who clear the RMO get to write the INMO (Indian National Maths Olympiad).


http://netlinks.no-ip.org/forum/index.php

you can talk about computers!

and much more!

Now POST one thing you like about yourself. [/u]

are you bullying me mister Power Ranger?!

tsk tsk

read a calc book for more information


Out of $ 22$ students surveyed on ice cream flavors, 12 liked chocolate, 5 liked only strawberry, and 6 liked vanilla. If 3 liked chocolate and vanilla, how many students did not like any of these flavors?

By PIE, $ 12\plus{}6\minus{}3\equal{}15$ students liked chocolate or vanilla (or both), and $ 5$ liked only strawberry. I suppose the problem assumes that there are no other students that liked strawberry and some other flavor, so the answer is most plausibly $ 22\minus{}15\minus{}5\equal{}\boxed2$ students not mentioned in the problem (that is, they don't like any flavors).


trigonometry geometry perimeter circumcircle inradius inequalities algebra

Given triangle ABC.
Prove that:
$cosAcosB+cosBcosC+cosCcosA = \frac{p^{2}-4R^{2}+r^{2}}{4R^{2}}$

I assume that $p$ is the perimeter, $R$ is the circumradius, and $r$ is the inradius?

p is a half of perimeter
$p = \frac{a+b+c}{2}$

[quote="bibobeo"]Given triangle ABC. Prove that: $\cos A\cosh+\cos B\cos C+\cos C\cos A = \frac{p^{2}-4R^{2}+r^{2}}{4R^{2}}$[/quote][hide="Hint"][color=darkblue]$\sum\cos A=1+\frac{r}{R}\ \ \ ,\ \ \sum\cos^{2}A=\frac{1}{2}\cdot\sum(1+\cos 2A)$ $\mathrm{\ \ \ ,\ \ \ }\sum\cos 2A=-1-4\cos A\cos B\cos C$.
The power $p_{w}H$ of the orthocenter $H$ w.r.t. the circumcircle $w=C(O,R)$ is given by the relation $p_{w}H=-8R^{2}\cos A\cos B\cos C$.
The power $p_{w}G$ of the centroid $G$ w.r.t. the circle $w$ is given by the relation $p_{w}G=-\frac{a^{2}+b^{2}+c^{2}}{9}$.
$HO^{2}=R^{2}+p_{w}H=R^{2}-8R^{2}\cos A\cos B\cos C$ and $GO^{2}=R^{2}+p_{w}(G)=R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}$. Therefore,
$HO=3\cdot GO$ $\implies$ $HO^{2}=9\cdot GO^{2}$ $\implies$ $R^{2}-8R^{2}\prod\cos A=9(R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9})$ $\implies$
$\boxed{\ 8R^{2}\cos A\cos B\cos C=a^{2}+b^{2}+c^{2}-8R^{2}\ }$ and $ab+bc+ca=p^{2}+r^{2}+4Rr$.

[b]Remark.[/b] $\{\begin{array}{c}\cos A+\cos B+\cos C=1+\frac{r}{R}\\\\ \cos A\cos B+\cos B\cos C+\cos C\cos A=\frac{p^{2}++r^{2}-4R^{2}}{4R^{2}}\\\\ \cos A\cos B\cos C=\frac{p^{2}-(2R+r)^{2}}{4R^{2}}\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}\cos A\\\ \cos B\\\ \cos C\end{array}$ are the roots of

the equation $\boxed{\ 4R^{2}\cdot X^{3}-4R(R+r)\cdot X^{2}+(p^{2}+r^{2}-4R^{2})\cdot X+(2R+r)^{2}-p^{2}=0\ }$.[/color][/hide]

[quote="bibobeo"][color=darkred]Let be given a triangle $ABC.$ Prove that:
\[\cos A\cos B+\cos B\cos C+\cos C\cos A = \frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}\]
[/color][/quote]
[color=darkblue][b]Proof. [/b]I'll use some well-known relations: $\left\{\begin{array}{c}p=R\sum\sin A\\ r=4R\prod\sin\frac{A}{2}=R\left(\sum\cos A-1\right)\end{array}\right\|$

We have:
$\frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}=\frac{\left(\sum\sin A\right)^{2}+\left(\sum\cos A-1\right)^{2}}{4}-1=\frac{\sum\sin B\sin C+\sum\cos B\cos C-\sum\cos A}{2}\\ =\frac{\sum\left(\sin B\sin C-\cos B\cos C\right)-\sum\cos A}{2}+\sum\cos B\cos C=\sum\cos B\cos C$ [/color]

[color=darkblue][u][b]The shortest method.[/b][/u] $\{\begin{array}{c}ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ \cos A=-\cos (B+C)=-\cos B\cos C+\sin B\sin C\\\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ \cos A+\cos B+\cos C=1+\frac{r}{R}\end{array}$ $\implies$ $\sum\cos B\cos C=$

$\sum\sin B\sin C-\sum\cos A=$ $\sum\frac{bc}{4R^{2}}-(1+\frac{r}{R})=$ $\frac{p^{2}+r^{2}+4Rr}{4R^{2}}-\frac{R+r}{R}$ $\implies$ $\boxed{\mathcal \sum\cos B\cos C=\frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}}$.

[b]Consequence.[/b] $\frac{(R-2r)(7R+2r)}{4R^{2}}+\frac{3}{4}\le \cos A\cos B+\cos B\cos C+\cos C\cos A\le(\frac{R+r}{r\sqrt 3})^{2}$.
[hide]You can use above identity and the remarkable chain of inequalities $3r(4R+r)\le p^{2}\le\frac{(4R+r)^{2}}{3}$.[/hide][/color]

another way:
assume the polynomial $p(x)$ as $p(x)=4R^{2}x^{3}-4R(R+r)x^{2}+(p^{2}+r^{2}-4R^{2})x+(2R+r)^{2}-p^{2}$
now one can see that $cosA,cosB,cosC$ are the roots of $p(x)$ so by [b]viet identity[/b](I dont know the exact spelling) we get the result,and also some other useful identities such as $cosA+cosB+cosC=\frac{R+r}{R}$ and $cosA.cosB.cosC=\frac{p^{2}-(2R+r)^{2}}{4R^{2}}$ and $cos^{2}A+cos^{2}B+cos^{2}C=\frac{6R^{2}+4Rr+r^{2}-p^{2}}{2R^{2}}$...  :wink:

wow, you guys have really amazing solutions.
[Edited]

to BaBaK Ghalebi: How do you come up with that p(x)? It's so wonderful.

[color=darkblue][b][u]Another short proof.[/u][/b] $\cos B\cos C+\cos A=\sin B\sin C=\frac{S}{aR}$ $\implies$ $\sum\cos B\cos C=\frac{S}{R}\cdot\sum \frac{1}{a}-\left(1+\frac{r}{R}\right)$ $\implies$

$\sum\cos B\cos C=\frac{S}{R}\cdot \frac{ab+bc+ca}{4RS}-\frac{R+r}{R}$ $\implies$ $\frac{p^{2}+r^{2}+4Rr}{4R^{2}}-\frac{R+r}{R}$ a.s.o.[/color]

[quote="bibobeo"]to BaBaK Ghalebi: How do you know that $cosA, cosB, cosC$ are the roots of $p(x)$?  :maybe:[/quote]
[color=darkblue]I"ll answer in Babak Ghalebi's stead because I have now some times. You can prove easily that

$(X-x_{1})(X-x_{2})(X-x_{3})=X^{3}-s_{1}X^{2}+s_{2}X-s_{3}$, where $\{\begin{array}{c}s_{1}=x_{1}+x_{2}+x_{3}\\\\ s_{2}=x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}\\\\ s_{3}=x_{1}x_{2}x_{3}\end{array}$.

See [i][b]again[/b][/i] the my first reply (namely in the end) ![/color]

thank you virgil nicula for your answer...  :)
[quote]are the roots of  so by viet identity[/quote]
from [b]viet identity[/b] in my post,I ment the same thing that virgil nicula said in his last post...
always think about this idea,its very useful,and also powerful in solving the dentities of triangle...
and also there are some other polynomials and more identities which are very useful,and will help you to solve many kinds of problems,you should have some of them in mind,it could be useful someday...  :wink:
I dont know from where you can find them in your country,but I can post some of them here,for every one who is interested...  :maybe:
just let me know...

oh that's very kind of you, BaBaK Ghalebi. I'd like to know.

Btw, I understand what Virgil Nicula said. However, in your proof, you assumed $p(x)$, then you said "one can see that ..." and then you gave out some identities that come from Viet.

Virgil Nicula said having all 3 identities $\Longleftrightarrow$ $cosA, cosB, cosC$ are the roots of $p(x)$

Look again in your proof, you still have to prove $cosA, cosB, cosC$ are the roots of $p(x)$ first, then you can use Viet to give out those amazing identities.

Am I right?  :maybe: It really confuses me.

actually the idea of using polynomials in identities and the way of getting to that huge polynomial is this:

first they thought about [b]viet[/b],then they tried to get a polynomial of degree $3$ that the roots are the length of the sides
of $\triangle ABC$,$p(x)=x^{3}-2px^{2}+(p^{2}+r^{2}+4Rr)x-4Rrp$ how?like this:

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

[hide]ok heres the approach:we want to use [b]viet[/b] so first of all we need to compute $a+b+c,ab+bc+ca,abc$...

first we know that:

$a+b+c=2p$

$S=\frac{abc}{4R}=pr$

so we get that $abc=4Rpr$ now lets compute $ab+bc+ca$ in form of $R,r,p$ we know that:

$S=\sqrt{p(p-a)(p-b)(p-c)}=pr$ ([b]heron[/b])

so we have $(p-a)(p-b)(p-c)=pr^{2}$

therefor we get that $p^{3}-p^{2}(a+b+c)+p(ab+bc+ca)-abc=pr^{2}$

but we have computed $a+b+c$ and $abc$ before,so lets put them in the last equation,we`ll get that $ab+bc+ca=p^{2}+r^{2}+4Rr$ so now we have computed $ab+bc+ca$ in the form of $p,r,R$ as we wanted to...

[u]and now,acording to [b]viet[/b] we get the folowing identity:[/u]

$p(x)=x^{3}-2px^{2}+(p^{2}+r^{2}+4Rr)x-4Rrp$[/hide]

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

now by the law of $\sin$ we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ so we can put $2R\sin A$
istead of $a$ in the computed formulas above to get some similar identities for $\sin A,\sin B,\sin C$ (I mean formulas for
$\sin A+\sin B+\sin C$ and $\sin A\sin B+\sin B\sin C+\sin C\sin A$ and $\sin A\sin B\sin C$) and again with [b]viet[/b] we
get another polynomial...

