no, i don't mean problem solving strategies; more like when to guess, leave blank, etc.

i'm aiming for 100 or more on the 12 (yeah, i'm not very good...)

the strategy i developed for myself:

answer at least 15 questions that i'm confident i got right

leave 7 questions blank that i can't solve and/or have no time to solve them

guess on 3 questions that i know i can't get

if all goes well, this plan gives me a 100.5.

does this sound like a good plan?

I wouldn't do that... if you got one wrong and didn't guess any right, you would automatically disqualify yourself from AIME. Personally, I wouldn't keep anything in mind about leaving $ x$ questions blank, etc.: just prepare as best you can and answer the ones you can.

Don't try to game the test. You should always answer as many as possible just in case you get one or two wrong. If you can eliminate two answers, guess.

Honestly, if your goal is truly to just make AIME, set your plan to be little bit higher than the exact 100. By doing this, you will be ready for any unexpected mistakes.

Even if you phrase strategies in terms of answering so-and-so many questions and leaving so-and-so many blank, the real element of strategy is [i]time management[/i].

So what's a time management strategy consistent with trying for 100 points? As you've indicated, answering 18 questions and making 3 mistakes would suffice; so would answering 20 questions and making 4 mistakes. So let's say you're committed to doing about 20 questions - and you're comfortable with the idea that maybe you're never even going to read through some of the questions on the last page. (Suffice it to say that someone whose ambition is a 130 would adopt a very different strategy.)

Don't set question by question timetables, and don't constantly check your watch or the clock. Once in a while, look up to see where you are with time, but don't make yourself crazy. [i]Don't hurry![/i] Make very sure that you've read [i]exactly[/i] what the problem says. Take the time to write out intermediate steps in sufficient detail and with sufficient care on your scratch paper. (And remember that you'll need to look at that scratch work as you check through your answers - don't be so cryptic that you can't yourself figure out what you said or why.) Breathe. Don't be afraid to abandon - at least temporarily - a problem that has you stumped. Don't get frantic but do move onwards.

How many guesses? I'm going to suggest that you not guess at all, but that in some cases you be bold - make an assumption that you're pretty sure is right but that you can't fully prove to your satisfaction right now.

And what if it goes wrong anyway? You just can't find enough question that you're confident of to have a reasonable hope of it adding up to 100? It was [i]way[/i] harder than the practice rounds you attempted? That could mean any of several things. One of the things it could mean is that the test [i]is[/i] harder than standard - and because of that the scores may run so low that the AIME cutoff isn't 100 but 96 or 94.5 or something like that. This scenario has happened. The AMC isn't the ETS; they don't pre-test or statistically validate their questions in advance, and they don't know if they have the difficulty calibrated they way they like until the scores come in. So don't throw away any points at the end, because you don't know whether they might matter.


Could anybody (perhaps Darij) tell me What About construct the geometry basen in isometries????

Thanks
José Carlos

With my knowledge, we can construct the geometry with invariants of group on space.
Through vector space with inner product (Euclidean space) we can define a isometric transformations group.
And F.Klein saw geometry as: the study of invariants under a group of transformations.


Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:

$\textbf{(A)}\ 4\text{ mph}\qquad
\textbf{(B)}\ 8\text{ mph} \qquad
\textbf{(C)}\ 12\text{ mph} \qquad
\textbf{(D)}\ 16\text{ mph} \qquad
\textbf{(E)}\ 20\text{ mph}$

let the time they traveled be $ t_1$, and the rate of B be $ r_1$. Because A traveled to Poughkeepsie, then 12 miles back, he traveled 72 miles. B traveled 60-12 = 48 miles.
$ 72 \equal{} t ( r\plus{}4)$
$ 48 \equal{} tr$
so r $ \equal{} 8$ and the rate of a is $ r \plus{}4 \equal{} 12 \boxed {C}$

Minor correction: B is the faster one, so it's just $ \boxed{\textbf{(B)}\ 8\,\text{mph}}$. Also, more detail on solving: $ t\equal{}\frac{72}{r\plus{}4}\equal{}\frac{48}r\equal{}\frac{24}4\equal{}6\implies\frac1r\equal{}\frac18\implies r\equal{}8$.


How many 3-element subsets of the set $\{3,3^2,3^3,\ldots,3^{1000}\}$ consist of three numbers which form a geometric sequence?

This can be rephrased as: How many three-element subsets of $\{1,2,\dots,1000\}$ form an arithmetic sequence.

Let the elements be $a,b,c$ where $a<b<c.$ We must have $a+c=2b$, hence $a$ and $c$ must be of the same parity.

1) Among $500$ even numbers in the given set, for every even $c$ we have ${c\over 2}-1$ even numbers smaller than it. Choosing one of them as $a$ will automatically determine $b$. The number of such choices for all possible $c$ is $\sum_{k=1}^{500}(k-1)=\sum_{k=0}^{499}k=\frac{499\cdot 500}{2}$

2) Among $500$ odd numbers in the given set, for every odd $c$ we have $\frac{c-1}{2}$ odd numbers smaller than it. Similarly as in the previous case, we find that the number of choices is $\sum_{k=1}^{500}(k-1)=\sum_{k=0}^{499}k=\frac{499\cdot 500}{2}$

Therefore, the total number of choices is $499\cdot 500=249500$

[quote="jhcreinhardt"]Another Solution

[hide]Take this and turn it into the amt. of arithmetic sequences from (1,2,3...1000), like Farenhajt did.  With common difference 1, there are [(1,2,3) (2,3,4) ... (998,999,1000)] as solutions, a total of 998.  Common difference of 2, there are [(1,3,5) ... (996,998,1000)], making 996.  This continues on toward the maximum common difference of 499, which only has two solutions.  To find all of the solutions, we find the summation of the first 499 even numbers, which is [b]249500[/b][/hide][/quote]That was my solution. :)


Did anyone else get something like this?

If you did, please post it so i can see i am not the only doomed one

:(

What do you men? I have gotten 13.37, 5 on the dot.... what do you mean?

ur a rare breed

nothing to worry about, its just that u got 23 secs on the spot.

so may i say...

NICE!  :D

rating update: my rating is still 1068. :harhar:

do u mean your 2.3 (or is that a 23?) time?

If that is indeed what u mean, then its not weird. u just happened to get 23.000 (or w/e that time is, its hard to see)

i once got a prob in 8 secs exact, no .000001 or etc

jain, this was from june, plz don't revive an old thread

Please do not respond to this anymore.

It is a very old thread.

I know this post is hypocritical.


Let $ a, b$ and $ c$ be non-negative numbers such that $ ab\plus{}ac\plus{}bc\equal{}3$.

Prove that $ \sqrt{a\plus{}3}\plus{}\sqrt{b\plus{}3}\plus{}\sqrt{c\plus{}3} \ge 6$.

[quote="mathwizarddude"]Let $ a, b$ and $ c$ be non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3$.

Prove that $ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge 6$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196058

And what is the best constant of the inequality:
$ \sum \sqrt{a\plus{}k} \geq 3\sqrt{1\plus{}k}$
When $ ab\plus{}bc\plus{}ca\equal{}3$?  :)


Let $ A \equal{} \{(a_1,\dots,a_8)|a_i\in\mathbb{N}$ , $ 1\leq a_i\leq i \plus{} 1$ for each $ i \equal{} 1,2\dots,8\}$.A subset $ X\subset A$ is called sparse if for each two distinct elements $ (a_1,\dots,a_8)$,$ (b_1,\dots,b_8)\in X$,there exist at least three indices $ i$,such that $ a_i\neq b_i$.
   Find the maximal possible number of elements in a sparse subset of set $ A$.

A subset of $ A$ with the desired property cannot have 8 elements. In a $ 8$ element set,since $ a_6$ can be filled in 7 ways, by pigeon principle we can find two distinct elements $ (a_1,a_2,\ldots,a_8)$ and $ (b_1,b_2,\ldots,b_n)$ we can have only $ a_7\neq b_7$ and $ a_8\neq b_8$.

A construct of $ X$ with 7 elements:

$ 1,1,1,1,1,1,1,1$
$ 1,1,1,1,1,2,2,2$
.
.
.
.
$ 1,1,1,1,1,7,7,7$

Maybe I misundustood the problem, but if you add:
$ 1,2,1,1,1,7,8,9$ you have eight members in $ X$

Yes, we can add the 8 th element in this case. Then there is something wrong with what I wrote.

[quote="I"]... we can have only $ a_7\neq b_7$ and $ a_8\neq b_8$.[/quote]

I see here is where I slipped. First 6 elements can be filled in $ 2.3.4.5.6.7\equal{}5040$ ways. In a set with $ 5041$ elements,by pigeon hole principle we can find $ (a_1,\ldots,a_8)$ and $ (b_1,\ldots,b_8)$ so that $ a_i\equal{}b_i\quad \forall 1\leq i\leq 6$
qed ?

Clearly you prove that the number is lower than 5041, however is it possible to find a 5040-example?

Actually,no clear proof was given here,so here is my proof,i don't know official,but maybe they are the same:
Answer:$ 7!$
Solution:
In fact,them most hard part of the problem is to construct correct example,when the right answer achieved and to prove that it is correct indeed.
But at first we will proof that $ |X|\leq 7!$.
Let $ X$ be sparse subset of $ A$ and $ \overline{x}$ be a sequence of natural numbers $ (a_1,a_2,\dots,a_n)$,if i write $ \overline{x}\in X$,that means
$ \overline{x} \equal{} (a_1,a_2,\dots,a_8)$.
Consider $ X' \equal{} \{(a_1,a_2,\dots,a_6)$: such that for some $ a_7,a_8\in\mathbb{N}$,element $ (a_1,a_2,\dots,a_8)\in X)\}$
Let $ \overline{x},\overline{y}\in X'$,then $ \overline{x}\neq\overline{y}$,otherwise there will be less than $ 3$ indices,in $ x \equal{} \overline{x}(x_7,x_8),y \equal{} \overline{y}(y_7,y_8)\in X$,such that $ x_i\neq y_i$,contradiction.Hence $ \overline{x}\neq\overline{y}$.
So $ |X| \equal{} |X'|\leq 2\cdot 3\cdot 4\dots\cdot 7 \equal{} 7!$.
Let $ A' \equal{} \{(a_1,a_2\dots,a_6)$,$ 1\leq a_i\leq i \plus{} 1,a_i\in\mathbb{Z}\}$.Thus $ |A'| \equal{} 7!$.
Now for each $ \overline{a'}\in A'$,we construct $ \overline{a} \equal{} \overline{a'}(a_7,a_8)$,such that
$ a_7 \equal{} a_1 \plus{} a_2 \plus{} \dots \plus{} a_6 (mod 7)$,and
$ a_8 \equal{} \sum_{i \equal{} 1}^{6} i\cdot a_i (mod 7)$.
The last step is to prove that $ X \equal{} \{\overline{a}: \overline{a'}\in A'\}$-is a sparse subset of $ A$.
Consider two elements of $ X$,$ \overline{a},\overline{b}$.
It is easy to prove that there are at least three $ 3$ indices,such that $ a_i\neq b_i$,just consider three cases,
1. $ \overline{a'},\overline{b'}$ have three diffrent indices.
2. $ \overline{a'},\overline{b'}$ have two diffrent indices.
3. $ \overline{a'},\overline{b'}$ have one diffrent indice.
So the problem is proved.


LaTeX
real analysis
real analysis unsolved

sorry if this isn't clear, but how do I show that (∑k^n/k!,k=0,∞)(∑(-1)^k k^n/k!,k=0,∞) is an integer (-1)k kn/k!) is an integer?

Nope. At this stage I have trouble understanding what is asked :P. I see the product $ \Bigl(\sum_{k=0}^\infty\frac{k^{n}}{k!}\Bigr)\Bigl(\sum_{k=0}^\infty(-1)^{k}\frac{k^{n}}{k!}\Bigr)$ but the rest of your sentence [i]"is an integer $ (-1)k kn/k!$) is an integer"[/i] just violates the English grammar and the parity principle for parentheses. Should it be just [b]is an integer[/b] :??

By the way, the rest of us will greatly appreciate if you spend half an hour learning basic $ \text{\LaTeX}$ :)

$ k^{n}$ can be written as the integer combination of $ k_{a}=k(k-1)\cdots (k-a+1)$. It is easy to prove that $ \sum_{k=0}^{\infty}\frac{k_{a}}{k!}=e$ and $ \sum_{k=0}^{\infty}\frac{(-1)^{k}k_{a}}{k!}=\frac{(-1)^{a}}{e}$. So $ \sum_{k=0}^{\infty}\frac{k^{n}}{k!}=\alpha e$, $ \sum_{k=0}^{\infty}\frac{(-1)^{k}k_{a}}{k!}=\frac{\beta}{e}$, where $ \alpha$ and $ \beta$ are integers.


[quote="kunny"]Let $x,\ y,\ z$ are distinct numbers such that $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}.$ Find the common value of of this equation.[/quote]
[hide]$x=y=z$?
[/hide]

[quote="kunny"]Let $x,\ y,\ z$ are [b]distinct[/b] numbers such that $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}.$ Find the common value of of this equation.[/quote]

I don't think $x=y=z$ works.

[quote="kunny"]Let $x,\ y,\ z$ are distinct numbers such that $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}.$ Find the common value of of this equation.[/quote]

[hide]$x+\frac{1}{y}=y+\frac{1}{z}$

$x=\frac{y^{2}z-z+y}{yz}$

$y+\frac{1}{z}=z+\frac{1}{x}$

$x=\frac{z}{yz-z^{2}+1}$

$\frac{y^{2}z-z+y}{yz}=\frac{z}{yz-z^{2}+1}$

$\Leftrightarrow (y-z)(1+z(y-1))(1+z(y+1))=0$

Since $y\not= z$, $y=-\frac{1}{z}\pm 1$

Sub it into $y+\frac{1}{z}$ to get $\pm 1$ and that's the common values.

The corresponding solutions are $\{-2,1,-\frac{1}{2}\}$ and $\{2,-1,\frac{1}{2}\}$[/hide]


This is Hlawka inequality, no doubt it was seen before here (what wasn't ?)

