geometry
circumcircle
Triangle
concurrency
IMO Shortlist

Let $ ABC$ be an acute-angled triangle. The lines $ L_{A}$, $ L_{B}$ and $ L_{C}$ are constructed through the vertices $ A$, $ B$ and $ C$ respectively according the following prescription: Let $ H$ be the foot of the altitude drawn from the vertex $ A$ to the side $ BC$; let $ S_{A}$ be the circle with diameter $ AH$; let $ S_{A}$ meet the sides $ AB$ and $ AC$ at $ M$ and $ N$ respectively, where $ M$ and $ N$ are distinct from $ A$; then let $ L_{A}$ be the line through $ A$ perpendicular to $ MN$. The lines $ L_{B}$ and $ L_{C}$ are constructed similarly. Prove that the lines $ L_{A}$, $ L_{B}$ and $ L_{C}$ are concurrent.

I posted a solution to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38024]a very similar problem[/url] a few days ago. The lines $L_{A}$, $L_{B}$ and $L_{C}$ are the isogonal conjugate lines of the altitudes of the triangle. Hence they concur at the isogonal conjugate point of the orthocentre, which is the circumcentre.

For the sake of completeness, an angle chase solution:

Let O be the circumcenter of triangle ABC. We will prove that the lines $L_{A}$, $L_{B}$, $L_{C}$ are concurrent at the point O.

We work with directed angles modulo 180°. Since the points A, H, M, N all lie on the circle $S_{A}$, we have < AMN = < AHN. Since the point N lies on the circle $S_{A}$ with diameter AH, we have < ANH = 90°, so that, in the right-angled triangle ANH, we have < AHN = 90° - < NAH. But since < AHC = 90°, in the right-angled triangle AHC we have < HCA = 90° - < CAH. Thus, < AMN = < AHN = 90° - < NAH = 90° - < CAH = < HCA = < BCA. On the other hand, the point O is the circumcenter of triangle ABC, thus the center of a circle through the points A, B, C; hence, by the central angle theorem, < OAB = 90° - < BCA. Thus,

  < (AO; MN) = < (AO; AB) + < (AB; MN) = < OAB + < AMN = (90° - < BCA) + < BCA = 90°,

so that $AO\perp MN$. Thus, the line $L_{A}$, being defined as the perpendicular to MN through A, must coincide with the line AO, and therefore pass through the point O. Similarly, the lines $L_{B}$ and $L_{C}$ pass through O, and therefore the lines $L_{A}$, $L_{B}$, $L_{C}$ are concurrent at the point O. Problem solved.

  darij

My solution is the same as Arne's. If we use the properties of isogonal conjugates like that in an olympiad, do we have to prove or justify them?

We claim that the desired intersection point is $O$, the circumcenter of $\triangle ABC$.  Let $X$ be the feet from $A$ to $MN$. Note that $\angle HAC=90^{\circ}-\angle C.$ Hence, $\angle AMX=\angle HNA=90^{\circ}-\angle HAN=\angle C.$  Therefore, $$\angle MAX=90^{\circ}-\angle C=\angle BAO$$ so $O$ lies on $AX$, which is $L_A.$ We can similarly show that $O$ lies on $L_B$ and $L_C.$ $\blacksquare$


[hide="Important Note"] Please think that each proper noun listed (like Sunny, or Snow) is a country. And that this is a magical land and every day is the exact weather. So, if my birthday was Sep 19th, it would always be rainy on Sep 19th[/hide]

In the middle of the summer in Sunny, everyone dresses black.

In the middle of a blizzard in Snow, everyone dresses bright yellow.

In the middle of fog, Fog always dresses in a gray color.

Why do these people dress up as a certain color?

Please think outside the box. All the better if you do!

is it a holiday? like the day the country was founded?

your on the right track...

yet your train and your passengers are gonna die :p

okay is it a holiday?
are the required to wear this clothing?
are they able to make different colour clothing?
does the weather affect their clothing?

funcia...

YOUR SOOOO CLOSE!

but too bad. think outside the box..

[hide="hint"] the people in the country are wise[/hide][/hide]

[quote="funcia"]does it have anything to do with the sun?(OK I was just wondering)
And ARE they able to wear different clothing?
is is enforced by law to wear that clothing?

too bad i can't think outside the box but that makes me so happy about a story i once wrote...[/quote]

anything to do with the sun? : no

are they able to wear different clothing: yes, at certain times of day

is it enforced by law to wear that clothing: by some weaker laws.

AND YOU DIDNT READ MY HINT lol  :rotfl:

yes i did give me another because i didn't get what you were getting at...
so i thought perhaps it had to do with the sun absorption in different colour fabric...
ok do certain colour clothes come in particular styles?

they are WISE, but they are not old

they are STUPID in some ways, but are still wise.

think abotu thsoe two sentences.

and no, certain color clothes do NOT come in particular styles

ok then is it a holiday?
are they saving money by not buying new clothes?
 :)

actually, they are quite rich.

and it has nothing to do with holidays. trust me.

hint 2: daily events are done in an orderly fashion

is it so they can distinguish between the different citizens of different cities?

good i will change them sometime but as for now i like to express what i am singing
lalalala
(trust me: i had to get this avatar it isn't me it's amy lee)
Do they wear this clothing to tell each other what the weather is and therefore not need a weatherman?

nah, they have tvs and considered to be the most accurate country

Do the clothes "represent" something?
[hide="Like..."] I don't know... a big sports game every day and people are wearing the team uniforms to cheer, there are different sports for fog, sun, and blizzards, and as it happens the sports teams all have the same colored uniforms?[/hide]

 :P

That is thinking way too hard, right?
Do these people have a special color for other conditions, like rain, hail, or ?
Are these clothes normal clothes, like those I would get if I just went to the nearest store and bought some? (I happen to live on Earth.)

 :maybe:

Oh I know! IT'S ALL A COINCIDENCE!  :rotfl:

Are the actual colors in the question important? What if I changed the question to:

[quote]In the middle of the summer in Sunny, everyone dresses in hot pink.
In the middle of a blizzard in Snow, everyone dresses in lime green.
In the middle of fog in Fog, everyone dresses in blood red.[/quote]

Also, don't insult Amy Lee :)

[hide="Not Sure"]
They're wearing in some kind of relation to the colors of the [url=http://en.wikipedia.org/wiki/Mood_ring]mood ring[/url]?
[/hide]

If I visited one of the countries during the appropriate weather, would I...

feel inclined to wear this color? (reasons other than everybody wearing the same color except for you)
be manually forced (i.e. police) (EDIT: Sorry, my bad, I meant e.g.) to wear this color?
possibly wear this color? (EDIT: Sorry I mean possibly NOT wear this color)

Is it possible for at least 1 member of Homo Sapiens enter one of the countries during the weather and not wear the color?

Will anything happen if I removed "the middle of" from the problem?

Are the weathers corresponding to the days determined by some algorithm, or random? (Pseudo-random = random for this one.)

Do the habitants speak English?

Hey wait... are there more than two people in every country?

are there people in every country? Cause if yes then everybody = nobody.

do the countries exist?  :|

Does their clothing change color?

Are there only those color materials available? (e.g., animals with black fur migrate to Sun in the summer?)

Do the people always wear those colors, even when it isn't that season?

hey D3m0n Shad0w, could you answer the questions?
i've really wanted to see the result..

I am very interested in this puzzle.  Can you provide the solution please?  I guess that Sunny is in the Southern Hemisphere.

Or maybe there never was an answer... :D

Guess I never should have revived it in the first place. Well, it's the kind of thing you do if you post some and forget completely about AoPS for ~ 2 years...

In Sunny, people wear black sunglasses.
     In Snow...not about this one yet.
     And you said "in the middle of fog, Fog always dresses in a gray color". That would be implied, since fog IS gray (at least generically).

[quote="D3m0n Shad0w"]they're kids. under 13[/quote]
When I saw this clue, I immediately thought about:
[hide]That show where a bunch of kids start their own town... I think it's called KidNation[/hide]

I think KidNation was created after I made up this topic.

[quote="madshock"]so you will never answer it?!?!?!
Please tell us![/quote]
I too am anxious to know the answer, but the original poster, D3m0n Shad0w, already gave his reply regarding this situation a couple posts above:
[quote="D3m0n Shad0w"]I am very sorry, but this topic is a bit over 2 years old, and, well, I have forgotten the answer. Sorry!  :D[/quote]


A bag contains $a$ blue balls, $b$ red balls and $c$ white balls. You will take out a ball at random and replace into the bag looking at the color of the ball. Repeat this operation $n$ times repeatedly.Find the expectation value of the number of colors that are taken out.

kunny,

I'm trying to figure out what exactly the question is asking.  Could you try to restate it?

Thanks.

I have just edited, JBL.

How about this time? :)

do you mean for us to find an expression that will tell us how many times we expect to draw a red, blue, or white ball after $n$ repetitions?

That would be $\frac{a n}{a+b+c}$ right? (and b*n and c*n)

Sorry for my poor English, I have just edited again. :blush:

kunny

Ah!  I would say it like this:

A bag contains $a$ blue balls, $b$ red balls and $c$ white balls.  You will take a ball at random from the bag, observe its color, and replace it into the bag.  Repeat this operation $n$ times.  What is the expected value of the number of colors that you have seen?

So, in particular, the answer is at least 1 (for $n \geq 1$) and at most 3.  And, once one has solved this question, it should be generalized to $m$ colors.  (This is very similar to a question I've heard phrased as the stamp-collector's problem: you can buy a new stamp each day, and you need at least one of each of 5 stamps which appear with a certain probability distribution.  What is the expected number of days until you achieve your goal?)

Thank you for your restating to this problem! :)
Finally, is the answer is $1, 2,$ and $3?$ :maybe:

So, let me do the case where we have only 2 colors, appearing $a_{1}$ and $a_{2}$ times, respectively:

[hide]There are $(a_{1}+a_{2})^{n}$ ways to choose $n$ balls from our collection.  Both colors appear, so long as we are not in any of the $a_{1}^{n}$ combinations where only color 1 appears, or the $a_{2}^{n}$ combinations in which only color 2 appears.  Thus, our expected value is $\frac{2((a_{1}+a_{2})^{n}-a_{1}^{n}-a_{2}^{n})+1(a_{1}^{n}+a_{2}^{n})}{(a_{1}+a_{2})^{n}}= 2-\frac{a_{1}^{n}+a_{2}^{n}}{(a_{1}+a_{2})^{n}}$.[/hide]

[quote="kunny"]A bag contains $ a$ blue balls, $ b$ red balls and $ c$ white balls. You will take out a ball at random and replace into the bag looking at the color of the ball. Repeat this operation $ n$ times repeatedly.Find the expectation value of the number of colors that are taken out.[/quote]

[color=red]Last Edit[/color]

[color=blue]A bag contains $ a$ blue balls, $ b$ red balls and $ c$ white balls. You will take out a ball at random and replace into the bag looking at the color of the ball. Repeat this operation $ n$ times repeatedly. Denote $ E_n$ as the expectation value of the number of colors that are taken out, prove that we have $ E_n\equal{}3\minus{}\left(\frac{a\plus{}b}{N}\right)^n\minus{}\left(\frac{b\plus{}c}{N}\right)^n\minus{}\left(\frac{c\plus{}a}{N}\right)^n$.[/color]


Determine all integers k for which there exist infinitely the pairs of integers (a, b, c) satisfying
the following equation
$ (a^2\minus{}k)(b^2\minus{}k)\equal{}c^2\minus{}k$
 :D


A convex pentagon has interior angles with measures $ x\plus{}1$, $ 2x$, $ 3x$, $ 4x$, and $ 5x\minus{}1$ degrees.  What is the measure of the largest angle?

We get $ 15x\equal{}540 \implies x\equal{}36 \implies 5x\minus{}1\equal{}\boxed{179}$.


Let be ABC a triangle with area 1. If x=1/AB, y=1/AC and z=1/BC, find the minimum of expression x^4 + y^4 + z^4.

[b]Flavian, learn quickly LaTeX.[/b]

Denote the area $[ABC]$ of the triangle $ABC$.
$\boxed {\ \min_{[ABC]=1} {\left( \frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right) }=\ ?\ }$ Like you how I wrote your proposed problem ?.

This problem is equivalently with the following problem: ascertain
$\min {\left\{\left|\ \sum \frac{1}{(x+y)^4}\right|\ xyz(x+y+z)=1,\ x>0,\ y>0,\ z>0\ \right\} }.$

Yes! I hope that I will learn quickly Latex.  Thank you for translate the problem! :)

Can somebody post a solution? I made the problem with the help of something which is difficult, so my solution is very long!:(   I want see a nice solution! :D

Is 9/16. The equality is for an equilateral triangle. :)

[quote="eagerkill"]i think it need for a+b+c=const
wiht a,b,c is leght of straight[/quote]

hmmm... But the triangle has the angles less than 90. I didn't try to see what's the minimum if an angle is 90, but I don't think that is difficult. Maybe in that case u have right. But the problem is correct :)

This problem is another version of Iran 's famous inequality in 1996.
 Let $x, y, z$ be positive. Then
  $(\sum yz) (\sum \frac{1}{(y+z)^2}) \geq \frac{9}{4}$.
 If you replace $x = cotgA$, $y=cotgB$, $z=cotgC$ in the above inequality, then we obtain the minimum value of the expression.


modular arithmetic
number theory unsolved
number theory

Let n be a positive integer such that $ 1\plus{}2^n\plus{}4^n$ is prime. Prove that n must be a power of three.

