(1) When acetone is treated with conc. H2SO4 and distilled, what compound do we get.

(2) Which reagent(s) can be used to convert acetaldehyde into 2,2-dihydroxymethyl-1,3-propandiol?

(3) When tert-butyl trifluoromethyl ketone is subjected to Baeyer Villiger oxidation (using per acid, say PBA), where is the O inserted? The tert-butyl is the bulkier of the 2, but the trifuoromethyl carbon is very highly electropositive, so which will have higher migratory aptitude?

(4) What is the reaction, if any of chloroform or chloral with water, individually?

(5) Explain what could be the intermediates in the following 3 reactions?
In the reaction involving SeO2, can the O be inserted in between the 2 carbonyls? Referring to the mechanism, it would involve anegative charge on the caronyl. Also what is the final step of the sequnce?
In the second reaction, what coulbe the role of quinone?

[quote="hell_ever"]
(2) Which reagent(s) can be used to convert acetaldehyde into 2,2-dihydroxymethyl-1,3-propandiol?

[/quote]

treat acetaldehyde with excess of HCHO! (first 3 moles--->aldol condensation....next canizzaro  :D )

EDIT: Stupid 10 minute rule!  :mad: ...i am editing this post and posting ans for first question!

1 ans- 1,3,5-trimethyl benzene(mesitylene )

5) the first intermediate is a 1,2 -cyclopentanediketone(well, my IUPAC naming sux..........i think it is 2-oxocyclopentanone right?)

The second intermediate is the anhydride because when 1,2-diketones are subjected to baeyer-villager, anhydrides are formed........next hydrolysis ofc to form the corresponding dicarboxylic acid......next no idea! :(

EDITING again!  :mad:  for 4th quest., isnt it chloral hydrate?(extensive intramolecular H-bonding....hence very stable)  :maybe:

[quote="Da Vinci"]for 5 C, i think its just mixed aldol condensation(after all sulphur and oxygen have similiar properties)(only one possibility of carbanion.......so easy  :D ) and maybe next step a $ SH_2$ grp is removed just like dehydration![/quote]

Do you mean intermolecular aldol condensation? Why should it be mixed? :?:

[quote="Da Vinci"]3) hey! the migratory aptitude depends on the electron releasing ability (the higher the ER ability, greater the migratory apti.) (i.e, the ease of migration of substituents  depends on their ability to stabilize a positive charge in the transition state)

Is $ CF_3$ grp an electron releasing grp?  :|  :?:[/quote]

It is a very powerful e withdrawing group. But when a group migrates, doent it migrate with its pair of electrons? So shouldnt in fact the group be a powerful e withdrawing group rather than e releasing?

EDIT: It is indeed a stupid rule.  :P  There is possible formation of ammonium salt  and then amide after step 4 in 5a.... :maybe:

EDIT: Thanks for all the help. It was really very helpful.  :)

EDIT 3: (phew)
[quote="Da Vinci"]For 5(b)

Quinone acts as a very good hydrogen acceptor (guess why!  :wink: )

The resulting oxidised product is[/quote]

I am not thinking very well lately, so can you explain the shift of the double bond?

[quote="hell_ever"]

I am not thinking very well lately, so can you explain the shift of the double bond?[/quote]

i am afraid i cant!  :oops:
Its an experimentally observed fact that $ \beta \minus{} \gamma$ unsaturated alcohols form $ \alpha \minus{} \beta$ unsaturated ketones when subjected to oppeneur oxidation!  :P

[quote="hell_ever"]
So shouldnt in fact the group be a powerful e withdrawing group rather than e releasing?

[/quote]

If its $ e^\minus{}$ withdrawing how can it stabilise an electron deficient oxygen?  :|

Also, if you believe that electron withdrawing ability increases the mig. apti.,(which i think isnt the case) why did you have a doubt between t-butyl and $ CF_3$ grps?
clearly, t-butyl is an ERG!  :maybe:

[quote="hell_ever"]
Do you mean intermolecular aldol condensation? Why should it be mixed?

[/quote]

sorry da!  :D
whatever i said doesnt make sense to me also now!  :o i must have typed that in sleep!  :P
lemme try it again!

which round in chem olympiad are u both aiming for ???


I noticed that you created an article about the Asymptote package cse5.

Shouldn't that be included here: [[Asymptote: Macros and Packages]]?

No. CSE5 is far more vast and it needs its own place. In plus it was developed by a member of this community so it's more important than other packages :P
I'm just trying to make that resource actually useful. In particular unless you are doing 10 asy pictures a day, even if you have it installed, etc. know the basic principles it still takes one a lot of time to create a picture when he needs one. That's why I made those to entries, and I will continue to add to them as soon as I have the time.

Also, should not

Asymptote: * and LaTeX: *

be moved to

Asymptote:* and LaTeX:* (eliminating the space)

for consistence with other namespaces?


geometry
rectangle
probability
analytic geometry
absolute value
probability and stats

X and Y are two independent Uniformly distributed random variables. X is Uniformly distributed between 0 and 1. Y is Uniformly distributed between 0 and 0.5.

Find P(|X-Y| < 0.5).

The absolute value sign confuses me. Any ideas?

Thanks!
-PD

Any ideas? DRAW THE PICTURE!!!

This is a rectangle, and the distribution is such that probability is proportional to area. Figure out what the set you want looks like. Ask whether it might be more convenient to compute the probability of the complement. (It is.) Use what you know about the areas of triangles. Don't compute any integrals.

I guess the brute force method was inevitable. Thanks, Dr. Merryfield!

-PD

What I'm wondering actually is whether the areas where one variable or the other, is non-existent, counts as well into the probability.

We know X is a square from 0 to 1, both x and y-coordinate-wise, and Y is a rectangle with 0<x<0.5 and 0<y<2. Both random variables overlap from 0<x<0.5 and 0<y<1.

So if the extra areas where either X or Y is non-existent are included, then P(|X-Y| < 0.5) is 0.5.

If we're only concerned about the rectangle where both X and Y R.V.'s overlap, then P(|X-Y| < 0.5) is 0.875.

I'll think about this more later. I've got other finals to study for.

-PD

Whoa, there. Geometrically, $ X$ is not a square and $ Y$ is not a square. They're one-dimensional, not two dimensional. $ X$ lies in the interval $ [0,1]$ and $ Y$ lies in the interval $ [0,.5].$ Representing them together and simultaneously gives us the rectangle $ [0,1]\times[0,.5].$

Now: $ X\equal{}Y$ is a line from the origin cutting across that rectangle. $ |X\minus{}Y|<.5$ is a stripe centered on that line. We consider the complement, $ |X\minus{}Y|\ge .5.$ That lies either above the line $ y\equal{}x\plus{}.5$ or below the line $ y\equal{}x\minus{}.5.$

The line $ y\equal{}x\plus{}.5$ just barely touches the upper left corner of our rectangle; the area outside it is zero.

The line $ y\equal{}x\minus{}.5$ extends from the middle of the bottom edge to the upper right corner. Outside it lies a triangle. That triangle constitutes exactly $ \frac14$ of the area of the rectangle. Hence $ P(|X\minus{}Y|\ge.5)\equal{}.25,$ from which $ P(|X\minus{}Y|<.5)\equal{}.75.$


limit
trigonometry
calculus
calculus computations

Compute: $\displaystyle \lim_{x \to \infty} \frac{ 2x \sin^{3} x+1}{x^{2}\cos^{2}x+1}$.

thank you!

This can be written as

$\lim_{x\to\infty}\frac{2x\sin^3 x +1}{x^2(1-\sin^2 x )+1}$

$-1 \le \sin^n x \le 1$

So as $x\to\infty$, the expression oscillates depending on the divisibility of $\infty$, so the limit does not exist.

Let me know if this makes sense.


symmetry
real analysis
number theory proposed
number theory

It took me a week to find a proof of this.

Let p=8n+3 be a prime number congruent to 3 mod 8. The set
{2,4,6,...,4n}
contains exactly n squares mod p.

It is equavalent to $\sum_{1\le k<p/4} (k/p)=0$, were $(k/p)$ is lejander symvol. It is easy to prove.
I know better fact for prime p=3(mod 8): $\sum_{3p/10<kp/3} (k/p) =3h[1+(3/p)-(5/p]$, were h is number of classes for field $Q(\sqrt{-p})$.

Let $p = 11$ $( p \equiv 3 ( \mbox{mod } 8 ) )$.
Let $n = 2$, then the number of square $\mbox{mod } p$ in $\{ 2, 4, 6, 8 \}$ should be $2$ (according to the statement).
Homever we have
\[ \left( \frac{2}{11} \right) = - 1, \left( \frac{4}{11} \right) = 1, \left( \frac{6}{11} \right) = - 1, \left( \frac{8}{11} \right) = - 1  \]
Hence there is only $1$ square $\mbox{mod } p$ in $\{ 2, 4, 6, 8 \}$.
Are you sure the statement is correct ?

Notation: $\left ( \frac{x}{p} \right )$ is the [i]Legendre symbol[/i].

[quote="Rust"]It is equivalent to $\sum_{1\le k<p/4} (k/p)=0$, were $(k/p)$ is the Legendre symbol.[/quote]True.
[quote]It is easy to prove.[/quote]Well, I know that (2|p)=(-1|p)=-1, but I could not find an easy symmetry argument for the theorem. I had to look at the orbits of the permutation $x\mapsto 2x$ and it was not trivial, at least for me :)

Here is a sketch of a proof:
http://www3.telus.net/ldh/math/bigsmall.pdf

(edit) A correspondent sent me this reference:

V.-A. Lebesgue,
Demonstration de quelques theoremes relatifs aux residus et aux non-residus quadratiques, J. Math. Pures Appl., 7 (1842) 137-159.


Let $ a,b,c\ge0$ such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ \frac1{a^3\plus{}b^2\plus{}c}\plus{}\frac1{b^3\plus{}c^2\plus{}a}\plus{}\frac1{c^3\plus{}a^2\plus{}b}\le1$

by Caushy shwartz :
$ (a^3 \plus{} b^2 \plus{} c)(\frac {1}{a} \plus{} 1 \plus{} c)\geq(a \plus{} b \plus{} c)^2$

$ \iff$  $ \sum\frac {1}{a^3 \plus{} b^2 \plus{} c}\leq\frac {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} a \plus{} b \plus{} c}{(a \plus{} b \plus{} c)^2}$

but : $ \frac {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} a \plus{} b \plus{} c}{(a \plus{} b \plus{} c)^2}\leq\frac  {a \plus{} b \plus{} c}{9}$
$ \equal{} 1/3\leq1$

I think you are wrong mehdi cherif, because from your observation with cauchy Schwartz it should have been :

$ \sum \frac{1}{a^3\plus{}b^2\plus{}c}\leq \frac{\sum \frac{1}{a}\plus{}\sum a\plus{}3}{(a\plus{}b\plus{}c)^2}\equal{}\frac{\sum \frac{1}{a}\plus{}6}{9} \leq 1$.the last one is equivalent to :
$ \sum \frac{1}{a}\leq 3$ which is not true.Actually the reverse holds : $ \sum \frac{1}{a} \geq \frac{9}{a\plus{}b\plus{}c}\equal{}3$.

