ratio
limit
algebra
functional equation
geometric series
algebra unsolved

Is there a simple form of:
$f(x)=x+\frac{x^{1/2}}{2}+\frac{x^{1/4}}{4}+\frac{x^{1/8}}{8}+...+\frac{x^{1/2^m}}{2^m}+...$
which is a solution to the following functional equation:
$f(x)=x+\frac{1}{2}f(\sqrt{x})$.

INCORRECT:

This is mearly a geometric series with first term $x$ and common ratio $\frac{\sqrt{x}}{2}$.
So, $f(x)=\frac{x}{1-\sqrt{x}/2}=\frac{2x\sqrt{x}-4x}{x-4}$

Sorry, I misread the series as $x+\frac{x^{1/2}}{2}+\frac{x^{2/2}}{2}+\frac{x^{3/2}}{2}+...$

It isn't a geometric series. Sorry.

I don't think that the series is necessarily a valid solution.
It depends on $\displaystyle\lim_{m\rightarrow\1}\frac{1}{m}f(x^{1/2^m})$.
The proposed solution assumes that the above limit is $0$, but it might not be.  We know that $f(1)=2$ from the original equation, but $f(x)$ as $x$ approaches $1$ is unknown.
However, if it is given that $f$ is continuous, then the series is valid.  This would actually solve a second problem as well:
There seemed to be unconnected "chains" of numbers such as $...2^{1/4},2^{1/2},2^{1},2^{2},2^{4}...$ and $...3^{1/4},3^{1/2},3^{1},3^{2},3^{4}...$
Unless we say that $f$ is continuous, there is no connection between these two chains of numbers.


geometric sequence
number theory proposed
number theory

show that if $a,b$ are natural numbers with $b>1$
then $ab+1$ and $ab^{3}+1$ cannot be both perfect squares :wink:

[quote="Albanian Eagle"]well I can say that the last step in your proof is wrong because $k \not| \ b_{3}$
but you're very near to the end :wink:[/quote]
Thank you for your replay ... actually I thought about this problem for 5 min that's why I wrote such stupid things  :blush: It is obvious that $b$ becomes bigger and we can see that actually $b_{3}>k^{2}-1$ so $b=b_{2}.$ Am I right?

yes now you're right ! :lol:
with this auxilliary problem we can easily show that
four triangular numbers can not be in geometric progression
can you guess how ?


Αν $ n>1$ φυσικος να αποδειξετε οτι $ n$ δεν διαιρει τον $ 2^n\minus{}1$

[hide]estw oti iparxei n fysikos pou isxyei ayti i sinthiki
thewrw ton mikrotero prwto diaireti tou n,estw p,
tote
$ 2^n \equiv 1 {modp}$
kai tin taksi(ord) tou 2 modulo p,estw d
tote
$ 2^d \equiv 1 {modp}$
omws
apo fermat little theorem isxyei
$ 2^{p\minus{}1} \equiv1 {modp}$
ara $ d|p\minus{}1,d|n \equal{}> d<p \equal{}> d\equal{}1$
dld p|1 atopo.[/hide]

Βαζω μια λυση ιδια με αυτη του Αθηναιου γραμμενη με διαφορετικο τροπο:

[hide]
Εστω οτι υπαρχει τετοιος θετικος ακεραιος $ n$ και εστω $ p$ ο ελαχιστος $ > 1$ διαιρετης του ο οποιος ειναι πρωτος. Το συνολο $ \{1,2,...,p \minus{} 1 (modp)\}$ ειναι ομαδα ως προς τον πολλαπλασιασμο διοτι καθε στοιχειο του ειναι σχετικα πρωτο με τον $ p$ και επομενος εχει αντιστροφο $ (modp)$. Η ομαδα αυτη εχει πληθος στοιχειων $ p \minus{} 1$ και εστω $ x$ η ταξη του 2 το οποιο ειναι στοιχειο της, τοτε $ 2^x \equiv 1(modp)$ και απο το Θεωρημα του Lagrange εχω $ x|p \minus{} 1$. Επηδη $ x$ ειναι ο ελαχιστος θετικος ακεραιος τετοιος ωστε $ 2^x \equiv 1(modp)$, εχω $ 2^n \equiv 2^{kx \plus{} m} \equiv (2^x)^k2^m \equiv 2^m \equiv 1(modp) \Rightarrow m \equal{} 0$ αφου $ 0 \leq m < x$, επομενος ο x διαιρει τον n. Ομως απο $ x|p \minus{} 1 < p \equal{} min\{y > 1  /  y|n\}$ εχω τελικα $ x \equal{} 1$ και $ 2^x \equiv 2^1 \equiv 2 (modp) \Rightarrow 2 \equiv 1 (modp) \Rightarrow p \equal{} 1$ το οποιο ειναι ατοπο. Αρα δεν υπαρχει αριθμος $ n$ που να διαιρει τον $ 2^n \minus{} 1$ και η αποδειξη ολοκληρωθηκε.[/hide]


Hi all,

May I know how what is the equivalent latex command of "Justify" (Command which is normally found next to the tab for "Aligned Left") in Microsoft word?

Thanks!  :)

Cheers,
Ruifang

Hi Ruifang,

justifying is the default behavior, so a command for it is not really necessary. If you use commands to change that you should limit the scope (of \raggedright, \raggedleft) or use the corresponding environments (flushleft, flushright).
Concerning such issues the package [url=http://ctan.org/pkg/ragged2e]ragged2e[/url] is recommendable. It provides a command \justifying that solves this question.

Stefan


linear algebra
matrix
induction
superior algebra
superior algebra unsolved

If $ A\in M_n(K)$ is nilpotent, the $ A$ is similar to a matrix all of whose entries are 0 except for certain entries 1 on the superdiagonal.

try induction on $ n$, or use jordan decomposition.


Are the earlier AIMEs more or less difficult than the more recent ones?

The AIMEs from the 80's are, on average, much less difficult than the AIMEs from the 90's. The AIMEs in the last few years have varied quite a bit in difficulty. This has been discussed MANY times as well; try using the Search function to find out about it.

The AIME that I think was the Hardest was 1994 AIME and the easiest is probably a tie between 2003 AIME 1, 2005 AIME 1, and 2006 AIME 2.

But the decade that was hardest was 90s and easiest was 80s.

[quote="Xantos C. Guin"][quote="mathgeniuse^ln(x)"]The AIME that I think was the Hardest was 1994 AIME and the easiest is probably a tie between 2003 AIME 1, 2005 AIME 1, and 2006 AIME 2.
[/quote]

The easiest AIME's were in the early 1980's. NOT in this decade.

I got a 13 on the 1984 AIME after 2 hours.

The hardest AIME's were in the 90's

I got a 4 oond the 1994 AIME

The AIME's in this decade are of medium difficulty.

I usually get somewhere between 7 and 9.[/quote]

Well, I only took the 1989 and I got an 11 on that one.  So that is why I said that.  I got a 5 on the 1994 and a 12 on 2003, 2005, and 2006.  So that is why I said my cases.  But I haven't taken the 1984 yet so that is probably when I will change my hierarchy.

paladin is right.

I realize that you're new but it's even more effective to learn to use search button at early stage. There are MANY people who do not even bother to use it (and I mean some experienced people as well).

For instance, here are top difficult AIMEs:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=difficulty+aime&t=60178

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=difficulty+aime&t=10995

So, use search button and the topic is LOCKED.


$ a$, $ b$ and $ c$ are positive reals such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \le 3$. Prove the inequality: \[ \sqrt{a(b\plus{}c)}\plus{}\sqrt{b(c\plus{}a)}\plus{}\sqrt{c(a\plus{}b)}\le 3\sqrt 2\]

[quote="April"]$ a$, $ b$ and $ c$ are positive reals such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \le 3$. Prove the inequality:
\[ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)}\le 3\sqrt 2
\]
[/quote]
[hide="solution1"]by Cauchy-S inequality we have $ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)} \leq 3 \sqrt{\frac{\sum a(b\plus{}c)}{3}} \leq 3\sqrt{\frac{2\sum a^2}{3}} \leq 3\sqrt{2}$[/hide]

[hide="solution2"]by AM-GM we have $ \sqrt{a(b\plus{}c)} \leq \frac{1}{\sqrt{2}}\frac{2a\plus{}b\plus{}c}{2}$
setting up 2 other similar inequality then add them up we obtain :
$ \sum \sqrt{a(b\plus{}c)} \leq \frac{1}{2\sqrt{2}}(4\sum a) \leq \frac{1}{2\sqrt{2}}(4\sqrt{3(\sum a^2)}) \leq \frac{12}{2\sqrt{2}}\equal{}3\sqrt{2}$[/hide]


abstract algebra
superior algebra
superior algebra solved

Let $ R \equal{} k[X,Y]$, where $ k$ is a field.

Give an example of a finitely generated, free $ R$-module $ L$ which has a submodule $ H$ which is not free.

Edit: Sorry, this is trivial. Take $ L \equal{} k[X,Y]$ and $ H \equal{} (X, Y)$.

[quote="-oo-"]in fact, "submodules of all free $ R$-modules are free" is equivalent to "$ R$ is PID".[/quote]

can you prove it?

"=>" should be in most good books (see also http://www.mathlinks.ro/viewtopic.php?t=304313 ), "<=" is rather trivial.

"<=" is not trivial for free modules of infinite rank, we have to well-order the basis etc ;)

Looks more like I used the wrong directions for my naming...


vector
logarithms
linear algebra
number theory
prime factorization
superior algebra
superior algebra solved

Is the set of all real numbers $\mathbb{R}$ a finite dimensional vectorspace over the the rationals?  Not sure where this question fits; in this section or the linear algebra section.... I posed myself this question when reading the first pages of field theory. However, i'm not sure it has to do something with it.

Consider the numbers of the form $\sqrt n$.

A simple application of cardinality does the job: any vector space over $\mathbb{Q}$ with finite or countable dimension is countable. Since $\mathbb{R}$ isn't, $\mathbb{R}$ has uncountable dimension. With a little more work, we can show that the dimension of $\mathbb{R}$ has the same cardinality as $\mathbb{R}$: the cardinality of $\bigoplus_{i\in S}X$ is the maximum of the cardinalities of $S$ and $X$ as long as both are infinite.

Now, what about the dimension of the set of algebraic numbers as a vector space over $\mathbb{Q}?$

The set of algebraic numbers is countable. Therefore, by the theorem jmerry just quoted, the dimension of the vector space can be no more than countable.

But on the other hand, there's the argument suggested by mecrazywong, above. In fact, let's narrow it to this: can we prove that the set $\sqrt{p},\ p$ prime, is independent? If we can do that, then we show that the algebraic numbers are infinite dimensional over the rationals.

But the two arguments together and the dimension is countably infinite.

[quote="ZetaX"]Showing that the $\sqrt{p}$ are independent is not the easiest thing and normally requires ideas of Kummer theory.
A more direct argument is to consider the $2^{k}$-th roots of $2$.[/quote]
Yes, that's what I did too. But then R as vectorspace over Q being infinite-dimensional seemed contradictionary.

it's very easy to check that $\{\ln(p) : p \text{~prime~}\}$ is linear independant in $\mathbb{R}$ over $\mathbb{Q}$. it follows directly from the unique prime factorization of integers.

Yes, but $\{\ln p\}$ doesn't answer my question about the dimension of the algebraic numbers. For that, I'll stick with ZetaX's version.

Mmmm,

How can one prove that $\mathbb{R}^{n}\cong \mathbb{R}$ as vectorspaces over $\mathbb{Q}$.

(2500th post!)

Choose a basis $X$ for $\mathbb{R}$ over $\mathbb{Q}$. $X\times X\times\cdots \times X=X^{n}$ ($n$ terms) is a basis for $\mathbb{R}^{n}$ over $\mathbb{Q}$. Since $X$ is infinite, $X^{n}$ has the same cardinality as $X$, and we there is a bijection between $X$ and $X^{n}$. This bijection of bases induces an isomorphism.

Yes, i'm sure:

Let $dim_\mathbb{Q}(\mathbb{R})=\kappa$, then $\mathbb{R}$ is (after choosing a basis) isomorphic to the set $\mathbb{Q}^{(\kappa)}$.

Then we have:
$\kappa\leq |\mathbb{R}|=\left|\mathbb{Q}^{(\kappa)}\right|$
$\leq \left|\dot{\bigcup_{X\in P_{fin}(\kappa)}}\mathbb{Q}^{X}\right|=\left|\dot{\bigcup_{X\in P_{fin}(\kappa)}}\mathbb{Q}\right|$
${=|P_{fin}(\kappa)}|\cdot|\mathbb{Q}|=\kappa\cdot|\mathbb{Q}|=\kappa$
where $P_{fin}(\kappa)$ is the set of all finite subsets of $\kappa$.

Therefore $\kappa=|\mathbb{R}|=|2^{\mathbb{N}}|=2^{\aleph_{0}}$

[quote="amfulger"][quote="Gockel"]No, it isn't. In fact the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is $|\mathbb{R}|=2^{\aleph_{0}}$.[/quote]
Are you sure about the $|\mathbb{R}|=2^{\aleph_{0}}$ part?[/quote]

I definitely agree to $|\mathbb R|=|dim_{\mathbb Q}\mathbb R|$.
As I knew, $|\mathbb R|$ was denoted $c$. My problem is whether $c=2^{\aleph_{0}}$ or not. It relates to the continuum hypothesis.

[quote="amfulger"]As I knew, $|\mathbb R|$ was denoted $c$. My problem is whether $c=2^{\aleph_{0}}$ or not. It relates to the continuum hypothesis.[/quote]

Huh? Isn't $c=2^{\aleph_{0}}$ a rather easy-to-prove theorem? (Roughly speaking, an irrational real number uniquely corresponds to an infinite sequence of integers from $\left\{0;\ 1\right\}$, namely its binary representation, and each such sequence uniquely corresponds to a sequence of integers; actually, both steps here have to be precised and we have to care about rational reals which, fortunately, are countable.) As far as I remember, the CH states that there are no cardinals between $\aleph_{0}$ and $c$, what is a much different statement...

As for the initial problem, it is obvious that $\mathbb{R}$ is not a finite dimensional vector space over $\mathbb{Q}$, because if it was one, then we would have a bijection between $\mathbb{R}$ and $\mathbb{Q}^{n}$ for some $n\in\mathbb{N}$, but this is impossible since $\mathbb{R}$ is uncountable, while $\mathbb{Q}$ and therefore also $\mathbb{Q}^{n}$ is countable. No great theory of infinities required.

On the other hand, I have questions regarding the solutions posted by some of you. There are some assertions mentioned which I would like to have proven:

[b]1.[/b] If a set $A$ is infinite, then $\left|A\right|=\left|A^{n}\right|$ for any $n\in\mathbb{N}$.
[b]2.[/b] If a set $A$ is infinite, and $B$ is the set of all finite subsets of $A$, then $\left|A\right|=\left|B\right|$.

