Hi!

I'm quite stumped on this problem. Help on it would be greatly appreciated. Here it is:

Cities $C_1$, $C_2$,...,  $C_N$ are served by airlines $A_1$, $A_2$,..., $A_n$. There is direct non-stop service between any two cities(by at least one airline), and all airlines provide service in both directions. If $N\geq2^n+1$, prove that at least one of the airlines can offer a round trip with an odd number of landings.

Suppose the contrary. Then the graph $G_i,\ i=\overline{1,n}$ formed by our $N$ vertices (cities) and having as edges the services of the airline $A_i$ is bipartite, for all $i\in\overline{1,n}$, with bipartition $U_i,V_i$. To each vertex $C_k$ we can then associate a binary string $x_k$ with $n$ letters as follows: the component $i$ of $x_k$ is $0$ if $C_k$ belongs to $U_i$, and $1$ if it belongs to $V_i$.

The hypothesis that each two cities are connected by at least one airline service basically says that all binary strings $x_k,\ k=\overline{1,N}$ are distinct. Since there are exactly $2^n<N$ binary strings with $n$ letters, we have our contradiction.


If $ a,b,c$ are real numbers, then

$ \left(a^2\plus{}b^2\plus{}c^2\plus{}5bc\plus{}5ca\plus{}5ab\right)^2\geq 12(a\plus{}b\plus{}c)^2(bc\plus{}ca\plus{}ab),$

with equality if and only if $ a\equal{}b\equal{}c.$

This is very easy. :lol:

$ \left(a^2 \plus{} b^2 \plus{} c^2 \plus{} 5bc \plus{} 5ca \plus{} 5ab\right)^2\geq 12(a \plus{} b \plus{} c)^2(bc \plus{} ca \plus{} ab)$

$ \Longleftrightarrow \{(a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ca)\}^2\geq 0$.

The equality holds when $ (a \plus{} b \plus{} c)^2 \equal{} 3(ab \plus{} bc \plus{} ca)\Longleftrightarrow (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 0$.

Therefore we have real numbers $ a \equal{} b \equal{} c$.


Find all positive integer $ x,y,z$ such that
$ (x^2\plus{}2)(y^2\plus{}3)(z^2\plus{}4)\equal{}60xyz$

[hide="Wrong Solution hidden later"]We have $ x\geq 1,y\geq 1,z\geq 1$ because $ x,y,z$ positive integers.
$ x^2 \plus{} 2\geq 3x\iff (x \minus{} 1)(x \minus{} 2)\geq 0$ which is obvious
$ y^2 \plus{} 3\geq 4x\iff (y \minus{} 1)(x \minus{} 3)\geq 0$ which is obvious
$ z^2 \plus{} 4\geq 5z\iff (y \minus{} 1)(x \minus{} 4)\geq 0$ which is obvious
Multiplying we have $ (x^2 \plus{} 2)(y^2 \plus{} 3)(z^2 \plus{} 4) \geq 60xyz$
Equality holds iff $ (x,y,z) \equal{} (1,1,1)$[/hide]

[quote="JZambrano201"]Akashnil, your solution is correct, but I think you made a mistake with some variables when you factored the quadratics.[/quote]

I agree that there are errors in the variables, but I do not think the solution is correct:

[quote="Akashnil"]
$ x^2 \plus{} 2\geq 3x\iff (x \minus{} 1)(x \minus{} 2)\geq 0$ which is obvious
$ y^2 \plus{} 3\geq 4x\iff (y \minus{} 1)(x \minus{} 3)\geq 0$ which is obvious
$ z^2 \plus{} 4\geq 5z\iff (y \minus{} 1)(x \minus{} 4)\geq 0$ which is obvious[/quote]

These are just typographical errors; they should be:

$ x^2 \plus{} 2\geq 3x\iff (x \minus{} 1)(x \minus{} 2)\geq 0$ which is obvious
$ y^2 \plus{} 3\geq 4y\iff (y \minus{} 1)(y \minus{} 3)\geq 0$ which is obvious
$ z^2 \plus{} 4\geq 5z\iff (z \minus{} 1)(z \minus{} 4)\geq 0$ which is obvious

However, the last line regarding equality is not correct: we can choose $ x\in \{1,2\}$, $ y\in \{1,3\}$, $ z\in \{1,4\}$, and any of the $ 8$ resulting ordered triples is a solution. That is:

$ \boxed{\{(x,y,z)|x\in \{1,2\},y\in \{1,3\},z\in \{1,4\}\}}$

Those inequalities are both not obvious and not correct. Any of $ y \equal{} 2, z \equal{} 2, z \equal{} 3$ will break them.

[hide="Hint"]The right side is divisible by $ 5$. This is a very strong condition in this problem.[/hide]

[quote="MellowMelon"]Those inequalities are both not obvious and not correct. Any of $ y \equal{} 2, z \equal{} 2, z \equal{} 3$ will break them.[/quote]

Oops... well the inequalities are true for all positive integers $ x,y,z$ except for the ones you mentioned.  :roll:

[hide="my solution"]
Since $ x^2 \plus{} 2$ and $ y^2 \plus{} 3$ are not divisible by 5 (3 and 2 are not quadratic residues modulo 5), $ z^2 \plus{} 4$ must be divisible by 5, so $ z \equal{} 5z_1 \pm 1$ for some $ z_1$.

Also, it is easy to see that x, y, and z aren't very large; LHS $ > x^2y^2z^2$, which is greater than the RHS if $ xyz > 60$.
So testing with $ z \equal{} 1, 4, 6, 9$ gives the 8 ordered triples $ {(1,1,1),(1,3,1),(2,1,1),(2,3,1),(1,1,4),(1,3,4),(2,1,4),(2,3,4)}$.[/hide]


geometry
geometric transformation
reflection
rotation
puzzles

How many different designs can be formed by painting 3 squares(1x1) of a 5x5 grid with black?

If a design can be formed by rotating another one by several times,then these are considered to be the same designs, but reflection is not allowed.

For instance,

BBWWW          WWWWB        WWWWW         WWWWW
WWWBW         WWWWB         WWWWW        WBWWW
WWWWW        WWWWW        WWWWW         WWWWW
WWWWW        WWWBW         WBWWW         BWWWW
WWWWW        WWWWW        WWWBB          BWWWW

are all the same designs  but

BBWWW                     WWWBB
WWWBW                    WBWWW
WWWWW       and       WWWWW
WWWWW                   WWWWW
WWWWW                   WWWWW

are considered to be different.

Does this stop being a puzzle and belong to more of a straightforward combinatorics section instead?

[quote="roadnottaken"]Before allowing for rotation, there are 25*24*23 different ways.  Because you can rotate the box four times before it comes back to the original position, there are 25*24*23/4 or 3450 possible ways.[/quote]It's not correct because in some cases, when you rotate it four times, it can appear only 2 different designs.

Oops! :blush: I used permutation instead of combination.  The answer should be divided by six.  575

OHO also raised a point that I hadn't considered.  In order to account for this, you would need to add back in the times that rotations only give two answers, which occur when one black square is in the center, and the other two are reflections of each other.  This happens in, after rotation, six places, yielding an answer of 581.

I think you should add 3 not 6.
575+3=578

I say six because there are six distinct positions where one of the three could be if the second mirrors it and the third is in the center.  I've denoted the possible positions as X, and the reflections (where the third point would be) as Y.  Each X is a distinct point, even after rotation.

X X X w w
X X X w w
w w b w w
w w Y Y Y
w w Y Y Y

The reason I am adding these points back in is because Their 180 degree rotation has already been eliminated by the use of combination.  Eliminating them twice results in an artificially depressed answer.

I agree that there is 6.
But I think we should add half of it.

We subtracted each one one time too many, not half a time too many.

Unfortunately, what I think is likely to happen here, unless someone else joins the discussion, is that the two of us will just keep arguing over whether it is three or six, and we won't get anywhere.

If we add 6, that means none of that six designs are counted in 575.
But they are counted.

The designs are counted, but their reversals are not.  However, the reversals [i]are[/i] subtracted when we divide by four.  The reason we add six, is to avoid subtracting these patterns twice.

What are their reversals? I think their revesals don't need to be counted in the final answer because they are repeated.

Before dividing by $4$, these 6 are counted twice and the others are counted 4 times. So we may add 2 times of thost 6 to make them counted 4 times like the others. We can add 12 before dividing.

It is $\frac{\frac{25\cdot24\cdot23}{3\cdot2\cdot1}+12}{4}=578$

You're right.

This would not have extended nearly so long if you had explained it more clearly originally.


Problema data la clasa a IX-a , o scriu din minte :

Fie $a,b \ge 0$, astfel incât :
$a \le 1$

$a+b \le 10$.

Aratati cã :
         $\sqrt{a}+\sqrt{b}\le 4$.
                                            Liviu Vlaicu  , Zalau
    Eu am o solutie dar e cam urata ... as vrea sa vad si alte idei .

Cauchy-Schwartz:
$(\sqrt{a}+\sqrt{b})^{2}=\left(1\cdot \sqrt{a}+\sqrt{3}\cdot \sqrt{\frac{b}{3}}\right)^{2}\le (1+3)\left(a+\frac{b}{3}\right)=4\cdot \frac{2a+a+b}{3}\le 4\cdot \frac{2+10}{3}=16.$

Multumesc . Mult mai eleganta este solutia dvs . Se poate pleca si de la :

Sunt, de fapt multe abordări posibile. Jensen, de pildă:
\[\frac{1}{4}\sqrt{a}+\frac{3}{4}\sqrt{\frac{b}{9}}\le\sqrt{\frac{1}{4}\cdot a+\frac{3}{4}\cdot \frac{b}{9}}=\sqrt{\frac{3a+b}{12}}\le 1.\]
În 1977, Vasile Cârtoaje a propus la baraj problema următoare: $a,b,c,d$ pozitive, cu $a\le 1,a+b \le 5,a+b+c\le 14, a+b+c+d\le 30.$ Atunci
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10.\]

asta iese perfect cu prima metoda propusa de dumneavoastra  :


a & b is equal to a to the b power.

What is 2 & 3?
What is 3 & 2?
What is (2 & 2) & 3?

Challenge:

Which is greater, 2 & 1,000,000 or 1,000,000 & 2?

*gasps*
[hide="to shinwoo"]
3. Think about it.
[/hide]

whoa....

I missed it...

then it would be the opposite one...

is there any better way to find it out?

 :blush:

Shinwoo: [hide]Try comparing $2^n$ to $n^2$ for n = 1, 2, 3, 4, 5, 6, 7, and 8 (or until you get bored).  See if you notice a pattern appearing.[/hide]


1. Solve for $x$: $4^{\log_2 x}= 8-x^2$

2. Find the ordered pair of positive integers $(x,y)$ for which $\log_{10} (\log_{10} (xy))=\log_{10} (\log_{10} x)) +\log_{10} (\log_{10} y))$.

3. Solve for $x$: $\log_2 3x =\log_3 2x$

[quote="frt"]2. Find the ordered pair of positive integers $(x,y)$ for which $\log_{10} (\log_{10} (xy))=\log_{10} (\log_{10} x)) +\log_{10} (\log_{10} y))$.[/quote][hide="Answer to 2"]$\log_{10} (\log_{10} x)) +\log_{10} (\log_{10} y)) = \log_{10} (\log_{10} (x) \log_{10} (y))$ getting rid of the logs on each side, $\log_{10} (xy) = \log_{10} (x) \log_{10} (y) \Rightarrow \log x + \log y = \log (x) \log (y)$ setting $a= \log x$ and $b= \log y$, $a+b=ab$ for which the only solutions are $(0,0)$ and $(2,2)$ so $(x,y)$ are $\{(1,1), (100, 100)\}$ but $(1,1)$ is extraneous as it gives a $\log 0$ , so $\boxed{(100,100)}$[/hide]

You are correct, that is an invalid step in changing the base.
[hide="Idea"]$\log_3 2x = \frac{\log_2 2x}{\log_2 3}$[/hide]

[quote="frt"][quote="Hokkage"][quote="Apprentice"]3:
[hide]$log_{2}3x = log_{3}2x$
$\frac{1}{log_{3}2x} = log_{3}2x$
$1= log_{3}2x \dot log_{3}2x$
$\frac{1}{2} = log_{3}2x$
$x = \frac{\sqrt{3}}{2}$
[/hide][/quote][hide="Mistake?"]I don't think that $1= log_{3}2x \dot log_{3}2x \Rightarrow \frac{1}{2} = log_{3}2x$ is a valid step. Instead it should be $1= log_{3}2x \dot log_{3}2x \Rightarrow \pm 1 = log_{3}2x$ giving $2x = 3, \frac13 \Rightarrow x= \frac32 , \frac16$. But it seems that $\frac32$ doesn't work, so $\boxed{\frac16}$[/hide][/quote]
$\frac16$ is correct, but I don't think $\log_2 3x=\frac{1}{\log_3 2x}$ is true. Instead, it should be $\log_2 3x=\frac{1}{\log_{3x} 2}$... :?[/quote]

see the table and explanation at:
http://oakroadsystems.com/math/loglaws.htm
Oh wait... uh.. nevermind.... wow just look at my signature!!!


A bag contains 4 balls. Each ball is a single color, and 3 distinct colors are represented in the bag. What is the probability that the first 2 balls selected without replacement are the same color?
Express your answer as a common fraction.

There are $ 2$ balls that have the same color, so our answer is $ \frac{2}{4} \cdot \frac{1}{3} \equal{} \frac{1}{6}$.


[quote="joejia"]What is the units digit of the product of the first 100 prime numbers?[/quote]

[hide]2 * 5 = 10, Thus the units digit will be

[b] 0 [/b] [/hide]

[hide]
Since you have 2, and 5 being multipied, 2*5=10 which has a units of 0. 0*anything=0 so the answer (for the units digit!!!) is 0.
[/hide]

[hide]You have the digits 2 and 5 so it makes 0.[/hide]

[quote="ajai"]how did you solve the problem?[/quote][hide="Hello, ajai please click to see how to solve"]You have to know first few prime numbers: 2,3,5,7,...
If start multiply, 2x3=6
6x5=30
30x7=210
etc
Units digit of product is always 0[/hide]

[hide]
Let $P$ be the product of the first $100$ prime numbers. Let $p$ be the product of the first $100$ prime numbers excluding $2$ and $5$.
$P=2\cdot 3 \cdot 5 \cdot \cdots=p\cdot 2\cdot 5=p\cdot 10$
Therefore, the product is a multiple of 10, and the unit's digit of the product is $0$.[/hide]


probability
combinatorics unsolved
combinatorics

Two numbers are chosen at random from the set - {1,2,3,4.....,188,189}.Find the probability that the sum of the squares of these two chosen numbers is also a perfect square.

One of the chosen numbers is of the form $d\cdot(a^{2}-b^{2})$ so you can easy get all the possibilities. However, the computations are boring...Is there a nice way to solve the problem?