[quote="cnyd"]$ n \equal{} 3^{p}q$  $ (q,3) \equal{} 1$

if $ q\equiv 1\pmod{3}$  $ \implies$  $ 2^{3^{p}} \equal{} x$  $ \implies$  $ x^{2q} \plus{} x^{q} \plus{} 1\equiv 0\pmod{x^{2} \plus{} x \plus{} 1}$

similalry  $ q\equiv \minus{} 1\pmod{3}$ $ \blacksquare$[/quote]

Nice :)


For $a,b,c$ sidelenghts of a triangle prove that

$\sum \frac{a}{b}+5 \geq \frac{8}{3}\sum \frac{a+b}{b+c}$

I write it as

$[\frac{1}{ca}-\frac{8}{3(b+c)(a+b)}](a-c)(b-c)+[\frac{1}{ab}-\frac{8}{3(b+c)(a+c)}](a-b)^{2}\geq 0$

WLOG we can assume $a\geq b \geq c$

It is easy to prove that $[\frac{1}{ca}-\frac{8}{3(b+c)(a+b)}]$ is positive but I cannot prove that $[\frac{1}{ab}-\frac{8}{3(b+c)(a+c)}]$ is positive.

Can anyone interested in this problem help me please?

Thank you very much.

As far as I know, this one belongs to 10maths_tp0609. The solution is also not very hard.  :lol:

Can you please post your solution 10maths_tp0609 ?

Thank you very much.

Have you a solution, 10maths_tp0609 ?

Thank you very much.

I cannot find it  :blush:  :blush:


[hide] 100% this and that plus 50% this and that is 150% this and that[/hide]

[quote="i_like_pie"]This and that and half of this and that is what percent of this and that?[/quote]
[hide]
there are several ways of interpreting this, because there are 3 and's, so i do not know the grouping.
for $t$=this, $f$=that:
1. $(t+f)+(\frac{t}{2}+f)$
2. $(t+f)+\frac{(t+f)}{2}$
3. $t+(f+\frac{t}{2}+f)$
4. $t+(f+\frac{(t+f)}{2})$
5. $(t+f+\frac{t}{2})+f$
etc
it would be less confusing if it is in "a, b, c, and d" form instead of "a and b and c and d"

the only ways that have an actual ratio are ways 2 and 4, which give a ratio of $\frac{3}{2}$
however, if "this and that" is counted as one variable, then it would be
$t$=this and that
$t+\frac{t}{2}$
$\frac{3}{2}t$
it is a ratio of $\frac{3}{2}$
[/hide]


Assume that $a,b,c,d$ are positive real numbers such that
\[(a+c)(b+d)=4abcd. \]
Find the greatest value of the real $q$ for which
\[abcd\geq 1+q\left(\frac{(a-c)^{2}}{ac}+\frac{(b-d)^{2}}{bd}\right). \]

[color=blue]Edit: Fixed a typing error.[/color]

For $a=c$, $b=d$, $a<>b$the condition $(a+c)(b+d)=4abcd$ <=> $ab=1$, our inequality <=> $1=a^{2}b^{2}\geq 1+2q(\frac{(a-b)^{2}}{ab})$ $\Rightarrow $ $q\leq 0$.
For $q=0$ the inequality $abcd\geq 1$ is obvious by $A(m)\geq G(m)$ $Q-E-D$  :wink: .
 :wink:
$Note$
We have $4abcd(ac+bd)=(ac+bd)(ab+bc+cd+da)=\sum{acd(a+c)=\sum{cd(a^{2}+b^{2})}}$ $\Rightarrow $
$RHS=1+q\frac{\sum{cd(a-b)^{2}}}{abcd}=1+q\frac{-8abcd+\sum{cd(a^{2}+b^{2})}}{abcd}=1+q\frac{-8abcd+4abcd(ac+bd)}{abcd}=1+4q(ac+bd-2)$ $\Rightarrow $
<=> $abcd\geq 1+4q(ac+bd-2)$


probability
ratio
number theory
relatively prime

Let $p$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before one encounters a run of 2 tails.  Given that $p$ can be written in the form $m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.

From the diagram that I have attached, we can see all ways that we don't lose when we start with H or T.  All lines except (1) and (6) will continue since they have not yet gotten the requirements to stop.

The probability of getting (1) (5 heads in a row) is obviously 1/32.

The probabilities of getting (2), (3), (4), and (5) are 1/4, 1/8, 1/16, and 1/32, respectively.  After each of (2), (3), (4), and (5), we will need to have 5 heads to stop (1/32), or we can have each of the cases (2), (3), (4), and (5) repeated again (1/2+1/4+1/16+1/32=15/32).  (We are only looking at cases where we don't lose, ie, get TT).

Let (2), (3), (4), (5) be put together as one (2345) with probability of 15/32.

We can either win by:
getting the 5 heads at the beginning (1).
(2345) followed by (1).
(2345) followed by (2345) followed by (1).
...
(2345) followed by ... (2345) follwed by (1).

So, we can win (get 5 heads) after however many of these iterations we want, so we have an infinite series with first term 1/32 and common ratio 15/32.
Summing it, we get $\frac{\frac{1}{32}}{1-\frac{15}{32}}=\frac{1}{17}$.

Then we see that we have the exact situation if we had flipped a T for the first flip, which has a probability of 1/2 times what we got above because of the first T flip.  Therefore, our answer is $\frac{1}2 \cdot \frac{1}{17}+\frac{1}{17}=\frac{3}{34}$.

[hide="Different"]
Call a string of heads and tails "good" if it contains neither $5$ consecutive heads nor $2$ consecutive tails. We seek the probability that repeatedly flipping a coin generates a good string directly followed by $5$ consecutive heads. Let $t_k$ denote the number of good strings of length $k$ ($k\ge 0$). Let $a_k, b_k, c_k, d_k, e_k$ be the number of good strings beginning with $T$, $HT$, $HHT$, $HHHT$, and $HHHHT$, respectively.  Then

$t_k=a_k+b_k+c_k+d_k+e_k$

$a_{k+1}=b_k+c_k+d_k+e_k$

$b_{k+1}=a_k$

$c_{k+1}=b_k$

$d_{k+1}=c_k$

$e_{k+1}=d_k$

Combining the last four equations gives $b_k=a_{k-1}$, $c_k=a_{k-2}$, $d_k=a_{k-3}$, $e_k=a_{k-3}$. Substituting into the first two equations,

$t_k=a_k+a_{k-1}+a_{k-2}+a_{k-3}+a_{k-4}$

$a_{k+1}=a_{k-1}+a_{k-2}+a_{k-3}+a_{k-4}$

Combining these to get

$t_{k+1}=t_{k-1}+t_{k-2}+t_{k-3}+t_{k-4}$

We can also verify that $t_0=1, t_1=1, t_2=1, t_3=2, t_4=3$. The probability that repeatedly flipping a coin generates a good string of length $k$, directly followed by $5$ consecutive heads is $\frac{t_k}{2^{k+5}}$. Summing over all $k\ge 0$, we want

$\sum_{k=0}^{\infty}\frac{t_k}{2^{k+5}}$

There is a clever way to compute this sum which takes advantage of the recursion we derived for $t_k$. The details are left to the reader, but the idea is as follows. For simplicity factor out the $2^5$ from the denominators, and let the remaining sum be $S$.  By grouping terms with the same denominator, we see that

$S+2S+4S+8S=(14t_0+6t_1+2t_2)+\sum_{k=0}^{\infty}\frac{t_k+t_{k+1}+t_{k+2}+t_{k+3}}{2^k}$

$15S=22+\sum_{k=0}^{\infty}\frac{t_{k+5}}{2^k}$

$15S=22+(32S-(32t_0+16t_1+8t_2+4t_3+2t_4))$

$15S=22+(32S-70)$

$S=\frac{48}{17}$

The actual sum is then $\frac{1}{2^5}(\frac{48}{17})=\frac{3}{34}$. The answer is $\boxed{37}$.
[/hide]

In problems like these, we will always just look at the end cases and not worry about the intermediate ones. Let H represent a heads, and T represent a tails. So our game can end in
1. HHHHHH --> 1/32
2. THHHHH --> 1/64
3. TT --> 1/4
4. HTT --> 1/8
5. HHTT --> 1/16
6. HHHTT --> 1/32
6. HHHHTT --> 1/64

So the total is 2/64 + 1/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64= 34/64. Out of these only 1/32 + 1/64 result in a winning case. Thus we have 3/34 => 037.


Is this a question or some sort of crazy challenge

[hide]I think you would be right if there was only 9 digits in the decimal system [/hide]

[hide]When there are 2n digits in the palindrome:

First number must not be 0.

A1A2A3...AnAnAn-1An-2...A1.

A1 has 9 choices, every other one has 10. So there are 9 * 10n-1 in total.

When there are 2n+1 digits in the palindrome:

First number must not be 0.

A1A2A3...AnAn+1AnAn-1An-2...A1.

A1 has 9 choices, every other one has 10. So there are 9 * 10n in total.[/hide]


If I'm not mistaken, one can prove it by induction. :)

marshell, - why?

[quote="sinajackson"]Let $ x_{1},x_{2},\cdots,x_{n}$ be nonnegative reals, Prove that:

$ \prod^{n}_{i=1}\frac{1-x_{i}}{1+x_{i}}\geq\frac{1-(x_{1}+x_{2}+\cdots+x_{n})}{1+(x_{1}+x_{2}+\cdots+x_{n})}$ :)[/quote]

A proof by induction.
1. For n=1 we have equality.
2. Assume it's true that
$ \prod^{n}_{i=1}\frac{1-x_{i}}{1+x_{i}}\geq\frac{1-(x_{1}+x_{2}+\cdots+x_{n})}{1+(x_{1}+x_{2}+\cdots+x_{n})}$
Then we must prove that
$ \prod^{n+1}_{i=1}\frac{1-x_{i}}{1+x_{i}}\geq\frac{1-(x_{1}+x_{2}+\cdots+x_{n}+x_{n+1})}{1+(x_{1}+x_{2}+\cdots+x_{n}+x_{n+1})}$
We have from our assumption
$ \prod^{n+1}_{i=1}\frac{1-x_{i}}{1+x_{i}}\geq \frac{1-x_{n+1}}{1+x_{n+1}}\cdot \frac{1-(x_{1}+x_{2}+\cdots+x_{n})}{1+(x_{1}+x_{2}+\cdots+x_{n})}$
Let $ a = x_{1}+x_{2}+\cdots+x_{n}$ and $ b = x_{n+1}$. Then it's enough to prove that
$ \frac{1-b}{1+b}\cdot \frac{1-a}{1+a}\geq \frac{1-(a+b)}{1+(a+b)}$
Which is equivalent to $ a^{2}b+b^{2}a \geq 0$.
Equality holds iff n=1 or x_1 = ...= x_n = 0.

[color=red]
edit: This proof works only if x_i<=1 for all i.[/color]

Sorry, the condition $ x_{1}+x_{2}+...+x_{n}\le 1$ is too good but this inequality need some conditions

[quote="kar"]$ \prod^{n+1}_{i=1}\frac{1-x_{i}}{1+x_{i}}\geq \frac{1-x_{n+1}}{1+x_{n+1}}\cdot \frac{1-(x_{1}+x_{2}+\cdots+x_{n})}{1+(x_{1}+x_{2}+\cdots+x_{n})}$
[/quote]

This requests that $ \frac{1-x_{n+1}}{1+x_{n+1}}\ge 0$ so $ x_{n+1}\le 1$. So the conditions are $ x_{i}\le 1\;\;\forall i$

Oh! I see now, you are right, marshell, thank you.
Hmm..is it possible that the inequality is true without these conditions? (if we use another proof?)


calculus
integration
derivative
calculus computations

How to make up a differential equation that does not possess any real solutions ? :)

Theorem (Cauchy-Peano): Let $I=[t_{0},t_{0}+\beta]$ and let $\Omega$ be the closed ball of radius $r$ centered at $x_{0}\in\mathbb{R}^{n}$. If $f(t,x)$ is continuous on $I\times\Omega$, there exists a solution $X(t)$ to the integral equation $X(t)=x_{0}+\int_{t_{0}}^{t}f(s,X(s))\,ds$, which is continuous on $I_\alpha=[t_{0},t_{0}+\alpha]$ for some $\alpha$ (We can choose $\alpha=\min(\beta,\frac{r}{M})$, where $M=\max_{I\times\Omega}|f(t,x)|$).
By the FTC, $X$ also satisfies the differential equation $X'(t)=f(t,X(t))$ with initial condition $X(t_{0})=x_{0}$.