Can somebody show it, please..
and I`ll really be thankful if somebody show this inequality, too:

$|x_1+x_2+..+x_n|\leq |x_1|+|x_2|+..+|x_n|$

In this inequality can we set (without loss of generality) that $|a|\geq |b|\geq |c|$?


function
trigonometry
limit
calculus
derivative
calculus computations

i was asked to evaluate sinhx.sinh(e^-x) as x approaches infinity, since e^-x approaches zero faster than x approaches infinty, the answer should be zero, but how do i show it  mathematically?

but it is  product which i dont know how to arrangne into f(x)/g(x) since the functions are sinh, do i used cosech??

More concretely: write down the definition of $ \sinh(x)$, and then estimate $ \sinh(e^{\minus{}x})$ using an approximation of $ \sinh$ near zero.

The answer is neither zero nor $ \infty$.

why isnt it zero, just to clarify we were asked to evalute  $ \sinh x\sinh e^{\minus{}x}$ as x approaches infinty

[quote="calculus_jy"]why isnt it zero, just to clarify we were asked to evalute  $ \sinh x\sinh e^{ \minus{} x}$ as x approaches infinty[/quote]

as  j merry said  first  write  the  sinh  terms  according to the definiton.  then u will automatically  see the answer

Yeah, same here.  I had to apply the definition in exponentials before applying L'Hopital.

$ {\frac {e^x - e^{ - x}}_{2} }\frac {e^{e^{ - x}} - e^{ - e^{ - x}}}_{2}$ does not give a sound intermediate form for evalution of the limit

Why not?  Multiply it out and do some more work.

That's not very useful. We're looking at large values of $ x$; what's big and what's small?
First, that $ \frac{e^x\minus{}e^{\minus{}x}}{2}$ is basically just $ \frac{e^x}{2}$. Second, that $ \sinh(e^{\minus{}x})$ is the hyperbolic sine of something close to zero. What does $ \sinh(t)$ look like for small $ t$?

Lol.  Overcomplicated solution is overcomplicated.

[hide]
$ \lim_{x \to \infty} (\sinh \,x) (\sinh \,e^{ \minus{} x})$

$ \equal{} \frac {1}{4} \lim_{x \to \infty} \left(e^x \minus{} e^{ \minus{} x}\right) \left(e^{e^{ \minus{} x}} \minus{} e^{ \minus{} e^{ \minus{} x}}\right)$

$ \equal{} \frac {1}{4} \lim_{x \to \infty} e^x \left(e^{e^{ \minus{} x}} \minus{} e^{ \minus{} e^{ \minus{} x}}\right) \plus{} 0$

$ \equal{} \frac {1}{4} \lim_{x \to \infty} \frac {e^{e^{ \minus{} x}} \minus{} e^{ \minus{} e^{ \minus{} x}}}{e^{ \minus{} x}}$
[/hide]

L'Hopital's rule is not the only tool. I'd rather not calculate that chain rule derivative at all, using instead that $ \sinh(t)\equal{}t\plus{}O(t^3)$ near zero. We have $ \lim_{x\to\infty}(\frac12 e^x\plus{}\text{small})(e^{\minus{}x}\plus{}\text{small})\equal{}\lim_{x\to\infty}\frac12\plus{}\text{small}\equal{}\frac12.$
Overcomplicated? It took me about a second, in my head.

[quote="jmerry"]Overcomplicated? It took me about a second, in my head.[/quote]

Referring to my own.

i get how to derive the limit 1/2 but i still dont understand what is wrong with the following statement
since e^-x approaches 0 faster than x approaches infinity, sinhe^-x approaches 0 faster than sinhx approaches infinity, so the product of the limit should be zero,

i am so confused, can somehow explain in detail what went wrong with that and how do we know that the limit 1/2 is definitly correct when just above you have explained to me various ways to do it, how can you gurantee that there is no other way which will give a different limit


algorithm
pigeonhole principle
induction
inequalities
combinatorics unsolved
combinatorics

1) Let $S$ be a subset of $\{ 1,2,\cdots, 2046 \}$ with $21$ elements. Show that there exist distinct $a,b,c$ in $S$  such that $bc < 2a^{2}< 4bc$

2) Let $n$ be a positive integer. Find all $n$ - tuples of positive integers $(a_{1},a_{2}, \cdots, a_{n})$ such that $a_{1}+a_{2}+\cdots+a_{n}= 2n$ but such that no subset of the $a_{i}$ has sum equal to $n$

Please help me as soon as possible, because combinatoric is my weakest subject and I want to study more about combin. :)
I'm sorry if one of the questions has been posted before...  :maybe:

Um..  Did I do something wrong ? Can anybody help me ? Please, I just want to know whether anybody can solve these problems..

[quote="RDeepMath91"]Um..  Did I do something wrong ?[/quote]

Yes: You waited only 13 hours after you posted to start complaining that no one was solving your problems.

Both of these strike me more as number-theoretic then combinatorial, anyhow.  The second one, as far as I can tell, has only the "obvious" solutions -- the problem, of course, is to prove that there aren't any others.

Actually they look pretty combinatorial to me. Hmmm ok so for the second problem, I think you can just do this:

WLOG $a_{1}\le a_{2}\le \cdots \le a_{n}$. We use the following method to try to construct a subset of these numbers that sums to n. You start with two empty buckets each with capacity n liters, and starting from i = n and going down, we pour $a_{i}$ liters of water into any bucket that won't overflow.

The problem asks us to find $a_{i}$ so that this is not possible. Suppose we are unable to carry out this algorithm for $a_{k}$. Then this means that $a_{k}> 1+a_{1}+a_{2}+\ldots+a_{k-1}\ge k$, or else by pigeonhole one of the buckets would have had at least $a_{k}$ liters unfilled. However, note that we now have

$a_{n}+a_{n-1}+\ldots+a_{k}\ge (n-k+1)a_{k}\ge (n-k+1)(k+1)$, and

$a_{k-1}+a_{k-2}+\ldots+a_{1}\ge k-1$.

We thus have

$2n = a_{1}+a_{2}+\ldots a_{n}\ge (n-k+1)(k+1)+k-1 = nk-k^{2}+n+k$
$n+k^{2}\ge nk+k$
$(k-n)(k-1) \ge 0$

This means that either k = 1 or k = n. If k = 1, then we have $a_{1}> 1 \implies a_{1}= a_{2}= \cdots = a_{n}= 2$, which works if n is odd. If k = n, then we have $a_{n}> n \implies a_{n}= n+1, a_{1}= a_{2}= \cdots a_{n-1}= 1$, which always works.

Very nice job Probability.  Heres what I believe as a possible solution for the first part, but the bound of $2046$ seems very small:

We proceed by contradiction.  Suppose there is a subset $T\subseteq \{1,2,\dots,2046\}$ of cardinality $21$ which does not contain three distinct elements with the forementioned property.  We show such a subset cannot exist.

First, suppose we have two positive integers $x$ and $y$.  Also suppose that there is some integer between $x$ and $x+y$ such that these three numbers do not satisfy the given property.  Then either $(x+1)^{2}\le \frac{x(x+y)}{2}$ or $(x+y-1)^{2}\ge 2x(x+y)$.  These yield $y\ge x+2$ or $y\ge x+4+\frac{2}{x}$.  Hence $y\ge x+2$.

Now, if $T=\{t_{1},t_{2},\dots,t_{21}\}$ and $t_{1}<t_{2}<\cdots<t_{21}$, by above, we have that $t_{i+2}-t_{i}\ge t_{i}+2$ for all $i$.  Thus by induction, we have $t_{2k+1}\ge 2^{k}t_{1}+2^{k+1}-2\ge 3\cdot 2^{k}-2$ for all $k\in\mathbb{N}$.  But then $t_{21}\ge 3\cdot 2^{10}-2=3070$, which is obviously a contradiction.  This gives the desired result.

The curious thing about this problem is that $21$ is highly unlikely to be best possible.  I am pretty sure that even the bound $t_{5}\ge 10$ is not optimal.

I'm sorry 'coz I forgot to open this topic again..
Thx a lot, probability..  A very good solution.. ^^
But I'm still confused with the answer of my first question..
Why either $(x+1)^{2}< \frac{x(x+y)}{2}$ and $(x+y-1)^{2}\geq 2x(x+y)$ ?  :maybe:

Since these three numbers do not satisfy the given inequality, we must have $a^{2}\le\frac{bc}{2}$ or $a^{2}\ge 2bc$ for some $x\le a\le x+y$.  If you pick the first $a$ for which these inequalities will hold, you get the ones I wrote earlier...


Prove that it is impossible for three consecutive squares to sum to another perfect square.

$n^2+(n+1)^2+(n+2)^2=m^2\Longrightarrow 3n^2+6n+5=m^2$

Squares are either 0 or 1 $\pmod{3}$, but the left side is 2 $\pmod{3}$ so there are no solutions.


linear algebra
matrix
algebra
polynomial
quadratics
linear algebra unsolved

Find all matrices $ A \in M_n(\mathbb{R})$ such that $ A^2\plus{}Id\equal{}0$. Couldn't manage to solve it 8-) Can anyone give me a hint on how to approach it?

what can you say about $ n$ ? (odd, even) about the eigenvalues ?  Can you find such a matrix in $ M_{2n}(\mathbb{R})$ ?

We have the minimal polynomial: $ x^2\plus{}1.$ We must have that $ n$ is even and the characteristic polynomial is $ (x^2\plus{}1)^{n/2}.$ Find everything that is similar to the following matrix:
\[ \begin{pmatrix}0&\minus{}1&&&\\1&0&&&\\
&&0&\minus{}1&\\&&1&0&\\&&&&\ddots\end{pmatrix}\]

Thanks Alekk and Kent for your replies..

Kent: I cant see why the similar matrices to the one you gave are all the soutions. I see that all the solutions are those matrices similar over $ \mathbb{C}$ to the diagonal matrix with n/2 "i" and n/2 "-i" in the diagonal.. The question is: how do I know that there is no other (real) matrix which is similar to yours over $ \mathbb{C}$ but is not similar to yours over $ \mathbb{R}$? In the case such a matrix existed it would also be a solution...

There's an argument [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28984]here[/url]; it's different in detail, but the idea is the same. Since the minimal polynomial $ x^2\plus{}1$ is irreducible, we can construct the basis we want by choosing $ v_1$ arbitrary, $ v_2\equal{}Av_1$, $ v_3$ arbitrary outside the span of the preceding, $ v_4\equal{}Av_3$, and so on.

There's also the variant of Jordan form over $ \mathbb{R}$, where all irreducible polynomials are linear or quadratic; our blocks either look like $ \begin{bmatrix}a&1&0&\cdots&0\\0&a&1&\cdots&0\\0&0&a&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&a\end{bmatrix}$ for real roots $ a$ or $ \begin{bmatrix}a&\minus{}b&1&0&\cdots&0&0\\b&a&0&1&\cdots&0&0\\ 0&0&a&\minus{}b&\cdots&0&0\\ 0&0&b&a&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\cdots&a&\minus{}b\\ 0&0&0&0&\cdots&b&a\end{bmatrix}$ for complex roots $ a\pm bi$.

In this case, all of our blocks correspond to $ \pm i$, and we have nothing above the diagonal because the minimal polynomial is $ x^2\plus{}1$ rather than some higher power of it.

How do you show that if $ v\plus{}w\equal{}0$ with $ v \in <v_1,v_2>$ and $ w \in <v_3,v_4>$ then $ v\equal{}w\equal{}0$?


Given the nine-sided regular polygon $ A_1A_2A_3A_4A_5A_6A_7A_8A_9$,
how many distinct equilateral triangles in the plane of the polygon
have at least two vertices in the set $ \{A_1, A_2, \ldots A_9\}$?

Any pair of vertices determines two equilateral triangles, so $ 2 \binom{9}{2}\equal{}72$.  However, we have triple-counted $ A_1A_4A_7$, $ A_2A_5A_8$ and $ A_3A_6A_9$, so we must subtract each of them out twice.  $ 72\minus{}6\equal{}\boxed{66}$.

Hopefully this will suffice:

[hide="Explanation"]Consider triangle $A_1A_4A_7$. As AIME15 stated in his explanation, for each pair of vertices, we determine two equilateral triangles. Thus, segments $A_1A_4$, $A_4A_7$, and $A_7A_1$ each contribute 2 triangles towards the sum. However, note that the triangle $A_1A_4A_7$ is counted three times, once for each pair of vertices. Thus, we must subtract two of these three (we still want to keep one), so we subtract 2 from the sum. Similarly, because of $A_2A_5A_8$ and $A_3A_6A_9$, we must subtract $2+2=4$ more, so our answer is $2\dbinom{9}{2}- 6= \boxed{66}$.[/hide]

thnx guys this one kind of confused me (sort of just could not rap mind around it after 72 was incorrect)

SassmAN

[quote=mathyoga]I don't understand why $ A_{3}A_{6}A_{9} $, $ A_{2}A_{5}A_{8} $, and $ A_{1}A_{4}A_{7} $ have been triple counted. Could someone please explain?[/quote]

 These have all three vertices on the polygon, so we divide by 3
Sorry for bumping

[quote=Idefix]I don't get it. If the triangles are distinct, then does that mean that they are not congruent?

Idefix[/quote]

No, they can be congruent. Being distinct means they are in different positions, even if they are congruent.