I was totally clueless on this.....but I am trying

There are 142 such instances (so says Mathematica).


limit
Gauss
function
integration
number theory
real analysis
real analysis solved

Let (an)n>=1 be a sequence with terms in [0,1) s.t. \suman<00. Does it follow that \pi1/(1-an)<00?

I'm not sure what your bounds on an are, but assuming they are reasonable, the answer to your question is yes. Actually, you can prove an even stronger result:

[u]Theorem:[/u] If 0 \leq  an < 1, then \sum an converges if and only if \pi (1-an) converges.

A proof can be found in G.E. Andrew's book [u]Number Theory[/u] (Appendix B). But it is a good exercise to try to prove it.

P.S. Note that an infinite product is said to converge if it has a non-zero finite limit.

\prod (1-a_n) converges ==> \lim 1-a_n = 1 ==> \lim a_n = 0
\sum a_n coverges ==> a_n = 0

Under the condition a_n = 0, we have

\lim (ln(1-an) / (-an)) = 1,

hence \sum ln(1-a_n) and \sum (-a_n) both converge or diverge, thus \prod (1-a_n) and \sum a_n both converge or diverge.

more theorems: (the proof is left as exercises  :D )

Theorem 1
Suppose a_n > -1, n = 1,2,...
The following three statement are equivalent:
  - \prod (1+a_n) converges absolutely
  - \prod (1+|a_n|) converges
  - \sum a_n converges

Theorem 2
Suppose \sum a_n converges, then

\prod (1+a_n) converges <==> \sum a_n^2 converges

Answer grobber's question, then we can talk about other things later. I believe  \pi 1/(1-ai) diverges for every ai\in(0,1) for which \sum ai converges.

For gauss: The bounds are [0,1), as you thought they were, it was just a typo. \pi(1-an)<1 so it's <00 anyway, so the only interesting property is that it's >0, which is equivalent to \pi1/(1-an) being <00, so \pi(1-an) converges iff \pi1/(1-an) converges, so the other part, \pi1/(1-an) converges => \suman converges isn't that hard. We just write 1/(1-an) as 1+an+an2+.. and we can see that the product is > the sum, so if the product is <00 then so is the sum. I have the feeling that the same approach might work for the converse (the thing I posted), but it's a bit more elaborate. If L=\suman then we should be able to get an upper bound for the product in terms of L by writing 1/(1-an) as a series (just like I did above).

For 3x: I'm sure they're both convergent, because I found the property as being equivalent to the following Voytch Yarnk (2001) problem:

Let f: (0,00)->(0,00) be a stricly decreasing function s.t. \int{0->x}f(t)dt is finite. Show that xf(x)->0 as x->00.

I don't really have the time right now to show you how I reached the posted problem, but it's not hard. So here's what I thought: if the Voytch problem is Ok (it's been posted and solved here on the forum, I myself had a different proof, but I was now trying to solve it in another way) and it's equivalent to that implication I posted in my first post on this topic, then that implication must also hold,m right? I just couldn't prove it.


Consider the sequence $ a_1 \equal{} 1$, $ a_2 \equal{} 13$, $ a_3 \equal{} 135$, $ \ldots$ of positive integers. The $ k^{\mathrm{th}}$ term $ a_k$ is defined by appending the digits of the $ k^{\mathrm{th}}$ odd integer to the preceding term $ a_{k\minus{}1}$. For example, since the $ 5^{\mathrm{th}}$ term is $ a_5 \equal{} 13579$, then the $ 6^{\mathrm{th}}$ term is $ a_6 \equal{} 1357911$. Note that $ a_3$ is the first term of the sequence divisible by 9. What is the value of $ m$ such that $ a_m$ is the $ 23^{\mathrm{rd}}$ term of the sequence that is divisible by 9?

If you notice a pattern, you can see that every third term $ a_m$ is divisible by $ 9$.  Thus, $ m \equal{} 23 \times 3 \implies \boxed{69}$.


logarithms
integration
trigonometry
calculus
real analysis
real analysis unsolved

hello
can you help me with this problem. i didn't get it in class and i need little help
i think it's easy one but ....
however
here is the problem:

y''' = 2(y'' - 1) ctg(x)

thanks in advance

Solve the homogeneous equation first.

$y''' = 2y''\cot(x)$

By inspection, I think we should have $ln(y'') = 2\ln(\sin(x)) + C$, so $y'' = k\sin(x)^2$.

Again by inspection, $y'' = 1$ should be a particular solution. So our solution is $y'' = 1 + k\sin(x)^2$. I'm afraid I have guests coming, so I can't find $y$ right now, but it should have two more arbitrary constants in it, since we started with a third-degree equation.

You are right, Xevarion.

Rewrite $y'''=2(y''-1)\cot x$ as $(y''-1)'=2\cot x \cdot (y''-1),$

we have $y''-1=Ke^{\int 2\cot x\ dx}=K\sin ^2 x,$ where $K$ is a arbiterary constant.

Solve this we obtain $y=A\sin 2x+Bx^2+Cx+D.$ where $\ A,B,C$ and $D$ are integral constants.


[quote="nguyenvanhuyen"]For $ a, b, c$ are positive real numbers, prove that
$ 2\left ( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right ) \ge \frac {3a \plus{} b}{a \plus{} b} \plus{} \frac {3b \plus{} c}{b \plus{} c} \plus{} \frac {3c \plus{} a}{c \plus{} a}$[/quote]
From AM-GM:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 3$
By Cauchy-Schwarz we have:
$ \frac{1}{a\plus{}b} \le \frac{1}{4}(\frac{1}{a}\plus{}\frac{1}{b}$
So
$ \sum \frac{3a\plus{}b}{a\plus{}b}\equal{}3\plus{}\sum \frac{2a}{a\plus{}b} \le 3\plus{}\frac{1}{4}(6\plus{}2 (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \le 2(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})$


induction
advanced fields
advanced fields unsolved

Let X and Y be well-ordered sets and $ f_{1}$ and $ f_{2}$ are mappings from X to Y such that, for both of them, $ f_{i}(x)$ coincides with the smallest element in A := {$ {y\in Y: y \neq f_{i}(w)}$ for some w < x}. Show that $ f_{1} \equal{} f_{2}$.

I will appreciate a hint on this one. I think I need to proceed by contradiction.

I am still confused as to how to apply the hint.
Let $ y_{0}$ be the smallest element in $ A : \equal{}${$ {y \in Y: y \neq f_{i}(z)}$ for some $ z < x$}.
Choose $ x \in X$. Then for all $ x_{0} < x$, we have $ f_{i}(x) \equal{} y_{0}$ so that $ f_{i}(x) \equal{} y_{0}$ for all x.

I haven't seen an argument using transfinite induction before this post so I must be doing this wrong.

Define $ x_0\equal{}\min\{x\in X \mid f_1(x)\neq f_2(x)\}$, then $ f_1(x)\equal{}f_2(x)$ for every $ x<x_0$. Now derive the contradiction $ f_1(x_0)\equal{}f_2(x_0)$.


integration
real analysis
real analysis unsolved

Let $f \in C^4([0,\pi]), f(0) = f(\pi) = \int_0^{\pi} f(x) \cdot \sin(x) dx = 0$
Then $\int_0^{\pi} f^{12}(x) dx \geq 4 \cdot \int_0^{\pi} f^2(x) dx$
When equality holds ?
[isn't it cute  :roll: ]

[quote="mareleG"]Let $f \in C^4([0,\pi]), f(0) = f(\pi) = \int_0^{\pi} f(x) \cdot \sin(x) dx = 0$
Then $\int_0^{\pi} f^{12}(x) dx \geq 4 \cdot \int_0^{\pi} f^2(x) dx$.
[/quote]The estimate ought to be homogeneous in $f$.

As stated, if $g = \lambda f$, then $\int_0^{\pi} g^{12}(x) dx \geq 4 \lambda^{10}\cdot \int_0^{\pi} g^2(x) dx$.

Perhaps you intended an estimate like $\left(\int_0^{\pi} f^{12}(x) dx\right)^{1/6} \geq c \cdot \int_0^{\pi} f^2(x) dx$ which is no doubt correct for some $c>0$.


Find all function $f: \mathbb{R}\to\mathbb{R}$ satifying two following conditions:
a) $f(x^{2}-y)=xf(x)-f(y)$  $\forall x,y\in\mathbb{R}$
b) $xf(x)>0$  $\forall x\neq 0$

[hide="hint"]For $x=0$ then $f(y)=-f(-y)$
Therefore $f(x^{2}-y)>f(-y)\forall x\neq 0$, so $f$ is increasing!
Now, you need prove $f(x+y)=f(x)+f(y)\forall x,y\in\mathbb{R}$.[/hide]

Let $z=-y,x=\sqrt t , t>0$, then
(1) $f(z+t)=f(z)+\sqrt t f(\sqrt t)$.
z=0 give $f(t)=\sqrt t f(\sqrt t )$. Therefore
(2) $f(z+t)=f(z)+f(t) \ \forall z,t\in R$
If t<0 we use $f(-t)=-f(t)$.
Because f increase $f(x)=ax, \ a=f(1)>0$.

(*)  $f(x^{2}-y)=xf(x)-f(y)$ for all $x,y\in\mathbb{R}$

$x=y=0\Rightarrow f(0)=-f(0)\Rightarrow f(0)=0$
$y=0\Rightarrow f(x^{2})=xf(x)$ for all $x\in\mathbb{R}$
$\Rightarrow f(x^{2}-y)+f(y)=xf(x)=f(x^{2})$ for all $x,y\in\mathbb{R}$
$(x,y)=(\sqrt{a+b},b)$ gives $f(a)+f(b)=f(a+b)$ for all $a,b\geq 0$.

Then for all $x\geq 0$ we get
$(x+1)(f(x)+f(1))=(x+1)f(x+1)=f((x+1)^{2})=f(x^{2}+x+x+1)\\ \hspace*{1.45in}=f(x^{2})+f(x)+f(x)+f(1)=xf(x)+2f(x)+f(1)$
and $f(x)=f(1)x$.
$x=0$ in (*) gives $f(-y)=-f(y)$ for all $y\in\mathbb{R}$.
So $f(x)=ax$ for all $x\in\mathbb{R}$ with $a: =f(1)\in\mathbb{R}$.
There is no need for b) or any monotony.

[quote="N.T.TUAN"][quote="olorin"]
There is [color=red]no need [/color]for b) or any monotony.[/quote]
No, You sloved another problem( nice than). I think so. But don't said ''no need''  :wink:[/quote]
Well, $f(1)>0$ instead of b) would have been enough.


incircle of triangle $ABC$ touches BC,AC,AB at $D,E,F$ .$I$ is incenter of $ABC$. on lines $ID,IE,IF$ choose $A',B',C'$ with equal distance from $I$ prove that $AA',BB',CC'$ are concurrent.

It's called Kariya's Theorem. It's mentioned in these threads (although there's no complete proof there):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Kariya&t=16402[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Kariya&t=30114[/url]

and you can also find the exact statement and related properties here:

[url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Kariya_p.html[/url]


A good one on ineqs.

Assume that $a\geq b\geq c$. Then $\frac{1}{b+c}\geq\frac{1}{a+c}\geq\frac{1}{a+b}$

so Cebasev's inequality gives

$\sum\frac{a^3}{(b+c)^2}\geq\frac{1}{3}\cdot (\sum a^3) \cdot (\sum\frac{1}{(a+b)^2})$ where $\sum$ denotes cyclic sums.

Morover from Power mean ineq we get $\frac{\sum a^3}{3}\geq(\frac{a+b+c}{3})^3$ or $\sum a^3\geq\frac{(a+b+c)^3}{9}$

For $\sum\frac{1}{(a+b)^2}$ we have $3\sum\frac{1}{(a+b)^2}\geq(\sum\frac{1}{a+b})^2\geq(\frac{9}{2(a+b+c)})^2$ or

$\sum\frac{1}{(a+b)^2}\geq\frac{27}{4(a+b+c)^2}$

so finally we have

$\sum\frac{a^3}{(b+c)^2}\geq \frac{1}{3}\cdot (\sum a^3) \cdot (\sum\frac{1}{(a+b)^2})\geq\frac{1}{3} \cdot \frac{(a+b+c)^3}{9} \cdot \frac{27}{4(a+b+c)^2}
=
\frac{a+b+c}{4}$

Another way is to see that

$\frac{8a^3}{(b+c)^2}+(b+c)+(b+c)\geq3\sqrt[3]{\frac{8a^3}{(b+c)^2} \cdot (b+c) \cdot (b+c)}=6a$

so $\frac{a^3}{(b+c)^2}\geq\frac{3a-b-c}{4}$

and finally $\sum\frac{a^3}{(b+c)^2}\geq\sum\frac{3a-b-c}{4}=\frac{a+b+c}{4}$

I think both solutions are nice.The first one is easier to find and requires recalling a lot of well-known ineqs,famous ineqs!!

 :)

[quote="silouan"]$a,b,c$ are real numbers.Prove that
$\sum\frac{a^3}{(b+c)^2}\geq\frac{a+b+c}{4}$ :)[/quote]

Cauchy:

$\sum\frac{a^3}{(b+c)^2} \sum a \ge \Big( \sum \frac{a^2}{b+c} \Big)^2 \ge \frac{\Big( \sum a \Big)^4}{ \Big( 2 \sum a \Big) ^2 }$

$\sum\frac{a^3}{(b+c)^2}\geq\frac{a+b+c}{4}$

Often you can divide, get fractions in the denom, and use Jensen. :D

$ LHS\equal{} \sum \frac{a}{(\frac{b}{a}\plus{}\frac{c}{a})^2}$
We can use Jensen on $ \frac{1}{x^2}$, we get
$ LHS \ge (a\plus{}b\plus{}c) \cdot \frac{1}{(\frac{2(a\plus{}b\plus{}c)}{a\plus{}b\plus{}c})^2}\equal{}\frac{a\plus{}b\plus{}c}{4}$  :D

[quote="dgreenb801"]Often you can divide, get fractions in the denom, and use Jensen. :D

$ LHS \equal{} \sum \frac {a}{(\frac {b}{a} \plus{} \frac {c}{a})^2}$
We can use Jensen on $ \frac {1}{x^2}$, we get
$ LHS \ge (a \plus{} b \plus{} c) \cdot \frac {1}{(\frac {2(a \plus{} b \plus{} c)}{a \plus{} b \plus{} c})^2} \equal{} \frac {a \plus{} b \plus{} c}{4}$  :D[/quote]

May be stupid, I can't get your works.
 :blush:
Would you explain it in details,please?