The AC (axiom of choice) can be used. Note that I am able to prove that for an infinite set $A$, we have $\left|A\right|=\left|A\times B\right|$, where $B$ is a finite set. (Already this took me 4 pages...)

  Darij

All these questions depend a lot on the different constructions of cardinals. For instance, in our "cours de logique" the teacher presented von Newmann's approach: construction of ordinals and then proof that any cardinal is in bijection with a unique ordinal (Zermelo's theorem). After such introduction, proving that $A^{n}=A$ for any cardinal is really not difficult at all.

[quote="darij grinberg"][quote="amfulger"]As I knew, $|\mathbb R|$ was denoted $c$. My problem is whether $c=2^{\aleph_{0}}$ or not. It relates to the continuum hypothesis.[/quote]

Huh? Isn't $c=2^{\aleph_{0}}$ a rather easy-to-prove theorem? (Roughly speaking, an irrational real number uniquely corresponds to an infinite sequence of integers from $\left\{0;\ 1\right\}$, namely its binary representation, and each such sequence uniquely corresponds to a sequence of integers; actually, both steps here have to be precised and we have to care about rational reals which, fortunately, are countable.) As far as I remember, the CH states that there are no cardinals between $\aleph_{0}$ and $c$, what is a much different statement...

[/quote]

You are right. I rushed into it.


$X+Y+Z+W+T=16$

1. How many non-negative integer solutions are there?

2. How many positive integer solutions are there?

3. How many positive odd number integer solutions are there?

What about if you place an extra restriction on part 2, such as W must also be less than 10? Is there a general technique for it?

[quote="Farenhajt"][quote="neelnal"][quote="Farenhajt"][hide="Answers"]1. ${20\choose 4}=4845$

2. ${15\choose 4}=1365$

3. None, as five odd numbers can't add up to an even number.[/hide][/quote]

I am a bit confused. Order does matter here I thought.
Can you explain a bit more please?

I am noob  :P .[/quote]

Imagine 16 X's and 4 vertical bars, and put them in any order - for example, XXX|X|XXXXXXX|XX|XXX. Then all what's left of the first bar denote by $X$, between the first and the second by $Y$, etc. (If two vertical bars stand next to one another, they define a zero, of course.) You get all the possible solutions of the equation, and the number of the orderings is $\frac{20!}{16!4!}={20\choose 4}$

For (2), substitute $X=X'+1, Y=Y'+1$ etc. and then apply (1) to $X'+Y'+Z'+W'+T'=11$[/quote]

You can solve (2) without making a substitution.  Place the 16 X's in a line.  There are 15 spaces in between lines, and 4 must be occupied by dividers, so there are $\binom{15}{4}$ solutions.

The answers are correct though they are different from how the book solved them but it does't really matter.

[hide]1. ${20\choose 4}=4845$
2. ${15\choose 4}=1365$
3. 0, since the sum of 5 odd numbers cannot be even.[/hide]


calculus
integration
real analysis
real analysis solved

Evaluate the following definite integral.

\[\int^1_{-1} \frac{x^2}{1+e^x} dx\]

Cute, but quick if you see it. The answer is $\frac13.$ The key is the identity $\frac1{1+e^x}+\frac1{1+e^{-x}}=1$.

Let $u=-x$. Then $\int_{-1}^1\frac{x^2}{1+e^x}dx= \int_{-1}^1\frac{u^2}{1+e^{-u}}du= \int_{-1}^1\frac{x^2}{1+e^{-x}}dx$

So, $\int_{-1}^1\frac{x^2}{1+e^x}dx+ \int_{-1}^1\frac{x^2}{1+e^{-x}}dx = \int_{-1}^1x^2dx = \frac23.$

Divide by two to get the answer.

You are right ,Kent Merryfield.

Alternative solution:

Let $f(x)=\int^x_{-x} \frac{t^2}{1+e^t} dt$,since $ f'(x)=\frac{x^2}{1+e^x}+\frac{x^2}{1+e^{-x}}=x^2$ ,by integrating,we obtain $f(x)=\frac{1}{3}x^3+Const.$

Since $f(0)=0,$yielding$ f(x)=\frac{1}{3}x^3.$ Therefore the answer is $f(1)=\frac{1}{3}$.

kunny


logarithms
real analysis
real analysis unsolved

Assume  $x_n=\frac{1^{ia}+2^{ia}+\dots+n^{ia}}{n}$ and $a\ne 0$ and $i^2=-1$ Prove that, the set of all limitic points of $x_n$ is a circle with radius $\{1+α^2}^{-\frac{1}{2}}$

Is it $x_n=1^{i a}+2^{ia}+\dots+n^{ia}$ where $a\in\R\setminus \{0\}$, $i^2=-1$? Also do you define $k^{ia}$ by $k^{i a}=e^{i a\ln k}$? If so, the statement is wrong: $x_n$ tend to infinity, at least for small $a\ne 0$. If not, what is the correct formulation? :?


function
calculus
derivative
factorial
algebra proposed
algebra

Let $f\in\mathbb{R}[x]$, $\deg f=n$, and $f(x)\geq 0\ \forall x\in\mathbb{R}$.  Show that $\sum_{k=0}^{n}f^{k}(x)\geq 0\ \forall x\in\mathbb{R}$. ($f^{i}(x)$ denotes the $i-th$ derivative and $f^{0}(x)=f(x)$).  Also, please provide a proof using discriminants iff possible. :)

Masoud Zargar

Any idea on how to show that ? Obviously $\sum_{k=0}^{n}{\frac{f^{(k)}(x)}{k!}}\geq 0$ by the Taylor formula, but I couldn't drop the factorials (even though the result seems to be true to me).

My friend's simple solution :
Let $g(x)=\sum_{i=0}^{n}f^{(k)}(x)$. We just have to show that $g(x)\geq 0$ at its extremal points. But in this points we have $g'(x)=0$. Note that $g(x)=f(x)+g'(x)$. So at this points $f(x)=g(x)\geq 0$.
And we are done.

This was the intended way to solve it in a book, namely "Problem Solving Through Problems."  However, are there any other unique/beautiful ways that you are aware of?

Masoud Zargar


topology
advanced fields
advanced fields unsolved

I have this problem from Hatcher's book (chapter 2, section 2, problem 35)

Use the Mayer–Vietoris sequence to show that a nonorientable closed surface,
or more generally a finite simplicial complex $ X$ for which $ H_1(X)$ contains torsion,
cannot be embedded as a subspace of $ \mathbb{R}^3$ in such a way as to have a neighborhood
homeomorphic to the mapping cylinder of some map from a closed orientable surface
to $ X$.

Any hint to start with..

Say the closed orientable surface is $ Y$. Then get a covering $ \mathbb{R}^3 \equal{} U \cup V$ where $ U \cap V$ deformation retracts to $ Y$, and $ U$ deformation retracts to $ X$.

Then

$ H_2(\mathbb{R}^3) \to H_1(U \cap V) \to H_1(U) \oplus H_1(V) \to H_1(\mathbb{R}^3)$

is exact, giving an isomorphism

$ H_1(Y) \equal{} H_1(U \cap V) \equal{} H_1(U) \oplus H_1(V) \equal{} H_1(X) \oplus H_1(V)$.

But then $ H_1(Y)$ cannot be torsion-free, since $ H_1(X)$ has torsion.

I don't see why we can cover $ \mathbb{R}^3\equal{}U \cup V$ such that $ U\cap V \equal{}Y$ ?

Let $ U$ be the neighborhood homeomorphic to the mapping cylinder of $ Y \to X$. Let $ V$ be, say, the complement of $ X$ (if it is closed), such that $ U \cap V$ does not contain any part of the mapping cylinder that lies in $ X$ (that's the only constraint here, and its not a big one). Then $ U \cap V$ will def. retract to $ Y$.


Let $ x_{1},x_{2}$ are solutions of $ x^2\minus{}6x\plus{}1\equal{}0$. Prove :
$ \forall n\in \mathbb{Z}, x_{1}^n\plus{}x_{2}^n \in \mathbb{Z}$ and $ x_{1}^n\plus{}x_{2}^n \ne 5K$

A kind of lame approach would be to find the roots and use the Binomial Theorem. This ends up working out because all the terms involving square roots cancel, leaving only integers. Examining the sum mod 5 can show that it's not 0, but there's probably a better solution.

[quote="new_member"]Using Viete's relations we get:

    $ x_1 \plus{} x_2 \equal{} 6\in N$ and $ x_1x_2 \equal{} 1\in N$.

Using induction on N we can assume that the number $ x_1^n \plus{} x_2^n$ is an integer and we will prove that $ x_1^{n \plus{} 1} \plus{} x_2^{n \plus{} 1}$
is also an integer.

Well,notice that the number $ (x_1 \plus{} x_2)(x_1^n \plus{} x_2^n)$ would be an integer and hence we have that:

    $ (x_1 \plus{} x_2)(x_1^n \plus{} x_2^n) \equal{} (x_1^{n \plus{} 1} \plus{} x_2^{n \plus{} 1}) \plus{} (x_1x_2^n \plus{} x_2x_1^n)$.

  But notice that the number $ x_1x_2^n \plus{} x_2x_1^n$ is indeed an integer since:

   $ x_1x_2^n \plus{} x_2x_1^n \equal{} x_1x_2(x_1^{n \minus{} 1} \plus{} x_2^{n \minus{} 1}) \equal{} ... \equal{} (x_1x_2)^{n \minus{} 1}(x_1 \plus{} x_2) \equal{} 1^{n \minus{} 1}6 \equal{} 6$.
Is it correct?[/quote]

Why $ x_{1}^n \plus{} x_{2}^n \ne 5K, K\in \mathbb{Z}$???


OOPs I waas dividing the top and bottom instead of multiplying!
4*9/30=36/30=6/5

[hide]
$\frac{\frac{9}{3}\cdot\frac{4}{3}}{\frac{5}{3}\cdot\frac{6}{3}}
$
first multiply the top, you get 4
second multiply the bottom and you get 10/4
so 4/1*4/10=16/10 or
$\boxed{\boxed{\boxed{\boxed{\frac{8}{5}}}}}$[/hide]

[quote="236factorial"]What is $\displaystyle \frac{\frac{9}{3}\cdot\frac{4}{3}}{\frac{5}{3}\cdot\frac{6}{3}}$?[/quote]

$\frac65$

[quote="math92"]second multiply the bottom and you get 10/4
[/quote]

But this is incorrect,.  ;)

yeah, it should be 10/3
then 4/1*10/3=6/5

[quote="236factorial"]What is $\displaystyle \frac{\frac{9}{3}\cdot\frac{4}{3}}{\frac{5}{3}\cdot\frac{6}{3}}$?[/quote]
6/5


inequalities
trigonometry
inequalities open

Determine the least real number $ M$ such that the inequality

$ \sqrt {x^2 \plus{} y^2} \plus{} 2\sqrt {y^2 \plus{} z^2} \plus{} 3\sqrt {z^2 \plus{} x^2}\leqq M\sqrt {x^2 \plus{} y^2 \plus{} z^2}$

holds for all real numbers $ x,y$ and $ z$.

For given positive numbers $ a,\ b$ and $ c$, the following inequality holds by Cauchy-Schwarz inequality.
\[ a\sqrt {x^2 \plus{} y^2} \plus{} b\sqrt {y^2 \plus{} z^2} \plus{} c\sqrt {z^2 \plus{} x^2}\leq \sqrt {2(a^2 \plus{} b^2 \plus{} c^2)}\sqrt {x^2 \plus{} y^2 \plus{} z^2}.
\]

[quote="kunny"]For given positive numbers $ p,\ q$ and $ r$, the following inequality holds by Cauchy-Schwarz inequality.
\[ p\sqrt {x^2 \plus{} y^2} \plus{} q\sqrt {y^2 \plus{} z^2} \plus{} r\sqrt {z^2 \plus{} x^2}\leq \sqrt {2(p^2 \plus{} q^2 \plus{} r^2)}\sqrt {x^2 \plus{} y^2 \plus{} z^2}.
\]
[/quote]

Yes,but this equal holds only when (if $ x^2\plus{}y^2\plus{}z^2\neq 0$)
$ p\equal{}k\sqrt {x^2 \plus{} y^2},q\equal{}k\sqrt {y^2 \plus{} z^2},r\equal{} k\sqrt {z^2 \plus{} x^2}$ (where $ k$ is a real constant,$ k> 0$).

$ \Longleftrightarrow x^2\equal{}\frac{r^2\plus{}p^2\minus{}q^2}{2k^2},y^2\equal{}\frac{p^2\plus{}q^2\minus{}r^2}{2k^2},z^2\equal{}\frac{q^2\plus{}r^2\minus{}p^2}{2k^2}$

Therefore,the condition to exist reals $ x,y,z$ holding the equal is

$ p^2\plus{}q^2\geqq r^2, q^2\plus{}r^2\geqq p^2, r^2\plus{}p^2\geqq q^2$.

On this problem,however,if Cauchy's inequality is used directly,$ p\equal{}1,q\equal{}2,r\equal{}3\Longrightarrow p^2\plus{}q^2<r^2$

How do you make use of it? :?:

If $ \{x^2,y^2,z^2\} \equal{} \{a,b,c\}$ with $ a\geqq b\geqq c$ we have

$ \sqrt{x^2\plus{}y^2} \plus{} 2\sqrt{y^2\plus{}z^2} \plus{} 3\sqrt{z^2\plus{}x^2} \leqq \sqrt{b\plus{}c} \plus{} 2\sqrt{c\plus{}a} \plus{} 3\sqrt{a\plus{}b}$ by rearrangement, so it's left to maximize

$ f(a,b,c)\equal{}\sqrt{b\plus{}c} \plus{} 2\sqrt{c\plus{}a} \plus{} 3\sqrt{a\plus{}b}$

with $ a,b,c\geqq0$ and $ a\plus{}b\plus{}c\equal{}1$.

Well, we will prove that $ f(a,b,c)\leqq f(a,b\plus{}c,0)$, id est,

$ 2\sqrt{c\plus{}a} \plus{} 3\sqrt{a\plus{}b} \leqq 2\sqrt{a} \plus{} 3\sqrt{a\plus{}b\plus{}c}$.