[quote="can_hang2007"][quote="dduclam"]Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac1{a^3 \plus{} b^2 \plus{} c} \plus{} \frac1{b^3 \plus{} c^2 \plus{} a} \plus{} \frac1{c^3 \plus{} a^2 \plus{} b}\le1$[/quote]
Very nice, my friend. :)
My solution is the following :)
By the Cauchy Schwarz Inequality, we have
\[ \sum \frac {1}{a^3 \plus{} b^2 \plus{} c} \equal{} \sum \frac {a \plus{} b^2 \plus{} c^3}{(a^3 \plus{} b^2 \plus{} c)(a \plus{} b^2 \plus{} c^3)} \le \sum \frac {a \plus{} b^2 \plus{} c^3}{(a^2 \plus{} b^2 \plus{} c^2)^2} \equal{} \frac {3 \plus{} \sum a^2 \plus{} \sum a^3}{(\sum a^2)^2}
\]
It suffices to prove that
\[ 3 \plus{} \sum a^2 \plus{} \sum a^3 \le (\sum a^2)^2
\]
Setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, we can easily rewrite the inequality as
\[ 3(r \minus{} 1) \plus{} (q \minus{} 3) \minus{} 4(q \minus{} 3)^2 \le 0
\]
which is true since $ r \le 1,q \le 3$. :)[/quote]

Thank you,Can  :)
Above problem  appeared when I solve [url=http://www.mathlinks.ro/viewtopic.php?t=195375]this problem[/url].

The similar ones:
$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ P\equal{}\frac1{a^2 \plus{} b^2 \plus{} c} \plus{} \frac1{b^2 \plus{} c^2 \plus{} a} \plus{} \frac1{c^2 \plus{} a^2 \plus{} b}\le1$

$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ Q\equal{}\frac a{a^3 \plus{} b^2 \plus{} c} \plus{} \frac b{b^3 \plus{} c^2 \plus{} a} \plus{} \frac c{c^3 \plus{} a^2 \plus{} b}\le1$

What do you think about problems:[b]find the minimum value of P and Q[/b]? I think,they're hard problems  :)

[quote="dduclam"]The similar ones:
$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ P \equal{} \frac1{a^2 \plus{} b^2 \plus{} c} \plus{} \frac1{b^2 \plus{} c^2 \plus{} a} \plus{} \frac1{c^2 \plus{} a^2 \plus{} b}\le1$
[/quote]
It is equivalent to
 $ \sum_{sym}(6a^6 \minus{} 4a^5b \plus{} 56a^4b^2 \plus{} 18a^3b^3 \minus{} 21a^4bc \minus{} 43a^3b^2c \minus{} 12a^2b^2c^2)\geq0,$ which true by Muirhead. :wink:

[quote="dduclam"]
$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ Q \equal{} \frac a{a^3 \plus{} b^2 \plus{} c} \plus{} \frac b{b^3 \plus{} c^2 \plus{} a} \plus{} \frac c{c^3 \plus{} a^2 \plus{} b}\le1$[/quote]
Applying Cauchy - Schwarz inquality, we have
$ (a^3\plus{}b^2\plus{}c)(\frac{1}{a}\plus{}1\plus{}c)\geq (a\plus{}b\plus{}c)^2$
$ \Rightarrow \sum\frac{a}{a^3\plus{}b^2\plus{}c}\leq\sum\frac{1\plus{}a\plus{}ac}{(a\plus{}b\plus{}c)^2}\leq 1$ (by hypothesis $ a\plus{}b\plus{}c\equal{}1$)

[quote="dduclam"]$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ P \equal{} \frac1{a^2 \plus{} b^2 \plus{} c} \plus{} \frac1{b^2 \plus{} c^2 \plus{} a} \plus{} \frac1{c^2 \plus{} a^2 \plus{} b}\le1$[/quote]
By the same way,
$ P\leq\sum\frac{2\plus{}c}{(a\plus{}b\plus{}c)^2}\leq 1$

[quote="Sudoku"][quote="dduclam"]$ \bigstar$Let $ a,b,c\ge0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ P \equal{} \frac1{a^2 \plus{} b^2 \plus{} c} \plus{} \frac1{b^2 \plus{} c^2 \plus{} a} \plus{} \frac1{c^2 \plus{} a^2 \plus{} b}\le1$[/quote]
By the same way,
$ P\leq\sum\frac {2 \plus{} c}{(a \plus{} b \plus{} c)^2}\leq 1$[/quote]
yes just by C.S :) .


A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 1 through 4.  If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins $1.  If the number chosen appears on the bottom of both dice, then the player wins $2.  If the number chosen does not appear on the bottom of either of the dice, the player loses $1.  What is the expected return to the player, in dollars, for one roll of the dice?
(source: AMC 10B 2007)

Thanks everyone

[hide="Solution"]The player's choice is independent of the dice rolls, so the player correctly chooses the die roll $ \displaystyle\frac{1}{4}$ and incorrectly does so $ \displaystyle\frac{3}{4}$.

We can therefore easily compute the expected value: $ \displaystyle\frac{9}{16}(\minus{}1) \plus{} \displaystyle\frac{6}{16}(1) \plus{} \displaystyle\frac{1}{16}(2) \equal{} \boxed{\minus{}\displaystyle\frac{1}{16}}$ dollars.

[/hide]


function
calculus
derivative
linear algebra
matrix
vector
quadratics

I know for a function of two variables: f(x,y), there exists the second derivative test to find minimums, maximums, or saddle points for the function f(x,y).

So for example, z=x^4+y^4-4xy+1 has a saddle point at (0,0) and a local minimum at (1,1) and (-1,-1) using 2nd derivative test.

How would you find minimums, maximums, or saddle points for functions of three variables f(x,y,z)?

For example, w = "some function with 3 variables x, y, and z"

It is exactly the same test. The gradient should vanish and you should look at whether the second derivative matrix is positive definite (minimum), negative definite (maximum) or indefinite (saddle).  It works for any number of variables.

Expanding out what fedja said: For your function $w(x,y,z)$, at a critical point (a point where the gradient vanishes), consider this matrix:

$H=\begin{bmatrix} \frac{\partial^2w}{\partial x^2} & \frac{\partial^2w}{\partial x\partial y} & \frac{\partial^2w}{\partial x\partial z} \\ \frac{\partial^2w}{\partial x\partial y} & \frac{\partial^2w}{\partial y^2} & \frac{\partial^2w}{\partial y\partial z} \\ \frac{\partial^2w}{\partial x\partial z} & \frac{\partial^2w}{\partial y\partial z} & \frac{\partial^2w}{\partial z^2} \end{bmatrix}.$

Second derivative tests are about Taylor polynomials. Let $u$ be the incremental vector
$u=\begin{bmatrix} \Delta x & \Delta y & \Delta z\end{bmatrix}^T.$

Then $w(x+\Delta x,y+\Delta y, z+\Delta z) = w(x,y,z)+\nabla w\cdot u + \frac{u^THu}2 + o(||u||^2).$

Classifying the critical point is a matter of classifying the quadratic form $u^THu.$ $H$ is a symmetric matrix, so it has real eigenvalues. We don't need to know the actual values of the eigenvalues, just their signs. There is a test available. Start in the upper left corner and calculate determinants of all the sizes starting in that corner. That is, compute

$\frac{\partial^2w}{\partial x^2},\, \det\begin{bmatrix} \frac{\partial^2w}{\partial x^2} & \frac{\partial^2w}{\partial x\partial y} \\ \frac{\partial^2w}{\partial x\partial y} & \frac{\partial^2w}{\partial y^2}\end{bmatrix}$ and $\det H.$

If all three of these are positive, then $H$ is positive definite (all three eigenvalues are positive) and the critical point is a minimum.

If these three determinants have the strict alternating sign pattern $-,\,+,\, -,$ then $H$ is negative definite (all three eigenvalues are negative) and the critical point is a maximum.

If all three determinants are nonzero but there is some other sign pattern, then there is at least one positive eigenvalue and at least one negative eigenvalue. In that case, the critical point is a saddle point.

If $\det H = 0$ then there is at least one zero eigenvalue. If the other two eigenvalues are one positive and one negative, then we have a saddle point. But if the other eigenvalues are of the same sign, then the second derivative test is inconclusive.

[Edit: I forgot to divide the quadratic terms of the Taylor polynomial by 2. Fixed now.]

Just to add: the use of three variables and a $3\times3$ matrix in my post was for the sake of illustration. As fedja said, this all works in any number of variables.

Also: the last paragraph is somewhat incomplete, as I didn't show how to determine the signs of the eigenvalues in the case of a singular matrix. For that, consult a good linear algebra text.


quadratics
Gauss
number theory unsolved
number theory

(a/p) = (a/p) sounds very complicated  :wink:

did you mean $ (\frac{a}{p})\equal{}(\frac{a}{q})$ ?
if this is you question .... am asking if someone has a proof without using the reciprocity low :)

That question [b]IS[/b] the quadratic reciprocity law. Thus the proof doesn't use quadratic reciprocity. :D

:blush:  oups !!! i didn't notice that ....by the way i heared that gaus have gave may proofs for the reciprocity law... so can some one give me as many proofs as he knows ?
for me i only know two proofs : the first uses a lemma established by gauss and the second uses the gauss sums.

[quote="MagnusZ"][quote="MagnusZ"]If p and q are odd primes satisfying p=q+4a for some a, establish that (a/p)=(a/p)

(a/p) is Legendre symbol[/quote]

sorry, (a/p)=(b/p) :)[/quote]

(a/p)=(a/q)  :P


According to Maple this integral cannot be expressed by elementary functions.

Well, it's not even defined everywhere...


counting
distinguishability
rotation
factorial
symmetry
MATHCOUNTS
ceiling function

How many ways are there to arrange 5 white chips and 6 blue chips in a circle?

[quote="shaggy75"]is it 11! ?[/quote]

No, because the blue chips are indistinguishable. It should be
$ \frac{11!}{5!\cdot 6!}$
to count the repetitions.

That should be easy.
[hide]I'm pretty sure that this is the answer:  11!/5!/6!/11=[b]42[/b] ways.  If objects are put around a circle, then you have to divide by the number of objects on the circle, I think.  [/hide]

The Great One, please [u][i][b][color=red][size=200]stop[/size][/color][/b][/i][/u] spamming!  I beg of you! :surrender:

[hide]

Are they distinguishable or not?

If the chips are [b]distinguishable[/b], then the answer would be $ \frac{11!}{11}$. This is because there are 11 distinguishable chips. If the people were standing in a line, then there are $ 11!$ ways to arrange the people. But in a circle, you must account for rotation. There are 11 times you can rotate any arrangement. Therefore, we must divide the $ 11!$ by $ 11$. So, the answer is $ 10!$.

However, if the chips are indistinugishable, then the answer would be $ \frac{11!}{6!\times 5!\times 11}$ since you divide by the factorial of the number of the two pieces as well as 11[/hide]

No, I seriously doubt $ \frac{10!}{5!6!}$ is correct because some things like vishalarul said are not repeated exactly 11 times.  Try it with an easier case and you'll see why.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159364 .  And stop spamming.

