geometry
3D geometry
induction
algebra
polynomial
limit
system of equations

what is general formula for

1^4+2^4+3^4.............

if anything is there

can anybody please post the derivation

$\binom{k+1}{5}-\binom{k}{5}=\binom{k}{4}$ for all $k\in\mathbb{N}$.
Sum up this equations to get $\binom{n+4}{5}=\binom{n+4}{5}-\binom{4}{5}=\sum_{k=4}^{n+3}\binom{k}{4}=\sum_{k=1}^{n}\binom{k+3}{4}$ for all $n\in\mathbb{N}$.
Now multiply this with $4!$ to get:
${n(n+1)(n+2)(n+3)(n+4)\over 5}=\sum_{k=1}^{n}k(k+1)(k+2)(k+3)\\ \hspace*{2in}=\sum_{k=1}^{n}(k^{4}+6k^{3}+11k^{2}+6k)$
Now use formulas for $\sum k$,$\sum k^{2}$ and $\sum k^{3}$ to get $\sum k^{4}$.

There is an explicit general formula for $\sum k^{m}$ using Bernoulli numbers, but can't remember it right now.

can we do without using any combinatorics

using algebra

however what is the simple formula to calculate 1^4+2^4+3^4.......

please do the derivation using algebra

i am not familiar with combinatorics

What do you mean by using algebra, do you know of any way of deriving it for cubes using only algebra?

Do you mean method of solving system of equations for coefficients in the induction step?

[quote="olorin"]
There is an explicit general formula for $\sum k^{m}$ using Bernoulli numbers, but can't remember it right now.[/quote]

$\sum_{j=1}^{n}j^{k}= \sum_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!$   :)

But there is one with Bernoulli numbers. I should have just wikied it. :blush:
[url]http://en.wikipedia.org/wiki/Bernoulli_number[/url].

[quote="boxedexe"]That's the formula using Stiling Numbers of the second kind, not Bernoulli numbers.[/quote]

did i say that was the formula using Bernoulli numbers?  :wink:  :wink:

wow...

Logically, when you quote someone referring to Bernoulli numbers, and you immediately answer them without any explanation, then people think that you are answering to the exact question (or something related to it). :wink: Sorry if you felt insulted because of that.


Ed takes five 100-point tests in his algebra class. He scores 87, 85 and 87 points on the first three tests. If the scores of his last two tests differ by three points and he has a $ 90\%$ average on his five tests, what was his highest test score?

87 is 3 under 90
85 is 5 under 90
87 is 3 under 90

In order to maintain an average of 90, the last two test scores must have been 3+5+3=11 over 90. Since one of them is higher by three, the higher one is (11+3)/2=7 over 90, whichi is 97

We could also do:

If the average is 90, then the sum of his 5 test scores is $ 90 \cdot 5 \equal{} 450$. We subtract the sum of his other 3 test scores to get $ 191$. We then have:

$ x \plus{} x \plus{} 3 \equal{} 191$

$ 2x \plus{} 3 \equal{} 191$

$ 2x \equal{} 188$

$ x \equal{} 94$

So the higher test score is $ 94 \plus{} 3 \equal{} 97$.


logarithms
real analysis
real analysis unsolved

Show that for any positive sequence $ \{a_n\}_{n\equal{}1}^{\infty}$,we have $ \limsup_{n\rightarrow\infty} (\frac{1\plus{}a_{n\plus{}1}}{a_n})^n>e$.

To prove $ \limsup_{n\rightarrow\infty} (\frac {1 \plus{} a_{n \plus{} 1}}{a_n})^n\geq e$ is trivial,but I strongly guess the equlity can never occur. :)

still no one? :( Now I have prove it if $ \{a_{n\plus{}1}\}$ does not tend to $ \infty$

[quote="Ravi B"]When $ a_n \equal{} n \ln n$, I believe the lim sup is $ e$.[/quote]
If $ a_n \equal{} n \left(\ln n\right)^k$ with $ k > 0$, then the [b]limit[/b] is $ e$.


Are there any schools near Chapel Hill, NC that would allow me to take the B test there? Just curious.

You could always move down into the lower part of South Carolina and join my school  :D .  Then we would win EVERY (local) competition.


graph theory
combinatorics unsolved
combinatorics

There are $100$ cities in a country, some of them being joined by roads. Any four cities are connected to each other by at least two roads. Assume that there is no path passing through every city exactly once. Prove that there are two cities such that every other city is connected to at least one of them.

I dont remember the exact solution,but as far as I remember the fbegining was as wollow...
[hide]first of all name the graph $G$...
in the case that $G$ isnt connected,the problem is easy to solve...
so assume that $G$ is connected,now name the vertices of $G$ from $1$ to $100$
now if any $2$ vertex were connected the we would have a hamiltonian path,so there exist $2$ vertices that are not connecte, WLOG name them $1$ and $(($,now if every other vertex was connected to at least one of $1,99$ then the problem is solved,so there exists a vertex that is not connected to $1$ and not to $99$,again WLOG name it $100$...
so any other vertex $i$,must be connected to at least $2$ of the $1,99,100$ beacause if not,then vetices $1,i,99,100$ are $4$ vertices with less than $2$ edges...
so every $i$ is connected to at least tow of $1,99,100$ so it must be connected to at least one of the vertices $1,99$...
so every other vertices (other than $100,99,1$) must be connected to $1$ or $99$ or both of them...
now take the longest path between $1,99$ (such path exist since we assumed that the graph is connected...)
so this path must cover every vertices between $1,...,99$...
now WLOG name these vertices in increasing order $1,2,...,99$
now we prove that if $i$ is connected to $99$ then so is $i+1$,becuaes otherwise it must be connected to $1$ so the cycle $1,...,i,99,98,...,i+1,1$ is a hamiltonian cycle between $99$ vertices of $G$ so somhow $G$ would have a hamiiltonian path...
this contradiction shows that if $i$ is connected to $99$ then so is $i+1$[/hide]

I dont understand this step:

so this path must cover every vertices between $1,...,99$...

Why this path cover every vertices and does vertex $100$ belong to this path(I think it should)?   :maybe:

I think this will work:
take the longest path in $G$,and name its ends $1,99$ now we may assume that $1,99$ are not connected,we will prove this:
proof:assume that $1,99$ are connected,so the longest path is actually a cycle,now in this cycle,if there were two un-connected vertices like $i,j$ we would take these vertices as the ends of the path...so every 2 vertices of this cycle are connected,so we have a complete graph,now for any other vertex not in this cycle (there exist such vertex) like $k$,we see that $k$ could not be connected to any vertex from this cycle...(otherwise we would have a longer path,which is a contradiction)
this contradiction shows that the ends of the longest path in this graph are not connected...
name these ends $1,99$ now if every other vertex was connected to at least one of these 2 vertex,then the problem is solved,so there exists a vertex $100$ such that it isnt connected to $1,99$ we want to gain a contradiction:
so there isnt any edge between vertices $1,99,100$,now for any other vertex like $i$ we get that $i$ must be connected to at least one of the vertices $1,99$...
and also because $1,99$ are the ends of the longest path,so their neighbours must be in this path,and from the last thing that I said,we get that every vertex except $100$ is in the path...
now WLOG name the vertices of this path in increasing order $1,...,99$ now note that if $i$ was connected to $99$ as shown in the last post,$i+1$ must be connected to $99$ to... otherwise we would have a cycle along vertices $1,...,99$ and we kknew that $G$ was connected so vertex number $100$ has a neighbour along vertices $1,...,99$ let it be $t$ so the path $100,t,t+1,...,99,1$ will be a longer path and also a hamiltonian path,which is a contradiction...
so if $i$ was connected to $99$ then is also $i+1$...
now we just have to prove that this path is a cycle,because if we show this then as showed above,we can show that the graphb has a hamiltonian path which is a contradiction...


If $ a\equal{}\minus{}2$, the largest number in the set $ \left \{ \minus{}3a,4a,\frac{24}{a},a^2,1 \right \}$ is

\[ \textbf{(A)}\ \minus{}3a \qquad
\textbf{(B)}\ 4a \qquad
\textbf{(C)}\ \frac{24}{a} \qquad
\textbf{(D)}\ a^2 \qquad
\textbf{(E)}\ 1
\]

Since $ a$ is negative, you'd need the scalar in front to be negative as well to maximze direction.  So it is $ \minus{}3a$.  $ a^{2}$ is also positive, but that is only 4, whereas $ \minus{}3a\equal{}6$.

[hide="I dont really understand what scalar and direction are, so..."]
\begin{align*}
-3a&=6 \\ 4a&=-8 \\ \frac{24}{a}&=-12 \\ a^2&=4 \\ 1&=1
\end{align*}

Clearly the largest is $ -3a$, or $ \boxed{\textbf{(A)}}$.
[/hide]

A scalar is a coefficient, and direction is either positive or negative.

Scalar doesn't have a direction; vector is the one which has directions.

Every number has a magnitude and a direction.  Magnitude is size.  So if I ask whether 6 or 10 has a bigger magnitude, it's 10.  Also, if I ask whether -10 or 3 has a bigger magnitude, it's 10, despite the fact that it's negative.  Magnitude doesn't care whether it's negative or positive, just how big.  Direction cares about negative and positive.  So when you ask which number is the largest, you need to take both magnitude and direction into account.  So $ \left|\minus{}10\right|>\left|2\right|$, but because 2 has positive direction, it is larger.

[quote=KrrishPradhani]I dont really understand what scalar and direction are, so...[/quote]

Seems you copied the name of the hide tag in @Post 3. Try rereading the thread.


Note: I'm looking only for solutions by AM-GM.  I'm sure that these inequalities are easily proven by stronger inequalities than AM-GM.

1. Given $ x,y,z>0$, prove that $ \frac{1}{3x\plus{}y}\plus{}\frac{1}{3y\plus{}z}\plus{}\frac{1}{3z\plus{}x}\ge\frac{1}{2x\plus{}y\plus{}z}\plus{}\frac{1}{2y\plus{}z\plus{}x}\plus{}\frac{1}{2z\plus{}x\plus{}y}$.

2. Given $ a,b,c>0$ and $ abc\equal{}1$, prove that $ \frac{ab}{ab\plus{}a^5\plus{}b^5}\plus{}\frac{bc}{bc\plus{}b^5\plus{}c^5}\plus{}\frac{ca}{ca\plus{}c^5\plus{}a^5}\le 1$.

Well, here are the answers for you expectation.
The first problem, I can only give you a hint
$ \frac{1}{3x\plus{}y}\plus{}\frac{1}{2z\plus{}x\plus{}y} \ge \frac{4}{4x\plus{}2y\plus{}2z}\equal{}\frac{2}{2x\plus{}y\plus{}z}$
and after that, you sum up all the inequalities.

For the second problem,
we have $ a^5\plus{}b^5 \ge ab(a^3\plus{}b^3)$ and $ a^3\plus{}b^3 \ge ab(a\plus{}b)\equal{}\frac{a\plus{}b}{c}$( because $ abc\equal{}1$)
It follows
$ \frac{ab}{a^5\plus{}b^5\plus{}ab} \le \frac{ab}{ab\plus{}ab(a^3\plus{}b^3)}\equal{}\frac{1}{a^3\plus{}b^3\plus{}1}\le \frac{1}{\frac{a\plus{}b}{c}\plus{}1}\equal{}\frac{c}{a\plus{}b\plus{}c}$
Finally, you keep summing up all the similar inequalities as shown.
Have fun!

See number two here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247194


Three real numbers are chosen between 0 and 10.  What is the probability that their sum is less than 20??

I've seen this kind of problem many times before, but never knew how to set it up....thanks in advance!

That problem was discussed [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=20506]here[/url] a little bit.  blahblahblah's solution uses calculus but its basically the same as finding volume.


induction
linear algebra
linear algebra unsolved

Let $ A,B \in M_n(R)$ such that $ Rank(AB\minus{}BA)\equal{}1,(A\plus{}B)^n\equal{}0,A^n\equal{}0$ Prove that $ B^n\equal{}0$

$ rank(AB\minus{}BA)\leq{1}$ implies that $ A,B$ are simultaneously triangularizable......

[quote="loup blanc"]$ rank(AB \minus{} BA)\leq{1}$ implies that $ A,B$ are simultaneously triangularizable......[/quote]
 I don't known prove this  simultaneously triangularizable . Please Prove .... Thanks !

QuyBac,

I thought the result that loup blanc mentioned was on this forum, but I can't find it.

I know the result it true in $ \mathcal{M}_n(\mathbb{C})$, not so sure in $ \mathcal{M}_n(\mathbb{R})$, but it doesn't matter for the inital question, you show that the matrices are nilpotent in $ \mathcal{M}_n(\mathbb{C})$, then they are nilpotent in $ \mathcal{M}_n(\mathbb{R})$.

To prove that $ \mathrm{rank}(AB \minus{} BA)\leqslant 1$ implies that $ A,B$ are simultaneously triangularizable, you start to show that that $ B(\ker A) \subset \ker A$ or $ B(\mathrm{Im} A) \subset \mathrm{Im} A$, then you proceed by induction over $ n$.

loup blanc,

Are you sure that if $ A,B \in \mathcal{M}_n(\mathbb{R})$, $ \mathrm{rank}(AB \minus{} BA)\leqslant 1$ implies that $ A,B$ are simultaneously triangularizable in $ \mathcal{M}_n(\mathbb{R})$?

Hi JC_math,
of course the field must be algebraically closed. Here if $ A,B\in\mathcal{M}_n(K)$ then we prove the result  in  $ \mathcal{M}_n(\bar{K})$ where $ \bar{K}$ is the algebraical closure of $ K$.
References:
1) Laffey /  Simultaneous triangularization of matrices low rank cases and non derogatory case. Linear and multilinear algebra 6 (1978) p. 269-305.
About the simult. Triang., the bible is:
2) Radjavi, Rosenthal /  Simultaneous triangularization, Springer, 2000

Hi loup blanc

Radjavi and Rosenthal's book is very nice, and a good source for problems (that I used a few times :wink: )

[quote="othman"]i think you don't need such a theorem to solve that problem the proof is more elementary....well i think i have such a proof i will check my proof and i will post it[/quote]
 I will wait a proof of you !
Thanks!  :roll:


The repeating reciprocal digits function ($RRD(n)$ for short) is a discrete function that returns the number of digits it takes for $\frac{1}{n}$ to repeat and returns $0$ if it doesn't repeat.  For example,
$\frac{1}{2}= .5$, so $RRD(2) = 0$.
$\frac{1}{3}= .3333 \hdots$, so $RRD(3) = 1$.
$\frac{1}{37}= .027027027 \hdots$, so $RRD(37) = 3$.

