algebra
polynomial
Ring Theory
geometry
superior algebra
superior algebra unsolved

Problem:
Theorem: If F is a field, then every ideal of F[x] is principal.
Use the above theorem to prove the equivalence between these two theorems:

Fundamental Theorem of Algebra: Every nonconstant polynomial in C[x] has a zero in C.

Nullstellansatz for C[x]: Let f_1(x), ...,f_r(x) in C[x] and suppose the every alpha in C that is a zero of all r of these polynomials is also a zero of a polynomial g(x) in C[x]. Then some power of g(x) is in the smallest ideal of C[x] that contains the r polynomials f_1(x),..., f_r(x).

The attempt at a solution:
Lets prove the FTA implies that Nullstellansatz first. Say p(x) generates the ideal. I think its pretty clear that all the alpha's will be zeroes of p(x), right? So, what does that mean about the relationship between p(x) and powers of g(x). All the alphas are zeroes of both...but why does that necessarily mean p(x)q(x)=g(x)^n for some q(x)...

the problem is to deduce a nontrivial classical result (the nullstellensatz for algebraically closed fields) from a trivial boring result that polynomial rings are PID? ...

the one direction can be found in every introduction in (classical) algebraic geometry, or here:

http://planetmath.org/encyclopedia/ProofOfHilbertsNullstellensatz.html
http://planetmath.org/?op=getobj&from=objects&id=7313

for the other one: assume that the nullstellensatz holds in $ K[x]$. assume there is some $ f \in K[x]$ with no roots. but then the nullstellensatz applies to $ g \equal{} 1$ and we have $ 1^k \in (f)$ for some $ k$, i.e. $ f$ is constant. thus $ K$ is algebraically closed.

[quote="-oo-"]the problem is to deduce a nontrivial classical result (the nullstellensatz for algebraically closed fields) from a trivial boring result that polynomial rings are PID? ...[/quote]
No it is not, read again. The problem is to show the equivalence of two results, and the second one is only a very special case of the Hilbert Nullstellensatz which can be proven much simpler.

@FibonacciFan: use that $ p(\alpha)\equal{}0$ means $ p(x)\equal{}(x\minus{}\alpha)q(x)$ for some polynomial $ q$.

oh yes, only one indeterminate is boring, too. *g*


Let $ (u_n)$ be the sequence of integers defined by $ u_1\equal{}39, u_2\equal{}45$ and $ u_{n\plus{}2}\equal{}u_{n\plus{}1}^2\minus{}u_n$. Prove that $ 1986$ divides infinitely many terms of the sequence.

Can we use the same technique used [url=http://www.mathlinks.ro/viewtopic.php?t=150826]here[/url]?

it is the same problem than [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=150819[/url]

Lemma 1) $ 2|u_n \iff n \equal{} 3 \plus{} 3k$ for $ k \in \mathbb{N}_0$
Lemma 2) $ 3|u_n$ for all $ n$

Lemma 3) $ 331|u_n \iff n \equal{} a \plus{} 449k$ for $ k \in \mathbb{N}_0$ and $ a \in \{3,116,226,339\}$
After several calculations mod $ 331$, $ u_{447} \equal{} 39$ and $ u_{448} \equal{} 45$ and the only $ u_k \equal{} 0$ with $ k < 448$ are $ u_3 \equal{} u_{116} \equal{} u_{226} \equal{} u_{339} \equal{} 0$


trigonometry
function
conics
ellipse
calculus
calculus computations

How about something like this? :)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=236293#p236293[/url]

It's not exactly what you mean but try  [url=http://www.peda.com/grafeq/gallery.html]this[/url]

[quote="FREAK"]How can you create a cartoon face using tignometry graphs? You need to use at least 3 different types of trig functions and equations that when graphed gives a cartoon face.[/quote]

you can just use three circles (eyes + face) and a parametrically restricted ellipse (mouth)...

Another face here: http://www.niksula.cs.hut.fi/~hkankaan/Homepages/face.html


calculus
integration
logarithms
trigonometry
real analysis
real analysis solved

Find a reccurence relation for the integral:

$\displaystyle I_{n} = \int \sqrt{x^{2}+a} \cdot \ln^{n} (x+\sqrt{x^{2} + a}) dx,$ where $n \in N,$ $n \geq 2 $ and $a > 0.$

I can't understand the meaning[b] n[/b] of $ln ^n.....$

It's simply the $n$'th power of $\ln x$, where $\ln$ is the natural logarithm.

Thank you, grobber!

Now,I will post only answer.

Set $x=\frac{e^t-ae^{-t}}{2}$

The answer may be

\[2^{n+2}I_n-\frac{a}{n+1}u^{n+1}=u^n\frac{e^u-e^{-u}}{2}-nu^{n-1}\frac{e^u+e^{-u}}{2}+n(n-1)\left(2^nI_{n-2}-\frac{a}{n-1}u^{n-1}\right)\]

where $u=2\ln (x+\sqrt{x^2+a)}$

By the way,how can I type [b]hyperbolic sine[/b],[b]hyperbolic cosine [/b]in Tex?

kunny

oops! I made a mistake in calculation.

I will try again!

kunny

[quote="kunny"]

By the way,how can I type [b]hyperbolic sine[/b],[b]hyperbolic cosine [/b]in Tex?
[/quote]

Try $\sinh$ and $\cosh$. These are standard abbreviations.

[quote="Lagrangia"][quote="3X.lich"]Lagrangia: where are your '[i]cheers!!![/i]? I miss them  :blush:  :blush:  ;)
Sorry for posting a useless reply  :P[/quote]

They'll be back .. just that I've been very busy in the last months....   ;)[/quote]
Shame on me  :blush:  :blush:  :blush:


What is ''a circle orthogonal to a line''?  :blush: I know only ''a circle orthogonal to a circle'' .

The same thing.  The two are perpendicular at the meeting points.  Therefore the center of the circle lies on the line.  After all, a line can be seen as just a specific kind of circle.  This is the meaning I have seen and heard.


There is a triangle $ ABC$ with the property that there exists points $ M$ on $ BC$,$ N$ on $ AC$ and $ P$ on $ AB$ which satisfy
the following conditions:
$ (1)$ $ AM$,$ BN$,$ CP$ are concurrent
$ (2)$ $ AM\equal{}BN\equal{}CP$.
Prove that:
$ a)$ The triangle $ ABC$ is acuteangled.
$ b)$ If $ AB\equal{}AC$ and $ AB\geq BC$,then there exist points $ M$ on $ BC$,$ N$ on $ AC$, $ P$ on $ AB$ which satisfy $ (1)$ and $ (2)$.

This problem was given at a contest for juniors. :D

The condition b) implies ABC acute angled as well, doesn't it?

Best regards,
sunken rock

[color=darkblue]I was surprised to find an easy solution ...

[b]a)[/b] Suppose that triangle is $ A$-obtuse. Call $ AM \equal{} BN \equal{} CP \equal{} a$. Then we have $ a \equal{} BN > AB$ and $ a \equal{} CP > AC$. But one of the angles $ AMC$ and $ AMB$ is obtuse $ \Longrightarrow$ $ a \equal{} AM > AB$ or $ a \equal{} AM > AC$, contradiction.

[b]b)[/b] Draw altitudes $ CC_1$ and $ AM$. On the line segment $ AC_1$ take point $ P$ such that $ \sin{\angle{BPC}} \equal{} 2\sin{\frac {\angle{A}}{2}}$.
It is clear that such point $ P$ exists because $ \sin{\angle{BPC}} \equal{} 2\sin{\frac {\angle{A}}{2}} \equal{} \dfrac{\sin{\angle{A}}}{\cos{\frac {\angle{A}}{2}}} > \sin{\angle{A}} \Longrightarrow \angle{BPC} > \angle{A}$.
We'll prove that $ AM \equal{} CP$.

$ \left\| \begin{array}{cc} AM \equal{} \dfrac{BC\cdot \tan{B}}{2} \equal{} \dfrac{BC \cdot \sin{\angle{B}}}{2\cos{\angle{B}}} \equal{} \dfrac{BC \cdot \sin{\angle{B}}}{2\sin{\frac {\angle{A}}{2}}} \\
 \\
\dfrac{CP}{\sin{\angle{B}}} \equal{} \dfrac{BC}{\sin{\angle{BPC}}} \equal{} \dfrac{BC}{2\sin{\frac {\angle{A}}{2}}}\Leftrightarrow CP \equal{} \dfrac{BC \cdot \sin{\angle{B}}}{2\sin{\frac {\angle{A}}{2}}} \end{array} \right\| \Longrightarrow AM \equal{} CP$

Denote $ CP \cap AM \equal{} \{O\}$ and $ BO \cap AC \equal{} \{N\}$. Then $ AM \equal{} CP \equal{} BN$ and we are done!
[/color]


Occasionally, I'll write problems and say, "No, I'm not going to use them in the contest." Instead of throwing them away, I'll post here as "Proposal" problems. They can be made by me or donated from other users or slight modification from existing AMC/AHSME/other contest problems. I hope that by doing this, more people can get interested in participating in my contest (so far 40 people registered!!).

For positive real numbers $ x,y$, the following equation is true:
\[ \log_{10} (2x) \plus{} \log_{10} (y) \equal{} 2
\]

\[ \log_{10} (x^2) \minus{} \log_{10} (2y) \equal{} 4
\]
If $ x \plus{} y \equal{} \frac {m}{n},$ then find $ m \plus{} n$.

Level: Easy question in Hidden Discoveries Preliminary Round (first round equivalent to AMC-12)

[hide="Hint"]Know your rules of logs?
[hide="Answer"]By basic rules of logs, the first equation can be manipulated:
\begin{eqnarray*}
\log_{10} 2x + \log_{10} y&=&2\\
\log_{10} 2xy&=&2\\
2xy&=&10^2\\
2xy&=&100\\
xy&=&50
\end{eqnarray*}
Similar manipulation on the second equation yields $ \dfrac{x^2}{2y}=10000$.
Solve the first equation for $ y$: $ xy=50 \iff y=\dfrac{50}{x}$
Substitute into the second equation and solve for $ x$:
\begin{eqnarray*}
\dfrac{x^2}{2y} &=& 10000\\
\dfrac{x^2}{2\cdot\dfrac{50}{x}} &=& 10000\\
\dfrac{x^3}{100} &=& 10000\\
x^3 &=& 1000000\\
x &=& 100
\end{eqnarray*}
Now you can plug in $ x$ into either equation to find $ y$: $ y=\dfrac{50}{x} \rightarrow y=\dfrac{50}{100} = \dfrac{1}{2}$.
$ x+y=100+\dfrac{1}{2}=\dfrac{201}{2}$, so the final answer is $ \boxed{203}$.
[/hide][/hide]

Correct. You could've multiplied $ (2xy)$ and $ \left(\frac {x^2}{2y}\right)$ to solve for $ x$ first though.

Did you like this problem?  :)

[quote="mathemonster"]yea its nice bt EZ   :D[/quote]

Yes it is pretty easy. If you noticed, I did put its level to be easy question in the first round. The hard questions in the first round and problems in the second round and third round will never be this easy.  :)

It's good that you enjoyed it though!

[hide]
$ \log{2x}\plus{}\log{x^2}\plus{}\log{y}\minus{}\log{2y}\equal{}2\plus{}4$

$ \log{x^3}\equal{}6$

$ x^3\equal{}10^6$

$ x\equal{}100$

plugging in, we also get $ y\equal{}\frac{1}{2}$

so $ x\plus{}y \equal{} \frac{201}{2}$

and $ 201\plus{}2\equal{} \boxed{203}$[/hide]

Please restrict posts in the AMC forum to topics directly related to the AMC/AIME/USAMO or actual problems from those contests.  Other problems should go in the appropriate middle/high school forum.

Sorry, DPatrick.
Let me ask one question for MathLinkers except you.