Also I feel so proud of myself that I’ve done this problem while doing amc12  :rotfl: lmao

the equilateral triangle made it tricky for me

[quote=Singh96]isn't the problem just asking to determine distinct equilateral triangles that have vertices on nonagon? how are there any other equilateral triangles other than {a1, a4, a7}, {a2, a5, a8}, {a3, a6, a9}? Can someone provide clarity?[/quote]

Only 2 vertices need to be on the nonagon

[hide=Solution Progress]WLOG, let $A_1=e^0,A_2=e^{2\pi i/9},A_3=e^{\pi i/3},A_4=e^{4\pi i/9},A_5=e^{5\pi i/9},A_6=e^{2\pi i/3},A_7=e^{7\pi i/3},A_8=e^{8\pi i/9},$ and $A_9=e^{\pi i}.$

We want to find the number of equilateral triangles that have all their vertices at any three $A_i.$ In order for this to happen, we must have that, for $k,l,m\in\{1,2,3,4,5,6,7,8,9\},$ the complex number $A_m$ is the result of rotating $A_k$ about $A_l$ by an angle of $\pm\frac\pi3$ counterclockwise. In exponential form, these numbers are $e^{m\pi i/9},e^{k\pi i/9},$ and $e^{l\pi i/9}.$

Since $e^{m\pi i/9}$ is the result of rotating $e^{k\pi i/9}$ about $e^{l\pi i/9}$ by an angle of $\pm\frac\pi3$ counterclockwise, we must have either $$e^{m\pi i/9}=e^{\pi i/3}\left(e^{k\pi i/9}-e^{l\pi i/9}\right)+e^{l\pi i/9}\iff e^{m\pi i/9}=e^{\pi i/3+k\pi i/9}-e^{\pi i/3+l\pi i/9}+e^{l\pi i/9}$$ or $$e^{m\pi i/9}=e^{-\pi i/3}\left(e^{k\pi i/9}-e^{l\pi i/9}\right)+e^{l\pi i/9}\iff e^{m\pi i/9}=e^{-\pi i/3+k\pi i/9}-e^{-\pi i/3+l\pi i/9}+e^{l\pi i/9}.$$

@lamboreghini You are overthinking this problem by a lot.


modular arithmetic
quadratics
symmetry
number theory
relatively prime

Let $ a,b,c,x,y,z \in \mathbb{Z}$ such that
$ a^2 \plus{} b^2 \equal{} x^2$
$ b^2 \plus{} c^2 \equal{} y^2$
$ a^2 \plus{} c^2 \equal{} z^2$
Prove that $ 518400 \mid abcxyz$.

$ 518400 \equal{} 2^8 3^4 5^2$

we have three sets of pythagorean triples.  In any triple, at least one of the legs must be even (this results from analysis in mod 4).  Suppose $ a$ is even.  Then we still need either $ b$ or $ c$ to be even.  Without loss of generality, let $ b$ be even.  then so is $ x$.  Now $ c$, $ y$ and $ z$ may still be odd, so some of our evens are actually multiples of higher powers of 2.

in fact, we can rewrite our first condition as $ (a/2)^2\plus{}(b/2)^2 \equal{} (x/2)^2$ and $ a/2$ or $ b/2$ must be even.  So far we have shown that $ 2^4 | abcxyz$

Anyway, I would continue in this way, analyzing the equations in mods 3, 4, and 5 and see what comes up.  I'll work more on this later.

[quote="facis"]$ 518400 \equal{} 2^8 3^4 5^2$

we have three sets of pythagorean triples.  In any triple, at least one of the legs must be even (this results from analysis in mod 4).  Suppose $ a$ is even.  Then we still need either $ b$ or $ c$ to be even.  Without loss of generality, let $ b$ be even.  then so is $ x$.  Now $ c$, $ y$ and $ z$ may still be odd, so some of our evens are actually multiples of higher powers of 2.

in fact, we can rewrite our first condition as $ (a/2)^2 \plus{} (b/2)^2 \equal{} (x/2)^2$ and $ a/2$ or $ b/2$ must be even.  So far we have shown that $ 2^4 | abcxyz$

Anyway, I would continue in this way, analyzing the equations in mods 3, 4, and 5 and see what comes up.  I'll work more on this later.[/quote]

i think it's really hard.

I think mod 8 will actually work better than mod 4.  anyway, haven't had enough time to work on it yet.  Probably will figure it out over the weekend.

[quote="JBL"]What's the source of this problem?  (This was just posted by a different user in pre-olympiad -- are you cheating off the same test, or what?)[/quote]

I saw it in another site, but without solution. If it's here, give me the link please.

got it.

First, note $ 518400 \equal{} 2^8*3^4*5^2$.

lemma: if $ p,q$ are relatively prime and $ p|a$ and $ q|a$, then $ pq|a$
Proof - the statement is a direct consequence of Chinese remainder theorem.

So it is sufficient to prove these three statements separately: $ 2^8|abcxyz$, $ 3^4|abcxyz$, $ 5^2|abcxyz$

[hide="twos"]
Assume that, in any of the equations, both of the variables on the left side are odd.  Then the right side, a perfect square, is equivalent to $ 2 \pmod{4}$ which is a quadratic nonresidue (odds squared give 1 mod 4 and evens squared give 0 mod 4).  So this is a contradiction.

Thus at least one of the variables on the left hand side of each equation is even.

In $ a^2\plus{}b^2\equal{}x^2$, let $ a$ be even.  Then in $ b^2\plus{}c^2\equal{}y^2$, $ b$ or $ c$ must be even.  Without loss of generality, let $ b$ be even. (if we let $ c$ be even, we could proceed in the same way except analyze $ a^2\plus{}c^2 \equal{} z^2$ instead).

Now $ a^2\plus{}b^2\equal{}x^2$ and both variables on the left are even, so $ 4|x^2$.  We can now rewrite this as $ (a/2)^2\plus{}(b/2)^2 \equal{} (x/2)^2$, where $ a/2$, $ b/2$, and $ c/2$ are integers.  Thus either $ 2|a/2$ or $ 2|b/2$.  Without loss of generality, let $ 4|a$.

if $ b \equiv 2 \pmod{4}$ and $ c$ is odd, then $ y^2 \equal{} b^2 \plus{} c^2 \equiv 4\plus{}1 \equiv 5 \pmod{8}$, but  $ 5$ is a quadratic nonresidue mod $ 8$, contradiction.  So either $ 4|b$ or $ c$ is even.

[hide="Case 1: c is even"]
Then $ y$ and $ z$ are also even.  We then have the equation $ (b/2)^2\plus{}(b/2)^2 \equal{} (y/2)^2$ so either $ 4|b$ or $ 4|c$.  In either case we know that $ 8|bc$ and earlier had $ 4|a$ so $ 32|abc$ and $ 8|xyz$ since $ x,y,z$ are even so $ 2^8|abcxyz$ as desired.
[/hide]

[hide="Case 2: 4|b"]
Then we have $ 4|x$ and $ (a/4)^2\plus{}(b/4)^2 \equal{} (x/4)^2$ so one of $ a/4, b/4$ is even.  Without loss of generality, $ 2|a/4$, ie $ 8|a$.

If $ a/4 \equiv 2 \pmod{4}$ and $ b/4$ is odd, then $ (a/4)^2 \plus{} (b/4)^2 \equiv 4\plus{}1\equal{}5 \pmod{8}$ which is a quadratic nonresidue, contradiction.  So either $ 16|a$ and $ 4\parallel{}b$ or $ 8|a$ and $ 8|b$.  In both cases, $ 2^6|ab$ and so $ 2^8|abx$ and thus $ 2^8|abcxyz$ as desired.
[/hide]

[/hide]

[hide="threes"]
(The quadratic residues in mod $ 3$ are $ 0$ and $ 1$.)

Suppose both $ a$ and $ b$ are not divisible by $ 3$; that is, $ a,b \equiv \pm 1 \pmod{3}$.  Then $ x^2 \equiv 2 \pmod{3}$ which is impossible since $ 2$ is not a quadratic residue mod $ 3$.  Thus one of $ a$ and $ b$ is divisible by $ 3$.

Without loss of generality, let $ 3|a$.  identical reasoning with the equation $ b^2\plus{}c^2\equal{}y^2$ tells us that either $ 3|b$ or $ 3|c$.  By symmetry, we can choose, without loss of generality, to let $ 3|b$.

now $ a^2\plus{}b^2 \equal{} x^2$, so $ 3|x$.  we now have the equation $ (a/3)^2\plus{}(b/3)^2\equal{}(x/3)^2$, and again if neither of the quantities squared on the left hand side are divisible by $ 3$, $ (x/3)^2 \equiv 2 \pmod{3}$ which is impossible.  One of $ a/3$ and $ b/3$ is a multiple of $ 3$, so $ 3^4|abx$

Thus $ 3^4|abcxyz$ as desired.
[/hide]

[hide="fives"]
The quadratic residues in mod $ 5$ are $ 0, 1, 4$.

Addition of our quadratic residues:
$ 0\plus{}0\equal{}0$
$ 0\plus{}1\equal{}1$
$ 0\plus{}4\equal{}4$
$ 1\plus{}4\equal{}0$
$ 1\plus{}1\equal{}2$
$ 4\plus{}4\equal{}8 \equiv 3 \pmod{5}$

Except for the last two combinations, the sum of two quadratic residues is another quadratic residue.  Examining all the valid cases, we see that at least one of $ a$, $ b$, or $ x$ is divisible by $ 5$.

[hide="5|x"] If $ 5|x$, we must have $ a\equiv 1$, $ b \equiv 4 \pmod{5}$ (or vice-versa).  Then we have $ 1\plus{}c^2 \equiv z^2$ and $ 4\plus{}c^2 \equiv y^2$.  If $ c^2 \equiv 1,4 \pmod{5}$ then either $ z^2$ or $ y^2$ is equivalent to a quadratic nonresidue, which is impossible.  Hence $ c^2 \equiv  0 \pmod{5}$.  Hence $ 5|c$ so $ 25|cx$ and $ 25|abcxyz$ as desired.[/hide]

[hide="5|a or b"] Without loss of generality, assume $ 5|a$.  If $ 5|b$, then we are done.
So assume $ 5$ does not divide $ b$.  But in $ b^2\plus{}c^2 \equal{} y^2$, we must have that one of the variables is divisible by $ 5$.  Thus either $ c$ or $ y$ is divisible by $ 5$, and $ 25|acy$ so $ 25|abcxyz$ as desired.
[/hide]

[/hide]


modular arithmetic
number theory proposed
number theory

Let p be a prime which is greater than 3 and $f(x)=ax^2+bx+c$ with a,b,c are positive integers,(a,p)=1.Suppose that there exits p consecutive integers $x_
{i}$,i=1,p such that $f(x_{i})$ is a square modulo p.
Prove that:$b^2-4ac$ is divisibled by p.

I think it's true, here is the reason why,
[I mean SQ = square modulo $p$]

As the hypothesis said, $f(x)$ is SQ modulo $p$ in $\{1,2,...,p\}$
we have in the set ${1,2,...,p-1}$ has $\frac{p-1}{2}$ SQ and the left is non SQ. So at first there are two distince mod p x,y such that
$f(x)\equiv f(y)\pmod{p}$ then $a(x+y)\equiv-b\pmod{p}$ but the last equation has unique solution. From the fact that we got $p$ SQs which $p=2\frac{p-1}{2}+1$ then if there are three $u,v,w \in\{1,2,...,p\}$ distinct such that $u+v=v+w=u+w\pmod{p}$ this contract! So there is one x such that $f(x)\equiv0\pmod{p}$ (take a look at the set $\{0,1,...,p-1\}$) . Note that the number of such $x$ is odd cause in case the number is even then the the number of the lefts is odd this contract to the above arguments!

Now

$4af(x)=(2ax+b)^2-\Delta$ where $\Delta=b^2-4ac$ then it implies $\Delta$ is SQ in case it is not divisible by $p$. But the equation
$(2az+b)^2\equiv\Delta\pmod{p}$ has exactly two distince roots in case  $\Delta$ is not divisible ie... $f(x)=a(x-x_1)(x-x_2)$. So if $\Delta$ is not divisible by $p$ then we have exactly two $x$ that $f(x)$ is divisible by $p$. This can not cause the number of such $x$ is odd (two is even? :D )
So $\Delta $ is divisible by $p$.


The floors of a 4-floor house can be coloured in eight ways using the colours blue and yellow, in the case you keep to these rules: No two neighbouring floors may be blue, but they may be yellow. How many colourings are there for a 10-floor and 20-floor house ?

let $a_n$ denote the number of colouring with $n$-th floor blue.
let $b_n$ denote the number of colouring with $n$-th floor yellow.
we have
\[a_n = b_{n-1},\quad b_n = a_{n-1}+b_{n-1}\]
thus
\[b_n = b_{n-2}+b_{n-1}\]
 :D
We come over to a Fibonacci sequence...  :D

Yes, you are right. The answer is the $n+2$th Fibonacci number. But your proof does not seem very detailed.

well, i'd better do some explanations.

$a_n = b_{n-1}$ is because we need $n$-th floor being blue, that requires ($n-1$)th floor being yellow.

$b_n=a_{n-1}+b_{n-1}$ is because that, we need $n$-th floor being yellow, ($n-1$)-th has no constraints: it can be either yellow ($b_{n-1}$) or blue ($a_{n-1}$).

you should check both equalities for initial values.

and the answer goes to $a_n+b_n = b_{n-1} + b_n = b_{n+1}$, and we know that $\{b_n\}$ is a Fibonacci sequence, thus the answer is the ($n+2$)th item of Fibonacci sequence.


function
limit
logarithms
algorithm
real analysis
real analysis solved

Is there a real function $f$ that tends to infinity as x tends to infinity, but such that $\frac{f(x)}{ln_n(x)}$ tends to 0 for all values of n? Here $ln_n$ is the n-th composition of the logarithm.

Yes, for instance the inverse of the Ackermann function.

[quote="grobber"]Bu what does "$f$ tends to infinity as $x$ tends to infinity" mean?[/quote]
It means $\lim_{x\to\infty} f(x) = \infty$.

[quote="alekk"]but the Ackerman function is define over the integer isn't it ?[/quote]
Connect $f(n)$ and $f(n+1)$ by a line, which extends $f$ to $\mathbb{R}$.

OK  :D  but now, how do you prove the stament with the ackerman function ?

Sorry about that stupid question. I understood it as some sort of comparison between $f$ and the idenitity function, for some reason, even though I use that formulation quite oftem myself :?. Sorry :blush:. Also, I think there should be other examples :).