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

[hide]from $a+b+c=2p$ we get that $\sin A+\sin B+\sin C=\frac{p}{R}$ and from $abc=4Rrp$ we get that $\sin A\sin B\sin C=\frac{pr}{2R^{2}}$ and from $ab+bc+ca=p^{2}+r^{2}+4Rr$ we get that $\sin A\sin B+\sin B\sin C+\sin C\sin A=\frac{p^{2}+r^{2}+4Rr}{4R^{2}}$ so by [b]viet[/b] we would have the folowing polynomial:

$q(x)=x^{3}-\frac{p}{R}x^{2}+\frac{p^{2}+r^{2}+4Rr}{4R^{2}}x-\frac{pr}{2R^{2}}$[/hide]

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

we can do the same trick to get THE polynomial for $\cos A,\cos B,\cos C$...  :lol:
did you get the idea?

[color=darkblue][u][b]Example :[/b][/u] $\{\begin{array}{c}s_{1}=\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=\frac{r_{a}+r_{b}+r_{c}}{p}=\frac{4R+r}{p}\\\\ s_{2}=\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=\frac{r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}}{p^{2}}=1\\\\ s_{3}=\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}=\frac{r}{p-a}\cdot\frac{r}{p-b}\cdot\frac{r}{p-c}=\frac{r^{3}}{pr^{2}}=\frac{r}{p}\end{array}$

Therefore, the equation $t(X)\equiv pX^{3}-(4R+r)X^{2}+pX-r=0$ has the roots $\{\ \tan\frac{A}{2}\ ,\ \tan\frac{B}{2}\ ,\ \tan\frac{C}{2}\ \}$.

I used the well-known identities : $\|\begin{array}{c}(p-a)(p-b)(p-c)=pr^{2}\\\\ \|\begin{array}{c}r_{a}+r_{b}+r_{c}=4R+r\\\\ r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}\\\\ r_{a}r_{b}r_{c}=p^{2}r\end{array}\end{array}$

[b][u]Remark.[/u][/b] The equation $r(X)=X^{3}-(4R+r)X^{2}+p^{2}X-p^{2}r=0$ has the roots $\{\ r_{a}\ ,\ r_{b}\ ,\ r_{c}\ \}$.

[b][u]Exercise.[/u][/b] Prove that $\boxed{\ \triangle ABC\mathrm{\ is\ right\ }\Longleftrightarrow p=2R+r\ }$. Indeed, $\triangle ABC$ is right $\Longleftrightarrow$ $90^{\circ}\in\{\ A\ ,\ B\ ,\ C\ \}$ $\Longleftrightarrow$

$1\in \{\ \tan\frac{A}{2}\ ,\ \tan\frac{B}{2}\ ,\ \tan\frac{C}{2}\ \}$ $\Longleftrightarrow$ $t(1)=0$ $\Longleftrightarrow$ $p-(4R+r)+p-r=0$ $\Longleftrightarrow$ $p=2R+r$

[b][u]Another exercises.[/u][/b] $\{\begin{array}{c}\boxed{\ 60^{\circ}\in\{\ A\ ,\ B\ ,\ C\ \}\Longleftrightarrow t(\frac{\sqrt 3}{3})=0\Longleftrightarrow p=(R+r)\sqrt 3\ }\\\\ \boxed{\ 30^{\circ}\in\{\ A\ ,\ B\ ,\ C\ \}\Longleftrightarrow t(2-\sqrt 3)=0\Longleftrightarrow p=R+r(2+\sqrt 3)\ }\end{array}$[/color]

@Virgil Nicula: are $r_{a},r_{b},r_{c}$ the exradii? (radii of the excircles - see [url]http://mathworld.wolfram.com/Excircles.html[/url])

the 'well-known' identities can probably be shown if one knows that $r_{a}= \frac{K}{s-a}$ where $K$ is the area, $s$ is semiperimeter.

How do you prove $\sum(r_{a}) = 4R+r$?
I have really long prove...

Following Teki-Teki, so I prove
$S^{2}(\sum(\frac{1}{p-a}) = abc+\frac{S^{2}}{p}$
then use Heron, and blahblah ...

Is there any other way?

And btw, do you have any other special identities like those?  :D

This is somewhat beside the point, but the relation between the roots and coefficients of a polynomial are named after [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Viete.html]François Viète[/url] and so presumably should be referred to as "Viète's identities."

[quote][color=darkred]Let be given a triangle $ABC.$ Prove that: \[r_{a}+r_{b}+r_{c}=4R+r\] [/color][/quote]
[color=darkblue][b]Proof.[/b] \begin{eqnarray*}r_{a}&=&p\tan\frac{A}{2}=R\left(\sum\sin A\right)\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\\ &=&R4\prod\cos\frac{A}{2}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}= 4R\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\\ \Longrightarrow-1+\frac{r_{a}}{R}&=&-1+4\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\\ &=&-1+2\sin\frac{A}{2}\left(\cos\frac{B-C}{2}+\cos\frac{B+C}{2}\right)\\&=&-1+2\cos\frac{B+C}{2}\cos\frac{B-C}{2}+2\sin^{2}\frac{A}{2}\\&=&\cos B+\cos C-\cos A\\ \Longrightarrow-3+\sum\frac{r_{a}}{R}&=&\sum\cos A=1+\frac{r}{R}\\ \Longrightarrow \sum r_{a}&=&4R+r\end{eqnarray*}[/color]

[quote="bibobeo"]ok, I think I've understood everything. Thank you very much Virgil Nicula and BaBaK Ghalebi.[/quote]
below,I`ll post some polynomials with their identities,so you would use them...

and as said before from the $\sin$ law you can compute $\si8n A,\sin B,\sin C$ and come up with another useful polynomial and useful identities from the angles of the triangle:
the polynomial $p(x)=x^{3}-\frac{p}{R}x^{2}+\frac{p^{2}+r^{2}+4Rr}{4R^{2}}x-\frac{pr}{2R^{2}}$ has roots $\sin A,\sin B,\sin C$ (proved before...) so we get that:
-------------------------------------------
$1)\sin A+\sin B+\sin C=\frac{p}{R}$

$2)\sin A\sin B\sin C=\frac{pr}{2R^{2}}$

$3)\sin A\sin B+\sin B\sin C+\sin C\sin A=\frac{p^{2}+r^{2}+4Rr}{4R^{2}}$

$4)\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}=\frac{p^{2}+r^{2}+4Rr}{2pr}$
-------------------------------------------
again by [b]viete`s identity[/b]...  :)

[quote="bibobeo"][color=darkred]How do you prove $\boxed{\sum r_{a}= 4R+r}$ ? I have really long prove ...  Following Teki-Teki, so I prove
$S^{2}\cdot \sum \frac{1}{p-a}= abc+\frac{S^{2}}{p}(*)$. Then use Heron, and blahblah ...  Is there any other way ? And b.t.w., do you have any [b]other special identities[/b] like those ?[/color][/quote]

[color=indigo][u][b]Oops ! Here has been finished ! Indeed, the relation $(*)$ means $S\sum r_{a}=4RS+Sr$, i.e. $r_{a}+r_{b}+r_{c}=4R+r$ !!!![/b][/u][/color]

[color=darkblue][b][u]Lemma[/u][/b] ("[b]other special identities[/b]"). Let $ABC$ be a triangle. The its incircle $C(I,r)$ touches the sides in the points $D\in BC$, $E\in CA$, $F\in AB$.

Denote the its $A$- exincircle $C(I_{a},r_{a})$ a.s.o. Then $\{\begin{array}{c}I_{a}I_{b}I_{c}\sim DEF\ ,\ \frac{I_{b}I_{c}}{2R}=\frac{EF}{r}\ \ ,\ \ \boxed{\ [I_{a}I_{b}I_{c}]=2pR\ }\\\\ \frac{[DEF]}{r^{2}}=\frac{[ABC]}{2Rr}=\frac{[I_{a}I_{b}I_{c}]}{4R^{2}}\ \ ;\ \ [DEF]\cdot [I_{a}I_{b}I_{c}]=[ABC]^{2}\end{array}$.

[b]Remark.[/b] When I"ll have some times, I"ll present the its prove. Soon !

Here is [u][b]a short proof[/b][/u] of the identity $\boxed{\ r_{a}+r_{b}+r_{c}=4R+r\ }$.

$S=(p-a)r_{a}=(p-b)r_{b}=(p-c)r_{c}=pr$ $\Longrightarrow$ $p(r_{a}-r)=ar_{a}$ a.s.o. $\Longrightarrow$ $p(\sum r_{a}-3r)=\sum ar_{a}$.
$ar_{a}=2\cdot [BI_{a}C]$ a.s.o. $\Longrightarrow$ $\sum ar_{a}=2([I_{a}I_{b}I_{c}]-[ABC])=4pR-2pr$ $\Longrightarrow$ $\sum ar_{a}=2p(2R-r)$.
Therefore, $p(\sum r_{a}-3r)=2p(2R-r)$ $\Longrightarrow$ $\sum r_{a}-3r=2(2R-r)$ $\implies$ $\sum r_{a}=4R+r$.

[b][u]Another shorter proof[/u][/b] (own ? ; I don't think !) of the same identity.

$\{\begin{array}{c}\sum (p-b)(p-c)=\sum \frac{S^{2}}{r_{b}r_{c}}=\frac{S^{2}(r_{a}+r_{b}+r_{c})}{r_{a}r_{b}r_{c}}=r(r_{a}+r_{b}+r_{c})\\\\ \sum (p-b)(p-c)=\sum p^{2}-p\sum (b+c)+\sum bc=3p^{2}-4p^{2}+(p^{2}+r^{2}+4Rr)=r(4R+r)\end{array}\|$ $\implies$ $r_{a}+r_{b}+r_{c}=4R+r$.[/color]

Thanks a lot, I've learnt a lot from you, BaBaK Ghalebi, Virgil Nicula and other people too.


linear algebra
matrix
linear algebra unsolved

For a matrix $ A\in M_n(C)$, we know $ A$ and its transpose $ A^T$ are similar. Is it true that $ A$ and  $ A^T$ are unitarily similar?  :maybe:

It would be very nice if such a result was true, but it is not  :wink:
One can find a $ 3 \times 3$ matrix $ A$ such that $ A$ and $ A^T$ are not unitarily similar.

A couple of people asked me for an example via private message, so I thought I'd write one here:
$ A\equal{}\begin{pmatrix}1 & 4 &0\\0& 2 & 5\\ 0 &0& 3\end{pmatrix}$.
[hide="Hint "]To prove it, use Specht's theorem[/hide]

A simpler method than Specht's theorem:
Let $ A$ be the considered matrix. $ 2$ normed eigenvectors of $ A$ are:$ u_1\equal{}[1,0,0]^T,u_2\equal{}1/\sqrt{17}[4,1,0]^T$ and $ u_1.u_2\equal{}4/\sqrt{17}$.
The normed eigenvectors of $ A^T$ are $ v_1\equal{}1/\sqrt{26}[0,\minus{}1,5]^T,v_2\equal{}[0,0,1]^T,v_3\equal{}1/\sqrt{117}[1,\minus{}4,10]^T$. Then for all $ i<j$,  $ v_i.v_j\not\equal{}\pm{u}_1.u_2$.  QED

For any matrix $ A$, does there always exist a nonsingular matrix $ S$ such that $ A^T\equal{}S^*AS$?


