MATHCOUNTS geometry rectangle ARML email

I have heard of trading "pins" at MathCounts Nationals.  I have never been to Nationals before, so I have no idea what these are.  Can anyone tell me?

I agree. This is my school's first year at MATHCOUNTS  . . . I have no way of knowing.

Every state has it own design and each kid on the team gets some pins (decent number like 50 or so) - numbers could vary from state to state.  These are used as ice breakers.  While you wait in the airport waiting for your ride, you can exchange your pins with mathletes from other states.   Kids then decorate their hats/shirts or whatever with their collection.

Last year images of some state pins were posted on this forum.

Naga

Many mathletes try to collect one from every state.  Unfortunately, every year there's always a state or two that forgets to bring pins or doesn't have enough.  I'm still missing one or two, even over two years and a monopoly on the prized Illinois pin last year :(.  So remember to bring your pins, and lots of them!  Oh yeah and make sure you don't let out too many too fast, so demand remains high.  Maybe you should study some economics, maybe figure out the optimal strategy :P.

[quote="Naga"]
Last year images of some state pins were posted on this forum.
[/quote]

Do you know where, by any chance?

[quote="kchande"]In a really old post, I think

Thanks[/quote]

That would be: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=pins&t=12207]here[/url]

Theoretically if each state gave 57 pins to each student, then each student could get one of each state. But that will never happen.  :(

I remember wholesaling off my pins :P . NJ pins became worthless soon afterwards :P .

oops! I thought it meant how many pins do you have, like from your state.  :blush:  yea the 41- 56 is me.

The first year I went as a coach the 4 kids pooled their resources to collect a full set.  Although there was one state that didn't have any pins, they managed to collect 2 or 3 pins from some states that brought different variations from their state.  So I ended up with 70+ pins from 56 of the teams.  In the past couple years I managed to get that last one.  Now I have a cup full of them from different years, and different events(I was at a teacher thing in DC where teachers traded pins, and we were even more cut throat, it wa pretty funny watching adults bargain with each other for pins.)

I went two years ago. I acquired 43 pins (counting my own) :D. It was pretty fun trying to find all the pins, but some teams ran out too fast and I couldn't get them all. Oh well. Everyone should bring pins, though; it makes for a much more interesting time (I think we were all running around during dinner looking for them :P).

Ah, yes. Ze pinz! The spirit of mathcounts. I remember my first year going out and trading 4 ct pins for 4 of another state's and doing that for about 20 or so states. The CT pin was soon the most worthless pin. I did get 49 that year though. I did slightly better last year, getting 52. I still have all the pins, lol. Btw, someone tell me after Nats if CT still has the cruddy plastic pin that says "We're full of surprises!" on it.

SO where do you get these pins?  Coordinator? Coach?

Or as my coach puts it "to mingle".  I agree with sirorange, the point of the pins is to interact with other people.  My coach was pretty supportive of us getting to know other people.  Unfortunately our team was...socially challenged when we were not going crazy.

Getting all the pins is fun, but meeting people and trading fairly is more important.
I met several people in getting a full set and got a full set with only 1 for 1 trades and a free pin from the Idaho coach (thanks).

Two of my teammates traded all their pins in a few hours and I soon became the only one with NJ pins except for my roomate, but he lost his so I got to trade NJ pins at high prices, like two Alaska pins and two Whyoming pins.

Socially challenged... not me! Though they should have everyone try to get everyone else's email addresses. I met more people that way than trading pins. Pin-trading you are only guaranteed to meet 56ish other people. Email-address-collecting you have to meet all of them. Just my opinion though.

My team had a number who were adverse to being particularly social/mixing with other states. As long as they don't mind other people being social and mixing with other states, who cares?

Billy

[quote="solafidefarms"]

My team had a number who were adverse to being particularly social/mixing with other states. As long as they don't mind other people being social and mixing with other states, who cares?

Billy[/quote]

Our team was socially active only when we were going crazy.  Well...I sometimes I tried to talk to some random people.  I asked the Massachusetts team about their blue headbands during the GM tour.

[quote="solafidefarms"]
Though they should have everyone try to get everyone else's email addresses

Billy[/quote]

Oh so that's why Mathgeek fowarded that email from you to me.  You wanted my email adress(I guess..I dunno really).  I remember you can to our table during breakfast saturday morning asking for our email adresses.  I told you it was on my aops account and Asod didn't want to.  Ugh..I remember that morning.  Mathgeek thought it would be funny if he put sugar, salt, pepper and all sorts of condiments into my orange juice.

Well at the tour I put mustard in your tea cause you stupidly WAY oversugared it. :lol:

Oh so you were that guy walking around asking for everyones email or aops usrn. That list of ppl is really long...

I didn't way oversugar it!!  I like my tea...bitter at the top and super-super-super-super sweet at the bottom.

[quote]Our team was socially active only when we were going crazy.  Well...I sometimes I tried to talk to some random people.  I asked the Massachusetts team about their blue headbands during the GM tour.  [/quote]

No offense or anything, but when you asked that to us it sounded malicious, which is kinda why we didn't respond. We wore them for team spirit of course.

[quote]No offense or anything, but when you asked that to us it sounded malicious, which is kinda why we didn't respond. We wore them for team spirit of course.[/quote]

Malicious?  Why did it sound malicious?  I was just curious.

[quote]I don't know, it just did. Also, someone before said we looked like Nazis, and I thought it might've been you guys as you asked us later.[/quote]

Well.....I didn't say it.  If it was on the Utah Team it was either Justin or Asod.  Most likely Asod because I don't remember Justin saying that and I was walking with him.  If I remember we were talking about the glass bridge.  And then I decided to ask you guys about the head bands.

Justin and Asod are obsessed with Nazis....

That's not very healthy. Although if anybody is obsessed with [i]despising[/i] Nazis it's me - I wrote and gave a speech about the horrors of Holocaust denial and the neo-Nazis who perpetrate it.
If you want me to post it, I will  :)

send it to Asod--If you want his email address, PM me.

Asod isn't a nazi.  He actually doesn't talk about them as much as you do mathgeek.


Contradiction
USA(J)MO
USAMO
induction
pigeonhole principle
grid
combinatorics

We say that a black square is row-happy if it's the only black square in its row, and column-happy if it's the only black square in its column. Note that each black square must be at least one of the two.

We'll find the maximum $n$ for which we can color $n$ squares without leaving a right triangle. There can be at most $999$ row-happy black squares in the best case. If there are more than $1000$, two of them would be in the same row, contradiction. If there are exactly $1000$, there must be one of them in each row and each row could not contain any other squares, so there would be exactly $1000$ black squares in total. Analogously, in the best case there can be at most $999$ column-happy black squares. Since each square is at least one of the two, there can be at most $1998$ black squares in total. It's easy to find an example for $1998$: color the bottom row and the leftmost column, but leaving the bottom-left corner white.

So we can color $1998$ squares without leaving a right triangle, and this is the maximum. Thus the required $n$ is $1999$.

Suppose that there are 1999 or more colored squares.
If a colored square shares both a row and a column with another colored square, then we have the required right triangle.
So in order for there to be no such right triangle, each colored square must be alone in either its row or column.
We say that a colored square A "forbids" a square B if A shares a row with another colored square and shares a column with B, or vice versa.
Thus if A forbids B, then B cannot be colored.
A square can only be forbidden by two colored squares, one in its row and one in its column. If it is forbidden by three, then two of them must share a row or column and thus one of them cannot be colored.
Each of the 1999 colored squares forbids 999 other squares, and so at least M = int(1999*999/2) squares are forbidden.
M = int((2000-1)(1000-1)/2) = int((2000000-3000+1)/2) = 1000000-1500
So at most 1500 squares are not forbidden. However, there are 1999 colored squares, so at least 499 of them must be forbidden. Contradiction.

Consider a board with 1999 colored squares such that each square in the leftmost column and the bottom row are all colored. Now uncolor the left-bottom corner, leaving 1998 colored squares. Clearly no colored square shares both a row and a column with another colored square, so we have a setup with 1998 colored squares such that no such right triangle can be formed.

So n=1999.

[hide="Solution"]
Say we have an $ n \times n$ board for $ n \ge 2$. We wish to prove that if $ 2n \minus{} 1$ squares are colored, then we have a right triangle. We use induction.

For $ n \equal{} 2$, it's easy to see that if we color $ 3$ squares, we must have a right triangle.

Now assume the hypothesis for $ n$.

Note that if we exchange any two rows or columns in a board that has a colored right triangle, it will still have a colored right triangle.

Say that we have an $ n \plus{} 1 \times n \plus{} 1$ board for $ n \ge 2$.

Since there are only $ 2n \plus{} 1$ colored squares, there must be a column that has no more than one colored square. Switch this column with the rightmost one.

Similarly, there must a row with no more than one colored square. Switch this with the bottom row. If this row contained the one colored square in the rightmost column, then we only have one colored square in the union of the rightmost column and bottom row. Otherwise, we still have no more than two colored squares there.

This means that in the $ n \times n$ board obtained by deleting the rightmost column and bottom row, we have at least $ 2n \minus{} 1$ colored squares. By the induction hypothesis, there must be a colored right triangle there. By repeating the row and column exchanges, we see that there must have been a colored right triangle in the original $ n \plus{} 1 \times n \plus{} 1$ board. This completes the induction.

For $ n \equal{} 1000$, we thus have that if $ 1999$ (or more) squares are colored, there must be a colored right triangle. It's easy to show that $ 1998$ might not have a colored right triangle (fill one column and one row, but not their intersection).

[/hide]

Note that switching columns or rows will not change whether there is a right triangle or not, as squares which share the same row or column will still share the same row or column.

Now suppose it was possible to color a $ nxn$ board with $ 2n\minus{}1$ squares. Since we have $ n$ rows, at least one row contains two black squares. Similarly, at least one column contains two black squares.

Call the black squares that share a row $ R_1$ and $ R_2$, and call the black squares the share a column $ C_1$ and $ C_2$. Note that $ C_1$ or $ C_2$ cannot share a column with $ R_1$ or $ R_2$ or else there would be a triangle. Similarly, $ R_1$ and $ R_2$ cannot share a row with $ C_1$ or $ C_2$. Therefore we have four distinct squares. Shuffles the rows and columns in such a way that $ R_1$ and $ R_2$ lie in the first two squares on the first row, and $ C_1$ and $ C_2$ lie on the bottom two squares of the rightmost column.

Note that any black square in the column of $ R_1$ or $ R_2$ or the row of $ C_1$ or $ C_2$ would form a triangle. So essentially we use four squares to cut out the leftmost two columns and bottom two rows. We now have $ 2(n\minus{}2)\minus{}1$ squares to color a $ (n\minus{}2)X(n\minus{}2)$ board. Continue this operation until we must color a $ 2X2$ board with $ 2(2)\minus{}1 \equal{} 3$ squares. Contradiction.

[quote="MithsApprentice"]Find the smallest positive integer $ n$ such that if $ n$ squares of a $ 1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.[/quote]

 I think I seen a very similar problem in st. petesburg´s olympiad.But i don´t remenber the year...
But not only there, I think that i´ve seen it more then one time.Anyway, here is my solution:(it is really wweird, i don´t know if the correctors would accept this,but i had to post it fast)

1998 clearly can´t be.i´ll prove that it is 1999.Suppose for absurd that it is not 1999.
we´ll have a columm with 2 painted squares.if we have another painjted square in one of these lines ,absurd.so we can "remove" these 2 lines from the board. But , if we have another painted one in this columm,we´l have to eliminate another line.So , to maximize the number of possible painted ones,
we´ll have to have the other painted ones in a different columm. we also have to color in such a way that we remove the least possible number of new squares.so the next 2 painted ones will have to be in the same line,but not in the same collum as the first 2 .Continuing in such manner, we see that
we will have one last coloring to make .But then we´ll have a triangle.Absurd.
So n=1999

[hide="Solution"]
We claim that the smallest $n$ is $1999$.

First, we show that $n=1998$ is not sufficient. Color the whole left-most column of the board except the bottom-left corner, and color the bottom row of the board (except the bottom-left corner). We then have colored $1998$ squares, but there are no three squares satisfying the condition.

Assume that the result does not hold for $n=1999$. By the Pigeonhole Principle, there is at least one column containing at least $\lceil 1999/1000\rceil=2$ colored squares. Let $C_i$, $1\le i\le 1000$ denote the number of colored squares in column $i$. WLOG, let $C_i\ge 2$, for $1\le i\le k$, so $C_i\le 1$ for $k+1\le i\le 1000$.

Note that given any two squares in columns $x$ and $y$, $1\le x,y\le k$, they cannot be in the same row. Otherwise, those two squares along with another square in column $x$ would violate our assumption. Hence, there are no two squares in columns $1$ to $k$ inclusive that occupy the same row. Since there are $1000$ rows, it follows that $C_1+C_2+...+C_k\le 1000$.

If $C_1+C_2+...+C_k=1000$, every row contains a colored square. But we still have only colored $1000$ squares, so consider a square $S$ not in columns $1$ through $k$ that is colored. It must be in the same row as some square in column $i$, $1\le i\le k$, so taking square $S$ with the two squares in column $i$ contradicts the assumption. Therefore, $C_1+C_2+...+C_k\le 999$.

