induction
pigeonhole principle
combinatorics proposed
combinatorics

Let $ n \geq 3$ be a positive integer and let $ m \geq 2^{n\minus{}1}\plus{}1$. Prove that for each family of nonzero distinct subsets $ (A_j)_{j \in \overline{1, m}}$ of $ \{1, 2, ..., n\}$ there exist $ i$, $ j$, $ k$ such that $ A_i \cup A_j \equal{} A_k$.

$ i, j, k$ are pairwisely distinct, right?

Consider a non empty set $ A\in F$ where $ F$ is the family of non empty subsets of $ \{1,2,\dots,n\}$ and $ |F|\equal{}m\geq 2^{n\minus{}1}\plus{}1$.
Let $ F'\equal{}\{B_j|B_j\equal{}A\cup A_j \quad \forall A_j\neq A  \quad and \quad A_j\in F \quad and \quad j\equal{}1,2,\dots,m\}$
Since each $ B_j$ is distinct, $ |F'|\geq2^{n\minus{}1}$
$ \Rightarrow |F\cap F'|\geq1$ and we are done.

Is Tomcio's predict correct? Otherwise, it just an easy induction.

It seems like a straightforward inductive solution should work even if you require pairwise disjointness, but it turns out there's a booby trap to watch out for.  To start with, here's a solution with a hole in it:

The case $ n \equal{} 3$ can be handled by a quick case analysis (alternatively, we can start our induction at $ n\equal{}2$, where there is only one possible collection of $ 3$ sets).   Now let us assume that the result is true for subsets of an $ n \minus{} 1$ element set, and we are given a collection of $ 2^{n \minus{} 1} \plus{} 1$ subsets of an $ n$ element set.  By pigeonhole either there are $ 2^{n \minus{} 2} \plus{} 1$ sets not containing $ n$ or $ 2^{n \minus{} 2} \plus{} 1$ sets containing $ n$.

In the former case, we know by the inductive hypothesis that three of the sets not containing $ n$ satisfy $ A_i \cup A_j \equal{} A_k$, so we are done.  In the latter case, we just apply the hypothesis to the (at least $ 2^{n \minus{} 1} \plus{} 1$) sets formed by taking each set containing $ n$ and dropping $ n$.

[b]Why doesn't this work?[/b]
[hide]
The problem stated that the sets were nonempty, and it's a necessary assumption (e.g. if $ n \equal{} 4$ and you take all the subsets of cardinality $ 0, 1,$ and $ 3$).  If our original collection contained the singleton set $ \{n\}$, we can't drop $ n$ from it like we did in the second case of the induction.
[/hide]

[b]How can we fix this?[/b]
[hide]
By the above argument, we're done unless our collection contains the singleton $ \{n\}$.  So let us assume that this is the case.   Now we divide the $ 2^n$ possible subsets into $ 2^{n \minus{} 1}$ pairs $ \{X, X \cup \{n\}\}$, where $ X \subseteq \{1, \dots, n \minus{} 1\}$.  By pigeonhole our collection must have both elements of some pair, and we can take our sets to consist of the pair and the singleton we're assuming to be present.
[/hide]

It's ok but this solution is very easy for a 3rd question of Romanian TST.

$ F$ is the family of non empty subsets of $ [n]\equal{}\{1,\dots,n\}$ and $ |F|\geq 2^{n\minus{}1}\plus{}1$.We have $ A,B\in F$ so that $ A\cup B\equal{}[n]$ and $ A\cap B\equal{}\emptyset$. If $ [n]$ is contained in $ F$ we are done, else ( we know $ \emptyset \notin F$) we have another pair $ C,D\in F$ so that $ C\cup D\equal{}[n]$ and $ C\cap D\equal{}\emptyset$. Consider the four disjoint sets $ A\cap C, A \cap D, B\cap C$ and $ B\cap D$. Among many forbidden pairs we cannot have for example $ A\cap C, A \cap D$ both present in $ F$. Hence we have another pair $ E,G\in F$ so that $ E\cup G\equal{}[n]$ and $ E\cap G\equal{}\emptyset$. By recursively employing the same process we empty out all such pairs in $ F$ which are of course finite in number and hence the contradiction.

[quote="srikanth"]By recursively employing the same process we empty out all such pairs in $ F$ which are of course finite in number and hence the contradiction.[/quote]

I think you need to go into more detail here.   For example,

If $ A \equal{} \{1\}, C \equal{} \{2\}$, then $ A \cap C \equal{} \emptyset$, so $ (A \cap C, A \cap D)$ doesn't lead to a new forbidden pair.  Of course, in this particular example we're not stuck yet (we can look at $ (B \cap C, B \cap D)$ instead).   However, it may be possible that what eventually will happen is that all the forbidden pairs you generate in your new step will overlap with previously generated pairs in such a way, and your process will halt before you actually forbid all the pairs from $ F$.


calculus
integration
trigonometry
Euler
calculus computations

OK fellas, another  one...  :D

show

$ \int_0^\infty\;\frac{x^m\;\;\textbf dx}{x^2\plus{}2ax\cos\,a\plus{}a^2}\;\;\equal{}\;\;\boxed{\frac{\pi\;a^{m\minus{}1}}{\sin\,a}\;\frac{\sin\,ma}{\sin\,m\pi}}$

(where $ 0<m<1,\quad 0<a<\pi$ )

do [b]not[/b] use techniques from complex analysis.

This is the well-know Euler integral, but I don't know how to solve it without use complex techniques.


MagnusRex wrote:1. The school store has construction paper in 8 different colors. Angelo needs to buy five sheets for an art project. How many different selections are possible?

2. How many of the integers from 100 to 999 have at least one digit that is a zero?

[hide]2. 170 integers[/hide]

I don't know from the question, is there enough supply for him to pick 2 colors the same, 3, etc.?

If not, then another question to try to answer would be

'The school store has construction paper in 8 different colors, with each color in unlimited supply. Angelo needs to buy five sheets for an art project. How many different selections are possible? '

For the first one, I don't think it's 8C5 or ordered repetition (8^5). Like, if he picked RGGBY it would be the same as picking RBGYG.

And for the second one, I didn't quite get your answer, but very very close...do your arithmetic again.

so wat's the answer for #1? i wanna know....

That's actually number 2, not number 1.  Unanswered, post your solution, not just your answer!  (And the same for everyone else!)

Whoops. Oh well, that works out well, since no one has really tried thinking about number 1. If you draw all the papers simultaneously, it's different because arrangements are the same. This isn't the same as asking how many 5 letter words can be formed or how many groups of 5 people can I choose from 8 people.


linear algebra
matrix
algebra
polynomial
absolute value
linear algebra unsolved

Assume $A=(a_{ij})_{1\leq i,j\leq n}$ is a $n\times n$ matrix with complex entries such that $A^k=I_n$, for
some integer $k\geq 1$. Prove that $\sum_{1\leq i,j\leq n}|a_{ij}|^2\geq n$.

I really doubt that. $A$ is simply a conjugate of a diagonal matrix having roots of unity on its diagonal. Conjugation can change it a lot if you choose the invertible matrix by which you perform this conjugation properly. A simple counterexample:

Take $a+b=1$ and then take $A=\begin{pmatrix}a&b\\1&-1\end{pmatrix}\cdot\begin{pmatrix}i&0\\0&-i\end{pmatrix}\cdot\begin{pmatrix}1&b\\1&-a\end{pmatrix}$. In other words, you obtain $A$ through conjugation of $\begin{pmatrix}i&0\\0&-i\end{pmatrix}$ by $\begin{pmatrix}a&b\\1&-1\end{pmatrix}$. The $(1,1)$ entry in $A$ is $i(a-b)$, which can be chosen to have absolute value as large as you want.

Am I missing something? :?

P.S.

The result is true if you also know that $A$ is normal, for example. In that case, you even have equality.

what is "complex entire" ? It is clearly false for complex number, for integer numbers .. so what is it ?

Probably complex entries (coefficients complexes)

of course not: any upper triangular matrix, with distinct roots of the unity on the diagonal, satisfies $A^k=I$.
But one can put anything above the diagonal ...

Maybe the correct problem is to prove that $\sum_{1\leq i,j\leq n}|a_{ij}|^2\geq n^{1/k}$ for $A^k = I$

Actually, $\ge n$ should work. Schur's Theorem states that a matrix can be unitarily upper-triagonalized. We perform this operation on $A$ and end up with an upper triangular matrix $UAU^*=(t_{ij})$ (with $U$ unitary) whose diagonal entries are roots of unity. This means that $\sum_{i,j}|t_{ij}|^2\ge n$.  All that's left now is to notice that unitary similarity preserves the norm $\||A\||_2=\sqrt{\sum_{i,j}|a_{ij}|^2}$. Indeed, we have $\||A\||_2=\mbox{Tr}(AA^*)=\mbox{Tr}(UAA^*U^*)=\||UAU^*\||_2^2$ for any matrix $A$ and any unitary matrix $U$.

I believe this is what cimabue meant (the sign $\le$ should be changed to $\ge$). I don't know why I didn't see it earlier, when I posted the message above :).

"an original mistakes is more worthy that a trivial truth", Dostoïevski.  :D

For any A complex matrix we have (1) $||A||^2\geq \sum_{k=1}^{n}|b_k|^2$

where $b_k$ are the roots its characteristic polynomial

 $||A||=\sqrt{\sum_{i,j}|a_{ij}|^2}$

equality in (1) iff $AA^{*}=A^{*}A$


[quote="kritikos"]Μπορεί καποιος να μου θυμησει το θεωρημα του καραθεωδωρη στα διανυσματα??????[/quote]
Το θεώρημα και η απόδειξη του με μιγαδικούς βρίσκονται στο attachment

[quote="meinstein"]Πάντως Μητσάρα λίγο δυσκολο να πέσει σε Θαλή :wink:
Του ειχαμε κανει μια απόδειξη περυσι στο μαθημα κ χωρίς μιγαδικούς!
Αυτά παθαίνει οποιος μιλάει την ώρα του μαθήματος :rotfl:[/quote]
Από το παράρτημα της Ε.Μ.Ε. του Ν. Ημαθίας στη Βέροια, έχουν έκδοθεί μέχρι σήμερα (μόνο) τέσσερα τεύχη (δυστυχώς), του μαθηματικού περιοδικού ''ΑΠΟΛΛΩΝΙΟΣ''.

Στο 2ο τεύχος, σελίδα 28, υπάρχει απόδειξη χωρίς μιγαδικούς, του [b][size=100]Γιάννη Απλακίδη[/size][/b].

Στο 3ο τεύχος, σελίδα 163, υπάρχει άλλη απόδειξη επίσης χωρίς μιγαδικούς, του [b][size=100]Θεόφιλου Χρυσοστομίδη[/size][/b].

Κώστας Βήττας.


[i]I've seen a lot of repeated questions lately, so I'm making a FAQ to which we can point people so we don't have to answer the same question again and again. If you have any questions to add/comments/whatever feel free to reply. tokenadult wrote a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]comprehensive description of the AMC series[/url], but it is a little out of date and aimed more at parents. This will grow as I add more.[/i]

[size=200][b]Frequently Asked Questions[/b][/size][list]
[*][i]I'm new to the AMC series. How does it work? What are the different tests?[/i]
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]here[/url].

[*][i]All the problems seem really difficult. What should I do to improve?[/i]
The most important thing to do is just solve problems. Use the old AMCs in the Contests section above; the more you do, the better you will get. You can also work through books like the ones published by Art of Problem Solving; these help and are available through the bookstore. If you're having trouble with a problem on an old test in the Contests section, you can click the number and find the original discussion. If you don't understand the solution, feel free to revive the thread and ask.
It's a good idea to do contests under timed conditions. After you're done, read the solutions to any problems you got wrong or even found mildly difficult.
If you're experienced with MathCounts/AMC-style tests and want to move on to olympiad-style ones, the only solution is to work on progressively harder problems. Easier olympiads like Canada or older USAMOs and harder short-answer contests like AIME are good stepping stones.
I see a ton of questions that ask this in various forms. The only answer is to do more problems, and preferably ones on the edge of your skill level. There is no silver bullet that will make you instantly better.

[*][i]What books can/should I buy for a more comprehensive look?[/i]
The books that I see most frequently recommended are the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts_ProbSolve.php]original Art of Problem Solving[/url], [url=http://www.amazon.com/dp/0471789011/]the Art and Craft of Problem Solving[/url], and [url=http://www.amazon.com/dp/0387982191/]Problem Solving Strategies[/url], in order of difficulty.

[*][i]I'm a freshman, sophomore, or below. Which test should I take, the AMC 10 or the AMC 12?[/i]
It depends on your goals and current knowledge. If you haven't started Algebra 2 at least, you should take the AMC 10. Otherwise, you have to decide whether to aim for the USAJMO or the full USAMO. The USAMO is much tougher to make, as you have to face the more difficult AMC 12 test. The USAJMO, on the other hand, is easier to make, and if that's your goal take the AMC 10.

[*][i]I have a question about qualification for the tests or their administration. Where can I find the answer?[/i]
Look on the [url=http://www.unl.edu/amc]AMC website[/url]. Specifically, look at the [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml]Contest FAQ[/url] and the teacher's manuals for the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-AMC_1012A,2009.pdf]AMC 10/12 (2009)[/url], [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-AIME,2008.pdf]AIME (2008)[/url], and [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-USAMO,2008.pdf]USAMO (2008)[/url].

[*][i]You didn't answer my question.[/i]
Look on the AMC website, and if you still can't find it, make a new topic on the forums to ask.

[*][i]I've got a question about a problem on a past test - where can I ask it?[/i]
Find it in the [url=http://www.artofproblemsolving.com/Forum/resources.php]Resources section[/url] and click on the problem number - that's a link to the thread which originally contained the problem. Ask the question there. It doesn't matter if you're reviving an old thread; we just want to keep all discussion on a problem in a single place.[/list]

The "Which test should I take" answer is now outdated.  For 10th graders and below looking to make USAMO it is almost certainly better to take the AMC 12 now, and/or the AMC 10 for a chance at the USAJMO (Jr. math olympiad).