If $x,y$ are positive integers and $n$ nonnegative such that $\frac{xy}{x+y}>n$, then prove that

$\frac{xy}{x+y}\geq n+\frac{1}{n^{2}+2n+2}$

Oh,It is a interesting lemma for a problem of Bulgrian( :?: I always forgot some English word) 2004(Maybe).
we have $(x-n)(y-n)>n^{2},assume that (x-n)(y-n)=n^{2}+c$
then $x-n+y-n\leq n^{2}+c+1$
then $\frac{xy}{x+y}=n+\frac{c}{xy}\geq n+\frac{c}{n^{2}+2n+c+1}\geq n+\frac{1}{n^{2}+2n+2}$ :)

[quote="Hawk Tiger"]Oh,It is a interesting lemma for a problem of Bulgrian( :?: I always forgot some English word) 2004(Maybe).[/quote]

Yes, indeed  :)  (It's 2005)

Anyway, Hawk Tiger, do you have the problems of Bulgarian spring/winter tournaments this year? I haven't seen yet.

Sorry ,I don't have.I also want to have them.The Bulgarian problem is very interesting.Could some one who have the problems upload it or send them to me by E-mail?Thanks

It looks like we have no hope, Hawk Tiger  :D


I'm not sure what the closed form is for the following summations:

               n             n
            Sigma (i ( Sigma 3 ) )
              i=3         j=i+1

*Sigma represents the sigma symbol usually present in a summation

Any help would be greatly appreciated.

Sorry, your notation is quite unclear.  Do you mean

$ \sum_{i\equal{}3}^{n} i \left( \sum_{j\equal{}i\plus{}1}^{n} 3 \right)$

?

Yes i do, haha sorry...I didn't check how the forum formatted my text.  I just submitted without looking it over.

The inner summation is quite easy; do that first, and then evaluate what's left.

[hide="Possibly?"]
$ \sum_{i\equal{}3}^{n} \left( i \left( 3 \left(n\minus{}i \right) \right) \right)$
$ \sum_{i\equal{}3}^{n} \left( \minus{}3 i^{2} \plus{}3in \right)$
$ \frac{\minus{}n(n\plus{}1)(2n\plus{}1) \plus{} 3n(n\minus{}2)(n\plus{}3)\plus{}30}{2}$
$ \frac{n^3\minus{}19n\plus{}30}{2}$.
[/hide]

if it was as easy for me to just do it, I never would have posted in this forum.  I also followed that link of yours and it didn't shed to much light on my situation.

t0r gave you a suggestion on how to proceed.  Why don't you follow his suggestion as far as it leads you and post the results here -- then t0r or others may provide further suggestions to move you along.

You might consider learning a little bit of [[LaTeX]] (see, e.g. that link or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 or use the quote button on what AstroPhys wrote to figure out some of the basics) so that you can write math more readably.  (You've already noticed that the forum automatically eliminates spaces, which makes ASCII formatting more difficult.)

j = i + 1; (i = 3, so j = 4)
j = 4 -> n
3 + 3 + 3 + 3 +....+ 3n

is that right?  3n is than the closed form?

It's not at all clear to me what you did (and it doesn't give you the right answer).

The inner sum is $ \sum_{j \equal{} i \plus{} 1}^n 3$.  Can you evaluate this sum first?  (Then we can worry about the outer sum later.)

That's exactly what I tried doing.  The numbers I wrote correspond only to the inner sum.  That was the answer I got for the inner sum.

You did it for $ i \equal{} 3$.  But $ i$ isn't fixed (in the outer sum, it runs from $ 3$ to $ n$, so takes $ n \minus{} 2$ different values) -- your expression for the inner sum should end up in terms of $ i$ as well as $ n$.

I don't know, 3n+i ???  I was hoping to get the solution, a detailed step by step on how to get to that solution in order to apply these steps to the actual problem I need to solve for my studies.  All this back and forth hasn't helped me at all and has wasted alot of time.  The problem I am trying to solve is alot more difficult, but i cant attempt it until I understand closed forms of summations, along with double summations like the example I gave.  So can anyone help me, or not?

Hint
[hide]
The key is to understand the meaning of this summation:
For $ j\equal{}i\plus{}1, i\plus{}2, ..., n$, add $ 3$.
You can regard the $ 3$ as a constant function of $ j$, which takes the value from $ i\plus{}1$ to $ n$ in the summation.
Or this way:
In the sequence $ 3,3,3...,3,...$, sum the terms from $ i\plus{}1^{th}$ to $ n^{th}$.
You can use the same way to do the outer summation.
[/hide]

[hide="HINTS"]
$ \sum_{i\equal{}x}^{y} k$ for some constant $ k$, is $ k \left( y \minus{} x \plus{}1 \right)$.

Sum of the first $ n$ integers
$ \frac{n\left( n \plus{}1 \right)}{2}$.

The sum of the first $ n^2$ integers
$ \frac{n \left( n \plus{} 1 \right) \left( 2n\plus{}1 \right) }{6}$.
[/hide]

[hide="ANOTHER HINT"]
$ \sum_{i\equal{}j}^{l} ik\equal{}k \sum_{i\equal{}j}^{l} i$ For some constant $ k$.
[/hide]


calculus
integration
calculus computations

show, for $ n>m$, :D

$ \int_0^1\;\dfrac{x^m\plus{}\dfrac{1}{x^m}}{x^n\plus{}\dfrac{1}{x^n}}\;\;\dfrac{dx}{x}\;\equal{}\;\boxed{\dfrac{\pi}{2n}\cdot\sec\left(\dfrac{m\pi}{2n}\right)}$

Indeed you have many problems such that, misan.


inequalities
triangle inequality
inequalities proposed

Even the other way, are you sure it's true?

$\sum\frac{2-y-z}{2-x}=\sum\frac{y+z-2}{x-2}=\sum(1+\frac{y+z-x}{x-2})=3+\sum\frac{x+y-z}{z-2}$

The inequality holds iff $\frac{x+y-z}{z-2}+\frac{y+z-x}{x-2}+\frac{z+x-y}{y-2}\le 0$
That sum can be positive (for example, if $x,y,z>2$, and the three satisfy the triangle inequality), negative (for example, un Sung-yoon Kim's example), or zero if all of the numbers are 0...so I think you might have mistyped the left side if anything.


limit
linear algebra
matrix
calculus
calculus computations

Find all numbers $ a,b,c$ such that:
$ \displaystyle\lim_{x\to \pm \infty} {\frac{a(2x^3\minus{}x^2)\plus{}b(x^3\plus{}5x^2\minus{}1)\minus{}c(3x^3\plus{}x^2)}{a(5x^4\minus{}x)\minus{}bx^4\plus{}c(4x^4\plus{}1)\plus{}2x^2\plus{}5x}}\equal{}1$

It is not hard. Consider degree to find out $ a,b,c$

You have:

$ 5a\minus{}b\plus{}4c\equal{}0$

$ 2a\plus{}b\minus{}3c\equal{}0$

$ \minus{}a\plus{}5b\minus{}c\equal{}2$.

It's easy system.

$ \begin{pmatrix}5&\minus{}1&4\\2&1&\minus{}3\\\minus{}1&5&\minus{}1 \end{pmatrix} \begin{pmatrix}a\\b\\c \end{pmatrix} \equal{}\begin{pmatrix}0\\0\\2\end{pmatrix}$

This system has only one solution because

$ \begin{vmatrix}5&\minus{}1&4\\2&1&\minus{}3\\\minus{}1&5&\minus{}1 \end{vmatrix} \equal{} 5 \cdot 14 \plus{} 1 \cdot (\minus{}5) \plus{} 4 \cdot 11 \equal{} 109 \neq 0$.

Therefore:

$ a \equal{} \frac{\begin{vmatrix}0&\minus{}1&4\\0&1&\minus{}3\\2&5&\minus{}1 \end{vmatrix}}{109} \equal{} \minus{} \frac{2}{109}$

$ b \equal{} \frac{\begin{vmatrix}5&0&4\\2&0&\minus{}3\\\minus{}1&2&\minus{}1 \end{vmatrix}}{109} \equal{} \frac{46}{109}$

$ c \equal{} \frac{\begin{vmatrix}5&\minus{}1&0\\2&1&0\\\minus{}1&5&2 \end{vmatrix}}{109} \equal{} \frac{14}{109}$.


Let L be a six digit (base ten) integer whose last digit is not 0, and let M be the number formed by moving this last digit to the front. For example, if L=354268, then M=835426. Find all possible L such that L=3M. What is the sum of all the digits of all the possible L ?

I have been bugged by this problem for a while. Can anyone please explain in detail on how to solve it. Thanks in advance.

$ L \equal{} \overline{abcdef} \equal{} 10A \plus{} f$, $ M \equal{} \overline{fabcde} \equal{} 10^5f \plus{} A$. From $ L\equal{}3M$ we get $ (3\cdot 10^5 \minus{} 1)f \equal{} 7A$, so $ 299999f \equal{} 7A$, or $ 42857f \equal{} A \equal{} \overline{abcde}$. Therefore

1. $ f\equal{}1$, so $ L \equal{} 428571 \equal{} 3\cdot 142857 \equal{} 3M$;
2. $ f\equal{}2$, so $ L \equal{} 857142 \equal{} 3\cdot 285714 \equal{} 3M$.

Well, clearly I denoted $ A \equal{} \overline{abcde}$. Now $ L \equal{} \overline{abcdef} \equal{} A\cdot 10 \plus{} f$, no?


Two different integers from 1 through 20 inclusive are chosen at random. What is the probability that both numbers are prime? Express your answer as a common fraction.

First let's find how many prime numbers there are from $ 1\sim 20$.

We have: $ 2, 3, 5, 7, 11, 13, 17, 19$

That's $ 8$ numbers in total. So the probability of our first number being prime is $ \frac{8}{20}$, or $ \frac{2}{5}$.

Now we want to find the probability of our second number being prime. There are $ 8$ primes, but these have to be two different primes, so we only have $ 7$ "valid" primes. And, for the same reason, it's not out of $ 20$, but out of $ 19$. So the probability of our second number being prime is $ \frac{7}{19}$.

Since we now have both probabilities, we just multiply them! So: $ \frac{2}{5} \times \frac{7}{19} \implies  \boxed {\frac{14}{95}}$


One candle burns completely in EXACTLY one hour. There are two candles and a box of matches in a completely dark room, but you can still see the table with all the items on it (don't ask why). Find a way to burn out both candles in EXACTLY 45 minutes.

NOTICE:

No counting time.