$a)$  Compute or describe all integers $n$ for which $RRD(n) = 3$.
$b)$  Compute or describe all integers $n$ for which $RRD(n) = 4$.
$c)$  Compute $RRD(2007)$.
$d)$  Prove that $RRD(n) = 2007$ has no solutions.
$e)$  Compute the smallest number $k > 2007$ for which $RRD(n) = k$ has one or more solutions.
$f)$  Describe the set of integers $k$ for which $RRD(n) = k$ has one or more solutions.

[hide="a"]
Suppose $RRD(n) = 3$, then we can write $\frac{1}{n}= 0.abcabcabc...$ (where a, b and c denote digits).
Now $\frac{1}{999}= 0.001001001$, so $\frac{1}{n}= 0.abcabcabc... = \frac{abc}{999}$.
Since we have to be able to write $\frac{abc}{999}$ as $\frac{1}{n}$, abc has to divide 999 (and $n = \frac{999}{abc}$.
So all divisors of 999 will be ok for abc, and thus for n, except for those with $a=b=c$ (because then $RRD(n) = 1$).

Now we can easily find all divisors of 999 by its prime factorisation ($999 = 3^{3}*37$).

Remark: as originally stated, $RRD(37) = 3$.
[/hide]


$ a,b,c\in \mathbb{R}_\plus{}$ such that $ a\plus{}b\plus{}c\equal{}1$, prove that

$ \frac{1}{a^2\plus{}b}\plus{}\frac{1}{b^2\plus{}c}\plus{}\frac{1}{c^2\plus{}a}\geq\frac{27}{4}$

[quote="tobeno_1"]$ a,b,c\in \mathbb{R}_ \plus{}$ such that $ a \plus{} b \plus{} c \equal{} 1$, prove that

$ \frac {1}{a^2 \plus{} b} \plus{} \frac {1}{b^2 \plus{} c} \plus{} \frac {1}{c^2 \plus{} a}\geq\frac {27}{4}$[/quote]
It's wrong. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88792


Does anyone know of a good site online that gos through the basics of geometry (that is, the stuff normally taught in a high school geometry course)? I need to brush up on some of the very basics (ie when triangles are congruent, when they are similar, what happens in cases like AAS where the triangle is not congruent, etc...).  :mrgreen:  :mrgreen:  :mrgreen:

cant you just read through the parts in aops again?

I only have AoPS 2, not 1 (which I really don't need since I know most of the stuff in it).

[quote="Fierytycoon"]Does anyone know of a good site online that gos through the basics of geometry (that is, the stuff normally taught in a high school geometry course)?[/quote]

Hi, Fiery, prompted by your question, and by the fact that my son has a geometry final coming up in about a month, I just Googled "geometry review" and found quite a few interesting sites, with varying levels of depth. Next I'll try Googling "geometry end of course examination" or "geometry lecture notes" or other such search terms.

Hope this helps!

tokenadult:

Well, what a surprise! I had thought that googling up such a general topic like that would yield a boatload of useless and poorly-made websites (having had many such previous experiences), but I actually found a few decent resources in the first 20 matches or so. Thanks.


Hello everybody,

I am sad to inform you that Murray Klamkin (1921-2004) passed away on August 6th 2004 because of an intestinal tumour and heart failure.

He was arguably the most well-known mathematics problem-solver in the whole world. He left his mark in every major journal which had a problem section (American Mathematical Monthly, Crux Mathematicorum, SIAM Review,....etc).

We will all miss him.

Unfortunately, I did not read this message before. Yeah, we all will miss him in the problem-solving community. How comes that you get to know from it ? Is there any obituary for him ? I am really interested in it. He made so many valuable contributions !!!  :)

BTW what do you think are the mos well-known people in the problem-solving world ? I am thinking about Pl Erds and Arthur Engel. Any comments requested.  :)

Well, Pal Erdos was not really a problem-solver but a professional mathematician. There is no doubt that he was one of the greatest mathematicians in the 20th century.

Now, concerning problem-solving, there are -fortunately- several very talented people. I will just give an example with names of problem-solvers specialized in  geometry:

Chris Fisher (Canada), Michael Lambrou (Greece), Da

Oops! sorry for this error:  here is the list of names:

Chris Fisher (Canada), Michael Lambrou (Greece), Dan Branzei (Romania), Francisco Bellot-Rosado (Spain), Sergey Ivanov (Russia), Michel Bataille (France), Waldemar Pompe (Poland), Istvan Reiman (Hungary), Toshio Seimiya (Japan),...etc

And, in few years, I will be happy to include the name of Darij Grinberg

I feel flattered... but why??!! Just because I'm posting day and night on ML? I don't know what I will do in a few years, but there are many people about whom I know well that I will always remain under their level. Why haven't you included the name of Christian Reiher in the list (he is actually doing well with solving [i]open[/i] problems)? And what about Andrey Badzyan, Wolfgang Burmeister, Titu Andreescu, Valentin Vornicu, Reid Barton? I think they have and will have solved much more and much harder problems than me, maybe with the difference that they don't post every of their solutions into a maths forum...

  Darij

Darij:  all the people you cited are very talented, but I am talking about people who spend 90% of their time thinking about geometry. Furthermore, you are the one who has already published very important results (sometimes with the collaboration of professional mathematician) in a specialized geometric journal. Finally, do not forget that you are "still" 16 years old. CONGRATULATIONS!

I still assert that Pl Erds is a great problem solver and even greater mathematician. I read some books about famous mathematicians and other scientists including P. Erds, S. Ramanujan, K. Gdel, A. Turing, L. Euler, C.F. Gauss, A. Wiles, A. Einstein, W. Heisenberg, E. Noether, J. Nash, B. Gates etc. Maybe it is just me or do I love reading biographies...  :)

I read a biography of Erdos, but I also meet him personaly two times (in September 1995 and May 1996). I can tell you that he is most known as a problem-proposer rather than a problem-solver!!

[quote="Kantor"]
Chris Fisher (Canada), Michael Lambrou (Greece), Dan Branzei (Romania), Francisco Bellot-Rosado (Spain), Sergey Ivanov (Russia), Michel Bataille (France), Waldemar Pompe (Poland), Istvan Reiman (Hungary), Toshio Seimiya (Japan),...etc [/quote]

I see you often read Crux Mathematicorum and papers by the Forum Geometricorum.  :D

Does anyone know of a biography or obituary for Murray Klamkin?

[quote="orl"]Well I see people discuss other things in this thread now. But does anyone have any reports or articles about the life of Murray Klamkin ? Does anybody know a nice web site ?  :)[/quote]

Hi EveryOne,

When I read the first post in this thread I wanted to confirm that Klamkin did indeed pass away before I wrote anything.  I had trouble finding any death notices, etc., but I finally came across one this evening on the Mathematical Association of Americas website  [What follows if from < [url]http://www.maa.org/news/inmemoriam.html[/url] >]:

[quote][b]Murray Klamkin[/b], a long time member of the MAA and one of the original organizers, leaders, and coaches for the American Mathematics Competitions program, died on August 6, 2004 at the age of 83. Klamkin received a Certificate of Merit from the Association in 1978. He was a member of MAA since 1948. [/quote]

 :(

I'm a novice problem solver, but I have heard about many of the contributions made by Klamkin (for example, his "Klamkin Quickies" in Crux, etc.).  I was not fortunate enough to meet this man.   Even though I only new Klamkin as a name mentioned in periodicals, I think we should honor his memory by continuing what he seem to have liked best --- solving math problems!

By the way, I would bet that in the issues of some of next month's mathematical related periodicals, there will be eulogies, etc., about Klamkin.

Yours truly,

mathfreak

Oh...I wasn't aware of Murray's dead...
You are right : We will miss him.

Pierre.

I am very sorry for his death and I think his beautiful problems published

on AMM,CRUX,Mathematics Magazine,.... will miss every people keen on

problem solving.

[b]Curiosity:[/b]
In a personal letter, I appreciate  from period 1982-1986 (??) , Murray Klamkin has informed me that he had the possibility to study papers written in Romanian.The reason was that his wife has the ancestors in Romania - [Bucharest].


calculus integration function Putnam algebra domain logarithms

evaluate(find a formula -- not value in terms of functions and known constatnts)

$ \int_{0}^{1}  \frac{ln(1\plus{}x^2)}{(1\plus{}x)} .dx$

........

Well, it's not [i]exactly[/i] the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].

A similar trick might work: $ \ln(x^2 \plus{} 1) \equal{} \int_0^{x^2}\frac1{1 \plus{} t}\,dt$, so the integral becomes $ \int_0^1\int_0^{x^2}\frac1{1 \plus{} x}\cdot\frac1{1 \plus{} t}\,dt$.
Substituting $ t \equal{} x^2y,dt \equal{} x^2\,dy$ to make the domain a square, we have
$ \int_0^1\int_0^1 \frac1{1 \plus{} x}\cdot\frac {x^2}{1 \plus{} x^2y}\,dy\,dx \equal{} \int_0^1\int_0^1 \frac1{1\plus{}y}\left(\frac1{1\plus{}x}\plus{}\frac{yx}{1\plus{}yx^2}\minus{}\frac{y}{1\plus{}yx^2}\right) \,dx\,dy$
$ \equal{}\int_0^1 \frac1{1\plus{}y}\cdot\left[\ln(1\plus{}x)\plus{} \frac12\ln(1\plus{}yx^2)\minus{}\sqrt{y}\arctan(\sqrt{y}x)\right]_{x\equal{}0}^{x\equal{}1}\,dy$
$ \equal{} \int_0^1 \frac{\ln 2}{1\plus{}y}\plus{}\frac{\ln(1\plus{}y)}{2(1\plus{}y)}\minus{} \frac{\sqrt{y}\arctan(\sqrt{y})}{1\plus{}y}\,dy$

All three terms have elementary antiderivatives.
For the second: $ \int_0^1 \frac{\ln (1\plus{}y)}{1\plus{}y}\,dy\equal{}\int_0^{\ln 2}u\,du\equal{}\frac12(\ln 2)^2$ by the substitution $ u\equal{}\ln(1\plus{}y)$.
For the third: $ \int_0^1 \arctan{\sqrt{y}} \cdot\frac{\sqrt{y}}{1\plus{}y^2}\,dy\equal{}\int_0^{\frac{\pi}{4}}2t\tan^2 t\,dt$ by the substitution $ y\equal{}\tan^2 t$. We can find that $ \frac{d}{dt}\left(t^2\plus{}2t\tan t\plus{}2\ln(\cos t)\right)\equal{}2t\tan^2 t$ by integration by parts; the integral of this piece is $ 1\plus{}\frac{\pi}{2}\plus{}2\ln\frac1{\sqrt{2}}$

Putting the parts together, the integral is $ \frac54(\ln 2)^2\plus{}1\plus{}\frac{\pi}{2}\minus{}\ln 2$.

I like the Putnam problem better.

[quote="jmerry"]
$ \int_0^1\int_0^1 \frac1{1 \plus{} x}\cdot\frac {x^2}{1 \plus{} x^2y}\,dy\,dx \equal{} \int_0^1\int_0^1 \frac1{1 \plus{} y}\left(\frac1{1 \plus{} x} \plus{} \frac {yx}{1 \plus{} yx^2} \minus{} \frac {y}{1 \plus{} yx^2}\right) \,dx\,dy$
...
Putting the parts together, the integral is $ \frac54(\ln 2)^2 \plus{} 1 \plus{} \frac {\pi}{2} \minus{} \ln 2$.
.[/quote]
This is not correct??? the $ y$ on RHS should not be there with $ x$ i.e it should be:-....
$ \int_0^1\int_0^1 \frac1{1 \plus{} x}\cdot\frac {x^2}{1 \plus{} x^2y}\,dy\,dx \equal{} \int_0^1\int_0^1 \frac1{1 \plus{} y}\left(\frac1{1 \plus{} x} \plus{} \frac {x}{1 \plus{} yx^2} \minus{} \frac {1}{1 \plus{} yx^2}\right) \,dx\,dy$
and the furthur antiderivatives are not trivial....

I was hesitating the same but I didn't post 'cause I thought may be I was mistaken.

 :ninja:

OK, I see it- I thought the zeros of the denominator were at $ \pm\sqrt{\minus{}y}$, not $ \pm\sqrt{\minus{}1/y}$.

New version: the second and third terms are $ \int_0^1 \frac{\ln(1\plus{}y)}{2y(1\plus{}y)}\,dy$ and $ \minus{}\int_0^1 \frac{\arctan(\sqrt{y})}{\sqrt{y}(1\plus{}y)}\,dy$. That last term is easier after the change- we don't need to integrate by parts, and the value is $ \left(\frac{\pi}{4}\right)^2$.

The other term can be broken up by partial fractions; there's $ \minus{}\frac12\int_0^1\frac{\ln(1\plus{}y)}{1\plus{}y}\,dy$, which we know how to do, and $ \frac12\int_0^1\frac{\ln(1\plus{}y)}{y}\,dy$, which is a well-known function without an elementary antiderivative. For this term, we can use the power series of the logarithm; we get $ \frac12\int_0^1 1\minus{}\frac{y}{2}\plus{}\frac{y^2}{3}\minus{}\frac{y^3}{4}\plus{}\cdots\,dy\equal{} \frac12\left(1\minus{}\frac14\plus{}\frac19\minus{}\frac1{16}\plus{}\cdots\right)\equal{}\frac{\pi^2}{24}$

The original integral is $ \frac34(\ln 2)^2\minus{}\frac{\pi^2}{48}.$


Prove that if $ 0<a<b,$ then $ \frac{a\plus{}b}{2}\minus{}\sqrt{ab}<\frac{(b\minus{}a)^2}{8a}.$

Rearrangeing the terms it euivalent with $ {(\sqrt{b}\minus{}\sqrt{a})}^{2} > 0$, which is trivial.

i think he just means the trivial inequality:

x^2 >0 is true for all real x.

