geometry
geometric transformation
rotation

what do you mean by maps itself? do you mean like this is a regular polygon and if you rotate it 13 degrees it appears the same? What I've got now isn't very good at the moment:

[hide]
Finding sum of interior angles
(n-2)*180
(n-2)*180=13/n
n(n^2 - 2n) = 13 / 180
...............
180 n^3 = 13 + 360 n^2
[/hide]

But I don't know how to get further than this. I havn't done square root simplification, so if someone could show me how to do it, that'll be great :P

P.s. to the admins, check where my sig goes to first. You'll end up hot and bothered  :lol:

[quote="Detour"]
But I don't know how to get further than this. I havn't done square root simplification, so if someone could show me how to do it, that'll be great :P[/quote]
So If you have like $\sqrt{8}$ you can get it to $2\sqrt{2}$ because $\sqrt{8}=\sqrt{4}\times\sqrt{2}=2\times\sqrt{2}=2\sqrt{2}$
Square roots mulitply like regular numbers except there's a square root symbol. Find perfect square factors in a square root and simplify.
$\sqrt{162}=\sqrt{81}\times\sqrt{2}=9\times\sqrt{2}=9\sqrt{2}$

[quote="Treething"]A polygon, when rotated about its center, maps to itself if rotated 13 degrees. Prove that if it is rotated 69 degrees, it maps to itself.[/quote][hide] If a figure rotated $13^\circ$ maps to itself, from this position, we can rotate it again. This rotates it $26^\circ$, and it still maps to itself. Then we can continue to rotate it at $13^\circ$ each, and it will map to itself each time. [/hide]
[quote="Detour"][hide]
Finding sum of interior angles
(n-2)*180
(n-2)*180=13/n
n(n^2 - 2n) = 13 / 180
...............
180 n^3 = 13 + 360 n^2
[/hide][/quote]
I'm not sure exactly what you're doing, but there's an error going from the second to third line.

[quote="lotrgreengrapes7926"]Then we can continue to rotate it at $13^\circ$ each, and it will map to itself each time.[/quote]
So what you're saying is that rotating by $13^\circ$ or any mulitple of $13^\circ$ will make it map to itself? Well $69^\circ$ is not a multiple of $13^\circ$

[quote="Detour"]Finding sum of interior angles
(n-2)*180
(n-2)*180=13/n
n(n^2 - 2n) = 13 / 180
...............
180 n^3 = 13 + 360 n^2[/quote]

If the 2nd line is $180(n-2)=\frac{13}{n}$, then the 3rd line should be $n(\boxed{n-2})=\frac{13}{180}$, not ${n(\boxed{n^{2}-2n})=\frac{13}{180}}$

[quote="Detour"]bah, my bad :p
I'm trying to find out sum of interior angles so I can find out the number of edges, and then from there prove that it maps to itself at 69 degrees.

so

[hide]
n(n-2) = 13 / 180
n^2 - 2n = 13 / 180
n^2 = 13/180 + 2n

then what? I thought of making (13/180 + 2n) all square rooted..
then substituting.

Sorry, I'm terrible at maths compared to you guys :p
[/hide]

Oh, and I've done square root simplification for integers, but not with pronumerals (such as n).[/quote]
There's really no way to simplify $\sqrt{2n+\frac{13}{180}}$.
But you can solve $n^{2}= 13/180+2n$ as a quadratic.
I still have no idea what you're doing.  :roll: Oh, well.

well you can rotate the polygon by $13^{o}$ $33$ times to get $429^{o}$ and it will be the same. but if you rotate anything by $360^{o}$, it comes back to its initial state. $429-360=69$..

a similar problem is to construct a $1^{o}$ angle with just a $19^{o}$ angle.

[quote="Detour"]bah, my bad :p
I'm trying to find out sum of interior angles so I can find out the number of edges, and then from there prove that it maps to itself at 69 degrees.

so

[hide]
n(n-2) = 13 / 180
n^2 - 2n = 13 / 180
n^2 = 13/180 + 2n

then what? I thought of making (13/180 + 2n) all square rooted..
then substituting.

Sorry, I'm terrible at maths compared to you guys :p
[/hide]

Oh, and I've done square root simplification for integers, but not with pronumerals (such as n).[/quote]

Well you can't really find the number of sides because it does not say the polygon has to ge a regular polygon so using the formula for the sum of interior angles is not really a good idea.

minsoens, nice proof! Woot


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Does $R^n$ have an additive measurable subgroup of positive measure?

No, I don't think there are such proper subgroups. The idea is that if $A\subset\mathbb R^n$ is measurable and has positive measure, then $A+A$ contains a ball, so in our case the group would actuall have to coincide with the whole $\mathbb R^n$.

Suppose WLOG that the $A$ we're talking about is situated in the unit ball $\mathcal B$, and that it has a very large measure, covering almost the entire ball (i.e. if $m$ is the measure of the unit ball, then we take an $\varepsilon>0$, small enough for our purposes, and assume that $A\subset\mathcal B$ has measure $\ge m-\varepsilon$).

If we assume there are points $x_n$ arbitrarily close to $0$ which do not belong to $A+A$, then $A$ must be a subset of $\mathcal B-(x_n-A)$, for all $n$. If $n$ is large enough, however (and so $x_n$ is close enough to the origin), $x_n-A$ will be almost entirely situated in $\mathcal B$, so for some very small $\varepsilon'$ we will have two disjoint sets, one with measure $\ge m-\varepsilon$ (this is $A$), and one with measure $\ge m-\varepsilon-\varepsilon'$ (this is the part of $x_n-A$ that's situated in $\mathcal B$), both inside $\mathcal B$. If the $\varepsilon$ we chose at the beginning is small enough, and $\varepsilon'$ is also small enough, this is impossible.

You, that's what I also did. I find it very very nice this problem. Here is another one: find all closed subgroups in $R^n$.

Yes, this one is also nice. Where do you find them, man? :)

I hope the following is correct (it seems a bit "shaky" to me :)):

Let $0\le s\le n$ be the maximal positive integer with the following property: we can find a sequence $(u_n)_n$ of elements of our group $G\le \mathbb R^n$ with $\|u_n\|\to 0$ s.t. for all $N\in\mathbb N$ the vectors $(u_n)_{n\ge N}$ span an $s$-dimensional subspace of $\mathbb R^n$. In this case, this $s$-dimensional subspace $\mathcal S$ will have to be the same for large enough $N$ and for all such sequences. It follows easily from here that $\mathcal S$ is included in $G$.

Now let $\mathcal T$ be a complement of $\mathcal S$ in $\mathbb R^n$, and let $G_{\mathcal T}$ be the additive group we get by projecting $G$ onto $\mathcal T$. It's clear from the maximality of $s=\dim\mathcal S$ that $G_{\mathcal T}$ must be a discrete subgroup of $\mathbb R^{n-s}$, which means that it's a lattice spanned by at most $n-s$ vectors (this has been discussed before on the forum; there are two threads, and I believe I posted a proof in one of them, started by dickchimney).

It follows from all of this that $G$ is the direct sum of $0\le s\le n$ copies of $\mathbb R$ and $0\le t\le n$ copies of $\mathbb Z$, with $s+t\le n$.

Is it even remotely Ok? :)

Answer to first question: there are magnificient problems given at oral ENS and X for long time and the treasure of these problems includes this one also. To second question: it's good!!!

for the first one, one can use the fact that if $A$ has strictly positive measure, then $A-A$ contains a neighboorhood of $0$.

yes, I admit I haven't read his solution  :blush:  I thought it was something different because it seems too long :)


algorithm
number theory
greatest common divisor
Euclidean algorithm

How can I solve Diophantine Equations using the Euclidean Algorithm?

hello, see here:
http://mathworld.wolfram.com/DiophantineEquation.html
Sonnhard.

It's easy with linear Diophantine equations (of form $ Ax\plus{}By\equal{}C$).
You use the extended Euclidean algorithm on $ A$ and $ B$ to find a solution $ (x_0,y_0)$ such that $ Ax_0\plus{}By_0\equal{}\text{gcd}(A,B)$.
Then you multiply both sides by $ C/\text{gcd}(A,B)$ to get one solution to the equation (call it $ (x_1,y_1)$).
Now all the solutions are of the form:
$ x\equal{}x_1\plus{}\frac{B}{\text{gcd}(A,B)}k$
$ y\equal{}y_1\minus{}\frac{A}{\text{gcd}(A,B)}k$

If I remember correctly.

Other Diophantine equations require more guesswork (look into Pell's equation, and other forms like that).


ratio
inequalities
arithmetic sequence
algebra
partial fractions
rearrangement inequality

If $\{a_{n}\}_{n\geq0}$ is an arithmetic sequence where the first term ant it's ratio are pozitive, then $\frac{1}{a_{1}a_{2}}+\frac{1}{a_{3}a_{4}}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_{0}a_{2n}}$ for any $n\in\mathbb{N^{*}}$.

[quote]
...is an arithmetic sequence where the first term ant it's ratio ...
[/quote]
Do you mean geometric sequence?

He was probably confused, as it said ratio
maybe you should change it to difference?

[quote="seamusoboyle"]Beuatiful question! Ive got a nice solution with partial fractions, telescoping and rearrangement inequality, but its a bit late in the night over here and im knackered, so im going to hit the hay. Ill post it tomorrow if someone dosnt get to it before me.[/quote]

Hi seamusoboyle

Your method is right.  And I leave it to you to post the solution.

We'll prove that

$\sum^{2n-1}_{i=0} \frac {1}{a_ia_{i+1}}=\frac {2n}{a_0a_{2n}}$

Let $a=a_0$, $d$ be the common difference, and $n$ be the number of terms.  We have

$\frac {1}{(a+id)(a+(i+1)d)}=\frac {1}{d(a+id)}-\frac {1}{d(a+(i+1)d)}$

Sum this (it telescopes).  We are left to prove that

$\frac {1}{da}-\frac {1}{d(a+2nd)}=\frac {2n}{a(a+2nd)}$

Multiply through by $da(a+2nd)$  We are left to prove that

$(a+2nd)-a=2nd$

which is obvious.

---

Now, this is an identity, and it shows that the original inequality is true iff

$\frac {1}{a_0a_1}>\frac {n}{a_0a_{2n}}$, which is equivalent to

$\frac {1}{a+d}>\frac {n}{a+2nd}$ which reduces to $a+2nd>na+nd\implies a+nd>na$

So, say that that last inequality does not hold (say, for $a=10$, $n=3$, $d=1$)

Then the LHS is $5/176$, but the RHS is $3/160$, which shows that the problem is false.

blahblahblah, your sum has 2n terms, the sum in the original problem has n terms.

However, if you split your sum into 2 parts, you can easily prove the problem is actually correct.

So can anyone post a solution? I solved it fot 3-4 times but it's not working. It can be solved by induction.

Thanks for the observation, fuzzylogic - I can't believe I missed that the first time around :).  Slizzel, I have done almost all of the work towards a solution - see if you can finish it from my first identity.

[hide]
$\frac {1}{a_ia_{i+1}}>\frac {1}{a_{i+1}a_{i+2}}$
[/hide]

Blahblahblah, unfortunately that wont work, it was the first one i tried...
Here we go. Let the first term be $a$ and the difference $d$

Notice that:

$\frac{1}{(a_n)(a_{n+1})}= \frac{1}{(a+nd)(a+(n+1)d)}=\frac{1}{d(a_n)}-\frac{1}{d(a_{n+1})}$

When you split it up into partial fractions.

Now notice when you add up:

$\frac{1}{a_0a_1}+\frac{1}{a_1a_2}+\frac{1}{a_2a_3} \cdots \frac{1}{a_{2n-1}a_{2n}}$

It will all telescope down to $\frac{1}{da_0}-\frac{1}{da_{2n}}=\frac{2dn}{a_0a_{2n}}$

Were going to replace all the terms that we dont want in this sum with the ones already there, and by the rearrangement inequality the sum will be less.

This gives us the desired result

Hm why won't that work?

the sum in the given inequality is the "smaller half" of $\sum^{2n-1}_{i=0} \frac {1}{a_ia_{i+1}}$, therefore its less than half of the RHS

[quote="seamusoboyle"]Blahblahblah, unfortunately that wont work, it was the first one i tried...[/quote]

Then you did it wrong.  Take my identity, number the terms from 1 to 2n, and note that term 1 is greater than term 2, that term 3 is greater than term 4, and so on and so forth.

Oh, sorry. So if $\sum^{2n-1}_{i=0} \frac {1}{a_ia_{i+1}}=\frac {2n}{a_0a_{2n}}$ we have $\frac {2n}{a_0a_{2n}}<\frac{n}{a_{0}a_{2n}}$.  :?

In my book says it can be solved using induction. Anyone?


Could someone please solve this equation for me step by step please as i don't understand it. Thanks.

sin(2x)+sin(4x)+sin(6x) = (4(cos(x)))(cos(2x))(sin(3x))

Not only that question, but i've got a problem with a couple of more questions. I've posted all of them on my website here.

http://members.optusnet.com.au/gujju_style/

4cosx * cos2x * sin3x
= 2 * sin3x * (2cosx * cos 2x)
= 2 * sin3x * (cos3x + cosx)
= 2 * sin3x * cos3x + 2 * sin3x * cosx
= sin6x + 2 * sin3x * cosx
= sin6x + sin 4x + sin2x,
 so the solution is any real x. :rotfl:

Thanks for the reply, I really appreciate it.
If i'm not pushing my luck, could someone go through Question 3a and b on that website i pasted before with the questions.

http://members.optusnet.com.au/gujju_style/

Urgently, could anyone please help me? (Sorry to be pushy.) with the questions on that link please??