There is NOTHING else inside that room.

The candles must be burnt COMPLETELY, and the resulting time must be COMPLETELY EXACT.

Why are you posting this in so many different places?

well this is the first one IVE seen so far so anyways:

u could cut both of the candles into peices and burn each of them and thatll take a short time.

[quote="Athena"]well this is the first one IVE seen so far so anyways[/quote]

That's becuase you only go the Fun and Games section of AoPS and MATHLINKS.

oh wait i get it! first u light the candles. then u blow it out after 45 mins exactly! that's so smart right?

no because the CANDLES haf 2 be burned out not the flames...i think. so u could also break of a quarter of each of the candles and burn them separately from the big ones so u no its exactly 45 minutes. and if u dunno how 2 burn the smaller portions, just lite a separate match for them.

hey just listen to me! i'm always right! ALWAYS!!!!!!!!!!!!!!!!!!!!!!!!I'M NEVER WRONG!!!!!! HAHAHA!!!!!!!!! I WILL RULE THE WORLD ONE DAY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

You have no way of keeping track of the time, remember?

o yea and that 2. so my anser DOES work.

You got a ruler on hand?  Because the timing has to be EXACT.

but the problem never said that u had no way of measuring. only no way of keeping time. OOOOHHHHHH!!!!!!

[quote="marucho8d7"]One candle burns completely in EXACTLY one hour. There are two candles and a box of matches in a completely dark room, but you can still see the table with all the items on it (don't ask why). Find a way to burn out both candles in EXACTLY 45 minutes.

NOTICE:

No counting time.

[b]There is NOTHING else inside that room.[/b]

The candles must be burnt COMPLETELY, and [b]the resulting time must be COMPLETELY EXACT[/b].[/quote]

You were saying?

ok fine. u can't keep track of time BUT! get a super high-tech robot and let it hold the candles then after EXACTLY 45 minutes, make the robot self-destruct. there.  perfect answer. :D

Athena, you are not supposed to assume that the candle burns at a constant rate. The total time is one hour, but some parts of the candle may burn faster than others.

yea but then we would have no way of knowing which parts burn faster and how fast they burn.  but the whole thing is made of wax so...yea

No, but if you use the official solution, when you light from both ends, it will burn out completely in half an hour regardless of where the two flames meet, since the average speeds of the two flames must average to one cph (candle per hour). And since half an hour's worth of the other candle has been burnt out, the other half is half an hour's worth as well, no matter what size it is. And then you apply the same principle again.

But as for completely exact, there are always matters of human error--the time between you noticing that one candle has been burnt out and you lighting the second end of the second candle, for example--and candles themselves can't be that exact, so 'buy a clock regulated by a precision atomic clock in (someplaceorother)' is still a better idea.

did anyone at all read my preveious post? i gues u didn't my answer is perfectly correct so just read it

I did read your post, and I know it's 'correct'. I was just trying to explain why Athena's answer isn't correct and why the official solution is.

[quote="themonster"]Hey, marucho.  Was Mikey's solution correct and/or were you thinking of something different?[/quote]
mikey had the correct answer

ways such as measuring and breaking the candles are NOT the answers. Never. Know why? Because YOU CAN NEVER BE ACCURATE! I gave exact conditions in the question: EXACTLY one hour, EXACTLY 45 minutes, etc. And I forgot to mention that you are carrying NOTHING with you and that there is NOTHING else in the room

[quote="LynnelleYe"]But as for completely exact, there are always matters of human error--the time between you noticing that one candle has been burnt out and you lighting the second end of the second candle, for example--and candles themselves can't be that exact, so 'buy a clock regulated by a precision atomic clock in (someplaceorother)' is still a better idea.[/quote]

quite a point here. Try to forget ALL outside factors. gameworld, your answer is NOT correct. Read my previous post. I should give all the conditions at the start of the problem

[quote="Athena"]1) i could be carrying a ruler with me

2) it doesnt say that i cant measure in my head.

3) why would sumone wanna make a perfectly good 1-hour candle burn in 45 minutes?[/quote]

forget ALL outside factors. If you went to the Beginners section, you'll notice that I said: NO COUNTING IN YOUR HEAD. I so hate myself. Why didn't i write it in this problem?

[quote="gameworld7"]ok fine. u can't keep track of time BUT! get a super high-tech robot and let it hold the candles then after EXACTLY 45 minutes, make the robot self-destruct. there. perfect answer. [/quote]

get real!

Just ignore him, he never makes any sense.

[quote="RobinHood3000"]All good points, but quadruple-posting is generally discouraged.[/quote]

ill try my best to stop.

"Jig's up, kid. Why don't you just give?"...PUNCH! WHACK! BAM! ---Infamous Bully of Our School Who Beat People to Pieces

at least someone thinks my answer is correct AND!!  it is! and i never quadruple posted before.  my most is only triple. and u know, u could always edit it to one long post.


Source: MENSA

Express $ \frac{1}{2} $ using all of the digits 1 through 9 exactly once.

There are many solutions.  One is
$3+4+6+8-5-7-9+1/2$
Clearly, you can use 1 and 2 in 1/2.  Then you are left with 3 4 5 6 7 8 9.  You need to make 0 with these.  Well the sum is 42, so if you can find a set of them which add up to 21, you can make those positive, the others negative and the 1/2 positive, and you're done.  3 4 6 8 is one such set.

Well technically that is a solution, but I was looking for 1/2 to be expressed as the ratio of two integers, using all digits 1-9 once. Try again.

it is basically 1x/2x with 1-9 so 1+2+3...+9=45 so x clearly is 15 so have like

3+4+8/1+2+5+6+7+9

i think he might mean like a ratio of a 4 digit integer to a 5 digit integer where the digits 1-9 are all used once

Okay.  In that case,
(1+6+8)/(2+3+4+5+7+9)
You get 15 on the top and 30 on the bottom.

I think that alan has the answer we're looking for

Alan has the right answer. There's also another solution in that form.
Can you say how you found it?

Ah.  In that case
7692/15384
works as well, for instance.
You know that the unit's digit of the bottom number has to be even.  So there are clearly 4 possibilities for it.  For each one, the unit's digit of the other one can be 1 of two things.  Then from there you just form two numbers with one fewer digit each.  If the unit's digit of the top number is half the unit's digit of the second number, then the first 3 digits in the first number are half of the first 4 digits in the second, for example.


inequalities
geometry
circumcircle
inequalities unsolved

Let a, b, c be sides of a triangle with circumradius R. Prove that

${1/ab}+{1/bc}+{1/ca}\geq{1/R^2}$

Let $p=\frac{a+b+c}2$ be the semiperimeter and $r$ the inradius. $S$ is the area. The inequality is equivalent to $\frac {2p}{abc}\ge \frac 1{R^2}\iff \frac {p}{2RS}\ge \frac 1{R^2}\iff \frac 1r=\frac pS\ge \frac 2R$, which is well-known.

Similar:

Let [tex]K[/tex] be the area of the triangle, [tex]s[/tex] be the semiperimeter, [tex]R[/tex] be the circumradius, and [tex]r[/tex] be the inradius. We have [tex]K=rs=\frac{r(a+b+c)}{2}[/tex], thus [tex]a+b+c = \frac{2K}{r}[/tex]. We also have [tex]K = \frac{abc}{4R}[/tex], thus [tex]abc = 4KR[/tex].

We now have [tex]\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{abc} = \frac{\frac{2K}{r}}{4KR} = \frac{1}{2Rr}[/tex].

From the well-known [tex]2r \leq R[/tex], we find [tex]2Rr \leq R^2[/tex], and finally [tex]\frac{1}{2Rr} \geq \frac{1}{R^2}[/tex], and the inequality is proven.


trigonometry
geometry
AMC
AIME
MATHCOUNTS
logarithms
AMC 8

There really shouldn't be too much trig or stuff like that on here.  I ll allow some stuff if you give like a brief lesson beforehand, or it is something that beginners should know.  I think logs are fine.  They are pretty much just exponents and learning exponents is something that beginners should do.  Definitely keep the problems within precal reach and don't go into AIME and above level... easier AIME questions are ok.

Is it alright if I post a challenge set then, just for fun? In the beginner's forum of course. There won't be any trig, I think, but some logs.

I'd love to see it.  Logs are my favorite things to play with in math. :lol:

[quote="white_horse_king88"]This is directed more to the moderators:

If for example, I wanted to post a challenge set for the beginner's forum, what would be appropriate. Can I touch on some logs and trig? Or, is it all limited to algebra and geometry?[/quote]

I'm one of the several (four, right now, isn't it?) moderators here, and what I have been using as a benchmark for what's fair game here on Getting Started is that anything that would appear on the AMC 8 contest or on any level (chapter, state, or national) in Mathcounts is fair game for Getting Started. I think logarithms and trigonometry are both NOT on the list for those two contests, and the usual curriculum progression in most countries is such that problems with those topics better belong on the Intermediate Topics forum. ALL of the forums here tend to have challenge problems that are too hard, as the site owners agree (we have PMed one another about this issue), so when in doubt, post HIGHER if you have a new challenge problem. If you're not sure whether your problem fits Getting Started or Intermediate, post it on Intermediate, where readers will appreciate you posting a more approachable problem.

I've just spent the last six days driving more than 2,000 miles round trip to pick up my son from his summer program, so I'm a little behind reading forum messages. Thank you for asking, and thank you for working up challenge problems for the forum.


geometry
power of a point
geometry unsolved

Let $ ABC$ be an equilateral triangle anf $ D$ an  internal point of the side $ BC$.  A circle, tangent to $ BC$ at $ D$, cuts $ AB$ internally at $ M$ and $ N$,and $ AC$ internally at $ P$ and $ Q$.

Show that $ BD \plus{} AM\plus{}AN\equal{} CD \plus{}AP\plus{}AQ$.