But observe that $C_i\le 1$ for $k+1\le i\le 1000$, so $C_{k+1}+C_{k+2}+...+C_{1000}\le 1000-(k+1)+1=1000-k\le 999$. This implies $C_1+C_2+...+C_{1000}\le 999+999=1998$. But a total of $1999$ squares are colored, contradiction. Thus, when $1999$ squares are colored, there exists three squares whose centers form a right triangle with legs parallel to the edges of the board. // [/hide]

[hide="Solution"]
We generalize the problem to an $n \times n$ board. A simple construction exists with $2n-2$ colored squares and no right triangles: color the leftmost $n-1$ squares in the uppermost row and the bottommost $n-1$ squares in the rightmost column.  We now show by induction that if $2n-1$ squares are colored, there must be a right triangle.

The base case is obvious for $n=2,3$. The inductive step is $n \to n+2$. Assume the claim holds for an $n\times n$  chessboard.  Suppose that it is possible to color $2n+3$ squares of a $n+2 \times n+2$ chessboard and have no right triangles. By the pigeonhole principle, there must be two colored squares in the same row, say $s_1$ and $s_2$. If there is a colored square in the same column as either $s_1$ or $s_2$, then we have a right triangle, so assume there are no colored squares in either of those two columns.

Then there are $2n+1$  colored squares amongst the $n$ other columns. By the pigeonhole principle again, there must be two colored squares in the same column, say $s_3$ and $s_4$, which are distinct from $s_1$ and $s_2$. If there is a colored square in the same row as either $s_3$ or $s_4$, then we have a right triangle, so assume there are no colored squares in either of those two rows.

Delete the two columns occupied by $s_1$ and $s_2$ and the two rows occupied by $s_3$ and $s_4$ (in which there are no colored squares). We are left with an $n \times n$ chessboard with $2n-1$ colored squares. But the hypothesis  guarantees a colored right triangle in the $n \times n$ board, so we are done.

The answer to the original problem is then $2*1000-1=1999$
[/hide]

[hide=" Very badly written solution"]
I shall show that in general, for a $n \times n$ board, the answer is $2n-1$. It's easy to find a construction for $2n-2$- select any row, color $n-1$ of its cells, consider the column containing the uncolored cell of that row and color all of the other $n-1$ cells of that column.

Now, suppose $2n-1$ cells are colored with no right triangle being formed. Let $R_i$ and $C_i$ be the number of colored cells in the $i$th row and column respectively. Let $S_r$ be the sum of all $R_i$'s such that $R_i \geq 2$. Similarly define $S_c$.

[b]Claim:[/b] $S_r > n$ and $S_c > n$.
[i]Proof:[/i] If $S_r \leq n,$ there must exist at least $2n-1-n= n-1$ rows having no more than 1 colored cell. Hence, there must exist a row containing the remaining $2n-1-(n-1)=n$ colored cells. This, however, is impossible because then we get a right angled triangle for any cell not in that row. Similarly, $S_c > n.$ $\blacksquare$

Now, this proof is trivial. Consider an arbitrary row with 2 or more colored cells, let's call this row $r_0$. For any other row with 2 or more colored cells, project its colored cells in that row perpendicularly onto $r_0$. We get at least $n+1$ colored cells on $r_0$ by this process, so two of them must overlap, implying the existence of a right angled triangle with edges parallel to the edges of the chessboard (if two colored cells from rows $i$ and $j$ overlap, then the triangle with one of its vertices at the cell from row $i$ and other two vertices from row $j$, one of them being the one which overlapped with that cell, form a right triangle). $\blacksquare$ [/hide]

[hide="A Worse Written Solution"]
We will proof by induction that an $m \times n$ right triangle free grid can have at most $m+n-2$ black squares.

The base case for $m+n=4$ and $m+n=5$ is easy :)

Now in the $m \times n$ grid let the maximum number of black squares such that the grid is right triangle free be denoted by $b(m,n)$
Where $m$ denotes the number of rows and $n$ the number of columns

By PHP at least one row has at least $2$ black squares.
Obviously the respective columns (of these black squares) won't have any black squares (else a right triangle is formed)
Now remove these 2 columns,
we are left with $m \times (n-2)$ grid with $b(m,n)-2$ black squares
$\implies b(m,n)=b(m,n-2)+2=b(m-2,n)+2$ (by symmetry)
Thus by our induction hypothesis $b(m,n) \le m+n-4 +2 = m+n-2$

Thus the proof

The construction of such a case is easy as shown in the other proofs :)
[/hide]

[hide="solution"]
Suppose for contradiction that there is a configuration with $n=1999$ that does not form a triangle as described.  Note that every square must either be the only one in its row or column (otherwise, it, along with one in its row and one in its column, would form a triangle).  Then, we can define each black square by either the row or column that it is alone in.  Note that, because $n=1999$, either every column or every row has a corresponding black square.  WLOG, every column has a black square.  Then, however, there could be at most 1000 squares, because adding any more would cause some column to have more than one.  Thus, there is a contradiction.
Now simply note that $n=1998$ is easily achieved by coloring every square along one row and one column except their intersection.
Thus, $n=1999$ is the desired minimum.
[/hide]

[hide]Consider the bipartite graph $R \cup C$, where a black square is marked by connecting the two vertices. Then the existence of a right triangle is equivalent to the existence of a path with length $3$. Call a tree $\emph{oceanic}$ if all but one of its vertices are leaves. The existence of an example for $1998$ is evident; just combine two oceanic trees with $1000$ vertices each. Now we show that there must exist a path of length $3$ if there are $1999$ edges. Assume the graph is not connected. Then one can easily see that there are many connected components, and one has equal or more edges than vertices. Hence it would contain a cycle, and any cycle has length $\ge 4$, so we can remove an edge for the desired path. Otherwise, it is a tree. Assume that it is a tree. Then, if it not $\emph{oceanic}$, there is an easy way to get a path of length $3$. Start from a leaf, go to the root $(\ge 1)$ steps, go to a non-leaf ($\ge 1$ steps), and go to a leaf again ($\ge 1$ steps). Hence, it remains to see to the case where it is $\emph{oceanic}$. But then there are 1999 rows and 1 column, which is contradiction.[/hide]

[quote=MithsApprentice]Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.[/quote]

Let $f(m, n)$ be the answer to this problem for an $m \times n$ grid. I claim $\boxed{f(m, n)=m+n-1}$.

Induct on $k=m+n$; here base cases are clear. To see $f(m, n) \ge m+n-1$ take a row and a column and color all cells on them except for their intersection. WLOG, $m \ge n$. If each row has two colored cells, we get $2m>m+n-1$ colors. Else, pick that row with just one colored cell and look at the column containing it. Let there be $s \ge 1$ colored cells on this column. Then, none of the rows containing them can have a colored cell. Delete this column and these rows; "glue" together the rectangular pieces of the board so formed in the natural way to get a new board. For this board, $k \mapsto k-s-1$ so we may conclude that the original can have no more than $k-1$ cells. Our induction is complete. $\blacksquare$

[hide = Solution]
We use induction to find the smallest n such that no right triangle exists for the general k by k grid:

For k = 2 it's trivial that the maximum n for which no right triangle exists is 2(k-1) = 2.
For k = 3, we have to do a little more to prove that the above formula holds:

First 2(k-1) = 2(3-1) = 4.

Note the following invariants:

1) Swapping any row does not change the number of colored squares in any column or row.
2) Swapping any column does not change the number of colored squares in any column or row.

By pigeonhole, there is a row with at least two colored squares. First we swap the columns so that these two colored squares are in the first and second columns. Then we swap the row with the first row.

Also by pigeon hole, there is a column with at least two colored squares. If this column is among the first or second columns, a right triangle would exist. The column having a square in the first row would also create a right triangle. Assume otherwise. Then this column is the third column, with its colored squares in the second and third rows.

It follows that there cannot be any other squares in the second and third rows. The only square left now if the top-left corner, which clearly cannot be colored. Hence, the largest value of n for k = 3 is 4.

Now for the induction step:
Suppose the max n for $k \leq l$ is $2l-2$.

We prove the $n>2(l+1)-2$ implies the existence of a right triangle for the $k = l+1$ case. Suppose that at least $2l+1$ squares are colored.

Like the k = 3 case, we swap the rows such that $a_{11}, a_{12}$ are colored. We can also swap the squares so that $a_{n-1n}, a_{nn}$ are colored (otherwise a right triangle would exist by the same logic as the k=3 case). It follows that there are no other colored squares in the first, second columns and last, second to last rows. Then there must exist at least $2l-3$ colored squares among the top-right $l-1$ by $l-1$ grid, which contradicts our inductive hypothesis.

Construction: Take $a_{11}, a_{12}, ... a_{1n-1}, a_{2n}, a_{3n}, ... a_{nn}$

And we're done.
[/hide]

Construction by StanPZ28
We claim that $n=1999$ is the smallest such $n$ such that there is guaranteed to be a right triangle. For $n=1998$, take two lines, one vertical and one horizontal and color every square on those lines except for their intersection. Clearly there are no right triangles, and $1998$ colored squares. Hence, we see that $n\geq1998$.

Now we show that $1999$ squares is enough. Denote a square a [i]column square[/i] if the square has another colored square in the same column, and define a [i]row square[/i] analogously. Furthermore, denote a square to be [i]neither[/i] if it is neither a row nor a column square. If a square is simultaneously a row and a column square, then we are done, so assume that this is not the case.

Suppose that all the squares are either row or column squares. Let there be $r$ column and $c$ row squares respectively. Since $r+c=1999$, we have that one of $r$ or $c$ must be greater than or equal to $1000$, so we are done, since there must be some "overlap". Now suppose that there are $n$ neither squares. We now have that there are $n$ neither squares, taking up $k$ rows and $k$ columns. Hence we have that there are $1999-k$ rows/columns squares taking up $1000-k$ rows and $1000-k$ columns. Hence, there must be overlap, and we are done. $\square$

(There is a generalized form in the Pranav combinatorics book.)
Let a column with more than $1$ shaded square be bad.

Firstly find the max shaded squares where we can ever possibly make no right triangle form.
Take an $n\times n$ ($n\ge2$) grid and let the number of bad columns be $a$, then the good columns ($0/1$ shaded squares) is $n-a$.
If there are no bad columns ($a=0$) then the total number of shaded squares is at most $n-0=n$.

If there is at least $1$ bad column:
The shaded squares in a bad column can't share a row with another other square otherwise a right triangle forms. In particular the total number of shaded squares in the bad columns is at most $n$, otherwise $2$ (out of $n+1$) of these squares share a row.

If there were exactly $n$ of these squares none of the good columns would have any shaded squares since those $n$ squares definitely would take up all $n$ rows. Total shaded squares is $n$.

Otherwise there are at most $n-1$ of these squares. The other $n-a$ columns have at most $n-a\le n-1$ squares, giving a maximum of $2(n-1)$.
Clearly $2(n-1)\ge n$, thus, we have $2(n-1)$ shaded squares at most if no right triangle ever forms.
On the converse, if $m \ge 2n-1$ then we will always find a right angle triangle.

For $m = 2(n-1)$ we can construct an example where no right triangle appears, just shade the top row and left column except the top-left corner square.

We proved the minimum is $2n-1$, then setting $n=1000$, $2*1000-1=1999$.

The answer is $n = 1999$.

For a construction with $n = 1998$, take a punctured L as illustrated below (with $1000$ replaced by $4$): \[ \begin{bmatrix} 1 \\ 1 \\ 1 \\ & 1 & 1 & 1 \end{bmatrix}. \]

We now show that if there is no right triangle, there are at most $1998$ tokens (colored squares). In every column with more than two tokens, we have token emit a bidirectional horizontal death ray (laser) covering its entire row: the hypothesis is that the death ray won't hit any other tokens.

[asy] size(6cm); for (int i=1; i<=8; ++i) { 	if ((i-1)*(i-3)*(i-5) != 0) draw( (0.5,i)--(8.5,i), red+1, Arrows(TeXHead) ); 	for (int j=1; j<=8; ++j) { 		if ( (i-j)%2 == 0 ) 			filldraw(shift(i-0.5,j-0.5)*unitsquare, invisible, grey); 		else 			draw(shift(i-0.5,j-0.5)*unitsquare, grey); 	} }

draw( (3,2)--(3,7), blue); draw( (5,4)--(5,8), blue); void token(pair P) { fill(circle(P, 0.2), black); } token( (1,1) ); token( (2,1) ); token( (3,2) ); token( (3,6) ); token( (3,7) ); token( (4,1) ); token( (5,4) ); token( (5,8) ); token( (6,5) ); token( (7,3) ); token( (8,1) ); [/asy]

Assume there are $n$ tokens and that $n > 1000$. Then obviously some column has more than two tokens, so at most $999$ tokens don't emit a death ray (namely, any token in its own column). Thus there are at least $n-999$ death rays. On the other hand, we can have at most $999$ death rays total (since it would not be okay for the whole board to have death rays, as some row should have more than two tokens). Therefore, $n \le 999 + 999 = 1998$ as desired.

The answer is $n=1999$. To show that $n=1998$ fails, pick an arbitrary row, an arbitrary column, and color in both. Then uncolor their intersection, which clearly works.