This game is open to 8 people.
Basically, what you do is chose Rock, Paper or Scissors and pm it to me.
The person you are against at that time sends their choice to me as well.

$ \begin{tabular}{|c|c|c|c|c|c|c|c|c| |c|}
\hline Name& A & B & C & D & E & F & G & H & Wins\\
\hline A & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline B & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline C & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0\\
\hline D & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0\\
\hline E & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0\\
\hline F & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0\\
\hline G & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0\\
\hline H & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0\\
\hline \end{tabular}$

I am in.

[size=200]ZEEKY BOOGY DOOG[/size]

Y'all should play Rock, Paper, Scissors, Zuming instead.

$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c| |c|}
\hline Name& qwerty & hedge & ernie & math & 1\equal{}2 & cf & crazy & Catalyst & isabella & pacman &Wins\\
\hline qwerty & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline hedge & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline ernie & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline math & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline 1\equal{}2 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0\\
\hline cf & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0\\
\hline crazy & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0\\
\hline Catalyst & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0\\
\hline isabella & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0\\
\hline pacman & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & x\\
\hline \end{tabular}$

Sigh.

I suppose I'll add the two munchkins who posted above me. (I said 8 people.)

To the people who have sent me their thingamambob, please wait.

I don't think Catalyst joined? Eh, maybe?

$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c| |c|} \hline Name & qwerty & hedge & ernie & math & 1 \equal{} 2 & cf & crazy & isabella & pacman & Wins \\
\hline qwerty & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline hedge & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline ernie & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
\hline math & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0 & 0\\
\hline 1 \equal{} 2 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0 & 0\\
\hline cf & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0 & 0\\
\hline crazy & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0 & 0\\
\hline isabella & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 & 0\\
\hline pacman & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0\\
\hline \end{tabular}$

Yeah okay.

Current games:
qwerty vs hedge
ernie vs math
1=2 vs cf
crazy vs isabella

sorry pacman.

i'll play to even out the #s.

But i call dibs on rock

[quote="Dojo"]Yes. Pm them to me.[/quote]

bump

I will try keeping it alive, by posting so it is on the top

This is a great game. We should start another one. I wonder what happened to Dojo...


vector
linear algebra
matrix
linear algebra unsolved

V is a finite dimensional vector space and T is a linear map from V to V. Show that T is a scalar multiple of the identity if and only if TS = ST for every linear map S from V to V.

Call them matrices instead of linear maps.

Let $S$ have a $1$ in the $(i,j)$ position and zero otherwise. Assume $i\ne j.$

$ST$ is a matrix which is zero except for the $i$th row, and whose entries in the $i$th row are $(ST)_{ik}=t_{jk}.$

$TS$ is a matrix which is zero except for the $j$th column, and whose entries in the $j$th column are $(TS)_{kj}=t_{ki}.$

These two matrices must be equal. The designated column and row intersect in only one place:

$(ST)_{ij}=(TS)_{ij}=t_{jj}=t_{ii}.$

All the other entries are zero, so $t_{jk}=0$ for $j\ne k.$

Do this for all $i\ne j$ and we see the matrix $T$ must be zero except on the main diagonal and that all the entries on the main diagonal must be equal. That makes $T$ a scalar multiple of the identity. Since such a matrix commutes with all other matrices, we are done.


geometry
3D geometry
combinatorics unsolved
combinatorics

Each vertex of a cube is labelled with a different positive integer. Each edge of the cube is then assigned the average of the two numbers at its two endpoints.  Is it possible that the sum f the numbers at all vertices is equal to the sum of the numbers on all edges?

please explain to me with steps..  THZ A LOT!

I think that answer is o.k.

$ A$ is sum of the numbers at all vertices, $ B$ is sum of the numbers on all edges. We must have $ A=B$. When we count all numbers on edges, we count every single number at vertex 3 times. Because of that we must have: $ A=\frac{3A}{2}$, or: $ 2A=3A$. But that is impossible because $ A>0$.


[url]http://acm.tju.edu.cn/toj/showp2506.html[/url]

How to solve it? It is said that it is 2-SAT problem. can any one please clear me? :blush:[/url]


geometry
perimeter
conics
ellipse
geometry unsolved

Prove that among all possible triangles whose vertices are $3,5$ and $7$ apart from a given point $P$, the ones with the largest perimeter have $P$ as incentre.

Supose ABC is that triangle. Let $\Sigma$ the elipse of the points $Q$ such that $AQ+QB=AC+BC$, and $\Gamma$ the circle of centre $P$ and radius $7$. $C$ is the unique intersection point of $\Sigma$ and $\Gamma$ ( because the choice of $ABC$). In particular , there is a comun tangent $\iota$ to $\Sigma$ and $\Gamma$ passing by $C$. Let $X$ and $Y$ points in $\iota$ such that $C$ is betwen them. For property of the elipse we have $<XCA=<YCB$. Since $\iota$ is tangent to $\Gamma$ then $<ACP=<BCP$.

And what about this similar problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=116987 :?:

here is a very similar problem.

http://www.mathlinks.ro/Forum/viewtopic.php?t=114926

Yes.. This makes you wonder -- maybe there is a generalization for this... :maybe:

Yes, it can be generalized. It's not hard to see that the same procedure works for any fixed distances $ PA, PB, PC $.


group theory
abstract algebra
superior algebra
superior algebra unsolved

In the questions below: $G$ is a ﬁnite group; $H \leq G$ a subgroup of $G; |G : H |$ the index of $H$ in $G; |X |$ the number of elements of $X \subseteq G; Z (G)$ the center of $G; G'$ the commutator subgroup of $G; N_{G}(H )$ the normalizer of $H$ in $G; C_{G}(H )$ the centralizer of $H$ in $G$; and $S_{n}$ the $n$-th symmetric group.

Let $H \leq G$ and $a, b \in G.$ Prove that $|aH \cap Hb|$ is either zero or a divisor of $|H |.$

If $x\in aH\cap Hb$, then $aH\cap Hb = xH \cap Hx = (xHx^{-1}\cap H)x$ has the same size as $xHx^{-1}\cap H$, a subgroup of $H$.


What is the length of the longest altitude in triangle ABC if $a=7$, $b=8$, $c=9$?

not sure if this is the largest..

[hide]
$\boxed{\frac{24\sqrt{5}}{7}}$
[/hide]

[quote="mathgeniuse^ln(x)"]You need to find the area of the triangle, using Heron's Formula.

Then use the area of A=1/2*bh.  You know the height is maximum when the base is minimum and it goes from there.[/quote]

Yes, that is the right idea.

EDIT: By the way, SM4RT's answer is correct.

ya i did it that way and got the same answer..so iwas correct originally..


First i'll give the case with 2 integers  (naturals, actually).

Case 1:
If $a$ and $b$ are two co-prime positive integers and $k$ is a sum of non-negative multiples of $a$ and $b$.  What is the largest $N$ such that $N \neq k$ for any $k$?

[hide="case 2"]
 Once you've enjoyed my unelegant solution to the first problem, or come up with one yourself and critisized mine (which i welcome) i would invite you to help me with this next case.

If $a$, $b$ and $c$ are 3 co-prime positive integers and $k$ is a sum of non-negative multiples of $a$, $b$ and $c$.  What is the largest $N$ such that $N \neq k$ for any $k$?
[/hide]

[hide="solution to case 1"]

aaa
woodchuck
would chuck as much
wood as a
woodchuck could chuck if a
woodchuck could chuck
wood

[hide="real solution"]

Claim: $N=ab-a-b$

[hide="proof"]

Let $k$ be an integer such that $\exists$ non-negative integers $s$ and $t$ such that

\[k=sa+tb\]

WLOG (without loss of generality) let $a<b$.

Consider $b$ mod $a$, $2b$ mod $a$,..., $(a-1)b$ mod $a$.
Since $gcd(a,b)=1$ then $b,2b$ etc are distinct mod $a$ and are, of course, the congruence classes $1,2,(a-1)$ of Z/aZ in possibly some other order.

For $a \leq k < b$ we must have $k\equiv 0$ mod $a$,
for $b \leq k <2b$ we must have $k \equiv0$ or $k \equiv b$ mod $a$ and so on

We can achieve any $k$ with $b(a-1) \leq k$  because $k$ can now be any remainder class of $a$.

For $(a-2)b \leq k < (a-1)b$ we must have $k \not\equiv (a-1)b$ mod $a$.  Clearly $(a-2)b \leq N < (a-1)b$ and $N \equiv (a-1)b$ where $N=(a-1)b-a$ is the largest possibility.
Simplifying: $N=ab-a-b$.
[/hide]
[/hide]
[/hide]

Case 1 is indeed well-known (and due to Sylvester, I think). It's $ab-a-b$.

A related problem (but I don't think it can be of any real use here) is problem $3$ from IMO 1983: if $a,b,c$ are pairwisely coprime, then the largest integer which isn't of the form $xbc+yca+zab,\ a,y,z\ge 0$ is $2abc-ab-bc-ca$.

[quote="pbornsztein"]The problem for $n=3$ has been soved by Rdseth in 1978, but there is no easy formula in the general case (as it does exist for $n=2$).
As far as I know, for $n \geq 4$, only bounds are known.

O.J. Rdseth, 'On a linear diophantine problem of Frobenius', Journal fr die Reine und angewandt math., 1978, p. 170-178.

Pierre.[/quote]
My source is newer than you ;)

[quote="Myth"]Yes, case 1 is well known. About case 2: some years ago I saw review of scientific project, which aim was to solve your problem in common case. As I remember, author (great world-famous russian mathematician) stated that for three numbers $N\approx 2abc$, but I am not sure.[/quote]

???

clearly $N\leq ab-a-b$ for the $a,b,c$ case with $a<b<c$.

One more aops lover :( Don't mind it please.

I assumed $\gcd(a,b,c)=1$, not all numbers are pairwise coprime.

No more interesting things but :
http://mathworld.wolfram.com/CoinProblem.html

Pierre.


analytic geometry
trigonometry
integration
calculus
algebra
function
domain

Here is a problem:

A force in the xy plane is given by $\vec F = \frac{F_{0}}{r}(y \vec i-x \vec j)$ where $F_{0}$ is a constant and $r=\sqrt{x^{2}+y^{2}}$ Find the work done by this force on the particle that moves around a circle of radius 5m centered at the origin. Is the force conservative?

First let us take your defining formula $\vec{F}= F_{0}\left(\frac{y}{r}\hat{i}-\frac{x}{r}\hat{j}\right)$ and transform it to polar coordinates using the following parametric equations:

\begin{eqnarray}x &=& r\cos t \\ y &=& r\sin t\end{eqnarray}

We find $\vec{F}= F_{0}\left(\sin t\, \hat{i}-\cos t\, \hat{j}\right)$

Taking the gradients of (1) and (2) we find that

\begin{eqnarray}\hat{i}&=& \cos t\, \hat{r}-\sin t\, \hat{t}\\ \hat{j}&=& \sin t\, \hat{r}+\cos t\, \hat{t}\end{eqnarray}

Substituting these into the defining formula we obtain:

$\vec{F}= F_{0}\left(\sin t \cos t \,\hat{r}-\sin^{2}t \,\hat{t}-\cos t \sin t \,\hat{r}-\cos^{2}t \,\hat{t}\right).$

$\Rightarrow \vec{F}=-F_{0}\,\hat{t}.$

$W = \int_{0}^{2\pi}\vec{F}\cdot r\mathrm{d}\vec{t}= \left.-2\pi r F_{0}\right|_{r = 5}=-10\pi F_{0}.$

Conservative forces are forces for which $\nabla \times \vec{F}= 0.$

In this case, $\nabla \times \vec{F}=-\frac{\hat{k}}{r}F_{0}.$

Therefore, $\vec{F}$ is not a conservative force.

[quote]Conservative forces are forces for which $\nabla \times \vec{F}= 0.$[/quote]

Locally, yes - but globally, that's not the only condition.

For this example, $\nabla \times \vec{F}\ne 0$ anywhere, so it was not conservative in any region. But let's change to a different formula. Let

$\vec F = \frac{F_{0}}{r^{2}}(y \vec i-x \vec j)$

Then $\nabla \times \vec{F}=0$ everywhere except where $x=y=0.$

If we take the line integral for the work done over a closed path that does not wind around the origin, we get zero. But if we take the line integral on a circle around the origin (of any radius), we get a work of $-2\pi F_{0}.$

Hence this force would not be conservative over its whole domain.

The appropriate theorems are these:

If $\vec F$ is a conservative force over some domain, then $\nabla \times \vec{F}=0$ everywhere on that domain.

If $\nabla \times \vec{F}=0$ everywhere on some open domain, [b]and if that domain is simply-connected[/b], then the force is conservative.

The plane but not the origin, or, equivalently, $\mathbb{R}^{3}$ but not the $z$-axis, are not simply-connected domains.


What are we exactly looking for?

what's his goal? like how much is he willing to lose?

Your looking for how many days it takes him to get to his original wieght. Sorry that was a little ambigous.

[quote="A+MATH"]After eating way to many drumsticks at the Chinese new years party Frankinfueter has to go on a diet. He now wieghs 921 pounds. The amount of wieght he loses each day alternates between 12 and 9. If his original wieght was 111 how many days will he have to be on this diet?[/quote]

o_O

Are drumsticks even Chinese :P ?

[hide="solution"]
Every two days he loses 21 pounds.

921 - 111 = 810
810 / 21 = 38 + 12/21

So 76 days plus one more day, or [b]77 days[/b].
[/hide]

oo..

so the ambigious problem is solved

then the way Treething solved it is absoultley correct

Yup. That probably should have been stated better.