Squaring and doing little manipulation we get the equivalent one

$ 5c \geqq 12\sqrt{(a\plus{}b)(a\plus{}c)} \minus{} 12\sqrt{a(a\plus{}b\plus{}c)}$ which is

$ 5c \geqq 12\frac{(c\plus{}a)(a\plus{}b)\minus{}a(a\plus{}b\plus{}c)}{\sqrt{(a\plus{}b)(a\plus{}c)}\plus{}\sqrt{a(a\plus{}b\plus{}c)}}$ which is

$ 5c\sqrt{(a\plus{}b)(a\plus{}c)}\plus{}5c\sqrt{a(a\plus{}b\plus{}c)} \geqq 12bc$, but since $ a\geqq b$ and $ c\geqq 0$ we have

$ (a\plus{}b)(a\plus{}c)\geqq 2b^2$, $ a(a\plus{}b\plus{}c)\geqq 2b^2$ and $ 5c\sqrt{(a\plus{}b)(a\plus{}c)}\plus{}5c\sqrt{a(a\plus{}b\plus{}c)}\geqq 10\sqrt 2 bc \geqq 12 bc$ since $ 10\sqrt 2>12$ and $ c\geqq0$. So we found out that $ f(a,b,c)\leqq f(a,b\plus{}c,0)$ and thus, we can assume that $ c\equal{}0$ (since we are trying to get the maximum). In that case we have to maximize

$ f(a,b,c)\equal{}\sqrt{b} \plus{} 2\sqrt{a} \plus{} 3\sqrt{a\plus{}b}$ for $ a\plus{}b\equal{}1$.

AM-QM (or Cauchy or whatever) gives $ \sqrt{b} \plus{} 4\frac{\sqrt{a}}2 \leqq \sqrt 5\sqrt{b\plus{}4(\frac{\sqrt{a}}2)^2} \equal{} \sqrt{5}\sqrt{a\plus{}b}\equal{}\sqrt 5$ so $ f(a,b,c)\leqq 3\plus{}\sqrt 5$. Homogenizing (or the way it should be spelled) we have $ f(a,b,c)\leqq (3\plus{}\sqrt 5) \sqrt{a\plus{}b\plus{}c}$, with equality if and only if $ a: b: c\equal{}4: 1: 0$.

Back to the original we have that $ M\equal{}3\plus{}\sqrt 5$, and equality holds if and only if $ |x|: |y|: |z| \equal{} 2: 1: 0$.

Nice problem, Bovy11!

PS: Why do you use $ \geqq$ instead of $ \ge$? Svetoslav Savchev (a famous problem proposer, which nowadays is assisting latinamerican competitions) also uses $ \geqq$ and $ \leqq$; do you know him?

bimbo,thank you for posting nice solution :!:
I don't know him.Just because as a sign of inequality $ \geqq$ is more popular than $ \ge$ in fundamental educational places here in Japan.

My solution is by many calculations ,not elegant:

$ a \equal{} \sqrt {x^2 \plus{} y^2},b \equal{} \sqrt {y^2 \plus{} z^2},c \equal{} \sqrt {z^2 \plus{} x^2}$

We get $ a,b,c\geqq 0$ and $ x,y,z\in \mathbb R \Longleftrightarrow a^2 \plus{} b^2\geqq c^2,b^2 \plus{} c^2\geqq a^2,c^2 \plus{} a^2\geqq b^2 \cdots (1)$

When $ c \equal{} 0 \Longrightarrow z \equal{} x \equal{} 0 \Longrightarrow M\geqq 3$

When $ c\neq 0$, we can have $ \frac {a}{c} \equal{} r\cos\theta,\frac {b}{c} \equal{} r\sin\theta$ ($ r\geqq 0, 0\leqq\theta\leqq \frac {\pi}{2}$) and then

$ (1)\Longrightarrow r\geqq 1$

Now, $ \sqrt {x^2 \plus{} y^2} \plus{} 2\sqrt {y^2 \plus{} z^2} \plus{} 3\sqrt {z^2 \plus{} x^2}\leqq M\sqrt {x^2 \plus{} y^2 \plus{} z^2}$

$ \Longleftrightarrow \frac {\sqrt {2}(r\cos\theta \plus{} 2r\sin\theta \plus{} 3)}{\sqrt {r^2 \plus{} 1}}\leqq M$

So we should find the Maximum value of the left

$ \frac {\sqrt {2}(r\cos\theta \plus{} 2r\sin\theta \plus{} 3)}{\sqrt {r^2 \plus{} 1}}\equal{} \frac {\sqrt {2}(\sqrt {5}r\sin(\theta \plus{} \alpha ) \plus{} 3)}{\sqrt {r^2 \plus{} 1}}$  (where $ \tan\alpha \equal{} \frac {1}{2})$  $ \leqq \frac {\sqrt {2}(\sqrt {5}r \plus{} 3)}{\sqrt {r^2 \plus{} 1}}$

$ f(r) \equal{} \frac {\sqrt {2}(\sqrt {5}r \plus{} 3)}{\sqrt {r^2 \plus{} 1}}$ ,then we get $ f'(r) \equal{} \frac {\sqrt {2}(\sqrt {5} \minus{} 3r)}{(r^2 \plus{} 1)^\frac {3}{2}} < 0$ ($ r\geqq 1$)

So $ f(r)$ is strictly decreasing, thus $ \frac {\sqrt {2}(r\cos\theta \plus{} 2r\sin\theta \plus{} 3)}{\sqrt {r^2 \plus{} 1}}\leqq f(r)\leqq f(1) \equal{} 3 \plus{} \sqrt {5}$

This equation holds if and only if $ \cos\theta: \sin\theta \equal{} 1: 2, r \equal{} 1 \Longleftrightarrow a: b: c \equal{} 1: 2: \sqrt {5} \Longleftrightarrow y \equal{} 0$ and $ |x|: |z| \equal{} 1: 2$

So we get the least $ M \equal{} 3 \plus{} \sqrt {5}$

[quote="kunny"]For given positive numbers $ p,\ q$ and $ r$, the following inequality holds by Cauchy-Schwarz inequality.
\[ p\sqrt {x^2 \plus{} y^2} \plus{} q\sqrt {y^2 \plus{} z^2} \plus{} r\sqrt {z^2 \plus{} x^2}\leq \sqrt {2(p^2 \plus{} q^2 \plus{} r^2)}\sqrt {x^2 \plus{} y^2 \plus{} z^2}.
\]
[/quote]

I believe the inequality holds for all real numbers $ x,\ y,\ z$. :wink:


Prove  that   (1/sin(pi/7))=(1/sin(2pi/7))+(1/sin(3pi/7))
pi=perimeter/diameter

it is known that in a correct polygon with 7 sides A1A2A3A4A5A6A7(i don't know the name) the following equality holds,it's easy to prove.
1/A1A2=1/A1A3+1/A1A4
from the triangle A1A2A4 we get A1A2=2*R*sinA1A4A2 and A1A4A2=pi/7
from the triangle A1A3A5 we get A1A3=2*R*sinA1A3A5 and A1A3A5=2*pi/7
from the triangle A1A3A6 we get A1A4=2*R*sinA1A6A3 and A1A6A3=3*pi/7
and that's the needed equality.


Hi Ashrafmod.

Is the purpose of your problem to find a limit?


geometry
circumcircle
inradius
trigonometry
area of a triangle
Heron\u0027s formula
Pythagorean Theorem

A circle is inscribed in pentagon $ ABCDE$. $ AB\equal{}BC, CD\equal{}DE, AE\equal{}6, AC\equal{}8, CE\equal{}7$. Find the radius of the circle inscribed in pentagon and angle $ BCD$

Let $ a \equal{} CE \equal{} 7, c \equal{} EA \equal{} 6, e \equal{} AC \equal{} 8$ be sides and $ s \equal{} \frac{21}{2}$ semiperimeter of the $ \triangle ACE.$ By Heron's formula, its area $ S$ is

$ S \equal{} \sqrt{s(s \minus{} a)(s \minus{} c)(s \minus{} e)} \equal{} \frac{21 \sqrt{15}}{4}$

and its circumradius $ R$ is

$ R \equal{} \frac{ace}{4S} \equal{} \frac{16}{\sqrt{15}}.$

The perpendicular bisectors of $ AC$, $ CE$ meet at its circumcenter $ O.$ The perpendicular bisectors of these 2 segments are identical with the internal bisectors of the $ \angle ABC, \angle CDE$ of the pentagon $ ABCDE,$ therefore, $ O$ is the pentagon incenter. The pentagon incircle $ (O, r)$ is tangent to the chord $ EA \equal{} 6$ of the triangle circumcircle $ (O, R).$ Let $ M$ be the midpoint of $ EA.$ By Pythagorean theorem for the right angle $ \triangle AOM,$ the pentagon inradius is

$ r \equal{} OM \equal{} \sqrt{OA^2 \minus{} MA^2} \equal{} \sqrt{\frac{256 \minus{} 9 \cdot 15}{15}} \equal{} \frac{11}{\sqrt{15}}.$

The tangency points $ S, T$ of tangents to the circle $ (O, r)$ from $ B \in (O, R)$ are intersection of the circle $ (O, r)$ with a circle with diameter $ OC.$ From the right angle $ \triangle OTC,$

$ \sin \frac{\widehat{BCD}}{2} \equal{} \sin \widehat{OCD} \equal{} \frac{r}{R}.$

$ \cos \widehat{BCD} \equal{} 1 \minus{} 2 \sin^2 \widehat{OCD} \equal{} 1 \minus{} \frac{2r^2}{R^2} \equal{} 1 \minus{} \frac{2 \cdot 121}{256} \equal{} \frac{7}{128}.$


$x$ white marbles and $y$ black marbles are placed into a bag. 2 marbles are drawn without replacement, and the probability of choosing 2 of the same color is $\frac{1}{2}$.

1. Prove $x+y$ is a perfect square.
2. Prove $x$ and $y$ are consecutive triangular numbers.

[quote="AdorableAddict"]1 is a direct corollary of 2.  :lol:[/quote]
Yes, but I'm not quite sure how to prove 2.  :maybe:
[hide="Hint for 1"] $\frac{x(x-1)+y(y-1)}{(x+y)(x+y-1)}=\frac{1}{2}$ and manipulate. :D [/hide]

[hide="Full solution"]From

$\frac{{x\choose 2}{y\choose 0}+{x\choose 0}{y\choose 2}}{{x+y\choose 2}}={1\over 2}$

after some manipulation we get

$x^{2}-2xy+y^{2}-x-y=0$

which directly yields $x+y=(x-y)^{2}$, i.e. part 1. But it also gives

$x^{2}-(2y+1)x+y^{2}-y=0$

Solving this as a quadratic in $x$, we get

$x_{1,2}=\frac{2y+1\pm\sqrt{4y^{2}+4y+1-4y^{2}+4y}}{2}=\frac{2y+1\pm\sqrt{8y+1}}{2}$

Since $x$ is integer, we must have $8y+1=t^{2}$ where $t$ is integer. Obviously $t$ is odd, hence $t=2k+1$ for some integer $k$. That gives

$8y+1=(2k+1)^{2}=4k^{2}+4k+1\iff y=\frac{k(k+1)}{2}$

Plugging that back into the expression for $x$ and simplifying, we find

$x_{1}=\frac{(k+1)(k+2)}{2},x_{2}=\frac{(k-1)k}{2}$

which proves part 2.[/hide]


algebra
polynomial
number theory solved
number theory

Suppose that p is a prime, r is a positive integer and q is a prime such that
p> $q^{r-1}$ and q doesn't divide r but q divides p-1. Suppose that for some integers $a_1, ... a_r$ we have $ \sum a_i ^ { \frac {p-1}{q} } $=0 mod p. Prove that one of the $a_i $ is divisible by p.

This problem is stated wrong. The correct conditions are $p > r^{q-1}$, $q|(p-1)$, and q does not divide r. A counterexample to the original statement:
$p=683$, $q=11$, $r=3$; $a_1=1$, $a_2=1$, $a_3=10$

A full solution to the corrected problem will follow.

The solution:

Let $p$ and $q$ be primes and $r$ be a positive integer, with $p>r^{q-1}$ and $q|(p-1)$. Assume that $\displaystyle \sum_{i=1}^r a_i^{\frac{p-1}{q}} \equiv 0$ mod $p$ for some $a_i$, none of which are zero. Let $a$ be a primitive $q$-th root of 1 mod $p$. Then each $a_i \equiv a^{t_i}$ mod $p$ for some integers $0<t_i < q$. Let $\displaystyle f(x)=\sum_{i=1}^r x^{t_i}$. This is a polynomial, and $f(a)$ is divisible by $p$. Let $\displaystyle g(x)=\prod_{j=1}^{q-1} f(x^j)$. Then $g(a^j)$ is divisible by $p$ for each $j$, $1 \le j \le q-1$.
We have $\displaystyle g(x)=\sum_{k=0}^{(q-1)^2} b_k x^k$ for some positive integers $b_k$ with $\displaystyle \sum_{k=0}^{(q-1)^2} b_k =r^{q-1}$. Let
$\displaystyle h(x) = \sum_{k=0}^{q-1} \sum_{i=0}^{q-2} b_{k+qi}\cdot x^k$.
Then $h(x) = g(x) - v(x)(x^q-1)$ so $h(a^j)$ is divisible by $p$ for each $j$, $1 \le j \le q-1$. Define $c_k$ by $\displaystyle h(x) = \sum_{k=0}^{q-1} c_k x^k$. By construction, each $c_k$ is positive and $\displaystyle \sum_{k=0}^{q-1} c_k = r^{q-1}$
Now, considering the polynomials mod $p$, $1+x+x^2+ \dots +x^{q-1}$ and $h(x)$ are polynomials of equal degree with the same roots, so they are constant multiples of each other and all of $c_0, c_1, \dots c_{q-1}$ are congruent mod $p$. Their sum is less than $p$ by hypothesis, so they are equal and $q\cdot c_0=r^{q-1}$, which gives $q|r$.


calculus
integration
function
logarithms
complex analysis
trigonometry
topology

I am new to complex analysis. I got this question.
$\oint_{C}\frac{dz}{z}$
Where $C$ is a unit circle.

Now why cant I use the fundamental theorem of line integral? The function $f(z) = \ln z$ has property that $f'(z) = \frac{1}{z}$, right?

And so,
$\oint_{C}\frac{dz}{z}= 0$.

But that is not true! It should be $2\pi i$.
----

And please dont refer to the Theory of Residues because I want this to be an elementary answer. To why I cant use it.

There is no continuous logarithm defined on the entire unit circle. The F.T. has as a hypothesis the existence of a continuous antiderivative defined on some open set containing our curve- that isn't satisfied here.
If you look closer, the imaginary part of the logarithm is "$\theta$"- in order to define that all the way around the circle, you need a jump somewhere.

In this case, we can calculate the integral directly by parametrizing: $\int_{C}\frac1z\,dz=\int_{0}^{2\pi}\frac1{e^{it}}\cdot ie^{it}\,dt=\int_{0}^{2\pi}i\,dt=2\pi i$.