Now I will show that $n=1999$ works. Now identify each cell in the chessboard as an ordered pair $(i,j)$ where $i$ represents the column number and $j$ the row number, so the bottom left is $(1,1)$ and the top right is $(1000,1000)$. We can see that such a right triangle exists if and only if there exists some $(i,j)$ such that column $i$ contains at least two colored squares and row $j$ contains at least two colored squares.
Suppose FTSOC that there exists some coloring of $1999$ squares and no right triangles are formed. View the coloring column-by-column, so we first color some squares in the first column, then the next, and so on. At some point in time, call a row [i]solitary[/i] if there can be at most one colored square in it, otherwise a right triangle is formed. Hence every time we color a column, we create some (possibly zero) solitary rows. Since we cannot color cells lying in already solitary rows, it is clear that no new solitary columns are created by coloring a column with a single square or no squares, and if $k \geq 2$ columns are colored then $k$ new solitary columns are created: one for each square colored in that column. Thus after some number of columns are colored, with $k$ cells being filled in, the total number of solitary rows is equal to:
$$k-\text{the number of columns with 1 square colored},$$
hence after all the columns are colored there are:
$$1999-\text{the number of columns with 1 square colored}.$$
Now note that any coloring without right triangles must have at most $998$ columns with one colored cell: $1000$ is impossible since then we only color $1000$ squares, and $999$ is impossible since it implies the row that doesn't have one colored cell has $1000$ cells, and a right triangle can be clearly formed. Hence after the columns are all colored we have at least $1001$ solitary rows, which is absurd since there are only $1000$ rows in the chessboard. Hence such a coloring does not exist, so every coloring of $1999$ squares contains a right triangle. $\blacksquare$

The answer is $n=1999$. To see that $n=1998$ doesn't work, color the 999 squares that are in the topmost row and rightmost column, but not the square in both. It is easy to see that then $n<1998$ doesn't work.

Now to prove that $n=1999$ works. For a coloring of 1999 squares, consider the columns where at least 1 square is colored and pick 1 square is each of those columns to be called interesting. Note that there are between at most 1000 interesting squares, each column has at most 1 interesting square, and columns without an interesting square have no colored squares. If all the interesting squares are in the same row, then we can chose any colored squared and it would form a right triangle with 2 interesting squares. So, assume not all the interesting squares are in the same row. For the other 999 (possibly more) colored squares that are not interesting, if any two are in the same row, we are done since each of these colored non interesting square is in the same column as an interesting square. The right triangle is then formed by the two colored non interesting squares in the same row and the interesting square in the same column as one of these 2 squares.
The other case is if the 999 colored non interesting squares are all in different rows. Since the interesting squares are in at least 2 rows, one of these 999 colored non interesting squares is in the same row as an interesting square, and by definition of interesting that specific colored non interesting square has to be in the same column as an interesting square as well, so we have the right triangle.

Oops why did this take me like $15$ minutes to solve, but $1$ hour to write up... I'm obviously inexperienced at Olympiad Combo. [s]Also why are all the other solutions so short and appealing, especially Evan's, unlike mine.[/s]

We claim $n = 1999$ is the answer.

Call a chessboard [i]good[/i] if it contains a right triangle and [i]bad[/i] otherwise. Observe that a chessboard is [i]good[/i] if and only if it contains a colored square that has another colored square within its own row and column. Now, we will try to construct [i]bad[/i] chessboards.

Let a row be [i]singular[/i] if it contains exactly $1$ colored square and [i]non-singular[/i] if it contains $2$ or more. Notice that in order for a chessboard to be [i]bad[/i], every single colored square within a [i]non-singular[/i] row must have $0$ additional colored squares in its own column.

[b]Claim:[/b] Suppose we are constructing a [i]bad[/i] chessboard. If our chessboard has at least $1$ [i]singular[/i] row, then the largest possible amount of colored squares within our [i]non-singular[/i] rows is $999$.

[i]Proof.[/i] Since colored squares within [i]non-singular[/i] rows must have their own column (distinct from all other colored squares), the result easily follows, as the colored squares located within [i]singular[/i] rows could all share a single column. $\square$

Similarly, we conclude that for boards containing $0$ [i]singular[/i] rows, the maximum possible amount of colored squares contained within the [i]non-singular[/i] rows is $1000$.

It's easy to see there exists a [i]bad[/i] chessboard for $1 \le n \le 1000$, as we cannot guarantee there will exist a [i]non-singular[/i] row (by the Pigeonhole Principle) for these values of $n$. If $1001 \le n \le 1998$, then we can construct [i]bad[/i] chessboards as follows:
    $\bullet$ Color the first $n - 999$ squares of row $1$ from column $1$ to column $n - 999$;
    $\bullet$ Color the last $999$ squares of column $1000$ (from row $2$ to row $1000$).

Obviously, the total number of colored squares (here) is $n$, and we've satisfied all conditions necessary to form a [i]bad[/i] chessboard, so we're done.

Now, we will utilize contradiction to prove there exist no [i]bad[/i] chessboards for $n = 1999$.

[b]Claim:[/b] If a [i]bad[/i] chessboard has $1999$ colored squares, then it must contain exactly $1000$ [i]singular[/i] rows.

[i]Proof.[/i] Suppose we have $0$ [i]singular[/i] rows within our chessboard. Now, we have $1999 > 1000$ colored squares left for the [i]non-singular[/i] rows of our board, implying a [i]bad[/i] chessboard cannot be constructed for this case (due to the corollary of our first claim).

Now, assume we have at least $1$ [i]singular[/i] row on our chessboard. Then, the largest possible amount of colored squares contained within our [i]non-singular[/i] rows is $999$, implying the number of colored squares within our [i]singular[/i] rows (for this case) is at least $1999 - 999 = 1000$. Because our chessboard only has $1000$ rows, however, we conclude that exactly $1000$ [i]singular[/i] rows are needed for this case. $\square$

The last claim implies there must exist $1000$ [i]singular[/i] rows on our board to (possibly) construct a [i]bad[/i] chessboard when $n = 1999$. This means the number of colored squares here is just $1 \cdot 1000 = 1000$. Obviously, we have a contradiction as $1000 \ne 1999$, so all chessboards with $n = 1999$ are [i]good[/i] and we're done. $\blacksquare$

[hide]The answer is 1999. For a construction for why 1998 doesn't work, refer to vEnhance's post.

We show that 1999 doesn't work by contradiction, so assume it does. Now let $a_i$ be the number of colored squares in column $i$. Note that if we swap rows $i,j$, it doesn't change the outcome (the total number colored and whether it works) so assume $a_1 \geq a_2 \geq \dots \geq a_{1000}$.

Now note that if a column has multiple colored squares, then no more colored squares can be in a common row with any of those squares. However, if a column has only 1 colored square, then other columns who have 1 colored square can have the colored square in the same row. Now we show doing this is optimal to use the least number of rows, and to do this, say WLOG that there are multiple rows who have at max 1 colored square in each column, then we can move all of those colored squares to the same row thereby conserving those other rows.

Now if $x$ is the number of columns with 1 colored square, we have that the number of rows that have a colored square is at least \[1+(1999-x)=2000-x\] but since $x\leq 1000$, $x=1000$. But then this implies that we only have $1000$ colored squares - a contradiction, so we are done. [/hide]

[hide=does this work ???]
Replace $1000$ by any arbitrary constant $x$. We claim that at least $2x-1$ squares need to be colored. Coloring all but one square of a column and coloring the remaining squares in the uncolored square's row is a valid counterexample for $2x-2$.

FTSOC assume an arrangement in which $2x-1$ squares are colored but the desired conditions not fulfilled exists. Note that by PHP, at least one column must have $2$ colored squares ; Let the column with the maximum colored squares have $m$ colored squares. Also let $n$ columns have one colored square. We need to color $2x-1-n-m$ more squares. However, the rest of the columns have more than or equal to $2$ colored squares (or none at all but it can be disregarded). Moreover, henceforth each "gap" (rows with no colored squares in the column with the largest number) must contain at most $1$ colored square. So we must have $2x-n-m-1 \leq x-m \implies n \geq x-1$. This forces $n=x-1$ so $m=n$. However this is a contradiction $\blacksquare$

We offer a stronger claim.

[b]Claim: [/b] For $a,b\geq 2$, the rectangle $a\times b$ can have at most $a+b-2$ colored cells while not having any right triangles.
[b]Proof: [/b]. First note that we can always achieve this by covering a full row and full column, minus the intersection.

Induction. The base case $2\times 2$ clearly yields at most 2 colored cells. Otherwise, consider the row with the maximum number of colored cells, let this be $M$. If $M=b$, then the maximum number of cells is $b\leq a+b-2$. If $M=b-1$, then we are forced to have at most $b-1+a-1 = a+b-2$ by filling out the final unfilled the row (This is the equality case).Otherwise, the other $b-M$ cells in the row, and the $a-1$ other cells in the column, all cannot be colored. Additionally, by the inductive hypothesis, the maximum number of cells in the remaining $(b-M)\cdot (a-1)$ rectangle is $(b-M)+(a-1)-2$ if $a-1\geq 2$, and $(b-M)\leq (b-M)+(a-1)-1$ if $a-1=1$.  Thus, the total is at most
\[M + (b-M) +(a-1)-1 = b+a-2\]
and we're done. $\square$.

This gives us an answer of $1000+1000-2+1 = \fbox{1999}$

[hide=Solution]
We claim that the smallest $n$ is $1999$. By picking one entire row and one entire column, then removing the intersection, there are no right triangles. Now, suppose that there are $1999$ colored squares, and there are no right triangles. Suppose that there are $a_i$ numbers in row $a_i$. Then, if $a_i\geq2$, then $a_i$ columns can only have one colored square. Therefore, $\sum_{a_i\geq2}a_i\leq1000$. Since there are $\sum_{i=1}^{1000}a_i$ colored squares, this means that the maximum number of colored squares occurs when as many of the $a_i$ are equal to $1$ as possible. This number is at most $999$. If there exists $a_i=1000$, then there are at most $1000$ colored squares. Therefore, there are at most $999+999=1998$ colored squares. Therefore, the smallest positive integer $n$ that satisfies the conditions is $\boxed{n=1999}$.

We claim that $n=1999$. To see that $n=1998$ fails, pick a cell and color all the cells lying on the same row and column of it, and do not color the picked cell. Thus, there are no right triangles.

Now, let $G= A \cup B$ be the bipartite graph of the chessboard, where $A$ is the set of rows and $B$ is the set of the columns. We connect $a \in A$ to $b \in B$ iff their intersection on the chessboard is colored. We will prove that if $|E(G)| \geq 1999$, then $G$ has a path of size $3$ $a-b-c-d$, which clearly implies that there is a right triangle.

Thus, suppose for the sake of the contradiction that $|E(G)| \geq 1999$ but $G$ has not a path of size $3$. Then, for every $a \in A$ with $deg (a) \geq 2$, its neighbors have degree exactly $1$. The same is true for $b \in B$ with $deg (b) \geq 2$. Therefore, $G$ is a disjoint union of trees, and $G$ has exactly $2000$ vertices, so if $G$ has more than a tree, there will be less than $2000-1=1999$ vertices. Hence, $G$ is a tree. However, this clearly implies that there will be a size of path $3$ in $G$, implying the desired result.

$\blacksquare$

The answer is $1999.$ To show that $1998$ does not work, simply consider the case where the first row and the first column is colored, but not their intersection. Define vertices $R=\{r_1,r_2,\dots, r_{1000}\}$ and $\{g_1,g_2,\dots, g_{1000}\}$ in a bipartite graph where $(r_i,g_j)$ is connected if and only if row $r_i$ and column $g_j$ is colored. Note that if there exists a path with two vertices in $R$ and two in $G$, then there is a $3$-path, implying that there is a right triangle.

Assume that we can draw $1999$ edges without creating a $3$-path. Clearly there must be no cycles, so our bipartite graph is a forest. Consider any tree. It may not have two vertices in $R$ and two in $G$, so WLOG assume it has only one vertex in $R$. That means that one the forest has at least two trees. Therefore, the number of vertices is at least two more than the number of edges, $2000\ge 1999+2$ contradiction.

e claim that the answer is 1999. A counterexample for 1998 is when a cell is shaded if and only if it is in the first row or first column but not both.\\

We will then show that any coloring of 1999 cells works. We will try to construct a case with no triangles and show that it is impossible.\\

Note that in order for there to be no triangles, if any two rows share a colored cell in the same position, then that shared position must be the only colored cell in both of those rows.\\

Call a row orz if it contains exactly one colored cell, and unorz otherwise. Note that there can be at most 999 orz rows, so there must be at least 1000 colored cells that are in an unorz row.\\

Case 1: There is at least one orz row. Then, we cannot have any unorz row that shares a colored cell in the same position as in a orz row (since otherwise there will be a triangle), so we have at most 999 columns available for unorz rows. However, since there are at least 1000 cells in unorz rows, we must have two colored cells in unorz rows that are in the same column, which causes a triangle.\\

Case 2: There are no orz rows. Then, all 1999 colored cells are in unorz rows. Therefore, clearly, there exist some two colored cells that are in the same column in two unorz rows, which causes a triangle, so we are done.

The answer is $\boxed{1999}$. A counterexample for $1998$ is just shading the first row and first column, but not the bottom left corner.

Suppose that a right triangle did not exist for $n = 1999$. Call a square good if it is in a column with two or more colored squares. Clearly, each good square is the only colored square in its row. Assume that there were $k$ good squares. Since there are only $1000$ rows, $0\le k\le 1000$. If $k = 0$, then each column has at most $1$ colored square, and if $k = 1000$, each row has at most $1$ colored square, both result in at most $1000$ colored squares, contradiction. If $0<k<1000$, then there are at most $999$ columns without good squares, which means at most $k + 999$ total squares. However, $k + 999<1999$, contradiction. Therefore a right triangle must exist.

