AMC
AIME
blogs
ratio
trigonometry
geometry
number theory

Okay guys the test is now ready. You can find it on my blog http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=196979

Enter your answers on http://mersenne.zapto.org/Mock_AIME/usercp.php, (courtesy of #H34N1). Please do not discuss the test with people who have not taken the test until Tuesday/Wednesday. Results will be posted then. Good luck!

Thanks for the work!

Though by tuesday/wednesday, the forum will be closed down, and people will be taking the II already so I don't see how much that will help.

Scores are automatic? I'm not sure how the site works. It won't be much help if we don't know our scores before the real thing (for those taking the II). Answers in 001 form instead of 1 right?

Scores are automated. :D

Yes put 001 instead of 1 or your score may not be accurate.

hm okay well I guess i'll try to post results by monday/tuesday, not necessarily in this part of the forum.

For #14, is that angle supposed to be read as:

[quote]...so that $ \angle OAB$ is acute...[/quote]

or is it correct the way it is?

Well, I think it's a positive integer! Looking at the question you will see if n=0 then it leads to,again, infinite number(2,3,4,5,....)

[quote="pianoforte"]Clarification for 9. Let n be the smallest integer...

Positive integer? Or non-negative integer (n=0 trivializes it though).[/quote]

It is obvious what it is supposed to mean. (smallest positive integer).

[quote="aznlord1337"]Um clarification on number 2. Does the question imply that n C k < n C k+1 < n C k+2 only, or does it allow for the reverse case (n C k+2 < n C k+1 < n C k).[/quote]

The first one...

[quote="aznlord1337"]Um clarification on number 2. Does the question imply that n C k < n C k+1 < n C k+2 only, or does it allow for the reverse case (n C k+2 < n C k+1 < n C k).[/quote]

Yes, are those terms respectively in the ratio 1:2:3?

As Altheman said, just the first one.

@#H34N1, it should be $ \angle OAB$ is acute.

i think numbers 1-10 were all pretty straightforward.

wow...i got a 7...

the test seemed easier than the ones i took in practice.

how do we know which ones we got right?

Well, this test isn't hard but I just got 6(although I click"submit" accidentally before typing answer for #14 and #15 :oops: ). The actual problem I didn't get after thinking for a while(around 30-45 minutes) is #11. This problem is tough, I think. Does anybody get #12? Applying some new geometry theorems is absolutely the best way to solve it, in my opinion. :roll:

Ouch, 9, fail. Thought I had 13--Hmm...I should have put more time in...

I only did the first 10 problems because I knew I wouldn't have any time tomorrow for them, and I got an 8.
Darn my stupid mistakes!

um i went to enter my answers and it says that i already took this test and that i got a 0.  :?:

That problem with the equal arcs was probably the hardest.

Yes, I have you marked down for just sending a blank answer list. I'm not sure how to fix this besides asking #H34N1, but you can just pm me your answers instead. I'll score it by hand for you.

If you ever need something cleared, I can do it for you (I think)

Later tonight the individual problems will be open to discussion on my blog since the forum will be closed. Also, answers will be posted.

Late bump: Does anyone have a solution to #[url=http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_1_2007-2008_Problems/Problem_11]11[/url]?

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3717871_2008_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Say you have $n$ bugs, each on a seperate vertex of a regular $n \text{-gon}$ with side length $1$. Starting simultaneously, each bug travels directly towards its neighbor counterclockwise. What is the length one of the bug's path before meeting the others in the center?

wlog 1 unit per second traveling...consider the velocity component going to the center...split into perpendicular vectors, that is
[let x=90(1-n/2)]

cos x velocity component to center

for a distance of 1/2tan x

for a time of 1/2*sin x/cos^2 x

and thus the length of path is r*t=1/2*sin x/cos^2 x

solution from mathworld.com
http://mathworld.wolfram.com/MiceProblem.html

i think this requires some calculus to solve...


LaTeX
inequalities
number theory unsolved
number theory

Let a1,...,an be integers such that for all i,  ai | a(i+1)+a(i-1)   (a(n+1)=a1, a0=an).
Show that the sum of these quotients (a(i+1)+a(i-1))/ai is less or equal to 3n.

Good luck and thanks.

Here's the problem again, in $ \text{\LaTeX}$: Let $ a_1, \ldots, a_n$ be positive integers and suppose that $ q_i = \frac{a_{i-1}+a_{i+1}}{a_i}$ is an integer for all $ 1 \leq i \leq n$, where the indices are taken modulo $ n$ (that is, $ a_0=a_n$ and $ a_{n+1}=a_1$). Prove that $ \sum q_i \leq 3n$.

We will actually improve this a tiny little bit  :) and show $ \sum q_i \leq 3n-1$. The reason we bother with such a minor improvement is that this is actually exactly tight for all $ n \geq 1$: Let $ a_i=i$ for $ i=1,\ldots,n$ and note that now $ q_1=n+2$, $ q_k=2$ for $ 1 < k < n$ and $ q_n=1$ so $ \sum q_i = 3n-1$.

[hide="Proof."]

If all the $ a_i$ are equal, then $ q_i=2$ for all $ i$ and $ \sum q_i = 2n \leq 3n-1$ as required. Otherwise, consider the maximum value $ a = \max a_i$. There may be several $ a_i$ that attain this maximum, but since not all $ a_i$ are equal there must be a $ k$ such that $ a_k=a$ while $ a_{k-1}$ and $ a_{k+1}$ are not both $ a$. Therefore $ a_{k-1}+a_{k+1} < 2a$ and in order for $ q_k$ to be an integer we must have $ a_k=a_{k-1}+a_{k+1}$, so $ q_k=1$.

Our goal is to show that we can throw away this $ a_k$. Indeed, consider $ a_{k-1}$ which divides $ a_{k-2}+a_k = a_{k-2} + a_{k-1} + a_{k+1}$. Whenever $ x|x+y$ we have $ x|y$, and so in fact $ a_{k-1}$ divides $ a_{k-2}+a_{k+1}$, which are precisely the new neighbors of $ a_{k-1}$ after we delete $ a_k$. The same applies to $ a_{k+1}$. So the truncated sequence still satisfies the condition $ q_i \in \mathbb{Z}$ for all $ i$, and it only remains to check how this move impacts the statement $ \sum q_i \leq 3n-1$.

Since $ n$ changed to $ n-1$, $ 3n-1$ decreases by 3. $ q_k=1$ itself disappears (losing 1), while $ q_{k-1}$ changes from $ \frac{a_{k-2}+a_{k-1}+a_{k+1}}{a_{k-1}}$ to $ \frac{a_{k-2}+a_{k+1}}{a_{k-1}}$, losing exactly 1 again. $ q_{k+1}$ also decreases by 1 so overall $ \sum q_i$ decreases by exactly 3. Therefore the inequality $ \sum q_i \leq 3n-1$ is satisfied before the deletion iff it is satisfied after the deletion.

Finally, starting with an arbitrary sequence $ a_1,\ldots, a_n$ we see that we can keep trimming it until all the terms are equal (which happens, at the latest, when we're left only with a single term...). Notice that the deletion argument above works perfectly well even when $ n=2$. Therefore $ q_i \leq 3n-1$ always holds.

Also, starting with $ n=1$ and $ a_1=1$ and doing the trimming in reverse, inserting the sum of two elements in between them at each stage, we obtain various sequences for which equality actually holds, including the example we mentioned above, $ a_i=i$.
[/hide]


trigonometry
vector
calculus
geometry
AMC
AIME
complex numbers
\/closed

What happened to Intermediate Trigonometry and Complex Numbers?

It was renamed to the Precalculus Class.

Yes, we are planning to have a new Precalculus class and textbook in late fall.  This will include the old Interm Trig/Complex Number class material along with new (for us) material on vectors and matrices.

AoPS is also writing a Calculus book which will be released in the fall right?
AoPS must be really busy!

What date would you say about when the books will be released?

If all goes as planned, that is correct.

So will the book be used for the next pre-calculus class?

Yes; this is why the class is starting in December rather than in October -- gives me a little more breathing room to finish the book.

wait so if I take alg 2 and precalculus courses through AoPS then I can take calculus in highschool?

That depends on your high school.

But our calculus class will teach you more ;)

What does the Precalculus book cover in terms of trig and complex numbers?
I have decent knowledge in these areas, I just fail at both of them.

Trig: basics, identities, applications to geometry.

Complex numbers: basics, trig & exponential form, roots of unity, applications to geometry.

Basically, anything you'd see up through the AIME and a bit beyond is covered.

[quote="AIME15"]Pre-Calculus is supposed to come out in Fall 2009, and Calculus is supposed to come out before the year's out.[/quote]

:wink:

If you looked at the Online Classes page, you'd see that there is a Calculus class.

There will be a textbook, too, around year-end.


Prove, non analytically, that 3 points determine a circle.

Not too sure about this...
[hide]Since every 3 noncollinear points determine a triangle, and every traingle can be inscribed in a circle, every three points determine a circle.[/hide]

[quote="t0rajir0u"]ragnarok:  You haven't shown that every triangle can be inscribed in a circle, which is equivalent to the original problem.

[hide="Simple"] Three points determine a circle if we can show that there exists a point equidistant to all three points.

Take two of the points and draw the locus of points equidistant from them, which is a line perpendicular to the line between them.  Now take the third point and one of the first two and draw the locus of points equidistant from them, which is another line perpendicular to the line between them.

These lines intersect at the center of a circle containing all three points if and only if the three points are not collinear. [/hide][/quote]
You also have to prove that the center point is equidistant from the first and third points. This is not too hard though (look for congruent triangles)

Er, I already did.  The intersection of the loci of points equidistant to two pairs out of three points is the loci of points equidistant to all three points.


[quote="guile"]1.	Show that $|x|-|y| \leq | x-y |$
2.	Show that $|y|-|x| \leq |x-y|$
3.	Show that for $\forall n \in Z^{+}, |x_{1}+|x_{2}+\ldots+x_{n}|\leq |x_{1}|+|x_{2}|+\ldots+|x_{n}|$[/quote]

All are based on the main ineq $| x |+|y| \geq |x+y|$

eg $|x-y|+|y| \geq |x-y+y|=|x|$.......

 :)

Can you please do it? I'm asking because I don't know if my solution is right.

I believe he just solved the first two.  For the last one, use basic induction. :)

Masoud Zargar


Three darts are thrown at the figure given, each landing in a different square. What is the probability that the squares they land in form a row, either horizontally, vertically, or diagonally? Express your answer as a common fraction.

FIGURE:

[url=http://mathforum.org/alejandre/magic.square/3x3grid.gif]This is the figure (I know its a picture of a magic square, but its also the same as the figure)[/url]

[hide]Not so sure but you can create 8 rows/diagonals/columns with three squares. $9C3=84$ so it is $\frac{8}{84}=\frac{2}{21}$.[/hide]

[quote="indianamath"][hide]it can be three vertical, three horizontal, or 2 diagonal.  3+3+2=8

$\frac{8}{9C3}=\frac{2}{21}$[/hide][/quote]

Oh My Goodness.
I had it all down, but somehow, it wrote it as $8C3$.
Don't ask  :blush:

[quote="egghead91"]Can you explain your solution? I'd like to know how you solved it.[/quote]

[hide]
Basically, if you count, there are eight ways to get 3 in a row (3 vertical, 3 horizontal, and 2 diagonal). To find the probability, you have to put the desired outcome over the total possible outcomes. The total possible outcomes can be found out by a function called nCr (or how many ways to choose r things out of n). When you do that, you find there are a total 84 ways to chose 3 boxes out of nine, so you get 8/84 which can be simplified as 2/21 [/hide]


rotation
calculus
integration
calculus computations

Let W be the region consisting of all the points ( x , y ) such that  -1 <= x <= 1,  4/(5 + 2x + X^2) <= y <= 1.

Compute the volume of the solid generated when the region W is rotated about the line y = 1 .

Well you know your limits of integration, so that's good.  You also know the radius of your circle, so that's good.  Then you use
$ V\equal{}\pi\int_{a}^{b}\left|f^{2}(x)\minus{}g^{2}(x)\right|\ dx$


The  initial  ineq  is  a  special case of this  one :
[tex]a,b,c>0[/tex]  then [tex]1+2abc+a^2+b^2+c^2\geq2(ab+bc+ca)[/tex] which  has been discust  here .

Yeah, but what about the one i've posted? ;)

[quote="IneqLover"]If $a$,$b$,$c$ are positive reals and $abc=1$
Prove that
    $a^2 +  b^2 + c^2 + 3 \geq 2ab + 2bc + 2ac$[/quote]
I give another solution:
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ which x,y,z are positive.
by calculating, the problem equals to
$x^4z^2+y^4x^2+z^4y^2+3x^2y^2z^2\ge2(x^3y^2z+xy^3z^2+x^2yz^3)$
which can be easily done by AM-GM...

Skyward_Sea,

How to do that by AM-GM?
I can't do it neh...

Why do people like to post apmo 04 no.5 again, again and again..

