geometry
circumcircle
symmetry
ratio
IMO Shortlist

Determine the smallest positive real number $ k$ with the following property. Let $ ABCD$ be a convex quadrilateral, and let points $ A_1$, $ B_1$, $ C_1$, and $ D_1$ lie on sides $ AB$, $ BC$, $ CD$, and $ DA$, respectively. Consider the areas of triangles $ AA_1D_1$, $ BB_1A_1$, $ CC_1B_1$ and $ DD_1C_1$; let $ S$ be the sum of the two smallest ones, and let $ S_1$ be the area of quadrilateral $ A_1B_1C_1D_1$. Then we always have $ kS_1\ge S$.

[i]Author: Zuming Feng and Oleg Golberg, USA[/i]

This is from IMO Shorlist 2007,problem G6

The answer is $k=1$. We can get $\frac{S}{S_1}$ arbitrarily close to $1$ by taking $AA_1BB_1CC_1DD_1$ arbitrarily close to a medial triangle configuration (e.g. approaching $D_1=A=A_1$).

[b]Lemma.[/b] Let $\triangle{XYZ}$ be a triangle and $X_1,Y_1,Z_1$ points on sides $YZ$, $ZX$, $XY$. Then $[X_1Y_1Z_1] \ge \min_{\text{cyc}}([XY_1Z_1])$.

[i]Proof.[/i] For convenience, take an affine transformation sending $\triangle{X_1Y_1Z_1}$ to an equilateral triangle, and construct its antimedial triangle $\triangle{X_2Y_2Z_2}$. Let $O$ be the circumcenter of $\triangle{X_1Y_1Z_1}$.

Suppose the ``border areas'' $[XY_1Z_1],\ldots$ are all strictly greater than $[X_1Y_1Z_1]$. Then $d(X,Y_1Z_1) > d(X_1,Y_1Z_1)$, so $X$ cannot lie inside $(OY_1X_2Z_1)$. Since $Y_1Z_1$ separates $X,X_1$, we must have $\angle{Y_1XZ_1}<60^\circ$. By symmetry, we obtain $\angle{X}+\angle{Y}+\angle{Z} < 3\cdot60^\circ=180^\circ$, contradiction.$\blacksquare$

We will now prove $S_1 \ge S$. Inspired by the degenerate case, we define $T = A_1C_1\cap B_1D_1$, $f(A) = [AA_1D_1]$, $g(A) = [TA_1D_1]$, and $h(A) = \max([B_1A_1D_1],[C_1A_1D_1])$; similarly define $f(X),g(X)$ for $X=B,C,D$.

[b]Case 1:[/b] There exists $X\in\{A,B,C,D\}$ such that $f(X) \le h(X)$. WLOG $f(A) \le h(A)$ and $[B_1A_1D_1] \le [C_1A_1D_1]$. Then
\begin{align*}
S_1-S &\ge [C_1A_1D_1]+[A_1B_1C_1]-f(A)-\min(f(B),f(C),f(D)) \\
&\ge [A_1B_1C_1]-\min(f(B),f(C),f(D)).
\end{align*}But $A_1BCDD_1$ is convex, so letting $B' = CB\cap D_1A_1$ and $D' = CD\cap A_1D_1$, we find $f(B) \le [B'A_1B_1]$, $f(C) = [CB_1C_1]$, $f(D) \le [D'C_1D_1] \le [D'C_1A_1]$. The lemma now immediately gives $S_1-S \ge0$, as desired.

[b]Case 2:[/b] $f(X)>h(X)>g(X)$ for every $X\in\{A,B,C,D\}$, so $d(A,A_1D_1)>d(T,A_1D_1)$, etc. Taking an affine transformation if necessary, WLOG assume $\angle{A_1TB_1}=90^\circ$. For convenience, we write $T=(0,0)$, $A_1=(-a,0)$, $B_1=(0,b)$, $C_1=(c,0)$, $D_1=(0,-d)$ for positive reals $a,b,c,d$.

Then in particular, either $x(A)<-a$ or $y(A)<-d$ must hold (analogous statements hold for $B,C,D$). WLOG suppose $y(A)<-d$. Then we easily deduce
\begin{align*}
y(D)>-d &\implies x(D)>c>x(C) \\
&\implies y(C)>b>y(B)\implies x(B)<-a<x(A).
\end{align*}A simple ratio calculation (note $A$ is bounded by $C_1D_1$, $D_1A_1$, and $x=-a$) yields
\[ [C_1A_1D_1] \le h(A) < f(A) \le \frac{a}{c}[A_1D_1C_1], \]whence $a>c$. But similarly, we deduce $b>d$, $c>a$, and $d>b$, contradiction.

[b]Comment.[/b] There are certainly other ways to prove the lemma. Most obviously, we can directly use barycentric coordinates or standard area calculations. If we want to avoid both affine transformations and calculation, we can compare $\triangle{X_1Y_1Z_1}$ with the medial triangle $\triangle{X'Y'Z'}$ of $\triangle{XYZ}$.

[b]Comment.[/b] The key is to look at degenerate cases (and realize the answer is $k=1$). "Simple case analysis" is one way to motivate this, but smoothing provides a bit more insight, both for why the degenerate case should be optimal and how to go about proving $S_1\ge S$. Indeed, suppose we have a configuration with $S/S_1$ maximal (or at least ``close'' to maximal). Then we can smooth to the case $f(A)\le f(B)=f(D)\le f(C)$ (otherwise, consider rotating $BC,CD$ about $B_1,C_1$ to increase $S$ while fixing $S_1$). Now if $f(A)<f(B)=f(D)$ and $B_1$ is closer to $AB$ than $D_1$ (or $C_1$ is closer to $AD$ than $A_1$), we can slide $A_1$ away from $A$ a bit (resp. $D_1$ away from $A$) to decrease $S_1$ while increasing $S$.

Hence an ``optimal'' configuration (with $S/S_1$ maximal) should have $d(D_1,AB) \le d(B_1,AB)$ (and $d(A_1,AD) \le d(C_1,AD)$). While we can't do much more obvious smoothing, this at least suggests we consider the case when at least one of $A_1,D_1$ coincides with $A$ (or just $f(A)$ is really small).