Let $ f(x)\equal{}Ax^2\plus{}Bx\plus{}C$ where $ A>0$. Let $ a_1 \geq a_2 \geq a_3$, $ b_1 \geq b_2 \geq b_3 \geq \frac{\minus{}B}{2A}$, $ a_1 \geq b_1$, $ a_1\plus{}a_2 \geq b_1\plus{}b_2$ and $ a_1\plus{}a_2\plus{}a_3 \geq b_1\plus{}b_2\plus{}b_3$. Prove that: $ f(a_1)\plus{}f(a_2)\plus{}f(a_3) \geq f(b_1)\plus{}f(b_2)\plus{}f(b_3)$

Could you give the solution more clearly? I can't find how to use Abel express???

is abel express same as abel's sum.
well looking at it the first thing that comes to my mind is majorization inequality but have to hind a suitable $ a_{4}$ and $ b_{4}$ such that $ \sum a_{i}\equal{} \sum b_{i}$


$ 50$ contestants participated at a mathematical competition with $ 8$ problmes. At the end, there were $ 171$ correct solutions. Prove that $ 3$ of the problems were solved by at least $ 3$ of the contestants.

I must miss something, but if $ n_i$ is the number of correct solutions of problem #$ i$, then $ n_1\plus{} \cdots \plus{} n_8 \equal{} 171$ and $ n_i \leq 50$ for each $ i$.
If at most two of the $ n_i$'s are greater than $ 2$ then $ n_1\plus{}\cdots n_8 \leq 50\plus{}50\plus{}6 \times 2 < 171$, a contradiction. :wink:

Pierre.

I interpreted the problem differently and I apologize if I am wrong. I thought that it asks for a proof that there are 3 problemes and 3 contestants such that each of the 3 contestants solved each of the 3 problems.... :maybe: I also apologize for not writing this from the begining but I wrote it in the way I found it.... :blush:

i agree with mon, in that the problem is asking that the *same 3 contestants* solved some 3 problems. it's the more challenging interpretation.

assume that there aren't three students that solved the same three problems. S is the set of 4-tuples such that a student and 3 problems that he/she solved. let's count the number of elements of S. each 3 problem set can be solved by at most 2 student. so S has at most C(8,3)x2=112 elements.
a student that solved A problems adds C(A,3) elements to S. to minimize the sum of C(A,3) we must take 21 students solved 4 and 29 students solved 3 questions.(it comes from this lemma if a-b>=2 C(a,3)+C(b,3)>C(a-1,3)+C(b+1,3) so the difference of any 2 is at most 1 in the minimum) so S has at least 21xC(4,3)+29xC(3,3)=113 elements. Contradiction

Nice technique,thank you :lol: .  Do you think it can be used to solve this problem, too? http://www.mathlinks.ro/Forum/viewtopic.php?t=178013
I've tried but either I don't use the given condition in the best way or the approach is not a good one :maybe:


In a triangle ABC, points P, Q lie respectively on the sides AB, AC.
The Cevians BP, CQ intersect at R.
Given that AP = p. AC and AQ = q. AB where 0<p<1 and 0<q<1.
evaluate the ratio between the area of the quadrilateral AQRP and
the area of the triangle ABC.

Let $ AC=x$ and $ AB=y$. We have $ AP=xp, PC=(1-p)x, AQ=qy, QB=(1-q)y$.
By putting weight $ 1$ to point $ A$, weigth $ \frac{q}{1-q}$ to point $ B$ and finaly weight $ \frac{p}{1-p}$ to point $ C$ we get that center of weights is point R. ($ AP \cdot q=BP \cdot \frac{q}{1-q}\Rightarrow$ center of weights belongs to $ QC$; by analogy we get that center of weights belongs to $ BP \Rightarrow$ it is $ R$)
Let $ S_{\Delta ABC}=s$. Then $ \frac{S_{\Delta AQC}}{S_{\Delta BQC}}=\frac{q}{1-q}\Rightarrow S_{\Delta AQC}=qs ,S_{\Delta BQC}=s(1-q)$.
We may rearrange weights in points $ A(1)$ and $ B(\frac{q}{1-q})$ to point $ Q(1+\frac{q}{1-q}) \Leftrightarrow Q(\frac{1}{1-q})$.
Now we have $ QR \cdot \frac{1}{1-q}=CR \cdot \frac{p}{1-p}$.
$ \frac{S_{\Delta QBR}}{S_{\Delta RBC}}=\frac{QR}{RC}$ and $ S_{\Delta QBR}+S_{\Delta RBC}=S_{\Delta BQC}=s(1-q) \Rightarrow S_{\Delta QBR}=\frac{sp(1-q)^{2}}{1-pq}$.
$ S_{\Delta ABP}=sp, S_{AQRP}=S_{\Delta ABP}-S_{\Delta QBR}=ps-\frac{sp(1-q)^{2}}{1-pq}=\frac{spq(2-p-q)}{1-pq}\Rightarrow \frac{S_{AQRP}}{S_{\Delta ABC}}=\frac{pq(2-p-q)}{1-pq}$.

[quote="vprasad_nalluri"]Join AR.
Let (XYZ) denote the area of the triangle formed by the points X, Y, Z .
Let (ABC) = Δ
First we evaluate the ratios BR : RP and CR : RQ
Apply Menelau’s Theorem to ΔAQC and transversal BRP:
AP.CR.QB = PC.RQ.(-BA),
p.AC.CR.(1-q).AB = (1-p).AC.RQ.AB,
p(1-q).CR = (1-p).RQ,
CR : RQ = (1-p) : p(1-q) from which it follows
RQ : CQ = p(1-q): (1-pq)
Next apply Menelau’s Theorem to ΔABP and transversal CRQ:
AQ. BR. PC = QB. RP. (-CA),
q.AB. BR. (1-p). AC = (1-q). AB. RP. AC,
q(1-p). BR = (1-q). RP,
BR:RP = (1-q) : q(1-p) from which it follows
RP : BP = q(1-p): (1-pq)
Next we note (ABP) = (½ )AP.BP.SinP and
                       (ARP) = (½ )AP.RP.SinP
So (ARP) : (ABP) = RP : BP = q(1-p): (1-pq)
Also (ABP) : (ABC) = AP/AC = p,
         (ABP) = pΔ
So (ARP) =  pq(1-p)Δ : (1-pq) Likewise,
     (AQR) = pq(1-q)Δ : (1-pq))
Adding, required area (of the quadrilateral AQRP)
       (AQRP) = (ARP) + (AQR)
                     = pq (2-p-q) Δ : (1-pq)
(AQRP):(ABC) = pq(2-p-q): (1-pq)
If P, Q are points of trisection we have the following
Table of values:
      p  ----------------> 1/3       2/3       1/3      2/3
      q  ----------------> 1/3       1/3       2/3      2/3
(AQRP):Δ            ------> 1/6       8/15      2/7     2/7[/quote]

[b] I have just put the image on the screen to do the study easier.[/b]

[u] Babis[/u]


Let $ A \in M_n(R)$ such that : $ e^{A\plus{}A^T}\equal{}e^{A^TA},A^{2009}\equal{}O$ Prove that $ A\equal{}O$

We replace the hypothesis $ A^{2009}\equal{}0$ with $ trace(A)\equal{}0$.
$ A\plus{}A^T$  and $ A^TA$ are symmetric then $ A\plus{}A^T\equal{}A^TA$. Thus $ trace(A^TA)\equal{}0$; moreover $ A^TA$ is nonnegative. Then $ A^TA\equal{}0$ and $ A\equal{}0$.


vector
inequalities
geometry unsolved
geometry

The vectors $ \vec{a},\vec{b},\vec{c}$ are given in a plane. Prove the inequality
$ |\vec{a}|\plus{}|\vec{b}|\plus{}|\vec{c}|\plus{}|\vec{a}\plus{}\vec{b}\plus{}\vec{c}|\geq |\vec{a}\plus{}\vec{b}|\plus{}|\vec{b}\plus{}\vec{c}|\plus{}|\vec{c}\plus{}\vec{a}|$

Please see here

 http://www.mathlinks.ro/Forum/viewtopic.php?highlight=karamata&t=16600

 Babis


real analysis
topology
continued fraction
real analysis unsolved

Construct a compact,uncountable set of irrationals in [0,1].
I meant that my reqd set will be a subset of [0,1] & will consist only of irrationals belonging to [0,1]

Enumerate the rationals. Remove an open interval of length $3^{-n}$ centered around the $n$th rational. This leaves a closed, hence compact, set of measure at least $\frac12$, which must therefore be uncountable. Since it obviously does not contain any rational numbers, we are done.

Could you please elaborate your proof.I didnot understand anything you have said

It's just a simple consequence of the fact that the rationals have measure zero- I find an open set of small measure containing all of them, and take the complement.

Sorry I dont know any measure heory.I only know when is a set said to have measure 0,i.e the defn thats all

Just because I feel like it, here's a completely different example. Choose a finite set of (at least two) positive integers $S$. Take all continued fractions $\frac1{a(1)+\frac1{a(2)+\frac1{a(3)+\dots}}}$ where $a(n)\in S$ for all $n$. Since the continued fraction of any rational number terminates, this set contains only irrational numbers.


modular arithmetic
number theory
divisibility tests

[hide]If all of them are bigger than 3 then
a,b,c can't divide by 3
=> $\ a^2\equiv1\mod3$
     $\ b^2\equiv1\mod3$
     $\ c^2\equiv1\mod3$
=> $\ a^2+b^2+c^2 can divide by 3$ and $\ a^2+b^2+c^2>3$
=> $\ a^2+b^2+c^2 is not a prime number$
So one of them is 3.We just have two case that is 2,3,5 and 3,5,7
But $\ 2^2+3^2+5^2=38$ is not a prime number so that is 3,5,7.[/hide]

Math is my life :rotfl:

Find all n(n belong to N) satisfying:
a) $\ n^4+4$ is a prime number
b) $\ n^3-n^2+n-1$ is a prime number
c)  $\ n^{1997}+n^{1975}+1$ is a prime number

Math is my life :|

You're good but you have a mistake because n belong to N.What about the question c?That's hard question.And,next time please hide your answer.

Math is my life :!:

[quote="M4RI0"][quote="José"][quote="TRAN THAI HUNG"]Let a,b,c be the prime connective numbers satisfying $\ a^2+b^2+c^2$ is a prime number.Find a, b, c.[/quote]

$3^2+5^2+7^2=83$[/quote]

And how did you get that :?:[/quote]

calculator ejejejejejej just 1º attempt

[quote="TRAN THAI HUNG"]Find all n(n belong to N) satisfying:
a) $\ n^4+4$ is a prime number
b) $\ n^3-n^2+n-1$ is a prime number
c)  $\ n^{1997}+n^{1975}+1$ is a prime number
[/quote]
[hide="Solution to part c"]
Note that $n^2+n+1$ divides $n^{1997}+n^{1975}+1$.

[hide="explanation"]
Since $(x^2+x+1)(x-1)=x^3-1$, $x^2+x+1=0$ gives the non-real roots to $x^3=1$.
Hence, $x^2=x^5=x^8=...=x^{3k+2}$ and $x^1=x^4=x^7=...=x^{3k+1}$.

Now consider $n^{1997}+n^{1975}+1=0$.
Since $1997\equiv 2\pmod{3}$ and $1975\equiv 1\pmod{3}$, $n^2+n+1$ divides $n^{1997}+n^{1975}+1$.
[/hide]

Therefore, if $n^{1997}+n^{1975}+1$ is prime, then we must have either $n^2+n+1=1$ or $n^2+n+1=n^{1997}+n^{1975}+1$.
The first case gives $n=0,-1$, but $n$ is a natural number, so we have no solution.
The second case simplifies to $n(n^{1996}+n^{1974}-n-1)=0$, and the only positive integer solution is $n=1$.

Hence the only possible value for $n$ that satisfies the given condition is $n=1$.
[/hide]

Edit: Fixed.

[quote="frt"][quote="TRAN THAI HUNG"]Find all n(n belong to N) satisfying:
a) $\ n^4+4$ is a prime number
b) $\ n^3-n^2+n-1$ is a prime number
c)  $\ n^{1997}+n^{1975}+1$ is a prime number
[/quote]
[hide="Solution to part c"]
Note that $n^2+n+1$ divides $n^{1997}+n^{1975}+1$.

[hide="explanation"]
Since $(x^2+x+1)(x-1)=x^3-1$, $x^2+x+1=0$ gives the non-real roots to $x^3=1$.
Hence, $x^2=x^5=x^8=...=x^{3k+2}$ and $x^1=x^4=x^7=...=x^{3k+1}$.

Now consider $n^{1997}+n^{1975}+1=0$.
Since $1997\equiv 2\pmod{3}$ and $1975\equiv 1\pmod{3}$, $n^2+n+1$ divides $n^{1997}+n^{1975}+1$.
[/hide]

Therefore, if $n^{1997}+n^{1975}+1$ is prime, then we must have either $n^2+n+1=1$ or $n^2+n+1=n^{1997}+n^{1975}+1$.
The first case gives $n=0,-1$, but $n$ is a natural number, so we have no solution.
The second case simplifies to $n(n^{1996}+n^{1974}-1)=0$, and obviously it does not have any positive integer solution.

Hence there is no possible value for $n$ that satisfies the given condition.
[/hide][/quote]

what about $n=1$?

[quote="TRAN THAI HUNG"][quote="TRAN THAI HUNG"]Oh, Why don't you try did problem?
Find the two natural numbers m,n satisfying:
$\ A=3^{3m^{2}+6n-61}+4$ is a prime number.[/quote]
I think that's a little hard.Try this first:
      Prove that with all the integer numbers n>1
a)$\ 2^{2^{2n+1}}+2$ is not a prime number.
b)$\ 2^{2^{4n+1}}+7$ is not a prime number.
c)$\ 19.8^n +17$ is not a prime number.[/quote]

Well, a) is obviously true, since you can factor a $2$.....