Well, that's really trivial. But i suppose that solution is not complete. Am i right?

Completing anderssonsw's solution

$ \frac {a \plus{} b}{2} \minus{} \sqrt {ab} < \frac {(b \minus{} a)^2}{8a} \iff \\
\frac {b \minus{} 2\sqrt {ab} \plus{} a}{2} < \frac {(b \minus{} a)^2}{8a} \iff \\
(\sqrt {b} \minus{} \sqrt {a})^2 < \frac {(b \minus{} a)^2}{4a} \iff \\
\frac {(b \minus{} a)^2 \minus{} 4a(\sqrt {b} \minus{} \sqrt {a})^2}{4a} > 0 \iff \\
\frac {(\sqrt {b} \minus{} \sqrt {a})^2[(\sqrt {b} \plus{} \sqrt {a})^2 \minus{} 4a]}{4a} > 0 \iff \\
\frac {(\sqrt {b} \minus{} \sqrt {a})^2(b \plus{} 2\sqrt {ab} \minus{} 3a)}{4a}$

Thus it suffices to prove

$ 2\sqrt {ab} \ge 3a \minus{} b$.

$ 2\sqrt {ab} > 2a > 3a \minus{} b \iff \sqrt {ab} > a$, which is trivial. Did I make any algebraic mistakes?

EDIT: Of course that's easier... You just need to remark that $ a \neq b$ (so the equality cannot be achieved).

Wouldn't it be easier to go

$ (\sqrt{b}\minus{}\sqrt{a})^2 \le \dfrac{(b\minus{}a)^2}{4a}$

$ \Rightarrow 2\sqrt{a}(\sqrt{b}\minus{}\sqrt{a}) \le b\minus{}a$  which we can do becasue $ b>a$

$ \sqrt{ab} \le \frac{a\plus{}b}{2}$

Which is true by AM-GM


[quote="zvi"]It is well known that a convex quadrilateral ABCD is tangential iff AB+CD=BC+AD.
What about concave quadrilateral? is it true that for a concave quadrilateral ABCD, there exists a circle touching AB, BC, CD & DA or their extensions iff AB+CD=BC+AD?
And what about self-intersecting quadrilaterals?[/quote]

it is also true for concave quadrilaterals.

See http://www.mathlinks.ro/Forum/viewtopic.php?p=359769#359769 , the proof given by nttu

Indeed, a beautiful proof, which applies also to a concave quadrilateral.


vector
geometry
parallelogram
Euler
geometry unsolved

$A,B,C,D$ are points on a circle in this order clockwise. If $AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$, find $\vec{AC}\cdot\vec{BD}$ in term of $AB$ and $BC$.

[quote="OHO"]$AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$, [/quote]

This holds if and only if quadrilateral $ABCD$ is a parallelogram (that follows from Eulers theorem), but because $ABCD$ is cyclic, we conclude that $ABCD$ is a rectangle.

Now, it's not hard to see that wanted value is $AB^{2}+BC^{2}$

Bye

If $ABCD$ is a quadrilateral with sidelengths $a,b,c,d$ and diagonal length $e$ and $f$ then $a^{2}+b^{2}+c^{2}+d^{2}= e^{2}+f^{2}+4m^{2}$ where $m$ is length of the segment which connects two midpoints of diagonals. So $m=0$ iff $ABCD$ is a parallelogram.


$x,y,z>0$.Prove that:
$\sum_{cyclic}\frac{yz}{(y+z)^2}+2\sum_{cyclic}\frac{yz}{(x+y)(x+z)} \le \frac{9}{4}$

[quote="arqady"]$\frac{9}{4}-\sum_{cyc}\frac{yz}{(y+z)^2}-2\sum_{cyc}\frac{yz}{(x+y)(x+z)}=\frac{(x-y)^2(x-z)^2(y-z)^2}{4(x+y)^2(x+z)^2(y+z)^2}.$ :)[/quote]
Nice solution!
Here's my solution:
Let $a=\frac{y}{x+y},b=\frac{z}{y+z},c=\frac{x}{z+x}$
Rewrite the original ineqiality as
$\sum b(1-b)+2\sum a(1-c)\leq \frac{9}{4}$
$\Leftrightarrow \sum a-\sum a^2 +2\sum a-2\sum ab\leq \frac{9}{4}$
$\Leftrightarrow (\sum a-\frac{3}{2})^2\geq 0$


vector
geometry
area of a triangle
geometry proposed

Let ABCDE be a convex pentagon, and let M, N, P, Q, R be the midpoints of its sides AB, BC, CD, DE, EA respectively.
If the lines AP, BQ, CR, DM all pass through the same point O, prove that the line EN also passes through this point O.

Something like vectors or affine transformations might work.

Just a generalization of Ceva 's theorem : Poncelet 's theorem.  :D

Here is my solution of the problem. It uses a generalization of the Ceva theorem indeed (by the way, I have never seen it being called Poncelet theorem, but who knows):

[color=blue][b]Theorem 1.[/b] Let n be a positive integer, and let $A_1A_2...A_{2n+1}$ be any (2n + 1)-gon. We work with indices cyclic modulo 2n + 1, so that $A_{2n+1+i}=A_i$ for any i. Let P be a point in the plane; for every index i, let the line $A_iP$ meet the line $A_{n+i}A_{n+1+i}$ at a point $X_i$. Then,

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=1$.

Hereby, we work with directed segments.[/color]

[i]Proof of Theorem 1.[/i] Let's use not only directed segments, but also directed areas, and let's denote the directed area of a triangle $P_1P_2P_3$ by $\left[P_1P_2P_3\right]$.

Let S and T be the orthogonal projections of the points $A_{n+i}$ and $A_{n+1+i}$ on the line $A_iP$. Then, the two lines $A_{n+i}S$ and $A_{n+1+i}T$ are parallel to each other (since they are both perpendicular to $A_iP$); if we direct these two lines in the same way, then Thales yields $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{A_{n+i}S}{A_{n+1+i}T}$. On the other hand, the segments $A_{n+i}S$ and $A_{n+1+i}T$ are the altitudes of the triangles $A_iPA_{n+i}$ and $A_iPA_{n+1+i}$ to the common side $A_iP$; since the area of a triangle equals $\pm\frac12\cdot\text{ side }\cdot\text{ corresponding altitude}$, where the $\pm$ sign is + or - depending on the orientation of the triangle, it follows that $\left[A_iPA_{n+i}\right]=\pm\frac12\cdot A_iP\cdot A_{n+i}S$ and $\left[A_iPA_{n+1+i}\right]=\pm\frac12\cdot A_iP\cdot A_{n+1+i}T$; note that the two $\pm$ signs are either both + or both - (since the lines $A_{n+i}S$ and $A_{n+1+i}T$ are directed in the same way). Hence, $\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}=\frac{A_{n+i}S}{A_{n+1+i}T}$. Comparing this with $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{A_{n+i}S}{A_{n+1+i}T}$, we obtain $\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}$. Thus,

$\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=-\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=-\frac{\left[A_iPA_{n+i}\right]}{\left[A_iPA_{n+1+i}\right]}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_i\right]}=\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]}$,

where in the last step, we have used the fact that $A_{n+\left(n+1+i\right)}=A_{2n+1+i}=A_{i}$ (since indices are cyclic modulo 2n + 1). This identity holds for every index i; multiplying all such identities, we get

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\prod_{i=1}^{2n+1}\frac{\left[A_iPA_{n+i}\right]}{\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]}=\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]\left/\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]\right.$.

But actually, $\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]$ and $\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]$ are the same product, just with a different order of factors. Hence,

$\prod_{i=1}^{2n+1}\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\prod_{i=1}^{2n+1}\left[A_iPA_{n+i}\right]\left/\prod_{i=1}^{2n+1}\left[A_{n+1+i}PA_{n+\left(n+1+i\right)}\right]\right.=1$,

and Theorem 1 is proven.

Now, let's reformulate the problem (note that the pentagon does not need to be convex):

[color=blue][b]Theorem 2.[/b] Let $A_1A_2A_3A_4A_5$ be an arbitrary pentagon, and let $M_1$, $M_2$, $M_3$, $M_4$, $M_5$ be the midpoints of its sides $A_3A_4$, $A_4A_5$, $A_5A_1$, $A_1A_2$, $A_2A_3$, respectively. If the lines $A_1M_1$, $A_2M_2$, $A_3M_3$ and $A_4M_4$ have a common point O, then the line $A_5M_5$ also passes through this point O.[/color]

[i]Proof of Theorem 2.[/i] Since the lines $A_1M_1$, $A_2M_2$, $A_3M_3$ and $A_4M_4$ pass through the point O, the points $M_1$, $M_2$, $M_3$, $M_4$ are the points where the lines $A_1O$, $A_2O$, $A_3O$, $A_4O$ meet the lines $A_3A_4$,  $A_4A_5$, $A_5A_1$, $A_1A_2$, respectively. Let the line $A_5O$ meet the line $A_2A_3$ at a point $N_5$. Then, by Theorem 1, we have

$\frac{A_3M_1}{M_1A_4}\cdot\frac{A_4M_2}{M_2A_5}\cdot\frac{A_5M_3}{M_3A_1}\cdot\frac{A_1M_4}{M_4A_2}\cdot\frac{A_2N_5}{N_5A_3}=1$.

But since $M_1$, $M_2$, $M_3$, $M_4$ are the midpoints of the sides $A_3A_4$, $A_4A_5$, $A_5A_1$, $A_1A_2$, we have $\frac{A_3M_1}{M_1A_4}=1$, $\frac{A_4M_2}{M_2A_5}=1$, $\frac{A_5M_3}{M_3A_1}=1$, $\frac{A_1M_4}{M_4A_2}=1$. Thus, we also have $\frac{A_2N_5}{N_5A_3}=1$, so the point $N_5$ must be the midpoint of the side $A_2A_3$. Thus, $N_5=M_5$. Now, from the definition of the point $N_5$, it is clear that the line $A_5N_5$ passes through the point O; hence, the line $A_5M_5$ passes through the point O, and Theorem 2 is proven.

  Darij

using affine transformations is not a good idea since we have $R^2$ and $4$ vectors to "control"

We can use area method to prove it just like the Ceva theorem


Given: Triangle ABD, Angle BAD = 90* altitude AC (C lying on segment BD)
AD = 6, BC = 5..

To Prove: AB and CD
Proof: NOW HERES WHERE I GET STUCK... plz som1 help i have no idea what to do..

Are you trying to find AB and CD, or prove AB = CD?

err im going to assume he means find $AB$ and $CD$, because $AB \not = CD$

[hide]
we know that in this triangle

$AB^2 = BC(BD)$

setting $CD = x$, we have $AB^2 = 5(5+x)$

then we also know $AD^2 = CD(BD) \Rightarrow 36=x(5+x) = x^2 + 5x$

so $x^2 + 5x - 36 = 0$, $x = 4 or -9$ but we can toss out the negative value of $x$, so $x=CD=4$

plugging that back into the equation for $AB$, we have $AB^2 = 5(5+4) = 45$

so $AB = 3\sqrt5$

the formulas i used (i.e. $AB^2 = BC(BD)$ ) can be derived using the similar triangles that you make when you draw the altitude, so if u arent familiar with them jacquz you can work them out that way

[/hide]

hah $\$*%# som1 hah  :lol: havent used this internet talk for a long time since joining MathLinks :D

this problem has been posted twice in same forum so jaquz please avoid from doing that in the future

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45195]http://www.mathlinks.ro/Forum/viewtopic.php?t=45195[/url]

Yeah, i set $AC=x$ and then used similar triangles and quadratic formula to solve...


For what value of $ x$ does $ \frac{0.\overline{09}}{x} \equal{} 11^{\minus{}1}$?

We know that $ {0.\overline{09}=\frac{1}{11},}$ so we have that $ \frac{\frac{1}{11}}{x}=\frac{1}{11}\implies\boxed{x=1}.$


inequalities
algebra
polynomial
calculus
derivative
inequalities proposed

[hide]
$ \frac{5}{2}\sum_{cyc}\sqrt{ab}+\frac{1}{2}\sum_{cyc}\sqrt{a}\geq 9$ by $ AM-GM$
so we need to prove
$ a^{2}+b^{2}+c^{2}+3\geq a+b+c+ab+bc+ca$
set $ a=x^{3},b=y^{3},c=z^{3}$ and manipulate the inequality into
$ \sum_{cyc}x^{6}+3x^{2}y^{2}z^{2}\geq \sum_{cyc}x^{4}yz+\sum_{cyc}x^{3}y^{3}$
which is true because $ LHS\geq \sum_{sym}x^{4}y^{2}$ by schurr and the rest is easy...
[/hide]

For proving $ x^{2}+y^{2}+z^{2}+3\geq x+y+z+xy+yz+zx$ you can mix variables, or simply use the Turkevici inequality:

\[ \sum a^{4}+2abcd\geq \sum_{sym}a^{2}b^{2}\] for \[ d=1, abc=1, x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}\]

ok, i think i found a solution, but i need a check up.
[hide]
we want to prove that
$ (a-b)^{2}+\frac{5ab+1}{a+b}\geq a+b+1$ which is equivalent to proving
$ (a-b)^{2}(a+b-1)+(a-1)(b-1)\geq 0$. now if both $ a,b$ are $ \geq 1$ it's obvious.
if both are $ \leq 1$ it's safe to assume that $ a+b-1\leq 0$ otherwise we are done. in this case we have
$ (1-a)(1-b)=1-a-b+ab\geq 1-a-b\geq (1-a-b)(a-b)^{2}$ because $ 1-a-b$ is positive and $ (a-b)^{2}\leq 1$ because $ 0\leq a,b\leq 1$
now the third case is $ a\geq 1\geq b$. assume at first that $ a+b-1\geq \frac{1}{4}$
now we have $ (a-b)^{2}(a+b-1)\geq \frac{1}{4}((a-1)+(1-b))^{2}\geq (a-1)(1-b)$
assume now that $ a+b-1<\frac{1}{4}$. look at the polynomial of variable $ b$
$ f(b)=b^{3}-b^{2}(a+1)+b(3a-a^{2}-1)+(a-1)^{2}(a+1)$.we can assume $ a\not=1$ because then it's trivial. we want to prove that $ f(b)\geq 0$ for all $ b\geq 0$. lets assume different. then $ f$ has at least one positive real root. since $ f(0)>0$ it has at least one negative root and we conclude that it has three real roots, one negative and two positive. we can also assume that the positive roots are different otherwise we are done. now we have
$ f(b)=(b+p)(b-q)(b-r)$ with $ p,q,r\geq 0$ comparing coefficients with $ b^{2}$ we have $ p-q-r=-a-1$ or $ q+r=1+a+p$. WLOG assume $ q>r$ so we have $ 2q>q+r=1+a+p>1$ so $ q>\frac{1}{2}$.
we assumed there exists $ b_{0}$ such that $ f(b_{0})<0$. if $ b_{0}>q$ then $ f$ would have another root greater than $ q$ which cannot be so $ q>b_{0}$.
if there exists $ \epsilon>0$ such that for all $ x$ from $ (q-\epsilon,q)$ we have $ f(x)>0$ then $ q$ would be a local minimum of $ f$ and it's derivative would be $ 0$ and $ q$ would be a double root of $ f$ and that isn't true(i think you can do this part without derivatives but it's just shorter for me).
now pick $ \epsilon$ such that $ \epsilon<\frac{1}{4}$ now we have that $ f(b_{0})<0$ where $ b_{0}>\frac{1}{4}$ but that's impossible because then we would have $ b_{0}+a-1>\frac{1}{4}$
[/hide]


How many square units are in the area of polygon $ ABCD$ with vertices $ A(\minus{}4,4), B(1,\minus{}4), C(1,\minus{}1)$ and $ D(\minus{}4,\minus{}1)$?