[quote="A+MATH"]After eating way to many drumsticks at the Chinese new years party Frankinfueter has to go on a diet. He now wieghs 921 pounds. The amount of wieght he loses each day alternates between 12 and 9. If his original wieght was 111 how many days will he have to be on this diet?[/quote]

[hide="solution"]he losses 21 pounds every 2 days

he needs to loose 810 pounds

it takes him 2 days, 38.5 times so 2*38.5=[b]77[/b][/hide]


4. Given is an integer $n\geq 1$. Find out the number of possible values of products $k \cdot m$, where $k,m$ are integers satisfying $n^{2}\leq k \leq m \leq (n+1)^{2}$.

we can use one nice lemma, from Indian MO! 98, i think.

doesn't exist four different positive integers a,b,c,d with $n^{2}\leq a,b,c,d < (n+1)^{2}$ and ad = bc.

to prove this, suppose that our afirmation isn't true.
make the substituition a = pq, d = rs, b = pr, c = qs.
suppose $a < b < c < d$.

we have that c > a implies $s\geq p+1$ and b>a implies $r\geq q+1$

so, we have:

$d = rs \geq (p+1)(q+1) = pq+p+q+1 \geq pq+2\sqrt{pq}+1 = (\sqrt{a}+1)^{2}\geq (n+1)^{2}$

impossible!

We have

(n+1)² - n² =2n+1

So;

k or m can have 2n + 3 values

Do

The product  k.m have (2n+3)² values

From what e.lopes said, the only values for which $ ad\equal{}bc$ are $ a\equal{}n^2, d\equal{}(n\plus{}1)^2, b\equal{}c\equal{}n(n\plus{}1)$.  So taking in consideration that $ k \le m$ the number of possibilities are:
$ (2n\plus{}2)\plus{}(2n\plus{}1)\plus{}...\plus{}2\plus{}1\minus{}1\equal{}2n^2\plus{}5n\plus{}2$.


Find Eric’s average weekly earnings if he earned $ 90 each week for 3 weeks and then$45 each week for 2 weeks.

Eric earned a total of $ 90 \times 3 \plus{} 45 \times 2 \equal{} 270 \plus{} 90 \equal{} 360$ dollars, over a time span of 5 weeks. Therefore, he earned an average of $ 360 \div 5 \equal{} 72$ dollars per week.


logarithms
number theory unsolved
number theory

For arithmatic progression a[size=59]n[/size]
log(a[size=59]1[/size]a[size=59]2/2[/size]a[size=59]n/n[/size])=(n^2 + n)^2
Find sigma(k=1 to infinity) of (n^2 + log a[size=59]k[/size])

I think you mean $ \sum_{k \equal{} 1}^{\infty} (k^2 \plus{} \log a_k)$, but that diverges.


floor function
inequalities
ceiling function
combinatorics proposed
combinatorics

Strange...because, if you know some proofs of Sperner's theorem, you generally know the equality cases....

Pierre.

[quote="abbas_mehrabian"]Let $A$ be a family of subsets of $\{1,2,\ldots,n\}$ such that no member of $A$ is contained in another. Sperner’s Theorem states that $|A|\leq{n\choose{\lfloor\frac{n}{2}\rfloor}}$. Find all the families for which  the equality holds.[/quote]
  your's solution
       what is Sperner's theorem ?

to pbornsztein:students are supposed to deduce when n is odd integer why there are just two families that equality occurs for them.

Here is a solution.

Firstly, we shall prove Sperner's Theorem.

[i]Sperner's Theorem:[/i]

Let $S$ be a set with $n$ elements, and let $\cal S$ be a set of its subsets, such that $A\nsubseteq B$ for each $A,B\in\cal S$. Then $|\cal S|\leqslant\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)$.

[i]Proof:[/i]

Let $s_{i}$ be the number of elements of $\cal S$ with $i$ elements. Obviously, $\sum_{i=0}^{n}s_{i}=|\cal S|$. Since \[\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)\geqslant \binom{n}{i}\quad(*)\] for each $i\in\{1,2,\dots,n\}$, it follows $\frac{s_{i}}{\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)}\leqslant\frac{s_{i}}{\binom{n}{i}}$, and therefore $\sum_{i=0}^{n}\frac{s_{i}}{\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)}\leqslant\sum_{i=0}^{n}\frac{s_{i}}{\binom{n}{i}}$. By the following lemma, RHS is $\leqslant 1$, and the theorem follows. $\square$

[i]Lemma (Lubell-Yamamoto-Meshalkin inequality):[/i]

With the former conditions and notation, the following holds:
\[\sum_{i=0}^{n}\frac{s_{i}}{\binom{n}{i}}\leqslant 1\]

[i]Proof:[/i]

By [i]chain[/i] we mean any array $(S_{i})_{i\geqslant 0}$ of subsets of $S$ such that $S_{i}\subset S_{i+1}$ for each $i$. Obviously, the maximal chain has $n+1$ elements, and there are $n!$ maximal chains. Fix an $\overline{S}\in{\cal S}$. There are $|S|!(n-|S|)!$ maximal chains that contain $\overline{S}$, because we start from empty set, add one element at the time till we get to $\overline{S}$ in $|S|!$ ways, and then add the rest $n-|S|$ elements in $(n-|S|)!$ ways. Also note that it is impossible that two elements $A,B$from $\cal S$ are in the same chain, because that would mean $A\subset B$ or $B\subset A$. Therefore:
\[\sum_{i=0}^{n}s_{i}i!(n-i!)=\sum_{\overline{S}\in\cal S}|S|!(n-|S|)!\leqslant n!\quad(**)\]
and the lemma follows by dividing with $n!$ (note that $\binom{n}{i}=\frac{n!}{i!(n-i)!}$). $\square$

Now we may return to the main problem.

Note that we have the equality iff there is equality in (*) and (**). For even $n$ it is easy: we note from (*) that the equality is attained iff all sets are of the size $\frac{n}{2}$. For odd $n$ it holds $\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)=\left(\begin{array}{c}n\\{\left\lceil\frac{n}{2}\right\rceil}\end{array}\right)$ and we shall prove that the equality holds iff all the elements of $\cal S$ are simultaneously either of cardinality $\left\lfloor\frac{n}{2}\right\rfloor$ or $\left\lceil\frac{n}{2}\right\rceil$.

[b]Proof:[/b]

Let the equality in Sperner's Theorem is obtained. Therefore, each element of $\cal S$ corresponds to one and only one chain, as described in the proof of Lemma.

(1): If a $\left\lfloor\frac{n}{2}\right\rfloor$-elements subset of $S$ is [u]not[/u] in $\cal S$, then when we add any of the remaining elements, we shall always get a subset which is in $\cal S$, because otherwise no element of $\cal S$ will correspond to the chain which begins with the elements of the "missing" set.

(2): More, if a $\left\lceil\frac{n}{2}\right\rceil$-elements subset of $S$ is in $\cal S$, it is obvious that there is no $\left\lfloor\frac{n}{2}\right\rfloor$-elements subsets of that set in $\cal S$.

Suppose that there is at least one $\left\lceil\frac{n}{2}\right\rceil$-elements set in $\cal S$, and that there is at least one $\left\lceil\frac{n}{2}\right\rceil$-elements set which is [u]not[/u] in $\cal S$. Call them $P_{1}$ and $P'$, respectively. Look at the following "train":

\[P_{1}\supset Q_{1}\subset P_{2}\supset Q_{2}\subset P_{3}\supset Q_{3}\subset\cdots\]

$P_{i}$'s are pairwise different $\left\lfloor\frac{n}{2}\right\rfloor$-elements subsets of $S$ [u]in[/u] $\cal S$, and $Q_{i}$'s are pairwise different $\left\lceil\frac{n}{2}\right\rceil$-elements subset of $S$ [u]not in[/u] $\cal S$. In order to form $P_{i+1}$ from $Q_{i}$, we shall choose an element from $P'\setminus\left(\bigcup_{j=1}^{i}P_{j}\right)$. This is possible by (1) and (2). However, it is obvious that, by this procedure, the train will "arrive" at $P'$. Contradiction, what completes the proof. $\blacksquare$

Not sure if this is the same as the above, but once we get that all subsets must be of size $\left \lfloor \frac{n}{2} \right \rfloor$ or $\left \lceil \frac{n}{2} \right \rceil$, we can finish by analyzing Hall's condition.

Indeed, we can clearly discard the case when $n$ is even as trivial, as let $n = 2k+1.$

Now consider the matching between subsets of $[n]$ of size $k$ and subsets of $[n]$ of size $k+1$, where we match $A$ ($|A| = k$) with $B$ ($|B| = k+1$) if $A \subset B.$ This matching is regular of degree $k+1$ and so Hall's condition is satisfied.

Let $S$ be the set of subsets which are of size $k$. If $|S| = 0$ or $\binom{2k+1}{k}$, we're done.

Else, let $N(S)$ be the set of neighbors of vertices of $S$ with size $k+1$, as in Hall's Theorem. It's clear that there are at most $\binom{2k+1}{k} - |N(S)|$ subsets chosen with size $k+1$, and so we just need to show that $|N(S)| > |S|.$

To do this, notice that the sets in $S$ have $|S| (k+1)$ edges coming from them, so for $|N(S)| \le |S|$ we require $|N(S)| = |S|$ and no other sets of size $k$ are adjacent to anything in $N(S).$ However, we may consider a set $C$ of size $k$ such that it differs from a set in $S$ in only one element (do this by starting with something in $S$ and switching out one element at a time until we exit $S$), and notice that $C$ is not in $S$ yet is adjacent to something in $N(S).$ This implies that $|N(S)| > |S|$ and we're done.

So the only possibilities for $n = 2k+1$ are the set of $k-$element subsets and the set of $(k+1)-$element subsets.

$\square$


In acute-angled triangle $ ABC$, the points $ D,E,F$ are on sides $ BC,CA,AB$, respectively such that $ \angle AFE \equal{} \angle BFD, \angle FDB \equal{} \angle EDC, \angle DEC \equal{} \angle FEA$. Prove that $ AD$ is perpendicular to $ BC$.

An easy and nice problem!  :P

We have that AB, BC and AC are external bisectors from triangle DEF.So, A is ex-incenter relative to side EF.So DA is an internal bisector and the angle formed by an internal bisector and an external bisector is 90º.The conclusion follows imediatelly.

Sorry for bad English !  :D

$ \triangle DEF$ is orthic triangle!!
This problem is a famous property of orthic triangle.

Let $ \angle AFE = \angle BFD=x, \angle FDB = \angle EDC=y, \angle DEC = \angle FEA=z$. Then from the triangles $ AFE, BDF, CED$ we get
\[ \{\begin{array}{cccc}&x+A+z=\pi
&y+B+x=\pi
&z+C+y=\pi  \end{array}\]
Adding the above equations we get $ x+y+z=\pi$. Therefore from the equations above we conclude that $ z=B, x=C, y=A$. Hence $ \triangle ABC\sim\triangle AEF\sim\triangle DBF\sim\triangle DEC$, and $ \angle FDE=\pi-2A, \angle DEF=\pi-2B, \angle EFD=\pi-2C$. Therefore triangle $ DEF$ is similar to the orthic triangle $ D'E'F'$ of $ ABC$, and $ DE\parallel D'E',EF\parallel E'F',FD\parallel F'D'$. Suppose that $ E'F'>EF$. Then we must have $ F'D'>FD$ and $ D'E'>DE$ and hence lines $ DE$ and $ EF$ must intersect outside $ \triangle ABC$, a contradiction. Similarly $ E'F'<EF$ is not possible either. Thus $ DEF$ coincides with $ D'E'F'$ and the conclusion follows.


algebra
function
domain
Ring Theory
number theory
relatively prime
number theory unsolved

a)Find all integers $x,y\in \mathbb{Z}$ such that $y^2=x^3-11$
b)Find all integers $x,y,z\in \mathbb{Z}$ such that $x^3+5y^3+25z^3-15xyz=\pm 1$.

For the first one, we can work in the euclidean domain $\mathbb{Z}\left[\frac{1+\sqrt{-11}}{2}\right]$.
We have $(x+\sqrt{-11})(x-\sqrt{-11})=y^3$. We can show that $x+\sqrt{-11}$ and $x-\sqrt{-11}$ are relatively prime, so: $x+\sqrt{-11}=\left(\frac{a+b\sqrt{-11}}{2}\right)^3$ with $(a,b)\in\mathbb{Z}^2$. Thus $8=b(3a^2-11b^2)$ and  we have $(x,y)=(4,3)$ or $(x,y)=(58,15)$.

_______________
blang (sorry for my french english!)

Sorry, you have to exchange $x$ and $y$...

_______________
blang (sorry for my french english!)

For b)
We can factor it as
\[\big(x+y\sqrt[3]5+z\sqrt[3]25\big)\cdot\big((x^2-5yz)+(5z^2-xy)\sqrt[3]5+(y^2-xz)\sqrt[3]25\big)=\pm 1.\]
Is there a divisor of $\pm 1$ in $Z[\sqrt[3]5]$, which is distinct from $\pm 1$?


I used many ways to figure out the right answer, but I came away with $40$ for all of them. For some reason, I cannot trust my answer. I will post my solutions when i verify with others.

f(10)+f(-6)=10^4+6^4+a(10^3-6^3)+b(10^2+6^2)+(10-6)c+2d.
which is
11296+784a+136b+4c+2d
we know from f(1)=10 that
1+a+b+c+d=10 so a+b+c+d=9.___________Lets call a+b+c+d, e.
we know that from f(2)=20 that
16+8a+4b+2c+d=20 so 8a+4b+2c+d=4.____Lets call 8a+4b+2c+d, f.
We know from f(3)=30 that
81+27a+9b+3c+d=30 so 27a+9b+3c+d=-51. Lets call 27a+9b+3c+d, g.

so 11296 784a+136b+4c+2d=11296+xg+yf+ze. where x is the amount of gs, y is the amount of f's, and z is the amount of e's.

