vector
linear algebra
matrix
Putnam
modular arithmetic
least common multiple
number theory

Hey,

This problem is from the VTRMC 2006.

3. Recall that the Fibonacci numbers $ F(n)$ are defined by $ F(0) \equal{} 0$, $ F(1) \equal{} 1$ and $ F(n) \equal{} F(n \minus{} 1) \plus{} F(n \minus{} 2)$ for $ n \geq 2$. Determine the last digit of $ F(2006)$ (e.g. the last digit of 2006 is 6).

As, I and a friend were working on this we noticed an interesting relationship when writing the Fibonacci numbers in "mod" notation.

Consider the following,

01 = 1 mod 10
01 = 1 mod 10
02 = 2 mod 10
03 = 3 mod 10
05 = 5 mod 10
08 = 6 mod 10
13 = 3 mod 10
21 = 1 mod 10
34 = 4 mod 10
55 = 5 mod 10
89 = 9 mod 10

Now, consider that between the first appearance and second apperance of $ 5 mod 10$, there is a difference of five terms. Following from this we see that the third appearance of $ 5 mod 10$ occurs at a difference 10 terms from the second appearance. Following this pattern we can create the following relationships.

$ F(55) \equal{} F(05) \plus{} 5({2}^{2})$

This is pretty much as far as we got, any ideas?

The Fibonacci numbers mod $ 5$ have period $ 20$, with the zeros coming every fifth term. The Fibonacci numbers mod $ 2$ have period $ 3$, with even numbers every third term. Combine the two, and you get $ 5$ mod $ 10$ whenever $ n$ is $ 5$ or $ 10$ mod $ 15$. Also, the period of the Fibonacci sequence mod $ 10$ is $ 60$.

$ F_{2006}\equiv F_{26}\equiv \minus{}F_{\minus{}4}\equiv \minus{}3\equiv 7$.

Note that there are only $ 100$ possible distinct ordered pairs $ (F_{2k}, F_{2k\plus{}1}) \bmod 10$, and note further that each ordered pair uniquely determines the one after it; after at most $ 100$ terms, then, the Fibonacci sequence must repeat $ \bmod 10$.  jmerry has demonstrated an idea that you can use to calculate a number $ P$ such that the period of the Fibonacci sequence $ \bmod n$ divides $ P$.

If we consider the Fibonacci numbers mod $ n$ for any $ n,$ they must be periodic. The longest possible period would be $ n^2 \minus{} 1,$ but the only $ n$ for which that maximum period is realized are $ n \equal{} 2$ and $ n \equal{} 3.$

The best way to explain this is to consider the sequence of $ 2$-vectors given by $ v_k \equal{} \begin{bmatrix}f_k \\
f_{k \plus{} 1}\end{bmatrix}.$ We then have that the Fibonacci recursion is a one-step linear recursion $ v_{k \plus{} 1} \equal{} Mv_k,$ where $ M \equal{} \begin{bmatrix}0 & 1 \\
1 & 1\end{bmatrix}.$ Note that since $ M$ is invertible and its inverse also has only integer entries, then the recursion can be run backwards just as easily as it can be run forwards.

If we work mod $ n,$ then there are exactly $ n^2$ possible states of the $ 2$-vector. [This is what t0rajir0u said.] However, one of those states is the zero vector; the zero vector can only give rise to more zero vectors, and since $ 1$ is a Fibonacci number, that's not what happens. Hence there are at most $ n^2 \minus{} 1$ states that can occur. After $ n^2 \minus{} 1$ repetitions - or earlier - we arrive at a state we've been at before; since subsequent iterations depend only on the current state of the vector, we must repeat ourselves. Since the iteration can be run backwards as well as forwards, we have that $ f_k$ is periodic mod $ n$ as $ k$ ranges from $ \minus{} \infty$ to $ \infty.$

For a composite $ n$ like $ 10,$ it makes sense to consider what happens with its prime factors.

Mod $ 2,$ the Fibonacci numbers have period $ 3.$ (The longest possible; $ 3 \equal{} 2^2 \minus{} 1.$) They go $ 0,1,1,0,1,1,0,1,1,\dots.$

Mod $ 5,$ the period turns out to be $ 20.$ I'll show the repeating block with some extra spaces to make it easier to read and see what is going on:

$ \begin{matrix}0,1,1,2,3, & 0,3,3,1,4, & 0,4,4,3,2, & 0,2,2,4,1\end{matrix}$

Then by the Chinese remainder theorem, the period mod $ 10$ is the least common multiple of $ 3$ and $ 20,$ namely $ 60$ (which is what jmerry said).

Although APSStudent's post doesn't immediately have anything to do with the Putnam, there was a recent Putnam problem which did involve a forward-and-backward periodic mod $ n$ linear recursion.

Let me post this now and look for the link - back in a few minutes.

Ah ... very recently. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122447]Putnam 2006 A3[/url].

With some number theory and knowledge of finite fields, we can say more about the periods, particularly mod a prime $ p$. $ p\equal{}2$ and $ p\equal{}5$ have already been mentioned; they're exceptions to the following rules. If $ p>2$ and $ p\equiv 1$ or $ p\equiv 4$ mod 5, the length $ l(p)$ of the period mod $ p$ divides $ p\minus{}1$. If $ p>2$ and $ p\equiv 2$ or $ p\equiv 3$, $ l(p)$ divides $ 2(p\plus{}1)$, and $ \frac{2(p\plus{}1)}{l(p)}$ is odd.

Isn't $ F_{2006}\equiv 3\pmod{10}?$

$ 2006\equiv6\pmod{20},$ so $ F_{2006}\equiv F_6\equiv 3\pmod{5}.$

$ 2006\equiv2\pmod{3},$ so $ F_{2006}\equiv F_2\equiv 1\pmod{2}.$

Ah, I see jmerry's mistake. He's correct up to $ \minus{}F_{\minus{}4},$ but $ F_{\minus{}4}\equal{}\minus{}3,$ so $ \minus{}F_{\minus{}4}\equal{}3.$

My friend, last year, actually found out the last digits of the first twenty or thirty fibonacci numbers, and this isn't hard at all. After that, the digits will repeat. And that's how she found $ F(2006)$!

OK, except it's sixty, not "twenty or thirty." Doing that on a computer, using a spreadsheet program, is quick and painless.  Doing that by hand requires doing 60 or more additions (mod 10) with no mistakes, which is riskier.

That's why several people above worked mod 2 and mod 5 rather than mod 10.

Yes, you are right, it's sixty! I realized "twenty or thirty" wasn't right after I submitted the post, and I was too lazy to edit the same. I'm sorry about that "mistake."

Recall that the Fibonacci numbers $F(n)$ are defined by $ F(0) = 0$, $ F(1) = 1$ and $ F(n) = F(n - 1) + F(n - 2)$ for $ n \geq 2$. Determine the last digit of $F(2006)$.


geometry
geometric transformation
reflection

The original problem states:

Some checkers placed on an $ n\times n$ checkerboard satisfy the following conditions:

a) Every square without a checker on it shares a side with a square with a checker on it.
b) Given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.

Prove that at least $ \frac {n^2 \minus{} 2}{3}$ checkers have been placed on the board.

Generalization 1: What is the minimum number of checkers that satisfy the conditions? The given bound is pretty weak.

Generalization 2:
Is the desired statement true if we remove condition b? (My testing suggests that it might be...) Or even stronger - what is the minimum number of checkers?

Generalization 2 seems false.

[code].X..
...X
X...
..X.[/code]

This pattern tesselated, particularly in $ 4k \times 4k$ boards, gives a configuration substantially below the given bound unless the board is small.

Still thinking about the first one.

If it just has the first restriction, I get the result that the minimum is:(n-1)/2 when n is odd and n/2 when n is even.Is it correct? :roll:

@ MellowMelon: Could you find the minimum if condition b is left out? It should tend to about $ \frac {n}{5}$ as $ n$ gets large, but this is more like an upper bound for the minimum.

@ ghjk: The minimum for a 5x5 is at most 11 :wink:

[code]
x O x O x
x O x O x
x O x O x
x O x O x
x O O O x
[/code]

Just glancing at your table, I think you're wrong aspiringchesslete! Since if I'm not mistaken, O means the square which is painted black and X is the square which is not painted(white). Because you have two X lines adjacently, your table must be wrong!
Here is my diagram:
0 X 0 X 0
X 0 X 0 X
0 X 0 X 0
X 0 X 0 X
0 X 0 X 0
There are 12 squares being painted on this chessboard. Now, you can easily prove that if O<=11, you won't have any ways to paint this table by examining the least number of painted squares in 2 consecutive rows in horizontal line must be >=5 :)

But each x still has an O beside it! (And mine also satisfies the second restriction!)

The second restriction part was just extra.

Note that each x in my diagram has an O beside it!!

[quote="ghjk"][quote="aspiringchesslete"]The original problem states:

Some checkers placed on an $ n\times n$ checkerboard satisfy the following conditions:

a) Every square without a checker on it shares a side with a square with a checker on it.
[/quote]
I don't mention the 2nd restriction! Do you really interpret this sentence carefully? It means:"Form both vertical lines and horizontal lines, each square must satisfy it" . Your diagram doesn't  reflect that, so how can it be true?[/quote]
Oh my goodness! You should read my statement carefully, not just glancing! I said:"Both vertical and horizontal lines" not just horizontal lines :huh:

ghjk, you are not interpreting the problem correctly. ;) Any square without a checker must be adjacent to [b]at least one[/b] square with a checker. If this is satisfied, it does not matter if two empty squares are adjacent to each other.

Some checkers placed on an n\times n checkerboard satisfy the following conditions:

a) Every square without a checker on it shares a side with a square with a checker on it.

Really? So why doesn't this sentence have:"at least"? If it's at least as you say, I will do it again because it leads to completely different conclusion(obviously not as easy as it was)

MellowMelon is right. (Sorry, I wasn't sure what I had explained wrong, but I see now)

So, given any square, as long as there is (at least!) ONE adjacent square with a checker on it, you are fine. (At least I think this was the interpretation on the actual USAMO...)

In general, the word "a" implies "at least one."

So, you could rewrite the sentence as:

Every square without at least one checker shares at least one side with at least one adjacent square with at least one checker on it.

Much wordier, and more annoying, but it means the same thing.


Find all function $f: R^+\to R^+$ s.t.:
$f(x)f(yf(x))=f(x+y)$
Here $R^+$ does not contain zero

I assume you mean x,y are positive reals in the functional equation.

===========

Suppose some $x$ has $f(x) = c > 1$.  Then try $y=\frac{x}{c-1}$ and $f(x)f(yf(x)) = f(x+y)$ becomes $(c-1)(f(\frac{cx}{c-1})) = 0$, contradiction.
---
In the following, we consider that the image of $f$ is $(0,1]$.

Suppose $f(x) = 1$ for some $x$.  It implies $f(y) = f(x+y)$ ; ie., $f(x)$ is periodic with period $x$.

Suppose $f(x) = f(x+y)$, ie., $f$ is not injective.  It implies $f(yf(x)) = 1$, so that $f(x)$ is periodic, with period $yf(x)$.

Now consider all functions with image in $(0,1)$ that are injective.  Then $f(x) > f(x+y)$ and $f(yf(x)) > f(x+y)$.  The first implies that $f$ is strictly decreasing; the second implies $yf(x) > x + y$.  It implies $y(f(x) - 1) > x$, which cannot be true.

Now consider all functions in general with image in $(0,1)$.  It must be not injective- from before we see it must be periodic, say with period $p$.  Let $f(p) = k$.  Then $f(p) f(pf(p)) = f(p + p) \Rightarrow kf(pk) = k \Rightarrow f(pk) = 1$, contradiction.

So we must have $f(p) = 1$ for some $p$.  Then $p$ is also the period of $f$.  Finally, set $y = p$.  Then $f(pf(x)) = 1$ for all $x$.  So take any $x$ with $f(x) = m < 1$.  Then $f(pm) = 1$ implies that $pm < p$ is a period of $f$, whereby we can make our period arbitrarily small.  Contradiction.  So $f(x) = 1$ always.

[quote="Singular"]
Now consider all functions with image in $(0,1)$ that are injective.  Then $f(x) > f(x+y)$ and $f(yf(x)) > f(x+y)$.  The first implies that $f$ is strictly decreasing; the second implies $yf(x) > x + y$.  It implies $y(f(x) - 1) > x$, which cannot be true.
[/quote]
How can it comes true ? :?  $f(x).f(yf(x)=f(x+y)$ that implies $f(yf(x)>f(x+y)$ but $f(x)$ is decrease so we get $yf(x)<x+y$ which is trivial

This looks like an old IMO question, not a British one.

