linear algebra
matrix
induction
algebra
polynomial
binomial theorem

The linear map $ \mathcal{A} : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined by
\[ {x \choose y} \mapsto {x' \choose y'} \equal{} \mathsf{A} {x \choose y} \equal{} {\minus{}5x \plus{} 9y \choose \minus{}4x \plus{} 7y}.\]Find the matrix $ \mathsf{B}$ of the map $ \mathcal{A}$ relative to the basis
\[ \left\{ {3 \choose 2}, {1 \choose 1} \right\},\]and interpret the map geometrically. Hence show that, for each positive integer $ n$,
\[ \mathsf{B}^n \minus{} \mathsf{I} \equal{} n(\mathsf{B \minus{} I}),\]where $ \mathsf{I} \equal{} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Hence evaluate $ \mathsf{A}^n$. Verify that $ \det (\mathsf{A}^n) \equal{} (\det \mathsf{A})^n$.

I have obtained $ \mathsf{A} \equal{} \begin{pmatrix} \minus{}5 & 9 \\ \minus{}4 & 7 \end{pmatrix}$.

Since $ \begin{pmatrix} \minus{}5 & 9 \\ \minus{}4 & 7 \end{pmatrix}{3 \choose 2} \equal{} {3 \choose 2}$ and $ \begin{pmatrix} \minus{}5 & 9 \\ \minus{}4 & 7 \end{pmatrix}{1 \choose 1} \equal{} {4 \choose 3} \equal{} {3 \choose 2} \plus{} {1 \choose 1}$, $ \mathsf{B} \equal{} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. How do I interpret this geometrically?

I think $ \mathsf{B}^n \minus{} \mathsf{I} \equal{} n(\mathsf{B \minus{} I})$ is due to the Cayley-Hamilton Theorem when we consider the characteristic equation of $ \mathsf{B}$, but I don't know how to link it to $ \mathsf{A}^n$ after that.

Any hints?

You do have $ A$ and $ B$ computed correctly. Note that what you've done can be summarized by saying that $ B = P^{ - 1}AP,$ where $ P = \begin{bmatrix}3 & 1 \\
2 & 1\end{bmatrix}.$ We have that $ A$ and $ B$ are similar matrices.

As for the statement about $ B^1 - I:$ since $ B$ has a particularly simple form (it's what is known as a Jordan block), why don't you just directly verify by induction that $ B^n = \begin{bmatrix}1 & n \\
0 & 1\end{bmatrix}.$ That would be straightforward.

Or else you could say this: let $ N = B - I = \begin{bmatrix}0 & 1 \\
0 & 0\end{bmatrix}.$ It's clear that $ N^2 = 0.$ Then $ B^n = (I + N)^n = I + nN + \binom{n}2N^2 + \cdots + N^n$ by the binomial theorem - but all but the first two terms of that sum are zero. $ B^n = I + nN,$ or $ B^b - I = nN = n(B - I).$

Now, what about $ A?$ Instead of computing directly, let's just use the similarity. $ A = PBP^{ - 1},$ so

$ A^n - I = (PBP^{ - 1})^n - I = PBP^{ - 1}PBP^{ - 1}P\cdots P^{ - 1}PBP^{ - 1} - I = PB^nP^{ - 1} - PP^{ - 1}$

$ = P(B^n - I)P^{ - 1} = P(n(B - I))P^{ - 1} = n(PBP^{ - 1} - PP^{ - 1}) = n(B^n - I).$

Cayley-Hamilton has nothing whatsoever to do with this. Since $ A$ and $ B$ are similar, they have the same characteristic polynomial: $ (x - 1)^2.$ We cannot tell from that characteristic polynomial whether $ A$ and $ B$ are diagonalizable or not, and in particular, we cannot obtain that identity for $ B^n - I$ just from the characteristic polynomial.

[quote="Kent Merryfield"]Or else you could say this: let $ N = B - I = \begin{bmatrix}0 & 1 \\
0 & 0\end{bmatrix}.$ It's clear that $ N^2 = 0.$ Then $ B^n = (I + N)^n = I + nN + \binom{n}2N^2 + \cdots + N^n$ by the binomial theorem - but all but the first two terms of that sum are zero. $ B^n = I + nN,$ or $ B^b - I = nN = n(B - I).$[/quote]
I guess this what I was looking for - thanks.

[quote="Kent Merryfield"]
Now, what about $ A?$ Instead of computing directly, let's just use the similarity. $ A = PBP^{ - 1},$ so

$ A^n - I = (PBP^{ - 1})^n - I = PBP^{ - 1}PBP^{ - 1}P\cdots P^{ - 1}PBP^{ - 1} - I = PB^nP^{ - 1} - PP^{ - 1}$

$ = P(B^n - I)P^{ - 1} = P(n(B - I))P^{ - 1} = n(PBP^{ - 1} - PP^{ - 1}) = n(A^n - I).$

Cayley-Hamilton has nothing whatsoever to do with this. Since $ A$ and $ B$ are similar, they have the same characteristic polynomial: $ (x - 1)^2.$ We cannot tell from that characteristic polynomial whether $ A$ and $ B$ are diagonalizable or not, and in particular, we cannot obtain that identity for $ B^n - I$ just from the characteristic polynomial.[/quote]

I recently discussed on this forum something about tangent lines to $ y = x^n$: say $ y = ax + b$ is a tangent line to $ y = x^n$ at the point $ x = 1$. We can write $ x^n  - (ax + b) = (x-1)^2 Q(x)$ since it has a double root. Apparently there's a connection between this "analogy" and the Cayley-Hamilton theorem?


What are member ratings for and what determines a member's rating?


What are the solutions of

2^u + 3^v = 5^w

I think $u,v,w\in Z_{+}$, then $u=v=w=1.$

If $u,v,w\in Z$, then $(2,0,1)$ also works.

u=v=w=1 is the trivial solution.

 any other solutions ? if not, how to prove it ?


group theory
abstract algebra
rotation
superior algebra
superior algebra unsolved

It appears to me that Hom(Q/Z,Q/Z)=Q/Z, but elsewhere i've heard that it is uncountable. Which is the case?

It's uncountable. Specifically, we can decompose $ \mathbb{Q}/\mathbb{Z}$ as the direct sum of the union of $ \frac1{p^n}\mathbb{Z}$ for each $ p$, and those get us the $ p$-adic integers. The full homomorphism group is $ \prod_p \mathbb{Z}_p$.

$ \mathbb{Q}/\mathbb{Z}$ wouldn't make sense anyway- multiplying by $ \frac12$ doesn't work, because it's not unique.

Sorry, but could you please explain in more detail?

First, the direct sum: we can send, say, $ \frac{3}{2^3}\plus{}\frac{2}{3}\plus{}\frac{11}{5^2}\equal{}\frac{225\plus{}400\plus{}264}{600}\equal{}\frac{289}{600}$, and decompose in the other direction with the Chinese Remainder Theorem. Each element is a sum of finitely many elements with prime-power denominators, which is precisely what we get from a direct sum of these prime groups $ G_p$.
Now, an element of prime-power order must be sent to an element of prime power order, so any homomorphism maps each $ G_p$ to $ G_p$, and the whole is the direct product of these rings $ \text{Hom}(G_p,G_p)$. If infinitely many of them have two or more elements, that's already enough to make the product uncountable.

Finally, we determine what $ \text{Hom}(G_p,G_p)$ is. Each subgroup $ \mathbb{Z}/p^n\mathbb{Z}$ is mapped to $ \mathbb{Z}/p^n\mathbb{Z}$, and that is canonically $ \mathbb{Z}/p^n\mathbb{Z}$ acting by multiplication. This gives a sequence $ (x_1,x_2,x_3,\cdots)$ with $ x_n\in \mathbb{Z}/p^n\mathbb{Z}$. Each $ x_n$ determines $ x_m$ for all smaller $ m$, which we can express with the compatibility relation $ x_m\equiv x_n\mod p^m.$ That's one of the standard ways to define the $ p$-adic integers, and we're done. As it happens, $ \mathbb{Z}_p$ is uncountable on its own.

I did misspeak in one place: we get a ring structure on this Hom, not a (multiplicative) group. The automorphism group would be the group of units $ \prod_p \mathbb{Z}_p^*.$

For some reason earlier I was making a really stupid mistake by forgetting what Q/Z was, any way:
Is what you're saying amounting to
\[ \text{Hom}(\mathbb{Q} / \mathbb{Z},\mathbb{Q} /\mathbb{Z}) \equal{} \text{Hom}(\oplus\mathbb{Z} /p^n\mathbb{Z},\oplus\mathbb{Z} /p^n\mathbb{Z}) \equal{} \oplus \text{Hom}(\mathbb{Z} /p^n\mathbb{Z},\mathbb{Z} /p^n\mathbb{Z})
\]

No. First, direct products and direct sums aren't the same thing; the product is much larger when there are infinitely many pieces. Second, it's a sum/product over $ p$, and a limit over $ n$; you need to notate that limit somehow. There are also two different types of limit here: the prime pieces of $ \mathbb{Q}/\mathbb{Z}$ are direct limits, and the $ p$-adic integers are inverse limits.


function
trigonometry
AMC
USA(J)MO
USAMO
number theory
relatively prime

A calculator is broken so that the only keys that still work are the sin, cos, tan, sin-1, cos-1, and tan-1 buttons. The display initially shows 0. Given any positive rational number q, show that pressing some finite sequence of buttons will yield q. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

Man this isn't so hard, but I just thought it was pretty KWAZY.

Good exercise for the young'uns.

[quote="lowest_reynolds"]this isn't so hard[/quote]

 :?:  It doesn't look so easy to me.  Would you like to post the solution?

[quote="sunchips"][quote="lowest_reynolds"][hide]

Let q = a/b, and (a/b)^2 = m/n for relatively prime a and b.

$\sqrt{\frac{m}{n}}$ can actually obtained for any m and n positive.

If m > n, then:

$\sqrt{\frac{m}{n}}= \tan\arccos\left(\sqrt{\frac{n}{m+n}}\right)$
$= \tan\arccos\sin\arctan\left(\sqrt{\frac{n}{m}\right)}$
$= \tan\arccos\sin\arctan\sin\arctan\left(\sqrt{\frac{n}{m-n}\right)}$

If m < n, then:

$\sqrt{\frac{m}{n}}= \sin\arctan\left(\sqrt{\frac{m}{n-m}}\right)$

Walk through the cases, getting simpler and simpler until m = n.

You'll eventually get to the square root of 1, or cos 0.

Now do this backwards  :D .

[/hide]

I emphasize, it wasn't hard; it was kwazy...[/quote]

Hey....i remember,. this was from USAMO some year right?[/quote]

Yeah...

and iirc it wasn't a 1 or 4 either ;)


geometry
incenter
trigonometry
projective geometry
geometry unsolved

Given triangle $ABC$, incenter $I$, incircle of triangle $IBC$ touch $IB,IC$ at $I_{a},I_{a}'$ resp similar we have $I_{b},I_{b}',I_{c},I_{c}'$ the lines $I_{b}I_{b}'\cap I_{c}I_{c}'=\{A'\}$ similarly we have $B',C'$ prove that two triangle $ABC,A'B'C'$ are perspective.

If I am not mistaken, I see by the drawing that this result is not true. I will check again more carefully, for a definite answer.

I have in mind an easy proof for the case when the points $A',$ $B',$ $C',$ are the incenters of the triangles $\bigtriangleup IBC,$  $\bigtriangleup ICA,$  $\bigtriangleup IAB$ respectively, instead of their Gergone points $($ as a direct application of Jacobi's theorem about the pairs of isogonal lines, through vertices of a given triangle $).$

kostas Vittas.

Dear Kostas, I redraw it by cabri and my cabri say it is true, but actually I don't have any solution, I see it accidentaly by cabri and now when see your post I check it, the computer still say true

Sorry dear gemath. I was inattentive because I have been drawn a wrong configuration for your problem. I thought the points $A',$ $B',$ $C',$ as the Gergone points of the triangles $\bigtriangleup IBC,$ $\bigtriangleup ICA,$ $\bigtriangleup IAB,$ respectively...  :(

Now I have been drawn the correct figure and I see $($ by computer $)$ that the problem is true.

About the solution, I think that it is enough to prove that the lines $AD,$ $BE,$ $CF,$ are concurrent, where $D,$ $E,$ $F,$ are the touch points of the incircles of $\bigtriangleup IBC,$ $\bigtriangleup ICA,$ $\bigtriangleup IAB,$ on $BC,$ $AC,$ $AB,$ respectively.

Unfortunately, I still have not any proof for this result and I am eager for any helpful idea.

Kostas Vittas.

[quote="vittasko"]About the solution, I think that it is enough to prove that the lines $AD,$ $BE,$ $CF,$ are concurrent, where $D,$ $E,$ $F,$ are the touch points of the incircles of $\bigtriangleup IBC,$ $\bigtriangleup ICA,$ $\bigtriangleup IAB,$ on $BC,$ $AC,$ $AB,$ respectively.[/quote]

More generally, there is a result by Floor van Lamoen:

[color=blue][b]Lamoen theorem.[/b] Let A', B', C' be three points in the plane of a triangle ABC such that < B'AC = < BAC', < C'BA = < CBA' and < A'CB = < ACB', where all angles are directed angles modulo 180°. Let X, Y, Z be the feet of the perpendiculars from the points A', B', C' to the lines BC, CA, AB. Then, the lines AX, BY, CZ concur.[/color]

[hide="Proof"][url=http://mathforum.org/kb/thread.jspa?forumID=125&threadID=348786&messageID=1069651#1069651]Aeons ago, I wrote in the geometry-college newsgroup:[/url]

[code]   THEOREM 1, JACOBI THEOREM.
   --------------------------

   On the sides of a triangle ABC we erect triangles
   BA'C, CB'A and AC'B (not necessarily outwards),
   for whose (directed) base angles the equations

       angle C'AB = angle CAB';
       angle A'BC = angle ABC';
       angle B'CA = angle BCA'

   hold. Then the lines AA', BB', CC' concur.