Lemma: in a triangle ABC with X,Y,Z on sides BC,CA,AB then (XYZ) $\ge$ (AYZ),(BXZ) or (CXY)
Proof: WLOG ABC is quadrilateral, now just computations.

The proof k=1 is easy with this result.WLOG rays AB and DC intersect at P, then apply lemma to triangle APD and we get S1 > (A1C1D1) $\ge$ (AA1D1), (DDAC1) or (PA1C1). In the last case, (PA1C1) > S so we are done. Otherwise, WLOG (A1C1D1) $ \ge$ (DC1D1) and we easily get (A1B1C1) > (A1BB1) or (B1CC1), so we get S1 > S.

For equality let D=A=D1, and A1, B1, C1 be midpoints of ABC.

We claim that the answer is $1$. First, we'll show that $S_1=S$ can be achieved. Let $ABC$ be a triangle, and $D$ be infinitely close to the midpoint of $AC$. Then, we can let $A_1B_1C_1D_1$ essentially be equivalent to the medial triangle of $\triangle ABCD$, giving an answer of $1$.

Now, we show that $S_1\ge S$ is true. We claim that for any triangle $XYZ$ and $X_1$ on $YZ$, $Y_1$ on $ZX$, and $Z_1$ on $XY$, \[[X_1Y_1Z_1]\ge \text{min}([XY_1Z_1],[YZ_1X_1],[ZX_1Y_1])\]
Assume otherwise, then apply the affine transform that will take $[X_1Y_1Z_1]$ to an equilateral triangle. Let $X'$, $Y'$ and $Z'$ be such that $\triangle X_1Y_1Z_1$ is the medial triangle of $\triangle X'Y'Z'$. Evidently, $X'$ is closer to $Y_1Z_1$ than $X$. Since $X'Y_1Z_1$ is equilateral, that means $X$ lies outside $(X'Y_1Z_1)$ so $\angle YXZ=\angle Y_1XZ_1>60^\circ$. Similarly, $\angle XZY,\angle ZYX>60^\circ$, contradiction because the sum of angles in a triangle adds to $180^\circ$. This proves our claim.

Now, let $E$ be $AB$ intersection with $CD$. Without loss of generality, $BC$ is closer to $E$ than $AD$. Let $T_1=[AA_1D_1]$, $T_2=[BB_1A_1]$, $T_3=[CC_1B_1]$, $T_4=[DD_1C_1]$ then by our claim,
\begin{align*}
S_1 &= [A_1C_1D_1] + [A_1B_1C_1] \\
&= \text{min}(T_1,T_2+T_3,T_4)+\text{min}(T_3,T_1+T_4,T_2) \\
&\ge S
\end{align*}
as desired.

The answer is $k=1$. For a construction, consider the degenerate case where $ABCD$ is a triangle $ABC$ with $D$ the midpoint of $\overline{BC}$, where $A_1B_1C_1D_1$ are the midpoints of their respective sides (hence forming the medial triangle of $ABC$.)

To show that this is minimal, consider the following:

[b][color=#f00]Claim:[/b][/color] For any triangle $DEF$ inscribed on the sides of a triangle $ABC$, $[DEF] > \operatorname{min}([AEF], [BDF], [CDE])$.

[i]Proof.[/i] Take an affine transform sending $DEF$ to an equilateral triangle. Assume for the sake of contradiction that the result is not true; then, let $XYZ$ be the image of $DEF$ upon a homothety of ratio $4$ at its center. It is equivalent to show that there cannot exist points $A, B, C$ outside $XYZ$ such that $DEF$ is inscribed in $ABC$.

However, notice that for any point $A$ outside $XYZ$, we always have $\angle EAF < 60^\circ$ (as the circle for which $\angle EAF = 60^\circ$ is tangent to triangle $XYZ$). Hence the angles of $ABC$ add up to less than $180^\circ$, contradiction. $\blacksquare$

Now WLOG $BC < AD$ and construct the point $E = \overline{AB} \cap \overline{CD}$. The lemma implies that $[B_1A_1C_1] < \operatorname{min}([BA_1B_1], [CB_1C_1])$ and $[A_1D_1C_1] < \operatorname{min}([DC_1D_1], [AA_1D_1])$, and summing yields the result.


probability function symmetry algebra polynomial logarithms inequalities

Recall that a graph A is rigid iff the only automorphism of A is the identity function from A onto A.  Let G_n be the set of simple graphs with node set {1,...,n}.  Let G be the union of G_n for all finite n.  Let R contained in G be the rigid of simple graphs and let R_n = R (intersect) G_n.  What I want to show is that lim(n->infinity) |R_n| / |G_n| = 1

I've been trying this for a long time, and have even been told that the proof should exist in standard literature. Does anyone know how the proof goes, or where I might be able to find it? thanks.

It is definitely correct (in virtually any probability context) that the probability of a nontrivial symmetry in a random structure tends to 0.

 Suppose that there is a cyclic permutation symmetry of order $k$ of some vertices of the graph and consider the orbit of one vertex.  The amount of information required to describe the cycle is $k log k$ and the amount needed to select the vertices in the orbit is $k (log (n))$.  To specify the connection pattern of one vertex, and therefore all the vertices in the orbit, to the rest of the graph requires $n$ bits of information.  Total is at most  $n + 2k (log(n))$.    To specify the connection of $k$ given vertices to the rest of the graph, however, requires $kn$ bits, which is higher for a nontrivial symmetry ($k > 1$).

This is a soft but rigorizable argument just expressing the obvious fact that "symmetry reduces entropy".   For uniformly selected random graphs it implies some estimate of how fast the probability of a symmetry tends to 0.

can you explain some of the terms used here? I am not familiar with the notion of the orbit of a vertex or cyclic permutation symmetry. I also don't understand how what you showed implies the result.

The basic idea: Choose a permutation. Estimate the probability that this permutation is a symmetry of the graph. Sum over all permutations, and show that the expected number of nontrivial symmetries tends to zero.