Similar to (but much easier than) USA TST 2011/8.

Interpret the chessboard as the lattice points with $x$ and $y$ coordinates between $1$ and $1000$, inclusive. We claim that the answer is $1999$. The points $(2,1),(3,1),\ldots,(1000,1),(1,2),(1,3),\ldots,(1,1000)$ show that this is a lower bound.

We claim that $n=1999$ satisfies the problem statement. Assume FTSOC that no colored point has another colored point with the same $x$-coordinate and another colored point with the same $y$-coordinate. If a colored point $P$ has a colored point with the same $x$-coordinate, then consider the projection of $P$ to the $y$-axis. Otherwise, consider the projection of $P$ to the $x$-axis. These $1999$ projections must be distinct: no two colored points with the same $x$-coordinate can be projected to the $x$-axis, and no two colored points with the same $y$-coordinate can be projected to the $y$-axis. Since there are $1999$ colored points, the set of projected points must contain all of $(1,0),(2,0),\ldots,(1000,0)$ or $(0,1),(0,2),\ldots,(0,1000)$. However, this means the $1000$ colored points that created these projections must be in pairwise distinct rows and columns, which means no other colored point can be added, a contradiction. $\square$

Rather dirty induction done twice.  :what?:

[b][color=pink]Claim:[/color][/b] For a general $m \times n$ grid with $m, n \ge 1$, the maximum number of colored squares that can be placed while this doesn't hold is at most $m + n - 1$.
[i]Proof.[/i] The base case of a $0 \times n$ or $n \times 0$ grid is immediate. Else, take the row/column with the maximum number of squares, say $C$.
If $C = 1$, then the bound follows immediately. Else, this rules out at least $C$ rows/columns. Induct down on the smaller grid. $\blacksquare$

[b][color=pink]Claim:[/color][/b] For a general $m \times n$ grid with $m, n \ge 2$, the maximum number of colored squares that can be placed while this doesn't hold is at most $m + n - 2$.
[i]Proof.[/i] The base case of a $2 \times n$ or $n \times 2$ grid is immediate. Else, take the row/column with the maximum number of squares, say $C$.
If $C = 1$, then the bound follows immediately. Else, this rules out $C + 1$ rows/columns (including the row they are in). Using the weaker claim finishes. $\blacksquare$
A construction of $n = 1998$ comes by coloring all of a row and a column besides their intersection. Thus, $n = 1999$ is the answer.


calculus
derivative
trigonometry
induction
integration
function
calculus computations

[quote="levi"]$For\ n\in N^*\ and\ a^2-4b<0,\ prove\ that:$
$\left(\frac{1}{x^2+ax+b}\right)^{(n)}=(-1)^n\cdot n!\cdot\frac{2}{\sqrt{4b-a^2}}\cdot\frac{\sin\left[ (n+1)arctan\frac{\sqrt{4b-a^2}}{2x+a}\right]}{\sqrt{(x^2+ax+b)^{n+1}}}.$[/quote]
Let $ r_1,r_2 $ be the complex roots of equation $x^2+ax+b=0 . $ Because
\[=\frac{1}{r_2-r_1}\left(\frac{1}{x-r_2}-\frac{1}{x-r_1}\right)\]
and $  \left(\frac{1}{x-\alpha}\right)^{(n)}=\frac{(-1)^{n} n!}{(x-\alpha)^{n+1}} $ , with notation $B:=x-r_1\; ,\; A:=x-r_2 $ we find
\[(*)\; \; \; \;
E:=\left( \frac{1}{x^2+ax+b} \right)^{(n)}=(-1)^n
n!\cdot\frac{B^{n+1}-A^{n+1}}{(B-A)(AB)^{n+1}}\; .
\]
Suppose $r_1=-\frac{a}{2}-i\frac{\sqrt{4b-a^2}}{2}\; ,\;
r_2=-\frac{a}{2}+i\frac{\sqrt{4b-a^2}}{2}\; ,\; $ that is
\[ B=
\left(x+\frac{a}{2}\right)\left(1+i\cdot\frac{\sqrt{4b-a^2}}{2x+a}\right)=\frac{2x+a}{2}\rho(\cos\phi+i\sin
\phi)\; ,\;\] \[ A=
\left(x+\frac{a}{2}\right)\left(1-i\cdot\frac{\sqrt{4b-a^2}}{2x+a}\right)=
\frac{2x+a}{2}\rho(\cos\phi-i\sin \phi)\; ,
\]
with $\rho:=\displaystyle\frac{2\sqrt{x^2+ax+b}}{|2x+a|}\; \; ,\; \;
\phi=\displaystyle\varepsilon\cdot
\arctan{\frac{\sqrt{4b-a^2}}{2x+a}}\; ,\; \varepsilon=\mbox{\rm
sign}(2x+a)\; . $ Therefore $ AB=x^2+ax+b \; \; ,\; $ and
\[\frac{B^{n+1}-A^{n+1}}{B-A}=\frac{2\varepsilon\cdot\sin{(n+1)\phi}}
{\sqrt{4b-a^2}}\sqrt{(ax^2+bx+c)^{n+1}}=\]
\[
=\frac{2\sin{(n+1)\varphi}} {\sqrt{4b-a^2}}\sqrt{(ax^2+bx+c)^{n+1}}
\; ,\] with $ \varphi:= \displaystyle
\arctan{\frac{\sqrt{4b-a^2}}{2x+a}}\; . $ In conclusion from (*)
\[E=\frac{2(-1)^nn!}{\sqrt{4b-a^2}}\cdot\frac{\sin{\left((n+1)\varphi\right)}}{\sqrt{(x^2+ax+b)^{n+1}}}
\]
as desired.

It is O.K. But without to use derivatives over the set of the complex numbers !

[quote="levi"]How ? I am tried and I didn't managed.[/quote]
Of course , many proofs are possible ! Let us try another one.
[b]Proof II:[/b] by denoting $ h= h(x)=\displaystyle ax^2+bx+c $ we have
\[(**)\;  \; h(x)\left(\frac{1}{h(x)}\right)^{(n+2)}
=-(n+2)(2x+a)\left(\frac{1}{h(x)}\right)^{(n+1)} -
(n+1)(n+2)\left(\frac{1}{h(x)}\right)^{(n)} ,
\]
where the superscript $ {}^{(m)} $ denotes the $m^{th} $ derivative with respect to variable $ x . $ Note that (**) is an easy exercise by induction. Let us try to solve the recurrence (**) . To do this write (**) in the equivalent form
\[(***)\; \; \; H_{n+2}=2\cdot\lambda\cdot H_{n+1}-H_n \; ,\; n\in {\mathbb N}, \]
where  $\; \; (\pounds)\; \; \; \; \; \; H_n:=\displaystyle
\frac{(-1)^n}{n!}\sqrt{\left(h(x)\right)^{n+2}}\left(\frac{1}{h(x)}\right)^{(n)}  $  and $  \lambda:=\displaystyle \frac{x+\frac{a}{2}}{\sqrt{h(x)}} \; .$ But (***) , being with constant coefficients it's easy to solve. One finds
\[ H_n=\frac{\sin{\left((n+1)\arccos{\lambda}\right)}}{\sqrt{1-\lambda^2}}
\]
and from $ (\pounds) $ we conclude that  \[
\left(\frac{1}{h(x)}\right)^{(n)}=\frac{(-1)^n n
!}{\sqrt{\left(h(x)\right)^{n+2}}}\cdot
\frac{\sin{\left((n+1)\arccos{\lambda}\right)}}{\sqrt{1-\lambda^2}}=
\]
\[=\frac{2(-1)^n n !}{\sqrt{\left(h(x)\right)^{n+1}}}\cdot
\frac{\sin{\left((n+1)\arccos{\lambda}\right)}}{\sqrt{4b-a^2}}\; .
\]
Because $ \;  \arccos{\lambda}=\left\{\begin{array}{lcl}
\displaystyle
\arctan\frac{\sqrt{1-\lambda^2}}{\lambda}  &,& 0<\lambda\le 1,\\
&&\\ \displaystyle
\pi+ \arctan\frac{\sqrt{1-\lambda^2}}{\lambda}  &,&-1 \le \lambda<0,\\
\end{array}\right. , $ you find the solution, i.e. an explicit form of    $ \displaystyle\left(\frac{1}{x^2+ax+b}\right)^{(n)} \; .$
[b]Remark:[/b] In my proof is considered  only  case $ 2x+a>0 .$ It remain to study case II , i.e. when  $ 2x+a<0 $. It's not difficult ! I consider that in the definition of $\lambda $ we must consider $\left|x+\frac{a}{2\right}| $ instead of $ x+\frac{a}{2} . $

[quote="flip2004"] More difficult questions[/quote]
Let us propose a bit more general questions:
[b]Problem  1. [/b][i] Find  $y^{(n)}(x)\equiv \displaystyle \frac{d^n}{dx^n}y(x)\; ,\; \; y(x): =(x^2+ax+b)^p$ where p is any real number.[/i]
[b]Answer : [/b] If we use the notation  $h(x)=x^2+ax+b$ , $h_1(x)= 2ax+b ,$ and ${z\choose k}=\frac{z(z-1)\cdots (z-k+1)}{k!}\; ,\; k\in {\mathbb N}, \; z\in {\mathbb C},$ then
\[ y^{(n)}(x)=n!\sum\limits_{k=0}^{\left[\frac{n}{2}\right]}{p\choose n-k}{n-k \choose k} \left(h(x)\right)^{p-n+k}\left(h_1(x)\right)^{n-2k}\; ,  \]
where $[\; \cdot \; ]$ denotes integral part.

[b]Problem  2.--(Hard !) [/b][i] Find  $y^{(n)}(x)\equiv \displaystyle \frac{d^n}{dx^n}y(x)\; , \; y(x): =(x^3+ax^2+bx+c)^p$ where p is any real number.[/i]
[b]Answer : [/b] Using the notation  \[ \left\{ \begin{array}{lcl} h&: =&h(x)=x^3+ax^2+bx+c \\ h&: =&h_1(x)= 3x^2+2ax+b \\ h_2&: =&h_2(x)=6x+2a \end{array}\right.  \] one finds
\[ y^{(n)}=n!\sum\limits_{k=0}^{\left[\frac{2n}{3}\right]} \frac{1}{2^k} {p\choose n-k}h^{p-n+k} \sum\limits_{j=0}^{\left[\frac{k}{2}\right]}4^{j}{n-k \choose k-j}{k-j\choose j}h_1^{n-2k+j}h_2^{k-2j}  \]

[b]Remarks:[/b]
[b]a)[/b] Such questions are not simple. They have applications also in Combinatorics and Special Functions (e.g. by finding [i] generating functions[/i]  of some sequences).
[b]b)[/b] Of interest is the following formula ( I found it in Schwatt [14] ,see (83) at page 13): Let $F=H(u) \; ,\; u=g(x)$.
 For $H, g$  admissible functions, we have the identity
\[ \begin{array}{|c|} \hline \\ \displaystyle\frac{d^kF}{dx^k} = \displaystyle\sum\limits_{\nu=1}^{k}\frac{(-1)^{\nu}}{\nu !}\frac{d^{\nu}F}{du^{\nu}}\displaystyle \sum\limits_{\alpha=1}^{\nu}\displaystyle (-1)^{\alpha}{\nu \choose \alpha}u^{\nu-\alpha} \frac{d^k u^{\alpha}}{dx^k}\\ \\ \hline \end{array}\; .  \]
[b]Examples:[/b]
 [b]i)[/b]  $H(t)=t^{-p} \; ,\; g(x)=x^2+ax+b$
[b]ii)[/b]  $H(t)=t^{-p} \; ,\; g(x)=x^3+ax^2+bx+c$
[b]iii)[/b]$H(t)=t^{-p} \; ,\; g(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\ldots +a_n$
However perhaps this formula was discovered earlier by Faa di Bruno [7]  (?) .

Some (incomplete) references, regarding above subject, which I found,  are:
   [b] Refernces[/b]
[1] L.F.A  Arbogast [1755--1803],[i] Du Calcul des Derivatives[/i],p.15
[2] P. E.  Appell [1855--1930] , [i]Sur les developpments en serie suivant les  inverses de polynomes donnes,[/i] Bulletin de la  S.M.France, 48  (1920)1-8. See also:
http://www.numdam.org/item?id=BSMF_1920__48__1_0
and also
http://www.numdam.org/item?id=BSMF_1923__51__189_0
[3] J. Bertrand [1822-1900] , [i]Traite de Calcul Differential et de Calcul Integral[/i], Gauthier-Villars,1864,vol.I, p.140. See also:
http://visualiseur.bnf.fr/Visualiseur?Destination=Gallica&O=NUMM-99558
[4] Czuber ,[i]Vorlesungen ueber Differential und Integralrechnung[/i], part I , p.361.
[5] U. Dini  [1845--1905] , [i]Lezioni di Analisi Infinitesimale,[/i]part I,p.361 . See also
http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=02930001&seq=1
[6] J. Edwards [1854--1931], [i]Differential Calculus  with applications and  numerous examples,[/i] Macmillan, London, 1886, pp.57 and 449. See also
http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath&idno=ACM2343
[7] F. Faa di Bruno [1825--1888] ,Quarterly Journal of Mathematics vol. I ,p.359.
[8] Fujisawa , Journal of the College of Science, Imperial University of Tokyo, 4,p.174.
[9] A. Genocchi [1817--1889], and G.Peano [1858--1932], [i] Calcolo Differenziale[/i], p.52. Also see
http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=02840002&seq=1
[10] E. Goursat  [1858--1936], [i] A Course in Mathematical Analysis[/i], vol.I ,p.34 . See
http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath&idno=ABA9351
[11] U.H.Meyer , [i]Sur les derivees d'une fonction de fonction,[/i] Grunerts Archiv der Mathematik und Physics, 9 (1847) 96--100.
[12] L. Saalschutz [1835--1913], [i]Vorlesungen ueber die Bernoullischen Zahlen[/i], 1893, p.67. See also
http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=00450002&seq=1
[13]O.X. Schlomilch [1823--1901] , Zeitschrift fuer Mathematik und Physik, vol.3,p.65.
[14] I.J. Schwatt ,[i]An introduction to the operations with series[/i],First edition 1924 , Second edition,reprinted by  Chelsea Publishing Company, New York, p. 1.
[15] O.Stolz [1842--1905] , [i] Grundzuege der Differential und Integralrechnung [/i], p.121. See also
http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=02870002&seq=1
[16] I. Todhunter [1820--1884],[i] Differential Calculus[/i], Macmillan,London ,1878 p.148.  See also
http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=03380002&seq=1
[17] Wiliamson ,[i] Differential Calculus [/i], p.88.
[18] Worpitzki , [i]Lehrbuch der Differential und Integralrechnung  [/i], vol. I , p.140.