[quote="socrates"]Another way is to see that

$ \frac {8a^3}{(b \plus{} c)^2} \plus{} (b \plus{} c) \plus{} (b \plus{} c)\geq3\sqrt [3]{\frac {8a^3}{(b \plus{} c)^2} \cdot (b \plus{} c) \cdot (b \plus{} c)} \equal{} 6a$

[/quote]
 :| but a,b,c are not positive numbers

[quote="probability1.01"]$ \Big( \sum \frac {a^2}{b \plus{} c} \Big)^2 \ge \frac {\Big( \sum a \Big)^4}{\Big( 2 \sum a \Big) ^2 }$

[/quote]
Excuse me, but how did you get this?

That is just Cauchy-Schwarz Inequality in Engel Form.

Note that if you move the denom over to the left and take the square root it is the same as
$ ((a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a))(\frac{c^2}{a\plus{}b}\plus{}\frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{a\plus{}c}) \ge (a\plus{}b\plus{}c)^2$, Engel form is what it looks like after dividing, it is helpful for you to get used to seeing it in the divided form right away.

[quote="dgreenb801"]Note that if you move the denom over to the left and take the square root it is the same as
$ ((a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a))(\frac {c^2}{a \plus{} b} \plus{} \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c}) \ge (a \plus{} b \plus{} c)^2$, Engel form is what it looks like after dividing, it is helpful for you to get used to seeing it in the divided form right away.[/quote]
Oh, I knew it's in the "divided form"; I just didn't realize that $ \sum 2a$ could be rewritten as $ (a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a)$, but thanks anyways.
(By the way, why is it called "Engel form"?)

[quote="probability1.01"][quote="silouan"]$ a,b,c$ are real numbers.Prove that
$ \sum\frac {a^3}{(b \plus{} c)^2}\geq\frac {a \plus{} b \plus{} c}{4}$ :)[/quote]

Cauchy:

$ \sum\frac {a^3}{(b \plus{} c)^2} \sum a \ge \Big( \sum \frac {a^2}{b \plus{} c} \Big)^2 \ge \frac {\Big( \sum a \Big)^4}{\Big( 2 \sum a \Big) ^2 }$

$ \sum\frac {a^3}{(b \plus{} c)^2}\geq\frac {a \plus{} b \plus{} c}{4}$[/quote]

I could not agree that..because $ a,b,c$ are real only , not surely positive ...

If $ \sum a$ are negative ,how can get$ \sum\frac {a^3}{(b \plus{} c)^2}\geq\frac {a \plus{} b \plus{} c}{4}$

 :roll:

An easy approach to present although you can spend some computing the right coeficients...

Set $ a \plus{} b \plus{} c \equal{} 3$ because of homogenity, then we have $ \frac{a^3}{(b\plus{}c)^2} \equal{} \frac{a^3}{(3\minus{}a)^2}$.

Now you can easily prove that:
$ \frac{a^3}{(3\minus{}a)^2} \ge a \minus{} \frac{3}{4} <\equal{}> 27(a\minus{}1)^{2} \ge 0$

Now just sum it all up:

$ \sum_{a,b,c}\frac{a^3}{(3\minus{}a)^2} \ge \sum_{a,b,c} (a \minus{} \frac{3}{4}) \equal{} \frac{3}{4} \equal{} \frac{a\plus{}b\plus{}c}{4}$

Just as we had to prove...

Also, the numbers are probably ment to be positive (altought a sufficient condition is $ a\plus{}b\plus{}c>0$ !) . You can try plugging in the values:

$ a : \equal{} \minus{}2$
$ b : \equal{} \minus{}1$
$ c : \equal{} \minus{}1$

And you get $ \minus{}2 \minus{} \frac{2}{9} \ge \minus{}1$

The mistake in my solution is that you can not assume their sum is 3 unless the sum of original numbers isn't positive. I actually think that my proof can be trivially extended so that:

If $ a\plus{}b\plus{}c>0$ then $ \sum\frac{a^3}{(b\plus{}c)^2}\geq\frac{a\plus{}b\plus{}c}{4}$
If $ a\plus{}b\plus{}c<0$ then $ \sum\frac{a^3}{(b\plus{}c)^2}\leq\frac{a\plus{}b\plus{}c}{4}$
If $ a\plus{}b\plus{}c\equal{}0$ then $ \sum\frac{a^3}{(b\plus{}c)^2}\equal{}\frac{a\plus{}b\plus{}c}{4}\equal{}0$

The last statement can be easily proved independantly of the others...

[quote="silouan"]$ a,b,c$ are real numbers.Prove that
$ \sum\frac {a^3}{(b \plus{} c)^2}\geq\frac {a \plus{} b \plus{} c}{4}$ :)[/quote]

Are you sure $ a,b,c$ are only  [color=red]real[/color] numbers?

As Zuza's proof , it cannot hold unless $ a,b,c$ are real positive.

 :D


there is no need to use exsisting result to prove the inequality.

[quote=quykhtn-qa1]Given$ a,b,c>0$ such that:$ abc\equal{}1$.Prove that:
$ \frac{a(1\plus{}a\plus{}ab)}{a\plus{}2b}\plus{}\frac{b(1\plus{}b\plus{}bc)}{b\plus{}2c}\plus{}\frac{c(1\plus{}c\plus{}ca)}{c\plus{}2a}\geq\ 3$
Good luck  :)  :)  :blush:[/quote]


geometry
3D geometry
sphere
calculus
integration
real analysis
real analysis unsolved

Calculate : $ \oint_{S_R}{\frac {xdydzdt \minus{} ydxdzdt \plus{} zdxdydt \minus{} tdxdydz}{(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^m}}$
 Here $ S_R$ stands for the $ \mathbb{R}^4$ sphere of radius $ R$ with center at the origin $ O(0,0,0,0)$ , $ m$ are positive integers .

I have many mistake in calculate integral this then general result for $ \mathbb{R}^{n}$ sphere of radius $ R$
 I hope that you can for me a answer
and here : http://www.mathlinks.ro/viewtopic.php?t=323813  :roll:


Express this ratio as a fraction in lowest terms:

6 feet : 3 feet 8 inches

Let's change everything to inches to compare.

$ 6$ feet : $ 3$ feet $ 8$ inches

$ 72$ inches : $ 44$ inches

$ 72$ : $ 44$

$ \boxed{\frac {18}{11}}$


calculus
integration
inequalities
trigonometry
real analysis
real analysis solved

Prove the following inequality.

\[1-\frac{1}{e}<\int^\frac{\pi}{2}_0 e^{-\sin x}dx<\frac{\pi}{2}\left(1-\frac{1}{e}\right)\]

For right side $\sin x \geq \frac{2}{\pi}x$ on $[0;\frac{\pi}{2}]$


prove :D

$ \sum_{k\equal{}1}^{n}\;\frac{1}{k^{2}}\;\equal{}\;\boxed{\sum_{j\equal{}1}^{\infty}\;\frac{(\minus{}1)^{j\plus{}1}}{j}\cdot\dbinom{n}{j}\cdot\textbf H_{j}}$

where $ \textbf H_{j}\;\equal{}\;\sum_{i\equal{}1}^{j}\;\frac{1}{i}$

Apply [url=http://mathworld.wolfram.com/EulersSeriesTransformation.html]Eulers' series transformation[/url] once to obtain the desired result.


percent
ratio
algebra
system of equations

These are problems from a skill drill in a some book.

The time limit is 3 minutes.

See if you can identify the "theme" of these problems

When you're ready.. problems are in spoiler.

[hide]1. Evaluate 1+2+3+4+5+7+8+9+10*0

2. Cassandra hikes uphill at a rate of 3mph. Reaching the peak, she hikes back downhill at a rate of 6mph. What is her average speed expressed in mph?

3. Solve the following system of equations for A:

2A-B-C=0

A+2B+3C=15

AC=2C

4. One hundred percent % can be expressed as 10. Find a.

5. There are 5 numbers, a, b, c, d, x. Bill adds 20% to each number. How much % did Bill add to the sum a+b+c+d+x?[/hide]

PM me with answers, or just write them below.

These are trick questions. Here the answers:  (I didn't finish yet)

[hide]

1. 1+2+3+4+5+6+7+8+9+10*0 = 1+2+3+4 +5+6+7+8+9 = 45

3. 2A-B-C = 0

    A+2B+3C = 15

    AC=2C

From the third equation, either c= 0 or a =2.  First try c=0:

2a-b = 0

a+2b = 15

a = 3

b = 6

c = 0.  This works.

4.  100% in decimal form is 1.0.  Thus a = 0.

5.  20% = 0.2

.2+a+.2+b+.2+c+.2+d+.2+x

= 1+a+b+c+d+x

100% = 1, so 100% is added.[/hide][/hide]

You misunderstood 5.

[hide].2*a+.2*b+.2*c+.2*d+.2*x=.2*(a+b+c+d+x), so the answer is 20%.[/hide]

I think there are two interpretations of #2. The normal way is to use the physics definition of "speed"; but if "trickiness" is involved, then ...

For number 3, how do you know that that solution is unique?

nr1337 wrote:These are trick questions. Here the answers:  (I didn't finish yet)
[hide]
1. 1+2+3+4+5+6+7+8+9+10*0 = 1+2+3+4 +5+6+7+8+9 = 45

3. 2A-B-C = 0
    A+2B+3C = 15
    AC=2C
From the third equation, either c= 0 or a =2.  First try c=0:
2a-b = 0
a+2b = 15

a = 3
b = 6
c = 0.  This works.

4.  100% in decimal form is 1.0.  Thus a = 0.

5.  20% = 0.2
.2+a+.2+b+.2+c+.2+d+.2+x
= 1+a+b+c+d+x
100% = 1, so 100% is added.[/hide]

You got 3 and 5. You interpreted #5 correctly, but u missed something on 1 and 4.

For #2, just use the normal definition of speed.

Never mind.

Edited, because I can't count.  Or read.

Mathdool -- there is a little x next to the edit button on your messages that are the double-post of this.  Click it to delete them.  (Actually, it will only be on the bottom one, but once you get rid of that, you can get rid of the top one, too.)

[quote="mathdool"]You got 3 and 5. You interpreted #5 correctly, but u missed something on 1 and 4.
For #2, just use the normal definition of speed.[/quote]

Are you sure about number 5?  20% is never used additively.  There has to be 20% of something.  If you add 20% to the price of a $20 item, the final price will be $24, not $20.20.

Ravi B wrote:Joel, are you talking about Problem 1? Spoiler:[hide] Because even with PEMDAS, nr1337 did something wrong on that problem.  Look carefully.[/hide]

Oops.  I can't read.

I personally would think that if left unspecified that the percent would be of the number.  I can see how it can be .2 but that is really awkward to say 20% and mean for it to be .2.

mathdool wrote:[hide]Magnara wrote:Are you sure about number 5?  20% is never used additively.  There has to be 20% of something.  If you add 20% to the price of a 24, not $20.20.

This is the trick. The problem never states 20% of the number, so it really means to add .2 to each number.[/hide]

To add 20% to a number still means you multiply by 1.2, you definitely don't add 0.2. There really is no such thing as a percentage if you don't specify what you are taking it of..

[hide]1. 1+2+3+4+5+7+8+9+0=39

because their is no 6

2.There is a formula for finding the average of two different speeds traveled over the same distance. (2*a*b)/(a+b) so I just plugged 3 and 6 in for a and b and got 4.

3. Can't A also equal 2, (with B=-1 and C=5)?

4-B-C=0

2+2B+3C=15

2C=2C

2+2B+3C+2(4-B-C)=15+2(0)

10+C=15

C=5

4-B-5=0

-B-1=0

-B=1

B=-1

4.Since a doesn't equal 0, I am going to say 2 because 10 :^2: is 100

5. I agree with nr1337

[/hide]

By the way, my observation in the last post that [hide]% = /100[/hide] gives me a different answer on number 4 than any of you, but I bet mathdool agrees with me that it is [hide]-2[/hide].

By the way, what is the average velocity of the hiker in problem 2?

You're right JBL #4 is -2.
Be careful with those tricky percents, 100%%=1%=.01=10^-2.

These problems do have awkward wordings, but the book included in a previous lesson that "1%=1/100=0.01"

[quote="mathdool"]These problems do have awkward wordings, but the book included in a previous lesson that "1%=1/100=0.01"[/quote]

What is the name of this book? I am with the people who have previously objected in this thread to the view that "20 percent" means ".2" (of some unit) unless we are first told that it is 20 percent OF THAT UNIT. Percentages are customarily applied to quantities that may consist of any number of the units being considered, and it does violence to English usage (my specialty; I'm the verbal guy here) to use the term "percent" in some nonstandard way.

So what book are these problems from? I'm curious.

I have always learned that a percentage is just a fraction.  I don't mean this comparison to be insulting, but to me the argument that a percent cannot represent a stand-alone number sounds like someone disputing the existence of rational or negative numbers.  "What do you mean, 'negative one?'  Negative numbers don't have any meaning except when applied to some other, larger number."  Or "What do you mean, 'one half?'  One half only has meaning when you take away some number of objects from a larger collection."

For the record, the definition of percent I have here is "as expressed in hundreths," or "one hundreth," neither of which seem to imply a requirement of a unit.

Oh, I agree that "1 percent" means "one-hundredth" of something, but the usage of such a term almost always requires defining what the "something" is. And that is what was poorly specified (IMHO) about the original problem. I wouldn't write a word problem in English that way. I seem not to be alone in this opinion, considering what else has been posted in the thread.

I'd still like to know the name of the book.

[quote="JBL"]In the spirit of friendly disagreement, I will stand up for mathdool on this one.  One only adds 20% of a number when one says exactly that: "20% [b][i]of[/i][/b] a."  This is common usage in all circumstances for rational multiples of a number -- the word "of" in "take half of 5" is used in exactly the same way, with the same meaning, as "times" is used in other circumstances, such as "take :pi: times five."  (We also happen to have special words for some multiplications by positive integers: double, twice, triple, thrice, quadruple, and so on.)  If the "of" isn't there, (except in the cases of special words: "double 5," for example), we aren't multiplying.  So, we reduce our question to asking, "Does 20% mean anything without the 'of?'"  I will argue that it does -- in every circumstance where it is used, the "%" symbol could be replaced with the symbols "/100" while retaining exactly the meaning of the phrase.  There is no particular reason that this should cease being true when we aren't multiplying 20% by something.[/quote]

I disagree; % is not an operator.  That is like saying the degree symbol, , is an operator.  It is, rather, a symbol for units.

interpreting #4 to mean 100% of 1% is rediculous... the question is incorrectly worded, because 100% is 1/1 or just 1... the question asks for just 100%...

I would also like to know what book this question comes from...

[quote="TripleM"][quote="blahblahblah"][quote="JBL"](snip)[/quote]

I disagree; % is not an operator.  That is like saying the degree symbol, , is an operator.  It is, rather, a symbol for units.[/quote]

I'll have to disagree with that one.. you can't say "10 degrees of x", that just doesn't make sense. % is very similar to an operator really..[/quote]

But a percent is not a real number, it requires conversion, just as a degree is not a real number.