Here's an estimate: A permutation which moves exactly $k$ vertices and leaves the other $n$ fixed moves at least $k(n-k)+\frac{k^2-2k}{2}=nk-\frac{k^2}{2}-k$ edges. Because that $k^2$ is annoying, we will estimate that $k^2\le nk$, so the number of edges moved is at least $\frac{nk}2-k.$ The probability that a random graph is fixed by such a permutation is at most $2^{-\text{\# edges moved}/2}\le 2^{\frac{2k-nk}{4}}$.
Now, let $D(k)$ be the number of permutations of $k$ elements which move all of them. The expected number of nontrivial permutations which fix a random graph is at most $\sum_{k=1}^n\binom{n}{k}D(k)\cdot2^{{\frac{2k-nk}{4}}}$

Estimating $D(k)<k!$ and $\binom nk<2^n$, this expected value is less than
$\sum_{k=5}^n\frac{k!\cdot2^n\cdot2^{\frac k2}}{2^{\frac{nk}{4}}}+\sum_{k=1}^4\frac{k!\cdot\binom nk\cdot2^{\frac k2}}{2^{\frac{nk}{4}}}$

The second sum clearly goes to zero since the numerator is polynomial in $n$ and the denominator is exponential. In the first sum, we have the estimate $k!<k^k=2^{k\log_2 k}.$ Since $k\log_2 k$ is much less than $k^2$ for large $k$, and $nk$ is larger than $k^2$, the denominator dominates the numerator, and the terms for large $k$ go to zero rapidly. Since each individual term goes to zero exponentially, and the new term added on at each step goes to zero even faster, the sum must go to zero.

The entropy idea is just a way to gloss over the details and immedately see the result.

[quote="bansh33"] I am not familiar with the notion of the orbit of a vertex or cyclic permutation symmetry. I also don't understand how what you showed implies the result.[/quote]

An automorphism is a permutation of the vertices (satisfying some conditions).  Suppose an automorphism of the graph is given, and that one of its cycles (of length $k > 1$) is $(v_1 v_2 ... v_k)$.  Then the set $\{v_1,...,v_k\}$ is the orbit under discussion.  The argument is that you need about $1/k$ as many bits to encode the edges of the graph that have one of the $v_i$ as endpoint.
This in turn implies that the number of graphs with symmetries is dramatically lower than the number without.

[quote="jmerry"]The basic idea: Choose a permutation. Estimate the probability that this permutation is a symmetry of the graph. Sum over all permutations, and show that the expected number of nontrivial symmetries tends to zero. [/quote]

That's a better method, but it is a priori solving a different problem.
Conceivably one could have a situation where the expected number of thingies has a different behavior than the probability that a thingie appears.
There are situations where the expected number goes to infinity and the probability goes to zero.

[quote] The entropy idea is just a way to gloss over the details and immedately see the result.[/quote]

The entropy idea is the same as a probabilistic solution in the (not always realistic) sense that probability arguments and counting arguments are equivalent.    But it should be clear that it is actually rigorous; if you prove that graphs with some special feature are described by $d = d(n)$ bits of information with ${N \choose 2} - d \to \infty$ then you have proved that the probability of a special feature goes to zero. (If the graphs are chosen uniformly.)

In this case one can describe a graph with symmetry by specifying the symmetry (a permutation of $1,2,..,n$; this requires $n \log_2 n$ bits) and one bit for each orbit (cycle) of the induced permutation of the edges (given the permutation one can run through the edges in some predetermined order and specify whether each edge is in the graph, but only if that information does not follow from previously specified edges).
For each induced cycle of length $K$ one saves $K-1$ bits compared to an asymmetric graph, so one must show that the number of saved bits is substantially larger than $n log n$. By the reasoning above the amount saved should be about $2n$ bits, and the only non-rigorous point above was the omission of the proof that this is the case (you have to look a bit more carefully at the structure of the induced permutation, as some statements are false for $n = 3$, for example).  The "bits of information" approach is otherwise fully rigorous.

Note that this argument fails for, e.g. random trees.  A tree always has automorphisms, and the number of bits to specify a tree is about $\log (n^n) = n \log n$ rather than $O(n^2)$, so the above argument fails.  But I think you can use the same reasoning to rule out automorphism groups that are "too large", even for trees.

It's possible that the expectation can go to infinity with the probability still going to 0.  However, it can't happen the other way around for discrete variables.

In fact, by Markov's inequality we have that for any nonnegative random variable $P(X \geq a) \leq E(X)/a$.  Since our variable only takes on integer values, we can take $a=1$ and get $P(X \neq 0) \leq E(X)$.  Therefore if the expected number of symmetries goes to 0, the probability there is a nonzero symmetry tends to 0 as well (and at least as quickly)

There is a more basic reason why the probability of a thingie ( = expectation of "there is at least one thingie") is $\leq$ the expectation of "number of thingies":  just from the definition of expectation and the fact that the number of thingies is a non-negative integer.


quadratics conics algebra polynomial inequalities logarithms function \/closed

Our Intermediate Algebra textbook is [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]now available![/url]

The text covers the non-trigonometric topics of Algebra 2 and Precalculus, as well as many problem solving strategies required for success on the AMC and AIME.  Subjects covered include complex numbers, quadratics and conic sections, polynomials, multivariable expressions, sequences and series, identities, inequalities, exponents and logarithms, piecewise-defined functions, functional equations, and much more.

The text is co-authored by Mathew Crawford and me, and Naoki Sato was the lead author of the solutions manual.  The text is 720 pages and the solutions manual is 336 pages.

To order, or to see the table of contents, index, and a couple excerpts, [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]click here![/url]

Wow, 720 pages; that must be a new record  :D .

I hope I can get it, but my parents aren't crazy about spending $ \$ 60$ for a math book.  :(

SWEET COVER!!!

Unfortunately, I have to wait until my dad gets home to order... :(

One of those copies will be mine soon, hopefully.

Geometry remains by far my biggest weakness, but I still want to sink my teeth into 720 pages of algebraic delight.  :)

[quote="jvenezuela716"]but I still want to sink my teeth into 720 pages of algebraic delight.  :)[/quote]It's more than 1000 if you add the Solutions Manual.

Hey how much will this help with olympiad/hard AIME algebra? It doesn't seem as advanced as the intermediate counting in terms of contests....