This integral is the foundation on which the theory of residues is built; we need an independent calculation of it in order to do any residue calculations.
Suppose we have a function $f$ with an isolated singularity at $a$. We can expand $f$ in a Laurent series near $a$: $f(z)=\sum_{n\in\mathbb{Z}}c_{n}(z-a)^{n}=\frac{c_{-1}}{z-a}+\sum_{n\neq-1}c_{n}(z-a)^{n}$. The second term has an antiderivative $\sum_{n\neq-1}\frac{c_{n}}{n+1}(z-a)^{n+1}$ defined and continuous on a punctured neighborhood of $a$; its integral around any closed curve near $a$ is zero. The first term does not have a global antiderivative, and we need the independent calculation of its value ($2\pi i$ times the winding number around $a$).
That's only a few details away from being the residue theorem.

Thank you for responding jmerry, but I do not understand. How is it that $f(z) = \ln z$ [b] is not [/b] continous? My problem is that $\ln z$ is continous on $\mathbb{C}-\{0\}$ and since we are integrating along $e^{i\theta}$ we never reach the origin, so we are always continous. Where am I wrong?

No, it isn't. Zero is the only point that has to be a discontinuity, but it can't be made into an isolated singularity- there has to be a whole line/curve of discontinuities, going between zero and $\infty$. The standard definition puts this "branch cut" on the negative real axis.

You don't hit the origin- but you have to cross the branch cut somewhere. Think of the imaginary part as a spiral staircase; if you go around once, you're higher than you started out.

Okay, so let me see if I have this right.
Lets say I define for $z\not = 0$: $\ln z = \ln |z|+i \arg z$ where $\arg z$ is the interval $(-\pi,\pi]$.
Then this function is holomorphic on $\mathbb{C}-(-\infty,0]$.

And the way I actually show this (the holomorphic region) is by the Cauchy-Riemann equations. Which I am lazy to check because it gets a little messy with the arctangent function $\tan^{-1}\frac{y}{x}$. Because I need to consider the case where $x=0$ and the case when $x>0$ and the case when $x<0$, ect.... So it gets messy if you know what I mean.

2nd question] You know of any good sites which explain branch cuts in good details with examples because my book does a terrible job.


geometry
incenter
ratio
geometric transformation
reflection
symmetry
circumcircle

Dear Mathlinkers,
let ABC be a rectangular triangle at A,
A' the foot of the A-altitude on BC,
J, K the incenters of the triangles AA'B, AA'C
and Fe the Feuerbach's point of ABC.
The circle with diameter JK passes through Fe.
Any reference? I haven't see thsi result in the litterature.
Sincerely
Jean-Louis

Let $ X,Y$ denote the tangency points of the incircle $ (I)$ and the $ A$-excircle $ (I_a)$ with $ BC.$ $ W$ denotes the foot of the angle bisector of $ \angle BAC.$ Since the midpoint $ M$ of $ BC$ is also midpoint of $ XY$ and the cross ratio $ (A',W,X,Y)$ is harmonic, it follows from Newton's theorem that $ MX^2 \equal{} MY^2 \equal{} MW \cdot MA'.$ Consequently $ A'$ and $ W$ are inverse points under the inversion with center $ M$ and power $ MX^2 \equal{} MY^2.$ Such inversion takes $ (I)$ and $ (I_a)$ into themselves and  the nine-point circle $ (N)$ of $ \triangle ABC$ into the common internal tangent $ n$ of $ (I),(I_a)$ different from the sideline $ BC,$ which is the reflection of $ BC$ about $ AW.$ Then  $ F_e' \equiv n \cap (I)$ is the image of $ F_e$ under the referred inversion.

From the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=299708]Incircles and parallel segments[/url], we know that the circumference $ (V)$ with diameter $JK$ goes through $ A', X$ and the triangle $ \triangle KXJ$ isosceles and right at $X,$ thus $ XV$ bisects $ \angle KXJ.$ Further, $ XK \parallel AB$  and $ XJ \parallel AC,$ i.e. angles $ \angle BAC$ and $ \angle KXJ$ have  parallel sides $\Longrightarrow$ their bisectors are parallel as well $ \Longrightarrow$ $ VX \parallel AW.$

Let $ S$ be the midpoint of $ IW,$ which is obviously by symmetry the center of the circle $ \odot( XWF_e').$ We have $ \angle SXI \equal{} \angle IXV$ due to $ VX \parallel IW.$ In other words, angles between circles $ (V)$ and $ (S) \equiv ( XWF_e')$ with $ (I)$ are equal but oppositely directed. Since the inversion preserves angles between curves, the point $ X$ is double and  $ A'$ and $ W$ are inverse images,  it follows that $ (S)$ is identical with the inverse image of $ (V)$ under the inversion. $ F_e'$ is common point of $ (I),(S),(n),$ then in the primitive figure, $ F_e$ must be common point of $ (I),(V),(N).$

Dear Luis and Mathlinkers,
I appreciate your proof requiring inversion.
Why not a synthetic one without inversion?
A new challenge...
Sincerely
Jean-Louis

Let $ L$ be the incenter of triangle $ AA'B$.
A well-known result:  $ I$ is the orthocenter of triangle $ ALK$.
$ \Rightarrow \angle IAK = \angle ILK = \angle KCB = \angle LAA'$.
Therefore the circumcenter $ H$ of triangle $ ALK$ lies on $ AA'$.
Since $ \angle LAK = 45^o$ then $ \angle LHK = 90^o$. We get $ AH = HK = DK = ID = r = AR$
But $ AK$ is the bisector of angle $ HAR$ then $ AHKR$ is a rhombus.
$ \Rightarrow RK = r.$
Let $ P$ be the intersection of $ BI$ and $ AC$. Construct a line which passes through $ P$ and parallel to $ RK$, it cuts $ KM$ at $ S$.
We will show that $ RS$ is a bisector of angle $ KRM$
$ \Leftrightarrow \frac {RK}{RM} = \frac {KS}{SM} = \frac {RP}{PM}$
$ \Leftrightarrow \frac {r}{RM} = \frac {r - RP}{RP} = \frac {r}{RP} - 1 (*)$
Note that $ r = \frac {b + c - a}{2}, RM = \frac {b}{2} - \frac {b + c - a}{2} = \frac {a - c}{2}$ and $ RP = AP - AR = \frac {bc}{c + a} - \frac {b + c - a}{2} = \frac {(a + c - b)(a - c)}{2(a + c)}$
So $ (*)\Leftrightarrow \frac {b + c - a}{a - c} + 1 = \frac {(c + a)(b + c - a)}{a + c - b}$
$ \Leftrightarrow a^2 = b^2 + c^2$ (right!)
We obtain $ \angle PSR = \angle PRS = 1/2\angle B$. Therefore $ PS$ is a tangency of $ (I)$.
Thus $ MS$ passes through $ Fe$. Similarly $ QL$ passes through $ Fe$ and $ \angle LFeK = \angle MFeQ = 90^o$ hence $ Fe$ lies on $ (LK)$

Dear Livetolove and Mathlinkers,
nice metric proof...
Why not a synthetic one without calculus?
A new possible challenge...
Sincerely
Jean-Louis

Let the incircle $ (I)$ of $ \triangle ABC$ touch $ BC, CA, AB$ at $ D, E , F.$ $ BI \perp AK, CI \perp AJ$ $ \Longrightarrow$ $ I$ is orthocenter of $ \triangle AJK$ and $ AI \perp JK.$ $ \angle KAA' \equal{} \frac{_1}{^2} \angle ABC \equal{} \frac{_1}{^2}(90^\circ \minus{} \angle BCA) \equal{} \angle IAJ$ $ \Longrightarrow$ $ AI, AA'$ are isogonals WRT the angle $ \angle KAJ$ $ \Longrightarrow$ circumcenter $ P$ of $ \triangle AJK$ is on $ AA'.$ Since $ \angle KAJ \equal{} 45^\circ$ $ \Longrightarrow$ $ \triangle PJK$ is isosceles right. Let $ (Q)$ be its circumcircle on diameter $ JK$; then $ A', D \in (Q).$ (This is true for any $ \triangle ABC$ and for any $ A' \in BC,$ well known and posted before). From the right angle $ \angle DA'P,$ $ DP$ is diameter of $ (Q)$ and $ JDKP$ is a square. Since $ JK \perp AI,$ it follows that $ DP \parallel AI$ and $ DP \perp EF.$ By the problem [url=http://www.mathlinks.ro/viewtopic.php?t=301033]3 circles concurrent at Feuerbach point[/url], the circle $ (Q)$ with diameter $ DP$ goes through the Feuerbach point $ Fe$ of $ \triangle ABC.$

________________________________________________________________________
At the time of this post, the problem referenced above is to be solved. I do have a synthetic proof.

i'll use the following lemma:
Lemma- $ ABC$ is a right angled triangle at $ A$ and $ H_a$ its projection on $ BC$. Let $ K,J$ be the incenters of $ AH_aC,AH_aB$.Let $ C',B'$ be the midpoints of $ AB,AC$.then $ C'J,B'K$ intersection lies on the circle with diameter $ JK$ and also on the circle with diameter $ B'C'$(The nine-point circle).
Let $ S$ be the intersection point mentioned then we have
$ \Delta AH_aB \sim \Delta AH_aC \implies \angle JC'A \equal{} \angle KB'C,\angle A'B'C \equal{} \angle A'C'A \equal{} 90^\circ \implies \angle A'C'J \equal{} \angle A'B'K \implies \mbox{C'SA'B' is concyclic thus} \angle JSK \equal{} \angle C'SB' \equal{} \angle C'A'B' \equal{} 90^\circ$
Done.
Also i will use this well-known property :Let $ I$ be the incenter and $ D$ its projection on $ BC$. then we have $ \angle IDJ \equal{} \angle B,\angle KDI \equal{} \angle C$.it has a synthetical proof.
Let $ T_c,T_b$ be the intersections of $ AJ,CK$ and $ AK,BJ$ respectively.Let $ X,L$ be the projections of $ I$ on $ AB,AC$ respectively.Let the line parallel to $ BC$ and passing through $ I$ intersects $ C'J,B'K$ at $ T$ and $ T'$ respectively.Let $ K',J'$ be the midpoints of $ JI,KI$ respectively.we have the following properties:
1-$ \angle JAK \equal{} 45^\circ$,
2-$ I$ is the orthocenter of $ \Delta AJK$ which implies the perpendicularity of $ CK,AJ$ at $ T_c$ and similarly for $ BJ,AK$ at $ T_b$,
4-$ T_c,T_b,B',C'$ are collinear,
5-$ IT_c \equal{} T_cJ,IT_b \equal{} T_bK$,
6-$ JD \equal{} DI \equal{} DK$,
7-$ X,T_c,D$ are collinear,Similarly for $ D,T_b,L$,
8-$ \angle T_cDT_b \equal{} \angle XDL \equal{} 45^\circ$,
9-$ DI \equal{} T_cT_b$,we can prove this by the fact that $ \Delta DJ'I \equal{} \Delta J'T_cT_b$.
10-$ TI \parallel{} T_cT_b$,$ T'I\parallel{}T_cT_b$.
11-$ DK' \equal{} K'X,\angle K'T_bD \equal{} \angle K'DT_b \equal{} 45^\circ \implies \Delta DT_bX \mbox{is right \minus{} angled isocele}$ and similarly for $ \Delta T_cDL$.
12-$ T_cKT_bX,JT_bLT_c$ are parallellograms.
13-we'll prove $ TT_c\parallel{}JT_b$.We must prove $ \frac {C'T}{TJ} \equal{} \frac {C'T_c}{T_cT_b}$.
but we have $ C'T_b\parallel{}TI \implies \frac {C'T}{TJ} \equal{} \frac {T_bI}{IJ}$ and also $ XT_c\parallel{}AT_b \implies \frac {C'T_c}{T_cT_b} \equal{} \frac {C'X}{XA}$.So we must prove the equality $ \frac {T_bI}{IJ} \equal{} \frac {C'X}{XA}$.
Let $ O$ be the projection of $ C'$ on $ AJ$.Let $ P \equal{} AJ \cap XT_b$.If we prove $ C'P$ is the angle bisector of $ B'C'A$ then we'll have $ \Delta C'OA \sim \Delta JKT_b$.
So we should only show that $ C'P \perp AK$.but it's known that $ AC' \equal{} C'T_b,AP \equal{} PT_b$ so both $ C',P$ lie on the perpendicular bisector of $ AT_b$.But we have $ \angle OC'P \equal{} \angle C'PX \equal{} 180^\circ \minus{} ((\angle B/2) \plus{} (90^\circ \plus{} \angle C/2)) \equal{} 45^\circ$.in this way we get: $ C'P$ is the angle bisector of $ AC'B'$ thus $ \frac {C'X}{XA} \equal{} \frac {OP}{PA}, \Delta C'OA \sim \Delta JKT_b,\angle OC'P \equal{} \angle IKT_b \equal{} 45^\circ, \implies \frac {OP}{PA} \equal{} \frac {T_bI}{IJ}$.
14-$ TT_c\parallel{}JT_b \implies TT_c \parallel{}IT_b$,Similarly for $ T'T_b$ and $ IT_c$.
By (10) and (9) we can concude that $ TT_cT_bI$ and $ T_cT_bT'I$ are parallellograms $ \implies T_bT_c \equal{} DI \equal{} IT \equal{} IT'$.But we have $ \angle TST' \equal{} 90^\circ \implies IS \equal{} IT \equal{} IT' \equal{} ID \implies \mbox{S is on the nine \minus{} point circle and the incenter thus } S \equal{} F_e$
DONE!

Dear Mathlinkers,
a file concerning my result has been open on a French site :
http://www.les-mathematiques.net/phorum/read.php?17,538526
Sincerely
Jean-Louis

Dear Mathlinkers,
a synthetic proof of this result can be see
on: http://perso.orange.fr/jl.ayme    vol. 5  Le théorème de Feuerbach-Ayme
Sincerely
Jean-Louis


factorial
function
combinatorics unsolved
combinatorics

Find an expression for $ \sum_{i \equal{} 1}^ni(i \plus{} 1)(i \plus{} 2)\dots (i \plus{} r \minus{} 1)$. I am guessing $ r$ is a positive number (in the question it doesn't even state what $ r$ is).
pg. 107 #52 An Introduction To Mathematical Thinking (William J. Gilbert and Scott A. Vanstone) Fun little book, easy read.

[hide="My solution"]$ r!\sum_{i \minus{} 0}^{n \minus{} 1}\binom{r \plus{} i}{i} \equal{} r!\binom{r \plus{} n}{n \minus{} 1}$[/hide]

[hide="Prove"]$ \sum_{i \minus{} 0}^{n \minus{} 1}\binom{r \plus{} i}{i} \equal{} \binom{r \plus{} n}{n \minus{} 1}$ Combinatorially is more fun ;P Then again I couldn't figure out an algebraic proof, so maybe I am being insecure =)[/hide]

That depends on what you mean by an "algebraic" proof.  The generating functions proof is essentially the identity $ \frac{1}{1 \minus{} x} \cdot \frac{1}{(1 \minus{} x)^r} \equal{} \frac{1}{(1 \minus{} x)^{r\plus{}1}}$.  There is also a purely manipulative proof by telescoping using Pascal's identity.