A derivation of the Kariya Theorem from this fact
is shown in Michael de Villiers' site

http://mzone.mweb.co.za/residents/profmd/kosnita.htm

This is usually proven by the trigonometric version
of the Ceva Theorem. Here is a synthetic proof,
assuming A', B', C' outside triangle ABC and without
using directed angles:

The parallel to BC through A' meets CA at O and AB
at P. The parallel to CA through B' meets AB at M
and BC at N. The parallel to AB through C' meets BC
at K and CA at L.

Let AA' meet BC at A", BB' meet CA at B", and CC'
meet AB at C". Then we have

    AC"   LC'    BA"   PA'     CB"   NB'
    --- = ---;   --- = ---;    --- = ---.      (1)
    C"B   C'K    A"C   A'O     B"A   B'M

Since angle A'BC = angle ABC' and angle ABK = angle
PBC, we have

   angle A'BP = angle PBC - angle A'BC
              = angle ABK - angle ABC' = angle C'BK.

Further, from parallels we get angle APO = angle ABC
= angle LKO, thus angle A'PB = angle C'KB. Thus,
triangles A'BP and C'BK are similar, and hence

    PA'   BA'
    --- = ---.                                 (2)
    KC'   BC'

Now let X, Y, Z be the feet of the perpendiculars
 from A', B', C' to BC, CA, AB. Since we have angle
A'BC = angle ABC', i. e. angle A'BX = angle C'BZ,
and also angle A'XB = angle C'ZB = 90Ã?Â°, the triangles
A'BX and C'BZ are similar. Consequently,

    BA'   XA'   BX
    --- = --- = --.                            (3)
    BC'   ZC'   BZ

Together with (2), this yields

    PA'   XA'
    --- = ---.
    KC'   ZC'

Analogously,

    LC'   ZC'               NB'   YB'
    --- = ---      and      --- = ---.
    MB'   YB'               OA'   XA'

Multiplication gives

    PA'   LC'   NB'   XA'   ZC'   YB'
    --- * --- * --- = --- * --- * ---,
    KC'   MB'   OA'   ZC'   YB'   XA'

i. e.

    LC'   PA'   NB'
    --- * --- * --- = 1.
    KC'   OA'   MB'

With oriented edges, this gets

    LC'   PA'   NB'
    --- * --- * --- = 1.
    C'K   A'O   B'M

After (1), thus we get

    AC"   BA"   CB"
    --- * --- * --- = 1.
    C"B   A"C   B"A

After the Ceva Theorem, the lines AA", BB", CC"
concur, i. e. the lines AA', BB', CC' concur. This
proves the Jacobi Theorem.

Note that from (3), we also conclude

    BX   XA'
    -- = ---,
    BZ   ZC'

and similarly,

    CY   YB'                 AZ   ZC'
    -- = ---       and       -- = ---.
    CX   XA'                 AY   YB'

Multiplying these equations, we get

    BX   CY   AZ   XA'   YB'   ZC'
    -- * -- * -- = --- * --- * ---,
    BZ   CX   AY   ZC'   XA'   YB'

i. e.

    AZ   BX   CY
    -- * -- * -- = 1.
    BZ   CX   AY

With oriented edges, this will be

    AZ   BX   CY
    -- * -- * -- = 1.
    ZB   XC   YA

After the Ceva theorem, this shows that the lines AX,
BY, CZ concur. So we get the following theorem of
Floor van Lamoen:

   THEOREM 2, LAMOEN THEOREM.
   --------------------------

   In the configuration of theorem 1, if X, Y, Z are
   the feet of the perpendiculars from A', B', C' to
   BC, CA, AB, then the lines AX, BY, CZ concur.

See for this theorem

http://home.wxs.nl/~lamoen/wiskunde/isogonaal.htm
[/code][/hide]

Exercise to you (as I don't have a minute of time at the moment): Figure out why this yields your problem.

  Darij

[quote="darij grinberg"]

More generally, there is a result by Floor van Lamoen:

[color=blue][b]Lamoen theorem.[/b] Let A', B', C' be three points in the plane of a triangle ABC such that < B'AC = < BAC', < C'BA = < CBA' and < A'CB = < ACB', where all angles are directed angles modulo 180°. Let X, Y, Z be the feet of the perpendiculars from the points A', B', C' to the lines BC, CA, AB. Then, the lines AX, BY, CZ concur.[/color]

...

Darij[/quote]

Thank you dear Darij for your advices.

Now I can prove the gemath's problem $($ finaly it is easy to prove the result I mentioned $).$

I will post the solution I have in mind tomorrow, because this moment I am very tired.

Thank you again, Kostas Vittas.

Let us to change the notations of gemath’s problem, as follows:

$I_{1},$ $I_{2},$ $I_{3},$ are the incenters of the triangles $\bigtriangleup IBC,$ $\bigtriangleup IAC,$ $\bigtriangleup IAB,$ respectively.

$D,$ $D_{1},$ $D_{2},$ are the touch points of the incircle $(I_{1})$ of $\bigtriangleup IBC,$ on $BC,$ $IC,$ $IB,$ respectively. Similarly we denote the points $E,$ $E_{1},$ $E_{2},$ on $AC,$ $IA,$ $IC,$ respectively and the points $F,$ $F_{1},$ $F_{2},$ on $AB,$ $IB,$ $IA,$ respectively.

So, let be the points $A'\equiv E_{1}E_{2}\cap F_{1}F_{2},$ $B'\equiv D_{1}D_{2}\cap F_{1}F_{2}$ and $C'\equiv D_{1}D_{2}\cap E_{1}E_{2}.$

Finaly, we denote the points $D'\equiv BC\cap D_{1}D_{2},$ $E'\equiv AC\cap E_{1}E_{2}$ and $F'\equiv AB\cap F_{1}F_{2}.$

$\bullet$ Because of $\angle I_{1}BD = \angle I_{3}BF$ $\Longrightarrow$ the right triangles $\bigtriangleup I_{1}DB,$ $\bigtriangleup I_{3}FB,$ are similar and so, we have that $\frac{DB}{FB}= \frac{I_{1}D}{I_{3}F}$ $,(1)$

Similarly from the similar right triangles $\bigtriangleup I_{2}EC,$ $\bigtriangleup I_{1}DC,$ $\Longrightarrow$ $\frac{EC}{DC}= \frac{I_{2}E}{I_{1}D}$ $,(2)$ and from $\bigtriangleup I_{3}FA,$ $\bigtriangleup I_{2}EA,$ $\Longrightarrow$ $\frac{FA}{EA}= \frac{I_{3}F}{I_{2}E}$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}= 1$ $,(4)$

From $(4)$ we conclude, by Ceva’s theorem, that the segment lines $AD,$ $BE,$ $CF,$ are concurrent at one point $($ I am surprised because of I ascertain that I have been proved the Floor van Lamoen’s theorem, 40 years ago...  :wink:  $)$

$\bullet$ As well known, the segment lines $ID,$ $BD_{1},$ $CD_{2},$ are concurrent at one point, as the Gergone’s point of the triangle $\bigtriangleup IBC.$

So, it is easy to show that the point $D'\equiv BC\cap D_{1}D_{2},$ is the harmonic conjugate of $D,$ with respect to $B,$ $C.$ Hence, we have that $\frac{D'B}{D'C}= \frac{DB}{DC}$ $,(5)$

Similarly for the points $E',$ $F'$ and then $\frac{E'C}{E'A}= \frac{EC}{EA}$ $,(6)$ and $\frac{F'A}{F'B}= \frac{FA}{FB}$ $,(7)$

From $(4),$ $(5),$ $(6),$ $(7)$ $\Longrightarrow$ $\frac{D'B}{D'C}\cdot \frac{E'C}{E'A}\cdot \frac{F'A}{F'B}= 1$ $,(8)$

From $(8),$ by Menelaus's theorem, we conclude that the points $D',$ $E',$ $F',$ are collinear. Hence, by Desarques’s theorem, the triangles $\bigtriangleup ABC,$ $\bigtriangleup A'B'C',$ are perspective and the proof is completed.

Kostas Vittas.

Very nice solution by synthetic geometry Kostas!
I have found a solution by trigonometry as following
Lemma:Triangle $ABC$ incircle touch $AB,AC$ at $C',B'$ resp line $B'C'$ cut line $BC$ at $A'$ then $\frac{A'B}{A'C}=\frac{\cot\frac{B}{2}}{\cot\frac{C}{2}}$ solution
by Menelaus's theorem we easily seen $\frac{A'B}{A'C}=\frac{B'AC'B}{B'CC'A}=\frac{C'B}{B'C}=\frac{r\cot\frac{B}{2}}{r\cot\frac{C}{2}}=\frac{\cot\frac{B}{2}}{\cot\frac{C}{2}}$
Now apply this lemma(by my notation) called $B'C'\cap BC=\{A''\}$ we have
$\frac{A''B}{A''C}=\frac{\cot\frac{B}{4}}{\cot\frac{C}{4}}$
similar $B'',C''$ and easily seen
$\frac{A''BB''CC''A}{A''CB''AC''B}=1$ thus they are colinear by Menelaus's theorem therefore $A'B'C'$ and $ABC$ are perspective!
By this lemma we can prove a similar problem for Morley's Triangle :)


Pascal\u0027s Triangle
combinatorics proposed
combinatorics

I know an algebraic proof for the statement but I'd like to know if anyone can find an elegant, PURELY combinatorial proof of the following statement (i.e. don't use binomial coefficient formula):

Prove that in Pascal's Triangle, rows $2^k$ (e.g. 1, 2, 4, 8...) consist of only even numbers except for the two 1's at either end of the row.  You may also want to find a combinatorial proof to show that rows not of the form $2^k$ have at least one odd number other than the 1's at the end.

I'd be interested in seeing combinatorial proof for this!

More generally, let $n= l 2^k$, where $l$ is odd, and consider the entry in Pascal's triangle which corresponds to the number of $m$ element subsets of $\{1,2, \dots n\}$.  We define a map on these subsets as follows:

Define the  [b]periodicity[/b] of a subset $S$ to be the smallest $j$  such that $x \in S \iff x+l 2^j \in S$.  (where addition is taken mod $2^k$).  Now consider the map defined on subsets which adds $l 2^{j-1}$ to each element of $S$, where $j$ is the periodicity of $S$ (assuming $j$ is positive).  By construction the square of this map is the identity, but the map itself is not the identity.  Therefore this map induces a matching on all the $m$ element subsets with positive periodicity, proving there are an even number of them.

We're left with the sets of $0$ periodicity, that is to say the sets such that $x \in S \iff x+l \in S$.  These sets correspond to the union of various conjugacy classes $mod l$ (if you have one number congruent to $a$ mod $l$, you must have every number congruent to $a$ mod $l$).  This leads to some immediate corollaries:

-Each of these conjugacy classes has size $2^k$.  Therefore if $2^k$ doesn't divide $m$, there are no sets of 0 periodicity, so the total number of subsets is even.  In particular, if $l=1$ then the coefficient is even unless $m=1$ or $m=2^k$.

-If $m$ is a multiple of $2^k$, then the number of 0 periodicity sets is the number of ways to choose $m/2^k$ conjugacy classes out of the $n/2^k$ classes available to choose from.  Therefore we have that $\binom{n}{m}$ and $\binom{l}{m/2^k}$ share the same parity.  In particular, if $m=2^k$, then this expression reduces to $l$, which is odd.  This means every $n$ which is not a power of 2 has an odd number on the inside of its row in Pascal's triangle.


inequalities
geometry
inequalities proposed

$Please\ you\ !$. I want to learn of the inequality $s^2+r^2\le 3R^2+8Rr$ if is true.

A reference (an address), if it is possibly !

Is there a saturated inequality with the form $s^2+r^2\le mR^2+nRr$, where $m,n>0\ ?$.

For example, it is known the inequality $ s^2+r^2\le 4R^2+6Rr$.

[b]Remark[/b]. $3R^2+8Rr\le  4R^2+6Rr$. \[Thanks.\]

I have found it is equivalent to showing that for all reals $x, y, z > 0$ we have

\[
\sum_{sym} 3x^4y^2-5x^4yz+3x^3y^3+2x^3y^2z-3x^2y^2z^2 \ge 0
\]

The general one is

\[
\sum_{sym} 3mx^4y^2 + (3m-24)x^4yz + 3mx^3y^3
\]\[
 + (18m+12n-144)x^3y^2z + (5m + 4n - 56)x^2y^2z^2 \ge 0
\]

Or, under the assumption $2m+n=14$,

\[
\sum_{sym} 3mx^4y^2 + (3m-24)x^4yz + 3mx^3y^3
\]\[
 + (24-6m)x^3y^2z - 3mx^2y^2z^2 \ge 0
\]

Which evidently leads to a proof of $R \ge 2r$ by Muirhead.  ;)

To answer your question, it is not true for the case $x \rightarrow \infty$ and $y=z=1$. The best bound possible is in fact that known one (which also follows by Muirhead, incidentally.) The mysterious $x, y, z$ are the lengths of tangents to the incircle from $A, B, C$ respectively.