But I'm sure it's a great Alg2/Precal book as all of your books are xD. I think my sis will do well with it :)

And whoa! 720 pages!!! That is 3 times the amount of the Geometry one, which is already pretty huge!! Wow... you guys went all out.  :D

Mucho bravo! Oh wait, no, did I just mix spanish with italian....?

I'll be to sure to get.... That will be mine...

[/quote]

SWEET I'm definitely gonna use it for AMC prep! :lol:

I LOVE IT! :w00t:  :w00t:  (read the subject!)
I don't have to pay a cent because i made i school library pay for them!
i read both the excerpts....
I was not expecting Mr.Crawford to work on it....

I can't wait to see the section titled "Brute Force."

I need improvement in economical brute forcing. :D

[quote="FantasyLover"]uhhh... I found a typo in page 414.

Problem 13.3 (a)

Find x if $ 5^{x^2 \minus{} 3} \equal{} 25^{2x}$

But in solution of Problem 13.3, page 415, it says $ 5^{x^2 \minus{} 3} \equal{} 25^{x}$

I think $ 5^{x^2 \minus{} 3} \equal{} 25^{x}$ should be the real problem.[/quote]

Thanks!  (Already on page 415 - you work fast!)  If you find more typos, please email them to me!

Will you be selling this books at the imo 2008 in spain, i mean all the books, that are available on art of problem solving?

We won't be selling any books at the IMO.  Europeans can buy AoPS books at Highperception.

[quote="Wichking"]Even Intermediate Algebra?[/quote]  Yes; they should have them soon.  We sent some their way a week or two ago.  You might want to ask them about it in a few days.

It'd be very fun (not to mention very challenging!) for me to try a new algebra book.  :lol:

1) Would you recommend this if we've already taken Intermediate Algebra?
2) What books will be coming out in the next 6-9 months?

Depends, if you aced the (former) Intermediate Algebra, then this book might not help you much (but it's certainly a good way to review stuff at any time). But the book contains more material and exercises than the one covered by the old Interm Algebra course.

I was working through the book and I found a typo on page 198:  in problem 7.1.5, i believe it should say $ p(x)\equal{}3x^3\minus{}20x^2\plus{}kx\plus{}12$ instead of $ 3x^3\minus{}20^2\plus{}kx\plus{}12$.

I sure hope this is the right place to report typos.  :maybe:

Thanks!  You can also email typos to books@artofproblemsolving.com.

[quote="suma_milli"]Hi,  I think I found a typo on page 224.  On problem 7.6.4 I think it should be 64a + 16b + 4c + d = 256 not 64a + 16b + 4b + d = 256.[/quote]

I have looked around a bit, but have not found an online section for "errata" for all the books.    This seems like it would be a pretty helpful thing to have in general.  Does it exist?....Singularity5

[quote="rrusczyk"]There is a URL in the Foreword of the book with an Errata & Links page.[/quote]

thanks very much.  I had not noticed that before and I checked a few of the other books and found that they also had errata URLs, but I wonder why you do not just make these more directly and more clearly available from the website itself?  Just a suggestion....


abstract algebra
induction
analytic geometry
Ring Theory
superior algebra
superior algebra unsolved

$A$ is noetherian ring and $f\in A[[X]], f=\sum a_{n}X^{n}$.
Prove $f$ nilpotent is equivalent to $a_{n}$ nilpotent $\forall n$.

I have proved $(\Rightarrow)$ but I don't know for $(\Leftarrow)$

$\Rightarrow$

We can prove that $a_{n}$ is nilpotent by induction on $n$. If, say, $f^{k}=0$, then $a_{0}^{k}=0$ because $a_{0}^{k}$ is the free term of $f^{k}$. Then, for $n\ge 1$, notice that the coefficient of $x^{nk}$ in $f^{k}$ equals $a_{n}^{k}+b$ with $b\in(a_{0},\ldots,a_{n-1})$ (the ideal of $A$ generated by $a_{0},\ldots,a_{n-1}$). Since this coefficient is zero, we have $a_{n}^{k}\in(a_{0},\ldots,a_{n-1})$, and since this ideal is nilpotent (because it's generated by a finite family of nilpotent elements, by the induction hypothesis), $a_{n}$ must be nilpotent as well.

$\Leftarrow$

Because $A$ is Noetherian, the ideal $\frak a=(a_{0},a_{1},\ldots)$ is in fact generated by finitely many of the $a_{i}$. Suppose it's equal to $(a_{0},\ldots,a_{t})$. Since $a_{0},\ldots,a_{t}$ are nilpotent, there is a $k$ such that $\frak a^{k}=0$. The coefficients of $f^{k}$ are elements of $\frak a^{k}=0$, so we're done.

Thank you very much grobber, I like your solution.

Do you know a counterexample for $\Leftarrow$, i.e. where $A$ is not noetherian?

$\prod_{k=0}^\infty \mathbb{Z}/2^{k}\mathbb{Z}$ (or the sum of these rings) is one:
Let $a_{k}$ be the element that has $0$ on all coordinates except on position $k$, where it is supposed to have a $2$.
All $a_{k}$ are nilpotent, but the corresponding power series $S=\sum_{k=0}^\infty a_{k}x^{k}$ is not:
The coefficient of $x^{m(m+1)}$ in $S^{m}$ has on coordinate $m+1$ exactly the value represented by $2^{m}$, which is not $0$.


Several figures can be made by attaching two equilateral triangles
to the regular pentagon $ ABCDE$ in two of the five positions
shown. How many non-congruent figures can be constructed in this
way?

[asy]pair A,B,C,D,I;
A=(10,0);
B=(0,0);
C=(-3.1,9.5);
D=(5,15.4);
I=(13.1,9.5);
draw(A--B--C--D--I--cycle,linewidth(0.7));
draw(A--B--(5,-8.7)--cycle,dashed);
draw(A--I--(19.8,2.1)--cycle,dashed);
draw(C--B--(-9.8,2.1)--cycle,dashed);
draw(I--D--(14.15,19.4)--cycle,dashed);
draw(D--C--(-4.15,19.4)--cycle,dashed);
label("$A$",A,SE);
label("$B$",B,SW);
label("$C$",C,W);
label("$E$",I,E);
label("$D$",D,N);[/asy]

First pick one side to attach a triangle to.
Then, there are only 2 possibilities for the placement of the second triangle.