Sketch the graph of $ y \equal{} \log_{2}x^2$.

[hide]@kunny: simplify the log to become $ y\equal{}2\log_2{x}$. Change of base it, and sketch:

[img]http://myfiles.zapto.org/sketch.gif[/img][/hide]

"Sketch" usually implies "without computer/calculator assistance."

[hide="Try..."] Graphing the inverse function instead. [/hide]

[quote="#H34N1"][hide]@kunny: simplify the log to become $ y \equal{} 2\log_2{x}$[/quote].

Regrettably that's wrong. You should take into consideration the condition for antilogarithm. :(

[quote="kunny"][quote="#H34N1"][hide]@kunny: simplify the log to become $ y \equal{} 2\log_2{x}$[/hide][/quote].

Regrettably that's wrong. You should take into consideration the condition for antilogarithm. :([/quote]
[hide="Is it..."]$ y\equal{}2\log_2{|x|}$?[/hide]

Yes, that's right. The condition is $ x\neq 0$.

hey dude its qiute easy since the graph is y=x so a best fit is a must!

ans to kunny!


The expression $84x^{2}+233x+143$ factors into $(ax+b)(cx+d)$. Find $(a+b)(c+d)$.


How can i calculate the oxidation number for these molecules?
$ Hg_{2}HPO_{4}$
$ Ce_{2}(C_{2}O_{4})_{3}$
$ BrN_{3}$
$ Ce_{2}S_{3}$

Note molecules don't have oxidation numbers, it is the atoms that have oxidation numbers.

BanishedTraitor is right. The atoms is a species, not the species itself, are atributed oxidation numbers. To calculate the oxidation numbers (nox) in the atoms of a species we should bear in mind the following:

- For neutral species, the sum of nox for all atoms must be zero.
- For charged species, the sum of nox for all atoms must equal the charge of the species.
- Hydrogen has nox = +1 when combined with non-metal/metalloid elements, and nox = -1 when combined with metalic elements.
- Oxygen has nox = -2, except when present as peroxyde anion (where nox(O) = -1), or when combined with fluorine (where nox(O) = +2).
- Except in some unusual compounds, the alcaline and alcaline-earth elements have their group oxidation numbers.

Of course, in ionic compounds (as your case) the calculation can be facilitated if we can recognize some of the ions: for example, in Hg2HPO4 there is the hydrogenphosphate anion, $ HPO_{4}^{2\minus{}}$, and so the charge of Hg2 cation must be 2+. In your second, third, and fourth examples you have, respectively, the oxalate anion, the azide anion, and the sulphide anion. With this information in hand, and the rules above, try to compute the nox.

Thank you for your reply. My question was referred to the nox of the [i]atoms[/i] in that molecules. I was just wondering on how to do in that "special" cases. As an example i consider the molecule $ MnBr_{2}$. Consulting a periodic table where are annoted the nox of the atoms i can form the equation x+2y=0 and with substitution can obtain for Mn(+2) and for Br(-1) and so also for $ Bi_{2}(SO_{4})_{3}$ etc.. For the molecules i reported in my later post that method can't apply because for example it gives two solutions among which i can't choose. Using ions i think the problem doesn't exist but... to use this method is it necessary to memorize as much ions as possible or there is another way?

[quote="Humpty_Dumpty"]For the molecules i reported in my later post that method can't apply because for example it gives two solutions among which i can't choose.[/quote]

That's because there are many atoms that can assume several oxidation numbers, for example, nitrogen and phosphorus.

[quote="Humpty_Dumpty"]Using ions i think the problem doesn't exist but... to use this method is it necessary to memorize as much ions as possible or there is another way?[/quote]

Having some experience in the recognition of the most common ions is a very good help in nox calculations, and I think it wouldn't hurt if you spend a little time checking some of those ions. For example, in your first ionic compound, $ Hg_{2}HPO_{4}$, you cannot compute the nox without considering any recognizable ion, because both mercury and phosphorus can assume several oxidation numbers in their compounds. However, with my experience I imediatly recognize HPO4 as the hydrogenphosphate ion, $ HPO_{4}^{2\minus{}}$, and Hg2 as the mercurous ion, $ Hg_{2}^{2\plus{}}$. With this information in hand, we find nox(Hg) = +1, and nox(P) = +5.
In your second example, $ Ce_{2}(C_{2}O_{4})_{3}$, cerium and carbon can have several nox. However, with some knowledge we imediatly recognize C2O4 as the oxalate anion, $ C_{2}O_{4}^{2\minus{}}$, from which imediatly follows that nox(Ce) = +3.
In your third example, I recognize the azide anion, $ N_{3}^{\minus{}}$, and in your last example there is the sulphide ion $ S^{2\minus{}}$.


group theory
superior algebra
superior algebra unsolved

I have some problems trying tosolve this problem using the very fundamentals of group theory (i am not even allowed to use the fact that if e is the identity of a group and a,b are elements of this group, then if ab=e then ba=e)..I really need some help..Thanks..

[b]'' Let G be a group with identity e, and take an element a of G. If b,c are elements of G with ba=e and ac=e, show that b=c. Deduce that each element of G has a unique inverse. ''[/b]

Solve the first equation for $ b$ and the second equation for $ c$.

$ b\equal{}b \cdot e \equal{}b(ac)\equal{}(ba)c\equal{}e \cdot c \equal{}c$


The medians of a right triangle which are drawn from the vertices of the acute angles are 5 and $\sqrt{40}$.  The value of the hypotenuse is:
\[ \text{(A)}\ 10 \qquad \text{(B)}\ 2\sqrt{40} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ 2\sqrt{13} \qquad \text{(E) none of these} \]

[hide]I basically did it the same way as chess64 but let the side lengths of the original triangle be $2x$ and $2y$ to avoid fractions in my solution:

$x^2+4y^2=25$
$4x^2+y^2=40$

Solving, we get x=3, y=2. So the triangle has sides of length 4 and 6 and a hypotenuse of (D)$2\sqrt{13}$.[/hide]

[color=blue]I got the answer as 2sqrt13 [/color][/code]


geometry
AMC
AIME
number theory
relatively prime
power of a point

Source: Mildorf Geometry Lecture

$ABCDE$ is a cyclic pentagon with $BC = CD = DE$. The diagonals $AC$ and $BE$ intersect at $M$. $N$ is the foot of the altitude from $M$ to $AB$. We have $MA = 25$;$MD = 113$, and $MN = 15$. The area of triangle $ABE$ can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine the remainder obtained when $m + n$ is divided by $1000$

what are cyclic pentagons and relative prime numbers? :huh:

a cyclic pentagon is a pentagon that can be inscribed in a circle.

two numbers are relatively prime if the only number that divides both is $1$

Cyclic quadrilaterals are pretty bad, but cyclic pentagons?  :rotfl:

I'm bad at those kinds of problems, when given a polygon and a point inside it, and lengths of vertices of the polygon to the point inside...  :?

this most likely should be in intermediate as it is an aime problem, also not that many people in getting started know ptolemy

[quote="236factorial"]Ptolemy works for cyclic pentagons also? (or is the pentagon broken down into cyclic quadrilaterals somehow in the process of solving the problem?)[/quote]

I'm pretty sure it's broken down.

Moved.  Next time, however, trying identifying the source of the problem.

cyclic pentagon $\to$ Triangle+cyclic quad

[quote="keeyounglee"]wow.. nobody's solved this one yet?

[hide]Pythagoras...drawing lines...some simple trig...ptolemy[/hide][/quote]

The solution may not be especially high tech, but it's very difficult nonetheless. On this forum I have seen just one solution besides my own.

Ptolemy's theorem says that the product of the diagonals of a cylcic quadrilateral equals the sum of the products of opposite sides.

Yeah, it's in the "Cyclic Quadrilaterals" section of AoPS 2. Intersting stuff  :) But try to prove it yourself; it's not really that hard. Just try to use a combination of arc measurements and power of a point formulae.


Prove the inequality
\[\sum_{i<j}  \frac{a_ia_j}{a_i \plus{} a_j} \le \frac{n}{2(a_1 \plus{} a_2 \plus{}\cdots \plus{} a_n)}\sum_{i<j} a_ia_j\]
for all positive real numbers $ a_1, a_2,\ldots , a_n$.

[quote="mr.danh"]Prove the inequality
$ \sum_{i < j} \frac {a_ia_j}{a_i \plus{} a_j} \le \frac {n}{2(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)}\sum_{i < j} a_ia_j$
for all positive real numbers $ a_1, a_2, . . . , a_n$.[/quote]
The general one can be proved by the same way as the case $ n\equal{}3$,  :wink: .
\[ \frac{{ab}}{{a \plus{} b}} \plus{} \frac{{bc}}{{b \plus{} c}} \plus{} \frac{{ca}}{{c \plus{} a}} \le \frac{{3\left( {ab \plus{} bc \plus{} ca} \right)}}{{2\left( {a \plus{} b \plus{} c} \right)}}
\]
And to proved this inequality we use this $ \frac{1}{a} \plus{} \frac{1}{b} \ge \frac{4}{{a \plus{} b}}$, :).

[quote="BaBaK Ghalebi"][quote="Honey_S"]
And to proved this inequality we use this $ \frac {1}{a} \plus{} \frac {1}{b} \ge \frac {4}{{a \plus{} b}}$, :).[/quote]
I think you're wrong... :)[/quote]
Of course ... not, my friend,  :) .
To make it more clearly for you
\[ 2\left( {\sum a } \right)\left( {\sum {\frac{{ab}}{{a \plus{} b}}} } \right) \equal{} 2\sum {ab}  \plus{} 2abc\sum {\frac{1}{{a \plus{} b}}}  \le 3\sum {ab}
\]

Besides that, thanks you for a look at my solution here http://www.mathlinks.ro/viewtopic.php?t=224946&sid=2905796b9149d15b146dcde414121b86
 :)

yeah,you're right...
just a mistyped in your last post,it should be $ 3\sum ab$ instead of $ 3abc$.


Let $x,$ $y$ and $z$ are real numbers such that $x^{2}+y^{2}+z^{2}=1.$ Prove that $\sqrt{1-2xy}+\sqrt{1-2xz}+\sqrt{1-2yz}\geq\sqrt3.$

[hide]
WLOG let $x,y,z$ be nonnegative, since replacing each variable with its absolute value retains the condition and does not increase the left hand side. After squaring both sides, it's equivalent to
\[\sum_\text{cyc}\sqrt{(1-2xy)(1-2xz)}\geq xy+yz+zx\]
But after homogenizing, we find that $1-2xy=z^{2}+(x-y)^{2}$, etc. so that
\[\sqrt{z^{2}+(x-y)^{2})(y^{2}+(x-z)^{2})}\geq \sqrt{y^{2}z^{2}}=yz\]
so the desired inequality follows.
[/hide]

OK, scorpius119. Try it: :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156002

The following inequality is also true.
Let $x$, $y$ and $z$ be real numbers such that $x^2+y^2+z^2=1$. Prove that:
\[\sqrt{1-2xy}+\sqrt{1-2xz}+\sqrt{1-2yz}\geq\sqrt{5-2(xy+xz+yz)}\]


modular arithmetic
combinatorics unsolved
combinatorics

There are $ 3n, n \in \mathbb{Z}^\plus{}$ girl students who took part in a summer camp. There were three girl students to be on duty every day. When the summer camp ended, it was found that any two of the $ 3n$ students had just one time to be on duty on the same day.

(1) When $ n\equal{}3,$ is there any arrangement satisfying the requirement above. Prove yor conclusion.

(2) Prove that $ n$ is an odd number.

These are Steiner triple systems, see [url]http://mathworld.wolfram.com/SteinerTripleSystem.html[/url]. There exists one if and only if $N \equiv 1,3 \pmod{6}$, hence when $N=3n$ for $n \equiv 1 \pmod{2}$.