[hide="Solution"]
Let $ s$ be a side of the triangle. By Power of a Point,
$ AN \cdot AM\equal{}AQ \cdot AP$
$ (s\minus{}BM)(s\minus{}BN)\equal{}(s\minus{}CP)(s\minus{}CQ)$.
$ BM \cdot BN \minus{} CP \cdot CQ\equal{}s(BM\plus{}BN\minus{}CP\minus{}CQ)$
But by power of a point, $ BM \cdot BN\equal{}BD^2$ and $ CP \cdot CQ\equal{}CD^2$. Thus, the above simplifies to
$ BD^2\minus{}CD^2\equal{}s(CP\plus{}CQ\minus{}BM\minus{}BN)$
$ (BD\plus{}CD)(BD\minus{}CD)\equal{}s(BM\plus{}BN\minus{}CP\minus{}CQ)$
But $ BD\plus{}CD\equal{}s$. Thus,
$ BD\minus{}CD\equal{}BM\plus{}BN\minus{}CP\minus{}CQ$
But $ BM\equal{}s\minus{}AM$, $ BN\equal{}s\minus{}AN$, $ CP\equal{}s\minus{}AP$, $ CQ\equal{}s\minus{}AQ$. Substituting,
$ BD\minus{}CD\equal{}s\minus{}AM\plus{}s\minus{}AN\minus{}s\plus{}AP\minus{}s\plus{}AQ$, or
$ BD\plus{}AM\plus{}AN\equal{}CD\plus{}AP\plus{}AQ$.[/hide]


geometry
3D geometry
number theory unsolved
number theory

Find all primes p, for which p^2-p+1 is cube

[quote="venatrix"]There are no primes that satisfy the problem

in fact we have $p^{2}-p=p(p-1)=k^{3}-1=(k-1)(k^{2}+k+1)$

Since $(p,p-1)=1$ then we can have two cases

$\left\{\begin{array}{c}{a1}p=k-1p-1=k^{2}+k+1\end{array}$

which leads to $k-2=k^{2}+k+1\Rightarrow k^{2}=-3$. Impossible

$\left\{\begin{array}{c}{a2}p-1=k-1p=k^{2}+k+1\end{array}$

which leads to $0=k^{2}+1$ impossible.[/quote]

No, unfortunately, your solution isn't all right!
In fact, the solution is 19.

Here:
[hide]
Firstly it's important to see that $k<p$. If not, you would have $p(p-1)=(k-1)(k^{2}+k+1)\leq (k-1)k$, which is obviously false
We have $p|(k-1)(k^{2}+k+1)$ and because $p$ is prime
$p|(k-1)$ or $p|(k^{2}+k+1)$
$I.$ $p|k-1\implies p\leq k-1<p-1$ which isn't true
So $p|k^{2}+k+1$. This means that there exists a positive integer $n$ such that $pn=k^{2}+k+1$ (1)
Further $p(p-1)=(k-1)pn$ and $p-1=(k-1)n\implies p=kn-n+1\implies pn=kn^{2}-n^{2}+n$ (2)
From (1)(2) $\implies k^{2}+k(1-n^{2})+1+n^{2}-n=0$
So $d=(n^{2}-1)^{2}-4(n^{2}-n+1)$ has to be a perfect square!
But for $n>3 \implies (n^{2}-3)^{2}<d<(n^{2}-2)^{2}$ so $d$ cannot be a perfect square for $n>3$.
For $n=3 \implies p=19$
[/hide]
I thinks this is a problem from 2005 BalkanMO.

[quote="venatrix"]There are no primes that satisfy the problem

in fact we have $p^{2}-p=p(p-1)=k^{3}-1=(k-1)(k^{2}+k+1)$

Since $(p,p-1)=1$ then we can have two cases

$\left\{\begin{array}{c}{a1}p=k-1p-1=k^{2}+k+1\end{array}$

which leads to $k-2=k^{2}+k+1\Rightarrow k^{2}=-3$. Impossible

$\left\{\begin{array}{c}{a2}p-1=k-1p=k^{2}+k+1\end{array}$

which leads to $0=k^{2}+1$ impossible.[/quote]
By the way, you said that if $ab=cd$ and $(a,b)=1$ then $a=c$ or $a=d$
Nope. For example $a=14, b=15, c=7, d=30$ Even if $(c,d=1$ that isn't true for all integers!


I need  a hand in solving this..

4^(x) - 2^(x+1)= 3


linear algebra
matrix
vector
inequalities

For an $ n\times n$ matrix $ A$, let the $ \textit{adjoint}$ of $ A$ be the $ n\times n$ matrix whose $ (i,j)$ entry is $ (\minus{}1)^{i\plus{}j}\text{det }(A(i|j))$, where $ A(i|j)$ is $ A$ with the $ i^{th}$ row and $ j^{th}$ column removed. If $ \text{rk }(A)\equal{}n\minus{}1$, show that $ \text{rk }(\text{adj }A)\equal{}1$.

I'm not sure where to go with this. The problem before this in the book that I did solve is to show that if the rank of $ A$ is less than $ n\minus{}1$, then the rank of the adjoint is zero. I'm not sure if I have to use this or not...

Perhaps the most famous and important theorem concerning that adjoint (or adjugate) is this:
\[ A(\text{adj}\,A) \equal{} (\det A)I.\]
Now, we are supposing that $ \text{rank}\,A \equal{} n \minus{} 1,$ which means that $ \det A \equal{} 0.$ Hence $ A(\text{adj}\,A) \equal{} 0.$ Since that is true column by column, then each column of $ \text{adj}\,A$ belongs to $ \text{nul}\,A,$ but by the rank-nullity theorem, $ \text{nul}\,A$ is a one-dimensional subspace.

If every column of $ \text{adj}\,A$ lies in $ \text{nul}\,A,$ then there are exactly two possibilities:

(1) $ \text{col}(\text{adj}\,A) \equal{} \text{nul}\,A$ and is one-dimensional, in which case we're done, or

(2) $ \text{adj}\,A \equal{} 0.$

So what we need to do is rule out the second case.

Can you do that? Can you prove that if $ A$ has rank $ n \minus{} 1,$ that for some $ i$ and some $ j,$ we have $ \det(A(i|j))\ne 0?$ (To use your notation.)

Thanks for your response. :)

Here's what I wrote up, basically using what you hinted at:

[hide="Solution."]
=====
$ \textbf{LEMMA: }$ If $ \{\vec{v_i}\}_{i \equal{} 1}^{n \minus{} 1} \subseteq \mathbb{F}^n$ is a linearly independent set of vectors, then $ \exists p\in \mathbb{Z}$, $ 1\leq p\leq n$, such that the set of vectors $ \{\vec{v_i}'\}_{i \equal{} 1}^{n \minus{} 1}\subseteq \mathbb{F}^{n \minus{} 1}$ formed by removing the $ p^{th}$ element of each $ \vec{v_i}$ is also linearly independent.

$ \textit{Proof: }$ Consider the $ (n \minus{} 1)\times n$ matrix $ \textbf{A}$ whose rows are the row vectors $ \{\vec{v_i}\}_{i \equal{} 1}^{n \minus{} 1}$. Further let $ R(\textbf{A})$ be the reduced row echelon form of $ \textbf{A}$. If the rows of $ \textbf{A}$ are linearly independent, then $ R(\textbf{A})$ does not have a row of zeroes. However, since $ R(\textbf{A})$ has more columns than rows, there must also exist a column $ p$ with no leading entries. Thus if we remove this column from $ R(\textbf{A})$ to form $ R(\textbf{A}')$, we see that $ R(\textbf{A}')$ is an $ (n \minus{} 1)\times (n \minus{} 1)$ matrix in reduced row echelon form with no row of zeroes. If we perform the same operations on $ R(\textbf{A}')$ that convert $ R(\textbf{A})$ to $ \textbf{A}$, we get a matrix $ \textbf{A}'$ that is $ \textbf{A}$ with its $ p^{th}$ column removed, so that the row vectors of $ \textbf{A}'$ are $ \{\vec{v_i}'\}_{i \equal{} 1}^{n \minus{} 1}$. We have furthermore shown that these vectors are linearly independent, as desired. $ \square$

We have that the adjoint of $ \textbf{A}$ satisfies $ \textbf{A}\text{adj }\textbf{A} \equal{} (\text{det }\textbf{A})\textbf{I}_n$. Since $ \text{rk }\textbf{A} \equal{} n \minus{} 1 < n$, $ \text{det }\textbf{A} \equal{} 0$, so that $ \textbf{A}\text{adj }\textbf{A} \equal{} \textbf{0}$.

Consider any column vector $ \vec{v}$ of $ \text{adj }\textbf{A}$. We have $ \textbf{A}\vec{v} \equal{} \vec{0}$, so that $ \vec{v}$ is in the null space of $ \textbf{A}$. Thus the column space of $ \text{adj }\textbf{A}$ is a subspace of the null space of $ \textbf{A}$, so that $ \text{dim }(\text{col }(\text{adj }A)) \equal{} \text{rk }(\text{adj }\textbf{A})\leq \text{dim }(\text{null }\textbf{A}) \equal{} n \minus{} (\text{rk }\textbf{A}) \equal{} n \minus{} (n \minus{} 1) \equal{} 1$. It thus suffices to show that $ \text{rk }(\text{adj }\textbf{A})\neq 0$, or that $ \text{adj }\textbf{A}$ has at least one non-zero entry.

Since $ \text{rk }\textbf{A} \equal{} n \minus{} 1$, there exists a subset of $ n \minus{} 1$ rows of $ \textbf{A}$ that is linearly independent. The lemma above guarantees that we can remove corresponding elements from each of these row vectors and still have a linearly independent set of vectors. Note that our initial choice of $ n \minus{} 1$ rows corresponds to removing exactly one row of $ \textbf{A}$ and removing elements from the row vectors corresponds to removing exactly one column of $ \textbf{A}$, so that the remaining submatrix is a maximal submatrix with linearly independent rows. Thus the determinant of this submatrix is non-zero. But some entry of the adjoint of $ \textbf{A}$ is a non-zero multiple of this determinant, so that there is necessarily a non-zero entry of the adjoint. Thus the rank of the adjoint is necessarily non-zero, from which it follows that $ \text{rk }(\text{adj }\textbf{A}) \equal{} 1$, as desired. $ \blacksquare$
=====
[/hide]

[quote="jvenezuela716"]I did solve is to show that if the rank of $ A$ is less than $ n \minus{} 1$, then the rank of the adjoint is zero. [/quote]
So there are exactly three possibilities:

(1) $ \text{rank}\,A\equal{}n$ and $ \text{rank}\,\text{adj}\,A\equal{}n$. The invertible case.

(2) $ \text{rank}\,A\equal{}n\minus{}1$ and $ \text{rank}\,\text{adj}\,A\equal{}1.$

(3) $ \text{rank}\,A\le n\minus{}2$ and $ \text{adj}\,A\equal{}0.$

A simpler way of dealing with the case when $\text{rank}(A)=n-1$ :

Obviously, we have : $A\cdot\text{adj}(A)=O_n$ . By [u][b]Sylvester[/b]'s inequality[/u] we obtain :

$\text{rank}(A\cdot\text{adj}(A))\ge\text{rank}(A)+\text{rank}(\text{adj}(A))-n\implies$

$\text{rank}(O_n)\ge n-1+\text{rank}(\text{adj}(A))-n\implies\text{rank}(\text{adj}(A))\le 1$ .

Since $\text{rank}(\text{adj}(A))\ne 0$ it follows that : $\boxed{\ \text{rank}(\text{adj}(A))=1\ }$ .


algebra
polynomial
algorithm
greatest common divisor

1 and 3 are fairly trivial.

