Find the mean of all even numbers that can be formed using exactly three of the digits 2,3,5, and 7.

Analyzing the possibilities, we see that in order for the number to be even, the units digit must be $ 2$, as it is the only even digit out of the four digits provided. Therefore, we have $ 3$, $ 5$, and $ 7$ to choose two from for our two leftover digits.

Out of all of the possibilities for the hundreds and tens digits, we know that "$ 55$" would be the mean as $ 5$ itself is the mean between $ 3$, $ 5$, and $ 7$.

Thus, putting all our pieces together, the mean of all of the even possibilities would be $ \fbox{552}$.


I'll appreciate a person who gives me an idea about this problem:

Find a discontinuous function which preserves connectedness and compactness, and what is the weakest condition that when it's added, makes the function be continuous?

Any function from a totally disconnected space to a finite set (with any topology) satisfies this property. For a noncontinuous example, use the $ p$-adic integers and the set $ \{0,1,2,\cdots,p \minus{} 1\}$ with the discrete topology. Our function is the first (lowest power of $ p$) nonzero digit: $ f(1) \equal{} 1,f(p) \equal{} 1,f(p \plus{} 2) \equal{} 2$ and so on, with $ f(0) \equal{} 0$. The sequence $ p^n$ converges to zero, with $ f(p^n) \equal{} 1\neq f(0)$.

Now, is that really the question you wanted to ask?

Yes, almost it is. Thank you so much. For the second part, as I mentioned, I want to find a [b]weakest condition[/b], I named it $ P$, such that for any function which preserves connectedness and compactness and satisfying $ P$ implies that the function should be continuous.


let $a_i\in\mathbb{R}^+$ for $i=1,\dots,n;\;n\in\mathbb{N}^+$
and $m=\frac 1n \sum_{i=1}^{n} a_i$

proof:

$\sum_{i=1}^{n} \frac{\prod_{j=1}^{n} a_j - a_i}{\sum_{j=1}^{n} a_j}\cdot a_i \le m^n - m$

its one of my first creation, so tell me if im wrong:.
peeta

no one?

its way easy if u see $\prod_{j=1}^{n} a_j$ as p and $\sum_{j=1}^{n} a_j$ as s.

peeta

Consider $0<a_i<1$ s.t. $\sum a_i <1$, take n is even, then i think it's wrong.

well i dont know, whether the inequality is valid $a_i$ but let me show you proof of it:

$p:=\prod_i a_i\\
s:=\sum_i a_i\\
\\
\sum_{i=1}^{n} \frac{\prod_{j=1}^{n} a_j - a_i}{\sum_{j=1}^{n} a_j}\cdot a_i = \sum_{i=1}^{n} \frac{p\cdot a_i-a_i^2}{s}=\frac ps \sum_i a_i -\frac 1s \sum_i a_1^2=p-\frac{\sum_i a_i^2}{\sum_i a_i} \\\le m^n - m= \left(\frac sn \right)^n - \frac{\sum_i a_i}{n}$
since we know that $\sqrt[n]{p}\le \frac sn$ we only have to prove $\frac{\sum_i a_i^2}{\sum_i a_i}\ge \frac{\sum_i a_i}{n}$ which can be done, after a few transformations, wtih Jensen:
$\frac{\sum_i a_i^2}{\sum_i a_i}\ge \frac{\sum_i a_i}{n}\\
\sum_i a_i^2\cdot n \ge \left(\sum_i a_i\right)^2$
let $f(a_i)=n^2\cdot a_i^2$ we know $f$ is convex for $a_i\ge 0$ so:
$\frac 1n\cdot\sum_i f(a_i) \ge  f(\frac 1n\cdot\sum_i a_i)$
which comes out to be
$\sum_i a_i^2\cdot n \ge \left(\sum_i a_i\right)^2\blacksquare$

actually i did this problem with majorization which is too much for this because i anyway only use the fact that $(a_1;a_2;\dots;a_n)\succ (\frac sn;\dots;\frac sn)$, $s$ being $\sum_i a_i$ and that a $\phi(a)=\sum_i f(a_i)$ is convex if $f(a_i)$ is convex for all $a_i$.

im interested in that topic, so might someone post some links to threads where there are soloutions/prooves done with majorization.

thx peeta

btw tell me if the proof above isnt ok. ill be glad to hear, so i might learn from it.

i just skimmed your proof, it looks fine to me.
My posted counterexample is wrong, I misread the question, sorry for that.


modular arithmetic
combinatorics unsolved
combinatorics

Is this problem very hard?

A table $1\times n$ is given.
Two players play the game. In a move they can eat two consecutive
cells. A player who can't make a move - loses. Which player has a
winning strategy?

Of course, the answer depends on $n$. The case n odd is more interesting.

for n is even, the first player will win by first take the middle, then play symmetrically as the second player.
for n is odd, two cases:
case1: n = 1(mod 4) the second player will win. suppose after first step, the table splits into two parts with x, y cells respectively. x+y+2 = n. one of x and y may possibly be 0. WLG, assume x is even and y is odd.
if x = 0(mod 4), then y = 3(mod 4). second player then eat 2 and 3 of y. Thus both x and y are even now. Afterwards, if first player eats the middle of x or y, second player eats the other middle. Otherwise, second player always play symmetrically as first player in x or y.
if x = 2(mod 4), then y = 1(mod 4). second player eats the middle of x. Afterwards second player plays symmetrically as the first player within x and plays recusively within y by treating y as new n.
case2: n = 3(mod 4) the first player will win by just eating 1 and 2 of n.

[quote="grobber"]I did not read your explanation for $m\equiv 1\pmod 4$, but there must be something wrong in there. A simple case check should show you that if $n=13$ and the first player removes the cells $4,5$, then he wins. $n=13$ is the first $n\equiv 1\pmod 4$ for which the first player wins, and it's sort of large, so I can see why it would be hard to find. I was trying to compute the [url=http://mathworld.wolfram.com/Nim-Value.html]Sprague-Grundy numbers[/url] for the game, and that's how I found it.

P.S.

Sorry if it turns out I was wrong, but the counterexample seems Ok.[/quote]
oops, u r right. sorry about that :blush:


function
calculus
derivative
geometry
geometric transformation
calculus computations

3 levels of integrated functions. Don't know what else i should call it in english, but here is what the problem is all about:

$(\sqrt{x^4+1}+1)^5$

As far as i know of this function is made up of 3 functions. Any idea how to solve this?

The way to say this in English would be to call it a "composition" of functions.

Understand the chain rule as operating from the outside in. As you assemble the value, what is the last thing you do? In this case, you take the 5th power of something. Take care of the derivative of that first: the derivative of the 5th power of something is 5 times the 4th power of that thing:

$\frac{d}{dx}(\sqrt{x^4+1}+1)^5=5(\sqrt{x^4+1}+1)^4\cdot\frac{d}{dx} (\sqrt{x^4+1}+1)$

Now we find $\frac{d}{dx} (\sqrt{x^4+1}+1).$ It's a sum, but the second term is a constant and has derivative zero. The first term is a square root, and we know that $\frac{d}{du}\sqrt{u}=\frac1{2\sqrt{u}}.$ So

$\frac{d}{dx} (\sqrt{x^4+1}+1)=\frac1{2\sqrt{x^4+1}}\cdot\frac{d}{dx}(x^4+1) =\frac1{2\sqrt{x^4+1}}\cdot(4x^3).$

So we have

$\frac{d}{dx}(\sqrt{x^4+1}+1)^5=5(\sqrt{x^4+1}+1)^4\cdot\frac1{2\sqrt{x^4+1}} \cdot(4x^3)$

(Followed by a little algebraic work to "clean up" the answer.

$\frac{d}{dx}(x^2)=2x$ ... right?

Just translation again. We right it like this:

$(x^2)'=2x$

Will try and get used to the standard method though. Hope it won't give me too many problems remembering it in the future.

The problem was i started from the other end of the composition. I worked on $\sqrt{x^4+1}+1$ and then took a look $5x^4$. This resulted in some nasty work with the rule of division for differentiation. Kinda hard to explain, but i was way off course :D Let me try and do some analysis of this solution tomorrow (busy tonight) by putting it into $y=\frac{d}{dx}f(x)(X-x)+f(x)$ and see if i get a tangent. Will get back to this thread in case i do something wrong later and can't figure out what i wrong. Could you help me clean up this mess? Then i would be really happy - it ain't my strongest side.

Thx for your help (so far).

$\frac{d}{dx}h(x)=5(\sqrt{x^4+1}+1)^4\cdot\frac1{2\sqrt{x^4+1}} \cdot(4x^3)$

$\frac{d}{dx}h(x)=\frac{20x^3(\sqrt{x^4+1}+1)^4}{\sqrt{x^4+1}}$

$\frac{d}{dx}h(x)=\frac{20x^3(\sqrt{x^4+1}+1)^4}{\sqrt{x^4+1}}$

This is the only cleaning i can do. Don't know if that is what you've got too or if this is even more cleanable.


A convex hexagon $ ABCDEF$ satisfies $ AB\equal{}BC, CD\equal{}DE, EF\equal{}FA$ and: $ \angle ABC\plus{}\angle CDE\plus{}\angle EFA \equal{} 360^{\circ}$. Prove that the perpendiculars from $ A,C$ and $ E$ to $ FB,BD$ and $ DF$ respectively are concurrent.

Consider $ \triangle FBD$ and points $ A, C, E,$ such that $ AB \equal{} BC,\ CD \equal{} DE,\ EF \equal{} FA.$ Then

$ FA^2 \minus{} BA^2 \plus{} BC^2 \minus{} DC^2 \plus{} DE^2 \minus{} FE^2 \equal{}$ $ FA^2 \minus{} BC^2 \plus{} BC^2 \minus{} DE^2 \plus{} DE^2 \minus{} FA^2 \equal{} 0$

By Carnot theorem, perpendiculars to $ FB, BD, DF$ from $ A, C, E$ are concurrent. Conditions that $ ABCDEF$ is convex and/or $ \angle ABC\plus{}\angle CDE\plus{}\angle EFA \equal{} 360^{\circ}$ are not necessary.


See http://www.mathlinks.ro/Forum/viewtopic.php?t=52359 and http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf  .

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108230[/url]
This is analysis.  Why do you double post?


Prove that for any integer $ a, a^4 \equiv 0$ or $ 1$ (mod 5).

[hide]For one solution, you can just take the case where a is 0,1,2,3, and 4 (mod 5), or:from Fermat's Little Theorem, $ a^4\equiv1\pmod{5}$ unless gcd(a,5) is not 1, which only happens when a is 5. For a=5=0, we have $ a^4$ is 0 (mod 5), and the others are 1.
So all are 0 or 1.[/hide]


geometry
3D geometry
vector
analytic geometry
LaTeX
symmetry
geometric transformation

Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal.  Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP=5, PB=15, BQ=15,$ and $CR=10.$  What is the area of the polygon that is the intersection of plane $PQR$ and the cube?

[hide]Let A be (0,0,0) B be (0,0,20) C be (20,0,20) and D be (20,20,20)

Slogging through lots of vectors:

Coordinates of P,Q,R are:
(0,0,5)
(15,0,20)
(20,10,20)

The two vectors are (I dunno how to latex)

<15,0,15>
<20,10,15>

cross determinant is:
<-150,75,150>

Plane is:
-2x+y+2z=10

Calculating the intersection of the plane and the edges of the cube (for example setting x as 0 and y as 0 would give the intersection of the edge with endpoints of (0,0,0) and (0,10,0)), you get new points:

(20,20,15)
(0,10,0)
(5,20,0)

If you just draw any convex hexagon with points ABCDEF, you can find the area by calculating the area of ABC, ACD, ADE, and AEF.

By doing all 4 cross determinants and dividing by 2 we get 525[/hide]

Can anybody find a more elegant way...I don't like my way

[hide="Well..."]First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel.  Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane $PQR$, so they cannot be skew.  Therefore, they are parallel.

Let the cube's vertices be $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$, with $A$, $B$, and $C$ on the bottom face as before, $H$ being the other bottom vertex, $D$ directly above $C$, $E$ above $B$, $F$ above $A$, and $G$ above $H$.

Clearly, the next vertex of the intersection (starting with $P$, $Q$, $R$) will be somewhere on $DG$.  Let it be $X$, and have a distance of $x$ from D, and a distance of $20-x$ from $G$.

Then, the next vertex will be somewhere on $FG$.  It must be parallel to $PQ$, so this implies that it has a distance of $20-x$ from $G$, and thus a distance of $x$ from $F$.