The best inequality of this type is
$ms^2 \le (m+1)R^2-2(m^2-5m+2)Rr+(4-m)r^2$, $0\le m \le 1$,
with equality for an equilateral triangle, and also for an isosceles triangle having the sides $(a,b,b)$ with $a/b=1-m$.


We consider the series (x_n)n>=1 defined like so: x_1>0, x_(n+1)=x_n*a^(-x_n) for any n>=1 where a>1 and it is fixed.

Prove that the series: (y_n)n>=2, y_n=(x_1+..+x_(n-1))/ln(n) converges and find out the limit of y_n when n--> oo.

bye bye

It?s easy to see that the sequence x_n is decreasing. Let L be its limit (it?s convergent since it?s decreasing and all its terms are positive). Then L=L/a^L. If L>0 we get a contradiction, so L=0, so                      x_n->0.

Let?s write (y_n)n>=1, y_n=(x_1+..+x_n)/lnn. You can see that its limit will be the same as the limit of thesequence you denoted as y_n because we only added x_n/lnn, which tends to 0. In order to use Cesaro-Stolz we should first prove that x_1+..+x_n tends to infinity. This is easy, just write what each term means: x_1+..+x_n=x_1*(1+1/a^x_1+..+1/a^(x_1+..+x_(n-1))). Then assume that x_1+..+x_n tends to L<infinity (it?s an increasing sequence, so it has a limit) and you will get a contradiction. Try it, it?s not hard.

Let a_n=x_1+..+x_n. Now we should find the limit of x_n/(ln(n+1)-lnn) so we can use Cesaro-Stolz. Observe that (1/n)/ (ln(n+1)-lnn) tends to 1 as n tends to infinity, so we should actually calculate the limit of x_n/(1/n)=n*x_n=n*x_1/a^a_(n-1). Let?s try to calculate lim a^a_n-a^a_(n-1)= a^a_(n-1)* (a^x_n ? 1). Because (a^x_n ? 1)/x_n -> lna (because x_n->0) the limit we are trying to calc is lim
lna* x_n*a^a_(n-1). By calculating this according to the definition of x_n we get x_1*lna. So we found that lim a^a_n-a^a_(n-1) = lna*x_1. We now apply Cesaro-Stolz and we get lim n/a^a_(n-1) = 1/(lna*x_1), so lim n*x_1/a^a_(n-1) = 1/lna, so lim n*x_n=1/lna, so the limit we were looking for at the beginning is 1/lna. I hope I haven?t made any mistakes, but even if I have, this is the main idea.


geometry
incenter
circumcircle
power of a point
radical axis
angle bisector
geometry proposed

$A$ is a point on the exterior of $w=(O,R)$ and $M$ is the point of intersection of $w$ and the ray $OA$
Let $B,C \in w$ and $D \in OA \cap BC$
Then the following sentences are equivalent:

$(1)\ O,A,B,C$ lie on the same circle
$(2)\ D$ is inverse of $A$ w.r.t. $w$
$(3)\ M$ is the incenter of $\triangle ABC$

[color=blue][b][u][i]Note:[/i][/u][/b] I suppose that it is well known, but I want to use this as lemma[/color]

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97522]See the case (2) =>(3)[/b][/url]

[hide="My Solution"]
$(1) \Rightarrow (2)$
This is quite easy because the inverse of the chord $BC$ is the arc $BC$ (which doesn't contain $O$).
Hence the inverse of $A$ is the point $(OA \cap BC$, which is the point $D$.
[hide="But, anyway here is a proof:"]$D$ is on the radical axis of the two circles $\Rightarrow$
$OD \cdot DA = R^{2}-OD^{2}$
$OD \cdot (OD+DA) = R^{2}$
$OD \cdot OA = R^{2}$[/hide]

$(2) \Rightarrow (3)$
Let $MN$ be the diameter of $w$.
The points $A,D$ are harmonic conjugates of $M,N \Rightarrow$
$BM$ is the angle bisector of $\angle ABC$, and
$CM$ is the angle bisector of $\angle ACB$
Hence $M$ is the incenter of $\triangle ABC$.

$(3) \Rightarrow (1)$
On the circumcircle $k$ of $\triangle ABC$ let $O'$ be the midpoint of the arc $BC$ (which doesn't contain the point $A$)
According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65422]this topic[/url], $M$ is on the circle $z=(O',O'B)$
The circle $z$ passes through the points $B,C,M$, hence $z\equiv w \Rightarrow O' \equiv O \Rightarrow$
$O \in k$[/hide]

these are some nice result of pole and polar

i really enjoyed it, bravo :wink:


algorithm algebra polynomial calculus integration function domain

I have questions about polynomial over integral domain $ K$. Does $ K[x]$ form a euclidean ring? If not, what is the necessary and sufficient condition for two polynomials $ P(x)$ and $ Q(x)$ over $ K$ to have the greatest common divisor over $ K[x]$?

actually problem occurs while  i though about:

Suppose that $ P(x)$ and $ Q(x)$ are in $ K[x]$, then by extending $ K$ to a field of Quotient $ L$ we obtain $ d(x)$ which is the greatest common divisor over $ L$. Thus we have
\[ P(x)\equal{} d(x) \lambda(x) \qquad  Q(x)\equal{}d(x) \mu(x)\]

by Gauss's Lemma  $ P(x)\equal{}d'(x) \lambda'(x)$ and $ Q(x)\equal{}d'(x) \mu'(x)$ with $ d'(x),\lambda'(x),\mu'(x)$ are all in $ K[x]$. Does $ d'(x)$ necessary to be the greatest common divisor of $ K[x]$ ? if so, does we can always obtain the gcd by euclidean algorithm in integral domain?

Thx.

[quote="Beginner"]I have questions about polynomial over integral domain $ K$. Does $ K[x]$ form a euclidean ring?[/quote]

No - try $ K \equal{} \mathbb{Q}\left[y\right]$. Then, $ K\left[x\right]\cong\mathbb{Q}\left[x,y\right]$; this is a unique factorization domain, but not an Euclidean ring (and not even a PID).

[quote="Beginner"]Suppose that $ P(x)$ and $ Q(x)$ are in $ K[x]$, then by extending $ K$ to a field of Quotient $ L$ we obtain $ d(x)$ which is the greatest common divisor over $ L$. Thus we have
\[ P(x) \equal{} d(x) \lambda(x) \qquad Q(x) \equal{} d(x) \mu(x)
\]
by Gauss's Lemma  $ P(x) \equal{} d'(x) \lambda'(x)$ and $ Q(x) \equal{} d'(x) \mu'(x)$ with $ d'(x),\lambda'(x),\mu'(x)$ are all in $ K[x]$.[/quote]

The Gauss lemma that you are using holds only if $ K$ is a unique factorization domain.

[quote="Beginner"]Does $ d'(x)$ necessary to be the greatest common divisor of $ K[x]$ ?[/quote]

Yes, but only if you first divide the polynomials $ \lambda^{\prime}\left(x\right)$ and $ \mu^{\prime}\left(x\right)$ by the greatest common divisor of their coefficients (here I mean: of the coefficients of both of them; not just of the coefficients of one of them) and multiply the polynomial $ d^{\prime}\left(x\right)$ by this greatest common divisor. Here is the reason why this is required:

$ K \equal{} \mathbb{Q}\left[y\right]$, $ P\left(x\right) \equal{} yx^2$ and $ Q\left(x\right) \equal{} yx\left(x \plus{} 1\right)$. You compute the greatest common divisor of $ P\left(x\right)$ and $ Q\left(x\right)$ over $ L$ and obtain $ x$ (you could equally have obtained $ yx$, but this is equivalent to $ x$ [i]over $ L$[/i] because $ y$ is invertible in $ L$; hence, there is nothing that would keep you from obtaining $ x$ instead of $ yx$). So you get $ d^{\prime}\left(x\right) \equal{} x$, $ \lambda^{\prime}\left(x\right) \equal{} yx$ and $ \mu^{\prime}\left(x\right) \equal{} y\left(x \plus{} 1\right)$. But $ x$ is not the greatest common divisor of the elements $ P\left(x\right) \equal{} yx^2$ and $ Q\left(x\right) \equal{} yx\left(x \plus{} 1\right)$ of the ring $ K\left[x\right]$. You have to divide the polynomials $ \lambda^{\prime}\left(x\right) \equal{} yx$ and $ \mu^{\prime}\left(x\right) \equal{} y\left(x \plus{} 1\right)$ by the greatest common divisor of their coefficients (which is $ \text{gcd}\left(y,y,y\right) \equal{} y$) and multiply the polynomial $ d^{\prime}\left(x\right)$ by this greatest common divisor to obtain the actual greatest common divisor of the elements $ P\left(x\right) \equal{} yx^2$ and $ Q\left(x\right) \equal{} yx\left(x \plus{} 1\right)$ of the ring $ K\left[x\right]$ (which is $ yx$).

  darij

An even simpler non-PID example: $ \mathbb{Z}[x]$. The ideal $ (2,x)$ is not principal.


$ x \sqrt{x} \equal{} x^1 \cdot x^{ \frac{1}{2} } \equal{} x^{ \frac{3}{2} }$.  Then use the Power Rule.


According to the standard convension for exponentiation,
$2^{2^{2^{2}}}=2^{(2^{(2^{2})})}=2^{16}=65,536$
If the order in which the exponentiations are performed is changed, how many other values are possible?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5

[hide="I only found"]
1 other possibility - $2^{8}$

but I'm probably wrong
[/hide]

[hide]how did you get that? :huh: i find two, $16^{2}$ and $4^{4}$which are both equivalent to 256.[/hide]

How do you get 2^8? I didn't get any other possibilities.

edit: nvm, I see how fireemblem got his, but how do you get 2^8?

The question asks for "how many other values are possible?"
It doesn't matter that 2 different combinations obtain the same value. The answer is B.

yes, i know. that's why i said, "which are both equivalent to 256" giving only one other value.

[hide]2^8 is the other distinct value so the answer is 1.

$1$[/hide]


[color=red]Im looking for past MATHCOUNTS tests from nats. Does anyone know where I can find any? I tried Google but it didn't help. Ive contacted a couple people and I am awaiting responses........ :?:  [/color]

You probably need to order them from the MATHCOUNTS store.

For National, order from [url=http://www.mathcounts.org]Mathcounts[/url].  They will give you material (Must pay, of course!).  For countdown, they give you PowerPoinT files.

If u have been/are on/probably will be on the nats team, u can get from the state/coach of the state team.  I almost have a comprehensive set dating back a couple of yrs.  Of course, there's always the option of buying them online, but try seeing if u can get them from ur state coordinater, I believe they have complete sets.

is it possible if you could give them to us. do you have a website?

[quote="mamathbug"]Check if your coach has ever gone to nationals...I had mine make copies of every test they had...so he has a database (it took him a while though :-P)...[/quote]

huh wait do you go to clarke?

No...Alan Chou goes there, right? I've heard of him...

I believe you guys have a similar program (someone was telling me about what he heard from alan...)...where you go for like 4 hours on Saturdays...

though good guess :!: ma + math + bug

If anyone wants...I [b]might[/b] be able to get them from him...

I said something about a program?????

Anyways, about the topic... this kind of thing has been posted several times already, and for nats, either the Mathcounts store or get them from your coach.


I'm opening a topic for bug reports and suggestions concerning moderating.

[b]SOLVED NOW[/b]

  Darij

I know about that it's in process of being solved.

problem solved. topic closed (for good :D).


The City of Harmony is 455 miles from the city of Dissension.  At 12:00 noon Paul Haymaker leaves Harmony traveling at 60 miles per hour toward Dissension.  Simultaneously, Nick Plougham starts from Dissension heading toward Harmony, managing only 45 miles per hour.  When will they meet?

I got answer 4:20 but I'm not sure whether it's right or not.  So, can anyone like solve the question with explanation and post here?  I hope I can post what I did but I don't have time now, so please can anyone?

Thanks!

Here's my solution:

[hide]Let x be the time that they were traveling.  Paul distance from his starting point is 60x and Nick's position from NIck's starting point is 45x.  Nicks distance from Paul's starting point is 455-45x.  We want to find x such that 60x = 455 - 45x.  Solving this equation, we find x = 89/21, so they meet 89/21 hours after they left, or approximately at 4:15[/hide]

Actually you are right silverfalcon. nr1337, i think you divided by 5 wrong.

[hide]Combining them together, they travel at a total of 105 miles per hour together. They need to travel together for 455 miles, so the number of hours is 455/105 = 13/4 = 4 and 1/3. Multiply that 1/3 of an hour part by 60 for the number of minutes and get 20 minutes. Thus it takes them 4 hours and 20 minutes. 12:00 + 4 hours and 20 minutes = 4:20 PM.[/hide]

[quote="nr1337"]Oops.  Now I can't read, OR divide.[/quote]
i can relate to this

The cities are 455 km apart, and their speeds add up to 60+45=105 kph.  Look at it this way: the distance the cars travel after a given amount of time is d=rt, where r is the sum of their speeds because they are both covering distance at the same time.  Therefore, d=rt, 455km=105kph x Time.  Solving for time, we get Time = 455/105 or 13/3 hours=4 1/3  hours=4 hours and 20 minutes.  So your answer was right.


geometry
analytic geometry
complex numbers

Let $G$ be the centroid of $\triangle ABC$. Show that $AB^{2}+BC^{2}+CA^{2}=3(GA^{2}+GB^{2}+GC^{2})$.

[hide]

Let the intersection of line  with  be . Then you have

So . Add the other two equalities up cyclically and you get the desired relation.