The key to this problem is to recognize the trapezoid with bases AD and BC and perpendicular leg CD. Since AD=5, BC=3, and CD=5, the answer is (1/2)(5+3)(5)=[b]20[/b].

Hmm...the review says its $25$ but I got $20$ too.

I agree with 20

you can use the shoelace method (see [url=http://mathworld.wolfram.com/PolygonArea.html]a reference site[/url])with the points in clockwise order (ACBD) as follows

-4      4
1       -1
1        -4
-4       -1
-4       4

Left products in order: 4,-1,16,4
Right products in order: 4,-4,-1,-16

left sum: 23
right sum: -17
23--17=40
40/2=[b][i][u]20[/u][/i][/b]

I took a different approach but still got the same answer as previous posters:

Draw a very simple graph, just enough to get and idea of where the points. As you can see the graph is divided into 2 triangles which has a very clear base and a very clear height. Triangle 1: (-4,4)(-4,-1)(1,-1) triangle 2: (1,-1)(1,-4)(-4,-1)

Simple mind calc: triangle 1: 12.5units^2     triangle 2: 7.5units^2

Add these up and you get 20, however the computer said the answer is 25, going to report this problem now.


function
induction
algebra
functional equation
algebra unsolved

Find all continuous functions $f: \mathbb R \to \mathbb R$ so that $ f(x+f(x))=f(x)$ for all $ x \in \mathbb R$.

If we assume the uniform continuity on $ \mathbb{R}$ (not just the continuity), then I can prove that $ f$ is constant... :wink:

So, can we deduce the uniform continuity from the functional equation (which is just to prove that $ f$ has limit in $ \plus{}\infty$ and $ \minus{}\infty$)?

Pierre.

[quote="duytungct"]Find all continuous function $ f: R \to R$ so that $ f(x \plus{} f(x)) \equal{} f(x)$ for all $ x \in R$[/quote]

$ 1)$ First, it is easy to show with induction that $ f(x\plus{}nf(x))\equal{}f(x)$ $ \forall x\in \mathbb{R}$ and $ \forall n\in\mathbb{N}$

$ 2)$ Constant functions $ f(x)\equal{}c$ are obviously solutions.

$ 3)$ Assume it exists $ f(x)$ non-constant continuous solution. Then
It exists $ x_0<x_1$ such that $ f(x_0)\neq f(x_1)$
Let $ A\equal{}\{x>x_0$ such that $ f(x)\equal{}f(x_1)\}$
$ A\neq_\emptyset$ (since $ x_1\in A$) and $ x>x_0$ $ \forall x\in A$. Let then $ x_2\equal{}\inf(A)$.

Since $ f(x)$ is continuous, $ f(x_2)\equal{}f(x_1)$, $ x_2>x_0$ and $ \boxed{f(x)\neq f(x_1)\forall x\in[x_0,x_2)}$

Since $ f(x_2)\neq f(x_0)$, $ \exists n\neq p\in\mathbb{N}$ such that either $ x_0\plus{}nf(x_0)<x_2\plus{}pf(x_2)<x_2\plus{}nf(x_2)$, either $ x_0\plus{}nf(x_0)>x_2\plus{}pf(x_2)>x_2\plus{}pf(x_2)$
Since $ f(x)$ is continuous, so is $ x\plus{}nf(x)$, so $ \exists x_3\in(x_0,x_2)$ such that $ x_3\plus{}nf(x_3)\equal{}x_2\plus{}pf(x_2)$, so $ f(x_3\plus{}nf(x_3))\equal{}f(x_2\plus{}pf(x_2))$, and so $ f(x_3)\equal{}f(x_2)\equal{}f(x_1)$

But this is a contradiction since we know (see boxed statement above)  that $ f(x)\neq f(x_1)$ $ \forall x\in[x_0,x_2)$

So the only solutions are constant functions.

Ok, I got it!

Let $ f$ be a solution.
Clearly, $ f$ cannot be injective.

Let $ g(x)\equal{}f(x)\plus{}x$.
Let $ g^n$ denotes the $ n^{th}$ iterate of $ g$.

It is easy to verify that $ g$ is continuous and $ g^2(x)\equal{}2g(x)\minus{}x$ for all $ x$.
Thus, $ g$ is injective.
Since $ g$ is continuous and injective, we deduce that $ g$ is strictly monotone. But, if $ g$ is decreasing then $ f(x)\equal{}g(x)\minus{}x$ is (strictly) decreasing too, so that $ f$ would be injective, a contradiction.
Thus, $ g$ is increasing.

A straightforward induction leads to $ g^n(x) \equal{} ng(x)\minus{}(n\minus{}1)x \equal{} nf(x)\plus{}x$ for all $ x$ and positive integer $ n$.

Let $ a,b$ such that $ a<b$.
Since for each positive integer $ n$, the function $ g^n$ is increasing, we deduce that $ g^n(a) < g^n(b)$, which leads to $ f(a)\minus{}f(b) < \frac {b\minus{}a}{n}$ for all $ n>0$.
Taking limit on $ n$, leads to $ f(a)\minus{}f(b) \leq 0$, so that $ f$ is non-decreasing.

A straightforward induction leads to $ f(x\plus{}nf(x)) \equal{} f(x)$ for all $ x$ and non-negative integer $ n$
Since $ f$ is non-decreasing, it follows that $ f$ is constant on each intervall with endpoints $ x$ and $ x\plus{}nf(x)$ for each $ x$ $ n >0$.

Assume that there exists $ a$ such that $ f(a) > 0$ and $ c$ such that $ f(c)\equal{}0$.
Let $ M$ be the sup of the set $ S$ of the $ x$ such that $ f(x) \equal{}0$ (for all such $ x$, we have $ x < a$ since $ f$ is non-decreasing).
Using a sequence of element of $ S$ which has limit $ M$ and continuity of $ f$, we clearly deduce that $ f(M)\equal{}0$.
For each $ e>0$, we have $ f(M\plus{}e)>0$ so that $ f$ is constant on $ [M\plus{}e,\plus{}\infty[$. Clearly, when $ e$ varies, the constant value is always the same, say $ y$. Thus, for each $ e>0$, we have $ 0<y \equal{} f(M\plus{}e)$. Taking limit on $ e$ and using continuity, we deduce that $ f(M)\equal{}y >0$. A contradiction.

Assume there exist $ b$ such that $ f(b) < 0$ and $ c$ such that $ f(c)\equal{}0.$
Thus, from above $ f$ is constant on $ ]\minus{}\infty , b]$, and we get the same contradiction as above by considering $ m \equal{} \inf S$.

Therefore $ f$ is the zero function, or $ f$ has always the same sign on $ \mathbb{R}$.
In the latter case, if $ f(a) > 0$ for all $ a$, then $ f$ is constant on each $ [a, \plus{} \infty[$ for all $ a$, so that $ f$ is constant.
If $ f(b) < 0$ for all $ b$ then $ f$ is constant on each $ ]\minus{} \infty , b]$ for all $ b$, so that $ f$ is constant.

Thus, in each case $ f$ is constant.

Conversly, if $ f$ is constant then $ f$ is clearly a solution.

Pierre.

Aaargh... a bit too slow in writing... :oops:

Pierre.

That's a bit too complicated, Pierre.  :D  Once you know that $ g(g(x))\equal{}2g(x)\minus{}x$ you easily deduce that $ g$ is bijective (by continuity and monotony), so that relation $ g^n(x)\equal{}ng(x)\minus{}(n\minus{}1)x$ actually holds for all integers $ n$. Next write $ g^n(x)<g^n(y)$ for all integers $ n$ if $ x<y$ and you're done.


linear algebra
matrix
superior algebra
superior algebra unsolved

Hi,

I have the following set M = {$ (a,b): a,b \in R , a^2 \minus{} b^2 \neq 0$}

where R is the set of real numbers.

I need to define * on M by $ (a,b) * (c,d) \equal{} (ac \plus{} bd, ad \plus{} bc)$

namely, that M is closed under *, show that * is associative on M and find an identity element for * on M.

(1) For closure, I have shown that if $ (a,b) \in R$, then $ (a,b) * (c,d) \equal{} (ac \plus{} bd, ad \plus{} bc) \in R$

I am not sure if I am doing the right thing, but I think if $ (a,b) \in R$, then $ ac \plus{} bd \in R$ and $ ad \plus{} bc \in R$, thus $ (a,b) * (c,d) \equal{} (ac \plus{} bd, ad \plus{} bc) \in R$, which means that M is closed under *.

(2) As for the identity, as R is under multiplication, the identity element = 1, such that $ (a,b)*1\equal{}(a,b)$ but I am not sure how to prove it in the given context.

(3) I also need to find the inverses and show that (M, *) is a group.

So far I have done the following:

$ \dfrac{1}{(ac \plus{} bd)}$ = $ \dfrac{(ac \minus{} bd)}{(ac \plus{} bd)*(ac \minus{} bd)}$ = $ \dfrac{(ac \minus{} bd)}{(a^2)*(c^2) \minus{} (b^2)*(d^2)}$

We know that $ a^2 \minus{} b^2 \neq 0$

So, $ \dfrac{(ac \minus{} bd)}{(a^2)*(c^2) \minus{} (b^2)*(d^2)}$ = $ \dfrac{ac}{(a^2)*(c^2) \minus{} (b^2)*(d^2)} \minus{} \dfrac{bd}{(a^2)*(c^2) \minus{} (b^2)*(d^2)}$

This brings me to the conclusion that inverses $ \in R$. (I have not done the second bit, but it is a similar approach).

I am wondering if I am going the right thing here.

Any advice would be appreciated.

Thank you!

Please be more careful with the distinction between $ \mathbb{R}$ and $ M$.

I would be suspicious of this; it looks like we're extending $ \mathbb{R}$ by $ \sqrt{1}$. That's a problem, because both roots are already real numbers.

The identity element is $ (1,0)$. "$ 1$" is not an element of $ M$.

You have not completed the argument for closure- is the "norm" $ a^2\minus{}b^2$ still nonzero for a product? You haven't used the whole definition of $ M$.

Your inverse argument is nonsense. We don't care about inverting the real number $ ac\plus{}bd$; we want the inverse of $ (a,b)$.

The inverse of $ (a, b)$ is some element $ (a', b') \in M$ with the property that $ (a, b) * (a', b') \equal{} (1, 0)$.  Work from here.

Edit:  jmerry, it looks like this is (edit: the multiplicative group of units of) $ \mathbb{R}[X] / (X^2 \minus{} 1)$.

Not exactly- that's the ring, if we include the zero-divisors. With the norm condition, this is the group of units in that ring.

Thank you for your help.

I have tried to show that M is closed under *, and have got the following:

$ (a,b)*(c,d) \equal{} (ac\plus{}bd, ad\plus{}bc)$

So, if we square these, we get:

$ (ac\plus{}bd)^2 \equal{} a^2*c^2 \plus{} b^2*d^2 \plus{} 2acbd$
$ (ad\plus{}bc)^2 \equal{} a^2*d^2 \plus{} b^2*c^2 \plus{} 2adbc$

As $ a^2 \minus{} b^2 \neq 0$  then $ a^2 \neq b^2$. So,

$ a^2(c^2 \minus{} d^2) \minus{} b^2(c^2 \minus{} d^2) \neq 0$

(from the following expansion)

$ a^2*c^2 \plus{} b^2*d^2 \plus{} 2acbd \neq a^2*d^2 \plus{} b^2*c^2 \plus{} 2adbc$
$ a^2*c^2 \plus{} b^2*d^2 \minus{} a^2*d^2 \minus{} b^2*c^2 \neq 0$
$ a^2(c^2 \minus{} d^2) \minus{} b^2(c^2 \minus{} d^2) \neq 0$

$ a^2 \minus{} b^2 \neq 0$

which shows that M is closed under *.

Not sure if this is the correct argument for closure...

Also, with regard to inverses:

as $ (a,b)*(a',b')\equal{}(1,0)$

So, $ (a,b)*(a',b')\equal{}(aa' \plus{} bb', ab' \plus{}ba')$

$ a' \equal{} \frac{1}{a}$

$ b' \equal{} \minus{}b$

$ aa' \plus{} bb' \equal{} 1 \minus{} b^2$

$ ab' \plus{}ba' \equal{} a*(\minus{}b) \plus{} \frac{b}{a} \equal{} b \minus{} a^2*b \equal{} b*(1 \minus{} a^2)$

$ (a,b)*(a',b')\equal{}(1 \minus{} b^2, b(1 \minus{} a^2)) \equal{} (1,0)$

$ 1 \minus{} b^2 \equal{} 1, b\equal{}0$

$ b*(1 \minus{} a^2) \equal{} 0, b\equal{}0, a^2 \equal{} 1, a \equal{} \sqrt{1}$

So, $ b\equal{}0, a\equal{}1, a\equal{}\minus{}1$

and $ b'\equal{}0, a'\equal{}1, a'\equal{}\minus{}1$

Again, not sure if this is sufficient to show that M is a group under *.