That's nontrivial to prove; it requires something like the Arzela-Ascoli theorem. In essence, we always have a solution if we can at least solve for the highest derivative (since higher-order equations are equivalent to first-order systems.


[quote="gus135791"]a,b,c are all positive real numbers. Prove this:
\[ \frac {a \plus{} b \plus{} c}{\sqrt {ab \plus{} bc \plus{} ca}}\geq\sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 2bc}}
\]
[/quote]
$ \left(\sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 2bc}}\right)^2\leq\sum_{cyc}\frac {a}{a^2 \plus{} 2bc}\sum_{cyc}a\leq\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} ac \plus{} bc}.$ :wink:
The last inequality we can prove by arqady-Schur (  :D ) method:
$ \sum_{cyc}\frac {a}{a^2 \plus{} 2bc}\sum_{cyc}a\leq\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} ac \plus{} bc}\Leftrightarrow\sum_{cyc}\frac {a}{a^2 \plus{} 2bc}\leq\sum_{cyc}\frac {a }{ab \plus{} ac \plus{} bc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a(a \minus{} b)(a \minus{} c)}{a^2 \plus{} 2bc}\geq0.$
Let $ a\geq b\geq c.$ Hence, $ \frac {c(c \minus{} a)(c \minus{} b)}{c^2 \plus{} 2ab}\geq0$ and
$ \frac {a(a \minus{} b)(a \minus{} c)}{a^2 \plus{} 2bc} \plus{} \frac {b(b \minus{} c)(b \minus{} a)}{b^2 \plus{} 2ac} \equal{} (a \minus{} b)\left(\frac {a(a \minus{} c)}{a^2 \plus{} 2bc} \minus{} \frac {b(b \minus{} c)}{b^2 \plus{} 2ac}\right) \equal{}$
$ \equal{} \frac {(a \minus{} b)^2c(2a^2 \plus{} 3ab \plus{} 2b^2 \minus{} 2ac \minus{} 2bc)}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)}\geq0.$
Vornicu-Schur does not work here. :wink:

please Mr.arqady, what is Vornicu-Schur ?

[quote="arqady"]See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49427[/quote]
thank's a lot!

[quote="gus135791"]$ a,b,c$ are all positive real numbers. Prove this:

$ \frac {a \plus{} b \plus{} c}{\sqrt {bc \plus{} ca \plus{} ab}}\geq\sum\frac {a}{\sqrt {a^2 \plus{} 2bc}}$[/quote][quote="arqady"]$ \left(\sum\frac {a}{\sqrt {a^2 \plus{} 2bc}}\right)^2\leq\sum\frac {a}{a^2 \plus{} 2bc}\sum{a}\leq\frac {(a \plus{} b \plus{} c)^2}{bc \plus{} ca \plus{} ab}.$ :wink:

The last inequality we can prove by[/quote]$ \frac {a \plus{} b \plus{} c}{bc \plus{} ca \plus{} ab} \minus{} \sum\frac {a}{a^2 \plus{} 2bc}$

$ \equal{} \frac {abc\left[(4a^2 \minus{} 2b^2 \minus{} 2c^2 \plus{} 2bc \minus{} ca\minus{} ab )^2 \plus{} 3(b \minus{} c)^2(2b \plus{} 2c \minus{} a)^2\right]}{4(bc \plus{} ca \plus{} ab)(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}\geq0.$

[quote="Ji Chen"]$ \frac {a \plus{} b \plus{} c}{bc \plus{} ca \plus{} ab} \minus{} \sum\frac {a}{a^2 \plus{} 2bc}$

$ \equal{} \frac {abc\left[(4a^2 \minus{} 2b^2 \minus{} 2c^2 \plus{} 2bc \minus{} ca \minus{} ab )^2 \plus{} 3(b \minus{} c)^2(2b \plus{} 2c \minus{} a)^2\right]}{4(bc \plus{} ca \plus{} ab)(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}\geq0.$[/quote]
See also here: http://can-hang2007.blogspot.com/2010/03/inequality-122-unknown-author.html


In how many ways can two slashes be inserted between the letters of MATHCOUNTS to create three groups of letters?  Each ``group'' of letters must contain at least one letter.

There are $ 10$ letters, so there are $ 9$ spaces in between the letters.  We need to put the $ 2$ slashes in, so there are $ \binom{9}{2}\equal{}\boxed{36}$ ways to do this


geometry
geometric transformation
reflection
geometry theorems

Let a pint $D$ be on side $AB$ of a triangle $ABC$. What is isogonal conjugate to D wrt. $\triangle ABC$?

[quote="Centy"]The isogonal conjugate of any point on the triangle $ABC$ is the vertex opposite that line.

In your case it's $C$.[/quote]

And since isogonal conjugation is involutive, it means that the isogonal conjugate of $C$ is ?   ;)

Actually, I think that isogonal conjugation is not defined on the triangle sides...   :?

If P is a point on the side BC of triangle ABC such that $P\neq B$ and $P\neq C$, then the isogonal conjugate of the point P with respect to triangle ABC is the vertex A. In fact, the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA all concur at the point A - in fact, two of these reflections are the lines CA and AB, and the third one is a line through A.

On the other hand, if P is one of the vertices of triangle ABC, then the isogonal conjugate of P is not well-defined. For instance, if P = A, then the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA don't have a unique point of intersection - one of these reflections is not defined (because the line AP = AA is not defined), and the other two fall together to the line BC. The vertices of triangle ABC are "singular points" of the isogonal conjugation. One could say that the isogonal conjugate of the vertex A with respect to triangle ABC is an undefined point on the line BC; of course, this is far from being strict, but it helps you visualize the situation.

You encounter a similar problem if you work with inversion in the [i]projective[/i] plane. In the Möbius plane, the image of the center of a circle under inversion with respect to this circle is the point at infinity. In the projective plane, however, there is not one point at infinity but a whole line at infinity, and the image of the center of a circle under inversion with respect to this circle can be considered as an undefined point on the line at infinity. Strictly speaking, of course, it's not defined at all.

  Darij


3.  S = {1, 2, 3, 5, 8, 13, 21, 34}. Find ∑ max(A) where the sum is taken over all 28 two-element subsets A of S.

[hide="Solution from Kalva"]
3.   sum = 7·34 + 6·21 + 5·13 + 4·8 + 3·5 + 2·3 + 1·2 = 484
[/hide]

I don't know what the steps are behind that.

Because in a subset ${a,b}={b,a}$, only one of them needs to be counted, so just let $a$ be less than $b$.  We are summing over max(A) so that means $b$.  Now go through the elements of $S$, letting each element be $b$.  $34$ can be $b$ when combined with each of the $7$ lesser elements, $21$ with the $6$ lesser ones,...,until $1$ cannot be the max of any two element subset.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=508860#p508860

enjoy  :)


Gauss
complex numbers
absolute value
algebra unsolved
algebra

If $ z_{1},z_{2}$ and $ z_{3}$ are three non-collinear complex numbers and $ z$ is a variable complex number satisfying the equation

$ \frac{1}{z\minus{}z_{1}}\plus{}\frac{1}{z\minus{}z_{2}}\plus{}\frac{1}{z\minus{}z_{3}}\equal{}0$

then prove that $ z$ always lies inside the triangle formed by $ z_{1},z_{2}$ and $ z_{3}$.

This generalizes.  See the proof of the [url=http://en.wikipedia.org/wiki/Gauss-Lucas_theorem]Gauss-Lucas theorem[/url] given here.

Thanks t0rajir0u. But tell me how to interpret this.

On solving the given equation i got

$ z\equal{}\frac{z_{1}\plus{}z_{2}\plus{}z_{3}\plus{}\sqrt{{z_{1}}^2\plus{}{z_{2}}^2\plus{}{z_{3}}^2\minus{}z_{1}z_{2}\minus{}z_{2}z_{3}\minus{}z_{3}z_{1}}}{3}$

and

$ z\equal{}\frac{z_{1}\plus{}z_{2}\plus{}z_{3}\minus{}\sqrt{{z_{1}}^2\plus{}{z_{2}}^2\plus{}{z_{3}}^2\minus{}z_{1}z_{2}\minus{}z_{2}z_{3}\minus{}z_{3}z_{1}}}{3}$

$ \frac{z_{1}\plus{}z_{2}\plus{}z_{3}}{3}$ denotes the centroid of the triangle formed by $ z_{1},z_{2},z_{3}$ but how does one

interpret $ \frac{\sqrt{{z_{1}}^2\plus{}{z_{2}}^2\plus{}{z_{3}}^2\minus{}z_{1}z_{2}\minus{}z_{2}z_{3}\minus{}z_{3}z_{1}}}{3}$ geometrically.

$ z_1^2 \plus{} z_2^2 \plus{} z_3^2 \equal{} z_1 z_2 \plus{} z_2 z_3 \plus{} z_3 z_1$ if and only if $ z_1, z_2, z_3$ forms an equilateral triangle, so one way to interpret it is as a measure of the extent to which $ z_1, z_2, z_3$ fails to be equilateral.  If you'd like, you could write their difference as $ (z_1 \plus{} z_2 \omega \plus{} z_3 \omega^2)(z_1 \plus{} z_2 \omega^2 \plus{} z_3 \omega)$.

What do you mean by "measure of the extent to which $ z_{1},z_{2},z_{3}$ fails to be equilateral"

I mean if it has small absolute value then $ z_1, z_2, z_3$ is very close to equilateral and if it has large absolute value then $ z_1, z_2, z_3$ is very far from equilateral.  This is a heuristic and the factorization I gave can make this more precise.


[quote="quanghuyhungnguyen"]Find the smallest positive integer n such that n^n isnot a factor of 2009!=1.2.3....2009[/quote]

Looking first for prime solutions, we look for $ p$ such that $ p > v_p(2009!) > \frac {2009}p \minus{} 1$ and so $ p > 44$. The first next prime is $ p \equal{} 47$ and we indeed have $ 47 > v_{47}(2009!) \equal{} 42$. The question is now to see if there is a non prime number less than $ 47$ verifying the requirement.

Let $ p < \frac {47}2 < 24$ be the prime factor of the required composite number whose power does not fit. And so let this number be $ 47 > N \equal{} ap^n$  with $ p\not | a$.
So we need $ nap^n > v_p(2009!) > \frac {2009}p \minus{} 1 > \frac {1985}p$

So we have $ 47 > N > \frac {1985}{pn}$ so $ n > \frac {1985}{47p}$ and we want at least $ 47 > N \ge p^n > p^{\frac {1985}{47p}}$, so $ \ln(47) > \frac {1985\ln(p)}{47p}$ and so $ \frac {\ln(p)}p < 0.091162185 < \frac {\ln(24)}{24}$  and so $ p > 24$, which does not match the requirement.

Hence the answer : $ 47$


linear algebra
matrix
superior algebra
superior algebra solved

We have A in M_2 with real entries so that det(A)=1. Prove that det(A-A^(-1))=4.

Mamma mia traditore ! What about  A= I ?

well, seems to me if A=I_2 then the problem is trivial, I think you should also try the case when A<>I_2

cheers! :D

Yes, Lagrangia and Diogene, you are should consider A=I_2 and then A<>I_2

at least this is my opinion!

well my problem goes like this:
we have X in a 2nd ordered matrix with real entries (M_2(R)) with det(X)=a<>0
and det(X^2+I_2)=0 then we must prove that we have det(X-a^n*X^(-1))=4.

try it!

I have managed to prove that det(X)=1. is this true?

If det(X 2 +I 2)=0 then the eigenvalues of X are i and -i. So det X=i*(-i)=1. Then det(X-X -1)=det(X 2 -I2)=P(1)*P(-1)=4, where P(x)=x 2 +1.

Nice solutione pcalin! Do you have another problem, simmial to this one?

I would like to test my capabilities with eigenvectors :)

I have a file with over 70 problems of this type. unfortunately it's in romanian and I don't plan to translate it soon :(

[quote="Vialli"] A in M_2 with real entries so that det(A)=1.
Prove that det(A-A^(-1))=4.[/quote]

det(A+A^(-1))+det(A-A^(-1)) = 2(detA+detA^(-1))= 4

det(A+A^(-1)) = det(A+I) with A-trA.A+I=0

det(A+A^(-1))=(trA) so

det(A-A^(-1)) = 4- (trA)

Omarjee


Does sqrt(a^2)<sqrt(b^2) imply a<b?

Is my solution OK?

Case 1: a<b
Case 2: -a<-b or a>b
By Case 2 no.

Or how about this:

Suppose a<0 and b<0.
Then sqrt(a^2)=-a and sqrt(b^2)=-b.
Which results in -a<-b or a>b.
Therefore no.