Adding up corresponding coefficients, we get
784=27x+8y+z and
136=9z+4y+x and
4=3z+2y+x and
2=x+y+z.
Solving these, using some ugly substitution which will not be posted here, we get
x=z=64, y=-126.

So plugging these in
11296+64g-126f+64e is f(10)+f(-6).

We know g=-51 from the above, f is 4 from the above, and e is 9.
Plugging them back in we get
8104? (check this last step)
somethings wrong here...

ah that looks much better than my solution


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $R$ be a commutative noetherian ring, $p$ be a prime ideal of $R$ and $k(p) = R_{p}/pR_{p}$ be a residue field of a local ring $R_{p}$. Can somebody explain me why
\[\mbox{Hom}_{R_{p}}(k(p), E_{R_{p}}(k(p))) \simeq k(p)\]
where $E_{R_{p}}(k(p))$ denotes an injective hull of an $R_{p}$-module $k(p)$. Thx David

Let $(R,m,k)$ be a local ring. Let $E$ be an essential extension of $k$ as $R$-modules i.e. $k$ is included in $E$ and every nonzero submodule of $E$ intersects $k$ nontrivially.
We want $Hom_{R}(k,E))\simeq k$ as $R$ modules.

Let $f: k\to E$ be an $R$-module homomorphism. It is a monomorphism (injection) or the $0$ map because $k$ is a field.
If $f$ is not the zero map, then $f(k)$ is a nonzero submodule of $E$, so it intersects $k$ nontrivially. Since $k$ has no nontrivial sub $R$-module other than itself, it follows that $k\subseteq f(k)$. Look at $k$ as $R/m$.
Let $r\in R$ such that $f(\hat r)=\hat 1\in k$. Then since $Rf(r)=f(Rr)=f(k)$, we get $f(k)\subseteq k$, so $f(k)=k$. It is obvious to see that $f\leftrightarrow\hat r$ is a bijective correspondence between nonzero morphisms from $k$ to $E$ and nonzero mod $m$ classes in $R$.

From here it is easy to prove the isomorphism (also as modules not just as sets)


A resistance $ R_{1}$ is connected in series to a capacitor $ C$, and another resistance $ R_{2}$ is connected in series to a solenoid $ L$. These two pairs are connected in parallel, and then connected to alternating current of voltage $ V$ and frequency $ f$. What is the power generated?

Are you referring to the power given by the source to the circuit, or to the power dissipated on the circuit.

The equivalent impedance of R1 and C will be $ Z_{1C}\equal{}\left(R_{1}\minus{}\frac{j}{\omega C}\right)\,\,\Omega$, and the equivalent impedance of R2 and L will be $ Z_{2L}\equal{} (R_{2}\plus{}\omega L j)\,\,\Omega$, where $ \omega \equal{} 2\pi f$. The total impedance of the circuit will then be $ Z_{eq}\equal{}\frac{Z_{1C}Z_{2L}}{Z_{1C}\plus{}Z_{2L}}\,\,\Omega$. The current I can then be found from

$ I \equal{}\frac{V}{Z_{eq}}$.

Finally, the average power given by the source is $ P \equal{}\frac{1}{2}VI$, where both V and I are expressed as phasors.

[quote="Carcul"]Are you referring to the power given by the source to the circuit, or to the power dissipated on the circuit.

Finally, the average power given by the source is $ P \equal{}\frac{1}{2}VI$, where both V and I are expressed as phasors.[/quote]

How are the two different, could you tell me?

Also, why is $ P\equal{}\frac{1}{2}VI$? Shouldn't it be $ P\equal{}VI\cos\phi$, where $ \phi$ is the angle difference between $ V$ and $ I$?

[quote="Radio"]Shouldn't it be $ P \equal{} VI\cos\phi$, where $ \phi$ is the angle difference between $ V$ and $ I$?[/quote]

That's true, if you use the effective (or rms values) for V and I. Otherwise, $ P \equal{}\frac{1}{2}V_{m}I_{m}\cos\phi$, where $ V_{m}$ and $ I_{m}$ are the voltage and current amplitudes, which is the same to what I have written (expressing I as the conjugate and then taking the real part).

I think there's another way. You can just calculate the powers on each of the two resistors, and add them  :)

[quote="Carcul"]Yes you can, but for that you have to calculate the current through each resistor (applying the current division principle, for instance), which I think will require more work.[/quote]

Actually I think it requires less  :D

At least you don't have to go through that ugly $ Z\equal{}\frac{Z_{1}Z_{2}}{Z_{1}\plus{}Z_{2}}$


if n is natural and [b]1+2^n+4^n [/b]is prime prove that:
n=3^k

http://www.mathlinks.ro/Forum/viewtopic.php?t=5467 (but there are more of them)


Find the number of zeros at the end of the number \[ 4^{5^6} \plus{} 6^{5^4}.\]

Actually there are 5 zeroes.

With a [b]very[/b] bashy solution, you can show that $ 4^{5^6} \equiv 890624 \pmod {10^6}$ and $ 6^{5^4}\equiv 909376 \pmod {10^6}$, so $ 4^{5^6}\plus{}6^{5^4}\equiv 800000 \pmod {10^6}$, so there are 5 zeroes at the end of the number.


Solve the equation: $ \frac{3^x\plus{}1}{2^{x\minus{}1}}\minus{}\frac{2^x\plus{}1}{3^{x\minus{}1}}\equal{}1.$

[quote="moldovan"]Solve the equation: $ \frac {3^x \plus{} 1}{2^{x \minus{} 1}} \minus{} \frac {2^x \plus{} 1}{3^{x \minus{} 1}} \equal{} 1.$[/quote]

Multiplying by $ 6^x$, we get $ 2(3^x \plus{} 1)3^x \minus{} 3(2^x \plus{} 1)2^x \minus{} 6^x \equal{} 0$

Which may be written $ (3^x \plus{} 2^x \plus{} 1)(2\cdot 3^x \minus{} 3\cdot 2^x) \equal{} 0$

And so $ 3^{x \minus{} 1} \equal{} 2^{x \minus{} 1}$ and so the unique solution $ x \equal{} 1$


Prove that the divisibility rule for $2^{n}$ is if the first n digits are divisible by $2^{n}$(right to left), then the whole number is divisible bo $2^{n}$.

Prove that the divisibility rule for $5^{n}$ is if the first n digits are divisible by $5^{n}$(right to left), then the whole number is divisible bo $5^{n}$.

[hide="major hint for second one"]You should use the same method as the first one.[/hide]

[hide="solutions"][hide="the first one"]Let's say that the number is $a_{1}a_{2}a_{3}\cdot\cdot\cdot a_{x}$ where $a_{b}$ is the bth digit and $2^{n}$ is where n is less than x. We can also write $a_{1}a_{2}a_{3}\cdot\cdot\cdot a_{x}$ as $a_{1}*10^{x-1}+a_{2}*10^{x-2}+\cdot\cdot\cdot+a_{x-1}*10+a_{x}$. We know that each 10 has one 2 in it's prime factorization. Since we know that, we can cancel out $a_{x}*10^{x-1}$ because that is divisible by $2^{n}$. We can also cancel $a_{x-1}*10^{x-2}$ and so on until we reach $a_{x-n+1}*10^{x-n}$. We can also cancel out that, but no other terms. That leaves n terms to deal with. We now have the first n digits we wanted.[/hide]

[hide="the second one"]Let's say that the number is $a_{1}a_{2}a_{3}\cdot\cdot\cdot a_{x}$ where $a_{b}$ is the bth digit and $5^{n}$ is where n is less than x. We can also write $a_{1}a_{2}a_{3}\cdot\cdot\cdot a_{x}$ as $a_{1}*10^{x-1}+a_{2}*10^{x-2}+\cdot\cdot\cdot+a_{x-1}*10+a_{x}$. We know that each 10 has one 5 in it's prime factorization. Since we know that, we can cancel out $a_{x}*10^{x-1}$ because that is divisible by $5^{n}$. We can also cancel $a_{x-1}*10^{x-2}$ and so on until we reach $a_{x-n+1}*10^{x-n}$. We can also cancel out that, but no other terms. That leaves n terms to deal with. We now have the first n digits we wanted.[/hide][/hide]

I figured out the second one just today.

Ok. Now try a follow-up nutz_for2.718281828

What is the divisibility rule for $99\ldots9$ ($n$ $9$'s)? What about $100\ldots01$ ($n$ $0$'s)?


trigonometry
geometry
vector
calculus
integration
absolute value

Hi!

I was wondering about the magnetic flux law:
$ \Phi_B\equal{}BA\cos{\theta}$

B-magnetic field, A-area of the current loop

If the angle is measured with respect to the normal to the surface of the loop, then why can't the law be changed to:
$ \Phi_B\equal{}BA\sin{\theta}$

or some variation of that that will account for the -/+ changes with respect to cosine and sine?

Thanks.

$ \Phi_B \equal{} BA_{\text{effective}} \\
 \\
\text{See the diagram below, the effective area crossed by }\vec{B}\text{ is } |B|\cos\theta \text{ where }\theta\text{ is the angle between the areal vector and }\vec{B} \\
\text{If you choose the other angle of the respective right angled triangle as your }\theta\text{, the effective area crossed by }\vec{B}\text{ is indeed }|B\parallel{}A|\sin\theta$

Magnetic flux over a surface $ S$ is defined as:

\[ \Phi_m\equal{}\iint_S \mathbf{B}\,d\mathbf{A}\]

That is the surface integral of the magnetic field over the given surface. Here $ d\mathbf{A}$ is the vector normal to the surface, with absolute value of a surface element.\\

As we see the surface integral is basically the integral of an inner product, so the cosine of the angle between the surface normal vector and the magnetic field arises. Hence we can define some sort of "effective surface" as Rajiv suggested.

So the angle $ \theta$ in the "magnetic flux law" is the angle enclosed by the vector normal to the surface and the magnetic field. If however you choose $ \theta$ to be the, other angle, i.e. $ \frac{\pi}{2}\minus{}\theta$ then the above "law" changes appropriately.

As yeahnigga suggested, it's not a law but the definition of flux. Its concept is just to tell how much we have of perpendicular component (of the vector) "through" the surface.


The fractions $\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7},$ and $\frac{6}{7}$ are all represented by the same six digits in their decimal equivalents. Only the order of the digits is changed. in which of the repeating decimals do the third and fourth digit form a two-digit number which is one-half of hte number formed by the fifth and sixth digit? Write the fraction corresponding to this decimal.

Yes. I can do that..

but I wanted to ask for a faster way...

You could have narrowed it down to only the fractions with even numerators, I think.

Well, the subsequent ones are all the rotations of the first one that start with the right digit.  That's an easy way to find all of them.

If you don't want to list them, you could
(1) realize that in the decimal equivalents, the only "pairs" of digits that appear are 14, 28, 57.
(2) the third and fourth numbers have to be 14, fifth and sixth have to be 28.
(3) You're left with 57 as the first and second numbers.
(4) So, $.571428=\frac{4}{7}$

when you have a repeating decimal, how do you make it into a fraction?

@now a ranger:

I'm going to do it with just a simple fraction, $\frac{1}{3}$.

$\frac{1}{3}=0.\overline{3}$.

To change it, let $x=0.\overline{3}$.

Then, $10x=3.\overline{3}$

$10x-x=3.\overline{3}-0.\overline{3}=3$

$9x=3$

$x=\frac{3}{9}=\frac{1}{3}$

and if its something like .43 with the 3 or both repeating, then u'd multiply by 90, then if it was a 3digit decimal (unless they show like.222 repeating) then u multiply by 900

have a happy thanksgiving!
jorian hoover

basically, if you have a repeating decimal, $.a_{1}a_{2}\ldots a_{n}a_{1}a_{2}\ldots a_{n}\ldots$ you can express it as $\frac{a_{1}a_{2}\ldots a_{n}}{10^{n}-1}$  You can derive any other kind of repeating decimal from this.

what does it mean "only the order of the digits are changed" mean, because it doesn't make too much sense  :maybe:

-jorian

(thank you)

If you notice, the order of the digits stays the same. The only change is the first digit (as I said before, it increases as the numerator gets bigger).


Take the polynomial $P(x) = \sum_ {cyc} x (x-a)(x-b)(x-c)(x-d) - \frac 1{12} \sum_{sym} (x-a)(x-b)(x-c)$

Now we have, for our relations that $P(a)=P(b)=P(c)=P(d)=P(e)=0$ and thus because $deg[P] = 5$ we have that $P(x)= k(x-a)(x-b)(x-c)(x-d)(x-e)$.

Now, comparing the $5$th grade coefficients we have $k=5$.
Comparing the $4$th grade we have $a+b+c+d+e=0$
Comparing the $3$rd grade we have $ab+ac+ad+ae+bc+bd+be+cd+ce+de = -5$
Comparing the $2$nd grade we have $abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=0$
Comparing the $1$st grade we have $abcd+abce+abde+acde+bcde = \frac{15}4$
finally with $x=0$ we get $abcd=0$.

Thus, for Viète, $a,b,c,d,e$ are roots of the polynomial $x^5 - 5x^3 + \frac{15}4 x$.

So we have $a^3+b^3+c^3+d^3+e^3 = a^5+b^5+c^5+d^5+e^5 =0$ and so it's done.

ps this is because the roots are ${\pm \sqrt{\frac{5 \pm \sqrt{10} }2}}$ and $0$.