[/hide]

[b]Solution.[/b] Let $a\cdot b=\frac{1}{2}(\overline{a}b+a\overline{b})$ be the real product of the complex numbers $a,b$.  Then $a\cdot a=|a|^{2}$. Denote by lowercase letters as the complex coordinate of the capitalized point.   Let the circumcentre of $\triangle ABC$ be the origin of the complex plane.  Using $g=\frac{a+b+c}{3}$ as the centroid of the triangle, we eventually obtain, after simplification (left to the reader), $6R^{2}-2\sum_{cyc}a\cdot b=6R^{2}-2\sum_{cyc}a\cdot b$, which is clearly true. $\blacksquare$


analytic geometry
geometry
circumcircle
number theory
relatively prime
AMC

Let $ w$, $ w_{1}$, and $ w_{2}$ be the circles centered at $ (0,0)$, $ \left(\frac{1}{2},0\right)$, and $ \left(-\frac{1}{2},0\right)$ with radii $ 1$, $ \frac{1}{2}$, and $ \frac{1}{2}$, respectively. The line $ \mathcal{L}: y=a\left(x-\frac{1}{2}\right)$ for a nonzero, real number $ a$, intersects $ w$ at $ A$ and $ B$. The circumcircle of $ O: =(0,0)$, $ A$, and $ B$ intersects $ w_{1}$ and $ w_{2}$ again at $ A'$ and $ B'$, respectively. Given that $ a=\frac{8}{3}$ and $ [A'OB']=\frac{p}{q}$ where $ p$ and $ q$ are relatively prime positive integers, find $ p+q$.

[hide="hint"]There are a lot of circles...maybe an inversion would straighten them out...[/hide]


function
geometry
geometric transformation
rotation
vector

Have you ever passed through a situation when you have a glass of water full and want to move this glass between 2 different points? Generally people tend to move the glass slowly because they don’t want to get wet. But there is a way to move this full-glass of water more fast?

The statement of the problem is the following:

You have a cylindrical (or any kind of form) glass of height $ H$ and radius $ R$. You fill it with water until the height of $ H\minus{}h$. You want to move this glass between 2 horizontal points that is at distance $ L$. What is the inclination that the glass make with the vertical ($ z$ axis) in function of time? What is the position of the (center of mass or whatever specific point you want) of the glass in function of time?

Notes: You can disconsider any kind of perturbation in the surface of the liquid, since $ h$ is small in relation a $ H$, so the amplitude of any wave in the glass can be neglectable. So, by this, the surface of the water inside the glass would every be flat. Otherwise the problem would be very hard.
You can disconsider any force due the rotation of the glass.

[hide="Hint:"] if the acceleration is only due the gravity, the surface would be perpendicular to this acceleration, do not matter the orientation of the glass. In similar way, the surface will every be perpendicular to the vector $ g\minus{}a$ where $ a$ is the acceleration of the body. The signal negative is used because the acceleration of the water experiment is opposite to $ a$

So the movement will not be only horizontal. While you are accelerating the glass, you will rotate it.

(you can put a glass of water upside down without getting wet, if $ g\minus{}a\equal{}0$)
[/hide]

You want a really hard problem that's somewhat similar, go to pha.jhu.edu.  Click on Undergraduate and Course Homepage.  Go Fall 2007 semester and click on Classical Mechanics I 171.105.  It is question 5 on the final.  Hard.


linear algebra
matrix
algebra
polynomial
real analysis
real analysis unsolved

How can I solve the following system: D stands for differential operator notation

$ Dx\equal{}y, Dy\equal{}z, Dz\equal{}x$.

Thanks

In other words, $ \begin{bmatrix}x \\
y \\
z\end{bmatrix}' \equal{} \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\
y \\
z\end{bmatrix}.$

Note that the characteristic polynomial of that $ 3\times3$ matrix is $ \lambda^3 \minus{} 1,$ which has roots $ 1,\frac { \minus{} 1\pm i\sqrt {3}}2.$

We could also combine the equations to get that $ x''' \minus{} x \equal{} 0,$ which would lead us to the same place.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179254]this topic[/url], and especially the part that I responded to, for more.


superior algebra
superior algebra solved

Consider the sequence of matrices

$M_1=\left(\begin{array}{c}1\end{array}\right)$

$M_2=\left(\begin{array}{cc}1&2\\4&3\end{array}\right)$

$M_3=\left(\begin{array}{ccc}7&8&9\\6&1&2\\5&4&3\end{array}\right)$

$M_4=\left(\begin{array}{cccc}7&8&9&10\\6&1&2&11\\5&4&3&12\\16&15&14&13\end{array}\right)$

etc ....$M_n=...$

Find $trace(M_n)$


[hide="Solution"]Let the two points at which the square touches the other two squares be $ C$ and $ D$ from left to right, and let the vertex of the square diametrically opposite $ B$ be $ B'$. Since $ m\angle CB'B\equal{}m\angle DB'B$ by symmetry and $ m\angle CB'B\plus{}m\angle DB'B\equal{}90^\circ$, $ m\angle CB'B\equal{}m\angle DB'B\equal{}45^\circ$, so because $ CD\equal{}1$, the length of the foot from $ B'$ to $ \overline{CD}$ is $ \frac12$, and the foot from $ B'$ to the bottommost line is $ 1\minus{}\frac12\equal{}\frac12$. Finally, since $ BB'\equal{}\sqrt2$, the total distance from $ B$ to the base is $ \boxed{\textbf{(D)}\ \sqrt2\plus{}\frac12}$.[/hide]


umm...sorry if someone has posted this before but it i couldnt find the thread.  its also awkward that noone has posted this yet though.

"Indivisual" is spelled individual. :)

I got first place, I must have been totally lucky. There's no way I could have done that any other time. Now I have state on saturday, and I'll probably completely fail that, like everything else I've done this month. :(  :(

well i got 1st in the golden gate chapter :D
got a nice sharp trophy

if mathcounts had a good lawyer:
"mathcounts is not responsible for any trophy-related injuries..." lol :rotfl:

[quote="Ubemaya"]I'm pretty sure we're not allowed to discuss this yet.[/quote]

ur allowed to discuss ranking not scores

hey daemon, i live in cal too.  northern pwns southern though

[size=75][color=red]You [b]may[/b] discuss your own rank.  You [b]may not[/b] discuss your own score, until we tell you otherwise.

If people start posting scores, their posts will be deleted and/or this thread will be locked.  -- moderator rcv[/color][/size]

Wow, all six of us have gotten 1st so far.

i got second
but then again, im known for literally getting better scores on nationals

hey ppl who got 1st place
what were ur scores?   [size=150][color=red]<-- DO NOT POST YOUR SCORES!  DO NOT ASK ANYONE TO POST THEIR SCORES![/color][/size] [size=75][color=red]-- moderator rcv[/color][/size]

DO NOT POST YOUR SCORES

it's not allowed... and I can understand the mods' frustrations.

Mod: Hey guys, don't post your scores.
2 posts later: Hey guys, what're your scores?

[color=red]As of now, you may discuss problems and post your scores.[/color]

dude, this is gay. there should sooooo be a 6th place. i got 1st after cd though.

i got a 26, horrible. dan got a 29.

-jorian

[quote="jhredsox"]how again is that bombing out?

-jorian[/quote]
Well my score went down from last year so I was really disappointed. Deprovement usually is considered bad.

WHAT STATE R U FROM!!!!!! man, i wish i could've moved there! too bad it's too late now :( . but seriously, which state r u from?

I know, I wish I lived in like Alaska of or Guam or something.

[quote="acchessor"]I got first at my chapter and I got . . .
[hide]a 25 (15 Sprint, 10 Target).  I guess you would have wanted to be at my chapter.  Also, on the sheet my coach has, it says that the average score of students was around 11.  I won first at Countdown too.  Still no word on how I finished at state.  Didn't finish in top ten though ...[/hide][/quote]

thank you for changing the format. it'll probably save ppl minutes of worthless scrolling which can be spent on better things.

umm...lol, it's only 3 lines long!

6th to 1st in cd!!!

jorian

[quote="jhredsox"]umm...lol, it's only 3 lines long!
jorian[/quote]

Before instead of hiding he skipped like 30 lines.

oh, lol, i didn't know he did that :maybe:

sorry,
jorian

Hmm, I didn't change it.  Thanks to whoever did.

I live in Nebraska, Northeast Chapter to be exact.  I probably got the same score at state and finished 30th or something.  My coach didn't get the results yet, though (they ran out of ink of the printers).

I got 8th at Oakland Chapter (Awful, I know).
I did really bad at state, 22nd.

Bryoda! Edit your score out, you can't release it yet

Ya, too bad he also put his score in a million other places..

Actually, now that I re-reaad that:

I think that's his ranking, not his score. He just said "36" instead of "36th", which threw me off.

i got a score of 27, i found out, at my chapter, and it was both the best in  my school and it got me 3rd in the comp.

[quote="Bryoda"]I got 13 in Lincoln chapt. in PA. But my team got first so i got 2 go to states :lol: . I got 36 at state, not bad for 13 at chapter.   [color=red]<snip>[/color]  :mad: .[/quote]

I got 20-some in Fox River Valley in WI.  But my team got first so i got to go to states :lol: . I got 4th at state (3rd Written, CD Official), not bad for 20-something at chapter.

 :rotfl:


4.9: The total horizontal force exerted between the tires of a 1500-kg automobile and the ground is 980 N. If the car starts from rest, how far will it go in 5.0 s.

For this problem, I knew F=ma   so since 980 i thought it was 980=1500(a) but that was not correct.  I found a and did multiply it by 5 but did not give me the answer which was 8.2

Somebody please help me! (Please use solution techniques using Physics-B level.

The next problem is 4.7

A German Inter City Express train intially at rest accelerates to a speed of 200 km/h in 3 min 20s. What steady net force must be exerted by the engine if the total mass of the train is 8.63 x 10^5.

For this problem i have to find a, which is  delta v/ delta t. (Derivative, but since we're doing physics B, calculus is not required.)

so I try fidning a which is       200 /  (100/9)

(100/9  is 3min 20s in hours I believe so.)

but that gave me a complete weird a and it got really confusing.

The answer is 2.40x10^5 N by the way.

Can someone please help me. I need help badly :([/code]

For the second one you want to find force, which is in N, so convert the time into seconds and the velocity into m/s. Do that and you should get the right answer. I hope that helps.

i tried doing it but it doesnt work...
can you give me a solution?

3 min 20 sec is 200 seconds.
200 km / h simplifies to $\frac{500}{9}$ meters per second.
Change in velocity is $\frac{500}{9}$,
Change in time is 200
delta v over delta t = $\frac{\frac{500}{9}}{200}$
dv / dt = $\frac{5}{18}$
F = ma
$F ={8.63 \cdot 10^{5}}\cdot \frac{5}{18}$
$F = 2.39722 \cdot 10^{5}$ newtons
Force is roughly $2.40 \cdot 10^{5}$ newtons, which is the given answer.

No calculus involved  :lol:


linear algebra
matrix
induction
modular arithmetic
algebra
polynomial
calculus

We can generalize this: if $m$ divides $n$ then $F_m$ divides $F_n$ (I'll use $F_m$ for the $m$-th Fibonacci number, that is, $F_0=0$, $F_1=1$ and $F_{j+2} = F_{j+1} + F_j$ for $j \geq 0$).

Consider the matrix $A_n = \left[\begin{matrix} 1& 1\cr 1& 0\cr\end{matrix}\right]^n$. One can show by induction that $A_n=\left[\begin{matrix} F_{n+1}& F_n\cr F_n& F_{n-1}\cr\end{matrix}\right]$.

Looking these matrices $\bmod{F_m}$ and having in mind that $F_{m+1} = F_m + F_{m-1}$, we have $A_m \equiv \left[\begin{matrix} F_{m-1}& 0\cr 0& F_{m-1}\cr\end{matrix}\right] = F_{m-1}I \pmod{F_m}$. Thus $A_{km} = A_m^k\equiv F_{m-1}^kI\pmod{F_m}\iff \left[\begin{matrix} F_{km+1}& F_{km}\cr F_{km}& F_{km-1}\cr\end{matrix}\right]\equiv \left[\begin{matrix} F_{m-1}^k& 0\cr 0& F_{m-1}^k\cr\end{matrix}\right]\pmod{F_m}$

Thus $F_{km}$ is divisible by $F_m$.

[quote="random_mathematician"]Prove, for all n, that:

f(2n) is divisible by f(n)

where f(n) means f subscript n with fibonacci number n starting with f(1)=f(2)=1[/quote]didnt you already post this in intermediate? (read Sambitnayaks post!)

yeh, scrambled, thanx, but i needed a less general form: just for f(2n) instead of f(an) is divisible by f(n).

Thanx!!

i wanted a more elegant and shorter solution.

Here is another solution:

The formula for the nth Fibonacci is: $F_n = \frac{\phi^n}{\sqrt5}-\frac{\varphi^n}{\sqrt5}$ when $\phi=\frac{1+\sqrt5}{2} $ and $\varphi=\frac{1-\sqrt5}{2} $ (This can be shown but the proof is too long and I'm very lazy)
Now $F_{kn}=F_{n}\cdot{Q}$ and we want to show that Q is an integer.
$\frac{\phi^{kn}}{\sqrt5}-\frac{\varphi^{kn}}{\sqrt5}=(\frac{\phi^n}{\sqrt5}-\frac{\varphi^n}{\sqrt5}).Q$
$\frac{\phi^{kn}-\varphi^{kn}}{\phi^n-\varphi^n}=Q$
$\frac{(\phi^n)^k-(\varphi^n)^k}{\phi^n-\varphi^n}=Q$
$a^k-b^k$ is always divisible by $a-b$ so, Q is an integer.