Thus, there are $ \boxed{2}$ non-congruent figures that can be made this way.

There are $ 4$, but $ 2$ are the same under rotation.

its not clear. it does not say that the verticies are indistinguishable.

[quote=SugarHighOffense]I don't understand I counted and got $7$[/quote]

You probably overcounted

May I ask why this is the 10th problem of the AMC 12? The answer is 2 by 2 solutions:

1) Just count them.
We pick any first triangle, then we have two options: the one adjacent to the first triangle or the on after the triangle that is adjacent.

2) Do permutations.
5 choose 2 is the case without any restrictions. 5 c 2 = 10. Divide by 5 because there are 5 vertices.
10 / 5 = 2.

Wait why wouldn't it be 5 choices then... since there are 5 places in which the equilateral triangles can be placed in?

I don't understand at all! I got $20$ and $40$!


logarithms
limit
calculus
calculus computations

Let $ m$ be a positive integer, and here $ e$ is the base of the natural logarithm.

Evaluate:

$ \lim_{x \to 1}\frac{e^{(x^m)} \minus{} e}{ln(x)}$

answer is         [b]me[/b]

try using the l hospital`s rule

[quote="b555"]answer is         [b]me[/b]

[/quote]

You get me.

That is the answer to the question in my headline (to which this problem
I made up and worked out beforehand).


Is there a topological structure induced in measure space? If not why? Please explain!

~~Thanks ~~

No.  In general, a measurable space structure is much weaker than a topological space structure.  However, a topological structure induces a measurable space structure called the [url=http://en.wikipedia.org/wiki/Borel_algebra]Borel algebra[/url].

Thanks!

Is it so that countable unions of elements of $ \sigma$-algebra fail to produce arbitrary unions in topological spaces?

No, it's just that the two are very different structures.  Closure under complement is a property that neither open or closed subsets of a topological space share.


Find all positive integers $ n$ such that $ \sum_{i\equal{}1}^n \gcd(i,n)$ has at least three factors.

I'm not really seeing a pattern here... anyone have a hint? Just considering the prime case, it's not really clear whether $ 2p\minus{}1$ is composite as far as I can tell.

I'm not seeing any pattern, either. I wonder if I'm interpreting "at least three factors" correctly - I assume $ m$ satisfies this condition iff $ \#\{d : d|m, 1 < d< m\} \geq 3$.

I think "at least three factors" just means the number is composite.

If so, that's a mighty strange way of saying "composite".

OK, then assume what is meant is at least 5 factors - at least 3 proper ones. That seems to make more sense.

If the question is "for which $ n$ is it true that $ f(n) \equal{} \sum_{i\equal{}1}^n \gcd(i,n)$ is composite", the I have a solution of sorts - meaning, a solution that depends on which primes $ p$ are such that $ 2p\minus{}1$ is prime, which seems to be a prerequisite for this problem as worthawholebean observed.

[hide]
I'll answer the opposite question: $ f(n)$ is [i]prime [/i] iff $ n$ is prime and $ 2n\minus{}1$ is prime as well. It's hard to see why the question was the opposite of that (when is $ f(n)$ composite?), so perhaps we're still misinterpreting the question.

Anyway, to prove the above claim, note that:

* If $ n\equal{}p$ is prime, then $ f(p) \equal{} 2p\minus{}1$.

* If $ n\equal{}p^k$ when $ k>1$, then $ f(p^k)$ is divisible by $ p$. This is because $ \gcd(i,p^k)$ is divisible by $ p$ exactly when $ i$ is divisible by $ p$, and is $ 1$ otherwise. There are thus $ p^{k\minus{}1}$ summands that are divisible by $ p$, and $ p^k\minus{}p^{k\minus{}1} \equal{} p^{k\minus{}1}(p\minus{}1)$ summands that are 1, so the sum is divisible by $ p$.

* If $ m$ and $ n$ are relatively prime, then $ f(mn) \equal{} f(m)f(n)$. This is obvious since $ \gcd(i,mn) \equal{} \gcd(i,m)\gcd(i,n)$ in this case (such functions are usually called "multiplicative" or "weakly multiplicative" in number theory).

Therefore, if $ n$ is composite, then $ f(n)$ is composite as well. If $ n$ is prime, it all depends on whether or not $ 2n\minus{}1$ is prime as well, which I don't think anybody knows much about (it's a bit like Sophie Germain primes, which are mysterious).

[/hide]


Gauss
ceiling function
function
floor function
algebra
polynomial
sum of roots

Find the sum of the real roots satisfying the following equation with respect to $ x$:

$ [x]\plus{}[2x]\plus{}[3x]\plus{}[4x]\plus{}[5x]\plus{}[6x]\plus{}[7x]\plus{}[8x]\plus{}[9x]\equal{}44x$.

Note that for real number $ r$, denote the integer which is larger than or equal to $ r$ by $ [r]$.

[hide]
So the guass sign is the same as the ceiling function...

Note that $ 1 \plus{} 2 \plus{} 3... \plus{} 9 \equal{} \frac {9(10)}{2} \equal{} 45$
Since $ \sum\lceil x \rceil \ge \sum x \longrightarrow x\le 0$

Since the sum of guass functions will be an intiger, $ 44x$ must also be an intigier and let that intiger be $ k$.
Now $ x$ must be a rational number in the form $ \frac {k}{44}$

to make it easier (for me ) we will consider numbers $ x$ as postive and change the guass sign to floor function..