[quote="Simo_the_Wolf"]Thus, for Viète, $a,b,c,d,e$ are roots of the polynomial $x^{5}-5x^{3}+\frac{15}4 x$.[/quote]
Really, very nice solution !
Prehaps of interest is to see also  http://www.mathlinks.ro/Forum/viewtopic.php?t=97565


If $ a,b,c,d\in\mathbb{R}$, prove that $ (a^2\plus{}b^2)\sqrt{c^2\plus{}d^2}\ge2abc\plus{}d(a^2\minus{}b^2)$

[quote="IW@IT"]If $ a,b,c,d\in\mathbb{R}$, prove that $ (a^2 \plus{} b^2)\sqrt {c^2 \plus{} d^2}\ge2abc \plus{} d(a^2 \minus{} b^2)$[/quote]
By Cauchy inequality, we have:
$ LHS\equal{}\sqrt{(4a^2b^2\plus{}(a^2\minus{}b^2)^2)(c^2\plus{}d^2)}\ge 2abc\plus{}d(a^2\minus{}b^2)$


A uniform circular disc,weight W radius a is suspended by three equal vertical strings attached at three points on the circumference of the disc separated by equal intervals.
A weight w may be placed anywhere on the disc within a circular region with same centre as that of the disc and radius ra (r<1).

Prove the strings will not break if they can support a tension of  ( 2rw + w + W)/3.

Let A, B, C be the attachement points, forming an equilateral triangle $\triangle ABC$, with circumcenter = orthocenter = centroid O. The disk without any additional weight has the center of mass at O and the forces on the strings A, B, C are equivalent to putting mass points with weights $\frac W 3$ at A, B, C. An object with a weight w is added at a point P  = ra < a. The center of mass of the disk plus the object is now at a point G such that

$OG \cdot W = (OP - OG) \cdot w,\ \ OG = \frac{OP \cdot w}{W + w} = \frac{arw}{W + w}$

Let the cevians through the new center of mass G cut the opposite sides of the equilateral triangle $\triangle ABC$ at points $D \in BC, E \in CA, F \in AB$. The new center of mass can be achived by putting weights $w_A, w_B, w_C$ at points A, B, C, such that $w_A + w_B + w_C = W + w$ and $BD \cdot w_B = CD \cdot w_C$, $CE \cdot w_C = AE \cdot w_A$, $AF \cdot w_A = BF \cdot w_B$, which in principle allows to calculate the weights $w_A, w_B, w_C$. Because of the 3-fold rotational symmetry around the point O, we can limit ourselves to the point P in the $120^\circ$ plane sector with the vertex O and symmetrical with respect to the ray OA. The string A then has to carry the largest weight. The point G is on a $120^\circ$ circular arc of a circle with center O and radius $OG = \frac{arw}{W + w}$. Let $T$ be the intersection of the segment OA with this circle and $t$ a tangent to this circle at T. Since OT is a radius of this circle and OA the A-altitude of the equlateral triangle $\triangle ABC$, $t \perp OT \equiv OA \perp BC$, this tangent is parallel to BC. Let the cevian AD through G meet the tangent $t$ at a point G' and let K be the foot of the A-altitude. Obviously,

$\frac{w_A}{w_B + w_C} = \frac{w_A}{W + w - w_A} = \frac{DG}{AG} \le \frac{DG'}{AG'} = \frac{KT}{AT}$

$AT \cdot w_A \le KT \cdot (W + w - w_A)$

$w_A \le \frac{KT}{AT + KT}\ (W + w) = \frac{KT}{AD}\ (W + w)$

Thus we see that for a fixed r < 1, the weight $w_A$ will be maximum when the point G is identical with the point T, i.e., when the new center of mass G is on the altitude segment OA and the additional weight w at P is also on OA. The maximum weight $w_{A0}$ is easily calculated:

$w_A \le w_{A0} = \frac{KT}{AD}\ (W + w) = \frac{KO + OT}{AD}\ (W + w) =$

$= \left(\frac a 2 + \frac{arw}{W + w}\right) \cdot \frac{W + w}{\frac 3 2\ a} = \frac{W + w + 2rw}{2(W + w)} \cdot \frac 2 3\ (W + w) = \frac 1 3\ (W + w + 2rw)$


geometry
Linear Function
function
Inequality
algebra
IMO Shortlist
IMO Longlist

A linear binomial $l(z) = Az + B$ with complex coefficients $A$ and $B$ is given. It is known that the maximal value of $|l(z)|$ on the segment $-1 \leq x \leq 1$ $(y = 0)$ of the real line in the complex plane $z = x + iy$ is equal to $M.$ Prove that for every $z$
\[|l(z)| \leq M \rho,\]
where $\rho$ is the sum of distances from the point $P=z$ to the points $Q_1: z = 1$ and $Q_3: z = -1.$

[b]Darij Grinberg's translation from Morozova/Petrakov[/b]:

Let $z$ denote a variable complex number. Given the linear function $l\left( z\right) =Az+B$ with complex coefficients $A$ and $B.$ Given that the maximum of $\left| l\left( z\right) \right| $ on the interval $-1\leq x\leq 1$ (with $y=0$) of the real number axis of the complex plane $ z=x+iy$ is $M.$ Show that for any $z,$ the inequality $\left| l\left( z\right) \right| \leq M\rho $ holds, where $\rho $ is the sum of the distances from the point $P=z$ to the points $Q_{1}$ (corresponding to the complex numer $1$) and $Q_{3}$ (corresponding to the complex number $-1$).

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


trigonometry
calculus
calculus computations

Write the Taylor series of $ \sin x$ at the point $ a\equal{}\frac{\pi}{4}$ using $ \sum$ notation.

Let's see.  $ f(x)\plus{}f'(x)(x\minus{}a)\plus{}f''(x)\frac{(x\minus{}a)^{2}}{2}\plus{}f^{3}(x)\frac{(x\minus{}a)^{3}}{6}\plus{}\cdots$
So plugging in $ a\equal{}\frac{\pi}{4}$,
$ \sin{x}\plus{}\cos{x}(x\minus{}\frac{\pi}{4})\minus{}\sin{x}\frac{(x\minus{}\frac{\pi}{4})^{2}}{2}\plus{}\cdots$
and I guess it would continue like that.  It's late for me right now so I'm not really in the mood to write it in summation form.

Those terms should be $ f(a) \plus{} f'(a)(x \minus{} a) \plus{} f''(a) \frac{(x \minus{} a)^2}{2!} \plus{} ...$.  Clearly the expression you wrote is not a Taylor series.

Is mlok teaching that particular course this semester? This seems like it might have been a question on his latest quiz or test. So how did the class do on this question? (I don't think JRav would have gotten all that much partial credit.)

I am indeed teaching Calculus 2, but we don't have a test on series until next week. This is a textbook exercise. The series begins with
\[ \frac{\sqrt{2}}{2}\plus{}\frac{\sqrt{2}}{2}(x\minus{}\pi/4)
\minus{}\frac{\sqrt{2}}{2}\frac{(x\minus{}\pi/4)}{2!}\minus{}\frac{\sqrt{2}}{2}\frac{(x\minus{}\pi/4)^3}{3!}
\plus{}\frac{\sqrt{2}}{2}\frac{(x\minus{}\pi/4)^4}{4!}\]
but how do we write in the form $ \sum_{n\equal{}0}^{\infty}c_n (x\minus{}\pi/4)^n$?

Since $ \frac {d^{n}}{dx^{n}}(\sin(x))) \equal{} \sin(x \plus{} \frac {n\pi}{2})$ we get that

$ \sin(x) \equal{} \sum\limits_{n \equal{} 0}^{\infty}\sin\left(\frac {\pi}{4} \plus{} \frac {n\pi}{2}\right)\frac {(x \minus{} \pi/4)^{n}}{n!}$

$ \equal{} \frac {\sqrt {2}}{2}\sum\limits_{p \equal{} 0}^{\infty}( \minus{} 1)^{p}\frac {(x \minus{} \pi/4)^{2p}}{(2p)!} \plus{} \frac {\sqrt {2}}{2}\sum\limits_{p \equal{} 0}^{\infty}( \minus{} 1)^{p}\frac {(x \minus{} \pi/4)^{2p \plus{} 1}}{(2p \plus{} 1)!}$.

Hope I made no computational mistakes.

[quote="keira_khtn"]Hoping that here is what you are refering to:
$ \sin x \equal{} \sum_{n \equal{} 0}^{\infty}( \minus{} 1)^{p(n)}\frac {\sqrt {2}}{2}\frac {(x \minus{} \pi/4)^n}{n!}$ where $ p(n) \equal{} 2$ for $ n \equal{} 0,1(\mod 4)$ and $ p(n) \equal{} 1$ for $ n \equal{} 2,3(\mod 4)$[/quote]

Of course, we can write explicitly $ p(n) \equal{} \frac{n(n\minus{}1)}{2}$.

[quote="t0rajir0u"]we can write explicitly $ p(n) \equal{} \frac {n(n \minus{} 1)}{2}$.[/quote]
That's the answer given in the textbook. Which led me to wonder if the authors really expected calculus students to come up with $ (\minus{}1)^{n(n\minus{}1)/2}$ on their own. :maybe: Looks wildly optimistic to me.

I like the idea used by [b]didilica[/b]: think of $ (\sin x)'$ as $ \sin(x\plus{}\pi/2)$ rather than $ \cos x$. But I would not expect to see it in calculus, either.

I tell people that the summation notation isn't even all that important. Something like the first line of your post (#5 on this topic), or maybe that with one more term just to be sure that we've really established what the pattern is, is a sufficient answer as far as I'm concerned.

Yes, the sigma notation is compact, and spreading the terms out is an extravagant use of paper - but I accept the meaning as the same.

If the question were to write a computer (or calculator, or spreadsheet) program that adds up terms until some condition involving the size of the terms is met - well, that might well cause us to represent the terms in yet other ways. (My version of the program would probably try to calculate each term recursively from the previous term, rather than from scratch.)

I have the same opinion as yours.$ p(n)\equal{}$... is just a bit more formular,it made the sigma better-looking for nothing!


What fraction is equal to six times two-sevenths?  Express your answer as a common fraction.


calculus
integration
number theory solved
number theory

Determine whether the number of positive integral solutions (a,b,c,d,e) of the equation a^3+b^5+c^7+d^11=e^13 is zero, a finite positive number, or infinite.

It looks like the other one, but it's not the same type of question.

Now, first see that if (a,b,c,d,e) is a solution in positive integers then for eaxh positive integer k, (a*n 5*7*11*13 , b*n 3*7*11*13 ,  c*n 3*5*11*13 , d*n 3*5*7*13 , e*n 3*5*7*11 ) is a solution.
Thus, to prove that the number of solutions is infinite, we just have to find one.
And (2 5*7*11 , 2 3*7*11 ,  2 3*5*11 , 2 3*5*7 , 2 3*5*7*11+2 ) does the job...

Pierre.


[quote="akhrif"]can someone tell me why  $a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geq 0.$?[/quote]
Look here[url=http://planetmath.org/encyclopedia/SchursInequality.html]planetmath.org/encyclopedia/SchursInequality.html[/url]

[quote="akhrif"]niice  :D  thanks[/quote]
You'r wellcom ;) ,but next time try to serch,even in Mathlinks ther are a lot of topics with prove of Shur's ineq.!


I had the correct answer and procedure to solve number 5, and I got the right answer. However, I used $ C_1$ and $ C_2$ to define circles and centers of these circles interchangeably throughout the solution, and my diagram indicated what I ment. Although I believe that I could have made this a bit more clear, i got a 2/5. Should I appeal?

You don't have much to lose as this is the last round and you won't need the appeals for a later round. The worst thing that could happen is that they bump you down a point. I'd appeal if I were you.

I think that they give you 2 appeals at any given moment for any round. When I appealed for round 2, I was told I had one protest left, but then when my protest went through, I was told I had 2 left. Same thing for Round 3 (at least for me).

The way it is supposed to work is that you have two appeals to use. You use up an appeal every time you file one, and you are "refunded" the appeal if it goes in your favor. So, if you have 2 appeals, then protest your score on Problem #3, you are left with 1 appeal. If they accept your appeal, you get your appeal back and now have 2 appeals. If it gets denied, you only have 1 appeal left.

Thanks for the replies, I still have two appeals. I was actually wondering if similar errors got the same amounts of points docked off for the past rounds or thins round. In other words, do you think I have a chance of getting 1-2 points back?

You can't protest past rounds, but you can try to protest problems this round if you like. Do this only if you think you deserve a point though...


[quote="Ignite168"]$\frac{12}{x}+\frac{5}{y}=25$
$\frac{2}{x}-\frac{15}{y}=-18$

Solve the system of equations.

[hide="Minor Hint"]If two numbers are equal, their reciprocals are equal[/hide][/quote]
I've seen many like this. Here is how you should do it if you have no clue how to proceed.
[hide]Substitute $a=\frac{1}{x}$ and $b=\frac{1}{y}$ and chug away. Just remember to plug the numbers back in at the end.[/hide]

[quote="mysmartmouth"]I don't get nice numbers, so if someone wants to check my work...

[hide="Solution"]Let $a=\frac{1}{x}$ and $b=\frac{1}{y}$.

$12a+5b = 25$
$2a-15b=-18$

$36a+15b = 75$
$2a-15b=-18$

$38a = 57$
$a=\frac{3}{2}$
$b=\frac{7}{5}$

Substituting back into original equations:

$x=\frac{2}{3}$
$y=\frac{5}{7}$[/hide][/quote]

That is correct.

I don't see why you need to substitute a and b though.