Heh, all of you are wrong. It's
[hide]
Infinity!
Because you can put the chips different distances away from each other, and make a bigger or smaller circle! So, positive_infinity ways! :lol:
[/hide]
as for the "real" answer...
[hide]
For putting them in a line, it's 11!/6!/5! ways. So, for a circle, it's 11!/6!/5!/11 or 10!/6!/5!.
[quote]This is more complicated than just dividing by 11, since for some sequences, there are less than 11 repetitions. For example,
abababababa
has less than 11 repetitions.
[/quote]
Infact, abababababa rotates to something like ababab[b]aa[/b]baba.
No sequence has less than 11 repetitions, because:
say the 11 are: 0 1 2 3 4 5 6 7 8 9 10, each number representing either white or blue, and the new arrangement shifts it by x to get
( 1+x 2+x 3+x....... 11+x ) mod 11. Then, for an arrangement to have less than 11 repetitions, if f(n) equals the color of chip n,
f(1) = f(1+x)
f(1+x) = f(1+2x)
.
.
.
This loop must pass through all 11 chips before returning to f(1)
(example: 1 = 6 = 0 = 5 = 10 = 4 = 9 = 3 = 8 = 2 = 7 = 1)
because 11 is prime, which means to have <11 repetitions, every chip would need to be the same color.
So, the final answer is 10!/6!/5! (unless it really is positive_infinity.)
[/hide][/quote]

[quote="abacadaea"]i think(like half of you said, while everyone else was posting bogus stuff) $ \frac{10!}{5!6!}$ is correct answer. you have \binom{11} {6} ways to arrange blue chips. for the remaining spaces its all the same and doesn't matter.[/quote]

No actually its not 10!/5!6!, try it with a easier case...

try it with 3 red chips for example

If you list out, theres only 1 way

however, using your formula its 2!/3! which is a fraction!!!! :rotfl:  :rotfl:

Um, actually, with three chips, there would be 3!/3=2 ways, not one.  But I see your point.  However, the problem deals with two different types of chips, so I guess that you would have to put that into consideration.  If you're so confident with your answer, why don't you post it and explain? :P  I'm still 75% sure that my answer is correct, but I think that the Great One should post an answer.

OR, anyone who has the solutions book for AoPS book 1 can post the solution to this problem, seeing as how that's where the great one got the problem from. Does a circle consisting of 11 chips have reflectional symmetry? If it does, then the answer to this problem is $ 21$. However, if a circle consisting of 11 chips doesn't have reflectional symmetry, then the answer to this problem is $ 42$. Does anyone know whether or not a circle consisting of 11 chips has reflectional symmetry? :?:

Many people seem to think that the answer is $ 42$. It is not. You could see this from my second post in this thread why $ 42$ is not the right answer.
Please, this problem is much, much more complicated than counting for rotations. If any of you actually bothered to click t0rajir0u's link, you would see from JBL's post:
[hide][quote="JBL"]Indeed, the problem does not succumb easily to a simple formula you know.  The entry in the OEIS (shifted by a few terms) for the sequence is [url=http://www.research.att.com/~njas/sequences/A007997]here[/url].  The formula (where 12 is replaced by $ n$) is [hide]$ \left\lceil\frac{(n+1)(n+2)}{6}\right\rceil$.[/hide]  "Intelligent listing" in some form or another isn't a terrible way to solve the given instance.

Additional thoughts, for those who've read the answer already:
[hide]The answer is also equivalent to $ \left\lceil\frac{1}{n+3}\binom{n+3}{3}\right\rceil$.  This is interesting because it's exactly what you would expect if you didn't have to worry about the symmetries that Kunny pointed out.  Moreover, investigating the same question when we have $ n$ of one type and 0, 1 or 2 of the other type also gives the same result, i.e. the number of arrangements is in all of these cases $ \left\lceil\frac{1}{m+n}\binom{m+n}{m}\right\rceil$.  So, the question is: is this always the case?  Do there exist $ m$ and $ n$ such that in placing $ m$ things of one color and $ n$ of another around a circle, the number of possible arrangements is [i]not[/i] $ \left\lceil\frac{1}{m+n}\binom{m+n}{m}\right\rceil$?  What about the corresponding question for larger numbers of colors?[/hide][/quote][/hide]

This problem doesn't belong in the MathCounts forum. It is too difficult.

Yes, I clicked on the link, but if we assumed that JBL's generalization was correct, we get $ \frac{1}{11}*\dbinom{11}{5}=\boxed{42}$ as the answer, which you are saying to be wrong. Note that he asks us if the generalization is correct, and if there are any cases that prove it wrong; he is not saying that it is right. he asks us to explore further for ourselves.

[quote="erdogankerem123"]Yes, I clicked on the link, but if we assumed that JBL's generalization was correct, we get $ \frac{1}{11}*\dbinom{11}{5}=\boxed{42}$ as the answer. Note that he asks us if the generalization is correct, and if there are any cases that prove it wrong; he is not saying that it is right. he asks us to explore further for ourselves.[/quote]

I get a different answer...
$ \left\lceil\frac{(n+1)(n+2)}{6}\right\rceil = \left\lceil\frac{(11+1)(11+2)}{6}\right\rceil = 31$.

I thought that 31 was the right answer...

Well, that was the first formula he gave, which was rather specific to the question he was answering. That question involved "three things of same kind and twelve things of other kind around a circle." totaling 15 objects. That formula was the second formula "$ \left\lceil\frac{1}{m+n}\binom{m+n}{m}\right\rceil$", only with 3 as $ m$. Therefore, in our case, I think we should use the second formula, if it is right (the thing we were supposed to try and figure out.)

Like I said earlier, the great one got this problem from AoPS book 1. Anyone who has the solutions manual to AoPS book 1 could easily just look up the solution to this question and end this discussion.
[quote="vishalarul"][quote="The great one"]Both answers are wrong...

First, read, its a circle,
second counting repetitoins in a circle is a little different because

say this is a circle

aaaaabbbbb

is the same as

bbbbbaaaaa

when put in a circle.[/quote]

So,
aaaaaabbbbb
aaaaabbbbba
aaaabbbbbaa
aaabbbbbaaa
aabbbbbaaaa
abbbbbaaaaa
bbbbbaaaaaa
bbbbaaaaaab
bbbaaaaaabb
bbaaaaaabbb
baaaaaabbbb

are all the same.

This is more complicated than just dividing by 11, since for some sequences, there are less than 11 repetitions. For example,
abababababa
has less than 11 repetitions.[/quote]
vishalarul, all of those arrangements are the same because remember, we are arranging the chips in a CIRCLE. a CIRCLE has ROTATIONAL SYMMETRY. those arrangements however are indeed different when you arrange the chips in a row.

umm, when did he say that they were [b] not [/b] the same? I don't think that he meant it to be sarcasm, only as a statement.

Okay, never mind, now that I think about it, vishalarul probably just meant that you had to divide 11! by more than just 11, 5!, and 6!. The only other thing that I could think of dividing 11! by is 2, and that's only if the circle of 11 chips has reflective symmetry.


Triangle ABC is inscribed inside a circle with center O. A circle with center I is inscribed in triangle ABC. AI is drawn and extended to intersect the larger circle at D. Show that D is the circumcenter of triangle CIB.

I know this is dumb, but what's circumcenter?

this is from the merriam-webster unabridged vsion, nee the websters third new international dictionary:

circumcenter: the center of the circle circumscribing a triangle or a regular polygon of more than three sides

that definition was as useless as looking at the word, don't trust those dictionaries.

Maybe you'll trust mathworld :) : http://mathworld.wolfram.com/Circumcenter.html


All the positive integers with an initial digit of 2 are written down in succession. What is the 1978th digit thus written?

almost, but you missed one little catch
It's the 1978th digit written, not the 1978th '2' written

[hide]
1 digit numbers:
1

2 digit numbers
2 x 10

3 digit numbers
3 x 100

so far, that's 321 digits. The next ones digits will all be from numbers in the 2000s

$ \frac{1978-321}{4}= 414.25$

that means it's the 2nd digit of the 414th four digit number that starts with '2'

$ 2\boxed{4}13$

so the answer' is 4
[/hide]

[quote="igiul"]:maybe:  newish to me. Sounds like something from an... AMC?
[hide]
If my assumption is correct, we are searching for the actual digit within the "string" of numbers, formed by a concatenation of concessive positive integer "strings", beginning with $ 2$. I find subtraction would be the best way to do this. finding how many terms are included, with remainders of course, in $ 1978$ digits.

$ 8$ - 1 digit  $ \Rightarrow 1978-8 = 1970$ digits now remaining.
$ 90$- 2 digit $ \Rightarrow 1970-90=1880$ digits remaining.

To  find the number of 3 digit terms, so that we may hone in on our missingnomer; $ \frac{1880}{3}= 626 r2$
The Six-Hundred AndTwenty-Sixth $ 3$ digit number is $ 725$, then $ r2$, bringing us to the $ 1978$th digit,

$ 2$.[/hide]I hope that's right.[/quote]

Actually you got very lucky.  Yes your answer was right but your solution was wrong.  The answer is 2, but the 2 is the 2 in the number 2414.

hmmm show me how you got there.
I have checked a billion times, there must be something wrong with my brain, agh!

I find that there are $ 2888$ digits from 2-999, and that the digit in question is the 9 in 696 or something.

ah, silly mistake

.25 meant first digit, not second

[quote="erdogankerem123"]yes, I think you're right, Anthony Cai, but you didn't have to necessarily calculate the exact number that the two was in. Instead, you can simply see how many more digits you need to go once you hit 1000, which is 1657, then divide by four and see that the remainder is 1, so its the first digit of the next number, which of course is 2. Of course, if you calculated it out, it would only by 1 step more (saves only a few milliseconds)[/quote]

I don't understand why you need to subtract the first 1000...where does that number come from? Does it mean the 1000th digit? Or the first 1000 digits?

There is one 1 digit number that starts with 2,
10 2-digit numbers,
100 3-digit numbers,
and 1000 4-digit numbers..

I understand the dividing by 4 part, since at some point the numbers will all have 4 digits..., after 2000, but...

I'm not sure where your 1000 came from...please explain? :(

Yeah, I guess it was kind of confusing the way I worded it. It actually meant "the four-digit numbers" when I said 1000, (which I just used because it was the first 3-dig. number), and actually you never hit 1000, since that doesn't have first dig. two, but oh well. Sorry for the confusion.

[quote="erdogankerem123"]hmm... I think you may have misunderstood the problem. It doesn't mean that you're counting all positive integers' digits starting from 2, but rather that you're counting the digits of the integers that have a digit two in the front
(2,20,21,22...)[/quote]

 :rotfl: wow GG, this one isn't hard at all.