$f(x)=\frac{1}{cx+1}$ is another solution ( where c is a constant )

Here is my solution for the case $0<f(x)<1 \forall x$
We have $f(x)f(y)=f(x)f(\frac{y}{f(x)}f(x))=f(x+\frac{y}{f(x)})$
and $f(x)f(y)=f(y)f(\frac{x}{f(y)}f(y))=f(y+\frac{x}{f(y)})$
so $f(x+\frac{y}{f(x)})=f(y+\frac{x}{f(y)})$
but $f(x)$ is strictly decrease so $x+\frac{y}{f(x)}=y+\frac{x}{f(y)}$
Put $y=1$ we get $x+\frac{1}{f(x)}=1+\frac{x}{f(1)}$ that implies $f(x)=\frac{1}{cx+1}$ with $c=\frac{1}{f(1)}-1>0=const$

It is really well-know.The answer is$f(x)=\frac{a}{x-ax+a}$ where a=f(1).
Case 0<f(x)<1 can be treated as follow:put a=f(1) so
$f(x+1)=af(ax)=f(ax+x+1-ax)=f(ax)f(f(ax)(x+1-ax))$=>$a=f(f(ax)(x+1-ax))$.Since f is strictly decresing we have f(ax)(x+1-ax)=1 then$f(x)=\frac{a}{x-ax+a}$
It would be interesting if we find $f: R*\to R*$ where $R*=R^{+}$ and 0.The answer is
a)$f(x)=\frac{a}{x-ax+a}$ where a=f(1)
b)f(x)=0 if $x\geq\ a$ $a\geq\ 0$
   $f(x)=\frac{a}{a-x}$ for other case

Yes ,dhthstn's solution is also my one... :D.And it's funny that i don't remember the relation exactly then I have solved another problem.But its solution is similar to the above problem:$f(x)f(xf(y))=f(x+y)$ has unique solution:$f(x)=1$ for all $x>0$


Let $ a$ and $ b$ be rational numbers such that  $ \frac{2}{2a\plus{}b}\equal{}\frac{2a}{b}\plus{}\frac{b}{2a}\minus{}1$ (here $ a\ne0, b\ne0, b\ne\minus{}2a$).  Prove that $ 1\minus{}2ab$ is a square of rational number.

[quote="rogue"]Let $ a$ and $ b$ be rational numbers such that  $ \frac {2}{2a \plus{} b} \equal{} \frac {2a}{b} \plus{} \frac {b}{2a} \minus{} 1$ (here $ a\ne0, b\ne0, b\ne \minus{} 2a$).  Prove that $ 1 \minus{} 2ab$ is a square of rational number.[/quote]
[hide="Solution"]
Let $ c \equal{} 2a$.  Then, we get that $ \frac {2}{c \plus{} b} \equal{} \frac {c}{b} \plus{} \frac {b}{c} \minus{} 1$, so $ \frac {2}{b \plus{} c} \equal{} \frac {b^2 \plus{} c^2 \minus{} bc}{bc}$.  This means that $ b^3 \plus{} c^3 \equal{} 2bc$, so $ (c \plus{} b)[(c \plus{} b)^2 \minus{} 3cb] \equal{} 2bc$.  Let $ x \equal{} b \plus{} c$ and $ ab \equal{} y$, so $ x$ and $ y$ are rational.  This then turns into $ x(x^2 \minus{} 3y) \equal{} 2y$, so $ y(3x \plus{} 2) \equal{} x^3$, which means that $ y \equal{} \frac {x^3}{3x \plus{} 2}$.  Yet, notice that $ b$ and $ c$ are the soultions of $ k^2 \minus{} kx \plus{} y$, so since $ b$ and $ c$ are rational, the discriminant of this equation is a rational square.  This means that $ x^2 \minus{} 4y$ is a rational square.  Plugging this into our equation for $ y$, we find that $ \frac {2x^2 \minus{} x^3}{3x \plus{} 2}$ is a rational square.  Factoring out rational square $ x^2$, we see that $ \frac {2 \minus{} x}{3x \plus{} 2}$ is a rational square.  Now, we compute $ 1 \minus{} 2ab \equal{} 1 \minus{} bc \equal{} 1 \minus{} y$.  This is $ \frac {3x \plus{} 2 \minus{} x^3}{3x \plus{} 2} \equal{} \frac {(x \plus{} 1)^2(2 \minus{} x)}{3x \plus{} 2}$.  Yet, $ \frac {2 \minus{} x}{3x \plus{} 2}$ is a rational square, so $ 1 \minus{} 2ab$ is the square of a rational number, as desired. [/hide]


Prove (or disprove) that if $ a$ and $ b$ are irrational numbers, $ a^{b}$ could be a rational number.

Do they have to be real? If not...

[hide]$ e^{i\pi}=-1$[/hide]

Yeah, both $ a$ and $ b$ are real numbers.

[hide="Quick"] $ \sqrt{2}^{ \log_{ \sqrt{2}}3}= 3$.

It remains only to prove that $ \log_{ \sqrt{2}}3$ is irrational.  Well, if it were rational (say $ \frac{p}{q}$) then

$ 9^{q}= 2^{p}$

Contradiction. [/hide]

I prefer this solution, which I believe I learned on AoPS once upon a time (do a search!):
[hide]Consider $ {\sqrt{2}}^{\sqrt{2}}$.  If it's rational, we're done.  If it's irrational, then $ \left({\sqrt{2}}^{\sqrt{2}}\right)^{\sqrt{2}}={\sqrt{2}}^{2}= 2$ is rational, and we're done.[/hide]

[quote="JBL"]If it's irrational, then $ {{\sqrt{2}}^{\sqrt{2}}}^{\sqrt{2}}={\sqrt{2}}^{2}= 2$[/quote]

I think you mean $ \left( \sqrt{2}^{\sqrt{2}}\right)^{ \sqrt{2}}$.  I was trying to remember this solution, but had forgotten it  :P

$ e^{\ln 2}= 2$

NOTE: Yes, I realize we need to show $ \ln 2$ is irrational. I am sure it has been proven in diophantine approximation. So I accepted it by faith.


They are having a face-off (no other team is competing), best 4 out of 7. Yankees is winning by a point after 3 matches. There were no ties.
a.) How many more matches maximum do they have to do in order to decide the victor?
b.) What is the probability that the Red Sox win if they had the same chance of winning?
c.) If the Red Sox are as twice as strong as the Yankees what is the probability of them winning?

OK this isn't related to your problem, but it reminds me of a real nice problem.. vaguely remember it may have come up at this forum ages ago, but I'll post it again here. Possibly a bit too hard for mathcounts, but hey its the same best of 7 thing :)

Two teams playing a best of 7, both have a 50-50 chance of winning each time.. is it more likely to end after 6 games or 7 games? (they don't play any more after a winner has been decided, ie someone gets to 4.)

SPOILER![hide]4. The most games would be if the teams alternated victories: so their win-loss record would go

2-1---2-2----3-2-----3-3-----4-3 and it's over. That is 4 games, not counting the record at the beginning.[/hide]

[hide]The redsox need to win 3 to win and the yankees need to win 2 to win, so the chance of the redsox winning is 1/4.[/hide]

[hide]If the redsox are twice as strong, than their probability of winning is 2/3 since they should split the next 3 games 2-1 redsox, making sieres 3-3. and the rest if obvious.[/hide]

Yup yup, Go Sox! Reverse the curse!

*after watching the final game*
...the curse is still there...


What do you like the most, or consider the most important, about Mathematics?

1) [b]Theory[/b]: Theorems, and there dependence on each other, whereon we build theories fascinates me, and is what true Mathematics is all about.
2) [b]Problem[/b]: My opinion is that mathematics is all about the problems and developing thinking-strategies in order to attack problems. The theorems which we use for this are of less interest.

Motivations and comments may be given

If you voted, also mention your favorite area in mathematics. Maybe some pattern appaers :)

You may have a biased sample here: this website is not called http://www.artoftheoremproving.com  :wink:

[url=http://www.dpmms.cam.ac.uk/~wtg10/2cultures.pdf]The Two Cultures of Mathematics[/url] by Gowers is a standard reading on this topic. I would not say that "theorems ... are of less interest" to me -- it's just that I tend to judge the importance of a particular theorem by considering the problems that is allows us to solve.

[b]Message to the moderators[/b]:

If I edited my original message something weird happend with the poll. Could somebody please be so kind to fix this for me and delete this reply.

Thanks!


Let $ a \ge b \ge c > 0$. Prove that:
$ 2a \plus{} 3b \plus{} 5c \minus{} \frac{8}{3}\left( {\sqrt {ab}  \plus{} \sqrt {bc}  \plus{} \sqrt {ca} } \right) \le \frac{1}{3}\left( {\frac{{a^2 }}{b} \plus{} \frac{{b^2 }}{c} \plus{} 4\frac{{c^2 }}{a}} \right)$
:lol:

You can easily regroup the thing as:
$ (\sqrt{a}\minus{}\sqrt{b})^2\left(\frac{(\sqrt{a}\plus{}\sqrt{b})^2}{b}\minus{}4\right)\plus{}(\sqrt{b}\minus{}\sqrt{c})^2\left(\frac{(\sqrt{b}\plus{}\sqrt{c})^2}{c}\minus{}4\right)\plus{}(\sqrt{a}\minus{}\sqrt{c})^2\left(\frac{4(\sqrt{c}\plus{}\sqrt{a})^2}{a}\minus{}4\right)\geq 0$

Ps. Amazing title !!  :roll:


Hello, some of you might have seen my posts on the middle school boards. My name is Brian, I'm from around Cleveland, and I'd like to add a little traffic to this forum. A little bit about me and competitive math: I'll be helping represent our beautiful state at this year's national MathCounts competition, I tied for first this year in the AMC-8 with 25 pts, I took the AMC-10 but don't really remember my score...(or I'm just too embarassed to say it. Take your pick.)

I'll try and get a couple other people to join here.

Congratulations! I might see you on April 16  and on April 23 in Columbus.

Can you tell me who is in the team and where they are from. Thanks.

Ah! So you are the Gyan who our coach invited to our practices! I'm sorry, I didn't make that connection right away.

I don't like to put a person's full name on a public forum without their permission, but I will say we have me (Brian) who goes to Hawken School in Cleveland, Henry who goes to...shoot I'm blanking out here...I'll figure it out later and edit this post...in Columbus, Ashvin at Sycamore in Cincinatti, and Tong at Mason in Cincinatti.

I'm looking forward to being formally introduced to you at our next practice. We had our first one today.

Brian

[i]Moderators Edit: I deleted Gyan's last name[/i]

Thanks. For the info.
I agree about not  putting the full name in the public forum, so you may lile to edit  your message (including my last name :)

Thanks. See you at the next practice. I  think both students from  Sycoamore went to National last year too.

Oh lord, I'm really sorry. :blush: I hate it when I contradict myself like that. Luckily this forum doesn't get much traffic, I guess :blush:


superior algebra
superior algebra solved

Let $X$ be a set which is different from $\varnothing.$ Let $P(X)$ be the set off all the subsets of $X.$ If we denote with $S$ the number of invertible elements from the monoids: $( P(X), \cap)$ and $(P(X), \cup)$ then compute $S$.

and also.. which are these invertible elements? :?

1

The only invertible element is the identity- the whole set for intersection, the empty set for union.

and also.. a stupid question.. what is the sum of these invertible elements?

okay I forgot to count.. the sume of the number of the invertible elements is 2...


We take a sideways photograph of the wheel of a cycle going past us. Since the exposition time of the camera is finite, the spokes of the wheel arent clear on the picture, they are washed. But there are some points on the picture which are not washed, we can see them clearly. Where are these points?

can we assume it moves without slipping??

Actually one could have asked more; where the spokes are clear [i]and where the tire is clear[/i].

@[b]Thaakisfox[/b]
I've seen this one before. Where did you get it? I have a solution for it
but I'll post it later.

[hide=" answer"]
see figure

[b]Spokes: [/b]all points with $\sin{\varphi}=\frac{r}{R}$
That is a circle with radius $R/2$ centered on the midpoint of line between the hub and the contact point with the ground.

[b]tires:[/b] The contact point with the ground and the opposite point on the top
[/hide]


[quote="mdk"]If each person, in a group of n people, is a friend of at least half the people in the group, then it is possible to seat the n people in a circle so that everyone sits next to friends only.[/quote]

[hide="Hint"]
1)Phrase it as a graph theory problem.
2)Your looking for a [b]hamiltonian cycle[/b]
3)Since each vertex has degree at least half the number of vertices it means it hamiltonian.
[/hide]

Listen, mdk, your behavior is passing from annoying to outrageous.  Not only has this problem been posted on the forum before, but it was posted on the forum less than a week ago, by [i]you[/i].

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=358840811&t=159829

(I can't believe I'm using my 7000th post on such nonsense.)


geometry
trigonometry
trig identities
Law of Sines

Could someone please help me with the following problem?  Thanks in advance!

[hide]The answer I got was $ E$. Draw perpendicular from $ E$ to $ AD$ and label intersection $ R$. Draw Perpendicular from $ C$ to $ AB$ and label intersection $ S$. Then $ (ER)(AD) \equal{} \frac{(CS)(AB)}{2}$. So $ \frac{AD}{AB} \equal{} \frac{CS}{2ER}$.

WLOG, let $ ER \equal{} 1$ and let $ CS \equal{} y$. Now we need to find $ \frac{y}{2}$. Then $ AD$ = $ 1 \plus{} \frac{\sqrt{3}}{3}$ (to see this, look at the 45-45 and the 30-60 triangles that make up $ AED$). Similarly, $ AB \equal{} y \plus{} y\sqrt{3}$ . Therefore, $ \frac{y^2(1\plus{}\sqrt{3})}{2} \equal{} \frac{\sqrt{3}}{3}$.