I realized that the proof of the formula is important so, here it goes:

We want to find a sequence $F_n$ such that:
$F_n=F_{n-1}+F_{n-2}$, $F_0=0$and $F_1=1$.
Now, if a sequence $[A_n]$ is such that $a_n=a_{n-1}+a_{n-2}$ then, a sequence $k\cdot{[A_n]}$ too. And if two sequences $[A_n]$ and $[B_n]$ are like that, then the sequence $[A_n] + [B_n]$ is too.

We will look for an exponential succesion, so it has to verify:
$x^n=x^{n-1}+x^{n-2}$, one solution is 0, then dividing both sides by $x^{n-2}$ we have: $x^2-x-1=0$. The roots of this equation are $\phi=\frac{1+\sqrt5}{2}$ and $\varphi=\frac{1-\sqrt5}{2}$, so a succesion $F_n$ of the form $F_n=a.\phi^n+b.\varphi^n$ verifies the recurrence. Now, we have to look for a and b such that $F_0=0$ and $F_1=1$.

So, a and b have to verify
$a.\phi^0+b.\varphi^0=0$ and,
$a.\phi+b\varphi=1$
This gives $a=\frac{1}{\sqrt5}$ and $b=-\frac{1}{\sqrt5}$.

Sorry if the english is bad. Bye.

mcampo -- $a^k - b^k$ is divisible by (a - b) in the sense of "there exists a polynomial in a and b which is the quotient" -- you still have to prove that that polynomial will have an integral value for the particular values of a and b!  It's as if I said " $\sqrt{3} - \sqrt{2}$ divides $\sqrt{3}^2 - \sqrt{2}^2$, so the quotient must be an integer."

The particular case in which we wish to prove that $F_n$ divides $F_{2n}$ follows from mcampo's considerations: he proved that $F_n = \frac{1}{\sqrt5}(\phi^n - \varphi^n)$. So $F_{2n}=\frac{1}{\sqrt5}(\phi^{2n} - \varphi^{2n})=\frac{1}{\sqrt5}(\phi^n - \varphi^n)(\phi^n + \varphi^n) = F_nG_n$, where $G_n=\phi^n + \varphi^n$.

A simple induction proves that $G_n$ is always an integer. Indeed, $G_0 = 2$ and $G_1 = \phi + \varphi = 1$, since $\phi$ and $\varphi$ are the roots of the quadratic $x^2-x-1=0$. Similarly to mcampo's proof one can show that $G_{k+2}=G_{k+1} + G_k$ for all nonnegative integer $k$, so $G_n$ is always an integer.

Hence $F_{2n} = F_nG_n$ and $F_n$ (and $G_n$) divides $F_{2n}$.

Btw, my proof above with matrices shows in particular that $F_{2n} = F_n(F_{n-1}+F_{n+1})$.

:blush: You are right Joel, I forgot that I wasn't talking about integers, I will try to fix it and then if I can do it I'll post it.

I have found a nicer and shorter solution for the particular case:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18208url]Here[/url] is posted a proof of the fact that $F_{n+k}=F_{k-1}.F_{n}+F_{k}.F_{n+1}$. It is a short proof and uses induction. Now, for $k=n$ we have:
$F_{2n}=F_{n-1}.F_{n}+F_{n}.F_{n+1}$
$F_{2n}=F_{n}.(F_{n-1}+F_{n+1})$

Well, its a fibonacci identity so it mandates a counting proof.

Recall that if we define $f_n$ to be the number of tilings of a 1xn strip with squares and dominoes, then $f_n = F_{n+1}$ where $F_n$ represents the fibonacci numbers.

So we need to show $ f_{n-1} | f_{2n-1} $.

We look at a particular tiling and call it "splittable" between two cells on the board if there is not a domino over those two cells (imagine breaking a candy bar).

Then, we count tilings of a (2n-1) board by splittabillity between cells n-1 and n.

If it is splittable, then we have $f_{n-1}f_{n}$ tilings.

If it is not splittable, then we have $f_{n-2}f_{n-1}$ tilings.

Therefore, $f_{2n-1} = f_{n-1}(f_n + f_{n-2})$, so our claim is proven.


function
integration
real analysis
advanced fields
advanced fields unsolved

Let $g: [a,b] \rightarrow \mathbb{R}$ be a nondecreasing absolutely continuous function and $f: [g(a),g(b)] \rightarrow \mathbb{R}$ be a continuous function. Show that $\int_{g(a)}^{g(b)} f(t) d\lambda(t) = \int_{a}^{b} f(g(t))g'(t) d\lambda(t)$.
Hint: Consider the function $F$ defined by $F(x)=\int_{g(a)}^{g(x)} f(t) d\lambda(t)$.
($\lambda$ is the Lebesgue measure).

By absolute continuity of $g$ and of the set function $E \rightarrow \int_{E} f \ dm$ (defined for $E$ Lebesgue-measurable), one can easily show that $F$ is absolutely continous as well. Whence $F(x)= \int_{[a,x]} {F'(t) dt} = \int_{[a,x]}{f(g(t))g'(t)}$ (note that we used the continuity of $f$ in order to obtain its primitive and to be able to get $F'(t)=f(g(t))g'(t)$). Now taking $x=b$ gives the result.


topology
function
advanced fields
advanced fields unsolved

Hi

a topological space is compact when every open covering can be reduced to a finite open covering
a topological space is sequence compact if every sequence (no nets in general, just sequences) have a subsequence that converges

now my syllabus states that compactness does not imply sequence compact
but compactness and first countability axiom (every point has countable neighbourhood filter) do imply sequence compactness

My claim is that the first countability axiom is not necessary, here is my proof :

suppose the space is compact
suppose it isn't sequence compact. Let $x_{n}$ be a sequence that has no converging subsequence.
That means that for every point $p$ in the space, I can find an open set containing $p$ that contains $x_{i}$ only for a finite number ofpossibilities of $i$.
So for every point I have such an open set, together they cover the full space, I can reduce this to a finite covering of the full space with open sets

but all of these open sets contain $x_{i}$ for a finite number of $i$ , so that means that for $i$ sufficiently large, $x_{i}$ has nowhere to go anymore---> contradiction

Please tell me, did I use countabillity somewhere implicitly?  Is my claim and or proof correct?

i haven't really taken a look at your "proof", but it must be wrong, since there are certainly examples of spaces which are compact but not sequentially compact. such as $[0,1]^{[0,1]}$ with the topology of pointwise convergence. one can construct a certain sequence of functions $f_n: [0,1]\rightarrow[0,1]$ whose definition involves base-2 expansions and which has no convergent subsequence...but i'm too lazy to write out the details...

[quote="fredbel6"]Let $x_{n}$ be a sequence that has no converging subsequence.
That means that for every point $p$ in the space, I can find an open set containing $p$ that contains $x_{i}$ only for a finite number ofpossibilities of $i$. [/quote]

That's not true, and that's why this argument doesn't work. An example:

Let $X$ be the space of functions from $C$ to $\{0,1\}$ (the two-element set) with the product topology, where $C$ is the set of infinite binary strings. Convergence of a sequence is pointwise convergence.
Now, let $f_n(x)$ be the $n$-th bit of $x$. No subsequence $f_{n(k)}$ of the $f_n$ can converge, since we can simply choose $x_0$ so that $f_{n(k)}(x_0)=\left\{ \begin{array}{ccc} 1 & \text{if} & k\text{ is odd}\\ 0 & \text{if} & k\text{ is even}\end{array}\right.$.

Despite $f_n$ having no convergent subsequence, there are many points $p$ for which every neighborhood of $p$ contains infinitely many $f_n$. Unfortunately, they're very hard to imagine or write down.

The negation of the statement that I highlighted: For some point $p$ and every open neighborhood $U$ of $p$, infinitely many $x_i$ are in $U$. That's exactly the definition of "$x$ is a limit point of $\{x_i\}$." That's [b]not[/b] equivalent to "there exists a subsequence converging to $p$"; these infinite sets could be completely different for each $U$.

I am beginning to realise the difficulty :

jmerry, it would really help if you gave me a very precise definition of that limit point you speak of (when you use infinitely many $x_{i}$ do you really mean infinitely many, or infinitely many indices, if the sequence sometimes repeats an element, it gets subtle)

Also, could you give me a sketch of the proof in the case it IS true, when also supposing first countability
my rough guess would be that you would create a sequence of descending filters?
then i will really understand the difference

thanks for all the replies

My statement assumed no repeated values, and should be taken as "infinitely many indices. In the general case, that's not the definition of a limit point. At the very least, such points are in the closure of the set. I think I shouldn't use the term "limit point" in spaces like this anyway.

A proof in the first-countable case:
Let $x_i$ be a sequence and $p$ be a point. Suppose that every neighborhood of $p$ contains $x_i$ for infinitely many $i$.
Let $U_i$ be a countable neighborhood base at $p$; every open set containing $p$ contains some $U_i$. For each $i$, let $V_i=U_1\cap U_2\cap\cdots \cap U_i$. The $V_i$ are also a countable neighborhood base, and they are a descending chain under inclusion. By hypothesis, each $V_i$ contains infinitely many $x_i$; choose $i(1)$ so that $x_{i(1)}\in V_1$, $i(2)$ such that $i(1)\neq i(2)$ and $x_{i(2)}\in V_2$, etc. We have $x_{i(n)}\in V_n$, and the $x_{i(n)}$ are distinct.
Given any neighborhood $U$ of $p$, $V_N\subset U$ for some $N$. Since $x_{i(n)}\in V_n\subset V_N$ when $n\ge N$, the new sequence converges to $p$. We can rearrange the $i(n)$ to be increasing (or modify the process at the start to make them increasing), so that the rearrangement is a subsequence convergent to $p$ (Any rearrangement of a convergent sequence is convergent to the same limit).
By the argument you already gave, this property must hold for some point $p$ if the space is compact, and therefore the space is sequentially compact.

I cannot understand what you are arguing about. There’s the theorem: A topological space $X$ is compact if and only if each filter in $X$ has osculation point. Now if you take a sequence $(x_n)$ in a compact space $X$ and the elementary filter $\mathfrak F$ associated with $(x_n)$ (the filter of finite complements to elements of $(x_n)$) then by the above theorem $\mathfrak F$ has an osculation point. This is equivalent to the fact that $(x_n)$ has a convergent subsequence. Thus each compact topological space is sequentially compact. But reverse if false: (in general case) sequence compactness doesn’t lead to compactness.


our teacher gave us this problem for us to do. I get the basic idea but can't do it. :(  I need some help.

PROBLEM:

Using [b]coordinate[/b] geometry, prove that the midpoints joined in order of an isoceles trapezoid forms a rhombus.

Note: the teacher didn't want us to use specific coordinate such as (4,3) but wanted us to use coordinates like (a,b)

[quote="binonunquineist"]Um, okay.

[hide]I never took a geometry class so I don't know any exact format of a proof, but I think you can move the coordinate axes so that the origin lies in the exact center of the isoceles trapezoid. Then, the vertices of the trapezoid would be (-a,k),(a,k),(-b,-k), and (b,-k). Using the midpoint formula (m,n) [midpoint] = ((x1+x2)/2,(y1+y2)/2), we have the midpoints of the edges (0,k),((a+b)/2,0),(0,-k),((-(a+b))/2,0). If we use the distance formula d=sqrt((x1-x2)^2+(y1-y2)^2), we can calculate that the distance between any two adjacent vertices is sqrt((a+b)^2+k^2), and all four equal distances satisfies the definition of a rhombus.[/hide][/quote]

It will be more easy if one chooses the origin at the mid point of one of the parallel sides (generally the larger side)

[hide]Using coordinate geometry is an easier method to prove (in my opinion) than other methods. You just assign values to points and use formulas to get the specific distances you want.[/hide]


I've gotten how to solve systems so if anyone could tell me how one would solve this system of equations, I would really appreciate it!

F+3R = 10
R+2F = 8

[hide]You are given
$F+3R=10$ and $R+2F=8$

Multiply the first equation by 2 gives you:

$2F+6R=20$

Then subtract the two equations to get:

$5R=12$

Therefore $R=\frac{12}{5}$

Then substitute R into the original first equation to get:

$F+3*\frac{12}{5}=10$

Then solve for F.

$F=\frac{14}{5}$[/hide]

[hide]
Another way of doing it... (which might not be as useful in this case)

F+3R=10
F=10-3R

substitute into the second equation.
R+2(10-3R)=8
R+20-6R=8
-5R=-12
R=12/5

F=10-3R
F=10-3(12/5)
F=14/5

In some systems this method might be easier.[/hide]

I got another way

[hide]Multiply the first equation by 4 and the second one by 5 to get
$4F+12R=5R+10F$

Subtract R to get
$4F+7R=10F$

Then you get
$7R=6F$

Now plug it in to the equations:
Multiply the first equation by 6 and the second one by 7
Substitute 7R for 6F in the first equation to get
$7R+18R=60$
$25R=60$
$R=2.4$

Substitute 6F for 7R in the second equation to get
$6F+14F=56$
$20F=56$
$F=2.8$

To check yourself, plug the two numbers into the equations like this:

For the first one:
$F+3R=10$
$(2.8)+3(2.4)=10$
$2.8+7.2=10$
$10=10$

Now for the second one:
$R+2F=8$
$(2.4)+2(2.8)=8$
$2.4+5.6=8$
$8=8$[/hide]


Let ED be a chord of ABC (D lies on AB and E lies on AC) parallel to BC. If F = BE/\CD, let G be the point, a part from F, common to c(BDF) and c(CEF). Prove that AG is symmedian of ABC.