P.S.
you've no idea how much confidence you just gave back to me.  :lol:


In chess, a knight moves in an L-shaped manner - two spaces in one direction and one space in a direction perpendicular to the first direction - as shown. Beginning in and including the upper-left corner (marked $ \cdot$), what is the most number of squares in a $ 4\times4$ checkerboard that a knight can visit without visiting any square more than once?
[asy]draw((0,0)--(4,0));
draw((0,1)--(4,1));
draw((0,2)--(4,2));
draw((0,3)--(4,3));
draw((0,4)--(4,4));
draw((0,0)--(0,4));
draw((1,0)--(1,4));
draw((2,0)--(2,4));
draw((3,0)--(3,4));
draw((4,0)--(4,4));
dot((.5,3.5));
draw((.5,3.5)--(.5,1.5)--(1.5,1.5),EndArrow);[/asy]

I remember my old team getting 15. And the answer is 15, too. But I forgot how to do it (there's no specialized way, just guess and check) and even if I did, I don't know how to program Asymptote to put it here.

Is there like a systematic guess-and-check that would get you closer to the answer?

Here:

The hard part is showing that the knight can't touch all $16$ squares.

I'm a bit surprised that this is a FTW problem, as proving that you cant get 16 is a graph theory problem

Well this was a State Team round in the $1990's$ (I think).


geometry
parallelogram
incenter
angle bisector
geometry proposed

Let $ABCD$ be a paralleogram and $E$ be an arbitrary point on the diagonal $[AC]$. $P$ and $Q$ are the incenters of $\bigtriangleup AEB$ and $\bigtriangleup CED$. Find the locus of the midpoint of the segment $[PQ]$.

Please prove that.
[quote="Charlesww"]A segment which is parallel to the angle bisector of $\angle BAC$, the two extreme points are the midpoint of $A$ & the incentre of $\Delta ACD$ and  the midpoint of $C$ & the incentre of $\Delta CAB$[/quote]


geometry
rectangle
geometric transformation
rotation
vector
parallelogram
calculus

Let a rectangle $ ABCD$ and the foot $ E$ of $ B$ on $ AC$.If $ M$ is the midpoint of $ EA$ and $ N$ is the midpoint of $ CD$ , prove that $ \angle BMN \equal{} 90$.

[b] Comment[/b]

 [i]I'd like to see as many as possible solution to this simple and nice problem.[/i]

[hide]Notice that $ \triangle{ABE} \sim \triangle{DBC}$.  Hence, $ \triangle{MBE} \sim \triangle{NBC}$.  It then follows that $ \angle{BME} \cong \angle{BNC}$, so $ BCNM$ is cyclic.  The conclusion follows, because $ m\angle{BCN}\equal{}90^{\circ}$.[/hide]

The solution above is the obvious one... For the sake of adding another solution, I'll give the complex number solution.

Suppose that $ C\equal{}0$ and $ E\equal{}a$ where $ a$ is a positive real number. Suppose that $ B\equal{}a\plus{}bi$ where $ b$ is a positive real number. Then $ |AC\parallel{}EC|\equal{}|BC|^2$ implies that \[ A\equal{}\frac{a^2\plus{}b^2}{a}\] $ AC$ and $ BD$ have common midpoint, so\[ D\plus{}a\plus{}bi\equal{}a\plus{}\frac{b^2}{a}\plus{}0\implies D\equal{}\frac{b^2}{a}\minus{}bi\] Then $ N$ is the midpoint of $ DC$, so\[ N\equal{}\frac{b^2}{2a}\minus{}\frac{b}{2}\cdot i\] $ M$ is the midpoint of $ AE$ so \[ M\equal{}a\plus{}\frac{b^2}{2a}\] Now we may easily verify that \[ \frac{(B\minus{}M)\cdot i}{N\minus{}M}\equal{}\frac{\left( a\plus{}bi\minus{}a\minus{}\frac{b^2}{2a}\right)\cdot i }{ \frac{b^2}{2a}\minus{}\frac{b}{2}\cdot i\minus{}a\minus{}\frac{b^2}{2a} }\equal{}\frac{\left( \frac{b^2}{2a}\minus{}bi\right)\cdot i }{ a\plus{}\frac{b}{2}\cdot i }\equal{}\frac{b}{a}\] In other words, a $ 90$ degree counterclockwise rotation maps the vector $ B\minus{}M$ parallel to the $ N\minus{}M$ vector since $ \frac{b}{a}$ is real. Hence $ BM\perp NM$, as desired.

Let s.s.=spiral similarity. Notice that $ AEB\sim DCB$ so a s.s. about $ B$ maps $ DC$ to $ AE$. Since $ N$ is the midpoint of $ DC$ and $ M$ is the midpoint of $ AE$, so a s.s. about $ B$ maps $ CN$ to $ EM$. But then a s.s. about $ B$ maps $ CE$ to $ MN$. But then $ \frac{\pi}{2}\equal{}\angle CEB\equal{}\angle NMB$, as desired.

[hide]Let $ K$ be the midpoint of $ BE$. $ MK$ is midline in $ \Delta ABE$, so $ MK\equal{}\frac {AB}{2}\equal{}CN$, and $ MK \parallel{} CN \Rightarrow MKCN$ is a parallelogram, or $ CK \parallel{} MN.$
In $ \Delta CBM \; BE \perp CM, \; MK \perp BC \Rightarrow K$ is orthocenter, it follows $ CK \perp BM.$ [/hide]

Take $P$ midpoint of $\overline {AB}$; obviously $PM$ is midline of $\Delta ABE\implies PM\parallel BE\bot AC\implies M$ belongs to the circle of diameter $CP$, i.e. $BCNMP$ is cyclic, $BN$ another diameter, done.

Other solution, by power of $A$ w.r.t. $\odot BCNP$: $AP\cdot AB=AC\cdot AM'\implies M'\equiv M$ (here we called $M'$ the second intersection of the circle with $AC$, and the implication came from $AB^2=AC\cdot AE$).

Best regards,
sunken rock

with   vectors:

$\begin{gathered}
  (2\overrightarrow {BM) \cdot } (2\overrightarrow {MN} ) = (\overrightarrow {BA}  + \overrightarrow {BE} ) \cdot (2\overrightarrow {MA}  + 2\overrightarrow {AN} ) =  \hfill \\
   \hfill \\
   = (\overrightarrow {BA}  + \overrightarrow {BE} ) \cdot (\overrightarrow {EA}  + \overrightarrow {AD}  + \overrightarrow {AC} ) =  \hfill \\
   \hfill \\
  \overrightarrow {EA} ^2  - \overrightarrow {AB} ^2  + \overrightarrow {BE} ^2  = 0 \hfill \\
\end{gathered} $

(See: Book of Vector Calculus
  HIGH SCHOOL class B
Tsikaloudaki. Ask.11, 37 page 214)

[b]Remark[/b]: [color=#800000][i]the problem is valid for $\frac{AM}{ME}=\frac{ND}{CN}$ as well[/i].[/color]

Best regards,
sunken rock

[quote="Virgil Nicula"][quote="sunken rock"][b]Remark[/b]: [color=#800000][i]the problem is valid for $\frac{AM}{ME}=\frac{ND}{CN}$ as well[/i].[/color][/quote][b]Nice ![/b] I have one larger extension than yours :lol:
[color=darkred][b][u]An easy extension[/u].[/b] Let a convex cyclical quadrilateral $ ABCD$ , $\{E,M\}\subset (AC)$ and  $N\in (CD)$
so that $\widehat{AEB}\equiv\widehat {DCB}$ and $\frac {MA}{ME}=\frac {ND}{NC}$ . Prove that $BCNM$ is a cyclical quadrilateral.[/color][/quote]

Very nice!
From $\Delta ABE\sim\Delta DBC$ and $BM, BN$ homologues, we easily get $\Delta BMN\sim\Delta BEC\implies\angle MBN=\angle EBC=\angle MCN$, done.

Best regards,
sunken rock

Let K, M' ,    the intersection points of the circle (B, C, N) with the
AB, AC , respectively. Then:

$N\hat M'B = 90^o \mathop {}\limits^{} ,\mathop {}\limits^{} A\hat M'K = A\hat BC = 90^o \mathop {}\limits^{} and\mathop {}\limits^{} AK = KB$

 So:

$\left. \begin{gathered}
  AK = KB \hfill \\
  A\hat M'K = 90^o  \hfill \\
\end{gathered}  \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} \left. \begin{gathered}
  AK = KB \hfill \\
  M'K//EB \hfill \\
\end{gathered}  \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} AM' = M'E\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} M' \equiv M$


geometry
incenter
circumcircle
inradius
vector
geometry unsolved

Let $ ABC$ be a non-isosceles acute triangle and $ H; I; G$ be its orthocenter, incenter, centroid , respectively,
Now prove that $ \angle GIH >90^{o}$

Let be O the circumcenter of ABC, and E the midpoint of OH. Let R, r the circumradius and inradius, respectively. It is well know that OH = 3OG, OI^2 = R^2 - 2rR, IE = (R/2) -r.
From this we have
IH = 2IE - IO
IG = (2IE + IO)/3
Then the scalar product of the vectors IH . IG equals (4.IE^2 - IO^2)/3 = - 2r (R-2r)/3 <0 and therefore cos (GIH) <0.

This problem was proposed but not used in the IMO 1990, Beijeing, China, proposed by France (FR1)


Visit my blog and everything is way huge, however my font size is down to like 10/12 that range. It was fine yesterday...

For some reason your blog font size was missing. I've made it 8 now (you can change it to whatever you like though).


[hide]Let $x=\cos\theta+i\sin\theta$

Then we know that

$x+\frac{1}{x}=2\cos\theta$.

So $\cos{\theta}=-0.5$

Now, we also know that

$x^{999}=\cos{999\theta}+i\sin{999\theta}$
and
$x^{-999}=\cos{999\theta}-i\sin{999\theta}$

So, those two add up to $2\cos{999\theta}$, which equals to 2, in all cases, since $\cos{999\theta}=1$, for all values of $\theta$ satisfying the original equation for $\cos{\theta}=-0.5$.

$2$[/hide]

[quote="ashrafmod"]If $x+\frac{1}{x}=-1$ find $x^{999}+\frac{1}{x^{999}}$[/quote]
Assume $x^{n}+\frac{1}{x^{n}}=\alpha.$  Denote by $T_{n}(x)$ the Chebychev polynomial of degree $n .$ We note that
\[(*)\; \; \; \left\{ \begin{array}{c}T_{n}(y)=\frac{\left(y+\sqrt{y^{2}-1}\right)^{n}+\left(y-\sqrt{y^{2}-1}\right)^{n}}{2}=\\ \\ =\cos{\left(n\cdot \arccos{y}\right) }\; ,\; \; \mbox{ when}\; \; |y|\le 1\; . \end{array}\right. \]
From $(*)$ we give \[T_{n}\left(\frac{A+B}{2\sqrt{AB}}\right)=\frac{A^{n}+B^{n}}{2\left(AB\right)^\frac{n}{2}}\] which imply ( $A\leadsto x\; ,\; \; B: \leadsto \frac{1}{x}$ )
\[\begin{array}{c}A^{n}+B^{n}=x^{n}+\frac{1}{x^{n}}=2\cdot T_{n}\left(\frac{\alpha}{2}\right)\\ \end{array}\; . \]
For instance, when $\alpha=-1 \; ,\;$  because $T_{n}(-z)=(-1)^{n}T_{n}(z)$ , one finds
\[\begin{array}{|c|}\hline \\ x^{n}+\frac{1}{x^{n}}=2\cdot T_{n}\left(-\frac{1}{2}\right)=2(-1)^{n}T_{n}\left(\frac{1}{2}\right)=2(-1)^{n}\cos\left(\frac{n\pi}{3}\right) \\ \\ \hline \end{array}\; , \]
or (e.g. using  first equality from $(*)$ )
\[\begin{array}{|c|}\hline \hline \\ x^{n}+\frac{1}{x^{n}}=2(-1)^{n}\cdot T_{n}\left(\frac{1}{2}\right)= =(-1)^{n}\frac{\left(1+i\sqrt{3}\right)^{n}+\left(1-i\sqrt{3}\right)^{n}}{2^{n}}\\ \\ \hline\hline \end{array}\; . \]
Finally ($n\leadsto 999$)
\[\begin{array}{|c|}\hline \hline \\ x^{999}+\frac{1}{x^{999}}=-2 \frac{\left(1+i\sqrt{3}\right)^{999}+\left(1-i\sqrt{3}\right)^{999}}{2^{999}}=-2\cdot \cos{\left(\frac{999\cdot \pi}{3}\right)}=\\ \\ =-2\cdot\cos{(333\cdot \pi)}=1 \\ \\ \\ \hline\hline \end{array}\; . \]
[b][color=red]Happy New Year !   [/color][/b]

Another way to look at it is to use the recursion:

$(a^{n+1}+b^{n+1})=(a+b)(a^{n}+b^{n})-(ab)(a^{n-1}+b^{n-1})$

take $a=x$ and $b=\frac{1}{x}$ for this particular problem.

