Find the integer solutions of the equation: \[x^{3}+8=7\sqrt{8x+1}\]

First $8x+1\geq 0,$ integer $x=0,1,2,\cdots.$
Substitute $x=0,1,2$ into the given equation, we can get integer soultion.
But for $x=3,$ this is O.K.
On the condition of $x\geq 0,$ we have
$x^{3}+8=7\sqrt{x+1}\Longleftrightarrow (x^{3}+8)^{2}=49(8x+1)$
$\Longleftrightarrow x^{6}+16x^{3}-392x+15=0$
$\Longleftrightarrow (x-3)(x^{5}+3x^{4}+9x^{3}+43x^{2}+129x-5)=0.$
Ovbiously for $x=4,\ 5,\ 6,\ \cdots,$we have $x^{5}+3x^{4}+9x^{3}+43x^{2}+129x-5>0$
Thus desired integer solution is $x=3.$


I will post it here on  the 23rd. Anybody who wants to participate, say so. Also, if someone would like to do a countdown round, I'll help on that.

What time would work for you?

Billy

I'll do it mostly but im not perfectly sure

I think I can. What level will the questions be? I'll help with the countdown round. Last time was fun. :lol:

There WILL be some toough problems. :? :/ But there will also be a few easy problems.

I will post the problems both in PDF and as plain-text. We will do the problems on the honor system - you will have to get your answers in by 50 minutes after I post on the sprint, and 10 minutes for the targets

I would really like to do it, but I have to be in Vancouver, BC for some intermural competition with the Oregon and Washington MathCounts teams (I am on the WA team) and the British Columbia team for some other Canadian competition  . . . could you post it on another date, possibly the 24th?

[quote="solafidefarms"]Let me sum up everything I have said:

 I will post the problems both in PDF and as Word document. We will do the problems on the honor system - you will have to get your answers in by 50 minutes after I post on the sprint, and 10 minutes for the targets.

Here is the schedule:

7:00: Sprint posted.
7:50: Must have your answers in(note: the extra time does not mean that you have 50 minutes to solve, it means that you don't have to type your answers during the test); target #1 posted
8:00: Must have your answers in for target #1; target #2 posted
8:10: Must have your answers in for target #2; target #3 posted
8:20: Must have your answers in for target #3; target $4 posted
8:30: Must have your answers in for target #4

Yes, you will PM me the answers.

I will be posting the problems in a new thread, however, I will delete that thread and move the problems over to this one after it is all done.

And the last question: I will be posting rankings hopefully before 9 EST, but.... don't count on it. If I don't get it done tonight, it'll be at least Monday before I post scores.

Have fun!

Billy[/quote]

EDIT: Sorry I cannot compete in this competition. I'm a bit busy at the moment.

The results are in!

Firstly, the prolegomenous opening remarks:

Thank you all for participating in my mock competition. It was great fun to write, though admittedly I should have checked the problems before posting. I am quite sorry for the 3 mistakes in the sprint round, and also for my cunctation in giving the scores. I had not even done all my problems till this afternoon! (I said it would be prolegomenous!)

[hide="The rankings FOR SPRINT SCORES ONLY"]
Tarquin: 18
jb05: 17.5
nat mc: 16
Xantos C. Guin
chess64
GoBraves: .5 point![/hide]

[hide="The rankings BASED ON TARGET SCORE ONLY"]
nat mc: 10 pts.
jb05: 6 pts.
Xantos, chess, and Tarquin: 4 points apiece
[/hide]

[hide="the final rankings"]
Nat Mc!!!!
jb05
Tarquin
Xantos C Guin
chess64[/hide]

Congratulations, nat mc!

Thank you all again for participating! :lol:

Billy

NOTE: I didnt get all the problems wrong...I didnt even compete, I just pointed out a mistake so Billy gave me .5 points.

And nice job Tarquin on the sprint round for STARTING them with 10 min. to spare and still getting most right.

YAY!! I won!  Where did the questions go? Can we still look t them?

Hmm. That puts me in the top third, so i will get at least 76th... Probably better. Cool. It would have been quite a bit more enlightening if we had more than 5 people perticipating!

[quote="mathgeek2006"][quote="solafidefarms"]Here:http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34508[/quote]
Could you put that in a URL?  (It's not right now)[/quote]

how does it matter....copy paste it

[quote="mathgeek2006"][quote="solafidefarms"]Here:http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34508[/quote]
Could you put that in a URL?  (It's not right now)[/quote]

[hide="I hope this makes you happy..."]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34508  :P [/hide]

Yeah, I'm too [hide]lazy [/hide]to copy-paste...

Solafidefarms can I still turn in my answers? I forgot about the competition until today.
[hide]
I didnt cheat
[/hide]

Sure.

I'll have to go back through though and find my answer key, but I'll check it.

Billy


Easy one dude but it seems terrifying at first sight :rotfl:  :rotfl:
Nice one though,just solved it.

Does this problem come from a book or some contest?
Could someone tell me it`s source :?:


symmetry
linear algebra
matrix
combinatorics unsolved
combinatorics

In a very big party, for any A and B there exists only one person Z such that A and Z, and B and Z are friends.

Show that there exists at least one person such that he is friend of all other people.

[Edited! A typo in the part "for any A,B,..."]

[quote="grobber"]Are you sure you've seen it worded precisely like this? As far as I know, the friendship theorem states that the conclusion holds if [b]any two[/b] people have a unique common friend (i.e. it doesn't matter whether the two are friends to each other or not), so your hypothesis is weaker.[/quote]
you are right, grobber. in fact without that the statement is false.eg: take the Petersen graph.

[quote="seshadri"]you are right, grobber. in fact without that the statement is false.eg: take the Petersen graph.[/quote]

Actually, why not just take the graph formed by the sides of a pentagon?

Anyway, a proof of the friendship theorem can be found at http://matholymp.com/TUTORIALS/Friendship.pdf .

  darij

actually the proof is very much difficult and need linear algebra !!!  :blush:

The combinatorical proof is very nice, but how is the proof using linear algebra?

Here's a proof, but it uses both combinatorial and algebraic stuff :).

Let $x,y$ be two vertices of the graph $G$ which are not connected to each other. They have exactly one common neighbor $z$. Now, $z,x$ must have a common neighbor $u_x$, and $z,y$ must have a common neighbor $u_y$. Let $p_x$ be any neighbor of $x$ different from $u_x,z$. $p_x$ and $y$ must have a common neighbor $p_y$, which is different from $u_y,z$, because $p_y=u_y$ implies that $z,u_y$ have ar least two common neighbors, $y$ and $p_x$, while $p_y=z$ implies that $z,x$ have at least two common neighbors, $u_x$ and $p_x$. Also notice that given different neighbors $p_x,\bar p_x\ne u_x,z$ of $x$, the corresponding points $p_y,\bar p_y$ are different, for otherwise $p_x,\bar p_x$ would have at least two common neighbors, $x$ and $p_y=\bar p_y$. This means that we have a one-to-one application from the neighbors of $x$ diferent from $u_x,z$ to those of $y$ different from $u_y,z$, defined by $p_x\mapsto p_y$, as bove. The degree of $x$ is thus $\le$ that of $y$. By symmetry, the degrees of $x$ and $y$ must be equal. This shows that two vertices which are not connected to each other have equal degrees.

We now look at the connected components of the complement $\overline G$ of $G$. If there is a component with only one vertex, we're done: that vertex is connected to every other vertex in $G$. If, on the other hand, every component has at least two vertices, then there can be only one component, for any two distinct vertices in a component would be connected to any two distinct vertices in any other component, contradicting the hypothesis. This means that we now have to check the case when $\overline G$ is connected. From the paragraph above it follows that in this case $G$ is regular, every vertex having the same degree which we'll denote by $d$.

Let $A$ be the adjacency matrix of $G$. By the above, $A^tA$ equals the matrix havin $d$ on the main diagonal and $1$ everywhere else, meaning that $A^2=A^tA$ has $d-1$ as an eigenvalue with multiplicity $n-1$ and $d+n-1$ as a simple eigenvalue. In turn, this means that the eigenvalues of $A$ are $\varepsilon_i\sqrt{d-1},\ i=\overline{1,n-1},\ \varepsilon_n\sqrt{d+n-1}$, where the $\varepsilon$s are either $1$ or $-1$. By replacing $A$ with $-A$ with necessary, we may assume WLOG that $\varepsilon_n=1$.

The trace of $A$ equals zero, but at the same time it’s of the form $-t\sqrt{d-1}+\sqrt{d+n-1}$, where $t$ is some (positive) integer. From here we get $\frac n{d-1}=t^2-1$. The important thing here is that $d-1|n\ (*)$. On the other hand, if we consider $A$ to be the incidence matrix of a [url=http://mathworld.wolfram.com/BlockDesign.html]block design[/url], this block design satisfies the axioms of a [url=http://en.wikipedia.org/wiki/Projective_plane]projective plane[/url], so $n=d^2-d+1$ (to see this, just use the formulae at the top of the page explaining what a block design is, with $\lambda=1$, and $b=v=n,\ r=k=d$), meaning that $d-1\not |n$. This contradicts $(*)$ (well, except for $d=2,n=3$, but in this trivial case it's easy to see that the graph, which is a triangle, does indeed contain a vertex which is connected to all the others).

Grober can you please explain more about the block design argument?

Is any incident matrix can be considered as block design?

i don't understand the explaination from the wolfram :blush:  :oops: \

thanks

Incidentally, with a little more work we can show that the graph $G$ of friendships forms a "pinwheel", i.e. a bunch of disjoint triangles all connected to a single vertex.

Lemma:  $G$ is a connected union of disjoint triangles.

Proof:  $G$ is obviously connected from the problem condition.  For any edge $ab$, $a$ and $b$ are both connected to some vertex $c$, so $ab$ is part of triangle $abc$. If two triangles share an edge $ab$, then $a$ and $b$ are connected to two vertices.

The lemma plus the above implies that $G$ is a pinwheel.

If you dont know eigenvalues and other stuff here you can find nice elementary proof . https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=2ahUKEwiX3czd7s7cAhUMFCwKHZD8CA8QFjAEegQIBRAB&url=https%3A%2F%2Fwww.researchgate.net%2Fpublication%2F268313555_The_Friendship_Theorem&usg=AOvVaw17SEnQ4GKI-EvMpGmOPMU0
EDIT: If link is not working type on google (PDF) The Friendship Theorem - ResearchGate


Andrea and Lauren are $ 20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $ 1$ kilometer per minute. After $ 5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?

$ \textbf{(A)}\ 20 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 55 \qquad
\textbf{(D)}\ 65 \qquad
\textbf{(E)}\ 80$

[hide]
Let $ r_a$ be Andrea's speed and $ r_l$ be Lauren's speed.  We know that $ r_a\plus{}r_l\equal{}1$ and $ r_a\equal{}3r_l$, so $ r_a\equal{}\frac{3}{4}$ and $ r_l\equal{}\frac{1}{4}$.

After $ 5$ minutes, they are separated by $ 15$ kilometers.  Only Lauren bikes after this, and the time she takes is

$ \frac{15}{r_l}\equal{}\frac{15}{\frac{1}{4}}\equal{}60$.

So the total time is $ 65$ minutes.  Answer is $ D$. [/hide]

Please correct me if I'm wrong.  Here's how I did it:

[hide="Solution"]
Let $ l$ be Lauren's rate in km/minute.  Then Andrea's rate is $ 3l$, and since the sum of their distances traveled in one minute is $ 1$, we have $ 3l \plus{} l \equal{} 4l \equal{} 1 \Longrightarrow l \equal{} \dfrac{1}{4}$.  Since they have traveled $ 5$ km before Andrea gets a flat tire, Lauren has $ 15$ km left to go.  She does this in $ \dfrac{15}{1/4} \equal{} 60$ minutes, so the total time is $ 60 \plus{} 5 \equal{} 65$ minutes, or $ \boxed{D}$.
[/hide]
Seems a little too straightforward for #20, don't you think?

EDIT: Ugh, beaten by Euler.

no this was #20 and yea it was MAAAAD easy

[quote="ak909090"]I think this was like number 10...number 20 was something else.[/quote]It was #20.


invariant
linear algebra
matrix
algebra
polynomial
vector
complex numbers

Are there matrices such that all nontrivial elements of their commutant are invertible (let's say, over the complex numbers)? If so, how do they like?

(Something stupid was here.)

If $A$ has this property, then it must be invertible, and $\lambda I+A$ must be either invertible or zero for all $\lambda$. It cannot be zero, since scalar matrices don't work. Therefore, $A$ has empty spectrum. [i]A forteriori[/i], $A$ does not exist.

Ok, this was trivial. Now, passing to the real numbers? What about arbitrary fields?

If $A$ commutes with $B$, then the range of $B$ is an invariant subspace for $A$. Hence, any matrix without proper invariant subspaces has a very special commutant. For instance, $\begin{pmatrix}1&1\\-1&1\end{pmatrix}$. Is the converse true? Maybe. :maybe:

[quote="mlok"]Hence, any matrix without proper invariant subspaces has a very special commutant.[/quote]
Is this a technical term? Are we going to add it alongside perverse sheathes and wonderful compactifications and friendly numbers? :D

The full solution, over an arbitrary field:

Let $A$ be an $n\times n$ matrix.

If the minimal polynomial of $A$ factors as $p\cdot q$ for two nontrivial polynomials $p,q$, we have $0=p(A)\cdot q(A)$, and neither $p(A)$ nor $q(A)$ are zero. Both of these matrices commute with $A$, and neither is invertible.