[quote="aRb"]$ (a,b)*(a',b') \equal{} (aa' \plus{} bb', ab' \plus{} ba')$

$ a' \equal{} \frac {1}{a}$

$ b' \equal{} \minus{} b$[/quote]

I have no idea what you're trying to do here.  You're solving the system $ aa' \plus{} bb' \equal{} 1, ab' \plus{} ba' \equal{} 0$.  Try again.

[quote="aRb"]$ a^2*c^2 \plus{} b^2*d^2 \plus{} 2acbd \neq a^2*d^2 \plus{} b^2*c^2 \plus{} 2adbc$
$ a^2*c^2 \plus{} b^2*d^2 \minus{} a^2*d^2 \minus{} b^2*c^2 \neq 0$
$ a^2(c^2 \minus{} d^2) \minus{} b^2(c^2 \minus{} d^2) \neq 0$

$ a^2 \minus{} b^2 \neq 0$[/quote]

This is correct, but I'm quite confused by your last line.  What you need to do is factor the last line:

$ (a^2\minus{}b^2)(c^2\minus{}d^2)$.  Since you already know that neither of these factors can be zero, their product cannot be zero either.  Other than that, your closure argument is 100% correct.

This might help if you are familiar with matrices: $ M$ corresponds to

$ G \equal{} \left\{\left(\begin{array}{cc}a&b\\b&a\end{array}\right) : a^2\minus{}b^2 \neq 0\right\} \subset GL_2(\mathbb{R})$.

Then you just need to show $ *$ corresponds to matrix multiplication, and that $ G$ is actually a subgroup.

Thank you all!

$ aa' \plus{} bb' \equal{} 1$
$ ab' \plus{} ba' \equal{} 0$
$ a(a'\plus{}b') \plus{} b(a'\plus{}b') \equal{} 1$

So, $ a' \plus{} b' \equal{} \dfrac{1}{a\plus{}b}$

$ a' \equal{} \dfrac{1\minus{}ab'\minus{}bb'}{a\plus{}b}$

$ b' \equal{} \dfrac{1\minus{}aa'\minus{}a'b}{a\plus{}b}$

Would this be sufficient to show that M is a group under * (together with showing closure and the identity element)?

You need to express $ a', b'$ solely in terms of $ a, b$.  This is a system of two linear equations in two variables.


Hi all --

I've been doing past AMC 12 exams and have managed to get just barely qualifying scores (around 109.5). Do the exams generally stay around the same difficulty from year to year, or have they been getting easier/harder lately? I'm just hoping that this year's exam won't be too much more difficult than 2008's or 2007's.

Thanks,
Arachnotron

Yes, they're usually the same. You can tell by doing the old AMC 12 exams that can be found on this site. But if a certain year's AMC 12 is unusually difficult, then less than 5% of the participants will score above a 100, so the cutoff will be lowered. (For example, the cutoffs for AMC 12 A and B were lowered to 97.5.)

On a side note, they do get a little harder, so a 2008 test might not be harder than a 2007, but should be harder than one from 1970.

in 1970, they had AHSME, not AMC 12, and i am pretty sure AHSME is harder than AMC 12... i think

The last three problems on the 1970 AHSME were:

Find the sum of all the digits of all the numerals in the sequence $ 1,2,3,4,\ldots,10000$

What is the greatest integer that will divide 13511, 13903, and 14589 and leave the same remainder?

A retiring employee receives an annual pension proportional to the square root of the number of years of his service.  Had he served $ a$ years more, his pension would have been p dollars greater, whereas, had he served $ b$ years more, $ b \neq a$, his pension would have been $ q$ dollars greater than the original annual pension.  Find his annual pnesion in terms of a, b, p, and q

Those are clearly easier than the last 3 problems on a recent AMC 12.

Did the removal of the ability to use a calculator affect the question difficulty at all?

The calculations needed on recent tests do seem a bit easier. But some questions are harder; on old tests, for instance, you could approximate something and then compare to the answer choices, removing the need to manipulate trigonometric expressions, roots, etc. So it is harder IMO.

They've also changed the nature of the problems so calculators are not necessary.

seems to me that the 2008 test was somewhat harder than the others. how hard do you guys think it would be compared to other tests to get 100+ on this year's AMC 12

Obviously, as the cut-off was lower, it must have been harder.  I, personally, didn't feel it was any harder, and achieved what was my normal score (90-100) with a 97.5 on the AMC 12 B.

Don't forget that the difference in the cutoff is also a result of the scoring change (although I don't know when exactly they changed it). Leaving a question blank used to get you 2.5 points, but now it gets you 1.5.

That scoring change has been in action for 2 years already, and 3 for the AMC that's coming up next month.

I'm so excited for February!

Let's not compare AHSME especially back in 1970. That's almost bad as comparing George Washington to George Bush because they have first name. It doesn't quite make sense to be honest.

The difficulty usually stays same since AMC 12 came on year 2000. The difficulty level between A and B are usually same but differs in each person's weaknesses. For instance, person named A is good at geometry but insanely horrible at number theory. The exam A for that year had decent geometry questions with somewhat difficult NT problems. The exam B for that year had harder geometry but easier NT problems. In this case, person A would find B easier because he's better at geometry. People who are not good at geometry will find A easier.

Also, AHSMEs underwent significant changes. For one thing, back when it first came, there were 50 problems. Then it went down to 30 problems. Then when AMC 12s first came (with 5 less problems), you were allowed to use calculators until they were recently prohibited by MAA. So you can't honestly judge exactness in difficulty because of changes. Nevertheless, AMC 12s are about "same" whereas AHSMEs are somewhat mixed because they have extra 5 problems (and longer time I believe).

And oh, back in 1980's, AHSME was pretty insane because AIME did not come out till 1983. I can be positive that later problems in 1980's AHSMEs were harder than early-medium AIME problems in 80's.

[quote="buzzer11"]I don't think that affected the cutoff...[/quote]
It's hard to imagine that it wouldn't. The cutoff is either at the 5th percentile or at 100 (whichever is lower), and the 5th percentile would be lowered by giving fewer points for the same answers. If you don't believe me, take a look at the AIME cutoff scores on the AMC site... They were solid 100's from 2002-2006, then went down in 2007 and 2008. The average score also dropped in 2007 and 2008.

But it was only two years ago when the cutoff was lowered. There was no change last year, was there?


geometry
circumcircle
perpendicular bisector
geometry unsolved

[color=darkblue]Let $ (O;R)$ is a circle, let $ B, C$ are two points lie on $ (O), (BC<2R)$. Let $ A$ is a change point and $ A$ lie on $ (O;R)$, let $ AA'$ is a diameter of $ (O;R)$. Let $ AH \perp BC, BE \perp AA', CF \perp AA'$, with $ H$ lie on $ BC$ and $ E, F$ lie on $ AA'$.

1) Prove that $ HE \perp AC$. [i](easy)[/i]

2) Prove that $ \triangle HEF$ and $ \triangle ABC$ are similar. [i](easy)[/i]

3) Let $ I$ is center point of circle $ (HEF)$. Prove that $ I$ is a fixed point.[/color]

Yes. Thank you very much. Can you give me the full solution?

Yes! Thank you very much.

$ \bullet$ $ \angle OAB \equal{}\angle IHE$ and $ \angle OBA \equal{} \angle IEH$. But $ OA\equal{}OB\equal{}R \Longrightarrow \angle OAB \equal{}\angle OBA  \Longrightarrow \angle IHE \equal{} \angle IEH  \Longrightarrow  IE\equal{}IH$

$ \bullet$ Similar we get $ \angle IHF \equal{} \angle IFH  \Longrightarrow  IH\equal{}IF$

Hence $ IH\equal{}IE\equal{}IF  \Longrightarrow  I$ is center point of circle $ (HEF)$.

Problem have done.

It's well known the fact that AO and AH are isogonal cevians in triangle ABC, hence <BAH = <OAC ( 1 ) and <OAB = <HAC ( 2 ).
From ABHE cyclic, <CHE = <OAB and, with (2), <CHE = <HAC ( 3 ), but <HAC = 90 degs - <C ( 4 ).
From AHFC cyclic, <EFH = <C ( 5 ) and <CAF = <CHF ( 6 ). From (4) and (5) we get I on BC.
(6), (3) and (1) dau <EHF = <A ( 7 ), from (5) and (7) the triangles HEF and ABC are similar.
AHFC cyclic, with AC as diameter shows that the perpendicular bisector of HF passes through the midpoint of AC, while from (1) and (6) we have FH and AB perpendicular, therefore the perpendicular bisector of FH is medial line in triangle ABC, passing through the midpoint of BC, which is the circumcenter of (EFH).

Best regards,
sunken rock


Some unit cubes are assembled to form a 6x6x6 large cube. A sphere is inscribed in the large cube. How many unit cubes have their interiors inside the sphere?

In any one of the eight corner 3x3x3 cubes, the sphere will pass through the interiors of all but four cubes. Due to symmetry, each of the eight 3x3x3 cubes act exactly the same, meaning the sphere passes through the interior of 184 cubes.


integration
calculus
function
real analysis
real analysis unsolved

Let $f: [-1, 1]\to [f(-1), f(1)]$ be a stricly increasing and $C^{2}[-1,1]$ function. Prove that for any $n\in\mathbb{N^{*}}$, there is $c\in (-1,1)$     such that

                  \[ \int_{f(-1)}^{f(1)}(f^{-1}(x))^{2n-1}dx=\frac{2}{2n+1}\cdot f''(c).  \]

We make the change of variables $x=f(t)$.

Then the integral at the left becomes  $\int_{-1}^{1}t^{2n-1} \cdot f'(t)dt$ .

$\\ \int_{-1}^{1}t^{2n-1}\cdot f'(t)dt = [ \frac{t^{2n}}{2n} \cdot f'(t)]_{-1}^{1} - \int_{-1}^{1}\frac{t^{2n}}{2n} \cdot f''(t)dt = \\ \\ \\ = \frac{f'(1) -f'(-1)}{2n} -f''(a) \cdot \int_{-1}^{1}\frac{t^{2n}}{2n}dt =\frac{f''(b)}{n} -f''(a) \cdot \int_{-1}^{1}\frac{t^{2n}}{2n}dt$

 for some $a,b \in (-1,1)$ from mean value theorem .

Now $\frac{f''(b)}{n} -f''(a) \cdot \int_{-1}^{1}\frac{t^{2n}}{2n}dt = \frac{f''(b)}{n} -\frac{2 \cdot f''(a)}{2n(2n+1)}$

and $\frac{f''(b)}{n} -\frac{2 \cdot f''(a)}{2n(2n+1)}=\frac{2}{2n+1} \left(\frac{(2n+1)f''(b)- f''(a) }{2n} \right)$.

If $a=b$ we take $c=a=b$ and we are done. Suppose wlog that $a <b$.
We consider the function $g(x) = 2n \cdot f''(x) -(2n+1) \cdot f''(b) + f''(a)$
Since $g(a) \cdot g(b) <0$and $g$ is continuous from Bolzano theorem we have that
there exist $c \in (a,b)$ such that $g(c)=0$ . Then

$f''(c)=\frac{(2n+1)f''(b)- f''(a) }{2n}$ and we are done .

...................................................................................................
But i can't see why $\frac{(2n+1)f''(b)- f''(a) }{2n}$ can be written as $f''(c)$ for some $c \in (-1,1)$ .  :?

Maybe the mistake is not there ???  :D

Edit : I think i found it. We consider the function $g(x) = 2n \cdot f''(x) -(2n+1) \cdot f''(b) + f''(a)$
and suppose that $a < b$ .

It is $g(a) \cdot g(b) <0$and $g$ is continuous therefore there is $c \in (a,b)$ such that $g(c)=0$ .  :)

Edit #2: I completed the solution.  :lol:

soarer in $f('t)=f'(0)+tf''(c)$

$c=c(t)$ depend with $t$

you cannot move it outside of the integral


Basically, turnout for last year's AoPS meeting was pretty pathetic (the official meeting, at least), but that was probably due to the weather.

So, who's up for a meeting?  Discussion of place/time can commence once I am sure that somebody else aside from me is coming.

Dude I'm coming.

So I guess it's never too early to plan for ARML? :D

[quote="13375P34K43V312"]Yeah I'll show up I guess[/quote]

[quote="roadnottaken"]Sure, I'll probably come.[/quote]

I was kinda hoping for more enthusiasm than that.

OMG I'm coming.

I remember last year, nr1337 and me were the only AoPSers from Ohio, and we were walking around and he was like "I think we should go back to the dorms because it's raining" and I was like "no, it's not raining too bad" and then it started raining pretty bad and no one was at the pool, and no one was even swimming even though it was like June and then I went to the song contest and everyone was there and then I felt bad because I couldn't go swimming.

I searched all over the place and found no pool last year (probably due to the weather), so maybe we should have it at an easier-to-find place

WE'RE DOING IT AGAIN? YEEE-HAW!

[There, is that sufficiently enthusiastic? ;)]

I seem to remember various frozen-looking people swimming, or at least trying to ...

Anyway, I'll come, tell me where.

[quote]... various frozen-looking people ...[/quote]
The words "frozen-looking" would never have been used in Las Vegas.

[quote="MysticTerminator"]OMG I AM PROBABLY NOT COMING

BTW IT'S ARML

DID I TELL YOU ABOUT THE MACHINES THERE[/quote]

what machines?

and yea i'll be there for sure... i couldn't go there last year cuz i, #1, couldn't find the place, and #2, had to eat my veggie burger

[quote="E^(pi*i)=-1"]Please don't ask Arnav about the machines.

I'll try to come.  Last year was my team's first year and we sort of spent the whole afternoon and evening being confused and trying to figure out where stuff was and where to eat dinner.   :rotfl:

Personally, I think meeting at the song contest preliminaries is the easiest thing to do.