One way to approach an inequality like this is to test it for a few examples. Consider for instance $ a \equal{} 2$ and $ b \equal{} \minus{}4$

Note that $ |a| \equal{} \sqrt{a^{2}}$.

What akech said.

All you need to disprove a conjecture is one counterexample.

Case 1: $ a> 0$ and $ b>0$ then $ a<b$

Case 2: $ a<0$ and $ b<0$, then $ a>b$

Case 3: $ a<0$ and $ b>0$ then $ a<b$

Case 4: $ a>0$ and $ b<0$ then $ a>b$

*cases 3 and 4 are obvious since one is negative and the other is positive.


This problem was on my regional math team geometry contest this weekend and is the only one I can't figure out. We're given answers but not solutions, so help please? Calculators were allowed.

In $ \Delta ACE$, $ B$ lies on $ AC$, and $ D$ lies on $ EC$. If  $ \frac {AB}{BC}\equal{}\frac {4}{5}$ and $ \frac {ED}{DC}\equal{}\frac {1}{7}$, then the ratio of the area of $ \Delta BDC$ to the area of quadrilateral $ ABDE$ can be expressed as a common fraction reduced to lowest terms. Find that common fraction.

[hide="ANSWER"]$ \frac {35}{37}$[/hide]

BC=5m, AB=4m, CD=7n, DE=n. Using the formula area=$ (1/2)ab \cdot sin C$, we find if the area of BCD is x, then the area of ACE is $ x \cdot \frac{9}{5} \cdot \frac{8}{7}\equal{} \frac{72}{35}x$. So the area of ABDE is $ \frac{72}{35}x\minus{}x \equal{}\frac{37}{35}x$. So the answer is $ \frac{35}{37}$.

Let $ \triangle{BCD}$ have area $ A$. Draw $ AD$.  The area of $ \triangle{ABD}$ is therefore $ \frac {4A}{5}$.  The area of $ \triangle{ADE}$ is $ \frac {1}{7}(A \plus{} \frac {4A}{5}) \equal{} \frac {9A}{35}$.

Our desired ratio is the sum of the areas of triangles $ ABD$ and $ ADE$ over the area of triangle $ BCD$.  This is just $ \frac {A}{\frac {9A}{35} \plus{} \frac {4A}{5}} \equal{} \boxed{\frac {35}{37}}$.


Mark earns $ \$8.00$ per hour for the first 8 hours he works. If he works more than 8 hours, he gets paid 1.5 times his hourly rate for the extra hours. One day, Mark worked from 7:30 a.m. to 6:30 p.m. and took an hour for lunch. His lunch hour is unpaid. How many dollars did he earn?

He worked a total of $ 10$ hours. For the first $ 8$ hours he earned $ 8\times{8}\equal{}\$64$. For the extra two hours, he earned $ 12\times{2}\equal{}\$24$. Thus, the total amount of money he earned is $ \boxed{\$88}$


trigonometry
inequalities
algebra proposed
algebra

That that we can choose the signs $+\Big\slash-$ such that \[\left( \pm \cos 1 \pm \cos 2 \pm \ldots \pm \cos 1988 \right)^{2}+\left( \pm \sin 1 \pm \sin 2 \pm \ldots \pm \sin 1988 \right)^{2}\geq 1988 .\]

Shoose signs $\pm cos k=|cos k, \ \pm sink=|sink|$. Then \[\sum_{k=1}^{n}|cos k|>\frac{2n}{\pi }-1, \ \sum_{k=1}^{n}|sin k|>\frac{2n}{\pi}-1\].

[quote="Rust"] Then
\[\sum_{k=1}^{n}|cos k|>\frac{2n}{\pi }-1, \ \sum_{k=1}^{n}|sin k|>\frac{2n}{\pi}-1 \]
.[/quote]
How do you get this?


college
calculus
MIT
derivative
integration
geometry
geometric transformation

in preparation for the amc and other math contests, there are thousands of books, there are math mentors, there is the art of problem solving =), and so much more.  i know that for physics there's much much less, but is there anything at all?  at stuyvesant they give a physics olympiad class, and i know they do the same in some schools in russia and china.  is there anything available for someone who goes to a regular public school?  just as in math, doing well on a physics olympiad requires more than what you learn in a basic physics class, even an AP class.  there has to be some sort of preparation one can do.  for anyone who's in such an olympiad class, could you maybe explain how your class works?  if there's any literature that can be used to prepare for a physics olympiad rather than just past problems, please let me know about it.  if anyone has any information, i'd appreciate any feedback.  thanks!

a few questions for schulmannerism...

first of all, do these books use calculus and does the usapho have any calculus?  (i'm guessing they do, which is fine since im in bc now, and i assume that even if the usapho doesnt have calculus, it could serve useful)

second of all, where would you find a 15 dollar edition of these books?  on amazon they are very pricey, about 100 dollars each.

third of all, i know that the physics olympiad includes more than just the electricity and magnetism and mechanics.  do you know of any books for the rest of the topics such as optics, thermodynamics, modern physics, relativity, and anything else?

thanks!  anyone else who knows about anything is welcome to reply as well.

About the calculus: the books I mentioned use calculus, as do most introductory physics textbooks.

I was bit too optimistic on the pricing.
For finding cheap books, http://www.ebay.com and http://www.bookfinder.com are your best friends.

Mechanics book:
ebay has it for 20 + 5.50 shipping. http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=2228&item=6957648040&rd=1&ssPageName=WD1V

E&M book:
While this is also published by McGraw Hill, it is not published by McGraw Hill's ultra-cheap Indian subsidary Tata McGraw Hill. The one I own is my dad's old physics textbook from college back in the 60's. Anyway, that book is extremely hardcore for an introductory book, there are others that you might like better.

Whatever you decide to buy, if anything, you should try to look at it at first--if you have a university library near you that would be a good place to look at it.

For the other physics topics, I am currently looking for good books on these subjects, so that I can read up on them before I get to the training camp. Actually, two days ago I borrowed bunch of books from a nearby university library, but I haven't started reading them yet.

hmm, i see that http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/21N3R7VTQFDRP/ref=cm_lm_lists/103-8691437-3955064
is the listmania on amazon for the textbooks mit uses.  the two books you recommended are the first two there.  seeing that, do you think that some of the other books there might be useful as well, such as the one on optics or waves?  or are the other ones not of the same olympiad caliber considering they don't go along with an mit "honors" course?  you might not know anything about it, but i just felt like mentioning it.  thanks.

wow... very interesting list... I think iwill add this on to t he resources topic.  :)

if you're just looking for a book on e and m without the olmypiad aspect to it, then there are many good books.  look for authors such as serway and jewett and giancoli.  i used giancoli when doing physics without calculus and serway and jewett when doing physics with calculus.  i was just interested now to take it to the olympiad level with more abstract books.

If anyone's physics teacher reads the magazine The Physics Teacher it has some great problems in the challenges for teachers and students section.

Actually the rpoblems can be found online at http://www.aapt.org/tpt

Aren't the free response on the first qualifying test mostly kinematics (mechanics)?

read all the volumes of :
Holiday Physics.
also for Calculus, i think the best bookwould be:
Differentiation and Integration and analytic Geometry by George Thomas. Vol I&II

Just read Halliday/Resnick Fundamentals of Physics or something like that....

It's more than enough subject training for the olympiads if you raed every chapter and do every odd problem at the end of the chapters.

Of course, that's just the start.  Then you have to start doing physics problems that you find online and they're ALL over the place.

the only book you ever need to buy is Halliday  :D.  The rest are written by like arrogant physicists who want to show off their proof-oriented intelligence (and end up making it too hard and pointless to understand) and are not 100% necessary for olympiaders...

First, you should master those basic concepts.'
Second, you MUST do a lot of exercises. A LOT OF!
That's the only way to do well in physics contests.

Are there good online resources available which have advanced theory as well as tough , interesting problems in all topics covered in the Physics Olympiad??

Also, can anybody recommend some good theory and problem resources and books(preferably available online) in specific topics of Physics (mechanics,thermodynamics,Optics,Electrodynamics etc.) which might be helpful for the Olympiad as well as for general interest?

Note :- No point recommending Official IPhO site since I already know of it.

Regarding problem books there is this "Irodov's Problems in General Physics".I don't know if it is so easily available in the US and europe but it is a marveleous book.It's the best book having problems for high school or undergrad level.Irodov's theory books in mechanics and electrodynamics are also wonderful having both simple and advanced theory(although the translation can be befuddling at times!).if you can get them GET THEM NOW!!

I know Azerbaijan

1)Gold
2)Bronze
3)Bronze
4)Mansion(4)
5)Mansion(4)

I suppose you mean mention  :)
France got 5 bronze.

[quote="shobber"]Anyone knows the result of China?[/quote]

I do not, but I may guess.

[quote="schulmannerism"]Find a textbook with lots of non-routine examples and problems, and read it to supplement your school physics course.
I used
[u]An Introduction to Mechanics[/u] by Kleppner and Kolenkow
and
[u]Berkeley Physics Course -- Volume 2: Electricity and Magnetism[/u]
The problems in these books are challenging and are on the level of physics contests.

The two textbooks I mentioned might not fit you very well, hopefully someone else can give other recommendations. When you find a book, you can probably find a foreign (usually Indian) edition for around $15 online.[/quote]

Whoa cowboy... let's not get carried away. There is no reason that any high school student should consider trying to learn from these books. 8.012 (honors mechanics) and 8.022 (honors E&M) are extremely difficult and rigorous classes that both require multi variable calc (and rightly so) as pre-requisites. Even the average MIT student would fail these classes.

In fact, I would strongly suggest not going by the list of books for those classes there, as many of them are supplemental and strictly for students to glance at if they're looking for some more advanced reading. That site doesn't specificy which is which.

[quote]It looks as if the E&M books above are not for beginners. What if I want to learn E&M completely and start from nothing? Any good books? Is Feynman's Lectures volume II good for me as a beginner?[/quote]

I wouldn't consider the Feynmann lectures for beginners either, any volume.

Someone mentioned that Halliday and Resnick is a good book. I know that it's fairly standard in many universities for freshman physics. MIT in their regular freshman physics uses Young and Freidman for Mechanics, but I personally don't like the book. I think Giancoli is better (the calculus version).

Here are some of my opinions on the IPhO questions:

The problems themselves do not require calculus beyond basic integration. If integration is required, they provide you with integral... so as far as solving the problems, you don't need it. Now, with that said, as far as I can see, the theory you have to know to answer the problems is so advanced that I've never really seen it taught in any book except ones that use advanced calc (by advanced I mean calc involving vectors and associated threorems e.g. Divergence theorem). So, in my opinion knowledge of calculus is not just optional, it's required. If you want to understand the underlying physics (which is essential for IPhO), then you'll definitely need to know the calculus.

If you think these books are very [i]advanced[/i] you're wrong, you probably haven't seen them. They're difficult, yes, but not advanced. I stick to my assertion that they're a good supplement to a course like AP Physics C, both for general understanding and for contests. The Kleppner and Kolenkow doesn't require any multivariable calculus, and the exercises and examples are much more interesting and challenging than those of any typical introductory book. The Purcell book is a bit more advanced--it does use a lot of 3-dimensional vector calc, which it teaches in chapter 2, but it's not nearly as advanced as an actual second-year E&M course like Griffiths. Rather, it covers the simpler material in a very deep way.

By the way, while the IPhO syllabus only requires basic calculus and only a little more physics than you'd learn in AP physics C, for example, it helps a heck of a lot to know more advanced physics and math. For example, the most difficult of the three questions on the mock IPhO test they gave us at the US Physics Team training camp was basically to derive a certain effect that occurs when polarized light enters a dielectric substance. They lead you through the derivation, so you technically didn't need to know anything advanced, but it sure would've helped to understand the theory of electromagnetic fields in matter. There were a lot of other problems that are well-known effects that you could probably find fully worked out in some advanced textbooks

[quote="schulmannerism"]If you think these books are very [i]advanced[/i] you're wrong, you probably haven't seen them. They're difficult, yes, but not advanced. I stick to my assertion that they're a good supplement to a course like AP Physics C, both for general understanding and for contests. The Kleppner and Kolenkow doesn't require any multivariable calculus, and the exercises and examples are much more interesting and challenging than those of any typical introductory book. The Purcell book is a bit more advanced--it does use a lot of 3-dimensional vector calc, which it teaches in chapter 2, but it's not nearly as advanced as an actual second-year E&M course like Griffiths. Rather, it covers the simpler material in a very deep way.[/quote]

First of all, advanced is all a matter of opinion and it's relative. I assumed you were speaking strictly for high school students. Neither of the ones you mentioned are at a Physics C level, in my opinion. Yes I have seen both of them and I took physics C. Now, no they are not as "advanced" as griffiths, but griffiths wasn't a consideration, now was it? I personally think griffiths is not advanced at all, in fact it lacks a lot of depth that could easily be covered at that level. I don't consider Griffiths "deep" in any way, shape or form. Here is where it's strictly relative and opinionated. In any event, Jackson [i]Classical Electrodynamics[/i] is a "deep" book that explains the theory without the handwaving of griffiths.