The conversion process is almost identical (dare I say, percents are isomorphic to degrees?) and so I submit that I am correct.

Guys. These are all trick questions. They're meant to be interpreted in non-standard ways.

[quote="tokenadult"]I'd still like to know the name of the book.[/quote]

This book is nothing special, probably none of you have ever heard of it...
"Enriched Mathematics 8"

My friendly reading recommendation would be the [url=http://www.singaporemath.com/supplement_secmath.htm#Gardiner%20Order]Maths Challenge[/url] books (volumes 1, 2, and 3) by Tony Gardiner. They are good stuff, and the problems are very mathematical.

I'll have to agree that a percent is usually, and should be, used as a ratio or a "part of" something, but this assumption is  what makes this a [u]trick[/u] question.  :twisted:

well I have an equivilent question then... How many fingers am I holding up? I think that's a trick question too...

[quote="Spoon"]well I have an equivilent question then... How many fingers am I holding up?[/quote]

Um, thirteen?

token, sadly, you are off by a few... but nice guess!

How about.. 800% fingers?
Seriously people, anyone you talk to that does math(s) will tell you that % cannot exist by itself.. (including all the lecturers at my uni..)

This brings back old memories for me. That was what I thought when I was in fifth grade and was taking a placement test to see which math class I would be in in sixth grade at a public middle school (that I ended up not attending, but that's a different story). They gave me a 40-question test, and I got 39 right...

[quote="TripleM"]How about.. 800% fingers?[/quote]

Good response to the thread. With the clue I just got, I'm going to change my guess to 675% fingers.  :wink:

Listen, when I say "one half," what does that mean?  I can't just have "one half" -- I need to have one half of something.  When we say "one half," we mean "one half of one," where one is defined by the Peano axioms or field axioms (if you're in a field where 2 does not equal 0) or whatever.  I see no reason whatsoever that we shouldn't treat percentages exactly the same.

[quote="tokenadult"][quote="JBL"]I see no reason whatsoever that we shouldn't treat percentages exactly the same.[/quote]

I can think of a reason: etymology. Long-established usage does treat percentages differently from "natural" numbers, and that is to be expected on historical grounds.[/quote]

Percents are different from "natural" numbers, but can still be converted to fractions or decimals...

This is kind of cruel to beginners; it confuses them about the proper way to treat a percent. I have to agree with those saying that a percent must be used in conjunction with a "true" number, not having a percent stand alone.

20 percent added to your income tax would mean adding 20% of your income tax. Besides, even if you just add a percent to an integer, what would be the point of adding .2? I guess that would mean adding 20 cents, which you could probably find on the street.

[quote="probability1.01"]But what happens when I say I add 20% to your income tax? (besides getting me beat up, of course :-))  Is that adding 20% of your income tax or just 20%?[/quote]

It would mean that you are starting with the 0% that I currently pay and adding 20% to result in 20% -- only afterwards would you multiply it by my earnings.  The implication is that you will add the percent to the percent, but in this case you don't even have any other options:  notice that you cannot add 20% to $0 -- the money I pay (or don't) has a unit in front of it, and the percent does not.  So you can't add them.


Jenna has $ \$1.75$ in nickels and dimes. She has twice as many dimes as nickels. How many dimes does Jenna have?

In 1 quarter, you have 2 dimes and 1 nickel.
1.75=7 quarters=14 dimes (and 7 nickels.)

You could also let the number of dimes be $ d$ and write the following:

$ 0.1d \plus{} 0.05(0.5d) \equal{} 1.75$

$ 0.1d \plus{} 0.025d \equal{} 1.75$

$ 0.125d \equal{} 1.75$

$ d \equal{} \boxed{14}$


ABCD is a convex quadilateral with 4 inequal sides and angles.. M and N are 2 points on AB such that AM=MN=NB, P and Q are 2 points on DC such that DQ=QP=PC, prove that
Area of AMCP = Area of MNPQ -= 1/3 area of ABCD

How does one prove this?

Let's denote the area of the polygon $ A_1 A_2 ...A_n$ by $ \left[A_1 A_2 ...A_n\right]$. We'll use the following obvious
[b]Lemma:[/b] If $ a$, $ B$, $ C$ and $ D$ are all on one line and $ P$ is an arbitrary point which is not on that line, then $ \frac{[ABP]}{[CDP]}\equal{}\frac{\overline{AB}}{\overline{AC}}$.
Then
$ [AMCP] \equal{} [ABCD]\minus{}[MBC]\minus{}[APD] \equal{} [ABCD]\minus{}\frac{2}{3} [ABC]\minus{} \frac{2}{3} [ACD] \equal{} [ABCD] \minus{} \frac{2}{3}[ABCD] \equal{} \frac{1}{3} [ABCD]$
Doesn't work for $ [MNPQ]$ though :(


A $\LaTeX$ question: How do I write something over/under an arrow: $\longrightarrow$ ? I'm trying to write chemical reactions, and I have no idea how to do this.

All help would be welcomed.

Good question.  I have done something similar using \atop_{x}_{y} or something but I think there is a better way...

You can use \stackrel as in $A \stackrel{\alpha}{\longrightarrow}B$ but if you want to type a lot of chemistry then it would be better to use the [url=http://texcatalogue.sarovar.org/entries/mhchem.html]mhchem package[/url]. It's very easy to use and allows you to typeset things like

[img]http://sixthform.info/steve/wordpress/latexrender/pictures/3258fb81487d8bbde0c867a0c8884a38.gif[/img]


if x+y+z=1
prove that:
$ 6\sum x^2y^2\plus{}54(xyz)^2 \geq 12(\sum x^2\plus{}\sum xy)xyz$

[quote="BG Yoda"]There is something strange  :huh:  The both sides of the inequality can be divided by 6.

Except this, in this form the inequality is easily killed by Muirhead.[/quote]
Can you post your solution?I think there is a solution use Cauchy and Schur.But I don't know ..........?

[quote="tuandokim"]if x+y+z=1
prove that:
$ 6\sum x^2y^2 \plus{} 54(xyz)^2 \geq 12(\sum x^2 \plus{} \sum xy)xyz$[/quote]

$ \Leftrightarrow \sum_{sym}x^4y^2 \plus{}\sum_{sym}x^3y^3\plus{}\sum_{sym}x^2y^2z^2\ge\sum_{sym}x^4yz\plus{}2\sum_{sym}x^3y^2z$
It is true because following inequalities :
$ \sum_{sym}x^3y^3\plus{}\sum_{sym}x^2y^2z^2\ge2\sum_{sym}x^3y^2z$(Schur)
$ \sum_{sym}x^4y^2\ge\sum_{sym}x^4yz$(Muirhead)

[quote="NextPeace"][quote="tuandokim"]if x+y+z=1

$ \sum_{sym}x^4y^2\ge\sum_{sym}x^4yz$(Muirhead)[/quote][/quote]
I think this is only Cauchy inq

It is an easy ineq ,and I think it is not new,because there are many ways to prove it,for exmaple
pqr,SS,MV or SOS(I havent checked this way)


[quote="MCrawford"]If the length of a diagonal of a square is $a + b$, then the area of the square is:

(A)	 $(a + b)^2$
(B)	 $\frac{1}{2}(a + b)^2$
(C)	 $a^2 + b^2$
(D)	 $\frac{1}{2}(a^2 + b^2)$
(E)	 none of these[/quote]
the answer is (A),but i think this is a little hard for here. :D

that's wrong
[hide]
answer is B
[/hide]

I don't think it's too hard for this forum.  :ninja:

EDIT:

[hide]
doubling a square, take the diagonal as the length of the side.

2xside^2=diagonal^2
side^2=1/2xdiagonal^2

area=1/2 (a+b)^2[/hide]

I agree. The answer is B, since the dioganal is also the hypotenuse of creating two triangles, making two 1/2*(a+b)^2

Thus, the answer is B :)

Yes I agree the answer is [hide]B[/hide]

The answer is B because the definition of the area of any square can be writen as "the diagonal squared divided by 2", which is the exact anwer B has.  :)

Solution:

$x\sqrt{2}=a+b\implies a=\frac{a+b}{\sqrt{2}}$
$\implies A=\left(\frac{a+b}{\sqrt{2}}\right)^2=\frac{(a+b)^2}{2}\implies\boxed{B}$.

Masoud Zargar


probability
expected value
probability and stats

This is a problem from a regional Information Entropy Exam monitored in January and I have received full permission to post it online.

Two points are randomly selected from inside the unit circle. Let $ X$ be a random variable representing the distance between them. $ (a)$ What is the expected value of $ X$? $ (b)$ What is the information entropy of $ X$ in nats?


When do we get some sort of verification that we got whatever score we got and whatever award we got? Aren't they supposed to be sending books to winners?

Two years ago, I think they sent out books in early June.  I didn't do USAMTS last year, though, so I don't know about last year.

mathnovice, I think you get a book. Last year they sent out the books in mid-June I think. It was a Hungarian problem book called C2K2 or something.

I just got a book and a certificate today.
That's how I know this website.


function
absolute value
calculus
calculus computations

Graph the function f(x) = |x - 2| + |x - 5|. Where is f discontinous? Where is it not differentiable?

$f(x)$ is continuous at all points and is not differentiable at $x=2$ and $x=5.$

And what can you say about maxima and minima for this function?

I just have one more question. How do you graph functions with absolute values?

Remark that the values of $x$ for which the inside of the absolute value becomes 0, you can classify 3 cases as follows.

1. $x<2$
2. $2\leq x <5$
3. $5\leq x$


It is easy to calculate $ x_{n}\equal{}\frac{(\sqrt 2\plus{}1)^{2n\plus{}1}\plus{}(\sqrt 2\minus{}1)^{2n\plus{}1}}{2\sqrt 2 }$, therefore
$ x_{n}\{x_{n}\sqrt 2\}\equal{}\frac{1\plus{}(\sqrt 2\minus{}1)^{4n\plus{}2}}{2\sqrt 2}$ and $ \lim \equal{}\frac{\sqrt 2}{4}$.


While I will not solve either of these, I should point out that:
- You have a typo in the first: the first "=" should be a "+"
- "In" should be "ln"; that is, lowercase of LN, it means "Natural Log"

DON'T LOOK!
I don't know how to make this hidden

(1)
$ 2log_3 (x \plus{} 4) \minus{} log_3 (9) \equal{} 2$
$ log_3 (\frac {x \plus{} 4}{3})^2 \equal{} 2$
$ \frac {x \plus{} 4}{3} \equal{} 3$
$ x \equal{} 5$

(2)
$ ln (\frac {x \plus{} 1}{x}) \equal{} 2$
$ x \plus{} 1 \equal{} e^{2}x$
$ x \equal{} \frac {1}{e^{2} \minus{} 1}$

[quote]I don't know how to make this hidden [/quote]

Click the hide button, type your stuff, then click it again. The stuff between the tags is hidden.

[quote="sharkman"]Solve each equation. Express irrational solutions in EXACT form and as a decimal rounded to 3 decimal places.

(1) 2log_3 (x = 4) - log_3 (9) = 2

(2) In(x + 1) - In x = 2[/quote]
[hide]1. log_3 (x+4)^2/9=2, so (x+4)^2/9=9, (x+4)^2=81, so x+4 can be 9 or -9. That produces x=-13 and x=5 as our solutions.

2. ln [(x+1)/x]=2, so (x+1)/x=e^2, x+1=(e^2)x, so x=1/(e^2-1)[/hide]

[hide]-13 can't be a solution since you can only take logs of positive numbers.[/hide]


consider sequence An of natural nos. defined by A1=1,A2=4 and
 A(n+2)=6A(n+1)-A(n)-2. afterwards.show this sequence doesnt consist of any prime no. :|

dear rust,i want to suggest you ,please think atleast once carefully about what you post in hurry.here note terms of An are alternately even and odd.

If A(n) even, then A(n+2) is even. But I wrong, when calculate A(3) (I calculate 6*1-4-2=0) A(3)=21 is odd.
Let $q=1+\sqrt 2$. It is easy to calculate
$A(n)=\frac{1}{4}[q^{2n-1}-q^{1-2n}+2]=\frac{1}{4}(q^{n}-q^{-n})(q^{n-1}+q^{-n+1})=B(n)C(n)$. Were $B(n)=\frac{1}{2}(q^{n}-q^{-n}), C(n)=\frac{1}{2}(q^{n-1}+q^{n-1})$.
If n is odd B(n) and C(n) are integer. If n is even $B(n)C(n)=2*\frac{B(n)}{\sqrt 2 }\frac{C(n)}{\sqrt 2 }$, were $\frac{B(n)}{\sqrt 2},\frac{C(n)}{\sqrt 2 }$ are integer.


Some of the diagrams don't show up.  For example, number 12 and bunch of others here: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006

When I go to the direct link inside the image tags, I get a 404.  I'm not sure if this should be posted here, but is there an issue, or just that someone needs to reupload the pictures?

Thanks.

oh, I don't know but maybe it's a problem with the pdf?

The diagrams should have been made with asymptote (see #4 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1052272#1052272]this post[/url]). Instead of re-linking them, someone has to draw them with asymptote, or else the same thing will happen again.


geometry perimeter trapezoid circumcircle trigonometry parallelogram

I found this problem in the book called problems, problems, problems.
In the diagram, $\Delta ABC$ (right-angled) has $AC=b$, $BC=a$, and $AB=c$. Initially the triangle is located in square WXYZ of side c, as shown. The triangle moves counterclockwise inside the square so that the vertices A and B are always on the sides of the square. This motion continues until the triangle returns to its initial position. Determine the total distance in terms of a and b travelled by the point C.