[hide="Solution"] Recalling that $ \minus{} \ln (1 \minus{} x) \equal{} \sum_{k\equal{}1}^{\infty} \frac{x^k}{k}$ we substitute $ x \equal{} \frac{1}{2}$ to obtain

$ \minus{} \ln \frac{1}{2} \equal{} \ln 2 \equal{} \sum_{k\equal{}1}^{\infty} \frac{1}{k \cdot 2^k}$

hence our desired sum is $ 2 \ln 2$. [/hide]
[hide="Discussion"] Integrate $ \frac{1}{1 \minus{} x} \equal{} \sum_{k\equal{}0}^{\infty} x^k$. [/hide]


geometry
circumcircle
symmetry
perpendicular bisector
geometry solved

Let $\Omega ,\Omega '$ be the Brocard points of a triangle $\triangle ABC$. Show that if the Brocard line $\Omega  \Omega '$ is parallel to $BC$, then $AB = AC$.

Condition of the problem is equalent to $OK \perp BC$
It means that $K$ lies on middle-perpendicular of $BC$ or $BK = CK$.

$BK= \frac {a^2+c^2}{a^2+b^2+c^2}*s_b$ where $s_b $–semedian from $B$.

Using $s_b = \frac {2ac}{a^2+c^2}*m_b$ where $m_b $–median from $B$.

$BK= \frac {2ac}{a^2+b^2+c^2}*m_b$

So condition of the problem is equalent to $c*m_b =  b*m_c$

$ c^4+c^2 a^2 = b^4+b^2 a^2 $  or $(b^2-c^2)(a^2+b^2+c^2) = 0$

It means  $b=c$.

This problem is actually easier than you might think...

[color=blue][b]Problem (slightly extended).[/b] Let $\Omega$ and $\Omega^{\prime}$ be the two Brocard points of a triangle ABC. Prove that $\Omega\Omega^{\prime}\parallel BC$ if and only if b = c.[/color]

[i]Solution.[/i] We start with Prowler's idea:

Let O be the circumcenter and K the symmedian point of triangle ABC. Then, by a well-known property of the Brocard points $\Omega$ and $\Omega^{\prime}$, the line $\Omega\Omega^{\prime}$ is perpendicular to the Brocard axis OK of triangle ABC. Thus, $\Omega\Omega^{\prime}\parallel BC$ holds if and only if $OK\perp BC$.

Now, the point O, being the circumcenter of triangle ABC, lies on the perpendicular bisector of the segment BC; hence, $OK\perp BC$ holds if and only if the point K also lies on the perpendicular bisector of the segment BC.

Now, one direction of the problem is clear: If b = c, then the triangle ABC is isosceles with base BC, and hence its symmedian point K must lie on the axis of symmetry of the triangle, i. e. on the perpendicular bisector of the segment BC, and from the above it follows that $\Omega\Omega^{\prime}\parallel BC$.

Remains to show the other direction: If $\Omega\Omega^{\prime}\parallel BC$, then b = c.

Well, as we showed above, $\Omega\Omega^{\prime}\parallel BC$ implies that the point K lies on the perpendicular bisector of the segment BC. But since the point K is the symmedian point of triangle ABC, it must lie on the line AD, where D is the point of intersection of the tangents to the circumcircle of triangle ABC at the points B and C. In other words, the point A lies on the line KD. But since the two tangents from a point to a circle are equal in length, we have DB = DC, and thus the point D lies on the perpendicular bisector of the segment BC. We also know that the point K lies on the perpendicular bisector of the segment BC. Hence, the line KD is the perpendicular bisector of the segment BC. Therefore, since the point A lies on this line, we get AB = AC, and thus b = c. This completes the solution of the problem.

By the way, some properties of triangles ABC satisfying $\Omega\Omega^{\prime}\perp BC$ (instead of $\Omega\Omega^{\prime}\parallel BC$) can be found in http://www.mathlinks.ro/Forum/viewtopic.php?t=23609 post #3.

  Darij


If $ 2n\plus{}1$ and $ 3n\plus{}1$ are perfect squares, prove that $ 5n\plus{}3$ is not prime..
thx


calculus
integration
limit
calculus computations

Let $ f:[0,1]\to\mathbb R$ with $f\in C^0$ and a sequence $(a_n)_{n\ge 1}$ given
by:
\[ a_n=n\int_0^1x^nf(x^n)\,dx.\]
Conpute: $\lim_{n\to +\infty}a_n$.

Changing variables, $a_n=\int_0^1u^{1/n}f(u)\,du.$

Since $u^{1/n}\to1$ a.e. on $[0,1],$ we can interchange limit and integral, so

$a_n\to\int_0^1f(u)\,du.$

(You don't really need the Dominated Convergence Theorem - you could use that the convergence is uniform outside a small interval of negligible importance.)

O.K. Kent. For my students who don't know about the uniform convergence the following prove:
Let
\[F:[0,1]\to\mathbb R,F(x)=\int_0^xf(t)\,dt\Leftrightarrow F'=f\]
Then
\[a_n=\int_0^1x(F(x^n))'\,dx=xF(x^n)\big|_0^1-\int_0^1F(x^n)\,dx\\
=F(1)-\int_0^1F(x^n)\,dx=\int_0^1f(x)\,dx-\int_0^1F(x^n)\,dx\]
But
\[\big|F(x)\big|=\bigg|\int_0^xf(t)\,dt\bigg|\le\int|f(t)|\,dt\le Mx,\quad M=\displaystyle\sup_{t\in[0,1]}|f(t)|\\
\quad d\\
\big |F(x^n)\big|\le Mx^n,\quad x\in[0,1]\\
\quad d\\
\bigg|\int_0^1F(x^n)\,dx\bigg|\le M\int_0^1x^n\,dx=\frac M{n+1}\to 0\]


Find all [i]monotonically increasing[/i] functions $f$ : $\mathbb{R} \to \mathbb{R}$ such that:

1- $f(0)=0$

2- $x(f(x)+f(\frac{1}{x})) = f(x)^2+1$ for all  $x$  non zero in $\mathbb{R}$

Note. A function $f: \mathbb{R} \to \mathbb{R}$ is called [b]monotonically increasing[/b] if for any two numbers x and y such that $x\geq y$, we have $f\left(x\right)\geq f\left(y\right)$.

Mostly the same solution here:

$x=t$ $\to$ $t(f+g)=f^2+1$,
$x=\frac{1}{t}$ $\to$ $\frac{1}{t}(g+f)=g^2+1$,
Where $f=f(t)$ and $g=f \left( \frac{1}{t} \right)$.

$t=\frac{f^2+1}{f+g}=\frac{f+g}{g^2+1}$.
$(f+g)^2=(f^2+1)(g^2+1)$ $\to$ $f^2+g^2+2fg=f^2g^2+f^2+g^2+1$ $\to$ $f^2g^2-2fg+1=0$ $\to$ $(fg-1)^2=0$.
$fg=1$ $\to$ $t=\frac{f^2+1}{f+\frac{1}{f}}=f$.
(I checked that there are no divisions by 0.)

So, the only solution is $f(t)=t$ for every $t\in\mathbb{R}$.

Neither of us have used that the function is monotonically increasing.

It's a joke  :D ! A bad Indication!  :rotfl:  :rotfl:  :rotfl:


i'm also having trouble with this problem. help (not the answer) would be appreciated.  :)

When the number $ n$ is written in base $ b$ its representation is the two-digit number $ AB$ where $ A=b-2$ and $ B=2$. What is the representation of $ n$ in base $ b-1$?

I stink at number theory but this is what i have so far. i highly doubt im right
[hide] so in base b, n= (b-2)^1 + 2^0...i assume this because its a two-digit number.. now find n in base 10..solving i got b-2+2=b so n=b base 10..so in base b-1, n= (b-1)^1+1^0????[/hide]
like i said, i think im wrong..this problem doesn't make sense to me!!?? :(

[hide]Well, lets take any 2 digit number in base b of the form AB.

n is actually B*b^0 + A*b^1.

So in our problem we have n = (b-2)(b) + 2 = b^2 - 2b -2.

Now we want to represent n as XY, where n = Y + X(b-1). Go on from here. I hope this is right.  :maybe:

What you're doing wrong is you're making AB A^1 + B^0, but it's actually A*b^1 + B*b^0[/hide]

Actually, $ n=b^{2}-2b+2$, and n does not have to be a two digit number in base b-1.
In fact, the answer should be
[hide]101
cuz $ n=(b-1)^{2}+1$
[/hide]

[hide="Suggestions for how to proceed on this kind of problem"]In general, a good way to get some insight into this kind of problem is to consider some easy special cases and then look for a pattern, then try to prove why the pattern holds to get the general result.  E.g., if b = 10, then AB = 82 in base 10 and 101 in base 9.  Try a few more values of b, find a pattern, then prove the general result.

Even if you believe you can prove the general result without first trying special cases--as klebian and nealth approached it--it's still a good idea to check your general result afterwards by confirming that it works at least for a few easy to compute special cases.

More generally, on harder problems, one of the most powerful problem solving techniques is to try a few special cases to get some insight into the problem.[/hide]

yeah, that strategy got me into trouble on duke math meet 2006 tiebreaker #1 - the answer in this case would have been in terms of $ b$.  thankfully the guy who was proctoring me told me that simple fact so i got it right with only two wrong answers.  i got 2nd place with only solving 1 problem of 3!  Yay!  How the **** did that happen?


geometry
perimeter
geometric transformation
reflection
radical axis
geometry unsolved

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14573[/img]

 :(

Draw a circle, the perimeter of which is bisected by each of the 3 given circles  :?:

Assume that a circle $ (P)$ is bisected by $ (O_1)$ at $ A, B$, by $ (O_2)$ at $ C, D$ and by $ (O3)$ at $ E, F.$ This means that $ AB, CD, EF$ are diameters of $ (P).$ Then powers of $ P$ to $ (O_1), (O_2), (O_3)$ are equal:

$ \overline{PA} \cdot \overline{PB} \equal{} \overline{PC} \cdot \overline{PD} \equal{} \overline{PE} \cdot \overline{PF}$

Therefore, the center $ P$ of $ (P)$ is the radical center of $ (O_1), (O_2), (O_3),$ trivially constructed for intersecting circles. Let $ \varrho$ be the radius of $ (P).$ Since $ (O_1), (O_2), (O_3)$ cut $ (P)$ in diametrically opposite points, inversion in $ (P)$ with negative power $ \minus{} \varrho^2$ (equivalent to ordinary inversion with positive power $ \varrho^2$, followed by reflection in the inversion center $ P$) carries $ (O_1), (O_2), (O_3)$ into themselves $ \Longrightarrow$ reflection $ (O_1')$ of $ (O_1)$ in $ P$ intersects $ (O_1)$ at $ A, B \in (P).$ (We could reflect any circle $ (O_1), (O_2), (O_3)$ in $ P$ with the same result.) Then $ \varrho \equal{} PA \equal{} PB$ is the radius of $ (P).$


geometry
3D geometry
analytic geometry
symmetry

Calculate the total resistance over an n-dimensional cube where each vertice has a resistance.

Masoud Zargar

Did you mean each edge has a resistance? The easiest case is when you have to calculate the resistance between an opposite pair of vertices, so I'll do that:)

For easy navigation let's place the cube in N-dimensional space such that all the vertices have coordinates $[x_1,x_2,\ldots,x_N]$ where $x_i \in \{0,1\}$.

Now we can heavily exploit the symmetry of the situation - the vertices can be separated into levels according to how many 1-coordinates they have (i.e.$\sum{x_i}$). The source $[0,0,\ldots,0]$ is in the 0th level and the sink is in the Nth level.

In each of those levels the vertices have equal potentials (again, due to the symmetry). Hence, we can join them to only one vertex. Now we only have a chain of nodes, it remains to figure out how many resistors joined $k$th and $(k+1)$th level. There are $N\choose k$ vertices in level $k$, and each of them joins to $N-k$ vertices of the next level, hence there are ${N\choose k}(N-k)$ resistors between vertices in $k$th and $(k+1)$th level, all of which become paralell when we collapse the levels into a single node.

The total resistance is thus

\[ R = R_0\sum_{k=0}^{N-1}{\frac{1}{{N\choose k}(N-k)}}  \]

where $R_0$ is the resistance of a single edge.


[hide]N=2 (Mod 3, 4, and 5).  LCM (3,4,5)=60. The desired solution will be 60+2=62. :D

[/hide]

[hide]if it's 3 less than the multiples of 5, it have to end in 2 or 7, b/c the multiples of 5 is 5 and 10

if it's 2 less than the multiplies of 4, it's can be 2, 6, 0, 4, 8b/c the multiplies of 4 is 4, 8, 12, 16, 20

if it's 1 less than the multiples of 3 then it can be 2, 5, 8, 1, 4, 7, 0

and the only number that all sets have in common is 2, so it would be the LCM of 3,4,5 + 2

60+2 = 62[/hide]

[hide]The number of students in Teresa's graduating class is more than 50 and fewer than 100 and is 1 less than a multiple of 3, 2 less than a multiple of 4, and 3 less than a multiple of 5. How many students are in Teresa's graduating class?

50<x<100

the multiples of 3, 4, and 5 have to be consecutive.

therefore it has to be like x,y3,y4,y5

the only number that can be y is 6

so it becomes x,63,64,65[/hide]

[hide][b]x=62[/b][/hide]

Many solutions, but here's the easiest one.
[hide]call the number of students S. so:
S leaves a remainder of 2 when divided by 3,4, and 5. so it is two more than a multiple of 3, 4, and 5. So S leaves a remainder of two when divided by 60
so S= [b]62[/b][/hide]


email
AMC
USA(J)MO
USAMO
summer program
Mathcamp
ARML

Sure, as long as the person seems sane online and doesn't require you to meet him/her in a dark alley or something.

[quote="hatsup123"]Would you ever try to meet a stranger that you met online?[/quote]

Astounding question indeed. And highly ironic.

Yes, I most certainly would. :-) And the few times I have, (including MOP) it was certainly not a regretable experience.

I would probably do. I have never done it, basically because most of the people that I have met online and that I chat frequently are either in Europe or in South America!

[quote="Jhyon8149"][quote="eryaman"][b]only[/b] in a public place, or where there are many other people around.[/quote]I agree. I mean, I wouldn't go to a bar or an alley or something like that. :?[/quote]
I would definitely do that too

I would. unless if they are like one of those bad people that try to lure you online... :(

I would, if this person lived near to me. I have met someone at ICQ who lives in the same town as I do, and one day we met for like 5 minutes in the town center, as we had both something to do there that day and thought it would be a good opportunity to see each other irl. I would only do this in a very public place, however, and I don't think I would want to go to a bar or the cinema or something like that at the first meeting, it could be a strange or unpleasant or enervating or importunate person in reality after all.