[quote="pavel kozlov"]Use the sines Ceva theorem for $ABC$ and the point $F$.[/quote]

Coul you, please, give more details.

PS

If you are interested in, I  have an ordinary synthetic  proof.


If $a,b,c,k$ are nonnegative real numbers, then we have the following inequality :
$(\frac{a}{b+c})^{k}+(\frac{b}{c+a})^{k}+(\frac{c}{a+b})^{k}\ge min(2;\frac{3}{2^{k}})$
Mixing variables !


calculus
integration
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find the integer part of number: $A=\sum_{n=1}^{10^9}n^{\frac{-2}{3}}$.

I can only tell that it is around $B=3*10^3$. (it could be B, B-1, B-2 or B-3)

You can do better than that. Use an integral from $1$ to $10^9$ to construct your estimate instead of an integral from $0$ to $10^9$.

$\int_{a-1}^bf(x)dx\leq\sum_{i=a}^bf(i)\leq\int_a^{b+1}f(x)dx$ => thats what I used

After some integrating and doing stuff I get this:

$3\sqrt[3]{10^9}\leq A\leq 3\sqrt[3]{10^9+1}-3$

$3*10^3\leq A\leq 3*10^3-3$

What should I do next(else)?

P.S. How do I put the nice looking mutliply sign(dot) instead of *? And how do I put the floor sign?

You've got all of the inequality signs pointing in the wrong direction.

Use $\sum_{n=1}^{10^9} n^{-\frac23}=1+\sum_{n=2}^{10^9} n^{-\frac23}$ to get $\sum_{n=1}^{10^9} n^{-\frac23}\le 1+\int_1^{10^9} x^{-\frac23}\,dx$.

Can you please tell me how can I derive that inequality. (I've found this one on wikipedia)

It's exactly what you were already using. All forms of the inequality come from the fact that $f(n)\ge \int_n^{n+1}f(x)\,dx\ge f(n+1)$ for decreasing functions $f$. Estimating $A$ with a sum of those terms, we get $A-\frac1{1000}=\sum_{n=1}^{10^9-1}n^{-\frac23} \ge\int_1^{10^9}x^{-\frac23}\,dx\ge\sum_{n=2}^{10^9}n^{-\frac23} = A-1$.


[quote="kdpk"]fe(2+)  +  Cr2O7 (2-)  -----------> Fe (3+) + Cr (3+)

9 (Fe (2+)  ------------> Fe (3+) + e-)

9e- + 14H(+) + Cr2O7 (2-)  ------------> Cr (3+) + 7H2O

=

9Fe(2+) -----> 9Fe(3+) + 9e-
9e- + 14H(+) + Cr2O7 (2-)  ------------> Cr (3+) + 7H2O

And now I just cancell out?

i cant balance the charges no matter what i do pls help  :blush:[/quote]

I think you incorrectly balanced the half-reactions.

The $Cr^{3+}$ term should have a coefficient of $2$...

Here are the correct half-reactions:

$Fe^{2+}\rightarrow Fe^{3+}+e^{-}$

$6 e^{-}+14 H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2 Cr^{3+}+7 H_{2}O$

Mutliply the oxidation half-reaction by $6$ and then add.  That should give the correctly-balanced equation.


calculus
integration
calculus computations

I wish you'd spend more time solving such straightforward problems yourself instead of flooding the forum with them. All you need to show is $\int_{a}^{b}f(x)dx\leq\int_{a}^{b}|f(x)|dx$ and $\int_{a}^{b}(-f(x))dx\leq\int_{a}^{b}|f(x)|dx$.


trigonometry
algebra
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number theory unsolved
number theory

[b]Suppose that $ N $ is a positive integer. Prove or disprove the statement : there  exists a positive integer $ n $   and systems  $ \{c(1),c(2),...,c(n)\} \;  , \; \{b_1,b_2,...,b_n\} $ of rational numbers  ,  such that
\[  \sqrt{N} = \sum\limits_{k=1}^n c(k)\sin{(\pi b_k)}   .
\]
 Investigate if it's possible to consider $\ds b_k=\ds \frac{2k}{m} $ for a certain integer $  m$.[/b]

Note: When $ N=5 $ it seems that it's true -[easy].

As far as I remember, this problem was published in Gazeta Matematica by Marcel Tena, but his proof is too complicated for me.

Harazi is it possible to send the solution :blush:

i think it's related to the fact that $\mathbb{Q}(\sqrt{N})$ is contained in the field $\mathbb{Q}(\zeta_{4N})$ where $\zeta_{4N}$ is a primitive $4N$-th root of unity.

proof of the thing above: indeed, we have for an arbritary monic polynomial $P(x)$ of degree $n$ with complex zeros $x_1,...,x_n$ that $\prod_{i\neq j}(x_j-x_i)=\prod_{i=1}^{n} P'(x_i)$(the left side is almost the discriminant).

assume that $n$ is odd. we will show that already taking $\zeta_n$ suffices.
let $P(x)=x^n-1$. then $\prod_{i\neq j}(x_j-x_i)=\left(\prod_{i=1}^{n} \zeta_n^{i} \prod_{i=1}^{\frac{n-1}{2}} (1-\zeta_n^{i})^n)\right)^2=\left(\prod_{i=1}^{\frac{n-1}{2}}(1-\zeta_n^{i})\right)^{2n}$. the right side of the above equality is $n^n$. it follows that for some $x\in \mathbb{Q}(\zeta_n)$, we have $x^{2n}=n^n$, thus $x^2=n$(multiply $x$ by some power of $\zeta_n$ if you need). thus $x=\sqrt{n}$(multiply $x$ by -1 if you need).

now, if $n$ is even, let $n=2m$ with $m$ odd(indeed, higher powers of 2 may be ignored). then $\mathbb{Q}(\zeta_8),\mathbb{Q}(\zeta_m)$ are inside $\mathbb{Q}(\zeta_{4n})$. therefore, $\sqrt{2}$ and $\sqrt{m}$ are, therefore $\sqrt{n}$.


[quote="hungkhtn"]Let $a_1,a_2,...,a_n$ be positive real numbers such that
\[ a_1+a_2+...+a_n=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}.  \][/quote]
Prove that $\sum_{i=1}^n\frac{1}{n-1+a_i}\le 1$ ;)
Can we compare $\sum_{i=1}^n\frac{1}{n-1+a_i^2}$ and $\sum_{i=1}^n\frac{1}{n-1+a_i}$? :P

[quote="libra_gold"] Jensen inequalitiy works,

$\sum f(a_i^2)\leq nf(\frac{1}{n}\sum a_i^2) \leq nf(1)=1$[/quote]
I think it's wrong. :(
[b][u]eg[/u]:[/b]
$f(x)=x^2$ has $f^{'' }(x)=2>0$ but $f(x)+f(y)\ge 2 f(\frac{x+y}{2})\forall x,y\in R$ ;)

[quote="silouan"]The problem first appeared in a MOP and it was posted two times in the forum.
See here
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tiks&t=64349[/quote]

I checked your links and they're with $\sum\frac{1}{n-1+a_i}$ and not with $\sum\frac{1}{n-1+a_i^2}$

but the first one is much harder than the second

[quote="libra_gold"]Because,

$n(\sum a_i^2)(\sum a_i^2)\geq (\sum a_i)^2(\sum a_i^2)=(\sum \frac{1}{a_i})^2(\sum a_i^2) \geq n^3$

i,e. $\sum a_i^2 \geq n$

As $f(x)=\frac{1}{n-1+x}, f'(x)<0, f''(x)>0$, Jensen inequalitiy works,

$\sum f(a_i^2)\leq nf(\frac{1}{n}\sum a_i^2) \leq nf(1)=1$[/quote]

It's not the solution. The sign for the last is reserved. Two problem can't compare. So try again, this problem is not easy at all.  ;)

Who can solve this very nice one? Two days has gone.  :lol:

[quote="hungkhtn"]Let $a_{1},a_{2},...,a_{n}$ be positive real numbers such that \[a_{1}+a_{2}+...+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}.\] Prove the following inequality \[\frac{1}{n-1+a_{1}^{2}}+\frac{1}{n-1+a_{2}^{2}}+...+\frac{1}{n-1+a_{n}^{2}}\le 1.\] [/quote]

Well. It's still unsolved. Does anyone remember the inequality? \[a_{1}+a_{2}+...+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}.\] Prove the following inequality \[\frac{1}{n-1+a_{1}}+\frac{1}{n-1+a_{2}}+...+\frac{1}{n-1+a_{n}}\le 1.\] In this inequality, the equaliy holds if $a_{1}=a_{2}=...=a_{n}=1$ and also for $n=1$, but in my inequality, the equality even holds for $n=2$ (then for all $a_{1},a_{2}$). Take care in this case   :), but dont worry because both of them have solutions which are quite the same.


Prove for positive $ x$

$ 2\sqrt x \ge 3 \minus{} \frac 1x$.

EDIT: Whoops, meant to post in Intermediate.

[hide]$ \sqrt{x}\plus{}\sqrt{x}\plus{}\frac{1}{x}\geq 3\sqrt[3]{\sqrt{x}\cdot \sqrt{x}\cdot \frac{1}{x}}\equal{}3.$[/hide]


this is an easy money one:

Bob walks to his room to afford a new computer. He checks his piggy bank, but notices on the table that he has left tons of stuff. Latr, he counts his money that was strewn all about:

7 quarters
14 nickels
21 dimes
54 pennies
$5.00 bill 2$10.00 bills

How much money does he have in his piggy bank?

"Easy" is subjective.

I am in general agreement with lingomaniac88.

It is ambiguous because "checks his piggy bank," for one, doesn't indicate that any money was counted in it.  We are not told in a definite way whether, in fact, there is any money in the piggy bank, because of interruption in the action.
"Money that was strewn all about" doesn't indicate whether there was any ties to the denominations to the piggy bank.
(To be fair, you never stated there were.)

[quote="lingomaniac88"][hide]It can't be determined.  It doesn't say how much is in the piggy bank, does it?[/hide][/quote]

yes, that is correct, because i never stated how much was really in the piggy bank. however, if bob's little sister had not smashed it... :D


function
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inequalities
integration
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Suppose that f(x) is a differentiable function and g(x) is a double

 differentiable function , such that

|f(x)|<=1

f'(x)=g(x)

Suppose also that f(0)^2+g(0)^2=9

Prove that there exists some 'c'  in the interval (-3,3) such that

g(c)*g"(c) <= 0

[quote="pankajsinha"]please HELP[/quote]
Let me first put the problem in LATEX just so that it is more readable.
[b]
Problem[/b]: Suppose that $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function , such that $|f(x)| \le 1$ and $f'(x)=g(x).$ Suppose also that $f(0)^{2}+g(0)^{2}=9.$
Prove that there exists some $c$ in the interval $(-3,3)$ such that $g(c) \cdot g''(c) \le 0.$

[b]Solution[/b]: We will essentially use proof by contradiction. There are two cases to consider. The reasoning in the second one is similar to that in the first; so I'll just focus on the first case.
[b]
Case 1[/b]: Let us assume, for the sake of contradiction, that $g(x) > 0$ and $g''(x) > 0$ for all $x \in (-3,3).$ Note that the second condition implies that $g(x)$ is a convex (concave-up) function; hence, by Jensen's inequality, we have $g(\frac{a+b}{2}) < \frac{g(a)+g(b)}{2}$ for all $a,b \in (-3,3).$ In particular, for $a=x$ and $b=-x,$ we have $2 \cdot g(0) < g(x)+g(-x)$ for all $x \in (-3,3)$ ... $(*)$

Now, $f'(x) = g(x)$

$\Rightarrow \int_{-3}^{3}f'(x) \, dx = \int_{-3}^{3}g(x) \, dx$

$\Rightarrow f(3)-f(-3) = \int_{0}^{3}(g(x)+g(-x)) \, dx$ ... [using the identity $\int_{-a}^{a}g(x) \, dx= \int_{0}^{a}(g(x)+g(-x)) \, dx$]

$\Rightarrow f(3)-f(-3) > 2 \int_{0}^{3}g(0) \, dx$ ... [using $(*)$]

$\Rightarrow f(3)-f(-3) > 6 \cdot g(0)$

$\Rightarrow 6 \cdot g(0) < |f(3)-f(-3)| \le |f(3)|+f(-3)| \le 1+1 = 2$

$\Rightarrow 0 < g(0) \le \frac13 .$

But the above condition combined with the fact that $|f(x)| \le 1$ implies that $f(0)^{2}+g(0)^{2}\le 1+\frac{1}{3^{2}}= \frac{10}{9},$ which contradicts the condition that $f(0)^{2}+g(0)^{2}= 9.$

[b]Case 2[/b]: If we assume that $g(x) < 0$ and $g''(x) < 0$ for all $x \in (-3,3),$ we arrive at a contradiction in a similar way as shown above.

Thus, $g(x)$ and $g''(x)$ cannot be both positive or both negative for $x \in (-3,3).$ This implies that $g(c) \cdot g''(c) \le 0$ for some $c \in (-3,3).$ And, we are done.


If $ Y_n$ are random variables with characteristic functions $ \varphi_n$, then the existence of a $ \delta>0$ such that $ \varphi_n\to 1$ on $ B(0,\delta)$ implies that the $ Y_n$ converge weakly to $ 0$.