[quote="rem"]Let $\angle{CBA}=\alpha$
Let $B$ be on $WZ$, $A$ be on $WX$. Have $\angle{ACB}+\angle{BWA}=90^{o}+90^{o}=180^{o}$, so $WACB$ is cyclic, so $\angle{AWC}=\angle{CBA}=\alpha$.
Threfore, if $A$, $B$ are on $WX$, $WZ$ respectively, then $\angle{XWC}=\alpha=const$. By analogy, if if $A$, $B$ are on $YX$, $WX$ respectively, then $\angle{YXC}=\alpha=const$, if $A$, $B$ are on $ZY$, $YX$ respectively, then $\angle{ZYC}=\alpha=const$, if $A$, $B$ are on $WZ$, $YZ$ respectively, then $\angle{WZC}=\alpha=const$.
Threfore the path of $C$ is the perimiter of $MNPK$ where $M,N,W$; $N,P,X$; $K,P,Y$; $M,K,Z$ are collinear, and $\angle{XWM}=\angle{YXN}=\angle{ZYP}=\angle{WZK}=\alpha$. Therefore $MNPK$ is a square. Using the fact that $WM=XN=CP=KW=a$,$MN=NP=PK=KM=b$, get $WN=XP=YK=ZM=|b-a|$ we will get the perimiter of $MNPK$, which is the length of path of $C$, equal to $4|b-a|$.[/quote]
$4|b-a|$? I got $8\sqrt{a^{2}+b^{2}}-4a-4b$. are they equal :?:

My bad, I forgot that the point $C$ goes a little further, and the length of each section is $(c-b)+(c-a)=2c-(a+b)$, and te total elngth is $8\sqrt{a^{2}+b^{2}}-4(a+b)$.

btw i have this book. It's really good for prepaing for Canadian competitions below olympiad level.

i'm hoping on getting a better mark on euclid and fermat next year, hope this helps. :)

[color=darkblue][b]Remark.[/b] Inside of the square $ABCD$ construct four right (in the vertices $M,N,P,R$ respectively) and equivalent triangles $ABM$, $BCN$, $CDP$, $DAR$. If $AB=BC=CD=DA=a$, $BM=CN=DP=AR=b$ and $AM=BN=CP=DR=c$ then the quadrilateral $MNPR$ is a square, $MN=NP=PR=RM=|b-c|$ and we obtain a prove (with a figure and using areas) of the [b][u]Pythagorus's theorem[/u][/b] : $(b-c)^{2}+4\cdot \frac{bc}{2}=a^{2}\ .$[/color]

[quote="rem"]You have to remember that if you want to sole the last problem you need to practise on problems on higher level. The problems-problems-problems book is great for prectising for the first few questions of the contest. :)

And I am sure you will do better on contests next year!  :)[/quote]
Thanks Alex :lol:
you'll ace the contests; there's no question about it.
Virgil Nicula, in what kind of questions would Pitagora's theorem be useful? Please give an example. :)

[color=darkblue][b][u]Examples (about perpendicularity).[/u][/b]

$\blacktriangleright 0.$ Let $A,B$ be two fixed points and a mobile point $L$ for which the difference $LA^{2}-LB^{2}$ is constant.
Ascertain the geometrical locus of the point $L\ .$

$\blacktriangleright 1.$ Let $ABCD$ be a convex quadrilateral. Prove that : $AC\perp BD\Longleftrightarrow AB^{2}+CD^{2}=BC^{2}+AD^{2}$.

$\blacktriangleright 2.\ \mathrm{For\ any\ four\ points\ A,B,C,D\ there\ is\ the\ equivalence\ : \ }$ $AB\perp CD\Longleftrightarrow$ ${AC^{2}-AD^{2}=BC^{2}-BD^{2}}\ .$ [b][u]Very important property ![/u][/b]

$\blacktriangleright 3.$ ([b]The Carnot's Lemma[/b]) Given are a triangle $ABC$ and the points $M\in BC$, $N\in CA$, $P\in AB$. Prove that the perpendiculars in the points $M,N,P$ on the side-lines $BC$, $CA$, $AB$ respectively are concurrently if and only if $MB^{2}-MC^{2}+NC^{2}-NA^{2}+PA^{2}-PB^{2}=0\ .$

$\blacktriangleright 4.$ Let $ABCDEF$ be a hexagon (not necessarily convex). Prove that the perpendiculars in the points $A,C,E$ on the lines $BF$, $BD$, $DF$ respectively are concurrently if and only if $AB^{2}+EF^{2}+CD^{2}=AF^{2}+DE^{2}+BC^{2}\ .$

$\blacktriangleright 5.$ The incircle $C(I,r)$ of the triangle $ABC$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$.
Denote the intersection $L\in EF\cap BC$. Prove that $LI\perp AD$.

$\blacktriangleright 6.$ Let $ABNM$ be a trapezoid so that $AM\parallel BN$ and $AB\perp BN$. Denote the middlepoint $I$ of the side $[AB]$.
Prove that $MA^{2}+NB^{2}=MN^{2}$ $\Longleftrightarrow$ $IM\perp AN$ $\Longleftrightarrow$ $IN\perp BM\ .$

$\blacktriangleright 7.$ Let $d_{1},d_{2}$ be two lines (parallely or concurrently) and let $C(I,r)$ be a circle which is tangently to the lines $d_{1},d_{2}$ in the points $A,B$ respectively. Prove that for any two points $M\in d_{1}$ and $N\in d_{2}$, $IN\perp MB\Longleftrightarrow IM\perp NA\ .$

$\blacktriangleright 8.$ Let $CA_{1}B_{1}$ and $CA_{2}B_{2}$ be two triangles so that $A_{1}C\perp A_{1}B_{1}$, $A_{2}C\perp A_{2}B_{2}$ and $CA_{1}=CA_{2}$.
Prove that $B_{2}A_{1}\perp B_{1}C\Longleftrightarrow B_{1}A_{2}\perp B_{2}C\ .$

$\blacktriangleright 9.$ Let $ABC$ be a triangle with the circumcircle $C(O,R)$ and the incircle $C(I,r)$.
Denote the centroid $G$ and the orthocenter $H$ for the triangle $ABC$. Prove that :

[b]9.0.[/b] $AB\perp AC\Longleftrightarrow b+c=2(R+r)\ .$

[b]9.1.[/b] $OA\perp OI\Longleftrightarrow r(b+c)=aR\ .$

[b]9.2.[/b] $IA\perp IO\Longleftrightarrow IG\parallel BC\Longleftrightarrow b+c=2a\Longleftrightarrow$ $\cos\frac{B-C}{2}=\sqrt{\frac{2r}{R}}\Longleftrightarrow p^{2}+9r^{2}=18Rr\ .$

[b]9.3.[/b] $GO\perp GI\Longleftrightarrow p^{2}=7r^{2}+10Rr\ .$

[b]9.4.[/b] $IG\perp BC\Longleftrightarrow b=c\ \vee\ b+c=3a\ .$

[b]9.5.[/b] $OI\perp AM\Longleftrightarrow \frac{1}{b}+\frac{1}{c}=\frac{2}{a}\Longleftrightarrow l_{a}=a\cos\frac{A}{2}$,
where the point $M$ is the middlepoint of the side $BC$ and $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}\ .$

[b]9.6.[/b] $OH\perp AM\Longleftrightarrow b^{2}+c^{2}=2a^{2}$.

$\blacktriangleright 10.$ Let $A,B,C,D$ be four fixed points . Ascertain the geometrical locus of the mobile point $L$ for which :

[b]10.1.[/b] $LA^{2}+LB^{2}=LC^{2}+LD^{2}\ .$

[b]10.2.[/b] $LA^{2}+LB^{2}+LC^{2}=3\cdot LD^{2}\ .$

$\blacktriangleright 11.$ Let $ABC$ be a isosceles triangle ($AB=AC$). Denote the middlepoints $D,E$ of the sides $[BC]$, $[AB]$ respectively.
Ascertain the geometrical locus $\Lambda$ of the point $L$ for which $LB^{2}+LC^{2}=2\cdot LA^{2}+BC^{2}$ and prove that $P\in AD\cap \Lambda\Longrightarrow EP\perp AC\ .$

$\blacktriangleright 12.$ Let $ABCD$ be a convex quadrilateral inscribed in a circle $w$. For a mobile point $M$ define
the second intersections $X,Y,Z,T$ of the lines $MA$, $MB$, $MC$, $MD$ respectively with the circle $w$.
Ascertain the geometrical locus of the point $M$ for which $\frac{MA}{MX}+\frac{MB}{MY}+\frac{MC}{MZ}+\frac{MD}{MT}=4\ .$

$\blacktriangleright 13.$ see http://www.mathlinks.ro/Forum/viewtopic.php?t=101957

$\blacktriangleright 14.$ Let $ABC$ be a triangle. Prove that there is a point $M$ for which $MA^{2}=MB^{2}+MC^{2}$ if and only if $m(\widehat{BAC})\le 90^{\circ}\ .$.

$\blacktriangleright 15.$ Given are a triangle $ABC$ and a line $d$. Denote the projections $M,N,P$ of the points $A,B,C$ respectively on the line $d$. Prove that the perpendiculars from the points $M,N,P$ on the lines $BC$, $CA$, $AB$ respectively are concurrently.

$\blacktriangleright 16.$ Let $ABC$ be a triangle. Prove that the perpendiculars from a point $M$ on the lines $MA$, $MB$, $MC$ meet the lines $BC$, $CA$, $AB$ respectively in three collinear poins.

$\blacktriangleright 17.$ Let $ABC$ be a triangle with the fixed vertex $A$ inscribed in the fixed circle $w$ so that $AB^{2}+AC^{2}-BC^{2}=2k^{2}$ (constant).

[b]17.1.[/b] Ascertain the geometrical locus of the middlepoint of the side $[BC]$.

[b]17.2.[/b] Acertain the geometrical locus of the projection of the point $A$ on the line $BC$.

[b]17.3.[/b] Ascertain the geometrical locus of the projection of the point $B$ on the line $AC$.

$\blacktriangleright 18.$ Let $ABC$ be a triangle and let $D$ be the middlepoint of the side $[BC]$.
Prove that for any point $M$ there is the equivalence $MO\perp AD\Longleftrightarrow 2\cdot MA^{2}=MB^{2}+MC^{2}$.
[b]Remark.[/b] The geometrical locus of the point $M$ for which $2MA^{2}=MB^{2}+MC^{2}$ is the perpendicular line from the circumcenter of the triangle $ABC$ on the line $AD$.

$\blacktriangleright 19.$ Let $ABCD$ be a parallelogram, where $AB=CD=a$, $AD=BC=b$, $AC=e$, $BD=f,\ \frac{a}{b}\in \left(\frac{1}{\sqrt 7},\ \sqrt 7\right)$.
Denote the middlepoints $M$, $N$ of the sides $CD$, $AD$ respectively. Prove that $BN\perp AM\Longleftrightarrow 3(e^{2}-f^{2})=8(b^{2}-a^{2})\ .$
[b]Remark.[/b] $e^{2}+f^{2}=2(a^{2}+b^{2})$ ([u]the Euler's relation[/u]). \[\boxed{\mathrm{\ a.\ s.\ o.\ }}\] [/color]

[color=darkred]In the my previous message are 20  proposed problem about perpendicularity ![/color]

yes, that's what the title said. I'm studing analytic perpendicularity right now (3D ofcourse :) ), but this will help a lot. thanks again.

[quote="JBL"][quote="Virgil Nicula"]Pitagora's theorem[/quote]

In English, "Pythagorus's Theorem" or "the Pythagorean theorem."[/quote]
yes, i realized that a little while ago. :blush:
The problems are interesting though;varing difficulty.

[quote="Virgil Nicula"]Above is a marathon about perpendicularity ![/quote]
Hi Virgil. I suggest you start a new topic about the marathon about perpendicularity.


calculus
integration
logarithms
Euler
function
calculus computations

For (vii) I get:  $ f(x,b)=\ln(bx)+\gamma$ where $ \gamma$ is the Euler gamma.

If the others are also not elementary, I may just abandon this problem.  :lol:

[hide="cheap way to go about it"]
(i)

$ f(x,b) = \frac{2-2b}{e^{-b}\sqrt{1-x^{2}}}$ for $ b \in [0,1]$

$ f(x,b) = 0$ for $ b > 1$

it's a continuous integrable function over (0, infinity) =D
[/hide]

Aren't these just Laplace transforms of functions evaluated at $ b=1$?

If, so maybe we can use the [url=http://en.wikipedia.org/wiki/Inverse_laplace_transform]Fourier-Melling Integral[/url].  :maybe:

Just a suggestion.

For (iii) I get:

$ J_{0}(bx)$ where $ J_{0}(y)$ is the zeroth order Bessel function.  At this point I am bowing out...   :D


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra solved

$A\in M_2(R)$ such that its caracteristic polynomial $p(x)=det(xI_2-A)$ is

irreducible in $R[X]$

1/ Prove that $E=R.I_2\oplus R.A$ is a sub-algebra of $M_2(R)$

2/ For any $B \in E-(0)$ prove that $det(B)>0$

3/ Let $C\in M_{2n}(R)$

$C=\left (\begin{array}{cccc}A_{11}&A_{12}&...&A_{1n}\\A_{21}&A_{22}&...&A_{2n}\\...&...&...&...\\A_{n1}&A_{2n}&...&A_{nn}\end{array}\right)$

with $A_{ij}\in E$

Prove that $det(C)\geq 0$

Note that $R$ is the real numbers here- this fails in any other field I can think of.

1. It is obvious that $E$ is an $R$ submodule of $M_2(R)$. It is a sub algebra because of Hamilton-Cayley for $A$.
2. Let $B=xI+yA$. Let $z,\bar z$ be the complex roots of $p$. Then $\det B=(x+yz)(x+y\bar z)=|x+yz|^2\geq 0$ Equality occurs iff $x+yz=0$ iff $x=y=0$.
3. If we replace $A$ by some $uA+vI_2$ we don't change $E$. And we can chose $u,v$ such as $A^2+I_2=0$.
A is similar to $\left(\begin{array}{cc} 0 & 1\\ -1 & 0\end{array}\right)$. Notice that even though this similarity changes $C$, it does not change $\det C$.
By interchanging some rows and the corresponding columns we obtain a matrix of the form $\left(\begin{array}{cc} X & Y\\ -Y & X\end{array}\right)$.
This is a $2n\times 2n$ matrix with real entries. Add the last n columns multiplied by $i=\sqrt{-1}$ to the first n columns, then substract the first n lines multiplied by i from the last n to get $\left(\begin{array}{cc} X+iY & Y \\ 0 & X-iY\end{array}\right)$. The determinant of the last matrix is $\det (X+iY) \det (X-iY)=|\det(X+iY)|^2\geq 0$.


This small problem comes from a test for students from our department of Mathematics. If function $ f$ has $ f'(x) \ne 0$ for all $ x \in (a,b)$, then $ f$ is a monotone function in $ (a,b)$.


geometry
search
trapezoid
analytic geometry

I don't think I understand your solution.  could you elaborate a little more please? :huh:

[quote="Virgil Nicula"]Denote $ S\in RQ\cap AB$ . The division $ (R,Q;G,S)$ is harmonically and $ HG\perp HS\Longleftrightarrow\widehat {QHG}\equiv\widehat {RHG}$ . See the lemma from [url=http://www.mathlinks.ro/viewtopic.php?search_id=1736179779&t=46146][color=red][b]here[/b][/color][/url] .[/quote]

$ (R,Q;G,S)$ is harmonical division iff $ \frac {GR}{GQ}\equal{}\frac {SR}{SQ}$ (the points $ G$ , $ S$ are conjugate w.r.t. the points $ R$ , $ Q$ ). Please, read the mentioned link.

[quote="mathwizarddude"]In triangle ABC, P is any point inside the triangle.  The line AP intersect side BC at Q and the line BP intersect side AC at R.  PC and QR intersect at G, and H is the projection of G on side AB (such that GH is perpendicular to AB).  Prove that angle QHG = angle RHG.