[i]Dear sir (flip2004)[/i] ! Thanks you for these elegant solutions, these admirable generalizations with a generously bibliography which together are a choice present. You are a kindly man. Respectfully and with friendship, [i]Virgil Nicula. [/i]


linear algebra
matrix
linear algebra solved

Let $A,B$-be hermit(symmetric) matrices, one of which,say $A$, is positively defined. Prove that $Spec(AB)$-real

Suppose that $ \lambda $ is an eigenvalue of of $ AB $. Then there exists
$ v \in \mathbb{C}^{n} $ such that $ ABv = \lambda v $. Then
$ (ABv,Bv) = \lambda (v,Bv) $. Now, because $ A,B $ are symmetric we have that
$ (v,Bv) , (ABv,Bv) \in \mathbb{R} $. Now, if  $ (v,Bv) = 0 $ then
$ (ABv,Bv) = 0 $ and because $ A $ is positive we get that $ Bv = 0 $, therefore
$ \lambda v = ABv = 0 $, so $ \lambda = 0 $, and hence real.
Now, if $ (v,Bv) \neq 0 $, then $ \lambda = \frac{(ABv,Bv)}{(v,Bv)} \in \mathbb{R} $.
So in any case $ \lambda $ is real.

$ A$ is positively defined show that the exist matrix $ C$ is hermit(symmetric) matrices and  positively defined such that : $ A\equal{}C^2$ since
        $ SP_c(AB)\equal{}SP_c(C^2B)$=$ SP_c(C(CB))\equal{}SP_c(CBC) \subset R$ by $ CBC$ is hermit(symmetric) matrices (use $ SP_c(MN)\equal{}SP_c(NM)$)


[hide="Solution"] Multiplying together,

$(x+y)^{2}(y+z)^{2}(z+x)^{2}= 8100 \implies$
$(x+y)(y+z)(z+x) = 90$

Dividing each of the original equations into the above,

$y+z = 15$
$z+x = 3$
$x+y = 2$

Adding the equations,

$2x+2y+2z = 20 \implies$
$x+y+z = \boxed{10}$

[b]Remark:[/b]  Originally I was going to suggest that you could do this problem easily by assuming that the quantities involved were all integer, but this solution by symmetry does not require that assumption, and it is algebraically very tidy. [/hide]

[quote="anyone"]Find $|x+y+z|$ given the following:
$(x+y)(x+z)=6$
$(x+y)(y+z)=30$
$(y+z)(x+z)=45$

$A.10$  $B.18$  $C.20$  $D.24$  $E.27$[/quote]
[hide]Is the answer A? If it is I'll post my solution later because its really long and I hate it when posting a long solution and then someone tells you its wrong.[/hide]


Heights of the acute triangle ABC intersect at the point O. It is known
that OC = AB. Find the angle ACB.

C=  45 degrees    use the formula  that  OC=2R COS C


inequalities
number theory
inequalities proposed

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {b \plus{} c}{a} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2?$

I don't think it is 2 the minimum. :wink:[/quote]
Yeah, the inequality must be (I think and I have proved it)
\[ \sum \frac{a}{b\plus{}c} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2\]

[quote="can_hang2007"]
Yeah, the inequality must be (I think and I have proved it)
\[ \sum \frac {a}{b \plus{} c} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2
\]
[/quote]

$ LHS \geq\ \sum \frac {a}{b \plus{} c} \plus{}\frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 2$
(obviously with Schur ineq)
 :)

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {b \plus{} c}{a} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2?$

I don't think it is 2 the minimum. :wink:

[url]http://reflections.awesomemath.org/2007_6/trig_substitution.pdf[/url][/quote]
We can easily find the minimum of
\[ P\equal{}\sum \frac{b\plus{}c}{a} \plus{}k \sqrt{\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\]
my friend ;) :)

[quote="can_hang2007"][quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {b \plus{} c}{a} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2?$

I don't think it is 2 the minimum. :wink:[/quote]
Yeah, the inequality must be (I think and I have proved it)
\[ \sum \frac {a}{b \plus{} c} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2
\]
[/quote]
Since I proved it before, I assure that if we replace 2 by k=<2 into both sides, then the new inequality is still true(proving it in a number theory way, not inequality way though).So we can conclude that 2 is the best constant for the can_hang2007's inequality(in fact, it is actually a Bui Viet Anh's inequality).In addition, the best constant for this inequality is square root(2) if we keep the RHS=2(can anybody check this result since I don't remember my proof is true or not(3 years ago already)) :oops:

[quote="can_hang2007"][quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {b + c}{a} + 2\sqrt {\frac {abc}{(a + b)(b + c)(c + a)}} \geq 2?$

I don't think it is 2 the minimum. :wink:

[url]http://reflections.awesomemath.org/2007_6/trig_substitution.pdf[/url][/quote]
We can easily find the minimum of
\[ P = \sum \frac {b + c}{a} + k \sqrt {\frac {abc}{(a + b)(b + c)(c + a)}}
\]
my friend ;) :)[/quote]

Do you use AM-GM this way?

$ \sum \frac {b + c}{a} + k \sqrt {\frac {abc}{(a + b)(b + c)(c + a)}}=
  \sum \frac {b + c}{a} + \frac {k}{2} \sqrt {\frac {abc}{(a + b)(b + c)(c + a)}}+ \frac {k}{2} \sqrt {\frac {abc}{(a + b)(b + c)(c + a)}} \geq 5K$

where $ K^5=k^2/4$

[quote="manlio"][quote="can_hang2007"][quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {b \plus{} c}{a} \plus{} 2\sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 2?$

I don't think it is 2 the minimum. :wink:

[url]http://reflections.awesomemath.org/2007_6/trig_substitution.pdf[/url][/quote]
We can easily find the minimum of
\[ P \equal{} \sum \frac {b \plus{} c}{a} \plus{} k \sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}
\]
my friend ;) :)[/quote]

Do you use AM-GM this way?

$ \sum \frac {b \plus{} c}{a} \plus{} k \sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \equal{} \sum \frac {b \plus{} c}{a} \plus{} \frac {k}{2} \sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \plus{} \frac {k}{2} \sqrt {\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \geq 5K$

where $ K^5 \equal{} k^2/4$[/quote]
No, my friend.
Put $ t\equal{}\sqrt{\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}} \ge 2\sqrt{2}$, then we have that
\[ P\equal{}t^2\plus{}\frac{k}{t} \minus{}2\]
;)


Is it so difficult or so uninteresting?  :D

[quote="Armo"]Let ABC is a triangle .Prove that $\sqrt{r_{a}r_{b}}+\sqrt{r_{b}r_{c}}+\sqrt{r_{c}r_{a}}\geq{9r}$ where $r_{a},r_{b},r_{c}$ are exradiis of ABC and r is inradii.[/quote]

I'm doing again what I should not do, namely posting solutions into the Pre-Olympiad section.

[hide="Click here to reveal the solution of the problem"]Actually, let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Then, we know that $r_br_c = s\left(s-a\right)$, $r_cr_a = s\left(s-b\right)$, $r_ar_b = s\left(s-c\right)$ and $r=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$. Denoting x = s - a, y = s - b, z = s - c, we have

  x + y + z = (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s,

and thus $r_br_c = s\left(s-a\right)=\left(x+y+z\right)x$ and similarly $r_cr_a = \left(x+y+z\right)y$ and $r_ar_b = \left(x+y+z\right)z$. Also,

$r=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}=\sqrt{\frac{xyz}{x+y+z}}$.

We have to prove the inequality

$\sqrt{r_br_c}+\sqrt{r_cr_a}+\sqrt{r_ar_b} \geq 9r$.

Using the formulas $r_br_c = \left(x+y+z\right)x$, $r_cr_a = \left(x+y+z\right)y$, $r_ar_b = \left(x+y+z\right)z$ and $r=\sqrt{\frac{xyz}{x+y+z}}$, this inequality simplifies to

$\sqrt{\left(x+y+z\right)x}+\sqrt{\left(x+y+z\right)y}+\sqrt{\left(x+y+z\right)z} \geq 9 \sqrt{\frac{xyz}{x+y+z}}$.

In other words:

$\sqrt{x+y+z} \cdot \left(\sqrt x + \sqrt y + \sqrt z\right) \geq 9 \sqrt{\frac{xyz}{x+y+z}}$.

Multiplying this with $\sqrt{x+y+z}$, we get

$\left(x+y+z\right) \cdot \left(\sqrt x + \sqrt y + \sqrt z\right) \geq 9 \sqrt{xyz}$.

But this is almost trivial: AM-GM, applied firstly to the numbers x, y, z and secondly to the numbers $\sqrt x$, $\sqrt y$, $\sqrt z$, yields

$x+y+z \geq 3\sqrt[3]{xyz}$;
$\sqrt x + \sqrt y + \sqrt z \geq 3\sqrt[3]{\sqrt x \cdot \sqrt y \cdot \sqrt z}$,

and multiplication of these two inequalities yields

$\left(x+y+z\right) \cdot \left(\sqrt x + \sqrt y + \sqrt z\right) \geq 9 \sqrt[3]{xyz\cdot \sqrt x \cdot \sqrt y \cdot \sqrt z}$
$= 9 \sqrt[3]{x^{\frac32}y^{\frac32}z^{\frac32}} = 9 x^{\frac12}y^{\frac12}z^{\frac12} = 9\sqrt{xyz}$,

what is exactly our inequality. $\blacksquare$[/hide]

  Darij

Sorry for the newb post but wut does it mean by

"exradii" and "inradii"

I tried searching the two up but no results...

and sorry again, but wutz a "terahedron"?  (its my pure laziness to search up what this is)

[quote]Sorry for the newb post but wut does it mean by

"exradii" and "inradii"

I tried searching the two up but no results...
[/quote]
Exradiis are the radiis of the excircles, and inradii is the radius of the incircle.


"How many equilateral triangles can be formed within the same plane using at least two vertices that are also vertices of a given regular hexagon?"
 - 2001 State Team Round

[hide="Answer"]26

[url=http://googleads.g.doubleclick.net/pagead/iclk?sa=l&ai=BkmHR_VKQSZvnB4r4jATby_nCBuqWlReG7KnlAcCNtwGA8QQQARgBIL_SqAwoAzgAUOKQ6ZL6_____wFgya6tiPSj-A-gAYjCiv8DsgENcG9kLmNvbW9qLmNvbboBCTcyOHg5MF9hc8gBAdoBFWh0dHA6Ly9wb2QuY29tb2ouY29tL6gDAcgDB-gDtAPoA_MC9QMAAAAEmAQA&num=1&adurl=http://www.artofproblemsolving.com&client=ca-pub-7709415994183804&nm=3]Click Here[/url]

[/hide]

You can have $ 12$ triangles with a side length equal to the side length of the hexagon. $ 6$ with the third vertice outside of the hexagon and $ 6$ without (Try It!).

You can have $ 8$ triangles with a side length equal to the distance between any two points on the hexagon separated by one point. $ 6$ with the third vertice outside of the hexagon and $ 2$ without (Try It!).

You can have $ 6$ triangles using the diameter of the hexgon. $ 3$ with the third vertice outside of the hexagon and $ 3$ without (Try It!).

$ 12 \plus{} 8 \plus{} 6 \equal{} \boxed{26}$


$ a,b,c > 0$ and $ a \sqrt {a} \plus{} b \sqrt {b} \plus{} c \sqrt {c} \equal{} 1$ Then prove

$ \boxed{a^2 \plus{} b^2 \plus{} c^2 \ge \frac {1}{\sqrt [3] {3}}}$

Just use the power mean inequality: $ \left(\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{3}\right)^{\frac{1}{2}}\geqslant \left(\frac{a^{\frac{3}{2}}\plus{} b^{\frac{3}{2}}\plus{} c^{\frac{3}{2}}}{3}\right)^{\frac{2}{3}}$.

Applying AM-GM inequality, we have
$ 3.a^2\plus{}\frac{1}{3.\sqrt[3]{3}}\ge\ a^{\frac{3}{2}}.\frac{4}{\sqrt[3]{3}}$
and the 2 permutations.