You all are very lucky. Back when I was in high school I didn't such a thing like IPhO existed! (or IMO) or any international competition.

In fact I've been lucky as this year it was the first year France had a true IPhO team :D

[quote="antony"]Anyway, I'd just say that China got 5 gold.

Just like Taiwan, which also got the overall winner (tie with a Hungarian) and is the winning country  :D[/quote]
Thanks, and can you give us the general ranking?

The results should be at [url=http://www.ipho2005.com/]www.ipho2005.com[/url] but they still aren't .. hope there will be something soon.

But anyway, I found this year's IPhO quite a lot easier than last few years. What do you think of the questions?


Here's the queston for good problem solvers to do for me:

What is the least positive integer n such that n :^2: - n is divisible by some but not all integers values of k when 1 :le: k :le: n?

5, if you keep trying numbers from n=1 and upwards then you eventually get to 5 which works:
(1^2)-1=0, divisible by all possible values of k (1), so no
(2^2)-2=2, divisible by all possible values of k (1, 2), so no
(3^2)-3=6, divisible by all possible values of k (1, 2, 3), so no
(4^2)-4=12, divisible by all possible values of k (1, 2, 3, 4), so no
(5^2)-5=20, divisible by 1, 2, 4, and 5, so this works

Thanks, but there aren't any specific method on doing that?

I don't see any real other way.

Factor as n(n-1). If you want it to be divisible by all numbers 1<= k <= n, then it's obviously a pretty small number. You don't see numbers like 500 being divisible by 1 through 500.


function
calculus
derivative
trigonometry
limit
calculus computations

Consider the functions h(x) and k(x) given below.  Determine if each function is differentiable at the given value of x.  If it is differentiable at the given value of x, then find the value of the derivative there.  If it is not differentiable at the given value of x then clearly explain why you have reached that conclusion, using correct limit notation.

h(x)= {x^2+5x, x (< or =) 1}, {3x+1, x > -1}

Is h(x) differentiable at x = -1 ?

k(x)= {sinx, x (< or =) pi/6}, {3/pi, x (> or =) pi/6}

Is k(x) differentiable at x = pi/4 ?

\[ \begin{gathered}
  h\left( x \right) \equal{} \left\{ \begin{gathered}
  x^2  \plus{} 5x, \quad x \leqslant 1, \hfill \\
  3x \plus{} 1, \quad x >  \minus{} 1. \hfill \\
\end{gathered}  \right. \hfill \\
  k\left( x \right) \equal{} \left\{ \begin{gathered}
  \sin x, \quad x \leqslant \frac{\pi }
{6}, \hfill \\
  \frac{3}
{\pi }, \quad x \geqslant \frac{\pi }
{6}. \hfill \\
\end{gathered}  \right. \hfill \\
\end{gathered}
\]
 :?:

The definition of $ h$ is incorrect as written. There are two conflicting definitions for $ \minus{}1<t\le1,$ and that 's not a function. My best guess is this:

$ h(x)\equal{}\begin{cases}x^2\plus{}5x,&x\le\minus{}1\\3x\plus{}1,&x>\minus{}1\end{cases}$

In this case, even though $ \lim_{x\to\minus{}1^\plus{}}h'(x)\equal{}3$ and $ \lim_{x\to\minus{}1^\plus{}}h'(x)\equal{}3,$ the function would fail to be differentiable at $ \minus{}1$ because it is not continuous at that point. (It has a jump discontinuity.)

$ k$ is differentiable at $ \frac{\pi}4$ and its derivative is zero there. since $ \frac{\pi}4>\frac{\pi}6$ and differentiation is local, we might as well be differentiating a constant function.


geometry AMC USA(J)MO USAMO trigonometry percent MATHCOUNTS

What do you think about the math taught in school?

waay too simple...i always do my math homework the day she assigns it because im really bored.

[quote="roadnottaken"]I wish my teacher would post the math homework at the beginning of class so I can work on it during the class, a somewhat more constructive use of my time than listening to the lecture.[/quote]

your teacher is experianced. my teacher leaves it on the board and we must copy it to promote good usage of our school agenda...

if you are really bored (not to mention deprived due to parents who won't let you do anything fun) I would do every single problem in the next lesson for math....yeah i know its boring but since my parents suck I can't think of anything else to do

In NC, the standard math courses are WAY too easy.  I barely learned anything in the Geometry course, even though I didn't know much geometry beforehand.

NC is not noted for its difficult courses...

[quote="bpms"]Meh, in my alg. 2 class I only go once a week and on test days.
As for this poll, we are AoPSers. Consider the sheer number of USAMO qualifiers that this forum has. I don't think any of them have problems with school math.[/quote]
Maybe..............

All I do in trig class is take tests, do my homework, talk, and play calculator games. It's like I just walk into the room and suffer for 45 minutes just to get the credit...
(I miss questions on tests because of my "skill" with arithmetic and keying on my TI-89)

I use trig as a study hall period and ask the Spanish 4 people how to do my Spanish 1 homework.

I learned more from reading the USAMO trig book (Andreescu, Feng) for an hour than I have in the whole semester of trig class.

I am three years accelerated in math classes, and I still wish I could just save the 45 min/day to come to do other stuff (like read AoPS books). School math is so easy, it's worthless.

Once my trig teacher was scolding at this student after class because he played calc games for 5 min, but she still smiled at me    :blush:
*To elaborate on the deficiency of my high school math dept.*:

My cousins from Taiwan learned all this stuff by the time they were in 9th grade, without acceleration.

All the people who are accelerated in math like me last year do about the same thing as me in their math classes.
How to convert fractions into percents is taught in regular class, and solving single variable, linear equations is taught in adv. algebra (third year math).

Even though no high school teachers care about our math club, run by a university professor and a middle school teacher (they do care to tell us to shut the door to the room when we're done), we have produced several state Mathcounts champions,  USAMO participants, and one MOPer in the past five years. The Siffe Award from my high school last year went to a middle school teacher.

In general, I hate my school math department  :mad:

[quote="now a ranger"]They make school math tuned down so students of all levels can do it. This makes it much more easier than what REAL math really is. Therefore, it seems like a joke to many of us who can do it on a higher level. '

I wonder if that stupid No Child Left Behind Act has anything to do with math being like that nowadays....  :o[/quote]

well the name shure gives away a lot

[quote="7h3.D3m0n.117"][quote="now a ranger"]They make school math tuned down so students of all levels can do it. This makes it much more easier than what REAL math really is. Therefore, it seems like a joke to many of us who can do it on a higher level. '

I wonder if that stupid No Child Left Behind Act has anything to do with math being like that nowadays....  :o[/quote]

well the name sure gives away a lot[/quote]
<editing for math people not left behind in grammar>

The choice I wanted wasn't on there.

i could be playing Need For Speed, writing a book, watching ESPN, and writing a program at the same time as doing my math problems.

SERIOUSLY...PRECALCULUS IS SO EASY!!! AT LEAST THE BOOK WE'RE DOING!!!

I was less than 1 point away from a B in 6th grade adv math last year because I made a 15, a 25, and a 40 on notebook checks!  :P

lol today i prove Cauchy-Schwarz Inequality at math class :clap2:

In morocco, school maths is [b][u]TRIVIAL[/u] [/b]untill 1st year of high school(that's why we don't perform well on imo :D ), difficulties starts from the 2nd high school year espesially if you choose Mathematics Class, there are some notions, little bit tricky like Set Theory, Number Theory, combinatorics, and real difficulties start at "classes preparatoires" (after high school): Real, Complex analysis, linear algebra, analytical geometry almost everything and the programs are huge  :lol: .

I think this poll may be slightly biased... :P
EDIT: Lol I fail....
EDIT: Yes, I am a pro.

[quote="fairview13"]Very nice. How did you find this topic, anyway? xD[/quote]
 pi37 is a pro.

[quote="akkaren"][quote="now a ranger"]I think I like geometry slightly more than algebra because geometry is more Visual, whereas Algebra 2 requires immense calculation and knowledge of many definitions.
[/quote]

What type of Algebra II are you taking? I'd swap the descriptions you gave of algebra and geometry...

Actually, I'd say that algebra is more visual, but I really mean that it's more spatial. Geometry is clunky and dry, often on paper, which makes it harder to manipulate conceptually, whereas algebra and calculus get easier because you can feel where they're going...

Which I'm sure sounds like nonsense.

As for the original question, though, I really think it depends on where you go to school...

I've always heard that trig is the hardest math you can take before college, but I never took it...[/quote]

how can you say algebra is visual, there's not a lot to visualize, in geometry use, but you can't really visualize abstract math, what do you see when you visualize algebra????

You "see" what is called "algebraic geometry" (simplified: rings etc. as geometric objects) and "geometric algebra" (simplified: groups of symmetry and such) :P

Algebra can actually be very visual if you convert it into geometry :P

edit: 1300th post

[s]Good job! Don't post again today! Tomorrow, give an url on your blog showing which post what's your 1300th post! [/s]
In my school, if math is hard for one person, the whole class gets horribly easy math! AoPS saved me!  :omighty:  :omighty:  :omighty:

I agree that AOPS is very helpful, especially because teachers at school teach you a bunch of theorems and stuff but don't actually show you WHY they work... AOPS has great videos that explain everything.

I'm just really waiting for the Calculus, Vectors and Advanced Functions Courses (In Ontario, Canada, those are the math courses for grade 12).
At least those will be somewhat of a challenge...


Three darts thrown at this dart board land in three different regions. If the order
that the three darts are thrown does not matter, how many
combinations of three different regions from the 10 possible
regions would result in a sum of exactly 50 points?

[asy]import graph;
draw((0,0)--(0,20));
draw((0,0)--(19,6.2));
draw((0,0)--(-19,6.2));
draw((0,0)--(11.8,-16.2));
draw((0,0)--(-11.8,-16.2));
draw(Circle((0,0),10));
draw(Circle((0,0),20));
label("5",(-10,10),NW);
label("20",(10,10),NE);
label("15",(-10,-8),W);
label("10",(10,-8),E);
label("20",(0,-14),S);
label("40",(-3.3,4),NW);
label("35",(3.3,4),NE);
label("30",(-4,-3),W);
label("40",(4,-3),E);
label("45",(0,-6),S);[/asy]

You want to find the number of ways to add to 50. We have:

15+30+5

35+5+10

20+10+20

So we have $ \boxed{3}$ ways.


Six people sit around a circular table. In how many ways can three pairs of people shake hands so that no arms cross (nobody shakes hands with two people at the same time)?

Eum.. is this from 2003? I thought we got in in our training stage 2003 (around april)?

Maybe I'm misunderstanding this, but right now it looks pretty straightforward :?

Number the people $1,2,3,4,5,6$ in a clockwise direction. Clearly, $x$ can't shake hands with $k+2$, because then $k+1$ won't be able to shake kands with anybody without crossing his hand with the pair $k-k+2$. If $1$ shakes hands with $4$ then $2-3$ amd $5-6$ are the other two pairs. If $1-2$ then $2-3,4-5$ or $2-5,3-4$. If $1-2$ then $3-4,5-6$ or $3-6,4-5$. So there are five ways.

I'm sure there's something I'm getting wrong, because it can't possibly be that easy, so could you please clarify it?

The problem is wrong. It is supposed to be for 8,10 or 12 people; and if I remember well, it can be generalised for 2n. :)

There's a recurrence relation: $a_{2n}=a_{2n-2}+a_{2}a_{2n-4}+\ldots+a_{2n-4}a_2+a_{2n-2}$, I think. From the case $2n=6$ we easily deduce the values for $8,10,12$, but I'm not sure about the general formula.

[There was a mistake; I've edited it, and I hope it's Ok now]

Then $a_{2n} = \frac 1 {n+1} C_{2n}^n$, the $n^{th}$ Catalan number.

Pierre.

did you just look that up in sloane's or is there a nice way to deduce it? :)

Just use the recurrence relation given by Grobber.
This is a well-known relation which caracterizes the Catalan Numbers (if the first terms are well-chosen).

Pierre.


Polar-rectangular Equation conversion
Convert to rectangular form:

$ r^2\equal{}\sin(2\theta)$
$ r\equal{}\frac{5}{1\minus{}2\sin(\theta)}$

W8 what is rectangular form and if i new what it was i would think that it should go into the highschool intermediate section.

By rectangular he means as a point in the Cartesian plane.

Any solutions? Should I post this in high school intermediate topics?