[quote="TRAN THAI HUNG"]Oh, I'm sorry it just a mistake and I have edited it. :oops:

Math is my life :!:[/quote]

OK... I thought it was too easy then

[quote="TRAN THAI HUNG"][quote="TRAN THAI HUNG"]Oh, Why don't you try did problem?
Find the two natural numbers m,n satisfying:
$\ A=3^{3m^{2}+6n-61}+4$ is a prime number.[/quote]
I think that's a little hard.Try this first:
      Prove that with all the integer numbers n>1
a)$\ 2^{2^{2n+1}}+3$ is not a prime number.
b)$\ 2^{2^{4n+1}}+7$ is not a prime number.
c)$\ 19.8^n +17$ is not a prime number.[/quote]
Here are answers to a)-c):[hide]
a) $\ 2^{2^{2n+1}}+3$ is divisible by 7;
b) $\ 2^{2^{4n+1}}+7$ is divisible by 11;
c) $\ 19.8^n +17$:
case n=2.k: divisibility by 3,
case n=4.k+1: divisibility by 13,
case n=4.k+3: divisibility by 5
sorry that the solution is not shown here, you can check the answers by induction.
[/hide]
and here for the main problem:
[hide]
$\ A=3^{3m^{2}+6n-61}+4$ - obviously that A is not divisible by 3.
$\ 3.A=3^{3m^{2}+6n-60}+12$
$\ 3.A=27^{m^{2}+2n-10}+12$
3.A is divisible by 13, hence A is divisible by 13.
equation A = 13 gives:
$\ m^{2}+2n=21$
answers: (m,n)= (1,10) or (3,6)
[/hide]

[quote="TRAN THAI HUNG"]You're right.And here is another one.Prove that:
$\ N=\frac{5^{125}-1}{5^{25}-1}$ is not a prime number[/quote]

[hide]
numerator and denominator are both divisible by 4, so the fraction, or $N$ must be divisible by 4.[/hide]

[quote="xxreddevilzxx"]
numerator and denominator are both divisible by 4, so the fraction, or $N$ must be divisible by 4. [/quote]
Not clear, why that makes the fraction  non prime.  In FACT  $N$ is NOT divisible by 4.

(If you want to know the smallest prime factor of $N$ happens to be:
 $3597751$
[hide]Rest of the factors being
 $28707251$ and $767186663625251$  and $24687045214139234043375683501$

Hint:  (But it will give away the answer) [hide] One can  factor $1+x^2+x^3+x^4$ if  $x$ can be written as $5n^2$[/hide] [/hide]

[quote="TRAN THAI HUNG"]Good idea,You're right :P.I have this problem , it doesn't belong to prime number but it's really interesting.
Let n be a integer satisfying that $\ a_n=\ (\frac{3+\sqrt{5}}{2})^n+\ (\frac{3-\sqrt{5}}{2})^n -2$
Find n satisfying $\ a_n$ is a square of another number.

Math is my life(':harhar:')[/quote]
[hide="is it right?"]
n is even number:
if n=2k:
$\ a_{2k}=\ (\frac{3+\sqrt{5}}{2})^{2k}+\ (\frac{3-\sqrt{5}}{2})^{2k} -2$
$\ a_{2k}=\ ( (\frac{3+\sqrt{5}}{2})^k-\ (\frac{3-\sqrt{5}}{2})^k )^2$
[/hide]

:juggle: No, n is a odd number but try to find the solution.

Math is my life :!:


The first and ninth terms of an arithmetic sequence are $ \frac23$ and $ \frac45$, respectively. What is the fifth term?


ratio
geometric series
geometric sequence

The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is 15, and the sum of the squares of the terms of this series is 45.  The first term of the series is
\[ \text{(A)}\ 12 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 2 \]

[quote="tarquin"]This is strange.. I keep getting 5. Can someone tell me where I mess up?
[hide]$a= 15(1-r)$ so $a^2 = 225(1 - 2r + r^2)$. By substitution, $225(r^2 - 2r + 1) = 45(1-r^2)$. Solving, $r$ can be $\frac 23$ or 1. Putting $\frac 23$ in, I can only get $5$ for the first term..[/hide][/quote]I am also getting five, but using a different solution method.  Can someone tell me the flaw in this method?[hide]We can express the geometric series as $\frac{a}{1-r}=15$.  Writing out the second as $a^{2}+a^{2}r^{2}+a^{2}r^{4}...$ we can see that this is an infinite geometric sequence with first term $a^{2}$ and common ratio $r^{2}$.  So we can write $\frac{a^{2}}{1-r^{2}}=45$.

Dividing this equation by $\frac{a}{1-r}=15$, $\frac{a^{2}}{1-r^{2}}\times\frac{1-r}{a}=3$, so $\frac{a}{1+r}=3$.
Take the inverses of the two equations we have found ($\frac{a}{1-r}=15$ and $\frac{a}{1+r}=3$) to find $\frac{1+r}{a}=\frac{1}{3}$ and $\frac{1-r}{a}=\frac{1}{15}$.  Add the equations together and we get $\frac{2}{a}=\frac{6}{15}$, so $a=5$.
[/hide]

Well, as you can see both C and D have the value of 3. So perhaps it was just a typing error?

One of them is probably supposed to be 5. :)


let$ a\plus{}b\plus{}c\equal{}3$ and $ a,b,c>0$ prove that

 $ \sum\frac{a^2}{a\plus{}2b^2}\geq1$

$ \sum \frac{a^2}{a\plus{}2b^2}\equal{}\sum \frac{a^4}{a^3\plus{}2a^2b^2} \geq \frac{(a^2\plus{}b^2\plus{}c^2)^2}{(a^3\plus{}b^3\plus{}c^3)\plus{}2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}\geq 1$.So it's enough to prove that :

$ a^4\plus{}b^4\plus{}c^4 \geq a^3\plus{}b^3\plus{}c^3$ $ \Leftrightarrow$ $ 3(a^4\plus{}b^4\plus{}c^4) \geq (a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c)$ ,which reduces to :

$ 2\sum a^4 \geq \sum (a^3b\plus{}ab^3)$ , which is just Muirhead.Or you can also prove this last inequality by AM-GM.

[quote="Svejk"]$ \sum \frac {a^2}{a \plus{} 2b^2} \equal{} \sum \frac {a^4}{a^3 \plus{} 2a^2b^2} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{(a^3 \plus{} b^3 \plus{} c^3) \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\geq 1$.So it's enough to prove that :

$ a^4 \plus{} b^4 \plus{} c^4 \geq a^3 \plus{} b^3 \plus{} c^3$ $ \Leftrightarrow$ $ 3(a^4 \plus{} b^4 \plus{} c^4) \geq (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)$ ,which reduces to :

$ 2\sum a^4 \geq \sum (a^3b \plus{} ab^3)$ , which is just Muirhead.Or you can also prove this last inequality by AM-GM.[/quote]
Can you explain more how did you find the first inequality!.

It's only Cauchy-Schwartz inequality ;) : I believe that you are familiar with the following one :
For $ a_k,b_k \in \mathbb{R}$ we have (this is Cauchy Schwartz) :
\[ (a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)(b_1^2\plus{}b_2^2\plus{}...\plus{}b_n^2) \geq (a_1b_1\plus{}a_2b_2\plus{}...\plus{}a_nb_n)^2\]

Now if you make the substitution : $ a_k: \equal{} \frac{a_k}{\sqrt{x_k}}$ and $ b_k: \equal{}\sqrt{x_k}$ ,where $ x_k>0$ $ k\equal{}\overline{1,n}$ we get :

\[ \left( \frac{a_1^2}{x_1}\plus{}\frac{a_2^2}{x_2}\plus{}...\plus{}\frac{a_n^2}{x_n}\right)(x_1\plus{}x_2\plus{}...\plus{}x_n) \geq (a_1\plus{}a_2\plus{}..\plus{}a_n)^2\]
which is equivalent to  :

\[ \boxed{\frac{a_1^2}{x_1}\plus{}\frac{a_2^2}{x_2}\plus{}...\plus{}\frac{a_n^2}{x_n} \geq \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)^2}{x_1\plus{}x_2\plus{}...\plus{}x_n}}\]
for $ a_k \in \mathbb{R} ,x_k>0 ,k\equal{}\overline{1,n}$

[quote="Svejk"][quote="rachid"]ohhhh,sorry i'm really stupid :lol: .[/quote]

It doesn't mean that if you haven't seen this trick before you are stupid.It's very usefull and you would better have it in mind for the future inequalities ;) .Good luck ![/quote]
No,because i usually use it :blush: .

or you can say $ \sum \frac {a^2}{a \plus{} 2b^2} > \equal{}  \frac {(a \plus{} b \plus{} c)^2}{a \plus{} b \plus{} c \plus{} 2\sum b^2} > \equal{}  \frac {9}{3 \plus{} 2*9/3} \equal{} 1$
only andreescu

Athinaios: Your last inequality is false, check it again  :wink:

[quote="Svejk"]$ \sum \frac {a^2}{a \plus{} 2b^2} \equal{} \sum \frac {a^4}{a^3 \plus{} 2a^2b^2} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{(a^3 \plus{} b^3 \plus{} c^3) \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\geq 1$.So it's enough to prove that :

$ a^4 \plus{} b^4 \plus{} c^4 \geq a^3 \plus{} b^3 \plus{} c^3$ $ \Leftrightarrow$ $ 3(a^4 \plus{} b^4 \plus{} c^4) \geq (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)$ ,which reduces to :

$ 2\sum a^4 \geq \sum (a^3b \plus{} ab^3)$ , which is just Muirhead.Or you can also prove this last inequality by AM-GM.[/quote]

We can prove the original inequality by using CS but without multiplying by $ a^2$

[quote="arqady"][quote="M.A"]let$ a \plus{} b \plus{} c \equal{} 3$ and $ a,b,c > 0$ prove that

 $ \sum\frac {a^2}{a \plus{} 2b^2}\geq1$[/quote]
The following inequality is true too:
For positive $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 3$ prove that:
\[ \frac {a^2}{2a \plus{} b^2} \plus{} \frac {b^2}{2b \plus{} c^2} \plus{} \frac {c^2}{2c \plus{} a^2}\geq1
\]
[/quote]

 $ \sum\frac {a^2}{2a \plus{} b^2} \equal{} \sum\frac {a^4}{2a^3 \plus{} a^2b^2}\geq\frac {(\sum a^2)^2}{2\sum a^3 \plus{} \sum a^2b^2}$ so we must prove that $ \sum a^4 \plus{} \sum a^2b^2\geq2\sum a^3$ or $ \sum a^4 \plus{} 3\sum a^2b^2 \geq 2\sum a^3b \plus{} a^3c$  and i think the last inequality is true

Of course the last is was true ;) .Because :
<=>$ \sum  (a\minus{}b)^4 \ge 0$ ;)

[quote="rachid"]I don't think [b]FOURRIER[/b],or it's better to show us your solution,thank's.[/quote]

homogenize then use CS  :wink:

I think, the following inequality harder:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201362
 :wink:


Do you work better in contests or in practice?

I work better in contests, because there is pressure to work hard.

Same, during practice I just kind of get lazy (except AIME practice) and miss easy questions. For instance, I missed a Sprint Round question on the school test, but got perfect scores on the chapter and state tests (obviously not on nationals though)

During practice.  I tend to choke severely at contests (state MathCounts place: 36, 45)

There ought to be an 'about the same' option. I do slightly better in practice with MathCounts, but as far as I can tell I perform similarly in practice and for real on AMC-8 and -10.

Well,

I haven't participate in a contest since 1998, but somehow I enjoy more the problem solving now when I have the oportunity and the calm to think about the problem and try to find not only a solution, but a good and elegant solution, even when I spend three of four hours in a single problem, but before I had to do it at least four times that fast, and made silly mistakes!

Best,

I think it varies for me.  On Mathcounts state in 7th grade, I did significantly better than my practice as I got a 38.  On other contests sometimes I do better than normal.  However, on Mathcounts chapter rounds, I do absolutely horrible, my score was worse than at state for 2 years even though state is supposed to be harder.  So for me, it really depends.

[quote="peter"]How can I add options to the poll?[/quote]

Maybe editing your original post will work.

I've added the extra option for you.  In general, you would be able to do it by editing your original post.

when i click edit on the original post, there is no place to add poll options

only mods/admins can create/edit/delete polls

I do best when I practice, which is really annoying!!!!!!!!!!

Take the AMC contests, for every problem, I always end up doing something screwy, but as soon as the test is over, I go ahead and solve most of the problems that I did not know how to solve during the test time.

it depends on what im doing
for mathcounts, i always bomb the real test, last year scoring in our chapter with a 31 ( :( ) but on AMC 10, i did about the same (140.5) as in the contest


For each real number $ x$, let $ f(x)$ be the minimum of the numbers $ 4x\plus{}1$, $ x\plus{}2$, and $ \minus{}2x\plus{}4$.  Find the maximum value of $ f(x)$.

Thanks in advance!

[hide="hint"]Break this problem into separate cases for the possible values of $ f(x)$.[/hide]

[hide="solution"]We shall break this problem into 3 cases; where each of the expressions is the smallest.

[b]CASE 1:[/b]  $ 4x \plus{} 1$ is the minimum
$ 4x \plus{} 1\leq x \plus{} 2\Rightarrow x\leq\frac13$
$ 4x \plus{} 1\leq \minus{} 2x \plus{} 4\Rightarrow x\leq\frac12$
Thus, $ f(x) \equal{} 4x \plus{} 1$ whenever $ x\leq\frac13$, so the maximum value in this case is $ 4\cdot\frac13 \plus{} 1 \equal{} \frac73$

[b]CASE 2:[/b]  $ x \plus{} 2$ is the minimum
$ x \plus{} 2\leq 4x \plus{} 1\Rightarrow x\geq\frac13$
$ x \plus{} 2\leq \minus{} 2x \plus{} 4\Rightarrow x\leq\frac23$
Thus, $ f(x) \equal{} x \plus{} 2$ when $ \frac13\leq x\leq\frac23$, so the maximum value in this case is $ \frac23 \plus{} 2 \equal{} \frac83$

[b]CASE 3:[/b]  $ \minus{} 2x \plus{} 4$ is the minimum
$ \minus{} 2x \plus{} 4\leq 4x \plus{} 1\Rightarrow x\geq\frac12$
$ \minus{} 2x \plus{} 4\leq x \plus{} 2\Rightarrow x\geq\frac23$
Thus, $ f(x) \equal{} \minus{} 2x \plus{} 4$ whenever $ x\geq\frac23$, so the maximum value in this case is $ \minus{} 2\cdot\frac23 \plus{} 4 \equal{} \frac83$

It follows that the maximum value of $ f(x)$ is $ \frac83$[/hide]

Algebra is cool and all, but the quickest way to do this problem is undoubtedly to draw a picture.  Unfortunately, I don't have any Asymptote skills, so someone else will have to do the dirty work.