And once again, please do not ask Arnav about the machines.[/quote]

eeh, okay

then i'll ask you

what are the machines?  :|

Last year I offered a game of ultimate frisbee meeting next to the outside pool stadium.
It rained.
But a couple of people on my team played anyway.
It was fun. I recommend it. Don't wear socks or shoes, though.
One big game of ultimate frisbee.

Any discussion about the machines will be deleted.  PM Arnav if you really must know.  But it adds nothing to this discussion (it's not even really that humorous).

Anyway, I'm definitely up for a game of frisbee.

Frisbee would be fun  :)

But where exactly will we meet? And what time? And some place better than some swimming pool...

I guess i'll come.  And by the way, 2006 ARML song contest ftw.  You know SC shoulda won but they rigged the preliminaries.

[quote="lingomaniac88"]So, if this occurs, where would it be?[/quote]

Edit: Sorry, someone pointed out by PM (not_trig) that the information I received was incorrect

I met a lot of people at the song contest last year.  That wouldn't be a bad spot.  Or at least in that building since it's a pretty easy place to find.  I believe the name of the building is Findlay Commons?  In any case, there's a basketball court right outside of that building and some fields nearby.  I'll probably play as much frisbee/basketball as I can fit in my spare time there.

So meet in Findlay Commons near the registration/gameroom/big screen (i.e. the side closer to the b-ball courts, unless I'm not remembering correctly) at some time on Friday?  I would set a time but I'm unsure of when the SC team will be going to get dinner on Friday.  Perhaps an hour or so before the song contest would work out well since the song contest prelims were (and should be again) in that local neighborhood.

[quote="joml88"]So meet in Findlay Commons near the registration/gameroom/big screen (i.e. the side closer to the b-ball courts, unless I'm not remembering correctly) at some time on Friday? ... Perhaps an hour or so before the song contest would work out well since the song contest prelims were (and should be again) in that local neighborhood.[/quote]

Well, I object to this meeting time---I'll be giving a talk "an hour or so" before the song contest, so I wouldn't be able to go to the gathering!

Why not meet right after the song contest prelims?  It'd be easy to announce/remind people that way, too.

Well, I got back to Penn state after dinner very late.  I think that Findlay Commons sometime on Saturday (preferably during breakfast or lunch) would be a good time and place, since we know that we'll all be there.

OMG GUYS I JUST FOUND OUT THERE'S NO CHIPOTLE NEAR PENN STATE, I DON'T KNOW WHAT I'M GOING TO DO!!!!!!!!!!!!!

The song contest sounds good.

This was pre-empted by the "Join sides...", correct?


Four prime numbers are randomly selected without replacement from the first ten prime numbers. What is the probability that the sum of the four selected numbers is odd? Express your answer as a common fraction.

the sum is prime if and only if one of the numbers is 2 (figure out why this is true youself)

therefore, out of the ten numbers, we have to choose a 2, so we need to choose three numbers out of the nine numbers remaining
9C3 = $ \frac{9\times8\times7}{3\times2} \equal{} 3\times4\times7 \equal{} 84$

there are a total of 10C4 ways to choose 4 numbers out of 10
10C4 = $ \frac{10\times9\times8\times7}{4\times3\times2}\equal{}10\times3\times7 \equal{} 210$

so our probability is $ \frac{84}{210}\equal{}\frac{42}{105}\equal{}\frac{14}{35}\equal{}\boxed{\frac{2}{5}}$

hope i didn't make a mistake

As already stated, the sum is odd only if 2 is one of the primes selected.  What is the probability that 2 is one of the first 4 primes selected out of 10?  $\frac{4}{10} = \frac{2}{5}$.  Easy peasy.

[font=TIMES NEW ROMAN][hide=Logical Thinking]We can first think about how we can obtain an odd number through addition. We know that for two numbers, whenever we add one odd number and one even number we get an odd number as a result.

So for four numbers, we need an odd number and an even number, as well as either two odd numbers or two even numbers.

Fortunately, there is only one even prime number: two. So we know that one of the numbers that is chosen must be two in order for the two numbers to sum to an odd number. This means that our chosen numbers must include two, and three other odd numbers in order to work.

So we can solve this problem by supposing we have to choose three numbers from nine - the number two must always be included, so we just choose from prime numbers two to ten.

This means that there are $\tbinom{9}{3}=84$ combinations that work. There are $\tbinom{10}{4}=210$ total possible combinations, so the probability of getting working one is $\boxed{\tfrac{84}{210}\leadsto\tfrac{2}{5}}.$
$$\definecolor{color}{rgb}{0.1, 0.8, 0.1} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

Another way that I thought about it was that you must get $2$ in order to fit the solution. You have a $\frac{1}{10}$ chance of getting it. Since you choose a number $4$ times, you just do $\frac{1}{10} \times 4$. This results in $\frac{2}{5}$


factorial
number theory unsolved
number theory

Find the first digit of 100 factorial

Sriram

An easy computer program shows it's 9, but I can't prove it yet.

I don't believe that there are the proof.

Well, I think it has something to do with Wilson's theorem. Applying it we obtain:
$100! \equiv -1$ $mod$ $101$ as 101 is a prime, but I can not come to more concrete conclusions.


Find all complex roots.  Leave your answer in polar form with the arguments in DEGREES.

QUESTION:

The complex CUBE roots of -8.

[hide="hint"]
Rewrite $ \minus{}8$ in polar as $ 8 cis 180^\circ$
Then apply DeMoivre's
[/hide]


If $BC=12\sqrt3$ and Circles $O$ and $P$ are congruent and tangent, what is the radius of the circle?

[hide="Answer"]$\triangle OAB \sim \triangle OPC$
$\bigcirc O = \bigcirc P \Rightarrow 2(OA)=OP=2(PC)$
$2(OB)=OC=8\sqrt{3}$
$(PC)^2+(8\sqrt{3})^2=(2(PC))^2 \Rightarrow 3(PC)^2=192 \Rightarrow \boxed{PC=8}$[/hide]

How do you make a box around the answer?

[quote="JesusFreak197"][hide="Answer"]$\triangle OAB \sim \triangle OPC$
$\bigcirc O = \bigcirc P \Rightarrow 2(OA)=OP=2(PC)$
$2(OB)=OC=8\sqrt{3}$
$(PC)^2+(8\sqrt{3})^2=(2(PC))^2 \Rightarrow 3(PC)^2=192 \Rightarrow PC=8$[/hide]

How do you make a box around the answer?[/quote]

rather than using the pythagorean (sp?) theorem i used 30-60-90. it's a good alternative in this case
 :roll:

[quote="JesusFreak197"][hide="Answer"]$\triangle OAB \sim \triangle OPC$
$\bigcirc O = \bigcirc P \Rightarrow 2(OA)=OP=2(PC)$
$2(OB)=OC=8\sqrt{3}$
$(PC)^2+(8\sqrt{3})^2=(2(PC))^2 \Rightarrow 3(PC)^2=192 \Rightarrow PC=8$[/hide]

How do you make a box around the answer?[/quote]
correct
This is how you box it:
[code]\boxed{...}[/code]


I've seen my brother's blog on AoPS and I'm just wondering how i make one because he forgot.

It should be under profile. I believe there is a new blog button.

awwwww....

It's not in profile some reason.. :(

Could you please provide a screenshot so I can see what you see when you go to profile? Thanks.

Due to a spam epidemic a couple months ago, there is a minimum post requirement before making a blog. You probably haven't reached that level yet.


algebra
polynomial
advanced fields
advanced fields open

Is linear programming strongly polynomial?  ;)

Unsolved problem as of a few years ago.   If solved it would have been big news, I think,


Consider the inequality $x^{3}+y^{3}+z^{3}\ge 3xyz, x, y, z \in \mathbb{R}^{+}$.

Find as many ingenious proofs of this inequality as you can.  There are several standard ones, but can you find a few that are not so standard?

[hide="For example..."]
$x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z)( (x-y)^{2}+(y-z)^{2}+(z-x)^{2}) \ge 0$ [/hide]

:)  Good luck!

Two obvious ones are Power mean and Holders.

Yep, or also rearranging :
We may assume wlog $x \geq y \geq z$. Then we have $x^{2}\geq y^{2}\geq z^{2}$, so :
$x \cdot x^{2}+y \cdot y^{2}+z \cdot z^{2}\geq x \cdot y^{2}+y \cdot z^{2}+z \cdot x^{2}= xy \cdot y+yz \cdot z+zx \cdot x$.
But, $xy \geq zx \geq yz$, so :
$xy \cdot y+yz \cdot z+zx \cdot x \geq xy \cdot z+yz \cdot x+zx \cdot y = 3xyz$.

Here's a similar one to T0rajir0u's:
[hide]
from a well known factoring trick:

$x^{3}+y^{3}+z^{3}\ge 3xyz \implies (x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz) \ge 0$

Now, $x+y+z$ is always positive, so we have to prove that:

$x^{2}+y^{2}+z^{2}\ge xy+yz+xz$

WOLOG $x \ge y \ge z$, then the above follows straight from rearrangement.
[/hide]

also, AM-GM on $x^{3}, y^{3}, z^{3}$ is pretty obvious...


If $ a^2\plus{}b^2\plus{}c^2\equal{}3$ and $ a,b,c>0$

prove

$ \frac{3\minus{}a^2}{b\plus{}c}\plus{}\frac{3\minus{}b^2}{c\plus{}a}\plus{}\frac{3\minus{}c^2}{a\plus{}b} \geq a\plus{}b\plus{}c$

[hide="Solution"]From the given condition, we have $ \frac{3\minus{}a^{2}}{b\plus{}c}\plus{}\frac{3\minus{}b^{2}}{c\plus{}a}\plus{}\frac{3\minus{}c^{2}}{a\plus{}b}\equal{}\frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}$.

Hence, it suffices to prove that $ \sum_{\text{cyc}}\frac{a^2\plus{}b^2}{a\plus{}b}\ge \sum_{\text{cyc}}\frac{a\plus{}b}{2}$, which follows from AM-GM Inequality.

We are done.[/hide]

[quote]
\[ \sum\limits_{cyc} {\frac {{3 \minus{} a^2 }} {{b \plus{} c}}} \equal{} \sum\limits_{cyc} {\frac {{b^2 \plus{} c^2 }} {{b \plus{} c}} \equal{} } \sum\limits_{cyc} {\frac {{\left( {b \plus{} c} \right)^2 \minus{} 2bc}} {{b \plus{} c}} \equal{} \sum\limits_{cyc} {b \plus{} c \minus{} \frac {{2bc}} {{b \plus{} c}}} } \geqslant a \plus{} b \plus{} c\]
[/quote]

[quote]
\[ b \plus{} c \minus{} \frac {{2bc}} {{b \plus{} c}} \plus{} a \plus{} b \minus{} \frac {{2ab}} {{a \plus{} b}} \plus{} c \plus{} a \minus{} \frac {{2ca}} {{c \plus{} a}} \geqslant a \plus{} b \plus{} c\]
[/quote]

[quote]
\[ a \plus{} b \plus{} c \equal{} \frac {{\left( {a \plus{} b} \right) \plus{} \left( {b \plus{} c} \right) \plus{} \left( {c \plus{} a} \right)}} {2} \geqslant \frac {{2bc}} {{b \plus{} c}} \plus{} \frac {{2ab}} {{a \plus{} b}} \plus{} \frac {{2ca}} {{c \plus{} a}}\]
[/quote]
es directo ya que
[quote]\[ \sum\limits_{cyc} {\frac{{a \plus{} b}}
{2}}  \geqslant \sum\limits_{cyc} {\frac{2}
{{\frac{1}
{a} \plus{} \frac{1}
{b}}}}  \equal{} \sum\limits_{cyc} {\frac{{2ab}}
{{a \plus{} b}}}\][/quote]

hello, your inequality is equivalent to
$ \frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}\plus{}\frac{a^2\plus{}b^2}{a\plus{}b}\geq a\plus{}b\plus{}c$
and this is equivalent with
$ a(b^2\minus{}c^2)(b\minus{}c)\plus{}b(a^2\minus{}c^2)(a\minus{}c)\plus{}c(a^2\minus{}b^2)(a\minus{}b)\geq0$
and this is true.
Sonnhard.

$ \frac{b^{2}\plus{}c^{2}}{b\plus{}c}\plus{}\frac{c^{2}\plus{}a^{2}}{c\plus{}a}\plus{}\frac{a^{2}\plus{}b^{2}}{a\plus{}b}\geq a\plus{}b\plus{}c$

chevychev's inequality works , we have $ \sum \frac{a^{2}\plus{}b^{2}}{a\plus{}b} \ge \sum \frac{a\plus{}b}{2}\equal{}a\plus{}b\plus{}c$


parameterization
quadratics
LaTeX
algebra
quadratic formula

Solve the equation with the parameter $ a\in\mathbb{R}$ :

$ x \plus{} \sqrt {a \plus{} \sqrt {x}} \equal{} a$

This is a really annoying problem... I've done the multiple squaring and the existence conditions but I can't get anywhere, could anyone post a [i]solution[/i] if you can find the time?

[hide]Here's a "slick" solution:
Let $ t \equal{} sqrt(a \plus{} sqrt{x})$. We have: $ x \plus{} t \equal{} t^2 \minus{} \sqrt {x}$.
This is equivalent to: $ (t \minus{} 1/2)^2 \equal{} (\sqrt {x} \plus{} 1/2)^2$.
So either $ t \minus{} 1/2$ = $ \sqrt {x} \plus{} 1/2$ or $ 1/2 \minus{} t \equal{} \sqrt {x} \plus{} 1/2$.
Since $ t\geq 0$, we see that the 2nd cases occurs at only $ t \equal{} 0$ and $ x \equal{} 0$.
Now,we have left with the 1st case, which is: $ t\equal{}\sqrt{x}\plus{}1$.Replace $ t \equal{} \sqrt{a \plus{} \sqrt{x}}$ and squaring both sides with $ t\geq 1$, we get the quadratic equation:
$ x \plus{} \sqrt {x} \plus{} 1 \minus{} a \equal{} 0$
Using quadratic formula we get: $ x \equal{} ( \minus{} 1 \plus{} ( \minus{} )\sqrt {4a \minus{} 3})/2$. Since x must be greater than or equal to 0, $ x$ = $ ( \minus{} 1 \plus{} \ sqrt(4a \minus{} 3))/2$($ a\geq 3/4$). Now plugging back in this value of x into the equation to find any possibly essential restrictions of a.[/hide]

ghjk, thank you for your solution. I understand what you did and I don't see any mistakes, but could you recheck what you did? I have as answers:

$ \begin{cases}a<0 \text{ or }0<a<1\;\;\;\Rightarrow \emptyset \\
a=1\;\;\;\Rightarrow 0\\
a\geq 1\;\;\; \Rightarrow \frac{2a-1-\sqrt{4a+3}}{3} \end{cases}$

Thanks again.