For (2), set $ x^n\minus{}1\equal{} (x\minus{}1)f(x)\plus{}r$ where $ f$ is a polinomial of degree $ n\minus{}1$ with integer coefficients. Then $ r$ is constant and setting $ x\equal{}1$ yields $ 0\equal{}r\plus{}0\implies r\equal{}0.$

Or simply $ a^{n}\minus{}1 \equal{} (a\minus{}1)(a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{}.....\plus{}1)$

[hide="1"]Since all divisors of a number are natural numbers and less than or equal to the original number, all divisors of 1 re less than or equal to 1, but there are no naturals less than 1, so all of the divisors of 1 are 1 and only 1. Therefore, a=1.[/hide]

[hide="3"]We use the Euclidian Algorithm, then get (1,1)=1.[/hide]

[quote="guile"]1.) show that if $ a|1$, then $ a \equal{} 1$
2.) prove that if $ a \neq 1$ then $ (a \minus{} 1)|(a^{n} \minus{} 1)$
3.) prove that $ (1,n) \equal{} 1$ for all $ n$[/quote]

(3) By the definition of the greatest common divisor, $ (1,n)|1$. But (1) implies that this yields $ (1,n)\equal{}1$ ;)


2-4: How many positive two-digit numbers have a perfect square as the product of their digits?

[b]I found 17 but my answer was wrong.[/b]

2-5: For what rational number c do the equations x³ + cx² + 3 = 0 and x² + cx + 1 = 0 have a common solution?

[b]Why is the answer -10/3?[/b]

[b]I just don't have any methods on getting the answer:[/b]

If n is a positive integer, then n! is the product of all integers from 1 to n inclusive. The number of times 5 occurs as a prime factor of 32! is
(A) 7    (B) 6    (C) 5    (D) 21   (E) 1

If 90! = (90)(89)(88) . . . (2)(1), then the exponent of the highest power of 2 which will divide 90! is
(A) 86    (B) 45    (C) 90    (D) 75    (E) 85

2-4: They counted 0 as a perfect square.

2-5: You can use the quadratic formula for the second equation to solve for x, then put that in the cubic. If you keep track of your plus/minus cases (do the plus and minus cases separately if need be), many terms will cancel.

For the second-last question, we need to find the number of times 5 divides the product. 5 divides each of 5, 10, 15, 20, and 30 once, and divides 25 twice. 5+2=7, so (A) is the answer. If you are not seeing intuitively why this is the solution, you can write out the consecutive integers in the product, and start dividing out 5's. It should make sense.

For the last question: first we divide out all the multiples of 2, then all the multiples of 2^2 = 4, then all the multiples of 2^3 = 8, etc. This is similar to the process above. There are 45 multiples of 2, 22 multiples of 4, etc. up to 1 multiple of 64. Add these together. We needn't consider multiples of 2^7 = 128 or any higher power of 2 since they are too big to be in the list.

[quote="d.lam.86"]2-5: For what rational number c do the equations x³ + cx² + 3 = 0 and x² + cx + 1 = 0 have a common solution?

[b]Why is the answer -10/3?[/b][/quote]

Multiply $x^2+cx+1=0$ by $x$ to get $x^3+cx^2+x=0$, but we know that $x^3+cx^2+3=0$, so $x=3$. Thus $3^2+3c+1=0\implies c=-\frac{10}{3}$. :)

[quote="d.lam.86"]Thanks for the help!

BTW, why is 0 a perfect square? I know 0 × 0 is 0 but isn't there an exception to 0 being a perfect square?  :oops:[/quote]I don't think there is any consensus in the mathematical community. If the question does not specify whether 0 is a perfect square when it makes a difference, then there should be two possible answers.

[quote="rem"]I have a cool solution for 2nd problem
 [hide]From 2nd equation, x^2 + cx = -1.Subst in 2nd equation to get  - x + 3 = 0. Then x = 3 and solving for c, get c = -10/3. :D[/hide][/quote]

isnt it the same as the other solution?

No, you use substitution instead of multiplication by x.


calculus
integration
conics
parabola
geometry
function
calculus computations

Suppose a parabola with the axis as the $ y$ axis, concave up and touches the graph $ y\equal{}1\minus{}|x|$. Find the equation of the parabola such that the area of the region surrounded by the parabola and the $ x$ axis is maximal.

So, we need to find a function $ f(x) \equal{} a x^2 \plus{} b x \plus{} c$ with the mentioned properties. Firstly, since the function it's supposed to touch is symmetric with respect to the y-axis as well as the parabola (with respect to it's central axis), we conclude that $ b\equal{}0$. This gives us $ f(x) \equal{} a x^2 \plus{} c$. Now, we know that solving the equations $ f(x) \equal{} 1\minus{}x$ or $ f(x) \equal{} 1\plus{}x$ should yield only one solution, so we can use that to determine the connection between the two parameters. More precisely, we have the solutions $ x_{1,2} \equal{} \frac{\minus{}1 \pm \sqrt{1\plus{}4a\minus{}4ac}}{2a}$ solving with $ 1\minus{}x$, and this gives us $ 1\plus{}4a\equal{}4ac \Rightarrow c \equal{} 1 \plus{} \frac{1}{4a}$. Now, our function becomes $ f(x) \equal{} a x^2 \plus{} \frac{1}{4a}\plus{} 1$. The function we need to maximize is represented by the area under the parabola, so we need it's roots for the integrals boundaries and these are $ x_{1,2} \equal{}  \pm \sqrt{\minus{}\frac{1}{a}\left(1\plus{}\frac{1}{4a}\right)}$ but since the surface is symmetrical, we can just look at one side (double, of course). This is $ I(a)\equal{}\int_0^{\sqrt{\minus{}\frac{1}{a}\left(1\plus{}\frac{1}{4a}\right)}}\left(a x^2 \plus{} \frac{1}{4a}\plus{} 1\right)\,dx$ and after some easy calculation, we get $ I(a)\equal{}\minus{}\frac{1}{12 a^2}\left(\minus{}1\minus{}4a\right)^{\frac{3}{2}}$. All that we need to do now is to solve $ I'(a)\equal{}0$, and for the derivation we have $ I'(a) \equal{} \minus{}\frac{\sqrt{\minus{}1\minus{}4a}(1\plus{}a)}{6a^3}$. The solutions are obvious, $ a_1 \equal{} \minus{}1$ and $ a_2 \equal{} \minus{}\frac{1}{4}$ but the second one gives $ 0$ (the minimum value - when the parabola has it's maximum on the x-axis). So, the final answer is $ f(x)\equal{}\frac{3}{4}\minus{}x^2$ and it encloses the surface of $ I(\minus{}1) \equal{} \frac{\sqrt{3}}{4}$.

Hi [b]Kunny![/b]
[u][b] My solution:[/b][/u]
Let $y:=f(x)=ax^2+bx+c; (a \neq 0)$ .
+ Because "[i]Suppose a parabola with the axis as the  $y$- axis[/i]" $\Rightarrow f(\pm x_0)=0; \overset{Viete}{\rightarrow}
0=(x_0)+(-x_0)=-\frac{b}{a}\Rightarrow \boxed{b=0}$
$\Rightarrow y=a(x^2-x_0^2);(P);$
+ Because $(P)$  "[i]concave up and touches the graph [/i]$y=1-|x| ;(d)$
$\implies $
$\begin{cases}
 & \math{} ax^2-ax_0^2=1-|x|=1-x \\
 & \math{} 2ax= -1 \\
 & \math{} y''= 2a<0 \\
\end{cases}$      $\Rightarrow \frac{1+4a}{4a^2} = -x_0^2\Rightarrow \boxed{y=ax^2+\frac{1+4a}{4a^2}}$;
+ $S=2\int_{0}^{x_0}(ax^2+\frac{1+4a}{4a^2})\ dx \Rightarrow \frac{S}{2}=[\frac{ax^3}{3}+(\frac{1+4a}{4a^2})x]\mid_{0}^{x_0} $
$\Rightarrow \frac{S}{2}=\frac{ax_0^3}{3}+(\frac{1+4a}{4a^2})x_0 \Rightarrow \frac{S}{2}=(\frac{a-3}{3})x_0^3= -(\frac{\sqrt{-1-4a}}{2a})^3=g(a)$
Where $g(a)= -(\frac{\sqrt{-1-4a}}{2a})^3 \implies \max g(a) =  \max S = \frac{8}{3}.(\frac{\sqrt{3}}{2})^3 $
$\implies a =-1 \implies \boxed{y=-x^2+\frac{3}{4}}$.


i know the formulas for distinguisbible marbles and boxes (boxes ^ marbles), indistinguisible balls and boxes, indistinguisible balls but distinguisible boxes, but what is the formula for distinuishible balls and indistinguisible boxes?

[quote="akhil0422"]If you have $ m$ distinguishable marbles, and $ b$ indistinguishable boxes, isn't it just $ \dbinom{m}{b}$?

I don't really understand what you're saying: indistinguishability and distinguishability aren't really my strong point though.[/quote]

That is wrong.  If $ n$ is the number of balls and $ k$ is the number of boxes, then the formula is $ \binom{n\plus{}k\minus{}1}{k}$

that is also incorrect. that is the formula for indistinguisible balls and distinguishable boxes. The formula i am looking for is distinguisible balls and indistinguishable boxes.

Hmm..I'm pretty sure that's right..ask math154.  Also, refer to this topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244341

And click math154's link.

Your answer is incorrect. for instance, if there were 4 distinguishible balls and 2 indistinguishable boxes, there would be 8 different ways to place the balls into the boxes. 4-0 (1 way) 3-1 (4 ways) 2-2 (3 ways). This does not fit the formula of (n+k-1 nCr k-1). HOwever, if there were 4 indistinguishable balls and 2 distinguishable boxes, there would be 4 ways. (4+2-1 nCr 2-1)

For a full treatment, read all of the cases [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]here[/url]. As you can see, you have to specify whether duplicates are allowed or not.

@math154 it would be a good idea to put that on the wiki titled "Balls and Urns" so people don't always ask that question.


function
linear algebra
matrix
limit
inequalities
topology
linear algebra unsolved

Let $ M_{nxn}$ be the space of real $ nxn$ matrices . Regard it as a metric space with the distance function : $ d(A,B) \equal{} \sum_{i,j \equal{} 1}^{n}{|a_{ij} \minus{} b_{ij}|}$ ($ A \equal{} (a_{ij}),B \equal{} (b_{ij}))$)
Prove that the set of nilpotent matrices in $ M_{nxn}$ is a closed set .

The function $ p: \mathbb{R}\times\mathbb{R}^{n^2}\rightarrow\mathbb{R}\,,\,p(t,X)\equal{}\det(tI_n\minus{}X)$ is continuous, since its calculation consists only of elementary operations of the input variables. Remember, that a real matrix $ X$ is nilpotent, iff $ p(t,X)\equal{}t^n$ holds. Now, let $ X_k$ be a convergent series of nilpotent matrices with limit $ X$ with regard to the $ d$-distance. Then for each matrix entry we have $ X_{ij}\equal{}\lim_{k\rightarrow\infty}(X_k)_{ij}\Longrightarrow p(t,X)\equal{}\lim_{k\rightarrow\infty} p(t,X_k)\equal{}\lim_{k\rightarrow\infty}t^n\equal{}t^n$ and $ X$ must be nilpotent, too, i.e. the set of nilpotent matrices is closed.