[quote="Ravi B"]An example that is perhaps easier to prove than inverse Ackerman's: Define f(x) to be the number of times you need to take natural logarithm to bring x to at most 1.[/quote]

As you know this was the function from Tarjan's paper on Log* behavior in UNION-FIND.
Is there any natural computation problem leading to inverse-Ackermann behavior in the bounds
rather than Log*?

Yes, actually the exact running time of the union-find algorithm does involve the inverse Ackermann function.  Some algorithms books are content to show that the running time is bounded above using the iterated logarithm function I defined above (usually called $\log^*$).  But Tarjan, either in his original paper or a later one, showed that the true behavior was inverse Ackermann, both as an upper and lower bound.

As another example, I think Noga Alon and one of his colleagues showed that there exist linear-size superconcentrators with inverse Ackermann depth.

Hmm, fleeting_guest, I wonder if I know you.


How'd everyone do?

Here's my results:

asteroid       *  x  x  *  x  *  *  *  *  x  t
ontherun     *  *  *  *  *  *  *  t  t  t
twalk           *  *  *  *  t  t  t  t  t

The competition was interesting for me.  I thought my ontherun solution was going to be the worst, but it was the best, I thought my twalk solution was going to be the best, and it was the worst, and I used DP on asteroid contrary the the given net flow solution (I'm not that familiar with 'cover' problems).

asteroid    *  *  x  *  *  *  *  x  *  x  x
ontherun    *  *  *  *  *  x  *  x  x  x

eh... i'm a little rusty. haven't worked on usaco probs for a while now.

i just used greedy for the asteroid one (blast row/column with most asteroids)... and ontherun had a very flawed algorithm (simply chose closest point to go to) but it evidently worked for quite a few cases.

Ya it seems that the test cases this time lended themselves to flawed solutions.  Did you not do twalk?  I didn't have any optimizations (spent all my time coming up with solutions for the other two), so my basic recurrence algo was tooooo slow.

no, i didnt want to do twalk, it didnt look really fun... and i thought i'd have issues with the dictionary file so i didnt want to waste time coding that.

It was rather easy. I got a max score, but several other contestants did the same. See you in the next contest.  :P

Yea, I'm definetly not as great as a lot of other people here, but this is what I got.  And this was for the SIlver division too.

acrobat     *  *  *  x  x  *  *  x  *  x
skyline     *  *  *  *  *  *  *

Are the rankings for the contest up yet? Or am I missing something?  Usually they're out on the Thursday after the contest.  I have not received any email about them being posted yet.

[quote="PenguinMath"]ace.delos.com/NOV05results[/quote]

Thank you! Been looking for that.


[quote="shyong"]$z_1,z_2,z_3$ are three points on a unit circle in a complex plane . They can be written as $\alpha,\beta,\gamma\in\mathbb{C}$ respectively . Show that

$\frac{\alpha\beta\gamma}{(\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)}$ is real .[/quote]

[hide="this is nice"]just note that $\mid \alpha \mid ^2=\alpha \cdot \overline {\alpha}$
Take the conjugate and do some calculations!
[/hide]

 :)

[quote="shyong"]$z_1,z_2,z_3$ are three points on a unit circle in a complex plane . They can be written as $\alpha,\beta,\gamma\in\mathbb{C}$ respectively . Show that

$\frac{\alpha\beta\gamma}{(\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)}$ is real .[/quote]

let $\arg{\alpha}=a$, $\arg\beta=b$ and $\arg\gamma=c$

theorem: $\arg(\alpha+\beta)=\frac{a+b}{2}$

(consider the parallelogram formed by the numbers in the complex plane)

$\arg\left[ \frac{\alpha\beta\gamma}{ (\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)} \right]$
$=\arg\alpha+\arg\beta+\arg\gamma-\arg(\alpha+\beta)-\arg(\alpha+\gamma)-\arg(\beta+\gamma)$
$=a+b+c-\frac{a+b}{2}-\frac{b+c}{2}-\frac{c+a}{2}=0$

and since the argument is 0, the number lies on the real line

Apply only the following two simple properties :
$\boxed {\ z\in R\Longleftrightarrow z=\overline z\ }$ and $\boxed {\ |z|=1\Longleftrightarrow \overline z=\frac 1z\ }\ .$


geometry
geometric transformation
reflection

Here's one.

Dogs think they are human, cats think they are God.

"Quitting smoking is very easy. I have done it many times myself."
- [i]Mark Twain[/i]

What do you mean by appropriate?

[quote="bubka"]

What do you mean by appropriate?[/quote]

nothing involving sex, drugs, murder, or communism... basically nothing that I'd post in allcaps

Does tobacco count as a drug? If so I have already put in an inappropriate one.  :wink:  :rotfl:

Anyway, approppriate means only nothing involving sex or ethnicity. That will do for the time being.

[hide="Read iff you are a male human."]Mirrors reflect without talking and women talk without reflecting.[/hide]

There are no typos in this message.

"I have never said anything worthy of being quoted" --Nicole June DeVine

I know, not very funny...

"Sure!" - [i]George Bernard Shaw[/i]
(on being told by somebody that sugar is the only word in english which begins with a 'sh' sounding 's'.)

[quote="eryaman"][quote="Xantos C. Guin"]"It is better to keep your mouth shut and make people think that you are stupid than to open it and erase all doubt."

This was in someone's sig.[/quote]I think it's SplashD's signature...[/quote]

ccy says he had the quote in his sig first.


Γίνεται να δομήσουμε το $ \mathbb R$ ως $ \mathbb C$-διανυσματικό χώρο;Αν ναι πώς; :maybe:

Πολύ όμορφη άσκηση.  :)

[quote="Demetres"]Πολύ όμορφη άσκηση.  :)[/quote]

Πράγματι!

Ας συμβολίσουμε με $ z\circ x$ τον πολλαπλασιασμό ενός "βαθμωτού" $ z\in\mathbb{C}$ με ένα "διάνυσμα" $ x\in\mathbb{R}$.

Ένα λογικό πρώτο βήμα είναι να υποθέσουμε ότι $ z\circ x \equal{} zx$ (δηλ. το συνηθισμένο γινόμενο του $ \mathbb{R}$) για όλα τα πραγματικά $ z\in\mathbb{R}$. Τότε όμως, αν θέσουμε $ i\circ 1 \equal{} a\in\mathbb{R}$, θα έχουμε:
$ \minus{} 1 \equal{} ( \minus{} 1)\circ 1 \equal{} i^2\circ 1 \equal{} i\circ(i\circ 1) \equal{} i\circ a \equal{} i\circ(a\circ 1) \equal{} (ia)\circ 1 \equal{} a\circ(i\circ 1) \equal{} a\circ a \equal{} a^2$
πράγμα που δεν μπορεί να ισχύει αφού $ a\in\mathbb{R}$.

Έτσι, δεν μπορούμε να έχουμε $ x\circ y \equal{} xy$ για κάθε $ x,y\in\mathbb{R}$. Από την άλλη όμως, έχουμε οπωσδήποτε $ q\circ r \equal{} qr$ για όλα τα $ q,r\in\mathbb{Q}$.

Υπάρχει λοιπόν επιλογή $ i\circ 1 \in\mathbb{R}\setminus\mathbb{Q}$ που να δουλεύει?

Cheerio,

Durandal 1707

Νομίζω οτι πρεπει να βαλεις και αλλες συνθηκες στο ερώτημα σου. Κατ' αρχας τα δυο σύνολα έχουν την ιδια πληθικότητα και οτιδήποτε έχει την πληθικότητα των μιγαδικών μπορει να ταυτιστει με τους μιγαδικους και αρα να κληρονομήσει οποιαδηποτε δομή έχουν οι μιγαδικοί. Απο την αλλη είναι αρκετα ενδιαφερον το οτι οι μιγαδικοι και οι πραγματικοι δεν ειναι μονο ισομορφικες ομαδες αλλα και ισομορφικοι χωροι επι των ρητων. Οποτε και απο αυτες τις δυο ταυτισεις μπορουν να κληρονομησουν δομες επι των μιγαδικών. Αρα τετοιες δομες υπαρχουν αρκετες. Ενδεχομενως να θελεις να θεσεις πιο συγκεκριμενες συνθηκες οι οποιες να καθιστουν την υπαρξη μιας τετοιας δομης αδύνατη.

Το αρχικό ερώτημα (όπως το κατάλαβα εγώ τουλάχιστον) είναι να κρατήσουμε τη δομή του $ \mathbb{R}$ ως αβελιανή ομάδα (διανύσματα) και τη δομή του $ \mathbb{C}$ ως σώμα (βαθμωτά), και να βρούμε ένα πολ/σμό βαθμωτών με διανύσματα $ \circ: \mathbb{C}\times\mathbb{R}\to\mathbb{R}$ ώστε να ικανοποιούνται τα αξιώματα ενός δ.χ.

[quote="Jondrette"]Απο την αλλη είναι αρκετα ενδιαφερον το οτι οι μιγαδικοι και οι πραγματικοι δεν ειναι μονο ισομορφικες ομαδες αλλα και ισομορφικοι χωροι επι των ρητων. Οποτε και απο αυτες τις δυο ταυτισεις μπορουν να κληρονομησουν δομες επι των μιγαδικών.[/quote]

Αν θέλεις, γίνε λίγο πιο explicit... το ερώτημα είναι ενδιαφέρον αλλά δεν είναι απαραίτητα δεδομένο ότι είναι όλοι εξοικειωμένοι με Hamel bases ;)

Cheerio,

Durandal 1707

[quote="Durandal"]Το αρχικό ερώτημα (όπως το κατάλαβα εγώ τουλάχιστον) είναι να κρατήσουμε τη δομή του $ \mathbb{R}$ ως αβελιανή ομάδα (διανύσματα) και τη δομή του $ \mathbb{C}$ ως σώμα (βαθμωτά), και να βρούμε ένα πολ/σμό βαθμωτών με διανύσματα $ \circ: \mathbb{C}\times\mathbb{R}\to\mathbb{R}$ ώστε να ικανοποιούνται τα αξιώματα ενός δ.χ.
[/quote]
Αυτό ακριβώς εννοούσα.Sorry αν μπερδεύτηκε κάποιος. :blush:

Ειναι ακριβώς όπως τα λες Durandal. Το $ \mathbb{R}$ και το $ \mathbb{C}$ είναι ισομορφικές αβελιανές ομάδες ακριβώς επειδή είναι ισομορφικοί διανυσματικοί χώροι επί του $ \mathbb{Q}$ (θα επανέλθουμε σε αυτό στο τέλος). Με άλλα λογια το $ \mathbb{R}$ και το $ \mathbb{C}$, σαν αβελιανές ομάδες είναι το ίδιο και το αυτό. Οποιαδήποτε επιπλέον δομή έχει η μία ομάδα μπορεί να αποκτήσει και η άλλη. Αυτό απλά μέσω ενός ισομορφισμού. Με άλλα λόγια αν
$ \phi: \mathbb{R}\to\mathbb{C}$ ένας ισομορφισμός, τότε ορίζουμε βαθμωτό πολλαπλασιασμό στο $ \mathbb{R}$,
με τον τύπο:

\[ \lambda\cdot u\equal{}\phi^{\minus{}1}(\lambda\phi(u))\]

όπου $ u\in\mathbb{R}$ και $ \lambda\in\mathbb{C}$. Τώρα είναι εύκολο να αποδείξει κανείς ότι ο πολλαπλασιασμός αυτός ικανοποιεί όλες τις απαιτούμενες ιδιότητες.

Ο λόγος για τον οποίο το $ \mathbb{R}$ και $ \mathbb{C}$ είναι ισομορφικοι ρητοί διανυσματικοί χώροι, είναι επειδή
μια αλγεβρική βάση (με άλλα λόγια ένα μεγιστικό γραμμικά ανεξάρτητο σύνολο) ενός οποιουδήποτε από τους δύο χώρους έχει  κατά ανάγκη  πληθικότητα ίση με το χώρο. Η απόδειξη αυτού του ισχυρισμού ανάγεται στο ότι το σύνολο των πεπερασμένων υποσυνόλων ενός απειροσυνόλου έχει πληθικότητα ίση με το απειροσύνολο.

Στο βιβλίο του "I want to be a mathematician" ο Paul Halmos μιλάει για κάποιο φοιτητή που σε κάποιες εξετάσεις από το άγχος του διάβασε λανθασμένα την ερώτηση

Να δειχθεί ότι το $ \mathbb{C}$ είναι διανυσματικός χώρος πάνω από το $ \mathbb{R}$.

ως

Να δειχθεί ότι το $ \mathbb{R}$ είναι διανυσματικός χώρος πάνω από το $ \mathbb{C}$.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a local ring and let $ M$ be its maximal ideal. Its residue field is defined to be the field $ k \equal{} R/M$ and $ R$ is said to contain its residue field if there is an (injective) homomorphism $ i: k\rightarrow R$ such that $ i(r \plus{} M) \equal{} r (modM)$ for all $ r\in R$. Now let $ R$  be an Artinian local ring that contains its residue field. Show that $ R\cong k[x_1...x_n]/I$ for some ideal $ I$ such that $ (x_1...x_n)^m$ is contained in or is equal to $ I$

any hints or suggestions are greatly appreciated =)

I think he means for some $ m \in \mathbb{Z}$.

[hide="Hint Part 1"]For the first part, use the universal property you mentioned in your other post ;)[/hide]

[hide="Solution Part 1"]
With $ i$, we can make $ R$ into a $ k$-algebra. Let $ x_1,\cdots,x_n$ be generators of the ideal $ I$, which exist since $ R$ is Artinian and therefore Noetherian. Then any element of $ R$ can be expressed as an element of the image of $ i$ plus an element of $ I$, so $ x_1,\cdots,x_n$ generate $ R$ as a $ k$-algebra. By the universal property, we can view $ R$ as a quotient of $ k[x_1,\cdots,x_n]$.
[/hide]

Edit: I don't believe this proof is correct. This shows that $ I$ is finitely-generated as an ideal, i.e. as an $ R$-module, but that doesn't mean that $ x_1,\cdots,x_n$ generate it as a $ k$-algebra. However, I believe there's some standard argument why $ I$ is finitely generated as a $ k$-algebra?