Now, the next vertex (call it $Y$) will be somewhere on $AF$.  The segment must be parallel to $QR$, so $FY$ must have length $2x$,  and $AY$ must be $20-2x$.

Since $DX \parallel AP$, $DR \parallel AY$, and $RX \parallel PY$, we must have triangle $APY$ similar to triangle to $DXR$; therefore, \[\frac{AP}{DX}=\frac{AY}{DR}\]  \[\frac{5}{x}=\frac{20-2x}{10}\]  \[x^{2}-10x+25=0\]  \[x=5\] We can now find that the hexagon has side lengths $15\sqrt{2}$, $5\sqrt{5}$, $5\sqrt{5}$, $15\sqrt{2}$, $5\sqrt{5}$, and $5\sqrt{5}$.  Moreover, opposite angles of this must be equal (by symmetry), so segment $RY$ divides the hexagon into two isosceles trapezoids.  It is easy to find the length of $RY$ (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or $20\sqrt{2}$), so it is now easy to finish the problem.  I leave the rest as an excercise (translation: I'm too lazy to do the last calculations).[/hide]

Anyone has an easier solution for this problem?

Here's my solution:

Note that the intersection is going to be a hexagon (actually there's an aime #15 in one of the previous years that deals with the net of the portion of the cube cut by the hexagon, and I instantly remembered that problem and realized this is a hexagon).

Let the opposite edge of DC be AE, with midpoint M. From here, it suffices to find the area of the isosceles trapezoid MPCR (since PB=BQ and therefore by symmetry the plane is going to intersect AE at M, further, the trapezoid is going to be half the area of the hexagon by symmetry).

MR is equal to the diagonal of the face of the cube, which is $ 20\sqrt2$
To find the other base PQ of the trapezoid, use the pythagorean thm on $ PB,BQ$ to get $ 15\sqrt2$

$ RC = MP = \sqrt {5^2 + 10^2} = 5\sqrt5$

So the height of the trapezoid is ${ \sqrt {(5\sqrt {5})^2 - [0.5(20\sqrt2 - 15\sqrt2)]^2}} = \sqrt {5^3 - 25/2}$

So the answer is $ (20\sqrt2 + 15\sqrt2)(\sqrt {5^3 - 25/2}) = 35\sqrt {2\cdot5^3 - 5^2} = 35\cdot15 = 525$.


linear algebra
matrix
algebra
polynomial

Here's another way you can see this:

for a degree $ n$ polynomial $ f$, the polynomial $ f(x\plus{}n\plus{}1)$ is always a linear combination of $ f(x), f(x\plus{}1), \dots, f(x\plus{}n)$.


I'm looking for a code that shifts the characters of a string a set number of characters. I would lurk moar, but I'm not actually sure what to search on google.

Do you know where one could find a simple code that lets you input and decode strings using the simple letter shift mentioned above?

Do you understand the algorithm or do you just need code?  This is a [url=http://en.wikipedia.org/wiki/Caesar_cipher]Caesar cipher[/url].

in what lang?
me and  [b] one [\b] (the other dude on the team did almost nothing)of my friends on this for almost 6 months (in java) but here is the code [hide][url]http://www.challenge.nm.org/finalreports/[/url][/hide] scroll down to the Aspen Elementary section..


Radii OA, OB, and OC divide the circle into 3 congruent parts. Let D be a point on the arc AB such that the arc AD has half the measure of arc DB. find measure of angle OCD.

[quote="bigboy82892"]$160^{o}$!! Because a third of the circle is $120^{o}$ and a third of that is $40^{o}$!! Add and you get $160^{o}$!!![/quote]
Use \circ instead of o for degrees. :wink:

[quote="bigboy82892"]$160^{o}$!! Because a third of the circle is $120^{o}$ and a third of that is $40^{o}$!! Add and you get $160^{o}$!!![/quote]

huh?! i think it's incorrect.

$Arc AD$ is $40^\circ$.. therefore, if you extend $CO$ and make it hit the circle at $P$, $D$ is between $arc AP$.

and you get $arc DP=20^\circ$

$angle OCD=10^\circ$ ?


Amos is packing cans. He has a large supply of boxes that hold either 5, 7, or  11 cans. Each box used must be fully packed of cans. What is the largest number of cans that Amos cannot pack with the boxes, if Amos must use at least one of each box size?

5+7+11 is the minimum that we should have. 23 is 5n + 3. so we can have all combinations of this form by simply adding 5s. 23+7 = 5n, so we can have all such combinations (>=30) by simply adding 5s. 23+11 = 5n + 4 so we can have all such combos (>=34). 23+7+7 = 5n + 2, so we have all such combos (>=37). 23+7+11=5n+1. so we have all such combos >=41.
now we can form all numbers >=41. (as we have covered, 5n, 5n+1, 5n+2,5n+3, 5n+4) the last to be formed was 5n+1 at 41. hence a no. of the form 5n+1 is the max that cannot be formed. since we are able to form all other combinations for >=37 (except for 5n+1) the max number which cannot be formed using all 3 no.s is 36.


algebra polynomial Vieta inequalities geometry symmetry induction

Let's discuss the qualities of the six IMO'2007 questions!

Problem 4 was far too easy; problems like that are simply not appropriate for IMO.

Problem 1 was fairly easy. Pretty much every approach works out: Just draw some kind of diagram of this and the solution is obvious (as Goblin wrote). But in contrast to problem 4, one had to think a little bit.

Problem 5 was easy to medium, but unfortunately attackable by a standard trick. darij grinberg wrote about it: "An IMO problem should require maybe 75% brains and 25% preparation and not the other way round. I really wonder why we have the same style of Vieta jumping over and over on IMO's since the 1988 problem by Arthur Engel. But then again, the quality of number theory problems in the IMO is suffering an infinite descent..." Problems like that should not even make it to the shortlist.

Problem 2 was easy to medium, but darij grinberg observed that it is the converse of a South Africa 1999 problem. This is not the best choice for an IMO problem...

Problem 6 was hard. But vess observed that it is an easy consequence of the Combinatorial Nullstellensatz, a well-known theorem in the MathLinks community. This makes it trivial for MathLinkers, and hence totally inappropriate for IMO.

Problem 3 was very hard, and a perfect choice for IMO.

In summary the jury failed to select good medium problems. The gap in difficulty between problems 1,2,4,5 and problems 3,6 is far too big.  It would have been much better to replace the trivial problem 4 by a problem of medium difficulty, and to replace problem 6 by a slightly easier problem.

[quote="test20"]In summary the jury failed to select good medium problems. The gap in difficulty between problems 1,2,4,5 and problems 3,6 is far too big.[/quote]

This is exactly my opinion.
Besides, the two hardest problems were combinatorical ones. This is not pretty balanced IMHO. There are hard inequalities ("hard" meaning not just 5 pages of computation), hard number theory problems and hard geometry problems in the world.

As for problem 2 being the converse of a SAMO problem, this was a remark by myself 2 minutes after I had seen the problem. This has actually not much to do with the solution - it is rather different from the solution of the SAMO and requires some more thought. But still the problem is anything but original, and I really hoped for something better on the IMO.

  Darij

[quote="Mamat"]I'm sure there are nice inequalities in the shortlist. Couldn't they choose better one?

I liked the 2nd problem.[/quote]

Also as I said to Darij in Yahoo, I feel something unbalanced in the contest. Actually, two problem 2 and 4 are easy for IMO, kind like problem 5 appears many time in IMO. Problem 1 is not bad in general, but for IMO, it is not good. Problem 3 is perfect, just this one. Personally, I like the problem set of IMO 2006 more.

The only logical conclusion one could arrive to after IMO 2007 is that [b]Geometry is completely useless in Mathematical contests[/b]. To be more precise, [b]future competitors shouldn’t even practice geometry at all![/b] This fact is proven by (so called) “problems” like [i]IMO 2007 (4), IMO 2006 (1), IMO 2004 (1), IMO 2003 (4)[/i] which do nothing but foster a sense of humiliation to all people that enjoy solving Geometry problems (and on the other side, represent a free gift (i.e. 7 assured points) to all those that hate Geometry). [b]These kinds of questions shouldn’t even be shortlisted[/b] (they don't have anything to do with thinking).

Moreover, questions like IMO 2007(2) (despite the fact that it is only slightly harder than IMO 2007(4) )  are far too easy to be selected as second problems at IMO’s. In other words, this IMO proved to be a complete failure with respect to Geometry.  Unfortunately, the aforementioned examples illustrate the fact that nobody cares about Geometry anymore; one doesn’t even have to have any math skills in order to solve a Geometry problem at IMO. [b]This disgrace has to stop as soon as possible[/b], otherwise what’s the point of studying Geometry at all ?!

It's pretty weird how both 3 and 6 were combinatorics, this is very good for some countries and bad for others.

It's hard to comment on the choosing of the problems, but usually the leaders try to select the best problem set available from the ISL. I think that by next year we can judge whether this was the best way to do it (but for now I trust the leaders assessment).

[quote="vess"] Looking from the partial results, so far the highest score on that problem is a 5, in the Russian team, thus it may safely be said that, [b] so far[/b], it has only been solved by one (this) person![/quote]

Mladen Radojevic from Serbia (SRB1) solved it (7 points).

[quote="vess"]To be honest, I think that a random computer-generated choice would have selected better problems from the shortlist. There must be original, fresh problems in there that would have made much better choices then the [i]lazy[/i]  #5 and #6.[/quote]

I totally agree!

[quote="Goblin"][quote="vess"] Looking from the partial results, so far the highest score on that problem is a 5, in the Russian team, thus it may safely be said that, [b] so far[/b], it has only been solved by one (this) person![/quote]

Mladen Radojevic from Serbia (SRB1) solved it (7 points).
[/quote]

Good job!  :first:  The Serbian results on #3 haven't yet been posted, which is why I didn't know about this. I suppose that there will be other 7's as well, not only in the China team -- but still, the results posted so far show that the problem is one of the most difficult ever at IMO (despite the simple-looking solution(s)!). But in case there aren't other 7's, he (Mladen Radojevic) might be awarded a special prize for solving this! ... In any case, it's an achievement, congratulations! :)

[quote="vess"]I agree with most things said, in particular, that #3 is a terrific problem (at once one of the finest and one of the most difficult combinatorial problems ever at IMO!). Looking from the partial results, so far the highest score on that problem is a 5, in the Russian team, thus it may safely be said that, [b] so far[/b], it has only been solved by one (this) person! Of course, we have to wait for the results of the other contestants, most notably ITA6 and the Chinese team; but it seems that, statistically, this problem could surpass in difficulty the celebrated 1988/6!

Unfortunately, the other problems are totally a bad choice. I agree with what was said above, [b]except[/b] that #6 is a combinatorial problem. (If it were, I don't think that the Chinese total score on that problem would have been 3 points  :lol: -- an indication that, as suspected, the proposers don't have a combinatorial solution and only have the standard polynomial solution). This is not an original problem, and it shouldn't have made it even to the shortlist, in my humble opinion. The Committee should have had a better connaissance with Olympiad-style problems in the literature, and should have noted that, at the least, this is not a new problem. [b] They also should have attempted to solve the problem(s) on their own [/b] before selecting it (them) on the basis of solution/comments given in the shortlist.

To be honest, I think that a random computer-generated choice would have selected better problems from the shortlist. There must be original, fresh problems in there that would have made much better choices then the [i]lazy[/i]  #5 and #6.[/quote]

Personally, I realize your assessment. You are totally right !  :maybe:

[quote="marko avila"]Its totally unfair for #5 on an imo to be solved with a standard trick as vieta jumping. i mean , if you know vieta jumping you will most probably solve the problem . as i see the solution using the vieta jumping is just  standard computation using the method. its unjust because if someone doesnt know the trick , the question becomes pretty hard. also an imo problem should have various solutions, i have only seen one. this problem ,has nothing to do in an imo test. i think its wrong that the medals are decided simply by knowledge and not by talent. this is wrong. moreover as everybody can see, pretty much evereyone except a couple people have a positive score on problems 3 and 6. wich means that problem 5 decides in a way who wins gold medals. anyways , i saw this coming judging from vietnam national olympiads, it seems that number theory problems like 5 are pretty frecuent in vietnam olympiads. this test was horrible!!!  its just unfair!!![/quote]

Hi, dont be angry like that !!! We are just discussing !!!   :lol:

[quote="Valentin Vornicu"]It's hard to comment on the choosing of the problems, but usually the leaders try to select the best problem set available from the ISL.[/quote]
yeah, they usually do, but it seems that they failed *two years in a row* :(

One can and should judge an IMO test on several different points.