[quote="socrates"][quote="cezar lupu"]In the same way it can be proved his sister: $a^2+b^2+c^2+3\geq ab+bc+ca+a+b+c$ ;)  ;)[/quote]

How?/
I cannot reach a complete solution   :([/quote]
$\sum a^2 + 3 \ge 2(\sum ab)$ (proved)
$\sum a^2 + 3 \ge 2(\sum a)$ (amgm)

[quote="IneqLover"]If $ a$,$ b$,$ c$ are positive reals and $ abc \equal{} 1.$ Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \geq 2(bc \plus{} ca \plus{} ab).$[/quote]$ a^2 \plus{} b^2 \plus{} c^2 \minus{} 2(bc \plus{} ca \plus{} ab) \plus{} 3\equiv x^6 \plus{} y^6 \plus{} z^6 \minus{} 2\left(y^3z^3 \plus{} z^3x^3 \plus{} x^3y^3\right) \plus{} 3x^2y^2z^2$

$ \equal{} \left(x^2\plus{}y^2\plus{}z^2 \plus{} yz\plus{}zx\plus{}xy\right)\sum{x^2(x \minus{} y)(x \minus{} z)}\geq0,$

which is true by the Schur inequality.

See also : http://www.mathlinks.ro/viewtopic.php?t=18430

Let $ min(a,b,c)\equal{}c$
$ ab(a\plus{}b)>1 \Longrightarrow$
if we take $ a\plus{}b\equal{}x$ and $ ab\equal{}y$
$ (xy\minus{}1)^2 \geq (2y.\sqrt{y}\minus{}1)^2$
$ (2y\sqrt{y}\minus{}1)^2 \geq 4y^3\minus{}3y^2 \Longleftrightarrow$
$ (\sqrt{y}\minus{}1)^2(3y\plus{}2.\sqrt{y}\plus{}1)\geq0$ which is obviously true.
$ (xy\minus{}1)^2 \geq 4y^3\minus{}3y^2 \Longleftrightarrow$
$ a^2\plus{}b^2\plus{}\frac{1}{a^2b^2} \geq 2ab\plus{}\frac{2a\plus{}2b}{ab} \Longleftrightarrow$
$ a^2\plus{}b^2\plus{}c^2\plus{}3 \geq 2ab\plus{}2bc\plus{}2ac$.

[quote="IneqLover"]If $ a$,$ b$,$ c$ are positive reals and $ abc \equal{} 1$
Prove that
    $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \geq 2ab \plus{} 2bc \plus{} 2ac$[/quote]
WLOG,we assume that $ (a\minus{}1)(b\minus{}1) \geq 0$.Rewrite the inequality as $ (a\minus{}b)^2\plus{}(c\minus{}1)^2\plus{}2c(a\minus{}1)(b\minus{}1) \geq 0$ which is true.

Done !


Problem 1
In the shape below two squares are given such that the sides of the exterior square are equidistant from the sides of the small square, and the length of th segment $AB$ is equal to 18m. If the area of the small square is equal to $\alpha$ and the shaded area is equal to $\beta$, calculate $2\alpha$+$\beta$.

Problem 2
The surnames of four persons are Vasiliou, Iakovou, Nikolaou and Andreou. Their jobs are accountant, lawyer, dentist and doctor (not necessarily with the given order). The accountant and the lawyer work in their offices whereas the dentist and the doctor work on their surgeries. Free medical care is given to the accountant, who also keeps Mr Nikolaou accontant books. Mr Iakovou does not know Mr Vasiliou and moreover his doctor is living in the street where Mr Vasiliou office is situated. Mr Iakovou and th acconatant exchange services. Find the jobs of each person.

Problem 3
Two tangent circles $(K,R)$ and$(\Lambda,\rho)$ are external to one another and their point of tangentcy is $Z$. A straight line $(\epsilon)$ is tangent to the circles with center $K$ and $\Lambda$ at the points $A$ and $B$ respectively. A straight line $(\delta)$ is parallel to $(\epsilon)$ and is tangent to the circle with center $\Lambda$ at $\Gamma$. Prove that the points $A$, $Z$, $\Gamma$ are collinear.

Problem 4
Prove that

$\frac{1}{2007}<\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{2005}{2006}<\frac{1}{44}$

[hide]Problem 1: In the figure we have $BC,CD,DE,EB$ intersect $AP,PN,NM,MA$ at $T,R,U,S$ respectively. It's easy to prove that $URTS$ is a square and also $\alpha+\frac{\beta}{2}= 18^{2}$
As a consequence : $2\alpha+\beta = 2 \times 18^{2}= 648$

Problem3: First let's prove that $\angle AZB = 90$.
Let the tangent to both circles passing through $Z$ intersect $AB$ at  $C$. As a consequence:
$\angle CBZ = \angle CZB \Rightarrow \angle ABZ = \angle CZB$
$\angle CAZ = \angle AZC \Rightarrow \angle BAZ = \angle AZC$
$\Rightarrow \angle BAZ+\angle ABZ = \angle AZB \Rightarrow 180-\angle AZB = \angle AZB \Rightarrow \angle AZB = 90$
It's obvious that $B, \Lambda , \Gamma$ are on a straight line.

Now assume that $AZ$ intersects circle ( $\Lambda , \rho$ ) at $\Gamma^{'}$ . We have $\angle AZB = \angle BZ\Gamma^{'}= 90$ as a consequence $B\Gamma^{'}$ is a diameter of the circle ( $\Lambda , \rho$ ), so is $B\Gamma$, so $\Gamma = \Gamma^{'}$ and the proof is complete.[/hide]


algebra
polynomial
calculus
calculus computations

I need to be enlightened (and I bet pco would like it too): where from come all these ugly polynomials, with no reasonable roots? Is there a malign teacher who feeds them to the naive younger generation, or is it a mushrooming of occult origin?

hello, you have to solve the following polynomial
$ 3\,{x}^{4}\plus{}{x}^{5}\minus{}24\,{x}^{3}\minus{}28\,{x}^{2}\plus{}48\,x\plus{}100\equal{}0$
Good luck!
Sonnhard.


A game of solitaire is played with $R$ red cards, $W$ white cards, and $B$ blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of $R, W,$ and $B,$ the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

Unexpectedly nice. Here is a nice "local"-type solution.
The answer is $\min(BW,2WR,3RB)$. We can achieve $BW$ by removing $B$ blue cards (with $W$ penalty each), then removing the red cards with $0$ penalty each and then the white cards with $0$ penalty each. We can also achieve $2WR$ and $3RB$ with a similar procedure.
Consider the order in which cards are played. If a player first plays three red cards, then one white, then one red, then two blue, we can represent the order as $RRRWRB$. Define a "C-block" of plays as some consecutive (possibly zero-length) plays in which the same card C is played, so an R-block is red cards for example. Clearly, we can represent every card order as a list of blocks such that every red block is followed by a blue block, every blue block is followed by a white block, and every white block is followed by a red block (except for the last block) and no two consecutive blocks are zero-length. For instance, we can represent $RRRRRBBW$ as a 5-long R-block, a 2-long B-block, and then a 1-long W-block. We can represent $WBBR$ as a 1-long W-block, a 0-long R-block, a 2-long B-block, a 0-long W-block, and finally a 1-long R-block.
Consider some arbitrary play order $P$. I claim that we can find some play order $P'$ which results in at most as much penalty as $P$ and has fewer nonempty blocks than $P$, unless $P$ has at most 3 nonempty blocks. If $P$ has more than 4 nonempty blocks, it is clear that at least one of the block orders R-B-W-R, B-W-R-B, W-R-B-W occurs, where the first and fourth blocks are nonempty, while the second and third are not necessarily nonempty.
Focus on the R-B-W-R case. Suppose there are $b$ blue cards in the blue block, $w$ in the white block, $r_1$ in the first red block, and $r_2$ in the second red block. Now consider the following subcases:
[u]Subcase 1:[/u] $3b \leq 2w$. Consider the play order $P'$ formed by deleting the second red block and adding $r_2$ red cards to the first block. For example, if the blocks are $\boxed{RRRR}\boxed{B}\boxed{WW}\boxed{RR}$ before, then they will be $\boxed{RRRRRR}\boxed{B}\boxed{WW}$. Clearly this decreases the number of nonempty blocks. It is clear that this process increases the penalty by $3r_2b$, but also decreases it by $2r_2w$. Since $3b \leq 2w$, it follows that $2r_2w \geq 3r_2b$ and the penalty does not increase.
[u]Subcase 2:[/u] $3b > 2w$. Consider the play order $P'$ formed by deleting the first red block and adding $r_1$ red cards to the second block. Clearly this decreases the number of nonempty blocks. It is clear that this process increases the penalty by $2r_1w$, and decreases it by $3r_1b$. Since $3b>2w$, it follows that $3r_1b>2r_1w$ and the penalty decreases.
Combining these cases implies that there always exists such a $P'$ for the R-B-W-R case. The other two cases can be dealt with similarly, which implies the claim.

Repeating this process shows that any play order $P$ accumulates at least as much penalty as some play order with three or fewer nonempty blocks. Hence to find the minimum it suffices to only consider play orders with three or fewer nonempty blocks. Clearly if there are at most two nonzero numbers in the multiset $\{R,B,W\}$, the minimum is zero, which fits the expression. Otherwise, if all of $\{R,B,W\}$ are nonzero, we can easily derive the above expression by considering cases on the order in which blocks are removed, which completes the proof. $\blacksquare$

[quote=JSteinhardt]We claim (inductively) that the minimum is just going to be $min(BW,2WR,3RB)$. We'll start our induction with the case where one of the three quantities is zero, in which case we verify that we can indeed get away without any penalty by, for example, discarding blue if we are out of white.

Now, for the inductive step, let $f(R,W,B)$ be the minimum we seek. Note that
\[f(B,W,R) = min(W+f(B-1,W,R),2R+f(B,W-1,R),3B+f(B,W,B-1)) \]
By our inductive hypothesis, $f(B-1,W,R) = min((B-1)W,2WR,3R(B-1))$. In order for this to cause our inductive step not to hold, we would require that $W+min((B-1)W,2WR,3R(B-1)) < min(BW,2WR,3RB)$. It is evident that the first two entries in the $min$ expression cannot cause this to happen, so that we need only consider $W+3R(B-1) < min(BW,2WR,3RB)$. So $W+3R(B-1) < BW$, whence $3R < W$. But $W+3R(B-1) < 3RB$, so that $W < 3R$, a contradiction.

For the other two cases, we can get similar contradictions, so that our inductive step must hold, and so $f(B,W,R)$ is indeed $min(BW,2WR,3RB)$.

We now need only establish how many ways to do this. If one of these quantities is smaller, our induction and the fact that it is eventually zero guarantees that it will continue to be the smallest quantity as cards are discarded. (For example, if it is currently optimal to discard a blue card, it will continue to be so until we run out of blue cards.) Therefore, assuming that there is currently only one best choice of card to discard, this will continue to be so in the future, whence if $BW \neq 2WR \neq 3RB$, there is only $1$ optimal strategy.

Suppose, now, that $BW = 2WR$. It is thus optimal to discard either a $B$ or $W$ card. If we ever discard a blue card, then we will cause $BW < 2WR$, whence there is only one possible strategy from then on. However, if we discard a white card, then we will still have $BW = 2WR$, meaning that we continue to have the choice of discarding a white or blue card. Since we can discard a white card at most $W$ times, there are $W+1$ choices for how many $W$ cards to discard ($0$ to $W$), meaning that there are $W+1$ optimal strategies.

By similar logic, we get $R+1$ optimal strategies if $2WR = 3RB$, and $B+1$ optimal strategies if $3RB = BW$.

The final case, then, is if $BW = 2WR = 3RB$. In this case, if we first discard a white card, we are left with the $BW = 2WR$ case, and similarly for a blue and white card. The total number of optimal strategies in this case is then the sum of the optimal strategies in those cases, or, in other words, $B+W+R$.

To summarize:
The minimum penalty is $min(BW,2WR,3RB)$.
If $BW \neq 2WR \neq 3RB$, there is $1$ optimal strategy.
If $BW = 2WR < 3RB$, there are $W+1$ strategies.
If $2WR = 3RB < BW$, there are $R+1$ strategies.
If $3RB = BW < 2WR$, there are $B+1$ strategies.
If $BW = 2WR = 3RB$, there are $R+B+W$ strategies.[/quote]

You are using Induction of which quantity?

[b]1. Setup[/b]
For each order of playing, we write it out by repeatedly appending either $r$, $w$, or $b$ based on the color played in that move to the end of a sequence which starts off as empty. For example, if the player plays five reds and then three blues, then we say this player played sequence $\alpha=rrrrrbbb.$ We call a sequence $u$ a subsequence of another one $v$ if and only if we can remove some elements of $v$ and get $u$. Denote by $\#$ to be a function taking in a sequence and outputting a nonnegative integer that finds the number of times the sequence is a subsequence of $\alpha.$ Note that the penalty is equal to $\#(bw)+2\#(wr)+3\#(rb).$ Call a subsequence of consecutive elements of the same color a block. Furthermore, let a block of $k$ $r$'s be denoted by $r^k$.

[b]2. In the minimum-penalty played sequence $\alpha$, $b$ must be followed by $r$, $r$ by $w$, and $w$ by $b$.[/b]
Assume otherwise that we have a played sequence $\alpha$ for which two consecutive elements are $rb$, $bw$, or $wr$. Clearly, transposing these two will not cause change in the penalty induced by any other element. The penalty induced by the second transposed element ($b$, $w$, $r$, respectively) will also not change. However, the first transposed element's penalty will decrease because we are losing either an $(rb)$ subsequence, a $(bw)$ subsequence, or a $(wr)$ subsequence.