After some simplifying of radicals, we determine that $ y^2 \equal{} \frac{2}{\sqrt{3}}$, so $ y \equal{} \sqrt{\frac{2}{\sqrt{3}}}$, $ \frac{y}{2} \equal{} \sqrt{\frac{1}{2\sqrt{3}}}$ or $ \frac{1}{\sqrt[4]{12}}$.[/hide]

[hide="Law of Sines solution"]We have $ AE \cdot AD \equal{} 1/2 \cdot AC \cdot AB$.
Also, $ \frac {AE}{sin60} \equal{} \frac {AD}{sin 75}$ and $ \frac {AC}{sin 30} \equal{} \frac {AB}{sin 105}$
Combining the two, $ AD^2 \cdot \frac {sin 60}{sin 75} \equal{} AB^2 \cdot \frac {sin 30}{2 sin 105}$
So $ \frac {AD}{AB} \equal{} \sqrt {\frac {sin 30 \cdot sin 75}{2sin60 \cdot sin 105}}$, which simplifies to E.
[/hide]


trigonometry
quadratics
special factorizations
inequalities proposed
inequalities

Prove that:$3x^{2}+3y^{2}+z^{2}\geq 10$ where $x;y;z$ are real numbers satisfying $xy+yz+xz=5$

[hide]Use $x^{2}+y^{2}+z^{2}\geq 2xy\cos{A}+2yz\cos{B}+2zx\cos{C}$, where $A,B,C$ are three angles of a triangle.[/hide]

[quote="kunny"]$3x^{2}+3y^{2}+z^{2}-2(xy+yz+zx)$

$=3\left(x-\frac{y+z}{3}\right)^{2}+\frac{2}{3}(2y-z)^{2}\geq 0$[/quote]

ok...but how did you think of that...i see people completing the square in such random ways [or at least seemingly so], and i would like to know how i could think of such manipulations

Regard this equation as the quadratic equation with respect to $x.$
Rearrange this for $x.$ :)

[quote="kunny"]$3x^{2}+3y^{2}+z^{2}-2(xy+yz+zx)$

$=3\left(x-\frac{y+z}{3}\right)^{2}+\frac{2}{3}(2y-z)^{2}\geq 0$[/quote]
Nice solution  :wink: But with problem follow, your method not work

[b]Problem.[/b] Let be given positive reals $x,y,z$. Find minimum of $xa^{2}+yb^{2}+zc^{2}$, where $a,b,c$ are positive reals such that $ab+bc+ca=1$.

The conclusion is equivalent to $3x^{2}+3y^{2}+z^{2}\ge 2(xy+yz+zx)$This is equivalent to $(x-y)^{2}+(x\sqrt{2}-z\frac{1}{\sqrt{2}})^{2}+(y\sqrt{2}-z\frac{1}{\sqrt{2}})^{2}\ge 0$ which is true


I have a question in the form, "I have a question about Problem __, are we required to prove..."

Are these questions allowed to be posted? If yes, then I will reply with my question.

No, unfortunately clarification is not allowed to be given on problems during the contest.


There is supposed to be a "rate this blog" thing on everyone's blog (except one's own, of course!). Where is this, please?

[quote="mathnerd314"]There is supposed to be [...]. Where is this, please?[/quote]

This is a free web site, whose operators owe nothing to the users (outside of the paid-for classes).  If they ever get around to adding such a blog feature, wonderful, but it is hardly a matter of promises or entitlements.

A less (actually or potentially) offensive phrasing of the question would be "has it been implemented" or "if that has been installed, how do I use it?", or something like that.

At this moment this feature is not implemented. It might be in the future.

The blog rating is for the blogs listed under the Blogs tab near the top of the page (on the AoPS view); it is not just for the blogs that are part of the Forum.


geometry
projective geometry
geometry solved

Give an example of a finite projective plane which does not satisfy Desargues theorem.

As i remember, Desargues holds iff the plane is a plane over some field (well, over some skew field, which is the same thing in finite case). There are some finite projective planes not over fields, their constructions are crazy and may be found in web.

Can you post some "crazy" constructions?

The least of them has $9^2+9+1=91$ points. A conjecture is that any projective plane of $p^2+p+1$ points for prime $p$ is over field. See

http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html

or

http://www.math.virginia.edu/~ww9c/katie.pdf

(the last link contains constructions of 4 projective planes of order 9)


[quote=alex2008]Let be $ a,b,c\in \mathbb{R}_\plus{}$ such that $ a\plus{}b\plus{}c\equal{}2005$ . Find the minimum value of :

$ E\equal{}a^{2006}\plus{}b^{2006}\plus{}c^{2006}\plus{}$ $ \frac{(ab)^{2004}\plus{}(bc)^{2004}\plus{}(ca)^{2004}}{(abc)^{2004}}$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1308813p7005030]Moldova national olympiad 2006, grade 9, pr.1[/url]


mishe be man yek ketab ya jozveye khoob moareffi konid ke dar morede moadelate tabeii ya chand jomleii bashe va ham dars va ham masalee bashe chon man ketabe aghaye MEHDI SAFA kar mikonam vali khoob tozih nadadan age mishe yek ketabe khoob moareffi konid ke khoob tozih dade bashe va mesal haye kheili khafan va khob ham dashte bashee!!!!

Mitooni az marjae kamele ryzai baraye dars va az 101masale jabr fatemi baraye masale estefadeh koni.vali bazam behtarin keteb jabr be nazaram ketabe mehdi safae. :D

Tamae tavabe haghighie $f$ ra tori biabid ke :
$xf(y)+yf(x)=(x^{2}+y^{2})f(xy)$


algebra
polynomial
superior algebra
superior algebra solved

$A,B\in M_{3}(K)$ where K can be Q,R,C

Prove that

 $2det(A+xB)=2det(A)+\ ( det(A+B)-det(A-B)-2det(B) \ )x+$
$\ ( det(A+B)+det(A-B)-2det(A) \ )x^2+2det(B)x^3$

[quote="Moubinool"]where K can be Q,R,C[/quote]
No need to be that restrictive: the argument I'm about to give will work for any $K$ of characteristic not 2, and in characteristic 2, the theorem is trivially true (both sides being zero.)

Let $P(x)=2\det(A+xB) = \alpha+\beta x +\gamma x^2 +\delta x^3.$

By direct examination of the expression for determinant, the only way we can get $x^3$ is to use only elements of $B.$ Hence, $\delta = 2\det(B).$

Now evaluate the polynomial at $0, 1, -1$:

$P(0)=2\det(A)$. Hence, $\alpha=2\det(A).$
$P(1)=2\det(A+B).$
$P(-1)=2\det(A-B).$

Using our knowledge of $\alpha$ and $\delta,$ the last two equations become

$\beta+\gamma=2\det(A+B)-2\det(A)-2\det(B).$
$\beta-\gamma=-2\det(A-B)+2\det(A)-2\det(B).$

These equations can be easily solved (assuming the characteristic of the field is not 2) to give the coefficients asked for.

Yep, same sol here. Only..
[quote="Kent Merryfield"]Let $P(x)=\det(A+xB) = \alpha+\beta x +\gamma x^2 +\delta x^3.$[/quote]--> I think you meant $2\cdot\det(A+xB)$ (just for the detail)


Each of the numbers 1 to 10^6 is repeatedly replaced by its digital sum until we reach 10^6 one digit numbers. Will these have more 1's or 2's?

[hide]Let $ S(x)$ denote the digital sum of $ x$.  Since $ S(x)\equiv x\pmod 9$, the list of one-digit numbers will simply be the one-digit positive number that $ x$ is congruent to in $ \mod 9$.  Since $ 9|999\text{,999}$, the list from 1 to 999,999 contains each digit from 1 to 9 an equal amount of times.  The 1,000,000 adds another 1, so there are more 1s.[/hide]


function
real analysis
real analysis unsolved

Let $f: I\to\mathbb{R}$ be a function that has  primitives and \[f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}\] holds for every $x,y\in I$($I$ is an interval). Prove that $f$ is continuous.

I'd say $I$ must be an interval like $(a,b)$(ask a question about language,how do I call the two points of the start and end of an interval?),otherwise $f$ doesn't need to be continuous at the start and the end of the interval.
And what do you mean by "has primitives"?(Sorry for my English about maths)

The points are called endpoints, but $I$ can be any interval. A function has primitives if there exists a function $F$ such that $F'=f$, the function $F$ is called a primitive of $f$.

Harazi posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=primitive+convex&t=45524]here[/url] a rather ugly in appearance proof.


Let A,B\in M_n(C) s.t. AB+BA=O_n and detA=0.
Prove that det(A^3-B^3)=0.

Strange

A=0, B=I_n verify the statement and
det(A^3 -B^3) = det(-I_n)=/=0

sorry moubi.. this is the problem like i have it.. maybe we should try to solve it if A,B=/=O_n. I couldn't solve it.. but the official solution uses that A,B=/=0_n

Even if A,B =/= to 0_n and I_n
Here an example of such A,B  in  M_2(C)

AB+BA = 0, detA = 0

and det(A^3 - B^3) =/=0

See just above Posted: Sat Mar 06, 2004 1:45 am


function
geometry
calculus
calculus computations

Find the function f (x) such that for all L ≥ 0, the area under the graph of
y = f (x) and above the x-axis from x = 0 to x = L equals the arc length of
the graph from x = 0 to x = L.

The relevant ODE is $ \sqrt {1 \plus{} \left( f'\right )^2} \equal{} f$.  One family of solutions is $ f(x) \equal{} \cosh (x \plus{} c)$ for each $ c\in \mathbb{R}$. $ f(x)\equiv 1$ is also a solution.


I do not know if this is the correct forum to post this question so mods, please feel free to move this post when I posted this on the wrong forum.

I have another problem.

I have a set data where [b] X = a + b + c + d [/b]

Using a spreadsheet, I was able to derive the following values

[img]http://www.smhs-alumni.net/smhspics/chart.gif[/img]

Can you please tell me if there is an equation I can use to derive the value of a, b, c and d given a particular value of X?

Thank you very much.

Ive been trying to solve this for several days now but I couldn't

The outputs were generated by an algorithm which I do not know but Iam assured that produces very regular and predictable results.

I couldn't find what is regular and predictable in the following data set. Is somebody else here trying?


Let $ a_1,a_2,a_3,\dots$ be infinite sequence of positive integers satisfying the following conditon: for each prime number $ p$, there are only finite number of positive integers $ i$ such that $ p|a_i$. Prove that that sequence contains a sub-sequence $ a_{i_1},a_{i_2},a_{i_3},\dots$, with $ 1 \le i_1<i_2<i_3<\dots$, such that for each $ m \ne n$, $ \gcd(a_{i_m},a_{i_n})\equal{}1$.

If $ a_{i_1}$ is arbitrary fixed, choose $ a_{i_2}$ s.t. $ i_2: \equal{}\min\{y \in \mathbb{N}_0: \text{lpf}(a_{y})>\text{gpf}(a_{i_1})\}$, so also with $ a_{i_3}$ s.t. $ i_3: \equal{}\min\{z \in \mathbb{N}_0: \text{lpf}(a_{z})>\text{gpf}(a_{i_2})\}$, etc. []


integration
calculus
calculus computations

$ \int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}} cos(t) dt \equal{}[sin(t)]_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}} \equal{}2$

but if we put $ x\equal{}cos (t)$ it will be:
$ \int_{0}^{0} ... \equal{}0$

The problem is that the integrand $ \pm\frac x{\sqrt{1\minus{}x^2}}dx$ doesn't give you an integral of a closed differential form because of the sign change at $ x\equal{}1$, so you cannot reduce the integration path that goes from $ 0$ to $ 1$ to $ 0$ to just one point. Generally speaking, you should be quite careful when making non-monotone changes of variables.

[quote="fedja"]Generally speaking, you should be quite careful when making non-monotone changes of variables.[/quote]

So all changes of variables should be monotone?
Either those of the kind $ t\equal{}cos x$ and the kind $ x\equal{}cos t$

Here's another one:

$ \int_{ \minus{} 1}^{1} x^2 \equal{} 2/3$

But puting $ t \equal{} x^2$ it's 0

You can make a non-monotone change of variable sometimes but, as I said, then you should understand quite well what you are doing. So, just stick to the monotone mappings for now to avoid trouble (otherwise, you will end up "proving" that $ \int_a^b f(x)\,dx\equal{}0$ whenever $ f(a)\equal{}f(b)$).


There was a person who said his school has AMC 12 on Monday next week!! I don't think we should discuss problems until then.

I'm sorry to post this here but AMC forum was locked..so..

AMC12 doesn't count unless it's done today..

[quote="xxazurewrathxx"]Yeah, it'll be safe to discuss it tomorrow when aops is unlocked.[/quote]

Although I'm not sure that I want too.  :P  The other people are right, the AMC 12 must be administered today, so either the school is just doing it for fun on the Monday or they made a costly mistake.

Yeah, it's too bad for the person, I suppose, if they did make a costly mistake. I would feel really bad if I took the test only to find that it didn't count.  :ninja:

That happened to some schools that got snowed out for the A... even ones near where I live.

[quote="davidyko"]Wait, so they didn't know they had to take it on a certain day?[/quote]

No, they went into the test knowing that it wouldn't count because it got postponed.  Sorry about the confusion.


Suppose $ ABCDEF$ is a hexagon  in a circle $ O$.
Let $ BD$ be perpendicular to $ CF$.

Let $ G \equal{} BD\cap CF, H \equal{} AC\cap BE, I \equal{} AD\cap CE$. Let $ AI \equal{} CI$. Prove that

$ CH \equal{} AH \plus{} DE$ if and only if $ GH\cdot BD \equal{} BC\cdot DE$

Dear indybar,
Can you introduce your proof?

can you tell specificly where this prblem is from and add a solution to it?