First of all, the questions should separate the contestants sufficiently, especially around the medal cuts. Personally I think having average scores of around 4/2/1 (on both days) and many partial marks work rather well.
This means problem 1,2, and 5 are doing good, 4 is too easy (average 5,5) and 3 and 6 are much too hard (though on problem 3 many good countries haven't coordinated in Joseph's Meyers list). It is especially unfortunate that both 3 and 6 are too hard, as this implies there is no problem to separate "good silvers" from "bad golds". Problem 1 can cover a bit for problem 4 to separate between those deserving no medal and the bronze medallists.

Secondly there is the (subjective) question of beauty of problems and their solutions. I won't comment on the geometry problems here as I can (almost) only solve those by algebra which destroys all beauty immediately, but the geometry problems did not look horrible to me. Of the other problems I think problem 1 is pretty nice, problem 3 is very nice, problem 5 is ok (somewhat standard but not bad) and problem 6 would not have been my choice (the solution requires a method high school students should not be expected to know or come up with).

Thirdly there is the question of fairness. Here we want to have problems which are not solved by tricks you should have seen in order to be able to come up with them and that there is a balance in different subjects. As for the distribution of subjects I think it is perhaps not a good idea to to have a combinatorics and an algebraic combinatorics problem on 3 and 6. As for the problem of tricks being necessary, I think that the complaints about Vieta jumping on problem 5 are a bit rash as it can be done as well by infinite descent, and the proof I found using that method is simpler than the proof using Vieta jumping on the forum. And number theory will always be a subject in which you need to know some things in order to proof anything, and infinite descent is part of any decent IMO training program (and even some not so decent ones). However I am concerned that problem 6 can only be solved by students having seen something similar.

All in all not a perfect test, but surely not bad. If problem 6 was something else it could well have been a very good test. Considering it is impossible for the jury members to solve all shortlist problems (especially the candidates for problems 3 and 6) themselves in the short time they have, and the inherent difficulty in trying to judge the difficulty level of problems,  I'd say they did a rather good job.

Disclaimer: I reviewed the submitted problems of the Dutch team and already at that time was not very positive about problem 6, though I deemed it good enough to at least be reviewed by the problem selection committee. Sadly enough the same proposer also submitted three (not sure here, could be four) more problems which I all liked, but were not chosen, one of them would even have been a good candidate for problem 6.

[quote="shunka"]Problem 3 was solved by 3 (three) people. One from China (7 points), one from Serbia (7 points), and one from Hungary (6 points). That's it. Of these three, two are from countries that did not make it in top 10, and one got silver medal. I am sure that all three of them are extremely happy and proud about their success (and I would like to congratulate them wholeheartedly), but it is clear that the problem did not have any measurable effect on the competition. [/quote]

Actually I did not solve problem 3. Just got way to close to the solution. Maybe if I had had some more time I could have solved it...

[quote="Valentin Vornicu"]The same happened to a number of other students. In my opionion problem 3 should have been problem 6, and problem 3 should have been chosen a number theory question from the ISL. The Olympiad would have been a lot better this way.[/quote] Basically I totally agree with your point, Valentin. Just want to say that I heard (only rumor, but I trust the source) that in 2007 ISL there are few good candidates for number theory. That may explain the situation.

Also I heard that there is no other good and difficult algebra problem for problem 3 and 6 (of course, the definition of "good" problems varies even among some great mathematicians, so this is just an idea). If so, then it turns out that there are only two choices available - either like the IMO this year, or choose a geometric problem for problem 3. It could not be combinatorics geometry, so it must be pure geometry (pardon, I don't know if there is such a word).

But it is never easy to find a nice difficult geometry problem. Many nice geometry problem are moderately difficult, but in my opinion very few are as difficult as, say, IMO 2003-3. Also there is some danger to put such problem in problem 3, as there was for IMO 2000-6. It might be solvable by brute force trigonometry or coordinate computation.

And I want to say something for problem 6 of this year. We all know that there are three possibilities for one to solve problem 6.
1. He/She knows CN, and understands it well.
2. He/She knows knows some idea of such polynomial method. And with some nice thought the problem is then solved.
3. He/She knows have seen nothing about this, but work out the problem with great intuition.
I really have no idea about what distribution of the celebrated five solvers of the problem in IMO has in this (maybe someone has the idea? or should we do a survey on this :D)  But if most of them are of category 2 and 3, then maybe the problem 6 is not bad. After all, only a small percentage of the contestants are Mathlinkers.

I agree with Darij said that "IMO should require maybe 75% brains and 25% preparation and not the other way" (though Darij was not talking about this problem). And I think this problem is not totally bad in this point, since it encourages contestants to study some interesting deeper and advanced mathematics, rather than doing lots of problems or studying some strange mathematics Olympiad books. To learn stuff like CN is, in my opinion, much better than spending time learning some ugly inequalities tricks (this has no relation with the inequality book sold and distributed at IMO in Hanoi, of course!) or so. This may not be a good reason for it to appear in IMO, but only that I think this problem is not that bad.

Personally I hope they choose a harder geometry problem for problem 5 and move problem 5 to problem 2. But again the gap between problem 1 and 2 will be big then. It might be possible to choose a slightly easier combinatorics problem to replace problem 3. But I like this one. IMO should have some problems as (or more!) beautiful and incredible as some in St. Petersburg, All Russian's, BMO, RMO, ...etc.

Sorry for this post to be so long :D And we have to wait a year to see if the 2007 ISL is really like that  :maybe: I heard that Vietnamese tried a new "modern" style in choosing the ISL, maybe the shortlisted this year is just a little  strange comparing to those of earlier years.

[quote="Darkseer"]I heard that Vietnamese tried a new "modern" style in choosing the ISL, maybe the shortlisted this year is just a little  strange comparing to those of earlier years.[/quote]Oh, I definitely didn't mean that our host is to be blamed for some disadvantages of 2007 IMO. I am sorry if it sounds like so. I just guess that the committee didn't get used to the new style, so that the choice is limited.

edited: the earlier version of this post contained some description about the 2007 ISL. Although I think it won't really disclose anything important for next year's TST, it is still better to delete it.

[quote="Valentin Vornicu"][quote="Darkseer"]After all, only a small percentage of the contestants are Mathlinkers.[/quote]You are basing this statement on ...  :?: I'm sorry but I will have to contradict you.[/quote]No offense really  :roll: ...It is only my guess. And Valentin you're right I should not write that so confidently. But by Mathlinker (but again, sorry for the abuse of language) I mean a stricter definition, that is, for our discussion, I consider only those who have visited AoPS-Mathlinks and participated in the discussion or at least read others' ideas in the Olympiad section many times.

I just think that in most Asian country few contestants are Mathlinkers. And in some (but only some) European countries the contestants don't put so much time in preparing IMO, so I also guess (maybe I am wrong) that also not many of them are Mathlinkers. It seems to me that it might be strange if more than 30% of the contestants are Mathlinkers but only 5 contestants solves problem 6 (maybe it is not really strange, I don't know. But I only want to say that I think quite few contestants knows about CN when facing problem 6)

But after all, I believe you can have better idea in this point than me. So how do you think, Valentin?

I think that you have absolutely no basis for your statistics :) On the other hand I have clear statistics as of visits from countries from around the world, and I can definitely tell you that you are wrong. But that's not the point of this. I'm thinking about myself: I saw the CN postings some time ago, and I knew the idea behind it. However I didn't solve the problem when I thought about it (granted I only tried about 30 minutes  :) ). Just as multiplying the plane equations didn't automatically mean you've solved the problem (check the number of 1,2 points scores which is rather big).

As I have written above, from my experience (and I've been 2 times coordinator, 2 times contestant and 3 times observer at the IMO, so it's not quite small) and also from knowing this year's ISL (and part of the ILL - the Romanian one that is) I can tell you that there could have been much more better questions selected for the IMO, and that it [b]is not[/b] the Jury's fault I think. You, on the other hand, do not know many of these arguments, so your deductions are purely theoretical and I dare to say that they have nothing to do with the reality. In an year the ISL is going to be public and then I can better explain my arguments ;)

Oh Valentin if you have the shortlisted, then of course I can believe you in your judgement. I haven't seen the ISL nor any of the ILL (except the six problems which everyone has seen :D) myself. But I trust my source - I can only say that either I misunderstood him (as he could not disclose any useful detail of the ISL to me, it was likely to hapen) or there are just some differences in the way we judge a problem, which happens quite often.

And I presented my argument about problem 6 just to say that I think problem 6 is not that bad, as some have argued and complained so in several threads. I agree that you must have clear statistics - thank you for correcting me - yet I only want to present my idea. There could be error in that part of the argument - just skip it then.

Sure I understand and I appreciate you stating your point of view. I agree with some of your answers, but I disagree with the others and I've said my arguments on this ;) That's why we have these Forums, so that people can discuss their (different :) ) opinions on the subjects.

[quote="Valentin Vornicu"]I think that you have absolutely no basis for your statistics :) On the other hand I have clear statistics as of visits from countries from around the world, and I can definitely tell you that you are wrong.
[/quote]

As a person interested in statistics I am looking forward for seeing an [url=http://www.mathlinks.ro/viewtopic.php?t=3788]for the following thread.[/url] I also would not mind to get some distribution over the folks who are visiting. One of the available tools shows kind of cluster map, [url=http://www2.clustrmaps.com/counter/maps.php?url=http://homepage.mac.com/davidrh/]here is an example.[/url]

I suppose that I have no particular authority that makes my opinion valuable with respect to the quality of the 2007 IMO questions. But I would still like to voice my opinion.

I only looked at 3 of the problems thoroughly.

1. Apparently this was a fine problem. One of my friends solved it relatively quickly. I think this is well-placed.

2. I thought that this problem seemed too easy. I solved it in 45 mins. I had seen the converse before. Knowing that made the problem easier. I think there are better geometry problems out there.

3. no idea.

4. This problem is trash. I saw how to solve it practically instantly. In terms of my 'problem-solving algorithm,' this was just a "sine-area" bash. It took no thinking whatsoever. All necessary lengths were trivial computation. Of course there was a slightly nicer solution, but that does not change the fact that this problem was totally trivial.

5. This problem was also very bad. I learned the root-flipping idea a few days before at my camp. According to the algorithm that I developed for this type of problem, this was a mere exercise. IMO problems should require thinking, not this type of learning of techniques.

6. I don't know that technique...

I suppose that I was pretty negative overall. I didn't really attempt the hard combo problems. Again, I have no particular authority to say any of these things. I still feel like the geometry was lame. I cannot really solve IMO combo problems yet...so I cannot offer a general opinion of those. I would still hope that they could find problems that require elegant arguments rather than knowledge of 'random' techniques.

[quote="Altheman"]I would still hope that they could find problems that require elegant arguments rather than knowledge of 'random' techniques.[/quote]It's a matter of getting them into the ISL and then voted :) The problems exist ;)

Orlando, the stats are something like
Posts per day: 	532.09
Topics per day: 95.07
Users per day: 	19.40

I took part in IMO1981 as a high school student from a school with no real history or links with the "Olympiad World" and found that the problems and training material them were accessible to me, so if there are specialist techniques required to solve problems or which make them very easy, I think that makes them bad problems for IMO - a good high-school student looking at the solution should be able to say 'hey I could have done that' - the difficulty should be in seeing the way through the problem.

Problem 3 2007 seems to me to be a potentially good problem - undoubtedly difficult, but various possible ways in - but perhaps in retrospect a little too difficult. And based on comments the illusion of possible progress may have lured some candidates into wasting time.

Problem 6 by contrast seems to me to be poor. The 'answer' becomes obvious with a bit of testing. But there are some obvious ways of achieving it with parallel planes - some highly redundant and others uneconomical (very few points on a plane - easy to leave a spare point which needs just one plane to pick it up) and there is (reasonably obviously) no need for the planes concerned to be 'rational'. So there is not going to be any elementary solution based on proving that one, or a few, possible arrangements are extreme and it looks very much as though any solution is going to have to deal with all the (infinite number of) minimal configurations at once. I did try reducing any solution to some kind of standard form as a technique, but I haven't been able to make this work. All the coefficients for the planes can be made rational without difficulty.