[hide="Hint Part 2"]How can you turn the condition into a fact in $ R$? Relate it to ideals?[/hide]

For reference, one may find LkNsngth's other post at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258860.

Ahh thank you guys; I actually sort of drafted a solution last night assuming that universal property was true (which was why I asked you guys =P)

My solution is like this:

Because $ R$ a Noetherian ring (we know this because it is Artinian) containing $ k$ as a subfield, we know $ R$ is finitely generated over $ k$. Therefore it is the quotient of $ k[x_1,x_2...,x_n]/I$ for some ideal $ I$. Now if $ I$ doesnt contain $ (x_1,...x_n)^m$ for some $ m$, then we can create a descending chain of ideals $ (x_1,...x_n) \supset (x_1,...x_n)^2 \supset (x_1,...x_n)^3$ etc... contradicting the fact that $ R$ is Artinian.


Archipelago consists of $ n$ islands : $ I_1,I_2,...,I_n$ and $ a_1,a_2,...,a_n$ - number of the roads on each island.  $ a_1 \equal{} 55$, $ a_k \equal{} a_{k \minus{} 1} \plus{} (k \minus{} 1)$, ($ k \equal{} 2,3,...,n$)
a) Does there exist an island with 2008 roads?
b) Calculate $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n.$

a) It can be easily proven by induction that $ a_n \equal{} 55 \plus{} \frac {n(n \minus{} 1)} 2$, so $ a_{63} \equal{} 2008$.

b) Result is $ 55n \plus{} {n \plus{} 1 \choose 3}$, also by induction. The inspiration for this result came from Pascal's triangle.

a) is trivial. b) Our sum is 55n + 1 appears once in a_2, twice in a_3, etc. until n-1 times in a_n. Then 1 appears a total of n(n-1)/2 times, 2 appears a total of (n-1)(n-2)/2, summing over everything is n(n-1)/2+2(n-1)(n-2)/2+…

Need hint

a) The number of roads is $a_n=55+\binom n2.$

[i]Proof:[/i] We proceed by induction.

The base case is $n=1.$ We have $a_1=55+\binom12=55+0=55.$

Now, we assume $a_k=55+\binom k2.$ We have

\begin{align*}
a_{k+1}&=a_k+k\\
&=55+\binom k2+k\\
&=55+\frac{k(k-1)}{2}+k\\
&=55+\frac{k^2-k+2k}{2}\\
&=55+\frac{k^2+k}{2}\\
&=55+\frac{k(k+1)}{2}\\
&=55+\binom{k+1}{2}.
\end{align*}

Therefore, $a_n=55+\binom n2$ by induction.

b) The sum is

\begin{align*}
a_1+a_2+\cdots+a_n&=55+\left(55+\binom22\right)+\left(55+\binom32\right)+\cdots+\left(55+\binom n2\right)\\
&=55n+\sum_{k=2}^n\binom k2.
\end{align*}

By the [[Hockey-Stick Identity]], $\sum_{k=1}^n\binom k2=\binom{n+1}{3},$ so we have $a_1+a_2+\cdots+a_n=55n+\binom{n+1}{3}.$


for me, i don't think "Grey" is too annoying... it just seems like the "e" makes "grey" kinda scary
i really  :furious: "catsup" it seems like i should sound nothing like ketchup.

post other spelling variants and especially ones that you think are annoying

also, i can't find the poll's forum... did it get merged with Games and Fun Factory?

--my first G&FF topic

ok yeah... the poll glitched... only vote in the top 4 if you know what's good for you :D

Klebian... the poll isnt working!! is it cuz im putting 0 days? it says thats ok!
ill try 1 more time... then can u plz add it?

i don't (or rather, I wouldn't) pronounce catsup the same as ketchup

also, yes

Which is the correct spelling, grey or gray?

And is catsup even a real word?

catsup is a word..

i don't understand why they say memorise. it's memorize.
also they sometimes say "you should have practised." it's practiced.

[quote="stevenmeow"]hey Klebian... did u delete my poll?  :( i know it was there cuz i voted in it[/quote]no i didn't
maybe it's in RT? dunno
i don't even know what poll you are referring to

the poll i put for this thread...
i might just make it again

klebian:
can you plz edit the options? they totaly glitched out...
they are supposed to be Gray/Ketchup, Gray/Catsup, Grey/Ketchup, and Grey/Catsup (boo!)

the poll keeps disappearing...

I know quite a few:

colour --> color
armour --> armor
grey --> gray
practised --> practiced
memorised --> memorized

I think the words on the left column are used in Europe and the ones on the right are more commonly used in US. Thats why if you type in the first 3 on MS word, there will be a red squiggly line under it.

Yeah, I remember it being English/Canadian English and American English.

Go CANADA (even though I'm in the us)!!!


Ascertain the real numbers $x$, $y$ so that
$xy+\frac 1x+\frac 1y=3$
$x+y+\frac{1}{xy}\ge -\frac{15}{4}\ .$

There is no unique solution in x & y because these two are in effect:

1) they are required to be real numbers in the problem, instead of merely being integers AND

2) instead of there being both equations, one is an inequality, which allows for a REGION of solutions

No, the system has a finite number of solutions !

Denote $p=x+y,q=xy.$ Then

\begin{eqnarray}q+{p\over q} &=& 3\\ p+{1\over q} &\geqslant& -{15\over 4}\end{eqnarray}

From $(1)$ it follows $p=q(3-q)$, so $(2)$ becomes

\begin{eqnarray*}-q^2+3q+{15\over 4}+{1\over q}\geqslant 0 &\iff& \frac{4q^3-12q^2-15q-4}{q}\leqslant 0\\ &\iff& \frac{4q^3-16q^2+4q^2-16q+q-4}{9}\leqslant 0\\ &\iff& \frac{(q-4)(2q+1)^2}{q}\leqslant 0\\ &\iff& \frac{q-4}{q}\leqslant 0\\ &\iff& \frac{4}{q}\geqslant 1\\ &\iff& 0<q\leqslant 4\end{eqnarray*}

On the other hand, $x$ and $y$ are the solutions of the equation $t^2-pt+q=0$, which therefore must have a non-negative discriminant:

$p^2-4q\geqslant 0 \iff q^2(3-q)^2-4q\geqslant 0$

Since we've established that $q$ is positive, this gives

\begin{eqnarray*}q(9-6q+q^2)-4\geqslant 0 &\iff& q^3-6q^2+9q-4\geqslant 0\\ &\iff& q^3-4q^2-2q^2+8q+q-4\geqslant 0\\ &\iff& (q-4)(q-1)^2\geqslant 0\\ &\iff& q-4\geqslant 0\\ &\iff& q\geqslant 4\end{eqnarray*}

From $0<q\leqslant 4$ and $q\geqslant 4$ it follows $q=4.$ Then $p=q(3-q)=-4$

Therefore $x,y$ are the solutions of the equation $t^2+4t+4=0$, meaning that the [u]unique[/u] solution to the initial system is $x=y=-2.$

The set of the solutions is $: \ S=\left\{\ (-2,-2)\ ,\ (1,1)\ ,\ \left(-2,\frac 14\right)\ ,\ \left(\frac 14,-2\right)\ \right\}\ .$

Oh silly me!  :?  :blush:

I disregarded the cases $q=-{1\over 2}$ in $\frac{(q-4)(2q+1)^2}{q}\leqslant 0$ and $q=1$ in $(q-4)(q-1)^2\geqslant 0.$

Don't know what I was thinking...  :rotfl:


AMC
AMC 12
AMC 10
probability
USA(J)MO
USAMO
AIME

I'm annoyed right now.

I realize this may sound like whining, but my school forced me to take the AMC 12 and as a result I did not qualify for USAMO.

Thoughts on if this system of equalizing the tests for underclassmen should be changed?

[quote="Hamster1800"]If you can take the AMC12, you should take the AMC12. I don't know why your school forced you to take the AMC12, nor how you know that it directly caused you not making USAMO. Oddly enough my school (actually the high school right next to my middle school) last year forced me to take AMC10, as well as everyone else in 10th grade or below.[/quote]
Simple. Had I taken the 10, I would have gotten, in all probability, better than 123.5, and thus would have qualified, based on the scores of similar skill-level people in my school, such as Ignite168.

[quote="13375P34K43V312"]Well if you can take both, and you're still having some fun with the 10, take the 10 and the 12. I took 10A/12B this year, most likely will do the same next year, as a sophomore there's a chance I'll do 12A/12B.

But yeah, don't miss out on a chance to take the 12, however I'm a bit skeptical that they [i]forced[/i] you to do it.[/quote]
My school doesn't offer both and my parents won't let me go anywhere else to take it.

well then study harder
next year i'ma go take the amc12

[quote="Ignite168"][quote="jrshoch"]I ran into the same problem last year.......

Moral of the story:  Underclassmen seeking to make USAMO should always take AMC10[/quote]

Not if they believe they can get an easy perfect on the AMC10, in that case, they could probably set the floor and still get in if they don't (with a good AIME score)[/quote]

This is true, but I think the determining factor for taking the AMC12 should be that you feel that you can get 120+ on AMC12, since this will almost surely get you above the tiebreaker if you get the floor, although not necessarily, but if you can get 130+ then you should definitely take 12.

Last year was not the case, which is why they had the later expansion. Last year, a few people scored 130 on the AMC12 and were left out. In fact, jrshoch scored a point above the floor and above 120 on the AMC12 and had the bad luck to be cut out by the initial "tie-breaker," one of the obviously unfair cases that caused them to reverse their decision.

yeah....... i got 120.5 AMC 12A, 9 AIME I... 210.5 index, the tiebreaker cutoff was 212.

I was pissed  :lol:

I really don't think it's fair that the website says something completely different from the TM, advising you to take the 12, when actually the process favors the 10. They may have changed the TM since last year, but the website still says the same thing.

[quote="Ignite168"][quote="jrshoch"]I ran into the same problem last year.......

Moral of the story:  Underclassmen seeking to make USAMO should always take AMC10[/quote]

Not if they believe they can get an easy perfect on the AMC10, in that case, they could probably set the floor and still get in if they don't (with a good AIME score)[/quote]

I think this is a good reason to take the 10; getting perfect is always good motivation.


Does anyone know how to prepare for the Physics Olympiad, i.e. what books to read and classes to take? I calculus required on the exam? I am only taking calculus right now.

For that matter, does anyone know how one's school registers for a student to take the first round of the USAPhO?  Or the USABO?  Or the USNCO?  Or the USACO?

USACO is taken as an individual. You register on their site. The info for all those others is available on their respective websites. Find them with Google.

USABO registration is open. Check out the website. Link on the right: http://www.cee.org/usabo/index.shtml

USNCO is a little more prickly. http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?id=5a5185d6803211d5e6103fba9e800100 is a how-to-participate page with barely any info. Email c_hernandez@acs.org for info on the local coordinator for the USNCO. Wrote to them a week ago, still waiting for a reply...

The APhO I'm still trying to get info about. The website itself doesn't say too much but there's an application form on it (http://www.aapt.org/Contests/physicsteam.cfm) that's a year old, but will give you some idea of what to expect.

the thing about the chem olympiad is that 2 people at every school can take it. my school selects a 10 person team. this team participates at the state contest and afterwards, the highest 2 people in our school team are invited to nationals. im not sure if this system works everywhere else though.

Are the biology olympiads just pure memorization?

After all, what type of biology "problems" are there that you can solve with pencil and paper?

awww :( no fun genetic problems??? (in AP bio, we're doing that now, so much fun)

MIT OpenCourseWare has some taped physics lectures on E&M for anyone that's interested. I just watched Lecture #19. It was really good - right at my level (I've studied this stuff before), although I should review with a couple of the earlier lectures.

http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/


induction
number theory unsolved
number theory

For which values of N is it possible to write numbers from 1 to N in some order so that for any group of two or more consecutive numbers, the arithmetic mean of these numbers is not whole?

I think you're right.

$n$ must indeed be even, because otherwise $1+2+\ldots+n=\frac{n(n+1)}2$ is divisible by $n$. Conversely, if $n$ is even, then we can arrange the numbers $2,1,4,3,6,5,\ldots,2k,2k-1$. It should be fairly easy to check that this arrangement satisfies the condition (not sure; just wrote it in a hurry :D).

yes it does, prove it by induction as an exercise.


The fibonacci sequence is $1,1,2,3,5,8,13...$
What is the reminader when the $100th$ term of this sequence is divided by $4$?

[hide]Look at the Fibonacci numbers in terms of mod 4: 1,1,2,3,1,0,1,1,2,3,1,0,...They go in cycles of 6. 100/6 leaves remainder 4, so the 100th Fibonacci number leaves a remainder of 3 when divided by 4.[/hide]

[hide]$\boxed{3}$
The remainders are 1,1,2,3,1,0,1,1. Then it repeats after every 6 terms. $100\equiv 4 \bmod{6}$. The remainder is $3.$[/hide]

My ULTIMATE method of doing it last year:
[hide]Guess 3 and hope you get it right because you're running out of time.[/hide]

Well, it's either 0,1,2 or 3 so your chances of guessing and getting it right isn't that bad.

[quote="xxm0rg07hxx"]Don't tell me you actually guessed 3 and got it right.[/quote]
Fine, I won't tell you.


The integer $ n$ is a triangular number if and only if $ 8n\plus{}1$ is a perfect square.

By definition, \[ n \equal{} \frac{k(k\plus{}1)}{2}\]

for some positive integer $ k$. Note that when we substitute this expression for $ n$ in $ 8n \plus{} 1$, we get \[ 8 \left( \frac{k(k\plus{}1)}{2} \right) \plus{} 1 \implies 4k^2 \plus{} 4k \plus{} 1\] which factors as \[ (2k\plus{}1)(2k\plus{}1)\] This is a perfect square, so we're done. QED.