If the minimal polynomial $p$ of $A$ is irreducible of degree $k<n$, we have the following:
Choose $v_{1}$ nonzero arbitrarily, and let $v_{2}=Av_{1},v_{3}=Av_{2},\dots,v_{k}=Av_{k-1}$. We have $Av_{k}=A^{k}v_{1}=-c_{k-1}v_{k}-\cdots-c_{1}v_{2}-c_{0}v_{1}$, where the $c_{i}$ are the coefficients of $p$. Since $p$ is irreducible, these $k$ vectors are linearly independent; any dependence relation would give a polynomial factor of $p$ with lower degree. Now choose $v_{k+1}$ arbitrarily outside the span of the first $k$ vectors, and let $A^{i}v_{k+1}=v_{k+1+i}$ for $1\le i\le k-1$. The first $2k$ vectors are now linearly independent: any dependence relation can be expressed as $p_{1}(A)v_{1}=p_{2}(A)v_{k+1}$. Since $p_{2}$ has degree less than $k$, it is relatively prime to $p$ and we can find a polynomial $q$ such that $q\cdot p_{1}\equiv 1\mod p$. Then $q(A)p_{1}(A)v_{1}=v_{k+1}$, contradicting our choice.
Repeat this process until we have a basis for $\mathbb{R}^{n}$. In this basis, the matrix of $A$ is $\begin{bmatrix}M&0&\cdots&0\\0&M&\cdots&0\\ \vdots&\vdots&\ddots&\cdots\\0&0&\cdots&M\end{bmatrix}$, where $M=\begin{bmatrix}0&0&0&\cdots&0&-c_{0}\\ 1&0&0&\cdots&0&-c_{1}\\ 0&1&0&\cdots&0&-c_{2}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-c_{k-2}\\ 0&0&0&\cdots&1&-c_{k-1}\end{bmatrix}$. Then $\begin{bmatrix}I&0&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\cdots\\0&0&\cdots&0\end{bmatrix}$ is a non-invertible matrix that commutes with $A$.

This leaves only one case: $A$ has an irreducible minimal polynomial, which is equal to its characteristic polynomial. In this case, the commutant of $A$ is naturally isomorphic to the field of polynomials mod $p$; by the field structure, all except zero are invertible.

To prove this claim, note that the above construction generates one block given any starting vector, so we can translate to a new space:
Let $V$ be the algebra of polynomials mod $p$, and consider the linear transformation $A$ from $V$ to itself given by $A(q(x))=x\cdot q(x)$. In the standard basis, the matrix of $A$ is exactly the companion matrix we found earlier. Any linear transformation $B$ which commutes with $A$ commutes with multiplication in general: $B(x^{n})=x^{n}B(1)$, and $B(q(x))=q(x)\cdot B(1)$. The transformation $B(q(x))=q(x)\cdot\lambda(x)$ is exactly $\lambda(A)$, and every linear transformation which commutes with $A$ is a polynomial in $A$.


analytic geometry
combinatorial geometry

Is there a generalization of the shoestring for 3-d coordinates?

Not in the same form, since you'd need to use a triple scalar product instead of a cross product.

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

Edit:  If you're asking about how to find the volume of the [url=http://mathworld.wolfram.com/ConvexHull.html]convex hull[/url] of a set of points, I really doubt there's a formula, since finding the convex hull of an arbitrary set is NOT easy.


Together Susan and Sam have $ \$140$. When Sam gives Susan $ \$10$, they each have the same amount. How many dollars did Sam have originally?

At the end they each have 70, so sam starts w/ 70+10=80


group theory
abstract algebra
superior algebra
superior algebra solved

Show that if H is a subgroup of Sn, then either every member of H is even or exactly half the members are even.

Consider the composition $ H\hookrightarrow S_{n}\xrightarrow{\mathrm{sgn}}\{\pm1\}$. owk


Let $ p$ be a prime number ($ p > 2$) and $ n,k$ be positive integers such that $ n > kp$. Prove that $ (^n_{kp}) \minus{} (^m_k)$ is divisible by $ p$ (here, $ m \equal{} [\frac {n}{p}]$).

Other way is Lucas theory.
Let $ n \equal{} \sum_{i \equal{} 0}^t a_ip^{i}$
      $ k \equal{} \sum_{i \equal{} 0}^t b_i p^{i}$
Where $ a_i,b_i\in\{0,1,...,p \minus{} 1\}$
Then
$ C_{n}^k\equiv \prod_{i \equal{} 0}^t C_{a_i}^{b_i}(\mod p)$
Apply it for this problems .


geometry
incenter
angle bisector
geometry proposed

In the triangle $ABC$ we divide each of the angles $B$ and $C$ in three equal parts. Let the rays $d'$ and $d\"$ be the trisectors of the angle $B$ and $e'$ and $e\"$ be the trisectors of the angle $C$. (The notation counts the rays from $BC$ toward $AB$, respectively $AC$.) Let us denote by $N$ the intersection of $d'$ and $e'$ and by $M$ the intersection of $d\"$ and $e\"$.
 Prove that $MN$ is perpendicular on $BC$ if and only if the angle $B$ has the same measure as the angle $C$

The trisectors $d' \equiv BN,\ e' \equiv CN$ of the angles $\angle B, \angle C$ closer to the side BC are the internal bisectors of the angles $\angle CBM, \angle CMB$. Hence, N is the incenter of the triangle $\triangle BCM$ and MN is the internal bisector of the angle $\angle CMB$. It is perpendicular to the opposite side BC iff the triangle $\triangle BCM$ is isosceles with $\frac 2 3 \angle B = \angle CBM = \angle BCM = \frac 2 3 \angle C$ or $\angle B = \angle C$.

[u]Yetti , nice solution![/u]

[b] An other aproach (?) .[/b]

 Let $K$ the point where  $MN$ meets  $BC$ and $MN \perp BC$.  From the angle bisector teorem we take :

 $\frac{BK}{BM}= \frac{NK}{NM}= \frac{CK}{CM}$ . This means that  right angled triangles $MKB , MKC$ are similar , so $\angle MBK = \angle MCK$
 The other part of the problem is obvious.
 This moment I wonder if really the two triangles are similar , or I must delete the post.The  argument works for the converse however , which is very easy. I ....run in the class !!!

 Babis

[u]Babis[/u]


function
limit
real analysis
real analysis unsolved

Consider the signum  function which is defined as \[ sgn(t)=\left\{\begin{array}{c}1\;\;\;for\;\;t\geq 0\ 0\;\;\;for\;\;t<0\end{array}\right.\] The Fourier transform of this signal is usually comuted in Signals anf systems textbooks as a limit like \[ sgn(t)=\lim_{a\rightarrow 0}\left[e^{-at}u(t)-e^{at}u(-t)\right]\] for $ a>0$. Thus \[ F\left\{sgn(t)\right\}=\lim_{a\rightarrow}\left[\frac{1}{a+j\omega}-\frac{1}{a-j\omega}\right]=\frac{2}{j\omega}\] Now the above result should be same even if I apply series expansion to the above limit. That is,
\[ F\left\{sgn(t)\right\}=\lim_{a\rightarrow}\frac{-2}{a}\left[\frac{j\omega}{a}+\left(\frac{j\omega}{a}\right)^{3}+\left(\frac{j\omega}{a}\right)^{5}+\cdots\right]=\frac{2}{j\omega}\]Can I do that? If not why? Can anybody clarify me?

regards
bhupala.

You can't, because the series doesn't converge everywhere; it diverges near the point you care about.

By the way: begin{cases}(stuff)end{cases} works well for piecewise definitions, and & is the right tool for horizontal spacing in an array environment.

I would assume that you really want $ \text{sgn}(t) \equal{}\begin{cases}1, & t > 0\\ \minus{}1, & t < 0\end{cases}$

And I don't care how or whether it's defined at zero.

If a function is in $ L^{1}(\mathbb{R})$ (is absolutely integrable on the line), then we may define its Fourier transform pointwise by the usual integral.

If a function is in $ L^{2}(\mathbb{R})$ (is square integrable on the line), then we may define its Fourier transform as a member of $ L^{2}$ by using the Plancherel theorem to extend the definition from $ L^{1}\cap L^{2}.$

The signum function is not in $ L^{1}$ or in $ L^{2},$ so those definitions do not apply. The only sense in which the signum function does have a Fourier transform is as a tempered distribution. It will have a transform that is a tempered distribution; that distribution might not be given by a locally integrable function.

Pursuing that line of thought:

$ \text{sgn}(t) \equal{}\lim_{a\to0^{\plus{}}}[e^{\minus{}at}u(t)\minus{}e^{at}u(\minus{}t)]$

makes sense as the [i]weak[/i] limit (distributional limit).

Hence the following should make sense [i]as a weak limit[/i]:

$ \widehat{\text{sgn}}(\omega) \equal{}\lim_{a\to0^{\plus{}}}\left[\frac{1}{a\plus{}i\omega}\minus{}\frac{1}{a\minus{}i\omega}\right]$

$ \equal{}\lim_{a\to0^{\plus{}}}\frac{\minus{}2i\omega}{a^{2}\plus{}\omega^{2}}.$

(Note: as a mathematician, not an EE, I'd prefer not to write $ j.$)

But at this point, we do realize that we are dealing with a weak limit, and that the limit must be interpreted as a tempered distribution. It's not enough to say that the limit is the function $ \frac{2}{i\omega}.$ After all, that's not a locally integrable function. The true limit is the [b]principal value[/b] of $ \frac{2}{i\omega},$ considered as a tempered distribution.

Does that power series converge weakly? I'm not going to say that it doesn't but I can't see how to use it.

So now what happens for the Fourier transform of the signal,\[ \lim_{a\to 0}\left[e^{\minus{}at}u(t)\plus{}e^{at}u(\minus{}t)\right]\]
atleast in distributional sense?

regards,
bhupala.


probability
LaTeX
number theory proposed
number theory

If you have

x = :sqrt: 3/1 + 4/2 + 5/3 + ... (n+2)/n

what is the probability that the x is a rational number (n is of course a positive integer).

is the square root over everything or just the 3 because if it is just the 3, it will never be rational.

let p be a maximal prime number which is less tnan n.
Put A/B= 3/1 + 4/2 + 5/3 + ... (n+2)/n
B divisible by p, but A is not.  So x not rational

I thought that there would have been some probability.

[quote="spoudyal"]Square root is over everything. I just do not have time to learn latex.[/quote]I don't think you don't have 15 minutes to learn $\LaTeX$ ...  :?  :? ... use the link in my signature ;)

Sure if only 15 minutes. I installed the programs needed for it just now. I will know LATEX by Monday.


search
function
number theory
number theory proposed

Host topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:

To make it short: [b]ALWAYS USE YOUR COMMON SENSE IF POSTING![/b]
If you don't have common sense, don't post.

More specifically:

[u]For new threads:[/u]

a) [b]Good, meaningful title[/b]:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad.
That's in fact a requirement for being able to search old problems.

Examples:
[u]Bad titles:[/u]
 - "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
 - "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
 - "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
 [u]Good titles:[/u]
 - "On divisors of a³+2b³+4c³-6abc"
 - "Number of solutions to x²+y²=6z²"
 - "Fibonacci numbers are never squares"

b) [b]Use search function[/b]:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

c) [b]Good problem statement[/b]:
Some recent really bad post was:
[quote="tranthanhnam"]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question, no answer, no sense (and is by the way stupid).
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.

[u]For answers to already existing threads:[/u]

d) [b]Of any interest and with content[/b]:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) [b]Well written and checked answers[/b]:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.

To repeat it: [b]ALWAYS USE YOUR COMMON SENSE IF POSTING![/b]

Everything definitely out of range of common sense will be [u]locked or deleted[/u] (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this in http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .

Greetings ZetaX.

I have posted many new topics, but I have not received even one answer. You may have observed that my problems are almost entirely in the in the tradition of classical diophantine analysis. Did I not use my common sense when posting? Is this the right place to post these problems? Are my problems too classical? Are they too difficult? Are they not interesting? Or what else? And why?

Thanks.


equation
algebra
Perfect Square
exponential equations

[quote="cdymdcool"]4^x+4^1993+4^1996=perfect square
what are the three intefral values of x?[/quote]

Did you prove by a formal demostration that this are the only 3 or just factorize the expression as the square of a binomial???

...those seems to be the only ones but I'm not totally sure, short ago a similar problem was solved on the intermediate, and it is also similar to an easy IMO problem, 2006 problem 4.

 :)


Hi,

I do not find an archive for the "Problem of the day" section.
Is there an archive for the "Problem of the day" section and
if so, how do I find it?

Thank you!

Sebastien

The problem is to be found here: http://www.mathlinks.ro/ .

sebastien: no, but up to what I know, this is planed for the future.

Thank you, ZetaX.

If this is so, do you know how
to recover the graph theory "Problem of the day"
which appeared only a couple of days ago?
(I would estimate it to have appeared not
within the last week.)

It was about constructing a graph with many edges
but no cycle of length four.
I made a mental note to myself to read it later,
but know I cannot find it anymore, even when
searching with the search facility for various
combinations of
like "graph", "cycle", "length four", "length 4" and other terms.

This was the POTD about a week ago: http://www.mathlinks.ro/viewtopic.php?p=816424#816424
A couple of days later this was the POTD: http://www.mathlinks.ro/viewtopic.php?p=819057#819057

These two match the "description" you've made so I'm hoping it's one of them :)

Thank you very much!
The second was the post I was looking for.

While it's not finished yet, this is a start for such an archive:

http://www.mathlinks.ro/search.php?search_id=potd

A few more things are to be added. I'm waiting on suggestions on what to add. For now I have

- the date when the problem was potd
- solutions (if existing)
- the link to the contest it belongs to

Let me know if there is anything else I have forgotten.


Three circles of radius $ r$ cover a circle of radius $ 1$. Find the minimum possible value of $ r$.