[It also looks as though there might be some viable induction on dimension - the one-dimensional case being trivial, but this does not seem to work]

[quote]After all, only a small percentage of the contestants are Mathlinkers.[/quote]

 If you consider my country usually 4-5 participants are Mathlinkers :)

The geometry problem this time was really rather simple. It belongs to simple middle school geometry- it is certainly not worthy of inclusion in the IMO. More so when you consider the level of IMO geometry in the past.
How come they don't include solid geometry? :?:

[quote="rg13.1729"]How come they don't include solid geometry? :?:[/quote]

1) Most students of nowadays don't have a solid ( ;) ) foundation in solid geometry. For instance, solid geometry taught in Germany until the 12-13'th grades is confined to computing volumes of shapes like tetrahedra and spheres, and even this is done very informally and partly without proof. Things like the uniqueness of the common perpendicular to two skew lines or the theorem of the three perpendiculars are not taught until the last two grades of school, where they are considered in algebraic fashion. And IMO curriculum is supposed to be based on middle school rather than high school knowledge.

2) Obtaining marks for a problem should not depend on one's drawing abilites. Unfortunately, good drawings are hard to make in solid geometry, and the skills required for this are not quite mathematical.

3) Many facts in solid geometry - many more than in plane geometry - are much easier to prove knowing linear algebra than without. The IMO does not want to have its problems advantage students learning higher mathematics.

  darij


search
function
algebra
polynomial
MATHCOUNTS
complex numbers

To my knowledge, the TI-89 can calculate, simplify, or solve most things that you can throw at it.

A quick Google search returns [url]http://www.physicsforums.com/showthread.php?t=60283[/url]. It appears that you should use certain functions while dealing with complex numbers (ie. csolve() instead of solve(), and so forth). A quick peek at the manual  (here: [url]http://ti-89.org/uploads/ti-89-manual.pdf[/url]) seems to back this up.

[quote="math92"]Can the TI-89 titanium calculate and simplify expressions with the imaginary number, i?[/quote]

yes, as well as many other things. ;)

[quote="Glen Mann"]but would the 89 be worth getting? is it allowed on contests?[/quote]

the 83+ can do it...

just go to MODE and hit a+bi

the only good reason for the 89 is that i solves and holds [b]radical functions[/b]...the other TI counterparts display all radicals as decimals...for example the 89 can simplify

$\sqrt{12}= 2 \sqrt{3}$

where as the TI-83+ will show it as $3.4641016$...

[quote="bpms"][quote="7h3.D3m0n.117"][quote="Glen Mann"]I dont even have a graphing calculator but I bet they can come in handy on contests..[/quote]

Well the whole point of  a [b]graphing[/b] calculator is that it graphs functions...for that reason it is pointless in competitions...the only remarkable and worth mentioning item is that the TI can hold BASIC programs so you can calculate tedious forumlas easier (may or may not be allowed on contests)[/quote]
not true, you can find the roots of a polynomial within a few decimal places by graphing.
THis is incredibly useful on contests.[/quote]

Yeah but I've never seen a competition (maybe Olympiad though) that would require you to graph (certainly not MATHCOUNTS, but maybe AMC) polynomials. Besides, creating a BASIC program to find the roots of a polynomial is far better (an 11 degree polynomial can be hard to track down its roots  :D )


function
integration
trigonometry
limit
calculus
abstract algebra
real analysis

For $ f\in{L^{1}(\mathbb{R}})$ and $ n\in\mathbb{N}$, the function of $ \phi_{n}$ over $ \mathbb{R}$ is defined by $ \phi_{n}(x) \equal{}\int_{\minus{}\infty}^{\infty}\frac{\sin n(y\minus{}x)}{n(y\minus{}x)}f(y)\ dy$, provided that we interpret as $ \frac{\sin n(y\minus{}x)}{n(y\minus{}x)}\equal{} 1$ when $ y \equal{} x$.

Prove that $ \lim_{n\to\infty}\phi_{n}$ is uniformly converge to 0 over $ \mathbb{R}$.

Write $ f(x)\equal{}g(x)\plus{}h(x),$ where $ g$ is a "nice" function - let's go ahead and say $ C^{\infty},$ compactly supported, just for overkill - and $ \|h\|_{1}<\epsilon.$

We know that $ \int_{\minus{}\infty}^{\infty}\frac{\sin(n(x\minus{}y))}{x\minus{}y}\,g(y)\,dy$ converges uniformly to $ \pi g(x).$ My favorite argument for that would involve splitting that at $ x$ into two integrals and then integrating each half by parts, differentiating $ g$ and integrating the "sinc" function (or Dirichlet kernel, if that's what you want to call it.)

But you'll notice that what we actually have is $ \frac{1}n$ times that integral. Uniform convergence to $ \pi g$ implies uniform boundedness, and multiplying by $ \frac{1}n$ then gives that this part of the integral converges uniformly to zero.

As for the other portion, for all $ x,$

$ \left|\int_{\minus{}\infty}^{\infty}\frac{\sin(n(x\minus{}y))}{n(x\minus{}y)}\,h(y)\,dy\right|\le\int_{\minus{}\infty}^{\infty}1\cdot|h(y)|\,dy\equal{}\|h\|_{1}<\epsilon.$

Since this estimate is uniform on $ \mathbb{R},$ that finishes the argument.


2*(106+24b)^(-I-1) = (106-12b)^(-I-1)

2 = [(106+24b)/(106-12b)]^(I+1)

I don't understand the second line. How do we get from the first line to the second line?


Euler
modular arithmetic
function
Alcumus
number theory
totient function

[url]http://www.projecteureka.org/[/url] - this site is a compendium a problems. It is in a way very similar to this forum, however it is meant for the problem solver, hence a user can not see other ppl solutions unless herself solved the problem.

The discussions happen only after solving the problems, IMHO,  i think it is cool, as it force you to think. :wink:

Also I like the test section where problems can be grouped together to form tests, this is very useful for whoever prepares for contests.

similar website: projecteuler.net

Just thought I'd point it out.

[quote="uzurpatorul"]Yeah, in a sense it's similar to project euler, however project eureka has a broader audience (unlike euler's which is math problems that involves some sort of programming in order to get the solution) and it is user driven (anybody can submit interesting problems).[/quote]

Not to mention that most of the problems are easier. A few are hard, but none of them are really olympiad-level. Except, perhaps, this one:

[quote]Given that 37! = 13763753091226345046315979581abcdefgh0000000 (N! = 1x2x3...N),  determine, with a minimum of arithmetical effort, the digits a, b, c, d, e, f, g, and h.[/quote]
All I can figure out is that h is 0.

I've registered under the name "sdkudrgn88".

Actually even that is not of olympiad level. It just requires a lot of computations w/ modular arithmetic. Prime factorize $ 37!$ and then take $ \frac{37!}{10^7} \pmod{10^8}$. It will probably involve using the Carmichael function because it is obviously going to give you better results than the Totient function at this point.

I guess in time project eureka will have also Olympiad level problems, probably its community is not quite at that level  :wink:

It's a great site! Sort of a mix between Alcumus and a problem-based Wiki. I like it a lot.


Γεια  σε  όλους  και  καινούργιους  και  παλιούς .  Εχω  καιρώ  να  μπώ  αλλά  πρέπει  να  ξέρεται  πως   πιστεύω  το  εξής :  αξιζει  να  μπαίνει  κανείς  σε  κάποιο  φόρουμ   μονάχα  όταν 1) επικρατεί  ευημερία και  2) έχει  χρόνο .  Επειδή  αυτή  η  περίοδος  είναι  ακριβώς  έτσι  λέω  να  ξεκινήσω  με  ένα  ολοκληρωμα  απλό  έτσι  για  αρχή.......... :)  :)

        Λοιπόν ....  Να  υπολογιστεί  το ολοκλήρωμα

                  $\int\frac{x^{2}e^{Arctanx}dx}^{\sqrt{1+x^{2}}}$

Gia su desko+wellcome again...(an +ise pios palios ap m1...)
Gai to olokliroma..meta apo arketes praksis+an dn exo knai kapio la8os.. me mia entelos anor8odoksi me8odo..eftasa..sto..
$\frac{1}{2}e^{Arctan{x}}\sqrt{1+x^{2}}(x-1)$  (1)

+leo anor8odoksi gt den efarmosa kanena <kanona>oloklirosis aplos skeftika  antistrofa...diladi <ti 8a eprepe na ixa oste an paragogiza na eperna ayto>..+siga siga..kateliksa s ayto pu grafo pio pano,to opio malista an to paragogisis su dini tin arxiki posotita..tora..dn ksero mpori na iparxi+ali lisi i tipota periorismos..ala i (1) sigura ine mia lisi... :lol:

 :D

Ακριβώς  αυτή  είναι  η  λύση  ........ (βάλε  μονάχα  ένα  c  στο  τέλος  για  να  είμαστε  οκ  με  την  μαθηματική  μας  συνείδηση   :lol: )

  Κατα  τα  άλλα  το  θέμα  αυτό  είχε  μπεί  στην  Ρουμανική  ολυμπιάδα  το  1975  ...... για  τις  μεγάλες  τάξεις .......  Έτσι  όπως  το  είχε  λύσει  τότε  η  επιτροπή  έχει  αρκετό  ενδιαφέρον  από  μεθοδολογικής  άποψης ........ Θα  περιμένω  λίγο  ακόμη  μήπως  το  σκεφτεί  κανείς    και  μετά  θα  βάλω  την  λύση  που  είχαν  δώσει  τότε ......

  Υ. Σ.   Να  σου πω  την  αλήθεια   Ράλλη  και  εγώ  όταν  την  πρωτοέλυσα  με  μπόληκες  πράξεις  την  έβγαλα  ........ αν  δεν  σου  κάνει  κόπο  βάλε  την  λύση  σου  ..... μην  ξεχνάς  κάθε  λύση   εχει  πάντα  κάτι  να  πει  :)  :)


calculus
derivative
function
logarithms
integration
limit
complex analysis

Prove that $f(x)=e^{x}$ is its own derivative, and that it is the only function with this property (except of course $f(x)=ae^{x}$)

Note that $e^{x}= xln(e)$.
We get $lnf(x) = xlne=x$.
Differentiating with respect to x and applying the chain rule, we get
$\frac{1}{f(x)}*f^{'}(x) = 1$
and it follows that $f^{'}(x) = f(x) = e^{x}$.

The precise nature of this proof depends on how you're defining $e^{x}$.

[b]Definition 1:[/b]  $e^{x}= \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.  Then the result is obvious.

[b]Definition 2:[/b]  $e^{x}= \ln^{-1}(x)$, where $\ln x = \int_{1}^{x}\frac{dt}{t}$.  Here we can solve a first-order differential equation

$\frac{dy}{dx}= y \Leftrightarrow$
$\frac{dy}{y}= dx \Leftrightarrow$
$\int \frac{dy}{y}= \ln y = \int \, dx = x+C \Leftrightarrow$
$y = e^{x+C}$

You missed a bunch of solutions, t0rajir0u.  :lol:

Hm?  As far as I know, $e^{x+C}= e^{C}\cdot e^{x}= Ae^{x}$ is the only solution to that differential equation.

Strictly speaking, $e^{C}>0$, whereas $Ae^{x}$ with $A\le 0$ is also a solution. This is easy to fix, of course.

I was thinking of $e=\lim_{x\rightarrow\infty}(1+\frac{1}{n})^{n}$...

[quote="mlok"]This is easy to fix, of course.[/quote]
I guess, but you will still need either a uniqueness-theorem for ODEs or some careful analysis to show that there aren't any solutions that are 0 sometimes and non-zero at some other point.

This is why I think that separation of variables shouldn't be taught until you get to an ODE-course. I mean, the method breaks down in cases that are not really pathalogical.

t0rajir0u: The reason you miss solutions is that you clearly do not have the equivalence you have written between your steps, since $y$ may attain the value 0 or be negative (or maybe both?).

[quote="mlok"]Strictly speaking, $e^{C}>0$[/quote]

Oh, allow me the luxury of $C \in \mathbb{C}$  :P

[quote="Kalle"]t0rajir0u: The reason you miss solutions is that you clearly do not have the equivalence you have written between your steps, since $y$ may attain the value 0 or be negative (or maybe both?).[/quote]

I see your point.  Would the series substitution, then, be rigorous enough, or would I also have to deal with questions of convergence?

For uniqueness part I guess assuming f(x) is also a solution and investigating f(x)/e^x is enough.

Here is a nice proof that also happens to be the first one yet that doesn't assume the question (come on people, the Taylor Series is not a [b]definition[/b]; the integral for ln is not a [b]definition[/b]).