Interesting how for some problems you have so precise information and for the others you give nothing... but I don't care anyway. The problem is really nice, proposed by Constantin Caragea for a Gazeta Matematica contest. I think that among all problems posted in the latest days this one is the most interesting.

Indeed , we have a stronger ineq put in Old and New ineqs (problem 19d , p.10) . It is :
  $ abc(a+b+c) \leq \frac{1}{2}+2a^2b^2c^2 $


Three brothers and their wives visited a friend in hospital. Each person made just one visit, so that there were six visits in all. Some of the visits overlapped, so that each of the three brothers met the two other brothers' wives during a visit. Show that one brother must have met his own wife during a visit.

Can someone please help :)

[hide="Solution"]Call the brothers $ A,B,C$ and their wives $ a,b,c$. There are three moments you know must exist.
1. $ a,b,C$ all in simultaneously.
2. $ a,B,c$ all in simultaneously.
3. $ A,b,c$ all in simultaneously.
If any of the moments occurred at the same time, done. Else, order the three moments chronologically: first, middle, last. Then some wife was in there in the first and last of these moments, and thus the middle. Her husband was in the room in that middle moment also. Done.[/hide]


ratio
geometry
geometric transformation
homothety
circumcircle
perpendicular bisector

more like intermediate?
[hide="hint"]circumscribe a circle around CEFB.[/hide]

[quote="parijat"]Let D,E,F be the feet of altitude from A,B,C in a triangle ABC.Prove that the perpendicular bisector of EF also bisects BC.
[/b][/quote]
here is the outline:
ABC,AFE similar.Dilatate AFE with ratio -1,center A.Then use homothety with ABC and midpoint of AFE
does anyone have simple euclidean solution?
moon

easily show the midpoint of BC is the circumcenter of EFBC


inequalities
geometry
inequalities proposed

Yes ,it is nice! We use the Ravi substitutions $a=y+z,b=x+z,c=x+y$.We use also the well known formulas and we will have to  prove $ 4\sum xy\geq 4\sqrt{3xyz\sum x}+\frac{\sum xy}{2}-\frac{\sum\sqrt{xy(y+z)(z+x)}}{4}$.Now if we prove that $\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{x}}\geq 2\sqrt{3\sum_{cyc}x}$ we are done.Note that $ 2\sqrt{\frac{(x+y)(x+z)}{x}}=2\sqrt{\sum x+\frac{yz}{x}}\geq{\sqrt{3\sum x}+\frac{\sqrt{yz}}{\sqrt{x}}}$.We add up the similar ones so it remains to prove $\sum xy\geq \sqrt{3xyz\sum x}$ which is simple.(QED) :lol:

Yes thank you mateivld, that's a good proof ! My solution is using Finsler-Hadwiger's inequality for the triangle created by three excenter $J_{a},J_{b},J_{c}$  :)
A similar problem:
$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+\frac{(b^{2}-c^{2})^{2}+(c^{2}-a^{2})^{2}+(a^{2}-b^{2})^{2}}{a^{2}+b^{2}+c^{2}}$
hope you are interested in !! :)

This one is also nice.Let us denote $\sum a^{2}b^{2}=x$ and $4S=s$.We have to prove $(2x+s)^{2}\geq 3s(4x-s)$ so we have to prove $x^{2}-2xs+s^{2}\geq 0$ equivalent to $(x-s)^{2}\geq 0$.(QED) :lol:

[quote="gemath"]yes that a good and nice proof, too! I still have this problem:
$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+\sum (b-c)^{2}+16Rr(\sum (cos^{2}\frac{A}{2}-cos\frac{B}{2}cos\frac{C}{2})+\sum (sin\frac{A}{2}-sin(\frac{\pi}{4}-\frac{A}{2}))$
I see it isn't nice but I still hope you are interested in!  :)  :)[/quote]
Are you sure about
$sin\frac{A}{2}-sin(\frac{\pi}{4}-\frac{A}{2}))$?
First it is not homomorphism.
Second it is not equal to 0,when $A=B=C$

zhaobin sorry I type error the fact it is:
$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+\sum (b-c)^{2}+16Rr(\sum cos^{2}\frac{A}{2}-cos\frac{B}{2}cos\frac{C}{2})+16r^{2}(\sum sin(\frac{\pi}{4}-\frac{A}{4})-\sum (sin\frac{A}{2})$
Thank you!

[quote="gemath"]Yes thank you mateivld, that's a good proof ! My solution is using Finsler-Hadwiger's inequality for the triangle created by three excenter $J_{a},J_{b},J_{c}$  :)

Can you post it? I'm really interrested to see it. :)

Hi Cezar Lupu thank you for your interest! My solution relys on a property quite nice of triangle:
We know the famous ineq $a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S(1)$ and $\triangle J_{a}J_{b}J_{c}$(denote $a',b',c',S'$ is three side and area of $\triangle J_{a}J_{b}J_{c}$)has a property :
$a'^{2}+b'^{2}+c'^{2}-4\sqrt{3}S'=\frac{2R}{r}(a^{2}+b^{2}+c^{2}-4\sqrt{3}S-(b-c)^{2}-(c-a)^{2}-(a-b)^{2})$
through this properties we see if we use $(1)$ for $\triangle J_{a}J_{b}J_{c}$ we will have Finsler-Hadwiger's inequality thus what happens when we use Finsler-Hadwiger's inequality for $\triangle J_{a}J_{b}J_{c}$?? We will have
$0\le a'^{2}+b'^{2}+c'^{2}-4\sqrt{3}S'-\sum(b'-c')^{2}=\frac{2R}{r}(a^{2}+b^{2}+c^{2}-4\sqrt{3}S-(b-c)^{2}-(c-a)^{2}-(a-b)^{2})-\sum(b'-c')^{2}$
note $a'=4Rcos\frac{A}{2}$ then we have my ineq
by this method we will creat infinite ineq stronger Finsler-Hadwiger's inequality but it isn't nice, and $\triangle J_{a}J_{b}J_{c}$ has a nice property:
[b]If we have a ineq on $\triangle ABC$ and use it for $\triangle J_{a}J_{b}J_{c}$ we will get a stronger ineq[/b] The fact is quite plain because we see three angle of $\triangle J_{a}J_{b}J_{c}$ is $\frac{\pi}{2}-\frac{A}{2},\frac{\pi}{2}-\frac{B}{2},\frac{\pi}{2}-\frac{C}{2}$
they more "smooth" than three angle of $\triangle ABC$  :)  :)

[quote=gracen307][quote=sqing][quote=gemath]Given $\triangle ABC$ prove that
$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+(b-c)^{2}+(c-a)^{2}+(a-b)^{2}+16Rr(\sum cos^{2}\frac{A}{2}-cos\frac{B}{2}cos\frac{C}{2})$ :)[/quote]
See also here
http://www.mathematica.gr/forum/viewtopic.php?f=50&t=10139[/quote]
(1)We have
$$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+4R^2{\cdot}\sum{(sin{B}-sin({C})^2}+\frac{8}{9}{\cdot}Rr\sum{ (cos{B}-cos{C})^2}$$
where $4,\frac{8}{9}$ are the best.
(2)
$$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+4R^2{\cdot}\sum{(sin{B}-sin{C})^2}+8Rr{\cdot}\sum{ (cos\frac{B}{2}-cos\frac{C}{2})^2}+\frac{2}{5}r^2{\cdot}\sum{ (cos{B}-cos{C})^2}$$
(3)Find the maxmum k such that
 $$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+4R^2{\cdot}\sum{(sin{B}-sin{C})^2}+8Rr{\cdot}\sum{ (cos\frac{B}{2}-cos\frac{C}{2})^2}+kr^2{\cdot}\sum {(cos{B}-cos{C})^2}$$[/quote]
(4)prove or disprove that
$$a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S+4R^2{\cdot}\sum{(sin{B}-sin{C})^2}+8Rr{\cdot}\sum{ (cos\frac{B}{2}-cos\frac{C}{2})^2}+(8\sqrt{2}-4\sqrt{3}-4)Rr{\cdot}\sum {(sin{B}-sin{C})^2}$$


algebra
polynomial
complex numbers
Divisibility Theory

[quote="Peter"]Suppose that $4^{n}+2^{n}+1$ is prime for some positive integer $n$. Show that $n$ must be a power of $3$.[/quote]

Copying [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47701]my proof from MathLinks[/url]:

Consider the polynomial $\Phi_3\left(x\right)=x^2+x+1$. We will first show:

[color=blue][b]Lemma 1.[/b] If k is a positive integer not divisible by 3, then $\Phi_3\left(x^k\right)$, written as a polynomial of x, is divisible by the polynomial $\Phi_3\left(x\right)$ over $\mathbb{Z}\left[x\right]$; in other words, there exists a polynomial u with integer coefficients such that $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$.[/color]

[i]Proof of Lemma 1.[/i] The complex zeroes of the polynomial $\Phi_3\left(x\right)=x^2+x+1$ are the two primitive 3rd roots of unity, i. e. the two complex numbers z such that $z^3=1$ but $z\neq 1$ (each being a zero with multiplicity 1). Let u be one of these two zeros. Then, $u^3=1$ but $u\neq 1$. Now, since k is an integer, $\left(u^k\right)^3=u^{3k}=\left(u^3\right)^k=1^k=1$. On the other hand, since k is not divisible by 3, it is coprime with 3, so we can find integers $\alpha$ and $\beta$ such that $\alpha\cdot k+\beta\cdot 3=1$. Hence, if we had $u^k=1$, then we would have $u=u^1=u^{\alpha\cdot k+\beta\cdot 3}=u^{\alpha\cdot k}\cdot u^{\beta\cdot 3}=\left(u^k\right)^\alpha\cdot\left(u^3\right)^\beta=1^{\alpha}\cdot 1^{\beta}=1$, contradicting $u\neq 1$. Thus, we must have $u^k\neq 1$. [Well, I know this argumentation was overkill, but it's generalizable.] Thus, the number $u^k$ satisfies $\left(u^k\right)^3=1$ but $u^k\neq 1$. Hence, $u^k$ is also a primitive 3rd root of unity, i. e. a root of the polynomial $\Phi_3\left(x\right)$. In other words, $\Phi_3\left(u^k\right)=0$. But this can also be interpreted as follows: The number u is a root of the polynomial $\Phi_3\left(x^k\right)$. Hence, we have showed that, if u is a root of the polynomial $\Phi_3\left(x\right)$, then u is also a root of the polynomial $\Phi_3\left(x^k\right)$. Thus, the polynomial $\Phi_3\left(x^k\right)$ must be divisible by the polynomial $\Phi_3\left(x\right)$. Since both polynomials have integer coefficients and the leading coefficients are both 1, it follows that the quotient is a polynomial with integer coefficients, i. e. there exists a polynomial u with integer coefficients such that $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$. Lemma 1 is proven.

As a consequence of Lemma 1, if x and k are positive integers and k is not divisible by 3 and satisfies k > 1, then the number $\Phi_3\left(x\right)$ is a proper divisor of $\Phi_3\left(x^k\right)$. In fact, since x is an integer and $\Phi_3$ and u are polynomials with integer coefficients, the equation $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$ yields that the number $\Phi_3\left(x\right)$ divides the number $\Phi_3\left(x^k\right)$, and in fact it is a proper divisor since it is not equal to 1 (in fact, $\Phi_3\left(x\right)=x^2+x+1>1$) and not equal to $\Phi_3\left(x^k\right)$ (since $\Phi_3\left(x\right)=x^2+x+1<\left(x^k\right)^2+x^k+1=\Phi_3\left(x^k\right)$).

Now, if the number n is not a power of 3, then it has a prime divisor $\neq 3$. Let q be this prime divisor. Then, q is a positive integer, q > 1 (since q is a prime), and q is not divisible by 3 (since q is a prime $\neq 3$), and also, the number n can be written in the form n = qr, where r is some positive integer. Hence, by the above, the number $\Phi_3\left(2^r\right)$ is a proper divisor of $\Phi_3\left(\left(2^r\right)^q\right)$. Thus, $\Phi_3\left(\left(2^r\right)^q\right)$ cannot be a prime. But $\Phi_3\left(\left(2^r\right)^q\right)=\Phi_3\left(2^{qr}\right)=\Phi_3\left(2^n\right)=\left(2^n\right)^2+2^n+1=1^n+2^n+4^n$ must be a prime by the condition of the problem. Hence, n must be a power of 3.

That's all.

  darij

Here's my generalisation from MathLinks:

Let $\Phi_{n}(x)$ denote the $n$-th cyclotomic polynomial, defined on any way you like, e.g. by $x^{n}-1= \prod_{d|n}\Phi_{d}(x)$ for all $n$ or by $\Phi_{n}(x) = \prod_{m \in (\mathbb{Z}/n \mathbb{Z})^{*}}(x-\zeta_{n}^{m})$, with $\zeta_{n}= e^{\frac{2 \pi i}{n}}$ a $n$-th primitive root of unity.

[b]Generalisation:[/b]
[b]Theorem 1[/b]
Let $a \geq 2$ be an integer and $p$ be prime. Then $a^{(p-1)m}+a^{(p-2)m}+...+a^{2m}+a^{m}+1$ can only be prime if $m=p^{k}$ for some $k$.
[u]Proof:[/u]
This is the case $n$ prime of theorem 2.

Remark: this still holds for composite $n$, but then the term is never prime.