Some LaTeX suggestions for both of you:

1) Proofread!  There's a preview button so that you can look at your post before you make it.  This will prevent, for example, repeated ugly copies of $ sqrt$ appearing in your text, as well as an arrow disappearing from using \Rightarrpw.

2) \pm gives $ \pm$

3) For a longer arrow, use \Longrightarrow

4) When you're using the cases environment, use an ampresand (&) to separate the two parts:

\begin{cases}
a<0 \text{ or }0<a<1  &  \Rightarrow \emptyset \\
a=1  &  \Rightarrow x = 0\\
a\geq 1  &   \Rightarrow x = \frac{2a-1-\sqrt{4a+3}}{3}
\end{cases}

gives

\[ \begin{cases}
a<0 \text{ or }0<a<1  &  \Rightarrow \emptyset \\
a\equal{}1  &  \Rightarrow x \equal{} 0\\
a\geq 1  &   \Rightarrow x \equal{} \frac{2a\minus{}1\minus{}\sqrt{4a\plus{}3}}{3}
\end{cases}
\]

[quote="triplebig"]ghjk, thank you for your solution. I understand what you did and I don't see any mistakes, but could you recheck what you did? I have as answers:

$ \begin{cases}a < 0 \text{ or }0 < a < 1\;\;\;\Rightarrow \emptyset \\
a = 1\;\;\;\Rightarrow 0 \\
a\geq 1\;\;\; \Rightarrow \frac {2a - 1 - \sqrt {4a + 3}}{3} \end{cases}$

Thanks again.[/quote]
I think your root is wrong. Since I already showed my root in both cases above, you can see that it's completely different from your roots in the case $ a\geq 1$. If u did check my solution again, as well as I did, then u probably make mistakes somewhere in your solution.

These answers aren't mine, they are the answers from the list.

I will trust your solution, as I can't find any visible mistakes, thank you for helping out.

Since $ \frac{2a \minus{} 1 \minus{}\sqrt{4a \plus{} 3}}{3} < 0$ for all $ a$ such that it's defined, you can be quite confident that it isn't actually a root of the given equation  :wink:

Simillar problem:

Solve ${ x=a+\sqrt{a+\sqrt{x}}}\ (a>0)$.

1972 Tokyo Medical and Dental University entrance exam


1)Prove that for any $a,b,c>0$ and $k\ge 2$ we always have
\[\frac{a^{2}}{ka^{2}+bc}+\frac{b^{2}}{kb^{2}+ca}+\frac{c^{2}}{kc^{2}+ab}\le \frac{3}{k+1}\]
A related inequality:
2)Prove that for any $a,b,c>0$ we always have
\[\frac{a^{2}}{2b^{2}+ca}+\frac{b^{2}}{2c^{2}+ab}+\frac{c^{2}}{2a^{2}+bc}\ge 1 \]
An open question: Is the general of (2) true?

[hide="1"]
This is equivalent to
\[\sum_\text{cyc}\frac{bc}{ka^{2}+bc}\geq \frac{3}{k+1}\]
after multiplying by $k$ and subtracting each term from 1. We can prove this with Cauchy:
\[\left( \sum_\text{cyc}\frac{bc}{ka^{2}+bc}\right) \left( \sum_\text{cyc}bc(ka^{2}+bc)\right) \geq (ab+bc+ca)^{2}\]
\[\Rightarrow \left( \sum_\text{cyc}\frac{bc}{ka^{2}+bc}\right) \geq \frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+kabc(a+b+c)}\]
Showing that the right hand side is at least $\frac{3}{k+1}$ is equivalent to $(k-2)(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}-abc(a+b+c))\geq 0$, which is a simple AM-GM.
[/hide]

[hide="2"]
The inequality
\[\sum_\text{cyc}\frac{a^{2}}{kb^{2}+ca}\geq \frac{3}{k+1}\]
holds for any nonnegative $k$, since by Cauchy,
\[\left( \sum_\text{cyc}\frac{a^{2}}{kb^{2}+ca}\right) \left( \sum_\text{cyc}a^{2}(kb^{2}+ca)\right) \geq (a^{2}+b^{2}+c^{2})^{2}\]
\[\Rightarrow \sum_\text{cyc}\frac{a^{2}}{kb^{2}+ca}\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+a^{3}c+b^{3}a+c^{3}b}\]
Now we need to show the right hand side at least $\frac{3}{k+1}$; this follows from
\[k[a^{4}+b^{4}+c^{4}-a^{2}b^{2}-b^{2}c^{2}-c^{2}a^{2}]+[(a^{2}+b^{2}+c^{2})^{2}-3(a^{3}c+b^{3}a+c^{3}b)]\geq 0\]
But maybe there is a way to prove it without resorting to such a strong result as Vasc's inequality.
[/hide]

[quote="The soul of rock"]2)Prove that for any $a,b,c>0$ we always have
\[\frac{a^{2}}{2b^{2}+ca}+\frac{b^{2}}{2c^{2}+ab}+\frac{c^{2}}{2a^{2}+bc}\ge 1 \]
An open question: Is the general of (2) true?[/quote]
The sharper for (2):
\[\sum\frac{a^{2}}{2b^{2}+ca}\ge \frac{3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}\]
with equality with $a=b=c$ and $a=b,c=0$. :) My solution for it is too urgly.  :(

Sorry, my inequality is not true (I have some mistakes in calculations). :( But the orginal problem is true, and we can prove it easily by Cauchy Schwarz
\[\sum\frac{a^{2}}{2b^{2}+ca}\ge \frac{(a^{2}+b^{2}+c^{2})^{2}}{2\sum a^{2}b^{2}+\sum ab^{3}}\ge \frac{(a^{2}+b^{2}+c^{2})^{2}}{2\sum a^{2}b^{2}+\sum a^{4}}=1 \]
:)


limit
number theory solved
number theory

For each positive integer n, let $a_n$ denote the number of integer solutions (x,y) to the equation $n^2+x^2=y^2$ such that $x\ge n,y\ge n$.

a) Prove that for every M, $a_n > M$ is satisfied by at least one positive integer n.
b) Is it true that $\lim_{n->\infty}a_n = \infty$?

a) We can take the sequence $n_k=p_1p_2\ldots p_k$, where $\{p_k\}$ is the sequence of primes. For any $M$, we will eventually have $a_{n_k}>M$ for some $k$.

b) No, it doesn't tend to $\infty$. We can take $n_k=p_k^2$. There is at most one decomposition of $n_k$ as the difference of two squares satisfying those conditions.


I found this game site a while ago, and I come to realy love it!
[url]http://www.kongregate.com/?referrer=jackdillon[/url]
tell me what you guys think!
also, any flash game creators can publish there games on the site.

i know what that site is as well, and i have an account too

please dont post it as a referral link

[url]http://www.kongregate.com[/url]

there are cards and badges for people who get certain goals in games


Let $ (O)$ be a circle with diameter $ AB$.$ M$ is a point on $ OB$.$ CD$ is a chord through $ M$($ CD \neq AB$).Suppose that $ O_1$ and $ O_2$ be circumcenter of triangle of $ AMC$ and $ BDM$.Suppose that $ E$ is midpoint of $ O_1O_2$
Prove that $ E$ lies on $ AB$

$ \angle{O_2MB}\equal{}90^o \minus{} \angle{BDM}\equal{}90^o\minus{}\angle{BDC}\equal{}90^o\minus{}\angle{BAM}$. Hence $ O_2M \perp AC$.

Since $ OO_1 \perp AC$, so $ O_2M \parallel OO_1$. Similarly, we have $ OO_2 \parallel O_1M$. Therefore $ O_1OMO_2$ is a parallelogram. Hence the midpoint of $ O_1O_2$ is always on $ AB$.


I took a practice test SAT at Kaplan yesterday and I don't know how to solve this problem,it's a grid in.

Ambrose,working alone, can paint a certain room in 8 hours. Either one of Benedict or Charles,working alone,can paint a room in 4 hours.If all 3 people, working at these rates, work together to paint the room, what fraction of the room is painted by Ambrose?

charles and bob work $ 2$ times as fast as ambrose. Thus together they work 4 times a fast and will have done 4 times as much work as ambrose when they finish painting the room $ \implies$ ambrose does $ \frac{1}{5}$ of the work.


Two students set their digital watches to 10:00. One watch runs one minute per hour too slow, and the other watch runs 2 minutes per hour too fast. What time will the slow watch show when it is exactly one hour behind the fast watch?

The slow one falls behind a total of 3 minutes every real hour. So it takes 20 hours for the slow one to be behind the fast one by 1 hour. Now, since the slow one loses one minute per hour, it will be 19 hours and 40 minutes past 10:00 on the slow one when the fast one is 1 hour ahead. This is the same as 19:40 mod 12:00 = 7:40 later. After adding that to 10:00 we get 10:00 + 7:40 = 17:40 = $ \boxed{5: 40}$ mod 12:00.

Wait, if we were to use variables, how would we set the equation up?


$ P,Q \in R[x]$  of polynomials with real coefficients such that $ P(\frac{1}{2010}\plus{}x\plus{}Q^{4}(x))\equal{}Q(\frac{1}{2010}\plus{}x\plus{}P^{4}(x))$ $ x \in R$
Prove that $ P\equal{}Q$

_________________________________________
Azerbaijan Land of Fire  :ninja:

Where is the solution ???
Who can manage this problem ???


conics
ellipse
parabola
geometry
rectangle
parallelogram
cyclic quadrilateral

I was thinking about some problem and wondered: Can any given quadrilateral be inscribed in an ellipse?  For certain quadrilaterals, does there exist only one such ellipse?  Can someone give a proof/explanation as to what the case is, and why it is true or false?  Thanks  :)

I think the answer is yes.  The non-rigorous way of showing it would be to divide the quadrilateral into two triangles.  For example in ABCD consider ABC and DBC.  The problem is equivalent to stretching the altitudes of the triangles to BC to make a cyclic quadrilateral A'BCD'.  As you stretch A and D perpendicularly to BC, the angles A and D are monotonically increasing, and their sum is 360 and 0 at the two extremes.  Thus at some point the sum is 180.

hrm. the problem with projections is that it may not be an ellipse; it may be a parabola or hyperbola. The general property in this case is that 5 points (in general configuration) uniquely define a conic (so, in general, a quadrilateral would not uniquely define an ellipse).

[quote="s0mp"]you don't know whether that sum of opposite angles = 180 property holds for quads inscribed in an ellipse...maybe some sort of projection would help in this prob...[/quote]

:? read my post again.  If what MysticTerminator said is true, then it seems that a quadrilateral must be inscribable in an ellipse, if not uniquely.

You guys are thinking too deeply about this.  [hide]This problem asks if there is a unique ellipse determined by four coplanar points.[/hide]

Equation for ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$.

Since we have four variables, there should be just one ellipse (then again, they are squared, so that might introduce more solutions)?

[quote="MysticTerminator"]umm...Magnara, yea. How does that simplify anything I've said?[/quote]

Heh.  I didn't read that post.... :)

Paladin, don't forget that that's an ellipse with major and minor axes parallel to the x and y axis.

Yeah, for a rectangle, there are an infinite number of ellipses.  However, for, say, a more general parallelogram, i think it may be unique...dunno.
But yeah, the general equation for a conic is $ax^2+by^2+cx+dy+e=0$; five coefficients, five points, five equations, one solution, makes sense...

Hm, for the ellipse, doesn't a length of the minor and major axis, ie. 2 variables, define a ellipse.
I was thinking about this problem as a circle. 3 pts define a circle because it can define a radius, and that's all the circle needs.

Or maybe I'm wrong, but I'm just saying that I don't think the center/position of the ellipse doesn't really matter.

[quote="MysticTerminator"]K8: you forget the xy term; you didn't read my post ... it's not unique in general[/quote]
o ya
But it does appear that for any quadrilateral, there is at least one ellipse.


geometry
pigeonhole principle
algebra unsolved
algebra

Show that it is impossible to cover a unit square with five equal squares with side $s<\frac{1}{2}$.

[quote="jin"]maybe you mean $s<\frac{1}{2}$.[/quote]

Yes you're  right !!!

Thanks

Assume we can cover it with 5 squares. However look at a side of the unit square , since $s<\frac{1}{2}$  it must be covered by at least 3 of the 5 squares. Analogously the opposite side must be covered by at least 3 squares. Therefore at least one of these squares must cover both sides but this is impossible since its diagonal is less than $\frac{\sqrt{2}}{2}$ which is in turn less than 1. Therefore we have reached a contradiction and we are done.
Correct me if I'm wrong please.

Yes, unfortunately you're wrong, since for $s<1/2$, but large enough, it is easy to use two squares to cover a side. A start towards a solution is to look at the nine points given by the vertices of the unit square, the midpoints of its sides, and its centre. No square of side $s<1/2$ may cover three of them, since if collinear they stretch on a length of at least $1 > s\sqrt{2}$, while otherwise they determine a triangle of area at least $1/8 > s^2/2$. Therefore by pigeonhole principle, we need at least five small squares.

Ok now I see that my mistake was assuming we purposefully had to place the squares inside the unit square.
However your solution is not clear to me, you mention the fact that we can't cover those nine points simultaneously with the squares and then you prove that we can't do so with [b]4[/b] squares? Since if you mean it's not possible with 5 I have an easy counterexample for that.
The interesting thing about this problem is that it is so simple it should be possible to generalize.
I'm sorry if I misunderstood your proof.