We have $\frac{d}{dx}e^{x}=\lim_{h\rightarrow 0}\frac{e^{h+x}-e^{x}}{h}=e^{x}\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}$. But $e=\lim_{h\rightarrow 0}(1+h)^{1/h}$. Then we can imbed the limits: $\frac{d}{dx}e^{x}=e^{x}\lim_{h\rightarrow 0}\frac{(1+h)^{h/h}-1}{h}=e^{x}$, as desired.

As for uniqueness, see the post above.

[quote="Kalle"]I have a hard time telling if your post is some type of troll or not.

1) How do you define the exponential function?

2) If you don't think there is a problem with taking the logarithm of a negative number then you don't know enough to make this assertion. There is a big problem with this, discussed in complex analysis.

3) What t0rajir0u did is in a strict sense wrong. While he made an algebraic mistake and wrote $\ln y$ instead of $\ln |y|$ this is not really the heart of the issue. The problem is that $1/y$ blows up for $y=0$ and so t0rajir0us solution might miss solutions that assume both the value 0 and some other value. Introducing a complex logarithm definitely doesn't fix it, and furthermore the method of separation of variables stops making sense when talking about complex valued things.[/quote]

Sorry about the nature of my earlier post. I certainly don't mean to turn math threads into battlegrounds-there's already enough fighting on these forums. I consider the definition of $e^{x}$ to be an exponential with base $\lim_{h\rightarrow 0}(1+h)^{1/h}$, interpolated to the reals, as this seems to be the standard definition of e. Even if it isn't, the other definitions make the problem really trivial, and the author of the post made it clear that he wanted this definition. Sorry again if that appeared to be a troll. I was just stating in an admittedly too-mean way that the author's question as he stated it hadn't yet been answered.

Nevermind about any complex analysis stuff. I haven't really studied complex analysis formally, though I've worked with it a little on my own. Anyway, I think we all agree that Kahsi at least was right on the other direction.

But saying that $\ln x = \int \frac{1}{x}$ (or, equivalently, $\frac{d}{dx}\ln x = \frac{1}{x}$ assumes that $\frac{d}{dx}e^{x}= e^{x}$, right?

I've always thought of $e^{x}$ as a function that satisfies $e^{a}\cdot e^{b}= e^{a+b}$ and a few other nice properties, where $e$ is defined as above.

Sorry if that isn't rigorous enough; I really don't know much about complex analysis either :blush:

Maybe I misinterpreted this when I was first learning this stuff in calculus (as opposed to using natural logs in pre-calc stuff), but the impression I got was that the definition of exponential and logarithmic functions was primarily due to the integral $\int\frac{1}{x}dx$.  There was no way to evaluate this integral with any functions that already existed, so the natural log was defined as $\ln{x}=\int^{x}_{1}\frac{1}{t}dt$.  From this, $e$ was defined as the number such that $\ln{e}=1$, and using the integral definition of natural logs, the definition of $e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ was derived.

Assuming that I understood that correctly, I would consider the most fundamental definition of $e$ to be t0rajir0u's definition 2, using the integral definition of natural logs...

[quote="jb05"] I consider the definition of $e^{x}$ to be an exponential with base $\lim_{h\rightarrow 0}(1+h)^{1/h}$, [b]interpolated to the reals[/b][/quote]

Of course, the problem is this: how exactly is such an interpolation done?

I've seen enough analysis to suspect that a rigorous treatment of exponentiation begins with the [b]definition[/b] $\ln x = \int_{1}^{x}\frac{dt}{t}$ and works from there (and Kalle's statements seem to back me up on this).  Everything is defined totally rigorously in this way, and we can prove the familiar properties of logarithms and exponents from here.

Ask yourself this:  working from your definition, how on earth would you go about showing that $\ln x = \int_{1}^{x}\frac{dt}{t}$?

[quote="DanK"]definition of exponential and logarithmic functions was primarily due to the integral $\int\frac{1}{x}dx$.  There was no way to evaluate this integral with any functions that already existed, so the natural log was defined as $\ln{x}=\int^{x}_{1}\frac{1}{t}dt$.  From this, $e$ was defined as the number such that $\ln{e}=1$, and using the integral definition of natural logs, the definition of $e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ was derived.[/quote]

I actually suspect that both approaches were considered.  It wouldn't surprise me if one reason mathematicians considered $e$ such an interesting constant is that those two definitions turn out to be equivalent in that manner.

I guess defining the exponential like jb05 suggested is possible, although I haven't seen it done like that myself. It sounds like a difficult approach.

[quote="mathnerd314"]
I've always thought of $e^{x}$ as a function that satisfies $e^{a}\cdot e^{b}= e^{a+b}$ and a few other nice properties, where $e$ is defined as above.

Sorry if that isn't rigorous enough; I really don't know much about complex analysis either :blush:[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=equation+functional&t=86119 In my post here I outlined how to find all continuous solutions to the functional equation $e^{a}e^{b}= e^{a+b}$.

[quote="t0rajir0u"][quote="jb05"]the integral for ln is not a [b]definition[/b]).[/quote]

I'd say it's the only sensible way to define real-valued exponentiation.[/quote]
Hmm, t0r, your statement seems too strong to me.  For example, can't we define natural logarithms of postive numbers by
\[\ln x = \lim_{n \to \infty}n(x^{1/n}-1) \, , \]
and then define $e^{x}$ as the inverse function?  Alternatively, we can define $\ln$ as the inverse of the exponential function, where the exponential function is defined by
\[e^{x}= \lim_{n \to \infty}{(1+\frac{x}{n})}^{n}\, , \]
or equivalently by the power series written in your first post.  Probably I misunderstood what you were trying to say.

[quote="Ravi B"]Alternatively, we can define $\ln$ as the inverse of the exponential function, where the exponential function is defined by
\[e^{x}= \lim_{n \to \infty}{(1+\frac{x}{n})}^{n}\, , \][/quote]

Hmm.  I considered this approach, but couldn't figure out how to prove any properties this way  :P

You're right, of course, that other definitions can be just as rigorous, but I think the integral definition is the one that leads most readily to a proof of common properties.  Perhaps I'm mistaken.

[quote="t0rajir0u"]You're right, of course, that other definitions can be just as rigorous, but I think the integral definition is the one that leads most readily to a proof of common properties.  Perhaps I'm mistaken.[/quote]

That's the way I always understood it too...

That's basically what I was trying to establish: that the two definitions of e were indeed equivalent.

[quote="gauss202"]I think some people are uncomfortable with the idea that in mathematics things can be defined in more than one way. But you shouldn't be uncomfortable with this idea, its actually a very handy thing.

The real work comes in proving that all three of the definitions given for e are indeed equivalent. None of them are really any more "natural" than the others. Once you've established that they are equivalent (not an easy thing to do), then using whichever one is most convenient for the task is perfectly logical.[/quote]

I agree with you completely.  But even though one isn't more "natural" than another, there are some that can't be derived without another.  That was the only point I was trying to make...

[quote="DanK"]But even though one isn't more "natural" than another, there are some that can't be derived without another. [/quote]

I actually don't think this is true.  For example, the definition

$e^{x}= \lim_{n \to \infty}\left( 1+\frac{x}{n}\right)^{n}$

Is surprisingly workable, since

$\frac{d}{dx}e^{x}= \lim_{n \to \infty}n \left( 1+\frac{x}{n}\right)^{n-1}\cdot \frac{1}{n}= e^{x}$

And

$e^{a}e^{b}= \lim_{n \to \infty}\left( 1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^{2}}\right)^{n}= \lim_{n \to \infty}\left( 1+\frac{a+b}{n}\right)^{n}= e^{a+b}$

Not completely rigorous, but you get the idea.

Hmm, maybe.  e's a really interesting number in a lot of ways...


MATHCOUNTS
summer program
Mathcamp
\/closed

We need more active moderators in MATHCOUNTS.

There has been a lot of spam in there lately.

I agree. We had people who made multiple accounts and spammed by making non-Mathcounts related topics that were somewhat inappropriate...

I wholeheartedly agree.  The forum is not what it used to be...  or maybe this is what happens every summer.

No. Last summer, it was extremely clean. Now a days, since most of the mods are out of 8th, they don't visit as much. We need younger mods.

By extremely clean, I mean I only saw like 2 spam topics the whole summer last year, and those were promptly deleted.

I haven't ever seen a mod do something in the MathCounts forum except for Kyuanmathcount (I think i spelled it wrong).

Yes. Same here. solafidefarms used to do stuff like 2 years ago, but now that he's in like 11th (10th?) grade, I don't think he visits MC anymore.

It hasn't been this bad for some time.

i dont think there has been a mod in mathcounts for a long time... its definitely time to make new ones.

Kyyuanmathcount does pretty well...but he/she's only one person... :(

I volunteer also. I really havent seen much of the mods expect once or twice in MC.

Uhh... Guys that is not how mods are chosen.
[url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_forums_tutorial#Becoming_a_Moderator]Click[/url]

But to be safe, I volunteer too.

so will I.

but guys, its really up to the admins whether there are going to be new mods or not.

I hope they do ... it's really out of control there.  Also, more of you should practice the principle of "don't feed the trolls" -- when you respond to spam, you're probably just going to encourage more of it.

Hmm... well I think mods in the Mathcounts forum should be middle schoolers and active competitors, because they will probably look more in the forum (obviously, as they are part of the competition) than high schoolers.

Edit: Actually, I think there are way better candidates for mods than me, so I won't volunteer.

Where are the mods? Seriously? Now there's jimmy sun, spamming MC. What is seriously up?

This is what me and everybody else in this thread mean:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161482[/url]

Also, no one post at the thread i just gave you...

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=901693#901693

I'll volunteer to be a moderator, too.

You all can probably stop volunteering now  :wink:   It appears the administrators have noticed your plea and are taking some action, including some well-deserved bannings.

Ban [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=32044]this guy[/url] for posting [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=901794#901794]this[/url]. I mean ban his IP.

As to where the mods are, kyyuanmathcount is in Texas (AMP) and will return home tomorrow, LadyKn1ght is in Maine (MC), I believe that ends very soon, nebula42 is now inactive.

I'm not inactive, I just haven't posted much recently.  :P
(I went to 3 camps / events this summer, so that might be the reason you haven't seen me much.)
I have been trying to check the forum occasionally and delete spam, but there really is a lot of it...  :|

So guys, try to avoid making spammy posts to make the forum better (and make the job easier for the moderators  :) ).
For example, when responding to a problem, don't just post an answer, post your solution (or at least your work / ideas) as well.
Also, if your post isn't longer than one or two lines, don't post it unless you're sure it's not spam.

Another thing: if you haven't done so already, it would definitely be a good idea to read the
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19760]MATHCOUNTS Forum Guidelines[/url].

Hopefully things will get better soon, especially after the admins have banned several troublesome spammers.

Sorry for being inactive, I've been at Mathcamp the last 4 weeks (one week is left).

Yes, what Archimedes1 said. Ban the IP cause he's made multiple accounts saying that he's some random person.

Unfortunently you cant ban this IP.

Ill even tell you what my IP is now

71.183.34.54 go ahead ban it..

it doesnt work on me.

Well it looks like a lot of the mods will be back at the end of the summer, so maybe we can just pull through the summer without.

[quote="#H34N1"]It looks like Anthony Cai is using a proxy, either Tor, or some online thing[/quote]

Or he might just be using a dynamic IP address.

I was away too and now I've been busy working on my math meet. Sorry for being absent.

[And actually I'm not as inactive as you think; complain to me in a PM and posts get deleted.]


analytic geometry
geometry
trigonometry
rectangle
probability

24. Let $S$ be the set of all points $(x, y)$ in the coordinate plane such that ${0}\le{x}\le{\frac{\pi}{2}}$ and ${0}\le{y}\le{\frac{\pi}{2}}$. What is the area of the subset of $S$ for which ${\sin^2{x}-\sin{x}\sin{y}+\sin^2{y}}\le{\frac{3}{4}}$?

I hope this is in the right forum.  :)

I was one step away from solving this on the actual thing.   :(

(Sorry, I wrote this in a hurry :blush: )
[hide="(edited) Solution (not the official one, but way cooler)"]
Multiplying by $4\sin x +4 \sin y$, we obtain

$4\sin^3 x + 4\sin^3 y \le 3\sin x + 3 \sin y$

Recall that $\sin (3x) = 3 \sin x - 4 \sin^3 x$.  We can then rewrite

$0 \le \sin 3x + \sin 3y$
$\sin 3x \ge - \sin 3y$

Graph this however you want; the way I did it on the test, from $0 \le x, y \le \frac{\pi}{3}$ both $\sin 3x$ and $\sin 3y$ are positive, so this area is definitely included into our summation, and has area $\frac{\pi^2}{9}$.