[hide="The long algebraic way"]Find the common intersection points and test intervals:

1. $ 4x\plus{}1\equal{}x\plus{}2$ ==> $ x\equal{}\frac13$, so then $ 4x\plus{}1<x\plus{}2$ for $ x<\frac13$, reverse inequality sign otherwise.

2. $ x\plus{}2\equal{}\minus{}2x\plus{}4$ ==> $ x\equal{}\frac23$, so then $ x\plus{}2<\minus{}2x\plus{}4$ for $ x<\frac23$, reverse inequality sign otherwise.

3. $ \minus{}2x\plus{}4\equal{}4x\plus{}1$ ==> $ x\equal{}\frac12$, so then $ 4x\plus{}1<\minus{}2x\plus{}4$ for $ x<\frac12$, reverse inequality sign otherwise.

So for $ x<\frac13$, $ 4x\plus{}1<x\plus{}2<\minus{}2x\plus{}4$. For $ \frac13<x<\frac12$, $ x\plus{}2<4x\plus{}1<\minus{}2x\plus{}4$. For $ \frac12<x<\frac23$, $ x\plus{}2<\minus{}2x\plus{}4<4x\plus{}1$. For $ \frac23<x$, $ \minus{}2x\plus{}4<x\plus{}2<4x\plus{}1$. Thus,

\[ f(x)\equal{}\begin{cases}4x\plus{}1&\text{if }x<\frac13\\x\plus{}2&\text{if }\frac13\le x<\frac23\\\minus{}2x\plus{}4&\text{if }x\ge\frac23\\\end{cases}\]

Maximum occurs only on one of the horizontal lines or 'corners' of the graph of $ f$. Since $ f\left(\frac13\right)\equal{}\frac73$ and $ f\left(\frac23\right)\equal{}\frac83$, max is $ \frac83$.
[/hide]


How many strictly positive integer solutions (x, y, z) are there, such that
x + y + z = 100

$ \binom{102}{2}$ is the number of nonnegative solutions. This is smaller, but the same ideas work.

I think there are 5,151 combinations.

where x=0, y+z=100 | 101 combinations

where x=1, y+z=99 | 100 combinations

where x=2, y+z=98 | 99 combinations

therefore, x=n, y+z=101-n | 101-n combinations.

101
sum(x) = (101+1)(101/2)= 5151
x=0

That should cover all combinations becoz for each value of x, every possible value of y+z is computed, therefore for every value of y and z, x+y and x+z = 100 respectively.

Your thoughts?

[b]Lemma 1[/b] The number of positive integer solutions $ (a_1,a_2,...,a_n)$ to the equation $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} k$ where $ k\ge n$ is $ \binom{k \minus{} 1}{n \minus{} 1}$.
[b]Proof[/b] Consider $ k$ circles in one row. Then, if we put $ n \minus{} 1$ pluses between any two circles with at most one plus between two circles, then the groups of circles between the pluses, the start and the end will each contain at most one circle, will be disjoint, and in total the number of circles in the groups will be $ k$. So, for every valid placement of the pluses, we get a unique partition in groups, which will give us a unique solution in positive integers $ (a_1,a_2,...,a_n)$ with $ a_i$ being the number of circles in $ i$th group from left to right. The number of placements of pluses is $ \binom{k \minus{} 1}{n \minus{} 1}$ as there are $ n \minus{} 1$ pluses and $ k \minus{} 1$ spaces between consecutive circles. Then Lemma 1 follows.

Using lemma 1 for the given problem, the number of solutions will be $ \binom{100 \minus{} 1}{3 \minus{} 1} \equal{} \binom{99}{2}$ solutions.

Note that if we are looking for nonnegative integer solutions $ (a_1,a_2,....,a_n)$ we can reduce the problem to lemma 1 by adding $ 1$ to $ a_i$ for $ i \equal{} 1,2,...,n$ and then the number of required solutions will be equal to the number of positive solutions to $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} k \plus{} n$ so the number of solutions in this case will be $ \binom{n \plus{} k \minus{} 1}{n \minus{} 1}$.

How about if you could repeat numbers?

I tried this problem with n = 100 and k = 5 in the case where you repeat just two numbers.

In this case, there would be k - 2 pluses and n - x/2 - 1 places to position the pluses, where x = 1, 2, 3 ... n/2 - 3. Each instance of x gives a different combination, for example:
x = 1 corresponds to the case: 1 + 1 + 3 numbers that sum to 98.
x = 47 corresponds to the case 47 + 47 + 3 numbers that sum to 6.

Since each of these cases is a different combination, taking the sum would give total combinations if two numbers are repeated:

sum [ (n - 2x - 1) nCr (k - 2) ] where x = 1, 2, 3 ... n/2 - 3.

I tried to generalize this solution to any n and k and any amount of repeated numbers, but I'm having trouble with the end-points. For example, in the n = 100, k = 5, repeats = 2 case, x has to stop at n/2 - 3 because 6 is the smallest even number that can be expressed as the sum of three distinct numbers. Any suggestions?

Please ignore my last message.

Here's a more challenging problem: What if all k numbers had to be distinct?

In either case you could generate fns (Multinomial theorem)
 :wink:


geometry
3D geometry
real analysis
real analysis unsolved

Let $\{U_{1},U_{2},\dots \}$ be a cover of $\mathbb{R}^{n}$ by open sets. Prove that there is a cover $\{V_{1},V_{2},\dots \}$ such that $1.V_{j}\subset U_{j}$ for each $j$ $2.$each compact subset of $\mathbb{R}^{n}$ is disjoint from all but finitely many of the $V_{j}$.
--------
Here is a discussion of two students on that problem, what do you think about it?

Student$1$: A natural approach is to make the given covering minimal in the sense that if $U_{i}\subset \bigcup_{k=1}^{\alpha}U_{k}$ then we replace $U_{i}$ with the empty set.

Student$2$: Hum, how do you know that the second condition is satisfied? In my opinion this method works only in $\mathbb R$...

what solution is there to the original question?

A dyadic interval is an interval of the form $[(k-1)2^{m},k2^{m})$, where $k,m\in\mathbb Z$. A dyadic cube is a product of dyadic intervals with the same length $2^{m}$. Two dyadic cubes are either disjoint or one of them contains another. In order to cover $\mathbb R^{n}$ by open cubes, we rescale each dyadic cube by $1+\epsilon$ and make it open (let's call such cubes $Q_{j}$). For each $x\in\mathbb R^{n}$ choose $U_{i}$ that contains $x$ and then choose $Q_{j}$ so that $x\in Q_{j}\subset U_{i}$. We get a new open cover, for which the idea of Student 1 works.

See also: [url=http://en.wikipedia.org/wiki/Paracompact]Paracompact space[/url]

Do you also agree that the method of the first student works in $\mathbb R$?

If the sets are intervals (which is not much of a restriction).


What is the value of $ 39^{2} \plus{} 2 \cdot 39 \plus{} 1$?

hello, we get after the binomial formula
$ 39^2\plus{}2\cdot39\plus{}1\equal{}(39\plus{}1)^2\equal{}40^2\equal{}1600$.
Sonnhard.


function
number theory unsolved
number theory

let n be natural number and A be set of divisors of n and f:A...A  be fuction that for a,b element of A and if a divides b then f(a)divides f(b)prove that there is m in A subject that f(m)=m

only positive divisors or both pos and neg ?

I guess you excludes $1$ and $n$ from the set of divisors? because it highly trivializes the problem

Anyway, just consider the number $a$ from $A$ which divides the most of other divisors. Suppose $\forall _m f(m) \not = m$ so $f(a)$ has the same number of divisors as $a$. Analogically $f(f(a))$ has the same number of divisors as $f(a)$. Continuing this we see, as there is no $m$ for which $f(m)=m$, that for some $k$: $f_k(a)=a$ ($k$-th iteration). Contradiction.


[quote="jinmadaxia"]$ a,b,c>0$,and $ a^2\plus{}b^2\plus{}c^2\equal{}1$, proof
$ (1\minus{}ab)^{\minus{}1}\plus{}(1\minus{}bc)^{\minus{}1}\plus{}(1\minus{}ca)^{\minus{}1} \leq 9/2$[/quote]

It's the same as [url=http://www.mathlinks.ro/viewtopic.php?t=312356]this one[/url], there it's just for four variables...

And sorry, but is it that difficult to put dollar signs and write \leq instead of <=? It's not that hard I think...but it looks many times better...thanks :)

[quote="jinmadaxia"]a,b,c>0,and a^2+b^2+c^2=1,proof
(1-ab)^-1+(1-bc)^-1+(1-ca)^-1<=9/2[/quote]

$ \Longleftrightarrow \sum{\frac {a^2 \plus{} b^2 \plus{} c^2}{a^2 \plus{} b^2 \plus{} c^2 \minus{} ab}}\le\frac {9}{2}$

$ \Longleftrightarrow \sum{a^4(a \minus{} b)(a \minus{} c)} \plus{} 3abc\sum{a(a \minus{} b)(a \minus{} c)} \plus{}$ $ 2abc\sum{(b \plus{} c)(a \minus{} b)(a \minus{} c)} \plus{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$ $ \plus{} (a^2 \plus{} b^2 \plus{} c^2)\sum{(a \minus{} b)^4}\ge0$

true.

[quote="spanferkel"]See also here:
http://www.mathlinks.ro/viewtopic.php?t=109029[/quote]

thanks...
good topic...

Using my method from [url=http://www.mathlinks.ro/viewtopic.php?t=312356]this[/url]topic, we can easily prove the following:

If $ x_1, x_2, \ldots, x_n$ are $ n$ positive real numbers satisfying $ \sum_{i\equal{}1}^n x_i^2\equal{}1$, then the inequality
\[ \sum_{1\leq i <j \leq n} \frac{1}{1\minus{}x_ix_j}\leq \frac{n^2}{2}\]
holds.


trigonometry logarithms induction function quadratics algebra polynomial

[color=darkred]For $ n\in\mathcal N^*$ and $ \Delta\equiv a^2\minus{}4b<0$ prove that :[/color]

$ 1\blacktriangleright (\arctan)^{(n)}\equal{}(\minus{}1)^{n\minus{}1}(n\minus{}1)!\cdot \frac {\sin\left(n\cdot\arctan\frac 1x\right)}{\sqrt {\left(x^2\plus{}1\right)^n}}$ .

$ 2\blacktriangleright\left[\ln \left(x^2\plus{}1\right)\right]^{(n)}\equal{}2(\minus{}1)^n(n\minus{}1)!\cdot\frac {\cos\left(n\cdot \arctan\frac 1x\right)}{\sqrt {\left(x^2\plus{}1\right)^n}}$ .

$ 3\blacktriangleright\ \boxed {\ \left(\frac {1}{x^2\plus{}ax\plus{}b}\right)^{(n)}\equal{}\frac {2(\minus{}1)^nn!}{\sqrt {4b\minus{}a^2}}\cdot\frac {\sin\left[(n\plus{}1)\arctan\frac {\sqrt {4b\minus{}a^2}}{2x\plus{}a}\right]}{\sqrt {\left(x^2\plus{}ax\plus{}b\right)^{n\plus{}1}}}\ }$ .

Hmmm... Once the answers are given, a simple induction combined with some elementary trigonometry does the job. Another approach that does not require knowing the right hand side in advance is to differentiate a rational function with a quadratic polynomial in the denominator using its decomposition into simple fractions with complex coefficients. We'll get something like $ \frac 1{(x\minus{}z)^n}\plus{}\frac1{(x\minus{}\bar z)^n}$, which can be rewritten as $ \frac {2\cos[n\operatorname{arg}(x\minus{}z)]}{|x\minus{}z|^n}$ (this is if we need the real part; if we need the imaginary part, it'll be $ \sin$). This approach seems to take care of all three formulae in a uniform way.

at first I was frightened when I saw right side of equation $ ...(arctan \frac {1}{x})..$  hence my intuition decided all this to replace $ y \equal{} arctan \frac {1}{x}$ but I'm not sure will it work or not (now I have no time to check, but I think it will work because $ x \equal{} \frac {1}{tan y}$ and
$ 1 \plus{} x^{2} \equal{} 1 \plus{} \frac {1}{tan^{2}y } \equal{} \frac {1}{\sin^{2}}$ ...)

