floor function
group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Prove that there are at least $ 2^{\lfloor (\frac{n}{4})^2 \rfloor}$ different subgroups of $ S_{n}$.

I'm going to prove this in the case that $ n=4m$ is a multiple of $ 4$. I'm quite sure the method can be tweaked to give the required lower bound in other cases, too.

Let $ Z < S_n$ be the subgroup generated by $ (1,2), (3,4), (5,6), \ldots, (n-1, n)$. This is an abelian group, isomorphic to $ (\mathbb{Z}/2\mathbb{Z})^{2m}$. We will show that we can find enough subgroups of $ Z$ to satisfy the lower bound; there are, of course, lots and lots of other subgroups of $ S_n$ which we just ignore.

The group $ Z$ (like any elementary abelian $ p$-group) is actually a vector space - in our case, over the field $ \mbox{GF}(2)$ of $ 2$ elements, and the subgroups of $ Z$ are obviously also vector subspaces. So we need to count the number of vector subspaces of a vector space of dimension $ d$ over $ %Error. "mathbox" is a bad command.
{GF}(2)$; more generally, let $ \genfrac{[}{]}{0pt}{}{d}{k}_q$ be the number of subspaces of dimension $ k$ inside a vector space of dimension $ d$ over the field with $ q$ elements. We have:

$ \genfrac{[}{]}{0pt}{}{d}{k}_q = \frac{(q^d-1)(q^d-q)(q^d-q^2)\cdots(q^d-q^{k-1})}{(q^k-1)(q^k-q)(q^k-q^2)\cdots(q^k-q^{k-1})}$

This is the famous "Gaussian binomial coefficient" or "$ q$-binomial coefficient". If you've never seen it, here is a
[hide="proof of the formula."]
If $ V$ is a $ d$-dimensional vector space over $ F = \mbox{GF}(q)$, let us count the number of sequences $ (v_1, v_2, \ldots, v_r)$ of linearly independent vectors over $ F$. We can choose $ v_1$ in $ q^d-1$ ways since all vectors are eligible except for the $ 0$ vector. Then, we can choose $ v_2$ to be any vector except for the $ q$ multiples of $ v_1$, so it has $ q^d-q$ options.

Now $ v_1, v_2$ are linearly independent so they span the $ q^2$ vectors $ c_1v_1+c_2v_2$ as $ c_1,c_2$ range over $ F$. Thus $ v_3$ can be anything except for those $ q^2$ vectors, giving it $ q^d-q^2$ options. Continuing in this way, we find that the number of linearly independent sequences is just $ (q^d-1)(q^d-q)(q^d-q^2)\cdots(q^d-q^{r-1})$.

Now, to choose a $ k$-dimensional vector space $ W$ we can just choose an ordered basis $ (v_1, v_2, \ldots, v_k)$ for it, and then divide by the number of bases of $ W$ which is precisely the number of linearly independnent sequences $ (w_1, w_2, \ldots, w_k)$ in the $ k$-dimensional space $ W$. Overall we get that the number of $ k$-dimensional subspaces of $ v$ is the formula displayed above, which is very much analogous to the formula for the ordinary binomial coefficients.
[/hide]

Thus the total number of subspaces of a $ d$-dimensional space over $ \mbox{GF}(q)$ is
$ N(d,q) = \sum_{k=0}^d \genfrac{[}{]}{0pt}{}{d}{k}_q$.

To estimate this sum from below, we simply focus on just one of the summands. The best one to focus on is the largest, and it's not hard to guess that the largest number of subspaces occurs at dimension $ d/2$ (just like with subsets of a set). Thus, obviously,

$ N(2m,q) \geq \genfrac{[}{]}{0pt}{}{2m}{m}_q = \frac{(q^{2m}-1)(q^{2m}-q)\cdots(q^{2m}-q^{m-1})}{(q^m-1)(q^m-q)\cdots(q^m-q^{m-1})}$

and we take the easiest approach once again, replacing every term on the enumerator with $ q^{2m}-q^{m}$ which is obviously smaller than all of them, and replacing every term on the denominator with the largest one $ q^m-1$. The result is

$ N(2m,q) \geq \frac{(q^{2m}-q^{m})^m}{(q^m-1)^m} \geq q^{m^2}$

so in our case, with $ q=2$, the number of subgroups of $ Z$ is at least $ 2^{m^2}$, as needed. QED

It seems like we made a huge number of compromises here: we took a specific abelian subgroup $ Z$ of $ S_n$ and only considered its subgroups. Obviously there are many more copies of $ Z$ inside $ S_n$, many of which are totally disjoint from ours, and of course there are many subgroups of $ S_n$ which are not contained in any of those elementary abelian groups. In addition, while estimating the number of subgroups of $ Z$ we only looked at the mid-level dimension, and even there we made some pretty careless estimates.

So it would seem that $ S_{4m}$ should have [b]tons[/b] more subgroups than $ 2^{m^2}$. Indeed, it does, but surprisingly the exponent we got is in fact the correct order of magnitude: $ S_n$ has about $ 2^{O(n^2)}$ subgroups, if I remember correctly, though I couldn't quickly find a reference. So in a sense a good chunk of the subgroups of $ S_n$ are of this simplistic nature.

It was a bonus question for a problem set in Algebra [We were given about 14 in total]. Thank you for solving it. I will read your solution after the exams.


calculus
derivative
function
real analysis
real analysis solved

Let [tex]f \in C([0,1])[/tex] s.t. [tex]f[/tex] is differentiable on [tex](0,1)[/tex]. Prove that there is a sequence of real continuous functions which is pointwise convergent to [tex]f'[/tex] on [tex](0,1)[/tex].

What about $f_n(x) = n(f(x+\frac{1}{n})-f(x))$?

Your function is undefined on $[1-1/n,1]$  :?

[quote="grobber"]Can't that easily be repaired? :? Take it as liyi defined it on $[0,1-\frac 1n]$, and $\frac {f(x)-f(1)}{x-1}$ on the rest of $[0,1]$.[/quote]
I just meant that. I think this part is easy...  :blush: So I didn't write it out...
In fact, it is also okay if you define $f_n(x) = 0$ on $(1-\frac 1n,1)$.

But then the functions won't be continuous..  :?

oops...
I didn't notice that..
Then do it as grobber says...

Anyway, it is easy to smooth the "jumping" discontinuity..


There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?

[hide="solution"]For each case, (E.g. one white, then black, then white, then black, then black, then black.) each have a probability of $ \frac{1}{5^6}$ of happening. So we just need to find the number of cases. We divide this into casework:

[i]Case 1: one white and five black[/i]

We have wbbbbb. There are 6 arrangements of that total.

[i]Case 2: two white and four black[/i]

We have wwbbbb. There are $ \frac{6!}{4!*2!}\equal{}15$ arrangements of that total.

[i]Case 3: three white and three black[/i]

We have wwwbbb. There are $ \frac{6!}{(3!)^2}\equal{}20$ arrangements of that total.

[i]Case 4: four white and two black[/i]

We have wwwwbb. There are 15 arrangements of that total.

[i]Case 5: five white and one black[/i]

We have wwwwwb. There are 6 arrangements of that total.

$ 6\plus{}15\plus{}20\plus{}15\plus{}6\equal{}62$

So we put that over $ 5^6$:

$ \boxed{\frac{62}{5^6}}$[/hide]

That is assuming that you meant only black balls and white balls...

Or, to simplify your argument, we have two choices for each ball (black or white) except that we cannot choose all black or all white, so $ 2^6 \minus{} 2$.  (Another way to write this simplification is that you added $ \binom{6}{1} \plus{} \binom{6}{2} \plus{} \ldots \plus{} \binom{6}{5}$; this is the same as $ \sum_{n \equal{} 0}^6 \binom{6}{n} \minus{} \binom{6}{0} \minus{} \binom{6}{6}$.

I read the problem to mean, "what is the probability that there are some black and some white balls?" not "what is the probability that there are some black and some white and no other colored balls?"  Someone should solve that version, as well.


Not only did David Yang win the mathcounts written portion of the nationals competition, he also was one of the top 12 of the UASMO, and he's only in 7th grade...

here's the link: http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/2009usamo-highest.shtml

I agree. And he would have won countdown if he hadn't been in a rush to buzz in.

ya, lol. he was like half a second too late to answer

That's nice, but we already know that. I think this was a useless thread. I think this forum is for discussion of MC problems and contests not admiration.

[i]Added by admin -- yes.  Thread locked.[/i]


[hide="Solution"]Let x be the number of Jellybeans.

Thus, $\frac{4^2}{5^2}x=32$, and $x=50$.[/hide]

[hide=Solution]
Let $j$ be the # of jellybeans on the first day.

So: Beginning of first day: $j$ jellybeans

First day: $0.8j$ jellybeans.

Second day: $0.64j$ jellybeans left.

We now have $0.64j = 32$. Solving for $j$, we have $j = 50$. Thus, the correct choice is $(B)$.

@Darren the goat.... why would you bump A DECADE OLD post. We already knew the answer. Please don't do this in the future.

[hide][color=#783F04]Jenny ate $20\%$ of the jellybeans each day, thus each day the jellybeans remained will be $80\%$ of the last day. So after 2 days, it will be $64\%$. Answer is B because $50 \times 64\% = 32$[/color][/hide]

[hide=Solution]Since we know that Jenny ate $20$% of the jellybeans each day, we would know that she ate $64$% of the jellybeans after 2 days. Since the problems states that there were $32$ jellybeans at the [b]end[/b] of the 2nd day, we can set up an equation:
$0.64 \cdot x=20$
Solving this we get $x=50$, giving us answer choice [b][u]C[b][/u][/hide]

[quote=4everwise]Each day, Jenny ate $ 20\%$ of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, $ 32$ remained. How many jellybeans were in the jar originally?

$ \textbf{(A)}\  40\qquad \textbf{(B)}\  50 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\  60\qquad \textbf{(E)}\ 75$[/quote]

The problem is from 2000 AMC 10/12 #3

[hide=sol]
If there are $x$ jellybeans at the start of a day, then at the end of the day, there are $0.8x$ jellybeans. Therefore at the end of the second day there are $(0.8)^2 x = 0.64x$ jellybeans. We know that $0.64x = 32$ so $x = \frac{32}{0.64} = 50$ jellybeans. There were $50$ jellybeans in the jar originally so the answer is $\boxed{B}$

[quote=OlympusHero][b]Start Time: 3:12 PM (#1-2 Already Solved)[/b]

[hide=Solution]Let her starting amount be $x$. Then $0.8x$ remain after the first day and $0.64x = 32$ remain after the second day. We hence have $32 \cdot \frac{1}{0.64} = x \implies x = 50 \implies \boxed{\text{(B)}}$.[/hide][/quote]

Are you trying to speedrun this AMC 10?

Also, here's my [hide = Solution]Let the number of jellybeans that were in the jar originally be $x$. Then, since Jenny eats $20\%$ of the total jellybeans each day, and there were $32$ jellybeans remaining at the end of the second day, we have that $0.8(0.8x) = 32$. Therefore, the total number of jellybeans, $x$, that were in the jar originally was $0.8(0.8x) = 32 \Rightarrow 0.64x = 34 \Rightarrow \frac{64}{100}x = 32 \Rightarrow x = 32 \cdot \frac{100}{64} = \frac{100}{2} = \boxed{\textbf{(B)}\ 50}$.[/hide].

Let’s assume the number of jellybeans to be 100.
So by the end of second day number of jellybeans will be 64.
Now letting the original number of beans be x,
64/100= 32/x
So x=50


mấy anh chỉ cho em chỗ nào là box phương trình

Tớ ko rành chuyện này lắm! :lol:  Nhưng tớ nghĩ cậu có thể gửi bài vào đây:
http://www.mathlinks.ro/index.php?f=150


What is the positive difference between the 75th and 50th term of the sequence in which the $ n$th term is found by adding the first n positive integers?

75th term=75+74+73...+1
50th term=50+49+...1
Thus, the difference is 51+52...75
Using the arithmetic sequence formula, get (51+75)/2*25 terms=1575.


geometry
circumcircle
combinatorics proposed
combinatorics

Let P be a covex polygon with n sides  which have no 4 vertices concyclic.A circle is called "C" if it through 3 vertices of P
and contain all remaining vertices.A circle "C"  is called "B" if it through 3 consecutive ,and called "I"  it's not through any
two consecutive vertices of P.The number of circle "B" is f(P) ,the number of circle "I" is g(P).
Prove that : f(P)=2+g(P)   (for all convex polygon P ,and n>3.)

This problem seems to have something of the IMO style. Anybody has an idea?

I'll just outline the main ideas.

Try to show that the triangles having diagonals of the polygon as sides and whose circumcircles encompass the polygon form a [url=http://en.wikipedia.org/wiki/Polygon_triangulation]triangulation[/url] of the polygon. Then, associate a tree to this triangulation as follows: the vertices of the tree are the triangles themselves, and two vertices are connected iff the two corresponding triangles share a side. Notice that the triangles with property "B" (in Nbach's terminology) correspond to the vertices of the tree having degree $ 1$, while those with property "I" correspond to the vertices of degree $ 3$. The problem has now become:

Let $ T$ be a tree in which all vertices have degree $ \le 3$. Prove that the number of vertices of degree $ 1$ is larger than that of vertices of degree $ 3$ by two.