Also, if $r$ is a solution, then

$r+\frac{1}{r}=-1$
$r^{2}+r+1=0$
$r^{3}-1=0$

Then the equation obviously reduces... $(r^{3})^{333}+(r^{3})^{-333}=(1)^{333}+(1)^{-333}=1+1=2$

[quote="Altheman"]Another way to look at it is to use the recursion:...
Then the equation obviously reduces... $(r^{3})^{333}+(r^{3})^{-333}=(1)^{333}+(1)^{-333}=1+1=2$[/quote]
Inded, it's a more ellementary method ! Best wishes,proposer .


geometry
circumcircle
modular arithmetic
combinatorics unsolved
combinatorics

There is a frog in every vertex of a regular 2n-gon with circumcircle($n \geq 2$). At certain time, all frogs jump to the neighborhood vertices simultaneously (There can be more than one frog in one vertex). We call it as $\textsl{a way of jump}$. It turns out that there is $\textsl{a way of jump}$ with respect to 2n-gon, such that the line connecting any two distinct vertice having frogs on it after the jump, does not pass through the circumcentre of the 2n-gon. Find all possible values of $n$.

[b]Claim:[/b] $2n = 4 \pmod 8$.

I label the vertices with classes $\pmod{2n}$. Suppose the frogs follow the way of jump described (i.e.: after the jump no opposite vertices are both occupied).

[i][u]Sufficiency[/i][/u]

Consider the following w.o.j.: the frog in vertex $k$ jumps in $k+1$ if $k = 0,1 \pmod 4$, otherwise it jumps in $k-1$. This is easily seen as an example of a good way of jump. []

[i][u]Necessity[/i][/u]

[b]Subclaim[/b] [i]If, after the jump, a frog lands in $k$, then another frog must land also in $k+n-2$.[/i]

In fact, the opposite of $k$ is $k+n$ and must be empty. The frog starting in $k+n-1$ must now find itself in $k+n-2$. []

WLOG we can suppose that after the jump, vertex 0 is occupied. Now apply repeatedly the subclaim: every multiple of $n-2$ must be occupied and, this is equivalent to ask that every multiple of $\gcd(2n, n-2)$ must be occupied.

Two cases: (even) $2n = 0 \pmod 8$ and (odd) $2n = 2 \pmod 4$.

[b]Even case:[/b] $\gcd(2n, n-2) = 2$ [because $4 = 2n-2(n-2)$, but $4 \nmid n-2$]. So every even node must be occupied. In particular, $n$ is occupied. But then $0$ and $n$ are opposite vertices both occupied, and this is a contradiction. []

[b]Odd case:[/b] $\gcd(2n, n-2) = 1$ [$n-2$ is odd]. So every single node must be occupied and clearly this contradicts the hypotesys. []

We will show that $v_2(n)=1$.
Key observations:
[b]1.[/b] Any frog has only $2$ vertices to hop onto (since it has $2$ adjacent vertices). So only $2$ vertices are possible for a frog, and also a vertex can have only $2$ possible frogs.
[b]2.[/b] There will be $n$ vertices and $2n$ frogs so by [b]1[/b] each vertex must have exactly $2$ frogs.
Hence, we assign to each vertex a pair: $(x_i,x_{i+1})$ to the [hide=odd vertices]
Label the vertices $v_1, v_2, \cdots v_{2n}$. The the odd vertices are $v_1, v_3, v_5 \cdots v_{2n-1}$ and the rest are even.
This is made more clear by the diagrams below.
[/hide] and $(y_i, y_{i+1})$ to the even vertices, where $1 \le i \le n$ (and $x_{n+1}=x_1, y_{n+1}=y_1$).
[b]3.[/b] By [b]2[/b], no vertex can be occupied by only $1$ frog. Hence if a vertex $(a,b)$ is chosen, then the 2 vertices of the form $(a,x), (b,y)$ cannot be chosen.

[b]Contradiction when $n$ is odd[/b]
Start with $(x_1,x_2)$ and start moving clockwise. By [b]3[/b], $(x_2,x_3)$ cannot be chosen. Hence $(x_3, x_4)$ must be chosen. (since we can't leave out $x_3$). Hence $(x_4, x_5)$ cannot be chosen. Hence $(x_5, x_6)$ mst be chosen and so on. So we get the chain (of the vertices we choose):
$$(x_1, x_2) \implies (x_3, x_4) \implies (x_5, x_6) \implies \cdots \implies (x_n, x_1)$$
which contradicts our point [b]3[/b] as we cannot chose both $(x_1, x_2)$ and $(x_n, x_1)$. $\blacksquare$

[b]Note[/b]: Since $n$ is odd, the antipode of $(x_1, x_2)$ is an even vertex, and so won't bother our chain. Also, since $n$ is odd, hence $(x_n, x_1)$ is a member of our chain.
[hide=Diagram]
[asy]
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import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -180.5556147171591, xmax = 178.61017945740556, ymin = -38.903026657714214, ymax = 230.72568568296495;  /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);

draw((-39.241507915278135,25.78735718949745)--(-2.677024375985958,36.541617053995175)--(20.583068850171806,66.73406022838508)--(21.654206732294128,104.83219962344998)--(0.12725100604783535,136.28384089840648)--(-35.77523291545517,149.07552608819023)--(-72.33971645474733,138.3212662236925)--(-95.59980968090511,108.12882304930261)--(-96.67094756302744,70.03068365423769)--(-75.14399183678115,38.579042379281205)--cycle, linewidth(2) + rvwvcq);
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draw((0.12725100604783535,136.28384089840648)--(-35.77523291545517,149.07552608819023), linewidth(2) + rvwvcq);
draw((-35.77523291545517,149.07552608819023)--(-72.33971645474733,138.3212662236925), linewidth(2) + rvwvcq);
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draw((-75.14399183678115,38.579042379281205)--(-39.241507915278135,25.78735718949745), linewidth(2) + rvwvcq);
 /* dots and labels */
dot((-39.241507915278135,25.78735718949745),dotstyle);
label("$(y_3 , y_4)$", (-44.89336951430181,14.34440458440733), NE * labelscalefactor);
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label("$(x_6 , x_1)$", (-118.82929314985903,112.69953235647897), NE * labelscalefactor);
dot((-96.67094756302744,70.03068365423769),dotstyle);
label("$(y_4 , y_5)$", (-118.49013753685189,62.84365724442886), NE * labelscalefactor);
dot((-75.14399183678115,38.579042379281205),dotstyle);
label("$(x_4 , x_5)$", (-88.30528797921615,20.449205618535913), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]
[/hide]
The figure shows the case when $2n=10$.

[b]Contradiction when $v_2(n) \geq 2$[/b]
As above, we get
$$(x_1, x_2) \implies (x_3, x_4) \implies (x_5, x_6) \implies \cdots \implies (x_{n-1}, x_{n})$$
The issue here is that if the antipode of $(x_1, x_2)$ is of the form $(x_{n/2+1}, x_{n/2+2})$, then $n/2+1$ is odd since $n/2$ is even. Hence, the antipode is a part of this chain, a contradiction. $\blacksquare$

[hide=Diagram]
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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label("$(x_1 , x_2)$", (-66.95563726162416,397.3310125981661), NE * labelscalefactor);
label("$(y_4 , y_1)$", (-182.51608893635506,375.1876925167805), NE * labelscalefactor);
label("$(x_4 , x_1)$", (-237.18241038727567,276.9267096556319), NE * labelscalefactor);
dot((-167.89411094597546,204.48968653324079),dotstyle);
label("$(y_3 , y_4)$", (-210.195239038087,181.43364180465645), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]
[/hide]
The figure shows the case when $2n=8$.

[b]Construction when $v_2(n)=1$[/b]
It is evident by the above arguments that we must have $2|n$ and $2 \nmid n/2$, i.e. $v_2(n)=1$.
We now present our construction for $2n=12$, and in general, the idea is just to choose alternate pairs of vertices. The chosen vertices are marked.
[hide=Construction]
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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draw(shift((31.12849872998072,4540.19813182906))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
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draw(shift((-3104.6960378774274,3898.498909868954))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
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label("$(x_6 , x_1)$", (-3942.3666756701746,3871.0365004349123), NE * labelscalefactor);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]
[/hide]

Assume that jumping is over and the condition is satisfied..Label the vertices $A_1 \dots A_{2n}$ with $L$ or $R$ depending on the direction the frog on them has jumped(left or right).Note that at least half of the vertices are empty and every vertex has either 0,1 or 2 frogs,so by PHP,exactly $n$ vertices are empty and so each vertex contains either 0 or 2 frogs.This means that
 $$ \forall i \leq 2n :A_i \neq A_{i+2} \implies 4|2n \implies \text{n is even}$$
that's because if the neighbors of $A_i$ have same labels,then $A_i$ contains exactly one frog,So the labels are $RRLLRR \dots RRLL$ in this order.Note that if every block $RR$ is antipode of a $LL$,the condition would be satisfied,so $n=4k+2$ works and also,if every $RR$ is antipode of a $RR$ then the condition will not satisfied;meaning that $n$ is not divisible by 4,so answer is $n=4k+2$.

Claim: $n\equiv 2 \pmod 4$.

There are $2n$ frogs and $2n$ vertices or $n$ pairs of diametrically opposite points. If the frogs end up in more than $n$ vertices, then by PHP, we can find a pair of diametrically opposite vertices each with at least one frog.

Number the vertices: $1,2,\dots ,2n$ in anticlockwise direction.

We have $2n$ frogs, at most $n$ vertices and at each vertex, at most $2$ frogs (from the neighbours). Thus $n$ vertices have exactly $2$ frogs.