[b]3. If $\alpha$ contains exactly four consecutive blocks, we can decrease that number to three blocks while not increasing the penalty.[/b]
Suppose we have $\alpha=w^{a_1}b^{a_2}r^{a_3}w^{a_4}$. The penalty is $a_2a_4+2a_1a_3.$ Then, $w^{a_1+a_4}b^{a_2}r^{a_3}$ has penalty $2a_1a_3+2a_3a_4$ and $b^{a_2}r^{a_3}w^{a_1+a_4}$ has penalty $a_2a_1+a_2a_4.$ Depending on the relative sizes of $a_2$ and $2a_3$, one of these must not increase the penalty. The process for the other three cases are similar and I will not go into detail.

[b]4. One minimal-penalty $\alpha$ is in the form $w^{W}b^{B}r^{R}$ or any cyclic shifts.[/b]
Note that in $\alpha$ we can apply the previous step to any four consecutive blocks, and we will only affect the penalties of elements inside of that block, and the other blocks will not be affected. Therefore, if we continually apply the algorithm in the previous step, the number of blocks decreases while the penalty doesn't increase. Therefore, one of the minima must be the case where there are three consecutive blocks, in the form $w^{W}b^{B}r^{R}$ or any cyclic shifts. [b]The penalty here is either $BW,2WR,3RB$ so the minimum is $\min(BW,2WR,3RB).$[/b]

[b]5. The equality case.[/b]
In the final application of the algorithm, as described in step $4$, the equality case of step $3$ is either $W=3R$, $B=2R$, or $2W=3B$. Therefore, if none of these equalities are true, then the inequality in the final application in step $4$ is strict and so the only minimum case is the one with exactly three blocks. Note that in this case, $BW,2WR,3RB$ are pairwise distinct. If the minimum is $BW$, then the only minimum case is $b^Br^Rw^W.$ Similar goes for the other cases. The other cases can all be deduced by reverse application of the algorithm in step $4$ mindful of the equality case.

[hide=Equal case]JSteinhardt's inductive method summarizes the equality cases better than this one can.[/hide]
[hide=@above]Induction on the number of total cards.[/hide]

The answer is $\boxed{\min(BW, 2RW, 3BR)}$.  Each of these is achievable with $\textbf{BRW}$, $\textbf{WBR}$, and $\textbf{RWB}$, respectively, where $\textbf{B}, \textbf{R}, \textbf{W}$ are some number of (in this case, all) blue cards, red cards, and white cards.  We will prove this is minimal.

Consider a moveset that gives minimal penalty (we can assume this since there are finitely many possible movesets), which we will call a $\textit{Mondrian}$.  For convenience, we will let $B$, $W$, and $R$ denote a blue, white, and red card when describing a moveset.  Then notice that if there exists $BW$ in the moveset, we can switch both to strictly decrease the penalty, which would be a contradiction, so there do not exist $BW$ in any $\textit{Mondrian}$.  Similarly, neither does $WR$ nor $RB$.  Therefore, every $\textit{Mondrian}$ is of the form $X_1X_2X_3\dots X_n$, where each $X_i$ consists of $R\cdots RW\cdots WB\cdots B$ (where there are a positive number of each $R$, $W$, and $B$), and $X_1$ may start with $R$, $W$, or $B$ and $X_n$ may end with $R$, $W$, or $B$.

When $BW \neq 2RW \neq 3BR \neq BW$, notice that we can now move the first "block" of cards of the same color in the first $X_i$ to the "block" in the last $X_n$ (with the same color, so red goes into a red block) to strictly decrease the penalty of a $\textit{Mondrian}$, a contradiction.  (If this does not decrease it, do the opposite and move the last block to the first one, since either one is guaranteed to work).  Therefore, the $\textit{Mondrian}$ must be of the form we mentioned at the start.  There is obviously only $\boxed{1}$ way to achieve this.

By extremely similar logic, we can conclude that
$BW = 2RW \neq 3BR \implies \boxed{W+1}$,
$BW = 3BR \neq 2RW \implies \boxed{B+1}$,
$2RW = 3BR \neq BW \implies \boxed{R+1}$, and
$BW = 2RW = 3BR \implies \boxed{R+W+B}$, done.

Alright here we go... this sols different from above sols i think and it is legitimately solved
    We use a local technique (even though this is in DCY-process :rotfl: )

    The answer is min(BW,2RW,3RB), obtained by taking all of B then R then W (cyclically shifting for other minimums). Note that WB,RW,BR are more efficient versions of BW,WR,RB, hence we know the dealing of cards occurs in blocks $B_1B_2...B_n$, where each $B_k$ is a sequence of $R...RW...WB...B$, and henceforth suppose that we have obtained the minimal penalty config s.t. $n\ge 2$, with $BW\ne 2RW\ne 3RB$. Between any $B_k,B_{k+1}$, having $p,q,r$ and $u,v,w$ number of R,W,B in the respective blocks, we assume the penalty is already optimal, and perturb the penalty by combining only the Rs with each other into the same block $B_k$, and see if it's more efficient; it increases by 3(ur-qu). If this is not equal to 0, then either putting R in both blocks into $B_k$ or $B_{k+1}$ suffices, since they are negation of each other, so we would combine the blocks s.t. it increases by negative amount, contradiction; if they are 0, q=r; by similar reasoning on other ones, we deduce they are all =0, meaning combining blocks with their respective color doesn't change the optimality.

    We're now home free: ,We can WLOG combine them, and now apply this algorithm GLOBALLY to show that (at least one of the) minimum(s) is when it's one block of R...RW...WB...B.
    It remains to find the number of ways: There's one way if $BW\ne 2RW\ne 3RB\ne BW$, which is playing all of one type at once; if $BW=2RW\ne 3RB$ there's W+1 ways (erasing 0 to W of white cards, then playing out all blue cards, then all red cards, and finish with the rest of white cards. (Each white card discarded still preserves $BW = 2WR$.)

    For the second part, the motivation was that the R and B must now be in one block due to our above reasoning, but we can have white separated, because it holds the same equality as if it were R...RW...WB...B, but it wouldn't hold the equality as the same if we had done R...RW...WB...B with a blue not in the block of blues (easily checked, remember that we can WLOG combine the other blocks). We can apply similar reasoning in other cases to get 1,W+1,B+1,R+1,R+W+B ways in their respective equals.


009 	Nidhi Ramaraj 	10 	Bangalore, National Public School
014 	Pradeep George Mathias 	12 	Bangalore, National Public School, Indiranagar
071 	N Karthik Rajagopal 	12 	Chennai, DAV Boys Sr Sec School, Gopalapuram
072 	Hamsini Krishnakumar 	11 	Chennai, DAV Girls School, Gopalapuram
090 	S Shreyas 	12 	Chennai, PS Sr Sec School
093 	V Prashant 	11 	Chennai, PS Sr Sec School
096 	Gaurav C 	12 	Chennai, PSBB, KK Nagar
097 	Gokul Ravee Ramanathan 	11 	Chennai, PSBB, KK Nagar
098 	Sudarshan V 	12 	Chennai, PSBB, KK Nagar
121 	Aayush Singhal 	12 	Ghaziabad, Saraswati Bal Mandir
150 	Abhishek Ghosh 	11 	Durgapur, Hem Sheela Model School
152 	Tirtha Chatterjee 	11 	Durgapur, Hem Sheela Model School
184 	Bharat Agarwal 	10 	Kanpur, Seth Anandram Jaipuria School
228 	Siddharth Bhatia 	09 	Delhi Public School, Vasant Kunj
238 	Priyanshi Agrawal 	11 	Noida, Apeejay School
244 	Archit Vikrant Karandikar 	10 	Pune, Abhinava Vidyalaya
245 	Akshay Degwekar 	10 	Pune, DES Sec School
246 	Anand Degwekar 	10 	Pune, DES Sec School
248 	Vihang Mehta 	12 	Pune, Jog Junior College
271 	Siddharth Dagar 	12 	Bangalore, National Public School
272 	Pramod 	12 	Chennai, DAV Boys School, Gopalapuram
273 	B Srivatsan 	11 	Chennai, Sri Sankara Sr Sec School
274 	Nadeem Moidu 	12 	Doha, MES Indian School
275 	Samagra Agarwal 	10 	Kanpur, Seth Anandaram Jaipuria School
276 	Namit Dejappa Shetty 	12 	Thane, Smt Sulochanadevi Singhania School

Yeah!

Thanks to Madness for telling me. I owe him a treat. Congrats to Vihang (Hope we will have fun at the camp), Bruteforce (tell me who you are) and Pradeep.

Dei congrats da and i am telling everyone i know  :D  :D  :D

deo i forgot me da :rotfl:  :rotfl:  :rotfl:

Well after all not everyone can go to the TC's. That's gr8 da and so too for bruteforce (sorry for bugging u for ur identity), vihang and pradeep and ofc gaurav. I had told gaurav that his program would have exceeded memory but then............................ He's going to be mad wid joy :D  :D

@bruteforce: Do we got good vegetarian food in TISB?? (Sorry for showing off, but I am asking out of genuine concern)

Annd btw did u phone ramu and kvkk? And anyone told V Prashanth? That guy might never know  :D

Pm me wat they said to you da shreyas  :D

SHRE!!! Congrats man.. Now you have a legit reason to give me a treat... And yes. Someone has to inform Prashanth... Kid might not know..

Congrats to all the other people on board who got selected....

TRAINING CAMPS FTW :D

P. S - I am mad, dont mind me

And yes as shreyas says prashanth seems to have a golden touch  :) . Avanukku chinna vayasu laye semma talent  :D  :D

That guy is a prodigy of his own right.. Seriously, and really modest too... I like the kid, but its a pity he is so subdued in school.. You must know Rohit :)

[quote="Cyber-Shreyas"]Thanks to Madness for telling me. I owe him a treat. [\quote]

Dei bring ur purse full atleast 100 rupees for me exclusively da  :D  Maybe pizza hut  :P  :D

Come on SHRE!!!! Second person, you and all who head... Thala

How very conveniently inconvenient? :D
>>>>>>>>>><<<<<<<<<<<<  You know what this means, TREAT   >>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<

[quote="Cyber-Shreyas"]I've told the sirs about me and VP, so he will get to know soon.

That guy is awesome. His track record this year is exemplary, as chappli would say guru uchi'lla errukaru.

I am the 2nd person studying in our school currently who gotto TC........ Hopefully there will be more.[/quote]
And have u most importantly told Shankar and Yogashankar  :D  :D
And not to forget Deepak and Venkatesh  :)

I had kept a notify tool in my browser as i was too eager to know the result. So I got to know it as soon as it came.

And just like this Indian thread in aops , the camp was also dominated by chennaiites , who were mostly veg. So don't worry , shreyas . I am sure everyone will enjoy it.

I think there was a class wise cutoff this time , because there is no other way that so many can get in from class 10 and 11.

Congos to all selected.

@ brute force or shreyas: Can you give me instructions as to how to place a notify tool. Pleezeeeeee  :)

I use this add-on  on firefox https://addons.mozilla.org/en-US/firefox/addon/3149

[quote="bruteforce"]I had kept a notify tool in my browser as i was too eager to know the result. So I got to know it as soon as it came.

And just like this Indian thread in aops , the camp was also dominated by chennaiites , who were mostly veg. So don't worry , shreyas . I am sure everyone will enjoy it.

I think there was a class wise cutoff this time , because there is no other way that so many can get in from class 10 and 11.

Congaratulations to all selected. ( all the seniors have been selected , so u wont get any information about me when i say i got selected. )[/quote]

10 Chennaites this time. wow!! Dei Bruteforce, message me who you are now or else I will ask rohit to tell me.

COngrats To All THose who qualified,especially to Shreyas and all the Chennaities who qualified.
I expected more from Chennai. :wink:

[quote="Vihang"]Thanks and congratulations to all of those who got in.

4 students from Pune! Yay! This was the first time ZOI was conducted here ... Chennaiites beware![/quote]

Hey Vihang I guess int a[n] worked :D:D Meet you at the camp though you will be meeting my best friend before that. :D

Am I included in the treat?  :lol:  :huh:

and probably someone can also send some money by speed post so that i can eat in Pizza hut here :D

[size=200][color=red][b][i]CONGRATS TO ALL OF THEM WHO HAVE GOT SELECTED[/b][/i][/color][/size]

Well I didn't get selected. :(  Anyway, now I can wait for the chem olympiad results(with even more anxiety).

[color=blue][size=150]BEST OF LUCK SHREYAS[/size][/color] :bye:  :thumbup:

Come back with a GOLD medal or don't come back.

was there any  double meaning. to that :D ...

[quote="chemrock"]No I don't think so.
Just congrajulating him and wishing him[/quote]
that should have been congra[b]t[/b]ulating(English NV :P  :P  :P  :P )

[quote="Chapplikka"]Macha Shre, don spend more than rs.50 per person da.. I spent only rs.285 for more than 15 ppl in my second yr of IMOTC.. so for ur first yr, i would consider it illogical spendin so much money on treats.. But as I have been your biggest source of support over the last two years, I would definitely expect a treat from you, even though i dont generally take treats.. Even a rs.5 paneer roll from you in sakthi's kitchen would be equivalent to a candlelight dinner wit (ahem ahem) at taj... So even rs.5 is enough for me...[/quote]

50 per persona?? Yeah right, as if I'm frolicking in money.. Well unakku 50p dhan. :D Candlelight dunner with whom da? Miami beach babes a?

Pardesi: Don't worry da, one veg puff/roll reserved for you.