You can do it without it, but it's such a simple thing to do, and it reduces potential for careless error.

after using the substitution, just eliminate one of the variables  :D

[quote="jhredsox"]wow. what do u do if you have to subsitute but have no idea what to do?
-jorian[/quote]

No problem ever "requires" substitution, but sometimes it is almost impossible for a normal person to solve a problem without it.  A good example is one of the first problems from the AoPS Chapter 1 Vol 2 (Logarithms) problem set.  It involves letting y=log x and then finding a quadratic in y.  If you have the books, look for that problem.


vector
linear algebra
matrix
abstract algebra
linear algebra unsolved

find all linear maps $f: M_n(K) \to M_n(K)$ such that $f(XY)=f(X)f(Y)$ for all $X,Y \in M_n(K)$

[quote="jmerry"]Wrong question.[/quote]

<Edit>
You are right, I somehow missed that this is about endomorphisms of $M_n(k)$.  Thanks.

Here is a nice solution in a particular case ($f(I)$ non-zero) I've found yesterday evening, thinking at Moubinool.  :P
  So, assume that $f(I)$ is not 0 and let's prove that $f$ is an interiour authomorphism. I will use the notations of Moubinool: $VecA$ is the $n^2$ vector formed by writing one after the other the column vectors of $A$. Then it is not difficult to see that there is a matrix $K$ such that $Vec f(A)=K Vec A$ for all $A$. It follows that $ K (I*X)VecY=(I*f(X)) K VexY$ for all $Y$ (where $A*B$ is the Kronecker product of $A,B$). Thus $ K (I*X)= I*f(X) K$ for all $X$. Write $K=K_{ij}$ in block format where $K_{ij}$ are block $nxn$ matrices. It follows that $ K_{ij}X=f(X) K_{ij}$. Since $K$ is non-zero (because $ f(I)=K Vec I$ and $f(I)$ is not zero), we can find a matrix $K_{ij}$ which is non-zero. Suppose $K_{ij}$ is not invertible and take $v$ a non-zero vector in its kernel. Then we find that $Xv$ is in its kervel for all $X$. This means that all vectors are in its kernel and thus $K_{ij}=0$, contradiction. Thus $f$ is an interiour authomorphism.
   Assume now that $f(I)=0$. Then obviously $f$ is the zero function. Is there something wrong that I cannot see here?

VecX is the vector formed by puting in a $n^2$ vector the columns of the matrix $X$ in order. And $A*B$ is the Kronecker product of the two matrices, that is the famous matrix $a_{ij}B$.


group theory
abstract algebra
superior algebra
superior algebra solved

How can you show that if G is a finite group with just one Sylow p-subgroup then that subgroup must be normal?

The result you need is sylow II: any p-sylow subgroup are conjugate to each other. So if there is only one p-sylow subgroup, then its conjugation must be itself (otherwise it's another p-sylow group), thus it's normal

sylow II is much too strong.
If $H$ is $p$-Sylow in $G$, then $gHg^{-1}$ is $p$-Sylow in $G$ for all $g\in G$. By hypothesis there is only one $p$-Sylow subgroup, so $H=gHg^{-1}$ for all $g$, hence $H$ is normal.


In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms.  Then the common ratio is:
\[ \text{(A)}\ 1 \qquad \text{(B) about}\ \frac{\sqrt{5}}{2} \qquad \text{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \text{(D)}\ \frac{1-\sqrt{5}}{2} \qquad \text{(E)}\ \frac{2}{\sqrt{5}} \]

[hide]Answer (C).
t1=t2+t3, and tn=t1.q^(n-1), so:
               t1=t1q+t1q² = t1 (q+q²)
=> q²+q-1=0
=> q= answer (C)[/hide]

[quote="eryaman"][hide]We can write the sequence as $x$, $ax$, and $a^{2}x$.  So $a^2x+ax=x$.  Since $x\neq0$, we can therefore say $a^2+a-1=0$.  Solving for $a$ gives $a=\frac{\sqrt{5}-1}{2}$.[/hide][/quote]

Yes eryaman, that is exactly how I solved it  :)


I've been given such a task, which I can't solve. It may seen too easy to you, however, I really need help.

$\emptyset \neq X \subset R, \emptyset \neq Y \subset R, X+Y= \{u \in R: u=x+y, x \in X, y \in Y \}$
Prove that
$sup(X+Y) \le sup(X)+sup(Y)$
Give an example of X and Y, when
$sup(X+Y) \not= sup(X)+sup(Y)$

$z\in X+Y$ => there are $(x,y)\in X\times Y$ , $z=x+y$ => $z\le sup(X)+sup(Y)$ for all $z \in X+Y$ => $sup(X+Y)\le sup(X)+sup(Y)$
__________________________
$sup(X+Y)=sup(X)+sup(Y)$ for all X and Y

[u][size=150]for all [/size][/u]$a>0$:
there are $(x,y)\in X\times Y$ , $sup(X)-a/2<x\le sup(x)$
 $sup(Y)-a/2<y\le sup(Y)$
=> there is $z=x+y$ , $sup(X)+sup(Y)-a<z\le sup(X)+sup(Y)$ for all $a>0$
=> $sup(X+Y)=sup(X)+sup(Y)$

[quote="towersfreak2006"]Let \[x_{n}=(-1)^{n},\]  \[y_{n}=(-1)^{n+1}.\] $\sup x=\sup y =1$, but $\sup(x+y)=0$.[/quote]$X+Y$ is defined, in the original post by doctor_faust, as follows:
$\emptyset \neq X \subset R, \emptyset \neq Y \subset R, X+Y= \{u \in R: u=x+y, x \in X, y \in Y \}.$

This means we don't get the elements of $X+Y$ by adding just the corresponding terms of $\{x_{n}\}$ and $\{y_{n}\}$as you have done. In fact, in your example, $X = \{-1,1\}$ and $Y = \{-1,1\}$ (I am assuming the sequences are defined for $n \ge 1$ though it really doesn't matter even if we assume they are defined for $n \ge 0.$) So, $X+Y = \{-2,0, 2\}.$ Hence, $\sup (X+Y) = \sup X+\sup Y = 2.$

What aviateurpilot did was to show that $\sup (X+Y) = \sup X+\sup Y$ for the problem posed by doctor_faust.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114579]Here[/url] is a recent discussion on essentially the same problem.


can you plese post some good logarithm problems, please Im not good at them..because Ive just started. :?  :(

there's a good one here $\rightarrow$ http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43565]www.artofproblemsolving.com/Forum/viewtopic.php?t=43565

[quote="riddler"][quote="danne89"][quote="4everwise"]Try solving for x: $ln(3\sqrt{x})=\sqrt{ln x}$[/quote]
I'm very new to this too.
[hide][tex]ln(3\sqrt{x}=\sqrt{ln(x)}[/tex] <=>
[tex]ln(3\sqrt{x})^2 = ln(x)[/tex] <=>
[tex]ln(3\sqrt{x}^{ln(x)}=ln(x)[/tex] <=>
[tex]3\sqrt{x}ln(x)=ln(x)[/tex] <=>
[tex]3\sqrt{x} = 1[/tex] <=>
[tex] sqrt{x} = 1/3[/tex] <=>
[tex] x = 1/9[/tex][/hide][/quote]
sorry but I didnt understand how you got  $\displaystyle 3\sqrt{x}ln(x)=ln(x)$ from this $\displaystyle ln(3\sqrt{x}^{ln(x)}=ln(x)$
and how you got $\displaystyle ln(3\sqrt{x}^{ln(x)}=ln(x)$ from this$\displaystyle ln(3\sqrt{x})^2 = ln(x)$ :?  :([/quote]

For any log: log(x^y)=y * log(x) and log(x)^2 = log(x^x) (which logical follows from the other)

Anyway it's wrong. ln(3*1/3)=ln(1)=0 and not equal to ln(1/9), which is a negative number because 1/9<1. :(

[quote="danne89"][/quote][quote="riddler"][quote="danne89"][quote="4everwise"]Try solving for x: $ln(3\sqrt{x})=\sqrt{ln x}$[/quote]
I'm very new to this too.
[hide][tex]ln(3\sqrt{x}=\sqrt{ln(x)}[/tex] <=>
[tex]ln(3\sqrt{x})^2 = ln(x)[/tex] <=>
[tex]ln(3\sqrt{x}^{ln(x)}=ln(x)[/tex] <=>
[tex]3\sqrt{x}ln(x)=ln(x)[/tex] <=>
[tex]3\sqrt{x} = 1[/tex] <=>
[tex] sqrt{x} = 1/3[/tex] <=>
[tex] x = 1/9[/tex][/hide][/quote]
sorry but I didnt understand how you got  $\displaystyle 3\sqrt{x}ln(x)=ln(x)$ from this $\displaystyle ln(3\sqrt{x}^{ln(x)}=ln(x)$
and how you got $\displaystyle ln(3\sqrt{x}^{ln(x)}=ln(x)$ from this$\displaystyle ln(3\sqrt{x})^2 = ln(x)$ :?  :([/quote][quote="danne89"]

For any log: log(x^y)=y * log(x) and log(x)^2 = log(x^x) (which logical follows from the other)

Anyway it's wrong. ln(3*1/3)=ln(1)=0 and not equal to ln(1/9), which is a negative number because 1/9<1. :([/quote]
try to use $\text{\LaTeX}$ because its easier to follow what you are saying.

Try eliminating some of the quotes inside there, so your post isn't way too long. You don't need to quote everything, unless everything is actually relevant.

[quote="4everwise"]Try solving for x: $ln(3\sqrt{x})=\sqrt{ln x}$[/quote]

First of all, when x is 0 it obviously works.
Let's solve for this when x is not zero.
$\ln(3\sqrt{x})=\sqrt{\ln x}$
$\ln3 \cdot \ln \sqrt{x} = \sqrt{\ln x}$
$\ln3 \cdot \ln x^{1/2} = \sqrt{\ln x}$
$\ln3 \cdot \frac{\ln x}{2} = \sqrt{\ln x}$

Set $Y$ = $\ln x$
$\frac{\ln 3 \cdot Y}{2} = \sqrt{Y}$
Squaring both sides:
$\frac{(\ln 3)^2 \cdot Y^2}{4} = Y$
$(\ln 3)^2 \cdot Y^2 = 4Y$
${ln 3}^2 \cdot Y = 4$
$Y = \frac{4}{(\ln 3)^2}$
$\ln x = \frac{4}{(\ln 3)^2}$
$x \approx 27.498$

You made a mistake right at the beginning... $\ln (3\sqrt{x})=\ln 3 + \ln \sqrt{x}$, not $\ln 3\cdot \ln \sqrt{x}$.

[b]EDIT:[/b] To give the solution...

$\ln 3 +\frac{\ln x}{2}=\sqrt{\ln x} \Rightarrow Y=\ln x$
$(\ln 3 +\frac{Y}{2})^2=(\ln 3)^2+Y\cdot \ln 3+\frac{Y^2}{4}=Y$
$\frac{Y^2}{4}+Y(\ln (3) -1)+(\ln 3)^2=0$, which does not factor properly... while there is a solution, I don't think it can be found, since my TI-89 won't factor that equation.

[quote="JesusFreak197"]which does not factor properly... while there is a solution, I don't think it can be found, since my TI-89 won't factor that equation.[/quote]

But how about quadratic formula?

Err... non-real result. $(\ln 3 -1)^2<(\ln 3)^2$, so the square root is of a negative number. No wonder my calculator couldn't factor it.

I don't think a is neccessary here, but here is what i got:
\[\log 5^3 = b\]
\[3 \log 5 = b\]
\[\log 5 = \frac{b}{3}\]
\[\log 5 - \log 10 = \frac{b}{3} - \log 10\]
\[\log \frac{5}{10} = \frac{b}{3} - \log 10\]
\[\log \frac12 = \frac{b}{3} - 1\]

hmm?? i used both a and b.. sorry i dunno how to use latex.. so its gonna get quite messy

lg36 = 2lg6 = a
lg 125 = 3lg5 = b

lg (1/12) = lg (5/60)

hence, lg (5/60) = lg 5 - lg 60 , where -lg 60 = -lg 6 - 1

so lg (5/60) = lg (5/6) - 1

lg 5 = b/3 and lg 6 = a/2

substituting a and b, u get lg (1/12) = b/3 - a/2 - 1

shoot me if i am wrong.

Chrnox - you are correct  :10:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]Click Here for a LaTex tutorial[/url]

[quote="riddler"][quote="xenon_nightmare152"]If $\log36 = a$ and $\log 125 = b$ express $\log \frac {1}{12}$ in terms of $a$ and $b$

EDIT: $\frac {1}{12}$ not $\frac 12$   :blush: Sorry about that bornformath[/quote]
I think bornformath got it till $log\frac{1}{2}=\frac{b}{3}-1$
So I think Ill end it... multiplying $2$ both sides you get....
$log\frac{1}{4}=2(\frac{b}{3}-1)$ - Equation $\boxed1$
Now,
$log36=a$ whcih gives us
$log6=\frac{a}{2}$
$log6-log18=\frac{a}{2}-log18$
so $log\frac{1}{3}=\frac{a}{2}-log18$ -Equation$\boxed2$
On adding $\boxed1$ and $\boxed2$ you get
$log\frac{1}{12}=2(\frac{b}{3}-1)+\frac{a}{2}-log18$
Is that correct?????????
Any more cool log problems????[/quote]
what about me thats right isnt it???? :?  :(

Some random problems......
log_4 x^2=10
ln 3xy^2=56
Expand ln (x^2*y^2)/3z^2

I got the $1st$ one... I think...
[hide="1"]
$log4x^2=10$
$2log2x=10$
$log2x=5$
$log2x=log100000$
$2x=100000$
$x=50000$
[/hide]
Can somebody tell me what is $ln$????? :furious:  :mad:

I interpreted #1 differently

[hide]
$\log_4 x^2 = 10$

$2\log_4 x = 10$

 $\log_4 x = 5$

$x = 1024$

[/hide]

xenon_nightmare, Is ny solution to your sum wrong :?  :(  or right...