[quote=jgnr][hide](1) Label the girls $ 1,2,3,4,5,6,7,8,9$.

Day 1: 1,2,3
Day 2: 1,4,5
Day 3: 1,6,7
Day 4: 1,8,9
Day 5: 2,4,8
Day 6: 2,5,6
Day 7: 2,7,9
Day 8: 3,4,7
Day 9: 3,5,9
Day 10: 3,6,8
Day 11: 4,6,9
Day 12: 5,7,8

It is clear that this arrangement satisfies the given requirements.

(2) Assume that there are an even number of girls, namely $ A_1, A_2, A_3, \ldots, A_{2n}$. In one day, $ A_1$ must be with two of them. In another day $ A_1$ must be with another two. And so on. Finally $ A_1$ is already "out of partner". So the number of girls must be odd.[/hide][/quote]

I got the same exact triples in the order you gave  :D


Does anyone know the people that might make Nationals this year?

I heard that this year was a down year for Indiana, and their streak of top 5 finishes nationally might be in jeopardy...but thats just what I heard.

I know one person that should make Nationals, but does anyone know anybody else?  I heard rumor that there was another really good student that got perfect at chapter and also 144 on AMC, but I dont know who he is.

The Benjamins have another younger brother. :D

oh yea they are always gonna be good.

like I said people have told me a 7th grader in our district, but I have never heard of him before so its kind of weird...however, he got perfect at chapter so hes obviously good (and got top 10 state as 6th grader).

Hm perfect chapter doesn't really mean anything this year (like last year). In the hardest chapter in Massachusetts, the chapter winner (who had a perfect) came 17th at states.

Good luck to Indiana anyway. It should be interesting to see how they do this year when people say that they won't be at the very top again. Somewhat like Kansas in the NCAA.

^but he also got 144 on AMC 10 (not confirmed) and did get 8th last year as 6th grader.

Its just weird because hes in our district but no ones heard of him.  I dont think he goes to resurrection's dads classes.

Predictions =D

1. Mark Sellke
2. David Liang
3. Madelaine Snipes
4. Who knows? (yushi?

um yushi=2nd probly not 4th... who are the other people on ur list?

dude i'm good at predicting things. except for like a placement and a person. but w/e.

It was today?  who won.  hopefully yushi did.  i dont know anybody else.

eghh wait, reading what you said agian, resurrection, you predicted it ALL right?

Honey Creek 1st
Sycamore 2nd

No Honey Creek Kids made it(finished 5th and 6th).  1 kid from Sycamore(sibling of a past Nats member I think).  Mark Selke made it again, and I know there was a girl who made it too.

The competition went late Mr Fischer said, so they can add more when they get around.

Good luck Indiana team, sure I'll be hanging with you a bunch like last year,

Mr Frost
Massachusetts

It's Mr. Frost!! Do you know if Mr. Fish minds me posting the results? Mark told me to but I don't have permission yet so eh dunno.

nationals is in orlando this year. sdjf ksjfdbg lkdgf  i want to go -_-.
meh maybe if my parents hadn't made me start school early.

[quote="cognos599"][quote="frost13"] and I know there was a girl who made it too.[/quote]

yeah, the girl who came second goes to my school... does anyone know who came 4th? i know yushi came 3rd,...[/quote]
Wow... you and I must have REALLY sucked last year if she made it this year...

[quote="cognos599"][quote="MathAndKnowledge"]sucked[/quote]

wrong tense.  :)[/quote]

What is the correct tense?

Oh and by the way that's what Gytis said (sucked).

Yeah the guy who got first in chapter and 144 on AMC 10 is David Liang

He just moved here from China recently right-within 3 years?  Because that just shows how good Chinese schools are at math.


Hi I'm trying to get some experience with different methods of proofs.  For a question, I've done several methods, and one of which was proof by contrapositive.  I was wondering if anyone here may please check over my work.  That would be great, thank you!

Question:  Prove or disprove the following statement $P:$ "If the product of two integers is even, then both of the integers are even."

The contrapositive of this implication is: "If either $n$ or $m$ are odd then the product $nm$ is odd.
[u]Proof[/u]
Case 1: ($n$ and $m$) are odd then $n=2k+1$ and $m=2t+1$ for some integers $k,t$
The product $nm=(2k+1)(2t+1)$
$nm=4kt+2k+2t+1$
$nm=2(2kt+k+t)+1$
Since $2kt+k+t$ is an integer, $nm=2(2kt+k+t)+1$ is odd.

Case 2: (If either $n$ or $m$ are odd) say $n$ is odd and $m$ is even then $n=2a+1$ and $m=2b$ for some integers $a,b$.
The product $nm=(2a+1)(2b)$
$nm=4ab+2b$
$nm=2(2ab+b)$
Since $2ab+b$ is an integer, $nm=2(2ab+b)$ is even.

Since the two cases are not both true, the contrapositive is proved false and the initial implication was false. [b][i]Is this correct?[/i][/b]

that would be a valid proof.

however your first case was unneccessary, because it did not disprove anything. only case 2 provided a counterexample.

Wow that was fast, thank you tjhance for replying to my question!  I appreciate your help!

Yeah. If you are asked to prove something, and you found a counterexample, you can only write the counterexample to disprove it. :wink:

you can simply disprove the statement with this: $2=1\cdot2$. done :D


1) 1,3-Dimethyl-2-nitrobenzene + SnCl2/HCl/then HO- -----> A

2) A + ClCH2COCl -----> B

3) B + diethylamine -----> C

4) C + HCl -----> D

A- 2,6-dimethyl aniline
B- N-(2,6-dimethyl phenyl) 2-chloro ethanamide
C- N-(2,6-dimethyl phenyl) 2-(N,N-diethyl amino) ethanamide
D- Probably hydrolysis 2-(N,N-diethyl amino) ethanoic acid and A.

i think the last reaction is not hydrolysis but something similar to fries rearrangement. the acyl group migrates to ortho or para pos and you get aniline


modular arithmetic
number theory proposed
number theory

find all positive integers n such that $ \frac{2*10^{n}\minus{}4}{19}$ is a integer.

$ n \equal{} 18k \plus{} 17$ ,$ k$ is nonnegative integer

Exactly $ n\equal{}18k\minus{}1,k\in N$.

Isn't there a short way to solve it? without calculator

thanks

I calculate without calculator $ 10\equal{}2^{\minus{}1}\mod 19$ and 19 is prime.

I have a question. Using Rust's idea the problem is equivalen to $ 10^{n\plus{}1}\equal{}1\mod 19$, so $ n\equal{}17$ works, wherefrom $ n\equal{}17\plus{}18k$ comes?

[quote="asdrojas"][quote="boxedexe"]By Fermat's Little Theorem, $ 10^{18}\equiv 1\pmod{19}$, from which the answer follows.[/quote]

Yes i knwo that, but if $ 10^{18}\equiv 1\pmod{19}$, how can I know that $ 10^{9}\neq 1\pmod{19}$?[/quote]
I chek that $ 2^3\equal{}8,2^6\equal{}7\mod 19 \not \equal{}1\mod 19, 2^9\equal{}8*7\equal{}\minus{}1\mod 19$, therefore $ T\equal{}18$ is not divisor 18.


[quote="hanschinaman"]Suppose $a,b,c$ are all positive real numbers, show that
$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\leq \frac{3\sqrt{2}}{2}$[/quote]
see here: http://www.mathlinks.ro/viewtopic.php?t=148797


The area of some regular polygon is 1.5 times as large as the area of another polygon with sides 2 times as large as the side of the polygon. What is a possible pair of such polygons?

[quote="orangefronted"]The area of some regular polygon is 1.5 times as large as the area of another [regular] polygon with sides 2 times as large as the side of the polygon. What is a possible pair of such polygons?[/quote]

You must be missing the above word shown in brackets.

[hide]One pair that I could come up with is a regular hexagon with side length one half of an equilateral/equiangular triangle.

If you assign the length of the hexagon's sides the letter $ x$ then the triangle's side length would be $ 2x$.
The area of the hexagon would be $ \dfrac{3x^2\sqrt {3}}{2}$
The area of the triangle would be $ x^2\sqrt {3}$

The area of the hexagon is 1.5 times the area of the triangle.

If more explanation is needed, please reply.[/hide]

The area of some regular polygon is 1.5 times as large as the area of another [b]regular[/b] polygon with sides 2 times as large as the side of the polygon. What is a possible pair of such polygons?

@geotrig, is the hexagon/equilateral triangle pair the only possible pair?


What is the result when the following set's largest number is divided by its smallest number?
\[ \{0.47,\frac{1}{6},1\frac{1}{2},0.\bar{3}\}\]

First, write the terms that are fractions as decimals, to determine the largest and the smallest.

$ \{ 0.47, 0.1\overline{6}, 1.5, 0.\overline{3} \}$

Clearly, the largest is $ 1.5$ or ($ 3/2$) and the smallest is $ 0.1\overline{6}$ or ($ 1/6$), so the answer is $ \frac{3/2}{1/6} \equal{} \boxed{9}$


Εδώ αναλύστε τα θέματα και συζητείστε για τον ΕΥΚΛΕΙΔΗ  2007

Oups! Μόλις άνοιξα άλλο topic  :|

δεν πειραζει Στελλιο.πως τα ειδες ? για τη Β λυκειου?

Δεν τα πήγα όπως περίμενα.. Ίσως έφταιξε το άγχος, ίσως..

Πρέπει να έλυσα 2-3 ασκήσεις.. όχι όμως τρελά πράγματα..

Καλά αποτελέσματα όπως και 'χει στα υπόλοιπα παιδιά του forum  :lol:

νομιζω οτι εισαι πολυ καλα .!  :)

[b][color=red]1ο ΘΕΜΑ ( Β΄ΛΥΚΕΙΟΥ - Θ.2)[/color][/b]

 Θεωρούμε τόξο $AB = 90^{0}$ ενός κύκλου και προεκτείνουμε τη χορδή $AB$ κατά  τμήμα $BC = AB$.Έστω $CD$ το εφαπτόμενο τμήμα προς το τόξο και $K$ η προβολή του $A$ προς την ευθεία  $BD$.Να αποδείξετε ότι
$KB = 2KA$.

[i]Κατά τη διάρκεια  της επιτήρησης βρήκαμε με το συνάδελφο Θανάση Μπεληγιάννη την παρακάτω λύση.Δεν την έχουμε χτενίσει για τη μαζέψουμε ίσως λίγο , αλλά σας δίνουμε την πρώτη προεσέγγιση.Μας έφαγε λίγο χρόνο το γεγονός ότι $KA = KD$ και αυτό μας οδηγούσε στην $DB= AK$ , που την ψάχναμε αλλά δεν έβγαινε αμέσως.Πιθανόν τώρα που έχουμε τόσα στοιχεία , με λίγο ψάξιμο ακόμα να βρούμε και άλλο δρόμο.Πολές φορές η πιο ψύχραιμη μελέτη κάνει την πρώτη προσπάθεια να μοιάζει αστεία .Τέλος πάντων , ορίστε η 'εν θερμώ ' προσπάθεια.:[/i]
 [hide][b]Υπόδειξη[/b]
Αν $E$ είναι το αντιδιαμετρικό του $A$  και $O$ το κέντρο του κύκλου, τότε το $CE$ εφάπτεται στον κύκλο(αφού $BO // CE$), τα τρίγωνα $CEO, CDO$ είναι ίσα , από το εγγράψιμο τετράπλευρο $OAKB$ βρίσκουμε με λίγο παιγνιδάκι ότι τελικά η $OK$ είναι διχοτόμος της γωνίας $DOA$ και έτσι τελικά η γωνία $COD$ είναι ορθή.Άρα
   $\angle ECO = 90-\angle EOC = \angle KOA = \angle ABK$. Αυτό σημαίνει ότι τα τρίγωνα $OEC$ και $KAB$ είναι όμοια. Αλλά $EC = 2OB = 2OE$ , οπότε και  $\frac{KB}{KA}= \frac{EC}{EO}= 2$ , ΤΕΛΟΣ[/hide]

Και εγώ 3 θέματα έχω λύσει....