[quote="hatsup123"]Would you ever try to meet a stranger that you met online?[/quote]What do you mean by stranger? :) I met most of the people I know online (on this very site :P) and then actually met them in person. I just met with Pierre Bornstein and Moubi in Paris a few days ago :D No back alleys though :P

Don't let them lure you into one with a math book though!

[quote="shobber"]never... i don't know who they are therefore i am afraid to meet them...[/quote]
Well, it depends..if I met someone on AIM, I probably definitely never want to meet them. But if it's something like aOPS, maybe I would...if he or she lived close enough...

[quote="furious"]possibly if i met them on a math website (like this one :D ). for some reason, i don't see a math geek trying to abduct a person.
however, if it's some random person in a chatroom or the like, i most likely wouldn't (unless they promise me candy and a ride in their car to a remote location).[/quote]Well, I think that I wouldn't have much of a concern to meet someone from this forum personally, or at least those people who are quite active on it. It is just know you know a little more about them than a usual person who you meet online.

[quote="xxreddevilzxx"][quote="djimenez"]Well, I suppose that in the USAMO and in the IMO there are many AoPSers and MathLinkers. Now, on the other hand, we could do something.... What about Atlanta :D , or I am going to be in the West coast (Portland to be precise) at the end of June!! :P[/quote]

is that for mathcamp visit?[/quote]No, it is not. I am going to the MSRI Summer School in Mathematical Graphics. I didn't even know that there was a mathcamp on Oregon during those days!

[quote="MithsApprentice"]That's because there isn't MathCamp then. MathCamp is July/August.[/quote]OK, thanks for the information, but any way, by July 4th I'll be back in Atlanta (or at least that is the plan)

[quote="MithsApprentice"][quote="paladin8"]PS. Hey Miths, I'm less than 1000 posts behind you now! Haha, I'll catch up someday... hopefully...[/quote]

Hmm...less than 1000?...Since when?[/quote]

Grr, I was gone all day ;)

[quote="xxreddevilzxx"][quote="MithsApprentice"][quote="djimenez"]OK, thanks for the information, but any way, by July 4th I'll be back in Atlanta (or at least that is the plan)[/quote]

Well, MathCamp starts on the 3rd as far as I know. So you might not miss it completely.

And quit pulling excuses, paladin. Although, I'm sure you'll catch up soon enough. :P[/quote]

i think i will be able to catch up to paladin faster than he can to you... i started out 0 when he was around 350-400, and now we are about 100 off.[/quote]

spammer ;) i'll make sure i stay ahead of you :D

[quote="paladin8"][/quote][quote="xxreddevilzxx"][quote="MithsApprentice"][quote="djimenez"]OK, thanks for the information, but any way, by July 4th I'll be back in Atlanta (or at least that is the plan)[/quote]

Well, MathCamp starts on the 3rd as far as I know. So you might not miss it completely.

And quit pulling excuses, paladin. Although, I'm sure you'll catch up soon enough. :P[/quote]

i think i will be able to catch up to paladin faster than he can to you... i started out 0 when he was around 350-400, and now we are about 100 off.[/quote][quote="paladin8"]

spammer ;) i'll make sure i stay ahead of you :D[/quote]Well, under that definition probably you'll have to tell me spammer also :P , I am ahead of you and I have less than a fifth of the time you have in the forum :D . I hope my posts are not considered spam :?

Hey, if they counted active time, I would be owning you all :P I was just inactive for a year - I actually considered making a new account so my average posts/day would be higher :D

Why would having high post per day a good thing? I just like to have a high post number to show that I have contributed and participated in the site in any way.

BTW, the videos of rubik cubes are awesome. You gotta show me litoxxox. I can olny solve 2 sides at the most. ;)

Well djiminez was pointing out that he's been here for much less time and has more posts than me => his average posts/day is higher than mine. I was just explaining why mine is so low - when I came back this year it was about 0.5 I think; I've managed to work it up to about 2... *shrugs* but yeah, I guess it doesn't mean much. Let's just go by raw post count because then I'm a lot higher up :P

well... you joined about 2  months before me, so that explains.


Prove that:

$a^2 b^2 c^2  + a^2  + b^2  + c^2  + 2 \ge 2\left( {ab + bc + ca} \right),\forall a,b,c \in \left[ {0, + \infty } \right)$

As a quadratic in c, the discriminant is $4(a+b)^2 - 4(a^2b^2+1)(a^2+b^2+2-2ab)$

So we show that

$(a+b)^2 \le (a^2b^2+1)(a^2+b^2+2-2ab)$

which reduces to

$4ab \le 2 + a^2b^2(a^2+b^2+2-2ab)$

Then

$2 + a^2b^2(a^2+b^2+2-2ab) \ge 2 + 2a^2b^2 \ge 4ab$

so that's all.


#1:
Allison has sneezed exactly one million times in her life. Because of her age there has to have been at least one day when she  sneezed at least 101 times. What is the oldest, in days, Allison could be?

#2:
How many 4-digit #s have the property that  each of the 3 2-digit #s formed by consecutive digits is divisible either 19 or 31?

[hide]if its day 10000 (1000000/100), she's sneezed 100 times per day (on average), so on day $ \boxed{9999}$, there must have been a day where she sneezed more (101).[/hide]

[hide]1. easy. she could have sneezed one time the other days. so its 1,000,000-101+1=[b]999900[/b] erdogan i dont think your answer is correct because it never said anything about the average.

2. lets see...i dont know[/hide]

[quote="abacadaea"]1. easy. she could have sneezed one time the other days. so its 1,000,000-101+1=[b]999900[/b] erdogan i dont think your answer is correct because it never said anything about the average.

2. lets see...i dont know[/quote]

Actually you are wrong.  And i cant vote and that post is voted as spam so im the only one that agrees.

First dont say its easy when you get it wrong.  and erdogan got the answer right...
You want to max out on the other days.  If you have 999900 days, then guess what, you DONT HAVE to sneeze 101 times on a day.  Read the question.  It says "Because of her age there has to have been at least one day when she sneezed at least 101 times".  If her age was 999900 days, then she doesnt have to sneeze 101 times on a day, she can, but she doesnt HAVE too.  Otherwise she could be inifity years old because she can sneeze 0 times on a day.  So you want to maximize on 100 times a day and thats 10000 days.  However, 10000 days is the minimum you can have without having to sneeze atleast 101 times a day (again you can still sneeze 101 times in a day but YOU DONT HAVE TO).  So we take 1 less than that.

[quote="abacadaea"]but it what is the OLDEST she could be[/quote]

Omg but her age still has to fit the "for this age, she must have atleast one day of 101 sneezes"  If she was 999900 years old, she doesnt have to neccessarily have to have 1 day of 101 sneezes, she can but she DONEST HAVE TO.

Otherwise she can be $ \infty$ years old.  She could sneeze

0 times on day 1
0 times on day 2..

until

1,000,000 times on day $ \infty$

From that case, she can still sneeze 1,000,000 times and her age is the oldest as she can be ($ \infty$)\

But that age, just like 999900, doesnt fit the criteria of "Because of her age there has to have been at least one day when she sneezed at least 101 times".  For that age, there DOESNT HAVE TO BE ONE DAY

[quote="abacadaea"]im confused ill just stop[/quote]

Dude how did you get 999900 in the first place?  She doesnt have to even sneeze every day so if i were doing what you were doing, the answer to that would be infiity. You said in your original solution, "easy she couldnve sneezed 1 time on every other day."  Why 1 time?  Make it 0 times and she would be even older.  But that method is incorrect.

But like the future said, if she was 999900 days old, she doesnt have to sneeze atelast 101 times on a day.  She can, but she doesnt 100% have to.  However, she has to always (100%) of the time have atleast 1 day of 101+ sneezes, so its 9999.

Basically heres an expamle:
If its 999900 days old, she can go 500000 days of sneezing twice and 499900 days of sneezing 0 times for a total of 1,000,000 sneezes.  Now for that case, did she ever sneeze 101 times per day?  Nope.

But for 9999 days no matter how you make a case (there are lots and lots of cases), she will alwasy have atleast 1 day of 101+ sneezes.

oooooooooohhhhhhh i get it. so it has to be no matter what the cirmcumstances are.

Since no one answered #2, I will:

[hide]Let's look at all the possible two-digit sequences that might appear:

$ 19$
$ 38$
$ 57$
$ 95$
$ 31$
$ 62$
$ 93$

Let's split into cases, depending on what the first two digits are:
$ 19\rightarrow 93\rightarrow 38$
$ 19\rightarrow 93\rightarrow 31$
$ 19\rightarrow 95\rightarrow 57$
$ 38\rightarrow \text{nothing}$
$ 57\rightarrow 76\rightarrow 62$
$ 76\rightarrow 62\rightarrow \text{nothing}$
$ 95\rightarrow 57\rightarrow 76$
$ 31\rightarrow 19\rightarrow 93$
$ 31\rightarrow 19\rightarrow 95$
$ 62\rightarrow \text{nothing}$
$ 93\rightarrow 31\rightarrow 19$
$ 93 \rightarrow 38\rightarrow \text{nothing}$

So, we have $ \boxed{8}$ such $ 4$ digit numbers:
$ 1938$
$ 1931$
$ 1957$
$ 5762$
$ 9576$
$ 3193$
$ 3195$
$ 9319$

As an interesting side-note, we can obtain a number of any digit satisfying the conditions with these cycles:
$ 1931931931931\cdots$
$ 3193193193193\cdots$
$ 9319319319319\cdots$[/hide]

EDIT: What's wrong with me?

[quote="vishalarul"]Since no one answered #2, I will:

[hide]Let's look at all the possible two-digit sequences that might appear:

$ 19$
$ 38$
$ 57$
$ 95$
$ 31$
$ 62$
$ 93$

Let's split into cases, depending on what the first two digits are:
$ 19\rightarrow 93\rightarrow 38$
$ 19\rightarrow 93\rightarrow 31$
$ 19\rightarrow 95\rightarrow 57$
$ 38\rightarrow \text{nothing}$
$ 57\rightarrow 76\rightarrow 62$
$ 76\rightarrow 62\rightarrow \text{nothing}$
$ 95\rightarrow 57\rightarrow 76$
$ 31\rightarrow 19\rightarrow 93$
$ 31\rightarrow 19\rightarrow 95$
$ 62\rightarrow \text{nothing}$
$ 93\rightarrow 31\rightarrow 19$
$ 93 \rightarrow 38\rightarrow \text{nothing}$

So, we have $ \boxed{9}$ such $ 4$ digit numbers:
$ 1938$
$ 1931$
$ 5762$
$ 9576$
$ 3193$
$ 3195$
$ 9319$

As an interesting side-note, we can obtain a number of any digit satisfying the conditions with these cycles:
$ 1931931931931\cdots$
$ 3193193193193\cdots$
$ 9319319319319\cdots$[/hide][/quote]

I think theres only 8.  First you listed out 7 and said there was 9, but I think you meant 8 ( you mistpyed twice?) because in your previous work you had 8 that worked.

I was wondering if questions with infinity appear on Mathcounts and if there is a quick way to find out that a question's answer is surely infinity?

        Sorry I appologies for the spelling and grammar error.

No, actualy it occasionally appears, but it's not like limits, it's just basically geometric and arithmetic sequences.

[hide]the first 2 digits can be 19, 38, 57, 76, 85, 31, 62, 93, for 19 you can only have 3 being the next digit so we now have 193 and for the last digit you can have either 1 or 8 so that gives you 2 possibilities. for 38 you have to have a five next so you now have 385 and that gives 1 solution so we now have 3 total possibilities. for 57 the only possibility is 5762, for 76 there are none, for 85 you have 8576 for 31 you can have 3193, for 62 there are none, and for 93 there are 9319 and 9385. so total we have $ \fbox{8}$[/hide]


What common fraction is equivalent to $ .\overline{325}$?

Eh, when I put 325/999 on FTW, it says incorrect...

I believe they changed it so that a/b is acceptable (not sure if \frac{a}{b} is acceptable).

only for a very small portion of the problems, though...

On FTW, it says the answer is $ \frac{161}{395}$, but it should be $ \frac{325}{999}$.  :huh:

Correction MathChampion: FTW says it is 161/495, which is correct. But how come 325/999 works? Are there two answers? :(

[quote="GameBot"]What common fraction is equivalent to $ .3\overline{25}$?[/quote] is what it should be. there was a typo when entering the problem.