For complex numbers $ z\equal{}x\plus{}iy$ with $ x,y>0$

$ \left(\frac{d}{dx}\right)^n \left[ \frac12 \ln(x^2\plus{}y^2)\plus{}i \left(\frac{\pi}{2}\minus{}\arctan \frac{x}{y}\right) \right]$

$ \equal{}\left(\frac{d}{dx}\right)^n \ln(x\plus{}iy)\equal{}\left(\frac{d}{dz}\right)^n \ln z\equal{}\frac{(\minus{}1)^{n\minus{}1} \, (n\minus{}1)!}{z^n}$

$ \equal{}\frac{(\minus{}1)^{n\minus{}1} \, (n\minus{}1)!}{\sqrt{x^2\plus{}y^2}^n} \, \underbrace{\exp\left(\minus{}i\, n\arctan \frac{y}{x}\right)}_{\cos(...)\minus{}i\sin(...)}$

Compare imaginary and real part:

[b](1)[/b] $ \blacktriangleright \left(\frac{d}{dx}\right)^n \arctan \frac{x}{y}\equal{}\frac{(\minus{}1)^{n\minus{}1} \, (n\minus{}1)!}{\sqrt{x^2\plus{}y^2}^n}\, \sin\left(n\arctan \frac{y}{x} \right)$

[b](2)[/b] $ \blacktriangleright \frac12 \cdot \left(\frac{d}{dx}\right)^n \ln(x^2\plus{}y^2)\equal{}\frac{(\minus{}1)^{n\minus{}1} \, (n\minus{}1)!}{\sqrt{x^2\plus{}y^2}^n}\, \cos\left(n\arctan \frac{y}{x} \right)$

$ f(x)\equal{}\frac{1}{x^2\plus{}ax\plus{}b}\equal{}\frac{2}{\sqrt{4b\minus{}a^2}} \, \frac{d}{dx} \arctan \frac{2x\plus{}a}{\sqrt{4b\minus{}a^2}}$

$ f^{(n)}(x)\equal{}\frac{2}{\sqrt{4b\minus{}a^2}} \, \left(\frac{d}{dx}\right)^{n\plus{}1} \arctan \frac{2x\plus{}a}{\sqrt{4b\minus{}a^2}}$

$ \equal{}\frac{2}{\sqrt{4b\minus{}a^2}} \, \frac{(\minus{}1)^n \, n!}{\sqrt{\left(x\plus{}\frac{a}{2}\right)^2\plus{}\left(\frac{\sqrt{4b\minus{}a^2}}{2}\right)^2}^{n\plus{}1}} \sin\left( (n\plus{}1)\, \arctan\frac{\sqrt{4b\minus{}a^2}}{2x\plus{}a}\right)$


I am trying to find all solutions in the interval 0 deg<x<360 deg, for cosx-2sinxcosx=0.  It is making me very mad! :mad:  The answers are 30deg, 90 deg, 150 deg, and 270 deg, but I don't know how to get them.

$\cos x-2\sin x\cos x=\cos x(1-2\sin x)=0.$  By the zero product property either $\cos x=0$ or $\sin x=\frac 12.$

Think you can get it from there?


$ a_{1}\equal{}20$

$ a_{2}\equal{}30$

$ a_{n\plus{}2}\equal{}3a_{n\plus{}1}\minus{}a_{n}$

proof:n=3 is the [b]only[/b] answer when $ 1\plus{}5a_{n}a_{n\plus{}1}$ is a perfect square.

[quote="wxydx00"]$ a_{1}\equal{} 20$

$ a_{2}\equal{} 30$

$ a_{n\plus{}2}\equal{} 3a_{n\plus{}1}\minus{}a_{n}$

proof:n=3 is the [b]only[/b] answer when $ 1\plus{}5a_{n}a_{n\plus{}1}$ is a perfect square.[/quote]

$ 1\plus{}a_{n}a_{n\plus{}1}\equal{}(a_{n\plus{}1}\plus{}a_{n})^{2}\plus{}501$.

And if $ n\ge 250$ then $ n^{2}\plus{}501$ isn't a perfect square .And $ a_{n\plus{}1}>a_{n},n\in\mathbb{N}$ so $ a_{n}\plus{}a_{n\plus{}1}>250$ for $ n\ge 4$.

Try $ n\equal{}1,2,3$ we have $ n\equal{}3$ such that the condition  :wink: .

It was posted before ( by me  :D ) on N.T.TUAN'weblog :

http://trungtuan.wordpress.com/2007/08/19/two-lemmas-on-linear-sequences-of-order-two/.

By English, not Vietnamese  :roll:

[quote="Rust"]Exactly $ 1\plus{}a_{n}a_{n\plus{}1}\equal{} b_{2n\plus{}1}^{2}\plus{}501$, were $ b_{2n\plus{}1}\equal{} (15\minus{}5\sqrt 5 )(\frac{1\plus{}\sqrt 5}{2})^{2n\plus{}1}\plus{}(15\plus{}5\sqrt 5 )(\frac{1\minus{}\sqrt 5 }{2})^{2n\plus{}1}$ ($ a_{n}\equal{} b_{2n}$).[/quote]

We don't needed it, because from my express, we can prove this problem .


Let r are real number such that $ r \in [0;3]$. Prove that the inequality
$ \sum x^4\plus{}(1\plus{}r)\sum x^2yz \plus{}2r\sum x^2y^2\ge (1\plus{}r)\sum xy(x^2\plus{}y^2)$
is true for all real numbers x,y,z.
When does equality occurs? :D

Not any one can prove the inequality? Is it really hard? :lol:

This is my solution:
Setting $ r\equal{}1\plus{}2p$, we have $ p\in [\minus{}\frac 12;1]$.We need to prove that
$ \sum x^4\plus{}2(1\plus{}p)\sum x^2yz\plus{}2(1\plus{}2p)\sum x^2y^2\ge 2(1\plus{}p)\sum xy(x^2\plus{}y^2)$
Because $ p\in [\minus{}\frac 12;1]$, exist $ a,b,c\in R$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}1, ab\plus{}bc\plus{}ca\equal{}p$.
Thus, we need to prove
$ \sum a^2\sum x^4\plus{}2(\sum a^2\plus{}\sum ab)\sum x^2yz\plus{}2(\sum a^2\plus{}2\sum ab)\sum x^2y^2\minus{}2(\sum a^2\plus{}\sum ab)\ge 0$
This inequality can be written in the form
$ \sum_{cyc} [ax^2\plus{}by^2\plus{}cz^2\minus{}(a\plus{}b)xy\minus{}(b\plus{}c)yz\minus{}(c\plus{}a)xz]^2\ge 0$
which is true for all real numbers x,y,z.
So, a solution is nice for me. :winner_first: And you, an idea?
(I think that solution can wrong, so I very want an idea of you) :wink:


If you had to chose, which one would you work on?

I'd choose a National Sprint Round, because those are the problems that I want to focus on for this point on in time, for my third and final year at MATHCOUNTS. Unfortunately, I asked my coach today at math club and he couldn't find any other than stupid old school and chapter sprint problems.

Looking at the results, it seems as though team rounds are not very popular...

Sprints are usually better than Targets for practice (not that Targets are bad; they are essential, just not as essential). After all, they are worth more points...

Sprint rounds have more problems, so yeah....

[quote="Ihatepie"]national, it gives a better challenge.

(I'm training for AMC 10 now anyways)

Mathcounts Nats is harder than AMC 10 imo.[/quote]
Why do you think MC Nats is harder than AMC 10?
But AMC 10 is for high school kids, lmao.

I actually do better on amc10's than on nationals sprint because you get almost double the time to do 5 less problems. If I were given like 10-15 more minutes to do the sprints, then I would get a near perfect score every time. Sprint is sooo evil because afterwards you realize that you could do all the problems but the 40 minute time limit just kills you. :mad:

That's why I usually do better when I do Targets first--which is ALMOST never.

[quote="math154"]AMC10 gives you much more time per problem, and AMC10 is for high school, but it's more like, say, a MOEMS for high schoolers (except a bit harder). OK, I don't know if that makes sense, but MC Nats IS harder.[/quote]

To build on that, it's beginning to have the same difficulty as the AMC 12. Consider: 126 AMC 12 score and 33 Nats score.

I agree that the difficulties are more or less the same. However, AMC12 still has harder concepts than MC. It's the [b]speed[/b] attribute that makes MC nats hard; not necessarily the concepts.

Yes, I agree with most of you. I chose national sprint round only because there is more problems, it affects your individual score the most, and I just like to take it.

[quote="math154"]AMC10 gives you much more time per problem, and AMC10 is for high school, but it's more like, say, a MOEMS for high schoolers (except a bit harder). OK, I don't know if that makes sense, but MC Nats IS harder.[/quote]
Why would you say MC Nats is way harder? I still don't get it...

I didn't say substantially harder. Look at my last post.

AMC 10 has some really easy problems like MOEMS level, but Nationals at MC usually has mostly very difficult questions.

If you had to choose between these three, which one would you say was the hardest, and also which one would you rather work on;

National Sprint Round
AMC 10
AMC 12

This is precisely why I wrote a NATIONAL level mock test :D
Because (almost countably infinite) everybody voted either National sprint or National target :D

Good choice, man. That correlates with the popularity of Nats individual rounds.


[quote="aznmonkey1992"]if a is decreased by 10 % and b is increased by 10 %, they will be the same.
whats the ratio of a to b?[/quote]

[hide]
$\frac{9a}{10}=\frac{11b}{10}$

$9a=11b$

$9:11$[/hide]

Tread carefully, getting quite close to requiring brains here.

[quote="B4k4ka"][quote="aznmonkey1992"]if a is decreased by 10 % and b is increased by 10 %, they will be the same.
whats the ratio of a to b?[/quote]

[hide]
0.9a = 1.1b
9a = 11b

ratio is 11/9

[/hide][/quote]

I don't think that your answer is right, because 11/9 is b:a, but he asks for a:B.

[quote="math92"]where is .9 and 1.1 coming from?[/quote]

When a is decreased by 10%, .10 is taken away from the number (which we assume now is just 1, since that's 100%). 1- 0.10 = 0.9.

Similarly, when b is increased by 10%, .1 is added to 1 (which is the whole), giving us 1 + .1 = 1.1.


probability
geometry
geometric transformation
dilation
number theory
prime numbers

I don't know how to approach these, and some of them I'm getting the wrong answer.  I would appreciate it if someone could post some solutions.

24.) How many ordered triples of three prime numbers exist for which the sum of the members of the triple is 24?

27.) A bowl contains fewer than 50 marbles, and each is either red, green, or blue.  The probability of drawing a red marble is 2/5, and the probability of drawing a green marble is 3/7.  If two marbles are drawn without replacement, what is the probability that both are blue?  Express your answer as a common fraction.

28.) Brent likes to dilute his lemonade.  He starts with a full cup and drinks 1/3 of its contents.  He then fills the cup with water, stirs the contents, and again drinks 1/3 of its contents.  He repeats this process until he has consumed one cup of liquid.  What part of the original cup of lemonade remains in the cup?  Express your answer as a common fraction.

[hide="Correct Answers"]

24.) 15

27.) 3/119

28.) 8/27

[/hide]

[hide]27. Since there are only three types of marbles and the probability of drawing a red is 2/5 and the probability of drawing a green is 3/7 the probability of drawing a blue is 1-(2/5+3/7) or 6/35. Now the probability that both the marbles are blue is 6/35*5/34 or 3/119

28. He drinks 1/3 of the content leaving 2/3rds lemonade of the previous time. Since he never adds any more lemonade to the mixture we aren't concerned with the total amount of liquid. It takes him 3 turns to drink a whole glass so 2/3(2/3(1-1/3))  so now with that we get 8/27. [/hide]

[hide="hint for 24"]Consider the parity of the numbers.[/hide]

[hide="24"]
since the three numbers add up to 24, one must be even.
the only even prime is 2.
so now we need to find 2 primes that sum to 22
one pair is 11, 11. another is 17, 5. the last is 19, 3
however, the problem says ORDERED, so the triples must be counted for specific orderings.
for $(2, 11, 11)$ there are 3 orderings
for $(2, 5, 17)$ there are 6 orderings and
for $(2, 3, 19)$ there are 6 orderings
a total of $15$ ordered triplets(or whatever you call them)[/hide]


[quote="weatherbottle"]Source: Massachusetts Mathematics League March 2003

The remainder when $5(2x-1)^3+3(3x-2)^2 +k(x+5) + 7$ is divided by $x-1$ is 3. Find $k$.[/quote]

[hide]Is the answer -3?[/hide]

[hide]When expanded, $kx + 5k$ will only effect the first degree and zero degree terms of the expression. So...

Use synthetic division (with $1$) to find the relation between these terms.

You can set up the equations as

$y=$hole in synthetic division

$1+k=y$
$5k+y+14=3$

You can solve to get... $k=-2$ [$y=-1$][/hide]


I want you to say the name of the best movie that you have ever seen.(and if you can explain about the film,it will be so nice.) :D

[quote="José"][quote="jensen"]i will be happy to answer this question first.
One of the best movies that i have ever seen is  $Artificial$ $intelligence$.[/quote]

My TOP 3:

1)"9 reinas".
2)Truman Show.
4)Saw I; Saw II.[/quote]
2)Truman Show
i think it is very very good,with incredible playing of $JIM$ $CARREY$.
but i didn't see the others. :D

[quote="José"][quote="jensen"][quote="Chocoboom"]i agree completely with truman show and saw 1-2. best movies[/quote]
so i think i should see saw1,2. :D[/quote]

Jajaja...

About "9 reinas": It's an argentinian movie, considered the best film of the 90's... it's extremely cool... if you know spanish, try to watch it!!!!!![/quote]
if i don't know spanish,then what should i do? :?:


vector
linear algebra
matrix
linear algebra unsolved

Not quite related to http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tr&t=97713, but most likely classical, although I could't find it on this forum.

Suppose that $V$ is a vector space of dimension $n$ over a zero characteristic field and $T$ is an operator on $V$ s.t. $\forall k \in \{0,1,\dots , n-1 \}, \mbox{Tr}(T^{k})=0$, but  $\mbox{Tr}(T^{n})\not=0$. Prove that $T$ is diagonalizable.

What is the field is not of zero characteristic?

the complex case, other algebraicly closed fields must work the same:
the eigenvalues $x_{1}, \ldots, x_{n}$ satisfies without loss of generality
$x_{1}+x_{2}+\ldots+x_{n}= 0$
$x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}= 0$
$\ldots$
$x_{1}^{n}+x_{2}^{n}+\ldots+x_{n}^{n}= n$
Hence it is easy to compute the symetric sum $\sigma_{1}=x_{1}+x_{2}+\ldots+x_{n}$, $\sigma_{2}= x_{1}x_{2}+x_{1}x_{3}+\ldots+x_{n-1}x_{n}$, etc .. and see that the eigenvalues are the roots of $x^{n}-1=0$. Hence the eigenvalues are distinct, the matrix is diagonalizable.