[hide="For Starters"]
(a) $ r^2 \equal{} \sin{(2\theta)} \equal{} 2\sin{\theta}\cos{\theta}.$

Using $ r \equal{} \sqrt {x^2 \plus{} y^2}$, $ y \equal{} r\sin{\theta}$, $ x \equal{} r\cos{\theta}$, we obtain

$ x^2 \plus{} y^2 \equal{} 2\frac {y}{\sqrt {x^2 \plus{} y^2}}\cdot \frac {x}{\sqrt {x^2 \plus{} y^2}} \\
(x^2 \plus{} y^2)^2 \equal{} 2xy.$

I don't know how much you want it simplified so I'm leaving it there.

(b) $ r \equal{} \frac {5}{1 \minus{} 2\sin{\theta}}.$

Using identities similar to before, we have

$ \sqrt {x^2 \plus{} y^2}(\frac {\sqrt {x^2 \plus{} y^2} \minus{} 2y}{\sqrt {x^2 \plus{} y^2}}) \equal{} 5 \\
x^2 \plus{} y^2 \equal{} (2y \plus{} 5)^2.$

Again, I'll let you do the simplifying.
[/hide]

EDITED: thanks

Uh doesn't $ x\equal{}r\cos(\theta)$ and $ y\equal{}r\sin(\theta)$?


Can anyone give me any advice on how to acquire any sort of feedback on the problems in this book? I know that he says in the book that it is necessary to have "backburner" problems, but this is not as time practical as working diligently for as long as possible then receiving feedback on what was missing from the work. Has anyone found a solutions manual to this book, despite Wiley's decision to give it only to instructors at colleges and schools? If so, would you please consider pointing me in the right direction? I'd hate to go about reading this magnificent book in a vacuum.

I agree that this is the main fault of the book - a lack of a comprehensive solutions manual.  Even the one that has been produced only has solutions to roughly half of the problems.  I strongly suggest you use this forum as your de facto solutions manual - post the problems you are stuck on here (usually in pre-Olympiad or in one of the Olympiad sections), and you'll likely get a solution in fairly short order, if not 2 or 3.


[quote="rd5493"][quote="7h3.D3m0n.117"][quote="acchessor"]Hello.  When I was practicing for state (it's this Saturday, anyone here from Nebraska?), I kept running out of time on the last 5 or so problems on the Sprint round, and I practically missed the last few problems.  I think that I probably could solve these with a few minutes more, but how can I reduce the chance that I would run out of time on the test?[/quote]

Well spend less time double checking. I have the same problem (mainly due to me skipping questions because it would take too long to read them  :D ). So usually I do the short problems first. The last 5 problems are bound to be harder but just make sure you keep aware of your time and "quit" a problem if it is taking too long (then come back to it later).[/quote]

You too.  i thought it was just me.   :D that makes me feel so much better knowing that im not the only lazy procrastinator on this planet. :rotfl:[/quote]

Yeah by the time I get to the "question" part of the problem my mind zones out and I skip a couple of sentences (at one time I skipped "Express as a common fraction"...)

[quote="jka386"]YEah, if it takes me more than 15 seconds to read the problem, I usually skip it... idk if that's good or bad.[/quote]

whoa 15 seconds? thats a long time... :D  usually I would skip a problem if it included a diagram becaue that would be really annoying...

yeah... I didn't really get a close approximation... :maybe: about 6 sec.

approximations?  whos got time for those.  this work concept iststill relatively new to me

:wow:  :wow: :wow: :wow:
AHHHHHHHHHHHHHHH!!!!! THEYRE ATTACKING!!!  Please send help or reinforcements.  I dont know how much longer I can hold out.  (ive been told that this happens when you dont pay attention)

:ninja:  :ninja:  :ninja:     here are the special forces.

[quote="7h3.D3m0n.117"][/quote][quote="rd5493"][quote="7h3.D3m0n.117"][quote="acchessor"]Hello.  When I was practicing for state (it's this Saturday, anyone here from Nebraska?), I kept running out of time on the last 5 or so problems on the Sprint round, and I practically missed the last few problems.  I think that I probably could solve these with a few minutes more, but how can I reduce the chance that I would run out of time on the test?[/quote]

Well spend less time double checking. I have the same problem (mainly due to me skipping questions because it would take too long to read them  :D ). So usually I do the short problems first. The last 5 problems are bound to be harder but just make sure you keep aware of your time and "quit" a problem if it is taking too long (then come back to it later).[/quote]

It's usually safe to assume that it's going to be a common fraction. They rarely have you express answers otherwise.

You too.  i thought it was just me.   :D that makes me feel so much better knowing that im not the only lazy procrastinator on this planet. :rotfl:[/quote][quote="7h3.D3m0n.117"]

Yeah by the time I get to the "question" part of the problem my mind zones out and I skip a couple of sentences (at one time I skipped "Express as a common fraction"...)[/quote]


For $ a,b,c$ positive reals

$ \frac{a^{k}(a+b)}{a+c}+\frac{b^{k}(b+c)}{b+a}+\frac{c^{k}(c+a)}{c+b}\ge a^{k}+b^{k}+c^{k}$

for $ k=0,1,2$ (does not hold for $ k=3$)

Hello...here was something about it
[url]http://www.mathlinks.ro/viewtopic.php?p=895644#895644[/url]

[quote="manlio"]Yes, I know, in fact I said it was of Ductrung, but I don't remember if it was created by Ductrung or it is an old inequality of the forum.   :)[/quote]
To Manlio: for $ k=1$ use rearrangement or Cauchy-Schwarz.


Let $ a,b,c$ positive reals such that $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$. Prove that

$ \frac{b(a\plus{}c^2)\plus{}c^3}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{c(b\plus{}a^2)\plus{}a^3}{(b\plus{}a)(c\plus{}a)}\plus{}\frac{a(c\plus{}b^2)\plus{}b^3}{(c\plus{}b)(a\plus{}b)}\geq\frac{5}{4}$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$. Prove that

$ \frac {b(a \plus{} c^2) \plus{} c^3}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {c(b \plus{} a^2) \plus{} a^3}{(b \plus{} a)(c \plus{} a)} \plus{} \frac {a(c \plus{} b^2) \plus{} b^3}{(c \plus{} b)(a \plus{} b)}\geq\frac {5}{4}$[/quote]
I will prove it with the stronger condition $ a\plus{}b\plus{}c\equal{}1$. (The original inequality follows imediately from this proof since $ LHS$ is mixed of degree $ 0$ and degree $ 1$ and $ a\plus{}b\plus{}c \ge \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$)
Expanding:
$ \sum_{cyc} 4(b(a\plus{}c^2)\plus{}c^3)(a\plus{}b) \ge 5(a\plus{}b)(b\plus{}c)(c\plus{}a) \iff$
$ \sum_{cyc} 4(ab(a\plus{}b\plus{}c)\plus{}bc^2\plus{}c^3)(a\plus{}b) \ge 5(a\plus{}b\plus{}c)(a\plus{}b)(b\plus{}c)(c\plus{}a) \iff$
$ \sum_{cyc} 8a^3b\plus{}8ab^3\plus{}12a^2b^2\plus{}12a^2bc \ge \sum_{cyc} 5a^3b\plus{}5ab^3\plus{}10a^2b^2\plus{}20ab^2c \iff$
$ \sum_{cyc} 3a^3b\plus{}3ab^3\plus{}2a^2b^2 \ge \sum_{cyc} 8ab^2c$
This is true by muirhead.

It is easy to prove that $ \sum\frac{ab}{(a\plus{}c)(b\plus{}c)}\ge\frac34$ (clear the denominator and use AM-GM). It is also easy to prove that $ \sum\frac{c^2}{a\plus{}c}\ge1$ (use Cauchy Schwarz). Add the two inequalities, and we are done.


Given three positive real numbers $x,y,z$ such that $xy+yz+zx\leq8$, find the minimum value of
\[ f=\frac1x+\frac1y+\frac4z. \]

Nice one.
Rewrite
$E=\frac 1 {xyz}+\frac 3 z$
then we can solve it by $x=y$

We have the general problem of it...

See an other nice and general here  :

http://www.mathlinks.ro/Forum/viewtopic.php?t=59376

The question of mine has posted but noone solved!


number theory
least common multiple
floor function
number theory proposed

Given a positive integer $ n>3$, prove that the least common multiple of the products $ x_1x_2\cdots x_k$ ($ k\geq 1$) whose factors $ x_i$ are positive integers with $ x_1\plus{}x_2\plus{}\cdots\plus{}x_k\le n$, is less than $ n!$.

Obviously
\[ \textrm{lcm }\equal{} \prod_p p^{k_p}, \ k_p \equal{} \left \lfloor \frac np \right \rfloor \]
and $ n! \equal{} \prod_p p^{m_p}, \ m_p \equal{} \left \lfloor \frac np \right \rfloor \plus{} \left \lfloor \frac {n}{p^2}\right \rfloor \plus{} \cdots \equal{} \sum_k \left \lfloor \frac {n}{p^k}\right \rfloor \ge k_p$. Therefore $ \textrm{lcm } < n!$ for $ n\ge 4 \equal{} 2^2$.


1/Find all functions $f: [0,+\infty) \to [0,+\infty)$ such that:
$f(x^{2}+y^{2}+xy)+f(\frac{x^{2}+y^{2}}{2}-xy)=f(\frac{3x^{2}+3y^{2}}{2}) \forall x,y \geq 0$
2/Find all functions $f,g,h: R \to R$ such that: $f(x+y^{3})+g(x^{3}+y)=h(xy) \forall x,y \in R$

:blush:  :blush:
I made a silly thing there

[quote="aviateurpilot"]1)$y=-x$ => $f(3x^{2})=f(\frac{3}{2}x^{2})$
[/quote]
Read the problem carefully :lol:


ratio
function
logarithms
number theory open
number theory

some numerical results suggest the following:

for every fixed $ c>0$, for sufficiently large $ n$ will be $ \frac{p_{n\plus{}1}}{p_n}<1\plus{}c$

where $ p_n$ is n-th prome

Could you prove or disprove my conjecture ?

The distribution of primes is a goal of the Riemann-Zeta function and the Prime Number Theorem.  See http://en.wikipedia.org/wiki/Prime_number_theorem

you suggest something like this ?

$ p_n\sim n\ln n$ and $ p_{n \plus{} 1}\sim (n \plus{} 1)\ln(n \plus{} 1)$ so
$ \frac {p_{n \plus{} 1}}{p_n}\sim \frac {(n \plus{} 1)\ln(n \plus{} 1)}{n\ln n}\to 1$
and so for every fixed $ c > 0$ it will be $ < 1 \plus{} c$ if $ n$ suff. large...

is it a correct proof ?
I'm not sure if I could divide those two asymptotic formulas......

Erdös proved precisely this using Selberg's fundamental formula after Selberg gave a lecture on it.
This eventually led to an elementary proof of the prime number theorem, as well as a bit of a dispute.


geometry
rectangle
combinatorics proposed
combinatorics
combinatorics unsolved

We have an 8x8 chessboard with 64 squares. Then we have 3x1 dominoes which cover exactly 3 squares. Such dominoes can only be moved parallel to the borders of the chessboard and also only if the passing squares are free. If no dominoes can be moved, then the position is called stable.
a. Find the smalles number of covered squares neccessary for a stable position.
b. Prove: There exist a stable position with only one square uncovered.
c. Find all Squares which are uncoverd in at least one position of b).

a) I think the smallest number is $42$ , isn't it ?

I would guess that the answer is $21$, achieved by placing 7 trominos along a diagonal.

I was thinking an argument along the following lines would work:  You start with a single tromino, which fits in a $2 \times 2$ square.  You need to place a tromino to block its movement; there must be one which increases the sides of the minimal rectangle containing the trominos by at most 2, or else two trominos that increase the sides by at most 4.   By the end, the minimal rectangle must by the entire $8 \times 8$ board, or else a tromino nearest an open edge can be moved.  This is a total increase of 12, so we need 7 trominos.

I think Deedlit is using L-shaped trominoes. The problem states they're 3x1, so I think they are I-shaped trominoes.

Bleah, I'm not paying attention...

With I trominos, I can't do better than 16.

Hi, as I said we need at least $14$ trominoes since I could do it  :)

c)If we paint the board in three colours and rotate it 4 times with the same colouring, we easily get that uncovered squares can be squares with the coordinates (3,3), (3,5), (5,3), (5,5) if the downest square has the coordinates (1,1).

b) Easy. If we don't cover square (3,3) we can cover the uppest subtable 8x3, under it the table 5x3 containing the rightest column, an the rest can be tiled by 4 rectangles 2x3 (3x2).


Hey  i need someone to help me... here's my situation:
i took the sat on october 9 and I am applying early to a university. my guidence counselor is going to send the app before the oct. sat scores arrive, and I didn't send the scores when registering for the SAT!
can i make something in collegeboard.com to send them? i can't see anything..
I can send my previous Sat scores.. so i think i'll have to wait until OCT 24 when the scores are in to send them ? (the deadline of the early app is nov 1, will the scores be there before the deadline? do they send the scores by email or by mail?

thanks!

hey manual. :)

well usually as far as i know, the application deadline does not apply to the sat scores.
you can send in your application first then send in your sat scores (the sat scores must be mailed streight from Collegeboard).

you can even take the sat again and resend your score, if it was higher.