I don't get the last 2 steps. Can you please elaborate


[quote="vuthanhtu_hd"][b]Problem.[/b] (Vu Thanh Tu)Let $ a,b,c > 0$ .Prove that
$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]

we can prove that

$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{\frac{a\plus{}b\plus{}c}{3}}$

Assume that $ a \equal{} max\{a, b, c\}$
Because $ \sum (b \plus{} c \minus{} a)^2 \equal{} 3\sum a^2 \minus{} 2\sum ab$, so we only need to prove:
$ (\frac {1}{b} \minus{} \frac {1}{a})(c \plus{} a \minus{} b)^2 \plus{} (\frac {1}{c} \minus{} \frac {1}{a})(a \plus{} b \minus{} c)^2 \ge 0$
(obviously trues because $ a \equal{} max \{a, b, c\}$)
[quote="kuing"]

we can prove that

$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{\frac {a \plus{} b \plus{} c}{3}}$[/quote]
I think it's not trues, kuing  :maybe:

to prove
\[ \frac{{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac{{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac{{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac{{3(a^2  \plus{} b^2  \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\frac{{a \plus{} b \plus{} c}}{3}}}.\]

by CS, we have

\[ \left( {a(b \plus{} c \minus{} a)^2  \plus{} b(c \plus{} a \minus{} b)^2  \plus{} c(a \plus{} b \minus{} c)^2 } \right)\left( {\frac{{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac{{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac{{(a \plus{} b \minus{} c)^2 }}{c}} \right)\]
\[ \ge \left( {(b \plus{} c \minus{} a)^2  \plus{} (c \plus{} a \minus{} b)^2  \plus{} (a \plus{} b \minus{} c)^2 } \right)^2\]
\[ \equal{} \left( {3(a^2  \plus{} b^2  \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)} \right)^2 ,\]
so we get

\[ \frac{{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac{{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac{{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac{{\left( {3(a^2  \plus{} b^2  \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)} \right)^2 }}{{a(b \plus{} c \minus{} a)^2  \plus{} b(c \plus{} a \minus{} b)^2  \plus{} c(a \plus{} b \minus{} c)^2 }},\]
so we just need to prove that

\[ \frac{{3(a^2  \plus{} b^2  \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{a(b \plus{} c \minus{} a)^2  \plus{} b(c \plus{} a \minus{} b)^2  \plus{} c(a \plus{} b \minus{} c)^2 }} \ge \frac{3}{{a \plus{} b \plus{} c}}\]
\[ \iff a^2 b \plus{} ab^2  \plus{} b^2 c \plus{} bc^2  \plus{} c^2 a \plus{} ca^2  \ge 6abc,\]
true.

by Chebyshev Ineq we have

\[ \frac{{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac{{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac{{(a \plus{} b \minus{} c)^2 }}{c}\]
\[ \ge \frac{1}{3}\left( {(b \plus{} c \minus{} a)^2  \plus{} (c \plus{} a \minus{} b)^2  \plus{} (a \plus{} b \minus{} c)^2 } \right)\left( {\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}} \right)\]
\[ \ge \frac{1}{3}\left( {(b \plus{} c \minus{} a)^2  \plus{} (c \plus{} a \minus{} b)^2  \plus{} (a \plus{} b \minus{} c)^2 } \right)\frac{9}{{a \plus{} b \plus{} c}}\]
\[ \equal{} \frac{{3(a^2  \plus{} b^2  \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\frac{{a \plus{} b \plus{} c}}{3}}}\]

 :lol:

[quote="vuthanhtu_hd"][b]Problem.[/b]Let $ a,b,c > 0$ .Prove that
$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sum \frac{(b\plus{}c\minus{}a)^2}{a} \ge \frac{(a\plus{}b\plus{}c)[3(a^2\plus{}b^2\plus{}c^2)\minus{}2(ab\plus{}bc\plus{}ca)]}{ab\plus{}bc\plus{}ca}.$

[quote="can_hang2007"][quote="vuthanhtu_hd"][b]Problem.[/b]Let $ a,b,c > 0$ .Prove that
$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sum \frac {(b \plus{} c \minus{} a)^2}{a} \ge \frac {(a \plus{} b \plus{} c)[3(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)]}{ab \plus{} bc \plus{} ca}.$[/quote]

yes...

by Chebyshev Ineq we have
\[ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c}\]
\[ \ge \frac {1}{3}\left( {(b \plus{} c \minus{} a)^2 \plus{} (c \plus{} a \minus{} b)^2 \plus{} (a \plus{} b \minus{} c)^2 } \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)\]
\[ \equal{} \left(3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)\right)\frac{(ab\plus{}bc\plus{}ca)^2}{3abc(ab\plus{}bc\plus{}ca)}\]
\[ \ge \left(3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)\right)\frac{3abc(a\plus{}b\plus{}c)}{3abc(ab\plus{}bc\plus{}ca)}\]
\[ \equal{} \frac {(a \plus{} b \plus{} c)[3(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)]}{ab \plus{} bc \plus{} ca}.\]

 :)

[quote="can_hang2007"][quote="vuthanhtu_hd"][b]Problem.[/b]Let $ a,b,c > 0$ .Prove that
$ \frac {{(b + c - a)^2 }}{a} + \frac {{(c + a - b)^2 }}{b} + \frac {{(a + b - c)^2 }}{c} \ge \frac {{3(a^2 + b^2 + c^2 ) - 2(ab + bc + ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sum \frac {(b + c - a)^2}{a} \ge \frac {(a + b + c)[3(a^2 + b^2 + c^2) - 2(ab + bc + ca)]}{ab + bc + ca}.$[/quote]

stronger is:

$ \sum\frac{(b+c-a)^2}\geq \frac{(a+b+c)(4\sum{a^2}-3\sum{bc}){ab+ca+bc}.$

<=>
${ 4\sum(b^2c^2(a-b)(a-c)}+\sum{abc(b+c)(a-b)(a-c)}+(b-c)^2(a-c)^2(a-b)^2\geq 0.$
But by CS？
BQ

[quote="can_hang2007"][quote="vuthanhtu_hd"][b]Problem.[/b]Let $ a,b,c > 0$ .Prove that
$ \frac {{(b \plus{} c \minus{} a)^2 }}{a} \plus{} \frac {{(c \plus{} a \minus{} b)^2 }}{b} \plus{} \frac {{(a \plus{} b \minus{} c)^2 }}{c} \ge \frac {{3(a^2 \plus{} b^2 \plus{} c^2 ) \minus{} 2(ab \plus{} bc \plus{} ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sum \frac {(b \plus{} c \minus{} a)^2}{a} \ge \frac {(a \plus{} b \plus{} c)[3(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)]}{ab \plus{} bc \plus{} ca}.$[/quote]
See my proofs here: http://can-hang2007.blogspot.com/2010/01/inequality-43-v-t-tu-v-q-b-can.html

[quote="can_hang2007"][quote="can_hang2007"][quote="vuthanhtu_hd"][b]Problem.[/b]Let $ a,b,c > 0$ .Prove that
$ \frac {{(b + c - a)^2 }}{a} + \frac {{(c + a - b)^2 }}{b} + \frac {{(a + b - c)^2 }}{c} \ge \frac {{3(a^2 + b^2 + c^2 ) - 2(ab + bc + ca)}}{{\max \{ a;b;c\} }}$

 :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \sum \frac {(b + c - a)^2}{a} \ge \frac {(a + b + c)[3(a^2 + b^2 + c^2) - 2(ab + bc + ca)]}{ab + bc + ca}.$[/quote]
See my proofs here: http://can-hang2007.blogspot.com/2010/01/inequality-43-v-t-tu-v-q-b-can.html[/quote]

stronger is:

$ \sum\frac {(b + c - a)^2}{a}\geq \frac {(a + b + c)(4\sum{a^2} - 3\sum{bc})}{ab + ca + bc}.$

<=>
${ 4\sum(b^2c^2(a - b)(a - c)} + \sum{abc(b + c)(a - b)(a - c)} + (b - c)^2(a - c)^2(a - b)^2\geq 0.$
But by CS？
BQ

[quote="can_hang2007"][quote="xzlbq"]stronger is:

$ \sum\frac {(b + c - a)^2}{a}\geq \frac {(a + b + c)(4\sum{a^2} - 3\sum{bc})}{ab + ca + bc}.$

<=>
${ 4\sum(b^2c^2(a - b)(a - c)} + \sum{abc(b + c)(a - b)(a - c)} + (b - c)^2(a - c)^2(a - b)^2\geq 0.$
But by CS？
BQ[/quote]
If you want a solution by Cauchy-Schwarz, see here: http://can-hang2007.blogspot.com/2010/01/inequality-44-b-q-liu.html[/quote]

thank you!
BQ

4-variables is:
$ 4\,{\frac {\left( a \plus{} b \plus{} c \plus{} d \right) \left( {a}^{2} \plus{} {b}^{2} \plus{} {c}^{2} \plus{} {d} ^{2} \right) }{ab \plus{} bc \plus{} cd \plus{} ad}}\leq {\frac {\left( a \plus{} b \plus{} c \minus{} d \right) ^{2}} {d}} \plus{} {\frac {\left( b \plus{} c \plus{} d \minus{} a \right) ^{2}}{a}} \plus{} {\frac {\left( c \plus{} d \plus{} a \minus{} b \right) ^{2}}{b}} \plus{} {\frac {\left( d \plus{} a \plus{} b \minus{} c \right) ^{2}}{c}}.$

[size=200]or[/size]

$ {\frac { \left( a\plus{}b\plus{}c\plus{}d \right)  \left( {a}^{2}\plus{}{b}^{2}\plus{}{c}^{2}\plus{}{d}^{2
} \right) }{ab\plus{}bc\plus{}cd\plus{}ad}}\leq {\frac { \left( a\plus{}b\minus{}c \right) ^{2}}{c}}\plus{}
{\frac { \left( b\plus{}c\minus{}d \right) ^{2}}{d}}\plus{}{\frac { \left( c\plus{}d\minus{}a \right)
^{2}}{a}}\plus{}{\frac { \left( d\plus{}a\minus{}b \right) ^{2}}{b}}.$
BQ

Why are you afraid of 4 -variables?
Is 4 -variables of the C-S as a good grasp？

$ 0\leq {\frac { \left( y\plus{}z\minus{}x \right) ^{2}}{x}}\plus{}{\frac { \left( z\plus{}x\minus{}y
 \right) ^{2}}{y}}\plus{}{\frac { \left( x\plus{}y\minus{}z \right) ^{2}}{z}}\minus{}2\,{\frac {
 \left( y\plus{}z\minus{}x \right) ^{2}}{y\plus{}z}}\minus{}2\,{\frac { \left( z\plus{}x\minus{}y \right) ^{2
}}{z\plus{}x}}\minus{}2\,{\frac { \left( x\plus{}y\minus{}z \right) ^{2}}{x\plus{}y}}.$

$ 3/8\,{\frac { \left( a\plus{}b\plus{}c\minus{}d \right) ^{2}}{a\plus{}b\plus{}c}}\plus{}3/8\,{\frac {
 \left( b\plus{}c\plus{}d\minus{}a \right) ^{2}}{b\plus{}c\plus{}d}}\plus{}3/8\,{\frac { \left( c\plus{}d\plus{}a\minus{}b
 \right) ^{2}}{a\plus{}c\plus{}d}}\plus{}3/8\,{\frac { \left( d\plus{}a\plus{}b\minus{}c \right) ^{2}}{a\plus{}b\plus{}
d}}\leq {\frac { \left( a\plus{}b\minus{}c \right) ^{2}}{a\plus{}b}}\plus{}{\frac { \left( b\plus{}c\minus{}
d \right) ^{2}}{b\plus{}c}}\plus{}{\frac { \left( c\plus{}d\minus{}a \right) ^{2}}{c\plus{}d}}\plus{}{
\frac { \left( d\plus{}a\minus{}b \right) ^{2}}{a\plus{}d}}.$

Large-scale march to the 4-variables.

So that c-s method，
Just as if 3-variables!
BQ


geometry
incenter
geometry proposed
U sviti matematyky

A circle $ \omega_1$ touches sides of angle $ A$ at points $ B$ and $ C.$ A straight line $ AD$ intersects $ \omega_1$ at points $ D$ and $ Q,$ $ AD<AQ.$ The circle $ \omega_2$ with center $ A$ and radius $ AB$ intersects $ AQ$ at a point $ I$ and intersects some line passing through the point $ D$ at points $ M$ and $ P.$ Prove that $ I$ is the incenter of triangle $ MPQ.$

Denote BD∩⊙A=J  then we have ∠ABJ=∠AJB=∠BQD=∠AQB    =>  A,B,Q,J are concyclic     =>   BD*DJ=AD*DQ   =>   DM*DP=BD*DJ=AD*DQ  =>   A,M,Q,P are concyclic    =>  ∠MQA=∠PQA  MA=AI=AP  =>  I is the incenter of △MQP


[quote="mathemagician1729"]8 identical spheres each of diameter 20 fit tightly into a cube of side length 40 so that each sphere just touches three of the faces of the cube. What is the radius of the largest sphere that will fit in the central space, just touching all eight other spheres, to the nearest tenth?[/quote]
[hide]
Each sphere has a radius of 10.  Picture a sphere inside each of the
eight smaller, congruent, nonoverlapping cubes, where the larger cube is
divided into them.   Each smaller cube has a side length of 20.  One
vertex (corner) of each smaller cube meets (coincides with) the center
of the larger cube.

The radius of the center sphere will equal the distance from a smaller
cube's center to the center of the larger cube (which is where a
vertex from each of the eight smaller cubes meet), minus the radius
of one of the eight larger spheres.

The distance from the center of a smaller cube to one of its vertices is
$ d \equal{} \sqrt {x^2 \plus{} y^2 \plus{} z^2}$, where $ x, y$, and $ z$ can be the
side lengths of a rectangular box, of which each of the smaller cubes are
also rectangular boxes.  Here, $ x \equal{} y \equal{} z \equal{} 10$, because the distance
from the center of a smaller cube to one of its faces is half of $ 20$.

Then, the distance from the center of a smaller cube to the center of
the larger cube is
$ \sqrt {100 \plus{} 100 \plus{} 100}$
$ \equal{} \sqrt {300}$
$ \equal{} 10\sqrt {3}$.

Then the radius of the central sphere is
$ (10 \sqrt {3} \minus{} 10)$ $ units$.

This is approximately equal to $ 7.3$ $ units$, to the nearest tenth.

 [/hide]


geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

the following equality is the key:
  $\frac{n-k}{k+1}\binom{n}{k}=\binom{n}{k+1}$.
 the general term is $\frac{2k+1}{n+k+1}\binom{2n}{n+k}$.

Now, $\big{(}1-\frac{2k+1}{n+k+1}\big{)}\binom{2n}{n+k}\ = (\frac{n-k}{n+k+1})\binom{2n}{n+k+1}=$(by the first equality)$=\binom{2n}{n+k+2}$. so that the general term is simply \[\binom{2n}{n+k}-\binom{2n}{n+k+1}\].
so now the entire sum telescopes to $\binom{2n}{n}$.

Well, I think you meant
$\big{(}1-\frac{2k+1}{n+k+1}\big{)}\binom{2n}{n+k}\ = \binom{2n}{n+k+1}$

Anyway, does anybody have a "combinatorics" solution?

I have combinatorial solution.

Think about the number of the shortest root from $(0,0)$ to $(n,n)$.

Then it can be easily to know that

First term, $\displaystyle \binom{2n}{n}\frac{1}{n+1}$ is the number of the shortest root that are not bigger than $y=x$. (This is the Catalan Number)

Second term, $\displaystyle \binom{2n}{n-1}\frac{3}{n+2}$ is the number of the shortest roots that are not bigger than $y=x+1$ starting from $(0,0)$ and bigger than $y=x$ at least one time.

Remainders can be noticed similarly.

And the total number of the shortest roots is $\displaystyle \binom{2n}{n} $, cuz we have $n$ rightarrow, and $n$ uparrow.

so, we get $\displaystyle \binom{2n}{n}\frac{1}{n+1} + \binom{2n}{n-1}\frac{3}{n+2} + ... + \binom{2n}{0}\frac{2n+1}{2n+1} = \binom{2n}{n} $

$Q.E.D.$

[quote="Chang Woo-JIn"]
Second term, $\displaystyle \binom{2n}{n-1}\frac{3}{n+2}$ is the number of the shortest roots that are not bigger than $y=x+1$ starting from $(0,1)$
[/quote]

to $(n,n)$ ? Why?

It was my mistake.

$\displaystyle \displaystyle \binom{2n}{n-1}\frac{3}{n+2}$ can be proved by using reflection

like $(0,0)$ reflect to $y=x+1$. This technique is used in the proof of Catalan Number.

Hmm... how to reflect and get $\binom{2n}{n-1}\frac{3}{n+2}$? I can't see...