I'm don't know who moved my post, but if I did I would have private messaged that person, so I apologize for making this public.

I posted a question only a few hours ago, and it was apparently moved to the high school intermediate level forum.

With all due respect, this is a question that was assigned to a 400 level college class.  I'm sorry if you think it's too simplistic for you to bother having it in your forum, or if you think it's below you to have it here.

I respect that you want to keep your forums clean, but please keep in mind that just because you (and your genius level intellect) think it may be a simple problem, does not mean that it is easy for everyone.

[url]http://www.mathlinks.ro/viewtopic.php?t=187826&sid=b457e0747df626c726c7bb7c2237dbea[/url]

It was me.

I've no clue what a "400 level college class" is, but this problem can be done with intermediate methods. And it doesn't require much work, too. Thus it fits there. Possibly, Pre-Olympiad fits also, but I considered it better there.
Compared to the other problems in the Number Theory forum, it's really easy. It's simple diversion of difficulty. If the criterium for a problem to be kept here is "it's not easy for some person", then we must keep problems like "solve x^2=4", too. So this is no argument.

And yes, I want my forums to be clean  :wink:

Just to fill you in, I'm in my final semester as a mathematics major, and 400 level classes are "upper level" or "senior level" classes (typically the more difficult).

I realize now that the answer is not that difficult, but every now and again, I come across problems, that at first glance, cause a bit of "writer's block", and I just can't seem to get the ball rolling.

Thank you for your respectful tone. :wink:

The main question is:
What else should I do¿
As it looks, the only alternative is doing nothing. And it's not like your post is the only one ever moved, I move about 3 threads a day, I think (maybe I should start counting). Additional duties of a moderator are: deleting multiposts, searching/merging/linking reposts, stopping trolls and flames, helping on general questions, renaming threads, et cetera (and I'm far away from doing everything that should be done). Most of this isn't that much fun, and I've real world stuff (like university) to do, too.
Writing a personal message to every user in these cases would even make the time needed 5 times that much. Thus this is only the last resort if some user acts to extreme.

About 90% of the work I do is caused by users that in some sense act not "good". So in fact, most of this work could be avoided, and if that would happen, I would truly have more time to answer each remaining thing personally.

In this special case, it sounds like you feel discriminated or considered stupid because I moved it. Or that I think that the question is stupid. That's simply not true. But please understand that it does not fit here as this subforum is for problems around olympiad level or beyond, and these are normally really more advanced.


projective geometry
Olimpiada de matematicas

Aca les dejo los examenes de la OMCC 2006 realizada en Panama del 30 de julio al 5 de agosto.

¿Alguien tiene los resultados por países?

Los resultados se encuentran en la pagina oficial de la VIII centro, los resultados de la delegacion mexicana fueron:

Oro:
Paul y Jose Daniel

Plata:
Andres Emilsson

Vamos Mexico!!!!

Los problemas de la OMCC 2006 ya me salieron, pero con excepcion del 6, la vdd ya he hecho varias cosas, y todas me resultan contradictorias. Me gustaria saber si alguien lo ha hecho, para k me de cuenta de n k ando fallando.
Si kieren k escriba lo k he hecho asta aora, k me digan y con gusto lo pongo!!

Y no mas como comentario, creo k esta OMCC no fue dificil!!

Espero sus soluciones  :oops:

EXELENTE PRIMER PERFECTO EN OMCC FELICIDADES DANIEL!!!!

[quote="conejita"]
Si kieren k escriba lo k he hecho asta aora, k me digan y con gusto lo pongo!![/quote]

Los que podrías hacer es abrir 6 topics, cada uno con un problema, y ahi compartir la solución

seria bueno q alguien ponga la soluciones , especialmente para el 5 y la 6 en mi caso  :blush: ,

[quote]Sean $ \Gamma$ y $ \Gamma'$ dos circunferencias de igual radio con centros $ O$ y $ O',$ respectivamente.  $ \Gamma$ y $ \Gamma'$ se cortan en dos puntos y $ A$ es uno de ellos. Se escoge un punto $ B$ cualquiera en $ \Gamma.$ Sea $ C$ el otro punto de corte de la recta $ AB$ con $ \Gamma'$ y $ D$ un punto en  $ \Gamma'$ tal que $ OBDO'$ es un paralelogramo. Demuestre que la longitud de $ CD$ es constante, es decir, no depende de la elección de $ B.$[/quote]
Sea $ E \equiv BD \cap \Gamma,$ diferente de $ B.$ Basta observar que como $ \Gamma \cong \Gamma',$ entonces por evidente simetría si $ BD\parallel OO',$ los arcos $ AE$ y $ AD$ en $ \Gamma$ y $ \Gamma'$ son congruentes $\Longrightarrow$ $ \angle ABE  \equal{} \angle ACD.$ Así el triángulo $ \triangle BDC$ es isósceles con ápice $ D$ $ \Longrightarrow$ $ CD \equal{} BD \equal{} OO' \equal{} \text{const}.$

[quote]Sea $ ABCD$ un cuadrilátero convexo. Sea $ I$ el punto de intersección de las diagonales $ AC$ y $ BD.$ Sean $ E, H, F$ y $ G$ puntos sobre los segmentos $ AB, BC, CD$ y $ DA,$ respectivamente, tales que $ EF$ y $ GH$ se cortan en $ I.$ Sea $ M$ el punto de intersección de $ EG$ y $ AC$ y $ N$ el punto de intersección de $ HF$ y $ AC.$ Demuestre que $ \frac {_{AM}}{^{IM}} \cdot \frac {_{IN}}{^{CN}} \equal{} \frac {_{IA}}{^{IC}}.$[/quote]
Note que los triángulos $ \triangle AEG$ y $ \triangle CFH$ son perspectivos a través del perspector $ I,$ entonces por teorema de Desargues las intersecciones $ X \equiv AD \cap BC,$  $ Y \equiv AB \cap DC$ y $ P \equiv EG \cap HF$ son colineales. Considérese $ Q \equiv AC \cap XY.$ Luego por teorema de Menelao en los triángulos $ \triangle MQP$ y $ \triangle NQP$ cortados por las rectas $ AB$ y $ CD,$ tenemos

$ \frac {YQ}{YP} \cdot \frac {PE}{EM} \cdot \frac {MA}{QA} \equal{} 1 \ , \ \frac {YQ}{YP} \cdot \frac {PF}{NF} \cdot \frac {NC}{QC} \equal{} 1$

$ \Longrightarrow  \frac {MA}{NC} \cdot \frac {PE}{EM} \cdot \frac {NF}{PF} \equal{} \frac {QA}{QC} \ (*)$

Pero por teorema de Menelao en  $ \triangle MNP$ cortado por $ \overline{FIE},$ se tiene

$ \frac {PE}{EM} \cdot \frac {NF}{PF} \equal{} \frac {IN}{IM}.$  Combinando con $ (*)$ resulta

$ \frac {MA}{NC} \cdot \frac {IN}{IM} \equal{} \frac {QA}{QC}.$  Luego como $ (A,C,I,Q) \equal{} \minus{} 1,$ se sigue que

$\frac {AM}{IM} \cdot \frac {IN}{CN} \equal{} \frac {QA}{QC} \equal{} \frac {IA}{IC}.$

El problema anterior puede ser generalizado de la siguiente forma

[b]Proposición:[/b] $ABCD$ es un cuadrilátero convexo y $J$ es un punto en la diagonal $AC.$ Una recta por $J$ corta a $AB,CD$ en $P,Q$ y otra recta por $J$ corta a $BC,DA$ en $R,S.$ $AC$ corta a $PS,RQ$ en $M,N.$ Entonces $\frac{_{IA}}{^{IC}}=\frac{_{AM}}{^{JM}} \cdot \frac{_{JN}}{^{CN}}.$

Proyéctese la recta a través de $AD \cap BC$ y $AB \cap DC$ al infinito. Denotando los puntos projectados con subíndice cero, $ABCD$ con intercepto $I$ de diagonales se transforma en un paralelogramo $A_0B_0C_0D_0$ con centro $I_0$ y claramente $\triangle A_0P_0S_0$ y $\triangle C_0Q_0R_0$ son homotéticos a través de  $J_0.$ Como $I_0A_0=I_0C_0$ y $\frac{_{J_0M_0}}{^{J_0N_0}}=\frac{_{A_0M_0}}{^{C_0N_0}},$ se sigue que

$ \left (\frac{I_0A_0}{I_0C_0} \cdot \frac{M_0C_0}{M_0A_0} \right) \cdot \left (\frac{J_0M_0}{J_0N_0} \cdot \frac{C_0N_0}{C_0M_0} \right)=(A_0,C_0,I_0,M_0) \cdot (M_0,N_0,J_0,C_0) =1$

Así, $(A,C,I,M) \cdot (M,N,J,C) =1 \Longrightarrow \frac{IA}{IC} \cdot \frac{CN}{JN} \cdot \frac{JM}{MA}=1.$


geometry
geometric transformation
reflection
function

[quote="Altheman"]question: give the equation of the reflection of y = 2x-5 into y=3x
i am not sure if i got this right or not, and looking from the graph, it does not seem to be right, i would like a second opinion, my work is as follows,

[hide]consider point line distance formula for $ax+by+c=0$ and a point (x,y)
if we have 2 functions, and they reflect over a line, the distance from one line to to line it is being reflected into is always the same as the distance from the reflection and the line it is being reflected into for it to be a reflection,
so for the first function, it can be an ordered pair, (x,2x-5), and that distance from that point to $3x-y+0=0$ is the same as another linear function if it is to be an inverse, let our other function be $y=mx+b$ or (x,mx+b)
so setting them equal, we get
$\frac{|3x-y|}{\sqrt{3^2+1^2}} = \frac{|3x-y|}{\sqrt{3^2+1^2}}$ simplifying, and putting in our 2 functions,
$|3x-(2x-5)| = |3x -(mx+b)|$
$|x+5|=|(3-m)x-b|$, note that the previous function where m=2, and b=-5 satisfies this, but we need 2 other values of m and b for this, by inspection, we see that m=4, and b=5 also works, so the answer is
$\boxed{y=4x+5}$, but as i said, looking at the graph on my calc does not seem to show this? my answer seems logical?, is the graph on the calc just unreliable, or what?[/hide][/quote]

If a line $L_1$ is reflection of the line $L_3$ in the line $L_2$, then image of every point on $L_1$ w.r.t $L_2$ lies on $L_3$ and vice versa.  And $L_2$ is one of the bisectors of angle  between $L_1$ and $L_3$.
The lines $L_1  \equiv {\rm   2x-y - 5=0}$ and ${\rm L}_{\rm 2}  \equiv {\rm 3x-y=0}$meet at (-5, -15)
pick some point, say (5, 5) on the line $L_1  \equiv {\rm   2x-y - 5=0}$
Find the image of this point in the line ${\rm L}_{\rm 2}  \equiv {\rm 3x-y=0}$
it gives us (-1, 7).

The required line is nothing but the line joining the points (-5, -15) and (-1, 7)
and the answer is $11x-2y+25=0$


I guess it is 2,2-dimethyl bromocyclopentane...

No it isnt it. There s another methyl shift to produce a more stable tert carbocation.
So it is 1-bromo-1,2-dimethylcyclopentane.
Ashwath plz no your question.

2-bromo 1,1-dimethyl cyclopentane is the answer....

Yeah there is one more rearrangement. Raghavendra is right. So much for my paltry question...


is it possible to cut a square into several obtuse triangles?

if you can, can you give a proof for it?

i meant is it possible to dissect a square into only obtuse triangles?

Since any square can be dissected into four right triangles and that any right triangle can be dissected into three obtuse triangles, it is possible to dissect a square into obtuse triangles. A right triangle can be dissected into obtuse triangles by connecting the orthocenter to the vertices.

[quote="maximocapcom"] A right triangle can be dissected into obtuse triangles by connecting the orthocenter to the vertices.[/quote]
But, the orthocenter, here does not lie inside the (right) triangle :(

[hide](Of course, you can dissect into  2 (non right triangles)  and one which is not obtuse  can then be dissected into obtuse triangles :) )
[/hide]
(BTW,  A related problem is to dissect a square into all acute triangles) The answer is yes (Not too hard) and of course, a acute triangle can be cut into obtuse triangle by the above method)

(Hint: The question has been asked and answered  in AoPS before )

(Peek only if you want to see the answer)

[hide] [url=http://www.ics.uci.edu/~eppstein/junkyard/acute-square/]click here[/url][/hide]

Posted and discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=obtuse&t=105119

[quote="thealchemist"]Just curious,
if you knew the solution why did you post it as "can't do it"? :([/quote]

His original post is from January 2006. He solved it a few days ago. There's nothing wrong with that.


[quote="moogra"]Simplify the following polynomial:

(X – A)(X – B)(X – C)(X – D) . . . (X – Z)

is therea easy way to do this?[/quote]

Look at all of the seperate factors. Is there one that will make this problem much easier?  :D  [hide](This is really a trick question)[/hide]

[quote="moogra"]Simplify the following polynomial:

(X – A)(X – B)(X – C)(X – D) . . . (X – Z)

is therea easy way to do this?[/quote]
[hide="Easy"]The answer is $0$ because one of the parts is $(X-X).$[/hide]

[hide="My Solution"]

We know that $(X-X)$ is also in there, and x-x=0, so the whole thing is 0.[/hide]

[hide="hint"]Look at the last 5 terms or so[/hide]
[hide="answer"]$x-x$ is one of the factors...[/hide]
Can some mod please move this to puzzles and brainteasers (although I don't think moogra knew that this was a trick question)?

Solution. [hide]0 because of (x-x).[/hide]

Seems every high school does this question!

Actually, there was just a problem exactly like this in Puzzles and Brainteasers :D

right thanks. so its just when x-x=0 then 0*the rest is 0. ok and yeah i got this from a whitehouse.gov site...


Prove that in any acute angle triangle ABC we have:

sum(1/(sinA/2))>=2(sum(1/cos((A-B)/4))).

In the left member is it  \sum (1/(sin(A/2)) ?

Yaap .. soory about that! the ineq is like so:

Prove that in any acute angle triangle ABC we have:

sum(1/(sin(A/2)))>=2(sum(1/cos((A-B)/4))).