I eventually solved this.  The condition implies that the sequence of measures associated with the ch.f's is tight, so it suffices to look at weakly convergent subsequences, but the condition implies that any weakly convergent subsequence converges to 0 since the limit of the ch.f's is $ 1\plus{}o(t^2)$ in a neighborhood of $ 0$.


calculus
integration
algebra
polynomial
limit
trigonometry
real analysis

Let $ {T_n}$ be the sequence of Tchebycheff polynomials of first type $ T_n(x)\equal{}\cos(n\arccos(x))$ , $ x \in [\minus{}1,1]$
Prove that $ \lim_{n\rightarrow\infty }\int T_n(x)\frac {sinx}{x}d\mu(x)\equal{}0$
where $ d\mu\equal{}\frac {dx}{\sqrt{1\minus{}x^2}}$

The $ T_n$'s are nearly orthononormal with respect to $ \mu$ and $ \frac {\sin x}{x} \in L^2(\mu)$, so this follows from elementary Fourier stuff.


what type of open problems are there in the fields of cryptography and combinatorics? Any good resources?(ok, maybe this should have been in the other section)

A nice resource for unsolved problems can be found at the MathWorld website

http://mathworld.wolfram.com/UnsolvedProblems.html

http://mathworld.wolfram.com/Problem.html

http://mathworld.wolfram.com/topics/MathematicalProblems.html

Be sure to check out the bibliographies at the bottom of each page, as well as his more general bibliographies found here.. http://www.ericweisstein.com/encyclopedias/books/MathematicsProblems.htm


calculus
integration
logarithms
trigonometry
search
derivative
calculus computations

How could you know that the answer will come out by substitute $x=\frac{a-t}{1+at}$ :maybe:

[quote="perfect_radio"]I saw that $1+ax$, which made me think about [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=603056389&t=124541]this topic[/url]. It was a kind of pure luck that it actually worked :whistling:[/quote]

so, exactly what music r u whistling over there.... is it "fear of the dark" or "7th son of a 7th son" ? LOL !   :rotfl:

[quote="perfect_radio"]One possible approach is the following:

Make the substitution $x = \frac{a-t}{1+at}, \, dx =-\frac{1+a^{2}}{\left( 1+at \right)^{2}}\, dt$, so that
\[\int_{0}^{a}\frac{\ln (1+ax)}{1+x^{2}}\, dx = \int_{0}^{a}\frac{\ln \left( \frac{1+a^{2}}{1+at}\right)}{1+t^{2}}\, dt . \]
Now, if $I$ is your integral, then $I = \int_{0}^{a}\frac{\ln \left( 1+a^{2}\right)}{1+t^{2}}\, dt-I$. This gives the expected answer.

PS: I'd like to thank ciprian and didilica, who made this solution possible.[/quote]

that substitution is rather interesting, yes.  :)

the method i used was leibnitz's rule for differentiation under the integral sign:

$\frac{\partial}{\partial a}\int_{f(a)}^{g(a)}h(a,x)dx\;=\;\int_{f(a)}^{g(a)}\frac{\partial}{\partial a}\left(h(a,x)\right) dx\;+\;\left( \frac{\partial g}{\partial a}h(a,g(a))-\frac{\partial f}{\partial a}h(a,f(a))\right)$


[color=green]Use the side-angle inequality to show that in triangle ABC if the internal bisector of angle A meets BC at D and AB>AC, then DB>DC.[/color]

Thanks very much guys!!! :lol:

If AD is internal bisecter of angle A then $DB = \frac{AB}{AC}.DC$. Therefor if $AB>AC$ then $DB>DC$
Do you think so?

But how do you know that DB = (AB/AC)*DC?


Where can i find AMC 8 previous years papers - with answers and just question papers

Try [url=http://www.math.ksu.edu/main/handbook/ProblemSets]this link[/url] for the past tests.

You can purchase the rest from the [url=http://www.unl.edu/amc/d-publication/publication.shtml]AMC Ordering Page[/url].

I also have some of the copies of the later ones, like 2006 onwards. If you need them, you can PM me.

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=42]Here[/url] is another place you can look, too.


complex numbers
number theory unsolved
number theory

if (a^2+b^2)/1+ab is an integer prove that its a perfect square
 (who can solve this problem in complex numbers method )
 thank you :lol:

It is from IMO 1988,problem A3 in PEN. :wink:


inequalities
vector
inequalities unsolved

Let $ \sum\limits_{i\equal{}1}^na_i\equal{}\sum\limits_{i\equal{}1}^nb_i\equal{}1$, $ a_i>0,
b_i>0, i\equal{}1,2,\cdots,n$ and $ \lambda_1\geq
\lambda_2\geq\cdots\geq\lambda_n\geq 0$. Then whether the following
inequality holds?\\
$ \frac{\sum\limits_{i\equal{}1}^n(1\plus{}\lambda_i)^{m\plus{}1}a_i}{\sum\limits_{i\equal{}1}^n(1\plus{}\lambda_i)^{m}b_i}
\leq1\plus{}\frac{\sum\limits_{i\equal{}1}^n\lambda_i^{m\plus{}1}a_i}{\sum\limits_{i\equal{}1}^n\lambda_i^{m}b_i}$,
where $ m\in\mathbb{N}$.

This holds for $ m \equal{} 0$ and $ m \equal{} 1$, but not further.

[hide="Argument why it cannot hold for m greater than 1"]
In fact, first we notice that nothing will change if we replace the condition $ \lambda_1\geq\lambda_2\geq ...\geq\lambda_n\geq 0$ by the weaker condition $ \lambda_1,\lambda_2, ...,\lambda_n\geq 0$ (since everything else in your problem is symmetric). Since $ \sum_{i \equal{} 1}^na_i \equal{} 1$, your inequality rewrites as

$ \frac {\sum\limits_{i \equal{} 1}^n(1 \plus{} \lambda_i)^{m \plus{} 1}a_i}{\sum\limits_{i \equal{} 1}^n(1 \plus{} \lambda_i)^{m}b_i} \leq \sum_{i \equal{} 1}^n a_i \plus{} \frac {\sum\limits_{i \equal{} 1}^n\lambda_i^{m \plus{} 1}a_i}{\sum\limits_{i \equal{} 1}^n\lambda_i^{m}b_i}$.

This is homogeneous in the $ a_i$, so we can drop the condition $ \sum_{i \equal{} 1}^n a_i \equal{} 1$. Besides, this is continuous in the $ a_i$, so the condition $ \left(a_1,a_2,...,a_n\right)\in\left(\mathbb{R}_{ > 0}\right)^n$ can be replaced by $ \left(a_1,a_2,...,a_n\right)\in\left(\mathbb{R}_{\geq 0}\right)^n$. Finally, this inequality is [i]linear[/i] in the $ a_i$, so it holds for all $ \left(a_1,a_2,...,a_n\right)\in\left(\mathbb{R}_{\geq 0}\right)^n$ if and only if it holds for all $ \left(a_1,a_2,...,a_n\right)\in S$, where $ S$ is the standard basis of the vector space $ \mathbb{R}^n$ (this is the basis consisting of all possible vectors $ \left(0,0,...,0,0,1,0,0,...,0\right)$). So your inequality is always true if and only if

$ \frac {(1 \plus{} \lambda_j)^{m \plus{} 1}}{\sum\limits_{i \equal{} 1}^n(1 \plus{} \lambda_i)^{m}b_i} \leq 1 \plus{} \frac {\lambda_j^{m \plus{} 1}}{\sum\limits_{i \equal{} 1}^n\lambda_i^{m}b_i}$ for every $ j\in\left\{1,2,...,n\right\}$ is always true.

This rewrites as

$ \frac {(1 \plus{} q)^{m \plus{} 1}}{\sum\limits_{i \equal{} 1}^n(1 \plus{} \lambda_i)^{m}b_i} \leq 1 \plus{} \frac {q^{m \plus{} 1}}{\sum\limits_{i \equal{} 1}^n\lambda_i^{m}b_i}$ for every $ q\in\left\{\lambda_1,\lambda_2,...,\lambda_n\right\}$.

It can easily be seen that "for every $ q\in\left\{\lambda_1,\lambda_2,...,\lambda_n\right\}$" can be replaced by "for every $ q\in\mathbb{R}_{\geq 0}$ (in fact, if you have some $ q\in\mathbb{R}_{\geq 0}$ which is not among the $ \lambda_i$, you can add variables $ b_{n \plus{} 1} \equal{} 0$ and $ \lambda_{n \plus{} 1} \equal{} q$, and then $ q$ will be in the set $ \left\{\lambda_1,\lambda_2,...,\lambda_{n \plus{} 1}\right\}$ but otherwise nothing will change). Now, this already fails for $ b_1 \equal{} b_2 \equal{} \frac12$ and $ m \equal{} 2$, as can easily be checked on a CAS; it is more or less equivalent to Minkowski's inequality for the $ m$-norm (weighted version) holding with a $ \leq$ instead of a $ \geq$. Of course, this does hold for $ m \equal{} 1$ (since the $ 1$-norm satisfies Minkowski with an equality sign; we are talking about positive vectors here), but this is false for any $ m > 1$.
[/hide]

To give a concrete counterexample: $ m \equal{} 2$, $ \lambda_1 \equal{} 2$, $ \lambda_2 \equal{} 4$, $ \lambda_3 \equal{} 3$, $ a_1 \equal{} a_2 \equal{} 0$, $ a_3 \equal{} 1$, $ b_1 \equal{} b_2 \equal{} \frac12$, $ b_3 \equal{} 0$. Unfortunately, the inequality still won't hold if we replace $ \geq$ by $ \leq$.
For $ m \equal{} 1$, the inequality is quite an easy AM-GM and Cauchy-Schwarz application, and for $ m \equal{} 0$ it is an identity.

I believe your inequality holds for all real $ m\in\left[0,1\right]$, but as my above argument shows, it is merely a combination of

$ \sqrt [m]{\sum_{i \equal{} 1}^n\lambda_i^m b_i} \plus{} \sqrt [m]{\sum_{i \equal{} 1}^n b_i}\leq\sqrt [m]{\sum_{i \equal{} 1}^n\left(1 \plus{} \lambda_i\right)^mb_i}$

(which follows from the weighted Minkowski inequality) with Lemma 2 in [url=http://www.mathlinks.ro/viewtopic.php?p=161525#161525]http://www.mathlinks.ro/Forum/viewtopic.php?t=20547 post #10[/url] (applied to $ n \equal{} 2$; this is merely a particular case of Hölder).

  darij


A woman arrives at an airport to meet her husband sometime between 4:00 and 5:00 and stays there for 10 mins.  The husband arrives at the airport sometime between 4:00 and 5:00 and stays there for 20 mins to meet his wife.  What are the chances that the man and the woman will meet eachother?

I'm a beginner at geometric probability, so I think I'm wrong but is this the answer?

[hide="Solution?"]
The husband must arrive in a specific time period starting 20 minutes before the wife arrives and ending right as the wife leaves.  This is 30 minutes out of 60, so it would seem to be 1/2.  But if the wife comes right at the beginning, it's only 1/6.  What do I do next?
[/hide]

I did a fast calculation and i think this is the correct answer:

[hide="Answer"]$1-\frac{41}{72}=\boxed{\frac{31}{72}}$[/hide]

[quote="SM4RT"]
$1-\frac{49}{72}=\boxed{\frac{23}{72}}$[/quote]
Why?  Can you explain?

hmmm I have a different solution:

[hide]
Let us draw out a 60x60 square, where one side is the wife's arrival times and the other the husband's. The area of this square is the value of all...

ok well before I get into all of this (as it will probably require a picture etc. etc.) I should probably find out if my solution is right  :P :

$\frac{4}{9}$
[/hide]

EDIT: hmmm, I think I drew out the lines wrong for what areas to calculate...? I'll try again another time.

EDIT2: Yes I just re-drew the picture and get the same answer as is the general consensus here.

hmm i get 31/72..I just drew a graph.  The graph was bounded by the coordinates(up to 60) (0,0) (20,0) (60,40) (60,60) (50,60) and (0,10) but im not sure..im not experienced with this...

JUST KIDDING, I just realized that I already subtracted from 1 on my paper....>.>.

Stupid messy paper. Another reason to write neater and use a clean sheet of paper.

I agree with Art of Owna.  However, I cannot be certain either.  Here was my approach:

[hide]Draw a square with dimensions 60 by 60, then draw the diagonal.  Then draw shorter lines parallel to the diagonal.  If you can picture the left and top sides, the first will be connecting the points 20 unit up from the bottom-left vertex and 20 units left from the top-right vertex.  This forms an isosceles right triangle with side lengths 40 and area 800.  The area between the shorter line and the diagonal represents the possible arrival time of the husband.  Now do the same thing except the points will be 10 units right from the bottom-left vertex and 20 units down from the top-right vertex.  This forms another triangle with area 1250, the area between this line and the diagonal represents the possible arrival of the wife with respect to the husband.  Now subtract this area from the total area (complementary counting).  The result is $3600-800-1250=1550$.  The probability is $1550/3600=31/72$[/hide]

EDIT:  Zellex, I got Art of Owna's answer, however I subtracted for complementary counting.  I have a feeling that one of us found the wrong areas, and I'm pretty sure that my answer is right.

huh, what do you mean subtract from 1?  I got 41/72 and I subtracted that from 1 to get 31/72..not the other way around..