Denotes vertices of P  are $ a_{1},a_{2},...,a_{n}$ in the clock wise. A segment  $ [a_{k},a_{k+s}]$ is a sequence of points $ {a_{k}, a_{k+1},...,a_{k+s}}$ in the clock wise where indices modulo $ n$ .When $ s=1$ ,we call this segment is empty.Eg: $ [a_{n},a_{1}]$ is empty.With a triangle $ ABC$,we write 3 vertices in the cock wise is $ abc$ .It's mean that $ b \in[a;c]$
Call $ [a,b]; [b,c]; [c,a]$ are segment of triangle $ ABC$. Prove 3 lemma
Lemma $ 1$ : If $ abc;klm$ are triangles of kind "C" then the intersection of their interiors is empty.Or clearly,They can only have common vertice or side.
Lemma $ 2$: Suppose that $ [a,b]$ is empty or there exist $ d$ such that $ adb$ is a triangle of kind "C" and [b,a] isn't  empty ,then there exist unique $ c\in [b,a]$ such that $ abc$ is a triangle of kind "C".
Lemma $ 3$ :All triangles of kind "C" pave polygon P.


I was wondering this 2 years ago I think. But I was too lazy to, and I still am. :P

I would...[hide]graph[/hide][/hide]

Counting would work as well.
if fact, I think all multiples of 1.5 up to 132 can be expressed since 6*22 is 132, and the others can be subatituted with amounts of 1.5 since 4 1.5s make a 6

6*23+1.5*3
6*24+1.5*2, 6*24+1.5*3
are the only 3 that cannot be represented

[quote="TachyonPulse"]Suppose that we have  $ 0 \le x \le 150$. How many values of $ x$ can be expressed in the form $ 6m \plus{} 1.5n$  for nonnegative integers $ m,n$ such that $ m \plus{} n \le 25$?[/quote] Basically Rob Roobiks solution works. You can get any multiple of 1.5 except a few at the top.

First, I would multiply the equation $ 0\le 6m\plus{}1.5n\le 150$
by $ \frac{2}{3}$ to get
$ 0\le 4m\plus{}1n\le 100$

Note that the equation $ m\plus{}n\le 25$ does not change because the proportion of the number of $ m$ 's and $ n$ 's stays the same.

It is evident that the highest number that we can get is $ 100$.
First, we make $ {m\equal{}0, 1, 2, 3, ..., 25}$ .

We can obtain the every multiple of four by making $ n\equal{}0$
We can obtain the numbers inbetween the multiples of four by making $ n\equal{}1,2,3$ .

So we can make the integers $ 0, 1, 2, 3, ..., 91$ by making the pairs
$ (0,0)(0,1)(0,2)(0,3)$
$ (1,0)(1,1)(1,2)(1,3)$
$ (2,0)(2,1)(2,2)(2,3)$
$ (3,0)...$

We only have to stop when we get to the pair $ (22,3)$, which equals $ 91$, because $ m\plus{}n$ has finally reached $ 25$.
Now we just count the pairs where $ m\plus{}n\le 25$ and
$ 23\le m \le 25$ .
When $ m\equal{}23$, $ n$ can equal $ 0$, $ 1$, or $ 2$.
When $ m\equal{}24$, $ n$ can equal $ 0$, or $ 1$.
When $ m\equal{}25$, $ n$ can equal $ 0$.

So, we have four values of $ n$ for each value of
$ m\equal{}0, 1, 2, ..., 22$ and we have $ 3$, $ 2$, and $ 1$ value of n when $ m\equal{}23, 24, 25$, respectively, which equals
$ 4*23\plus{}3\plus{}2\plus{}1 \equal{} 92\plus{}6 \equal{} \boxed{98}$

From which AMC exam was the problem taken?

None of them. This problem is basically asking what the possible scores are on the AMC exam.

[quote="thegathering"]a very good website to study math: [url]http://diendantoanhoc.net/forum/index.php?[/url][/quote]

This is just a Vietnamese math forum, so I'm guess it's spam. Mods can delete these posts.

[quote="thegathering"]where are you come from?
Tôi là người Việt!!!!!!!!!!!!!!!![/quote]

Did you mean "Where are you from?"  ---> "I'm vietnamese"


A magician asks a participant  to randomly arrange 10 "face down" cards labelled 1 to 10. The participant does not show the arrangement of these cards to the magician, but does show them to the magicians assistant. The assistant looks at the 10 cards and flips over 6 of the cards in a [b]certain ordered manner[/b] to reveal their values to the magician. The magician immediately identifies the values of the four remaining cards.

How does the trick work?

Well this is not an elegant method, but you could do this:

WLOG, cards 1-4 are the first 4, in some order.  The assistant flips cards 5 and 6 first.  This does not matter.  Then, the assistant flips cards 7-10 in the same order as numerically are cards 1-4.

I'm pretty sure I misunderstood the problem, but...

That's what I came up with also. Maybe the assistant is supposed to flip 4 and have the magician determine the other 6?

I wonder if there is some similar problem that can take advantage of the extremely unusual (unique?) fact that 10!=7!6!. Anyway, zabelman's problem is definitely solvable (i.e. there is an algorithm for doing it).

I am sorry. The problem may not be well stated.

[quote]The assistant looks at the 10 cards and flips over 6 of the cards in a [b]certain ordered manner[/b] to reveal their values to the magician[/quote]

The assistant can only flip the cards either from left to right or from right to left.

I guess I confused a lot of people here. The following non-trivial problem appears in some books which have a section on the Pigeonhole Principle (Paul Zeitz? not sure, I don't have a copy here)
[quote]Given any sequence of mn+1 real numbers, some subsequence of (m+1) numbers is increasing or some subsequence of (n+1) numbers is decreasing. (Anybody interested in giving a proof?)[/quote]
Applying this to m=n=3, no matter how the cards are arranged, there is an  increasing or a decreasing subsequence of 4 cards. The rest of the trick is obvious.


[quote="black_hole123"]Some one has the USA TST 2005 or 2006 problems can post there?[/quote]

Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79839]here[/url] for 2004. For 2006 and 2006, please continue the discussion [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47106]there.[/url] Also check the [url=http://www.mathlinks.ro/Forum/resources.php?c=182&cid=35]resources[/url] for previous years. Topic locked.


Coloring
algebra
partition
Ramsey Theory
IMO Shortlist
combinatorics

Prove that the set of positive integers cannot be partitioned into three nonempty subsets such that, for any two integers $x,y$ taken from two different subsets, the number $x^2-xy+y^2$ belongs to the third subset.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Assume we have such a partition $ \mathbb{N}\equal{}A \cup B\cup C$ in nonempty subsets,
such that $ 1\in A$ and the minimal $ n\in \mathbb{N}\setminus A$ lies in $ B$. Then $ n\ge 2$.

1) $ x,y,x\plus{}y$ can't lie in all three different subsets for any $ x,y\in \mathbb{N}$:
Assume $ x\in A, y\in B, x\plus{}y\in C$, then by assumption
$ z: \equal{}x^2\minus{}x(x\plus{}y)\plus{}(x\plus{}y)^2\equal{}y^2\minus{}y(x\plus{}y)\plus{}(x\plus{}y)^2$ lies in $ B$ and in $ A$.
Contradiction!

2) $ n\ge 3$:
If $ n\equal{}2$, then $ 1\in A,2\in B$ gives $ 3\equal{}1^2\minus{}1\cdot 2\plus{}2^2\in C$.
But then $ 3\equal{}1\plus{}2$ contradicts 1).

3) We find $ k,l\in C$ with $ 0<l\minus{}k<n$:
By 2) we have $ 1,2\in A, n\in B$, so for $ n>3$ we can take $ k: \equal{}n^2\minus{}2n\plus{}2^2,l: \equal{}n^2\minus{}n\plus{}1^2\in C$
with $ l\minus{}k\equal{}n\minus{}3$, so $ 0<l\minus{}k<n$.
For $ n\equal{}3$ we have $ 1,2\in A$, $ 3\in B$, so $ 7\equal{}3^2\minus{}3\cdot 1\plus{}1^2\in C$.
Then by 1) we have $ 4\equal{}1\plus{}3\not\in C$ and $ 4\equal{}7\minus{}3\not \in A$.
So $ 4\in B$ and we can take $ k: \equal{} 4^2\minus{}4\cdot 2\plus{}2^2\equal{}12\in C$ and $ l: \equal{} 4^2\minus{}4\cdot 1\plus{}1^2\equal{}13\in C$.

Finally choose $ (k,l)\in C^2$ with $ 0<l\minus{}k<n$ and $ k$ minimal.
By minimal choice of $ n$ we have $ l>n$, so $ m: \equal{}l\minus{}n\in \mathbb{N}$.
Also $ k\minus{}m\equal{}n\minus{}(l\minus{}k)$ and $ 0<l\minus{}k<n$ gives $ 0<k\minus{}m<n$,
and again by minimal choice of $ n$ we have $ k\minus{}m\in A$.
Together with $ n\in B$ and $ k,l\in C$ this gives $ m\equal{}l\minus{}n\not\in A$ and $ m\equal{}k\minus{}(k\minus{}m)\not \in B$ using 1).
So we find $ (m,k)\in C^2$ with $ 0<k\minus{}m<n$, which contradicts the minimal choice of $ k$.

We can finish it in another way: let $a$ be smallest which is in set which not consists 1. Assume $1 \in A, a\in B$. We choose partition with smallest possible value of $a$. Now we see that $a^2-a+1$ is in $C$. So $a^2-a\not\in B$ (because $1, a^2-a,a^2-a+1$ aren't in three different subsets, like olorin showed). Now we make bijection $ka \rightarrow k$ and because $a \in B, a^2-a\not\in B$ we get contradiction.

If $x$, $y$, and $x+y$ appear on different sets, then \[A=x^2+(x+y)^2-x(x+y)=y^2+(x+y)^2-y(x+y)\] means that $A$ is in the same set as both $x$ and $y$, contradiction. Thus, $x+y$ must be on the same set as either $x$ or $y$. Let our sets be $S_1$, $S_2$ and $S_3$. WLOG, let $1\in S_1$. Let $m$ be the smallest integer not in $S_1$, and let it be in $S_2$.

Suppose $m=2$ then $1^2-1\cdot 2+2^2=3\in S_3$, contradiction. Suppose $m=3$ then $7\in S_3$, and following that, from the sum rule we found previously, $4\in S_2$, $5,6\in S_1$ and then $9,12,13\in S_3$ and so there is no place to put $8$.

Thus, $m\ge 4$. Note that $m^2-m+1$ and $m^2-2m+4$ are in $S_3$. Let $k$ be the maximum positive integer such that $m^2-km+1$ and $m^2-(k+1)m+4$ are both in $S_3$. Note that $(m^2-km+1)-(m)$ must not be in $S_1$ and $(m^2-(k+1)m+4)-(3)$ must not be in $S_2$. They are the same number, so $m^2-(k+1)m+1$ must be in $S_3$. Similarly, now, $(m^2-(k+1)m+1)-(m-3)$ must not be in $S_2$ and $(m^2-(k+1)m+4)-(m)$ must not be in $S_1$, and since they are the same number, $m^2-(k+2)m+4$ must be in $S_3$. Contradiction. We are done.

[hide=Post #5 by darij grinberg]
[url=https://aops.com/community/user/432]darij grinberg[/url] [url=https://aops.com/community/p2728758](view original)[/url]
@RaleD, what do you mean by "make bijection $ka \rightarrow k$"?

@olorin: nice proof, but you the (WLOG) assumption that $1\in A$ and that the minimal $n\in\mathbb N\setminus A$ lies in $B$ should be done after step 1.
[/hide]

What RaleD means, I think, is that if $a$ is the smallest which CAN be in a set without $1$. (Meaning that any smaller number cannot result in a valid partitioning) I guess the idea is that $x,y,x+y$ not being in three different sets is a homogeneous condition so we can multiply everything by $a$ so theoretically $a$ and $a^2-a$ must still be in the same set.

I don't think it works and the reason is that while $x,y,x+y$ is a homogeneous condition, the proof also uses the fact that $a^2-a+1$ is in the other set, which is not homogeneous and will probably change the proof after multiply by another $a$.


(I have a short and amazing (I think) solution by complex number but I still want to know a solution in pure geometry  :blush:  ( and I hope that solution is natural  :) )
for quadrilateral ABCD with $ AD \equal{} CD$, $ \angle DAB \equal{} \angle ABC$. A line through D and midpoint of BC intersect AB at E. prove that:$ \angle BEC \equal{} \angle DAC$

Let $ T$ be the midpoint of $ BC$ and  $ P\equiv DA \cap CB.$ By Menelaus' theorem for $ \triangle PTD$ cut by the line $ \overline{ABE},$ keeping in mind that $ \triangle PAB$ is isosceles due to $ \angle DAB \equal{} \angle ABC,$ we have

$ \frac {\overline{ET}}{\overline{ED}} \cdot \frac {\overline{DA}}{\overline{AP}} \cdot \frac {\overline{PB}}{\overline{BT}} \equal{} 1 \ \Longrightarrow \ \frac {\overline{ET}}{\overline{ED}} \equal{} \frac {\overline{TC}}{\overline{DC}}$

$ \Longrightarrow$ $ CE$ is the external bisector of $ \angle DCB$ $ \Longrightarrow$ $ \angle BCE \equal{} 90^{\circ} \minus{} \frac {_1}{^2}\angle BCD.$

By simple angle chasing we have:

$ \angle BEC \equal{} \angle DAB \minus{} (90^{\circ} \minus{} \frac {_1}{^2}\angle BCD) \equal{} \angle DAB \minus{} 90^{\circ} \plus{} \frac {_1}{^2}\angle BCD$

$ \angle DAC \equal{} 90^{\circ} \minus{} \frac {_1}{^2}(360^{\circ} \minus{} 2\angle DAB \minus{} \angle BCD) \equal{} \angle DAB \minus{} 90^{\circ} \plus{} \frac {_1}{^2}\angle BCD$

 $ \Longrightarrow \angle DAC \equal{} \angle BEC,$ as desired.


vector
Functional Analysis
real analysis
real analysis unsolved

Let $ H$ be a Hilbert space and $ T\in K(H)$ a compact operator. Show that $ \mbox{im} (T)$ is closed iff $ T$ has finite rank.