$ r \equal{} \frac {\sqrt {3}}{2}$? Any smaller and some point on the circumference would be uncovered.

(I don't have a proof for this, and the fact that the total area is $ \frac{9}{4}$ makes it a very wasteful tiling, but I can't find a better case.

yes, that's the correct point (to consider the circumference)
[hide]First, we must prove that any part of the circumference that a circle covers is continuous (not 2 disjoint arcs that don't connect together, just one continuous part of the arc. This has nothing to do with the calculus definition)
Well this is true for any size of circle, but we only need to prove it for circles smaller than the unit circle given.
The unit circle's circumference is covered at all if and only if the circles intersect. If they are tangent then only 1 point is covered. Otherwise, there are 2 intersections. Then one of the 2 arcs created by the 2 points is covered. (in fact, the minor arc because the second circle is smaller)
Ah, clearly, some part of the circumference is covered, and both arcs can't be covered.
And there can't be any other possible region of the circumference covered because this requires more intersections.
yay
then by pigeonhole, one circle must cover at least $ \frac {1}{3}$ of the circumference. There are 2 poitns on this arc $ \sqrt {3}$ units apart, so the smallest possible radius is $ \frac {\sqrt {3}}{2}$[/hide]

Inscribe an equilateral triangle in the circle, and draw three circles that have the sides of this triangle as diameters. Clearly they cover the original circle and their radius is $ \sqrt{3}/2$. Now suppose that $ r<\sqrt{3}/2$. If a circle is not tangent with the unit circle or interior to it it's intersection with the unit circle is a chord whose lenght is less than $ \sqrt{3}$, and therefore the arc delimited is less than $ 120^\circ$. We conclude that we must have an arc that is not covered, contradiction.


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

Consider a quadrilateral ABCD inscribed in a circle such that AB is the diameter.Draw tangent to the circle at A and B. Join mid point E of CD with A and B. Draw perpendicular from the mid point of AD to AE and extend the perpendicular to meet the tangent at A in  M. Similarly Draw perpendicular from the mid point of BC to BE and extend the perpendicular to meet the tangent at B in N.
Now Prove that MN is parallel to CD.

Circumcenter $ O$ of $ ABCD$ is midpoint of $ AB.$ Let $ K, L$ be midpoints of $ DA, BC.$

$ AO^2 \minus{} BO^2 \plus{} BL^2 \minus{} EL^2 \plus{} EK^2 \minus{} AK^2 \equal{} \frac {_1}{^4}(BC^2 \minus{} BD^2 \plus{} CA^2 \minus{} DA^2) \equal{} 0$

By Carnot theorem for $ \triangle ABE,$ perpendiculars from $ O, L, K$ to $ AB, BE, EA$ are concurrent at a point $ P.$ Let $ Q$ be midpoint of $ MN.$ Let $ F$ be reflection of $ E$ in $ O.$ $ AEBF$ is parallelogram. $ \triangle PMN \sim \triangle AEF$ are similar, because their corresponding sides $ PM \perp AE, PN \perp (AF \parallel BE)$ and corresponding medians $ PQ \perp AO$ are perpendicular. It follows that $ MN \perp EF.$ Since $ EF \perp CD$ as its perpendicular bisector, $ MN \parallel CD.$


linear algebra
matrix
Extremal combinatorics
combinatorics
IMO Shortlist

For an ${n\times n}$ matrix $A$, let $X_{i}$ be the set of entries in row $i$, and $Y_{j}$ the set of entries in column $j$, ${1\leq i,j\leq n}$. We say that $A$ is [i]golden[/i] if ${X_{1},\dots ,X_{n},Y_{1},\dots ,Y_{n}}$ are distinct sets. Find the least integer $n$ such that there exists a ${2004\times 2004}$ golden matrix with entries in the set ${\{1,2,\dots ,n\}}$.

Can they be multisets? I mean, can we repeat elements in the same column and row? or must they be distinct?, and if we can repeat them, what equality condition should we use??? :(

That's problem 3 of the 2nd German TST 2005. I guess it's from the ISL 2005. LaTeX writeup:

[color=blue]For an $n\times n$ matrix A, let $X_i$ be the set of all entries in the row i, and $Y_j$ be the set of all entries in the column j. We call the matrix A a [b]golden matrix[/b] if all sets $X_1$, $X_2$, ..., $X_n$, $Y_1$, $Y_2$, ..., $Y_n$ are pairwisely distinct. Find the smallest integer N such that there exists a golden $2004\times 2004$ matrix whose entries are numbers from the set {1; 2; ...; N}.[/color]

As for Pascual's question, two entries in a row or in a column can be equal, but the sets are sets, not multisets, so that equal elements are counted only once (hence, not all of the sets $X_1$, $X_2$, ..., $X_n$, $Y_1$, $Y_2$, ..., $Y_n$ must necessarily have 2005 elements), and two such sets are equal if they are equal as sets (just take the formal point of view).

In our test, the greatest score for this problem were 4 marks. When we were told the solution, I tried to follow at the beginning, but then I fell asleep since it was just too ugly and boring...

  darij

Thanks  :)

You can post one complete solution? ;/

ooppps i am very sorry but now i am starting to believe that the answer will be $(log_2 n)+2$. It is easy to prove one needs at least that, because there are $2^m$ subsets of $(1,2,3,4...,m)$. The only hard case is when $n$ is a power of $2$. For example when $n=2^t$, we need $t+2$ objects instead of $t+1$, because if it were possible with $t+1$, then all subsets should be in some row or some column, so there would be a row with $1,1,1.....,1$, but since there are only $2^t$ subsets containing $1$, there will be also a column with $1,1,1...,1$ (simple piggeonholes).

Until now, i proved that $f(2n)=f(n)+1$, we only have to prove left that: $f(2n+1)=f(n)+1$, and we will be done!

$f(n)$ is the minimun number!

Does someone knows a solution????

Until now i have found $f(2)=3$, $f(3)=4$, $f(n+1)\leq f(n)+1$, $f(2^n)=n+2$, $f(6)=5$. I want to prove that we always have $f(3.2^n)=n+4$ it is clearly enough to prove it for arbitrry increasing sequence of $n$. I think also one could prove that $f$ is monotonically increasing. I also think $f(2n+1)\leq f(n)+1$
 (not proved) and $f(2n)\leq f(n)+1$ (already proved) if one proves this facts the answer then will easily follows.

pascual, try to prove $f(2004)=13$ directly. actually, you showed that $12\leq f(2004)\leq 13$ already. now, it is some work to exclude the case $f(2004)=12$, but this can be done by considering some tons of cases - in the TST, i still had 2 or 3 hours for this step and couldn't succeed, but it is quite clear that $f(2004)>12$ since 2004 is quite near to the next power of 2, 2048.
our delegation leader than gave a complete proof to us, but i can't really remember it. maybe i will try to reconstruct the proof if you really want me to.

Solution-
I will only show that $n=12$ is not possible. My construction for $13$ is too hard to put into words.
So if possible let $A$ be a golden $2004\times 2004$ matrix with entries in $\{1,2,\cdots,12\}$.
 Let the row sets and the column sets be $X_1,X_2,\cdots X_{2004}$ and $Y_1,Y_2,\cdots,Y_{2004}$ respectively and let $\mathbb{H}=\{X_1,X_2,\cdots X_{2004}, Y_1,Y_2,\cdots,Y_{2004}\}$
[b]Observation[/b] - $X_i^c\cap Y_j^c\notin \mathbb{H}$
For if it was a $Y_k$ then it must have a element in common with $X_i$ and similar for other case.

Now total number of subsets of  $\{1,2,\cdots,12\}$ are $4096$ and hence if no subset in $\mathbb{H}$ had three elements we would have $4008=|\mathbb{H}|\leq 4096$- $ {12}\choose {3}$$=3876$, a contradiction.
Hence there is atleast one subset having exactly three elements . WLOG let it be $X_1=\{1,2,3\}$.
Now for each $1\leq i\leq 2004$ define $f(Y_i)=Y_i^c\cap X_1^c$. Note that $f(Y_i)\notin\mathbb{H} $ by observation.
For some $i$ let $S=\{Y_j| f(Y_i)=f(Y_j)\}$ we claim that $|S| \leq 7$
From this we have that the set $\{f(Y_i)| 1\leq i\leq 2004\}$ will have atleast $\frac{2004}{7}$ elements . Hence we would have atleast $286$ sets which are not part of $\mathbb{H}$, but this is a contradiction as $4096-4008=88<286$
Hence we only need to show that $|S|\leq 7$, for this consider a given set $B$ in the image of $f$.
To get corresponding $Y_i$ we can add any of $\{\phi\},\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\}$ to $B$ and then take it's complement, and note that these are the only possible ways and there are $7$ possibilities.
Hence proved.

The answer is $n=13$. We shall call a golden matrix with entries from $\{1,2,...,k\}$ a $k$-golden matrix.
We will first show that there exists a $13$-golden matrix. We will do so by proving the following claim:
[color=#00f]CLAIM 1.[/color] Let $p,q$ be integers such that $p<q\leq 2p$, suppose there exists a $p\times p$ $k-$ golden matrix, then there exists a $q\times q$ $(k+1)-$golden matrix.
[color=#00f]Proof. [/color]
Let $A$ be the $p\times p$ $k-$ golden matrix, let $q=p+m$ where $m\leq p$. Let $B$ be the matrix formed by the first $m$ columns of $A$, and let $C$ be the matrix formed by the first $m$ rows of $A$. Let $D$ be the $m\times m$ matrix with all entries being $n+1$, then it is straightforward to check that the block matrix
$$\begin{pmatrix}
A&B\\
C&D
\end{pmatrix}$$
is a $q\times q$ $(k+1)-$golden matrix. $\blacksquare$
Now $\begin{pmatrix} 1&1\\2&3\end{pmatrix}$ is a $2\times 2$ $3$-golden matrix. Consider the sequence $$(a_1,...,a_{11})=(2,4,8,16,32,63,126,251,501,1002,2004)$$
By the CLAIM 1 there exist a $a_i\times a_i$ $(i+2)-$golden matrix. This proves the existence of a $2004\times 2004$ $13$-golden matrix.

Now we show that there does not exists a $2004\times 2004$ $12-$ golden matrix. Suppose on the contrary that such matrix exists. Let $S=\{1,2,...,12\}$. Let $S_i$ be the family of subsets of $S$ with $i$ elements. Let $X=X_1\cup...\cup X_{2004}$, define $Y$ similarly and let $E=X\cup Y$. Notice that for each $i,j$ we have $X_i\cup Y_j\neq \emptyset$. Let $F=2^S\setminus E$
[color=#00f]CLAIM 2. [/color]
(i) $|F\cup S_0|=1$
(ii) $|F\cup S_1|\geq 11$
(iii) $|F\cup S_2|\geq 36$
[color=#00f]Proof. [/color]
(i) is obvious since the only element in $S_0$ is the emptyset, which does not belong to $E$. Now suppose on the contrary that $|F\cup S_1|\geq 2$, WLOG assume $\{1\},\{2\}\in E$, notice that they must both belong to either $X$ or $Y$. Suppose they belong to $X$, then there are only $2^{10}=1024$ elements in $2^S$ which are not disjoint with both of them, hence there are only $1024$ possibilities for elements in $Y$, contradicting the fact that $|Y|=2004$. This proves $(ii)$
Now suppose on the contrary that $F|\cup S_2|\leq 35$, WLOG assume $\{1,2\}\in X$, if none of the $2$ element subsets of $\{3,4,...,12\}$ are in $E$ then we are done, hence assume $\{3,4\}\in X$. If all of the 2 element subsets of $\{5,6,...,12\}$ are in $F$ then we are done, hence assume $\{5,6\}\in X$. Now by inclusion-exclusion principle, there are exactly
$$3\times 2^{10}-3\times 2^8+2^6=2368$$
elements of $2^S$ which are disjoint to at least one of $\{1,2\},\{3,4\}$ and $\{5,6\}$. Therefore all of them can't be in $Y$. Therefore there are only $4096-2368=1728$ possibilities for $Y$, contradiction. $\blacksquare$

Now we further divide into two cases
[color=#f00]CASE I:[/color] $S_2\cup E\neq \emptyset$
WLOG assume $\{1,2\}\in X$
[color=#00f]CLAIM 3: [/color]For each $i=3,4,5,6$ we have $S_i\cup Y=\emptyset$
[color=#00f]Proof.[/color]
Suppose there exists $A\in S_3\cup Y$. Since $|A\cap \{1,2\}|\geq 1$ we have $|A\cup \{1,2\}|\leq 4$, hence $|S\setminus (A\cup \{1,2\})|\geq 8$. Let $Z=S\setminus(A\cup \{1,2\})$, for each subset of $Z$, it is disjoint to both $A$ and $\{1,2\}$, hence it can't be in $E$, therefore $|F|\geq 256$, $|E|\leq 4096-256<2\times 2004$, contradiction. For similar reasoning we have $S_4\cup Y=\emptyset$.
Now suppose $A\in S_5\cup Y$, let $B=S\setminus A$ then consider all the 3 and 4 element subsets of $B$, they do not belong to $X$ since they are disjoint with $A$, they also do not belong to $Y$ since $S_3\cup Y= S_4\cup Y=\emptyset$, hence they belong to $F$, therefore,
$$|F|\geq 1+11+36+\binom{7}{3}+\binom{7}{4}>4096-2\times 2004=88$$
contradiction.
For $i=6$, we can use the similar reasoning to obtain
$$|F|\geq 1+11+36+\binom{6}{3}+\binom{6}{4}+\binom{6}{5}=89$$
contradiction. $\blacksquare$
Hence there are at most $4096-\sum_{i=3}^6\binom{12}{i}=1665$ sets in $Y$, contradiction.
[color=#f00]CASE II: [/color]$S_2\cup E=\emptyset$
If $S_3\cup E=\emptyset$ then we are done. Now assume $\{1,2,3\}\in X$, using the same reasoning we have $S_3\cup Y=\emptyset$
If $S_4\cup Y=\emptyset$, then $|F|\geq 1+11+\binom{12}{2}+\binom{8}{3}>88$, contradiction.
Similarly $S_5\cup Y=S_6\cup Y=\emptyset$, using the exact reasoning as in CASE I we obtain a contradiction.