Μου άρεσε πάρα πολύ το [b]3ο θέμα[/b] , που ήταν και το
δυσκολότερο κατα τη γνώμη μου .
Το έλυσα με αντίστροφη συνάρτηση .

An mporeite valte ta themata tou Likeiou na ta sxoliasoun oloi kai na ta doume osoi den eimastan ston eukleidi logw ilikias.

Alexandros

Το γράψιμο Αλέξανδρε , είναι το αδύνατο μου σημείο .
Τα έβαλα στο συνημμένο .

Pragmati, i lisi sou me antistrofi sinartisi einai poli omorfi.

Euxaristw gia to post me tis ekfwniseis

Alexandros

Παιδιά γεια σας .Έλυσα 3 θέματα . Δεν έλυσα την ανισότητα  :mad:  :mad:  . Μπερδεύτηκα μάλλον επειδή ήθελε ακεραίους .Ο Νίκος ο Ράπανος βρήκε την εξής πολύ ωραία λύση.

Από AM-GM gia a fores to a^3 kai b fores to b^3 einai

$\frac{a^{3}+a^{3}+....+a^{3}+b^{3}+...+b^{3}}{a+b}\geq\sqrt[a+b]{a^{3a}b^{3b}}$

telos .

Ελπίζω να περάσω  :(  με τρία θέματα

Sigoura tha peraseis Silouane me tria themata!!

H lisi tou Nikou einai akrivws idia me tin diki mou pou esteila sto proigoumeno post.

Oson afora to prwto thema me tous logarithmous tis 3hs Likeiou

$2(1-y)=\ldots$ (allagi vasis, apo 150 sto 50) $= \frac{2}{\log_{50}{3}+1}$
kai $1-x-y=\frac{1-\log_{50}{2}}{\log_{50}{3}+1}$

Ara

$\frac{1-x-y}{2(1-y)}=\frac{1-\log_{50}{2}}{2}=\log_{50}{5}$

kai telika $A=50^{\log_{50}{5}}=5$

Alexandros

Καλά δε νομίζω με 3 θέματα να μην περάσει κανείς δεδομένου ότι ήταν σχετικά δύσκολα τα θέματα...Ετσι δεν είναι;

Edit

Με προλάβατε... :D

Πιστεύετε ότι με 2-2,5 θέματα παιρνάει κάποιος ?Που νομίζετε θα είναι η βάση?

Δεν ξέρω πόσο θα είναι η βάση.. στη 3η γυμνασίου πχ είχα λύσει 3.5 και δεν πέρασα.. Ελπίζω να πέσουν για να περάσουμε  :P

Πάντως τα θέματα ήταν αρκετά δύσκολα..

ηταν ενας καλος ευκλειδης ,απο αποψη δυσκολιας .με δυο + μισο θεματα περνας σιγουρα.ακομα και με δυο θεματα .(ακομη και λιγοτερο απο 2 θεματα!!!!
να ξερεις Σιλουαν και υπολοιποι οτι περνας και με βαθμο ''κατω απο τη βαση'' αρκει να εχεις δειξει στο γραπτο σου οχι ξερεις  μαθηματικα.ειμαι σιγουρος οτι περνας μην αγχονεσαι αδικα λοιπον

[u]Β  Λυκείου[/u]
Πολύ ωραία τα θέματα...Και οι δύο γεωμετρίες πολύ ωραίες και η ανισότητα!
Η ανισότητα όμως μήπως ισχύει για όλους τους θετικούς πραγματικούς???Αν θέλετε μπορείτε να το κοιτάξετε.Θέλω επίσης να βάλω τη λύση μου στο δεύτερο θέμα που είναι ελαφρώς διαφοροποιημένη από του κύριου Στεργίου.

[u]Γ Λυκείου[/u]
Τα δύο πρώτα εύκολα.Το 4ο πανέμορφο θέμα.Χρησιμοποιώ οτι Β,Ι,Ο,Γ βρίσκονται πάνω σε κύκλο κέντρου Μ (μέσο του τόξου ΒΓ)
.... ΙΑ+ΙΒ+ΙΓ<=3R....
Το 3ο θέμα λιγουλάκι δύσκολο.Το έλυσα μαζί με έναν καθηγητή,αλλά μας έφαγε αρκετή ώρα...

Θα τα πούμε πιο αναλυτικά στη συνέχεια!

α) Είχα την εντύπωση ότι αυτή την ώρα θα έχουν κρεμάσει τις λύσεις στην ΕΜΕ.Αλλά φαίνεται ότι μας ....κρέμασαν.Βέβαια τις λύσεις τις είχα στο παράρτημα και δεν τις πήρα , ούτε και τις κοίταξα, αφού θα τις έβλεπα εδώ(στο  site της  ΕΜΕ).
β) Μη σκέφταστε τον Ευκλείδη. Πέρασε ! Ετοιμαστείτε για τον Αρχιμήδη.Όσο και να είναι οι βάσεις , εσεις - όλοι -  θα είστε εκεί !!!

                 [u]Μπάμπης[/u]

let's hope  :(  .
 .......

Εγώ πάντως δεν είμαι σίγουρος για να είμαι ειλικρινής .

Silouan,

Den iparxei periptwsi na meineis ap'eksw ap'ton Arximidi. Tha eisai stin Athina! Eimai sigouros! Kai me 2 themata teleia tha exeis perasei!

Min agxwnesai loipon kai sinexise tin kali douleia kai to diavasma! (Eksallou erxetai kai 2os diagwnismos telos Martiou opws eipame  :)  Den teleiwnoun edw fetos oi diagwnismoi! )

Alexandros

[quote="cretanman"]Silouan,
...(Eksallou erxetai kai 2os diagwnismos telos Martiou opws eipame  :)  Den teleiwnoun edw fetos oi diagwnismoi! )
...[/quote]
τι διαγωνισμος(plz μ η με βαλετε να ψαξω ολα τα posts ειμαι πολυ καινουριος στο forum και... :blush: ...βαριεμαι :blush: ...να τα διαβαζω ολα)
αν μπορουσατε να με καθοδηγησετε λιγο...θα σας ειμουν ευγνωμων...

darksaga

Για το 3ο θέμα της 3ης Λυκείου..

Είναι προφανής η λύση $x=1$..

Όμως η συνάρτηση είναι 1-1, άρα είναι και μοναδική.

Έχω κάπου λάθος;

[quote="kostas"]darksaga κοιτα εδω
http://www.mathlinks.ro/Forum/viewtopic.php?t=124494[/quote]tnx  alot :)

mostel

οι δύο συναρτήσεις (1ου και 2ου μέλους) είναι αντίστροφες
και επειδή είναι και γν.αύξουσες οι λύσεις της βρίσκονται στη διχοτόμο ψ = χ .
Αν λύσεις την f(x) = x , βρίσκεις την χ = 1 και χ = (ρίζα3 μείον 1)/2  και
χ = (- ρίζα3 μείον 1)/2 .

Σωστά Νίκο   :|

Πολύ καλή σκέψη ! Μπράβο!  :lol:

niko to x=-(riza3 -1)/2 aporriptetai afou prepei x >=1-3

Nick

το σκέφτηκα , αλλά τελικά προτίμησα να την κρατήσω .
Ειναι περιττής τάξης το ριζικό .
Δεν το είδα τυπικά το θέμα .

niko exeis dikio einai ena typiko thema orismou gi auto to anefera..
kai gia na sou po tin alitheia to thymamai auto apo tote pou enas mathimatikos sto gymnasio mou ekopse mia monada gia ena tetoio <<lathos>>...tespa emena me vriskeis symfono!


Let $ a,\ b,\ c$ be psitive numbers with $ abc\equal{}1$ for some positive constant numbers $ p,\ q$.

Prove that $ \sqrt{pa^{2}\plus{}q}\plus{}\sqrt{pb^{2}\plus{}q}\plus{}\sqrt{pc^{2}\plus{}q}\leq\sqrt{p\plus{}q}(a\plus{}b\plus{}c)$.

very nice, kunny  :wink: I will try it  :wink: we have a result of Jichen:
Let $ a,b,c\ge 0,abc \equal{} 1$ and $ 0\le k\le k_{0}$  Find the best $ k_{0}$ constant with such that the following inequality is correct:
\[ \sum\sqrt{ka^{2}\plus{}1}\ge\sqrt{k\plus{}1}({a\plus{}b\plus{}c})\]
I think it is not very hard  :wink: I will try ...

If $ x,y,z$ be positive real numbers such that $ xyz \equal{} 1$, then

$ \sqrt {23x^{2} \plus{} 2} \plus{} \sqrt {23y^{2} \plus{} 2} \plus{} \sqrt {23z^{2} \plus{} 2}\leq5(x \plus{} y \plus{} z)$,

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1$.

[hide="Links."]$ p \equal{} 8,q \equal{} 1$: http://www.mathlinks.ro/Forum/viewtopic.php?t=3301

$ p \equal{} 9,q \equal{} 4$: http://www.mathlinks.ro/Forum/viewtopic.php?t=165390

$ p \equal{} 3,q \equal{} 4$: http://www.mathlinks.ro/Forum/viewtopic.php?t=157600[/hide]
By the way, if $ x,y,z$ be nonnegative real numbers such that $ yz \plus{} zx \plus{} xy \equal{} 3$, then

$ \sqrt {1 \plus{} kx^{2}} \plus{} \sqrt {1 \plus{} ky^{2}} \plus{} \sqrt {1 \plus{} kz^{2}}\leq(x \plus{} y \plus{} z)\sqrt {1 \plus{} k}$

holds if and only if $ 0\leq k \leq \frac {11}{16}$; with equality if $ x \equal{} 0,y \equal{} z \equal{} \sqrt {3},k \equal{} \frac {11}{16}$.


linear algebra
matrix
inequalities
advanced fields
advanced fields unsolved

I know that any matrix A on $C^{nxn}$ can be written as $UTU^{*}$ where T is upper triangular and U is unitary (Schur factorization) but how
can I prove that A is normal iff it can be written as $UTU^{*}$ where L is diagonal.
If I make the comparision explicitly I get an expression I don't know how to get out of...

thanx.

Let
$B=U^{-1}AU=\left(
\begin{array}{cccc}
b_{11} & b_{12} & \cdots & b_{1n}\\
0 & b_{22} & \cdots & b_{2n}\\
\cdots & \cdots & \cdots & \cdots\\
0 & 0 & \cdots & b_{nn}
\end{array}
\right)
$
is an upper-triangle matrix.
Since $A^HA= AA^H$, it holds that $B^HB=BB^H$,
which is
$\left(
\begin{array}{cccc}
\overline{b_{11}} & 0 & \cdots & 0\\
\overline{b_{12}} & \overline{b_{22}} & \cdots & 0\\
\cdots & \cdots & \cdots & \cdots\\
\overline{b_{1n}} & \overline{b_{2n}} & \cdots & \overline{b_{nn}}
\end{array}
\right)\left(
\begin{array}{cccc}
b_{11} & b_{12} & \cdots & b_{1n}\\
0 & b_{22} & \cdots & b_{2n}\\
\cdots & \cdots & \cdots & \cdots\\
0 & 0 & \cdots & b_{nn}
\end{array}
\right)=\left(
\begin{array}{cccc}
b_{11} & b_{12} & \cdots & b_{1n}\\
0 & b_{22} & \cdots & b_{2n}\\
\cdots & \cdots & \cdots & \cdots\\
0 & 0 & \cdots & b_{nn}
\end{array}
\right)\left(
\begin{array}{cccc}
\overline{b_{11}} & 0 & \cdots & 0\\
\overline{b_{12}} & \overline{b_{22}} & \cdots & 0\\
\cdots & \cdots & \cdots & \cdots\\
\overline{b_{1n}} & \overline{b_{2n}} & \cdots & \overline{b_{nn}}
\end{array}
\right)$
Comparing the diagonal elements, we obtain the conclusion..

thanx a lot!!
But how do you know that $B^HB=BB^H$...