~amir

Post #2 is correct. The scores arrive separately from the College Board.


Hello,
I was wondering what the best books to self-study from for the respective olympiads (USAMO, USAPHO, USNCO, USABO) [I know... I know... there is no way I'm going to study for all of these, but just for reference]. For the USAPHO, I'd like to know what book would be the best to start with also.

USAMO-
Art and Craft of Problem Solving (Zeitz), Problem Solving Strategies (Engel), AoPS Vol. 2?

USABO-
Biology (Campbell)?

USNCO-
?

USAPHO-
?

See the subpages at

http://olympiads.win.tue.nl/index.html

for links specific to each Olympiad with reading suggestions.

Would you guys say that Physics B or Physics C would help you more on the Physics Olympiad? Thanks.


Points $ A_1,A_2, ... ,A_n$ inside a circle and points $ B_1,B_2,...,B_n$ on its boundary are positioned so that the segments $ A_1B_1,A_2B_2, ... ,A_nB_n$ do not intersect. A bug can go from point $ A_i$ to $ A_j$ if the segment $ A_iA_j$ does not intersect any segment $ A_kB_k$, $ k \neq i, j$. Prove that the bug can go from any point $ A_p$ to any point $ A_q$ in a finite number of steps.

A point may [i]see[/i] another point if the segment between them is not intersected by an existing chord (seeing is mutual).

Suppose a point $ X$ (connected by chord to a point $ W$) may see a point $ Y$; then we claim that $ X$ also sees the other endpoint $ Z$ of the chord containing $ Y$. The region defined by $ \overline{WX}, \overline{YZ}$ and the non-overlapping arcs that connect them is convex. Thus if a chord intersects the line $ \overline{XZ}$ it’s endpoints must be on both arcs, but then $ X$ may not see $ Y$, contradiction.

So it is not important if the bug jumps to an $ A$ or a $ B$ (if it jumps to a $ B_k$, instead change it so that it jumps to the corresponding $ A_k$). Now any endpoint $ A_i$ also sees at least one other endpoint of another chord (consider the point closest to $ A_i$, traveling along the circle).

Suppose from $ A_iB_i$ we may see $ A_jB_j$, and now we delete $ A_jB_j$. Suppose further that [only after this deletion] $ A_iB_i$ may see $ A_kB_k$. Then we claim that $ A_jB_j$ saw $ A_kB_k$ before it was deleted, which follows similarly from the convexity argument above.

Then we just continuously delete chords (which corresponds to the bug jumps), and since there is a finite number of chords eventually any chord may be seen from any starting point.


a_1=1. a_2=1. a_n=sqrt(a_(n-1)^2+a_(n-2)^2)

What is a_100000000000 (hopefully this is 1 trillion)?

Edit: Wow this was a lot easier then I thought. Well, not that much easier.

Are you sure your recursion is stated correctly?  You have two $a_{n-1}$ terms.  In any case, substitute $b_{n}=a_{n}^{2}$.

[hide="solution"]The sequence is simply the fibbonacci sequence with each term square rooted since the recursive function involves squaring, then adding, then square rooting once again.

$\sqrt{1},\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{8}...$

Plug in n = 1 trillion into the closed definition of the fibbonacci numbers and then take the square root.

$\sqrt{{\frac{1}{\sqrt{5}}[{(\frac{1+\sqrt{5}}{2})^{n}}-{(\frac{1-\sqrt{5}}{2})^{n}}]}}$

with n = 1 trillion

I'm new to Latex
How do you make the parentheses taller?
[/hide]

[quote="chickendude"]I'm new to Latex
How do you make the parentheses taller?
[/quote]

[code] \left(\frac{1}{2}\right) [/code]

will produce $\left(\frac{1}{2}\right)$.


abstract algebra
limit
group theory
superior algebra
superior algebra unsolved

Is $M=\mathbb{Z}^{\mathbb{N}}$ a free $\mathbb{Z}$ module?
----------------------------------------------------------------------------------
To explain a bit more:

$\mathbb{Z}^{\mathbb{N}}$ is $\mathbb{Z} \times \mathbb{Z} \times \ldots \times \mathbb{Z}$  a countable number of times.

The elements of M are $(x_n)_{n \in \mathbb{N}}$, with $x_n \in \mathbb{Z}$.
M has a group structure:
$(x_n)_{n\in\mathbb{N}}+(y_n)_{n\in\mathbb{N}}=(x_n+y_n)_{n\in\mathbb{N}}$.
$(x_n=0)_{n\in\mathbb{N}}$ is the neutral element of M.

We define $n\cdot x=x+\ldots +x$ (n times) for n in N and (-1)x=-x for all x in M.
Thus we can define zx for all $z \in \mathbb{Z}$ and $x \in M$. This means that M is a $\mathbb{Z}$ module

M is a free $\mathbb{Z}$ module, if it has a basis. That is if there exists B a subset of M such as:
1. for all x in M there exists a natural n, integers $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$ in B such as $x=\sum_{k=1}^n a_k\cdot b_k$.
2. for all n, integers $a_1,\ldots,a_n$ and distinct elements of B, $b_1,\ldots,b_n$, the condition $\sum_{k=1}^n a_k\cdot b_k=0$, implies $a_k=0$ for all $k=\overline{1,n}$.

The following proof is due to "Kaplanski - Infinite Abelian Groups".
M is not free.

Definition: Let R be a ring and I be a set.
$R^I=\{(x_i)_{i\in I}| x_i\in R\ \forall i\in I\}$ is called the direct product (or complete direct sum) of R idexed by I.
$R^{(I)}=\{(x_i)_{i\in I}| x_i\in R\ \forall i\in I$ and only finitely many $x_i$'s aren't 0} is called the direct sum of R indexed by I.

Suppose M is free. Let B be a basis for M.
Suppose B is countable. Then it is easy to see that the $\mathbb{Z}$ module M generated by B is countable. But $|M|=|\mathbb{Z}|^{|\mathbb{N}|}=\aleph_{0}^{\aleph_0}=\aleph_c$. So M is uncountable. We get a contradiction.
So B is uncountable.

Let p be a prime number in $\mathbb{N}$.
Let $v:\mathbb{Z}\to\mathbb{N}\cup \{\infty\}$ defined by $p^v(n)|n$ and $p^{v(n)+1}\nmid n$ for all $n\in \mathbb{Z}^*$. Also, $v(0)=\infty$.

Let $\displaystyle S=\{(x_n)_{n\in\mathbb{N}}\in M| \lim_{n\to\infty}v(x_n)=\infty \}$.

Let $f:M\to M$ defined by $f((x_n)_{n\in \mathbb{N}})=(p^n\cdot x_n)_{n\in \mathbb{N}}$. f is an injective morphism of M into S. The image of M by f must be an uncountable generated (cannot be countable generated) free submodule of S, so S cannot be countable generated. From this we should infer that S/pS cannot be countable generated as an $\mathbb{Z}/p\mathbb{Z}$ module.

On the other hand, we prove that $S/pS\simeq \mathbb{Z}^{(\mathbb{N})}$, which is free and countable generated.
By the definition of S, for all $(x_n)_{n\in\mathbb{N}}\in S$ there exists $N_x$ such that for $n>N_x,\ p|x_n$. So for $n>N_x$, $\hat{x}_n=\hat{0}$. So S/pS contains only finite combinations of $\hat{e}_k$'s ($\hat{e}_k$ is the element of M/pM that has $\hat{1}$ on the $k^{th}$ place and $\hat{0}$ in the rest). Obviously $e_k \in S\ \forall k\in\mathbb{N}$, so $\hat{e}_k\in S/pS\ \forall k\in\mathbb{N}$.
This proves that S/pS is isomorphic to the required direct sum which happens to be countable generated.

The problem is over if we prove the implication: S not countable generated as a $\mathbb{Z}$ module implies S/pS not countable generated as a $\mathbb{Z}/p\mathbb{Z}$ module. In Kaplanski's book, this is given as "manifestly" type implication.

[quote="amfulger"]...From this we should infer that S/pS cannot be countable generated as an $ \mathbb{Z}/p\mathbb{Z}$ module.
[/quote]

why?

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=189984]here[/url] and the various links.
It's only the second topic from the top- why didn't you see it?

I actually reached this thread from that link, and I started to go over the proof and got stuck at that place... I'm sorry if I'm missing something obvious.

Oops- I didn't check the date on the original post.

Basically, if you have an uncountably generated free module, the generators are distinct mod $ p$, and an uncountable abelian group isn't countably generated.


I am kind of learning Algebra 2 over the summer, and I came across a lesson about continuous compounding and [i]e[/i]. I was wondering where else e would be used and if there is a proof or something like that for $ A\equal{}Pe^{rt}$[/i]

You've probably gone through compounding a finite number of times periodically. The idea is that as you compound it more and more rapidly (ie, the limit as the period goes to 0), that approaches your continuously compounded interest formula.

Check out the definition of [url=http://en.wikipedia.org/wiki/Exponential_function]the exponential function[/url].

In fact, 'e' is a very special number, largely because it is the base of the exponential function and the natural logarithm, both of which are very special functions. You'll see in that link above that there are many varied definitions, properties, and identities satisfied by the exponential function. It then makes sense that the number is is ubiquitous in mathematics.


Last weekend I was watching leaves fall from a tree. As the wind picked up and the leaves fell faster I noticed an interesting pattern.

The first minute only one leaf fell. During each passing minute, the number of leaves that fell was equal to the total on the ground plus one.

1.If this pattern continues how many leaves fell during the 12th minute?The 24th minute?

2. Write an expression to represent how many leaves would fall during the nth minute.

This looks like it is out of your textbook....remember don't post homework here unless you maybe try and attempt the problem and ask why it's wrong, etc.

BTW, the pattern is 1, 3, 7, 15, 31...

I don't know if you would call this a formula but $ a_n \equal{} a_{n\minus{}1} \plus{} (a_{n\minus{}1} \plus{} 1 )$

Really? I thought the patter would be 1, 2, 4, 8, 16. Its because the leaf that fell before plus one. I dont get why is 1, 3, 7, etc.

argh..you are right.  i gave the total number of leaves on the ground instead of how many fall each time.

You serious? I was right? Im stupid at math and I knew it. (You did to but still)

 So i was wondering is the 12th- 2048 and the 24th- 8388608?


search
number theory
prime numbers
number theory unsolved

Let $ a,a \plus{} k,a \plus{} 2k$ be prime numbers $ >3$ .Prove that $ k \vdots 6 ,(k \in N)$

You forget, that $ a>3$. $ a\equal{}3,k\equal{}2$ give $ 3,5,7$ are primes.

[quote="sukivel"]Let $ a,a \plus{} k,a \plus{} 2k$ be prime numbers .Prove that $ k \vdots 6 ,(k \in N)$[/quote]
Is this problem quite trivial,or maybe  i made some mistake :maybe:
If $ k$ is odd then,$ a \plus{} k$ is even contradiction.Hence $ 2|k$.
As Rust said,$ a > 3$,so it is obvious that $ 3$ doesn't divide any of the $ a,a \plus{} k,a \plus{} 2k$.There is at least $ 2$ numbers among them,which difference is diviible by $ 3$.But differences are $ 2k,k$.Thus we are done.

Yes , The prolem is repaired that:
[quote="sukivel"]Let $ a,a \plus{} k,a \plus{} 2k$ be prime numbers $ > 3$ .Prove that $ k \vdots 6 ,(k \in N)$[/quote]

General problem: If $ a,d$ are positive integers such that $ a,a \plus{} d,a \plus{} 2d, \dots,a \plus{} (n \minus{} 1)d$ are prime numbers and $ p$ is the product of all primes $ \le n$, then $ p\mid d$.

The proof is quite easy. You can see my posts here: http://www.mathlinks.ro/viewtopic.php?search_id=1212712511&t=165462

Yes Rust I can see that it's true if $ a>n$ but what I [i]can't[/i] see is "[i]why[/i]?". :maybe:

I thought about this for long but without any luck. :( Can someone help? Sorry if this is a stupid question.

It is obviosly, if $ p\le n$ and $ p\not |d$, then for (at least ) one k $ p|a_k\equal{}a\plus{}(k\minus{}1)d$.


This problem I had posted in this forum before and the propose problem for VMEO.

Oh, I don't know that. I created it yesterday afternoon. Also in this afternoon, I created several others of similar kind. I like this one the most.

Could you show me the link, Hung?

See here

http://www.mathlinks.ro/Forum/viewtopic.php?t=76441

i can find almost no difference between them

[quote="pvthuan"]Thank you for the insights and inspirations you give us in the art of inequalities.[/quote]I also thank you for your appreciation.