PS: $\left(\begin{matrix}0 &0 &1 \\1 &0 &0 \\0 &1 & 0 \end{matrix}\right)$ is not diagonalizable over $\mathbb{R}$


USAMO
trigonometry
iscosceles triangle
Law of Cosines
Law of Sines
reflection
geometry

hey,i got the soln to the geometry problem.in the triangle reflect M across AC.(that's the hint)well,i'm in a hurry.so i'll post the full soln later.for the time being here's the final equation  that i got
 2cos10cos(60+x)=sinx  where x= angle MCB.evidently x=20 is a soln.i havent bothered to find the others(if any).therefore it turns out AB=BC.
Regards
vineet

yeah ... i know we aren't suppose to know this, but :D ...
consider the function f(x)=2 cos10 cos60+x- sin x
f'(x)=-2cos 10 sin 60+x - cos x and because x can be at most 80, we get that f'(x)<0 for x \in (0,80). Thus it is one-to-one, so it can only have one solution ... that is the one you found vineet.

waiting for the rest of the solution ... :)
[hide="Remark"]"This problem has been given to me by a young promising Romanian student, Gabriel Kreindler" was the text that started this post.
This was the first problem ever posted and discussed on MathLinks.[/hide]

well,the soln i 'll give here is different from my original solution as it doesnt need the reflection tool.apply "trig ceva" with angle MCB as "x"
we get sin20.sinx.sin40 = sin10.sin(80-x).sinsin30 which is equivalent to
sin (80-x) = 4sinx.sin40.cos10 now the identity 2sina.cosb=sin(a-b) + sin(a+b) yields sin(80-x) = 2sinx.(sin 30 + sin50) which implies
 sin(80-x) = sinx +2sinxcos40  therefore 2sinxcos40 = 2sin(40-x)cos 40.
rest is obvious.
my previous soln was a hybrid of pure geometry and trigonometry (more of the latter).actually i got a third soln  without trigonometry.if neone's willing to know abt it,i 'll post it.
regards
vineeet

yes vineet i'm quite curious ... i haven't gave this problem much tought, but anyway i'd like to see a purely geometric solution ( i'm in for the nicer solutions :) .. although generally i give ugly ones :D )

ya here is the solution using pure geo.
reflect the point A across BM and name the point D.than triangle AMD is isosceles with vertex angle 2 (180 - <BMA) = 2(<MAB + <ABM) = 60. and hence is equilateral.also <DBA = 2<MBA = 40 and since <BAC = 50 we have DB perpendicular to AC.let E be the intersection of DB with CM.then
<MED = 180-<CED=180 - (90-<ACE) = 90 +30 =120.therefore<MED +<DAM = 180.therfore quad. AMED is cyclic.therfore <DEA =<DMA = 60.we note that <DEC=<DEA=60.since AC perpendicular to DE theforewe see  that A and C are symmetric across the line DE which implies BA=BC.
i hope there iz nothing wrong with this proof
bye

ups ... i was going through some of my materials today and i happened to find this problem as being question no.4 on the 25th USAMO, held on May 2nd, 1996 :)

[hide="Nothing but Laws of Sines and Cosines and trig"]Assume all angles are in degrees. By the Law of Sines on $\triangle AMB$, $\frac{AM}{\sin20}=\frac{AB}{\sin{150}}$, so $AM=2AB\sin20$. By the Law of Sines on $\triangle AMC$ (:)), $\frac{AM}{\sin30}=\frac{AC}{\sin110}$, so $AM=\frac{AC}{2\sin110}$. Combining these, $2AB\sin20=\frac{AC}{2\sin110}$, so $\frac{AC}{AB}=4\sin20\sin110=4\sin20\sin70=4\sin20\cos20=2\sin40$, since $\sin2\theta=2\sin\theta\cos\theta$. By the Law of Cosines on $\triangle ABC$, $BC^2=AB^2+AC^2-2ABAC\cos50=AB^2+AC^2-ABAC(2\sin40)$, which is (by substitution)$AB^2+AC^2-ABAC\frac{AC}{AB}=AB^2+AC^2-AC^2=AB^2$, so $BC=AB$, as desired.[/hide]

[hide="Here is a solution similarly with the Vineet-s."]
[size=100][color=darkblue]Denote $m(\widehat {CBM})=x$. Thus, $m(\widehat {BCM})=80^{\circ}-x$.

I shall aplly the [u]Ceva's theorem[/u] ([u]in the trigonometrical form[/u]):[/color]

$\sin 10^{\circ}\sin x\sin 30^{\circ}=$ $\sin 40^{\circ}\sin 20^{\circ}\sin (80^{\circ}-x)$ $\Longleftrightarrow$

$\sin 10^{\circ}\sin x=$ $2\sin 40^{\circ}\sin 20^{\circ}\cos (10^{\circ}+x)$ $\Longleftrightarrow$

$\sin 10^{\circ}\sin x=$ $\left(\cos 20^{\circ} -\frac 12\right)\cos (10^{\circ}+x)$ $\Longleftrightarrow$

$2\sin 10^{\circ}\sin x=$ $2\cos 20^{\circ}\cos (10^{\circ}+x)-\cos (10^{\circ}+x)$ $\Longleftrightarrow$

$\cos (10^{\circ}-x)-\cos (10^{\circ}+x)=$ $\cos (30^{\circ}+x)+\cos (10^{\circ}-x)-\cos (10^{\circ}+x)$

$\Longleftrightarrow$ $\cos (30^{\circ}+x)=0$ $\Longleftrightarrow$ $x=60^{\circ}$ $\Longleftrightarrow$ $A=C=50^{\circ}$.

[color=darkred][b]Very nice the Vineet's pure syntetical solution[/b] ! Now and I am finding an another pure (i.e. non metrical) syntetical solution for this nice problem.[/color][/size][/hide][size=200]\[ \mathrm {A\ happy\ New\ Year,\ Mathlinks\ !} \][/size]

we put :    x=<MCB  tHEN : <MBC=80-x
then we use  SEVA theorem and write this:
(sin<MAB/sin<MAC).(sin<MCA/sin<MCB).(sin<MBC/sin<MBA)=1
====>   (sin10/sin40).(sin30/sinx).(sin(80-x)/sin20)=(sin(80-x))/4sin40cos10sinx
therefor we understand this:  sin(80-x)=2sinx.(2sin40cos10)=2sinx.(sin50+sin30)=sinx.(2cos40+1)
====>so:  2sinx.cos40=sin(80-x)-sinx=2sin(40-x)cos40
at the end we understand that  :sinx=sin(40-x)====> x=40-x ====>
<ACB=30+x=50 ===> x=20 ===><ACB=<BAC=50
THEN WE UNDERSTAND THAT THIS triangle is isosceles.
sorry i dont know how i use LATEX ;)

A solution with a 18 edges regular polygon without trigonometry .

[img]http://img107.imageshack.us/img107/5717/fig1al4.jpg[/img]

Domi

Sorry for my poor english  :blush:

Solution with an incenter:
Reflection of AB in AM cuts CM at D. Triangle ADC is D-isosceles with <DAC = <ACD = 30°. Perpendicular bisector of AC through D cuts AB at B'.
AM, DM bisect <DAB=20°, <B'DA = 120° => B'M bisects <AB'D = 40° => <MB'A = 20° => B' is identical with B => triangle ABC is B-isosceles.

There is a definite connection of this problem with Virgil's
   slicing problem III  (What happened to  I and II though ?)

   http://www.mathlinks.ro/Forum/viewtopic.php?t=179544

   Both problems deal with triangles having internal angles
   50,60,70.

   It seems that arith. prog. of angles is the key to solution.

   Thanks.
   M.T.

Take D the circumcenter of triangle AMC, see AMD equilateral, AD and BM perpendicular, there4 D is the reflection of A across BM, giving <ABD=40 deg, which makes BD and AC perpendicular, then remember AD=DC (D-circumcenter), this makes BD perpendicular bisector of AC and AB=BC.

Best regards,
sunken rock

That's an awesome solution, Domi, with the 18 sided regular polygon. How did you figure it out? That is a good method to create hard problems too.
I just found this website which discusses it: [url]http://www.cut-the-knot.org/triangle/TrigCeva.shtml[/url].
Using the applet on the website, here is another problem I created in case anyone is interested: In triangle ABC, angle ACB=30, angle ABC=110, and M is a point outside the triangle. Angle ACM=30 and angle MBC= 40. Find angle MAC. (You can use the applet to solve it).

[quote=pinkpig][quote=ThinkThink]Isn't this the oldest thread in AoPS?[/quote]

The oldest thread on AoPS got deleted. This is the second oldest.[/quote]

Really? What was the oldest thread about?

Nice problem(third USAMO geo ive solved lol)

Let $\angle{MBC}=\theta,\angle{MCB}=(80-\theta)$, every angle here is in degree.

Using trig Ceva, we can get the equation that $\frac{\sin20}{\sin40}*\frac{\sin10}{\sin20}*\frac{\sin\theta}{\sin(80-\theta)}=1$.

Now we look at three different cases, $AB=AC;AC=BC;AB=BC$

$(1)AC=AB$, $\theta=45$ and this cant work.
$(2)AC=BC$, $\theta=30,\angle{MCB}=50$. this cant work as well since $\frac{\sin10}{\sin20}*\frac{\sin50}{\sin20}\neq 1$
$(3),AB=BC,\theta=60$, we have to prove that $4\sin{20} \sin{20} \sin{80}=\sin{60}$

Write that $4\sin{(30-10)}\sin{(30+10)}\cos{10}=4(\frac{\cos^2{10}}{4}-\frac{3\sin^2{10}}{4})\cos{10}=4\cos^3{10}-3\cos{10}=\cos{30}=\sin{60}$ as desired

My solution using a Trigonometric Trick:

[color=#f00][b]A Trigonometric Trick:[/b][/color] The relation $x+y = a+b < \pi $ is given, provided that $x$, $y$, $a$, $b$ are positive angle measures. If $\dfrac{\sin x}{\sin y} = \dfrac{\sin a}{\sin b}$ then, $x=a$ and $y=b$. (Since it is easy to prove, I will not write it here. I can add the proof if requested.)

Let $\angle MBC = x$, $\angle MCB = y$. $x+y = 80^\circ$. By trigonometric Ceva, $$ \sin x \cdot \sin 30^\circ \cdot  \sin 10^\circ =  \sin y  \cdot \sin 40^\circ \cdot  \sin 20^\circ .$$
Easily we can prove that $$ \sin 20^\circ \cdot  \sin 40^\circ \sin 80^\circ = \dfrac{\sqrt{3}}{8} $$

Hence, $$ \dfrac{\sin x}{\sin y} =   \dfrac{\sin 60^\circ}{\sin 20^\circ}  $$

By [b][color=#f00]the trigonometric trick[/color][/b], $x=60^\circ$ and $y=20^\circ$. Thus, the triangle $ABC$ is isosceles.

(Lokman GÖKÇE)

[quote=ThinkThink][quote=veranikaia]I don't know. It could be.. it took place long before I was born.[/quote]

lol same[/quote]
we don't know. the first post might be the welcome to aops or sth, but it's deleted
This is the oldest thread still on aops today

Lol, looks like the mindset of constructing a bunch of points to make random isosceles triangles worked. Why are there so many trig solutions :noo:

By angle chasing, the following constructions are possible:
[list]
[*] $P$ on $\overline{AB}$ satisfying $AP=PM=MB$
[*] $Q$ on $\overline{AC}$ such that $\triangle MPQ$ is equilateral
[*] $R$ on $\overline{CM}$ such that $\triangle MQR$ is equilateral
[*] $S=\overline{AC} \cap \overline{BR}$ such that $QR=QS$ and $SC=SR$.
[/list]
Since $\angle MBR=\angle MRB=\angle CRS=\angle RCS$, $BMSC$ is cyclic. Thus, $\angle BCM=\angle BSM=\frac{\angle MQR}{2}=30^\circ$, so $\angle ACB=70^\circ$ and $\angle ABC=70^\circ$, as desired.

[quote=CyclicISLscelesTrapezoid]Why are there so many trig solutions :noo:
[/quote]

The AMSP book has a trig solution, and most people doing olympiad problems have the AMSP books.

Let $\angle MBC=x$, then $\angle MCB=80^\circ-x$. By Trig Ceva:
$$\frac{\sin\angle MBA}{\sin\angle MBC}\cdot\frac{\sin\angle MCB}{\sin\angle MCA}\cdot\frac{\sin\angle MAC}{\sin\angle MAB}=\frac{\sin20^\circ}{\sin x}\cdot\frac{\sin(80^\circ-x)}{\sin30^\circ}\cdot\frac{\sin40^\circ}{\sin10^\circ}=1.$$
Let $f(x)=\frac{\sin20^\circ}{\sin x}\cdot\frac{\sin(80^\circ-x)}{\sin30^\circ}\cdot\frac{\sin40^\circ}{\sin10^\circ}$, we notice that $f'(x)>0$ so at most one value of $x$ works.
But since a triangle exists with $x=60^\circ$, the triangle is isosceles with $\angle BCA=\angle BAC$.