Of course, a finite dimensional vector space is closed, so <-- is obvious. For the other direction, I read a proof in some book using the singular value decomposition. However, we haven't covered that yet so I would not like to use it. Does someone know a different way?

The result holds true even in Banach space.

[b]Proof.[/b] Let $ X, Y$ be Banach spaces $ T: X \to Y$ be a compact operator, $ T(X)$ be closed. Then $ T(X)$ itself is a Banach space, so we can consider $ T$ as surjective operator acting from $ X$ to $ T(X)$. Then the Open map theorem says that $ T(B_X)$ contains an open ball $ rB_{T(X)}$ of the space $ T(X)$. On the other hand according to the definition of compact operator $ T(B_X)$ is pre-compact, so $ B$ is precompact. Then the unit ball $ B_{T(X)}$ of $ T(X)$ is pre-compact as well.  But due to the famous F. Riesz's theorem if the unit ball of a normed space is  pre-compact then the space is finite-dimensional.

[quote="BlueVelvet"]Thank you, volodya. I didn't know the Lemma by Riesz, so I looked it up. Can I ask what you mean by $ B$ in "so $ B$ is precompact"?[/quote]

I meant $ rB_{T(X)}$. Sorry for the misprint!

No need to apologize  :lol:

I hate to ask this, but could you/someone take a look at [url=http://www.mathlinks.ro/viewtopic.php?t=313524]my other topic[/url] about compact operators? I placed in Advanced Fields, but now I think it should have been here, in Real Analysis.

[quote="BlueVelvet"]could you/someone take a look at [url=http://www.mathlinks.ro/viewtopic.php?t=313524]my other topic[/url] about compact operators? [/quote]

Well, I have had a look at your other post. The solution is basically correct. There are only few things that could be improved (in my opinion). From (i) you can deduce that $ T$ possesses an orthonormal basis consisting of eigenvectors with corresponding eigenvalues -- zeros and $ \frac2k$. For such a diagonal operator self-ajointness is equivalent to the fact that all the eigenvalues are real and compactness -- to the fact that the sequence of eigenvalues tends to zero. This solves (ii). Your solution of (iii) is OK.


Hello! I sent my solutions via email in the early AM on Oct. 4th, and I still haven't gotten an email or a check mark. I knew from previous posts that confirmations could take a while, but now I'm starting to think my solutions might not have been received. I'm looking at the Sent Mail record in my email program, and it looks like I sent it correctly. Is there anything that can be done?

Thanks,
Aaron

vRusczyk has PM'ed you with some details.

Hey, thanks for helping me straighten this out! I'd better get started on Round 2... looks pretty tough...


algebra
polynomial
calculus
topology
function
binomial coefficients
Ring Theory

Consider a polynomial which has form:
$(1+x+x^{2}+...+x^{p})^{n}$
where $p, n$ are two positive.
THe question is: can we calculate concretly the coefficient $a_{k}$ of this polynomial, for some k? Or for which cas, we could determine $a_{k}$?

Yes, of course this is possible as some type of sum. In many combinatorial applications it will be actually enough to find the coefficient $a_{k}$ in the series $\left( \sum_{k=0}^{\infty}x^{k}\right)^{n}= \frac{1}{(1-x)^{n}}= \sum_{k=0}^{\infty}\binom{-n}{k}(-x)^{k}= \sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{k}$
This will work if $k$ is not too big. Also note that this is not really calculus - the sums should be viewed as completely formal sums where we just identify coefficients, no convergence or anything to worry about. However, the fact that an identity holds for series in terms of calculus makes it hold in terms of formal sums too since equal sums means equal coefficients (although there is some obvious but tedious checking for example that multiplication of power series in calculus is the same as formal multiplication. I guess in a really good presentation you introduce some topology in R where every power series converges).

For the original question, there's no universal way to do it which is also computationally efficient- you're going to get sums involving binomial coefficients, probably in several layers. That's why you work with special cases- Kalle gave the case in which $k$ is at most $p$.

[quote="Kalle"]...where every power series converges[/quote]
That would be the ring of formal power series, and you lose the ability to evaluate the series as a function there. You can't have everything at once- that's why we move back and forth between the ring of formal power series and the ring of (real) analytic (germs) functions near zero.
Evaluation at a point should be a ring homomorphism; the only homomorphism from the whole ring of formal power series to $\mathbb{R}$ is the constant term (evaluation at zero).

There are several reasonable topologies for the ring of formal power series; in the strongest, $\sum_{n}a(m,n)x^{n}$ converges to $\sum_{n}b(n)x^{n}$ if and only if $a(m,n)=b(n)$ whenever $m>M(n)$ for some sequence $M(n)$. Weaker topologies use the topology of the coefficients as well; for series with coefficients in $\mathbb{R}$, we could say that $\sum_{n}a(m,n)x^{n}$ converges to $\sum_{n}b(n)x^{n}$ if and only if $a(m,n)$ converges to $b(n)$ for each $n$. Uniform convergence of analytic functions on some compact set is slightly weaker than this second topology, and of course only applies to the subset of series which converge on that set.


function
integration
inequalities
algebra
polynomial
real analysis
real analysis theorems

There are many different paths.

The first issue is notation - real or complex, over what period, and where does the $2\pi$ go? I'll make these choices:

$f$ a function on $\mathbb{T}_P$ - that is, periodic of period $P.$

For $k\in\mathbb{Z},\,\hat{f}(k)=\frac1P\int_0^Pf(x)e^{-\frac{2\pi ikx}P}\,dx.$

Here's an utterly formal "argument": suppose $f$ and $g$ are functions for which Fourier inversion works (and we agree not to look to closely at justifications for interchanging sums and integrals).

That is, suppose that $f(x)=\sum_{k=-\infty}^{\infty}\had{f}(k)e^{\frac{2\pi ikx}P}$ and similarly for $g.$

Then $\sum_{-\infty}^{\infty}\hat{f}(k)\overline{\hat{g}(k)}= \sum_{k=-\infty}^{\infty} \left(\frac1P\int_0^Pf(x)e^{-\frac{2\pi ikx}P}\,dx\right)\overline{\hat{g}(k)}$

$=\frac1P\int_0^Pf(x)\overline{\sum_{-\infty}^{\infty}\hat{g}(k) e^{\frac{2\pi ikx}P}} =\frac1P\int_0^Pf(x)\overline{g(x)}\,dx$

If we now let $f(x)=g(x),$ we get

$\sum_{-\infty}^{\infty}|\hat{f}(k)|^2=\frac1P\int_0^P|f(x)|^2\,dx.$

If you want more rigor than that, I would travel the following path: The functions $e^{\frac{2\pi ikx}P}$ are orthonormal in $L^2(\mathbb{T}_P).$ (With measure $\frac{dx}P.)$  The general theory of expansions on orthonormal sets now leads to Bessel's inequality:

$\sum_{-\infty}^{\infty}|\hat{f}(k)|^2\le\frac1P\int_0^P|f(x)|^2\,dx.$

with equality if the orthonormal set can be shown to be a complete orthonormal set.  (That is, the only function orthogonal to all of them is the zero function.) One way to show an orthogonal set is complete is to show that the finite linear combinations of that set (in this case, the trigonometric polynomials) are dense in the appropriate Hilbert space.

For this particular case, we can accomplish this constructively. The Cesaro means of the Fourier series of a continuous function converge uniformly to that function; the same proof applied to $L^2$ functions shows convergence in $L^2.$ This proof works by analyzing the convolution with the Féjer kernel.


[quote="etaratoris"]Let $ x,y,z >0$ . Find the minimum of $ \frac{x^5\plus{}y^5\plus{}z^5\plus{}3}{x\plus{}y\plus{}z}$[/quote]
Edited to display property. Don't use colons in mathematical expressions, since they are used in emoticons.

The final answer is not exactly like that but I like your idea!!

By 5th-order mean > AM,
$ x^5 \plus{} y^5 \plus{} z^5 \geq 3((x \plus{} y \plus{} z)/3)^5 \equal{} (1/3)^4 \cdot (x \plus{} y \plus{} z)^5$.
Equity's at x=y=z.

By AM>GM,
$ (x^5 \plus{} y^5 \plus{} z^5) \plus{} 3/4 \plus{} 3/4 \plus{} 3/4 \plus{} 3/4 \geq 5 \cdot [(3/4)^4 \cdot (x^5 \plus{} y^5 \plus{} z^5)]^{1/5} \geq 5 \cdot (3/4)^{4/5} \cdot (1/3)^{4/5} \cdot (x \plus{} y \plus{} z) \equal{} 5\cdot (1/4)^{4/5}(x \plus{} y \plus{} z)$.
Equity's at $ x^5 \plus{} y^5 \plus{} z^5 \equal{} 3/4$ and therefore minimum $ \equal{} 5 \cdot (1/4)^{4/5}$ and occurs at $ x \equal{} y \equal{} z \equal{} (1/4)^{1/5}$.

[quote="ghjk"][quote="jmerry"][quote="etaratoris"]Let $ x,y,z > 0$ . Find the minimum of $ \frac {x^5 \plus{} y^5 \plus{} z^5 \plus{} 3}{x \plus{} y \plus{} z}$[/quote]
Edited to display property. Don't use colons in mathematical expressions, since they are used in emoticons.[/quote]
Here's my solution:
Applying AM-GM we have: x^5+y^5+z^5+3/4+3/4+3/4+3/4>=5.5th-root[(3/4)^4.(x^5+y^5+z^5)>=5.5th-root[(3/16)^4].(x+y+z)
Therefore, we easily see that A=(x^5+y^5+z^5+3)/(x+y+z)>=[b]5.5th-root[(3/16)^4][/b]
The equality occurs when x=y=z=5th-root(3/4)
PS: We can solve the generalized problem by using the same method above:
Find minimum of: A=(x^k+y^k+z^k+l)/(x+y+z) with l and k are positive integers. :)[/quote]

[quote="etaratoris"]Here is a more elegant solution

x^5 +1+1+1+1 >= 5x (Cauchy)
so x^5>=5x-4 and by posing x={4^(1/5)}x we find x^5 +1 >=1,25{4^(1/5)}x
The equality for x=1/{4^(1/5)}
Therefore minimum=1,25{4^(1/5)} for x=y=z=1/{4^(1/5)}[/quote]
In fact, this is the same idea as the way I used Cauchy-Schwartz to solve it. By the way, nobody minds to try solving my generalized inequality posted above? :o


analytic geometry
number theory
greatest common divisor
factorial
vector
calculus
integration

Show that if we have a coordinate sistem in the plane there doesnt exist a regular pentagon ABCDE such that A,B,C,D and E have both coordinates integer.

Moreover, it is well known that if n =/= 4,
there does not exist a regular n-gon, the vertices of which are all lattice points.

yes, but I think he means without using that property ;)

well I think simply applying it for tan 72 (which is not a rational number) does the job as well :?

This is quite easy...

Assume that all the coordinates are integers.  Let the coordinates be A(a1, a2), B(b1, b2), C(c1, c2), D(d1, d2), E, (e1, e2).  WLOG assume that gcd(a1, a2, b1, b2, etc.) = 1 (otherwise, simply divide by the gcd, and the resulting pentagon A'B'C'D'E' is regular and has all integer coordinates.)  Also, we can assume that a1, a2 are both even (otherwise, just translate the entire pentagon such that this holds).

If b1, b2 are both even, then 4|AB2.  It's easy to see that c1, c2 are both even (otherwise 4 does not divide BC2).  By a similar argument d1, d2, e1, e2 are all even, which contradicts the earlier assumption.

If b1, b2 are both odd, then AB2 = 2 mod 4.  This implies that c1, c2 are both even => d1, d2 are both odd => e1, e2 are both even => a1, a2 are both odd.  Contradiction.

If b1 + b2 is odd => c1 + c2 is even => d1 + d2 is odd => e1 + e2 is even => a1 + a2 is odd.  Contradiction.

So for all cases you will arrive at a contradiction.  This method works for all regular n-gons where n is odd.

I knew a solution of this using the fact that Z[i] is a factorial ring, but I guess there's no point in posting it, since it proves the general property (that for n=/=4 there are no latticial regual n-gons), and we don't need that much here.

whatever way we solve it, we all agree it was easy :D

Here is an very nice and elementary way to solve a more general problem :

Let's give (O,u,v) be a system of coordinates (u and v are ANY two not collinear vectors) and let's L be the lattice form with the points with integral coordinates in this system.
Let P be a regular (with respect to the usual euclidean distance) convex n-gon inscribed in the lattice (that is each of its vertices is an element of L).
Then n = 3, 4 or 6.

Proof :
Lemma :
If P is a regular convex n-gon with n > 6. Let d be the common length of its sides and R be its circumradius. Then d < R.

Proof : From the law of cosines, we have d 2  = 2R 2 ( 1 - cos(2 \pi /n) ) < R 2 .