Then from $\frac{\pi}{3} \le x, y \le \frac{\pi}{2}$ both $\sin 3x$ and $\sin 3y$ are negative, so this area is definitely not included in our summation.

There are two rectangles of undetermined signage left over where $\sin 3x, \sin 3y$ have opposite signs.  This area is symmetrical, since probability distributions are symmetric, so half of this remaining area is included in our summation, which has area $\frac{\pi^2}{18}$.

Summing, we find our total area is $\boxed{ \frac{\pi^2}{6} }$.
[/hide]

I think you mean $\frac{\pi^2}{8}$ because $\frac{\pi^2}{16}$ is not a choice and is a quarter of the area. But, thanks t0rajir0u!! :D

There are definitely a number of typos in that solution, although it looks like it might work out anyhow -- in particular, after multiplying through you created a whole bunch of differences where there should be sums.

Pre-Olympiad is too hard for this problem (Intermediate would be better) but it's already been posted anyway; in fact the whole test can be found [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2006]here[/url] and click the $\boxed {24}$ to get to #24.


geometry
inequalities
logarithms
geometric transformation
complex analysis
geometry theorems

Inversion is a strong tool to solve some kinds of problems. But I don't know very much about this method. Can everyone post some interesting problems about this ?
Thanks

There is an article of Kin Li in Math.Excalibur:

http://www.math.ust.hk/excalibur/
http://www.math.ust.hk/excalibur/v9_n2.pdf

In fact, inversion as a geometric transformation and as a method of solving problems is discussed very nicely by Alexander Bogomolny on the website http://www.cut-the-knot.org/Curriculum/Geometry/SymmetryInCircle.shtml and some other pages which are linked there.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=56658 and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70654]Kiran Kedlaya's Geometry Unbound[/url].

  Darij

Check out the papers here:
[url=http://www.math.utah.edu/mathcircle/notes/01-02.html]www.math.utah.edu/mathcircle/notes/01-02.html[/url]

i thinks before that u should learn about it and its theorems.i think if we post the theorems of inversion,it will help

u better. ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=51865

these links are some problems about inversion.

but i only could find my proofs for inversion. :oops: i will try and find the other inversion problems. ;)

 http://www.mathlinks.ro/Forum/viewtopic.php?t=63420

http://www.mathlinks.ro/Forum/viewtopic.php?t=68589

 http://www.mathlinks.ro/Forum/viewtopic.php?t=66365

 http://www.mathlinks.ro/Forum/viewtopic.php?t=74481

 http://www.mathlinks.ro/Forum/viewtopic.php?t=54036

 http://www.mathlinks.ro/Forum/viewtopic.php?t=75187

 http://www.mathlinks.ro/Forum/viewtopic.php?t=39186

 http://www.mathlinks.ro/Forum/viewtopic.php?t=76632

 http://www.mathlinks.ro/Forum/viewtopic.php?t=28928

 http://www.mathlinks.ro/Forum/viewtopic.php?t=60787

Some problems to be solved by invesion:

1. Circles s1 and s2 are tangent externally in point K and tangent internally to the circle S in respectively A1 and A2. A line through K tangent to both circles s1 and s2 intersects S in point P. The line PA1 intersects s1 in point B1 and the line PA2 intersects s2 in point B2. Prove that line B1B2 is tangent to both s1 and s2. (it can be done without inversion, but inversion is very quick here)

2. Prove Ptolemy's inequality ($AB*CD+BC*AD \geq AC*BD$ in a quadrilateral) using inversion.

3. Given points A,B and a circle O construct a circle passing through A and B and tangent to O.

4. Given a point A and cirlces O1, O2 construct a circle passing through A and tangent to both O1 and O2.

What do  you want to know about it? The definition of inversion is that if $C$ is an $n$-dimensional ball $\mathbb{R}^{n}$ with radius $r$ and center $O$ and $P$ a point in $\mathbb{R}^{n}$, then the inversion point of $P$ with respect to $C$ is a point $P'$ in ray $OP$ such that $OP\cdot OP'=r^{2}$. Please ask more specific questions if you need more information.

the most important fomula of invension is
If $A'=I_{k}(A),B'=I_{k}(B)$ then
$A'B'=\frac{kAB}{IA.IB}$

I would like to see some examples about inversion in olimpic problems.And the properties about quaht the inversion do with circles, lines, intersecting circles...

thanks

circles, tangents, etc. : very good representation using inversion

[quote="boxedexe"]Logarithms is to Algebra as Inversion is to Geometry.[/quote]

Haha.. doesn't tell you how to use Inversion though..

[quote="weiquan"]Hmm.. coming to think of it, what about Mobius transformations? What are some useful properties of this mapping? I have not been able to find a nice book or website on this :([/quote]

What are [i]Mobius Transformations[/i]?

[quote="weiquan"][quote="sunchips"]What are [i]Mobius Transformations[/i]?[/quote]

Haha.. it is a superset of inversion, defined over the complex plane. Can anyone help?[/quote]

Hmm... does anyone know of any resources I can obtain on using Mobius Transformations in geometry? I really cannot find anything other than what they are :(

There are loads by grobber, for example these:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=106650
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=97487
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=41258
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=34049
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=32379
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=invert&t=23756

The best description of Möbius transformations I've ever read is in the book [i]Visual Complex Analysis[/i], of Tristan Needham. It's not that elementary, though. I don't know whether this transformations have applications in elementary geometry... :maybe:

If it not already sent, I think this is also a nice pdf text on inversion !

 http://www.imomath.com/tekstkut/inversion_ddj.pdf

[u] Babis[/u]

Do there a book that explain about inversion easily to understand ?

 :)  :)  :)

Try [i]Problems in Plane Geometry[/i] by I.F.Sharygin. It has a short, but instructive, appendix on inversion.


Let
$P(x)=\Sigma^{3}_{i=1}{a_{i}x^{p_{i}}(x+1)^{q_{i}}}$
where a_i are real and p_i,q_i are positive integers
Prove that P(x)=1 have at most 14 positive roots

It is false. For example if $p_{i}+q_{i}<14 \ \forall i$ then $deg P(x)<14$.

I dont think there is anything wrong with that >You arre asked to prove that it has $at most$ 14 positive roots.Possibly you have misinterpreted it as $at least$.

[quote="Skyward_Sea"]Nobody? So hard?[/quote]
I wouldn't call it hard, but awfully tedious.

First note, that if $T(x)=x^{m}(x+1)^{n}f(x)$ with $m,n\in\mathbb{Z}$ and $f\in\mathbb{R}[x]$ of degree $d$,
then $T'(x)=x^{m-1}(x+1)^{n-1}g(x)$ with
$g(x)=m(x+1)f(x)+nxf(x)+x(x+1)f'(x)\in\mathbb{R}[x]$ of degree $\leq d+1$.

Now assume $P(x)-1=\sum_{i=1}^{3}a_{i}x^{p_{i}}(x+1)^{q_{i}}-1$ has $15$ positive roots for some $p_{i},q_{i}\in\mathbb{Z},a_{i}\in\mathbb{R},i=1,2,3$.
Then $P'(x)$ has $14$ positive roots, using the mean value theorem (MVT), and
$P'(x)=\sum_{i=1}^{3}x^{p_{i}-1}(x+1)^{q_{i}-1}\alpha_{i}(x)$  for some $\alpha_{i}\in\mathbb{R}[x]$ of degree $\leq 1$ for $i=1,2,3$.
Set $r_{i}=p_{i}-p_{3},s_{i}=q_{i}-q_{3}\in\mathbb{Z}$ for $i=1,2$.
Then $Q(x): ={P'(x)\over x^{p_{3}-1}(x+1)^{q_{3}-1}}=\sum_{i=1}^{2}x^{r_{i}}(x+1)^{s_{i}}\alpha_{i}(x)+\alpha_{3}(x)$ has also $14$ positive roots,
and $Q''(x)$ has $12$ positive roots, again using MVT.
$Q''(x)=\sum_{i=1}^{2}x^{r_{i}-2}(x+1)^{s_{i}-2}\beta_{i}(x)$ for some $\beta_{i}\in\mathbb{R}[x]$ of degree $\leq 3$ for $i=1,2$.
Set $t=r_{1}-r_{2},u=s_{1}-s_{2}\in\mathbb{Z}$.
Then $R(x): ={Q''(x)\over x^{r_{2}-2}(x+1)^{s_{2}-2}}=x^{t}(x+1)^{u}\beta_{1}(x)+\beta_{2}(x)$ has also $12$ positive roots,
and $R^{(4)}(x)$ has $8$ positive roots, using MVT again.
$R^{(4)}(x)=x^{t-4}(x+1)^{u-4}\gamma(x)$ for some $\gamma\in\mathbb{R}[x]$ of degree $\leq 7$.
But now $\gamma$ has also $8$ positive roots.

This gives $\gamma\equiv 0\equiv R^{(4)}(x)$ and $R\in\mathbb{R}[x]$ of degree $\leq 3$ with $12$ positive roots,
so $R\equiv 0\equiv Q''(x)$ and $Q\in\mathbb{R}[x]$ of degree $\leq 1$ with $14$ positive roots,
so $Q\equiv 0\equiv P'(x)$ and $P(x)-1$ is constant with $15$ positive roots,
and so finally $P\equiv 1$.

For example $P(x)=(x+1)^{2}-x^{2}-2x\equiv 1$ or $P(x)=x^{2}+(x+1)-x(x+1)\equiv 1$.

If $p_{i}\in\mathbb{N}$ for $i=1,2,3$, then $P(0)=0$ and $P\not\equiv 1$.


superior algebra
superior algebra unsolved

Let $ S$ be a set with atleast two elements and let $ G$ be a group acting transitively on $ S$. For $ x \in G$ let $ f(x)$ denote the number of elements of $ S$ that are fixed by $ x$.  Prove that $ \displaystyle \sum_{x \in G} f(x) \equal{} \ G$ where
$ |G|$ denotes the cardinality of $ G$

Consider the set $ T \equal{} \{(x,s) \in G \times S \; : \: x \cdot s \equal{} s \}$. If we fix x and sum, we get $ |T| \equal{} \sum_{x \in G} f(x)$. If we fix $ s$ and sum, we get $ |T| \equal{} \sum_{s \in S} |Stab_G(s)| \equal{} \sum_{s \in S} \frac{|G|}{|Orb_{G}(s)} \equal{} \sum_{s \in S} \frac{|G|}{|S|} \equal{} |G|$.
Thus $ \sum_{x \in G} f(x) \equal{} |T| \equal{} |G|$.

This is also just the Burnside Lemma in the case that the action has 1 orbit  :wink:


Let a, b, c be non-negative real numbers such that $abc=1$

Prove that :

$\frac{ac+ab^{2}}{a+b}+\frac{ba+bc^{2}}{b+c}+\frac{cb+ca^{2}}{c+a}\geq \frac{9}{a^{2}+b^{2}+c^{2}}$

$\sum_{cyc}\frac{ac+ab^{2}}{a+b}=\sum_{cyc}\frac{ac^{2}+b}{ac+bc}\geq \frac{(\sum_{cyc}\sqrt{ac^{2}+b})^{2}}{2(ab+bc+ac)}$ (by Cauchy).
It's enough to prove that \[\frac{(\sum_{cyc}\sqrt{ac^{2}+b})^{2}}{2(ab+bc+ac)}\geq \frac{9}{a^{2}+b^{2}+c^{2}}\]
It's obvious that $\frac{1}{ab+bc+ac}\geq \frac{1}{a^{2}+b^{2}+c^{2}}$ ,so it's enough to prove that $\sum_{cyc}\sqrt{ac^{2}+b}\geq 3\sqrt{2}$ which it's true because $\sum_{cyc}\sqrt{\frac{ac^{2}+b}{2}}\geq \sum_{cyc}\sqrt[4]{abc^{2}}\geq 3\sqrt[12]{a^{4}b^{4}c^{4}}=3$ using AM-GM and the fact that $abc=1$

her is an other prouff (i think) i’ve found this solution .let  $s=\sum_{cyc}\frac{ac+ab^{2}}{a+b}$  now and by cauchy’s formula we have
$2(a+b+c)s\geq(\sum_{cyc}\sqrt{ac+ab^{2}})^{2}$, we claim that via $AM-GM$ we have  $\sum_{cyc}\sqrt{ac+ab^{2}}\geq\sum_{cyc}\sqrt{2}\sqrt[4]{ab}\geq3\sqrt{2}$ so we get
$2(a+b+c)s\geq(3\sqrt{2})^{2}=18$  it coms than $(a+b+c)s\geq9$.($1$)
now and since $a+b+c\geq3$ (by $AM-GM$) , and from  $a^{2}+1\geq2a$  we’ve got
$a^{2}+b^{2}+c^{2}\geq2(a+b+c)-3\geq(a+b+c)$ so $a^{2}+b^{2}+c^{2}\geq(a+b+c)$ so
$(a^{2}+b^{2}+c^{2})s\geq(a+b+c)s\geq9$ see ($1$)  and we are done
Vive $Maroc$.


quadratics
number theory proposed
number theory

Let p is prime, suth that $p=2^{l+1}m+1,l\ge 1$, m - is odd.
Prove, that equation \[2=x^{2^{l}}(mod \ p)\] has not solution (2 is not $2^{l}$ degree residue). For examle l=1 give p=5(mod 8) and 2 is not quadratic residue.

hallo,
i think i have one solution.
we have $\frac{p-1}{2}=2^{l}m$. so from fermats little theorem we get, if $(x,p)=1$ (otherwise we have $x \equiv 0 \bmod{p}$):
$x^{2^{l}m}\equiv \pm 1 \not\equiv 2 \bmod{p}$

I am sorry this problem is not correct for l>1. I want nesessary condition for prime $p=1+2^{l+1}m, m-odd$, when exist solution:
$2=x^{2^{k}},k=1,2,..,l+1$.


calculus
integration
algebra
partial fractions
calculus computations

If you want an exact indefinite integral, you're going to have to find the zeros of that irreducible cubic (all three are real). After that, you use partial fractions, and the answer will be a very ugly linear combination of logarithms.