This completes the proof.


Find all polygons P such that for any two vertices A and B of P, there exists a third vertex (C) of P such that ABC is an equilateral triangle.

It is strange that nobody wants to try this nice problem :s

Only 16 views of this pos till now! Why?

Is P assumed to be convex?

In that case, considering the biggest equilateral triangle formed with vertices from P, we prove that P is itself an equilateral triangle.

Pierre.

Here's a problem from the first Romanian TST in 2000:

We have a convex polygon P_1P_2..P_n in the plane s.t. for all i=/=j there is a k s.t. angle P_iP_kP_j=60 deg. Show that n=3 (so the polygon is,  obviously in this case, an equilateral triangle).

It's not the same, but it's similar and, I think, stronger.


geometry
search
videos
cyclic quadrilateral
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, A' the midpoint of BC, 1 the incircle of ABC, D, E, F the points of contact of 1 with BC, CA, AB, I the center of 1, A" the second point of intersection of DI with 1, U the meetpoint of DI with BC, X the meetpoint of A'A" with BC, and V the symmetric of U wrt X.
Prove synthetically that AV // A'A".
Sincerely
Jean-Louis

You must explain what is meetpoint.

I think it midpoint, and intersection of two line.
But it doesn't and so I search many website to find what is mean.
It is not first time and I only saw 'MEETPOINT' in your post.

P.S.
Maybe $ A, A', A''$ are on same line is useful for solving your problem.

Dear Mathlinkers,
I saw the term "meetpoint" i.e. "the point of intersection of two lines" in Mathlinks and in another site as an abreviation...
Sincerely
Jean-Louis

I think so, too -_-..

but if you say $ V$ is meetpoint of $ DI$ with $ BC$, then because $ D$ is tangent point of incircle with $ BC$ so $ V\equal{}D$
But I think it's not what you want to say.

Thank you for answer^^

Actually it would be $ U \equal{} D$, besides $ X \equal{} A'$.

Hello jayme !

I think your problem is wrong. see picture ! :maybe:

I think that $ A'I$ and $ AV$ are parallels !

If we can proof that $ A$ ,$ A"$ and $ V$ are collinears, then we conclude that $ A'I$ and $ AV$ are parallels ! :roll:

In fact, I've made a video inspired from your problem (in French, sorry, but Maths are international !):

[url]http://www.dailymotion.com/video/edit/x8jcub_59-un-peu-de-trigo-pour-deux-droite_lifestyle[/url]

Thanks for the idea, jayme !

Mihaï-Ioan.

Dear Mathlinkers,
sorry, sorry for my typos... I think that now it is corect...
let ABC be a triangle, A' the midpoint of BC, 1 the incircle of ABC, D, E, F the points of contact of 1 with BC, CA, AB, I the center of 1, A" the second point of intersection of AI with 1, U the meetpoint of DA" with EF, X the meetpoint of A'A" with EF, and V the symmetric of U wrt X.
Prove synthetically that AV // A'A".
Sorry
Sincerely
Jean-Louis

Hello jayme !

It's the good picture ? That's quite different !

Dear Mathlinkers,
yes...
Sincerely
Jean-Louis

Parallel to $ EF$ through $ A''$ cuts $ CA, AB$ at $ E', F'$; $ A''$ is midpoint of $ E'F'.$ Tangential quadrilateral $ BCE'F'$ and its contact cyclic quadrilateral $ DEA''F$ have the same diagonal intersection $ U.$ Newton line $ A'A''$ of $ AE'UF'$ with 2 diagonals $ E'F'$ and $ BC$ cuts its 3rd diagonal $ AU$ at its midpoint $ M$ $ \Longrightarrow$ the Newton line $ (A'A'' \equiv XM) \parallel AV$ is midline of $ \triangle AUV.$

Dear Mathlinkers,
a synthetic proof of this result has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


number theory
prime numbers
number theory unsolved

A positive integer $ n$ is called $ good$ if it can be uniquely written simultaneously as $ a_1\plus{}a_2\plus{}...\plus{}a_k$ and as $ a_1 a_2...a_k$, where $ a_i$ are positive integers and $ k \ge 2$. (For example, $ 10$ is good because $ 10\equal{}5\plus{}2\plus{}1\plus{}1\plus{}1\equal{}5 \cdot 2 \cdot 1 \cdot 1 \cdot 1$ is a unique expression of this form). Find, in terms of prime numbers, all good natural numbers.

[quote="moldovan"]A positive integer $ n$ is called $ good$ if it can be uniquely written simultaneously as $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k$ and as $ a_1 a_2...a_k$, where $ a_i$ are positive integers and $ k \ge 2$. (For example, $ 10$ is good because $ 10 \equal{} 5 \plus{} 2 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 5 \cdot 2 \cdot 1 \cdot 1 \cdot 1$ is a unique expression of this form). Find, in terms of prime numbers, all good natural numbers.[/quote]

Clearly primes are not good natural numbers.

So let $ n\equal{}ij$, where $ i,j \neq 1$ are positive integers. Then $ n\equal{}i\plus{}j\plus{}1\plus{}1\plus{}...\plus{}1$ (there are $ (n\minus{}i\minus{}j)$ (can be zero) $ "1"$s )
$ \equal{}i \cdot j \cdot 1 \cdot 1 \cdot ... \cdot 1$. And let this kind of expression be $ A(i,j)$.

Then unique expression for $ n$ $ \Rightarrow$ $ n$ can be uniquely expressed as product of two positive integers(both $ \neq 1$) $ \Rightarrow n\equal{}p^2$ or $ pq$, where $ p,q$ are primes , which clearly works.

So $ n\equal{}p^2$ or $ pq$ are all good natural numbers.

[quote]Then unique expression for $ n$ $ \Rightarrow$ $ n$ can be uniquely expressed as product of two positive integers[/quote]
Why ?Can anyone explain. . .I'm not getting :(

Actually it should be:
$n=p_1.p_2......p_i.1.1....1$ and $n=p^i$ are good numbers.
Motivation for proof:
1. Observing that prime numbers are not actually good numbers, since $p_i+1>p_i$.
2. Proving that it works for all numbers that is not prime, since for any $A=p_i.p_j.p_k...$,we have that
$p_i.p_j.p_k... >p_i+p_j+p_k....$, so we can easily insert $1$ as it's a multiplicative identity to get the desired result.


what trick are u talking about? i see that there are multiple 29/37 's that you can simplify, but am i missing something here? :huh:

[quote="LawOfSigns"]Solve for x. Express your answer as a common fraction in lowest terms.
(29/37) + (8/37)2 =(29/37)2+ x

Not sure what the trick here is...[/quote]
Not much of a trick anywhere. Just multiple everything by 37 to get rid of the denominators, then you can do the rest.

[quote="LawOfSigns"]Solve for x. Express your answer as a common fraction in lowest terms.
(29/37) + (8/37)2 =(29/37)2+ x

Not sure what the trick here is...[/quote]

That is unusual... Why would someone write ''2'' after the fraction rather than before the fraction?

I thought it was a square(that is, ^2), and there is indeed a little trick if that's the case.

In that case:

[hide]

To see the trick, let your first number $a$. Then you can easily see that

$a+(1-a)^{2}=a^{2}-a+1$
$a^{2}+(1-a)=a^{2}-a+1$

Thus $a+(1-a)^{2}=a^{2}+(1-a)$.

So in this case, you can substitute $\frac{29}{37}$ for $a$, which results in $x=1-a=\frac{8}{37}$.

Whatever number you have as $a$, you can subtract it from 1 and get the answer $1-a$.[/hide]

Sorry if that was not what you were trying to ask.

Yeah, I meant ^2

I'm  not easily seeing how you did that though.


inequalities
trigonometry
inequalities unsolved

Let $ a, b, c$ be the sidelenghts of triangle $ ABC$ and $ R$ circumradius. Prove the inequality
\[ \frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}\leq R\sqrt{3}.\]

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\leq \sqrt {3}R$

$ \Longleftrightarrow \sqrt{3}(\sin A\plus{}\sin B\plus{}\sin C)\plus{}\cos 2A\plus{}\cos 2B\plus{}\cos 2C\geq 3$.

Does anybody have solution of this problem, or trigonometric inequality above  :roll:

[quote="mkrnic"]Let $ a, b, c$ be the sidelenghts of triangle $ ABC$ and $ R$ circumradius. Prove the inequality
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\leq R\sqrt {3}.
\]
[/quote]
This problem is so easy.Let $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y$
$ R\sqrt3\equal{}\frac{4\sqrt3S}{abc}$
The ineq becomes:
$ 4(x\plus{}y\plus{}z)\sqrt{3xyz(x\plus{}y\plus{}z)}$≥$ (x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)(x\plus{}y)(y\plus{}z)(z\plus{}x)$
and it is not difficult to prove the last ineq.

[quote="mkrnic"]Let $ a, b, c$ be the sidelenghts of triangle $ ABC$ and $ R$ circumradius. Prove the inequality
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\leq R\sqrt {3}.
\]
[/quote]

Try : $ a\equal{}b$ and $ c \rightarrow 0$  :wink:

Must be : $ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\leq 2R$ $ <\equal{}> \sin^2 \alpha\plus{}\sin^2 \beta\plus{}\sin^2 \gamma \le \sin \alpha\plus{}\sin \beta\plus{}\sin \gamma$

Equality for $ \alpha \equal{} \beta\equal{}\frac{\pi}{2}$


Determinati termenul general al sirului $a_0=1,a_{n+1}=3a_n+\sqrt {8a_n^2  + 1},n \in N.$

See here

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27863


symmetry
rotation
geometry
geometric transformation
counting
distinguishability
search

Hi, I couldn't count all of the color combinations for this problem.  Can someone please help me?

For both of the solutions, let us denote the six positions as _ _ _ _ _ _ so that the ends can be connected to form a circle as given. Also, let the two colors be A and B.

[hide="Symmetry"]
We will divide in to cases by the minimum number of times it takes to rotate a certain orientation back to its original position. For example, in ABABAB, it takes two rotations to get it back to an indistinguishable rotation. Note that in this solution, we will find the number of cases where all rotations have been overcounted, which will make the symmetry useful. That is, in ABABAB, we will say that there are $ 2$ cases, and later divide by $ 2$ to get the $ 1$ case that there really is.

[b]Case:[/b] one rotation. The possibilities are AAAAAA and BBBBBB, so this case gives $ 2\cdot1 \equal{} 2$.
[b]Case:[/b] two rotations. The only possibility is ABABAB. This case gives $ 1\cdot2 \equal{} 2$ ways.
[b]Case:[/b] three rotations. Both AABAAB and BBABBA work. This gives us $ 2\cdot3 \equal{} 6$ ways.
[b]Case:[/b] four or five rotations. Assume that such a number $ n$ would be the minimum number of rotations needed to get back to the starting position. Then with $ 6 \minus{} n < 4$ more rotations, we would be able to get back to the original position again. But $ 6 \minus{} n$ is less than $ n$, which leads to a contradiction. Another way to look at this is to see that if $ n$ rotations works, then it is necessary and sufficient for $ n|6$, and $ 4,5$ don't satisfy this. So there are $ 0$ for both of these cases.
[b]Case:[/b] six rotations. The ones we haven't counted yet are of number $ 2^6 \minus{} 2 \minus{} 2 \minus{} 6 \minus{} 0 \minus{} 0 \equal{} 54$, where the $ 2^6$ is there because we are considering all rotations first in the first part of this solution, and thus there are $ 2^6$ colorings.

Now, since the actual problem does not consider visually indistinguishable rotations as distinguishable, the answer is $ \frac21 \plus{} \frac22 \plus{} \frac63 \plus{} \frac04 \plus{} \frac05 \plus{} \frac {54}6 \equal{} 2 \plus{} 1 \plus{} 2 \plus{} 0 \plus{} 0 \plus{} 9 \equal{} \boxed{14}$.
[/hide]

This technique of symmetry can also be used on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=299447215&t=248222]2008 State Team #10[/url] (the relevant solution is in my first post there).


I have never used Asymptote so it would be nice to see a complete example to draw the following type of diagrams:

On $ x$-axis, there are all integers $ 1,\ldots, 10$ and on $ y$-axis, there are integers, say $ 50,\ldots,300$. Then there are data points $ (i,y_i)$ where $ 50\leq y_i\leq 300$ and line segments from $ (i,y_i)$ to $ (i \plus{} 1,y_{i \plus{} 1})$ for all $ i\in \{ 1,\ldots,9\}$.

Also, I would like to know is the line segment diagram correct term for the diagram of that type.

Is this what you were looking for?