A diagram would help a lot.  :) [/quote]
[hide="Solution"]
Let $ S$ be the intersection between $ CP$ at $ AB$.  Furthermore, let the perpindiculars from $ Q$, $ R$, and $ C$ to $ AB$ meet $ AB$ at $ X$ and $ Y$, and $ Z$ respectively.  Then, if $ [ABC]$ represents the area of $ \triangle ABC$, we have that\[ \frac {RG}{GQ} \equal{} \frac {[GRC]}{[GCQ]} \equal{} \frac {\frac {RC}{AC}\cdot [GAC]}{\frac {QC}{BC}\cdot [BGC]} \equal{} \frac {RC}{CQ}\cdot \frac {BC}{AC}\cdot \frac {[AGC]}{[BGC]} \equal{} \frac {BC}{AC}\cdot \frac {RC}{QC}\cdot \frac {SA}{SB}\]Yet, Ceva's Theorem on concurrent cevians $ AQ$, $ CS$, and $ BR$ gives that $ \frac {SA}{SB}\cdot \frac {BQ}{QC}\cdot \frac {CR}{AR} \equal{} 1$, so $ \frac {SA}{SB} \equal{} \frac {QC}{BQ}\cdot \frac {AR}{CR}$.  Substituting this in, we find that\[ \frac {RG}{GQ} \equal{} \frac {BC}{AC}\cdot \frac {RC}{QC}\cdot \frac {QC}{BQ}\cdot \frac {AR}{CR} \equal{} \frac {BC}{BQ}\cdot \frac {AR}{AC}\]Since $ QX\perp AB$, $ CZ\perp AB$, and $ RY\perp AB$, we have that $ QX\parallel CZ\parallel YR$, which implies that $ \triangle AYR\sim \triangle AZC$ and $ \triangle BXQ\sim \triangle BZC$, so $ \frac {YR}{ZC} \equal{} \frac {AR}{AC}$ and $ \frac {XQ}{ZC} \equal{} \frac{BQ}{BC}$.  It follows that $ \frac {YR}{XQ} \equal{} \frac {AR}{AC}\cdot \frac {BC}{CQ} \equal{} \frac {RG}{GQ}$.  Yet, since $ RYXQ$ is a trapezoid with transversal $ GH$, we have that $ \frac {RG}{GQ} \equal{} \frac {HY}{HX}$.  Thus, $ \frac {YR}{XQ} \equal{} \frac {HY}{HX}$.  Since $ \angle HYR \equal{} 90 \equal{} \angle HXQ$, we have that $ \triangle HYR\sim \triangle HXQ$, so $ \angle HQX \equal{} \angle HRY\implies \angle QHG \equal{} \angle RHG$ because of the fact that $ QX\parallel GH\parallel RY$. [/hide]

great solutions - also try using analytic geometry to prove it

also try to use coordinates geometry to prove it please...


floor function
inequalities
number theory unsolved
number theory

For any real number $ p\geq 1$ consider the set of all real numbers $ x$ with $ p<x<(2\plus{}\sqrt{p\plus{}\frac{1}{4}})^2$.

Prove that from any such set one can select four mutually distinct natural numbers $ a,b,c,d$ such that $ ab\equal{}cd$.

[url=http://www.mathlinks.ro/viewtopic.php?t=150848]This related problem[/url] is of great help. At least it gives us an idea where to start from.
For $ p=1$ we have $ 2\cdot4=1\cdot8$. Assume $ p>1$.
Note that if we decrease $ p$ until $ \lfloor p\rfloor$ (without making $ p$ an integer) decreases the upper bound, hence it suffices to prove the following problem:

[i]There are $ 4$ mutually distinct natural numbers $ a,b,c,d$ in the interval $ I_p=\left[p,\left(2+\sqrt{p-\frac34}\right)^2\right)$ so that $ ab=cd$, where $ p$ is an integer.[/i]
[hide="Note"] In the original problem, the numbers $ a,b,c,d$ will be strictly greater than $ p$. That's because if $ p$ is an integer in the original problem we apply the above reformulation for $ I_{p+1}$[/hide]
Analyze $ p=2,3$ separately. Assume now $ p\ge4$. Let $ (M-1)^2\le p<M^2$, where $ M\in\mathbb{Z}$. Because $ p\ge4$, $ M\ge3$.

If $ (M-1)^2<p$ then let's show that $ (M+1)^2<\left(2+\sqrt{p-\frac34}\right)^2$. Obviously, we have to show that $ M+1\le2+\sqrt{(M-1)^2+1-\frac34}$, which is immediate. But then $ M^2-1, M^2+2M\in I_p$ and we have $ (M^2-1)(M^2+2M)=(M^2+M)(M^2+M-2)$ and we are done.

If $ (M-1)^2=p$ then let $ M-1=N$ and we would like to use the identity $ N^2(N^2+3N+2)=(N^2+N)(N^2+2N)$. All we need is $ N^2+3N+2=M^2+M\in I_p$, or
$ M^2+M\le\left(2+\sqrt{M^2-2M}\right)^2$. Multiplying by $ 4$ and weakening tightly the inequality, we obtain $ 2M-3\le 2\sqrt{M^2-2M}$, which holds for $ M\ge3$.


inequalities
calculus
derivative
integration
real analysis
real analysis unsolved

Let $ f: \left[ {0,1} \right] \to \Re$ have a continuous derivative, $ f(0) = 0$ and $ f'(x) > 0$ for $ x \in (0,1)$. To prove that
$ \int\limits_0^1 {\frac{{(f(x))^3 }}{{(f'(x))^2 }}dx \ge \frac{1}{9}} \int\limits_0^1 {(1 - x)^2 f(x)dx}$.

Can the constant $ \frac{1}{9}$ should be replaced with bigger?

I'm getting that this inequality is wrong.

We can muliply $ f$ by any constant, so wlog we can assume $ f(1) = 1$.

Let $ g$ be the inverse of $ f$. Then (changing $ x = g(y)$)
\[ \int\limits_{0}^{1}{\frac {{(f(x))^{3}}}{{(f'(x))^{2}}}dx = \int_0^1 y^3 g'(y) dy = 3\int_0^1 y^2 (1 - g(y))dy = 3\int_0^1 y^2 g_1(y) dy,}\]
where $ g_1$ is the inverse of $ f_1(x) = f(1 - x)$.

So we should compare this to $ \frac19 \int_0^1 x^2 f_1(x) dx$.

If we take $ f_1(x)\approx 1/n$ then $ g_1(x)\approx 1_{[0,1/n]}(x)$ then the first integral is approximately $ 3/n^3$ while the second is $ 1/(27n)$, which is definitely bigger for $ n$ large enough.

Are you sure that inequality is wrong? Here's the original, [url=http://www.fmi.uni-sofia.bg/Members/babev/studentski-olimpiadi-po-matematika/fmi/oly08in2.pdf/at_download/file]Problem 6[/url]

[quote="ferry2"]Are you sure that inequality is wrong?[/quote]

I would be grateful if you find a flaw in my argument.

[quote="ferry2"] Here's the original, [url=http://www.fmi.uni-sofia.bg/Members/babev/studentski-olimpiadi-po-matematika/fmi/oly08in2.pdf/at_download/file]Problem 6[/url][/quote]
"studentski-olimpiadi" - this was an olympiad for students in Bulgaria?
Could someone translate other problems, please ?
Thanks in advance

Apply Holder's inequality:
$ \displaystyle \left(\int_0^1 (1 - x)^3 f'(x) dx\right)^2\left(\int_0^1 \frac {f(x)^3}{(f'(x))^2}\right) \ge \left(\int_0^1 (1 - x)^2f(x)dx\right)^3$
Apply integration by part to $ \int_0^1 (1 - x)^3 f'(x) dx$, we can see that it is actually the same as $ 3\int_0^1 (1 - x)^2f(x)dx$. Thus the inequality is true.

Equality holds iff $ (1 - x)^3f'(x)$ and $ f^3(x)/(f'(x))^2$ are linearly dependent.

[quote="zhoraster"]
Let $ g$ be the inverse of $ f$. Then (changing $ x = g(y)$)
\[ \int\limits_{0}^{1}{\frac {{(f(x))^{3}}}{{(f'(x))^{2}}}dx = \int_0^1 y^3 g'(y) dy = 3\int_0^1 y^2 (1 - g(y))dy = 3\int_0^1 y^2 g_1(y) dy,}\]
[/quote]

The first equality sign is wrong. It's $ g'(y)^3$ instead.

[quote="Soarer"]
The first equality sign is wrong. It's $ g'(y)^3$ instead.[/quote] Yes, thank you  :blush:

[quote="Soarer"]
Equality holds iff $ (1 \minus{} x)^2f'(x)^{2/3}$ and $ \frac {f(x)}{f'(x)^{2/3}}$ are linearly dependent.
[/quote]
Should it say that $ (1\minus{}x)^3f'(x)$ and $ f^3(x)/(f'(x))^2$ are linearly dependent?

[quote="zhoraster"]
Should it say that $ (1 \minus{} x)^3f'(x)$ and $ f^3(x)/(f'(x))^2$ are linearly dependent?[/quote]

oops, my bad  :blush: edited

Anyway, the solution is $ f(t) \equal{} c(1\minus{}t)^a$. Which means that the equality is impossible. Moreover, it might be a sign that the inequality is true with a bigger constant.


Positive integers $x>1$ and $y$  satisfy an equation $2x^2-1=y^{15}$. Prove that 5 divides $x$.

Are there any integer solutions to this equation?  :?  Anyway, suppose what they want is not true. Of course, 2 divides $y+1$ and if we write the equation as $ ((y^5+1)/2)( y^{10}-y^5+1)=x^2$ we observe that $ ((y^5+1)/2, y^{10}-y^5+1)$ is a divisor of 3. If it is 1, then $y^{10}-y^5+1$ is a perfect square, impossible for y at least 2 (^we can put it between two consecutive squares). So, suppose it is 3. Then we can write $ y^5+1=6A^2$ and $y^{10}-y^5+1=3B^2$ with $x=3AB$. Since $ y^5-y$ is a multiple of 3, it follows that $ y+1$ is a multiple of 6. Thus, if we write $ ((y+1)/6)(y^4-y^3+y^2-y+1)=A^2$ we see that since 5 does not divide $A$, we have $ ((y+1)/6, y^4-y^3+y^2-y+1)=1$ and thus $ y^4-y^3+y^2-y+1$ is a square. Again, this is impossible by the fact that $ (2y^2-y)^2<4(y^4-y^3+y^2-y+1)<(2y^2-y+2)^2$.

Since the statement of the original question is false (as pointed out by there having no integer solutions), arent such claims such as $5|x$ logically true ?

[quote=ck204]See here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=399294&hilit=do+there+exists+marathon[/url]
No solution exists..[/quote]

Sorry but the solution is not true....

by mod 5
$2x^2 -1 \equiv  y^3 \mod 5$
$x^2 \equiv (y^3+1)*-2 \mod 5$
notice (-2) is not  quadratic residue mod 5 so does $(y^3+1)$
so $(y^3+1)^2 \equiv 4 \mod 5$
$(y^3+1) \equiv  2,-2\mod 5$
$y^3 \equiv 1 ,-3 \mod 5$
$5 \nmid x$


[quote="Interval"]An air traffic controller sees 2 aircraft flying at the same altitude on his screen.  Aircraft A is headed due west at 150  mph.  Aircraft B is 15 miles due north of Aircraft A and heading south at 400 mph.  How close will the two aircraft come to each other?[/quote]

[hide]they are closest to eachother when A is exactly west of B (i dont know how to explain besides related rates but perhaps someone else does?)  it takes $\frac{15}{400}=\frac{3}{80}$ hours for B to be east of A.  A travels $\frac{3}{80}\cdot 150=\frac{45}{8}$ miles.[/hide]

[quote="jli"][quote="Walk Around The River"][quote="Interval"]An air traffic controller sees 2 aircraft flying at the same altitude on his screen.  Aircraft A is headed due west at 150  mph.  Aircraft B is 15 miles due north of Aircraft A and heading south at 400 mph.  How close will the two aircraft come to each other?[/quote]

You never specified that Aircraft A and Aircraft B were the two that the controller saw on his screen.  :wink:[/quote]

wow  :huh: ... these problems arent supposed to trick you but its a good habit to find these mistakes i guess. :lol:[/quote]

for some reason i look for flaws.  i think tis because in mu alpha theta [big thing in florida], if i find a flaw, i can put E and i get the points and dont even have to do the problem.  :P

I will search for the question again and see if there is a mistype.


[quote="Inequalities Master"]Let $ a,b,c,d\geq0$.Prove that:
$ \frac {(1 \plus{} a^3)(1 \plus{} b^3)(1 \plus{} c^3)(1 \plus{} d^3)}{(1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)(1 \plus{} d^2)}\geq \frac {1 \plus{} abcd}{2}$.[/quote]
Nice inequality! :lol:
$ \prod_{cyc}\frac{1\plus{}a^3}{1\plus{}a^2}\geq\sqrt{\prod_{cyc}(1\minus{}a\plus{}a^2)}\geq\sqrt[4]{\prod_{cyc}\frac{1\plus{}a^4}{2}}\geq\frac{1\plus{}abcd}{2}.$

Can you show all the solution for this problem?

[quote="BaBaK Ghalebi"]it holds for $ 5$ variables too:

$ \frac{(1\plus{}a^3)(1\plus{}b^3)(1\plus{}c^3)(1\plus{}d^3)(1\plus{}e^3)}{(1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)(1\plus{}d^2)(1\plus{}e^2)}\geq\frac{1\plus{}abcde}2$[/quote]
Because $\frac{1+a^3}{1+a^2}\geq\sqrt[5]{\frac{1+a^5}{2}}$ and Holder.


it is easy when slove in geometric way.
and it is equal to find the sum of the distance of a point in a fixed line to the other fixed point.
 :lol:

Yes, and also if you want an inequality ?Use Minkowsky's. :)


On a standard six-faced die, the numbers on opposite faces always add to seven.  Dena rolls a pair of fair, standard six-faced dice.  She then takes the product of four numbers: the two numbers on the faces that are showing on top of the dice and the two numbers on the faces which are hidden on the bottom of the dice.  What is the greatest possible product of these four numbers?

The possible pairs between the one facing up and the one hiddem are $ (6,1)$, $ (5,2)$, and $ (4,3)$. We want to maximize the product, so we take the $ (4,3)$ pair. Thus we have $ 4\times{3}\times{4}\times{3}\equal{}\boxed{144}$.


modular arithmetic
induction
strong induction
combinatorics proposed
combinatorics

Let $ k,t\geq 1$ be coprime integers. Starting with the numbers $ 1,2,...,n$ in the increasing order, one can change places of any two integers that differ by $ k$ or $ t$. Show that every permutation can be obtained if and only if $ n\geq k+t-1$.