Now we have $ \sum_{i \equal{} 1}^9\left \lfloor \frac {ik}{44} \right \rfloor \equal{} k$

If $ k \le 4 \mod 44$ then $ \sum_{i \equal{} 1}^9\left \lfloor \frac {ik}{44} \right \rfloor \equal{} 45\left \lfloor \frac {k}{44} \right \rfloor$

From this we see that we need solutions to $ \left \lfloor \frac {k}{44} \right \rfloor \equal{} \frac {k}{45}$
This occurs at multiples of $ 45$ from $ 45 \rightarrow 43(45)$

So the sum of roots is; $ \frac { \minus{} 1}{44} \cdot 44 (1 \plus{} 2 \plus{} 3... \plus{} 43) \equal{} \minus{}\frac {43(44)}{2} \equal{} \minus{}946$  :)

[/hide]

I also think the meaning of Gauss sign is equivalent to ceiling in this topic.....but I know the ordinary JMO problem says its floor,so...
[hide="solution"]
$ LHS$ is integer,so the roots $ x$ can be wrriten as $ x \equal{} m \plus{} \frac {n}{44}$ $ (m\in\mathbb{Z}, n \equal{} 0,1,2,...43)$
Put this in the equation we have
$ 45m \plus{} [\frac {n}{44}] \plus{} [\frac {2n}{44}] \plus{} \cdots [\frac {9n}{44}] \equal{} 44m \plus{} n$
hence $ x \equal{} m \plus{} \frac {n}{44} \equal{} \frac {45n}{44} \minus{} ([\frac {n}{44}] \plus{} [\frac {2n}{44}] \plus{} \cdots [\frac {9n}{44}])$,hence the sum of root is:

$ \sum_{n \equal{} 1}^{43}\left(\frac {45n}{44} \minus{} ([\frac {n}{44}] \plus{} [\frac {2n}{44}] \plus{} \cdots [\frac {9n}{44}])\right)$

Now,we wanna get the value of $ \sum_{n \equal{} 1}^{43}[\frac {kn}{44}] \equal{} \sum_{n \equal{} 1}^{21}\left([\frac {kn}{44}] \plus{} [\frac {k(44 \minus{} n)}{44}]\right) \plus{} [\frac {k}{2}]$ for $ k \equal{} 1,2,...,9$.
notice that $ [\frac {kn}{44}] \plus{} [\frac {k(44 \minus{} n)}{44}] \equal{} [\frac {kn}{44}] \plus{} [ \minus{} \frac { \minus{} n}{44}] \plus{} k \equal{} k \minus{} 1$ if $ \frac {kn}{44}$ is not an integer $ \Longleftarrow n\ne 11$,
hence we have
$ \sum_{n \equal{} 1}^{43}[\frac {kn}{44}] \equal{} \sum_{n \equal{} 1}^{21}\left([\frac {kn}{44}] \plus{} [\frac {k(44 \minus{} n)}{44}]\right) \equal{} 20(k \minus{} 1) \plus{} [\frac {k}{2}] \plus{} [\frac {k}{4}] \plus{} [\frac {3k}{4}]$.

therefore we have the sum
$ \sum_{n \equal{} 1}^{43}\left(\frac {45n}{44} \minus{} ([\frac {n}{44}] \plus{} [\frac {2n}{44}] \plus{} \cdots [\frac {9n}{44}])\right)$
$ \equal{} \frac {45}{44}\times \frac {43\times 44}{2} \minus{} \sum_{k \equal{} 1}^{9}\left(20(k \minus{} 1) \plus{} [\frac {k}{2}] \plus{} [\frac {k}{4}] \plus{} [\frac {3k}{4}]\right) \equal{} \boxed{\frac {379}{2}}$[/hide]


Four vehicles are attempting to go through an elliptical arch of height 12 ft and width 10 ft. These vehicles are: a car 6 ft wide and 6 ft tall; a Cadillac 9 ft wide and 6 ft tall; a truck 7 ft wide and 8 ft tall; and an SUV 8 ft wide and 8 ft tall. Which vehicles can fit through the arch?

I assume that 12ft is the maximum height and that 10ft is the maximum width and that all vehicles pass through the middle.

[hide="solution"]The equation for this ellipse is $ \frac{x^{2}}{5^{2}}\plus{}\frac{y^{2}}{12^{2}}\equal{}1$.

So for the car that is 6 feet wide it passes at three feet from the center on either side, meaning $ \frac{3^{2}}{25}\plus{}\frac{y^{2}}{12^{2}}\equal{}1$. Solve for $ y$, $ \frac{9}{25}\plus{}\frac{y^{2}}{12^{2}}\equal{}1\rightarrow\left(\frac{y}{12}\right)^{2}\equal{}\frac{16}{25}\rightarrow\frac{y}{12}\equal{}\frac{4}{5}\rightarrow y\equal{}\frac{48}{5}\rightarrow y\equal{}9.6\text{ft}$
So the car can pass.

The Cadillac is 9ft wide or 4.5ft on each side, so $ \frac{(4.5)^{2}}{25}\plus{}\frac{y^{2}}{12^{2}}\equal{}1\rightarrow\left(\frac{9/2}{5}\right)^{2}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\left(\frac{9}{10}\right)^{2}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\frac{81}{100}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\left(\frac{y}{12}\right)^{2}\equal{}\frac{19}{100}\rightarrow\frac{y}{12}\equal{}\frac{\sqrt{19}}{10}\rightarrow y\equal{}\frac{6\sqrt{19}}{5}\rightarrow y\equal{}5.2307\text{ft}$
So the Cadillac can not get through.

The truck is 7ft wide or 3.5ft on each side, so $ \left(\frac{7/2}{5}\right)^{2}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\left(\frac{7}{10}\right)^{2}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\frac{49}{100}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\left(\frac{y}{12}\right)^{2}\equal{}\frac{51}{100}\rightarrow\frac{y}{12}\equal{}\frac{\sqrt{51}}{10}\rightarrow y\equal{}\frac{6\sqrt{51}}{5}\rightarrow y\equal{}8.5697\text{ft}$
So the truck can get through.

The SUV is 8ft wide or 4ft on each side so, $ \frac{4^{2}}{25}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\frac{16}{25}\plus{}\left(\frac{y}{12}\right)^{2}\equal{}1\rightarrow\left(\frac{y}{12}\right)^{2}\equal{}\frac{9}{25}\rightarrow\frac{y}{12}\equal{}\frac{3}{5}\rightarrow$
$ y\equal{}\frac{36}{5}\rightarrow y\equal{}7.2\text{ft}$
So the SUV cannot get through.

Wow, that was tiring. To sum it up:
The car and truck may pass through the arch, and the Cadillac and SUV may not.

The morale of this story is: buy generic cars, they can pass under arches :wink: .

[/hide]


(1) Find the 3 digit number that is equal to 25 times the sum of it's digits.

(2) Find the 3 digit number that is equal to 49 times the sum of it's digits.