[b]Generalisation of generalisation:[/b]
[b]Theorem 2[/b]
Let $a \geq 2$ and $n \geq 1$ be integers. Then for $\Phi_{n}(a^{m})$ to be prime, it is necessary that $m$ has only prime divisors also dividing $n$, except of the case $a=2, n=1$ where it is only necessary for $m$ to be prime.
[u]Proof:[/u]
The case $n=1$ is well know stuff from elementary results on Mersenne numbers. So let $n \geq 2$.
When $m$ is not of the desired type, see theorem 3 for a factorisation of $\Phi_{n}(a^{m})$.
It's enough to show that $\Phi_{mn}(a)>1$ and $\Phi_{kn}(a)>1$ now, since then $\Phi_{n}(a^{m})$ will surely not be prime.
For this it suffices to show that $\Phi_{w}(b)$ is greater than one for all $w,b \geq 2$. This is easy when looking at the roots of $\Phi_{w}(x)$: they all are on the unit circle and $\neq 1$ (the latter by $n \neq 1$), so their distance to $b$ is $>1$, so also the product of those distances, giving what we wanted.

[b]Generalisation of generalisation of generalisation:[/b]
[b]Theorem 3[/b]
Let $a \geq 2$ and $n \geq 1$ be integers. Then the polynomial $\Phi_{n}(x^{m})$ is irreducible iff $m$ is only divisible by prime divisors also dividing $n$; when $q|m=q^{v}k$ with $q\nmid k,n$, two nontrivial factors are explicitely given as $\Phi_{mn}(x)$ and $\Phi_{kn}(x)$.

[u]Proof:[/u]
If $q$ is some prime dividing $m=q^{v}k$ ($q \nmid k$) but not $n$, we see that the two factors proposed have only roots $\Phi_{n}(x^{m})$ has, and they have no common roots (all this may be hard to see at first). Thus these are indeed different factors. Another way to get these factors is by using the following Lemma sometimes.

You can stop reading all that generalisations, since only the already proven part is needed for the proofs of theorem 1 and 2.
But here goes the other direction:

We need a important property of cyclotomic polynomials (only the first part is needed but the other is good to know):
[b]Lemma:[/b]
Let $n$ be any positive integer and $q$ be prime. Then:
- $\Phi_{n}(x^{q}) = \Phi_{nq}(x)$ iff $q|n$
- $\Phi_{n}(x^{q}) = \Phi_{nq}(x) \cdot \Phi_{n}(x)$ iff $q \nmid n$
This is essentially done by comparing the roots of both sides using that $\Phi_{w}(x) = \prod_{a \in (\mathbb{Z}/w \mathbb{Z})^{*}}(x-\zeta_{w}^{a})$.

Back to the problem:
If $m$ has only such prime factors, by repeated use of the Lemma, part 1, we get that $\Phi_{n}(x^{m})=\Phi_{mn}(x)$. And the cyclotomic polynomials are irreducible as was shown in http://www.problem-solving.be/pen/viewtopic.php?t=585 .

Let  $ 3^k \parallel{} n$, write $ n \equal{} 3^k(3q\plus{}r)$ where $ r \equal{} 1,2$
It's easy to see that $ x^2\plus{}x\plus{}1 | x^{6q\plus{}2r}\plus{}x^{3q\plus{}r}\plus{}1$ for all $ q \in N_0 ,r \in \{1,2\}$
thus $ (2^{3^k})^2 \plus{}2^{3^k}\plus{}1 | 4^n\plus{}2^n\plus{}1 \equal{} (2^{3^k})^{6q\plus{}2r}\plus{}(2^{3^k})^{3q\plus{}r}\plus{}1$
since $ 4^n\plus{}2^n\plus{}1$ is prime and $ (2^{3^k})^2 \plus{}2^{3^k}\plus{}1>1$
Hence, $ 3q\plus{}r\equal{}1$ so $ r\equal{}1$ and $ q\equal{}0$.
thus $ n\equal{}3^k$ , a power of $ 3$


If $a,b,c>0$ then we have:

$ | \sum \frac{a-b}{a+b} | <1$.

cheers! :D :D :cool:

Use the fact that |sum (a-b)/(a+b)|=| prod (a-b)/(a+b)| and the later is of course smaller than 1.


vector
advanced fields
advanced fields unsolved

First of all, has there ever been presented a well-order on $ \mathbb{C}$? I searched, but didn't find anything.

Then, how can one find a total order on $ \mathbb{C}$ with the supremum proprety? That is, how can one define a bijection $ \phi: \mathbb{R}\to \mathbb{C}$ such that $ \phi(x)<\phi(y) \iff x<y$?

There is no "constructive" way to describe a well ordering on $ \mathbb C$, you need the axiom of choice (well, only for this cardiniality).

And simply take $ \psi: \mathbb C \to \mathbb R$ to be any bijection and then define for $ a,b \in \mathbb C$ that $ a<b : \iff \psi(a) < \psi(b)$. Then $ \phi=\psi^{-1}$.

For example, which $ \psi$ would satisfy those conditions?

We know there exists one, that suffices ;)
Using the axiom of choice, we can find bases of $ \mathbb R , \mathbb C$ as vector spaces over $ \mathbb Q$. Then both bases must have same cardinality, giving such a bijection (the map is even $ \mathbb Q$-linear then!).

Without it's a bit more complicated, one way would be to find a bijection between $ ]0,1]$ and $ \mathbb N^{\mathbb N}$ (sequences of positive integers). Then splitting that sequence into even and odd entries gives you a bijection between $ ]0,1]$ and $ ]0,1]^{2}$, so all you have to do is to find a bijection between $ ]0,1]$ and $ \mathbb R$, being much easier than the previous.

So it's difficult to constructively find such a bijection. [url=http://mathforum.org/library/drmath/view/61864.html]Here[/url] is discussed the supremum problem, which made me ask this.


geometry
circumcircle
incenter
parallelogram
rhombus
ratio
perpendicular bisector

$ \odot O$ is the Circumcircle of quadrilateral $ ABCD$; $ AC,BD$ intersect at $ E$.
Given $ G,K$ separately are incenter and excenter of $ \triangle ABE$; similar definition $ H,L$;$ I,M$;$ J,N$. Please prove:

(1): quadrilateral $ GJML$ and quadrilateral $ HINK$ both have a Circumcircle $ \odot P$; $ \odot Q$.
(2): Given $ F$ is an arbitrary point on $ \odot O$, the radius of $ \odot P$ and $ \odot Q$ are $ R_1$ and $ R_2$, then $ FP^2 \plus{} FQ^2 \equal{} R_1^2 \plus{} R_2^2$.

$ \angle BGM \equal{} \frac {_1}{^2}\angle BAC \plus{} 90^\circ \equal{} \frac {_1}{^2}\angle BDC \plus{} 90^\circ \equal{} \angle BDM$ $ \Longrightarrow$ $ BGDM$ is cyclic with a circumcircle $ (P_1)$ and similarly, $ BJDL$ is cyclic with a circumcircle $ (P_2)$ $ \Longrightarrow$ $ \overline{EG}\cdot \overline{EM} \equal{} \overline{EB}\cdot \overline{ED} \equal{} \overline{EJ}\cdot \overline{EL}$ $ \Longrightarrow$ $ GJML$ is cyclic with a circumcircle $ (P).$ In exactly the same way, $ BIDK, BHDN$ are cyclic with circumcircles $ (Q_1), (Q_2)$ $ \Longrightarrow$ $ \overline{EI}\cdot \overline{EK} \equal{} \overline{EB}\cdot \overline{ED} \equal{} \overline{EH}\cdot \overline{EN}$ $ \Longrightarrow$ $ HINK$ is also cyclic with a circumcircle $ (Q).$ Powers of $ E$ to $ (O), (P), (Q)$ are equal.

Let $ U \equiv BC \cap DA,$ $ V \equiv AB \cap CD.$ Let arbitrary line $ u$ through $ U$ meet $ AB, CD$ at $ U_1, U_2$ and let arbitrary line $ v$ through $ V$ meet $ BC, DA$ at $ V_1, V_2.$ By repeated Menelaus theorem, $ U_1V_1, U_2V_2, AC$ are concurrent and $ V_1U_2, V_2U_1, BD$ are also concurrent. If $ U_1V_1U_2V_2$ happens to be a parallelogram with $ U_1V_1 \parallel U_2V_2$ and $ V_1U_2 \parallel V_2U_1,$ then $ U_1V_1 \parallel U_2V_2 \parallel AC$ and $ V_1U_2 \parallel V_2U_1 \parallel BD.$ Assume now that $ u, v$ are internal angle bisectors of $ \angle BUC, \angle AVB$ intersecting at $ W.$ By a simple angle chase for the cyclic $ ABCD,$ $ \angle UWV$ is right, hence $ W$ is common midpoint of $ U_1U_2, V_1V_2,$ $ \Longrightarrow$ $ U_1V_1U_2V_2$ is a rhombus with $ U_1V_1 \parallel U_2V_2 \parallel AC,$ $ V_1U_2 \parallel V_2U_1 \parallel BD.$ Consequently, $ u, v$ are parallel to the internal angle bisectors $ k \equiv KGEIM, l \equiv LHEJN$ of $ \angle AEB, \angle BEC.$

$ o \equiv P_1P_2OQ_2Q_1$ is perpendicular bisector of $ BD.$ Perpendicular bisectors of $ AB, BG$ and the angle bisector $ k$ concur at $ X$ on the circumcircle $ \odot(ABE),$ while perpendicular bisectors of $ CD, DI$ and the angle bisector $ k$ concur at $ Y$ on the circumcircle $ \odot(CDE).$ $ \angle OXP_1 \equal{} \angle ABG \equal{} \frac {_1}{^2}\angle ABC \equal{} \frac {_1}{^2} (180^\circ \minus{} \angle P_1OX)$ $ \Longrightarrow$ $ \triangle XOP_1$ is isosceles with $ OX \equal{} OP_1$ and similarly, $ \triangle YOQ_1$ is isosceles with $ OY \equal{} OQ_1.$ By similar angle chase, $ \angle OXY \equal{} \angle AVW \equal{} \frac {_1}{^2} \angle AVD \equal{} \frac {_1}{^2}(180^\circ \minus{} \angle YOX)$ $ \Longrightarrow$ $ \triangle XOY$ is also isosceles with $ OX \equal{} OY.$ Combining, $ \overline{OP_1} \equal{} \minus{} \overline{OQ_1}.$ In exactly the same way, $ \overline{OP_2} \equal{} \minus{} \overline{OQ_2}$ and combining, $ \overline{P_1P_2} \equal{} \minus{} \overline{Q_1Q_2}.$

$ P$ is intersection of perpendicular bisectors of $ GM,$ $ JL$ through $ P_1, P_2,$ which are parallel to $ l, k,$ respectively, while $ Q$ is intersection of perpendicular bisectors of $ IK, HN$ through $ Q_1, Q_2,$ which are also parallel to $ l, k,$ respectively. Consequently, the right $ \triangle PP_1P_2 \cong \triangle QQ_1Q_2$ are centrally congruent with similarity center $ O$ and coefficient $ \minus{} 1$ $ \Longrightarrow$ $ \overline{OP} \equal{} \minus{} \overline{OQ}.$ Centers of $ (O), (P), (Q)$ are therefore collinear and their pairwise radical axes parallel, but their radical axes all go through $ E,$ so that these 3 circles are coaxal. If $ F \in (O),$ then powers of $ F$ to $ (P), (Q)$ are in the ratio $ \frac {FP^2 \minus{} R_1^2}{FQ^2 \minus{} R_2^2} \equal{} \frac {\overline{OP}}{\overline{OQ}} \equal{} \minus{} 1$ $ \Longrightarrow$ $ FP^2 \plus{} FQ^2 \equal{} R_1^2 \plus{} R_2^2.$


Find all functions $f: \mathbb{R}\mapsto \mathbb{R}$ such that
$f(f(x)^{2}+f(y))=xf(x)+y$ for all $x,y$.

Let $x=y=0 \Longrightarrow f(f(0)^{2}+f(0))=0$
So there is a $t \in R$ such that $f(t)=0$
If $f(x)=f(y)$ then $f(f(a)^{2}+f(x))=f(f(a)^{2}+f(y))$
$af(a)+x=af(a)+y \Longrightarrow x=y$.So $f$ is injective.
$x=y=t \Longrightarrow f(f(t)^{2}+f(t))=tf(t)+t \Longrightarrow f(0)=t \Longrightarrow f(f(0))=f(t)=o$
$x=f(0) \Longrightarrow f(f(f(0))^{2}+f(y))=f(0)f(f(0))+y \Longrightarrow f(f(y))=y$
$x=0 \Longrightarrow f(f(0)^{2}+f(y))=y=f(f(y)) \Longrightarrow f(0)=0$
$y=0 \Longrightarrow f(f(x)^{2})=xf(x) \Longrightarrow f(f(f(x))^{2})=f(x)f(f(x)) \Longrightarrow f(x^{2})=xf(x) \Longrightarrow f(x^{2})=f(f(x)^{2}) \Longrightarrow f(x)^{2}=x^{2}$
Let $x,y$ be nonzero reals, then
$f(f(x)^{2}+f(y))=xf(x)+y$
$(f(f(x)^{2}+f(y)))^{2}=(xf(x)+y)^{2}$
$(f(x)^{2}+f(y))^{2}=(xf(x)+y)^{2}$
$f(x)^{4}+2f(x)^{2}f(y)+f(y)^{2}=x^{2}f(x)^{2}+2xyf(x)+y^{2}$
$f(x)f(y)=xy$ $(*)$
We have $f(1)^{2}=1$
If $f(1)=1$ then take $y=1$ in $(*)$ $\Longrightarrow f(x)=x$
If $f(1)=-1$ then take $y=1$ in $(*)$ $\Longrightarrow f(x)=-x$


In a $ n$-gon $ A_1A_2A_3...A_n$,the smallest interior angle is $ \angle A_1\equal{}120^\circ$ and other interior angles satisfy $ \angle A_{k\plus{}1}\equal{}\angle A_k\plus{}5^\circ$ $ (k\equal{}1,2,3,...,n\minus{}1)$,determin the possible $ n$.