Here's a sketch:

Let $ a$ and $ b$ be the two coprime integers and w.l.o.g. $ a<b$.  First we show that for any $ n\ge a+b-1$ we can obtain all permutations.  If we can show that any arbitrary pair of integers $ i,j$ can be swapped without affecting any of the other numbers, we are done.  Because $ a$ and $ b$ are coprime, we know that the set $ \{b,2b,3b,\dots, ab\}\equiv\{1,2,\dots, a\}\pmod{a}$.  From this fact we can obtain a finite sequence of integers $ \le n$ that spans all congruences mod $ b$ by starting the sequence with $ b$, subtracting as many multiples of $ a$ still leaving us with a number greater than $ 0$, and then adding $ b$ and repeating.  Because $ a$ will be the final number in the sequence, we can always add $ b$ without getting bigger than $ a+b-1$.  Here is an example for $ a=6,b=11$:

$ 11,5,16,10,4,15,9,3,14,8,2,13,7,1,12,6$.

Hence, for any two numbers $ i$ and $ j$ in our permutation, we can obtain a sequence $ i=a_{1},a_{2},\dots,a_{k-1},a_{k}=j$ so that $ a_{m}$ and $ a_{m+1}$ differ either by $ a$ or $ b$.  Through the use of a strong induction on $ k$ and a little work, we can show that $ i$ and $ j$ can be swapped without changing the order of any of the numbers in between. (Use the key inductive step that we can always move from $ a_{1},a_{2},\dots,a_{k}\to a_{k},a_{k-1},\dots,a_{2},a_{1}$ for all $ k$).  This shows that we can always do it for $ n\ge a+b-1$.

Now, to show we cannot for all $ n<a+b-1$, we proceed by contradiction.  If we could, then we could have a sequence like the one above so that $ b=b_{1},b_{2},\dots,b_{r}=a$ (after all, the elements of the sequence $ (b_{r})$ are merely the numbers that are used to be swapped).  Because $ (a,b)=1$ we know $ b$ does not equal $ a$ modulo $ a$, meaning the sequence must contain another congruency class modulo $ a$.  From $ b$ the only different congruence mod $ a$ we could reach would be $ 2b$, because trying to subtract $ b$ from any element in that congruency modulo $ a$ would require negative numbers in the permutation.  From there the only new congruence class we can reach is $ 3b$, and we will not reach the congruency class with $ a$ until we finally reach $ ab$.  Hence we would hit all of the congruences modulo $ a$ in these steps.  But the number $ a-1$ would then have to appear in the sequence, and this number could not be increased by $ b$ since $ a+b-1>n$.  Thus $ a-1\equiv ab\pmod{a}$, which implies that $ -1\equiv 0\pmod{a}$.  Contradiction.

[quote="zgorkster"]

 Through the use of a strong induction on $ k$ and a little work, we can show that $ i$ and $ j$ can be swapped without changing the order of any of the numbers in between. (Use the key inductive step that we can always move from $ a_{1},a_{2},\dots,a_{k}\to a_{k},a_{k-1},\dots,a_{2},a_{1}$ for all $ k$).  [/quote]

Can you show the induction in detail?  :)

Okay.  We show by induction that for all $ k\in\mathbb{N}$, the sequence $ a_{1},a_{2},\dots,a_{k}$ where $ |a_{i}-a_{i+1}|\in\{a,b\}$ for all $ 1\le i\le k-1$ can be rearranged into the sequence $ a_{k},a_{k-1},\dots,a_{1}$.  If we can show this, then it is trivial to show that for any $ k$,

$ a_{1},a_{2},\dots,a_{k}\to a_{k},a_{k-1},\dots,a_{2},a_{1}\to a_{k},a_{2},\dots,a_{k-1},a_{1}$

as desired.  Now, to show the induction, the base cases are trivial.  So suppose it is true for all integers less than $ k$.  Then we have the sequence of moves:

$ a_{1},\dots,a_{k}\to a_{k-1},\dots,a_{1},a_{k}\to a_{k},a_{k-2},\dots,a_{1},a_{k-1}\to\cdots$
$ \to a_{k},a_{k-1},a_{k-2},\dots,a_{1},a_{2}\to a_{k},a_{k-1},\dots,a_{1}$.

This finishes the induction.


number theory
prime numbers
combinatorics proposed
combinatorics

We say $m \circ n$ for natural m,n $\Longleftrightarrow$
nth number of binary representation of m is 1 or mth number of binary representation of n is 1.
and we say $m \bullet n$ if and only if $m,n$ doesn't have the relation $\circ$
We say $A \subset \mathbb{N}$ is golden $\Longleftrightarrow$
$\forall U,V \subset A$ that are finite and arenot empty and $U \cap V = \emptyset$,There exist $z \in A$ that $\forall x \in U,y \in V$ we have $z \circ x ,z \bullet y$
Suppose $\mathbb{P}$ is set of prime numbers.Prove if $\mathbb{P}=P_1 \cup ... \cup P_k$ and $P_i \cap P_j = \emptyset$ then one of $P_1,...,P_k$ is golden.

Hi, Omid Hatami!
Are you sure that statement is correct?
Maybe I am wrong, but it seems I disprove it.

Yes I'm sure that problem is correct. It was our exam last week.

yes the main idea of this problem was given from the paper from erdos about random graph (just a liittle manipulation) :D

Of course the official solution is easy using Dirichlet theorem in number theory

If you say that problem is correct, it means that I still don't understand statement.

Well I mean that we ahve a relation and $A \subset \mathbb{N}$ we say $A$ is golden if and only if for evey $U,V \subset A$ that $U \cap V= \emptyset$ then thre exist $z \in A$ that z has relation with every $x \in U$ and does'nt have relation with every $y\in V$

Thank you, but the same is written above  :?

Well I don't realize where is the problem. Explain more that what don't you understand.

Let $P_1=\{5,7\}$. Set $P_1$ is not golden (U={5}, V={7}). Let $P_2=P\setminus P_1$.
Suppose $P_2$ is golden. Consider $U=\{13\}\cup \{a\in P_2\,|,13\circ a\}$, $V=\{11\}$.
We see $13\bullet 11$ and $2\not\in U$. For all remaining prime numbers we have $p \bullet p$, since binary representation of $p$ has 0 on p-th place. So if $z\circ x\ \forall x\in U$ then $z\circ 13$ $\Rightarrow$ $z\in U$, but $z\bullet z$ -- contradiction.
What is wrong?

Excuse me I didn't say that $U,V$ must be finite but I thought I had said.

I am very angry because both Omid and Sam-n said that statement is correct... rrrrrr...  ;) :)
Now problem become very clear and really simple if we are allowed to use Dirichlet theorem.

oh myth i think it is part of default (being finite )  :D anyway sorry


Prove that if $x, y, z $ are positive reals and $x^2+y^2+z^2=xyz$  then:

$\ xy+yz+zx+9 \ge \ 4(x+y+z)$

By AM-GM and Cauchy-Schwarz,
\[xy+yz+zx+9 \ge 4\sqrt{3xyz} = 4 \sqrt{3(x^2+y^2+z^2)} \ge 4(x+y+z).\]


Hi guys, again in the tradition of this topic forum members at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24485]IMO 2005[/url], it would be also nice to know who of participates in the [url=http://www.ioi2005.pl/about/]IOI in Nowy Sącz, Poland, 18-25 August 2005[/url]. Thus if you are a contestant, e.g. in the IOI Team for 2005, a deputy, observer or a team leader, or just a visitator then please leave a short message here.  :)

Well! It is a great fortune for me that I am going to post the first post in both IMO2005 and IOI2005 pages :D

Yes I am going to participate there and it is the first time for our country as well! :lol:

Slovak IOI team 2005:

Frantisek Simancik
Jakub Zavodny (= kubus)
Michal Burger
Peter Peresini

The first three of us are also going to IMO this year.

Ok, probably I'll also go to IOI 2005 :D
The Moldavian IOI team is (not 100%, some changes are possible)
1. Filip Simion (me :D )
2. Ciubatii Dumitru
3. Vanovschi Vladimir
4. Grecu Mircea

Hope to see some of you there :)

The Indian IOI team:

Ks hitij Bansal
Arnold Joseph Noronha
Preyas Popat
Harpreet Singh (<--That's me)

See you at the IOI.

Ok I forgot to write our team. Here it is:

1. Md. mahbubul Hasan (me)
2. Shahriar Rouf
3. Khobaib Chowdhury
4. M. Tanvir Kaykobad

See you there :)

Anyone has some sample questions? I don't know anything about their exams.

Good luck everyone!

For sample questions, try the USACO training pages: http://ace.delos.com/usacogate/

IOI 2005 practice tasks are at [url=http://www.ioi2005.pl/competition/tasks.php]www.ioi2005.pl/competition/tasks.php[/url], but I don't think the problems at the IOI are going to be any different from the normal IOI style .. from what I know about polish OI, they're going to be really nice and quite tough. Really looking forward to it!

well it seems to me that the sample tasks given at the site are very tough. so can't we discuss those here at this forum?

Did anyone think of the first problem; Alice and Bob?
It's really nice. I just read it and will try to solve it after my exams finish.

Very nice tasks. This is how example tasks should be, not like the last years "1+1" tasks. :D

well, so can we discuss those problems over here?

Kang Ji-hoon, Seoul Science High School
Lee Hu-yeon, Hansung Science High School
Yeo Yun-dong, Hansung Science High School
Jo Gang-won, Taegue Science High School

will participate IOI2005.

Serbia & Montenegro team:

1) Bojan Basic;
2) Stevan Joncic;
3) Igor Kabiljo;
4) Ivan Labat.


On my profile still appears the name of the blog that I created time ago, but I decided to delete it.

I want to create a new one, but I can't.

Can you fix the problem?

This was discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158633


Find all pairs of functions $ f,g: \mathbb{R}\rightarrow\mathbb{R}$ such that, for any $ x,y\in\mathbb{R}$:

$ f(xg(y\plus{}1))\plus{}y\equal{}xf(y)\plus{}f(x\plus{}g(y))$, and $ f(0)\plus{}g(0)\equal{}0$.

I've tried some basic replacements but until now the only thing I got is [hide="this"]$ f(g(y))\equal{}f(0)\plus{}y$, and $ f(1\plus{}g(0))\equal{}1$[/hide]

[color=red]The solutions are $ f(x) \equal{} x$ and $ g(x) \equal{} x$ for all $ x\in\mathbb{R}$.[/color]

We suppose that exist $ \alpha$ such that $ g(\alpha) \equal{} 0$, and we put $ y \equal{} \alpha \minus{} 1$.

We obtain: $ f(0) \plus{} \alpha \minus{} 1 \equal{} xf(\alpha \minus{} 1) \plus{} f(x \plus{} g(\alpha \minus{} 1))$, this is equal to:

$ f(x \plus{} g(\alpha \minus{} 1)) \equal{} \minus{} f(\alpha \minus{} 1)x \plus{} f(0) \plus{} \alpha \minus{} 1$

This is in the form: $ f(x \plus{} c) \equal{} mx \plus{} n$ with $ c,m,n\in\mathbb{R}$

So $ f(x) \equal{} px \plus{} q$ with $ p,q\in\mathbb{R}$.

$ x \equal{} 0$

the original equation become $ f(0) \plus{} y \equal{} f(g(y))$ and so $ LHS \equal{} q \plus{} y$, $ RHS \equal{} p\cdot g(y) \plus{} q$, $ LHS \equal{} RHS$ then $ y \equal{} p\cdot g(y)\Longrightarrow g(x) \equal{} \frac {x}{p}$.

So the original equation is $ xy \plus{} x \plus{} q \plus{} y \equal{} xpy \plus{} xq \plus{} px \plus{} y \plus{} q$
That become $ y \plus{} 1 \equal{} py \plus{} q \plus{} p$

If we substitute $ y \equal{} \minus{} 1$ then $ q \equal{} 0$.

If we substitute $ y \equal{} 0$ then $ p \plus{} q \equal{} 1$.

So $ p \equal{} 1$ and $ q \equal{} 0$.

[color=blue]The last thing to do is to  prove that existe an $ \alpha$ such that $ g(\alpha) \equal{} 0$.[/color]

With $ x \equal{} 0$ we obtain $ f(0) \plus{} y \equal{} f(g(0))$ so $ f$ is surjective and $ g$ is injective.

If we put $ xg(y \plus{} 1) \equal{} x \plus{} g(y)$. We obtain $ \displaystyle x \equal{} \frac {g(y)}{g(y \plus{} 1) \minus{} 1}$ and we substitute this $ x$ in the original

equation we obtain

$ y \equal{} \frac {g(y)f(y)}{g(y \plus{} 1) \minus{} 1}$

If $ y \equal{} 0$,we have

$ 0 \equal{} \frac {g(0)f(0)}{g(1) \minus{} 1}$.

From this we can easily conclude using that $ f$ is surjective, $ g$ is injective and $ f(0) \plus{} g(0) \equal{} 0$ that $ \alpha \equal{} 0$.

some frivolous mistake.

$ f(0)\plus{}y \equal{} f(g(0))$ must be corrected $ f(0)\plus{}y \equal{} f(g(y))$

and I think you can't put $ x \equal{}\frac{g(y)}{g(y\plus{}1)\minus{}1}$
because if you put x=0 then, g(1)=1 so g(1)-1=0.
And it is enough to show just f is surjective.
maybe f(g(-f(0)))=0 right?


group theory
abstract algebra
superior algebra
superior algebra unsolved

1. Let $ G$ act transitively on $ A$ and assume that $ G$ is finite. Define an action of $ G$ on the set $ A\times A$ by putting $ (a,b).g\equal{}(a.g;b.g)$. Let $ a\in A$. Show that $ G$ has the same number of orbits on $ A\times A$ that $ G_a$ has on $ A$.

2. An action og $ G$ on $ A$ is double transitive if $ G$ is transitive on the set of those odered pairs $ (a,b)$ with $ a\ne b$ (with the action as in the problem 1). Show that $ G$ is doubly transitive on $ A$ if $ \frac{1}{|G|}\sum_{g\in G}\varphi(g)^2\equal{}2$ where $ \varphi$ is the associated permutation character. Do not assume $ G$is transitive.

[quote="kidmath"]1. Let $ G$ act transitively on $ A$ and assume that $ G$ is finite. Define an action of $ G$ on the set $ A\times A$ by putting $ (a,b).g \equal{} (a.g;b.g)$. Let $ a\in A$. Show that $ G$ has the same number of orbits on $ A\times A$ that $ G_a$ has on $ A$.[/quote]

Assuming that $ G_a$ denotes the isotropy subgroup of $ a\in A$, consider the $ G_a$-equivariant map $ i_a: A\rightarrow A\times A$, $ b\mapsto (a,b)$ inducing the (well-defined) map $ \Phi_a: A/G_a \rightarrow (A\times A)/G$, $ G_ab\mapsto G(a,b)$.

$ \Phi_a$ is checked to be injective, in general, and surjective by the assumption that $ G$ acts transitively on $ A$.