Subtrack 3 in your inequality it's my inequality.

hi,
here is my solution :

$A\geq \frac{(a+b)(a+c)}{a^2+ab}+\frac{(b+c)(b+a)}{b^2+ab}+\frac{(c+a)(c+b)}{ca+ab}$
$A\geq\frac{a+c}{a}+\frac{b+c}{b}+\frac{c+a}{a}$

$A\geq 5$
 :D .

[quote="pvthuan"]Another one is
\[ \frac x{(x+y)(x+z)}+\frac y{(y+z)(y+x)}+\frac z{(z+x)(z+y)}\geq\frac2{x+y+z}.  \][/quote]

I don't know, maybe I misunderstanding something, but your inequality is equivalent with

                     \[ x(y+z)+y(z+x)+z(x+y)\geq\frac{2(x+y)(y+z)(z+x)}{x+y+z}\Leftrightarrow \]
                     \[ 2(x+y+z)(xy+yz+zx)\geq 2(x+y)(y+z)(z+x)\Leftrightarrow \]
                     \[ (x+y+z)(xy+yz+zx)\geq (x+y)(y+z)(z+x)\Leftrightarrow \]
                     \[ (x+y+z)(xy+yz+zx)\geq (x+y+z)(xy+yz+zx)-xyz\Leftrightarrow \]
                     \[ xyz\geq 0. \] :)

[quote="pvthuan"]The coming May 17 is the second anniversary of Vasc's joining Mathlinks. The following inequality is dedicated to him. Thank you for the insights and inspirations you give us in the art of inequalities.

Problem 85. Let $a,b,c$ be real nonnegative numbers, prove that
\[ \frac{(a+b)(a+c)}{a^2+bc}+\frac{(b+c)(b+a)}{b^2+ca}+\frac{(c+a)(c+b)}{c^2+ab}\geq5.  \][/quote]

Here is the summarised proof in my book.
Suppose that $a\geq b\geq c$. The desired inequality is equivalent to
\[ \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq2,  \]
which is written as
\[ \frac{(a-b)(a-c)}{a^2+bc}+\frac{(b-c)(a-c)}{c^2+ab}\leq\frac{b(a+c)}{b^2+ac},  \]
the left hand side of which is \[ (a-c)\left\{\frac{a-b}{a^2+bc}+\frac{b-c}{c^2+ab}\right\}\leq (a-c)\left\{\frac{a-b}{a^2}+\frac{b-c}{ab}\right\}\leq a\frac{2ab-b^2-ac}{a^2b}.  \]
It is sufficient to prove that
\[ (2ab-b^2-ac)(b^2+ac)\leq ab^2(a+c).  \]
After some manipulations, this comes downs to
\[ (a-b)^2b^2+a^2c^2\geq 2(a-b)bac.  \]

 :lol:

What is your solution, Hung?

Today is exactly the day you came here in this forum. Any other solution for this, Vasc?

Indeed pvthuan, I got over this two year resistance test. :(
Very nice your solution to this problem.
I have not an own nice solution. Only an idea: to use the Minh's idea and to show that
$\sum\frac{2a^2(b+c)^2}{(a+b)^2(a+c)^2} \geq 1$.
I don't know whether this inequality holds.  ;)

there is a proof for $\frac{1}{2}\le k \le 1$

Hi here is my solution I wish i did not do any mistake
Lets open upside parantesens then it becomes as
 $\frac{ac}{a^2+bc}+\frac{ab}{a^2+bc}+\frac{ab}{b^2+ca}+\frac{bc}{b^2+ca}+\frac{ac}{ab+c^2}+\frac{cb}{ab+c^2}\geq2$ then send numerators down the denominators then it becomes as
$\frac{1}{\frac{a}{c}+\frac{b}{a}}+\frac{1}{\frac{b}{a}+\frac{c}{b}}+\frac{1}{\frac{b}{c}+\frac{a}{b}}+\frac{1}{\frac{b}{c}+\frac{c}{a}}+\frac{1}{\frac{a}{c}+\frac{c}{b}}\geq2$ now take $\frac{a}{b}=x \frac{b}{c}=y\frac{c}{a}=z$ with $xyz=1$ then it becomes
 $\frac{xy+1}{x+y}+\frac{yz+1}{y+z}+\frac{zx+1}{z+x}\geq2$
after multiplying denominators with numerators it becoms
$(xy)^2+(yz)^2+(zx)^2+3(x+y+z)+3(xy+yz+zx)+(x^2+y^2+z^2)\geq2(\sum_{sym}x^2y +2)$ as we know $xyz=1$ so using it we will get:
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} +2(x+y+z) +x^2+y^2+z^2+2( xy+yz+zx)+ (x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq\frac{y}{x}+\frac{z}{x}+1+\frac{x}{y}+\frac{z}{y}+1+\frac{x}{z}+\frac{y}{z}+1-1$ then
$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2+(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+x+y+z)\geq2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(x+y+z)-2$ so it is obvious

That means we have no equality.  :)

Since for non-negative [b]x,y,z[/b]  we have: $x+y+z+\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 6$.

Davron

[quote="zhaobin"]What a nice inequality :)
but can we prove:
$\left(\frac{a(b+c)}{a^2+bc}\right)^k+\left(\frac{b(c+a)}{b^2+ca}\right)^k+\left(\frac{c(a+b)}{c^2+ab}\right)^k \geq2$
where $0< k \le 1$?[/quote] Yes. It suffices to prove the statement:
Let $x,y,z$ be non-negative numbers such that $x+y+z \geq 2$ and $xy+yz+zx \geq 1$. Then
$x^k+y^k+z^k \geq 2$ for all $0<k\leq1$.

[quote="ductrung"][quote="Vasc"]$\sum\frac{2a^2(b+c)^2}{(a+b)^2(a+c)^2} \geq 1$.
I don't know whether this inequality holds.  ;)[/quote]

It is since we have
$2\sum_{cycl}a^2(b+c)^4 \ge 32a^2b^2c^2+(a+b)^2(b+c)^2(c+a)^2$[/quote]

Yes, last ineq is very easy to prove by SOS  ;)


What is the smaller natural number that has 5 factors?

[hide="Answer"]I'm assuming five distinct factors. You can find distinct factors by prime factorizing a number; say $n=a^xb^yc^z$. The number of distinct factors of $n$ is $(x+1)(y+1)(z+1)$, because this is the number of different values for each of the prime factors (i.e., $a^0$ through $a^x$ is $(x+1)$ possible values of $a^m$). So, to get back to the problem, the smallest natural number with five distinct factors must have only one prime factor, as 5 is a prime number, so the number will be in the form $n^4$, and we want the smallest $n>1$, so our answer is $2^4=\boxed{16}$.[/hide]

[hide] using the counting principle  $x=a^b \cdot c^d $,   $x$ has $(b+1)(d+1)$ factors.  We also need to note that the smallest prime is 2, and that 5 is prime so were dealing with a number to the 4th power.  (  The fact that it has an odd number of factors, tell us it is a square number, but thats not important)

  Thus, we see our number is $2^4=\boxed{16}$.[/hide]

We know that the number $N$ has factor of $1$ and $N$ also, so we need three more factors. Well the integer $2$ is the smallest number which can be exponentiated to give us that. $2^1, 2^2, 2^3, 2^4$ will be $4$ more factors including $N$ itself which is $2^4=16$. To check this, $16=2^4$ and $4+1=5$ so it does indeed have five factors.


Prove or disprove that $ a^n\minus{}a^{n\minus{}1}\equal{}(a\minus{}1)^{n\minus{}1}$ for all naturals, integers, rationals, reals, and/or complexes.

$ a^{n}\minus{}a^{n\minus{}1}\equal{}a^{n\minus{}1}*(a\minus{}1)$

so you wat us to prove that $ (a\minus{}1)^{n\minus{}2}\equal{}a^{n\minus{}1}$ which is obviusly wrong (for integers for example $ 9^2>8$)


Hi guys ;)
I wonder,is it already posible to know the names of the team's guides?
I am very interesting to know hwo will be the guide of the Armenian team.
May someone from Mathlinkers have that information,please let us know.
Thank you! :)

It is not known yet. We are having a meeting soon and maybe I will know more then.

(Probably I'll be a guide of the Belgium team.)

[quote](Probably I'll be a guide of the Belgium team.)[/quote]

Nice! See you in Slovenia then, majaa!  :D  (I'm in Belgium's team)

Some news:

Turkey - Špela Hrvatin
Liechtenstein and Luxembourg - Janez Lev Kočevar
Netherlands - Brina Zupančič
Uzbekistan - Minca Šegula
Republic of Moldova - Matevž Kladnik
Latvia - Marjana Vuk
Ukraine - Andreja Hvalc
Pakistan - Maša Šabič
Mexico - Lea Cibrič
Belarus - Teja Šegula
Slovakia - Maja Alif ( that's me  :) )

[url=http://imo2006.dmfa.si/sub_Guides.html]Photos[/url]

Are the country guides going to appear in the internet site, or I shall wait till IMO begins. :(
Can you just name the guide of Iran? Thanks.

Sorry, but I know only the guides of these countries.  :(

does anybody knows the guide of Bosnia and Herzegowina team

A little mistake, majaa :) .
Turkey - Špela Peras.

Please can someone write the name of our guide ??

Sincerely
Davron

Davron, sorry, but we don´t know. We (guides) are in a few groupes and me and majaa are in the same. We know the states just for our group ...
Bye :)

[quote="master_yoda"]e ko sve govori srpski ili hrvatski od guidova????????? neko od njih ce valjda i nama da bude guid.[/quote]

Please English

[quote="master_yoda"]e ko sve govori srpski ili hrvatski od guidova????????? neko od njih ce valjda i nama da bude guid.[/quote]
   you are from Croatian or Serbian team?I think that they will give us(ex-Yu countries) guides regurarly without attention to who speaks our langauges and who not.
 dodatak:i ja sam u BiH timu pa se vidimo tamo  :lol:

[quote="master_yoda"]e ko sve govori srpski ili hrvatski od guidova????????? neko od njih ce valjda i nama da bude guid.[/quote]

When is the Serbian team arriving in Slovenia? The B&H team should be there on Monday (early in the morning).

P. S. Pozdravi srpskom timu od bh tima ;)

[quote="master_yoda"]serbian team will arrive on monday as well.

p.s. ja sam prilicno siguran da ce nam dati nekoga ko govori srpsko-hrvatski jer takvih sigurno ima.[/quote]
     ps.većina slovenaca i razumije ostale jezike iz ex-yu(bosanski,srpski,hrvatski) ali nisam siguran da li ga mogu koristiti kao maternji  :lol:

[quote="vedran6"]
p.s. ja sam prilicno siguran da ce nam dati nekoga ko govori srpsko-hrvatski jer takvih sigurno ima.[/quote]
     ps.većina slovenaca i razumije ostale jezike iz ex-yu(bosanski,srpski,hrvatski) ali nisam siguran da li ga mogu koristiti kao maternji  :lol:[/quote]
we understand srpsko-hrvatsko but I don´t think that u will get the guide who speaks your language. You will see :wink: .

Guys how I saw nobody knows the name of the Armenian team guide :maybe: ,but anyway tomorow I will find out it :lol: .
We will start to Slovenia tonight.
Good luck!!!!!
See you there!!!!

For those who are really impatient (since I guess most of you will meet their guide soon enough), the picture "Checking all the lists" on the official site http://imo2006.dmfa.si/images.html seems to show on the left hand side a list of guides and country codes. However, it is near impossible to read (but probably clear enough if you really want to).

I'm wondering what the list on the right hand side is though. It seems like a distribution in busses, and a set of times associated with each bus, but I don't know whereto (it can't be the examination room because no bus should leave there after 19.00).

[quote="Tasar"]..., the picture "Checking all the lists" on the official site http://imo2006.dmfa.si/images.html ...[/quote]

Maybe I'm blind, but I haven't found that picture... :ninja:

Added: false alarm, now I see these pictures, unfortunately Estonia is on the top of the list, so it isn't seen on the photo  :mad:

So Majaa, are you the team leader for our country as you expected (Belgium)? :lol:

What do you think?  :P  No, I'm not. I have the team from Slovakia.


How many 5-digit numbers are there with strictly growing digits from left to right?
How many 5-digit numbers are there with strictly decreasing digits from left to right?

I don't know if this is too easy for this forum, but it seemed about right. I can't figure it out :(

For the first question, Kalle, suppose you have any 5 distinct digits from the set {1, 2, ..., 9} - then there is exactly one way to arrange them such that they are increasing from the left to the right, do you see that? Therefore, the number you are looking for is equal to the number of subsets with 5 distinct elements of the set {1, 2, ..., 9}, and this number is ${9 \choose 5} = 126$.

Now, how does the answer to the previous question change when we consider decreasing digits? Can you figure that out?

Yes, then we can allow a 0 and we get $\binom{10}{5}$. I actually did figure out the first part too on my own for once. I guess I gave up on the problem too early and posted it here. I figured it out in the shower before reading this :) But thanks Arne, you keep helping me with everything that makes sleep difficult.