[quote="danne89"][quote="4everwise"]Try solving for x: $ln(3\sqrt{x})=\sqrt{ln x}$[/quote]
I'm very new to this too.
[hide][tex]ln(3\sqrt{x}=\sqrt{ln(x)}[/tex] <=>
[tex]ln(3\sqrt{x})^2 = ln(x)[/tex] <=>
[tex]ln(3\sqrt{x}^{ln(x)}=ln(x)[/tex] <=>
[tex]3\sqrt{x}ln(x)=ln(x)[/tex] <=>
[tex]3\sqrt{x} = 1[/tex] <=>
[tex] sqrt{x} = 1/3[/tex] <=>
[tex] x = 1/9[/tex][/hide][/quote]

danne89,

If you substitute $x =1/9$ into the original problem you get $ln(3\sqrt{1/9})=\sqrt{ln(1/9)}$

Which yields $ln(1) =\sqrt{-2ln(3)} \Rightarrow 0 = i \cdot \sqrt{2\cdot ln(3)}$

You might want to reconsider your solution, but the problem's a bit too complicated for Getting Started anyway.

riddler...you didn't get the answer that was expected but you could have just expressed it differently. I checked over your work and haven't found anything wrong...yet...
 :huh:

sorry i kept you waiting over that. in your previous posts you sounded as if you were ready to shoot me   :oops:

Pakman: I don't know how you got that, but $\sqrt{4-8\ln 3}$ is going to be imaginary.

[quote="10000th User"][quote="4everwise"]Try solving for x: $ln(3\sqrt{x})=\sqrt{ln x}$[/quote]
$3\sqrt{x}=e^{\sqrt{ln x}}$
$e^{\sqrt{ln x}}=e^ln x^{\frac{1}{2}}=\frac{1}{2}$
$\sqrt{x}=\frac{1}{6}$
$x=\frac{1}{6^2}$

But I plug in and it dont work. Help!  :help:[/quote]

$\sqrt{\ln{x}} \neq \ln{x^{\frac{1}{2}}}$

But what can do to problem? $(ln(3\sqrt{x}))^2=lnx$ dont work
$3\sqrt{x}=e^{\sqrt{lnx}}$ dont work too
Can you solve problem? I dont understand danny89 answer, JesusFreak197 answer, and Bornformath answer...
 :help:  :help:  :help:

If it can be answered, the answer will be imaginary; it will also not be very clean, as $(\ln 3)^2$ is not easy to simplify.

[b]EDIT:[/b] Okay, let me try this.

$\frac{Y^2}{4}+Y(\ln (3)-1)+(\ln 3)^2=0 \Rightarrow Y=2-2\ln 3\pm 2.188i=\ln x$

$x=\frac{e^{\pm i\cdot \sqrt{2\ln (3)-1}}\cdot e^2}{9}$

I think...


he probably won't have a chance to ever win a Super Bowl again as the starting quarterback, especially considering his age. Also, Green Bay is not that elite.

I thought he would retire earlier, but who knows, maybe he will be even more successful. I think a past Super Bowl ring is enough to be satisfied.

Boo Packers!!!

Go Bears!!!!

I like the Colts, but for the next week, boo Colts!!!


[quote="texas137"]it's part of the emailed school report:

"Students scoring at least 100 on the AMC 12 or students (in grades
   10 or below) scoring at least 110 on AMC 10A  or  120 on AMC 10B
   qualify for the AMERICAN INVITATIONAL MATHEMATICS EXAMINATION (AIME)
   which will be held on Tuesday, March 23, 2004. "[/quote]

This would mean that you have seen an emailed report about a B test, right? I thought it would be a few more days before an official cutoff would be set for the 10B, based on when the 10A cutoff was set.

[quote]I thought it would be a few more days before an official cutoff would be set for the 10B, based on when the 10A cutoff was set.[/quote]

That's what I thought, too. But yes, I've seen an emailed report sent this afternoon.

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHH .5  OFF NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

4 points off from the cutoff...oh well, I'll try harder next year.

wow.... where did you get the report? I mean, that was a fast one.

Just curious.

From AMC. They send out the reports in batches. To get it this past Friday, my program must have been in the first batch sent after the B exam. There will probably be another big group tomorrow. (After the A exam, our report came something like a week after other programs had theirs.)

I still haven't get my A results back, is that normal?(Although it's a good thing since I didn't make the AIME)

[quote="mathnovice"]Have any of you gotten your official scores back? (I mean, the AMC people have graded your exams and sent back the official results to your schools) I know what I should have on the AMCB but I haven't gotten the official score yet. Also, do we have to re-register for AIME if we qualified?[/quote]

To answer your questions, everyone (that is, every test administrator) who indicated an email address has received an A score report, and B score reports have been received by SOME administrators. But I don't think any of the paper score reports have reached anyone yet. The email notices are official; they're just not on paper. Some test administrators don't tell AMC about a contact email address, so they don't get the email notices. You should official scores on paper from an A exam pretty soon.

Registration for the AIME is automatic for those who qualify. The test administrator of the qualifying test (12A, 10A, 12B, or 10B) will recieve in the mail the AIME test packets, including instructions for the rather rigid rules about administering that test. I already know my schedule and location for my math team member's AIME, because the test administrator has already told me that by email. NONE of us here locally have seen paper score reports for either A tests or B tests, but we know who will be taking the AIME from our group. The AIME test kits will come with the same mail as the paper score reports from the previous test, I'm pretty sure.

Enjoy your suspense.

My teacher got our official scores back today.  They agree with texas137's 120 and 100 for the '10 and '12, respectively.

[quote="tokenadult"]You should official scores on paper from an A exam pretty soon.[/quote]

We got our paper report for the A on Friday, but you can never know how things will go with the mail...


geometry
circumcircle
inradius
geometry unsolved

[quote="mathservant"]Let R and r be the circumradius and inradius, respectively, of a triangle ABC. Let $h_a$, $h_b$, $h_c$ be the lengths of the altitudes of triangle ABC issuing from A, B, C, respectively, and let $w_a$, $w_b$, $w_c$ be the lengths of the interior angle bisectors of A, B, C, recpectively. Prove that

     $\frac{h_a}{w_a}+\frac{h_b}{w_b}+\frac{h_c}{w_c}\geq 1+\frac{4r}{R}$.[/quote]

This one is Crux Mathematicorum problem #3005, by Pham Van Thuan. The deadline for submitting solutions is over and the problem can now be discussed.

  Darij


function
LaTeX
integration
calculus
calculus computations

Problem 1:
What is the inverse laplace transform of $ F(s) = s$ ?

Problem 2:
Define $ g(t) = \text{ Dirac Delta Function }$. Then find the laplace transform of $ g'(t)$

Doubt:
What is the $ \text{\LaTeX}$ code for Laplace transform?

[quote="Euclidean Geometer"]Problem 1:
What is the inverse laplace transform of $ F(s) = s$ ?[/quote]

The inverse Laplace Transform of $ s$ is $ \delta^{\prime}(t)$.

[quote="Euclidean Geometer"]Problem 2:
Define $ g(t) = \text{ Dirac Delta Function }$. Then find the laplace transform of $ g'(t)$[/quote]

According to the [url=http://mathworld.wolfram.com/]MathWorld[/url] article about [url=http://mathworld.wolfram.com/DeltaFunction.html]the Delta Function[/url],

\[ \int f(x)\delta^{(n)}(x)\,\mathrm{d}x \equiv -\int \frac{d f}{d x} \delta^{(n-1)}(x)\,\mathrm{d}x\].

Given the above definition, the LT of $ \delta^{\prime}(t)$ is:

\[ \int_0^{\infty} e^{-st}\delta^{\prime}(t)\,\mathrm{d}t = s\int_0^{\infty} e^{-st}\delta (t)\,\mathrm{d}t = s\]

[quote="Euclidean Geometer"]Doubt:
What is the $ \text{\LaTeX}$ code for Laplace transform?[/quote]
I don't know that there is an embedded command for the Laplace Transform in $ \text{\LaTeX}$. You'd probably have to create one using the \newcommand command.

Just the $ \text{\LaTeX}$ remark: \mathcal{L} gives $ \mathcal{L}$.


I just got my entry form filled out, and won't be able to fax it in. Can I scan it and email it?

mail it. the postmark has to be by tomorrow.


a russian problem again huh? :) how do they make up these problems anyway ... ?
(rethoric question)

Probably spend 20 days thinking and then come up with a problem they dreamt of in the night before :) Really wierd , hehe :)

hii
after 20 mins of fruitless labour,i think the problem is pretty easy if induction  is used.i have divided it  into a few cases and and i am yet to fully rigorize the proof.i hope that i'm on the rite track.neway how did u solve the problem

I used invariants and I think it's quite ellegant but no time to post it now. :) Later !

I think I have a nice solution to this problem, and I didn't use neither induction, nor invariants.

First we must make the observation that for any $a,b>0$, we have $|a-b|=2\max(a,b) - (a+b)$.

So to show that  $\sum |a_i-b_i|=n^2$ means to show that $\sum 2\max(a_i,b_i) - \sum (a_i+b_i)=n^2$.

Since $\sum (a_i+b_i) = \sum a_i + \sum b_i = 1+2+\cdots+2n = n(2n+1)$, the problem is equivalent to showing that

$\sum 2\max(a_i,b_i) =n(3n+1)$.           (1)

The case $a_1>b_1$ is obvious, because in this case all the $a_i$'s are larger than all the $b_i$'s, so they must be $n+1, n+2, \ldots, 2n$. Therefore $\sum 2\max(a_i,b_i) =2(n+1+n+2+...+2n)=n(3n+1)$.

Likewise, the case $a_n<b_n$ means that all the $b_i$'s are larger than all the $a_i$'s, so the equality (1) holds.

So let's presume
\begin{align*}a_1 &< b_1 \\ a_2 &< b_2 \\ &\vdots \\ a_i &< b_i \\ b_{i+1}&<a_{i+1} \\ &\vdots \\ b_n &<a_n. \end{align*}

Therefore $\sum 2\max(a_i,b_i)=2(b_1+...+b_i+a_{i+1}+...+a_n)$.

Now I want to show that the numbers $b_1, \ldots, b_i, a_{i+1}, \ldots, a_n$ are exactly the numbers $n+1, n+2, \ldots, 2n$, and therefore their sum verifies (1).
It suffices to show that they are all greater than $n$, and because they are all different and their number is $n$, they must be exactly $n+1,\ldots,2n$.

Suppose we want to prove that $b_j > n$, with $1 \leq j \leq i$. There are $n-j+1$ of the $b_i$'s less than or equal to $b_j$ (namely $b_j, \ldots, b_n$).
Also, because $b_j>a_j$ there are $j$ of the $a_i$'s less than $b_j$. So there are at least $n+1$ numbers less to or equal to $b_j$. But that can only mean that $b_j>n$.
In the same manner we can show that for any $j$ with $i+1 \leq j \leq n$, we have $a_j>n$. And that proves (1).

indeed it is a nice solution - great job andrei! look up to the algebra section now ... there are two chinese problems that are still unsolved ... maybe you can help there too :)

if we put Mi=max(ai,bi) and mi=min(ai,bi), then the desired sum=sum(Mi-mi)
on the other hand it's easy to show that Mi>mj for every pair (i,j)
thus 1<=mi<=n and n+1<=Mi<=2n,so Sum=(n+1+n+2+....+2n)-(1+2+...+n)=n2

Claim: $\text{min} (a_i, b_i) \le n < \text{max}(a_i, b_i)$.

Proof: Suppose WLOG that $\text{max}(a_i, b_i) \le n$; then there must exist $\text{min}(a_j, b_j) > n$ by symmetry. Thus $a_i, b_i < a_j, b_j$; but this cannot be since the two sequences are oppositely ordered.

The claim shows that $$\sum_{i=1}^n |a_i-b_i| = 2n+(2n-1)+\cdots +(n+1) - n-(n-1)-\cdots -1 = n^2$$ as desired.

Let for some $k$, $k = b_i, k+1 = a_j$, so we can consider a new sets $A' = \{a_1' = a_1,\ldots , a_j' = k, \ldots , a_n' = a_n\}, B' = \{b_1' = b_1,\ldots , b_i' = k+1,\ldots , b_n' = b_n\}$ and see that $\sum |a_i' - b_i'| = \sum |a_i - b_i|$. So after some of these transpositions we get $\sum |a_i - b_i| = |(n+1) - 1| + |(n+2) - 2| +\ldots + |2n - n| = n^2$. Done


3x+5y=2xy - 1 whereas x and y are positive integers. Find all possible answers for x - y.

3/y + 5/x = 2 - 1/(xy) ; divide both sides by xy

Let 1/y = u and 1/x = v

3u + 5v = 2 - uv
3u + 5v + uv = 2
(u + 5)(v + 3) = 2 + 15 = 17

Using Diophantine eqn

(u + 5)(v + 3) = (17)(1)
So u = 12 and v = -2 implies y = 1/12 and x= -1/2

However it is given that x and y are positive.

Any comments?

You're looking at the inverses of $ x$ and $ y$ and then assuming that they're integers instead of $ x$ and $ y$.  Why didn't you apply the proper factoring to the original equation instead?  You get

$ 6x \plus{} 10y \equal{} 4xy \minus{} 2 \Leftrightarrow$
$ (2x \minus{} 5)(2y \minus{} 3) \equal{} 17$.

By the way, this is not quite Classroom Math.  A problem like this belongs in the Mathcounts forum or HSB.

[quote="t0rajir0u"]You're looking at the inverses of $ x$ and $ y$ and then assuming that they're integers instead of $ x$ and $ y$.[/quote]

I don't understand this but I do understand the factoring. What do you mean by him using inverses? Where did he?