[quote="chemrock"]No I don't think so.
Just congratulating him and wishing him

EDIT: Pardesi u did wrong NV first u said "CONGRAGULATING"[/quote]
any proof :P
@chappli i suppsoe u taught soumya tamil....
then please translate that.....
Bhartahiyar...ayarraar :P

If u ever go to the IOI, Each one pizza treat at Pizza Corner Da, if u win different medals there, wait let me think abt it..........  :D  :D  :D

[quote="bruteforce"]
I think there was a class wise cutoff this time , because there is no other way that so many can get in from class 10 and 11.
[/quote]

Yes, true!! They want to encourage younger people to get in, so as to start the training young. :D I guess the cutoff for class 12 must have been really high cause a lot of people here who did one and a half didn't make the cut.

Wow, as time goes, its gonna be dead difficult for class 12 newbies to make the cut. :(

Just found this in Wiki
[quote=IOITC]
Students who participate in the camp may be given direct admission to Chennai Mathematical Institute's B.Sc. (Hons) Mathematics programme. This is decided on a case-by-case basis by the admissions committee.
[/quote]

However this is what it says about IMOTC

The Chennai Mathematical Institute offers direct admission for its B.Sc. (Hons) Mathematics programme to students who have cleared INMO. This is somewhat different from its earlier policy of granting direct admission to students who have attended the IMOTC once and qualified as Seniors the next time. However, the policies are subject to change.

Just a note to other students who got selected to the camp :D

i know that school. a friend of mine is from there. buts its bloody expensive, and huge( some 350 acres or something like dat). am sure youll have fun there. good luck :D

[quote="Cyber-Shreyas"]This is really really weird. The first name from the list "Nidhi Ramraj" has just disappeared and the no.of.people selected has reduced to 24 from 25.

http://www.iarcs.org.in/inoi/2008/inoi2008/results_inoi2008.php

To view the original list, check the first post of this thread.

Bruteforce: Any idea???[/quote]
 i don't have any idea .
as long as ur name is there , u don't have to worry.

the different versions are available here http://www.iarcs.org.in/inoi/2008/inoi2008/

donno why they keep changing.

The edit was made on the fourth of march. Not that I'm worried but just curious. :D

Thanks Varun. I just saw the pics below the wiki page for TISB (they were posted by an IOITC guy).

[quote="Cyber-Shreyas"][quote="hell_ever"]hey, shreyas....

isnt the tc held at hbcse? i live quite close to it and pizza hut is even closer..... :wink:

(that was only for information  :D )[/quote]

Well, IMOTC, IPHOTC, ICHOTC, IBOTC and IAOTC are all held in HBCSE.

IOITC is held in Bangalore. In The International School. http://en.wikipedia.org/wiki/Tisb and is much later than the other camps.[/quote]

No...IAOTC usually held in Nehru Science Centre, E. Moses road
But this year in IIT Bombay....
regulated by HBCSE


function
real analysis
calculus
calculus computations

Suppose we have the indicator function $ I_{\mathbb{Q}}: [0,1] \to \mathbb{R}$ of rationals. To show that it is not Riemann integrable (without looking at upper and lower sums) I said that the Riemann sum can take on three possible general values:

$ \mathcal{R}(f,P) \equal{} \begin{cases} 0 \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} \notin \mathbb{Q} \ \text{for any} \ i \\
x_{i} \minus{} x_{i \minus{} 1} \ \ \text{if} \ x_{i}^{*} \in \mathbb{Q} \ \text{for exactly one} \ i \\
\geq \parallel{}P\parallel{}, \ \ \ \text{if} \ x_{i}^{*} \in \mathbb{Q} \ \text{for some} \ i \end{cases}$

Here $ \parallel{}P\parallel{}$ is the mesh of the partition. Because we know that $ \mathcal{R}(f,P) \equal{} \sum_{i \equal{} 1}^{n} f(x_{i}^{*})(x_{i} \minus{} x_{i \minus{} 1})$. So we can't bound $ |\mathcal{R}(f,P)|$? Because either $ \mathcal{R}(f,P) \equal{} 1$ or $ \mathcal{R}(f,P) \equal{} 0$. In other words, $ |\mathcal{R}(f,P)| < \epsilon$ and $ |\mathcal{R}(f,P)\minus{}1|< \epsilon$?

the real problem is that $ R(f,P,\{x_i^*\})$ is supposed to approach a limit as $ \parallel{}P\parallel{}\to 0$ regardless of your choices $ \{x_i^*\}$

but no matter how small the mesh is you can pick $ \{x_i^*\} \subset \mathbb{Q}$ and $ \{y_i^*\} \subset \mathbb{R}\setminus \mathbb{Q}$

and so $ R(f,P,\{x_i^*\})\equal{}1$, while $ R(f,P,\{y_i^*\})\equal{}0$

this would be so much easier if you just let us use the darboux condition -.-


A rectangle is formed. Its length is increased by 50\% and its width is cut in half in forming a second rectangle. Then each of the two new dimensions is decreased by 20\% to form a third rectangle. What percent of the original rectangle's area is the area of the third rectangle?

$ WLOG$ Let our rectangle have sides of length $ 10$. Its length is increased to $ 10 \cdot 1.5 \equal{} 15$ and its width decreased to $ \frac{10}{2} \equal{} 5$. Decrease both these dimensions by $ 20\%$ to $ 15 \cdot 0.8 \equal{} 12$ and $ 5 \cdot 0.8 \equal{} 4$. The area of our new rectangle is $ 12 \cdot 4 \equal{} 48$, compared to the original $ 10 \cdot 10 \equal{}100$. Thus our answer is $ 48\%$.


Prove that

cos(2*pi/5) + cos(4*pi/5) = -1/2

[hide="Hint"]\[\sum_{k=1}^{n}\cos \left ( \frac{2\pi}{n}k \right ) = 0 \ \ (\text{The integer that}\ n \ \text{is more than}\ 2) \][/hide]


Consider $n$ points $A_1, A_2, \ldots, A_n$ on a circle.  How many ways are there if we want to color these points by $p$ colors, so that each two neighbors points are colored with two different colors?

if $p=2$ then if $n$ is even you have $2$-colorings,else you have none.
 If $p\geq 3$ we can reduce this recursively as follows(this is quite an old idea):For $C_n$ denoting the circle graph on $n$ vertices, and $P_n$ denoting a path on $n$ vertices,it is easy to see that if $f(C_n,p)$ is the desired number,then $f(C_n,p)=f(P_n,p)-f(C_{n-1},p)$.this is easy since every proper coloring of $C_n$ leads to a proper coloring of $P_n$ as well(just delete an edge).conversely the only proper colorings of $P_n$ that extend to a proper coloring of $C_n$ are those for which the end vertices have different colors.The colorings of $P_n$ with the end vertices of the same color are equivalent to colorings of $C_{n-1}$ (where you 'identify' the end vertices as one single vertex).
finally $f(P_n,p)=p(p-1)^{n-1}$ and the rest is induction.

[quote="seshadri"]For $C_n$ denoting the circle graph on $n$ vertices, and $P_n$ denoting a path on $n$ vertices,it is easy to see...[/quote]

A stupid question - what is a circle graph/path?

$C_n$ is the cycle with $n$ vertices (you can draw it as a polygon with $n$ vertices), and $P_n$ is what you get from $C_n$ after you delete an edge (i.e. the path with $n$ vertices).

I think that this is a generalization of [url=https://artofproblemsolving.com/community/c5h1213135p6023626]2016 AIME II #12[/url]. Not a hard extension but pretty cool.


probability
invariant
function
geometry
geometric transformation
real analysis

The concept of an amenable group came up in my analysis class. I have looked at the [url=http://en.wikipedia.org/wiki/Amenable_group]Wikipedia article[/url] about it but I am afraid I don't really understand the concept. Here are my questions:

(1) My professor told us that the usual definition of amenability (which he didn't state) is equivalent to the statement:
[i]Suppose $X$ is a compact metric space and $G$ is a(n at most) countable group. $G$ is said to be amenable if for any continuous action of $G$ on $X$, there is a probability measure on $X$ which is invariant under the action. [/i]

What is the "usual" definition of amenability? (Wikipedia has several and I can't understand them.) Why is that equivalent to the probability measure one?

(2) What does amenability tell us about a group, in group-theoretic terms? All I can understand from the definition I have is with regards to the dynamical properties of its actions on a probability space...

The usual definition of amenability is given by Wikipedia: there exists a left invariant mean $\mu$, i.e. a positive left-invariant functional on $L^{\infty}(G)$ that sends $f\equiv 1$ to $1$. Since the linear span of characteristic functions is dense in $L^{\infty}$, a mean is determined by its values on sets. In this sense, $\mu$ is just a left-invariant probability "measure" -- the quotation marks are necessary because $\mu$ need not be sigma-additive, just finitely additive.

 For example, $\math R$ is an amenable group under addition, even though there is no translation-invariant finite measure on it. To construct a mean on $\mathbb R$, one begins with a sequence of measures $\mu_{n}(E)=\frac{1}{2n}|E\cap [-n,n]|$ (where $|\cdot|$ is the Lebesgue measure) and take their "limit" as $n\to\infty$. Note that the limit is $0$ if $|E|=0$, and it is always at most $1$. Of course, in general the limit of $\mu_{n}(E)$ does not exist, for which there is a workaround. The measure $\mu_{n}$ defines a linear functional of norm $1$ on $L^{\infty}(\mathbb R)$, that is, $\mu_{n}\in (L^{\infty})^{*}$. By the Banach-Alaoglu theorem there is a subsequence with a weak* limit. This limit is $\mu$.

This procedure works whenever there is a Folner sequence (such as $[-n,n]$) mentioned by Wikipedia. Roughly speaking, a Folner sequence is an exhaustion of a group by sets with relatively small boundaries.


geometry
geometric transformation
analytic geometry
vector
linear algebra
matrix
rotation

The following problem is confusing me, so any help would be greatly appreciated!

Consider the quadrilateral (namely Q) in R^3 formed by the points
(1, 0, 0), (2, 0, 0), (1, 1, 3), and (2, 1, 3).

What should the coordinates be for the figure R we get by rotating Q counterclockwise in the x-y plane by 45 degrees, then dilating it by a factor of 3/2, then translating it along the vector (-2, 1, -1)?

What matrix transforms Q into R?

Counterclockwise rotation by angle $\theta$ in the $xy$-plane is realized by the matrix $\begin{pmatrix}\cos\theta & \sin\theta & 0 \\-\sin\theta & \cos\theta & 0 \\ 0&0&1\end{pmatrix}$. Dilation by $3/2$ is simply $\begin{pmatrix}3/2 & 0 & 0 \\ 0 & 3/2 & 0 \\ 0&0&3/2\end{pmatrix}$. Translation along $\langle-2,1,-1 \rangle$ adds $-2$ to the $x$-coordinate, $1$ to the $y$-coordinate, etc. I don't know how one can represent translation by a matrix.  :maybe:

So would it be right if I did the following:

First, I took the matrix

[cosθ, sinθ, 0
-sinθ, cosθ, 0
0, 0, 1]

and I set all the θ equal to 45. This led to the matrix:

[1/√2, 1/√2, 0)
-1/√2, 1/√2, 0
0, 0, √2]

In order to do dilation by 3/2, I multiplied each entry by 3/2 retrieving:

[3/(2√2), 3/(2√2), 0
-3/(2√2), 3/(2√2), 0
0, 0, (3√2)/2]

Then to translate it along the vector (-2, 1, -1), I added -2 to the x coordinates, 1 to the y coordinates, and -1 to the z coordinates.

[3/(2√2) -2, 3/(2√2) +1, -1
-3/(2√2) - 2, 3/(2√2) + 1, -1
-2, 1, (3√2)/(2) - 1]

So the above was my final matrix (and I said that the rows of that matrix were the coordinates for the figure R.

No. That's definitely wrong. There is no way to include the translation with just a $3\times 3$ matrix for the transformation, and your attempt to do this is complete nonsense.

Then what is the procedure for doing this type of problem?

Okay, so I figured out how to find the coordinates.

What I did was I used the matrix

cosθ -sinθ 0
sinθ cosθ 0
0 0 1

Then substituted 45 for θ.

After this, multiplied the identity matrix for R^3 by 3/2 and then multiplied it by the matrix with 45 substituted for θ.

Then T(x,y,z) = (-2+3sqrt(2)x/4-3sqrt(2)y/4,1+3sqrt(2)x/4+3sqrt(2)y/4,-1+3z/2).

I substituted each of the quadrilateral points in for T(x, y, z) to come up with the four points and got:

(3/(2√2) - 2, 3/(2√2) + 1, -1),
(3/√2 - 2, 3/√2 + 1, -1),
(-2, 3/√2 + 1, 3.5),
(3/(2√2) - 2, 9/(2√2) + 1, 3.5)

However, I'm confused as to how to find the matrix that transforms Q into R.

[quote="SharkBites77"]... the matrix that transforms Q into R.[/quote]
There isn't one, or at least there isn't a $3\times 3$. See my first post for a way to incorporate the translation.

[quote="jmerry"]You can fit translation in by adding a dummy fourth coordinate: instead of the point $(x,y,z)^{T}$, we have $(x,y,z,1)^{T}$. The matrix of translation by $v$ is then $\begin{pmatrix}I&v\\0&1\end{pmatrix}$ in block form.

If you loosen the restrictions on the last row, this leads to projective geometry.[/quote]

Question - what does the v stand for in the matrix?

$v$ is the specific vector you're translating by, as a column vector.