Which is the best of k for this ineq is true:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}k.\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge \frac{3}{2}\plus{}k$

[quote="mathstarofvn"]Which is the best of k for this ineq is true:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} k.\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \ge \frac {3}{2} \plus{} k$[/quote]
$ k \le \frac{{\sqrt 3  \minus{} 1}}{2}$ :maybe:


[quote="kim77"]A bit difficult.[/quote]
[hide="Solution"]
Extend $ BM$ and $ CN$ to meet the extension of $ AD$ at $ X$ and $ Y$; also let $ P$ be a point on the extension of $ CB$ and $ Q$ be a point on the extension of $ DA$.  Now, let\[ \angle MBA \equal{} \alpha\implies \angle XAB \equal{} 180 \minus{} \angle ABP \equal{} 180 \minus{} 2\alpha\implies \angle BXA \equal{} 180 \minus{} \angle XAB \minus{} \angle XBA \equal{} \alpha \equal{} \angle XBA\implies AB \equal{} XA\] Similarly, $ CD \equal{} DY$.  We also notice that \[ \angle AMB \equal{} 180 \minus{} (\angle MAB \plus{} \angle MBA) \equal{} 180 \minus{} (\frac {\angle BAQ \plus{} \angle ABP}{2}) \equal{} 180 \minus{} \frac {180}{2} \equal{} 90\]  Thus, $ AM\perp BX$.  Since $ \triangle BXA$ is isoceles and $ AM\perp BX$, we have that $ MB \equal{} MX$.  Similarly, $ NY \equal{} NC$.  Hence, $ MN$ is the median of trapezoid $ BCYX$, so \[ MN \equal{} \frac {BC \plus{} XY}{2} \equal{} \frac {BC \plus{} AD \plus{} AX \plus{} DY}{2} \equal{} \frac {BC \plus{} AD \plus{} AB \plus{} CD}{2}
\]which completes our proof. [/hide]

Okay...solved in about 15 seconds haha...

let $ MN$ intersect $ AB$ at $ O$...then $ \angle BMO\equal{}\angle MBO$...so $ MO\equal{}BO$...likewise $ MO\equal{}AO$...so $ O$ is the circumcenter of that triangle, so $ MO\equal{}\frac{1}{2}\cdot AB$..Then Let $ P$ be the intersection of $ CD$ and $ MN$. It follows that $ OP$ is a mid line of the trapezoid, and $ PN$ is half of $ CD$. Adding up the parts finishes the problem...

It is nice...but easy. These problems should be in an easier forum.

[quote 'let MN intersect AB at O...then \angle BMO=\angle MBO...so MO=BO']
Altheman how did you prove that those angles are equal...?

I wasn't being that careful...We can prove it by saying let O and P be the midpoints of AB and CD, respectively. Then we note that O, P are the circumcenters of those triangles, as well as the fact that BC // OP // AD then we can use easy angle chasing to show that angle MOB= angle POB...


inequalities
trigonometry
inequalities unsolved

In any triangle  ABC  show that
[tex] \Sigma\frac{bc}{p-a} \geq (5-\frac{2r}{R})\cdot p  [/tex]

[quote="nickolas"]In any triangle  ABC  show that
$ \Sigma\frac{bc}{p-a} \geq (5-\frac{2r}{R})\cdot p  $[/quote]

This inequality is equivalent to
$ \Sigma\tan(\frac{A}{2})^2 \  + 8sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2}) \geq\ 2 $ (*)

treegoner has a very splendid solution to (*).

And fuzzylogic's general inequality can prove (*) too.

I remember a topic posted in Open Problems Section "A new solution to Jack Garfunkel inequality" where Treegoner wrote

[quote]But I get trouble with this equality
            \sum tg^2(A/2)+8sin(A/2)sin(B/2)sin(C/2)>=2
       I think it's right because I have checked it many times.
    Bye[/quote]

[quote="tranminhhoang"][...] This inequality is equivalent to

$ \sum\tan^2\frac{A}{2}\  + 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \geq\ 2 $ (*)

treegoner has a very splendid solution to (*). [/quote]

Thanks, Hoang for pointing out this.

[quote="manlio"]I remember a topic posted in Open Problems Section "A new solution to Jack Garfunkel inequality" where Treegoner wrote [...]
[/quote]

Thanks manlio, for pointing out the source of (*) at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3849

I have trouble to read non-LaTeXed post. But I got the general idea. To recap  treegoner's idea, (*) can be proved from the following:

[quote]Let $A, B, C$ be the angles of a triangle. Then for any reals $x, y, z$,

$x^2 + y^2 + z^2 \geq 2yz \cos{A} + 2zx \cos {B} + 2xy \cos{C}$.----(**)[/quote]

By taking $x=\tan\frac{A}{2}$, etc. in (**), (*) is reduced to an identity:

$2\sum \tan\frac{B}{2}\tan\frac{C}{2}\cos{A} + 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = 2$ which isn't very hard to verify.


Let $P(x)$ be a polynomail with integer coefficients that satisfies $P(17)=10$ and $P(24)=17$.  Given that $P(n)=n+3$ has two distinct integer solutions $n_{1}$ and $n_{2}$, find the product $n_{1}n_{2}$.

It's saying $P(x)$ and $x+3$ intersect at 2 distinct points in which $x$ is an integer.

[quote="junggi"]but if $P(n)=n+3$ then $P(17)=17+3=20$ :maybe:[/quote]

No, you mistunderstood - that's the [u]equation[/u], not the identity.

[hide="Solution."]
I'll use the well-known property of integer polynomials $a-b | P(a)-P(b)$ (it's really easy to prove through factorizations).
Let $m$ be one of the solutions of the equation $P(n)=n+3$.
We have that $m-24 | (m+3)-17 = m-14$ and that $m-17 | (m+3)-10 = m-7$.
Now, let's see for what numbers we have $k-10 | k$. If $k \geq 21$, then $k-10 > \frac{k}2$, so it can't divide $k$. If $k < 5$, then $k-10$ is bigger in absolute value than $k$, so again it's not possible. Now check the rest of the cases by hand; we get that the only solution are:
$k=20$: $10|20$;
$k=15$: $5|15$;
$k=12$: $2|12$;
$k=11$: $1|11$;
$k=9$: $-1|9$;
$k=8$: $-2|8$;
$k=5$: $-5|5$.
I hope I didn't miss any.

Both $m-14$ and $m-7$ fulfil this property, $k-10|k$. The only pairs with a difference of $(m-7)-(m-14) = 7$ are $(12,5)$ and $(15,8)$.
Thus, $n_{1}= 19$ and $n_{2}= 22$, so $n_{1}n_{2}= 418$.
[/hide]

[quote="perfect_radio"]
Both $m-14$ and $m-7$ fulfil this property, $k-10|k$. The only pairs with a difference of $(m-7)-(m-14) = 7$ are $(12,5)$ and $(15,8)$.
Thus, $n_{1}= 19$ and $n_{2}= 22$, so $n_{1}n_{2}= 418$.
[/hide][/quote]

I don't get this part  :wink: . Can you explain it? thanks.  :D

nice approach though

Let $a=m-14$ and $b=m-7$. Then we have that $a-10|a$ and $b-10|b$. Also, $b-a = 7$. Now just find the pairs of numbers with difference $7$. There are only two, so they're the ones we've been looking for. Is it clear now?


I read on the AMC website that for 2009, the AMC 10/12's will be held on February 10th and 25th. Does anyone know when AIME or USAMO will be held, or when that information will be released? Also, approximations would help too.

Thanks!

From the AMC12 Teacher's Manual:

[quote]Students who score 100 or above or finish in the top 5% on this AMC 12 or students who score 120 or above or finish in the top 1% on this AMC 10 will be invited to take the 27th annual American Invitational Mathematics Examination (AIME) on Tuesday, March 17 or Wednesday, April 1. [/quote]

USAMo is generally the Tuesday and Wednesday of the last or second-to-last week in April.

When in doubt, check [url=http://www.unl.edu/amc/whatswhen.shtml]the AMC website.[/url]

Elizabeth Claassen
Senior Accounting Clerk
MAA-AMC


Zuming Feng is going to come to the Bay Area for giving an 8-day math training sessions during the winter break.

I have signed up for both sessions. Anyone else?

Just looked it up cause I was curious...

@james4l: maybe you should have posted some more info?

I googled it:

http://ideamath.org/amc123.htm

I'm doing session 2. I want to do 1 as well, but i have some time conflicts  :(

I'm going!

My parents said they met ur parents james.

:D

i'm going to both sessions.

NOT JUST ZUMING, TITU IS COMING ALSO

Sorry to double-post, but do you know what kind of lunch they'll have there?

*smacks lips*

Zuming is way too beast for me :'( so I can't go.

Sry to revive.

Which "level" are you taking?  I'm taking the AIME level one, though it might be a tad bit hard.

Opinions on today's session?

I'm in the AIME class.  Today I learned a bunch of algebra, and I also learned that I fail uberly at AIME level algebra and I don't need to work on my AIME C&P.

Oh, I am taking sessions one and two.  I am in Amc10/12.  What classes are you guys taking? and what are your names?

...and the bald guy is AIME15's brother.

So, anyways, many AoPSers don't know what I look like, so the added info doesn't help...

[quote="AIME15"]I'm in the AIME class.  Won't say my name here, but I'm the guy next to the bald guy eating donut holes at first break, and short, and wearing a hoodie from cross country nats.  Maybe talking to james4l.[/quote]
Oh, I know who you are.  My brother knows you I think.

That's scary.  What's your brother's first name?

Cya guys in like 4 hours!

zephyredx first name?

I'll be coming for 2nd session.

Is there anything we need to bring? An emergency form, perhaps? I saw it on the website a while ago, but can't find in now. :(

Yeah, I think there was an emergency form.

Bring like...umm...i think we need ruler, protracter, but compass' are provided.  Or something like that.

Pencil would be useful also.

Well, it was a good camp. Too bad it is over.  You guys plan on returning in the spring?

If it's during my spring break, I'm definitely coming.


I edit a lot of LaTeX documents for myself in Kile, a LaTeX editor, and have a good amount of inline math. I like to have them all in display mode, so I currently append "\displaystyle" within the dollar signs. I was wondering if there was a way to configure Kile to automatically compile all inline math in display mode without me having to type "\displaystyle" every time, just like how AoPS does with LaTeX input..

ehh sorry if this seems a bit jumbled :| Thanks for any help.

\everymath{\displaystyle}
This is a tex command and works however the document is edited and compiled; Kile is only an editor so not really relevant to the problem.

ah :) I put it before \begin{document} and it works. Thanks stevem!


Let $ \overrightarrow a,\ \overrightarrow b,\ \overrightarrow c$ be three vectors such that $ \overrightarrow a\plus{}\overrightarrow b,\ \overrightarrow a\minus{}\overrightarrow b,\ \overrightarrow a\plus{}2\overrightarrow b\plus{}3\overrightarrow c$ are collinear. Prove that $ \overrightarrow a,\ \overrightarrow b,\ \overrightarrow c$ are collinear vectors.

ok, if (a+b) is collinear with (a-b), then those two vectors are collinear with (a+b) + (a-b) since the vectors point in the same direction. So (a+b) and (a-b) are collinear with 2a, which is collinear with a. So if a is collinear with (a+b), then b has to be pointing in the same direction as A. So a and b are collinear. Now if a and b are collinear with a+2b+3c, then vector c must be pointing in the same direction as a and b (a+2b is collinear with a and b). So a, b, and c are collinear vectors.


Units digits 3 and 7 repeat in cycles of 4 when raised to powers.

For powers of numbers with units digit 7: 7, 9, 3, 1, 7, 9, 3, 1, etc.

For powers of numbers with units digit 3: 3, 9, 7, 1, 3, 9, 7, 1, etc.

Now $ 27\equiv 3$ and $ 23\equiv 3$ in mod 4.

So that means $ 23^{27}$ has units digit 7 and $ 27^{23}$ has units digit 3. Their sum will then have units digit 0, therefore making it divisible by 10.

Thus $ 10|a$ where $ a\equal{}23^{27}\plus{}27^{23}$, completing our proof.


[quote="Michael Niland"]For all real x and $a_{r}$ ,   F(x) satisfies

F(x) =$\sum_{1}^{n}\frac{a_{r}}_{F(r)(x+r)}$

Prove F(-x)F(x) = $\sum_{1}^{n}\frac{2ra_{r}}_{r^{2}-x^{2}}$[/quote]

For all real x and $a_{r}$ ,   F(x) satisfies

F(x) =$\sum_{r=1}^{n}\frac{a_{r}}_{F(r)(x+r)}$

Prove F(-x)F(x) = $\sum_{r=1}^{n}\frac{2ra_{r}}_{r^{2}-x^{2}}$

Is this what you mean?

Reply to BoesFX

Yes.The summation is better the way you have written it.
My apologies for any confusion

Michael N.

if you inserted the word "where" immediately to the left of

"F(x) satisfies," then it would make sense, and a person would

find it more readable.


inequalities
conics
hyperbola
algebra unsolved
algebra

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 1

Let $p,q,r$ and $s$ be real numbers such that $p^{2}+q^{2}+r^{2}-s^{2}+4=0$. Find the maximum value of $3p+2q+r-4|s|$.

Use Cauchy-Schwarz inequality, $3p+2q+r\leq \sqrt{3^{2}+2^{2}+1^{2}}\sqrt{p^{2}+q^{2}+r^{2}}=\sqrt{14}\sqrt{s^{2}-4},$

$\therefore 3p+2q+r-4|s|\leq \sqrt{14}\sqrt{s^{2}-4}-4|s|.$

Assume that $\sqrt{14}\sqrt{s^{2}-4}-4|s|$ takes some value $k,$ there exists the real numbers $s$ such that $s^{2}\geq 4\Longleftrightarrow |s|\geq 2,$ that is to say, rewrite as $\sqrt{14}\sqrt{s^{2}-4}=4|s|+k.$
Consider on $s-t$ plane, the two graphs of the hyperbola $t=\sqrt{14}\sqrt{s^{2}-4}$ and the graph of broken line $t=4|s|+k.$ have the intersection points on the interval $|s|\geq 2.$

This condition can be obtained by the condition these graphs touch.