[asy]size(200, IgnoreAspect);
int numberOfPoints = 10;
pair[] points;
points.push((1,77));
points.push((2,103));
points.push((3,94));
points.push((4,126));
points.push((5,155));
points.push((6,216));
points.push((7,195));
points.push((8,255));
points.push((9,240));
points.push((10,277));

xaxis(0, 10, Ticks(true, true, 0, 0, 5, 1, 5, 3));
yaxis(0, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));

for(int i = 0; i < numberOfPoints - 1; ++i) {
draw(points[i]--points[i+1],green+linewidth(1));
}
for(int i = 0; i < numberOfPoints; ++i) {
dot(points[i],blue+linewidth(3));
}[/asy]

[code]size(200, IgnoreAspect);[/code]
This code sets the size of the image, and allows the y-axis to be a different scale than the x-axis

[code]int numberOfPoints = 10;
pair[] points;
points.push((1,77));
points.push((2,103));
points.push((3,94));
points.push((4,126));
points.push((5,155));
points.push((6,216));
points.push((7,195));
points.push((8,255));
points.push((9,240));
points.push((10,277));[/code]
This creates the an array of data points (like in a programming language). I just made up some values.

[code]xaxis(0, 10, Ticks(true, true, 0, 0, 5, 1, 5, 3));
yaxis(0, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));[/code]
This draws the x- and y-axes.

[code]for(int i = 0; i < numberOfPoints - 1; ++i) {
draw(points[i]--points[i+1],green+linewidth(1));
}
for(int i = 0; i < numberOfPoints; ++i) {
dot(points[i],blue+linewidth(3));
}[/code]
Finally, this draws the line segments (first for loop) and creates dots on the data points (second for loop).

[quote="MathWise"]Is this what you were looking for?
[/quote]
Almost. Thanks! But $ x$-axis should start from $ 1$ and $ y$-axis from $ 50$ instead of $ 0$. Also, can I name x-axis as "days" and y-axis as "passed tests"?

Is this better?

[asy]size(200, IgnoreAspect);
int numberOfPoints = 10;
pair[] points;
points.push((1,77));
points.push((2,103));
points.push((3,94));
points.push((4,126));
points.push((5,155));
points.push((6,216));
points.push((7,195));
points.push((8,255));
points.push((9,240));
points.push((10,277));

xaxis(Label("days", MidPoint), YEquals(50), 1, 10, Ticks(true, true, 0, 0, 1, 0, 3, 0));
yaxis(Label("passed tests", MidPoint), XEquals(1), 50, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));

for(int i = 0; i < numberOfPoints - 1; ++i) {
draw(points[i]--points[i+1],green+linewidth(1));
}
for(int i = 0; i < numberOfPoints; ++i) {
dot(points[i],blue+linewidth(3));
}[/asy]

Here is the code:
[code][asy]
size(200, IgnoreAspect);
int numberOfPoints = 10;
pair[] points;
points.push((1,77));
points.push((2,103));
points.push((3,94));
points.push((4,126));
points.push((5,155));
points.push((6,216));
points.push((7,195));
points.push((8,255));
points.push((9,240));
points.push((10,277));

xaxis(Label("days", MidPoint), YEquals(50), 1, 10, Ticks(true, true, 0, 0, 1, 0, 3, 0));
yaxis(Label("passed tests", MidPoint), XEquals(1), 50, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));

for(int i = 0; i < numberOfPoints - 1; ++i) {
draw(points[i]--points[i+1],green+linewidth(1));
}
for(int i = 0; i < numberOfPoints; ++i) {
dot(points[i],blue+linewidth(3));
}[/asy][/code]

[quote="MathWise"]Is this better?
[/quote]
At the moment yes. I have to ask if this is what my boss wants.


Let $ x,y,z>0$.

Prove that

(a) $ \sqrt{\frac{x}{y\plus{}2z}}\plus{}\sqrt{\frac{y}{x\plus{}2z}}\plus{}2\sqrt{\frac{z}{x\plus{}y\plus{}z}}>2$

(b) $ \sqrt[3]{\frac{x}{y\plus{}2z}}\plus{}\sqrt[3]{\frac{y}{x\plus{}2z}}\plus{}2\sqrt[3]{\frac{z}{x\plus{}y\plus{}z}}>2$

For a, $ \sqrt {\frac {x}{y \plus{} 2z}} \plus{} \sqrt {\frac {y}{x \plus{} 2z}}$ is minimized for a fixed $ x \plus{} y$ when $ x \equal{} y$ This is because the sum is $ \ge 2 \sqrt [4] {\frac {xy}{(x \plus{} 2z)(y \plus{} 2z)}}$ by AM-GM. Thus, we can let $ x \equal{} y \equal{} \frac {x \plus{} y}{2}$, and we have to show
$ \sqrt {\frac {x \plus{} y}{x \plus{} y \plus{} 4z}} \plus{} \sqrt {\frac {z}{x \plus{} y \plus{} z}} > 1$
Let $ \frac {z}{x \plus{} y} \equal{} a$, then we have to show
$ \frac {1}{\sqrt {4a \plus{} 1}} \plus{} \frac {1}{\sqrt {\frac {1}{a} \plus{} 1}} > 1$ which my calculator shows is true by the graph, I don't feel like expanding.
For b, $ \frac {1}{\sqrt [3] {4a \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {\frac {1}{a} \plus{} 1}} > 1$ is also true.

[quote="dgreenb801"]$ \frac {1}{\sqrt {4a \plus{} 1}} \plus{} \frac {1}{\sqrt {\frac {1}{a} \plus{} 1}} > 1$ which my calculator shows is true by the graph, I don't feel like expanding.
For b, $ \frac {1}{\sqrt [3] {4a \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {\frac {1}{a} \plus{} 1}} > 1$ is also true.[/quote]

Are there any ways not expanding?

Expanding actually isn't that hard, for a it gives $ 16a^2 \plus{} 11a \plus{} 4 > 0$, always true. I will try to find a way without expanding.

[quote="outback"]Let $ x,y,z > 0$.

Prove that

(a) $ \sqrt {\frac {x}{y \plus{} 2z}} \plus{} \sqrt {\frac {y}{x \plus{} 2z}} \plus{} 2\sqrt {\frac {z}{x \plus{} y \plus{} z}} > 2$

(b) $ \sqrt [3]{\frac {x}{y \plus{} 2z}} \plus{} \sqrt [3]{\frac {y}{x \plus{} 2z}} \plus{} 2\sqrt [3]{\frac {z}{x \plus{} y \plus{} z}} > 2$[/quote]
Hey outback,All of them are th cases of the genaral ineq
For all $ x,y,z,t > 0$ Prove that $ \sqrt {\frac {x}{y \plus{} z \plus{} t}} \plus{}\sqrt { \frac {y}{x \plus{} z \plus{} t}} \plus{}\sqrt { \frac {z}{y \plus{} x \plus{} t}} \plus{} \sqrt { \frac {t}{y \plus{} x \plus{} t}} > 2$(after let $ z \equal{} t$)
My hint is using AMGM

[quote=ghjk][quote="outback"]Let $ x,y,z > 0$.

Prove that

(a) $ \sqrt {\frac {x}{y \plus{} 2z}} \plus{} \sqrt {\frac {y}{x \plus{} 2z}} \plus{} 2\sqrt {\frac {z}{x \plus{} y \plus{} z}} > 2$

(b) $ \sqrt [3]{\frac {x}{y \plus{} 2z}} \plus{} \sqrt [3]{\frac {y}{x \plus{} 2z}} \plus{} 2\sqrt [3]{\frac {z}{x \plus{} y \plus{} z}} > 2$[/quote]
[hide]My approach for the 1st inequality:
By AM-GM we get: sqrt(x)(y+2z)=<x+y+2z/2
                          sqrt(y)(x+2z)=<y+x+2z/2
                          sqrt(z)(x+y+z)=<x+y+2z/2
So, we have: x/sqrt((x)(y+2z))+y/sqrt((x+2z)y)+2z/sqrt(z(x+y+z))>=2(x+y+2z)/(x+y+2z)=2(Q.E.D)
Easy to see that the equality doesn't occur.
The 2nd doesn't work quite well with the same method above.[/hide][/quote]
By AM-GM , we get:
$\sqrt{x(y+2z)}\leq \frac{x+y+2z}{2}$,$\sqrt{y(x+2z)}\leq \frac{x+y+2z}{2}$,$\sqrt{z(x+y+z)}\leq \frac{x+y+2z}{2}$ ,
So, we have: $ \sqrt{\frac{x}{y+2z}}+\sqrt{\frac{y}{x+2z}}+2\sqrt{\frac{z}{x+y+z}}>\frac{2x}{x+y+2z}+\frac{2y}{x+y+2z}+\frac{2z}{x+y+2z}=2$(Q.E.D)
Easy to see that the equality doesn't occur.
The 2nd doesn't work quite well with the same method above.


geometry
incenter
circumcircle
perimeter
geometry proposed

Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ . Denote

$ D\in BC$ for which $ AD\perp BC$ and $ AD \equal{} h_a$ . Prove that $ DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)$ .

[quote="pablo221092"] [color=darkred]Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ . Denote

$ D\in BC$ for which $ AD\perp BC$ and $ AD\equal{}h_a$ . Prove that $ \boxed {DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)}$ .[/color] [/quote]
[color=darkblue][b][u]Proof[/u].[/b] From the well-known identity $ \boxed{a\cdot XA^2\plus{}b\cdot XB^2\plus{}c\cdot XC^2\equal{}(a\plus{}b\plus{}c)\cdot XI^2\plus{}abc}$ for $ X: \equal{}D$ obtain

$ a\cdot DA^2\plus{}b\cdot DB^2\plus{}c\cdot DC^2\equal{}2s\cdot DI^2\plus{}abc$ . Therefore, $ \underline{2s\cdot DI}^2\equal{}ah_a^2\plus{}b\cdot\left(c^2\minus{}h_a^2\right)\plus{}c\left(b^2\minus{}h_a^2\right)\minus{}abc$

$ \equal{}2(s\minus{}a)bc\minus{}2(s\minus{}a)h_a^2\equal{}2(s\minus{}a)\left(2Rh_a\minus{}h_a^2\right)\equal{}$ $ \left(2sh_a\minus{}2ah_a\right)\left(2R\minus{}h_a\right)\equal{}$ $ \left(2sh_a\minus{}4sr\right)\left(2R\minus{}h_a\right)$

$ \equal{}\underline{2s\left(h_a\minus{}2r\right)\left(2R\minus{}h_a\right)}$ . In conclusion, $ 2s\cdot DI^2\equal{}2s\left(h_a\minus{}2r\right)\left(2R\minus{}h_a\right)$ , i.e. $ DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)$ .[/color]

Hello [b]Virgil Nicula[/b],
With your permission I put your statement, this better expressed. If you disagree let me know please.

Many thanks!!

[color=darkblue][b]Pablo[/b], I love [b]Spain[/b] and your songs. [b]Happy New Year ![/b]

Prove similarly the following interesting extension - [b]my present for you[/b].[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]Extension[/u].[/b] Let $ ABC$ be an acute triangle with the incircle $ w \equal{} C(I,r)$ and the circumcircle $ C(O,R)$ . Denote $ D\in BC$

so that $ AD\perp BC$ , $ X\in (AD)$ and $ AD \equal{} h_a$ , $ AX \equal{} x$ . Prove that $ \boxed {XI^2 \equal{} (2R \minus{} x)(h_a \minus{} 2r) \minus{} x(h_a \minus{} x)}$ .

[b]Particular cases.[/b] $ \left\|\begin{array}{ccccc} x \equal{} R & \implies & IX^2 \equal{} R(R \minus{} 2r) & \implies & IX \equal{} IO \\
 \\
x \equal{} 0 & \implies & IA^2 \equal{} 2R(h_a \minus{} r) & \implies & IA^2 \equal{} \frac {bc(s \minus{} a)}{s}\end{array}\right\|$ what are evidently. Thus we

verified the upper identity ! Also we can find easily the points $ \{X,Y\}\subset (AD)$ for which $ IX \equal{} IY \equal{} r$ , i.e. $ AD\cap w$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ P\in AD$ for which $ IP\perp AD$ . Apply the [u]generalized [b]Pytagoras[/b]' relation[/u] in $ \triangle AIX$ :

$ XI^2 \equal{} AX^2 \plus{} AI^2 \minus{} 2\cdot AX\cdot AP\equal{} x^2 \plus{} \frac {bc(s \minus{} a)}{s} \minus{} 2x(h_a \minus{} r)$ $ \equal{} sx^2 \plus{} bc(s \minus{} a) \minus{} xh_a(b \plus{} c)$ $ \implies$

$ s\cdot XI^2 \equal{} sx^2 \minus{} xh_a(b \plus{} c) \plus{} bc(s \minus{} a) \minus{} (s \minus{} a)h^2_a \plus{} (s \minus{} a)h^2_a \equal{}$ $ sx^2 \minus{} xh_a(b \plus{} c) \plus{} (s \minus{} a)h^2_a \plus{} (s \minus{} a)\left(bc \minus{} h^2_a\right) \equal{}$

$ \equal{} sx^2 \minus{} xh_a(b \plus{} c) \plus{} (s \minus{} a)h^2_a \plus{} h_a(s \minus{} a)\left(2R \minus{} h_a\right) \equal{}$ $ \left[sx \minus{} (s \minus{} a)h_a\right]\left(x \minus{} h_a\right) \plus{} h_a(s \minus{} a)\left(2R \minus{} h_a\right) \equal{}$

$ \equal{} h_a(s \minus{} a)\left(2R \minus{} h_a\right) \plus{} sx\left(x \minus{} h_a\right) \minus{} h_a(s \minus{} a)\left(x \minus{} h_a\right) \equal{}$ $ h_a(s \minus{} a)(2R \minus{} x) \minus{} sx\left(h_a \minus{} x\right)$.