Without loss of generality, vertex $1$ has $2$ frogs and frog from vertex $1$ jumped to vertex $2$. Notice that each vertex contains either two frogs or zero. Now vertex $2$ also has $2$ frogs or frog from vertex $3$ jumped to vertex $2$. As frog from vertex $2$ jumped to vertex $1$, frog from vertex $4$ jumped to vertex $5$ and so on. Also notice that frog from vertex $1$ did not jump to vertex $2n$ thus vertex $2n$ has $0$ frogs.

Concluding we get that vertex $k$ has two frogs if and only if $k\equiv 1,2\pmod 4$. We have that vertex $2n$ has $0$ frogs or $2n\equiv -1,0\pmod 4$. Thus $n$ is even, say $n=2m$. Mark 'X' on a vertex if it has 2 frogs and 'O' if it has 0.

If $n+1$ labelled vertex is marked $X$ then we get two diametrically opposite points with at least one frog on them. Thus we need $n+1\equiv 3,0\pmod 4$ or $2m\equiv 2,3\pmod 4$ or $m$ is odd. Thus $n = 4m+2$.

Construction for $n \equiv 2\pmod 4$:
The frogs on vertex $4k+1$ and $4k+2$ swap after jump. Frog on $4k+3$ jump on the left vertex or $4k+2$ and frog on vertex $4k+4$ jump on the right vertex or $4k+5$.

$n=4k+2$
First note that after the move each vertex has atmost 2 frogs since there are two neighbouring vertices.Call a vertex rich if it has frogs after the move and poor otherwise.Since no two opposite vertices are rich there are atmost $n$ vertices but since each vertex can accommodate atmost $2$,we conclude all rich vertices have $2$ frogs each

Claim $1$:$n$ is even.
Suppose not.
Consider a bipartite graph of rich and poor vertices.Draw an edge between $2$ vertices if they are adjacent.So every vertex has degree $2$.Notice that every poor vertex has to be connected to a rich vertex,since every frog is jumping from its place.Also a poor vertex cannot be connected to $2$ rich vertices which will imply that it is jumping at $2$ distinct places.Therefore every poor vertex is connected to exactly one poor vertex and every rich vertex is connected to exactly one rich vertex.But this means there are $\frac {n}{2}$ edges among poor vertices,contradiction to our assumption.

Claim $2$:$n\neq 4k$
Now a poor vertex $a_1$ is connected to a rich vertex $b_1$which is connected to rich vertex $b_2$ which is connected to poor vertex $a_2$.So they must lie in this order on the polygon(say in clockwise direction).Now to the right of $a_2$ there is a poor vertex $a_3$.So this process of formation of quadruple of $(a_{2k+1},a_{2k+2},b_{2k+1},b_{2k+2})$ will continue.But since the vertex numbers of opposite vertices are congruent mod $4$,we get that both are rich or both are poor.This is a contradiction.
For $n=4k+2$ the construction is simple.$4m+1$ and $4m+3$ jump to $4m+2$ and $4m+2$ and $4m+4$ jump on $4m+3$ where $m$ takes values from $0$ to $2k$

Note : I’ll be sometimes calling “inhabited” vertices R or red, similarly with “uninhabited” (blue)
We claim that $n=4k+2$ is the only possible value. Call a vertex “inhabited” if there’s at least one frog on it (after the algorithm is applied), and call a vertex “uninhabited” if not.

Claim 1: every vertex has 2 or 0 frogs on it
Proof : initially, there are $2n$ frogs. Now, $a_{1}+a_{2}+ \dots + a_{i} = 2n \leq 2i \leq 2n$. This occurs when $i=n, $ and there are two frogs on all inhabited vertices $\Box$

Claim 2: at least 2 inhabited vertices are adjacent.
Proof : The frog on any inhabited vertex must jump to an adjacent vertex.

claim 3: the vertices must occur as $RRBBRRBB \dots$
Proof : Let the vertices be labelled in the format $A_i$, such that the $i$s increase by one as we move clockwise. (Diagram below).  Let $A_1, A_2$ be adjacent R’s. $A_3$ should be $B$ as $A_2$s frog must have gone to $A_1$ and $A_3$s frog to $A_2$. $A_4$s frog going to $A_3$ isnt possible coz then it has only 1 frog. Similarly, $A_4$ is uninhabited too. Now $A_4$ s frog must have gone to $A_5$ so it is inhabited. Similarly $A_5$s frog went to $A_6$ so it is inhabited too. $A_{4n}$ and $A_{4n-1}$ have to be uninhabited coz $A_1$s frog went to  $A_2$. So basically, the string $RRBB$ is repeating. this implies the thing has $4k$ vertices (as if there are $4k+2$, it will end with $AAAA$, which as we said is not possible). Now, let $\text{number of vertices}=4t$. Note that $A_{2t+1}, A_{2t+2}$ are B. We see that $A_i$ is R if $i$ is $1,2 \mod 4$, B otherwise. So $2t+1 \equiv{3} \pmod{4} \implies t \equiv{1} \pmod{2}$. So number of vertices $=2n=4t=4(2k+1) \implies n \equiv{2} \pmod{4}$

Construction: label vertices as $a_i$s mod 4, that is $1,2,3,4,1,2,3,4, \dots$. Consider consecutive vertices $4,1,2,3$. Transfer frogs from $4,2$ to $1$, and from $1,3$ to $2$ for all $\blacksquare$

Solved with [size=150][color=#f00]None[/color][/size] since I am lonely and just got kicked out of my last Math group.
We claim is that the answer is just $n \equiv 2\pmod 4$.
Actually,one of our insights is that the simultaneous condition is not of any concern.We can just move the frogs one by one.

So,we just place a $1$ at each of the vertices and then $1'$ for a frog that has moved.
Now, we just start moving with the first frog and see that the diametrically opposite vertex just gets fixed and we circle it.
Now, the $1s$ adjacent to it have to move in a specific direction and thus we gain some insight.
With some casework by observing the patterns in the occurrence of $1's$ ,circles and empty points ,we are done.

The answer is $n \equiv 2 \pmod 4$.
All the congruences below are taken modulo $2n$.

After the frogs move, we color each vertex which has at least one frog.
[b]CLAIM:[/b] there are exactly $n$ colored vertices, each of them has exactly two frogs.

[i]Proof:[/i] First notice that the total number of diameters which connect the vertices of the 2n-gon is $n$, and each diameter has at most one colored vertex, hence $n$ is the maximum number of colored vertices. Also it is easy to see that each colored vertex has at most two frogs, hence the claim is proved.

Let $A_1A_2...A_{2n}$ be the 2n-gon, and take the indices modulo $2n$. Assume that $A_i$ is colored, then $A_i$ has two frogs from $A_{i-1}$ and $A_{i+1}$. Now it is easy to see that $A_{i-1}$ or $A_{i+1}$ is colored. Assume WLOG that $A_{i+1}$ is colored, then we can see a pattern of colored vertices, that is, each one of $A_{i+4k}$ and $A_{i+1+4k}$ is colored.

$\boxed A$ Since $A_{i-1}$ isn't colored, then:
$$i-1 \not\equiv{i+1+4k} \Rightarrow 2\not\equiv{4k} \Rightarrow 2\mid n$$
$\boxed B$ $A_{i+n}$ is the antipode of $A_i$, then:
$$i+n \not\equiv i+4k \Rightarrow n\not\equiv 4k \Rightarrow 4 \nmid n$$
Hence $v_2(n)=1$, and this actually works due to $A$ and $B$. $\blacksquare$


Let x,y,z be positive real numbers such that x^4+y^4+z^4=1.
Determine with proof the minimum value of x^3/1-x^8  + y^3/1-y^8  + z^3/1-z^8

let´s work with each term individually.
we have that $ \frac{x^4}{x(1\minus{}x^8)}$ attains it´s minimal value when it´s denominator is in its  maximum.so let´s see when $ x(1\minus{}x^8)$ is maximum.let $ k$ be this maximum.we have that :
$ 8x^8(1\minus{}x^8)^8\leq{8k^8}$.by AM-GM,we´ll get that $ 8x^8(1\minus{}x^8)\leq{(\frac{8}{9})^9}$,so we´ll
have that $ 8k^8\equal{}{(\frac{8}{9})^9}$.so $ k\equal{}\frac{8}{9\sqrt[4]{3}}$.now all you have to do is compute:
the min is obtained when the denominetor is $ k$.direct substitution yields that the minimun is
$ \dfrac{9\sqrt[4]{3}}{8}$. Done. :D

[quote="Stephen"][quote="v235711"]let´s work with each term individually.
we have that $ \frac {x^4}{x(1 \minus{} x^8)}$ attains it´s minimal value when it´s denominator is in its  maximum.so let´s see when $ x(1 \minus{} x^8)$ is maximum.let $ k$ be this maximum.we have that :
$ 8x^8(1 \minus{} x^8)^8\leq{8k^8}$.by AM-GM,we´ll get that $ 8x^8(1 \minus{} x^8)\leq{(\frac {8}{9})^9}$,so we´ll
have that $ 8k^8 \equal{} {(\frac {8}{9})^9}$.so $ k \equal{} \frac {8}{9\sqrt [4]{3}}$.now all you have to do is compute:
the min is obtained when the denominetor is $ k$.direct substitution yields that the minimun is
$ \dfrac{9\sqrt [4]{3}}{8}$. Done. :D[/quote]

'we have that $ \frac {x^4}{x(1 \minus{} x^8)}$ attains it´s minimal value when it´s denominator is in its  maximum.so let´s see when $ x(1 \minus{} x^8)$ is maximum.'

But why is it?[/quote]

i calculated this maximun denominator without interfering the value of the numerator.


OK I'm having trouble with this problem. Bare with me because I don't know how to type square roots

rationalize the denominator:

The square root of 9 / the square root of 20x^2y, this reduces to; 3 / the square root of 20x^2y. this is where I'm stuck, I don't know what to multiply it by.

Edit: Ok I looked at the answer in the back of my book and the answer is 3 times the square root of 5y / 10xy. Now you can get this answer by multiplying the equation by the square root of 5y / the square root of 5y, correct? if so why do you multiply it by that?

So, you were able to get to $ \frac{3}{\sqrt{20x^2y}}$.  Now how do you solve this?

I think the easiest way to do this is to start by simplifying the denominator before you rationalize it.  In other words:
$ \sqrt{20x^2y}\equal{}\sqrt{(4x^2)(5y)}\equal{}2x\sqrt{5y}$

Now, we have $ \frac{3}{2x\sqrt{5y}}$.  To rationalize the denominator, it should be clearer now that you should multiply by $ \frac{\sqrt{5y}}{\sqrt{5y}}$ to get:

$ \frac{3}{2x\sqrt{5y}}\cdot \frac{\sqrt{5y}}{\sqrt{5y}} \equal{} \frac{3\sqrt{5y}}{2x\cdot 5y} \equal{} \frac{3\sqrt{5y}}{10xy}$

Ok thanks that makes sense now. I couldn't find an example similar to this in the book.


inequalities
factorial
induction
number theory unsolved
number theory

If $ n \leq m$ s.t $ m,n \in \mathbb{Z}^\plus{}$ .