Zellex, I think you found the probability that tehy will NOT meet eachother

yes it is $\frac{31}{72}$ i made a small error...


number theory
greatest common divisor
number theory unsolved

@PhilG: $ 2^8\equal{}256$

The gcd of them all must divide 273 and 21, so it's 1 or 7

We'll prove it's 7.

Powers of 2 give 1,2,4 as residues modulo 7, 2 if the power is 3k+1, 4 if the power is 3k+2, 1 if the power is 3k

$ 2^{n\plus{}1},2^n,0$ all give different residues modulo 3, so every number in the list gives residue 2+4+1=7, or 0 modulo 7

Answer: gcd=7

[quote="Bugi"]Answer : gcd=7[/quote]

I think it's $ 21$  :(

if we put $ a_i \equal{} 2^{2^i}\plus{}2^{2^{i\minus{}1}}\plus{}1$
we can see that :
$ a_i\equal{}2^{2^i}\plus{}2^{2^{i\minus{}1}}\plus{}1\equal{}(2^{2^{i\minus{}1}}\plus{}1)^2\minus{}2^{2^{i\minus{}1}}\equal{}(2^{2^{i\minus{}1}}\plus{}1\minus{}2^{2^{i\minus{}2}})(2^{2^{i\minus{}1}}\plus{}1\plus{}2^{2^{i\minus{}2}})\equal{}(2^{2^{i\minus{}1}}\plus{}1\minus{}2^{2^{i\minus{}2}}).a_{i\minus{}1}$
so we get that $ a_n | a_{n\plus{}1}$ and it follows that the gcd of the terms of the sequence is $ a_2\equal{}21$.

Yes I forgot that 273 is divisible by 3. But the procedure is the same as with 7...


I'm sure this problem has a really easy solution, but it's got me stumped:

"The high school band corps has 8 flags - 4 are white, 3 are blue, and 1 is red. How many different ways can teh flags be lined up in a straight line?"

Thanks for any help y'all can provide...

PS - the answer choices given are:
a. 96
b. 144
c. 280
d. 1344
e. 3360


TN on a roll son!

3 guys  pwned the INOI and are thru to the camp.(they are only in 11th)

Hi,
I am from Chennai.I've made it through INOI.
Congrats to all who made it.Any idea where the camp will be held?
What has been the cut-off?

Congrats to chackjak and arjunarul...
make the best out of the camp...
best of luck...

hi da arjun . You on this forum too? This is anish

Congrats to all those selected!!  :coolspeak:

thanks everybody... but im genuinely surprised keshav u didnt make it..

Looking forward to the camp!

Congrats to all those who made it this year. The camp will be held in The Intl School Bangalore... probably in the end of June..

Don't worry you'll get mails from Prof.Madhav Mukund with all required details.

Can some post the questions of this years INOI??

And whats the difference between ZCO and ZIO? Does ZCO give another chance to qualify?

They started ZCO alternatively this year to allow students to qualify to write INOI(other than ZIO alone). ZCO was an online programming contest(on OPC server-3hrs)

Seems like I've qualified too.

Chronoz: The problems were something like this:

[list=1]
[*] Let B be a permutation of some set A. Find the minimum number of swaps required to get from A to B, given that you can only swap adjacent elements.

[*] In a game, there are two jars of beans P and Q. The number of beans in the jars P and Q at an instant of time are x and y, respectively. If the total number of beans is even, you move. Otherwise, your opponent moves. A value of f(x, y) is added to your score every instant, where f(x, y) = $ Ax^2 \plus{} By^2 \plus{} Cxy$; A, B, and C are constants. A player can do one of the following:
[list=1]
[*] Move a bean from P to Q.
[*] Remove a bean from Q.[/list]
The game ends when there are no beans left in both the jars. You always play to maximize your score, while your opponent always plays to minimize your score. Find the score you will end up with. [/list]

The problems weren't too interesting from an algorithmic standpoint. The first one required a little bit of insight for a O(n^2) solution, and the right data structure for a O(n lg n) one. The second one was straightforward DP.

Is there anything else about the training camp you would like to share with us?

yea, i got the n^2 solution for the first and the straightforward one for the second...

Chennai totally ruled the INOI this year

I did the first sum in around N*N/2.For the second sum I used Dynamic Initializing,similar to what you all have done.I only know what has been taught in our school in XI C++ and whatever I learnt in the OPC Server with the solutions(great site),so guess I will learn a lot in the camp.
Anything special to learn before the camp?

@Phinity:
Hi Anish,
I started using this site only two days ago,though I registered long ago.

Im Rishabh Kabra, from Dubai, nice to meet y'all.

@arjun: yeah, the OPC problems and solutions were pretty much my only source as well.
@manama: pls introduce (reveal) urself

@crackjak987
believe me... even i am genuinely surprised...

:D  ALL The BEST  :D
To all those who are selected for IOITC and will be writing the Asia Pacific Informatics Olympiad on may 9.... Get some prizes for INDIA. :!:

The APIO 09 will be over by the time you guys read this.. Please post the questions here and enlighten us lesser mortals...

We can't. There are others who are yet to write the APIO.

If the contest is over yet could someone post the questions
Please  :roll:

We have been told not to discuss till Monday.
Though,there is nothing to discuss for me. :rotfl: I spent most of my 5 hours trying to learn something useful about Linux(new to me).I just stopped short of playing games. :rotfl:

all the best to everyone going to IOITC.

All the Best at IOITC

Do post the final list of indian representatives at the IOI.
Please mention any good informatics sources provided at the camp :roll:

In fact, can you smart IOI TCers post your online camp mails here?

There wasn't much of an online camp really,

Please tell me something more about INOI.Why is there no Java? :huh:

http://en.wikipedia.org/wiki/Indian_National_Olympiad_in_Informatics
There is no Java in INOI because it is not there in IOI. A lot of reasons have been quoted for this
http://ioinformatics.org/newsletters/html/ioinews6.htm#Java
http://ioinformatics.org/admin/isc/iscmeetings/meet-09/java-proposal-zaf.pdf

:mad: Hey,That's unfair,that means I wont be able to participate as ICSE and ISC syllabus includes only Java!
Where could I get the syllabus?

The fact that we (in CBSE schools) have learnt c++ in school can only serve as a base for the lang. for e.g in my school they are still taking c++ (structures,classes or something) and if I were to wait for them to finish the syllabus I would have to wait till I pass XIIth!! Once you get the basic syntax ( condition and loop statements) it should be no trouble at all. You might want to check this as an intro into c++:
[url]http://cplusplus.com/doc/tutorial/[/url]
(the site also has reference for functions,libraries,etc. with quite some depth)

Thanks!I am havin problem with buffer statements,will this website help?

Really, if you KNOW java, C++ should take 30 mins. Obviously, the more advanced stuff requires time.


invariant
combinatorics unsolved
combinatorics

Two red points placed around a cycle. in each step we can :
1) Add a Red point somewhere and change the color of its neighbors (Red to Blue and Blue to Red)
2) Remove a Red point and change the color of its neighbors (Like 1).
Can we reach to the "two blue points around the cycle" ?

What daoes it mean two points around a cycle?

Does it mean "two beads on a string?"

Quite tough, it took me a long time to find this invariant...
1-"cut" the circle at a random place and write down in a line the colours of the points you encounter going, say, clockwise: e.g. RBRRRBBBR
2- write down the number of red points which lie between two subsequent blue points:  e.g., in the preceeding configuration,
1 3 0 0 1
3- sum and subtract alternately the numbers you have noted: e.g.
1-3+0-0+1=-1

Note that this number (let us call it i(X), where X is the R/B string) does not depend on the segment in which we have "cut" the circle, except from the fact that we can change its sign (e.g. getting 1 instead of -1 if we had started two positions later)
This also depends on the fact that the number of blue points on the circle is always even, another invariant which is easy to check.

Moreover, i(X) is invariant modulo 3 (apart from the sign: that is, it can be zero or non-zero mod 3, and this fact is invariant): in fact, for every possible move (X and Y denote any R/B string)
(as our invariant does not depend on cyclic permutations of the string, it suffices to check the case in which the move is at the beginning of the string)

RRX --> BRBX
2+i(X) --> i=-1+i(X)

BBX --> RRRX
i(X) --> 3+i(X)

RBX --> BRRX
1-i(X) --> -2-i(X)

BRX --> RRBX
-1-i(X) --> 2+i(X)
and in all four cases the quantity is preserved modulo 3.
In the starting configuration the invariant is nonzero (mod 3), while in the target it is zero: so, the target cannot be attained. QED.
some ideas I used in order to find this invariant...
1- apparently, sequences made up with 3k+1 and 3k+2 R's can be attained from RR, while 3k R's can only be attained starting from the BB
2- this is because we can "simplify" with a couple of steps a RRR sequence. Other  such sequences we can simplify are BB and BRBR
3- this way we can start from a long string and simplify it with successive steps to get a simpler one. So we can check for which sequences our candidate invariant must be equal or non-equal.4- Our invariant cannot involve only the sum of B and R's, since sequences such as BRBR and BBRR belong to the two different configurations. With a slightly longer counterexample we see it cannot either involve "couples" of successive colours.
5- based on similar problems, the invariant should be some kind of "linear" (in some sense)
... and then after much work and unsuccessful tries I found it.
thanks for proposing this problem, I like invariants :-)
--federico poloni


limit
trigonometry
calculus
calculus computations

$\textbf{Prove : }$  :rotfl:

$\lim_{n\to\infty}2^{n+1}\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_{\textbf{n}}}\;=\;\boxed{\pi}$

Sketch proof.
It can be shown that the limit is equal to
$\lim_{n\to+\infty}2^{n+2}\sin\left(\frac{\pi}{2^{n+2}}\right)=\pi$.

It seem the following thing and my question look really similar , but I cannot find the detail .

$\frac{\sqrt{2}}{2}\frac{\sqrt{2+\sqrt{2}}}{2}\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdots\;=\;\frac{2}{\pi}$

Could you show me

:maybe: $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$, $\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}$.


limit
trigonometry
calculus
geometry
algebra unsolved
algebra

Find:

$ \lim_{n \rightarrow \infty}  n \sin {\frac{2 \pi}{n}}$

Gee that was quick... whats your method?

Please dont use calculus!, but if you got it that fast i dont think you did...

Edit: OK i get it, that was too easy, if you want the solution i have, go without calculus... it has a lovely answer

It took me all of about 5 seconds - this is a common exercise in any calculus textbook.  Set $\ell=2\pi/n$; the limit we are looking for becomes

$\lim_{\ell\to 0} \frac {2\pi \sin\ell}{\ell}=2\pi\lim_{\ell\to 0} \frac { \sin\ell}{\ell}=2\pi$

Ok, Ok, it was a crap problem.

The solution i got, which is nice, and i suppose with blah's method is another proof of $ \lim_{n \rightarrow \infty} \frac{\sin n}{n}=1$

is thats its the area of an regular 'n-gon' insribed in a unit circle


calculus integration logarithms MATHCOUNTS trigonometry algebra polynomial

What is the first three digits of $2001^{2001}$?

I don't know the level of this problem as there seems to be a pattern with powers of two but then it starts getting messed up turning into 257 and 514 later on so if it's too difficult, please move it.

If you raise something that ends in 1 to any (integral) power, you should never get something that ends in 4 or 7...in fact it should always end in 1. This problem is pretty simple with modular arithmetic:

2001 = 1 (mod 1000)
2001^2001 = 1^2001 (mod 1000)

So the answer is 001.

isn't it the first 3, not the last three?

0->A
1->C
repeat a=2001
A+1->A
2001C->C
If C>1E90
then
C/1E10->C
end
disp c
end

if you are willing to wait...

2^2001*23^2001*29^2001 is the prime factorization, but I don't see its use for that...

[quote="Ignite168"]I think that's going to overflow before it gives the answer.[/quote]

0->A
1->C
repeat a=2001
A+1->A
2001C->C
[b]If C>1E90
then
C/1E10->C
end [/b]
disp c
end

[quote="binonunquineist"]I don't know if you people understand logarithms, but you can use logs to solve the problem.

[hide]First of all, you can express 2001^2001 in scientific notation, a.bc*10^n, where a, b, and c are digits, and n is an integer. We can write 2001^2001 as

10^(log(2001^2001))

Using the property log(a^b) = b*log(a) [I can prove it but I'm too lazy], we get

10^(2001*log(2001))

We can then seperate this expression into:

10^[integer part of 2001*log(2001)] * 10^[fractional part of 2001*log(2001)]

Notice that the first part of this expression is 10 raised to an integer, and the second part of this expression is 10 raised to a fractional power (thus is less than 10). So this number has been put in scientific notation! Grabbing our calculator, we take 2001 and multiply by the log of 2001. We get 6605.79542...

Subtract 6605 to get the fractional part. Now we are left with .79542...

Now take 10 to that power. You should get 6.24345...

So basically the number 2001^2001 expressed in scientific notation is simply 6.24*10^6605! (recall that 6605 was the integer part of 2001*log(2001))

Thus, our first three digits are [b]624.[/b][/hide][/quote] :what?: I understand logs but woah!

That is cool, but it took ehehheehee a few minutes to conprehend.  It helps that I already know log(a^b)=b(log(a))...

Thats a good solution, but I'm pretty sure there is a non-logarithm solution, this is MATHCOUNTS!

Supposedly, and I say supposedly because there are many problems that are easier using more advanced methods, you don't need trig, logarithms, or nearly anything in AoPS book 2, except:

One to one correspondances
Polynomial division/synthetic
Inverse functions

This is over ehehheehee's head.  This is ehehheehee's way of solving hard problems: 1. look at it. 2. look at it more. 3. brute force or look for a pattern (howerver long)