If you want to find a numerical estimate of the definite integral on some interval which does not contain any of the zeros of the denominator, any standard method of numerical integration should work.

[quote="eplag"]here's a question i'm having a big problem solving :(.  I did some research and someone told me i had to use a trapazoidal rule because this is a specialized integral. how would i go on solving this :?

$\int\frac{1}{x^{3} - 4x^{2} - 3x + 8}dx$[/quote]

eplag are u sure its not
$x^{3} - 4x^{2}$ [b]+[/b] $3x + 8$  :?

i can't seem to find the zeros


calculus
integration
logarithms
real analysis
real analysis unsolved

Calcutate: $ \int_{0}^{\plus{}\infty}{t^{\mu}e^{\minus{}\alpha t}\ln t \;dt},\; \alpha>0,\;\mu>\minus{}1.$

Note that $ \Gamma'(s) \equal{} \int_{0}^{\infty} t^{s} e^{\minus{}t} \log t \, dt$ by Leibniz integral rule.


I had to this problem in a math competition a couple days ago and we had 2 minutes and 30 seconds to solve it, I could solve it, but I'm wondering what is the fastest way.

The Problem:
100 people went to the party, the entrance fee for males is $ \$5$, the entrance fee for women is $ \$1$, and the entrance fee for children is 5 cents. If $ \$100$ were collected at the party, how many of each went(If at least one of each went)?

[hide="The answer"]
There were 19 men, 1 women, and 80 children.
[/hide]

If you know any sort of tricks for this problem please tell me.

First, I believe that you would write the equations:

Let x=#of men
Let y=# of women
Let z=# of children
$ 100\equal{} ($5)(x)+($ 1)(y)\plus{}($0.05)(z)
x+y+z=100
x,y,z>1

Then you would use reasoning to deduce that if at least one of each type came, z, the number of children has to be a multiple of 20.

Let us say that z is 20.
We now know that x+y=80

Substitute into the first equation to get that:
100=5(x)+1(80-x)+(.05)20
Solve to get x=19/4

X is not a whole number, so we will increment z by 20.

If z=40, then x+y=60
Substitute:
100=5(x)+1(60-x)+(.05)40...NOPE

z=60 then x+y=40
Substitute:
100=5(x)+1(40-x)+(.05)60...NOPE

z=80 then x+y=20
Substitute:
100=5(x)+1(20-x)+(.05)80...x=19

If x=19, then y=1

Therefore 19 men...1 woman...and 80 children attended the party.

120 SECONDS FLAT  :lol:

Well, I just guessed this problem, you know that the answer is in full dollar amount and since there have to be babies. The babies must come in group of 20s. You know that the money female and babies pay must be divisible by 5. You also know that the numbewr of male is less than 20. It just made sense that i should guess the number of the babies from the top, so either 60 or 80)
And you would come to a conclusion of 19 male, 1 female and 80 babies


Given $ a, b, c > 0$. Prove that: $ (1\plus{}a\plus{}b\plus{}c)(1\plus{}ab\plus{}bc\plus{}ca) \ge 4\sqrt{2}\sqrt{(a\plus{}bc)(b\plus{}ca)(c\plus{}ab)}$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ (1 \plus{} a \plus{} b \plus{} c)(1 \plus{} ab \plus{} bc \plus{} ca) \ge 4\sqrt {2}\sqrt {(a \plus{} bc)(b \plus{} ca)(c \plus{} ab)}$
 :)[/quote]
Note that
$ (1 \plus{} a \plus{} b \plus{} c)(1 \plus{} ab \plus{} bc \plus{} ca) \equal{} (1 \plus{} a)(1 \plus{} b)(1 \plus{} c) \plus{} (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 2\sqrt {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}.$
Therefore, it suffices to prove that
$ 8(a \plus{} bc)(b \plus{} ca)(c \plus{} ab) \le (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a).$
By the AM-GM Inequality, we have
$ 2\sqrt {(a \plus{} bc)(b \plus{} ca)} \le a \plus{} bc \plus{} b \plus{} ca \equal{} (c\plus{} 1)(a \plus{} b),$
$ 2\sqrt {(b \plus{} ca)(c \plus{} ab)} \le (a \plus{} 1)(b \plus{} c),$
$ 2\sqrt {(c \plus{} ab)(a \plus{} bc)} \le (b \plus{} 1)(c \plus{} a).$
Multiplying these inequalities, we get the result.

Yes, that's also the way I created it  :oops:

Let  $ a, b, c > 0$. Prove that$$ (a+b+c)(1+ab+bc+ca)\geq \sqrt{6(a+b)(b+c)(c+a)}$$


function
real analysis
topology
real analysis unsolved

[quote="vertigo74"]Let $ U,V$ be (open or closed) intervals in [b]R[/b], and let $ f: U \minus{} > V$ be a function which is strictly increasing and onto. Prove that $ f$ and $ f^{ \minus{} 1}$ are continuous.

I think I should be appealing to a theorem that says for every open set in V, if the inverse image of that set is open as well, then f is continuous. I still cannot come up with a good way to use that theorem.[/quote]

I think the fact that the function is onto is of critical importance in your proof, if you remove that condition then the statement is false. From Rudin: if f is a monotonically increasing function, then the left and right-side limits exist at every point. This might help too.


AMC
AMC 10
AMC 12
geometry
quadratics
logarithms
analytic geometry

I would like to propose an AMC problem marathon. Eligible problems must either be from the AMC10 or from the first 15 problems of the AMC12 (as per the High School Basics description).

Hopefully this will allow newer members to get some practice with solving as well as with some common methods of solving AMC problems.

Problem 1:
A circle passes through the three vertices of an isosceles triangle with side length 3 and base length 2. What is the area of this circle?

I suggest that you change the name into AMC Marathon so they don't get another idea and know this is a marathon :)

Area of a circle in which an isosceles triangle with side length 3 and base 2 is 81pi/32.

Don't forget to post a new problem, but i see you are new so that's ok.
Rules are post a solution and poast a new problem! :)
thanks

[hide="New Problem"]Let $ 4^a\equal{}5, 5^b\equal{}6, 6^c\equal{}7,$ and $ 7^d\equal{}8$ Find $ abcd$[/hide]

Please do not rely on looking up problems and/or memming them.

[quote="AwesomeToad"]Let $ 4^a \equal{} 5, 5^b \equal{} 6, 6^c \equal{} 7,$ and $ 7^d \equal{} 8$ Find $ abcd$[/quote]
$ 4^a \equal{} 5$, so applying log to both sides we get $ log_4(4^a) \equal{} log_4(5)$, or $ a \equal{} log_4(5)$. By the logarithmic change of base identity, this becomes $ a \equal{} log(5)/log(4)$.

Similarly we get $ b \equal{} log(6)/log(5)$, $ c \equal{} log(7)/log(6)$, and $ d \equal{} log(8)/log(7)$.

Now \[ abcd \equal{} (log(5)/log(4))(log(6)/log(5))(log(7)/log(6))(log(8)/log(7))\]
which simplifies nicely to $ log(8)/log(4)$.

As for another problem, here's one:

----
If in applying the quadratic formula to a quadratic equation

\[ f(x) \equiv ax^2 \plus{} bx \plus{} c \equal{} 0,\]

it happens that $ c \equal{} \frac{b^2}{4a}$, then the graph of $ y \equal{} f(x)$ will certainly:

$ \mathrm{(A) \ have\ a\ maximum  } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$ $ \mathrm{(C) \ be\ tangent\ to\ the\ x\minus{}axis} \qquad$ $ \mathrm{(D) \ be\ tangent\ to\ the\ y\minus{}axis} \qquad$ $ \mathrm{(E) \ lie\ in\ one\ quadrant\ only}$

I have a question; would log8/log 4 simplify to 3/2 if we change it to base 2?

[quote="dwks"]
As for another problem, here's one:

----
If in applying the quadratic formula to a quadratic equation
\[ f(x) \equiv ax^2 \plus{} bx \plus{} c \equal{} 0,\]
it happens that $ c \equal{} \frac {b^2}{4a}$, then the graph of $ y \equal{} f(x)$ will certainly:

$ \mathrm{(A) \ have\ a\ maximum } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$ $ \mathrm{(C) \ be\ tangent\ to\ the\ x \minus{} axis} \qquad$ $ \mathrm{(D) \ be\ tangent\ to\ the\ y \minus{} axis} \qquad$ $ \mathrm{(E) \ lie\ in\ one\ quadrant\ only}$[/quote]

Because the $ y$ coordinate of the vertex is obtained by using the formula $ y \equal{} c \minus{} \frac {b^2}{4a}$, we get that the y-coordinate of the vertex is 0, and so the graph is tangent to the x-axis.

[hide="NP"]
Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
(2002 AMC12A #10)
[/hide]

should we post the problems or solutions in this page?

You post a solution to the previous problem, then post your new problem which must be AMC10 or first 15 questions of AMC12

Question: Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
(2002 AMC12A #10)

Assume both cups have 24t 24t amount of coffee and cream respectively. Then by pouring half coffee (i.e. 12t into other cup, it has now 12t coffee and 24t cream).  Now half of second cup will have 6t coffee and 12t cream and that added to 12t coffee will now have 18t coffee and 12t cream.  So the ratio of cream is 12/30 = 4/10 = 2/5th.

Here is a new problem:
Find the largest divisor of 1001001001 that does not exceed 10000.

[quote="KirthiRaman"]Find the largest divisor of 1001001001 that does not exceed 10000.[/quote]KirthiRaman, this is an ARML problem.

Are ARML problems also welcomed in this marathon? You guys decide.

All participants of this marathon should post the source of your problem (AMC or AHSME, year and problem #). That way, we know where the problem comes from and give credit to the respective organizations for creating wonderful problems.

Now I understand that there are defined rules here.

Please disregard the question.  Here is an appropriate AMC question
If a,b,c, and d are integers such that a<2b, b<3c, and c<4d. If d<100, the largest possible value for a is?

(a) 2367
(b) 2375
(c) 2391
(d) 2399
(e) 2400

The largest possible value for $ d$ is $ 96$, so the largest possible value for $ c$ is $ 23$, so the largest possible value for $ b$ is $ 7$, so the largest possible value for $ a$ is ... 3?

EDIT: Sorry, I misread.

Why can't d have a value of 99?
Start from there.

The largest possible value of a is far from 3! At max:
d=99
so c=99*4-1=395
b=395*3-1184
a=1184*2-1=2367
A.

Remember teh rules and post a new problem.
Here's one:
The sum of two numbers is S. Suppose 3 is added to each number and then each
of the resulting numbers is doubled. What is the sum of the final two numbers?