Since $ h_a(s \minus{} a) \equal{} sh_a \minus{} ah_a \equal{} sh_a \minus{} 2sr \equal{} s\left(h_a \minus{} 2r\right)$ obtain $ XI^2 \equal{} (2R \minus{} x)(h_a \minus{} 2r) \minus{} x(h_a \minus{} x)$ .[/color]

nice problem :)
lemma.$ AI^2\equal{}\frac{bc}{P}(P\minus{}a)$
let $ K$ be a point on $ AD$ such that $ AK\equal{}2R$. then its easy to prove that $ KDII_a$ is cyclic and so $ KDI$ is similiar to $ KIA$ and so $ ID^2\equal{}\frac{IA^2}{IK^2}(KD^2)\equal{}\frac{IA^2}{AK}(DK)$. so we must prove that $ \frac{AI^2}{AK}\equal{}h_a\minus{}2r$ and its simple with lemma. :wink:

[b]MJ GEO[/b], What is P? perimeter, semiperimeter?  EDIT: P is semiperimeter.
Many thanks.

[b]Virgil Nicula[/b], many thanks. I think it is a great gift for my. I'll try! Happy new year!!

You don't try to prove the extension. It is a gift for you. See above my proof.
Maybe you wish to find another proof. This is your job ... :lol:

[b]Virgil Nicula[/b]In your proof (extension) I think where you says "P" did you mean "T" or vice versa.
Regards.

[quote=MJ GEO]then its easy to prove that $ KDII_a$ is cyclic [/quote]

I would like to know what $I_a$ is

Anyways nice result, looks fresh and classic

Law of cosine in triangles $ADI, AIO:$

$\displaystyle A{I^2} = {h_a}^2 + D{I^2} - 2{h_a} \cdot DI\cos \theta  = {h_a}^2 + D{I^2} - 2{h_a} \cdot DI \cdot \frac{r}{{DI}} \Leftrightarrow $ $\boxed{A{I^2} = {h_a}^2 + D{I^2} - 2{h_a} \cdot r}$ $(1)$

$\displaystyle O{I^2} = {R^2} + A{I^2} - 2R \cdot AI\cos \omega  \Leftrightarrow {R^2} - 2Rr = {R^2} + A{I^2} - 2R \cdot AI \cdot \frac{{{h_a}}}{{AZ}}.$  But, $\displaystyle \frac{{AI}}{{AZ}} = \frac{{{h_a} - r}}{{{h_a}}},$ thus:

$\displaystyle A{I^2} - 2R{h_a} + 4Rr = 0\mathop  \Leftrightarrow \limits^{(1)} {h_a}^2 + D{I^2} - 2{h_a} \cdot r = 2R{h_a} - 4Rr \Leftrightarrow $ $\boxed{DI^2=(2R-h_a)(h_a-2r)}$


It is very good problem from the bulgarian mathematician E. Stoyanov.

It is given a triangle $ ABC$ and a point $ T$ inside $ ABC$. $ l1, l2, l3$ are lines parallel to $ AB, BC, CA$ respectively. $ l1$ intersects $ AC$ and $ BC$ at the points $ M$ and $ N, l2$ intersects $ AB$ and $ AC$ at the points $ P$ and $ Q, l3$ intersects $ BC$ and $ AB$ at the points $ R$ and $ S$. Prove the following inequality:
$ MT.NT \plus{} PT.QT \plus{} RT.ST \leq R^{2}$

This problem has been posted before at [url]http://www.mathlinks.ro/viewtopic.php?t=118621[/url].


Joe has 6 coins and 7 dollar bills, where the coins could be nickels, dimes, pennies, and/or quarters and where the dollar bills could be worth $20,$10, $5, an/or$1, how many of nickels, dimes, pennies, quarters, $20 dollar bills,$10, $5, and$1 does Joe has if the sum of the money is $37.46?$

I got the cent part - 46c.
[hide]4 dimes, 1 nickel, 1 penny[/hide]
I couldn't get the dollar part - 37
 :?:

[hide]Try working backwards, he has 37 dollars. Therefore, he must have 2 1 dollar bills as that is the smallest bill. That leaves you with 5 bills which add up to 35 dollars. Hope that helps. [/hide]

[hide]2-10 dollar bills, 3-5 dollar bills, 2-1 dollar bills
4 dimes, 1 nickel, 1 penny [/hide]

[hide]4 dimes, 1 nickel, 1 penny.

2 dollar bills, 3 five dollar bills, 2 ten-dollar bills[/hide]


Square $ SQUA$ with midpoints $ M$ and $ N$ of sides $ SQ$ and $ SA$ respectively, has an area of 144 square units. What is the number of square units rounded to the nearest integer, in the rea of the largest circle that can be inscribed in pentagon $ MNAUQ$?

Didn't you already post a similar problem, or am I being stupid?

Isn't this the same problem you posted in another thread except without the diagram?

Sorry, I was left logged on as #H34N1 at school and my friend posted a few problems he had trouble with. I guess he didn't see that I had posted a similar problem. :roll:

And about a year ago...
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135169[/url]


inequalities
AMC
AIME
triangle inequality

While on the AMCs you don't really have to use inequalities to solve questions (other than for graphing inequalities or the triangle inequality), in a lot of solution posts to AIME problems in particular there is a very frequent use of inequalities. I am pretty bad with using inequalities in general, and I was wondering if someone wouldn't mind helping me understand when to use inequalities particularly in questions involving sequences.

Thanks.

Inequalities obviously aren't as powerful generally as identities, but they are often useful when it seems like you don't have enough information. I can't come up with an example off the top of my head, but I can vaguely remember more than a few problems where I've thought, "This condition feels particularly weak... maybe I can combine it with an inequality to strengthen it." You really need to know the equality cases, though, as if you can come up with an inequality that looks the same as a given equality, you have a whole bunch of extra information from the equality case. Say you are given
\[ \frac {x \plus{} 2y \plus{} 4z}{\sqrt [3]{xyz}} \equal{} 6
\]
for positive real $ x$, $ y$, and $ z$ and asked to find $ x/z$. By AM-GM,
\[ \frac {x \plus{} 2y \plus{} 4z}{3}\ge\sqrt [3]{8xyz}
\]
with equality iff $ x \equal{} 2y \equal{} 4z$. By our given equality, the sides are equal, so $ x \equal{} 4z$ and $ x/z \equal{} \boxed{4}$. You had a pretty weak condition and turned it into a stronger one using an equality case.

Edit: wow. oops.

Erm, isn't arithmetic mean greater than or equal to the geometric mean?

Yes. I am just not good at thinking sometimes.


How many non congruent quadrilaterals can be formed on this square dot grid if each vertex must coincide with a dot?
[asy]size(50);
for(int i = 0; i < 3; ++i){
 for(int j = 0; j < 3; ++j){
   dot((i,j));
 }
}[/asy]

Squares: 2
Rectangles: 1
Paralellograms: 2
Other: 2

Not sure, 7?

I found three squares...one can be at points (1,0), (2,1), (1,2), (0,1).

definitely more than 8:

not sure if this is right but, by careful casework, i have found $ \boxed{16}$:

label the 9 vertices (0,0), (1,0), (2,0)......(2,2) such that the lower left dot is at (0,0), the one to the right of it is at (1,0), etc.

for shortnes, 00 10 01 11 will mean (0,0), (1,0), (0,1), (1,1)

1.00 10 11 01
2.00 10 12 01
3.00 10 12 02
4.00 10 01 22
5.00 10 22 12
6.00 10 21 22
7.00 10 21 11
8.00 02 22 10
9.00 10 21 12
10.10 01 12 21
11.00 21 02 11 concave
12.00 20 02 22
13.00 12 11 21 concave
14.00 20 11 12 concave
15.00 21 12 11 concave
16.00 10 21 02

OK, so I dont really feel like checking right now that I did not typo, so someone should let me know if they think they see a typo. Also, I think I put the vertices in order of clockwise or counterclockwise around the quad, but again I dont feel like checking. BTW, I did casework by ummm:

I. convex quads
..A. at least one side of lenth one
....1. 4 sides length 1: $ 1$
....2. 2 sides length 1: $ 6$
....3. 1 side length 1: $ 3$
..B. no sides length one but at least one side length 2
....1. 4 sides length 2: $ 1$
....2. 2 sides length 2: $ 0$
....3. 1 side length 2: $ 0$
..C. no rational length sides: $ 1$
II. concave quads: $ 4$

I found that casework above to be easiest to not mess up (though maybe I have messed up). Hm, this is rather hard for a timed question.

To conclude....LET ME KNOW IF I MESSED UP

I would just organize by side length...
I will write these in the form $ a,b,c,d$ where $ a$ is adjacent to $ b$, etc.
$ 2\sqrt2,1,\sqrt2,1$-6 and 9
$ \sqrt5,1,\sqrt5,1$
$ \sqrt5,1,1,\sqrt5$-4 and 13
$ \sqrt5,1,2,2$-5 and 8
$ \sqrt5,1,\sqrt2,2$-11 and 14
$ \sqrt5,\sqrt2,1,\sqrt2$-15
$ \sqrt5,\sqrt2,1,2$-16
$ \sqrt5,\sqrt2,\sqrt2,1$
$ \sqrt5,\sqrt5,\sqrt2,\sqrt2$
$ 2,2,2,2$-12
$ 2,1,1,\sqrt2$-2
$ 2,1,2,1$-3
$ \sqrt2,1,\sqrt2,1$-7
$ \sqrt2,\sqrt2,\sqrt2,\sqrt2$-10
$ 1,1,1,1$-1

This gives me $ 15$...

EDIT: I also combined my results with hurdler's, and found out that I was missing some and that he was missing some and repeating some--here are the (hopefully) correct cases.

EDIT2: Does anyone have a good casework method? I thought mine was foolproof but it missed 2 similar cases.

Um, I just double checked, and did not see that I was repeating any.....

Although I realized that many of mine were not in clockwise order or counterclockwise order..and this order MATTERS since for example: number 13 and 15 use the same four points but are connnected in a different order. I have now edited my post above to correct the order of the points.

Math: are the small numbers not in LaTeX to the right of each of your cases corresponding to my counting? You have 16 listed, not 15.

Also, which ones were I missing or repeating? I still think that my casework posted above is quite fullproof, if you are careful.

If we break it into cases based on the number of corners used, we get:
4 corners -- 1 quadrilateral
3 corners -- 1 quad
2 corners-adjacent -- 5 quads (two concave using center and side point opposite from corners)
2 corners-opposite -- 3 quads
1 corner and center -- 4 quads  (two concave using both side points opposite from corner)
1 corner and no center -- 2 quads
0 corners -- 1 quad

17 quadrilaterals

Correction
If we break it into cases based on the number of corners used, we get:
4 corners -- 1 quadrilateral
3 corners -- 1 quad
2 corners-adjacent -- 5 quads (two concave using center and side point opposite from corners)
2 corners-opposite -- 3 quads
1 corner and center -- 4 quads (two concave using both side points opposite from corner)
1 corner and no center -- 1 quads  (a second possibility is congruent to a previous quad)
0 corners -- 1 quad

16 quadrilaterals

9C4, then subtract the ones with a straight line and one vertex because that is a triangle?

I think there has to be some other way than listing it all out. It may be helpful to look at the book "Competition Math For Middle School" from the bookstore, since I skimmed the probability chapter and saw stuff on this. I will post more when I get to that part in the book.


trigonometry
inequalities proposed
inequalities

Prove that if $ B\geq{C}$ and $ B$ and $ C$ are the angles of a triangle then
$ \sin{B}\minus{}\sin{C}\leq{\sin{(B\minus{}C)}}$

Try again,
your counterexample doesn't work

$ \sin (B\minus{}C)\equal{}2\sin \frac{B\minus{}C}{2}\cos \frac{B\minus{}C}{2}$ and $ \sin B\minus{}\sin C\equal{}2\sin \frac{B\minus{}C}{2}\cos \frac{B\plus{}C}{2}$, so we have
\[ \sin (B\minus{}C)\minus{}(\sin B\minus{}\sin C)\equal{}2\sin \frac{B\minus{}C}{2}(\cos \frac{B\minus{}C}{2}\minus{}\cos \frac{B\plus{}C}{2})\]
\[ \equal{}4\sin \frac{B\minus{}C}{2}\sin \frac{B}{2}\sin \frac{C}{2}\]
which is nonnegative.

same thing asked here:

[url]http://www.mathlinks.ro/viewtopic.php?p=1111914#1111914[/url]


Does there exist a function $f : Z^{+}$→$Z^{+}$ such that  for all positive integer $n$, we have
             $f(f(n)) = n^{2}-19n+99$ ?

I'm sorry about my poor English!!


induction
geometry
3D geometry
sphere
real analysis
real analysis unsolved

Prove that any uncountable subset of the Euclidean $ n$-space contains an countable subset with the property that the distances between different pairs of points are different (that is, for any points $ P_1 \not\equal{} P_2$ and $ Q_1\not\equal{} Q_2$ of this subset, $ \overline{P_1P_2}\equal{}\overline{Q_1Q_2}$ implies either $ P_1\equal{}Q_1$ and $ P_2\equal{}Q_2$, or $ P_1\equal{}Q_2$ and $ P_2\equal{}Q_1$). Show that a similar statement is not valid if the Euclidean $ n$-space is replaced with a (separable) Hilbert space.

I will solve only the first part.
We use induction on n.

n=1  Start with any 2 points. In each next step there is
at least one point free to chose because only finitely many points are
forbidden. So we can chose infinitely many points.

n->n+1  we try to chose points one by one. suppose at some step
we cannot chose the next one.
Then all the other points lie in the union of several hyperplanes
and n-spheres. Then a hyperplane or a sphere must contain
uncountably many points. In first case we are done by induction.
Suppose that a sphere contains uncountably many points. Then again
start choosing points on sphere one by one. If in some step we cannot
proceed then all the other points on the n-sphere must be contained
in the union of finitely many (n-1)-spheres. So at least one
(n-1)-sphere contains uncountably many points and we are done
by hypothesis.