[hide="bad solution"]The equation needed to be solved is 100x+100y+z=25(x+y+z)
which is equivalent to 75x-15y-24z=0.Setting x=1,2,3,...,9 and solving the diophantines that come up for positive and less than 10 values we can get the wanted number. :D [/hide] :D

$ \ a) 375
 \ \ b) 882$

I got the answers by trial-error, but certain things helped in case a), like the no. cannot be greater than 525, contains specific digits etc.
Hoping to see an algebraic solution  :maybe:

For a you can rather set values of z=0 and 5 and solve for the diophantine in the way i described.

225 also works for a). I broke it into cases where it ends in 00,25,50,and 75 (pretty much indirectly checking every number). For b), that fastest way is almost certainly to check all the multiples of 49 up to 1000.

[hide="a"]$ 100a \plus{} 10b \plus{} c \equal{} 25(a \plus{} b \plus{} c) \\
\Rightarrow 75a \equal{} 15b \plus{} 24c \\
\Rightarrow a \equal{} (5b \plus{} 8c)/25$
now $ a$ is a natural number,so
$ (5b \plus{} 8c)/25 \equal{} 25,50,75,100$
applying the constraint that $ 0 \leq b,c\leq9$
we get $ (b,c) \equal{} (5,0),(2,5)(7,5)$
The fourth equation gives no solution

so the three numbers are $ 150,225,375$[/hide]

[hide="b"]
$ 100a \plus{} 10b \plus{} c \equal{} 49(a \plus{} b \plus{} c) \\
\Rightarrow a \equal{} (13b \plus{} 16c)/17\\
\Rightarrow a\equal{}b\plus{}c\minus{}(4b\plus{}c)/17$

we just need solution of $ 4b\plus{}c\equal{}17,34$ which make $ a$ also an integer
we get $ (3,5) (4,1) (8,2)$
so the numbers are $ 735,441,882$

 [/hide]


Find all integer $n\ge 3$ such that
$2cos\frac{2\pi}{n}\in \mathbb{Z}$
P/s: It's my first post in this box! :D

See [color=blue]Forum index » Olympiad Section » Geometry » Geometry Proposed & Own Problems[/color]  :rotfl:

P.S: For your problem , because $-1\leq \cos{\frac{2\pi}{n}}\leq 1$, we need find $n$ such that $2\cos{\frac{2\pi}{n}}=-2,-1,0,1,2$.

So this is the corollary of this problem:
Find all integer $n$ such that $n\ge 3$ and exist an equilateral polygon with all integer points
P/s: It's in [b]Mathematics and Youth magazine, vol 307[/b]


inequalities
function
logarithms
calculus
derivative
algebra
binomial theorem

For real numbers $ x>y>0$, and positive integers $ n>k$, prove that

$ (x^k\minus{}y^k)^n<(x^n\minus{}y^n)^k$.

[quote]$ \begin{array}{c} x > y > 0 \\
 \\
a > b > 0\end{array}$ $ \implies$ $ \left(x^b \minus{} y^b\right)^a < \left(x^a \minus{} y^a\right)^b$ .[/quote]
[color=darkblue][b][u]Proof.[/u][/b] Consider the function $ f\ : \ (0,\infty )\rightarrow\mathbb R$ , where $ f(t) \equal{} \left(p^t \minus{} 1\right)^{\frac 1t}$ and $ p > 1$ .

Define $ x\mathrm {\ .s.s.\ }y$ $ \Longleftrightarrow$ $ x \equal{} y \equal{} 0$ or $ xy > 0$ . Prove easily that the function $ f$ is increasing.

Indeed, $ f'(x)\mathrm {\ .s.s.\ }\left[\frac {tp^t\ln p}{p^t \minus{} 1} \minus{} \ln \left(p^t \minus{} 1\right)\right]\mathrm {\ .s.s.\ }\left[ p^t\ln p^t \minus{} \left(p^t \minus{} 1\right)\ln\left(p^t \minus{} 1\right)\right] > 0$ .

Therefore, for $ p: \equal{} \frac xy > 1$ obtain $ f(b) < f(a)$ $ \implies$  $ \left[\left(\frac xy\right)^b \minus{} 1\right]^{\frac 1b} < \left[\left(\frac xy\right)^a \minus{} 1\right]^{\frac 1a}$ $ \implies$

$ \left(x^b \minus{} y^b\right)^{\frac 1b} < \left(x^a \minus{} y^a\right)^{\frac 1a}$ $ \implies$ $ \boxed {\ \left(x^b \minus{} y^b\right)^a < \left(x^a \minus{} y^a\right)^b\ }$ .[/color]

I don't know without derivatives ! Maybe you or an other user ...

[quote="minsoens"][quote="Virgil Nicula"]Consider the function $ f\ : \ (0,\infty )\rightarrow\mathbb R$ , where $ f(t) \equal{} \left(p^t \minus{} 1\right)^{\frac 1t}$ and $ p > 1$ .
Prove easily that this function is increasing.[/quote]Is there any way other than calculus? :?:[/quote]

Perhaps the Extended Binomial Theorem?

or taking a log?

$ (x^k\minus{}y^k)^n < (x^n\minus{}y^n)^k$

iff $ n\ln (x^k\minus{}y^k) < k\ln (x^n\minus{}y^n)$

iff $ n\ln x^k \plus{} n\ln (1\minus{}(y/x)^k) < k\ln x^n \plus{} k\ln (1\minus{}(y/x)^n)$

iff $ n\ln (1\minus{}(y/x)^k)<k\ln (1\minus{}(y/x)^n)$

iff $ (n\minus{}k)\ln (1\minus{}(y/x)^k)<k[\ln (1\minus{}(y/x)^n)\minus{}\ln (1\minus{}(y/x)^k)]$

Now LHS<0 because $ n\minus{}k>0$ and the thing inside Ln is less than 1.
Also 0<RHS since $ 1\minus{}(y/x)^n>1\minus{}(y/x)^k$ and Ln is an increasing function.
So the last inequality holds hence so is the first.