Okay, see I'm confused because I was told that in order to find the transformation matrix, all I have to do is multiply

cos45 -sin45 0
sin45 cos45 0
0 0 1

by

the inverse of
(3/(2sqrt2) - 2, 3/(2sqrt2) + 1, -1),
(3/sqrt2 - 2, 3/sqrt2 + 1, -1),
(-2, 3/sqrt2 + 1, 3.5),
(3/(2sqrt2) - 2, 9/(2sqrt2) + 1, 3.5)

But I'm not sure how to do this seeing that one is a 3 x 3 matrix and the other is a 4 x 3 matrix.


linear algebra
matrix
vector
inequalities

Let $ A$ be an $ m\times n$ matrix with rank $ m$ and $ B$ an $ n\times p$ matrix with rank $ n$. What is the rank of $ AB$?

The rank of $ AB$ is $ m.$

Note that $ A$ and $ B$ are both (possibly) wider than they are tall: $ m\le n\le p.$

The statement that $ A$ has rank equal to its number of rows is equivalent to saying that the rows of $ A$ are independent. Similarly, the rows of $ B$ are independent.

Saying that the rows of $ A$ are independent can be phrased this way: if $ x$ is a row vector ($ 1\times m$ in this case) such that $ xA\equal{}0,$ then $ x\equal{}0.$ Similarly, if $ yB\equal{}0$ then $ y\equal{}0.$

Now suppose $ x(AB)\equal{}0.$ We can write this as $ (xA)B\equal{}0,$ which implies that $ xA\equal{}0,$ which implies that $ x\equal{}0.$ Hence the rows of $ AB$ are independent and $ \operatorname{rank}AB\equal{}m.$

It is a corollary of:

Suppose that $ A \in M_{m \times n}$ and $ B \in M_{n \times p}$. Then:

$ rk(A) \plus{} rk(B) \minus{} n \leq rk(AB) \leq \min \{ rk(A), rk(B) \}$

mathangel , can you show me the proof of that corollary, please?

The left inequality is called Sylvester's inequality and it was discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20558]here[/url].


calculus
integration
function
algebra
polynomial
complex analysis
complex analysis unsolved

Let f be entire. If $ |f(z)| \leq M|z|^n$ for large $ |z|$, for a constant M, and for an integer n, show that f is a polynomial of degree $ \leq n$.

Thanks for your help.

since $ f$ is entire function so
$ f(z) \equal{} \sum_{m \equal{} 0}^{\infty}a_{m}z^{m}$ for all $ z\in D(0,R)$

where $ a_{m} \equal{} \frac {f^{n}(0)}{m!} \equal{} \frac {1}{2\pi i}\int_{C(0,R)}\frac {f(w)}{w^{m \plus{} 1}}dw$

use what you are given : $ |f(w)|\leq M|w|^{n}$, then let $ R\rightarrow\infty$ then $ |a_{m}| \equal{} 0 \forall m > n$

See the [url=http://www.mathematics.jhu.edu/kong/311f07/311midterm2sol.pdf]4th problem[/url].


If $x,y,z$ are positive reals prove that

$\dfrac{x^2}{x^2-(y-z)^2}+\dfrac{y^2}{y^2-(z-x)^2}+\dfrac{z^2}{z^2-(x-y)^2}\geq\dfrac{3}{4}$

try $1,2,4$ for which it is $-\frac 1{15}$ ...

Maybe you want $x$, $y$ and $z$ to be the sides of a triangle?

Yes. sorry boys.

I am wondering why always I forget one condition? Iam angry with myself. :blush:  :mad:

If $x$, $y$ and $z$ are the sides of a triangle it is completely trivial.

Then \[ \dfrac{x^2}{x^2 - (y - z)^2} \geq 1 \] and similarly for the other fractions, so the sum is at least 3.

And oh yes, equality can be attained.

Where did you get 3/4?

[quote="silouan"]The idea was a problem which said that if $x+y+z=0$ then

$\dfrac{x^2}{x^2-(y-z)^2}+\dfrac{y^2}{y^2-(z-x)^2}+\dfrac{z^2}{z^2-(x-y)^2}=\dfrac{3}{4}$[/quote][/quote]

it's good to have a try to generalize or extend a problem... but changing x+y+z=0 to x,y,z are positive reals or sides of a triangle simply does not make much sense..


A rhombus has an area of 108 square units. The lengths of its
diagonals have a ratio of 3 to 2. What is the length of the longest
diagonal, in units?

From the area formula for a rhombus, $ 216 \equal{} (d_{1})(d_{2})$. By utilizing the ratio given, we find that the longer diagonal has length $ \boxed{18}$.


tahala be in fekr kardin keh momkeneh oon chizi keh shoma mibinin hamoon chizi nabashe keh baghieh mibinan,

masalan rajeh be chi fekr kardi my darling?

tel me more...plz...

masalan momkene man ya rangi ro ghermez bebianam vali to hamoon ro abi bebini? :!:

shayad oon be ellat e koor rangi bashe.. :rotfl:

bargardim be bahse asli.
hatta man fekr mikonam in mozoo be rang ha mahdood nemishe,masalan momkene yeki dayera ro morabba bebine. :!:

adam har chiziro, harjoor keh bekhad mitooneh bebineh,

monteha bastegi dareh keh nazaresh kheir bashe ya na.

[quote="amir2"]man ein sag ba ashegh movafegham.
mohem niaate adame.[/quote]
are vali khob az in mozoo ke begzarim in masale kheili ajibe. :D


Given any positive integer n, show that we can sind a positive integer m such that mn uses all ten digits when n is written in the usual base ten.

The problem MP11 here is very similar:
[url]http://mathcentral.uregina.ca/mp/previous2000/[/url]

[hide="Full solution"]Consider the $ n$ numbers with the following digits written in the usual base ten:
$ N_{1}=1234567890$
$ N_{2}=12345678901234567890$
$ \dots$
$ N_{n}=12345678901234567890\dots1234567890$ ($ n$ times).

Now consider the $ n$ values of $ N_{i}\mod n$, $ (i=1,\dots,n)$.
There are two cases:
a) At least one of the values $ N_{i}\mod n$ is $ 0$.
Then $ m=\frac{N_{i}}{n}$ is integer and $ nm=N_{i}$, which uses all ten digits.
b) None of the values is $ 0$.
Then there are only $ n-1$ possible values left $ \mod n$.
Therefore, by the pigeonhole principle at least two values must be the same, let's say $ N_{i}\equiv N_{j}\mod n$, $ i>j$.
Then $ m=\frac{N_{i}-N_{j}}{n}$ is integer and $ nm=N_{i}-N_{j}$, which uses all ten digits.[/hide]


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Find the value of $ a$ for which $ \int_0^{\pi} \{ax(\pi ^ 2 \minus{} x^2) \minus{} \sin x\}^2dx$ is minimized.[/quote]

Is there a strategy to generally bound integrals? Or do we have only CS at our disposal??

Well I guess you could expand out the integral and evaluate  :o
But that is just ugly, so I would try using the Cauchy inequality.

[quote="kunny"]Find the value of $ a$ for which $ \int_0^{\pi} \{ax(\pi ^ 2 \minus{} x^2) \minus{} \sin x\}^2dx$ is minimized.[/quote]

can i do this ? :clap:

let $ \textbf f\;\equal{}\; \int_0^{\pi} \{ax(\pi ^ 2 \minus{} x^2) \minus{} \sin x\}^2 \; dx$

now, let us set

$ \dfrac { \partial \; \textbf f }{\partial a} \;\equal{}\; \int_0^{\pi} \dfrac { \partial  }{\partial a} \{ax(\pi ^ 2 \minus{} x^2) \minus{} \sin x\}^2 \; dx\;\equal{}\;0$

$ \Rightarrow \;  \int_0^{\pi} \;2 \{ax(\pi ^ 2 \minus{} x^2) \minus{} \sin x\} \cdot  x(\pi ^ 2 \minus{} x^2)  \; dx \;\equal{}\; 0$

$ \therefore a\;\equal{}\; \dfrac {  \int_0^{\pi} \sin x  \cdot  x(\pi ^ 2 \minus{} x^2) \; dx  } {  \int_0^{\pi}  x^2(\pi ^ 2 \minus{} x^2)^2 \; dx } \;\equal{}\; \dfrac {\pi^2 \cdot \textbf I_1 \minus{} \textbf I_3 } { \pi^4 \cdot \dfrac {\pi^3}{3} \minus{} 2\pi^2 \cdot \dfrac {\pi^5}{5} \plus{} \dfrac {\pi^7}{7} }$

 where we have,
 $ \textbf I_n \;\equal{}\;  \int_0^{\pi} x^n\; \sin x \; dx  \;\equal{}\; \left( n\,x^{n\minus{}1} \sin x \minus{} x^n \cos x \right) \bigg |_0^{\pi} \;\minus{}\;n(n\minus{}1) \cdot \textbf I_{n\minus{}2}$
$ \equal{}\; \pi^n \minus{} n(n\minus{}1) \cdot \textbf I_{n\minus{}2} \qquad \left(  \;\;\; \text{and} \;\;\; \textbf I_1 \;\equal{}\; \pi \;\;\;  \right)$

and so,
$ a \;\equal{}\; \dfrac { \pi^2\cdot \pi \minus{} ( \pi^3 \minus{} 6 \pi ) } { \pi^7 \left( \dfrac {1}{3}  \minus{} \dfrac {2}{5} \plus{} \dfrac {1}{7} \right) } \;\equal{}\; \boxed { \dfrac { 315 } { 4\; \pi^6 } }$


advanced fields
advanced fields theorems

What's the Yang-Mills Existence and Mass Gap Theory?  I'm not in high school or college yet, but I'm just curious.  Please explain in terms that stupid people can comprehend. :blush:

Is it the one you are looking for ? http://www.claymath.org/millennium/Yang-Mills_Theory/index.php


limit
logarithms
inequalities
real analysis
real analysis unsolved

[color=darkblue]Prove that $\boxed {\lim_{n\to\infty} \frac{1+\sqrt [n] 2+\sqrt [n] 3+\ldots +\sqrt [n] n-n}{\ln n}=1}$ ([b]Autor[/b] : [u]Virgil Nicula[/u]).[/color]

nice limit:

Aplying Lagrange th for $e^x$ we get.
$\frac{\sum_{k=1}{n} (e^{\frac{\ln k} n}-1)}{\ln n}= \frac{\sum e^{x_{k,n} } \frac{\ln k}{n}}{\ln n}$

since $\frac{\ln n}{n} \to 0$ we have that $\lim_n x_{k,n}=0,\forall k$ hence  our limit is equal with:
$\lim \frac{\sum \ln k}{n \ln n}=1$ from Cesaro-Stoltz.

(A more detailed proof is with $\epsilon,\delta$ but the ideea is the same).

[hide="Here is the author-s proof."]
$\left(\forall\right)x\in R\ ,\ e^x\ge x+1\ .$[hide="The proof of this inequality without derivatives."]From the Bernoulli's inequality $\left(1+\frac xn\right)^n\ge 1+x$ for $n\to\infty$ obtain $e^x\ge x+1\ .$ Is it nicely ? I think that "yes".[/hide] $k\in \overline {1,n},\ x: =\frac 1n\ln k\Longrightarrow \left(\forall\right)k\in \overline{1,n},\ \sqrt [n] k\ge 1+\frac{\ln k}{n}\Longrightarrow$
$n+\frac{\ln n!}{n}\le \sum_{k=1}^n \sqrt [n] k\le n\sqrt [n] n\Longrightarrow\frac{\ln n!}{n}\le \sum_{k=1}^n \sqrt [n] k -n\le n\left(\sqrt [n] n-1\right)\Longrightarrow$
$u_n\equiv \boxed {\frac{\ln n!}{n\ln n}\le \frac{\sum\limits_{k=1}^n \sqrt [n] k -n}{\ln n}\le \frac{n\left(\sqrt [n] n -1\right)}{\ln n}}\equiv v_n\ .$

$\lim_{n\to\infty} \frac{\ln (n+1)!-\ln n!}{(n+1)\ln (n+1)-n\ln n}=$$\lim_{n\to\infty} \frac{\ln (n+1)}{\ln \left(1+\frac 1n\right)^n+\ln (n+1)}=1\Longrightarrow \lim_{n\to\infty} u_n=1\ .$

$x_n\equiv\frac{\ln n}{n}\rightarrow 0\Longrightarrow v_n=\frac{e^{\frac{\ln n}{n}}-1}{\frac{\ln n}{n}}=\frac{e^{x_n}-1}{x_n}\rightarrow 1\ .$[/hide]

This problem was proposed today at arhimede contest at 11-th grade. If I were in the 11-th grade I woluld have been very lucky :D


analytic geometry
geometry
circumcircle
algorithm

Let $S$ be a finite subset of the Cartesian plane and $d$ be the diameter of $S$.

Prove that if $D$ is the[i][b] smallest[/b][/i] circular disc covering $S$ , then the radius of $D\leq \frac{\sqrt 3}3 d$.

Remark. [i]The diameter of a subset of S is defined as the maximum distance between any two points in the set.[/i]

Less than; there is a maximum value that $\textrm{radius}\left(D\right)$ can take on.

Is the diameter unique (i.e there exists 2 and only 2 points within S that define d)? I think it makes a difference to know this.

No, the problem should hold in either case.

It also shouldn't matter if the maximum is actually achived in $S$.  For instance, if you consider the set of points with coordinates $(x, 0)$ for $0 < x < 1$, that set has diameter 1 even though there are no two points in that set exactly 1 unit separated from each other.  That won't affect the truth of the statement here.