I hope that this is correct and understandable.

In fact, in an Euclidean space it is enough to assume that
the given set is infinite, i.e. it does not necessarily have to be
uncountable.


When I navigate to the [url=http://www.artofproblemsolving.com/Wiki/]AoPSWiki[/url], it gives an error message of "MediaWiki 1.11.0 / Please set up the wiki first. "  :(

We're in the process of upgrading the Wiki and a few other items on the site.  We should have it all running smoothly soon.

Have you looked at it? The logo isn't working and there's nothing of the AoPS template on the wiki, nor any way to get to any other part of the site. ...:(

[quote="Valentin Vornicu"]Go to preferences, skin -> AoPS Skin. We're trying to fix it now for the old users (new users of AoPSWiki will not have that problem).[/quote]

Apparently the monobook skin is there only for debugging purposes.


calculus
derivative
calculus computations

[img]http://img260.imageshack.us/img260/8335/30354062.jpg[/img]

if you explain the answer in deep, i'll be very appreciated.

$ e^x \equal{} \sum_{k \equal{} 0}^n\frac {x^k}{k!} \plus{} \text{error}$

The Lagrange (mean-value) form of the error term involves the $ n \plus{} 1$st derivative:

$ \text{error} \equal{} \frac {e^cx^{n \plus{} 1}}{(n \plus{} 1)!}$ where $ c$ is between $ x$ and $ 1.$

The worst case estimate would be $ x \equal{} 1,c \equal{} 1.$ ($ c$ won't be $ 1,$ but since we don't really have any better information about what it will be, that's the best we can do.)

We want this worst-case error to be less than $ 0.1.$ We can just trial-and-error this, since we know it won't take many terms. (I'll just use $ e < 3$ for simplicity.)

$ n \equal{} 2:$ the error is bounded by $ \frac {e}{3!}\approxeq .5$ Too big.

$ n \equal{} 3:$ the error is bounded by $ \frac {e}{4!} < \frac18.$ Close, but not quite.

$ n \equal{} 4:$ the error is bounded by $ \frac {e}{5!}\approxeq.025.$ That will work.

thank you Kent. I appreciate your effort


Let $P_{n}$ be the set of subsets of $\{1,2,...,n\}$. Let $c(n,m)$ be the number of functions $f: P_{n}\rightarrow \{1,2,...,m\}$ such that $f(A \cap B) = min\{ f(A), f(B)\}$. Prove that
\[c(n,m) = \sum_{j=1}^{m}j^{n}\]

It's easier than I expected!

[hide="Solution"] Let $S = \{ 1, 2, ... n \}$.  Let $s_{i}= \{ i \}, i = 1, 2, ... n$.  For a set $a \in P_{n}$, let $a^{-1}$ denote the complement of $a$ with respect to $S$.

Consider $f(S)$.  For any $A \in P_{n}$ we have

$f(S \cap A) = f(A) = \text{min}\{ f(S), f(A) \}$

So that we have $f(S) \ge f(A) \forall A \in P_{n}$.  In other words, if $f(S) = j$ then there are only $j$ possibilities $1, 2, ... j$ for any other $f(A)$.  This suggests that we group our functions $f$ according to their value of $f(S)$, which sets a maximum.

Given a value of $f(S) = j$, I claim that the values of $f(a_{i}^{-1}), i = 1, 2, ... n$ uniquely determine $f$.  There are $n$ values and they can take on $j$ different values for a total of $j^{n}$ possibilities as $j$ ranges from $1$ to $m$ - precisely our desired summation.

We have defined $f$ for subsets missing no members ($S$) and for subsets missing one member ($a_{i}^{-1}$).  The values of $f$ at subsets missing two members can be defined in terms of the subsets missing each of the two members; in other words,

$f( \{ x, y \}^{-1}) = \text{min}\{ f( a_{x}^{-1}), f( a_{y}^{-1}) \}$

Similarly, it's obvious from induction that for any $A \in P_{n}$ we have

$f( A^{-1}) = \text{min}\{ f(a_{x}^{-1}) \, | \, x \in A \}$

Because $A$ ranges across all of $P_{n}$, so does $A^{-1}$, and so every value in the domain of $f$ is uniquely defined.

Hence for a given value of $f(S) = j$ we have $j^{n}$ possible functions, so that we can conclude that across all possible values of $j$ there are

$c(n, m) = \sum_{j=1}^{m}j^{n}$

Possible functions $f : P_{n}\to \{ 1, 2, ... m \}$.  QED. [/hide]


calculus
integration
trigonometry
logarithms
calculus computations

[quote="misan"]$ \int\;\sqrt {\;\tan\;x\;}\;\textbf dx$[/quote]
this integral already was posted.
hint:
$ \tan x\equal{}y$
after $ \sqrt{y}\equal{}z$
and you have
$ \int \frac{2z^{2}}{1\plus{}z^{4}}dz$
and use this link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=137064

Make $ \tan x \equal{} t^2$, $ x' \equal{} \frac{2t}{1\plus{}t^4}$, to get

$ P \equal{} \int\sqrt{\tan x}\,dx \equal{} \int\frac{2t^2}{1\plus{}t^4}\,dt$.

Now make $ t \equal{} \frac{1}{u}$, $ t' \equal{} \minus{}\frac{1}{u^2}$, and so $ P \equal{} \minus{}2\int\frac{1}{1\plus{}u^4}\,du$. But:

$ \int\frac{1}{1\plus{}u^4}\,du \equal{} \int\frac{1}{(1\plus{}u^2)^2 \minus{} 2u^2}\,du \equal{} \int\frac{1}{[1\plus{}u^2 \minus{} \sqrt{2}u][1\plus{}u^2 \plus{} \sqrt{2}u]}\,du$

$ \equal{} \int\frac{\minus{}\frac{1}{2\sqrt{2}}u \plus{} \frac{1}{2}}{u^2 \minus{} \sqrt{2}u\plus{}1}\,du \plus{} \int\frac{\frac{1}{2\sqrt{2}}u \plus{} \frac{1}{2}}{u^2 \plus{} \sqrt{2}u \plus{} 1}\,du$.

Calculating each primitive we get as the final result

$ \int\frac{1}{1\plus{}u^4}\,du \equal{} \frac{1}{4\sqrt{2}}\ln\left(\frac{u^2 \plus{} \sqrt{2}u \plus{} 1}{u^2 \minus{} \sqrt{2}u \plus{} 1}\right) \plus{} \frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2}u}{1\minus{}u^2}\right)\plus{} C$.

Now making $ u^2 \equal{} \frac{1}{t^2}$ and then $ t \equal{} \sqrt{\tan x}$, we get

$ P \equal{} \int\sqrt{\tan x}\,dx \equal{} \frac{1}{4\sqrt{2}}\ln\left(\frac{\tan x \plus{} \sqrt{2 \tan x} \plus{} 1}{\tan x \minus{} \sqrt{2 \tan x} \plus{} 1}\right) \plus{} \frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2 \tan x}}{\tan x \minus{}1}\right) \plus{} C$.

You also can solve the original integral this way:

Supose $ \sin x >0$, then

$ \int \sqrt {\tan x} \; dx = \sqrt 2 \int \frac {\sin x \; dx}{\sqrt {2 \sin x \, cos x}} = \frac {\sqrt 2}{2} \int \left( \frac {\sin x + \cos x}{\sqrt{2 \sin x \, cos x}} + \frac {\sin x - \cos x}{\sqrt{2 \sin x \, cos x}} \right) \; dx =$

$ = \frac {1}{\sqrt 2} \left( \int \frac {d(\sin x - \cos x)}{\sqrt {1-(\sin x - \cos x)^2}} - \int \frac {d(\sin x + \cos x)}{\sqrt {(\sin x + \cos x)^2-1}} \right) =$

$ = \frac {1}{\sqrt 2} \left( \sin^{-1} (\sin x - \cos x) - \cosh^{-1} (\sin x + \cos x) \right) + C$


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \frac {a \plus{} b}{4c^2 \plus{} ab} \plus{} \frac {a \plus{} c}{4b^2 \plus{} ac} \plus{} \frac {b \plus{} c}{4a^2 \plus{} bc}\geq\frac {6(a \plus{} b \plus{} c)}{5(ab \plus{} ac \plus{} bc)}
\]

This type of inequality there is on my working table for two days.
I got the following inequalities of seven degree for any nonnegative numbers $ a,b,c$, no two of which are zero.

[b](A) [/b]If $ k\ge 2 \plus{} \sqrt 3$, then

$ \sum \frac {b \plus{} c}{ka^2 \plus{} bc} \ge \frac {6(a \plus{} b \plus{} c)}{(k \plus{} 1)(ab \plus{} bc \plus{} ca)}$;

[b](B)[/b]
$ \sum \frac { \minus{} a \plus{} b \plus{} c}{2a^2 \plus{} bc} \ge \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$;

[b](C)[/b] If $ \minus{} 2\le k\le 2(5 \plus{} 4\sqrt 2$, then

$ \sum \frac {ka^2 \plus{} b^2 \plus{} c^2}{a^2 \plus{} bc} \ge \frac {3(k \plus{} 2)}{2}$;

[b](D)[/b] If $ k \equal{} 2(1 \plus{} \sqrt 2)$, then

$ \sum \frac {a(2a^2 \minus{} b^2 \minus{} c^2)}{ka^2 \plus{} bc} \ge 0$,

with equality for $ a \equal{} b \equal{} c$, and again for $ \frac a{\sqrt 2} \equal{} b \equal{} c$ or any cyclic permutation.

I think the inequality [b](D)[/b] is nice and very hard, even for $ k\equal{}4$. The following are easier, but nice.

[b](E)[/b]
$ \sum \frac { \minus{} 2a \plus{} b \plus{} c)}{a^2 \plus{} bc} \ge 0$;

[b](F)[/b]
$ \sum \frac {a^2( b^2 \plus{} c^2)}{a^2 \plus{} bc} \le a^2 \plus{} b^2 \plus{} c^2$.

[quote="Vasc"]This type of inequality there is on my working table for two days.
[/quote]
I created it by my old inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171300

Are these inequalities solved with the SOS method or there is an easier (and more important - faster) one ?? :oops:


Country A has c% of the world's population and owns d% of the world wealth. Country B has e% of the world's population and f% wealth. Assume the citizens in country a shares the wealth of A equally, and b shares the wealth of B equally. find the ration of a citizen of A to a citizen of B.

sry about that it's the ratio of the wealth of a citizen in A and a citizen in B's wealth

[hide]In country A, each citizen will have $ \frac{d}{c}$ money, and for country B, each will have $ \frac{f}{e}$.
Simply divide to find the answer, $ \frac{de}{cf}$...[/hide] I think?

xpmath, i think you got the correct answer.

However, it may be more correct (if you're really picky) to say $ \frac{\frac{c}{100}}{\frac{d}{100}}$ as the fractions...

Yes I know, but since I was in a hurry, and because the 100's cancel out anyway, I didn't type that


There are 2 secondary pentyls possible from n-pentane.Can any one give me their common names?

The 2 sec- pentyls are:

$ \ce{CH3\minus{}CH2\minus{}CH2\minus{}CH\minus{}CH3}$ is 2-pentyl (common name: sec-pentyl)
and

$ \ce{CH3\minus{}CH(CH3)\minus{}CH\minus{}CH3}$ is 3-methyl-2-butyl (common name: i am not sure, but i think it doesn't have any)

How is the second one from n-pentane? It would have to have been derived from 2-methyl-butane.

I think you are looking for:
$ \ce{CH3\minus{}CH2\minus{}CH\minus{}CH2\minus{}CH3}$ (3-pentyl?)

I could be wrong though.

[quote="yanglunj"]How is the second one from n-pentane? It would have to have been derived from 2-methyl-butane.

I think you are looking for:
$ \ce{CH3 \minus{} CH2 \minus{} CH \minus{} CH2 \minus{} CH3}$ (3-pentyl?)

I could be wrong though.[/quote]

You are absolutely right..I guess I read the question as " sec-alkyls from pentane".So, I considered the isomers as well, unfortunately forgot the 3-pentyl....  :wallbash_red:

n-pentane is a structure of pentane.(straight chain)
n-pentane - CH3 - CH2-CH2-CH2- CH3

2 sec pentyls are- 1)   CH3 - CH-CH2-CH2-CH3
                            2)  CH3 - CH2-CH-CH2-CH3

Please write their[(1) &(2)] common names.

1) Sec-pentyl or 2-pentyl

2) Diethylmethyl or 3-pentyl


Let p be a prime number ,p>3.  Do there exists two prime numbers1< q,t<p such that:
           q :!=: 1(modp 2 ),t :!=: 1(modp 2 )

You probably mean $ q^{p-1}-1$ is not divisible by $ p^2$ and the same for the other. In this case, as far as I remember, the problem was posted by Valiowk and solved.

I think you posted a problem from Singapore, and Harazi noticed that the problem posted by you could be generalized by the problem in this thread. That's what I remember..


Just got this from a small maths competition in Australia.

The polynomial equation
$x^{3}-6x^{2}+5x-1 = 0$
has three real roots a, b and c. Determine the value of $a^{5}+b^{5}+c^{5}$

are newton's sums and vieta's sums the same?

No, they aren't the same, although they're related.

http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums
http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas

A more conventional way is to note that $a^{3}= 6a^{2}-5x+1$.  You can apply this repeatedly to express $a^{5}$ in terms of smaller powers of $a$, all the way down to a quadratic expression.  Then getting $a^{2}+b^{2}+c^{2}$ is not very hard, so you can finish that way.