[quote=Valentin Vornicu]Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.[/quote]

was this the first post on aops?

url is https://artofproblemsolving.com/community/c6h2

https://artofproblemsolving.com/community/c6h1 doesn't exist

Nice question! (Trig functions in degrees)
WLOG let $AB=1,$ angle chasing gives $\angle AMB=150,$ and $\angle AMC=110,$ using law of sines on $AMB,$ and $AMC,$ yields $AB\cdot \frac{\sin(20)}{\sin(150)}=AM=2\sin(20),$ now $AC=AM\cdot \frac{\sin(110)}{\sin(30)},$ note that $\sin(110)=\sin(90+20)=\cos(20),$ so this is $AM\cdot 2\cos(20)=4\sin(20)\cos(20)=2\sin(40),$ using law of cosines in $ABC,$ gives $BC^2=1^2+(2\sin(40))^2-4\sin(40)\cos(50)=1,$ hence $AB=BC,$ and triangle $ABC,$ is isosceles $\blacksquare$

Let $\angle MBC=\alpha$ and $\angle MCB=\beta$. Then $\alpha +\beta =80^\circ =60^\circ +20^\circ$, so it suffices to prove that $\frac{\sin \alpha}{\sin \beta}=\frac{\sin 60^\circ}{\sin 20^\circ}$. From the Trigonometric form of Ceva's Theorem we have $\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin \alpha}{\sin \beta}=1$, so it suffices to prove that $\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin 60^\circ}{\sin 20^\circ}=1$. Now using the identities $\sin x\sin y=\frac{\cos (x-y)-\cos (x+y)}{2}$ and $\cos x\sin y=\frac{\sin (x+y)-\sin (x-y)}{2}$ we obtain\begin{align*}\sin 10^\circ \sin 30^\circ & =\frac{\cos 20^\circ -\cos 40^\circ}{2}, \\
\sin 40^\circ \sin 20^\circ & =\frac{\cos 20^\circ -\cos 60^\circ}{2}, \\
\cos 20^\circ \sin 60^\circ & =\frac{\sin 80^\circ +\sin 40^\circ}{2}, \\
\cos 40^\circ \sin 60^\circ & =\frac{\sin 100^\circ +\sin 20^\circ}{2}=\frac{\sin 80^\circ +\sin 20^\circ}{2}, \\
\cos 20^\circ \sin 20^\circ & =\frac{\sin 40^\circ}{2}, \\
\cos 60^\circ \sin 20^\circ & =\frac{\sin 80^\circ -\sin 40^\circ}{2}.
\end{align*}Hence\begin{align*}\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin 60^\circ}{\sin 20^\circ} & =\frac{\cos 20^\circ -\cos 40^\circ}{\cos 20^\circ -\cos 60^\circ}\frac{\sin 60^\circ}{\sin 20^\circ} \\
& =\frac{\cos 20^\circ \sin 60^\circ -\cos 40^\circ \sin 60^\circ}{\cos 20^\circ \sin 20^\circ -\cos 60^\circ \sin 20^\circ} \\
& =\frac{(\sin 80^\circ +\sin 40^\circ )-(\sin 80^\circ +\sin 20^\circ )}{(\sin 40^\circ )-(\sin 80^\circ -\sin 40^\circ )} \\
& =\frac{\sin 40^\circ -\sin 20^\circ}{2\sin 40^\circ -\sin 80^\circ}.
\end{align*}Therefore it suffices to prove that $\sin 40^\circ -\sin 20^\circ =2\sin 40^\circ -\sin 80^\circ$, i.e. that $\sin 80^\circ =\sin 40^\circ +\sin 20^\circ$. We have\begin{align*}\sin 40^\circ +\sin 20^\circ & =2\sin \frac{40^\circ +20^\circ}{2}\cos \frac{40^\circ -20^\circ}{2} \\
& =2\sin 30^\circ \cos 10^\circ \\
& =\cos 10^\circ \\
& =\sin 80^\circ ,
\end{align*}as wanted.


This was a problem on the math team tryout I couldn't solve.

What is the sum

log(tan 1) + log (tan2).....log (tan89)

Log is base 10, and measurements are in degrees.

=log(tan1tan2...tan89)
=log(sin1...sin89/(cos1...cos89))

notice that cos(90-x)=sinx
hence the sum in equation is just:
=log(sin1...sin89/(sin89...sin1))
=log1
=0

Ah, its nice how every single problem like this cancels.  :D

this is like a mini-version of the more difficult trig series aime problems. heres one i remember : 100 *  :Sigma: x=1 to 44 (sin x) /  :Sigma: x=1 to 44 (cos x)

actaully
to solve that problem

theres a neat formula i learned over the summer that expresses
(sina+sinb+sinc+...)/(cosa+cosb+cosc+...)

If u use that.
This is a one step question.

im not sure what that formula is. to solve this problem, i used the identity sin x =  cos(:sqrt: (2)x -45 )


linear algebra
matrix
superior algebra
superior algebra solved

Show that if the matrix A is from M_2(R) so that there exists n odd, n natural, so that A^n=I_2 then A=aI_2 where a is from R\{0}.

How can this be true?
  Look here:
  Take A that has a_11=cos2pi/n=a_22, a_21=-a_12=sin 2pi/n.  THis one verifies A^n=I, but is not of the form aI.

This proble was proven false by Harazi, so I modified the title and I'll move it to solved.


geometry
circumcircle
ratio
parallelogram
analytic geometry
graphing lines
slope

Let S1 and S2 be two circumferences with centers O1 and O2 and common points M and N. Let t be the common tangent which is nearer to M, A and B be the points where t touches S1 and S2, respectively, C the point diametrally opposed to B and D the intersection of the line O1O2 with the perpendicular to AM trough B. Prove that M, D and C are collinear.

Very nice problem.... it can be solved with analytic geometry, or with a very sintetic proof using only angle chasing:
  Clearly you just have to show that the triangle ABM is similar to BCD. And to see that, let E be the intersection of MN and AB... E is in the radical axis of S1 and S2 so it is the midpoint of AB, and obviously O2 is the midpoint of BC, so it suffices to show that AEM is similar to BO2D. But with angle chasing it's easy to see that <EAM = <O2BD, and we know that EM is perpendicular to O1O2 so, if F is the point of intersection of MN and O1O2, EFO2B is cyclic, and we have <AEM = <BO2D. And the conclusion follows.

  RaMlaF...

[quote="RaMlaF"]  Clearly you just have to show that the triangle ABM is similar to BCD.

RaMlaF...[/quote]

I can easily see that if M,D and C are collinear then ABM ~ BCD. But, I'm sorry to not see so clearly the converse.

---edited

Ok! I have it!

Very very nice proof, RaMlaF!!

--- second edit

From my old notes:

Since MC is perp. to MB, to show M,D,C collinear, it suffices to show that DMB=90

Starting from this point, I will show an equivalent result: i.e. given that <DMB=90 then D, O1 and O2 are collinear.

Let S be the circumcircle of ABN with center O. Let E be the intersection between the perpendicular to BM trough A and the perpendicular to AM trough M. Is it possible to prove that F (the intersection between AE and BD) lies on S(*).
One can prove that, if r(S)/r(S1)=k, the triangles AME and BMD are similar with ratio k. Moreover is true that NA/NB=k then, since A,B,N and F are conciclic, <EAN=<DBN and therefore EAN and DBN are similar too, with the same ratio k. Then NE/ND=AE/BD=k and EM/DM=k (1).
Since EMDF is a parallelogram, then <EMD=<EFD. Moreover <END=<ANB=<AFB then <EMD=<END (2).
From (1) and (2), we have that EMD and END are similar. In this case, since they have a common side, are equal and symmetrically located respect to ED, this means that ED is the perpendicular bisector of MN, i.e. ED pass trough O1 and O2.

Ciao
Sprmnt21

(*) Is it true in general that:
(I) if you drawn two lines from A and B with the same slope respect to BM and AM, they meet on S.

There are a lot of interesting facts in the construction, one of these is the following:
(II) if you drawn two lines from A and B with the same slope respect to BM and AM, called A' the point where the line trough A meet S1 and called B' the point where the line trough B meet S2, then A', B' and M are collinear.

Another fact is:
(III) if R and S are the points common to AN and BM and to BN and AM respectively, then MN bisects RS.

One more fact is:
(IV) FN//ED.

And again:
(V) S is the geometric mean of S1 and S2.

Also it is:
(VI) for each F in S, the circumcircle of FNM meet AM and BM in the same points as FB and FA do.

And finally:
(VII) does exist a point F in S that, if U and V are the points common to AF and BM and to BF and AM respectively, then MN, RS and UV are concurrent.

well, your proof is just wonderfull!
  Now i'm going to prove (or at least try!) all those interestings facts.

another interesting idea for a diferent solution could be to consider the inversion with center $M$ and ratio $\frac{MN^2}{4}$...


geometry
ratio
trigonometry
trig identities
Law of Cosines

Alright, here is a neat problem that my students and I completed last week.  The authors labeled their question incorrectly and caused us much frustration, but I have fixed it for your enjoyment.

P and Q are the midpoints of the diagonals (BD and AC respectively) of a quadrilateral ABCD.  Prove that AB^2+BC^2+CD^2+DA^2 = AC^2+BD^2+4PQ^2.

Now, for the following problem, my students got a different value than me.  I am about to go check out why, but first let me post if for you guys.

For triangle ABC, P is the midpoint of [AB].  Q divides [AC] in the ratio 1:2.  R divides [BC] in the ratio 1:3.  What fraction of triangle ABC is triangle PQR (by area)?

oh goodie, you revived this forum  :wink:

Problem 2:

Let $\triangle{ABC}$ be equilateral with sidelength $4$. We then have $AP=PB=2$, $BR=1$, $RC=3$, $CQ=\frac{8}{3}$, $QA=\frac{4}{3}$.

$[\triangle{ABC}]=4\sqrt{3}$.

$[\triangle{PQR}]=[\triangle{ABC}]-[\triangle{APQ}]-[\triangle{PBR}]-[\triangle{RCQ}]$.

Using the sin-area formula and knowing $\sin{60}=\frac{\sqrt{3}}{2}$, we get that:

$[\triangle{PQR}]=4\sqrt{3}-\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{2}-2\sqrt{3}=\frac{5\sqrt{3}}{6}$.

Thus, $\frac{[\triangle{PQR}]}{[\triangle{ABC}]}=\frac{\frac{5\sqrt{3}}{6}}{4\sqrt{3}}=\boxed{\frac{5}{24}}$.

I'll do the first one later but I have a feeling you just use Law of Cosines a bunch of times and add everything up...

[quote="Iversonfan2007"]oh goodie, you revived this forum  :wink:
[/quote]
I can start posting problems here constantly after running AoPS Kings.  I seem to have found a passion for making problems.

whoa wait, dude are you from SC?  :maybe:

No, but I don't belong anywhere else either, and at least I have a friend in SC.


Well... no. 2 probably was one of the most difficult triangle geometry problems I have ever seen. Actually, my solution is 7 pages long and uses the Soddy circles theory.

Concerning no. 3, I have the solution for n >= 3 only (quite easy, hence I guess the case n = 2 is most difficult).

For no. 1, I guess f(x) = g(x) = 3/x is the only solution. No idea how to prove it.

How did you find the problems?

  Darij

I don't participate, but I do have an opinion. First of all, the first problem was fairly easy, but the second one was absolutely criminal. It is true that I have no relation with geometry, but this problem is the hell on earth! Problem 3 was beautiful and not very easy, but also not very difficult.

@ harazi: You say nr. 1 was easy. HOW HAVE YOU SOLVED IT?! I have worked for some hours and didn't get it!

I have posted my solution of nr. 2 and some ideas about the other problems at http://www.mathlinks.ro/Forum/viewtopic.php?t=17629 .

  Darij

It's my problem, so I think I have a solution to it. My solution was:
   Take a_1=2, b_1=4  and a_(n+1)=2/(1-1/b_n), b_(n+1)=2/(1-1/a_n). Then we prove by induction that for any n and any x we have x*min{f(x),g(x)}>a_n and x*max{f(x),g(x)}<b_n. This isn't hard, it follows immediately from the relations given. One can find the general term of a_n and b_n and can easily deduce that both sequences tend to 3. Thus, f(x)=g(x)=3/x.

Thanks! But calling this "simple"...

  Darij

the problem 3 was the easiest i think... it is in GM 12/2003 (gymnasium).

the fact that the problem appeared there at the gymnasium is just a mistake from the editors of the GM. the problem is not easy, as many people, who thought about it, can confirm. :)

I stumpled onto the possible source of #2 one time. The problem was used in the [url=http://science.up.ac.za/samo/s3q2003.pdf]South African MO 2003[/url]

Well, all "sources" of #2 have a common source: the ISL 2003...

  Darij

[quote="billzhao"]Could problem #3 be true if we limit ourselves to the Euclidean Plane instead of the space? Or is it possible to find a counter example? Also, might there be similar results for the other centres of the triangles? These are just some natural questions led from the problem.[/quote]
Yes it is (or at least according with my solution, if it is corect  :) ...). I hope Valentin posts the results soon so I can see if it is correct or not.  :D

I also found that this is true in the plane. About other centers, this isn't true for incenter and centroid: Divide the space(plane) into 2 halfspaces (halfplanes) (the boundry can belong to any of them) and color one of the halfplanes in one colour and the other one in another colour. Since the incenter and centroid always belong to the interior of the triangle, it follows that any monochromatic triangle has centroid/ incenter of the same colour as the vertices. The only center we haven't studied so far is the circumcenter. I didn't think about this one, but my first instinct (which is usually wrong) tells me that it should work because if it wasn't true, any circle couldn't contain more than 2 points of different colour from the center, which is probably imposible.

[post moved to http://www.mathlinks.ro/Forum/viewtopic.php?t=6004 - please delete this here]

hmm is there any proof of qn 3? I've been searching and can't find it, if there's any thread, please enlighten, thanks. the solution to qn 1 is brilliant!


calculus
integration
limit
calculus computations

I'm trying to calculate the following integral : $ \int_0^{ \plus{} \infty} \frac {1}{\sqrt {2\pi a^3}} \exp( \minus{} \frac {a^2}{2x}) \exp( \minus{} \lambda x) dx$ , we can see that $ \frac {\partial^2 I}{\partial a \lambda} \minus{} a \lambda I \equal{} 0$ so maybe $ I$ have the form $ \exp( \minus{} \frac{a^2\plus{}\lambda^2}{2})$ with some factor so that $ \lim I_{a \rightarrow 0^ \plus{} } \equal{} \plus{} \infty$   :(  ? any idea will be welcome  :idea:

hello, can you please correct your post, it is sennseless for me.
Sonnhard.

sorry for that, I have corrected the post but maybe It wasn't saved  :!:  Its done   :blush:
$ \lambda$ and $ a$ are strictly positives.

so no one could help ? Is it very easy or hard  :wink:  ?