[quote="JBL"]The question should say that $D$ is the [i]smallest[/i] disk covering the set.  (Obivously, we can increase the radius of $D$ as much as we want and still have it cover.)  I think this fixes it, yes?[/quote]

Yes, it should be the smallest disc.

[hide="proof"] Assume there exists 2 points $k_1$ and $k_2$ within $S$ that define $d$. The only other point that can possibly exist within $S$ without redefining $d$ is the point $k_3$ that is a distance $d$ away from both $k_1$ and $k_2$. You will now notice that these 3 points form an equilateral triangle with side measure $d$. Note that there exists one and only one such point $k_3$, for if there was another point (call it $k_4$) a distance $d$ away from both $k_1$ and $k_2$, then the distance between $k_3$ and $k_4$ would be greater than $d$, redefining it. Therefore, it is sufficient enough to find the smallest possible $D$ that contains the 3 points $k_1$, $k_2$, and $k_3$. Such a $D$ is the circumscribed circle formed from an equilateral triangle of side measure $d$, and the radius of such a circle is $\frac{\sqrt[]{3}}{3}$ $d$. Therefore, $Radius_D$ $\leq$ $\frac{\sqrt[]{3}}{3}$ $d$, $Q.E.D$ [/hide]

[quote="luimichael"]Let $S$ be a finite subset of the Cartesian plane and $d$ be the diameter of $S$.

Prove that if $D$ is a circular disc covering $S$ , then the radius of $D\leq \frac{\sqrt 3}3 d$.

Remark. [i]The diameter of a subset of S is defined as the maximum distance between any two points in the set.[/i][/quote]

[hide]
Let $A,B \in S$ such that the distance between $A$ and $B$ is $d$. Draw a circle of radius $d$ with $A$ as center. Do the same for $b$. All points in $S$ must belong in the intersection of these two circles in order to satisfy the definition of diameter. Thus, it suffices to find the radius of the circle that can cover the intersection. Connect $A$ and $B$ and let the intersection of the two circles on the top be $C$. $ABC$ is an equilateral triangle. The height of the radius of the minimal disk covering is the height of this equilateral triangle, which is $\frac{d}{\sqrt{3}}$. A circle with this radius can be situated on the midpoint of $\overline{AB}$ and that would suffice.

Sorry if this is unclear; I am too lazy to draw this in geometer's sketchpad.[/hide]

[quote="quantabe"]The only other point that can possibly exist within $S$ without redefining $d$ is the point $k_3$ that is a distance $d$ away from both $k_1$ and $k_2$.[/quote]
[quote="agolsme"]The height of the radius of the minimal disk covering is the height of this equilateral triangle, which is $\frac{d}{\sqrt{3}}$.[/quote]

Neither of these statements is true.

[quote="JBL"][quote="quantabe"]The only other point that can possibly exist within $S$ without redefining $d$ is the point $k_3$ that is a distance $d$ away from both $k_1$ and $k_2$.[/quote]

Neither of these statements is true.[/quote]

I didn't word it quite correctly. What i meant to say is that for a certain defined diameter $d$, if there are more than those 2 points within $S$ that make up $d$ then the "farthest" distance a POTENTIAL point within S could have would be the one a distance $d$ away from both $k_1$ and $k_2$.

If you still have problems with this, please explain why.

Certainly, two points at a distance $d$ define other potential points which are "as far away as possible."  However, it is not at all clear that putting a single point "as far away as possible" is the best way to go -- maybe, using 3 other points, I could make some sort of pentagon with a larger circumcircle than I'll get with your method.  When you pick your third point, you give up a lot of other potential points, so you have to justify that choice.  (The method you're trying is called the greedy algorithm -- taking at much at each step as possible.  While in many situations, including this one, it gives you the right answer, it doesn't always.  So using the greedy algorithm to get a maximum isn't a proof of a maximum -- you still have to prove you can't do better.)

[hide="Hint "]
Consider the three different cases separately:

Case 1. Only one point of S is lying on the boundary of D.

Case 2. Exactly two points of S are lying on the boundary of D.

Case 3. At least three points of S are lying on the boundary of D.
[/hide]

[hide="More hints"]
Case 1
It occurs if and only if S is a singleton set.

Case 2
Suppose $P_1$ and $P_2$ are the two points of S lying on the boundary of D.
Then $P_1P_2$ is in fact the diameter of D.

Case 3
If S has exactly 3 points, then the boundary of D will be the circumcircle of S.
Else S has more than 3 points, let $P_1$, $P_2$ and $P_3$ be the any three points of S on the boundary of D.
Then the three arcs seperated by these three points are all MINOR ARCS.
$\Delta P_1P_2P_3$ will provide enough information to complete the proof.[/hide]


LaTeX
algebra
polynomial
calculus
calculus computations

Lat $ x_1,..x_n\in R,x_i\#x_j$
Find $ S_k,k\ge n$
with $ S_k = \sum_{i = 1\to n} \frac {x_i^k}{(x_i - x_1)..(x_i - x_{i - 1})(x_i - x_{i + 1})..(x_i - x_n)}$
it is hard so who can help me??

Some useful LaTeX: \neq makes a not-equals sign.  \ldots makes the elipsis $ \ldots$, while \cdots makes the elipsis $ \cdots$.  \mathbb{R} makes $ \mathbb{R}$ (and this can be applied to other letters as well, e.g. $ \mathbb{Z}$).

The sum you've written is $ h_{k\minus{}n}({\bf x})$, the complete homogeneous symmetric polynomial of degree $ k\minus{}n$ in the variables $ x_1, \ldots, x_n$.  It's the sum of every degree-$ (k\minus{}n)$ monomial made up of the $ k$ variables.

[quote="JBL"]Some useful LaTeX: \neq makes a not-equals sign.  \ldots makes the elipsis $ \ldots$, while \cdots makes the elipsis $ \cdots$.  \mathbb{R} makes $ \mathbb{R}$ (and this can be applied to other letters as well, e.g. $ \mathbb{Z}$).

The sum you've written is $ h_{k \minus{} n}({\bf x})$, the complete homogeneous symmetric polynomial of degree $ k \minus{} n$ in the variables $ x_1, \ldots, x_n$.  It's the sum of every degree-$ (k \minus{} n)$ monomial made up of the $ k$ variables.[/quote]
Can you write your roof,I really need it,thank you


geometry
trigonometry
ratio
quadratics
inequalities
circumcircle
area of a triangle

Is it possible to define the three edges of a triangle where the three distances of the verteces from the orthocenter are given?

You have misunderstood the question: Not the altitudes of the triangle ABC are given, but the distances from the orthocenter to the vertices. This makes the problem much more complicated. Actually, there are algebraic ways to compute the sides of triangle ABC from these distances by solving cubic equations, but (in general) you cannot construct triangle ABC by compass and ruler, given these distances.

Here is a thread (in German) about this problem: http://groups.google.com/groups?hl=de&lr=&th=cc272176633c0839&seekm=1c5184ca.0405280359.1c94ab36%40posting.google.com&frame=off .

  Darij

[quote="mount"]Is it possible to define the three edges of a triangle where the three distances of the vertices from the orthocenter are given?[/quote]

Let $H$ be the orthocenter of the desired triangle $\triangle ABC$. Denote the triangle sides and angles as usual: $a = BC, b = CA, c = AB$, $\alpha = \measuredangle CAB, \beta = \measuredangle ABC, \gamma = \measuredangle BCA$. The given segments $AH, BH, CH$ are equal to $R \cos \alpha,  R \cos \beta,  R \cos \gamma$, respectively, where $R$ is the triangle circumradius. Denote

$x = \frac b c, y = \frac c a, z = \frac a b, xyz = 1$

$p = \frac {\cos \beta}{\cos \gamma}, q = \frac {\cos \gamma}{\cos \alpha},  r = \frac {\cos \alpha}{\cos \beta}, pqr = 1$

The ratios $p, q, r$ are known. If we could find the ratios $x, y, z$, we could construct a triangle $\triangle A'B'C'$ similar to the desired triangle $\triangle ABC$, find its segment $A'H'$ and then construct the desired triangle $\triangle ABC \sim \triangle A'B'C'$ with known coefficient of similarity $\frac {AH}{A'H'}$.

$a = b \cos \gamma + c \cos \beta$
$b = c \cos \alpha + a \cos \gamma$
$c = a \cos \beta + b \cos \alpha$

$x = \frac b c = \frac{c \cos \alpha + a \cos \gamma}{a \cos \beta + b \cos \alpha} = \frac{y + q}{\frac 1 z + \frac 1 r} = \frac{y + q}{xy + pq}$

By cyclic exchange of $x, y, z$ and $p, q, r$,

$y = \frac{z + r}{\frac 1 x + \frac 1 p} = \frac{\frac{1}{xy} + \frac{1}{pq}}{\frac 1 x + \frac 1 p} = \frac{xy + pq}{qy(x + p)}$

$z = \frac{1}{xy} = \frac{x + p}{\frac 1 y + \frac 1 q} = \frac{qy(x + p)}{y + q}$

Only 2 of the equations for $x, y, \frac{1}{xy}$ are independent. Using the first two equations for $x, y$,

$x(xy + pq) = y + q$
$qy^2(x + p) = xy + pq$

Calculating $y$ from the 1st equation, substituting to the 2nd one, reducing by $q$ and multiplying by $(1 - x^2)^2$,

$y = q\frac{px - 1}{1 - x^2}$

${q^3 \frac{(px - 1)^2}{(1 - x^2)^2}(x + p) = qx \frac{px - 1}{1 - x^2} + pq}$

${q^2 (px - 1)^2 (x + p) = (1 - x^2)[x(px - 1) + p(1 - x^2)]}$

${q^2 (px - 1)^2 (x + p) = (1 - x^2)(p - x)}$

This is a cubic equation for $x$. The equation has either one or three real roots and if at least one root is positive, the sides of the triangle $\triangle ABC$ can be defined, but not constructed with compass and straightedge.

This cubic equation will reduce to a quadratic equation, if at least 2 given segments $AH, BH, CH$ are equal to each other. If $BH = CH$, then $p = 1$, if $CH = AH$, then $q = 1$ and if $AH = BH$, then $r = \frac{1}{pq} = 1$ or $q = \frac 1 p$. In the 1st case, the cubic equation can be reduced by $1 + x \neq 0$, in the 2nd case, the absolute term of the cubic equation is zero and it can be reduced by $x \neq 0$ and in the 3rd case, the coefficient of the cubic term is equal to zero. Hence, an isosceles triangle can always be constructed, given 2 different distances of the vertices from the orthocenter.

[quote="Fiachra"]I believe so: in Heron's formula, we can divide across by $(ABC)^2$; on the RHS, this means replacing $a$ by $\frac{2}{h_a}$ etc. in every instance. Thus we find $(ABC)$, and then we can use $(ABC) = \frac{1}{2}a h_a$.[/quote]

The altitudes $h_a, h_b, h_c$ of a triangle $\triangle ABC$ are in the ratios

$h_a : h_b : h_c = \frac 1 a : \frac 1 b : \frac 1 c$

or

$a : b : c = \frac{1}{h_a} : \frac{1}{h_b} : \frac{1}{h_c}$

Hence, if reciprocals of the given segments $h_a, h_b, h_c$ satisfy the triangle inequality, it is possible to construct a triangle $\triangle A'B'C'$ with the sides $a' = \frac{1}{h_a}, b' = \frac{1}{h_b}, c' = \frac{1}{h_c}$ similar to the desired triangle $\triangle ABC$, find one of its altitudes $h_a'$ and then construct the desired triangle $\triangle ABC \sim \triangle A'B'C'$ with known similarity coefficient $\frac {h_a}{h_a'}$. But this is a different problem.

Thank you Yetti and Darij for your answers. You are very helpfull.

So I should guess now that this is the only way to find the circumradius R of this triangle? First compute the edges and after that the R, or there is a more direct way?

[i]Thanks!![/i]


function
calculus
integration
real analysis
real analysis unsolved

(a) Let $ E$ be a measurable subset of $ \mathbb{R}^{2}$ such that for almost every $ x\in\mathbb{R}$, $ \{y\mid (x,y)\in E\}$ has $ \mathbb{R}$-measure 0.  Show that $ E$ has measure 0, and that for almost every $ y\in\mathbb{R}$, $ \{x\mid (x,y)\in E\}$ has measure 0.  Let $ f(x,y)$ be nonnegative and measurable in $ \mathbb{R}^{2}$.
(b) Suppose that for almost every $ x\in\mathbb{R}$, $ f(x,y)$ is finite for almost every $ y$.  Show that for almost every $ y\in\mathbb{R}$, $ f(x,y)$ is finite for almost every $ x$.

I feel there should be some way to use Fubini's Theorem, Tonelli's Theorem, or some application of them.

Absolutely correct. Use Fubini on the characteristic function of $ E$ for (a). Take $ E\equal{}\{f\equal{}\plus{}\infty\}$ for (b).

I appreciate the help but I am having trouble setting up the integral, which seems silly.  Would it be $ m(E)\equal{}\iint_{\mathbb{R}^{2}}\chi_{E}\equal{}\int_{x\in E}\chi_{\{y\mid (x,y)\in E\}}\equal{}\int 0\equal{}0$ for the first one?  Is there a way to rearrange this to get that $ \{x\mid (x,y)\in E\}$ has measure 0?

For the second, I'm pretty sure Tonelli's Theorem is appropriate, but how do we set it up?

I guess the main question I'm having is setting up the boundaries/limits of integration.