P.S. A brute force solution might use the following inequality for matrices $ A,B$ and $ a\equal{}\max\{|A_{ij}|\},b\equal{}\max\{|B_{ij}|\}$

$ \begin{array}{l}\det (A\plus{}B)\minus{}\det (A)| \equal{} \left|\sum_{\sigma\in S_n}(\minus{}1)^{sgn(\sigma )}\sum_{I\subset\{1,\ldots,n\}}\prod_{i\in I}A_{i,\sigma(i)}\prod_{j\not\in I}B_{j,\sigma(j)}\right|\leq n![(a\plus{}b)^n\minus{}a^n]\end{array}$ and note, that $ d(A\plus{}B,A)\equal{}d(B,0)\leq n^2b$, which can be used to show that the non-nilpotent matrices form an open set.


Theo website của ĐCSVN:
         .   Lá cờ đỏ sao vàng lần đầu tiên xuất hiện trong khởi nghĩa Nam Kỳ , ngày 23/11/1940 .

          .  Tác giả là Nguyễn Hữu Tiến , một Đảng viên Cộng sản . Ông bị xử bắn ngày 28/8/1941 , cùng với nhiều chiến sỹ trong Khởi nghĩa Nam Kỳ :  Nguyễn Thị Minh Khai , Võ văn Tần , Hà Huy Tập ....

       Tôi không biết từ nguồn thông tin nào , mà analysis và tom-analog khẳng định rằng là cờ đỏ sao vàng thể hiện sự phụ thuộc vô điều kiện của VNDCCH với Trung Quốc , hay "lãnh tụ Bắc Việt dẫm phải cứt Mao Trạch Đông "  bởi vì lá cờ đỏ sao 5 sao của CHNDTrung Hoa chỉ chính thức được công nhận từ năm 1950 , sau khi nôi chiến của họ kết thúc . Trong khi đó Lá Cờ đỏ sao vàng của VNDCCH trở thành Quốc Kỳ từ tháng 5/1945 và đươc Hiến pháp công nhận năm 1946 .   Nếu ai đó không tin những thông tin do ĐCSVN cung cấp  , có thể dễ dàng thấy rằng Lá cờ đỏ Sao vàng đã được sử dụng , trễ nhất là trong Cách mạng Tháng Tám 1945 , qua những hình ảnh tài liệu .
         Vì vậy , xét về mặt "bản quyền" , VNDCCH đã nhanh chân hơn Trung Quốc rất nhiều năm . Không thể nói VNDCCH bắt chước Trung Quốc  . Câu hỏi cần đặt ra là  Trung Quốc có ý đồ gì không khi lấy là cờ có nhiều sao bao quanh như vậy ...
Thực sự , tôi cũng không tìm được chứng cứ  nào cho thấy Việt nam DCCH chịu sự chi phối của Trung Quốc hay Liên Xô trong giai đoạn này, nếu xét tương quan lực lượng và tình hình  lúc đó
       "Vào giữa thập niên 1940, thế lực chính trị của Liên Sô còn yếu. Lực lượng quân sự của Liên Sô đang dồn vào mặt trận Đông Âu, không có chút ảnh hưởng nào tại Đông Dương ngoài việc huấn luyện ý thức hệ cho các đảng viên đảng Cộng Sản Đông Dương. Trung Hoa còn thuộc chế độ Cộng Hòa của Trung Hoa Quốc Dân Đảng, từ Tôn Dật Tiên chuyển xuống Tưởng Giới Thạch, đang vất vả chống Nhật ngay trên lãnh thổ của họ và phải nhờ Hoa Kỳ trấn giữ các căn cứ chiến lược vùng đông nam Hoa Lục. Lực lượng Trung Cộng của Mao Trạch Đông chỉ mới phôi thai ở miệt tây bắc Trung Quốc, lúc ấy chưa hề bén mảng đến miền nam; mãi đến năm 1949, dành được Hoa Lục, mới bắt đầu ảnh hưởng vào vùng Đông Dương "

1. Tôi chỉ nêu lên câu hỏi cho sự tương đồng của 2 lá cờ Trung Quốc và Việt Nam. Tôi chưa hề khẳng định điều gì.

2. Những sử liệu từ ĐCSVN tôi đều đánh dấu chấm hỏi thật lớn.

3. Giả sử lá cờ đỏ sao vàng xuất hiện vào 1945. Điều này không thể phủ nhận sự tương đồng giữa hai lá cờ nói trên. Về mặt bản quyền (trước 4 năm), vẫn không thể suy luận, "Không thể nói VNDCCH bắt chước Trung Quốc."

4. Lá cờ đỏ sao vàng 1945 vẽ khác với lá cờ đỏ sao vàng hiện tại. Lá cờ hiện tại được ĐCSVN chính thức sử dụng vào năm 1955 (chỉ ở miền Bắc)(?) Lá cờ đỏ 5 sao vàng của ĐCSTQ được kéo lên năm 1949, vậy trước đó ĐCSTQ dùng lá cờ nào?

5. Nếu phân tích một cách vô tư, chúng ta có thể đưa ra 3 giả thuyết.
(a) Sự tương đồng của 2 lá cờ trên là sự ngẫu nhiên.
(b) Có sự ảnh hưởng lẫn nhau giữa 2 lá cờ; VN dựa theo TQ hoặc ngược lại.
(c) Vì hoàn cảnh địa lý, văn hóa, chính trị v.v...
Những chuyện thuần ngẫu nhiên không thể nói là không có, nhưng tỉ lệ có 2 lá cờ tương đồng như vậy rất thấp. Tôi cho rằng giả thuyết (a) không mấy chính xác. TQ cho dù có ý đồ gì với VN, cũng không cần thiết tạo ra 1 lá cờ tương đồng như vậy. Một đại cường quốc như TQ dĩ nhiên chỉ xem VN như một tiểu nhược quốc. Cho nên lá cờ tương đồng vì TQ có ý đồ với VN là không hợp lý. CSVN chịu ảnh hưởng của CSTQ, điều này hợp lý hơn. Có thể vì hoàn cảnh địa lý, văn hóa, chính trị (c)? Giả thuyết này cũng rất hợp lý, nhất là về khối cộng sản với cơn huyễn mộng thế giới đại đồng. Dĩ nhiên trong đó đàn em phải theo gương đàn anh. Có thể những lá cờ tương đồng này chỉ là kết quả đại đồng hóa của CS quốc tế?

Nếu muốn đào sâu vào vấn đề này để có 1 câu trả lời chính xác. Có lẽ phải tìm hiểu cặn kẽ lịch sử của 2 lá cờ, người vẽ, các bản vẽ, v.v... Tôi thật sự không có nhiều thời gian để làm chuyện này.

[img]http://i9.tinypic.com/2el7c5h.gif[/img]
1 bản vẽ khác (trong trên 3000 bản vẽ đã được đề xuất)


inequalities
function
integration
real analysis
real analysis unsolved

To be sincere, I have no idea how to prove the following: suppose that $\infty>p\geq 1$ and that $f,g$ are two functions in the Schwartz class of $R^{2n}$ such that $f+v\cdot\nabla_{x}(f)=g$ (here $f=f(x,v)$;, the same for $g$). Then $\parallel f\parallel _{L^{p}}\leq \parallel g\parallel _{L^{p}}$.

for $p=2$, maybe that Fourier transform with respect to $x$ works :
$\hat{g}(\xi, v) = \hat{f}(\xi, v) (1+iv.\xi)$
and Plancherel theorem shows that
$\|g\|_{L^{2}(\mathbb{R}^{2n})}^{2}= \int_{v \in \mathbb{R}^{n}}\|g(.,v)\|_{L^{2}(dx)}^{2}dv= \int_{v \in \mathbb{R}^{n}}\| \hat{g}(., v) \|_{L^{2}(d \xi)}^{2}dv \geq \int_{v \in \mathbb{R}^{n}}\| \hat{f}(., v) \|_{L^{2}(d \xi)}^{2}dv = \|f\|_{L^{2}(\mathbb{R}^{2n})}^{2}$
because $\| \hat{g}(., v) \|_{L^{2}(d \xi)}^{2}\geq \| \hat{f}(., v) \|_{L^{2}(d \xi)}^{2}$ since $|\hat{g}(\xi, v)|^{2}= |\hat{f}(\xi, v) |^{2}(1+(v.\xi)^{2})$
but I don't know what is happening for $p \neq 2$ ?!?

PS: what is the kinetic interpretation of this equation ?

[quote="alekk"]
but I don't know what is happening for $p \neq 2$ ?!?
[/quote]
Actually, alekk, you do :). There is one trivial observation you might not notice: if $|f|$ attains its maximum somewhere, then at this point $|g|\ge |f|$, so we have the $L^\infty$ inequality in addition to the $L^{2}$ one. Now the key words are complex interpolation and duality.

Fedja, did you mean Riesz - Thorin theorem? Could you explain a little bit more please.

Yes, I meant Riesz-Thorin, though it is a [b]huge[/b] overkill. I just had some reason to think that alekk would know a few things about Fourier multipliers by now, whence the remark. Of course, the problem is much simpler than that. First of all, we do not need $v$ as an argument: the inequality is true for each $v$ separately. Secondly, we do not need $\mathbb R^{n}$: when $v$ is fixed, the equation is completely determined by the values of $f$ and $g$ on lines in direction $v$ and it is enough to show that the inequality holds on each such line. Third, by scaling $f(x)\mapsto f(\alpha x)$, we see that we can assume that $|v|=1$. Thus, we have the following problem on the line: $f+f'=g$; prove that $\|f\|_{p}\le\|g\|_{p}$. But this equation allows to express $f$ in terms of $g$ quite easily: $f=g*h$ where $h(x)=e^{-x}\chi_{[0,+\infty)}(x)$. Since $\|h\|_{1}=1$, we get the desired conclusion immediately.


calculus
integration
logarithms
calculus computations

Evaluate $ \int_0^{\sqrt{2}\minus{}1} \frac{1\plus{}x^2}{1\minus{}x^2}\ln \left(\frac{1\plus{}x}{1\minus{}x}\right)\ dx$.

[hide]making the change of variables $ \frac{1\minus{}x}{1\plus{}x} \equal{} u$ we get

$ I \equal{} \int_1^{\sqrt{2}\minus{}1}\frac{\ln u}{u}\,du \minus{} 2 \int_1^{\sqrt{2}\minus{}1} \frac{\ln u}{(u\plus{}1)^2}\,du$.