Prove or disprove :

$ 2^n n! \leq \frac{(m\plus{}n)!}{(m\minus{}n)!} \leq (m^2\plus{}m)^n$.

 :D

$ \frac {(m \plus{} n)!}{(m \minus{} n)!} \equal{} (m \minus{} n \plus{} 1)\times....\times(m \minus{} n \plus{} 2n)\geq1\times2\times...\times2n\geq2\times4\times...\times2n \equal{} 2^n n!$
 and for $ 1\leq{k}\leq{n}$ we have:$ (m \minus{} n \plus{} k)(m \minus{}n\plus{} 2n \minus{} k \plus{} 1)\leq{m(m \plus{} 1)}$if multiple this inquality for $ k \equal{} 1,...,n$ we get
$ \frac {(m \plus{} n)!}{(m \minus{} n)!} \equal{} (m \minus{} n \plus{} 1)\times....\times(m \minus{} n \plus{} 2n)\leq{(m^2 \plus{} m)^n}$

[quote="ali666"]$ \frac {(m \plus{} n)!}{(m \minus{} n)!} \equal{} (m \minus{} n \plus{} 1)\times....\times(m \minus{} n \plus{} 2n)\geq1\times2\times...\times2n\geq2\times4\times...\times2n \equal{} 2^n n!$
 and for $ 1\leq{k}\leq{n}$ we have:$ (m \minus{} n \plus{} k)(m \minus{} n \plus{} 2n \minus{} k \plus{} 1)\leq{m(m \plus{} 1)}$if multiple this inquality for $ k \equal{} 1,...,n$ we get
$ \frac {(m \plus{} n)!}{(m \minus{} n)!} \equal{} (m \minus{} n \plus{} 1)\times....\times(m \minus{} n \plus{} 2n)\leq{(m^2 \plus{} m)^n}$[/quote]

Faster than I thought .....

Nice work dude  :wink:

Induction kills ti easily...especially left side...  :D


For how many positive integers $ n$ will $ \frac{60}{n}$ also be an integer?

factors of $ 60$. $ 60\equal{}2^2\cdot3\cdot5$, which has $ (2\plus{}1)(1\plus{}1)(1\plus{}1)\equal{}3\cdot2\cdot2\equal{}\boxed{12}$ factors. thus, the answer is $ 12$.


Is there a way to upload images to the forum without having them attached at the end of a post?  Sometimes when I'm solving a geometry problem, I make a diagram and want to include it next to my solution rather than at the end.  This would make it possible to hide a picture rather than having it display at the end.

(I believe this works: post the image as an attachment, click submit, edit, delete the attachment, and use the img tags to include it within the text body.  But it's a painful workaround...)

At this moment there is no way other than the one you suggested (which itself is a bug! and will be fixed - so don't do it anymore :) ) but maybe we'll figure out something in the future.

You can make an attachement in a future post and then edit your previous one to display the link.

I have same question with mathfanatic.

I see the images within the posts, not as the attached image. How do you do it? I saw those in this site so this IS possible. ;)

[quote="LynnelleYe"]Or you could sign up with a free direct-linkable host such as http://www.photobucket.com and upload your diagrams there...[/quote]

Yes... hot-linking works very well, except the pictures get deleted if you don't use your account often enough, i think....

Hmm I advise you not to do that because of many reasons. Mostly because there might be different issues with browsers. After all, up to one month ago there was no way to hide anything but text in spoilers, and everybody was happy :)

There will be a proper solution found soon, but I repeat, try [u][b]NOT[/b][/u]  to link images to external free-advertising site.


Chess problem marathon!!!!!!!!!

Post a problem after you solve it.

Compositions only, please.

Dont know how to post the picture, sorry.

W:Kg8 Qh7 Rc1 Re7 Na4 Nb7 Ba2 Bb2 Pb3 Pg7
B:Kd5 Qh1 Ra5 Rh5 Nd4 Bd1 Bg1 Pa7 Pa6 Pe3 Ph2

White mates in 2

By Godfrey Heathcote
[i]Hampstead and Highgate Express,[/i] 1905-06
(1st Prize)[/img]

Is this the position you're referring to?
[hide="If so, here is a possible solution"][list]1. Ng3+, Kc6
2. Qg6 1-0 (unless you count the knight block from black to e6 as an extra move)[/list][/hide]
A way to create a chess diagram and display it: [list]Go to [url=http://www.apronus.com/chess/wbeditor.php]www.apronus.com[/url],
Create a chess diagram (click "make diagram" to get a jpeg of the diagram after placing the pieces),
Save it to your desktop,
Add it as an attachment to your post.[/list]

Also, there used to be a chess marathon [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263&start=60]here[/url], although it hasn't been active since December 2008.

Sorry, that does count as 3 moves.

This is a chess composition, so the point is not to win; it is to do what the instructions say.

Hint: the KEY first move is not a check

Nice problem! I have not idea why I did not consider a [hide] capture[/hide] to be the first move until like after spending 30 minutes on this.

[hide="solution"]
1.Nxa5

if

1.... Nxb3
2.Rxd1#

if

1... Ne6
2.Rd7#

if black makes another move, then
2.Rc5#[/hide]

new problem (much easier):

White mates in 2:

W: Kb8, Qb3,Re1
B:Kd8, Re8, pd7 pe7 (yes this is pretty ridiculous but...)

hopefully I remembered this correctly..

sorry hurdler, but what about Nxa5 Bc2?

(this is still for the first porblem)

Here is a BIG hint.  [hide]1.Rcc7[/hide]

That big hint is more of a giveaway.
[hide="To put the first problem to rest"]After that move (1. Rcc7), white can just finish the next move with 2. Nc3++, with a few exceptions:

If 1. ... Ne2, 2. Qxh5++
If 1. ... Rc5, 2. Rxc5++
If 1. ... Rxa4, 2. Rc5++[/hide]

Geez i hope i remember Pattern's problem correctly...i saw ir once at home...but now some reason the image is a "x".

[hide]1...Qh8
2.Kh8 Rh2
3.Kh2 (or Nf3) Nf3
4.Kg1 Rh8[/hide]

Is this the right position i have in mind?

[quote="Patterns_34"][hide="At hurdler s Solution"]If 1. Nxa5, what about 1. ... Kd6 from black? That white rook at e7 would be hanging, and gives black an extra move (and even with 2. Ba3+, black can just block with the h-rook).[/hide]
[hide="As for hurdler s Problem"]We can just finish this in one move with 1. Qb6++ (++ is another way of saying mate).

But if we want to extend this to 2 moves,
1. Qe6.
[list]If black moves the rook, white finishes with 2. Qxe7++.
If 2. dxe6, Rd1++
If 2. d6 or d5, Qc8++
[/list][/hide]
[img]http://www.chessvideos.tv/bimg/zow0ceba8u8.png[/img]

A New Problem:
Black to move, and mate white (are we allowed to not specify how many moves to mate?).
[img]http://www.chessvideos.tv/bimg/4a32618rld0k0.png[/img][/quote]

1...rh2. Since black threatens mate on g2, white''s only choices are qf3 or kh2 (notice there are no checks for white). If qf3, obviously nf3 mate. If 2.kh2 rh8. AFter 3.nh5, just r takes night and mate on h1. Otherwise, kG1. Then, qg3+ and after fg3, nf3 +++ (lol it's checkmate (++) and double check (+) to give (+++)) .


Let $ \alpha\equal{}\{v_1,v_2\}$ be a basis for $ V$ and $ \beta^*\equal{}\{(v_1\plus{}v_2)^*,(v_1\minus{}v_2)^*\}$ where $ v_1^*,v_2^*$ is the canonical basis of the dual. Prove that $ \beta^*$ is basis of $ V^*$ and $ (v_1\plus{}v_2)^*\ne v_1^*\plus{}v_2^*.$

There's a flaw in the notation: $ *$ is not a meaningful operator when applied to individual vectors. It is only meaningful applied to bases, where we say that $ v_i^*$ is the linear functional which is $ 1$ at $ v_i$ and zero at $ v_j$ for $ j\neq i$.
By this definition, the $ *$ of a basis is automatically a basis for the dual space. Now, if we take $ w\equal{}v_1^*\plus{}v_2^*$ coming from the standard basis, we get that $ w(v_1)\equal{}1\plus{}0$ and $ w(v_2)\equal{}0\plus{}1$, so $ w(v_1\plus{}v_2)\equal{}2$ and $ w(v_1\minus{}v_2)\equal{}0$. That takes the wrong value at $ v_1\plus{}v_2$ to be $ (v_1\plus{}v_2)^*$ with respect to any basis.


AMC
AIME
function
AMC 12
AMC 12 B
AMC 10
AMC 10 B

Does anyone know the cutoff for the AMC 12 B... it was hard

It won't be determined until they score the tests.

Also: the initial inquiries were fine, but from this point forward, please don't ask us about the qualifying scores for the AIME.

We don't know.  We don't have any inside information.  When the AMC office knows, they will let us know.  For the A test, it took about a week.  It might take less time for the B, it might take more.  Please don't complain about it.  When we know, we'll know.  Absolutely do not contact the AMC to ask them about it.

[quote="AMCDirector"]We need to tabulate and verify a large number of answer forms in order to determine the 1% (on the 2007 AMC 10 B) and the 5% (on the 2007 AMC 12 B) levels.  Before we can do that, we have to receive, log-in, sort and open envelopes, and then run the answer forms through the scanners.  It will take us about two weeks before I can confidently report an AIME qualification cut-off score on the 2007 AMC 10 B and 2007 AMC 12 B.  As I write this on Thursday, 1 day after the contest date, we have received 22 packages out of an expected 1,500.  Please be patient.

I can tell you however that the cut-off score on the 2007 AMC 10 B will be no higher than 120, and on the 2007 AMC 12 B will be no higher than 100, just as we say on the contest cover page.

Steve Dunbar
Director, American Mathematics Competitions[/quote]

So if we get over $100$ on the 2007 AMC 12 B, we are sure to do the AIME?

100 or more, yeah, and 100 can't really be done, so sure.

[quote="usaha"]Can someone pm me the questions please because our test director took our booklets[/quote]

I'd have to request the same.  We actually had school cancelled on Wednesday, and only by chance did we manage to convince our principal to administer the test to us on Wednesday during the day.  Our test booklets are history though.

(or, when will AoPS post the problems in the contest resources section?)

what school do you go to? My school was cancelled but we got to convince our prinicipal to let us take it (i live in indiana

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135055]this post[/url].

[quote="usaha"]No i want the test, not the answers[/quote]

Yeah.. we're requesting questions, not answers.. since I didn't get to take any paper out of the testing room.

I also live in Indiana, where we all experienced such a wonderful fog day...

carmel? We got to convince the principal and like 40 out of trhe 100 people schelduled took it(im estimating)

If you want specific questions, I can post them for you, but I am not about to type out the entire test.

uh... 6-11 I don't remember what the questions were.

6.  Triangle $ABC$ has side lengths $AB=5$, $BC=6$, and $AC=7$.  Two bugs start simultaneously from $A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $D$.  What is $BD$?

(A) 1  (B) 2  (C) 3  (D) 4  (E) 5

7.  All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A=\angle B=90^{\circ}$.  What is the degree measure of $\angle E$?