For the first one, you've done half of the work. Just think now what will happen if you change the order of integration.

For (a), let $ E_x = \{y: (x,y)\in E\}$ and $ E^y = \{x: (x,y)\in E\}$. Then $ E = E_x\times E^y$
Hence
\begin{eqnarray*} (\mu\times \nu)(E) & = & \int_Ed(\mu\times \nu) \\
& = & \int_{E^y}\int_{E_x}d\nu\, d\mu \\
& = & \int_{E^y}0\, d\mu \, \, \, \, \, \text{because for a.e.} x, \nu(E_x) = 0 \\
& = & 0 \end{eqnarray*}
This is just what JRav has done.

Now for the second part of (a), observe that $ \int_{\mathbb{R}}\int_{E_x}d\nu\, d\mu = 0\, \,$ because it is given that for a.e.$ x,\, \, \nu(E_x) = 0$
But by Tonelli, $ \int_{\mathbb{R}}\int_{E_x}d\nu\, d\mu = \int_{E_x}\int_{\mathbb{R}}d\mu\, d\nu$
So for a.e.$ y$, $ \mu(E_y) = 0$.

Am I right?

[quote="limsup"]
But by Tonelli, $ \int_{\mathbb{R}}\int_{E_x}d\nu\, d\mu \equal{} \int_{E_x}\int_{\mathbb{R}}d\mu\, d\nu$
Am I right?[/quote]
Almost. It should really be $ \int_{\mathbb{R}}\int_{E_x}d\nu\, d\mu \equal{} \int_{{\color{red}\mathbb R}}\int_{{{\color{red}E^y}}}d\mu\, d\nu$. Also, nothing can be farther from the truth than the identity $ E \equal{} E_x\times E^y$ ;).

Thanks!  Along similar lines, if $ f$ and $ g$ are measurable on $ \mathbb{R}^{n}$, how would I show that $ h(x,y)\equal{}f(x)g(y)$ is measurable on $ \mathbb{R}^{n}$?  The problem is that if I assume that $ E_{1}$, $ E_{2}$ are measurable in $ \mathbb{R}^{n}$ and so $ E_{1}\times E_{2}$ is measurable in $ \mathbb{R}^{2n}$, and then construct approximating simple functions (which I can do), I am doing the problem backwards-I'm supposed to deduce the measurability fact from $ fg$ being measurable, not the other way around.

It suffices to show that the functions $ \left(x,y\right)\mapsto f\left(x\right)$ and $ \left(x,y\right)\mapsto g\left(y\right)$ are measurable. The preimage of a measurable set $ B$ under $ f$ is $ f^{\minus{}1}\left(B\right)\times\mathbb{R}^{n}$, which is measurable as a product of measurable sets. The same argument applies for $ g$.

[quote="fedja"]Also, nothing can be farther from the truth than the identity $ E \equal{} E_x\times E^y$ ;).[/quote]
I am wondering whether this is true :o

Whether which part is true, the identity or that fedja is saying it's not true?

Oh sorry JRav... I didn't really understand what fedja said at that time. Now I realize that he is saying the identity is completely false. Yes I do really agree that it's completely nonsense.
But for the problem, we can argue in the following way.
\begin{eqnarray*}(\mu\times\nu)(E)& = &\int_{\mathbb{R}}\nu(E_{x})\,d\mu,\,\,\,\,\,\text{(this is by a result)}\\ & = &\int_{\mathbb{R}}0\, d\mu,\,\,\,\,\,\text{(because for a.e.}x,\nu(E_{x}) = 0)\\ & = & 0\end{eqnarray*}


geometry
geometric transformation
reflection
inequalities
rotation
trigonometry
geometry solved

Let triangle $ABC$ be equilateral, with side $a = 2$ and center $O_1$. Inside $ABC$ we have another equilateral triange $DEF$, with side $b = 1$ and center $O_2$. Prove, that reflection of the point $O_1$ in the point $O_2$ belongs to both triangles.

Let $T$ be the equilateral triangle with side 1 and centre $O_1$, homothetic to the larger of the given triangles (call this $T_1$). Then $O_2$ must lie within $T$, otherwise its incircle will intersect $T_1$. But the reflection of $O_1$ through any point on $T$ is a point on $T_1$, as can be easily shown.

This is, of course, obnvious. You must prove that the reflection of $O_1$ in $O_2$ lies in both triangles, $ABC$ (this, you have shown) and $DEF$.

[quote="Fiachra"] $O_2$ must lie within $T$, otherwise its incircle will intersect $T_1$[/quote]

Can you prove it ?

I'll prove the initial problem for the case when two of the vertices of $DEF$ lie on two sides of $ABC$, and I'll let you guys figure out why it suffices :).

Assume $E,F$ belong to $CA,AB$ respectively. It's not hard to show that in this case, $AFO_2E$ is cyclic, $AO_2$ is the bisector of $\angle FAE=\angle BAC$, and this means that $A,O_2,O_1$ are collinear. If $T=FE\cap AO_1$, then the inequality we want is $AO_1-AO_2\le O_2T$, which is equivalent to $O_2T\ge O_2E+O_2F-O_2A\iff O_2T+O_2A\ge O_2E+O_2F$. This follows from AM-GM applied to $O_2T,O_2A$, taking into consideration the fact that $O_2T\cdot O_2A$ is constant and equals $O_2E^2$.

And now for the non-trivial part:

Suppose $a$ is the side closest to $O_2$ and let $K$ be the foot of the perpendicular from $O_1$ to $a$. Let $P$ be the reflection of $O_1$ in $O_2$, $B$ the point of intersection of $O_1 O_2$ and $a$ and $Q$ the midpoint of $BP$. Let $\theta = \angle KO_1 B$. Now if $O_2$ lies on $T$, then one of its sides must lie on $a$, and hence $P$ lies on that side. Otherwise, let us start with $T_2$ having a side parallel to $a$. Let $\alpha$ be the angle through which $T_2$ can be rotated about $O_2$ before intersecting $a$, and $\beta$ the angle through which it must be rotated to bring $P$ outside $T_2$. We will show that $\beta \geq \alpha$.

Let $d$ be the perpendicular distance from $O_2$ to $a$ and let $x$ be the distance from $Q$ to $a$. It is clear that $d=x+\frac{1}{2 \sqrt{3}}$. Suppose we rotate $T_2$ so that a vertex $D$ approaches $a$. When $D$ meets $a$, $O_2 D=\frac{1}{\sqrt{3}}$ and so $\cos (60-\alpha) = d \sqrt{3}=x \sqrt{3} + \frac{1}{2}$.

$\beta$ is also the angle through which $P$ must be rotated before meeting a side of $T_2$. The length of the perpendicular from $P$ to the side of $T_2$ (in its original position) is $x$. (Note that $Q$ is on this side.) We have $O_2 Q=\frac{1}{2 \cos \theta \sqrt{3}}$, $PQ=\frac{x}{\cos \theta}$ and $O_2 P = \frac{1}{2 \sqrt{3} \cos (\theta - \beta)}$. Hence $\frac{\cos (\theta - \beta)}{\cos \theta}=\frac{1}{1-2\sqrt{3} x}$.

Now since $\theta < 60$, $2 \cos (60 - \beta)=\cos \beta + \sin \beta \tan \60 \geq \cos \beta+ \sin \beta \tan \theta = \frac{\cos (\theta - \beta)}{\cos \theta} = \frac{1}{1-2\sqrt{3}x} \geq 1+2\sqrt{3}x = 2 \cos (60 - \alpha)$. Hence $\beta > \alpha$, as required.


what are the millenium problems.I have just heard of these things like Navier-Stokes Equation,Riemann hypothesis,etc. Someone pls send me the questions of the problems.

Please take a look. :)

[url=http://www.qeden.com/wiki/Millennium_Prize_Problems]www.qeden.com/wiki/Millennium_Prize_Problems[/url]

[url=http://www.claymath.org/millennium/]www.claymath.org/millennium/[/url]


linear algebra
matrix
quadratics
number theory
superior algebra
superior algebra unsolved

let A be a symmetric matrix (n dimensional) with integer entries
we call it congruent with another one : B , if and only if there is a matrix C, invertible and with an inverse with integer entries, such that

transpose(C)AC=B

in landau , elementary number theory, this fact is used :

if n=2/3, a quadratic form is congruent with the form with identity matrix if and only if it is positive definite and the determinant is 1
(it was a proof of the three squares theorem)

is this true for all n? are all positive definite integer matrices with determinant 1 congruent to the identity matrix?

thx, fred

Far from it.  For example, there are more than 80 million forms incongruent forms in 32 variables.  See "A course in arithmetic" by Jean-Pierre Serre, GTM 7, for an introduction.

Thanks, I posted this so long ago I even forgot!

I suppose there is no simple pair of integer matrices A and B, both positive and with determinant 1 and obviously noncongruent?


function
number theory unsolved
number theory

Problem 10 in sec 1.3 reads :" Prove that a positive integer of form 3k+2 has a prime factor  of same form"

I proved it by the following argument. Let us suppose 3k+2 does not have a prime factor of form 3k+2. Every factor has either the form 3k or 3k+1. Let's see how a products involving 3k and 3k+2 look,

(3k+1)(3k+1)=3k+1
(3k+1)(3k)=3k

Hence, if 3k+2 is decomposed into prime factors there must at least be one prime factor of form 3k+2. Similarly we have numbers 4k+3 and 6k+5 with the same property.

Here I go ahead and generalize,
Let us say that  the ordered pair $ (a,b)\in R$ iff $ ak\plus{}b$ has a prime factor of same form( $ R$ is a relation here, not to be confused with real numbers). Of course, $ 1\leq b<a$

(i) Do we have infinitely many ordered pairs in $ R$ ?
(ii) Can every number $ \geq 1$ be equal to $ a$ in atleast one ordered pair? (seems not, what about 5 ?) How do we find the subset of numbers that can be equal to $ a$ atleast in one ordered pair ?
(iii) For a given $ a$, do we have function (existence) that gives the list of all $ b$ so that $ (a,b)\in R$ ? Whats the function ?

If $ gcd(a,b)\equal{}1$, then $ 2,3,4,6$ are the only possible values of $ a$, because $ \varphi(a)\leq2$ (number of nonnegative integers less than $ a$ and coprime with $ a$). Suppose that $ \varphi(a)\geq3$ and $ gcd(a,b)\equal{}1$. There exists a number $ c$ coprime with $ a$ neither equal to $ b$ nor to $ 1$. There also exists a number $ d\equiv \frac{b}{c}(mod a)$. So $ ak\plus{}b\equal{}cd$, contradiction ($ d$ is also neither equal to $ b$ nor to $ 1$).
If there exists a prime $ p$ dividing $ gcd(a,b)$, then $ b\equal{}kp$ and $ b\neq k$ so $ b\equal{}p$. But $ (a\minus{}1)(a\minus{}p)\equal{}a(a\minus{}1\minus{}p)\plus{}p$ and $ a\minus{}1\not\equiv p(moda)$ so $ a\minus{}p\equiv p(moda)$ and $ a\equal{}2p$.
So $ R\equal{}\{(2p,p): p\minus{}$prime$ \}\cup\{(2,1),(3,2),(4,3),(6,5)\}$.


This question might have been asked before so please forgive me if it has, but I didn't see anything in the first couple search results so here goes.  What is the farthest an AoPS member has gotten in terms of AMC competitions/MATHCOUNTS?

I know many members have qualified for the USAMO, but has anyone been to MOP or IMO, AoPS ID's names would be helpful too.

at least three people (on the old AoPS, before the merge) whose ids I know, went to IMO (2 from US, 1 from New Zealand). After the merge, I expect we have dozens of IMOers representing various countries.

I'm not sure what you are asking here. If you are asking about members of the AoPS - MathLinks community which went to IMO, I think it is very big (I personally know 30 or so gold medalists at IMO, and many more Silver, Bronze medalists).

Actually some of the moderators and admins are former IMO medalists :P

There are also a couple of IMOers from US that took some AoPS Classes prior reaching the IMO, if the question was directed in that sense. :)

An old poll is relevant in this case: http://www.mathlinks.ro/Forum/viewtopic.php?t=2551

[quote="mna851"]I know many members have qualified for the USAMO, but has anyone been to MOP or IMO, AoPS ID's names would be helpful too.[/quote]

Some people may not wish to have their sn disclosed, so I'm not sure the admin could actually just post a list of who's done what. However, if you ask around or make a thread for "Whoever's been to MOP (or USAMO or IMO or etc) please post here" I'm sure you'll get responses. As Valentin said, there's probably more than 50 IMOers here (from various countries), 50 MOPpers, hundreds of USAMOers, and a lot more MC National Finalists than anyone could name.


Let $ n$ be a fixed positive integer. Define a relation $ R$ on the set $ Z$ of integers by, $ a R b \Leftrightarrow n|a\minus{}b.$ What does the symbol "|" indicate here? Thank you.

$ n|a\minus{}b$  means   n divides  (a-b)

[quote="turikachi"]$ n|a \minus{} b$  means   n divides  (a-b)[/quote]

ok. thank you. :-)


Add two to previous addition.

    1        sqrt: 1
+  3
_____
     4       sqrt:  2

+   5
_____
      9      sqrt:  3

+   7
_____
      16     sqrt:  4

+    9
_____
     25      sqrt: 5
ect.

Challenge:  formula to determine when a sequence of additions will reach an integer square?
Example problem: number of additions is 9.