Now, suppose that n > 6. Let P = A_1,...,A_n be an inscribed convex regular n-gon with minimal length d > 0. Let O be its circumcentre.
Let B_i such that vector OB_i = vector A_iA_(i+1) for i = 1,...,n (and with A_(n+1) = A_1).
It is easy to see that P' = B_1...B_n is an inscribed convex regular n-gon with side-length d' < R' = d (from the lemma), which contradicts the minimality of d.

Let n = 5. Suppose that A_1...A_5 be an inscribed convex regular pentagon. As above, we may assume that P has minimal length d > 0.
Let P' = B_1...B_5 be the convex pentagon whose vertices are the intersection ppoints of the diagonals of P. It is easy to see that P' is a convex regular inscribed (A_1A_5B_1A_2 is a parallelogram, and so on for the others...) pentagon with side-length d' < d, which contradicts the minimality of d.

So we are done....

It is easy to find u and v such that we may inscribe an equilateral triangle and a regular hexagon in the lattice. In another hand, it is a classical result, using the irrationality of  \sqrt 3 that the only convex regular n-gon which can be inscribed in the usual euclidean lattice is the square.
More generally, we may prove that the is no plane lattice into which we may inscribe both a square and an equilateral triangle.

Moreover, if we work in higher dimensions, we may prove that in |R  p with p > 2, we may inscribe a convex regular n-gon in Z p  if and only if n = 3, 4 or 6, and for n = 3 or 6 it can be done for p = 3.

Pierre.


geometry
trapezoid
circumcircle
geometry proposed

Let $ ABCD$ be a trapezoid such that $ AB \parallel CD$ and assume that there are points $ E$ on the line outside the segment $ BC$ and $ F$ on the segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Let $ I,J,K$ respectively be the intersection of line $ EF$ and line $ CD$,  the intersection of line $ EF$ and line $ AB$, and the midpoint of segment $ EF$. Prove that $ K$ is on the circumcircle of triangle $ CDJ$ if and only if $ I$ is on the circumcircle of triangle $ ABK$.
[i]Utari Wijayanti, Bandung[/i]

It's ISL 2008:   :)
[url]http://www.mathlinks.ro/viewtopic.php?p=1555917#1555917[/url]


[quote="mel"]Polynomials that have palindromic coefficients (like $ax^4 + bx^3 + cx^2 + bx + c = 0$) have a very special property, but my mind seems to have blanked, and I can't remember it.  :oops:  Can someone refresh my memory, please?[/quote]

[url=http://www.mathpages.com/home/kmath294.htm]Click for a paper on them[/url]

Thanks! I thought of the palindromic description on the spot, so I never thought to search for "palindromic polynomials." (Didn't know it was an actual term...:P)

i think you mean

$p(x)=ax^4+bx^3+cx^2+bx+a$

the roots and their inverses are roots too (think about the transformation to a polynomial with inverse roots)

Since the polynomial with roots that are reciprocals of $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ is $a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n$, if $a_i = a_{n-i}$, then the reciprocals of the roots are also roots...

not very useful, i know :P


[quote="Inspired By Nature"][quote="ZetaX"]I assume you speak from anyone [u]else[/u] reading them...
Mods can't.
Admins can do (as good as) everything, so as long there is not some cryptography, they could if they want, I think (I'm no admin, so I can't know for sure).[/quote]
I think they can. After all, they made aops/mathlinks. :)[/quote]

True. They have to control everything that happens. For privacy, write e-mails instead of PM's, altought I think admins aren't wasting their time only reading PM's.....

Yes, administrators can read "Private" Messages. :D On AoPS, pms are basically a special type of post, and as we all know, admins can do whatever they want with any post.

They can also edit your blog entries. (personal experience)

In my opinion, this topic is completely irrelevant. I just felt like posting. :P

[quote="MCrawford"][quote="Totally Zealous"]I was just wondering, can anyone read PMs? (like moderators or admins)[/quote]
Why?  Should we be reading yours?[/quote]
good point. :P
I doubt admins would read pm's for no good reason.

right MCrawford :)

just a privacy thing

Just to be clear, we would never go reading PMs unless somebody on the site was causing serious trouble.  We have about 1.7 billion higher priorities running the site alone.

But if anyone is concerned about privacy and our ability to read messages, the PM system is not the best option for communication.

To add on to this: the only time I have looked in the PMs in the database is when someone PM'ed me trying to sell mortgages or other spam.  Then I ban them, and go into the database and try to clean out the PMs they have sent - but I'm only then looking at the PMs sent by them.

Well I guess that's a good thing in that case

[quote="MCrawford"][quote="Totally Zealous"]I was just wondering, can anyone read PMs? (like moderators or admins)[/quote]

Why?  Should we be reading yours?[/quote]

No I was just wondering.
You can read mine if you want to.
All it is is a discussion of who dies at the end of the 7th harry potter book.

If anyone wants to know my guesses PM me.


calculus
integration
logarithms
calculus computations

can anybody help me to solve this integration, please:
\[ \int_0^1 \frac {\ln (1\plus{}x)}{1\plus{}x^2}\, dx\]
the messages gives is to use the series, there should be Taylor series or Maclaurin series, but i don't know how to use it.

You can see the solution here. :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=149755&search_id=1928257573#149755[/url]

thank you, kunny. a bundle of very wonderful answer. :P

[quote="kunny"]不用谢 :)[/quote]
 :lol: 你会中文，呵呵


Λοιπον η παρακατω ασκηση ειχε πεσει στους μεγαλους το 2002 και εχει ξαναμπει 2 φορες στο φορουμ αλλα δεν εχει δωθει σαφης λυση. Απο' τι καταλαβα θελει την αρχη της περιστεροφωλιας αλλα δεν μπορω να βγαλω ακρη.

Ενας μαθητης του μετσοβιου πολυτεχνειου διαβαζει μαθηματικα για 37 μερες υπο τους παρακατω κανονες:
1) Διαβαζει καθημερινα θετικο ακεραιο αριθμο ωρων μεγαλυτερο ή ισο του 1 και μικροτερο ή ισο του 12
2) Συνολικα και τις 37 μερες δεν διαβαζει περισσοτερο απο 60 ωρες

ΝΔΟ υπαρχουν καποιες διαδοχικες μερες κατα τις οποιες συνoλικα διαβασε 13 ωρες.

Το προβλημα επισης βρισκεται εδω http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2064786370&t=62046
και εδω http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1913267071&t=46281

Μπορειτε να με βοηθησετε?? :fool:

Χμμ..
Πιο συγκεκριμενα,
εσυ θελεις να βρεις ακεραιους $ m,z$ τετοιους ωστε να ισχυει
$ a_{m} \plus{} a_{m \plus{} 1} \plus{} .. \plus{} a_{z} \equal{} 13$
Για τα m,z δεν σε ενδιαφερει ποσο απεχουν αρκει να ισχυει $ m < z$.
ομως η παραπανω σχεση ειναι ισοδυναμη με
$ A_{z} \minus{} A_{m} \equal{} 13$, οπου $ A_{i} \equal{} a_1 \plus{} a_2 \plus{} .. \plus{} a_{i}$ και $ 1 \leq i \leq 37$
Τωρα αξιοποιωντα ςτα δεδομενα σου ισχυει,
$ 1\leq A_1 < A_2 < .. < A_{37} \leq 60$(1)
και προσθετοντας $ 13$,
$ 14 \leq A_{1} \plus{} 13 < .. < A_{37} \plus{} 13 \leq 73$(2)
οι σχεσεις ομως (1) και (2) δηλωνουν οτι εχεις 73 ακεραιους (φωλιες απο το 1 εως το 73) για να τοποθετησεις 74 ακεραιους(περιστερια που ειναι οι $ A_1,...,A_{37},A_1 \plus{} 13,...,A_{37} \plus{} 13$  ) :wink: επομενως υπαρχουν ακεραιοι m,z που θα ικανοποιουν το ζητουμενο.
Ελπιζω να βοηθησα σχετικα  :)

Καταλαβα. Η ισοδυναμια που γραφεις ειναι ολη η δουλεια. Ευχαριστω :)


trigonometry
inequalities
inequalities proposed

Let $ABC$ be an acute-angled triangle. Prove that:
\[ \frac{\cos{A}}{\cos{(B-C)}} + \frac{\cos{B}}{\cos{(C-A)}} + \frac{\cos{C}}{\cos{(A-B)}} \geq \frac{3}{2}.  \]

$\frac{\cos A}{\cos (B-C)}=-\frac{\cos (B+C)}{\cos (B-C)}=-\frac{1-\tan B\tan C}{1+\tan B\tan C}=1-\frac{2}{1+\tan B\tan C}$
So we just need to prove $\sum 1-\frac{2}{1+\tan B\tan C}\geq\frac{3}{2}$
$\sum\frac{1}{\tan B\tan C+1}\leq\frac{3}{4}$
Suppose $\frac{1}{x}=\tan A\tan B,\frac{1}{y}=\tan B\tan C,\frac{1}{z}=\tan C\tan A$
We know $x,y,z>0,x+y+z=1$
We want to prove $\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\leq\frac{3}{4}$
$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq\frac{9}{4}$
By Cauchy,easy.


Let $D$ be the midpoint of the base $AB$ of the isosceles acute triangle $ABC$. Choose point $E$ on segment $AB$, and let $O$ be the circumcenter of triangle $ACE$. Prove that the line through $D$ perpendicular to $DO$, the line through $E$ perpendicular to $BC$, and the line through $B$ parallel to $AC$ are concurrent.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=32810 .

  Darij

Let the line through $E$ perpendicular to $BC$ intersect $CD$ at $P$ and $BC$ at $Q.$ it is trivial to verify $P$ lies on $(ACE).$ $P$ is also the orthocenter of $BCE$ so $(ECQD)$ and $(BDPQ)$ are cyclic.

Let $(EQB)$ intersect the line through $B$ parallel to $AC$ at $R.$ Let $(ECQD)$ intersect the line through $D$ perpendicular to $DO$ at $J.$ By radical axis it suffices to show $J$ lies on $(DBR).$

Notice that $\angle EQB = \angle ERB = 90^\circ,$ and $\angle EBR = \angle BAC = \angle EBQ,$ thus $R$ is the reflection of $Q$ across $EB$ and $(DBR)$ is the reflection of $(BDPQ)$ across $AB,$ thus $\angle DRB = \angle BEC$ and it suffices to show $\angle CJD = \angle BJD.$

Let the circle with diameter $AB$ intersect $(ACE)$ at $K.$ Since $KA$ and $JD$ are both perpendicular to $DO,$ they are parallel. $\angle KCE = \angle KAE = \angle JDB = \angle JDE$ so $K$ lies on $JC.$ Notice that $\angle AKB$ is right and it easily follows that $DJ$ is the perpendicular bisector of $BK$ implying the result. $\square$


Below are the results for the Pythagorus Contest. I'll also compile the solutions soon and post them below:

Group A:

Shambles-5
Xantos C. Guin-5
dangvh-4

Group B:

dangvh-25
Shambles-25
Xantos C. Guin- 25
Zyloch -25

Group C:

dangvh-20
Xantos C. Guin- 25

Also, could you please comment on what you think I could do to improve the contest by either responding to the poll or posting below? Thanks.

Okay, I was going to post solutions, but I'm kinda busy at school right now and don't really have time to, not only post my solutions, but a compiled list of correct solutions that the contestants sent in.
So instead, the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33889]questions[/url], are up for grabs. Basically, if you know a way to solve a problem thats not already posted up here, post it.

I'll start off.

A1: Take a 6 by 6 by 6 cube and place it around the 4 by 4 by 4 cube thing so that there is always a distance of 1 between the faces of the two cubes. It is evident that any winning line will pass through exactly 2 of the unit cubes inside the 6 by 6 by 6 cube, but outside the 4 by 4 by 4 cube. Thus, since there are 6^3-4^3= 152 unit cubes inside the 6 cube but outside the 4 cube, making for 152/2=76 winning lines.

Blech... horrible explanation. At least I hope some of you understand it.


abstract algebra
topology
complex numbers
group theory

1. Compute the fundamental group of the space obtained from 2 two-dimensional tori S^1 x S^1 by identifying a circle S^1 x {0} in one torus with the corresponding circle S^1 x {0} in the other torus.

2. Let p and q be integers and let f: S^1 x S^1 ----> S^1 x S^1 be given by f(x,y)=(x^p,x^q) (for x and y complex numbers of magnitude 1).
Let X = D^2 x S^1 and Y = S^1 x D^2. Then both spaces have S^1 x S^1 as a boundary. Form the adjunction space by identifying the boundaries of X and Y using f as the attaching map.
Compute the homology of the resulting space.

Please help!  I have an exam tomorrow...

For (1), Van Kampen's theorem would suggest that the answer is

$ \langle a,b \rangle * \langle b,c \rangle$

where $ \langle a,b \rangle$ is the free abelian group on generators $ a,b$, and
$ G * H$ is the free product of $ G$ and $ H$.

Still looking at (2).


Let $ (O)$ is a circle, let $ B, C$ is two point lie on $ (O)$. Let $ (O')$ is a circle such that $ (O')$ contact with $ (O)$ at point $ A$ and $ (O')$ contact with $ BC$ at point $ D$. Prove that $ AD$ is bisector of angle $ \angle BAC$.

Hi every One. This problem is easy but I must solve it by many way. So please help me. Thanks you so much.