[quote="stevenmeow"]hint: express the average of the angles in terms of k and set it equal to the measure of an interior angle in a regular k-gon
solution:
[hide]the AM (average) of the angles can be represented as $ 117.5 \plus{} 2.5k$ so we get the equation
$ 180 \minus{} \frac {360}{k} \equal{} 117.5 \plus{} 2.5k$
$ 62.5 \minus{} 2.5k \equal{} \frac {360}{k}$
$ 125k \minus{} 5k^2 \equal{} 720$
$ k^2 \minus{} 25k \plus{} 144 \equal{} 0$
factor...
$ (k \minus{} 9)(k \minus{} 16) \equal{} 0$
$ k \equal{} 9,16$[/hide][/quote]

[hide]That is necessary condition,but not sufficient one at least in this problem.[/hide]

[hide]Another way would be to consider the measures of the exterior angles. These must add up to $ 360$, and it's just computing a partial sum of an arithmetic series to figure out what their sum is for each $ n$. The result is that $ n \equal{} 9$ is the only solution.[/hide]

[quote="stevenmeow"]but why can't the angles be more than 180?[/quote]

Try drawing such a $ 16$-gon,then you should notice.


Find all functions $f: \mathbb R^{+} \to \mathbb R^{+}$ such that
\[f(xf(x) \plus{} f(y)) \equal{} f^2(x) \plus{} y\]
for any $x, y>0$.

[quote="moriarti"]Find all functions $ f: R^ \plus{} \minus{} > R^ \plus{}$ such that

$ f(xf(x) \plus{} f(y)) \equal{} f^2(x) \plus{} y$

for any x, y>0.[/quote]
Let $ P(x,y)$ be the assertion $ f(xf(x)\plus{}f(y))\equal{}f(x)^2\plus{}y$
Let $ m\equal{}\inf f(\mathbb R^\plus{})$

$ f(y_1)\equal{}f(y_2)$ $ \implies$ (comparing $ P(x,y_1)$ and $ P(x,y_2)$) $ y_1\equal{}y_2$ and $ f(x)$ is injective.

$ \forall u>m^2$, $ \exists x$ such that $ m\leq f(x)<\sqrt u$. Then $ u\minus{}f(x)^2>0$ and $ P(x,u\minus{}f(x)^2)$ $ \implies$ $ f(\text{something})\equal{}u$ and so $ (m^2,\plus{}\infty)\subseteq f(\mathbb R^\plus{})$

$ P(x,f(y)^2)$ $ \implies$ $ f(xf(x)\plus{}f(f(y)^2))\equal{}f(x)^2\plus{}f(y)^2$
$ P(y,f(x)^2)$ $ \implies$ $ f(yf(y)\plus{}f(f(x)^2))\equal{}f(y)^2\plus{}f(x)^2$
So $ f(xf(x)\plus{}f(f(y)^2))\equal{}f(yf(y)\plus{}f(f(x)^2))$ and, since $ f(x)$ is injective, $ xf(x)\plus{}f(f(y)^2)\equal{}yf(y)\plus{}f(f(x)^2)$
So $ xf(x)\minus{}f(f(x)^2)\equal{}yf(y)\minus{}f(f(y)^2)$
So $ xf(x)\minus{}f(f(x)^2)\equal{}a$ for some real $ a$

Then $ P(x,y)$ may be written $ f(f(f(x)^2)\plus{}f(y)\plus{}a)\equal{}f(x)^2\plus{}y$ and, since $ (m^2,\plus{}\infty)\subseteq f(\mathbb R^\plus{})$ :

New assertion $ Q(x,y)$ : $ f(f(x)\plus{}f(y)\plus{}a)\equal{}x\plus{}y$ $ \forall x>m^4$,  $ \forall y>0$

Comparing then $ Q(x\plus{}z,y)$ and $ Q(x,y\plus{}z)$, we get $ f(f(x\plus{}z)\plus{}f(y)\plus{}a)\equal{}f(f(x)\plus{}f(y\plus{}z)\plus{}a)$ $ \forall x>m^4,y>0,z>0$ and, since $ f(x)$ is injective :

$ f(x\plus{}z)\plus{}f(y)\equal{}f(x)\plus{}f(y\plus{}z)$ $ \forall x>m^4,y>0,z>0$

So $ f(y\plus{}z)\minus{}f(y)\equal{}h(z)$ for some function $ h(z)$ independant of $ y$

But $ f(y\plus{}z)\equal{}f(y)\plus{}h(z)$ $ \forall y,z>0$ $ \implies$ $ f(y)\plus{}h(z)\equal{}f(z)\plus{}h(y)$ and so $ f(y)\minus{}h(y)\equal{}b$ for some real $ b$

So $ h(x)\equal{}f(x)\minus{}b$ and we got $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}b$ $ \forall x,y>0$

Let then $ g(x)\equal{}f(x)\minus{}b$ from $ \mathbb R\plus{}\to(\minus{}b,\plus{}\infty)$ : $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$
This is a Cauchy equation for a function which has a lower bound and so $ g(x)\equal{}cx$ (non continuous solutions have neither upper, neither lower bounds).

So $ f(x)\equal{}cx\plus{}b$ and, plugging this back in the original equation, we get $ c\equal{}1$ and $ b\equal{}0$

Hence the unique solution $ \boxed{f(x)\equal{}x}$


inequalities
trigonometry
trig identities
Law of Cosines

Show that for any triangle ABC  a^2 + b^2 +c^2 > \sqrt{3} max \{ |a^2 - b^2|, |b^2 -c^2|, |c^2 -a^2| \} .

 :maybe:

[hide]WLOG let $ a\ge b\ge c$. Then we have to show that $ a^2 \plus{} b^2 \plus{} c^2\ge\sqrt 3(a^2 \minus{} c^2)$, which is equivalent to $ b^2 \plus{} (\sqrt 3 \plus{} 1)c^2\ge (\sqrt 3 \minus{} 1)a^2$. Using the law of cosines this can be written as $ (2 \minus{} \sqrt 3)b^2 \plus{} 2c^2\ge 2(\sqrt 3 \minus{} 1)bc\cos A$. Since $ \cos A\le 1$, we'll be done if we can show that $ (2 \minus{} \sqrt 3)b^2 \plus{} 2c^2\ge 2(\sqrt 3 \minus{} 1)bc$. But this is just $ ((2 \minus{} \sqrt 3)b \minus{} \sqrt 2c)^2\ge 0$. Equality holds if $ b: c \equal{} \sqrt 2: 2 \minus{} \sqrt 3$ and $ \angle A \equal{} 0^\circ$, which is impossible. Hence equality never holds.[/hide]


Ever since treething started using "my head" everyone else started doing that too
Hey, who knows, maybe it makes you win mathcounts national competitions.

Anyway...
Monica and Chandler plan to meet for lunch. However, both have busy schedules. Monica only gets a 20 minute break, and Chandler only gets a 25 minute break. Monica plans to go to Central Perk between 11:30 and 12:30, while Chandler can only make it between 12 and 1. What is the probability that they meet?

[hide]That's what I got, but I think we both might have made a flaw somewhere in our drawing so I am not a 100% sure.[/hide]

I think it has to do with the wording.

I did the problem assuming they can go into the park during those times and leave after the range of time. Is that the right interpretation? :w00tb:

During the break, they warp (lol) to central perk (which is a coffeehouse)--watch friends, and stay there for all of the time that they can. It counts as meeting if one leaves at the exact same time that the other arrives.

Can they stay after their time range?

Like can Chandler go at 1 and stay until 1:25?

[hide="hmm.."]There are 41 different time periods Monica could be at Central Perk (11:30-11:50, 11:31-11:51, and so forth.) There are 36 different time periods for Chandler to be there, using the same logic. This yields 41*36= 1476 different time periods both could be there at. Using mainly brute force :) I got 496 ways they'll see each other, so $\frac{496}{1476}=\frac{124}{369}$.[/hide]

:spam: Any reason you posted the same thing twice?  :spam:

By the way, you're logic isn't correct, at least I don't think so.

this seems like a geometric probability problem, i'm pretty sure a timeline can be made, and then geometric probability can be used with a line....is that close [b]jclarkemathL314159[/b]?


[quote="Anonymous52111"]Need your help here, thanks.

Solve: f/7 - 3 = 1

Solve: 2x/5 - x/3 = 3[/quote]
[hide="first one"]$\frac{f}{7}-3 = 1$
add 3: $\frac{f}{7}=4$
multiply by 7 $f=28$[/hide]
[hide="second one"]$\frac{2x}{5}-\frac{x}{3}= 3$
change to common denominator (15) $\frac{6x}{15}-\frac{5x}{15}=3$
multiply by 15: $6x-5x=45$
$x=45$[/hide]

[quote="Anonymous52111"]Need your help here, thanks.

Solve: f/7 - 3 = 1

Solve: 2x/5 - x/3 = 3[/quote]
[hide="1"]Adding three to both sides you now have $\frac{f}{7}=4$. Multiplying by 7 and  you now have $f=28$.[/hide]
[hide="2"]Find a common deniminator, which is saying find the LCM of 5 and 3. Thats 15. So now we have $\frac{6x}{15}-\frac{5x}{15}=3$. Now you have $\frac{x}{15}=3$. Multiply by 15 to have $x=45$.[/hide]
EDIT: My hides looked exactly like quevvy, so I changed it. A coincidence?

[hide="Another way for the second one"]

Basically is the same:

\begin{eqnarray*}\frac{2}{5}x-\frac{1}{3}x &=& 3 \hfill \\ x\left({\frac{2}{5}-\frac{1}{3}}\right) &=& 3 \hfill \\ x \cdot \frac{1}{{15}}&=& 3 \hfill \\ x &=& 45 \hfill \\ \end{eqnarray*}[/hide]

[quote="Anonymous52111"]Need your help here, thanks.

Solve: f/7 - 3 = 1

Solve: 2x/5 - x/3 = 3[/quote]

[hide] First One:
$\frac{f}{7}=1+3$

$\frac{f}{7}=4$

$f={4}\times{7}$

$f=\boxed{28}$

Second One:
multiply each number or fraction by the LCM: 15

$6x-5x=45$

$x=\boxed{45}$
[/hide]

[quote="myspacesucks"]Do people have to repeat what others said already?  Just do it on pencil and paper if you have to do the problem, don't paste what the other guy wrote for your own reply or add something that another person  did exactly how you did, sheesh![/quote]
We don't know what the other person put until we're done if our replies. That's why everyone puts the hide. So other people can have a try on the question.


The three sides of a triangle have integer lengths. The perimeter is 8 units. What is the area?

Is the area sqroot8?I'm petty sure it aint....

[quote="Andreas"]$8$ is small so it is possible to find triples for which $a + b + c = 8$
Then it is possible to apply triangle inequality.
It should be $a = 2$, $b = 3$, $c = 3$ since the area is $20\sqrt{3}$.[/quote]
 :huh:  Im getting $\sqrt{8}$

[quote="236factorial"][quote="riddler"]
 :huh:  Im getting $\sqrt{8}$[/quote]

That's right, the sides have to be 2, 3, and 3, the semiperimeter is 4, applying herons we get $\sqrt8$.[/quote]

actually we get $2\sqrt2$ but it's the same thing. ;)

I'm going to show the work of doing heron's:
$A = \sqrt{s(s-a)(s-b)(s-c)}$
$A = \sqrt{4(2)(1)(1)}$
$A = \sqrt8$
$A = 2\sqrt2$

You guys got the right length.

I don't think Heron is necessary in here. This is pretty basic triangle so here's how "I" would do.

Fix $\triangle ABC$ with $AB = AC = 3$ and $BC = 2$. Since $\triangle ABC$ is isosceles, the altitude from $A$ is also median. Thus, we can apply Pythagorean Theorem to find out that height $AD$ (where $D$ is a midpoint of $BC$) is $\sqrt{3^2-1^2} = \sqrt{8}$. To find the area of the whole thing, we apply $A = b \cdot h \frac{1}{2}$. $b = 2$ and $h = \sqrt{8}$ so the area is $\sqrt{8} = 2 \sqrt 2$.

YAY!!Igot it right...Ithought my assumption of sides was muuging guesswork :rotfl:

you made assumptions? :huh:  :huh:  :huh:  :huh:  I just used,
$\text{In a triangle 2 sides are always greater than the third side}$

Same thing riddler,i thought  there was a better way to find out th esides,u know...more mathematically :) But that was the only thing possible :lol:

[hide] The side lengths are 2, 3, and 3. $ 3^2-1^1=8$, so $\sqrt{8}\cdot1=\sqrt{8}=\boxed{2\sqrt{2}}$
[/hide]