[hide="Converse"] Suppose that $ 8n\plus{}1$ is a perfect square. Clearly, $ 8n\plus{}1$ is odd for all positive integers $ n$. We may then write $ 8n\plus{}1\equal{}(2k\plus{}1)^2$, where $ k$ is a positive integer. Now,

$ 8n\plus{}1\equal{}(2k\plus{}1)^2$

$ 8n\plus{}1\equal{}4k^2\plus{}4k\plus{}1$

$ 2n\equal{}k(k\plus{}1)$

$ n\equal{}\frac{k(k\plus{}1)}{2}$.

But the RHS is simply the $ k$th triangluar number. So the "only if" statement in the original question is proven, as desired.[/hide]

$ n$ is a triangle number

                               $ n\equal{}\frac{r(r\plus{}1)}{2}$
                               $ 8n\plus{}1\equal{}4r(r\plus{}1)\plus{}1\equal{}4r^{2}\plus{}4r\plus{}1\equal{}(2r\plus{}1)(2r\plus{}1)$

[quote="yildobekir"]$ n$ is a triangle number

                               $ n \equal{} \frac {r(r \plus{} 1)}{2}$
                               $ 8n \plus{} 1 \equal{} 4r(r \plus{} 1) \plus{} 1 \equal{} 4r^{2} \plus{} 4r \plus{} 1 \equal{} (2r \plus{} 1)(2r \plus{} 1)$[/quote]

Again, this does not prove the converse which is what the poster was asking for.


Let $ ABCD$ be the quadrilateral. $ P \equal{} AC \cap BD$. $ S(ABP) \equal{} m, \ S(CDP) \equal{} 10m$ Find max $ S(ABCD)$.


geometry
circumcircle
parallelogram
geometric transformation
reflection
geometry proposed

Let $ ABC$ be a triangle and $ P$ is an arbitrary point, $ A'B'C'$ is pedal triangle of $ P$, through $ A',B',C'$ draw $ d_a\parallel PA,d_b\parallel PB,d_c\parallel PC$, $ H$ is orthocenter of triangle $ ABC$. $ d_a\cap HA \equal{} A'',d_b\cap HB \equal{} B'',d_c\cap HC \equal{} C''$, circumcircle of $ A'B'C'$ intersects nine-point circle of triangle $ ABC$ at $ F,F'$ prove that the circles with diameters $ (A'A''),(B'B''),(C'C'')$ are through $ F$ or $ F'$.

Note that, this is generalization of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=301033]3 circles concurrent at Feuerbach point[/url].

Let $ A_2,B_2,C_2, I, I'$ be the midpoints of $ BC, CA, AB, A'A'', AP$.
Note that $ AA''A'P$ is a parallelogram hence $ AA'\cap A''P\equal{}L$ so that $ L$ is the midpoint of $ AA'$.
Since $ I, I'$ are the midpoints of $ A'A'', AP$ then $ L$ is the midpoint of $ II'$ and $ II'\perp B_2C_2$.
Therefore $ (I',I'A)$ is the reflection of $ (I,IA')$ across the line $ B_2C_2$, which follows that $ (I)\cap (I')\equal{}\{M,N\}$ and $ M,N$ lie on $ B_2C_2$.
Applying the first [b]Fontene[/b]'s theorem we get $ MN, C'B', A'F'$ concur at a point $ V$.
$ \Rightarrow VF'.VA'\equal{}VC'.VB'\equal{}VM.VN$ thus $ M,N,A',F'$ are concyclic.
$ \Rightarrow \angle A'F'A''\equal{}90^o$.
We are done.


What is the value of $ \frac{(16\minus{}8)(16\plus{}8)}{(4\minus{}2)(4\plus{}2)}$?


Then the question is wrong, $RHS<2$ so n can only be 1

EDIT: :blush: typo

But the left hand side grows arbitrarily large!!!


Is this like a really easy ARML question or something?:

Find the ordered pair of positive integers (a, b) with a< b, for which

$ \sqrt{{1 \plus{} \sqrt{21 \plus{} 12\sqrt{3}}}} \equal{} \sqrt{a} \plus{} \sqrt{b}$

lawlz.....

Seriously, is that what you call an ARML problem? This seems more like a mathcounts problem. :|

Anyways, first you square both sides of the equation to get

$ 1\plus{}\sqrt{21\plus{}12\sqrt{3}}\equal{}a\plus{}2\sqrt{ab}\plus{}b$

Now you want to convert $ \sqrt{21\plus{}12\sqrt{3}}$ into a non-nested radical, which turns out to be $ 3\plus{}2\sqrt{3}$. (since when you square that, you get $ 21\plus{}12\sqrt{3}$.

This now gives you

$ 4\plus{}2\sqrt{3}\equal{}a\plus{}2\sqrt{ab}\plus{}b$

Using corresponding coefficients, we see that $ a\plus{}b\equal{}4$ and $ ab\equal{}3$. The only ordered pair that would work, with $ a<b$, is $ \boxed{(1, 3)}$.

I can't believe ARML puts this as a problem. Seriously, I feel this is a trivial mathcounts problem, supposing that the real ARML tests have harder, more interesting problems.

Maybe you should write a letter of complaint to the problem writers, asking why they would put such a TRIVIAL problem on ARML.

Or you could face the fact that if a contest had only problems that were "difficult", very few people would get anywhere with it. The purpose of a math competition isn't to show people how dumb they are.

That being said, instead of posting "trivial" ARML problems, maybe you should post the ones you find interesting. Gloating about how trivial you think a problem is does nobody any good.

hmm wow

letter of complaint= :rotfl:  :rotfl:  :rotfl: :rotfl:  :rotfl:  :rotfl:

According to Jeff Boyd, it's TRUE that harder math problems on contests really set the smartest kids apart from everyone else. That's what made bobby shen pwn.

EDIT: added another rotfl just to show off my humor in which that's the reason suma_milli likes me a lot......................yay gonna fit in six/six/seis rotfls lolz im more bored than that cuz the problem was trivial.

Yes, also calling things trivial........ is just bad in general...... though I seem to like to do this...

Hmm... I'll post some harder arml problems...

Also, arml isn't about the individual round (Despite socal's ability to beat everyone else on the individual round) as much as the team and power rounds...

Here is last years tiebreaker 1, which only 1 contestant could solve in ten minutes (and she solved it by drawing an accurate diagram!) I may have accidentally changed it a bit.

We have a triangle ABC. Consider the points P and Q that trisect BC. Draw AP and AQ. Similarly, consider the points that split AC into 7 identical parts and the connections between these points and B. Similarly, consider the points that split AB and the points that split it into nine identical parts and the lines connecting these points and C. How many regions do these lines split it up into?

And what? Its true that the harder problems seperate good people from almost good people (usamo does this pretty well) but was last years mathcounts harder than normal?

lol pythag you hypocrite. :P

lawlz and I knew how to solve already gaussy.  I just simplified $ \sqrt{21 \plus{} 12\sqrt{3}}$ and then simplified the whole thing afterwards.

@Aryth: I guess that's true..


quadratics
function
complex numbers
imaginary numbers

Let $f$ be the function defined by $f(x) = ax^2 - \sqrt{2}$ for some positive $a$.  If $f(f(\sqrt{2})) = \sqrt{2}$ then $a=$
\[ \text{(A)} \frac{2 - \sqrt{2}}{2} \ \ \ \text{(B)} \frac{1}{2}\ \ \ \text{(C)} 2 - \sqrt{2}\ \ \ \text{(D)} \frac{\sqrt{2}}{2}\ \ \ \text{(E)} \frac{2+\sqrt{2}}{2} \]

[quote="h_s_potter2002"]I didn't get the right answer, but when I plugged a into the equation, it worked.  Can someone tell me what I did wrong?

[hide]We can make $f(f(\sqrt{2})) = {{(2a^2-\sqrt{2})}^2}\cdot a - \sqrt{2} = {(4a^3-4\sqrt{2}a^2+2a-2\sqrt{2} = \sqrt{2}}$.  We can subtract the $\sqrt{2}$ from each side, and factor it, which makes it $(4a^2+2)(a-\sqrt{2}) = 0$.  If you make the first part equal to 0, you get an imaginary number, so the second part has to be 0, making $a = \sqrt{2}$.[/hide][/quote]

I did it the same way..... And obviously, I did something wrong. Can anyone help me see how I would solve that problem above?

[quote="TripleM"]Maybe theres a mistake in the question.. maybe you were meant to solve $f(f(\sqrt{2}))=-\sqrt{2}$ or something?[/quote]

I'm guessing you're probably right, but I don't remember which AHSME problem this is, so it's hard for me to look it up.

D if the question is f(f(sqrt(2))) = -sqrt(2)


Hi there!

 I thinks the title says it all. I cannot see Latex images only Latex codes on my computer. I feel sorry for my eyes. :)
 I would apreciate any help. thnks in advance. ;)

 Cezar

Same thing  happens to me. The "funny" is that when I'm Log out I can see the Latex images!

$x$

So above, you only see the text "x"? Or do you see a box with an X in it? Or something else?

Post a screenshot, because I know that's what V.V. will say. :)

Cezar solved his problem by restarting the computer :D

Seems to have something with the log in process.
Here goes a sample of my "Latex posting", maybe it will help somebody to help me.
$\alpha \, \beta \, \pi$

[quote="Immanuel Bonfils"]I finally managed to see Latex images properly. For some strange reason, BBCode was disabled in my profile. When I went to my profile -> Post Messages -> Show BBCode and enabled it, the problem was solved.[/quote] :lol: Yeah I totally forgot about that :)

[quote="Cezar Lupu"]Valentin, I have the same problem now. When I log out, I can see Latex images. :([/quote]Only when you log out? Have you tried Firefox or Opera9 ?

Valentin, but it is a little bit weird. Even when I logged in from Ukraine it didn't work. Strange indeed! :maybe:

[quote="Alexyang"]I can't see the LaTeX to. I only see a blank space but i know the TeX is there because when i put my mouse over it, i shows the equation.[/quote]
If you have a slow internet connection, you have to wait for the entire page to load before pressing any "hide" buttons (at least on my computer).

I have the same problem. Now, I have ti log out to see Latex images immediately after I post. :(


Let $\triangle ABC$ be a triangle and the points $X,Y\in (ABC)$. Is true the following implication ?

$XA+XB+XC\le 3R,\ YA+YB+YC\le 3R\Longrightarrow\left ( \forall \right) Z\in (XY)$ $,\ ZA+ZB+ZC\le 3R.$

The problem is trivialized by the following lemma:

[hide]In triangle ABC and point D on BC, $AD \le \frac{BD \cdot AC + CD \cdot AB}{BC}$.

Proof:

Let line L be perpendicular to AD and pass through A.  Let the respective feet of perpendiculars from B and C to L be X and Y.

$AD = \frac{BD \cdot YC + CD \cdot XB}{BC} \le \frac{BD \cdot AC + CD \cdot AB}{BC}$
[/hide]

And the lemma is a special case of Ptolemy's inequality...


calculus
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Support
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Hi, I am currently reviewing probabilities on my own and encountered a problem which i can't do. Hopefully someone here will be kind enough to offer me a solution. And yeah, i know this is a calculus forum, but why isn't there one for Probability?

Problem:

1. Let A1, A2, ..., An be n events. Show that if P(A1) = P(A2) = ... = P(An) = 1, then P(A1A2...An)=1.

2. Let A, B, and C be three events. Show that exactly two of these events will occur with probability P(AB) + P(AC) + P(BC) - 3P(ABC) [it's obvious when it's drawn with venn diagram, but i just dunno how to prove it formally]

Any help is greatly appreciated. I am not too bright, please elaborate your explaination whenever possible. :)

There aren't really enough probability and statistics questions to support a whole forum. I'd suggest that most statistics (as opposed to probability) questions should go into the "advanced topics" section; purely probability questions (including this one) can be dealt with either in "advanced topics" or here in analysis.

For problem #1, I'll repeat something I always tell students in a probability course: "always consider the complement."

The complement of $\bigcap_{j=1}^{n}A_{j}$ is simply $\bigcup_{j=1}^{n}A_{j}^{C}.$

Probability is "subadditive": the probability of a union is no greater than the sum of the probabilities. Hence

$P\left(\bigcup_{j=1}^{n}A_{j}^{C}\right)\le \sum_{j=1}^{n}P(A_{j}^{C}) =0.$

That in turns shows that $P\left(\bigcap_{j=1}^{m}A_{j}\right)=1.$

Note the full axioms for a probability space (essentially the same as the axioms for a measure space) include [b]countable[/b] additivity. With that, you can use the same proof to show that if $\{A_{j}\}$ is a sequence of events, $1\le j<\infty,$ such that $P(A_{j})=1$ for each $j,$ then

$P\left(\bigcap_{j=1}^{\infty}A_{j}\right)=1.$

For #2, I wouldn't have much to add: the Venn diagram points you to each of the eight possible triple intersections of these sets and their complements. You could simply account for what's going on, one such intersection at time.

Thank you for both of your help! I'll remember to always consider the complement from now on. Just out of curiosity, how do you remember things like "the intersect of the complement of An" is equal to "the union of An"? I mean this property makes perfect sense, but it does not occur to me so quickly.


[quote="anorithdialga"]Let the sum of a set of numbers be the sum of its
elements. Let $ S$ be a set of positive integers, none greater than
15. Suppose no two disjoint subsets of $ S$ have the same sum.
What is the largest sum a set $ S$ with these properties can have?[/quote]

anybody have thoughts on this problem? :roll:

The set {8, 11, 13, 14, 15} has the required property with sum 61.
[ no single element can be the sum of others because the max element is 15 and the minimum sum of two or more is 19,
  two elements can not be the sum of three since the total sum is odd,
  that two elements cannot be the sum of two others can be quickly checked noting that the fifth element would have to be be odd ]

It is easy to verify that this is the largest sum for a set of 5 or less elements with the property because there are just a few sets with larger sums.