Thus the desired maximum value is $-2\sqrt{2},$ when $(p,\ q,\ r,\ s)=\pm\left(\frac{6}{\sqrt{14}},\ \frac{4}{\sqrt{14}},\ \frac{2}{\sqrt{14}},\ 4\sqrt{2}\right).$

Alternative soluition: Let $x=|s|,$ we have $\sqrt{14}\sqrt{s^{2}-4}-4|s|=\sqrt{14}\sqrt{x^{2}-4}-4x=f(x)\ (x\geq 2).$

From $f'(x)=\frac{\sqrt{2}(32-x^{2})}{\sqrt{x^{2}-4}(\sqrt{7}x+2\sqrt{2}\sqrt{x^{2}-4})},$ the desired mximum value is $f(4\sqrt{2})=-2\sqrt{2}.$


We consider a set of positive integers $S$ and a positive integer $n$. We know that if $x,y\in S,x>y$ then $\frac{x^{n}y^{n}}{x^{2n}-y^{2n}}\in S$.
a) Is $S$ bounded?
b)What elements does $S$ have?

Asume that $S$ contains at least two different elements $x$ and $y$. Because $S$ contains only positive integeres it must be $x^{2n}-y^{2n}|x^{n}y^{n}$. Seting $a=x^{n}$ and $b=y^{n}$ yelds a^2-b^2|ab. There exists $k\in \mathbb{N}$ with $k(a^{2}-b^{2})=ab$. Let $t=\frac{a}{b}$. Then $1=k(t-\frac{1}{t})$ or $kt^{2}-t-k=0$ which has solutions $t_{1,2}=\frac{1\pm\sqrt{4k^{2}+1}}{2k}$. Because of positivity of $t$ it must be $t=\frac{1+\sqrt{4k^{2}+1}}{2k}$. Moreover because of rationality of $t$ $\sqrt{4k^{2}+1}$ is rational and it's rational iff it's natural number. So $4k^{2}+1=x^{2}$ for some $x\in \mathbb{N}$. So $1=(x+2k)(x-2k)$ or $k=0$, whch contradicts to $k\in \mathbb{N}$.

Hence solutions are $S=\emptyset$ or $S=\{s\}$ for arbitarly $s\in \mathbb{N}$.


geometry
incenter
circumcircle
symmetry
perpendicular bisector
cyclic quadrilateral
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, I the incenter, 1 the circumcircle, O the circumcenter, A' the second intersection point of AI and 1,
A" the intersection of the parallel to AI through O with BC, A*B*C* the tangentiel triangle of ABC and M the midpoint of B*C*.
Prove that A", A' and M are colinear.
Sincerely
Jean-Louis

Points $ O, A', A^*$ are collinear and $ OA'A^* \perp BC$ is the perpendicular bisector of $ (BC).$ Let the internal bisector $ AIA'$ of the angle $ \angle A$ cut $ BC$ at $ X$ and let a parallel to $ OB$ through $ A'$ cut $ BC$ at $ Y.$ $ BA'$ bisects the angle $ \angle CBA^* \equal{} \angle A,$ because $ \angle CBA' \equal{} \angle CAA' \equal{} \frac {\angle A}{2}.$ $ A^*A' \perp BC$ is the perpendicular bisector of $ (BC)$ and $ (YA' \parallel OB) \perp BA^*,$ because $ BA^*$ is a tangent of the circumcircle $ (O)$ at $ B.$ Therefore, $ A'$ is orthocenter of the $ \triangle A^*BY$ and $ BA' \perp YA^*$ its remaining altitude. Consequently, $ BA'$ cuts $ (YA^*)$ at its midpoint $ L.$ The $ \triangle A^*A''X$ with concurrent cevians $ A^*A', A''A', XA'$ is an image of the $ \triangle A^*BY$ with concurrent cevians (altitudes) $ A^*A', BA', YA'$ in an affine transformation (parallel projection), which keep all points of the line $ OA'A^*$ in place and which transforms every line perpendicular to $ OA'A^*$ into itself, (because $ O, A', A^*$ are fixed and $ OB \parallel YA', OA'' \parallel XA'$). It follows that $ A''A'$ cuts $ (XA^*)$ at its midpoint $ K.$

Let a tangent to $ (O)$ at $ A'$ cut $ C^*A^*, A^*B^*$ at $ D^*, E^*.$ $ D^*E^* \parallel BC$ and by symmetry, $ A'$ is the midpoint of $ (D^*E^*).$ The quadrilateral $ B^*C^*D^*E^*$ is tangential with incircle $ (O)$ and $ ABA'C$ is its contact cyclic quadrilateral, they have the same diagonal intersection $ X \equiv AA' \cap BC \cap B^*D^* \cap C^*E^*.$ Segments $ (A^*X), (E^*D^*), (B^*C^*)$ with midpoints $ K, A', M$ are the 3 diagonals of the complete quadrilateral $ (A^*D^*, D^*X, XE^*, E^*A^*).$ $ A''A'K$ is its Newton line and $ M \in A''A'K.$

Dear Mathlinkers,
a synthetic proof of this result has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis

My solution to your problem :
Let line IA intersects line EF at point N
Let P is feet of perpendicular from point I on A"A'
Let E' is Simmetry of point E wrt line IP
Let line IP intersects line DE at poin X
Let line IP intersects line EF at point Y
Easy to see that A" is Orthocenter of triangle IYA'
So IA" is Perpendicular to YA'
Let lines EA' intersects line IA" at point Z
Easy to see that angle IZE = 90 - ZA'Y = DA'E = DE'E = 90 - A'DE' , so angle ZA'Y = YXE' = EXY (Because XI is Perpendicular bissector to E'E)
So EXYA' is cyclic
So angle A'XP = A'EF
Angle IDE' = IEX = XE'I, so CXDIE is cyclic
Like the same let S is point of intersection of (IDF) and IF other then I , then angle PSA' = EFA' , so angle PSA' = NFA' = A'EF = A'XS , so PA' is perpendicular bissector of XS
Easy to see that angle CXI = BSI = IPA' = 90 , so M , P , A' are on same line , so MA"A' too . done


Is it true that calculators are not allowed in the AMC exams this year ?? I thought they were allowed...

Read the explanation in our newsletter:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/08AMC1012MathM.pdf

Read the Teacher's Manual, page 4, Item 4:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf

See also the reproduction of the cover of the AMC 10 and the AMC 12 in the Teachers' Manual, last pages:

Steve Dunbar
MAA Director, American Mathematics Competition


Let $S=\sum_{k=1}^{777}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}$. $S$ can be expressed as $\sqrt{p+\frac{m}{n}}-\frac{\sqrt{q}}{r}$, where $\gcd (m,n)=1$. Find $p+m+n+q+r$. $p,q,r,m,n\in\mathbb{N}$, $\frac{\sqrt{q}}{r}$ is in its simplest form, and $\frac{m}{n}<1$.

Masoud Zargar

[hide="Solution"]
\begin{eqnarray*} \frac{1}{ \sqrt{ 2k + \sqrt{4k^2 - 1} } } &=& \sqrt{ 2k - \sqrt{4k^2 - 1}} \\ &=& \frac{ \sqrt{(2k+1) - 2 \sqrt{(2k+1)(2k-1)} + (2k-1)} }{ \sqrt{2} } \\ &=& \frac{ \sqrt{2k+1} - \sqrt{2k-1} }{ \sqrt{2} } \\ \end{eqnarray*}

The sum telescopes.  $S = \frac{ \sqrt{1555} - 1}{ \sqrt{2} } = \sqrt{ 777 + \frac{1}{2} } - \frac{ \sqrt{2} }{2}$.

$p + q + r + m + n = \boxed{784}$.  I think.  (You should probably specify $p, q, r, m, n \in \mathbb{N}$ instead of $\mathbb{Z}$.)
[/hide]


invariant
abstract algebra
vector
superior algebra
superior algebra unsolved

Does anyone know of a proof of [url=http://en.wikipedia.org/wiki/Maschke%27s_theorem]Maschke's theorem[/url] that doesn't proceed by constructing a $ G$-invariant bilinear form?

Yes: It can be found in Cohomology of Number Fields in the chapter on spectral sequences: http://books.google.de/books?id=9WST1LnyCucC&dq=cohomology+of+number+fields&printsec=frontcover&source=bn&hl=de&ei=mOiMSrOyOJiJsAa08emwDA&sa=X&oi=book_result&ct=result&resnum=4#v=onepage&q=&f=false page 105.

It seems that the available pages change for different users ...

Look at the spectral sequence $ \mathrm{H}^p(G, \mathrm{Ext}_R^q(M,N)) \implies \mathrm{Ext}^{p \plus{} q}_{R[G]}(M,N)$, $ G$ a finite group, $ R$ a commutative ring and $ M,N$ $ R[G]$-modules. It is a special case of the Grothendieck spectral sequence.

Let $ M$ be projective as an $ R$-module. We want to show it is also projective as an $ R[G]$-module. Suppose the order of $ G$ is invertible in $ R$. Then $ \mathrm{H}^p(G, \minus{} ) \equal{} 0$ for $ p > 0$ as it is annihilated by $ G$ and an $ R$-module at the same time. So our spectral sequence degenerates in the $ E_2$-term, which yields $ \mathrm{Ext}^{p}_{R[G]}(M,N) \equal{} 0$ for $ p > 0$ and all $ N$, so $ M$ is projective.

[quote="t0rajir0u"]Sorry, page 105 isn't available in the preview.  Could you sketch the proof?  What I'm looking for is a proof that directly identifies a property group algebras have that more general algebras don't.[/quote]

The property is called semisimplicity, but I doubt this is what you are looking for, because semisimplicity is defined as "satisfies the assertion of Maschke's theorem" (unless you are reading the Wikipedia).

Anyway, most books prove Maschke without bilinear forms, instead constructing a projection:
Let $ \left(V,\rho\right)$ be a finite-dimensional representation of a finite group $ G$, and $ U$ be an invariant subspace. Let $ \pi: V\to U$ be any projection of $ V$ onto $ U$ as vector spaces (not necessarily as representations of $ G$). Then, it is easy to show that $ \frac {1}{\left|G\right|}\sum_{g\in G}\pi\circ\rho_g$ is a projection of $ V$ onto $ U$ as representations of $ G$.

Of course, $ \frac {1}{\left|G\right|}$ only makes sense here if the field in question has characteristic not dividing $ \left|G\right|$.

Is this really new to you??

  darij

Sorry, I worded my question poorly.  What I mean is something like this: suppose you are given a category of algebras defined in some way and you want to decide whether they're semisimple using some other easily checkable property of the category.  What sort of properties would a representation theorist look for?

Both the bilinear form proof and your proof of Maschke's theorem boil down to the nice properties of the averaging operator, but I don't really like "existence of averaging operators" as a structural property.  Is there a "better" structural property here?

myan, I'm afraid I can't follow your explanation.

Do you have an example of some algebras of which you would like to decide semisimplicity?

Myan's proof is for a much nicer property of group rings than semi-simplicity.

Using bilinear forms is an extremely important technique in representation theory.

Group algebras over infinite groups, for example.  Based on the usual proof of Maschke's theorem I would guess that the first class of groups to look at are compact groups, and then one replaces the average with the Haar integral.  Are there proofs in this setting which substitute a different property of compact groups for the existence of the Haar measure?

Oh, luckily that is an easy case: it is not semisimple.

Proposition 6.3 of Lam's First Course in Noncommutative Rings, page 81, has a proof.  If k is any nonzero ring and G is any infinite group, then the group ring kG is not semisimple.

To get the kind of algebras that are closer to semi-simple, normally instead of group rings (with finite sums) you take things like L^2(G) (with square-summable coefficients).  Once you do that, it becomes very natural to be using Haar integrals.


calculus
derivative
function
calculus computations

Hi
Here's a question that i came across....
f(x) is continuous and differentiable. f(x) takes values of the form +or-(I)^1/2 where I denotes set of whole numbers whenever x=a or b. otherwise f(x) takes real values. f(c)= -3/2 and |f(a)|<=|f(b)|.

then :

1. find the number of rational values of f(a)+f(b)+f(c)

2. find the number of values taken by (f(a))^2 +(f(b))^2 + (f(c))^2

3. find the number of possible triplets ( f(a), f(b), f(c)).
See the figure also.
Shyam

$ f(x)^2$ has derivative of twice the function on that graph. So $ f(x)^2$ increasing before $ c$ and decreasing afterward, so maximum at $ c$. We get $ f(a)^2\leq f(b)^2\leq f(c)^2\equal{}\frac{9}{4}$.

Since $ f(a)^2,f(b)^2$ are nonnegative integers, we get $ 0\leq f(a)^2\leq f(b)^2\leq 2$ giving 17 solutions for $ (f(a),f(b))$, namely $ f(a)\equal{}0, f(b)\equal{}0,\pm 1,\pm \sqrt{2}$, $ f(a)\equal{}\pm 1,f(b)\equal{}\pm 1,\pm \sqrt{2}$, and $ f(a)\equal{}\pm \sqrt{2},f(b)\equal{}\pm \sqrt{2}$.

The number of values for sum of squares is 5 (it can be $ \frac{9}{4}\plus{}n$ for $ n\equal{}0,1,2,3,4$).

For number of rational sums, $ f(a)\plus{}f(b)$ rational and $ f(a),f(b)$ give integers when squared, implying $ f(a),f(b)$ integers. So there are 3 sums (anywhere between -1 and 1 for $ f(a)\plus{}f(b)$).

I must say that that was a truly wonderful solution...thanks a lot!! :)


The Euler line of triangle $ ABC$ intersects $ BC,CA,AB$ in $ A_1,B_1,C_1$, then, the Euler lines of triangles $ ABC_1,BCA_1,CAB_1$ form a triangle congruent to ABC.
Can someone give a proof and post some properties of this configuration?

It is clear there is a typo, because A,B,C_1 are collinear and do not determine a triangle....Maybe should read ACA_1, ABB_1 and ACC_1? I do not yet know if the result is true.

A search in Google allow find some links related to the theorem of Gossard on Euler lines:
http://faculty.evansville.edu/ck6/bstud/gossard.html
http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html

There is the correct version of the theorem. This was discussed some years ago in others forums.
I will search some readable proof in old belgian journal Mathesis.