Actually I can't see how to do the condition "not bigger than $y=x+1$" and "bigger than $y=x$ at least one time" at the same time.


algebra
polynomial
Vieta
algebra unsolved

Find all integer a for which there exists distinct positive integers x and y such that (axy+1) divides (ax2+1)^2.
PS:Sorry but i want to show everyone this original problem(not a translated problem!)

I have no idea what you mean with x#y, but I don't see any problem putting $ x\equal{}y\equal{}1$ to have it for all $ a$... :?

By the way, it's not so useful to pm people with that. ;)

Is it trivial or super hard? Put $ y\equal{}ax^3\plus{}2x$ then $ axy\plus{}1\equal{}a^2x^4\plus{}2ax^2\plus{}1\equal{}(ax^2\plus{}1)^2$ :D

[quote="Arne"]Shouldn't $ x$ and $ y$ be distinct?[/quote]That wasn't there when I posted!  :(  [size=75](and that is probably what the disappeared x#y meant)[/size]

In any case the answer is still trivially for all $ a$, so I doubt we have the correct statement this time...

But how can you conclude the solution?It seems to be a particular case in this problem, right? But it's a very good step even though I haven't chosen any y like u!Anyway, thx for your helping!And i think i can solve the equation y=ax^3+2x by Vieta!
@PLURICOMPLEX:ARe u Vietnamese?? :)

Do you actually even understand your own problem? You ask for existence of $ x$ and $ y$, then don't be surprised we give you a particular choice for $ x$ and $ y$...  :maybe:

So, can you show me the final solution of this problem Prof.Peter?I just infer that when a>= -1 then we always find a pair of positive integers(x:y) satisfy this problem! But when a<-1 how can i prove it's unreasonable?I got it by applying Vieta jumping but it's too ugly because it's completely same as IMO 07 prob 5 when a<-1!

When $ a\leq \minus{} 1$ I found that $ a$ must be $ \minus{} 1$ and even addition that case, It's still not super hard :D

Yeah i got the final result is a>=-1!But i'm really curious about your solution, pluricomplex!Can you show me your solution?I'm very pleased if you shoe me your solution!
And I also proved the inverse  problem:
Problem:Find all integers a to exist a pair of distinct positive integer(x;y) to satisfy:(axy+1) is divisible by (ax^2+1) :)

In general, the problem is:
Find all integer $ a$ such that there exist $ x,y$ are distince positive integers such that $ axy\plus{}1\mid (ax^2\plus{}1)^2$.

[hide="Solution"]1) Any $ a\geq0$ satisfied because $ y\equal{}ax^3\plus{}2x$
2) $ a\equal{}\minus{}1$ then chose $ x\equal{}1$ and $ y\equal{}2$.
3) Assume $ a\geq2$ and $ x\neq y$ are positive integers such that $ axy\minus{}1\mid (ax^2\minus{}1)^2$. This implies $ axy\minus{}1\mid (x\minus{}y)^2$. So $ (x\minus{}y)^2\equal{}k(axy\minus{}1)$. Assume $ (x,y)$ are such positive integers with minimize $ x\plus{}y$. Assume $ x\geq y$. Because $ x^2\minus{}(2y\plus{}kay)x\plus{}k\plus{}y^2\equal{}0$ so $ x_1\equal{}2y\plus{}kay\minus{}x\equal{}\frac{k\plus{}y^2}{x}$ is positive integer and $ (x_1,y)$ is another root. So $ x_1\geq x$ which implies $ x^2\minus{}y^2\leq k\equal{}\frac{(x\minus{}y)^2}{axy\minus{}1}$ or $ x\plus{}y\leq \frac{x\minus{}y}{axy\minus{}1}$ so $ axy\minus{}1\leq\frac{x\minus{}y}{x\plus{}y}<1$ so $ axy<2$ which is impossible.[/hide]

It's very nice solution(similar to my solution!).But i think you made a mistake for your assumption cause if you assume a>=2 and there exists (x;y) satisfy the problem then you prove that it's unreasonbale! So, it means"when a>=2; there isn't exist a piar of positive intgers(x;y) like that!That's clearly false :maybe: ! And how about concerning my inverse problem? :)

when I wrote $ a\geq2$ I mean the case $ a\leq \minus{}2$, but by putting $ a_1\equal{}\minus{}a$ we can reduce to the last case as I showed.


[quote="CDP100"]$ 5(x^4 \plus{} y^4 \plus{} z^4) \plus{} 24(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 8[xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x)] \plus{} 12xyz(x \plus{} y \plus{} z)$[/quote]
Try $ x\equal{}y\equal{}z\equal{}\frac{1}{2}.$ :wink:

[quote="zaya_yc"][quote="CDP100"]$ 5(x^4 \plus{} y^4 \plus{} z^4) \plus{} 24(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 8[xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x)] \plus{} 12xyz(x \plus{} y \plus{} z)$[/quote]

$ 5(x^4 \plus{} y^4 \plus{} z^4) \plus{} 24(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 8[xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x)] \plus{} 12xyz(x \plus{} y \plus{} z)$
<->'t think you come up with this solution
$ 3(x^{4} \plus{} y^{4} \plus{} z^{4}) \plus{} (x^{2} \plus{} y^{2} \minus{} 4xy)^{2} \plus{} (x^{2} \plus{} z^{2} \minus{} 4xz)^{2} \plus{} (y^{2} \plus{} z^{2} \minus{} 4yz)^{2} \plus{} 6(xy \plus{} yz \plus{} zx)^{2}\geq 0$[/quote]

How about:3(x^4+y^4+z^4)+6(xy+yz+xz)<0? :)
PS: Try x=0,y=z=1/3 :wink:

[quote="CDP100"]I'm sorry! You're right. ;) It's not $ x \plus{} y$  -  it's  $ x^2 \plus{} y^2$.

$ 5(x^4 \plus{} y^4 \plus{} z^4) \plus{} 24(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 8[xy(x^2 \plus{} y^2) \plus{} yz(y^2 \plus{} z^2) \plus{} zx(z^2 \plus{} x^2)] \plus{} 12xyz(x \plus{} y \plus{} z)$

I hope there's no more mistakes :)

PS: Sorry for my bad English(maybe you are sick of this phrase, but I'm still improving my English  :oops: )[/quote]

$ 2(x^{2}\plus{}y^{2}\minus{}2xy)^{2}\plus{}2(x^{2}\plus{}z^{2}\minus{}2xz)^{2}\plus{}2(y^{2}\plus{}z^{2}\minus{}2xz)^{2}\plus{}x^{4}\plus{}y^{4}\plus{}z^{4}\plus{}12(x^{2}y^{2}\plus{}x^{2}z^{2}\plus{}y^{2}z^{2}\minus{}xy*yz\minus{}xy*zx\minus{}yz*zx)\geq 0$

$ x\equal{}y\equal{}z$

Good job, zaya_yc! One alternative solution:

$ LHS \minus{} RHS \equal{} (x \minus{} y)^4 \plus{} (y \minus{} z)^4 \plus{} (z \minus{} x)^4 \plus{} ( \minus{} x \plus{} y \plus{} z)^4 \plus{} (x \minus{} y \plus{} z)^4 \plus{} (x \plus{} y \minus{} z)^4$


geometry
inequalities
combinatorics unsolved
combinatorics

I'm not sure that it is combinatorics. But it is not algebra and it is not number theory. May be geometry? I don't know.

Problem. There are several blots $B_1, \ldots, B_n$ on a square $a \times a$ sheet of paper. Let $S(B_i)$ be an area of the blot $B_i$. It is known that for every $i$ the inequality $S(B_i) \leq 1$ holds. And also it is known that every line parallel to any side of the sheet cuts at most one blot. Prove that $\sum_{i=1}^{n}{S(B_i)} \leq a$.

Let $x_i,y_i$ be the respective projections of blot $B_i$ on some lines parallel to the each of the sides of the sheet.

If $x_iy_i \geq 1$ then $\sqrt {x_iy_i} \geq 1 \geq s(B_i).$
If $x_iy_i \leq 1$ then $\sqrt {x_iy_i } \geq x_iy_i \geq s(B_i)$.
Thus, in every case $\sqrt {x_iy_i } \geq s(B_i)$.

Now, any line parallel to the sides of the sheet intersects at most one blot, thus :
$x_1 + \cdots x_n \leq a$ and $y_1 + \cdots + y_n \leq a$.

Th desired conclusion follows now easily from AM/GM :
$s(B_1) + \cdots + s(B_n) \leq \sqrt {x_1y_1} + \cdots + \sqrt {x_ny_n} \leq \frac 1 2 (x_1+y_1) + \cdots + \frac 1 2 (x_n + y_n) = a.$

Pierre.


geometry
trigonometry
ratio
area of a triangle
Heron\u0027s formula
analytic geometry
similar triangles

"Given triangle ABC, points P,Q,R are taken on BC, CA, AB so that BP:PC = CQ:QA = AR:RB = m:n. Prove that area (PQR): area (ABC) = $ \frac{m^3 \plus{} n^3}{(m \plus{} n)^3}$"

[b]BTW - you're not supposed to use trigonometry (or heron's formula) or even analytic geometry[/b]

If you manage to find a solution or even a partial solution, please post a hint before posting the solution. Thanks very much!

[hide="a hint a requested"]
Join $ P,Q,R$ to each respective opposite vertice. Then use what they taught in geometry to find the ratio of areas...
[/hide]
Good Luck :)

okay. I give up, RobRoobiks, can you also post a solution?

[hide="Solution..."]
Soution: try harder

Took you exactly 16 minutes to give up
2 minutes after I post a hint
tsk tsk, you should try harder  :D
[/hide]

lol

I actually spent like 3 hours working on this problem before I posted it up here :)

But I finally figured it out though!!!!!!!!! YAY!

[hide="Hint"]
introduce the altitudes from A, B, C, P, Q, R  and use similar triangles to find the relative lengths of the altitudes from A, B, and C[/hide]

Altitutes?!
Joining BQ, we get
BQC:ABC = m:m+n
We also see that PQC:BQC = n:m+n
Combining the ratios, we get
$ \frac {nm}{(m\plus{}n)^2} \equal{} PQC: ABC$
Similarly we find ARQ and RBP to ABC are both $ \frac {nm}{(m\plus{}n)^2}$
Thus the ratio of RPQ: ABC
$ 1\minus{}\frac {3mn}{(m\plus{}n)^2}$
$ \frac {m^2\plus{}n^2\plus{}2mn\minus{}3mn}{(m\plus{}n)^2}$
$ \frac {m^2\plus{}n^2\minus{}mn}{(m\plus{}n)^2}$
$ \frac {m^3\plus{}n^3}{(m\plus{}n)^3}$

As desired

Hi theax! Maybe this would help:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314820

http://en.wikipedia.org/wiki/Routh%27s_theorem


Let $ A \in M_n{(\mathbb{C})} , Rank(A)\equal{}r$
Find linear independent roots number of equation :
1) $ AX\equal{}O$
2)$ AX\plus{}XA\equal{}O$
 Here unknown $ X \in M_n{(\mathbb{C})}$

1)
The solution space of $ AX \equal{} O$ is a $ n(n \minus{} r)$ dimensional subspace of $ M_n(C)$. And hence has $ nr$ independent solutions.
Consider $ X$ as a column of row vectors- $ [Y_1Y_2.......Y_n]$. $ AY_i \equal{} O$ is a vector space of $ n \minus{} r$ dimensions. Let the vectors $ {y_1,y_2,...,y_{n \minus{} r}}$ form a basis of the solution space of $ AY_i \equal{} O$.
Consider the vectors $ {[y_100...0],[0y_1000...0]....[00...y_1],[y_200...0],[0y_200..00],.......,[y_{n\minus{}r}00..0]...[000..y_{n \minus{} r}]}$ , each one belonging to $ M_n(C)$. These $ n(n \minus{} r)$ vectors form a basis for the solution space of $ AX \equal{} O$. Hence, that solution space is of dimension $ n(n \minus{} r)$.


This is from AMC 10A, Number 24

 Tina Randomly selects two distinct numbers from the set ${1,2,3,4,5}$ , and Sergio selects a number from the set ${1,2,...,10}$ . The probability that Sergio's number is larger than the sum of the two numbers chosen by tina is...

^It says two [b]distinct[/b] numbers. As such, only 11 out of the 25 listed are possible. We must first find the sums and frequency of each:
sum:            {3,  4,  5 , 6,  7,  8,  9}
frequency:   {1,  1,  2,  2,  2,  2,  1}
And then the chance that Sergio's number is greater:
(7/10) + (6/10) + 2(5/10) + 2(4/10) + 2(3/10) + 2(2/10) + (1/10) = 42/10 = 21/5
...and divide the result by 11, leaving the answer, [b]21/55[/b]

Unfortunately you are wrong, I am right.

[hide="Better solution if you want"]
[quote="AMC Solutions Pamphlet"]There are ten ways for Tina to select a pair of numbers. The sums 9, 8, 4, and 3 can be obtained in just one way, and the sums 7, 6, and 5 can each be obtained in two ways. The probability for each of Sergio's choices is 1/10. Considering his selections in decreasing order, the total probability of Sergio's choice being greater is \[ \left(\frac{1}{10}\right)\left(1+\frac{9}{10}+\frac{8}{10}+\frac{6}{10}+\frac{4}{10}+\frac{2}{10}+\frac{1}{10}+0+0+0\right) = \frac{2}{5}. \][/quote]
[/hide]

ugh.... I counted 2 ways to get 8 for some reason.... >_<


Recently in my algebra 2 class, we learned how to solve rational and quadratic (and higher degrees that can be factored completely) inequalities. The method was this.

1.) List out all the undefined values (i.e. in a rational equation, all the excluded values.)
2.) Treat the inequality as if it were an equation and solve for $ x$ (or whatever variable exists).
3.) Graph a number line containing all the results from steps 1 and 2 and then test each interval (and the bounds of the interval, provided they not unbounded) and make sure it satisfies the original inequality. If it does, use the appropriate symbols and darken the region with a line.

Although this method is quick and easy, I personally hate this method. There is hardly any mathematical intuition involved (other than solving the actual equation of course). So I decided to use casework after collection all algebraic terms to one side to solve the inequalities instead, regardless of how long it took me. It worked every time. Does anyone else feel the same way about it? (I hope I'm not the only one.) And also, are there any other methods (other than graphically viewing the inequality and using casework) that use pure mathematical processes?

The mathematical foundation behind this method is very intuitive - it is essentially the [url=http://en.wikipedia.org/wiki/Intermediate_value_theorem]intermediate value theorem[/url].  A continuous function cannot pass from positive to negative without going through zero, so the "boundaries" between regions where an inequality is true and regions where it is false occur either where the function defining it fails to be continuous or where equality occurs, so it is very natural to solve an inequality by finding the boundaries.