Since $U$ is unitary, $U^{-1}=U^H$ and $U^HU=UU^H=E$
$B^H B = (U^HAU)^H(U^HAU)=U^HA^H(U^H)^H U^HAU=U^HA^HU\ U^HAU=U^HA^HAU$
$BB^H = (U^HAU)(U^HAU)^H = U^HAUU^HA^H(U^H)^H = U^HAA^HU$
$A$ is normal, then $AA^H=A^HA$...


Let $p$ be prime such that $p\equiv\pm 3\pmod{5}$.

Assume that there exists an index $m$ such that $a_m b_m\equiv 0\pmod{p}$ and let it be the smallest such index. It is easy to check that $m\geq 2$.

Then for $n=0,1,\dots,m-1$, we have $a_n b_n\not \equiv 0\pmod{p}$ and thus they satisfy the congruences:
\[\begin{cases} a_n a_{n+1} \equiv 1 + a_n b_n \pmod{p} \\ b_n b_{n+1} \equiv 1 + a_n b_n \pmod{p} \end{cases}\]
Multiplying them, we get
\[c_n c_{n+1} \equiv (1+c_n)^2\pmod{p}\]
where $c_n = a_n b_n$.

For $n=m-1$, we have $c_{n+1}=c_m\equiv 0\pmod{p}$ and thus $c_{m-1}\equiv -1\pmod{p}.$

In turn, for $n=m-2$, we get:
\[- c_{m-2} \equiv (1+c_{m-2})^2\pmod{p}\]
that is equivalent to
\[(2c_{m-2} + 3)^2 \equiv 5 \pmod{p}.\]
However, this congruence is impossible since $5$ in not a square modulo $p$ (see [url=http://mathworld.wolfram.com/LegendreSymbol.html]formula (6)[/url]). The contradiction proves that there exists no such $m$ that $a_m b_m\equiv 0\pmod{p}$.


calculus derivative function inequalities integration limit induction

Let $f : \left[ 0,\infty \right) \to \mathbb R$ be a differentiable function such that $f(0)=0$ and such that there is a constant $M > 0$ for which $\left| f^\prime (x) \right| \leqslant M \left| f(x) \right|$ holds for all $x \in \left[ 0,\infty \right)$.

Prove that $f$ is identically null.

Let $g(x)=e^{-Mx}f(x).$ Then $g$ is also differentiable, and

$g'(x)=e^{-Mx}(f'(x)-Mf(x)).$

But since $|f'(x)|\le M|f(x)|,$ it follows that $g'(x)$ has the opposite sign to $f(x),$ hence to $g(x).$

Thus $g(x)g'(x)\le 0.$

Given that we also have that $g(0)=0,$ we can integrate this inequality to get that $[g(x)]^2\le 0.$

Hence $g(x)=0$ for all $x\ge 0$ and it follows that $f(x)=0.$

That's really nice. I would have used a more unattractive argument saying that $f$ must be 0 in a ball around 0, and then let $c$ be the infimum over all $x$ for which $f(x) \ne 0$, and then show that such a $c$ can't exist (because $f$ is 0 on $[0, c)$, and then use the derivative condition).

That was the argument I started to use. I hadn't yet decided whether or not it would work when I switched to the other argument.

Hmm... That exercise appears in my 11th grade book. There it is said that by proving the assertion we will get a proof of the fact that $f(x) = k f^\prime (x)$ has an "unique" solution. Kent's proof is about the same thing with the direct proof of the latter fact.

Anyway, I guess we can replace the "integrating" argument with one more 11th-gradish, so I guess it's OK. Thanks.

I found a solution that doesn`t use integration and it proves the generalisation of this $\left| f^\prime (x) \right| \leqslant M \left| f^{(n)}(x) \right|$ where $f^{(n)}=f\circ\ f \circ ... \circ f$

[quote="Xevarion"]$f$ must be 0 in a ball around 0[/quote]

I can't prove this...

BTW, spix, you can post your ideas.

The main idea is to prove that $f$ is $0$ on every interval of the form $[0,a],\forall a\in R$.
Let`s assume that there is $b$ such that $f$ is not $0$ on $[0,b]$ then there exist $A=\sup_{c\in[0,b]}f(c)$.
By Lagrange we have that $f(x)=xf'(c)\longrightarrow |f(x)|=x|f'(c)|\leq bM|f^{(n)}(c)|\leq bMA$ so $bMA\geq A$ which means that $bM \geq 1$.
Consider $\delta = (0,\alpha_1,\alpha_2,...,\alpha_{m-1},b)$ a divison of $[0,b]$ such that $M||\delta||<1$(*).
If $f$ is not identically null on $[0,\alpha_1]$ then taking $A_1$ the maximum value of $f$ on this interval and by the steps above $\alpha_1M\geq 1$ which contradicts (*).
After a finite number of steps the conclusion follows.

Here's a slight variation of spix's argument:

Let $\displaystyle x_0 \in \left( 0,\frac1{2M} \right]$. Then, by Lagrange's Theorem, we have
\[ \displaystyle \frac{f \left( x_0 \right) - f(0)}{x_0 - 0} = f^\prime \left( x_1 \right) \, \, \Longleftrightarrow \, \, f \left( x_0 \right) = x_0 f^\prime \left( x_1 \right) , \, \textrm{where} \, x_1 \in \left( 0,x_0 \right) ,  \]
so \[ \displaystyle \left| f \left( x_0 \right) \right| = x_0 \left| f^\prime \left( x_1 \right) \right| \leq \frac1{2M} \left| f^\prime \left( x_1 \right) \right| \leq \frac1{2M} \cdot M \left| f \left( x_1 \right) \right| = \frac12 \left| f \left( x_1 \right) \right| .  \]

We construct inductively a sequence $\displaystyle \left\{ x_n \right\}_{n \geq 0}$ such that $\displaystyle f \left( x_n \right) = x_n f^\prime \left( x_{n+1} \right)$, where $\displaystyle x_{n+1} \in \left( 0,x_n \right)$, and $\displaystyle \left| f \left( x_n \right) \right| \leq \frac12 \left| f \left( x_{n+1} \right) \right|$. Thus,
\[ \displaystyle \left| f \left( x_0 \right) \right| \leq \frac{1}{2^n} \left| f \left( x_n \right) \right| .  \]

$\displaystyle f$ is a differentiable function on $\displaystyle \left[ 0,x_0 \right]$, so it is also continuous, hence bounded, on that interval. We know that $\displaystyle x_n \in \left( 0,x_0 \right]$, for all $\displaystyle n \geq 0$, so $\displaystyle \left\{ \left| f \left( x_n \right) \right| \right\}_{n \geq 0}$ is a bounded sequence. Consequently,
\[ \displaystyle \lim_{n \to \infty} \frac{1}{2^n} \left| f \left( x_n \right) \right| = 0 .  \]
So, by taking the limit of $\displaystyle \left| f \left( x_0 \right) \right| \leq \frac{1}{2^n} \left| f \left( x_n \right) \right|$, we get that $\displaystyle \left| f \left( x_0 \right) \right| \leq 0$, so $\displaystyle f \left( x_0 \right) = 0$.

Since $\displaystyle x_0$ was an arbitrary number in $\displaystyle \left( 0,\frac1{2M} \right]$, we have $\displaystyle f \left( x \right) = 0, \, \forall x \in \left[ 0,\frac{1}{2M} \right]$.

Next, we shall prove by induction that $\displaystyle f(x) = 0, \, \forall x \in \left[ \frac{m}{2M}, \frac{m+1}{2M} \right]$, for all $\displaystyle m \geq 0$.

The induction base case $\displaystyle m=0$ was done above, so we'll turn our attention towards the induction step, i.e. passing from $\displaystyle m-1$ to $\displaystyle m$.

Consider the function $\displaystyle g : \left[ 0, \infty \right) \to \mathbb R$ defined by $\displaystyle g \left( x \right) = f \left( x + \frac{m}{2M} \right)$, for all $\displaystyle x \geq 0$. By the induction hypothesis, $\displaystyle g(0) = 0$. We also know that $\displaystyle \left| g^\prime (x) \right| \leq M \left| g(x) \right|$, for all $\displaystyle x \in \left[ 0,\infty \right)$. Using the same argument that we used to prove the base case, we see that $\displaystyle g \left( x \right) = 0$, $\displaystyle \forall x \in \left[ 0,\frac1{2M} \right]$. Thus, $\displaystyle f \left( x \right) = 0$, $\displaystyle \forall x \in \left[ \frac{m}{2M},\frac{m+1}{2M} \right]$.

Since $\displaystyle \bigcup_{m \geq 0} \left[ \frac{m}{2M}, \frac{m+1}{2M} \right] = \left[ 0, \infty \right)$, $\displaystyle f \equiv 0$.


function
algebra
polynomial
geometry
email
Support
number theory

I am not yet ACTIVE enough to volunteer to help, but I love number theory and your lovely collection of problems is bringing me much joy. Thank you. :D

Thank you, Ilthigore :D !  But, I'm sad. No volunteers...   :(

[quote="Peter VDD"]Do you need any special qualifications?

I am perhaps interested in doing it. How much time do you think it would cost me to keep up with it (if I want to do it good)?

I will only be able to start September 18 though, as I'm in exams until then. I hope that's no problem. But I thought I should reply now.[/quote]

Peter VDD, I deeply appreciate your help! Okay, there is no problem!
I'll contact you by email after September 18!

Dear math lover, it is not only important to use ML/AoPS to improve one's problem solving skills but also to offer something back to our community and friend Hojoo Lee. Writing up and collecting interesting problems will give you insight in classifying problems into topics and you may be easier for you to recognise problems in competitions and you should note that a community can be active only if everybody tries to contribute something!  :)

Thank you for your support, Orlando! As said (and confirmed in a pm with Hojoo), I will take over the work on PEN, and I will probably create an interactive way so people can help me with adding questions easily. No single person on this planet can beat an entire community each contributing a little bit. :)

[quote="ideahitme"]I believe that mathlinkers know PEN, which is one of my problem collection booklets.
There are two main reasons I developed this. One is that Number Theory is one of important  bridges of math olympiad and college mathematics. Furthermore, I wanted to offer a reference for proposers of Math Olympiad or for the IMO PSC. We have a duty to prevent problem recycling/re-discovering. So, I need to 'classify' the problems by topics. My booklet has many chapters:

3. Divisibility Theory I
4. Divisibility Theory II
5. Arithmetic in Zn
6. Primes and Composite Numbers
7. Rational and Irrational Numbers
8. Diophantine Equations I
9. Diophantine Equations II
10. Functions in Number Theory
11. Polynomials
12. Sequences of Integers
13. Combinatorial Number Theory
14. Additive Number Theory
15. The Geometry of Numbers
16. Miscellaneous Problems

Unfortunately, as you know, the project discontinued around 2004. However, I uploaded the TeX sourcee on my webpage at
http://myhome.naver.com/hojoolee/orange.html

I strongly feel that someone continue to develop my PEN more and more by collecting interesting NT problems in recent Olympiads.

So, if there are any ACTIVE (and young) volunteers, please email me at ultrametric@gmail.com ![/quote]

Hey Hojoo,don't you think that it would be nice if we try to collect also some nice solutions for the problems from this collection?

I will be glad to help you with that work,and I think it would be a very unique set of problems and solutions if we manage to write it :) .

And one more thing,I feel that we can do the same thing for other fileds, besides Number Theory,either:for example for Geometry :blush: .