Each of these integrals is easily evaluted and the final answer is

$ I \equal{} \frac12 \ln^2(\sqrt{2}\minus{}1) \minus{} \sqrt{2}(\sqrt{2}\minus{}1) \ln(\sqrt{2}\minus{}1) \minus{} \ln 2$.[/hide]

i got the same answer but i figured it was wrong since its so ugly =/

That's are correct, but what I wanted to post was $ \int_0^{\sqrt{2}\minus{}1} \frac{(1\plus{}x^2)^2}{1\minus{}x^2}\ln \left(\frac{1\plus{}x}{1\minus{}x}\right)\ dx$.


[quote="every"]Dear all member,
I want to see that problem in Mathlinks but I can't find links.Help me,please.[/quote]
You are Vietnamese,find it in http://diendantoanhoc.net/forum/
or                                        http://forum.mathscope.org/

But I want to see that problem in Mathlinks but I can't find links.Help me,please.


The sequence of real numbers $ x_1, x_2, x_3,...$ satisfies $ lim \to \infty (x_{2n} \plus{} x_{2n\plus{}1}) \equal{} 315$ and
$ lim \to \infty (x_{2n} \plus{} x_{2n\minus{}1}) \equal{} 2003$. Evaluate        $ lim \to \infty(x_{2n}/x_{2n\plus{}1})$

First of all, this problem does not belong in this forum, as it requires limits.

$ \text{\LaTeX}$ correction:

$ \lim_{n\rightarrow\infty}(x_{2n}+x_{2n+1})=315$
$ \lim_{n\rightarrow\infty}(x_{2n}+x_{2n-1})=2003$

Well, limits can be considered the foundations of calculus, but they can also be considered advanced algebra. Besides, limits were taught in Art of Problem Solving Volume 2, and thats supposed to be a precalculus text.

[quote="MathTwo"]Besides, limits were taught in Art of Problem Solving Volume 2, and thats supposed to be a precalculus text.[/quote]Really? How did the book define limits though? Did it use the sophisticated epsilon-delta definition (which I have yet to understand fully, especially proofs that require it)? I am asking because I don't have the book... in fact I don't have any reference book (i just learn problem solving techniques from solutions to the problems on AoPS), but whatever...

Guys!
Anyone havin slightest idea about the date of RMO 2009 in West Bengal :?:Which is the best reference book for RMO?
Is Hall and Knight a good book?Is Challenge and Thrill of Pre-College Mathematics good?

[quote="MathTwo"]I'm interested, where did you get this problem?[/quote]

hmmt shawty


logarithms
quadratics
algebra
quadratic formula

anyone know how the heck I can get the answer to $ 3^{x^2} \equal{} 6^{x \plus{} 4}$? u can't use conventional logarithms cuz x is squared............

$ 3^{x^2}\equal{}6^{x\plus{}4}$

[code]$3^{x^2}=6^{x+4}$[/code]

[hide="hint"]Try taking a log (any base will do) and using a few properties of logs.[/hide]

All I can say is that the answer I'm getting is really messy:

[hide]
$ 3^{x^2}\equal{}6^{x\plus{}4}$
$ \equal{}$
$ \log_{6}{3^{x^2}} \equal{} x\plus{}4$
$ \equal{}$
$ x^2\log_{6}{3} \equal{} x\plus{}4$
$ \equal{}$
$ (\log_{6}{3})x^2\minus{}x\minus{}4 \equal{} 0$
which is a lovely quadratic equation...
and then the quadratic formula, hence the "messy"
someone else can do that
[/hide]

i believe that that would probably be the best way to that problem, any complaints, counterarguments, disproofs, opposition, better methods...go for it.

[hide]Using the quadratic formula from ajai's partial solution, we find:
\[ x \equal{} \frac {1 \pm \sqrt {1 \plus{} 16\log_6 3}}{2\cdot \log_6 3}
\][/hide]


The $ 8\times 18$ rectangle $ ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $ y$?
[asy] unitsize(2mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$D$",(0,4),NW); label("$C$",(18,4),NE); label("$B$",(18,-4),SE); label("$A$",(0,-4),SW); label("$y$",(9,1)); [/asy]$ \textbf{(A) } 6\qquad \textbf{(B) } 7\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 10$

This question took me 30 minutes on the AMC.  Maybe if they said one of them went under a sliding transformation that would've helped me.  Oh well I guessed it right.

This also was on the ten... if you didn't realize that six was the only thing that worked, you could use a system of equations... I don't remember the problem at all so I can't post my proper solution. Can you put in the diagram chess?

I believe it looked like this.

Basically, I just saw that those 3 lengths must add up to 18 and all must equal $y$ for it to be a square, so $y = 18/3 = 6$.

(Let $X$ be that point between $A$ and $B$.)

Well, we now that its area must be $8\cdot18=12^2.$  So the side length must be 12.  We already know that $AD=8$ and $AB=18.$  Since the side length is 12, $AX=12$ and $BX=6.$  But then $BX$ and $y$ will be the same segment afterwards, so the answer is 6.

Yeah I used logic when I ran through it the first time and then to check I just did 8+4=18-y. Y=6. But that's not algebra it's logic.

I just slid it around in my head in order to get a square.   3y=18.....

Er...you guys are joking, right? This problem is trivial...

Slide the hexagon on the right so that it's on top of the first one, like so:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=561[/img]

OK the pic is off, but that's a square. :D Thus the side of the original rectangle is divided into three y's, so 3y=18. 18/3=6. Not too tricky.

Yeah see I never thought of sliding it.  *Shrugs*

Yeah I saw the sliding part but for some reason I divided 18 by 2 instead of 12 by 2 and got 9 instead of 6.. yeah, don't ask :P

Wasn't that bad if you saw how to slide it, heh.

[hide="Answer"]$8\times 18$ has area of $144$, so the square must be $12\times 12$. Therefore, if we label the points where the cut intersects the edges as $E$ and $F$, with $E$ on $AB$ and $F$ on $CD$, then $AE=CF=12$, and $BE=DF=6$. Therefore, since the cuts are perpendicular, $y=6\Rightarrow \boxed{A}$.[/hide]


A rectangular array of chairs is an arrangement of the chairs in rows and columns such that each row contains the same number of chairs as every other row and each column contains the same number of chairs as every other column.  If there must be at least two chairs in every row and column and all of the chairs in the room must be included, how many arrays are possible in a classroom containing $ 36$ chairs? Note that $ 12$ rows of $ 3$ chairs is different from $ 3$ rows of $ 12$ chairs.

We can list the pairs of numbers that multiply to $ 36$ with each pair having both numbers $ \ge2$.
$ \{2,28\},\{3,12\},\{4,9\},\{6,6\},\{9,4\},\{12,3\},\{18,2\}$ so there are $ \boxed{7}$ ways for the chairs to be arranged.

The first pair is actually 2,18 not 2,28


Post this in the Computer Science and Informatics section of the forum under Arts and Sciences.

lolz...

Niether did I. Is it just my computer or did he post a bunch of random characters?

I think you should post in the computers section.
OT: I think she failed the latex?

Look people, stop spamming. I think that he/she made clear that it was C++.

I tried compiling it, but my compiler lists several errors...

"
expected constructor, destructor, or type conversion before '(' token
expected unqualified-id before '{' token
"

There's no need to compile it... It's pretty easy to see that it's just a ton of random characters...

lol thats true, nothing is correct codding besides the stdio.....

hm technically its correct c (NOTE: not c++)
the main does not have to have a return type in c
the "char c;" can indeed be placed after the main
and variables can indeed be named "_"

its composed of several recursive main calls, many tertiary statements (?:), and a return statement enclosing it all
it is doing something with the string at the middle (there is actually a string in the middle) and at the bottom (decrypting?) and then prints
"On the twelfth day of Christmas my true love gave to me
twelve drummers drumming, eleven pipers piping, ten lords a-leaping,
nine ladies dancing, eight maids a-milking, seven swans a-swimming,
six geese a-laying, five gold rings;
four calling birds, three french hens, two turtle doves
and a partridge in a pear tree." (and all the other verses in that song)
even though its pretty impressive, i think the original post was still spam

This is the classic example of Obfuscated code, if you want to learn more about it, here is the wikipedia page: http://en.wikipedia.org/wiki/Obfuscated_code

If you want to see it reverse engineered into understandable code, here is the link: http://research.microsoft.com/en-us/um/people/tball/papers/XmasGift/

This topic should be moved to the Computer Science and Informatics section, and I think the original poster is a troll.

he/she just copied and pasted

on the other hand, that's pretty cool stuff

You mean the code?
As Eulers_Apprentice mentioned, it's obfuscated code for the twelve days of Christmas, and the link he gave says:
"Although unintelligible at first glance, it is a legal C program that when compiled and run will generate the 12 verses of The 12 Days of Christmas. It contains all the strings required for the poem in an encoded form inlined in the code. The code iterates through the 12 days displaying what it needs to."


Write a recursive equation that is equivalent to $f(x)=a(b^x)$ for $b>0$, $b\neq1$, and $x$ a nonnegative integer.

[quote="t0rajir0u"][hide="Solution"] It's trivial to see $f(k+1) = b f(k)$. [/hide] More interesting:  Write a recursive function equivalent to $f(k) = a^k + b^k$.  :)[/quote]

$a_{n+2}-(a+b)*a_{n+1}+(ab)*a_n=0$

with $a_0=2$ and $a_1=a+b$


Let $ n$ be integer.Prove that if expresion $ 2 \plus{} 2\sqrt {28n^2 \plus{} 1}$ is integer,then it's perfect square.

[i]I don't have the solution of this,It may be any typo on it.[/i]

It's in PEN.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150375[/url]


function
arithmetic sequence
real analysis
real analysis solved

Let $f : R \rightarrow R$ be continuous such that it attains both positive and negative values$\Gamma$and let$ n \geq 2$ be an integer. Show that there exists a strictly increasing arithmetic sequence
$a_1 <...< a_n$ such that$ f(a_1) +...+f(a_n) =0$

Let $a,b$ be reals such that $f(a) > 0$ and $f(b) < 0$.
Consider the function $g(t,x) = f(t+x) + f(t+2x) +...+f(t+nx)$.
Let $h(x) = g(a,x)$. Clearly $g$ is continuous. If there is $x>0$ such that $h(x) = 0$ we are done. Thus assume there is no such $x$. Then, using continuity, $h$ has the same sign on $[0; + \infty [$. Since it has limit $nf(a)$ when $x$ tends to 0, we deduce that $g(a,x) > 0$ for all $x > 0$.
In the same way, if there is no $x > 0$ such that $k(x) = g(b,x) = 0$ (otherwise we are done) then $g(b,x) < 0$ for all $x > 0$.
Now choose $x = 1$.
From above $g(t,1)$ is continuous function with respect to $t$, which is negative for $t=b$ and positive for $t=a$. The intermediate value theorem ensures that there exists $c$ such that $g(c,1) = 0$ and we are done.

Pierre.