$ AD \cap are BC \equal{} D' \Rightarrow \Delta AO'D \sim \Delta AOD' \Rightarrow O'D//OD \bot BC \Leftrightarrow AD$ is bisector of$ BAC$

it's easy to show many proof :P

Lemma: let $ BC$ be a chord of circle $ (O)$.A circle $ (I)$ such that $ (I)$ contact $ (O)$ at point A and contact with $ BC$ at point $ D$ then $ \dfrac{AB}{AC}\equal{}\dfrac{BD}{CD}$
 Dear thanhnam2902
 Seem Your figure is only considered in a case of that problem. :)  :)  See this figure

[quote="livetolovemath030894"]Lemma: let $ BC$ be a chord of circle $ (O)$.A circle $ (I)$ such that $ (I)$ contact $ (O)$ at point A and contact with $ BC$ at point $ D$ then $ \dfrac{AB}{AC} \equal{} \dfrac{BD}{CD}$
[/quote]

Please solve your lema? Dear livetolovemath030894!

Dear  inversionA007. Lemma which I posted above is the theorem generalized Ptolemaic
 Theorem generalized Ptolemaic: Let $ \triangle ABC$ and it's $ (O)$ circumcircle. $ (I)$ be a circle such that $ (I)$ contact with $ (O).AA',BB',CC'$ be tangent from $ A,B,C$ to $ (I)$ respectively.Then $ BC.AA' \equal{} AB.CC' \plus{} AC.BB'$
 You can prove it yourself because it's familiar to us :).In this case $ {C}\equiv{D}$ ($ D$ be tangency $ (I)$ with $ (O)$)

Dear Mathlinkers,
1. Let A' be the midpoint of the arch BC of the circle (O) which doesn't contain A and Ta' the tangent to (O) at A'.
2. Ta' is parallel to BC.
3. According to a special case of Reim's theorem, A, D and A' are collinear and we are done.
Sincerely
Jean-Louis


In my AP Chemistry class, we did a lab to determine the reactivity series for a few metals. We found (and our book confirmed) that iron was more reactive than copper. Yet our teacher took a solution of $FeCl_3$, added a square of Cu metal to it, left it alone for a long time (a little over an hour) and there was a reaction. Can someone explain to me, in some detail, why this reaction occurs? I'm not particularly spectacular at chemistry, so I would request that you not speak at an extremely high level.

Thanks.

Are you sure Copper is less reactive? Copper is located a few places to the right of Fe, which are both in the same period. That little chunk of chemicals (Fe, Mn, Tc, etc) comprises of some of the least reactive elements.  Elements to the right and left of them are more reactive because they either have few valence electrons in their outer shell or have a lot.  And also, as you go down a group, the elements become more reactive.

That is what our textbook said.

By the way, we might have a weird description of reactive. Our 'definition' of a metal being more reactive than another is that it is more prone to bond to other substances than the other metal.

Did that make any sense? I'm going on an English spree and feel tired.

A metal is [i]more reactive[/i] if it will replace another when both are present at $1M$ concentrations (this is related to the quantity $E^0$).  When they are at non-standard concentrations, we must consider $E$ (nonstandard), which is derived from the Nerst (Nernst?) equation.  Since the concentration of $\text{Cu}^{2+}$ ions was $0M$, the reaction will proceed to an equillibrium point (since iron is more reactive, the equillibrium will favor copper metal and iron ions).

When talking about a metal's reactivity, we are talking its tendency to form bondings with another substance. When you place copper in a Ferric Chloride solution, a Redox reaction takes place where the electrons in copper are transffered to the ferric ions in Ferric Chloride. Even though it is true that Iron is more reactive than copper, remember when that an ionic compound is dissolved it dissociates and form a positive and a negative ion in the solvent which in this case is water. Therefore, even though the reaction is slow and time consuming, it will take place eventually in tiny amounts.

The reduction potential for Fe3+ is significantly higher than that of that of Fe2+.  It's even higher than that of Cu2+.  So, the Fe3+ should oxidize the copper pretty readily.


linear algebra
matrix
trigonometry
induction
linear algebra unsolved

For arbitrary square matrix $ A$ define $ \sin A$ by means of series
\[ \sin A\equal{}\sum_{n\equal{}0}^{\plus{}\infty}{\frac{(\minus{}1)^{n}}{(2n\plus{}1)!}A^{2n\plus{}1}}.\]
Does there exist a $ 2\times 2$ matrix $ A$ such that $ \sin A \equal{}\begin{pmatrix}0.999 & 2007\\ 0 & 0.999\end{pmatrix}$ ?

Over $ \mathbb{C},$ one can develop the power series (centered at zero, standard branch) for $ \arcsin z.$ This power series has radius of convergence $ 1$ and the property that $ \sin(\arcsin z) \equal{} z.$

Now apply this power series to a $ 2\times 2$ matrix $ B$ and call the result $ \arcsin B.$ This makes sense whenever the power series converges. Since the radius of convergence in $ \mathbb{C}$ is $ 1,$ the matrix power series converges whenever the spectral radius of $ B$ is less than $ 1.$ The identity $ \sin(\arcsin B) \equal{} B$ is true for scalar multiples of the identity within the radius of convergence; by analytic continuation it should be true for all $ B$ with spectral radius less than $ 1.$

The spectral radius of the matrix given in the problem is $ 0.999.$ Therefore it should have an arcsine, and the answer to the question in the problem is "yes."

Spectral radius less than $ 1$ is a sufficient condition for having an arcsine. It's not a necessary condition.

Here's a cheap example:

$ A\equal{}\begin{bmatrix}0&\minus{}a\\a&0\end{bmatrix}$ for arbitrary real $ a.$

Then $ \sin A\equal{}\begin{bmatrix}0&\minus{}\sinh a\\ \sinh a&0\end{bmatrix}.$

This has spectral radius $ |\sinh a|,$ which can be made arbitrarily large.

Let $ B\equal{}\sin A\equal{}\left(\begin{array}{cc}0.999 & 2007\\0 & 0.999\end{array}\right)$, then for all $ n\geq 2$ we have the relationship $ B^{n}\equal{}n\beta^{n\minus{}1}B\minus{}(n\minus{}1)\beta^{n}1_{2}$ with $ \beta \equal{}0.999$ (this is easily shown by induction noticing $ B^{2}\equal{}trB\,B\minus{}detB\,1_{2}\equal{}2\beta B\minus{}\beta^{2}1_{2}$).

Now we consider the arcsin Taylor series expansion of $ B$:

$ A\equal{}\arcsin B\equal{}\sum_{n\equal{}0}^{\infty}\frac{B^{2n\plus{}1}}{(2n\plus{}1)2^{2n}}\,\left(2n\atop n\right)\equal{}B\plus{}\sum_{n\equal{}1}^{\infty}\frac{(2n\plus{}1)\beta^{2n}B\minus{}2n\beta^{2n\plus{}1}1_{2}}{(2n\plus{}1)2^{2n}}\,\left(2n\atop n\right)\equal{}\\ (B\minus{}\beta 1_{2})\sum_{n\equal{}0}^{\infty}\frac{\beta^{2n}}{2^{2n}}\,\left(2n\atop n\right)\plus{}1_{2}\sum_{n\equal{}1}^{\infty}\frac{\beta^{2n\plus{}1}}{(2n\plus{}1)2^{2n}}\,\left(2n\atop n\right)\equal{}\\ (B\minus{}\beta 1_{2})\,\frac{1}{\sqrt{1\minus{}\beta^{2}}}\plus{}1_{2}\arcsin\beta$

which determines the solution and from which we see that $ |\beta |<1$ was crucial.

just spotted a minor typo, which does not change the result: the second sum in the second row of equations starts at $ n\equal{}0$ instead of $ n\equal{}1$.


this an easy problem:
a person is standing at $A$, he wants to go to $B$,  he can move on the lines of the image below, HE CAN NOT MOVE ON A POINT TWICE. in how many ways can he go to $B$?

well....i'm assuming that you can only move up and to the right? if so, fill it in using pascal's triangle.

[hide]this is roughly what it will look like.....i think.

...............3......B (10)

...............3......7

1......2......3......4

A......1......1......1

thus, there are 10 ways to get from point A to point B
*the "......"s are there for spacing purpose only[/hide]


What is the last date to get their email?

I'm not sure, but when I contacted Perimeter last week, they said that "all decisions will be made by mid-June."

The email also looked mass-emailed...but I really don't know.

EDIT:  I believe John Schulmann (United States) also did this program.  I think his AoPs sn is schulmanerism or something like that.  Yeah...he probably knows more...

I just got my email this morning! Woot! I almost deleted it as spam since the title was all capitalized...  :oops:

So did I! I still can't believe it.
I'm one of the international students. I'll see you there, I suppose!  :)

Great stuff!

I'm an International as well, Jeremy Pham from the United States.  August session.  I've got a camera...hopefully ISSYP will be intellectually stimulating :).

The list of Canadian acceptances is up, for those of you interested...

Who are ya'll btw?

Mireia Crispin from Spain. July session.  :)

Oh shoot...I guess we won't see each other haha.  I've got something in July...

Apparently on the International List, I'm Jemery Pham :blush: .

I'm Tony Zhu from Vancouver, going in August... See you there!

Hmm...Yifan, you're on the list for the current ISSYP.  Might I ask, how's it going so far?  What's going on?

Did everyone else from the August session get their:

a) Follow-up email with all those trip forms and introductory letters?

b) Book?  (I haven't received mine yet :|).

Wonder if there's anything else I'm missing...  :?:

Hi everyone! I'm now writing from Waterloo, Ontario.. I'm one of the partecipants of ISSYP 1... It is really great! :D I'm Denis from Italy

It seems that the books are not going to be sent, since we didn't receive either. Nafessa said that is because of the large amount cost to ship.

I'm also going for the August session.  I'm from Ontario

Yes, Nafeesa also informed me that we are not in fact receiving books (leaving me quite disappointed).  Apparently that was last year's info that was forgotten to be removed from the e-mail.

 :( Arrgh, and now i have to write another essay for them, in a week, in the summer, when I avoid creative writing like the plague!!!

Well...I just got to Ontario.

Ah...I love creative writing opportunities like this.  One time, I wrote an essay proposing a fascist student government...it's amazing to see what types of responses freestyle topics (those "Write about yourself" prompt) can elicit!

Already sent mine  :rotfl: .

Can't wait until next week!  I wonder what the current people are up to...they're being mysteriously quiet about the program :roll: .

Yeah, it's fun though. We are just back from Toronto, spent a day there.
PS: I'm not sure that essay is necessary because I didn't write it.

Probably for them to match personalities and whatnot when pairing up roommates.

When are you guys arriving at Pearson International Airport? I need a ride and don't know when to show up.

My plane from Vancouver will be arriving at 2:30, so I guess you could arrive sometime between two and three.


Κατ' αρχάς να ευχαριστήσω όλα τα μέλη του φόρουμ για τις ευχές τους. Αφιερώνω το ολυμπιακό αυτό μετάλλιο στην κοινότητα αυτή που μου έμαθε τόσα και χωρίς τη βοήθεια της οποίας δε θα κατάφερνα ποτέ κάτι τέτοιο. Πριν χαιρετήσω για καλοκαίρι (γιατί λίγες διακοπές χρειάζονται) βάζω μια ανισότητα αφού δεν έπεσε στην IMO :)

$ \sum\frac{a\minus{}1}{a^2(a^4\plus{}1)}\leq0$

με $ abc\equal{}1$ και $ a,b,c$ θετικοί πραγματικοί.

Καλό καλοκαίρι!

[quote="Protonios"]Κατ' αρχάς να ευχαριστήσω όλα τα μέλη του φόρουμ για τις ευχές τους. Αφιερώνω το ολυμπιακό αυτό μετάλλιο στην κοινότητα αυτή που μου έμαθε τόσα και χωρίς τη βοήθεια της οποίας δε θα κατάφερνα ποτέ κάτι τέτοιο. Πριν χαιρετήσω για καλοκαίρι (γιατί λίγες διακοπές χρειάζονται) βάζω μια ανισότητα αφού δεν έπεσε στην IMO :)

$ \sum\frac {a \minus{} 1}{a^2(a^4 \plus{} 1)}\leq0$

με $ abc \equal{} 1$ και $ a,b,c$ θετικοί πραγματικοί.

Καλό καλοκαίρι![/quote]

Sigxaritiria Kosta gia to xalkino metalio

I Anisotita pou dineis isodinama grafetai:

$ LHS \equal{} \sum\frac {1}{a(a^4 \plus{} 1)}\leq \sum\frac{1}{a^2(a^4 \plus{} 1)} \equal{} RHS$

Apo tin Chebyshev exeis:

$ RHS \geq \frac{1}{3}(\sum{\frac{1}{a}})LHS \geq LHS$ apo AM-GM...

Mipos xanw kapou?

εστω $ f(x) = \frac {x - 1}{x^{2}(x^{4} + 1)} - \frac {1}{2}\ln x$, $ x$ στο $ (0, + \infty)$
Η $ f$ παρουσιαζει ολικο μεγιστο στο 1 το 0,
αρα $ f(a) + f(b) + f(c) \leq 0 \leftrightarrow \sum\frac {a - 1}{a^{2}(a^{4} + 1)} \leq \frac{1}{2} \ln (abc) = 0$. QED  :)
ΚΑΛΕΣ ΔΙΑΚΟΠΕΣ!!

[color=red]moderator edited [/color]


Consider the following two equilibria:

$N_{2}(g)+3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$

$H_{2}(g)+Br_{2}(g) \rightleftharpoons 2HBr(g)$.

For each one, discuss the effect of the addition of an inert gas at constant temperature and pressure.

Addition of an inert gas tells us a few things.