(A) 2S + 3 (B) 3S + 2 (C) 3S + 6 (D) 2S + 6 (E) 2S + 12

Okay, so let $ a \plus{} b \equal{} s$.  Then, we have $ (a \plus{} 3) \plus{} (b \plus{} 3)$, or $ a \plus{} b \plus{} 6$.  Then, substitute $ a \plus{} b$ with $ s$.

Answer: $ \boxed{\textbf{D}}$

New Problem:  In trapezoid $ ABCD$ with bases $ AB$ and $ CD$, we have $ AB \equal{} 52$, $ BC \equal{} 12$, $ CD \equal{} 39$, and $ DA \equal{} 5$. Find the area of trapezoid $ ABCD$.

[hide=" "]513642[/hide]


trigonometry
function
analytic geometry
graphing lines
slope

I know that $ sin x\equal{}sin (180\minus{}x)$ , but I was wondering what cos (180-x), tan (180-x), cot (180-x), csc(180-x), sec(180-x) and all of those with (360-x) were. Thanks in advance.

Just use addition formulas, and that's it...
For example:

$ \cos(\pi\minus{}x)\equal{}\cos\pi\cos x\plus{}\sin\pi\sin x\equal{}\minus{}\cos x$

But I don't know the addition formulas. I was trying to learn these first, then the addition formulas later.

So, what are they?

The addition formulas or your specific questions? I will assume you mean your specific questions. Realize that you only need to fugure out the sine and the cosine of each, since every other one of the functions can be derived from them.

$ \sin(180 \minus{} x) \equal{} \sin x$

$ \cos(180 \minus{} x) \equal{} \minus{} \cos x$

$ \sin(360 \minus{} x) \equal{} \minus{} \sin x$

$ \cos(360 \minus{} x) \equal{} \cos x$

I now draw your attention to something you may already know, but can be seen from the last two equations. If your familiar with the unit circle, you'll recognize that $ 0^{\circ}$ and $ 360^{\circ}$ are essentially "equivalent" (there is a term that describes it, but I can't think of it). So, I can rewrite the last two in this way:

$ \sin( \minus{} x) \equal{} \minus{} \sin x$

$ \cos( \minus{} x) \equal{} \cos x$

The addition formulas are overkill for these. Think visually. What do $ \sin{\theta}$ and $ \cos{\theta}$ represent visually? What does $ \tan{\theta}$ represent visually?

[hide]Hopefully you visualize $ \sin{\theta}$ and $ \cos{\theta}$ as the y-coordinate and x-coordinate, respectively, of a point on the unit circle at an angle $ \theta$ from the positive x-axis. And you can visualize $ \tan{\theta}$ as the slope of the ray pointing at an angle $ \theta$ from the positive x-axis.[/hide]
Now, if you can visualize where $ 180^o\minus{}\theta$ is in relation to $ \theta$, you can visualize how $ \sin{(180^o\minus{}\theta)}$, $ \cos{(180^o\minus{}\theta)}$, and $ \tan{(180^o\minus{}\theta)}$ relate to the values of $ \sin{\theta}$, $ \cos{\theta}$, and $ \tan{\theta}$.

The corresponding rules for $ \csc{(180\minus{}\theta)}$, $ \sec{(180\minus{}\theta)}$, and $ \cot{(180\minus{}\theta)}$ follow naturally once the previous three are understood.


geometry
cyclic quadrilateral
projective geometry
geometry unsolved

From a 2007 handout by Titu Andreescu, Ivan Borsenco, and Zach Abel:

Let ABCD be a quadrilateral and let E be the intersection of BC and AD, F be the intersection of AB and CD.  Consider O inside the quadrilateral ABCD and let the angles AOE and COF be equal.  Prove that <AOB + <COD = 180 degrees.

Sorry for my bad English.

Look at this figure, we can prove that the two statements about angles $ \iff$ P,Q,R,S are on a circle.

But $ O$ is an important point.

See here : http://bbs.cnool.net/topic_show.jsp?id=4538104&thesisid=494&flag=topic1

From cyclic quadrilateral $ OLEJ, OKFI, OLAI, OKCJ$ we have:
$ \angle LAO\equal{}\angle LIO, \angle LEO\equal{}\angle LJO$
$ \angle OCK\equal{}\angle OJK, \angle OFK\equal{}\angle OIK$
But $ \angle LAO\minus{}\angle LEO\equal{}\angle AOE\equal{}\angle COF\equal{}\angle OCK\minus{}\angle OFK$
$ \Rightarrow \angle LIO\minus{}\angle LJO\equal{}\angle OJK\minus{}\angle OIK$
$ \Rightarrow \angle LIO\plus{}\angle OIK\equal{}\angle OJK\plus{}\angle LJO$ or $ \angle LIK\equal{}\angle LJK$
So $ LIJK$ is cyclic quadrilateral.
$ \Rightarrow \angle LIJ\plus{}\angle LKJ\equal{}180^o$ or $ \angle AIJ\plus{}\angle DIJ\plus{}\angle LKB\plus{}\angle JKC\equal{}180^o$
$ \Leftrightarrow \angle AOL\plus{}\angle BOL\plus{}\angle DOJ\plus{}\angle COJ\equal{}180^o$ or $ \angle AOB\plus{}\angle COD\equal{}180^o$.

[quote="livetolovemath030894"]Dear kyyuanmathcount. So is The condition'' $ O$ inside $ ABCD$ '' unreasonable to stay together??  :)
 Sorry for my bad english  :)[/quote]

If I understand correctly, I think that it is possible to have O inside ABCD if we look at the diagrams in the last two post.

Thank you everyone for the solutions.  I am just wondering what is the motivation for dropping these four perpendiculars from O?

Also, is there another solution to this with projective geometry (that is the name of the handout from which I got this problem)?

Thanks


trigonometry
LaTeX
calculus
calculus computations

Determine for which $ x\in R$ the following partial sums are bounded.

$ \sum_{k \equal{} 1}^{n}\cos kx$ , $ \sum_{k \equal{} 1}^{n}\sin kx$.

Help me to solve this question.

Case 1 $ x\not \equiv 0(\mod 2\pi)$
$ \sum_{i \equal{} 1}^n\sin(kx) \equal{} \frac {x\sin(\frac {n \plus{} 1}{2})\sin(\frac {nx}{2})}{sin(\frac {x}{2})}$
Case 2 : $ x\equiv 0(\mod 2\pi)$ then $ \sum_{i \equal{} 1}^n\sin(nx) \equal{} 0$ .

1)$ \sum_{k \equal{} 1}^{n}\frac {\cos kx\sin\frac {x}{2}}{\sin\frac {x}{2}} \equal{} \sum_{k \equal{} 1}^{n}\frac {\sin(\frac {x}{2} \plus{} kx) \plus{} \sin(\frac {x}{2} \minus{} kx)}{2\sin\frac {x}{2}}$ .For $ x \equal{} 2\pi l$ , $ l\in Z$ the answer tends to infinity,right?How we can find that it is bounded for all $ x\neq 2\pi l$

2)$ \sum_{k \equal{} 1}^{n}\frac {\sin kx\sin\frac {x}{2}}{sin\frac {x}{2}} \equal{} \sum_{k \equal{} 1}^{n}\frac {\cos(\frac {x}{2} \plus{} kx) \minus{} \cos(\frac {x}{2} \minus{} kx)}{2\sin\frac {x}{2}}$ How did you find its sum?

Use \sin and \cos in your LaTeX, please.  (You'll have to add an extra space: \sinx gives nothing, while \sin x gives $ \sin x$.)

We focus on the cosinus case (the sinus case can be handled in analogy). Denote $ S_n(x)\equal{}\sum_{k\equal{}1}^n\cos kx$ and $ S_0(x)\equal{}0$

Case 1: $ x\in 2\pi\mathbb{Z}$

We trivially have $ S_n(x)\equal{}n$ is unbounded for $ n\rightarrow\infty$

Case 2: $ x\in 2\pi\mathbb{Q}\setminus2\pi\mathbb{Z}$

Write $ x\equal{}2\pi\frac{N}{M}$ with $ M>1,N\not\equal{}0$ and $ \gcd(N,M)\equal{}1$. First note, that the identity

$ S_M(x)\equal{}Re\left[\sum_{k\equal{}1}^M\exp\left(2\pi ik\frac{N}{M}\right)\right]\equal{}0$ holds. Further observe the property $ (k'\minus{}k)x\in 2\pi\mathbb{Z}\Longleftrightarrow k'\equiv k\bmod{M}$.

Decompose an arbitrary $ n\equal{}uM\plus{}r$ with $ 0\leq r<M$. Then the preparations imply $ S_n(x)\equal{}uS_M(x)\plus{}S_r(x)\equal{}S_r(x)$, i.e. $ S_n(x)$ assumes values in the finite set $ \{S_0(x),S_1(x),\ldots,S_{M\minus{}1}(x)\}$, in particular, it is bounded for $ n\rightarrow\infty$.

Case 3: $ x\in 2\pi\mathbb{R}\setminus2\pi\mathbb{Q}$

Direct calculation yields (note, that $ \sin x\not\equal{}0$ holds):

$ S_n(x)\equal{}\frac{1}{2\sin x}\sum_{k\equal{}1}^n[\sin(k\plus{}1)x\minus{}\sin (k\minus{}1)x]\equal{}\frac{\sin(n\plus{}1)x \plus{}\sin nx\minus{} \sin x}{2\sin x}$, i.e. $ |S_n(x)|\leq\frac{2\plus{}|\sin x|}{2|\sin x|}$ and $ S_n(x)$ is bounded for $ n\rightarrow\infty$.


Simplify for positive integer $ n>1$:
\[ \frac{(n\plus{}1)!}{n(n\minus{}1)!}\]

By definition (n-1)! is the product of the integers from 1 to n-1. Because this is multiplied by n, the bottom is equivalent to n!. The top is the product of the integers from 1 to n+1, so it can be written as n+1 multiplied by the product of the integers from 1 to n, or (n+1)n!, so our expression becomes $ \frac{(n\plus{}1)n!}{n!}\equal{}\boxed{n\plus{}1}$.


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

Well, I think this problem is pretty well-known...but I can't solve it...so....5 people participate at a conference...each one of them fall asleep 2 times. Anyway we choose two persons there is a moment when they were both asleep.Proove that there is a moment when three persons were asleep...

I wouldn't exactly call it a combinatorics problem, but rather a kind of Helly's theorem for a special non-convex figures (assumed to be intervals in this case). Anyway, this could be a possible road for a generalization to 2-D or to n-D. The problem is that I am so sick of problem solving that I don't even want to try a proof.

[[b]EDIT:[/b] The problem is [url=http://www.kalva.demon.co.uk/usa/usoln/usol862.html]15th USAMO 1986 problem 2[/url].]

  Darij

I think I got it :

Let $a,b,c,d,e$ be the persons.
For each of these 5 persons, there are two intervals of time during which this person is asleep. This gives 20 endpoints for these intervals. (1)

Now, suppose, for a contradiction, that no three persons are asleep together at some moment. (2)
Note that, it follows that we may assume that the endpoints are pairwise distincts.
Since any two persons are asleep together at some moment, it follows that the intervals associated to $a$ contains at least 4 endpoints (one for each other person). Moreover, from (2), if some endpoint from (say) $b$ is contained in an interval associated to $a$, then this endpoint cannot belong to any other interval.
The same is true for the others, so that all the intervals contain at least 20 endpoints.
Therefore, from (1), the intervals associated to $a$ must contain a total of exactly 4 endpoints, exactly one for each other person. In particular, no interval associated to $a$ can contain an interval associated to any other person (otherwise, it would contain two endpoint from this person).
The same is true for the other persons. It follows that each endpoint $p$ is contained exactly once in an other interval $I$ (which means $p$ is not defined as an endpoint from $I$).

But, considering these intervals along a line, the leftmost endpoint cannot belong to any other interval (and the same is true for the rightmost endpoint). A contradiction.

Pierre.

[quote="pbornsztein"]Note that, it follows that we may assume that the endpoints are pairwise distincts.[/quote]

I'm sorry but I don't understand why...  :?  Could you explain?

Thanks!
  Darij

We cannot have three intervals with a common endpoint otherwise at this moment three persons would be asleep which we have refused.
Therefore, if two intervals have a common endpoint, say $[x,y]$ and $[y,z]$ (as noticed, we cannot have $[x,y]$ and $[x,z]$ otherwise one of the interval would contained in the other, which is impossible from the above post). Then, we may choose $e>0$ sufficientely small and consider the intervals $[x,y+e]$ and $[y-e,z]$ without overlapping any other intervals (there are a finite number of intervals) and without change of situation.

Pierre.