$ a \in R$
($ for$ $ all$ $ n$ $ \in$ $ N*$) $ : lal\leq 1/n  \equal{}\equal{}>a\equal{}0$

[quote="greatestmaths"]$ a \in R$
($ for$ $ all$ $ n$ $ \in$ $ N*$) $ : lal\leq 1/n \equal{} \equal{} > a \equal{} 0$[/quote]
Oh yes because $ \lim_{n\to\infty}\frac {1}{n} \equal{} 0$ so $ |a| \equal{} 0\Rightarrow a \equal{} 0$

$ can$ $ you$ $ prove$ $ the$ $ exercice$ $ without$ $ use$  $ lim$
 :oops:

If $ a \not\equal{}0$ ,there exist $ N \in \mathbb{N}$ such that $ \forall n \geq N$, $ \mid a\mid >\frac1n$.


read the topic and tell us the date and month (don't worry i won't stalk u guys :D )

okay i'll be the first one to answer. 9/17/91

[quote="A+MATH"]7/23/1992  acordining to my math teacher in any group of 30 or more there must be atleast two people whose birthday is on the same month. We'll see about that :) .[/quote]

Shouldn't it be 13?

9/15/88

10/27/88

I was born in the same year as RobinHood3000 and my sister was born on the same day as him although she was born 2 years eariler....just felt like noting that.

I know this one kid who is exactly 1 month, 1 week, and 1 day older than me.  Creepy.  And my ex-girlfriend is exactly 5 months older than him, making her 6 months, 1 week, and 1 day older than me.

Me: 11/02/90
My sister(since she insists on putting hers.) 9/22/93

3-27-1991

it happens to be on easter next year in 2005 and the day the ppl who are going 2 yosemite leave. (thats wat it said on the green paper).  but it also happens to be on a sunday which is when i have chinese school. grr.......but im not sure cause after a unit, we have a 1-week rest so i hope its on that rest day even tho chances are.........not very big?

my sister wants to put hers too... her's is 5/5/89

u have a sister? shes older?.......WAT, sunnyray ur birthday was on the 8th? wait did i wish u a happy birthday already?

wow that took u awhile to find out i had a sister.  u shoulda known last year.  well its a good thing that you don't stalk me and listen to my conversations so much and JEFFERSON! don't u mention anything about "it" in this forum

Wow, it is scary, right now, in the poll resoults, the pick of AoPSer is almost ten years younger than me!!! I feel old!!!

I love my birthday, but if only I could have been born exactly 1 year  earlier!

8/8/89

born on friday may 13, 1988
and this year my birthday will again be on friday the 13th. cool huh?

[quote="GoBraves"]Wouldnt it suck if somebody was born on Feb. 29? Do any of you guys know anybody that was?[/quote]
yeah. my uncle was. he's 50-something years old but he has only had like 13 birthdays. that's unfortunate.

90!! wow 13 bday parties in 50 something years of living....now that would suck...

[quote="jackroland"]9/21/90
and I have a friend who has 9/11/90
9/11... harsh[/quote]

I too have a friend that was born 9/11, but he does not celebrate his birthday much.

7/7/88

14 April 1989
The anniversary of the day Titanic sank....


3(11-n) < 25 (.9)^n

The least value of n for which the inequility holds true is?

i know i can sovle it by graphing but i don't think i can solve such a complex inequality

3(11-n) < 25 (.9)^n

You could plug in integral values of n or graph it :P .

Hmm... n must be less or equal to than 11, as 25 (.9)^n can never be zero or less... that's all I can figure out just by glancing at it.

If that "25" was a "2", it'd be so easy... (the answer would be 11)

and ofcourse n can take only integer values which i forgot to mention.

I could well just plug in some values for n and see.
for example. substituting n = 9.
one the left of the equality we get 6. and on the right we get 9.68
so I know that n is between 1 and 9. continue doing so till i get the correct answer. but I was hoping that somebody would show me an easier way out of this.


abstract algebra
superior algebra
superior algebra unsolved

Let $ S$ and $ T$ be two sets, then the free groups $ F(S)$ and $ F(T)$ with basis $ S$ and $ T$ respectively are isomorphic iff $ S$ and $ T$ have the same cardinality (there is a bijection).

How can I show this elegantly? Can I use the universal property? What if $ S$ and $ T$ are infinite sets?

thanks

the one direction is trivial. for the other one, remark that the abelianization of $ F(S)$ is the free $ \mathbb{Z}$-module over $ S$ (just using the universal property), and free modules have a well-defined rank.


Each small square has sides of length 1 unit. How many distinct paths of length 6 units are there from A to B?
(look at the document for the diagram)
EDIT:i posted the document twice by accident and i don't know how to fix it.
[size=75][color=olive]
[Problem solved - click the posted attachments button on the Edit Post page - solafidefarms][/color][/size]

[hide="the answer"]
You need exactly $ 3 $ times going up and $ 3 $ times going right. That's like $ 3 $ heads and $ 3 $ tails. Hmmmmmmmmmm......... USE PASCALS TRIANGLE!!!!! look in the 7th row because there are six choices(remember the first row is row 0) and the 3rd number is ..............
[hide="the answer to the problem"]
[hide="maybe you should"]

$ 2 $

$ 0 $
Wow you are lazy. :D  :D

[/hide]
[/hide]
[/hide]

[hide="You COULD use"]pascal's triangle, or [hide="you could use"]binomial coefficient, which is [hide="..."]$\binom{6}{3}=\dfrac{6 \cdot 5 \cdot 4}{6}=20$[/hide][/hide][/hide]

Here is an extension to the problem.
My handwriting is messy so do the best you can to read it.

Try finding the number of paths in a 5 by 3 grid and then subtracting all the paths that go through the blank parts.

[quote="chess64"][quote="math92"]My handwriting is messy so do the best you can to read it.[/quote]

That handwriting is amazing![/quote]
really?
i think your handwriting is good.

PLEASE STOP SPAMMING ABOUT HANDWRITING!!!!!!

I should probably report this to a moderator, to lock or split this topic.

Sure, the end of that conversation: "oh, nvm" and some of the previous should not have been posted, and I'll delete that. It should have been done with PM. But it doesn't need splitting or locking; it isn't that off-topic. Nor is it really distracting.

[quote="biffanddoc"][quote="math92"]i was having a conversation with 236! one summer morning over pm's.  And i asked him if he had any nats test. He said that chess64 had given him one so he emailed the nats test to me.  And when i got it, it had handwriting on it from kchande.[/quote]What year is the nats test?[/quote]
i think its 1992 or 93
one of those two.


Fred's bike is 21 speed. If, for example, it has 3 main gears and seven subgears, he has a 21 speed bike. Let $ p$ be the main gears and let $ q$ be the subgears where the following equation has to be true : $ pq \equal{} 21$ Then, how many integer solutions are there  to the equation?

           [\hide]
      So first we prime factor $ 21$ to find that $ 21 \equal{} 3^1 * 7^1$ . Therefore we have $ 2 * 2 \equal{} 4$  integer factors of 21. So, the answer is $ 4$. [/hide][/hide]

In general...

[hide]
If you express the number in canonical prime factorization with each term of the form $ p_{i}^a_{i}$ then the formula is $ (a_{1}+1)(a_{2}+1)(......)(a_{n}+1)$ Where n is the number of different $ p_{i}$ 's.[/hide]

prime factorize 21 = 3 x 7
so the only solutions are (3 , 7) and (1 ,21)
each counts twice so [u]4[/u]


ratio
geometry
circumcircle
trigonometry
geometry unsolved

Since $MN \parallel AC$, the angle $\angle BNM = 90^\circ$ is right. The point M divides AB in the ratio $\frac{AM}{BM} = \frac{CN}{BN} = \frac{AC}{BC} = 2,$ hence CM bisects the right angle $\angle BCA$ and $\angle BCM = 45^\circ.$ Let BO meet CA at P. By Ceva's theorem, $\frac{AM}{BM} \cdot \frac{BN}{CN} \cdot \frac{CP}{AP} = 1,$ hence $\frac{CP}{AP} = 1$ and BP is the B-median of the triangle $\triangle ABC.$ From CA = 2 CB, we get CP = CB, so that the triangle $\triangle PBC$ is isosceles right, hence $AM \equiv AO \perp BP$ and the angle $\angle BOM = 90^\circ$ is right. Therefore, the quadrilateral BNOM is cyclic, with a circumcircle (Q) centered at the midpoint of Q of BM. Denote $r = \frac{BM}{2}$ its radius. From $MN \parallel AC,$ $\frac{MN}{BN} = 2.$ Since O is the center of masses $m_A = 1,\ m_B = 2,\ m_C = 1$ at A, B, C, $\frac{OB}{OM} = \frac{OC}{OM} = \frac{m_A + m_B}{m_C} = 3.$ Since $\angle NBO \equiv \angle CBO = 45^\circ,$ and (Q) is the circumcircle of the triangle $\triangle BNO$, $\frac{ON}{BM} = \frac{OB}{2r}= \sin 45^\circ = \frac{\sqrt 2}{2}.$ From the right triangle $\triangle BMO,$

$\frac{OM}{BM} = \cos \widehat{BMO} = \frac{1}{\sqrt{1 + \tan^2 \widehat{BMO}}} = \frac{1}{\sqrt{1 + \left(\frac{OB}{OM}\right)^2}} = \frac{1}{\sqrt{10}}$

$\frac{ON}{OM} = \frac{ON}{BM} \cdot \frac{BM}{OM} = \sqrt 5$

On the other hand, from KN = OK + OM,

$\frac{ON}{OM} = \frac{OK}{OM} + \frac{KN}{OM} = \frac{2 OK}{OM} + 1 = \sqrt 5\ \Longrightarrow\ \frac{OK}{OM} = \frac{-1 + \sqrt 5}{2} = \frac 1 g$

where $g = \frac{1 + \sqrt 5}{2}$ is the golden ratio. Then

$\frac{KN}{OK} = 1 + \frac{OM}{OK} = 1 + g = g^2$

Let the bisector of the angle $\angle ABC \equiv \angle MBN$ meet the circle (Q) again at a point S. Then

$\frac{2\tan \widehat{MBS}}{1 - \tan^2 \widehat{MBS}} = \tan \widehat{MBN} = \frac{MN}{BN} = 2$

$\tan^2 \widehat{MBS} + \tan \widehat{MBS} - 1 = 0$

$\frac{MS}{BS} = \tan \widehat{MBS} = \frac{-1 + \sqrt 5}{2} = \frac 1 g$

Let L be the foot of a normal from S to ON. Since $\angle LOS \equiv \angle NOS = \angle NBS = \angle MBS,$ the right triangles $\triangle SOL \sim \triangle MBS$ are similar and $\frac{OL}{SL} = \frac{BS}{MS} = g$ and $\tan \widehat{OSL} = g.$ The angle $\angle OSN = 180^\circ - \angle OMN \equiv 180^\circ - \angle CMN = 135^\circ,$ hence

$\frac{\tan \widehat{OSL} + \tan \widehat{NSL}}{1 - \tan \widehat{OSL} \tan \widehat{NSL}} = \tan \widehat{OMN} = -1$

Substituting $\tan \widehat{OSL} = g,$

${g + \tan \widehat{NSL} = g \cdot \tan \widehat{NSL} - 1\ \Longrightarrow\ \tan \widehat{NSL} = \frac{g + 1}{g - 1} } = g^3$

$\frac{LN}{OL} = \frac{\tan \widehat{NSL}}{\tan \widehat{OSL}} = g^2$

As a result, $\frac{KN}{OK} = \frac{LN}{OL} = g^2,$ so that the points $K \equiv L$ are identical. It follows that the points $T \equiv S$ are also identical, T lies on the circle (Q) with diameter BM, which means that the angle $\angle MTB = 90^\circ$ is right.

[quote="giancarlo"]A line parallel to the side $AC$ of a triangle $ABC$ with $\angle C = 90$ intersects
side $AB$ at $M$ and side $BC$ at $N$, so that $CN : BN = AC : BC = 2 : 1$.
The segments $CM$ and $AN$ meet at $O$. Let $K$ be a point on the segment
$ON$ such that $MO + OK = KN$. The bisector of $\angle ABC$ meets the line
through $K$ perpendicular to $AN$ at point $T$ . Determine $\angle MTB$.[/quote]

Let the side $CB=1$
At first we can find out
$AO*AN=\frac{3}{4}AN^2=\frac{3}{4}(AC^2+CN^2)=\frac{10}{3}= \frac{2}{3}AB^2=AM*AB$
So the points $O,M,B,N$ lie on a circle.
And finally let's note that the lines $BT$ and $KT$ pass through the middle of the segment $NOB$ of the circle
So $\angle MTB=\angle MNB=90^\circ$
V:)


group theory
abstract algebra
superior algebra
superior algebra solved

Is it true that if [tex]K\unlhd H\unlhd G[/tex] then [tex]K\unlhd G[/tex], where [tex]G[/tex] is a group?

Mmmm, don't understand any more what I wrote yesterday !

I'll try to find a counter-example !

How is that a counterexample?

I know that this result is false. I'll try to come up with a counterexample.

The statement is false.  Example:

Let [tex]S_4[/tex] be the group of permutations of four elements.

Then [tex]V_4 = \lbrace 1, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3) \rbrace[/tex] is a normal subgroup of [tex]S_4[/tex].

Moreover, the subgroup [tex]\langle (1 \ 2)(3 \ 4) \rangle[/tex] is normal in [tex]V_4[/tex] since it is of index 2.

However, this subgroup is not normal in [tex]S_4[/tex] since [tex](1 \ 3)(2 \ 4)[/tex] is a conjugate of [tex](1 \ 2)(3 \ 4)[/tex].

Hope this explains things well enough.[/tex]