1. It would not react with the species in the reaction

2. It may alter the partial pressure of each gaseous species

I think we should note that if adding an extra gas does not change the overall pressure then the overall volume must have increased.

Yes, that's correct. So what is the effect on the above equilibria of adding an inert gas, i. e., will each reaction proceed to the right, to the left, or remain unchanged?

Firstly, increasing the volume without altering the overall pressure by addition of an inert gas would mean that the partial pressure of its gaseous species is decreased. So a decrease in partial pressure would cause the equilibrium to shift to the side with more moles of gaseous species.

So I think:

[hide="Answer"] 1. Equilibrium shifts to the left

2. Equilibrium does not shift.[/hide]


geometry
parallelogram
trigonometry
geometry proposed

In a parallelogram $ ABCD$ with $ \angle A <90^o$,the circle with diameter $ AC$ meets the lines $ CB$ and $ CD$ again at $ E$ and $ F$,respectively,and the tangent to this circle at $ A$ meets $ BD$ at $ P$.Show that $ P,F,E$ are collinear

[hide="Hint"] Let M,N be intersection points of tangent in A with CD and BC. Try to prove that MNEF is cyclic. Then let S,R be intersection points of circle with diameter AC with BD. You can prove, that MNSR is cyclic.[/hide]

Let $ m(ACD)\equal{}\alpha$  ,  $ m(ACB)\equal{}\beta$ and $ m(BDC)\equal{}\gamma$.

First we will prove that      $ sin(\beta\minus{}\alpha).cos(\alpha\plus{}\beta).sin\gamma\equal{}sin\alpha.sin(\alpha\plus{}\beta).cos(\beta\plus{}\gamma)$ $ \Longleftrightarrow  cot\gamma\equal{}(\frac{sin(\beta\minus{}\alpha).cos(\alpha\plus{}\beta)}{sin(\alpha\plus{}\beta).sin\alpha}\plus{}sin\beta).\frac{1}{cos\beta}$ . Let $ K$ be the intersection of $ AC$ and $ BD$. $ |DK|\equal{}|BK|$

 $ \Longrightarrow \frac{sin\alpha}{sin\beta}\equal{}
{sin\gamma}.{sin(\alpha\plus{}\beta\plus{}\gamma)} \Longleftrightarrow cot\gamma\equal{}(\frac{sin\beta}{sin\alpha}\minus{}cos(\alpha\plus{}\beta)).\frac{1}{sin(\alpha\plus{}\beta)}\equal{}\frac{1}{cos\beta}.(\frac{sin\beta.cos\beta}{sin\alpha}\minus{}cos(\alpha\plus{}\beta).cos\beta)).\frac{1}{sin(\alpha\plus{}\beta)}\equal{}\frac{1}{cos\beta}.(\frac{sin\beta.cos\beta}{sin\alpha}\plus{}sin(\alpha\plus{}\beta).sin\beta\minus{}cos\alpha).\frac{1}{sin(\alpha\plus{}\beta)}\equal{}\frac{1}{cos\beta}.(\frac{sin(\beta\minus{}\alpha).cos(\alpha\plus{}\beta)}{sin\alpha}\plus{}sin(\alpha\plus{}\beta).sin\beta).\frac{1}{sin(\alpha\plus{}\beta)}\equal{}(\frac{sin(\beta\minus{}\alpha).cos(\alpha\plus{}\beta)}{sin(\alpha\plus{}\beta).sin\alpha}\plus{}sin\beta).\frac{1}{cos\beta}$

Now by Trigonometric Ceva in triangle $ APD$ for $ F$ we get easily result.

There is another solution without trigonometry
Let $ A'$ lie on circle such that $ PA'$ is tangent
Easy to see $ AEFA'$ ís conjugate quadrilateral as desire !


Passion Pink paint is produced by mixing 13 parts of red pigment per
quart of base paint. How many parts of red pigment are required to
mix $ 3\frac{1}{4}$ gallons of Passion Pink paint? Assume the pigment does not add to the volume of the mixture.

Drawing G-Man, we see that there are $ 13$ quarts of paint, so our answer is $ 13^2\equal{}\boxed{169}$.

Haha...I was just kidding, but here's a picture:
[img]http://illuminations.nctm.org/lessons/measureup/GallonMan.png[/img]

G stands for gallon, Q stands for quart, P stands for pint, and C stands for cup.


Singapore Mathematical Olympiads 2007

Forty points are placed equidistantly around a circle. How many triangles can be formed having their vertices as one of these points such that every vertice is separated from another vertice by at least 2 points?

[hide]This would be like having 40 O's around a circle and changing 3 of them to X's, with at least 2 points in between.

However, this is the same as having 34 O's around a circle and changing 3 of them to XOO's.

Hence, we want the number of distinct triangles we can make with 34 equally spaced points around a circle. Any three points will define a unique triangle, so the result is $ \binom{34}{3}= 5984$[/hide]

However, the official answer is $ 7040$?


geometry
calculus
integration
Putnam
function
calculus computations

Here's an interesting problem I saw on another site.

In the first quadrant, a horizontal line, y = c, cuts the curve $y = 8x - 27x^{3}$ into two regions under the

curve. Find the constant c such that the areas of the two regions are equal.

Does anyone have any ideas?.

There will be two places where the line intersects the curve. Let this point be, say, (a,c) and (b,c)

$8x-27x^{3}=0$

$x=\pm\frac{2\sqrt{6}}{9}$

The area of the entire region from 0 to $\frac{2\sqrt{6}}{9}$ is $\frac{16}{27}$.

Therefore, the area of one halved region would be 8/27.

I thought maybe the mean value theorem for integrals may come in handy, but to no avail.

For some reason, I'm stymied.

We must solve the equation:

$\displaystyle\int^{a}_{0}8x-27x^3dx = \frac{8}{27}$.

This gives us the equation:
$4x^2-\frac{27}{4}x^4 = \frac{8}{27}$

The only real positive solution to this equation is
$\displaystyle\frac{2\sqrt{-3(\sqrt{2}-2)}}{9}$

This is the x-value for the point at which the line y=c intercepts the function.

$f(\displaystyle\frac{2\sqrt{-3(\sqrt{2}-2)}}{9}) = \displaystyle\frac{16\sqrt{-3(\sqrt{2}-2)}}{9} - \displaystyle\frac{8(2-\sqrt{2})^{3/2}\sqrt{3}}{9}$

That last number is the value for c. (I think)

did this problem appear on an old putnam exam...because if so your wording is a litte off.

Thanks for the responses. This problem is a twist on a Putnam-esque problem.

Instead of finding a line which divides the region between the curve and the y-axis and the top of the curve and the

line into two equal regions, this one divides the area under the curve into two equal regions.

The answer I was given is c=.6107838722.

seesaw, you used a vertical line, not a horizontal one.

I do not think this question is suceptible to a nice, non-numerical attack.

[quote="JBL"]seesaw, you used a vertical line, not a horizontal one.
[/quote]

You're right.
 :blush:

I thought this was a fun problem. Here's my approach if anyone is interested. The approach I got from the person who showed me this problem is a little different and a little less messy. They used the top part of the graph above the line, but the answer is the same.

It is best to use some sort of technology for this problem, lest you have some mighty ornery algebra to contend with.

If we draw a horizontal line, y=c, through the graph of $8x-27x^{3}$, we see it intersects in 2 places. Call these (a,c) and (b,c)

[img]http://img118.imageshack.us/img118/3563/dividedregion1be.jpg[/img]

It is easily seen that the entire region is area 16/27. Therefore, we are looking for a line which divides the region into two subregions of area 8/27.

So we have:

$c=8a-27a^{3}$

$c=8b-27b^{3}$

Setting equal and solving for b(using technology, of course), we get:

$b=\pm\frac{(\sqrt{-3(81a^{2}-32)}-9a)}{18}$

Now, we can define our humongous functions with respect to 'a' only.

Using the bottom part of the graph below the line:

$\int_{0}^{a}(8x-27x^{3})dx+\int_{b}^{\frac{2\sqrt{6}}{9}}(8x-27x^{3})dx+c(b-a)=\frac{8}{27}$

Solving this monstrosity(with Maple's help) leads to a solution of [b]a=.0779462908[/b]

Subbing back into $8a-27a^{3}$, we get

[b]c=.6107838722[/b]


Given: triangle WZY is congruent to triangle WXY, and triangles WZY and XZY are isoceles.
Prove: WXYZ is a rhombus

okay, draw a typical-loooking parallelogram, and the draw the diagonals. the intersection of the diagonals is P, the top left corner of the parallelogram (p.) is W, the top right of the p. is X, the bottom left of the p. is Z, and the bottom right of the p. is Y.

sorry about the bad drawing...

one more..... and p.s the first correct answer is chosen as best

given: triangle LGK is congruent to triangle MJK and GHJK is a parallelogram
prove: GHJK is a rhombus

draw another typical parallegram, and then label the top left K. top right J, bottom left G, bottom right H, and intersection of the diagonals M. then draw a triangle with KG as its base. the tip of the triangle is labeled L.

that is about right... except l and k should be connected.

okay, now i need 2 2 column proofs. And please help me....!

Well, it is not too tough, I guess. The definition of a rhombus is that it is a quadrilateral with equal, parallel sides. I'm certainly not so sure, as it is hard without LatEx.

IF you guys actually TRIED to read it, it's not bad at all.  But since you insist...

GIVEN: $ \triangle WZY \cong \triangle WXY$, $ \triangle WZY$ and $ \triangle XZY$ are isosceles

PROVE: $ WXYZ$ is a rhombus.

[asy]dot((0,0));
dot((5,0));
dot((2,3));
dot((7,3));
draw((0,0)--(5,0)--(7,3)--(2,3)--cycle);
label("W",(2,3),NW);
label("X",(7,3),NE);
label("Y",(5,0),SE);
label("Z",(0,0),SW);
draw((5,0)--(2,3));
draw((0,0)--(7,3));
label("P",(3.5,1.5),N);
dot((3.5,1.5));[/asy]

second one:

GIVEN: $ \triangle LGK \cong \triangle MJK$, $ GHJK$ is a parallelogram

PROVE: $ GHJK$ is a rhombus

[asy]dot((0,0));
dot((5,0));
dot((2,3));
dot((7,3));
draw((0,0)--(5,0)--(7,3)--(2,3)--cycle);
label("K",(2,3),NW);
label("J",(7,3),NE);
label("H",(5,0),SE);
label("G",(0,0),SW);
draw((5,0)--(2,3));
draw((0,0)--(7,3));
label("M",(3.5,1.5),N);
dot((3.5,1.5));
dot((-2,1));
draw((-2,1)--(2,3));
draw((-2,1)--(0,0));
label("L",(-2,1),SW);[/asy]

Proofs:

[hide="1"]
From the given triangle congruence and CPCTC, we have WZ=WX.  Since ZY=XY, and the triangles WZY and XZY are isosceles, WX=XY=YZ=ZW and by the definition of rhombus, WXYZ is a rhombus.
[/hide]

[hide="2"]
From the triangle congruence we know that GK=JK, and because GHJK is a parallelogram it must then be a rhombus.
[/hide]

I won't put it into two-column format though, you should be able to do that yourself.


Newton's Second Law:

1. $F=ma$

2. $F= \frac{dp}{dt}$

How can you get from 1 to 2, vice versa?

Plus: how can I design an experiment to prove the Law of Conservation of Momentum?

${F=\frac{dp}{dt}=\frac{d(m v)}{dt}=v \frac{dm}{dt}+m \frac{dv}{dt}}$

${\frac{dm}{dt}=0\text{ },\text{ }\frac{dv}{dt}=a}$

${F=m \frac{dv}{dt}=m a}$

do you know derivatives?

How "scientific" does your experiment need to be? If you just need to show that momentum is passed from one object to another, you can just use one of those office toys with the steel balls attached to a rod--you know, where you lift a ball on one end up, let it fall, and it hits all the other ones and makes the ball on the other end go up.

However, if you need to prove definitively that no momentum is lost, you would need to move something with a known mass at a certain speed and make it collide with a second object of known mass. You would need a device to measure the speed of the second object after collision, and then you would need to plug all your data into the equations and show that the momenta of the two objects are equal (not taking friction into account).

Exactly, that's the hard part. How to measure the speed of the objects. I was hoping maybe there can be a way such that we measure something that is easily measured and deduce from the data, the momentum.

I'm not sure what you mean with the experiment. Theories (despite being labeled as a "law") are devised to conform and explain the observations obtained from experiments. It doesn't make much sense to "prove" a theory with an experiment.

if you want to verificate it you can make this experience

make a Brachistochrone curve http://en.wikipedia.org/wiki/Brachistochrone_curve

you can make this on wood. just google it. have some homemade ramps like that

put a ball of mass $m_{1}$ in one side and a ball with mass $m_{2}$ in the another side.

the balls will reach the low level of the curve in the same time.
Supposing a completelly elastic collision you can measure the velocity of each ball after and before they hints with the Height relative to the low level of the curve

Using the the Law of conservation of the energy you can deduce the law of conservation of the momentum.