I have located in Mathesis 1908 the result, it seems was comunicated by Cikot at least 8 years before Gossard, but without proof. The belgian journal Mathesis  includes a proof of a more general theorem, by J.Neuberg. If anybody is interested, I will translate the proof of Neuberg and send by email.


A club has 20 members, 12 boys and 8 girls.  Two of the members are chosen at random.  What is the probability that a boy and a girl are chosen?

I have noticed this for a lot of problems... I solve them by calculating them as the probabilities of multiple events, while the Alcumus solution involves counting combinations.

This problem I did 12/20 x 8/19 x 2.  It seems more intuitive this way to me.

yay admin!!! :D :D :D :D :D

[hide=my solution]Using complementary counting, we have $\frac{\binom{20}{2}-\binom{12}{2}-\binom{8}{2}}{\binom{20}{2}}=\boxed{\frac{48}{95}}$.[/hide]

[hide=Answer]There are 12 choose 1 ways (12) to choose a boy and 8 choose 1 ways (8) to choose a girl. There are 20 choose 2 ways (190) to choose 2 members. The amount of favorable outcomes is 12*8 = 96 and there are 190 possible outcomes. This makes 96/190 = [b]48/95[/b]. [/hide]


number theory
least common multiple
greatest common divisor
inequalities
limit
number theory proposed

If x and y are positive integers, L: LCM of x and y, G: GCD of x and y, then solve the following equations for x and y values:
1) xy=LG, L>x.
2) x^2+y^2 = L^2+G^2, L>x>G.

[quote="praneeth"]If x and y are positive integers, L: LCM of x and y, G: GCD of x and y, then solve the following equations for x and y values:
1) $ xy\equal{}LG, L>x.$ (1)
2) $ x^2\plus{}y^2 \equal{} L^2\plus{}G^2, L>x>G.$[/quote] (2)
The (1) is true for all  $ (x,y)$
Let $ d\equal{}\gcd(x,y)$ then exist a,b
$ x\equal{}da,y\equal{}db$ where $ \gcd(a,b)\equal{}1$ and $ a,b>1$
From (2) we have :
$ d^2(a^2\plus{}b^2)\equal{}(dab)^2\plus{}d^2$
$ \Longleftrightarrow a^2\plus{}b^2\equal{}(ab)^2\plus{}1$
$ (a^2\minus{}1)(b^2\minus{}1)\equal{}0$ (tradition !)
This equation has not solution.

The 1st equation I meant to be:

x^y=L^G; L>x

We have
$ d^2mn\equal{}(mn)^d$
$ d^2\equal{}(mn)^{d\minus{}1}$ (1)
Use inequalities to find out n.
Because
$ \lim_{x\to\infty}\frac{a^{x\minus{}1}}{x^2}\equal{}\plus{}\infty$ where $ a>1$
So exist $ n_0$ for  (1) is not true for all $ mn>n_0$
It is quite easy.

[quote="TTsphn"]We have
$ d^2mn \equal{} (mn)^d$
$ d^2 \equal{} (mn)^{d \minus{} 1}$ (1)
Use inequalities to find out n.
Because
$ \lim_{x\to\infty}\frac {a^{x \minus{} 1}}{x^2} \equal{} \plus{} \infty$ where $ a > 1$
So exist $ n_0$ for  (1) is not true for all $ mn > n_0$
It is quite easy.[/quote]

(2,4) is the only solution to 1st one


Hi,

I am working on creation of a wiki for Olympiads in India, at:

http://olympiadsindia.wiki-site.com

Currently, the wiki is in a very fledgeling state, and it needs the participation and support of a lot of interested people. The aim of the wiki is to provide information to Olympiad aspirants with an Indian focus and perspective. We would also like to incorporate regional information (information about the RMOs, Olympiad training and other Olympiad resources in different regions).

If you are interested in or involved with the Olympiad process in India, please check out the wiki.

You can log in or create an account at:

http://olympiadsindia.wiki-site.com/index.php/Special:Userlogin

You can also view a complete list of pages on the wiki at:

http://olmypiadsindia.wiki-site.com/index.php/Special:Allpages

Good idea, Vipul! I have joined too!
Sathej


Ronald's weight squared plus the Hamburglar's weight is two times Ronald's weight less than Ronald's weight plus the Hamburglar's weight squared. How much heavier is the Hamburglar than Ronald? Assume that the Hamburglar [b]is[/b] heavier than Ronald...

Thinking up problems takes a lot of work >___>

EDIT: This is the 30,000th post in the MathCounts forum. Go me.

[quote="slugbug321"][quote="Iron Fist"]Ronald's weight squared plus the Hamburglar's weight is two times Ronald's weight less than Ronald's weight plus the Hamburglar's weight squared. How much heavier is the Hamburglar than Ronald?

Thinking up problems takes a lot of work >___>

EDIT: This is the 30,000th post in the MathCounts forum. Go me.[/quote]

[hide]okay, so r=ronald and h=hamburglar.
r^2+h+2r=r+h^2.
r^2+h+r=h^2.
r^2+r=h(h-1)
r(r+1)=h(h-1)
r=h-1
h(h-1)=h(h-1)

The hamburglar's weight is one (unit?) more than Ronald's weight.[/hide][/quote]

In that case, Ronald would weigh 0 though meaning it might not be counted, there might be multiple solutions, not sure, haven't tried it. If there are, Iron Fist should specify their weights are both positive integers or something.

...well no duh their weight are positive integers...

When's the last time you saw somebody weigh less than 0 pounds?

[hide]What if Ronald was one pound and Hamburger was 2 pounds

$1^2+2+2=2^2+1$

1?[/hide]


What is the least number with 181 factors? Express your answer as a^b*c^d*e^f*...., where a, c, e, etc. are primes.

$ 2^{180}$

i think, because 181 is prime

TEST: $ \sqrt{181}\equal{}13.4....$
181/2=not integer
181/3=not integer
181/5=not integer
181/7=not integer
181/11=not integer
181/13=not integer

so 181 is prime

I think you need to prime factorize 180 since you have to add 1 to the exponents and multiply them together. But, there is no way I could do that....

unless...

[hide]180=2*2*3*3*5
so subtract 1 from all those

1,1,2,2,4
so I am going with 2^4*3^2*5^2*7*11=

277200
[/hide]

You're remembering the formula for the number of divisors wrong.  You actually do need to prime factorize $ 181$, not $ 180$.

we need the least number.  There's obviously no greatest number

[quote="pascal_1623"][quote="mz94"]$ 2^{180}$

i think, because 181 is prime

TEST: $ \sqrt {181} \equal{} 13.4....$
181/2=not integer
181/3=not integer
181/5=not integer
181/7=not integer
181/11=not integer
181/13=not integer

so 181 is prime[/quote]
do you actually need to check if 181 is prime? Since 2 is the smallest prime, any substitutes for 2 would mean a greater integer than 2^180[/quote]

Yes, we do. Take a number such as $ 39$ for example. If we assumed it was prime, then we would say that $ 2^{38}$ would be the smallest number with $ 39$ factors. However, if we discovered that 39 can be factored into $ 3 \cdot 13$, then we have to factor the number in the form $ a^2 \cdot b^{12}$. Obviously, we let the smallest prime number have the exponent of $ 12$ and the next smallest for the $ 2$, for a result of $ 2^{12} \cdot 3^2$. We can readily see that that is smaller than $ 2^{38}$ because $ 3^3 < 2^{26}$.


I have a problem that I cant seem to figure out.

f(x) = 3^(3^x)

I have taken the log of both sides to get 3^x=log(y/3)
but dont know where to go from here.

Please help

What are you trying to figure out?  The inverse of the function?

[hide="Then..."]$y = 3^{3^{x}}$
$\log_{3}y = 3^{x}$
$\log_{3}\log_{3}y = x$

And hence the inverse of the function is given by

$f^{-1}(x) = \log_{3}\log_{3}x$ [/hide]

[quote="dre1285"]How would I go about doing that in base 10?  The software I am using does not allow anything except base 10.[/quote]

Use the change of base formula; in this case, $\log_{3}x = \frac{ \log_{10}x}{\log_{10}3}$.

You need to use the change of base formula:

$\log_{a}b = \frac{\log b}{\log a}$

at least i think thats right...

edit: cincodemayo beat me to it!

the way I was approaching this problem is:

y = 3^(3^x)
log(y) = 3^x log(3)
log(y/3) = 3^x
log(y)-log(3) = 3^x
log((y-3)/3) = x
log(y)-log(3)-log(3) = x

am I completely wrong?

Wow, I can't believe i was so far off.  Thanks for the help.  Im going to have to tell all of my classmates about this place.


what i the formula for finding 1+4+7+...+103?

The formula for finding the Sum to the nth term for any arithmetic series is $ \frac{n(2a_{1}\plus{}(n\minus{}1)d)}{2}$, where $ a_{1}$ is the first term, $ n$ is the term number and $ d$ is the difference. Another version which you might find easier to use is $ \frac{a_{1}\plus{}a_{n}}{2}\cdot n$

so what would the $ a_n$ be in this sequence?

$ a_n$ is the last term of the sequence, which is $ 103$ in this case.

so its

\[ \frac{1\plus{}103}{2}\] times $ n$.....

what is n?

All you'll ever need is $ \sum_{k\equal{}0}^{n}k\equal{}\frac{n(n\plus{}1)}{2}$. Other formulas are derived from this.
It's much easier to come up with the answer like I did than it is to use weird formulas.

If you put it in sigma notation, $ \sum 1 \plus{} 3k$, you notice that $ 1 \plus{} 3k \equal{} 103$, so $ k \equal{} 34$. Because $ k$ ranges from $ 0$ to $ 34$ you have $ 35$ terms.

Alternatively, subtract 1 from each term, divide by 3, add one, and it becomes $ 1,2,3,...,35$.

[quote="DaReaper"]how do you figure out that there are 35 terms....[/quote]
In an arithmetic sequence, the nth term is derived to be $ a_{1}\plus{}(n\minus{}1)d$.(Can you derive this?)
Since the nth term is 103, we can plug and chug to find n.
$ 103\equal{}1\plus{}(n\minus{}1)3$
$ 103\equal{}1\plus{}3n\minus{}3$
$ 103\equal{}3n\minus{}2$
$ 105\equal{}3n$
$ n\equal{}35$


logarithms
limit
function
calculus
calculus computations

I don't know how to solve a problem like this.
Let us consider the Cauchy problem
$ y^' = y\log{(y + x)}$
$ y(0) = 1$
Prove that the maximal interval on which the solution is deﬁned is $ \mathbb{R}$.
Find $ \lim_{x \to \infty}\frac {y(x)}{e^x}$ and $ \lim_{x \to + \infty} \frac {y(x)}{x}$.

It's not immediately clear to me why $ y$ is well defined on all $ \mathbb{R}$, but if we assume it is then it's not hard to show $ \lim_{x\to \infty}\frac {y(x)}{e^x} \equal{} \infty$.

For $ x > 0$, $ y$ is a strictly increasing function, and in particular $ y' > y\log (x \plus{} 1)$ so $ y > \frac {(x \plus{} 1)^{x \plus{} 1}}{e^x}$ for $ x > 0$.  This estimate is good for small $ x$, but underestimates $ y$ dramatically for large $ x$ (the actual behavior for large $ x$ is something like $ e^{e^{x \minus{} c}}$); even so, it is already enough to show $ \lim_{x\to \infty}\frac {y(x)}{e^x} \equal{} \infty$.

A somewhat better but still rather crude lower bound:  Notice that $ y' > y\log y$, so the previous bound $ y(1) > \frac {4}{e}$ implies $ y > \left(\frac {4}{e}\right)^{e^{x \minus{} 1}}$ for $ x \geq 1$.


USAMO inequalities quadratics vector calculus induction

For me, this is the hardest and most beautiful inequality ever given in a contest. Absolutely incredible!

A very hard solution is published at kalva site.

I have studied it and I have found it very interesting but too much hard  for a common people like me.

wow: i love inequalities, but this one almost killed me - amazing  :D

I think I have found a simple proof of this famous inequality.

First, we will prove a handy lemma which shows that a certain quadratic form is nonnegative.

[b]Lemma 1:[/b] Let $r_1$, ..., $r_n$ be nonnegative numbers.  Let $x_1$, ..., $x_n$ be real numbers.  Then
\[
  \sum_{i, j} \min(r_i, r_j) x_i x_j \ge 0.
\]

[b]Proof:[/b] We may assume that $r_1 \le r_2 \le \ldots \le r_n$.  For convenience, define $r_0 = 0$.  The left-hand side of the inequality becomes
\[
  \sum_i r_i x_i^2 + 2 \sum_{i < j} r_i x_i x_j \, .
\]
But that expression is equivalent to the sum
\[
  \sum_i (r_i - r_{i-1}) \biggl( \sum_{j=i}^n x_j \biggr)^2 \, .
\]
Each term of that sum is nonnegative, so the whole sum is nonnegative. $\blacksquare$

To apply Lemma 1, we will choose the $r$ and $x$ vectors carefully.  Define
\[
  r_i = \frac{\max(a_i, b_i)}{\min(a_i, b_i)} - 1.
\]
(If the denominator is 0, set $r_i$ to be any nonnegative number.)  Define $x_i = \textrm{sgn}(a_i - b_i) \min(a_i, b_i)$.  We will next prove a lemma that connects our original problem to the quadratic form.

[b]Lemma 2:[/b] $\min(a_i b_j, a_j b_i) - \min(a_i a_j, b_i b_j) = \min(r_i, r_j) x_i x_j$.