One root of the cubic equation $ x^{3}\plus{}px^{2}\plus{}6x\plus{}q\equal{}0$, where $ p$ and $ q$ are real, is the complex number $ 5\minus{}i$.

Find the real root of the cubic equation.

Find the values of $ p$ and $ q$.

Is this possible? I think, finding roots to a third degree polynomial, then we only have the choices that:
(a) One root is real, the other two are complex.
(b) All roots are real.

so if $ 5\minus{}i$ is a root, then $ 5\plus{}i$ is also a root and that leaves the other root as the only real solution..

[quote="AndrewTom"]One root of the cubic equation $ x^{3} \plus{} px^{2} \plus{} 6x \plus{} q \equal{} 0$, where $ p$ and $ q$ are real, is the complex number $ 5 \minus{} i$.

Find the real root of the cubic equation.

Find the values of $ p$ and $ q$.[/quote]

So the expression is $ (x\minus{}a)(x\minus{}5\plus{}i)(x\minus{}5\minus{}i)$ $ \equal{}x^3\minus{}(10\plus{}a)x^2\plus{}(26\plus{}10a)x\minus{}26a$

So $ 26\plus{}10a\equal{}6$ $ \implies$ $ a\equal{}\minus{}2$ $ \implies$ $ q\equal{}52$ and $ p\equal{}\minus{}8$

Since $ 5 + i$ and $ 5 - i$ are roots $ (x - (5 - i))(x - (5 + i)) = (x^{2} - 10x + 26)$ is a factor of $ x^{3} + px^{2} + 6x + q$.

Now by Viete's formulas, $ (5 - i)(5 + i) + (5 - i)\alpha + (5 + i)\alpha = 6$, where $ \alpha$ is the real root.

Therefore $ \alpha = - 2$, so $ - 8 + 4p - 12 + q = 0$

$ 4p + q = 20$

$ (x + 2)(x^{2} - 10x + 26) = 0$

$ x^{3} - 8x \& ^{2} + 6x + 52 = 0$.

$ p = - 8, q = 52$.

I was doing the question and typing slowly and so was beaten by pco's much more elegant solutuion.


bache ha manie kalameye (algoritme havell _hakimi )chi mishe?

اين الگوريتم توي كتاب نظريه گراف وست هست و در مورد اينه كه يك دنباله گرافيكي است يا نه؟
در ضمن اگر اشتباه نكنم حكيمي ايرانيه


This problem was from 4everwise's mock AIME from 2006-2007:

Find the number of solutions, in degrees, to the equation $ \displaystyle \sin^{10}x \plus{} \cos^{10}x \equal{} \frac{29}{16}\cos^4 2x$, where $ \displaystyle 0^\circ \le x^\circ \le 2007^\circ$.


$ a,b>0.$  $ a\plus{}b\plus{}1\equal{}3ab$ prove that $ \frac{\frac{1}{a}\plus{}\frac{1}{b}}\geqslant2$   :)

hello, can you correct your formula please, in this form it is sensless for me.
Sonnhard.

sorry  :lol:  $ a,b>0.$  $ a\plus{}b\plus{}1\equal{}3ab$ prove that $ \frac{1}{a}\plus{}\frac{1}{b}\geqslant2$

[hide="easy"]
It sufices to prove that:
$ a+b \ge 2$.
But $ a+b \ge 2\sqrt{ab}$.
So it suffices to prove that $ ab \ge 1$.

But from the condition we have that:
$ 3ab \ge 3{\sqrt[3]{ab} \Longleftrightarrow ab \ge 1}$.[/hide]

[quote="Dimitris X"]
It sufices to prove that:
$ a \plus{} b \ge 2$.
[/quote]

If $ a\plus{}b\ge 2$ , also it is known that $ (a\plus{}b)\left(\frac{1}{a}\plus{}\frac{1}{b}\right)\ge 4$ , then it is not necessarily that $ \frac{1}{a}\plus{}\frac{1}{b}\ge 2$

[quote="Pain rinnegan"][quote="Dimitris X"]
It sufices to prove that:
$ a \plus{} b \ge 2$.
[/quote]

If $ a \plus{} b\ge 2$ , also it is known that $ (a \plus{} b)\left(\frac {1}{a} \plus{} \frac {1}{b}\right)\ge 4$ , then it is not necessarily that $ \frac {1}{a} \plus{} \frac {1}{b}\ge 2$[/quote]
 :huh: .I cant understand what you are saying.

The problem is to prove that:
$ \frac{1}{a}\plus{}\frac{1}{b} \ge 2 \Longleftrightarrow a\plus{}b \ge 2ab \Longleftrightarrow \frac{a\plus{}b}{2} \ge \frac{a\plus{}b\plus{}1}{3} \Longleftrightarrow 3(a\plus{}b)\ge 2(a\plus{}b\plus{}1) \Longleftrightarrow a\plus{}b \ge 2$.

I cant even understand your state.What you say is not true dear friend.... :wink:

Dimitris


Determine the minimum value of

\[\dfrac{\cot \left(\dfrac{\pi-A}{4} \right) + \cot \left(\dfrac{\pi-B}{4} \right) + \cot \left(\dfrac{\pi-C}{4} \right)}{\sin(A) + \sin(B) + \sin(C)}.\]

Assuming A,B,C are the angle of a triangle  ;)

I think 2 is the minimum value.

We want to prove  :Sigma: cot((:pi: - A)/4) - 2 sin(A) >= 0

But cot((:pi: - a)/4 - 2 sin(a) is convex so Jensen rules  :lol:

Minimum is reach for A=B=C=Pi/3


logarithms
function
algebra unsolved
algebra

find x so that 3^x+4^(1/x)=11 :P  :blush:

hmm.
It is clear that $x=2$ is the unique solution in $(1,\infty)$...
But there is another one in $(0,1)$.... I've no idea how to find the value of it...

Let $f(x)=3^x+4^{1/x}$. We see that $f'(x)=3^x\ln 3-\frac{4^{1/x}\ln 4}{x^2}$ is an increasing function for $x>0$. It follows that $f(x)$ is convex, thus $f(x)=11$ has at most two roots on $(0,\infty)$. We can easily find them from insight: $x_1=\log_3 2$ and $x_2=2$. And it is clear that there are no roots on $(-\infty,0)$.

[quote="Myth"]We can easily find them from insight: $x_1=\log_3 2$ and $x_2=2$. [/quote]
Wow..$x_1$ is not that easy to me  :(


He's not admitting to it, he's just saying what HE WOULD'VE done if he did it :wink: (kind of weird)

just a note:that same editor has come out with many contreversal books, like "juice" by jose cansecos, and some porn thing that was very contreversal :ninja:
-jorian

[quote="13375P34K43V312"]In his new book, "If I Did It, Here's What Happened," OJ Simpson confesses to the brutal murder of his wife and another woman. However, he was already found innocent of the crime, so he cannot be tried again for murder, due to double jeopardy. So basically he's a known murderer walking free. What's your opinion on this situation and the whole concept of double jeopardy?[/quote]

it's a hypothetical situation, he didn't say he did it (Who have you been listening to?)

That publisher is a weird one, she says she's doing it for payback.

[quote="fredbel6"]Is Simpson actually receiving money because of this book?!!!

It's extremely arrogant and :mad:  despicable.  It's also an insult to all those people who kept believing in his innocence.[/quote]

Not really, they were idiots to believe that in the first place. He is arrogant because he knows he is not as idiotic as those people who believed he was innocent and of course he'll receive money from the book. That's why he did it in the first place, he's just milking the story for money.

So where does that put us? Giving him free publicity and thus increasing his "worth"--if no one cared, he couldn't publish books and make money like this.

It's not like known murderers walking free is all that rare--usually just local, not so famous, though.

Double jeopardy's an important protection. Having a known murderer walk free is okay... as long as they don't repeat their crime. If they do, then we can get them that time around, so it's not as bad as it could be. With the alternative... someone could spend their whole life tied up in repeat trials, or be framed until the jury finally rules against them.

I dunno. Law's never perfect. I wonder if people have come up with mathematical models that would actually allow us to strike the best balance. It would still be an iffy prospect at best.

I really don't understand it either. It sounds like he had some big huge diabolical plot to kill his wife anyway and that is stuff that you have in the mind of a killer...THE F@*%ING GLOVE SHRUNK!!!!

My views a quite a bit different from what I read.  Allow me to share in case it offers a bit of comfort/enlightenment to some gentle readers.

If people really want to try him again, they can.  The principle of only trying him once applys for both governments, the state of California and the United States.

The real problem is that he is too politically powerful (not himself, but those with power who are willing to fight to keep him free because of his race).  This why he got off.  The elected prosecutor picked his worst, most incompentent team, didn't oppose the move of the case from the largely white area where OJ lived and where the crime was committed to the inner city which is largely black.

The thought may have been to avoid the race riots and certain deaths that a conviction, absent overwhelming belief that he was guilty held by those most likely to riot.  While millions of dollars were spent on the prosecution, no one dled, no riots occurred and no one thinks that the state will allow a potential wife killer to go free without prosecution.  Overall a great win for all the people by the elected prosecutor.

Judge Ito abused his position to keep the trial long and himself on TV.  The defense wanted a long complex trial to obscure the mountain of clear evidence.  The prosecution was incompetent.  For example after the trial was over, a network found a receipt for the brand of gloves that "didn't fit,"  they found footage of him wearing the gloves the month before the murder and they got the President of the company that made the gloves saying that the gloves found at the crime were the largest they had ever made.  If these didn't fit, then the receipt and the TV footage must have been planted by some great conspiracy.

There were a number of ways to present a clear case to the jury in less than a week from the mountain of evidence, but incompentence and greed of the prosecutors to be in a large, long, highly public case must have been too much.  (Another example the fibers found at the scene were from his Bronco -- only 3 sold in all of California.  This was the DNA evidence that was clear and that couldn't be tainted by detective behavior.)

Enought of this OJ was punished, no one else was killed and no race riots -- hard to argue that this wasn't a huge victory for the state, even as OJ walks off apparently in victory.

As Romans 12:19 says (Paul quoting God in Deut 32:35) "Vengence is mine; I will repay."  A short time into God's punishment, if OJ deserves puishemnt (as I am fully convinced he does), OJ will realize that he escaped nothing, he will have the full measure of punishment for what he did.  I am personally very comfortable leaving God the vengence he claims.

But that is really a lose-lose situation isn't it?

It would be better if he were to repent on Earth so that he would not have to face God's punishment.

[quote="Ignite168"]Saddam is also trying to get a book deal it seems, "If I killed 100,000 Kurds - Here's How It Happened"[/quote]

Wow,  :rotfl:, that was pretty good.

Although you're kinda getting off topic...

In my opinion, the strangest thing about all this is that people are actually buying this book.  Seriously, have they nothing better to do?

[quote="Ignite168"][quote="13375P34K43V312"][quote="bpms"]In my opinion, for his great crimes he will be reincarnated as a pig. That is all.
I can't believe that he had the gall to do this, isn't he rich anyways?[/quote]

Yeah, but after paying all those lawyers...apparently he got 4M for the book, not sure.[/quote]

I thought costs for counsel and court fees and such were transferred over to the losing side...[/quote]

I don't think so.

maybe he should be reincarnated into bpms :P

i am celebrating 500+ posts!

-jorian

Maybe we should stay remotely on topic?

Honestly, I don't see what's such a big deal about it. You haven't been directly affected by OJ, and if he commits another crime, he'll be punished for sure. What is the point of prison, if not to prevent prisoners from commiting crimes?

Don't pull any of that repentage on me, honestly, I believe that the prisoners that would truly repent for their sins (unless they're homosexual of course) deserve another chance. That and nobody wants to go through a lifetime of playing catcher, if you get my drift.

oh and hey jorian i'm celebrating 1855+ posts HOW ABOUT YOU?


Now I know that n ≡3 or 4 mod7 has no integer solution from n=x^3+y^3;n be an positive integer; and also does
n≡3,4,5 or6 mod 9.
From this,
If N  >=  2
           δ(N) =  (#{n є{1,2,…,N}: n≡ x^3+y^3mod N has a solution})/N
So I computed
           δ(7) = 5/7,    δ(9) = 5/9
If it is mod49. I try n≡ 3,4,10,11,17,18,24,25,31,32,38,39,45,46 mod49.There are 14 numbers that I think they make
n ≡ x^3+y^3 has no solution.Also n≡3,4,5,6,12,13,14,15,21,22,23,24mod27,there are 12 numbers.
    I computed   δ(7 2 ) = (49-14)/49=5/7,    δ(27) = (27-12)/27 =5/9
I'm not sure it is correct.

Well, there is the flaw that delta(7) = 2/7 != 5/7... I'm not sure exactly what you're looking for though.


[b]Problem 2[/b]

Let $ E$ be the set of all continuously differentiable real valued functions $ f$ on $ [0, 1]$ such that $ f(0) \equal{} 0$
and $ f(1) \equal{} 1$. Define

$ J(f) \equal{}\int^1_0 (1\plus{}x^{2})(f'(x))^{2}\,dx$

a) Show that $ J$ achieves its minimum value at some element of $ E$.
b) Calculate $ \min_{f \in E} J(f)$

[hide]i don't have much time left, i have to get back to the classes, so if someone could just type the rest of the problems.
cheers :)[/hide]

[i]Cauchy-Schwarz for the win[/i] :)

Let $ f \in E.$
\begin{eqnarray*} 1 = (f(1) - f(0))^2 & = & \left(\int_0^1 f'(x) dx \right)^2 \\
& \leq & \int_0^1 \frac {dx}{1 + x^2} \int_0^1 (1 + x^2)(f'(x))^2 dx = \frac {\pi}4 J(f), \end{eqnarray*}
so that
\[ \forall f \in E, \ \ J(f) \ge \frac 4{\pi}.
\]
Besides, we get an equality for $ f : x \mapsto \frac 4{\pi} \text{Arctan} x,$
hence
\[ \min_{f \in E} J(f) = \frac 4{\pi}.
\]