Suppose there were a set of $ n > 5$ elements with the property, and its smallest element were $ s$
The largest 5 elements in the set would also be a set with the property, so their sum would be 61 or less.
Then the sums  would range from s to at most $ s \plus{} 61 \plus{} 15(n\minus{}6)$, a range with at most $ 15n \minus{} 28$ possibilities.
A set of $ n$ elements has $ 2^n \minus{}1$ non-empty subsets, so if their sums are different, by pigeon hole principle we require
$ 2^n \minus{} 1 < 15n \minus{} 28$, but this is not possible for $ n > 5$

so the largest possible sum is 61.


Niki usually leaves her cell phone on. If her cell phone is on but she is
not actually using it, the battery will last for 24 hours. If she is using
it constantly, the battery will last for only 3 hours. Since the last
recharge, her phone has been on 9 hours, and during that time she has used
it for 60 minutes. If she doesn't talk any more but leaves the phone on,
how many more hours will the battery last?

$ \frac{1}{3}\plus{}\frac{8}{24}\plus{}\frac{x}{24}\equal{}1$, so $ x\equal{}8$hours.
$ \frac{1}{3}$ is battery spent on using the phone, $ \frac{8}{24}$ is time spent on leaving the phone on but not using.

[quote="GameBot"]Niki usually leaves her cell phone on. If her cell phone is on but she is
not actually using it, the battery will last for 24 hours. If she is using
it constantly, the battery will last for only 3 hours. Since the last
recharge, her phone has been on $ \boxed{9}$ hours, and during that time she has used
it for 60 minutes. If she doesn't talk any more but leaves the phone on,
how many more hours will the battery last?[/quote]
You read incorrectly bugi :D

Our equation is $ \frac{1}{3}\plus{}\frac{\boxed{9}}{24}\plus{}\frac{x}{24}\equal{}1$.  So $ x\equal{}7$.

You are not right. She has used it for an hour [b]during[/b] the time of 9 hours, so it wasn't used and was on for 9-1=8 hours.


I think that there is a typo or something here.

[hide]The answer is no. If we look at a square, we see that it is cyclic. For a square, AB+BC doesn't equal CD, because AB=BC=CD[/hide]

I forgot to mention that angle ABC = 120. I was a little dubious about this myself, but it is part of a problem on a worksheet, so...

[quote="Phelpedo"]Suppose we have a cyclic quadrilateral ABCD, with angle ABC=120. Does AB + BC = CD? If so, prove it.[/quote]
I don't think so. Imagine an angle $\angle ABC = 120^\circ$ that is inscribed in a circle. We can pick any point $D$ on arc $AC$ (the one not including point $B$), and the conditions given in the problem are satisfied, but the length of $CD$ obviously can vary greatly. Is there another condition of the problem that is missing?  :wink:


analytic geometry
graphing lines
slope
symmetry
algebra
system of equations

the midpoints of the sides of a triangle are (2,5) (4,2) and (1,1). Find the coordinates of the vertex that is in the first quadrant.

the answer is [hide](5,6)[/hide] but i have no clue how to get it...

[hide="hint"]Did you try the midpoint formula $ (\frac{x_{1} \plus{} x_{2}}{2}, \frac{y_{1} \plus{} y_{2}}{2})$, and just try solving by a system of equations?[/hide]

[quote="DaReaper"]the midpoints of the sides of a triangle are (2,5) (4,2) and (1,1). Find the coordinates of the vertex that is in the first quadrant. [/quote]
[hide="hint"]
I believe that when the smaller of two similar convex n-gons is inscribed within the other, there exists $ n$ pairs of parallel lines composed of one side from each n-gon.
[/hide][hide="solution"]
$ \frac{(5x\minus{}21)}{4}\equal{}\frac{(x\minus{}2)}{3}\Rightarrow x\equal{}5$
Thus, $ f(5)\equal{}y\equal{}6 \therefore (x,y)\equal{}(5,6)$
[/hide]

iguil, could you please explain your hint and solution to me? Which polygons are we concerned with in your hint and where does the equation come from in your solution?

yea, where did u get your equations from?

well you can solve it using slopes, just remeber something about parallels and midpoints of a triangle.

 :)

Here's an approach using algebra.  Let $ A$, $ B$, and $ C$ be the three vertices.  Then we can express the midpoint conditions with the following equations:
\begin{align*}
  \frac{A + B}{2} = (2, 5) \\
  \frac{A + C}{2} = (4, 2) \\
  \frac{B + C}{2} = (1, 1).
\end{align*}
Now solve these equations using ordinary algebra.  I'll let someone else finish.

[hide="solution"]I used the same approach Ravi B used, making (a,b) (c,d) and (e,f) the three vertices of the triangle. then i got the following equations set up:

(a+c)/2 = 2
(b+d)/2 = 5
(e+c)/2 = 4
(f+d)/2 = 2
(e+a)/2 = 1
(f+b)/2 = 1

I then solved for all variables and saw (c,d) was the only point with both x and y components positive, putting it in the 1st quadrants. Since c = 5 and d = 6 after solving for all variables, the answer is (5, 6) [/hide]

[hide="General Solution"] The points are (a,b); (c,d); (e,f).  The midpoints of the sides are $ \displaystyle(\frac {a + c}{2}, \frac {b + d}{2}); (\frac {a + e}{2}, \frac {b + f}{2}); (\frac {c + e}{2}, \frac {d + f}{2})$.
Let's say the midpoints are I, J, and K.  Let I(h, i), J(j, k) and K(l, m).  Then
\begin{eqnarray}\frac {a + c}{2}&=&h\\
\frac {a + e}{2}&=&j \\
\frac {c + e}{2}&=&l
\end{eqnarray}
Adding these 3 gives $ a + c + e = h + j + l$.  Subtracting twice equation (1) from this gives
$ e = j + l - h$
By symmetry, $ a = h + j - l$ and $ c = h + l - j$
Doing the same thing with b, d, and f, we get $ b = i + k - m$, $ d = i + m - k$, and $ f = k + m - i$.

What I do is I add the coordinates of the midpoints up, x's and y's, to get a single pair of coordinates.  Then I write twice the original coordinates underneath them.  Then I just subtract.
Example: $ (2, 5) (4, 2) (1, 1) \rightarrow{} (7, 8)$
            $ (4, 10) (8, 4) (2, 2)$
             $ (7 - 4, 8 - 10) (7 - 8, 8 - 4) (7 - 2, 8 - 2)$
            $ (3, - 2) ( - 1, 4) (5, 6)$
And these are the 3 coordinates of the triangle. [/hide]

Let $ ABC$ be the triangle and the midpoints be $ A'$, $ B'$, $ C'$ (with $ A$ opposite $ A'$). Then notice that $ A'B'AC'$ is a parallelogram.

[quote="DaReaper"]the midpoints of the sides of a triangle are (2,5) (4,2) and (1,1). Find the coordinates of the vertex that is in the first quadrant.

Let us consider that the triangle is ABC.
The mid points are D(2,5), E(4,2) and F(1,1)
D is mid point of AB, E is mid point of BC and F is mid point AC

To find the vertices, there is an easy technique.

A = D + F - E  ( Add two midpoints which are besides and subtract the opposite one)

   =  (2+1 - 4,5+1 - 2)
   = ( -1, 4)  Which is not in Q1

B = D + E - F

   =  (2+4-1, 5+2 - 1)

   =  (5,6) this lies in Q1

Therefore, this is your answer.

For your information,

C = E + F - D

   = (4+1 - 2, 2+1 - 5)

   = (3, -2) Which is not in Q1

B =

Wow my solution was quite unique.

I wrote the equation of the midpoint line that crosses through $ (1,1)$ and bisects the distance between the other two given coordinates at the points $ (3,3.5)$. If we call this line $ f$, then there exists a point $ x$ such that the slope between the points $ (2,5)$ and $ (x,f(x))$ is equivalent to the slope between the given points $ (1,1)$ and $ (4,2)$.
$ f(x)\equal{}\frac{5x\minus{}1}{4}$
Thus, $ \frac{f(x)\minus{}5}{x\minus{}2}\equal{}\frac{1}{3}\Rightarrow f(x)\equal{}\frac{x\plus{}13}{3}$
$ \frac{5x\minus{}1}{4}\equal{}\frac{x\plus{}13}{3}\Longrightarrow x\equal{}5$
$ \therefore f(5)\equal{}6\Rightarrow (5,6)$


group theory
Galois Theory
algebra unsolved
algebra

[b]Hi every One. I think every One known the Fermat Theorem. This theorem have done by Mr Andrew Wiles. Now I want see this solution. Who can help me?[/b]

This one?
[url]http://www4.tpgi.com.au/nanahcub/flt.pdf[/url]
But I think that the proof is very difficult  :lol:

[color=darkblue][b]Thank kunny very much and thank greenvert very much.[/b][/color]

Yes, it's very difficult. But I will try to read it.

Before you even begin to try to understand what they're talking about in that paper, you need to know about elliptic curves, have taken group theory and Galois Theory, representation theory, lots of number theory.  Then after that, you might be able to understand the definition of some words in the paper.

I think you would be better off studying some of the subjects that I've listed above than trying to read the paper.  I'm not trying to be rude, but I've taken all those classes and I still can't really understand anything.  I guess it's cool to look at it to see all of the crazy math that you might hopefully understand one day in the future.  But it probably takes a phd with a concentration in algebra to get the real gist of what's going on in that paper. Good luck.

kunny and greenvert you forgot an important part of the proof: in the same journal, pages 553-572. It's a joint work of Wiles and Richard Taylor (one of the best mathematicians in the world).
     This belongs to the mathematics story: there exists a gap in the pages 443-551 (thanhnam2902 will find where the problem stands). Wiles sees his error in 1993; he is near 40 old and he wants to obtain a Fields Medal which is reserved to the great mathematicians -before their 4Oth birthday-. Then he works with Taylor during one year but the full proof is obtained after the Field Medal symposium; the mathematical communauty thought that Wiles deserved the price but he did not obtain because a conspiracy of mathematicians (french and other...) prevented that.


help??

1. Oil spilled from a ruptured tanker spreads in a circle at a constant rate of 6mi/hr.  How fast is the radius of the spill increasing when the area is 9mi^2?

2. A 17ft. ladder is leaning against a wall.  if the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/sec, how fast will the top of the ladder be moving down the wall when it is 8 ft above the ground?

Thanks

[hide="1."]I dont understand what has a rate of 6mi/h.[/hide]

[hide="2."]$ x^{2}+y^{2}=17$ From Pythagoras.

$ 2xx'+2yy'=0$. We are given $ x'=5$ Now, use $ y=8$ and use first equation to find x. [hide="x="]$ 15$[/hide]
Now substitute in, $ 2(15)(5)+2(8)y'=0$

Solve for $ y'$. [hide="answer"]$ y'=\frac{75}{8}ft/sec$[/hide]
[/hide]

[quote="brenda"][quote="Jhonny Ruiz"][hide="1."]I dont understand what has a rate of 6mi/h.[/hide]

6 mi/h is the rate that the oil is moving as it spreads across the gound after it spilled out of the tanker.

Thanks for answering number 2![/quote][/quote]

This still doesn't help... what is the rate of?  Is it the rate in the change of the the circumference of the oil circle or of the radius or diameter or something else?

Based on previous problems in that exercise, I think they are looking for the rate of change in the area of the spreading oil, assuming that it is a circle and there is no "depth" increase in the puddle.

Thanks again.

The rate they are [b]looking for[/b] is the rate of change in the radius of the spill - that much is clear. It is less clear what is the meaning of the rate that is [b]given[/b]. My guess (based on common sense) is that you are given the rate of change of the [b]area[/b] of the spill, i.e. $ 6$ mi^2/hr. [After all, this is the rate that we can expect to remain constant.]

With this interpretation, we have $ A=\pi r^{2}$, and you told that $ A=9$ and $ dA/dt=6$. Go ahead and find $ dr/dt$.


floor function
logarithms
function
search
algebra

i feel uber stupid because I could've done this last year  :( :

find how many digits is 17^990 will have when expanded

don't feel stupid. I don't know the shortcut to doing this either.  :P

really? wouldnt that take like a gazillion digits, and calculators can't fit that many  :o

how can u use calc? i am JW no more than that  :?:

JW=Just wondering

gareen does it have to do with how the ----roots of 17 increase by size?

I can't go quite yet so...does this always work?

This works because:

The logarithm of a number $N$ tells you how many of powers of ten are in $N$, which essentially tells you how many digits are in that number (for example, if $5$ is the largest whole power of ten that goes into a number, the number is at least $10^{5}$ and is less than $10^{6}$ (or else the largest power of ten that goes into that number would be $6$ and not $5$) and so the number would have $6$ digits). If you round $\log{N}$ down to the nearest whole number (apply the greatest integer function), you will get the largest whole power of ten that goes into that number. For all $0 < N < 10$, $0 < \log{N}< 1$ and $\lfloor \log{N}\rfloor = 0$. For all $10 < N < 100$, $1 < \log{N}< 2$ and $\lfloor \log{N}\rfloor = 1$, etc. If you understand this pattern, you see that you simply add $1$ to the floor of the logarithm of a number to find how many digits that number has.

oh yeah i remember my friend telling me that thingy and now I forgot it....that question appears a few times in vestevia

The problem can be simplified to computing $\left\lfloor\log 17^{990}\right\rfloor+1$.

$\left\lfloor990\log17\right\rfloor+1$

$1219$

rnwang2 i have a feeling i should follow you everywhere so your math skills rub off me :roll:
but i will search your username that's a really nice way to do this!


A decorative arrangement of floor tiles forms concentric circles, as shown in the figure to the right. The smallest circle has a radius of 2 feet, and each successive circle has a radius 2 feet longer. All the lines shown intersect at the center and form 12 congruent central angles. What is the area of the shaded region? Express your answer in terms of $ \pi$.