(A) 90  (B) 108  (C) 120  (D) 144  (E) 150

8.  Tom's age is $T$ years, which is also the sum of the ages of his three children.  His age $N$ years ago was twice the sum of their ages then.  What is $T/N$?

(A) 2  (B) 3  (C) 4  (D) 5  (E) 6

9.  A function $f$ has the property that $f(3x-1)=x^{2}+x+1$ for all real numbers $x$.  What is $f(5)$?

(A) 7  (B) 13  (C) 31  (D) 111  (E) 211

10.  Some boys and girls are having a car wash to raise money for a class trip to China.  Initially 40% of the group are girls.  Shortly thereafter two girls leave and two boys arrive, and then 30% of the group are girls.  How many girls were initially in the group?

(A) 4  (B) 6  (C) 8  (D) 10  (E) 12

11.  The angles of quadrilateral $ABCD$satisfy $\angle A=2\angle B=3\angle C=4\angle D$.  What is the degree measure of $\angle A$, rounded to the nearest whole number?

(A) 125  (B) 144  (C) 153  (D) 173  (E) 180

Thanks a lot!  ...I guess it's good that I at least got the first 12 right..

[quote="moogra"]I thought the amc12b was easier than the amc12a. Maybe it would be 100 this time.[/quote]

I didn't find much difference.  Then again, I didn't try to do all of the problems.  The first 20 problems on both tests were similar difficulty, but this test seemed to have more algebra/counting problems that the A test.


Here's the link to the majority of Georgia competitions:
http://www.gctm.org/math_competitions.htm

I believe the next one would be Cobb County, which I'm pretty sure we're going to. Anyone else?

There's one in Alabama that GACS is going to soon

I might be coming to Cobb.  No guarantees whatsoever, though (auditions, auditions).  Then there's Duke (which is admittedly a North Carolina competition, but still), but again, I'm not sure if I'll be able to make it.  I'll be at UGA, though, which is the week after Duke.  (Be there, everybody!)

yeah i'm not sure about Duke; it's so far away!

when is duke? I'll be in NC in I think Charlotte Nov. 11-12, 2006 so JIC...

it is that weekend, here is the link for more info

http://www.math.duke.edu/dumu/contests/2006/index.html

Thanks! Too bad I don't think I'll make it then...

[quote="now a ranger"]are you all going to vestavia hills math tournament on December 9th, 2006?[/quote]
I'm no one, and I can't speak for anyone else, but I know Rockdale Magnet (recognize the name at all? I can't remember what place we got at Vestavia last year) is only sending 4 people this year. We took 8 or 12, maybe more, last year on a charter bus and everything... but this year they're kinda sneaking off in secret by car. Literally, in secret--apparently, if the whole team knew, too many people would ask if they could go, and they'd all be turned down.

Yeah... Vestavia... bad memories... Hope it all goes well for you, though.

yeah, I hope I can get into the top 10, score of 60+ at least.

[quote="cleansmyheart"]Well, it was supposed to be a secret. But everyone knows about it now--funny how that works, isn't it? Yes, Rockdale is taking four people, but they may not all be from Rockdale and we're going for fun because I begged our coach to let us go.[/quote]

Sorry, Amy, it probably wasn't my right to say the little I knew about it, but now a ranger just seems to be posting everywhere--thought I'd direct a message at him. Thanksgiving Break is boring. X.x


induction
geometry
3D geometry
algebra open
algebra

Prove by mathematical induction the formula for sum of the cubes of the first n positive integers:
1^3 + 2^3 + 3^3 + ... n^3 = (n(n+1)/2)^2. Write down explicitly the first five equations.

very simple :D
you can prove it yourself

(i don't think this is an open problem. please post it at Algebra Solved Problems)

For $n=1$, it is clearly true.

All you have to do is show that \[\left[\frac{n\left(n+1\right)}{2}\right]^{2}+\left(n+1\right)^{3}=\left[\frac{\left(n+1\right)\left(n+2\right)}{2}\right]^{2},\] which will complete the proof by induction.


Hi,

I have no idea about these problems.Can some one give me a hint.

http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=2612
http://www.spoj.pl/problems/LIAR/

Thanks.

The ACM one looks quite easy, just visualize the sentences as vertices of a graph and draw edges between them. There will be two types of edges - type 1 means that the vertices must have the same boolean value and type 2 means that they have the opposite boolean value. Rest is up to you.

The other one is actually even simpler, brute force should work all right. Suppose that person 1 says the truth - then you know exactly who is a liar and who's not, you only need to check that this setup works ($O(N^{2})$). Do this for every person. And then try the possibility that they're all liars. And that's it.

Your replies were really helpful.
Especially for Acm-Icpc problem After viewing it in terms of graph i could see components and cycles.I got it Accepted.

Now in this problem
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=3794
I was able to find maximal independent set size.I am not getting how to check
whether a unique solution exists or not.

Thanks a lot for the replies.

Well.... I do have an idea but I have this feeling that there exists a much simpler solution :).

So anyway, you could try to calculate for every person the max. size of the independent set if he is invited and the max. size if he isn't invited. If these two numbers are different for every person, the solution is unique, otherwise it isn't.

Hi ,
I used the approach that u told.It's giving TLE.I feel  it should pass because it takes O(n^2) to find MIS (when invited/no invited) on all nodes.

Any pointers regarding this.

Hm, finding the MIS for one person should take O(n), so the whole program should run in O(n^2). Is that what you meant?

Then it should definitely work. Perhaps you get TLE because your program gets stuck somewhere? I would try some random inputs and debugging, can't think of anythink else...

I figured the problem with my implementation.
Thanks 4 ur replies.. :lol:


induction
function
algorithm
topology
calculus
calculus computations

Show that a real solution exist for $ e^{\minus{}x}\equal{}x$ analytically.

What do you mean analytically? $ f(x) \equal{} e^{\minus{}x} \minus{} x$ is continuous, $ f(0) \equal{} 1 > 0$, $ f(1) \equal{} e^{\minus{}1} \minus{} 1 < 0$, so there exists a solution in $ (0,1)$. Since $ e^{\minus{}x}$ is decreasing, while $ x$ is increasing, moreover this solution is unique. How else?

Your answer is analytic opposed to the intuitive approach of the intersection of $ e^\minus{}x$ with $ x$. However, can you explain some more. Thanks

mavropnevma is relying on the Intermediate Value Theorem (for continuous functions).  Have you seen it before?  If not, you may wish to google it.

The Intermediate Value Theorem would have been my first choice, as well. But let's take a look at my second choice.

Define a sequence as follows: $ x_0\equal{}0,$ and after that, $ x_{n\plus{}1}\equal{}e^{\minus{}x_n}.$

Note that $ x_2\equal{}e^{\minus{}1}>0$ and I'm willing to argue (by induction) that $ x_2\le x_n\le x_1\equal{}1$ for all $ n\ge 2.$

That means we are iterating the function $ f(x)\equal{}e^{\minus{}x}$ on an interval $ [a,b]$ for some $ 0<a<b,$ and $ |f'(x)|\le |f'(a)|\equal{}e^{\minus{}a}<1$ for all $ x$ in $ [a,b].$ So the Banach Fixed Point Theorem applies and $ x_n$ converges to the unique fixed point of $ f(x),$ which is what you asked for.

This even gives a mindlessly simple (albeit inefficiently slow) numerical algorithm for approximating that root. Just run the recursive sequence $ x_n$ until the digits you're interested in stop changing.

I'm sorry, I meant to write "can you find more solutions". I had expected to see the intermediate value theorem. Although, I have never seen the Banach fixed point theorem. Can you explain each step of your solution kent? It implies that e^{-x} is a contraction mapping, correct?

For the Banach fixed point theorem to work, yes, I need a contraction mapping.

The first thing I need is a closed set (presumably a closed interval) such that $ f$ maps that set into itself. In this case, I claim that $ f(x)\equal{}e^{\minus{}x}$ maps $ [e^{\minus{}1},1]$ into $ [e^{\minus{}1},1].$ Since $ f$ is monotone (specifically, decreasing) it will suffice to show that the endpoints work. $ f(1)\equal{}e^{\minus{}1},$ so that end works. And $ f(e^{\minus{}1})<f(0)\equal{}1,$ so that end also works. On that interval, $ f$ is a contraction mapping: $ |f(a)\minus{}f(b)|\le K|a\minus{}b|$ for some $ K<1.$ But by the mean value theorem, $ f(a)\minus{}f(b)\equal{}f"(c)(a\minus{}b)$ for some $ c$ between $ a$ and $ b.$ Since $ |f'(x)|\le f(e^{\minus{}1})\equal{}e^{\minus{}e^{\minus{}1}}\equal{}K,$ and that $ K$ is smaller than $ 1.$ So we have a contraction mapping, and by the Banach theorem, we have a unique fixed point in that interval.

To show that that is the only global solution, we have to show that there also no solutions outside that interval. But if $ x>1,$ then $ e^{\minus{}x}<e^{\minus{}1},$ which is less than $ 1.$ So we can't have $ x\equal{}e^{\minus{}x}$ for $ x>1.$ And if $ x<e^{\minus{}1},$ then $ e^{\minus{}x}>e^{\minus{}e^{\minus{}1}}>e^{\minus{}1}$ (the latter since $ e^{\minus{}1}<1$), so you can't have $ x\equal{}e^{\minus{}x}$ for any of those, either.


Find real numbers $ x,$ such that $ x^2\minus{}\lfloor x\rfloor^2\equal{}1$

Hello,

I assume I did a mistake here, but I cannot find it  :blush: . :o

We need to prove $ x^{2} \minus{} \lfloor x\rfloor^{2} \equal{} 1$

And after that : $ \lfloor x\rfloor^{2} \equal{} x^{2} \minus{} 1$

Since $ \lfloor x\rfloor^{2}$ is an integer, so must $ x^{2} \minus{} 1$ be an integer.

And from there $ x^{2}$ must be an integer, so we can write:

$ x^{2} \equal{} x^{2} \minus{} 1$. So no solutions.  :blush:  :ninja:

$ \lfloor x \rfloor ^2 \neq \lfloor x^2 \rfloor$.

For example, $ \sqrt{2}$ is certainly a solution of the original equation.

[hide="Your mistake"]
$ x^2\in \mathbb Z$ doesn't imply $ x\in\mathbb Z$.[/hide]

[hide]
The solution is $ x \equal{} \sqrt {k^2 \plus{} 1}$ for any positive integer $ k$.

Let $ \lfloor x\rfloor \equal{} k$. We can write the equation as: $ x^2 \equal{} k^2 \plus{} 1$. Taking the square root of both sides yields $ x \equal{} \sqrt {k^2 \plus{} 1}$. Now we verify that the solution works. We have $ k < \sqrt {k^2 \plus{} 1} < \sqrt {k^2 \plus{} 2k \plus{} 1} \equal{} k \plus{} 1$, so $ \lfloor\sqrt {k^2 \plus{} 1}\rfloor \equal{} k$. Therefore:
\[ x^2 \minus{} \lfloor x\rfloor^2 \equal{} (k^2 \plus{} 1) \minus{} k^2 \equal{} 1.\][/hide]