     27      :(offset)
+   89      :(begin two plus previous addition)
_____
    116

+    91
______
     207

+     93
_______
      300

+      95
_______
       395

+       97
________
        492

+        99
________
         591

+       101
________
         692

+       103
_________
          795

+        105
_________
          900    sqrt: 30

Number of additions is nine.  What is a formula to determine when sum of additions will reach the square of an integer?

Any comments  or suggestions can be set to: atomicman631@aol.com
                                                     Subject: square formula

                                                                        Thanks
                                                                         Rob

Could you typeset this problem in $\LaTeX$?  I cant understand a thing.

Where did you get this problem?  It seems interesting.

Shouldn't this be moved to a (high school) math forum?

It is a simple perfect square formula.

(n+1)^2=n^2+2n+1


30 students in a class and they are paired up. what is the probability that margo is paired with irma? aren't there 435 pairs using n!/(n-r)!n!? then margo and irma could be one pair, so why isnt there answer 1/435? i think my reasoning's really wrong so please help!
thanks

Am I wrong or is the answer [hide]1/29[/hide] ?

First, let's just say what finite probability [i]is[/i]: the total number of "good" ways over the total number of ways.

Now, there are a lot of ways to go about this problem.  I'll do one, but I recommend that anyone else posts other methods, since this is not necessarily the cleanest.

Any pairing of people can be done by lining the 30 people up and then putting the first 2 into a pair, and then the second 2, and so on.  There are 30! line-ups.  However, we get the same result if we switch two people who are in a pair, so we have to divide by 2^15.  Also, if we switch the 15 pairs around, we get 15! lineups leading to the same pairing.  Thus, we have 30!/(15! * 2^15) total pairings.  (You can also get 30!/(2^15) if you consider C(30, 2), C(28, 2) and so on, picking one pair at a time.)

Now, how many ways can we put the two people in question together?  We have that they are one pair.  Then, the other 28 must be divided into 14 pairs.  Analogously to above, this can be done in 28!/(2^14 * 14!) ways.  So the probability is
[28!/(2^14 * 14!)] / [30!/(2^15 * 15!)] = 15 * 2 / (30 * 29) = 1/29.

Another way:  if we line them up, the first of our couple has to go somewhere.  What's the probability that the other one is in a position to be paired with the first?

Hm I thought of it like this :

Consider Irma's pair : (Irma, X) where X is unknown.
There are 29 possibilities for the unknown X.
Exactly one of them wil give us the good pair (Irma, Margo).
So the probability is 1/29.


algebra
polynomial
function
analytic geometry
Vieta
graphing lines
slope

Let $ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d\ ,\ x\in \mathcal R$ be a polynominal function, where $ a\ne 0\ .$
Let $ d$ be a line for which $ d\cap G_f \equal{} \{P_k\ |\ k\in \overline {1,3}\}$, where $ G_f$ is the graph of $ f\ .$
For any $ k\in\overline {1,3}$ the tangent to $ G_f$ in $ P_k$ meet again $ G_f$ in the point $ X_k\ .$ Prove :
$ 1\blacktriangleright$ The points $ X_k\ ,\ k\in\overline {1,3}$ are collinearly. and
$ 2\blacktriangleright$ The extremum points (if these exist) and the inflexion point of $ G_f$ are collinearly.

Very Interesiting Problem!

Solution for 1.

Since we can set the coefficient of $ x^2$ of $ f(x)$ as zero by pararell displacement for $ x$ direction, we will set $ b = 0$ follows.Suppose that $ P_k,\ X_k\ (k = 1,\ 2,\ 3)$ are mapped to $ P_k',\ X_k'$ and the line $ P_1P_2P_3$ denote by $ y = l_d(x)$, $ y = f(x)$ be mapped to $ y = l_0(x),\ y = f_0(x): = ax^3 + cx + d$ respectively. Let $ \alpha _k: \ x$ coordinate of $ P_k',\ \beta _k: \ x$ coordinate of $ X_k'$ and $ l_k': \　y = m_kx + n_k\equiv{l_0(x)}$ be the tangent at $ X_k'$. We can have $ f_0(x) - l_0(x) = a(x - \alpha _k)^2(x - \beta _k)$, i.e.  3 solutions of $ f_0(x) - l_0(x) = 0$ are $ \alpha ,\ \alpha ,\ \beta$, so Vieta's formula gives $ \alpha _k + \alpha _k + \beta _k = 0\Longleftrightarrow \beta _k = - 2\alpha _k\ \cdots \ (*)$. Now let $ m_{i,\ j}$ be the slope of the line $ P_i'P_j'$, W.L.O.G. $ \alpha _1 < \alpha _2 < \alpha _3$ and $ a > 0$, we have  $ m_{1, 2} = \frac {f_0(\alpha _2) - f_0(\alpha _1)}{\alpha _2 - \alpha _1} = \frac {\int_{\alpha _1}^{\alpha _2} f'_0(x)\ dx}{\alpha _2 - \alpha _1}$

$ = \displaystyle \frac { - 3a\int_{\alpha _1}^{\alpha _2} (x - \alpha _1)(x - \alpha _2)\ dx}{\alpha _2 - \alpha _1} = \frac {a}{2} (\alpha _2 - \alpha _1)^2$, similarly $ m_{2,\ 3} = \frac {a}{2}(\alpha _3 - \alpha _2)^2$. Thus 3 points $ P_k'$ are collinear $ \Longleftrightarrow m_{1,\ 2} = m_{2,\ 3}\Longleftrightarrow (\alpha _2 - \alpha _1)^2 = (\alpha _3 - \alpha _2)^2\ (\alpha _1 < \alpha _2 < \alpha _3)$, yielding $ 2\alpha _2 = \alpha _1 + \alpha _3$.　Consequently by $ (*)$, we obtain $ 2\beta _2 = \beta _1 + \beta _3$ which completes the proof.

Solution for 2:

For $ f_0(x) = ax^3 + cx + d\ (a\neq 0)$, we have $ f'_0(x) = 3ax^2 + c,\ f''_0(x) = 6ax$, yielding the point of local extrema are $ (\alpha _0,\ f(\alpha _0)),\ (\beta _0,\ f(\beta \0))\ (\alpha _0 < \beta _0)$, where $ \alpha _0,\ \beta _0$ are the $ x$ coordinates of the point of local extrema and the inflection point is $ (0,\ 0)$.  Consequently we obtain $ \alpha _0 + \beta _0 = 0\Longleftrightarrow 2\cdot 0 = \alpha _0 + \beta _0$, which completes the proof by the former problems argument.


Una palabra de longitud $n$ es una sucesion ordenada $x_1x_2...x_n$ donde $x_i$ es una letra del alfabeto $\{a,b,c\}$. Sea $A_n$ el conjunto de palabras de longitud $n$ que no contienen nigun bloque $x_ix_{i+1}$ de la forma $aa$ o $bb$ y sea $B_n$ el conjunto de palabras de longitud $n$ donde no hay ningun bloque $x_ix_{i+1}x_{i+2}$ que sea una permutacion de $\{a,b,c\}$. Demuestre que $|B_{n+1}|=3|A_n|$

i remember seeing this problem posted quite a few times on ML. can anyone help me find it?

Here

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=174213#p174213]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=174213#p174213[/url]

La solucion de Severius me ha parecido super!!
bueno y tambien la de leonardo muy interesante ...!!!

Carlos Bravo
Lima -Peru


A certain function $ f$ has the properties that $ f(3x) \equal{} 3f(x)$ for all positive real values of $ x$, and that $ f(x) \equal{} 1 \minus{} \mid x \minus{} 2 \mid$ for $ 1\leq x \leq 3$. Find the smallest $ x$ for which $ f(x) \equal{} f(2001)$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=171819828&t=77811
How did chess64 get f(x)=186?

What he do to get from the second to last line to the last line? What does the subscript "min" mean?

To answer your second question, the subscript min is short for minimum, so x_min is the smallest x, or the number you are looking for in this question.
As per the first question, he/she solved for x, but changed its name to x_min.

There's a really nice solution at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=714302[/url].


geometry
analytic geometry
linear algebra
matrix
geometric transformation
homothety

Let the incircle (with center $I$) of triangle $ABC$ touch the side $BC$ at $X$, and let $A'$ be the midpoint of this side. Prove the line $A'I$ (extended) bisects $AX$.

It's very easy with barycentric coordinates.

Let $W$ be the midpoint of $AX$ and $p = \frac{BC+CA+AB}2 = \frac{a+b+c}2$.
The homogenous barycentric coordinates are as follows: $A^\prime (0,1,1)$, $I (a,b,c)$, $A(1,0,0)$, $X(0,p-c,p-b)$.
The normalized barycentric coordinates of $X$ are $\left( 0,\frac{p-c}{a},\frac{p-b}a \right)$, so the normalized coordinates of $W$ are $\left( \frac12, \frac{p-c}{2a}, \frac{p-b}{2a}\right)$.
Thus, $W$ has homogenous barycentric coordinates $\left( a, p-c, p-b \right)$.
We want to prove that $A^\prime, I$ and $W$ are collinear. That's equivalent to
$ \, \left| \begin{matrix}0 & 1 & 1 \\ a & b & c \\ a & p-c & p-b \end{matrix}\right| = 0$
$ \, \Longleftrightarrow \, \left| \begin{matrix}0 & 0 & 1 \\ a & b-c & c \\ a & (p-c)-(p-b) & p-b \end{matrix}\right| = 0$
$ \, \Longleftrightarrow \, \left| \begin{matrix}0 & 0 & 1 \\ a & b-c & c \\ a & b-c & p-b \end{matrix}\right| = 0$
$ \, \Longleftrightarrow \, \left| \begin{matrix}a & b-c \\ a & b-c \end{matrix}\right| = 0$
$ \, \Longleftrightarrow \, a(b-c)-a(b-c) = 0$, which is obviously true.

Draw $B',C'$ on $AB,AC$ such that $B'C'//BC$ and tangent to Incircle $(I,r)$ at $K'$, if $K$ is the touch point of the excircle $(I_{a}, r_{a})$ at $BC$. So $K'$ is the image by homothety $h=-\frac{r}{r_{a}}$ with center $A$ besides $K'X$ is the diameter of $(I)$ and $A,K',K$ are collinears. Then if $A'$ is the midpoint of $BC$ also $A'$ is midpoint of $XK$, hence $A'I$ is the midline in the triangle $K'XK$, now $A'I$ passes through the midpoint of $AX$ too.


Click on the geogebra button (the rightmost button above the composition window). The graphing window will appear. If you want the full menu, double click on the graphing window. Make your picture, then click the Insert Picture button.


[quote="watermaximillion"]I got my mail notification today.[/quote]

Me too.

I will follow the trend and announce that, in fact, I did receive my notification today.

got it

(what exactly is the point of this thread, other than to make people who haven't gotten it yet freak out unnecessarily?)

Yeah...got it today too. Yeah, there's not too much of point.

[quote="nr1337"][quote="tenniskidperson3"]I may just be overly anxious, but has anyone received their mail notification?  I have received the email, but not the one in the mail.[/quote]

If you haven't gotten the mail invitation yet, you should probably start PANICKING.[/quote]

The only thing in the mail was the forms. You can get them online.

OK, that's all I really needed to know.  A mod can lock this now.


trigonometry
inequalities
geometry proposed
geometry

Let ABC be a triangle ; <B , <C are acute angles :
 $ \sin^2 B + \sin^2 C = \sqrt{\sin A} $
 $ \frac{a}{b+c} +  \frac{b}{a+c} +  \frac{c}{a+b} = \frac{5\sqrt2-4}{2} $
 Find <A , <B , <C ?

$ \sin^2 B \plus{} \sin^2 C \equal{} \sqrt{\sin A}$ $\Longrightarrow$ $ \frac{b^2 \plus{} c^2}{4R^2} \equal{} \sqrt{\frac{a}{2R}}$ $\Longrightarrow$

$ b^2 \plus{} c^2 \equal{} 2R \sqrt{2Ra} \ge 2Ra \ge a^2$ $\Longrightarrow$ $\measuredangle A \le 90^\circ$. Then

$ \sqrt{\sin A} \equal{}  \sin^2 B \plus{} \sin^2 C \ge \sin^2 B \plus{} \cos^2 B \equal{} 1$

$\Longrightarrow$ $\sin A = 1$ and $ \measuredangle A = 90^\circ$. Consequently, we can write the LHS of the 2nd equation as

$ \frac {1}{\frac b a \plus{} \frac c a} \plus{} \frac {\frac b a}{1 \plus{} \frac c a} \plus{} \frac {\frac c a}{1 \plus{} \frac b a} \equal{} $

$ = \frac{1}{2} \cdot \left( \frac {1}{\sin B \plus{} \cos B} + \frac {1}{\sin C \plus{} \cos C}\right) \plus{} \frac {\sin B}{1 \plus{} \cos B} \plus{} \frac {\sin C}{1 \plus{} \cos  C} \equal{}$

$ = \frac {\sqrt 2}{4} \cdot \left(\frac {1}{\sin (B \plus{} 45^\circ)} + \frac {1}{\sin (C \plus{} 45^\circ)}\right) \plus{} \left(\tan \frac {B}{2} \plus{} \tan \frac {C}{2}\right)$

Both bracketed terms have minima for $ \measuredangle B =  \measuredangle C = 45^\circ$ by Jensen inequality and their sum is equal to $\frac{\sqrt 2}{2} + 2(\sqrt 2 - 1) \equal{} \frac {5 \sqrt 2 \minus{} 4}{2}$.