[asy]fill((2,0)--(4,0)..scale(4)*dir(15).. scale(4)*dir(30)--scale(2)*dir(30)..scale(2)*dir(15)..cycle,gray(.7));
draw(Circle((0,0),2),linewidth(1));
draw(Circle((0,0),4),linewidth(1));
draw(Circle((0,0),6),linewidth(1));
draw(Circle((0,0),8),linewidth(1));
draw((0,0)--scale(8)*dir(30),linewidth(1));
draw((0,0)--scale(8)*dir(60),linewidth(1));
draw((0,0)--scale(8)*dir(90),linewidth(1));
draw((0,0)--scale(8)*dir(120),linewidth(1));
draw((0,0)--scale(8)*dir(150),linewidth(1));
draw((0,0)--scale(8)*dir(180),linewidth(1));
draw((0,0)--scale(8)*dir(210),linewidth(1));
draw((0,0)--scale(8)*dir(240),linewidth(1));
draw((0,0)--scale(8)*dir(270),linewidth(1));
draw((0,0)--scale(8)*dir(300),linewidth(1));
draw((0,0)--scale(8)*dir(330),linewidth(1));
draw((0,0)--scale(8)*dir(0),linewidth(1));[/asy]

We seek the area of the second most interior minus the most interior circle. The second most interior one has area $ \pi(4^2)\equal{}16\pi$. The most interior has area $ \pi(2^2)\equal{}4\pi$. Thus we have $ 16\pi\minus{}4\pi\equal{}12\pi$. But this is divided into twelfths, so $ \frac{12\pi}{12}\equal{}\boxed{\pi}$.


The acute triangle $ ABC$ is given and point $ M$ inside him. Points $ A_1, B_1, C_1$ are drawn on segments $ BC, AC, AB$ such that $ MA_1 \perp BC$,  $ MB_1 \perp AC$,  $ MC_1 \perp AB$. Prove that     the lines which connect midpoints of $ AM$ and $ B_1C_1$, $ BM$ and $ A_1C_1$, $ CM$ and $ A_1B_1$ are concurrent.

Proof:
We denote that O1,O2,O3,K,L are the midpoints of AM,BM,CM,C1B1,MB1 respectively .
It is easy to proof O1K⊥C1B1,O1O3⊥MB1.So we can say that O1,B1,L,K lies on one cycle.
We denote that O1K meets O2O3 on X, and Y,Z are the same.

O3X/O2X=(O1O3sin∠O3O1K)/(O1O2sin∠O2O1K)
O2Y/O1Y=(O2O3 sin∠O2O3Y)/(O1O3 sin∠O1O3Y)
O1Z/O3Z=(O1O2 sin∠O1O2Z)/(O2O3 sin∠O3O2Z)

If the problem is true then O3X/O2X * O2Y/O1Y * O1Z/O3Z=1(1)
If (1) is true then (sin∠O3O1K/ sin∠O2O1X) * (sin∠O2O3Y/ sin∠O1O3Y) * (sin∠O1O2Z/ sin∠O3O2Z)=1(2)

Notice that K,L,B1,O1 are on one cycle,so ∠O3O1K=∠C1B1M
And the same way we have: ∠O2O1K=∠MC1B1
sin∠O3O1K/ sin∠O2O1X=MC1/MB1
sin∠O2O3Y/ sin∠O1O3Y=MB1/MA1
sin∠O1O2Z/ sin∠O3O2Z=MA1/MC1

So (2) is true,thus we are done

It is an easy problem coming from the past.

Because of the quadrilateral $ AB_{1}MC_{1}$ is cyclic with diameter the segment $ AM,$ we have that $ XX'\perp B_{1}C_{1},$ where $ X,\ X',$ are the midpoints of the segments $ AM,\ B_{1}C_{1},$ respectively.

Similarly we have that $ YY'\perp A_{1}C_{1}$ and $ ZZ'\perp A_{1}B_{1}.$

So, because of the lines through vertices $ A_{1},\ B_{1},\ C_{1}$ of the triangle $ \bigtriangleup A_{1}B_{1}C_{1}$ and perpendicular to the sidelines $ YZ,\ ZX,\ XY$ of the triangle $ \bigtriangleup XYZ$ respectively, are concurrent at the point $ M,$ we conclude that the lines through the vertices $ X,\ Y,\ Z$ of the triangle $ \bigtriangleup XYZ$ and perpendicular to the sidelines $ B_{1}C_{1},\ A_{1}C_{1},\ A_{1}B_{1}$ of the triangle $ \bigtriangleup A_{1}B_{1}C_{1}$ respectively, are also concurrent at one point.
This is a well known theorem, easy proved by [b][size=100]Carnot theorem[/size][/b] and the triangles $ \bigtriangleup A_{1}B_{1}C_{1},$ $ \bigtriangleup XYZ,$ are named each other, [b][size=100]Orthologic triangles[/size][/b].

Hence, we conclude that the line segments $ XX',\ YY',\ ZZ',$ are concurrent at one point, so be it $ P$ and the proof is completed.

Kostas Vitrtas.

PS. The concurrency points $ M$ and $ P,$ are named as well, the [b][size=100]Orthologic centers[/size][/b], of the triangles $ \bigtriangleup A_{1}B_{1}C_{1},$ $ \bigtriangleup XYZ,$ respectively.


Hint : If $z_0$ is a zero of $g(z)$, then it is also a zero of $(z-1)g(z)$.

Abel方法也可以，就是麻烦点，不过本质上都一样的，老兄解得好啊！


function
inequalities
trigonometry
probability
probability and stats

for real-valued characteristic function $ f(t)$, prove that :
$ 1\minus{}f(tn) \leq n^{2}(1\minus{}f(t))$ for all $ n \in \mathbb{N} \cup \{0\}$

:arrow: You can find the proof in 2 books:
V.V.Petrov Sums of independent random variables
Lukasc E. Characteristic functions
There also russian variants of theses books.

can someone post solution? I can't find such books..
I know How to prove when $ t$ is near zero..

hm... I get that it's enough to prove inequality $ |\sin z |\geq \frac {| \sin nz |}{n}$ which seems to be true. Indeed. :)

Lazy bone  :)
[url]http://www.poiskknig.ru/cgi-bin/poisk.cgi?lang=ru&st=Petrov+V.V.&network=1[/url]-Petrov V.V.
[url]http://lib.org.by/info/M_Mathematics/MV_Probability/Lukacs%20E.%20Characteristic%20functions%20(1ed.,%201960)(KA)(T)(215s)_MV_.djvu[/url] - Lukacs E.

Ok. Let's show it by induction. for $ n \equal{} 1$ it's true
for $ n \plus{} 1$ (Suppose $ n|\sin z| \geq |\sin nz|$):
$ |\sin (n \plus{} 1)z| \equal{} |\sin nz \cos z \plus{} \cos nz \sin z| \leq |n \sin z \cos z| \plus{} |\cos nz \sin z | \leq |\sin z | (|n\cos z| \plus{} |\cos nz|) \leq |\sin z| (n \plus{} 1)$
Ok.
substitute $ z : \equal{} \frac {tx}{2}$ then we get :
$ 1 \minus{} cos tx \equal{} 2\sin^{2}tx/2\geq 2\frac {1}{n^{2}}\sin^{2}ntx/2 \equal{} \frac {1}{n^{2}} (1 \minus{} \cos ntx)$
now let's integrate with probability measure $ dF(x)$ we get
$ (1 \minus{} f(t))\geq \frac {1}{n^{2}}(1 \minus{} f(nt))$

I found this inequality only for the case $ n \equal{} 2$.    :(  in the case $ n \equal{} 2$ it's not so interesting.
what was really nontrivial: this inequality $ |\sin z| \geq \frac {1}{n}|\sin nz|$.
since for the real numbers $ n$ it's not true...
Thanks for the books  :)

it seems to me that your proof is ok but
$ 2\sin^2(\alpha)\equal{}1\minus{}\cos(2\alpha)$, but it doesn't influence the general result.


abstract algebra
number theory unsolved
number theory

If $ (a,m) \equal{} (b,m) \equal{} 1$ and if $ (\exp_m (a), \exp_m (b)) \equal{} 1$, prove that

$ \exp_m (ab) \equal{} \exp_m (a) \exp_m (b)$

Where the smallest positive integer $ f$ such that

$ a^f \equiv 1 (\text{mod }m)$ is called the exponent of $ a$ module $ m$, and is denoted by $ f \equal{} \exp_m (a)$

Suppose $ exp_m(a)\equal{}h$, $ exp_m(b)\equal{}k$, $ exp_m(ab)\equal{}d$. Then $ (ab)^{d}\equal{}1(modm)$ which implies $ (ab)^{hd}\equal{}1(modm)$ and hence $ b^{hd}\equal{}1(modm)$ implying k divides hd and since (h,k)=1, k divides d
Similarly, it could be proved that h divides d and hence, lcm(h,k)=hk divides d


Let $ABC$ be a triangle with $\angle B = 30^{\circ }$. We consider the closed disks of radius $\frac{AC}3$, centered in $A$, $B$, $C$. Does there exist an equilateral triangle with one vertex in each of the 3 disks?

[i]Radu Gologan, Dan Schwarz[/i]

Denote by $O$ the centre and by $R$ the radius of the circle circumscribed to $\triangle{ABC}$.
Since $B=30^\cdot$ we get that $AC=R$, hence $\triangle{OAC}$ is equilateral.
Now, it is enough to use the following Lemma (from a Romanian TST  2002  :) ):

[b]Lemma.[/b] If $X$ is a variable point on the disk centred at $A$ (with radius $R_1$) and $Y$ is a variable point on the disk centered at $C$ (with radius $R_2$), then the locus of points $Z$ such that $\triangle{XYZ}$ is equilateral, is the disk centred at $O$ and with radius $R_1+R_2$ (and, of course, its reflection in $AC$, but that's redundant here).

and the conclusion follows.

[quote]Now, it is enough to use the following Lemma (from a Romanian TST 2002  ):

Lemma. If  is a variable point on the disk centered at  (with radius ) and  is a variable point on the disk centered at  (with radius ), then the locus of points  such that  is equilateral, is the disk centered at  and with radius  (and, of course, its reflection in , but that's redundant here).[/quote]

Sorry, Sailor, could you give a demonstration to your lemma, please: I can't find the "Romanian TST 2002" on the Web  :blush:  , and I'm very intrested in this problem.  :)

Sorry for bumping such an old topic but can somebody please prove the stated lemma?? Thanks in advance..

Note: In my diagram, the triangle is $ABC$ in counterclockwise order.
Proof of the lemma:
Let the discs centered at $A,C$ be called $D_1,D_2$ respectively.
Let $Y$ be fixed and let $X$ vary on $D_2$. For each $X$, $Z$ is simply the $60^{\circ}$ counterclockwise rotation of $X$ with respect to $Y$ so, the locus of $Z$ as $X$ varies is simply the circle that results from rotating $D_1$ $60^{\circ}$ degrees counterclockwise about $Y$. Let this circle have center $P$, so $YAP$ is equilateral. Now if we let $Y$ vary as well, the locus of $P$ is simply the $60^{\circ}$ clockwise rotation of $D_2$ with respect to $A$, which is easily seen to be the circle centered at $O$ with radius $R_2$. Then the locus of $Z$ is just the union of all possible circles centered at $P$, which is clearly a circle centered at $O$ with radius $R_2+R_1$.

Without interfering with the above solutions, but the problem states 'with one vertex [b]in[/b] each of the discs...'.

Best regards,
sunken rock

Consider the circumcentre of $ABC$, $O$. Then $OAC$ is an equilateral triangle. Suppose in 3 the circles centered around $A,B,C$ there are points $J,K,L$ forming an equilateral triangle. Consider the spiral similarity sending $JKL$ to $JMC$, then that of $JMC$ to $AOC$. Then it is easy to see that $KM=LC<\frac{AC}{3}$, $OM=AJ<\frac{AC}{3}$, but $OK>\frac{2AC}{3}$, contradiction.

Since you revived this ...
[quote="sunken rock"]Without interfering with the above solutions, but the problem states 'with one vertex [b]in[/b] each of the discs...'.[/quote]
... but it also states "[b]closed[/b] disks".

There [b]does[/b] exist a (unique) such equilateral triangle. For a detailed analysis, see [b]RMC 2006[/b].

Ahhh. If we look at the inequalities established above we see that such a triangle is only possible when $AJ=BK=CL=\frac{AC}{3}$, and that $O,M,K$ collinear. Then using the same spiral similarities it is easy to determine that the triangle $JKL$ is determined by $AJ$ as reflection of $AC$ across $AB$ and similarly $CL$ reflection of $CA$ across $CB$, and by doing so we have a triangle that works.


AMC
USA(J)MO
USAMO
combinatorics proposed
combinatorics

Prove that if $ A={x_1,x_2,.....,x{}_2{}_k_+_1}$ is a set of positive integers such that if you remove any of the $ x_i$, the remaining elements can be split into two sets with equal sum then all $ x_i$ are equal.

All $ x_i$ have the same parity since every time we take one of them the sum of the remaining ones is even. Take the least $ x_i$, $ x_r$ and consider the new set $ B={y_1,.....,y{}_2{}_k_+_1}$ with $ y_i=x_i-x_r$, all the $ y_i$ are nonnegative integers. B also follows the condition that when removed every $ y_i$ the remaining elements can be partitioned into two sets with equal sum (property $ P$). So all $ y_i$ have the same parity, as $ y_r=0$ all $ y_i$ must be even. Then consider the set $ C_1={z_1,.....z{}_2{}_k_+_1}$ with $ z_i=y_i/2$ all $ z_i$ are integer since all $ y_i$ are even, and $ C_1$ also follows the property $ P$. As $ z_r=0$ all $ z_i$ must also be even. Then we can divide them by $ 2$ again and obtain another set $ C_2$ that also follows $ P$ if not all $ x_i$ were equal, then there must be a set $ C_t$ such that one of its elements is odd, but that is imposible since $ C_t$ follows $ P$ and the $ r$ element of $ C_t$ is cero.

[quote="Albanian Eagle"]he didn't say that the two sets must have equal number of elements. your solution works only for that case :wink:[/quote]

But if the two sets don't have to have equal numbers of elements, then it isn't true.  The set {1, 3, 5, 7, 9, 11, 13, 15, 17} is an easy counterexample.

Sorry!!!!

The two sets must have the same number of elements