Can anyone tell me mathematically or otherwise how Newton's Cradle demonstrates conservation of momentum?

http://www.iop.org/activity/education/Teaching_Resources/Teaching%20Advanced%20Physics/Mechanics/Momentum/page_3845.html

http://www.wfu.edu/physics/demolabs/demos/avimov/bychptr/chptr3_energy.htm

you can see the experiences on google


function
limit
real analysis
real analysis solved

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21463&postorder=asc
-------------------------------------------------------------------

$f:R_{+}\rightarrow R_{+}$ wich admit a primitive such that
$F(0)=0$
$q\geq 1$
$\lim_{x \rightarrow +\infty}f(x)F^q(x)=1$

Find an equivalent of $f(x)$ when $x$ tends to +oo

By l'Hospital $\lim_{x\to\infty}\frac{F^{q+1}(x)}x=q+1$.
Now $\lim_{x\to\infty}\frac{F^q(x)}{(x(q+1))^{\frac q{q+1}}}=1$ and
$\lim_{x\to\infty}f(x)F^q(x)\cdot\frac{(x(q+1))^{\frac q{q+1}}}{F^q(x)}=1$.
So $f(x)\sim \frac 1{(x(p+1))^{\frac q{q+1}}}$

There  is smth I don't get... what's the purpose of $F(0)=0$?


Solve in $ \mathbb{R}$:
$ e^{\sin(x\minus{}\frac{\pi}{4})}\equal{}\tan x$

[quote="joshisunny16"]x=pi/4 +arc(sin(ln(tan(x)))
  now value of tan can be frm -infinity to + infinity
   value of ln = 0 to infinity
    but arcsin will lie in 0 to pi/2
  so
x=pi/4+0   to pi/4+pi/2[/quote]
can u explain more :lol:
why don't you use latex?

srry but i dont know how to use it
but let me tell u arc means inverse
the first step I did was to take log(base e) on both the sides,
then (x-pi/4)=arcsin(ln x) where ln is log base e

hello, this equation has infinitey many solutions, i think you can solve this only by a numerical way.
Sonnhard.


This problem is too difficult for the Middle School Classroom Math forum.  I'm going to bump it up to a forum where others are likely to give you a good explanation.

[quote="lightfire22000"]I must have not paid attention in middle school because I'm a high school junior.  I've got a rules of divisibility question.  How come 9876543210 is divisible by 8.  210 is not divisible by 8.

There was this problem and the solution stated 9876543120 was the greatest integer multiple of 8 in which no digits repeated you see.

Thank you very much.[/quote]

Umm Are you talking about 9876543120 or 9876543210? You include both in your post.

Just fyi, the greatest multiple of 8 in which no digits are repeated is 9876543120, as 120 is a multiple of 8.

When you determine whether or not a number is dividisble by 8, you only need to look at the last 3 digits. For instance, 9876543120 is divisible by 8 because 120 is divisible by 8. However, 9876543210 is not divisible by 8 because 210 is not divisible by 8.

The reason why this rule works is because 1000 is divisible by 8.
In this case, $9876543120=9876543000+120=9876543\times 1000+120$, so we don't need to worry about $9876543000$.

In general, a number is divisible by $2^n$ if the last $n$ digits are divisible by $2^n$.

[quote="GoBraves"][quote="lightfire22000"]I must have not paid attention in middle school because I'm a high school junior.  I've got a rules of divisibility question.  How come 9876543210 is divisible by 8.  210 is not divisible by 8.

There was this problem and the solution stated 9876543120 was the greatest integer multiple of 8 in which no digits repeated you see.

Thank you very much.[/quote]

Something interesting happened while i was looking at this. I put 9876543210/8 in my TI-83 plus, and the result shown was 1234567901, which is incorrect. The actual answer is 1234567901.25. I think it showed this because it only shows the first 10 digits. So maybe you were using a TI-83 plus as well, and was misled like i was originally, but 9876543210 is not divisible by 8. You can check that by putting it in your computers calculator, which shows more digits.[/quote]

Or use exact mode :p

Now it makes sense.  I went to absolute mode and it helped me see.


There are $ 5$ pieces of cards in line.
Numbers from $ 0$ to $ 9$ are written to the card face up and face down.
What are $ 5$ numbers that it showed first at time of the following conditions?

$ 1: \text{uuuuu} \to \color{red}19\color{black}$
$ 2: \text{ddduu} \to 20$
$ 3: \text{ddudd} \to 35$
$ 4: \text{uudud} \to 11$
$ 5: \text{ududu} \to 31$
(u is face up, d is face down)

[url=http://en.wikipedia.org/wiki/Kakuro]Kakuro[/url] - Wikipedia

This problem is questions from "Takeshi Kitano presents Comaneci University Mathematics".
It is a Japanese TV program.

Sorry, my English is very poor.
I write the original in the Japanese as follows.

[quote="Japanese original"]一列に並んだ5枚のカードがあります。
カードの表と裏にはそれぞれ0～9の数字が1つずつ書かれています。
以下の条件のとき最初に見えていた5つの数字は何ですか？
[/quote]

[quote="Kouichi Nakagawa"]There are $ 5$ pieces of cards in line.
Numbers from $ 0$ to $ 9$ are written to the card face up and face down.
What are $ 5$ numbers that it showed first at time of the following conditions?

$ 1: \text{uuuuu} \to \color{red}19\color{black}$
$ 2: \text{ddduu} \to 20$
$ 3: \text{ddudd} \to 35$
$ 4: \text{uudud} \to 11$
$ 5: \text{ududu} \to 31$
(u is face up, d is face down)
[/quote]

Let $ uuuuu \equal{} abcde$ and $ ddddd \equal{} fghij$ and $ \{a,b,\cdots,j\} \equal{} \{1,2,\cdots,9\}$
$ 1: a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 19$
$ 2: f \plus{} g \plus{} h \plus{} d \plus{} e \equal{} 20$
$ 3: f \plus{} g \plus{} c \plus{} i \plus{} j \equal{} 35$
$ 4: a \plus{} b \plus{} h \plus{} d \plus{} j \equal{} 11$
$ 5: a \plus{} g \plus{} c \plus{} i \plus{} e \equal{} 31$

$ a \plus{} b \plus{} \cdots \plus{} j \equal{} 45$
then $ 1 \plus{} 3: c \minus{} h \equal{} 9$, then $ c \equal{} 9$ and $ h \equal{} 0$
$ 1: a \plus{} b \plus{} d \plus{} e \equal{} 10$
$ 2: f \plus{} g \plus{} d \plus{} e \equal{} 20$
$ 3: f \plus{} g \plus{} i \plus{} j \equal{} 26$
$ 4: a \plus{} b \plus{} d \plus{} j \equal{} 11$
$ 5: a \plus{} g \plus{} i \plus{} e \equal{} 22$
with $ a \plus{} b \plus{} d \plus{} e \plus{} f \plus{} g \plus{} i \plus{} j \equal{} 36$
from $ 1$ then $ \{a,b,d,e\} \equal{} \{1,2,3,4\}$
from $ 3$ then $ \{f,g,i,j\} \equal{} \{5,6,7,8\}$
from $ 4\minus{}1$ then $ j\minus{}e\equal{}1$ then $ j\equal{}5$ and $ e\equal{}4$
$ 2: f \plus{} g \plus{} d\equal{} 16$
$ 3: f \plus{} g \plus{} i \equal{} 21$
$ 5: a \plus{} g \plus{} i \equal{} 18$
with  $ \{a,b,d\} \equal{} \{1,2,3\}$ and $ \{f,g,i\} \equal{} \{6,7,8\}$
from $ 3\minus{}5$ then $ f\minus{}a\equal{}3$ then $ f\equal{}6$ and $ a\equal{}3$
$ 2:  g \plus{} d\equal{} 10$
$ 3:  g \plus{} i \equal{} 15$
with  $ \{b,d\} \equal{} \{1,2\}$ and $ \{g,i\} \equal{} \{7,8\}$
from $ 2$ then $ g\equal{}8$ and $ d\equal{}2$
from $ 3$ then $ i\equal{}7$ and thus $ b\equal{}1$

Solution:
$ \boxed{
1: \text{31924} \to \color{red}19\color{black},
2: \text{68024} \to 20,
3: \text{68975} \to 35,
4: \text{31025} \to 11,
5: \text{38974} \to 31
}$

Hi mszew :)
Your solution is perfect and the same as a model solution of the TV :clap:


algebra
polynomial
geometry
3D geometry
factorization
sum of cubes

Hi,

Can anyone help me with these two topics?

Cubic Formula and Newton's Sums? (I don't have the Art of Problem Solving 2)

I found the site for Cubic Formula:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Except I don't quite understand.

Is it formula I can use for any kind of polynomial (at least third-degree) because that would be awesome since I usually never get those questions.

And Newton's Sums, hmm... I don't understand it neither. One of the people in here told me for this problem, I have to use Newton's Sums but I don't quite understand this.

The polynomial \[p(x) = x^3+2x^2-5x+1\] has three different roots: a,b,c. (i.e., \[p(a) = p(b) = p(c) = 0)\]. What is then \[a^3+b^3+c^3\]

Thanks. :D

I can't believe I'm still awake just to check up if Mathlinks is back, and it is!!!
Anyways, I'm very tired and in short:

If $p(a)=p(b)=p(c)=0$, by Viete's coefficient formulas (this is something you can derive yourself)
$a+b+c=-2$,$ab+bc+ca=-5$,$abc=-1$.
What is $a^2+b^2+c^2$? Well, since $a+b+c=-2$, square both sides and get
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ and so we have
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(-2)^2-2(-5)=14$
Since $a,b,c$ are roots of $p(x)$
$a^3+2a^2-5a+1=0$
$b^3+2b^2-5b+1=0$
$c^3+2c^2-5c+1=0$
Add these three together and we have
$a^3+b^3+c^3+2(a^2+b^2+c^2)-5(a+b+c)+3=0$. Finally,
$a^3+b^3+c^3=-2(a^2+b^2+c^2)+5(a+b+c)-3=-2(14)+5(-2)-3=-41$

Now that you got the sum of cubes of roots, what is $a^4+b^4+c^4$? It's very easy and it should not grow mold in your head  :D


percent
geometry
geometric transformation
reflection
3D geometry
tetrahedron
conics

1. [4] Every day, the value of shares of Soap Bubble, Ltd. on the stock exchange either increases or decreases by n percent, where n is an integer and 0 < n < 100. Does there exist an n for which the price can take the same value twice?

(the price each day is calculated with infinite precision.)

2. [6] A billiard table has a polygonal shape, not necessarily convex, such that any two adjacent sides are perpendicular. From vertex A with an internal angle of 90o, a point-ball is launched. It moves along straight lines bouncing from the border according to the law "incidence angle is equal to reflection angle." If the ball hits any vertex it is absorbed. Prove that the ball cannot return to A.

3. [6] An orthogonal projection of a tetrahedron onto some plane has the maximal possible area. Prove that this plane is parallel to either one of the faces or two skew edges of the tetrahedron.

4. [7] Initially the number 2004! is written on the blackboard. Two players play a game according to the following rules:

- Let the number on the blackboard be B.
- The player, on his/her turn, may choose a number A that has no more than 20 distinct prime divisors, and A  B.
- B is replaced by B - A.

The player to write 0 on the board wins. Find the person who has the winning strategy and describe the strategy.

5. [7] A parabola y = x2 and a circle with exactly two common points A and B are plotted on the plane. At point A, both curves have the same tangent. Is it necessarily true that at poin tB both curves at the same tangent?

6. A magician and his assistant show the following trick.

A deck of 36 cards faced down, 9 of each suit, is placed in front of the magician. The magician calls the suit of the top card, the top card is revealed to everybody including the magician, and discarded. This procedure is repeated until the deck runs out. The magician attempts to guess the suits of as many of the cards possible. The card is rectangular, not square, in shape. The back of each card is identical but not symmetric, thus allowing the magician to distinguish the "orientation" of the top card each time. Before the show, the assistant knows the order of the cards, but the magician does not; the assistant, may switch the orientation of each card, but not the sequence. Is there an agreement regarding the orientation of the cards between the magician and the assistant that allows the magician to correctly guess

(a) [3] at least 19 cards?
(b) [5] at least 20 cards?

Ooh, number 3 I like.

In number 4, should it be A :le: B?  Because if A < B, I don't see how you could ever get to 0 ...

I'm not sure I understand number 6.

5:

[hide]Yes.  If the curves have the same tangent, the circle and parabola must both lie on the same side of that tangent line.  On a local scale (meaning near the point of intersection), the circle must be entirely on one side or the other of the parabola.  I'm trying to think of a non-calculus way to prove that statement, but I'm not sure how.  Maybe it has something to do with curvature, or because circles and parabolas don't have inflection points?  Maybe it's obvious and I'm just not thinking of it.  Either way, it's true.  In that case, if there were a non-tangential other point of intersection, the parabola would cross at that point from the exterior of the circle into the interior, which would necessetate (sp?) a third intersection point (since the parabola is unbounded, it would have to go off to infinity and thus cross the circle somewhere else).  So the second intersection point must also be a point of tangency.[/hide]

Necessitate.

Changed #4.

Let me explain #6 (most contestants were confused about this question as well):

- Magician does not know the sequence of the cards. He knows that there are 9 cards of each suit.

- The cards are rectangular in shape (think regular playing cards). The pattern printed on the back is not symmetric. If it helps, think of a large arrow printed on the card; you can rotate the card such that the arrow points in the opposite direction. (In other words, [hint] [hide]you can encode information using the arrows.[/hide])

- The assistant knows the sequence of the cards. He is not allowed to change the sequence, but he is allowed to rotate any number of cards.

- The magician is to guess the suit of each card. He can see the back of the card he is trying to guess. (Thus he is able to distinguish the two orientations of the card.)

- The magician and the assistant is to work out a system regarding the orientations of the cards such that the magician is guaranteed to guess correctly. Does there exist such a system with which the magician can guess

(a) [3] 19

(b) [5] 20

cards correctly?

This thread has now been reopened, as Osiris has informed me that the contest is now over.