[quote="Peter VDD"]Tiks, I am working on an online system to collect the solutions for the PEN problems. Patience now, I am very close to a release, hopefully the beta version will be there in a week or two.

I am very glad to see interest in the solutions, I hope we get a nice collection of solutions too. :)[/quote]

Please tell me if I can help,somehow  :) .
Here is my e-mail

tigran_sloyan € mail.ru

[moderator: just for case you don't want it to be spammed, I changed the @ into €]

 :wink:

[quote="Hawk Tiger"]I will be glad to help you with that work.And I have enough time.Please contact me by hawktiger89 % yahoo.com
I would like to write some solution or collect some new NT problem.
But, My English is not so good. :blush: But I think it is good enough to write the solution or the problem correctly :)[/quote]

The second author Peter is workng on it. We will let you know when it is possible that you can help us in various ways.

Thank you,
Hojoo Lee

The book is ready!! All is now on the website http://www.problem-solving.be/pen/ !!  :lol:

You can all help by solving the problems there (on the website) and by contributing new nice problems (on the website or by email to [img]http://lh5.google.com/image/ultrametric/RdG25j2x0AI/AAAAAAAABHM/TsiOhJJTjaw/petervandendriessche.jpg[/img]).


Can you solve this problem, I can't figure it out, I think it might be 28.

A farmers field is in the shape of a regular hexagon ABCDEF. The distance drom point A to point B is 420 feet. A fence post is laced at each vertex of the hexagon, and each side has 16 evenly spaced fence posts (counting the posts at the vertices). What is the distance from the center of one fence post to the center of an adjacent fence post on a side of the hexagonal field?

Thanks in advance!!

Well, the 16 posts divide the side into 15 congruent segments. If the 15 congruent segment add up to 420, the distance between each post is [b]420÷15=28[/b]


let $ABC$ be acute-angled  triangle prove the inequality:
$\frac{Cos(b)Cos(c)}{Cos(\frac{b-c}{2})}+\frac{Cos(a)Cos(b)}{Cos(\frac{a-b}{2})}+\frac{Cos(a)(Cos(c)}{Cos(\frac{a-c}{2})}\le \frac{3}{4}$

Is there any one interesting on this problem? :?

Ok,
Solution by Hojoo Lee:
Note that $A+B+C=\pi$.Using the $AM-GM$ inequality and $jensen$'s inequality ,we have :
$2\sum{\frac{Cos(B)Cos(C)}{Cos(\frac{B-C}{2})}}=
4\sum{\frac{Cos(B)Cos(C)}{Cos(B)+Cos(C)}Sin(\frac{A}{2})} \le
2\sum{\sqrt{Cos(B)Cos(C)}Sin(\frac{A}{2})} \le
 \sum{(Cos(B)Cos(C)+Sin^2\frac{A}{2})=
 \sum{Cos(B}Cos(C)}+\sum{\frac{1-Cos(A)}{2}}\le
 \frac{1}{3}\sum{Cos^2(A)}+\sum{\frac{1-Cos(A)}{2}}\le
Cos(\frac{A+B+C}{3})(\sum{Cos(A)})+\frac{1-Cos(A)}{2}=
\frac{1}{2}\sum{Cos(A)}+\frac{1-Cos(A)}{2}=\frac{3}{2}. $
***$\sum$ denote cyclic sum over A,B,C.
but i'am wating for another solutions.

Prove the inequality below:
3>2power(2power1/2) in two different ways! :ninja:

《Mathematics bulletin》(china beijing) NO.2(2012) Math problem 2050
 is the same question.


Seven different odd primes are given. Is it possible that for any two of them, the difference of their eight powers to be divisible by all the remaining ones ?

[i]Proposed by F. Petrov, K. Sukhov[/i]

Let p(1) < p(2) < p(3) < p(4) < p(5) < p(6) < p(7) be those primes.

p(2)^8-p(1)^8=[p(2)-p(1)][p(2)+p(1)][p(2)^2+p(1)^2][p(2)^4+p(2)^4]

p(i) > p(2) > p(2)-p(1) for every i, 2<i<8. Analogously, since all p(i) are odd and so are p(1) and p(2), all the p(i) dividing [p(1)^k+p(2)^k] must divide [p(1)^k+p(2)^k]/2 < p(2)^k < p(i1)p(i2)...p(ik), which means that at most k-1 of them divide [p(1)^k+p(2)^k] for each k. As a conclusion, p(i) divides p(2)^8-p(1)^8 for at most (1-1)+(2-1)+(4-1)=4<5 different values of i, and such a set cannot exist.

nice solution, Mimoide :) , but next time please use Latex.(just put dollar signs around the mathematical formulas and that would make your solutions easier to read)

It is not necessary that the primes be odd. Suppose that $2 = p_1 < p_2 < \cdots < p_7$ as others have done. Then, as others have done
\[ p_3 p_4 \cdots p_7 \mid (p_2^4 + 16)(p_2^2 + 4)(p_2 + 2) \]
From here we split into two cases.

[b] Case 1:[/b] Suppose that $p_3 \neq p_2 + 2$. Then
\[ p_3p_4p_5p_6p_7 \mid (p_2^4+16)(p_2^2+4) .\]
However, $p_2^2 + 4$ can only have one divisors among $p_3, \dots, p_7$ since
\[p_ip_j > (p_2 + 2)^2 > p_2^2 + 4. \]
Likewise $p_2^4 + 16$ can only have three prime divisors among the $p_i$, so this case is finished.

[b] Case 2:[/b] SUppose that $p_3 = p_2 + 2$. In this case we obtain that
\[ p_4 p_5 p_6 p_7 \mid (p_2^4+16)(p_2^2 + 4) . \]
Additionally, since $p_4 \neq p_3 + 2$ because there are no triplet primes (the only exception is if $(p_2,p_3,p_4)=(3,5,7)$ but $7 \nmid 5^8 - 3^8$), we also obtain $p_4 p_5 p_6 p_7 \mid (p_3^4 + 16)(p_3^2 + 4)$. Then
\begin{align*}
p_4 p_5 p_6 p_7 &\mid (p_2^4+16)(p_2^2 + 4) - (p_3^4 + 16)(p_3^2 + 4) \\
&= p_2^6 - p_3^6 + 4p_2^4 - 4p_3^4 + 16p_2^2-16p_3^2 \\
&= (p_2^2 - p_3^2)(16 + 4p_2^2 + 4p_3^2 + p_2^2p_3^2 + p_2^4 + p_3^4)
\end{align*}
None of the primes can divide $p_2^2 - p_3^2$ for size reasons, so they must divide the second factor. Taken modulo 3, we can see that that term is divisible by 3, so in fact we ahve
\[ p_4 p_5 p_6 p_7 \mid \frac{16 + 4p_2^2 + 4p_3^2 + p_2^2p_3^2 + p_2^4 + p_3^4}{3} < \frac{16 + 8p_3^2}{3} + p_3^4 < (2+p_3)^4\]
which is a contradiction.


Here's a simple sequence:-

$24,120,360,840.....$

Find the next term or even better generalise.

Well according to how I'm doing it,that's not right.

His next term or his generalization?
Because his next term is right.

Actually I was wrong about the generalization.
It should have been $a_{n+1}$ not $a_n$

That's what biff put for his next term and shady says its wrong...

o yeah :oops:  :oops:

how'd you get that?

[hide]The sequence seems to go $\times5$, $\times3$, $\times\frac{7}{3}$. $\frac{7}{3}=\frac{5+2}{3}$ so $\frac{3+2}{\frac{7}{3}}=\frac{15}{7}$ so you get $1800$ (I think). I'm probably wrong about the sequence.[/hide]


AMC
AIME
Putnam
USA(J)MO
USAMO
HMMT
AIME I

Heh, sorry for the title, but I figured it'd get some attention. Anyway, yeah, so I just realized that I am nowhere near as good as I would like to be as a problem solver. I was able to at least make AIME, but I fell flat on my arse and died there.

So could anyone share what they did to become "good" at problem solving? What books/problem sources did you use? How did you eventually learn to tackle problems that you couldn't before?

I have graduated high school, so I will not be participating in the AMC series again. And I doubt my skills will be enough for the putnam this year. I simply want to build my problem solving skills to help enrich myself. I enjoy doing problems and found the AMC/AIME to be a fun experience; ever since I encountered those types of problems, I have learned to love mathematics (school math is an insult). But I feel I can appreciate those types of problems (i.e. nonroutine) more once I can actually get the majority of them (I got slightly over 100 on AMC 12 and a 4 on AIME I 2009 -- I suck, basically).

I bought all of the CDs with all the AIMEs, AMCs, and USAMOs, and want to dedicate a significant portion of my waking hours to doing nonroutine mathematics. What resources -- if any -- should I use to supplement these materials? How do I go about USING what I have and whatever else I buy?

~Arachnotron

Find the problems whose difficulty is just barely outside your reach. Sit down and try your hardest to solve them. Have patience when you do this. At the olympiad level you should always give yourself at least an hour; I suppose for AIME problems around 15-20 minutes. Only when you have really knocked your brain against the problem for an extended period of time without success should you look at the solution. You'll learn a lot from this whether you solve the problem or not.

The answer to how to get better at problem solving is almost always the same: do problems.

Many people on this site will back up MellowMelon.
I for one, am one of those people. :)

I actually got some of the more advanced books, and read some of them, and tried to tackle the problems, as MellowMelon said. I found that I could do more and more over time, and now I surprise myself at what I can do frequently now.

I am also relative to most users on this site pretty bad. I am going into Grade 10 and this year did not do so well on American Math Contests. I took the AIME and got a 3, except I forgot to multiply one of my answers by 2 and on another I said that $ 2x\plus{}y\equal{}1000$ has 500 solutions for positive $ x,y$. Anyways, what I was wondering was where I should look for problems to practice on. Should I do old AIMEs or should I look for non-contest problems?

Thanks.

AoPS is not at all a representative sample of the population, and neither are AIME or even AMC takers. Just by getting to AIME you're already ahead of 95% of the people who took the AMC.

One of the problems I see with AoPS is that it becomes a sort of echo chamber - everyone thinks they are worse than everyone else, which really messes with the culture. Don't listen to people who say things like, "Oh, 10 on AIME, terrible..." They aren't worth listening to. It's fine if someone says they didn't meet their goals, but saying that you have "failed" by being in the top 1% instead of the top .1% is patently absurd. AoPS is still a great community, you just can't get swept up by that.

[quote="theprodigy"][quote="worthawholebean"]If you find yourself lacking in the specific tools needed to solve problems, a book's systematic treatment can be useful. Books like The Art of Problem Solving (Rusczyk/Lehoczky) and The Art and Craft of Problem Solving (Zeitz) are useful, but they aren't a panacea. The only solution is to do more problems.

[b]Also, a 4 on AIME is nothing to be ashamed of. That is still an incredibly impressive achievement. Don't belittle yourself[/b].[/quote]

problem is here it feels like everyone gets 6+ on AIME
but yeah 4 is above average...so you're better than most ppl who took it[/quote]

AoPS is full of exceptional people. I seriously think that if government wanted to hire a mathematician, they can find plenty from surfing on the web.

So, really... don't worry. Just enjoy. That's what mathematics is about --- challenging yourself and enjoying the moment.

[quote="1=2"]Do problems IN GENERAL. Mandelbrot, AIME, HMMT, PUMaC, whatever. They are good.[/quote]

I didn't think this year's PUMaC was particularly well written compared to last years.

[quote="worthawholebean"]If you find yourself lacking in the specific tools needed to solve problems, a book's systematic treatment can be useful. Books like The Art of Problem Solving (Rusczyk/Lehoczky) and The Art and Craft of Problem Solving (Zeitz) are useful, but they aren't a panacea. The only solution is to do more problems.