He set $ \dfrac{1}{y} \equal{} u$ and $ \dfrac{1}{x} \equal{} v$ and solved for $ u$ and $ v$.


[quote="modularmarc101"]A natural number with 3 digits is tricubic if it is equal to the sum of its cubed digits. Find all pairs of consecutive numbers such that both are tricubic.[/quote]

Let the two consecutive numbers be $ x_1\equal{}a_1b_1c_1$ as its decimal expression (sorry but I don't know how to add the straight line above the numbers) and $ x_2\equal{}a_2b_2c_2$, where $ x_2>x_1$

Clearly for $ c_1 \neq 9$, $ x_2\minus{}x_1\equal{}c_2^3\minus{}c_1^3\equal{}(c_1\plus{}1)^3\minus{}c_1^3\equal{}1$, yields $ c_1\equal{}0$

mod 10 the cubes we know $ y^3\plus{}z^3\equal{}0(mod 10)$ if and only if $ y\plus{}z\equal{}0(mod 10)$

So we try $ x_1\equal{}190,280,370,...,820,910$ and find that $ 370$ is the only solution in this case.

For $ c_1\equal{}9$ with $ b_1 \neq 9$ (or $ x_1>1000$), we have $ x_2\minus{}x_1\equal{}(b_1\plus{}1)^3\minus{}b_1^3\minus{}9^3\equal{}1$, impossible.

So $ 370$ is the only solution.

[quote="stephencheng"]sorry but I don't know how to add the straight line above the numbers[/quote]

Code \overline{anything} produces $ \overline{anything}$.


find, with proof, all positive integers $p,q,n$ such that $(p+q)^n=2(p^n+q^n)$

[hide="Hint"] Rational Root Theorem ;) [/hide]


Find all polynomials  $ P(x),Q(x)$ such that
\[P(Q(X))=Q(P(x)).\]

[b]sylvect[/b], the "Ukrainian" Vietnamese...
-mathmanman

Nota : I got t0r's proof saved and will all edit back when the deadline is over.

EDIT : here is [b]t0rajir0u[/b]'s proof :

First, it's pretty clear we need $ P(x), Q(x), R(x) \ge 0 \forall x$.  Suppose $ P(x), Q(x)$ share a factor $ x \minus{} r$.  Then $ R(r) \equal{} 0 \implies x \minus{} r | R(x)$.  Now let $ P(x) \equal{} (x \minus{} r)P'(x), Q(x) \equal{} (x \minus{} r)Q'(x), R(x) \equal{} (x \minus{} r) R'(x)$.  Then

$ \sqrt {P'(x)} \plus{} \sqrt {Q'(x)} \equal{} \sqrt {x \minus{} r} R'(x) \implies$
$ \sqrt {P'(r)} \plus{} \sqrt {Q'(r)} \equal{} 0 \implies$
$ P'(r) \equal{} Q'(r) \equal{} 0 \implies$
$ (x \minus{} r)^2 | P(x), Q(x)$,

so WLOG $ (P, Q) \equal{} 1$.  Then

$ 2 \sqrt {P(x) Q(x)} \equal{} R(x)^2 \minus{} P(x)^2 \minus{} Q(x)^2$

which implies $ P(x) Q(x)$ is the square of a polynomial - in particular, all of its roots have even multiplicity.  But then $ P(x), Q(x)$ must themselves be square polynomials.  Hence the solution set is

$ (P, Q, R) \equal{} (A^2, B^2, A \plus{} B)$

where $ A, B$ are polynomials non-negative everywhere.  In a more abstract setting, this boils down to showing that $ \mathbb{R}[x][ \sqrt {P}, \sqrt {Q}] : \mathbb{R}[x] \equal{} 1$, which is obvious since the given (and the square of the given) shows that it cannot be $ 4$ and if it were $ 2$ one of $ P, Q$ is a square, so this argument generalizes readily.

you are wrong
$ \sqrt {P(x)} \minus{} \sqrt{Q(x)} \equal{} R(x)$
not $ \sqrt {P(x)} \plus{} \sqrt{Q(x)} \equal{} R(x)$

Doesn't affect any detail of the solution, although you seem to have changed the problem beyond that and I'm not sure why.

Of course, that's because the n00b was trying to cheat...
http://www.mathlinks.ro/Forum/viewtopic.php?t=168836


limit
function
integration
trigonometry
algebra
domain
linear algebra

[color=violet]Some problems from [/color][color=orange][size=150]Competition for University Students[/size]
  of The Faculty of Mechanics and Mathematics of Kyiv National Taras Shevchenko University[/color]

[color=indigo][size=150]1996[/size][/color]

[b]1.[/b] Let $a,b,c\in\mathbb{C}.$ Find $\mathop{\overline{\lim}}_{n\to\infty}|a^{n}+b^{n}+c^{n}|^{1/n}.$

[b]2.[/b] Let function $f\in C([1,+\infty))$ be such that for every $x\ge1$ there exists a limit $\mathop{\lim}\limits_{A\to\infty}\int_{A}^{Ax}f(u)du=: \varphi(x),$ $\varphi(2)=1$ and function $\varphi$ is continuous at point $x=1.$ Find $\varphi(x).$

[b]3.[/b] The function $f\in C([0,+\infty))$ is such that $f(x)\int_{0}^{x}f^{2}(u)du\to1,\, x\to+\infty.$ Prove that $f(x)\sim\left(\dfrac{1}{3x}\right)^{1/3},\, x\to+\infty.$

[b]4. [/b] Find $\mathop{\sup}\limits_{\lambda}\left(\frac{\mathop{\sum}_{k=0}^{n-1}(x_{k+1}-x_{k})\sin (2\pi x_{k})}{\mathop{\sum}\limits_{k=0}^{n-1}(x_{k+1}-x_{k})^{2}}\right),$ where supremum is taken over all possible partitions of $[0,1]$ of the form $\lambda=\{0=x_{0}<x_{1}<\ldots<x_{n-1}<x_{n}=1\},\, n\ge1.$

[b]5.[/b] Let $\mathcal{D}$ be bounded connected domain with boundary $\partial\mathcal{D}$ and let $f(z),F(z)$ be functions analytical in $\overline{\mathcal{D}}.$ It is known that $F(z)\ne0$ and $\mathrm{Im}\frac{f(z)}{F(z)}\ne0$ for every $z\in\partial\mathcal{D}.$ Prove that functions $F(z)$ and $F(z)+f(z)$ have equal number of zeroes in $\mathcal{D}.$

[b]6.[/b] Let $A$ be linear operator in finite-dimensional space such that $A^{1996}+A^{998}+1996I=0.$ Prove that $A$ has basis which consists of eigenvectors.

[b]7.[/b] Let $A_{1}, A_{2},\ldots,A_{n+1}$ be $n\times n$ matrices. Prove that there exist numbers $\alpha_{1},\alpha_{2},\ldots,\alpha_{n+1}$ not all of which are zeroes such that the matrix $\alpha_{1}A_{1}+\ldots+\alpha_{n+1}A_{n+1}$ is singular.

[b]8.[/b] Let matrix $A$ be such that $\mathrm{tr}A=0.$ Prove that there exist positive integer $n$ and matrices $A_{1},\ldots,A_{n}$ such that $A=A_{1}+\ldots+A_{n}$ and $A_{i}^{2}=0,\, 1\le i\le n.$

[color=indigo][size=150]1997[/size][/color]

[b]1.[/b] Let $1\le k<n$ be positive integers. Consider all possible representation of $n$ as a sum of two or more positive integer summands (Two representations with differ by order of summands are assumed to be distinct). Prove that the number $k$ appears as a summand exactly $(n-k+3)2^{n-k-2}$ times in these representations.

[b]2.[/b] Prove that the field $\mathbb{Q}(x)$ of rational functions contains two subfields $F$ and $K$ such that $[\mathbb{Q}(x): F]<\infty.$ and $[\mathbb{Q}(x): K]<\infty.$ but $[\mathbb{Q}(x): (F\bigcap K)]=\infty.$

[b]3.[/b] Let the matrix $A\in M_{n}(\mathbb{C})$ has unique eigenvalue $a.$ Prove that $A$ commutes only with polynomials of $A$ if and only if $rank(A-aI)=n-1.$

[b]4. [/b] Let $a\in\mathbb{R}^{m}$ be a vector-column. Calculate $(1-a^{\mathrm{T}}(I+aa^{\mathrm{T}})a)^{-1}.$

[b]5.[/b] Let $f$ be positive non-increasing function on $[1,+\infty)$ such that $\int_{1}^{+\infty}xf(x)dx<\infty.$ Prove that the integral $\int_{1}^{+\infty}\dfrac{f(x)}{|\sin x|^{1-\frac{1}{x}}}dx$ is convergent.

[b]6.[/b] Find $\mathop{\lim}\limits_{n\to\infty}\mathop{\sum}_{j=1}^{n}\frac{j}{j^{2}+n^{2}}.$

[b]7.[/b] Let $I$ be the interval of length at least 2 and $f$ be twice differentiable function on $I$ such that $|f(x)|\le1$ and $|f''(x)|\le1,\, x\in I.$ Prove that $|f'(x)|\le2,\, x\in I.$

[hide="Solution to problem 1 (1996)"]
Let $|a| \ge |b| \ge |c|$

then the answer will be $|a|$

Just look to the argument $\phi$ between $a,b$ and $a,c$ when some equalities hold and for some $n$, $n \phi$ is almost multiple of $2 \pi$ by kroneckers approximation theorem and things like $lim_{n \to \infty} \sqrt[n]{r}=1$ for all fixed real $r>0$.

[/hide]

[hide="Solution to problem 1 (1997)"]

If the sum of numbers before $k$ is equal to $R$ and the sum of numbers after it is equal to $T$,
we have there are $2^{R-1}$ ways for the numbers before and $2^{T-1}$ ways for dividing $T$ in natural summands, except for $R,T=0$.
Because of $R+T= n-k$ we have $2^{n-k-2}$ ways for each place of $k$, which we count double when $R=0$ or $T=0$.
We have $n-k+1$ ways to chose the value for $R.$
Hence there are $(n-k+1+2)*2^{n-k+2}$  ways $k$ appear in the summands, as we had to prove.

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geometry
incenter
analytic geometry
ratio
area of a triangle
Heron\u0027s formula
angle bisector

Let $ (x_1, y_1) , (x_2, y_2), (x_3, y_3)$ form a triangle $ \bigtriangleup ABC$, determine the incenter point of  $ \bigtriangleup ABC$ in terms of   $ x_1, x_2, x_3, y_1, y_2, y_3$

Are $ (x_{1}, y_{1}) , (x_{2}, y_{2}), (x_{3}, y_{3})$ the coordinates of the apexes? If so, there should be ";" between x1 and x2.

If so, here is my solution.

Let  $ I (x;y)$ be the incenter, $ a$, $ b$, $ c$ - sides of the triangle. $ a \equal{} \sqrt{(x_{2}\minus{}x_{1})^2\plus{}(y_{2}\minus{}y_{1})^2}$, $ b \equal{} \sqrt{(x_{3}\minus{}x_{2})^2\plus{}(y_{3}\minus{}y_{2})^2}$, $ c \equal{} \sqrt{(x_{3}\minus{}x_{1})^2\plus{}(y_{3}\minus{}y_{1})^2}$,
Let $ S$ be the area of the triangle
Using Heron's formula we get $ 16S^2 \equal{} (a \plus{} b \plus{}c)(b\plus{}c \minus{} a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)$
$ S \equal{} pr$, where $ p$ - semiperimeter.

$ r^2 \equal{} \frac{(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)}{4(a\plus{}b\plus{}c)}$, $ r$ is the $ y$ coordinate of $ I$.

Now you should find $ x$.

I am not sure if my solution is suitable, because the $ y$ coordinate is expressed by a very long formula. Can anyone suggest a better solution?

From the given condition, we have the equations of the bisector of $ \angle ABC$ and $ \angle ACB$ (notice that the distance of any point on the bisector to its two sides of the angle is equal).
Then, let the two bisector intersect and we get the co-ordinate of the incenter
(we have 2 bisector of and angle, inside-bisector and outside-bisector, so we get 4 co-ordinates of the incenter, then we have to assess and choose the right one)

[quote="miwalin"][quote="skalgar"]If so, here is my solution.

Let  $ I (x;y)$ be the incenter, $ a$, $ b$, $ c$ - sides of the triangle. $ a \equal{} \sqrt {(x_{2} \minus{} x_{1})^2 \plus{} (y_{2} \minus{} y_{1})^2}$, $ b \equal{} \sqrt {(x_{3} \minus{} x_{2})^2 \plus{} (y_{3} \minus{} y_{2})^2}$, $ c \equal{} \sqrt {(x_{3} \minus{} x_{1})^2 \plus{} (y_{3} \minus{} y_{1})^2}$,
Let $ S$ be the area of the triangle
Using Heron's formula we get $ 16S^2 \equal{} (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)$
$ S \equal{} pr$, where $ p$ - semiperimeter.

$ r^2 \equal{} \frac {(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)}{4(a \plus{} b \plus{} c)}$, $ r$ is the $ y$ coordinate of $ I$.

Now you should find $ x$.[/quote]
$ r$ is not the $ y$ coordinate of $ I$, to be sure.[/quote]

HINT:

The angle bisector of a vertex of a triangle divides the opposite side of the triangle into the ratio of the adjoining sides

$ r$ is the $ y$ coordinate, provided $ A (0;0)$ If we draw the radius which intersects the tangent line in point where it touches the circumference, the radius is perpendicular to the tangent line. :)

[quote="pankajsinha"]

The angle bisector of a vertex of a triangle divides the opposite side of the triangle into the ratio of the adjoining sides[/quote]

How can it help us in calculating $ x$?