5.

[hide]No.

I have found a counter example some time ago.

When we plug y=x^2 into the general equation of a circle, we get a quartic. The idea is that if the intersection of the two curves is a triple root of the quartic (normally it would just be a double root since the two curves tangent to each other), then the curves are not tangent at the other intersection. Using this idea, we can easily find a set of counter-examples. Though, I don't have the time to reconstruct my counter-example right now.[/hide]

Number 4 seems interesting.  Clearly, whoever loses is the person who MUST pick a number with 20 or fewer factors.  So the first losing position is the smallest number with that property.  Hmm ... except that they're spread all out.  I'll keep thinking about this (going to mathjam now).

Main idea for number 4:

[hide]Let P be the product of the first 21 prime numbers. Then all losing positions (including the terminal position, 0) are multiples of P, and all the other numbers are winning positions.

(A winning position is a position such that if you have that number before your turn, then you can secure a win).[/hide]

One of the cases for 3:

[hide]There are 2 cases: the max area is achieved when we see the projection of one of the tetrahedron's faces or 2. The first case is easy to solve:

Let ABC be the face whose projection we're looking at. If the angle of the plane we're projecting the tetrahedron on and (ABC) is t, then the area of the projection is S(ABC)*cos(t), so it reaches it maximum when (ABC) is parallel to the plane.[/hide]


Find the positive difference between the solutions to $ \left|\frac{x}{2} \minus{} 6\right| \equal{} 20$.

We either have $ \frac{x}{2}\minus{}6\equal{}20$ or $ \frac{x}{2}\minus{}6\equal{}\minus{}20$, giving us the solutions $ x\equal{}52$ and $ x\equal{}\minus{}28$.  Their positive difference is $ 52\minus{}(\minus{}28)\equal{}\boxed{80}$

Or we could say for an equation $\left|ax+b\right| = c,$ the positive difference is $\frac{2c}{a}.$ Using this, we get $\frac{40}{\frac{1}{2}} = \boxed{80}.$

Also, a side note, the sum would be $-\frac{2b}{a}.$ In this the sum would be $\frac{12}{\frac{1}{2}} = 24.$


Find all function $f: Z \rightarrow Z$ satisfying $f(x+y+f(y))=f(x)+2y \forall x,y \in Z$

if $x=0$ and $y=0$, $f(f(0))=f(0)$.  if $x=f(0)$ and $y=0$, $f(f(0)+f(0))=f(0) \Rightarrow f(2f(0))=f(0)$, if $x=0$ and $y=f(0)$, $f(f(0) + f(f(0)))=f(0) + 2f(0)$
$\Rightarrow f(2f(0))=3f(0)$
$\Rightarrow 3f(0)=f(2f(0))=f(0)$
$\Rightarrow f(0)=0$ (this gives as an immediate result, $f(y+f(y))=2y$ if $x=0$)
if $x=-f(y)$, then $f(-f(y)+y+f(y))=f(-f(y))+2y$
$\Rightarrow f(y)=f(-f(y))+2y$
let's suppose there exist $p$ and $q$ such that $f(p)=f(q)$
$\Rightarrow f(-f(p))+2p=f(-(f(q))+2q$
$\Rightarrow p=q$
so it's an inyective function.
if $x=-y-f(y)$, then we have $0=f(-y-f(y))+2y$
so $2(-y)=f(-y-f(y))$ and we know that $2(-y)=f(-y+f(-y))$
so $f(-y+f(-y))=f(-y-f(y))$ $\Rightarrow -y+f(-y)=-y-f(y)$ $\Rightarrow f(y)=-f(-y)$
let's suppose we have $p$ and $q$ such that $f(p)-f(q)$ is even (let's say $2r$).
$f(p)=f(q)+2r=f(q+r+f(r))$
$\Rightarrow p=q+r+f(r)$
$\Rightarrow p-q=r+f(r)$
$\Rightarrow f(p-q)=f(r+f(r))=2r$
$\Rightarrow f(p-q)=f(p)-f(q)$
now if $f(p)-f(q)$ is not even, it still holds since one of $f(p)-f(p-q), f(q)-f(q-p)$ is even, and since we know $f(x)=-f(x)$, we can obtain $f(p)-f(q)=f(q-p)$ (this also assures that there are even numbers in the image of the function)
which can be written as $f(x)+f(y)=f(x+y)$ for every $x$ and $y$
we know that every function in $R$ that holds that, there exists a $\lambda$ such that
$f(x)=\lambda x$ for ever $x$ in $Q$
since this function is defined in $Z$, we have that $f(x)=\lambda x$ for a constant integer $\lambda$
we know that $f(x+y+f(y))=f(x)+2y$, and since y can be any integer, and we know that there exist even integer in the image, we can choose $f(x)$ to be one of such numbers.  then we can make $y$ vary, so every even number is in the image of the function.  so $2$ is in the image of the function.  so $|\lambda| \le 2$
if $\lambda=2$
$f(y+f(y))=2y$
$2(y+2y)=2y$
which is a contradiction
if $\lambda=-1$
$f(y+f(y))=2y$
$-(y-y)=2y$
which is a contradiction
if $\lambda=-2$
$f(x+y+f(y))=f(x)+2y$
$-2(x+y+-2y))=-2x+2y$
$-2x-2(-y)=2y-2x$
so the solutions are $f(x)=x$ and $f(x)=-2x$


ratio
geometric sequence
geometric series

Given that 1002004008016032 has a prime factor p > 250000, find it.

[quote="math154"]This is equal to

$ \frac2{10^3}S \equal{} 2\cdot10^{12} \plus{} 4\cdot10^{9} \plus{} 8\cdot10^6 \plus{} 16\cdot10^3 \plus{} 32 \plus{} \frac {64}{10^3}$, and

$ S\left(\frac {10^3 \minus{} 2}{10^3}\right) \equal{} 10^{15} \minus{} \frac {64}{10^3}$

.[/quote]

i don't really understand the 2nd line. Can you elaborate on that?

$ S$ is a finite [url=http://www.artofproblemsolving.com/Wiki/index.php/Geometric_series#Summing_a_Geometric_Sequence]geometric series[/url] with ratio $ \frac2{10^3}$.

So I multiplied $ S$ by $ \frac2{10^3}$, and then subtracted the resulting equation from the original equation to get the much simpler

$ S\left(\frac {10^3 \minus{} 2}{10^3}\right) \equal{} 10^{15} \minus{} \frac {64}{10^3}$.


inequalities
geometry
perimeter
inradius
circumcircle
inequalities unsolved

Prove that: $6\sqrt 3R^{2}+2pr\ge 5pR$ where $p,\,r,\,R$ are respectively the semi-perimeter, the inradius, circumradius of a triangle.

[color=purple]It's in Math & Youth 3-2007.You shouldn't discuss about it here [/color]


On their summer vacation, the Charen family flew from New York City to Raleigh and
then to Atlanta on their way to New Orleans. Returning from New Orleans, they flew
to St. Louis and then to Detroit on their way to New York City. According to the
mileage chart below, what was the total length of the familys round-trip?
\[ \textbf{Distance Between Cities (Miles)}
\]
$ \begin{tabular}{ c | c  c  c  c c c}
&Atlanta&Detroit&New Orleans&New York City&Raleigh&St. Louis\\ \hline
Atlanta&\minus{}&732&466&866&400&547 \\
Detroit&732&\minus{}&1077&606&707&542 \\
New Orleans&466&1077&\minus{}&1300&869&674 \\
New York City&866&606&1300&\minus{}&496&938 \\
Raleigh&400&707&869&496&\minus{}&828 \\
St. Louis&547&542&674&938&828&\minus{} \\
\end{tabular}$

Adding up the distances, $ 496\plus{}400\plus{}466\plus{}674\plus{}542\plus{}606\equal{}3184$, so the trip was $ \fbox{3184}$ miles long.


geometry
rectangle
combinatorics proposed
combinatorics

How many  1* 10sqrt2 rectangles can be cut from a 50*90 rectangle using cuts parallel to its edges[/quote][/list]


probability
AMC
USA(J)MO
USAMO
AIME
number theory
AMC 10

Hi, I'm a 9th grader entering Algebra 2/PreCalc. Do you think the AMC 12 is right for me?
Also, I'm using Introduction to Counting and Probability, Introduction to Number Theory, and AoPS Volume 2 to prepare? Is that enough? Do you think I can make it to AIME/maybe USAMO?

AoPS V2 should be enough to get you to the AIME, after that it's more about talent and experience. As for 10v12, you should try to find a way to take both (a nearby college or other school can do the B). If you can't do that, the 12 is a surer bet to get to the AIME; there's a bit more margin for error. Also, if you've actually read those books, Precalc is going to be a piece of cake...

If you're most of the way through that class (which you will be by February), you have the background material for the AMC 12. I would recommend taking the 12, or one of each if you have access to both dates.

if u want to make usamo, take the 10 - thats what im doing even though im taking BC calc next year. I am prepping by doing aops v1 (completed), Volume 2, Intro to Counting and Probability (Check), and doing a lot of practice.

Yep.  I took both but flunked the aime (6), so didnt take the usamo

[quote="Quickster94"]The general consensus is 12 is easier for AIME and 10 is easier for USAMO right?[/quote]

But also remember that since he's only freshman, it doesn't really matter which he does... He has the advantage of taking the use of floor value (as opposite to juniors and seniors going by index cutoff).

right, but sometimes they do use an index for amc-10, like this yr, i wish they just took every1 with a 6, then i would've made USAMO  :(


Compute the number of integers in the set $ {2000,2001,2002,...,2999}$ for which the units digit of the integer is a factor of the integer.

Thanks in advance!

[hide="Solution"]
If the units digit is 1: all of them works - 100
2: again, all of them works - 100
3: $ 2\plus{}a\plus{}b\equiv 0 \mod 3$
There are 2+5+8+9+6+3 of those - 33
4: hundreds digit can be anything, and the tens digit should be 0, 4, or 8 - 30
5: all of them works -100
6: $ 2\plus{}a\plus{}b\equiv 0\mod 3$
Again, 33
7: $ ab$ in $ 2ab7$ can be 03, 10, 17, 24, ..., 94. There are 14 of those.
8: $ ab8\equiv 0 \mod 8$ $ ab$ can be 04, 12, 20, ..., 92. There are 12 of those.
9: $ 2\plus{}a\plus{}b\equiv 0 \mod 9$
There are 8+3=11 of those.

Thus, the answer is $ 100\plus{}100\plus{}33\plus{}30\plus{}100\plus{}33\plus{}14\plus{}12\plus{}11\equal{}433$

Answer: 433[/hide]

[quote="FantasyLover"][hide="Solution"]
If the units digit is 1: all of them works - 100
2: again, all of them works - 100
3: $ 2 \plus{} a \plus{} b\equiv 0 \mod 3$
There are 2+5+8+9+6+3 of those - 33
4: hundreds digit can be anything, and the tens digit should be 0, 4, or 8 - 30
5: all of them works -100
6: $ 2 \plus{} a \plus{} b\equiv 0\mod 3$
Again, 33
7: $ ab$ in $ 2ab7$ can be 03, 10, 17, 24, ..., 94. There are 14 of those.
8: $ ab8\equiv 0 \mod 8$ $ ab$ can be 04, 12, 20, ..., 92. There are 12 of those.
9: $ 2 \plus{} a \plus{} b\equiv 0 \mod 9$
There are 8+3=11 of those.

Thus, the answer is $ 100 \plus{} 100 \plus{} 33 \plus{} 30 \plus{} 100 \plus{} 33 \plus{} 14 \plus{} 12 \plus{} 11 \equal{} 433$

Answer: 433[/hide][/quote]

Is there a way of doing this problem without using mod?

Er, in this case, mod is just a way of representing the remainders.

You can write like 'If you divide a+b+2 by 3, the remainder will be 0' instead of '$ a \plus{} b \plus{} 2\equiv 0 \mod 3$' if you want to.

Could you explain your solution to the cases where the units digit is 4, 7, 8, and 9, please?  (My pattern yields a different answer for those cases.)  Thanks!

[b]Octahedr0n[/b] — [b]FantasyLover[/b] undercounted by $ 20$ ($ ab$ can be $ 00, 02, ..., 98$), $ 0$ (you find the first $ 2ab7$ that works, and then you keep adding  $ 70$ until you reach a number that is greater than $ 2999$), $ 12$ (all $ ab$ that are divisible by $ 4$ count), $ 0$ (there are 8 integer solutions to $ 2 \plus{} a \plus{} b \equal{} 9, 0 \leq a, b \leq 9$, and 3 integer solutions to $ 2 \plus{} a \plus{} b \equal{} 18, 0 \leq a, b \leq 9$).

Um, the answer key says that the answer is 466.

metaphor, I'm having trouble understanding your post.  In which part of his solution did FantasyLover undercount?  Thanks!


For any positive integer $n$, let $a_{n}$ be the real number solution of $x^{3}+\frac{x}{n}=1$. Show that
(1) $a_{n+1}>a_{n}$;

(2) $\sum_{i=1}^{n}\frac{1}{(i+1)^{2}a_{i}}<a_{n}$.

(1) is trivial.
It can be easily shown that $a_{i}>\frac{i}{i+1}$. Then (2) immediately follows by using this fact.

[quote="mecrazywong"]$a_{i}>\frac{i}{i+1}$.[/quote]

$a_{n}= 1-1/{3n}+O(1/{n^{2}})$, which is consistent with that inequality. Maybe it can be improved to $a_{i}> \frac{3i}{3i+1}$ or something similar.