Also, a 4 on AIME is nothing to be ashamed of. That is still an incredibly impressive achievement. Don't belittle yourself.[/quote]

Check out: [url=http://www.overcomingbias.com/2009/07/stupider-than-you-realize.html]Stupider Than You Realize[/url]


Two jets leave St. Louis at the same time and fly in opposite directions. One is flying west, 50mi/h faster than the other. After two hours, the jets are 2500 miles apart. Find the speed of each jet.

This is the classical traveling problem that people hates so much..

It would be a good practice problem.

[hide="Solution"]Let the speed of the slower one be x.
The speed of the faster one is then x+50.
The joint speed is 2x+50.
We multiply it by two to get the speed per two hours
We set that equal to 2500.
4x+100=2500
4x=2400
x=600.
So the speeds are 600 and 650.
[/hide]

[quote="KBabe"]i don't get it....[/quote]

Imagine that the slower plane stood still, and the earth and the faster plane moved around it.
The earth would move x miles relative to the slow plane, and the fast plane moves x+50 miles relative to the earth.
So the speed that the fast plane moves away from the slow plane is 2x+50.

Why do I have feeling that I saw this before :lol:

All math problems look alike that's why! :P

Well , This is the problem, my old math teacher made up...
I though it would be good to prcatice with these because this is a forum for middle schoolers :D

[quote="shinwoo"]I like RC-7th 's answer

if you don't understand, KBabe, you could draw a function

it would be very helpful

it's really hard to draw two jets flying 600miles or 650miles perhour in any person's mind[/quote]

if she doesn't understand the explanation, I don't think she would know how to draw a function or even what a function is.

function....

hm...

how can I explain this...

if you go to the MATHCOUNT section, there is a part where Treething and I did some function problems..

You might want to take a look at that...

hm..

I might have to post some function problems..


Hi I wanted to get some more practice with the basics of mathematical logic, so I've done some questions involving negation.  I was hoping if someone here could again check over my work.  That would help me greatly.  Thank you again!

Question: Express the following statements in the form "for all ..., if ..., then ..." using symbols to represent variables.  Then write the negations, again using symbols. (Note: my book uses $\ni$ for "such that")

a)  An integer is odd or even.
$\forall n,$ if $n\in\mathbb{Z}$, then $n=2k\vee n=2k+1$ where $k\in\mathbb{Z}$
[b]Negation[/b] (The negation of an implication is $P\wedge\neg{Q}$)
$\exists n\ni n\in\mathbb{Z}\wedge(n\not= 2k\wedge n\not= 2k+1)$ where $k\in\mathbb{Z}$

b)  A function that is differentiable at 0 is continuous at 0.
$\forall f,$ if $f$ is differentiable at $0$, then $f$ is continuous at $0$.
[b]Negation[/b]
$\exists f\ni f$ is differentiable at $0$ $\wedge$ $f$ is not continous at $0$.

c)  1 is the smallest positve integer.
$\forall n\in\mathbb{Z}$, if $n$ is the smallest positive integer then $n=1$
[b]Negation[/b]
$\exists n\in\mathbb{Z}\ni$ $n$ is the smallest integer $\wedge$ $n\not=1$.

d)  When the product of two integers is even, then both integers are even.
$\forall n,m\in\mathbb{Z}$, if $nm=2k$ where $k\in\mathbb{Z}$ then $n=2s$, $m=2t$ where $s,t\in\mathbb{Z}$.
[b]Negation[/b]
$\exists, n,m\in\mathbb{Z}\ni nm=2k$ where $k\in\mathbb{Z}$ $\wedge$ $n\not= 2s$, $m\not= 2t$ where $s,t\in\mathbb{Z}$.

using a lot of symbols is annoying...they don't really make your solution shorter, and it is more difficult for graders when there are lots of symbols

That's quite... harsh!

Although you arguably have more and more positive feedback from graders than I do, I think it is important to know how to write in formal logic. That way, you don' get bogged down with english and trying to write an english paper and a math proof at the same time! It is very nice what you have done, mitochondria, although keep in mind that you are only doing exercises. You might want to post here a couple problems and solutions of your own (problems from olympiad)

Hi thank you for reviewing my question. I'm sorry to put so many of these logic questions here, but I just wanted to get some practice on writing mathematical statements with logic symbols and get some practice with logic proofs.  I just think that math is like a language and it might be helpful to learn its syntax and grammar.  I appreciate all of the feedback I am getting here, and I hope people won't mind if I put more logic problems in the future.  Anyways thank you again for the help.

oh sorry if i was a bit harsh...i did not mean to come off that way...

but in mathematical proofs, lots of symbols don't really have a places...unless you cannot speak english or something

also, what does $\ni$ mean?

As far as I know, $\ni$ means such that. Personally, I find the notation better than writing the exact same thing in English.

EDIT: I just noticed that he wrote the meaning in this post.  Also, you could use this in the way as $\in$ in for example $f: \mathbb{Z}\ni x\rightarrow [x]_{3}\in\mathbb{Z}_{3}$.


Can someone please help me with following two questions

1- Suppose $A$ is an $m \times n$ and there exist $n \times m$ matrices $C$ and $D$ such that $CA = I_{n}$ and $AD = I_{m}.$ Prove that  $m = n$ and $C = D.$

2- If the vectors $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ from $\mathbb{R}^{6}$, show that the vectors $a_{1}$, $a_{1}+a_{2}$, $a_{1}+a_{2}+a_{3}$, $a_{1}+a_{2}+a_{3}+a_{4}$, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ are also linearly independent.

hints and help would be really appreciated, thanks in advance!

1: What is $CAD$?

2: Let $b_{1}=a_{1},b_{2}=a_{1}+a_{2},\dots,b_{5}=a_{1}+a_{2}+\cdots+a_{5}$. Write the $a_{i}$ as linear combinations of the $b_{i}$.
You didn't get the statement right, by the way.

[quote="bylsals"]$C=I_{n}A^{-1}=A^{-1}$[/quote]
If this is a reference to the first problem, note that we do not know that $A$ is invertible and hence cannot invoke its inverse. In fact, at first we don't even know that $A$ is square.

(Follow jmerry's hint.)

1- is the excercise with whole Rank: from $CA = I_{n}$ and $AD = I_{m}$ follows whole column and row ranks, i.e. n=m. Therefore $C$ and $D$ are inversions of the same matrix.

2 - use determination.

assume that $m\geq n$
use the inqeuality:
$\mathrm{rank}(CA)\leq\mathrm{min}\{\mathrm{rank}A,\mathrm{rank}C\}$
it is quite easy to show that $\mathrm{rank}A=m$,
but $\mathrm{rank}A\leq n$, so $m\leq n$.
the other part is obvious.


This album is about large numbers. There. Have I bored you yet again?

Annoying animated scrolling zeroes. Use sparingly. If swallowed, please contact your local Poison Control Center immediately.


polynomial
algebra
system of equations
Discriminant
IMO
IMO 1968

Let $a,b,c$ be real numbers with $a$ non-zero. It is known that the real numbers $x_1,x_2,\ldots,x_n$ satisfy the $n$ equations:
\[  ax_1^2+bx_1+c = x_{2} \]\[ ax_2^2+bx_2 +c = x_3\]\[ \ldots \quad \ldots \quad \ldots \quad \ldots\]\[ ax_n^2+bx_n+c = x_1  \] Prove that the system has [b]zero[/b], [u]one[/u] or [i]more than one[/i] real solutions if $(b-1)^2-4ac$ is [b]negative[/b], equal to [u]zero[/u] or [i]positive[/i] respectively.

Consider the function
\[ f(x)=\left(\frac{2ax+b-1+\sqrt{\Delta}}{2ax+b-1-\sqrt{\Delta}}\right)^{\frac{1}{\sqrt{\Delta}}} \]
where $\Delta=(b-1)^2-4ac$.
If $x_i$ is an approximation to a root of this function, then $x_{i+1}$ is next approximation to the root of this function obtained by using the Newton-Raphson method.  So a solution of the system of equations corresponds to a failure of the Newton-Raphson method to find a root of this function, producing instead a cycle of length $n$.


induction
calculus
real analysis
real analysis unsolved

Hello there:

I have an induction problem I am not quite sure how to approach. I wasn't exactly sure where to post this. I hope

this is OK in the calculus forum, though, it's not technically calculus.

Let F(n)=fibonacci number and L(n)=Lucas number.

How can I prove by induction that F(2n)=F(n)*L(n)

I used F(2n)=F(n)*(F(n+1)+F(n-1)).

How can I transform the right side into, maybe, F(n+1)(F(n+2)+F(n))?.

I tried adding using F(2n)+F(2n+1)=F(2n+2) but then got hung up.

Any help is appreciated. Induction can be a booger.

$Fibbonaci: 1,1,2,3,5,8,...$

$Lucas: 1,3,4,7,11,...$

[quote="fredoniahead"]I used $F(2n)=F(n)(F(n+1)+F(n-1))$[/quote]

If it is true then we will use the other formulation of the mathematical induction (see [url=http://mathworld.wolfram.com/PrincipleofStrongInduction.html]here[/url])

We'll prove that $F(n+1)+F(n-1)=L(n), \ \ \ \forall n\geq 2$

For $n=2$ it is true: $F(3)+F(1)=2+1=3=L(2)$

We assume that it's true for all $n<k$ and we'll prove that is true also for $n=k$

so we want to prove that $F(k+1)+F(k-1)=L(k)$

Take the first part:

$S=F(k+1)+F(k-1)=$

$F(k)+F(k-1)+F(k-2)+F(k-3)=$

$[F(k)+F(k-2)]+[F(k-1)+F(k-3)]$

According to our assumption, we have:

$F(k)+F(k-2) = L(k-1)$

$F(k-1)+F(k-3) = L(k-2)$

$\Rightarrow$

$S=L(k-1)+L(k-2)=L(k)$

Hello and thanks for the fast response. This is not a homework problem. I am just trying to get better at induction.

May I ask, how  can I use your response to show F(2k)=F(k)*(F(k-1)+F(k+1))?.

It's probably obvious but I can't see it. I have been trying to figure this thing out for hours.

Wouldn't I have to add something to the left and right that would turn F(k)((F(k-1)+F(k+1))

into F(k+1)(F(k)+F(k+2))?.

Thanks a lot for your interest.

Back again.

I thought I had a breakthough.

 I tried $F_{2n}=F_{n-1}F_{n}+F_{n}F_{n+1}$

$F_{2n}+F_{2n+1}=F_{n-1}F_{n}+F_{n}F_{n+1}+F_{n}F_{n+1}+F_{n+1}F_{n+2}$

$F_{2n+2}=F_{n}(F_{n-1}+F_{n+1})+F_{n+1}(F_{n}+F_{n+2}).$

Of course, I tried some numbers and it didn't work. It looked so good. I think it's in the

$F_{2n+1}=F_{n+1}(F_{n}+F_{n+2})$

Hi again,fredoniahead

Well, I used this identity $F_{2n}=F_{n}(F_{n-1}+F_{n+1})$ because I saw it in your text.
I didn't know if it is true (by the way it is). I used it as a tool.

By the way, is a proof for this identity necessary? I mean, maybe this id is well known (you read it somewhere, right?)

About your proof,

[quote="fredoniahead"]I think it's in the

$F_{2n+1}=F_{n+1}(F_{n}+F_{n+2})$[/quote]

I agree with you, because $2n+1$ is odd

Thank you so very much. I went ahead and used:

$F_{2n}+F_{2n+1}=F_{n}(F_{n-1}+F_{n+1})+F_{2n+1}$

An identity is $F_{n}(F_{n-1}+F_{n+1})+F_{2n+1}=F_{n+1}(F_{n}+F_{n+2})$

Then we have $F_{2n+2}=F_{n+1}(F_{n}+F_{n+2})$. Which is what we want. I hope this will suffice,

otherwise, I am going to quit fooling with it.