[b]Proof:[/b] If we switch the values of $a_i$ and $b_i$, then both sides negate.  Thus we may assume $a_i \ge b_i$.  Similarly we may assume $a_j \ge b_j$.

If either $b_i$ or $b_j$ is $0$, then both sides are $0$.  Thus we may assume that $b_i$ and $b_j$ are positive.

By the definitions of $r$ and $x$, we have
\begin{eqnarray*}
  r_i & = & \frac{a_i}{b_i} - 1 \\
  r_j & = & \frac{a_j}{b_j} - 1 \\
  x_i & = & b_i \\
  x_j & = & b_j \, .
\end{eqnarray*}
Plugging those values in, we get
\begin{eqnarray*}
  \min(r_i, r_j) x_i x_j
    & = & \min\bigl(\frac{a_i}{b_i} - 1, \frac{a_j}{b_j} - 1\bigr) b_i b_j \\
    & = & \min(a_i b_j, a_j b_i) - b_i b_j \\
    & = & \min(a_i b_j, a_j b_i) - \min(a_i a_j, b_i b_j) \, .
\end{eqnarray*}
That completes the proof of Lemma 2.  $\blacksquare$

Finally, we can apply Lemma 2 and then Lemma 1 to get
\begin{eqnarray*}
  \sum_{i,j} \min(a_i b_j, a_j b_i) - \sum_{i, j} \min(a_i a_j, b_i b_j)
    & = & \sum_{i, j} \left[ \min(a_i b_j, a_j b_i) - \min(a_i a_j, b_i b_j) \right] \\
    & = & \sum_{i, j} \min(r_i, r_j) x_i x_j \\
    & \ge & 0 \, .
\end{eqnarray*}
Bringing one sum over to the right-hand side gives the desired inequality. :lol:

That's just great. I knew lemma 1 (to which I had a nice integral solution) and I also tried to use it, but I just could not find the good $r_i, x_j$. It is a fantastic solution, truly nice and beautiful, Ravi.  :first:

Somehow, I'm interested Ravi B how did you found that solution ? :)

Harazi, thanks for your kind words.  I think I see the integral solution of Lemma 1 (replacing the min with an integral), and it is cute.

Rafal, after many false starts, here's what I did.  I think the key idea was, instead of thinking of the pair $a_i$ and $b_i$ as two separate numbers, to think of them as consisting of three independent parts: their minimum, the ratio of their maximum to their minimum, and a bit indicating which is larger.  I gave a name to the first two parts, $m_i$ and $r_i$.  To handle the bit about which is larger, I split each of the original sums into four cases, depending on whether $a_i \ge b_i$ and/or $a_j \ge b_j$.  Each sum could then be expressed solely in terms of $m_i$ and $r_i$.  I subtracted the two sides, and then I realized I could collapse the four sums into one by using the sign function.  At that point, I had reduced the inequality to Lemma 1, and I knew I was home.

In the proof of Lemma 2, I should have pointed out that if either $a_i = b_i$ or $a_j = b_j$, then both sides are zero, so we can ignore those cases.

[quote="harazi"]I knew lemma 1 (to which I had a nice integral solution)[/quote]

How did you prove it with integrals?

Use the identity
\[ \min(r_i, r_j) = \int_0^\infty [t \le r_i] [t \le r_j] \, dt \, .  \]
Here $[S]$ means $1$ if $S$ is true, and $0$ otherwise.

There's one thing I don't understand about this inequality. Since $a_i$ and $b_j$ are said to be just nonnegative, then [b]why[/b] this inequality holds??? I mean, not looking into the solution, I just see the inequality and I don't know why it holds... Can anybody explain it to me?  :o

oops, I'm beginning to understand it.

Perhaps someone could write the integral proof of Lemma 1 for me, please.  :roll:

Well basically, you have the intuition that $a_{i}a_{j}$ and $b_{i}b_{j}$ are less "average" than $a_{i}b_{j}$ and $a_{j}b_{i}$, sort of like how you have $a^{2}+b^{2}\ge 2ab$. I think I have another solution--nowhere near as awesome as Ravi's, but I'll post it anyway for those who think the above one is too tricky to find.

We proceed by induction. The base case is trivial, as $\min(a_{1}^{2}, b_{1}^{2}) \le a_{1}b_{1}$, since $a_{1}b_{1}$ is the geometric mean of $a_{1}^{2}$ and $b_{1}^{2}$. Now we will prove the inequality for $n$ variables assuming that it holds for $n-1$ variables.

Consider $\sum \min(a_{i}b_{j}, b_{i}a_{j})-\sum \min(a_{i}a_{j}, b_{i}b_{j})$ as a function of $a_{i}$. It would suffice to prove the inequality when $a_{i}$ is set to a value such that this function is minimal (it is clear that it has a minimum, since it plateaus out when $a_{i}$ is sufficiently large, and it is continuous). This function is piecewise linear, so the minimum must occur at a "critical" point, i.e. when either $a_{i}b_{j}= b_{i}a_{j}$ or $a_{i}a_{j}= b_{i}b_{j}$ creates a cusp in the graph. Note that if the minimum occurs at $a_{i}= 0$, we immediately reduce the problem to $n-1$ variables and are done.

We also need the cusp to open upward. The cusp created when $a_{i}b_{j}= b_{i}a_{j}$ opens downward, which is not what we want, so our desired cusp must have been created when $a_{i}a_{j}= b_{i}b_{j}$ for some $j$. This applies to all $i, j$, so with the appropriate smoothing, we can assume that for each $i$, there exists $j$ with $a_{i}a_{j}= b_{i}b_{j}$.

WLOG, we will assume $\max \left( \frac{a_{1}}{b_{1}}, \frac{b_{1}}{a_{1}}\right) \ge \max \left( \frac{a_{i}}{b_{i}}, \frac{b_{i}}{a_{i}}\right)$ for all $i$, and that $a_{1}a_{n}= b_{1}b_{n}$. Also suppose $\frac{a_{1}}{b_{1}}= \frac{b_{n}}{a_{n}}= t \ge 1$ (we could always swap $(a_{1}, b_{1})$ with $(a_{n}, b_{n})$ to achieve this).

We again look to minimizing $\sum \min(a_{i}b_{j}, b_{i}a_{j})-\sum \min(a_{i}a_{j}, b_{i}b_{j})$. Note that for every $j$, $\min(a_{1}b_{j}, b_{1}a_{j}) = b_{1}a_{j}$, since $\frac{a_{1}}{b_{1}}\ge \frac{a_{j}}{b_{j}}$. Similarly, $\min(a_{1}a_{j}, b_{1}b_{j}) = b_{1}b_{j}$. These terms are the only ones involving either $a_{1}$ or $b_{1}$ in the sums, and they add up to

$b_{1}(a_{1}+2a_{2}+2a_{3}+\ldots+2a_{n}-b_{1}-2b_{2}-2b_{3}-\ldots-2b_{n})$.

In an analogous way, the terms involving $a_{n}$ or $b_{n}$ add up to

$a_{n}(b_{n}+2b_{n-1}+\ldots+2b_{1}-a_{n}-2a_{n-1}-\ldots-2a_{1})$.

Notice that $a_{1}$ and $b_{n}$ appear only in $a_{1}b_{1}+a_{n}b_{n}$, so if we decrease $a_{1}$ and $b_{n}$ by the same ratio in such a way that $\frac{a_{1}}{b_{1}}$ is still at least $\frac{a_{i}}{b_{i}}$ for any $i$, the entire sum is decreased. Then we can decrease them enough so that $\frac{a_{1}}{b_{1}}= \frac{a_{i}}{b_{i}}$ for some $i$, and WLOG $i = 2$.

However, we can replace $a_{1}$ and $a_{2}$ with $a_{1}+a_{2}$, and we can replace $b_{1}$ and $b_{2}$ with $b_{1}+b_{2}$. It is not difficult to verify that this does not change the sum in any way, but it reduces the number of variables to $n-1$, and so the inequality holds, and the induction works.

You might notice from this and the first solution that the ratio $\frac{a_{i}}{b_{i}}$ is important in this problem.

[quote=MathPanda1]What is $sgn(x)$ defined to be?[/quote]

See: https://en.wikipedia.org/wiki/Sign_function

Basically, $\text{sgn} (x) = -1 \text{ if } x<0$, $\text{sgn} (x)=0 \text{ if } x=0$, and $\text{sgn} (x) =1 \text{ if } x>0$.

I find the identity in post 5's solution impossible to motivate, so here's my solution, which shows how I derive a similar identity:

We will rewrite $\min (a_ib_j,a_jb_i) - \min (a_ia_j,b_ib_j)$ as $0.5(a_ib_j + a_jb_i - |a_ib_j-a_jb_i|) - 0.5( a_ia_j +b_ib_j - | a_ia_j-b_ib_j|)$, using the identity $\min (x,y) = 0.5(x+y-|x-y|)$. This last expression becomes $0.5( (b_j-a_j)(a_i-b_i) - |a_ib_j - a_jb_i| + |a_ia_j-b_ib_j| )$. We show the sum of all such terms is $\ge 0$.

Next, we notice that $|x| - |y|$ has magnitude $\min ( |x-y|, |x+y|)$, so $|x|-|y| = \pm \min (|x-y|, |x+y|)$. Of course, this does not help us find the value of $X=|a_ia_j-b_ib_j| - |a_ib_j-a_jb_i|$ unless we know the proper choice of sign, so we must determine when the previous quantity is nonnegative and when it is negative.

Notice $|a_ia_j-b_ib_j| \ge |a_ib_j-a_jb_i| \iff a_i^2a_j^2 + b_i^2b_j^2 \ge a_i^2b_j^2 + a_j^2b_i^2 \iff (a_i^2 - b_i^2)(a_j^2-b_j^2) \ge 0 \iff (a_i-b_i)(a_j-b_j)\ge 0$. Let's define $c_i = \text{sgn} (a_i-b_i)$. Then the sign of $X$ is just $c_ic_j$, so each of the original terms becomes equivalent to $0.5( (b_j-a_j)(a_i-b_i) +c_ic_j\min ( |a_ib_j -a_jb_i + a_ia_j-b_ib_j|, |a_ib_j -a_jb_i - a_ia_j +b_ib_j|) )$. We now define $d_i = a_i-b_i$ for convenience, noting $d_i = |a_i-b_i|c_i$, so this expression becomes $0.5( - d_id_j +c_ic_j \min ( |(a_j+b_j)d_i|, |(a_i+b_i)d_j|) = 0.5d_id_j \left(-1 + \min \left( \left|\dfrac{a_j+b_j}{d_j}\right|, \left|\dfrac{a_i+b_i}{d_i}\right|\right)\right)$, which equals $0.5d_id_j \min \left( \dfrac{a_j+b_j}{|a_j-b_j|} -1, \dfrac{ a_i+b_i}{|a_i-b_i|}-1 \right) = 0.5d_id_j \min (D_i,D_j)$ where $D_i = \dfrac{a_i+b_i}{|a_i-b_i|}-1 \ge 0$.

So now the problem is reduced to the following:

[b]Lemma:[/b] If $d_i\in \mathbb R$ and $D_i\in \mathbb R_{\ge 0}$ for $1\le i\le n$, then $\sum_{i,j} d_id_j \min (D_i,D_j) \ge 0$.

[b]Proof:[/b] Define $\lambda_i (x)$ to be $0$ if $x \not \in [0,D_i]$ and $1$ otherwise. Then the sum becomes $\sum_{i,j} d_id_j \int_0^{\infty} \lambda_i (t) \lambda_j (t) dt= \int_0^{\infty} \sum_{i,j} d_i\lambda_i (t) d_j\lambda_j (t) = \int_0^{\infty} \left( \sum_i d_i \lambda_i (t) \right)^2 \ge 0$, so we're done.

[quote=GetOffTheInternet][quote=fuzzylogic]Let $a_1, b_1, a_2, b_2, \dots , a_n, b_n$ be nonnegative real numbers. Prove that
\[
\sum_{i, j = 1}^{n} \min\{a_ia_j, b_ib_j\} \le \sum_{i, j = 1}^{n} \min\{a_ib_j, a_jb_i\}.
\][/quote]

Maybe I have misunderstood something, but it seems clearly false. Pick $\epsilon > 0$, but small. Let each of the $a_i$ be $\epsilon$ and each of the $b_i$ be $1$.

Then, independent of $i,\,j$ we have $\text{min}\{a_ia_j, b_ib_j\} = \text{min}\{\epsilon^2,1\} = \epsilon^2$ while independent of $i,\,j$ we have $\text{min}\{a_ib_j,b_ia_j\} = \epsilon$.

So the left hand sum is $n^2 \epsilon^2$, the right hand sum is $n^2 \epsilon$, and the inequality fails.[/quote]

why it fails? the inequality is still true!

In the following,$\int$ means intergrating from $0$ to $+\infty$.
We define $f(t,x)=1(\text{if}\ 0 \le x \le t),0(\text{otherwise})$.It is easy to see that $\int f(a,x)f(b,x)=\int cf(a/c,x)f(b/c,x)$ and $\min\{a,b\}=\int f(a,x)f(b,x)$.
Notice that
$$\begin{aligned}&\sum\limits_{1 \le i,j \le n}\min\{a_ib_j,b_ia_j\}\\=&\sum\limits_{1 \le i,j \le n}\int f(a_ib_j,x)f(b_ia_j,x)\\=&\sum\limits_{1 \le i,j \le n}\int b_ib_jf(a_i/b_i,x)f(a_j/b_j,x)\\=&\int (\sum\limits_{i=1}^n b_i f(a_i/b_i,x))^2\\=&\int \frac{1}{2}(\sum\limits_{i=1}^n b_i f(a_i/b_i,x))^2+\frac{1}{2}(\sum\limits_{i=1}^n a_i f(b_i/a_i,x))^2\\ \ge& \int (\sum\limits_{i=1}^n b_i f(a_i/b_i,x))(\sum\limits_{i=1}^n a_i f(b_i/a_i,x))\\=&\sum\limits_{1 \le i,j \le n}\int a_ib_jf(a_j/b_j,x)f(b_i/a_i,x)\\=&\sum\limits_{1 \le i,j \le n}\int f(a_ia_j,x)f(b_ib_j,x)\\=&\sum\limits_{1 \le i,j \le n}\min\{a_ia_j,b_ib_j\}\end{aligned}$$

Q.E.D.