[quote]2. An action og $ G$ on $ A$ is double transitive if $ G$ is transitive on the set of those odered pairs $ (a,b)$ with $ a\ne b$ (with the action as in the problem 1). [/quote]
Easy to check provided that the operation of $ G$ on $ A$ is assumed to be transitive. BTW: The reversed implication holds true,too.
[quote]Show that $ G$ is doubly transitive on $ A$ if $ \frac {1}{|G|}\sum_{g\in G}\varphi(g)^2 \equal{} 2$ where $ \varphi$ is the associated permutation character. Do not assume $ G$is transitive.[/quote]

Assuming that $ \varphi(g)$ means the cardinality of the fix point set $ Fix_A(g)\equal{}\{a\in A|a.g\equal{}g\}$ the claim follows from the general result \[ \sum_{g\in G} |Fix_A(g)|\equal{}|A/G|\cdot |G|\] for a operation of $ G$ on a set $ A$.

From the hypothesis we get that \[ |(A\times A)/G|\equal{}2 \,(*),\] i.e. the operation of $ G$ on $ A\times A$ has two orbits which must be the diagonal $ \Delta$ (which is $ G$-equivariantly isomorphic to $ A$) and its complement. Thus, g operates transitively on both $ A$ and $ A\times A$ (using (1) for the latter).

To see (*) note that $ Fix_{A\times A}(g)\equal{} Fix_A(g)\times Fix_A(g)$ such that
\[ 2|G|\equal{}\sum_{g\in G} |Fix_A(g)|^2\equal{}\sum_{g\in G}|Fix_{A\times A}(g)|\equal{}|(A\times A)/G|\cdot |G|\,.\]

... Done.  :P


trigonometry
inequalities proposed
inequalities

[quote="pvthuan"]Let be given a triangle $ABC$, find the maximum of
\[\cos^{k} \frac A2\cos^{m}\frac B2\cos^{l}\frac C2\]
where $k,m,l$ are positive numbers. Determine when equality holds.[/quote]                                                                                I think the equality holds iff   $ktg\frac A2 = mtg\frac B2 =l tg\frac C2 $ Am I true?

You wrote necessary extremum condition. I think you are right. But how can it help to find maximum? Is it possible in principle?

Hmm... I have made some calculations.
\[\max=\sqrt\frac{k^km^ml^l(k+m+l)^{k+m+l}}{(k^2+km+ml+lk)^k(m^2+km+ml+lk)^m(l^2+km+ml+lk)^l}.\]

Hope I didn't miscalculate...  :?


I really dont know how to do this...

We have twenty-seven 1 by 1 cubes. Each face of every cube is marked with a natural number so that two opposite faces (top and bottom, front and back, left and right) are always marked with an even number and an odd number where the even number is twice that of the odd number. The twenty-seven cubes are put together to form one 3 by 3 cube as shown. When two cubes are placed face-to-face, adjoining faces are always marked with an odd number and an even number where the even number is one greater than the odd number. Find the sum of all of the numbers on all of the faces of all the 1 by 1 cubes.

Picture attached....

Purple Comet!, 2009, Middle School Division, Question 15.

[hide="Problem and Solution"]
[quote][b][size=150]Problem 15[/size][/b]
We have twenty-seven 1 by 1 cubes. Each face of every cube is marked with a natural number so that two
opposite faces (top and bottom, front and back, left and right) are always marked with an even number
and an odd number where the even number is twice that of the odd number. The twenty-seven cubes are
put together to form one 3 by 3 cube as shown. When two cubes are placed face-to-face, adjoining faces are
always marked with an odd number and an even number where the even number is one greater than the
odd number. Find the sum of all of the numbers on all of the faces of all the 1 by 1 cubes.

Answer: 1377

Suppose one face of a 1 by 1 cube is on the outside of the 3 by 3 cube, and that face is marked with the
odd number n. Then the face opposite that face on the 1 by 1 cube is marked with 2n. The face on the
next cube adjacent to the face with 2n must be marked with 2n − 1. Its opposite face must be marked
4n − 2. The face adjacent to it must be marked 4n − 3, and the face opposite this face on its 1 by 1 cube
must be marked 8n − 6. For example, if a 1 appears on the side of the big cube, then 8 · 1 − 6 = 2 will
appear on the opposite side of the big cube. The sum of the six numbers starting with n and ending on the
opposite side of the big cube with 8n−6 is n +(2n) + (2n−1)+(4n−2)+(4n−3)+(8n−6) = 21n−12.
The diagram shows fourteen faces marked with either a 1 or 2 = 8(1) - 6, five faces marked with either a 3
or an 18 = 8(3) - 6, four faces marked with either a 5 or a 34 = 8(5) - 6, two faces marked with a 7, and
two faces marked with a 9. Thus, the 27 odd numbers appearing on the outside of the 3 by 3 cube add up
to 14 ·1+5·3+4·5+2·7+2· 9 = 81. For each of these odd numbers n on the outside of the 3 by 3 cube,
there are six other sides of 1 by 1 cubes adding up to 21n − 12. It follows that the sum of all the numbers
marked on all the sides of 1 by 1 cubes is 21 · 81 − 27 · 12 = 81(21 − 4) = 81 · 17 = 1377.[/quote][/hide]

Copied from the Purple Comet! Website at [url]http://www.purplecomet.org/home/resource/188/MiddleSchoolSolutions2009.pdf[/url], page 8. The diagram was not copied.
Hope that helps!

I just realized this, but the same problem is being discussed in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276652]this[/url] thread.

Moderators, please consider locking.


Given $n$ positive reals such that $\sum a^3=3$ and $\sum a^5=5$, prove that $\sum a >3/2$

[quote="mecrazywong"]By Cauchy Schwarz, $\sum a_i\sum a_i^5\ge(\sum a_i^3)^2$. Thus, $\sum a_i\ge9/5>3/2$ :?
Am I missing anything?[/quote]

No.  I didn't say it was hard  :D


Let $S = \lbrace 1, 2, \ldots, 15 \rbrace$.  Let $A_1, A_2, \ldots, A_n$ be $n$ subsets of $S$ which satisfy the following conditions:

[b]I.[/b] $|A_i| = 7, i = 1, 2, \ldots, n$;

[b]II.[/b] $|A_i \cap A_j| \leq 3, 1 \leq i < j \leq n$

[b]III.[/b] For any 3-element subset $M$ of $S$, there exists $A_k$ such
that $M \subset A_k$.

Find the smallest possible value of $n$.

I think the answer is 15,too :)
Here is my solution,and please tell me if it's wrong  :D
For $i=1,2,..,15$ ,let $x_i=|\{j\in[1,n]|i \in A_j\}|$
Then we have the following result:
1)$\sum_{i=1}^{15}x_i=7n$
2)$\sum_{i=1}^{15}\binom{x_i}{2} \leq 3 \binom{n}{2}$

From 2),we get $\sum x_i^2-\sum x_i \leq 3n(n-1)$
Then aplying AM-GM and 1),we can easily get $n \geq 15$
Now we will prove that $n=15$ satisfies the hypothesis,and we just have to check the 3 rd condition .

First,we notice that there does not exist 4 subsets that have 3 common elements (which is implied from the hypothesis)
So we just consider the following cases:

(*)Case 1:There exists unique $i,j,k$ (asumming they are 1,2,3)such that $|A_1 \cap A_2 \cap A_3|=\{1,2,3\}$.
Then the remaining 4 elements of each subset will get all the value in the set {4,5,6,...,15} (notice that 3+3x4=15)
We count the number of distiinct subset {x,y,z} where $x,y,z \in A_i$ The total is $3\binom{7}{3}-3+12\binom{7}{3}=522>\binom{15}{3}$
So the condition 3 is satisfied.
(*)Case 2:There exists only 2 subsets that have 3 common elements.
Like case 1,We count the number of distinct subset {x,y,z} where $x,y,z \in A_i$,the total is $2\binom{7}{3}-2+13\binom{7}{3}=523>\binom{15}{3}$
(*)Case 3:There does not exists indices i,j such that $|A_i \cap A_j|=3$
Like case 1 and 2,the total number of triples is $15\binom{7}{3}>\binom{15}{3}$
So we are done .

[quote="KDS"]From 2),we get $\sum x_i^2-\sum x_i \leq 3n(n-1)$
Then aplying AM-GM and 1),we can easily get $n \geq 15$[/quote]
Can anyone explain this step please? I get using AM-GM and (1) $\frac{49n^2}{15}-7n = (\sum_{i=1}^{15}x_i)^2/15 - \sum_{i=1}^{15}x_i \leq 3n^2-3n \implies n\leq 15?$

Another way of bounding $n$: For all 3 element subsets $M$ of $S$, let $d(M)$ be the number of $A_i$ containing M. Let a, b be the number of 3 element subsets $M$ of $S$ such that $d(M)=3,2$ respectively. Since $S\backslash M$ has 12 elements, $d(M)\le 4$. Note that the number of $(M,A_i)$ such that $M$ is contained in $A_i$ is
$$n\binom{7}{3} = 2a+b+\binom{15}{3} \hspace{2cm} (1).$$
Suppose next $\forall i \in S, i$ is contained in $x_i$ subsets, it follows that $\sum x_i = 7n.$ Each $A_i\cap A_j$ has size $\leq 2$ unless their intersection is some $M$ (of size 3), so $$n\binom{7}{3}- \binom{15}{3} + (n\binom{7}{3}- \binom{15}{3})/2 +2\binom{n}{2} \stackrel{(1)}{\ge} 3a+b+2\binom{n}{2} = \sum \binom{d(M)}{2} + 2\binom{n}{2}$$
$$\ge \text{ number of $(i,\{j,k\})$ pairs with $i\in A_j\cap A_k$} = \sum \binom{x_i}{2}$$
$$\ge  \frac{1}{2}((\sum_{i=1}^{n}x_i)^2/n - \sum_{i=1}^{n}x_i) =\frac{1}{2}(\frac{49n^2}{n}-7n),$$
so in shorter terms $$3(n\binom{7}{3}- \binom{15}{3})/2 +2\binom{n}{2} \ge \frac{1}{2}(\frac{49n^2}{n}-7n)$$. It follows from the inequality above that $n \geq 15$.

A third way: Consider $F$, the set of sets $A_i$ containing $x$ for an arbitrary $x\in S$. Since any two sets in $F$ after the exclusion of $x$ have intersection at most 2 but by (III) their union has size 14, it follows that each $x\in S$ is contained in $\ge 7$ sets $A_i$. Thus $3\binom{n}{2} \geq 15\binom{7}{2} \implies n\geq 15.$


If $ a<b$, what is the value of $ |a\minus{}b|\plus{}a\plus{}b$?

If $ a < b$, then $ |a \minus{} b| \equal{} b \minus{} a$.

$ b \minus{} a \plus{} a \plus{} b \equal{} \boxed{2b}$


inequalities
geometry
function
trigonometry
inequalities proposed

Prove that for real non-negative numbers $ a,b$ and $ c$ it is true that $ 27(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2\ge 64abc(a\plus{}b\plus{}c)^3$.

[hide="One way"]Let $ XYZ$ be the triangle with sides $ x \equal{} b \plus{} c$, $ y \equal{} c \plus{} a$ and $ z \equal{} a \plus{} b$.

The inequality is equivalent to $ \dfrac{9xyz}{x \plus{} y \plus{} z}\ge 4\sqrt {3}(XYZ)$ (where $ (XYZ)$ denotes the area of the triangle).

Which is true because $ \dfrac{9xyz}{x \plus{} y \plus{} z}\ge 2(xy \plus{} yz \plus{} zx) \minus{} x^2 \minus{} y^2 \minus{} z^2$ is equivalent to Schur's inequality

and $ 2(xy \plus{} yz \plus{} zx) \minus{} x^2 \minus{} y^2 \minus{} z^2\ge 4\sqrt {3}(XYZ)$ is Hadwiger-Finsler inequality
[/hide]

[hide="Another way"]Continuing with  the notation above.
Because $ f(w)\equal{}\sin(w), w\in[0,\pi]$ is a concave function we have that $ \sin(X)\plus{}\sin(Y)\plus{}\sin(Z)\le 3\sin\left(\dfrac{X\plus{}Y\plus{}Z}{3}\right)\equal{}\dfrac{3\sqrt{3}}{2}$. Multiplying the last inequality by $ 2R$ (where $ R$ stands for the circumradi) we get the equivalent inequality $ x\plus{}y\plus{}z\le 3\sqrt{3}R$.

Now
$ 4\sqrt{3}(XYZ)\equal{}2r\sqrt{3}(x\plus{}y\plus{}z)\le 18Rr\equal{}18\cdot\dfrac{xyz}{4(XYZ)}\dfrac{2(XYZ)}{x\plus{}y\plus{}z}\equal{}\dfrac{9xyz}{x\plus{}y\plus{}z}$
which proves what we wanted ($ r$ denotes the inradi). [/hide]

:)


It's sort of like Russian Roulette, except there is no chance involved! One pill is poison, one is not. You offer a pill. Your opponent choose whether or not to switch. Please pm me to sign up and which pill is poison (1 or 2).

Signups:

Crazychinesemaniac POSIONED A PILL!
Brut3Forc3 HAS NOT POISONED A PILL!!
Jjx1  POISONED A PILL!!
Budi713 POISONED A PILL!!
Dragon96 HAS POISONED A PILL!

So Russian Roulette with forced 6/0 splits? And each round has a guaranteed death?
Okay!

Sorry guys, I have to cancel this. My comp got a virus and I have little to no internet access.

I can take this over if people still want to play this.  (edit:  and if no one has a problem with it)

Edit #2:  Please PM me which pill you would like to have poisoned.


algebra
polynomial
geometry
3D geometry
algebra unsolved

Find all natural numbers "n" for which the polynomial " x^n  + 64 " can be written as a product of two nonconstant polynomials with integer coefficients ... Bulgaria 1998

Since $ 64$ is a perfect cube we can say that it is true for all positive integers $ n$ divisible by $ 3$.

Since $ 64\equal{}4*2^4$, $ x^{4n}\plus{}64$ can also be factored with the identity of sophie germain


function
inequalities proposed
inequalities

[quote="DCTPKTCSPHN"]Let $ a,b,c > 0$ such that $ : a + b + c = 3$.Prove that:
${ \frac {1}{a^2} + \frac {1}{b^2} + \frac {1}{c^2}\geq\ a^2 + b^2 + c^2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919

we can also use lagrange multipliers but when we will have to solve the system of the partial derrivatives we will have t introduce another function in 2 variables and use multipliers again ..finally we get the minimum of the function $ f(a,b,c,l)\equal{}LHS\minus{}RHS$,where $ l$ is the lagrange multiplier,  when a=b=c=1 and that min is $ 0$.. :wink:

I think this problem can be solved by [hide="uvw method"]http://www.mathlinks.ro/Forum/viewtopic.php?t=278791[/hide]