WOW!!!. I'll keep your post for future reference. Thanks alot. You were a big help.

I don't want to be a nuisance but do you know of anyway to prove

$F_{n}(F_{n+1}+F_{n-1})+F_{2n+1}=F_{n+1}(F_{n+2}+F_{n})$

I have been looking at it and taking ideas from your other proof.

If you don't feel like fooling with it, I understand.  It bugs me when I want

the solution and can't find it.

Take care

Thanks  alot. It's true that Fibonacci numbers seem to have endless properties and implications.

Take care and thanks for your expertise.

nice! :)

So we need again http://www.mathlinks.ro/Forum/viewtopic.php?t=45045


A circular ceiling fan rotates at a constant speed of 80 revolutions per minute. A point halfway between the center of the fan and the outer edge of the fan travels 97,968 inches in 15 minutes. How far does a point on the outer edge of the fan travel in 30 minutes?

Woah, no double posting.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323335


search
Functional Analysis
linear algebra
linear algebra unsolved

Let $ T$ be a bounded linear operator on a Hilbert space $ H$, and assume that $ \|T^n \| \leq 1$ for some natural number $ n$. Prove the existence of an invertible linear operator $ A$ on $ H$ such that $ \| ATA^{\minus{}1} \| \leq 1$.

[i]E. Druszt[/i]


What arithmetic symbol can we place between 2 and 3 to make a number greater then 2 but less than 3?

[i]from book "The Moscow puzzles" by Boris A. Kordemsky[/i]

[u]Please hide Your solution . Write inside Hide ... /Hide[/u]

[hide="Solution"]

A decimal point.

[/hide]

[hide]A decimal, dot, or period.  :D [/hide]


how do we can get the solutions of Hojo lee problems?(in number theory )

visit the [url=http://www.mathlinks.ro/index.php?f=461]PEN[/url] sub-forum...

Also some problems are solved in the PEN blog at [url]http://projectpen.wordpress.com[/url]


The sum of the digits of a number is 9. What is the probability it is prime?

[hide]
3,111,12,21

1/2

[/hide]

EDIT: If you put any number of 0's in those numbers, wouldn't you have infinite numbers that their digits sum is 3,

30,300,3000,10011,10101,3000000,3000000000000000000000...

[hide]
you could have
12
or 21
or 30
or 300
so there is an infinate amount of numbers so the answer is:
$\frac{1}{\infty}$
im wrong though.[/hide]

[quote="236factorial"]The sum of the digits of a number is 3. What is the probability it is prime?[/quote]

I think you should put a limit on the number of digits the number can have, that might makes this more solvable!

:) maybe you should put  :stretcher: [b]no zeros allowed[/b] on that problem. Then the answer would be $\frac{1}{2}$ :idea:

clever you......

[hide="hint"]
[hide="are you sure you need a hint?"]
what does it tell you if a number's digits have a sum of 3?
[/hide]
[/hide]

wait actually you forgot about the single prime number. Someone already said this but wouldn't it be $\frac 1{\infty}$ then?

i changed the problem to sum of 9. That should solve it. math92's answer is correct, but he didn't really do the right thng.

[quote="236factorial"]i changed the problem to sum of 9...quote]

[hide]Any number that's digit's sum is 9 is divisible by 9 so probability is 0[/hide]

[quote="236factorial"]nope guys, read the problem carefully  :P[/quote]

OHHHHHH... so you were trying to trick us...  :rotfl:
stupid me  :!:

EDIT: actually I think I mis read/ misunderstood the problem I was thinking what is the probablity that  the digits of the number are primes


Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.


A partition of n is a collection of one or more positive integers, with repetitions allowed, whose sum is n, An example of a partition of 7 is {3,2,1,1}. What is the probability that a randomly selected partition of 7 contains a prime number?

Hide your answers.

Billy

[hide]There are 15 partitions. Only the 6 1,  4 1 1 1, and 1 1 1 1 1 1 1 dont have a prime. so, there is a 12/15 or 4/5 chance[/hide]

[hide="hint"]Subtract the possibilities without a prime, and solve this yousing complementary probability.[/hide]
[hide]The only three that don't work are 61, 4111, and 1111111. THe probability of getting those is 1/5. 1-1/5=4/5.[/hide]


ARML
geometry
summer program
PROMYS
geometric transformation
reflection

worst power question set i've seen, ever!! i am serious!!

i liked last years' geometry problem 1000x better...

i was worried that nobody does ARML here anymore (graduated or whatever).

Why do you say that?  I dont remember last year's, was it the one with ceva's theorem?  I thought this was a little more interesting.  For the first question, someone knew the numbers off the top of his head... i dont know why anyone would memorize that, though.

Interesting...some questions were fun but the format was just odd. Qn 2 was too easy - almost depressingly so. 1 was odd - how did that person know?!

i dont know, that was really weird.  And the thing is, nobody wrote up a solution, so when time was running out, they werent able to justify it.  They just showed it worked, which wont get too much credit.

I dont know why anyone would memorize that, though.

10 was really easy too.  How about 9, did you get a solution for that?  we didnt.

Sometimes fractions can be reduced by cancelling out digits.  for example, in 16/64, you cancel out the 6 and end up with 1/4, which is correct.  Find the other two fractions that can be reduced in a similar manner, i.e. ab/bc = a/c, where a/c is a reduced fraction with a<c.

That doesn't explain why someone would memorize them.

we had half a solution for #9    we were pressed for time so we did not read the problem carefully.  i think it asked for all the solutions but we only determined one

At an appropriate time last year, I would have had those memorized because I designed a problem based on that principle.  I actually posted it here . . . 7ab / ab4 = 7 / 4.  So I had looked at all of them.

results are out:  Bulgarian math circle got 38, and some other team in the states
we got 37 :D
Stuyvesant got 36

hm, is there a website  where you can see the results?  I took a look at arml.com and didnt see it.

arml.com isn't updated.  However, they usually send it out to all of the schools that participated.

how was the second part of the contest??  i think its over by feb 19th so we are technically free to discuss the questions...

Why was the "ballot theorem" or whatever you want to call it on the ARML power part II ? Luckily I know the proof because we did it at PROMYS, but thats kind of an unfair question considering how difficult the proof is to come up with...

you got that ballot??  good job...  we think the formula is (N(B)-N(A))/(N(B)+N(A)) do you mind posting the proof?  i am curious as to how u do it.   btw for the last part in question 4, the answer was 4 right??

[quote="Nukular"]our school is tied with the Bulgarian math circle...

go AAST![/quote]

tied with Bulgarian math circle??  i recall that st. louis math circle was tied with them in round 1.  you are from that team??

how did anyone else do on the second set of power questions??

oh I thought we were in first... weird.

For shorthand I will say B  =  N(B) and A = N(A).

We ended up doing a proof similar to that of Catalan numbers. What we did is say that the number of voting sequences in which B beats A is the same as the number of lattice paths going up or right at each step from (1,0) to (B,A) such that the path never touches the line y=x. So there would be C(A+B-1, A) ways for this to happen without the condition. For every path that does cross y=x, reflect the path up to that point over y=x. This yields a bijection between paths that violate the condition and paths from (0,1) to (B,A).  There are C(A+B-1,A-1) such paths.

So the number of ways for B to beat A is C(A+B-1,A) - C(A+B-1,A-1). There are C(A+B, A) voting sequences in general, so the answer is

[C(A+B-1,A) - C(A+B-1, A-1)]/C(A+B,A). Now this simplifies algebraically to (B-A)/(B+A) which is the same thing you people got.

nice...  [quote]"oh I thought we were in first... weird. "[/quote]

umm i could have been the one thats mistaken...

anywayz thx a bunch for the proof

ballot theorem was question 2?  What exactly is that?  And how did you do part 2c?

ballot theorem was 5c.  i spent the entire time working on it so i have no idea what 2c is.

oh ok.  Hmm, we should have had the kid who went to promys work on that then.

who in your school went to promys? i might have met him there.

hm, no one from my school, but from my arml team.  The kid i was talking about was Steve Ehrlich (i think that's how you spell it).  And Alice Swanson went for the past few years, though i dont' know if she went this past yr.


What is the interest on $ \$240$ for 3 months at $ 8\%$ per year?

I'm assuming they want this simply.

240x.02=240/50=24/5=4.8

You also have to put 4.80, if you put 4.8, it will count it as wrong...i don't know why but it just happens


If the opposite faces of an hexahedron, are perspective quadrilaterals, then it’s diagonals are concurrent ( special case the cube ).

     Kostas Vittas.

Dear all my friends.

     I apologize for my typo mistake. I meant perspective (not projective) quadrilaterals. Now, my message has been corrected and sorry again.

     Kostas Vittas.

This problem is a direct application of the follows theorem of concurrent segments ( three segments are called concurrent if their segment lines are concurrent. Three parallel segments are concurrent at infinity ).

     THEOREM – Let  AA', BB', CC' be three concurrent segments, at a point P. We denote as A'', the intersection point of BC', B'C and as B'', the intersection point of AC', A'C and as C'', the intersection point of AB', A'B. Prove that:

a) The lines AA'', BB'', CC'', are concurrent at a point, so be it, M.
b) The lines A'A'',B'B'',C'C'',are concurrent at a point, so be it, N.
c) The line connecting the points M, N, passes through the point P.

     This theorem is true, for every configuration of the given segments ant their edges, in the plane, or in 3D space and there are many it's interesting applications. A simple proof of this theorem, has been posted in Hyacinthos forum.

     http://groups.yahoo.com/group/Hyacinthos/message/13310

      Kostas Vittas.

A simple construction ( helping to prove ) of the configuration of three concurrent segments, is the follows:

     Inwardly to a given triangle ABC, we consider a fixed point P and we denote as A', B', C', three arbitrary points, on the segments AP, BP, CP, respectively ( In my drawing AB = 11.3, BC = 15.7, CA = 13.9, AP = 4.5, BP = 9.6, AA' = 2.7, BB' = 4.2, CC' = 5.4 ).

     So, connecting the corresponded points as the theorem states and canceling the sidelines of ABC, we have also an image of the hexahedron’s concurrency, parallel projection on the plane.

     Kostas Vittas.

Dear all my friends.

I present a figure of the configuration as the problem states and I hope that it is enough to stimulate your interest, because it is a useful and not hard problem.

Kostas Vittas.


calculus
integration
limit
probability
calculus computations

In my opinion this question could be reduced to question:
Where     $ n \int_{0}^{1}\int_{0}^{1}\dots\int_{0}^{1}ln(x_1\plus{}x_2\plus{}\dots\plus{}x_{n\minus{}1}\plus{}1)dx_1dx_2\dots dx_{n\minus{}1}$  is going?
And this is infinity, because it is asymptotic equal with $ n^n ln(n)$  (or something about :) ).

$ 2/3$, of course: the integrand is bounded by $ 1$ and the numerator and the denominator have limits $ 1/3$ and $ 1/2$ in probability if you divide them by $ n$, so an appropriate version of the dominated convergence theorem finishes the problem off in no time. Let's make life a bit harder: is $ V_n$ greater or less than $ 2/3$ for large $ n$? ;) If it is still not hard enough, then approximately by how much? ;)


[hide="Answer"]Using synthetic division, we see that $k=4$, and we have $3x^2-9x^2+4x-12=(x-3)(3x^2+4)$, so the answer is $C$.[/hide]

[quote="joejia"]If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then it is also divisible by:
(A) $3x^2-x+4$
(B) $3x^2-4$
(C) $3x^2+4$
(D) $3x-4$
(E) $3x+4$[/quote]

Since it factors, it can be asumed that $(x-3)$ is a factor, so therefore $3x^2+4$ is the answer.

take out x from the first 2 terms and the last 2, the first would leave you with $3x^2(x-3)$ and the other should have $(x-3)$ so $4$ can come out, so combine those two and you get $3x^2+4$.