My first impression of this inequality is related to completing the square using the Cauchy Inequality, let me go for that for an hour and I will reply to this with my solution (if I can get the solution).

The splendid integral solutions are impossible to think of, and maybe the sol below is much more natural.

We only need $\sum_{i, j = 1}^{n} \frac{a_ib_j + a_jb_i - |a_ib_j-a_jb_i|}{2} - \sum_{i, j = 1}^{n} \frac{a_ia_j +b_ib_j - | a_ia_j-b_ib_j|}{2} \ge 0$,
which is $\sum_{i, j = 1}^{n} |a_ia_j-b_ib_j| -\sum_{i, j = 1}^{n} |a_ib_j-a_jb_i| \ge (\sum_{i = 1}^{n} (a_i-b_i))^2$.
We will prove $\sum_{i, j = 1}^{n} |a_ia_j-b_ib_j| -\sum_{i, j = 1}^{n} |a_ib_j-a_jb_i| \ge (\sum_{i = 1}^{n} |a_i-b_i|)^2$.

For $1 \le i \le n$, mark points $C_i(a_i, b_i)$  and $C_i'(b_i, a_i)$ on a $x-y$ coordinate. Let $l: y=x$. Draw $2n$ vectors $\overrightarrow{OC_i}$ and $\overrightarrow{OC_i'}$. Relabel all the vectors to be $v_1, v_2, \dots, v_{2n}$ such that the lines they belong to rotate clockwise from ${l}$ (if any two lines coincide, label them at any order). Then $v_i$ is in the first quadrant and $v_i, v_{2n+1-i}$ are symmetric wrt ${l}$.
Let  $v_1+v_2+\dots+v_n=\overrightarrow{OX}$ and $\blacksquare_{WXYZ}$ is a square where $W, Y$ is on ${l}$.
Let ${P}$ be the polygon with side vectors $v_1, v_2, \cdots, v_{2n}, -v_1, -v_2, \cdots, -v_{2n}$. ${P}$ is not self-intersecting, its sides are all counterclockwise, and ${P}$ is both axially (wrt ${l}$) and centrally symmetric. $X, Z$ is on ${P}$.
[img]https://cdn.artofproblemsolving.com/attachments/f/2/49519ed70327db62e750ad8cf8bd16f51dcff4.png[/img]
Since $|a_ia_j-b_ib_j|=v_i \times v_{2n+1-j}=v_j \times v_{2n+1-i}$, $-|a_ib_j-a_jb_i|=\begin{cases} v_i \times v_j & i<j \\ 0 & i=j \\ v_{2n+1-i} \times  v_{2n+1-j} & i>j \end{cases}$
and $WX=\sum_{i = 1}^{n} |a_i-b_i|$, we just need $\sum_{1 \le i<j \le 2n} v_i \times v_j \ge XW^2$, which is $S_P \ge S_{WXYZ}$.
Recall that all vectors are in the first quadrant and have positive slope, $P$ covers $\blacksquare_{WXYZ}$. Done.


Let a,b,c be three real numbers such that 0 \leq a \leq b \leq c \leq 3.
Prove that

(a – b)(a^2 – 9) + (a – c)(b^2 – 9) + (b – c)(c^2 – 9) ≤ 36.

Hi  Manlio....i  would  like  you  to  post some  inequalitis  from  italian  mo.....thanks!!
nickolas

I don't think this solution is correct !!!!!

Has anyone found a different solution????

Thank you very much !!!!!!

manlio, can you check this once again?
It is almost same with my solution; LHS≤24sqrt(2) Equality when a=0, b= sqrt(2), c=2sqrt(2) :)

Sorry, I don't understand what to do. :?

[quote="sqing"]Let $a,b,c $ be three real numbers such that $0 \leq a \leq b \leq c \leq 3$. Prove that\[(a - b)(a^2 - 9) + (a -c)(b^2 - 9) + (b - c)(c^2 -9) \le 36.\][/quote]

it is very easy to prove that for this condition we have

$f(a,b,c)=(a - b)(a^2 - 9) + (a -c)(b^2 - 9) + (b - c)(c^2 -9) \le 24 \sqrt 2$

just notice that  $f(a,b,c) - f(0,b,c)=a(a^2-ab+b^2-18)\le 0$


Well, the CMO is coming up. It will be held on [b]Wed. March 30, 2005[/b], [b]9:00 am to 12:00 noon local time[/b].

Good luck to all contestants on your final week of preparation.

Please feel free to leave any comments here, such as past experiences, or predictions.

If you are not aiming to go to IMO, if you only can solve two or three questions, make sure you solve them completely and explain lucidly.

Make sure that you write your solutions in the appropriate places (you'll know what I mean if you take the CMO, assuming they don't change the format).

[quote="billzhao"]Please feel free to leave any comments here, such as past experiences, or predictions.[/quote]

Read problems carefully, because it's easy to misinterpret problems.
Even experts like Michael made some mistakes because of that (I think), let alone myself.  :blush:

[quote="weileaf"][quote]Is it true that every year there is always one question which can be solved using AM-GM? 'Cause I think I heard this rumour from somewhere...[/quote]

if you see a inequality question in CMO, 90% of the time, you will be able to solve it using AM-GM. But then, I think chances are , you can also solved it with Cauchy....[/quote]

Yeah, but AM-GM is stronger than Cauchy anyways.

[quote="Lucky707"]Where do you guys learn all these inequalities anyway?[/quote]Where else? In the [url=http://www.mathlinks.ro/Forum/index.php?f=32]Inequalities Forum[/url] :)

[quote="11th dimension"]I have a question, in CMO if they say natural number, does it include 0?
In theory, I believe 0 is a natural number, but in many contests, it is excluded.[/quote]

Whole numbers include 0, natural numbers do not.  Please correct me if I'm wrong.

Natural numbers [i]usually[/i] excludes 0. But if they'll probably say positive integers or non-negative integers to be more precise.

Notationwise, it's conventional (at least in Canada) to have $\mathbb{N} = \mathbb{N^*}= \{1, 2, \ldots\}$, $\mathbb{N}_0 = \{0, 1, 2, \ldots\}$. But they are likely to define anything ambiguous, as they did in 2002.

CMO will start in less than 12 hours for many of us  :? , good luck everyone! :)

My sincere wishes to every participant, good luck!  :)

Okay, well, people are still writing in BC right now, so I'm not going to say anything other than... well, that was interesting.  I think I'll edit this post later to make it more exciting.

[quote="geoffreynham"]well, that was interesting.[/quote]

Very.

[quote="Aidin Kashigar"]i want to see a solution for question 4. it didn't seem like the average inequality.[/quote]

I have also provided a solution (purely algebraic).  It uses the substitution $u=a+b-c$, which is something you should always consider in triangles.


prove that any integer can be written in the form x^2 + y^2 - 5z^2

[hide="Solution"]
Let $ x,y,z$ be integers, then let:
(I) $ x=2z$ and $ y=z+1$,  hence we have:
$ x^{2}+y^{2}-5z^{2}=2z+1$ so we can get here all odd integers.
(II) $ x=2z+1$, $ y=z-1$
$ x^{2}+y^{2}-5z^{2}=2(z+1)$ so we can get here all even integers.
[/hide]


limit
search
calculus
calculus computations

Two questions: Should the first one be: $\lim_{n\to\infty}\sum_{i=1}^{n}\frac{(-1)^{i}}{i}$ ?

Should the $n$ and $k$ on the last term of the second one be switched?

For 2) search on this box for my name :D

1) ln(1/2)=-ln(2)=-1+1/2-1/3+1/4+...
2) We can divide by a_k. Then we see that  (a_i/a_k)^n->1 if a_i=a_k
else (a_i/a_k)^n->0. So, lim=lim ((a_k)*(s_n)^(1/n)), where s_n<=k.
That is lim=a_k.


(a) Solve $sin2x=cos3x$
(b) Solve $sin3x=cos2x$
(c) Find the exact value of $cos22.5+sin22.5$

[hide="a"]
$ 90-2x=3x $
$ 18=x $

use same method for b...
[/hide]

[/hide]

I meant answer in general form... (not just $x$ between 0 and 90 degrees)

[quote="deej21"]Well isn't $\sin\theta=\cos(90-\theta)$ and vice versa true for all values of $\theta$, not just $0<\theta<90$?[/quote]
He means that we should add something like $2k\pi$ to it.

3
[hide]
Using half angle formulas
$\frac{\sqrt{2+\sqrt 2} + \sqrt{2 - \sqrt 2}}{2}$
Simplifying top:
$\sqrt{\left(\sqrt{2+\sqrt 2} + \sqrt{2 - \sqrt 2}\right)^2}$
$\sqrt{4+2\sqrt 2}$
answer:
$\frac{\sqrt{4+2\sqrt 2}}{2}$
[/hide]

It might be possible to break that down more... I'm no good with sqrts inside sqrts... I'm pretty surprised I managed to simplify the top the way I did.

This stuff is all in the Art of Problem Solving books.  :lol:


A collection of subsets of $\{1, 2, \cdots , n\}$ has the property that each pair of subsets has at least one element in common. Prove that there are at most $2^{n-1}$ subsets in the collection.

[hide]Every subset that you put into the collection automatically excludes its complement subset from also being included in the collection. Because every subset has a unique complement, at most half of the subsets can be included. There are $2^{n}$ total subsets, so at most $2^{n-1}$ of them can go into the collection.

Note: The complement of a subset are the elements of the universal set that are [i]not[/i] in the subset. For example the complement of $\{1,2,3\}$ in this problem is $\{4,5,6,\ldots,n\}$
[/hide]

This would be a place to use pidgeonhole.


graph theory
advanced fields
advanced fields unsolved

Hi guys ;

Does anybody know a special program for graph theory ? liked to sketch bipartite , I know I can sketch using other program , but I asked if there is a special program or some applet on internet .

 :)

this might be what you're looking for

free
http://pigale.sourceforge.net/

free trial
http://www.geocities.com/pechv_ru/

also others mentioned in [url=http://echochamber.me/viewtopic.php?f=17&t=45732]this[/url] thread at xkcd


Let $e$ denote the highest power of 2 that divides $n!$ and $b$ the number of $1$'s in the binary representation of $n$. Prove that $n=e+b$

Then show that the number of odd terms in the $n$th row of Pascal's Triangle is $b$.

are you sure? :huh:
$(5!)_2=1111000$ and $b=4$, $e=3$(no. of final zeros) but $4+3\neq 5$

[quote="JohnConstantine"]are you sure? :huh:
... $b=4$...[/quote]

$b$ is the number of $1$'s in $n$, not in $n!$. In your example it should be $2$ ($5_{10}=101_{2}$), which gives $e+b=n$.


I have 18 distinguishable socks in my drawer: 8 white, 6 brown, and 4 blue.  In how many ways can I choose a pair of socks, provided that I get two socks of the same color?

Separate the three categories of color. We choose $ 2$ from each quantity of a certain color, then add up our results. There are $ \binom{8}{2}\plus{}\binom{6}{2}\plus{}\binom{4}{2}\equal{}28\plus{}15\plus{}6\equal{}\fbox{49}$ ways to pick a pair of socks that have matching colors.


p is prime number and"m,n"are integers,find all(p,m,n):
p=m^2+n^2 and (m^3+n^3)-4\p is integer

Do you mean (m^3+n^3-4)/p because if it would be trivial if it was (n^3+m^3) - 4/p

[hide="Solution"]Write $ \frac {m^3 \plus{} n^3 \minus{} 4}{n^2 \plus{} m^2} \equal{} m \plus{} n \minus{} \frac {mn(m \plus{} n) \plus{} 4}{m^2 \plus{} n^2}$. If $ m$ and $ n$ are both even, then $ p$ must have a factor of $ 4$ so it cannot be prime. If $ m$ and $ n$ are both $ odd$, then the numerator is even so $ p$ must have a factor of $ 2$. This is only possible if $ m^2 \plus{} n^2 \equal{} 2\implies{m, n \equal{} \pm 1}$. The numerator is also even if one of $ m$ and $ n$ is odd in which case there are no solutions. Thus, the only two solutions are $ (m, n, p) \equal{} (\pm 1, \pm 1, 2)$.[/hide]

if one of"m"and"n"is odd and the other one is even,the equation have an answer(maybe answers)
like this:(m,n,p)=(2,1,5)

why you don't answer this question?

barobache haye iran,la'aghal shoma ye javabi bedin :wink:


Here is a problem that I came up with. I am not sure how hard this problem is; I am not a good judge of the difficulties of inequality problems. I hope that it is appropriately placed.

If $ x$, $ y$, and $ z$ are positive real numbers, then

\[ \frac{x}{y^{2}(xy+z^{2})}+\frac{y}{z^{2}(yz+x^{2})}+\frac{z}{x^{2}(zx+y^{2})}\ge \frac{3}{2xyz}\]

[quote="Altheman"]That is essentially my solution. I am suprised that algebraic manipulation was so straightforward for you. I am probably just inexperienced at inequalities.

Is there anything in particular that signaled you to solve the problem in the way that you did?[/quote]

no.. just i realized that the cyclic sum of $ x(xy+z^{2})$ is simple, so i applied cauchy directly. though, i think this is really nice!