[url=http://www.mathlinks.ro/viewtopic.php?t=164224]here[/url] is a non-calculus solution by me

[quote="Virgil Nicula"]
Indeed, $ f'(x)\mathrm {\ .s.s.\ }\left[\frac {tp^t\ln p}{p^t \minus{} 1} \minus{} \ln \left(p^t \minus{} 1\right)\right]\mathrm {\ .s.s.\ }\left[ p^t\ln p^t \minus{} \left(p^t \minus{} 1\right)\ln\left(p^t \minus{} 1\right)\right] > 0$ .
[/quote]

I'm sorry I'm not too good at differentiation. Please show a bit more steps how to find $ f'(x)$?

[quote="nguyentrang"][quote="Virgil Nicula"]Indeed, $ f'(x)\mathrm {\ .s.s.\ }\left[\frac {tp^t\ln p}{p^t \minus{} 1} \minus{} \ln \left(p^t \minus{} 1\right)\right]\mathrm {\ .s.s.\ }\left[ p^t\ln p^t \minus{} \left(p^t \minus{} 1\right)\ln\left(p^t \minus{} 1\right)\right] > 0$ .
[/quote]I'm sorry I'm not too good at differentiation. Please show a bit more steps how to find $ f'(x)$?[/quote]we can write the function $ \displaystyle f(t)\equal{}(p^t\minus{}1)^{\frac{1}{t}}\equal{}e^{\frac{1}{t}\ln{(p^t\minus{}1)}}$
$ f'(t)\equal{}\left(\minus{}\frac{1}{t^2}\ln{(p^t\minus{}1)}\plus{}\frac{p^t\ln{p}}{t(p^t\minus{}1)}\right)e^{\frac{1}{t}\ln{(p^t\minus{}1)}}$


Ring Theory
superior algebra
superior algebra unsolved

my apologises, i dunno what i was thinking , it is very easy to prove that it can be a zero divisor

[quote="Adriana"]Im Stuck thinking about this:

Let $R$ be a ring with $1$ and $S$ a ring.   Let $f: R \to S$ be a ring homomorphism.  Is $f(1)$ always the identity element in $f(S)$?.

I know if $f(1)$ is not a zero divisor in $f(S)$, then it is the identity element in $f(S)$ (not too hard to prove), so can $f(1)$ be a zero divisor?.  Obviously in that case $f(1)$ cannot be the identity element.

Adriana

P.S.: There's many examples where $f(1)$ is not the identiy element in $S$ so that is not a problem in this case :P[/quote]

No. Let $R$ be a ring with 1 and $S$ be a ring without 1. Then $\phi: R\to S$ defined as the trivial homomorphism certianly is a homomorphism. But yet your statement is false.

Even in your example it is true, since $f(1)$ ($=0$) is the identity element of the subring $f(R)={0}$ (trivially)

[quote="Gockel"]f(1) can be a zerodivisor:
Let $R=\mathbb{Z}/(6), S=\mathbb{Z}/(2), f(k)=3k$. Here f(1)=3 is a zerodivisor in S.[/quote]

It's not true, $f(1)=1$ is not a zero divisor in $S$(actually its a unity).   My question was about the ring $f(S)$ , and in that case $f(1)$ is [b]always[/b] a identity element for $f(S)$ , $f(1)f(a)=f(1 \cdot a)=f(a)=f(a\cdot 1)=f(a)f(1)$.   If $f$ is the trivial homomorphism , then $f(S)$ is just the set $\{0\}$ and $f(1)=0$ can be thought as unity.   If $f$ is not the trivial , then $f(S)$ has more than one element and [b]the unity[/b] $f(1)$ [b]cannot[/b] be a zero divisor, if this is the case, then there exists $f(a) \neq 0$ in $f(S)$ such that $f(1)f(a)=0$, which is equivalent to $f(a)=0$ , a clear contradiction.

Adriana

[quote="Gockel"]Oh, I made a mistake... $S: =\mathbb{Z}/(6)$ then f(1)=3 is a zerodivisor in S.

What do you mean with f(S) ? As f is a homomorphism from R into S, so f(S) is not defined... Or was your question about f(R) ? This one is really trivial, because the above calculation shows, that f(1) is always a identity for f(R).[/quote]

yes, you are right , i made a mistake too; it was $f(R)$ , and yes it's trivial.  If I am working in $S$ we can take the  ring homomorphism $f: Z_{2}\to Z_{6}$ defined by $f(0)=0$ and $f(1)=3$.  In this case $f(1)=3$ is a zero divisor in $S=Z_{6}$ ($2 \cdot 3 = 0$).

Adriana


geometry
perimeter
inequalities proposed
inequalities

Let's consider a convex hexagon $ABCDEF$. Let $A_1, B_1, C_1, D_1, E_1, F_1 $ be midpoints of the sides $AB, BC, CD, DE, EF, FA$ respectively. Denote by $p$ and $p_1$, respectively, the perimeter of the hexagon $ABCDEF$ and hexagon $A_1 B_1 C_1 D_1 E_1 F_1 $. Suppose that all inner angles of hexagon $A_1 B_1 C_1 D_1 E_1 F_1 $ are equal. Prove that $p\geq \frac{2\sqrt{3}}{3} p_1 $. When does equality hold?

Put $ AA_1=a ; BB_1=b ; CC_1 =c ;...; FF_1 =f ; A_1B_1 =x ; B_1C_1 =y , C_1D_1 =z ; D_1E_1 =t , E_1F_1 =u ; F_1A_1 =v  $ .
 First , we can prove that : $ a+b+c \geq A_1C_1 \geq \frac{\sqrt3}{2}(x+y) $
 Similarly , we have another 5 ineqs .
 Add them all , we have : $ a+b+c+d+e+f \geq \frac{\sqrt3}{2}(x+y+z+t+u+v) \iff p \geq \frac{2.\sqrt3}{3}p_1 $  ;)


I just noticed that the USAMO list was posted and I checked......

No One From Longfellow Made USAMO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Indeed. Longfellow has now fallen to second place among all middle schools, with 8 USAMO qualifiers in the last ten years (Miller in San Jose is first with 9).
TJ, on the other hand, bounced back with 18 qualifiers, most of any school this year.

Don't complain too much...

I heard about miller, but san jose?

Never heard about that middle school.

I said [u]in[/u] San Jose. That's where Miller is.

BTW if thats true, our middle+elementary school got 2 USAJMO qualifyers this year


calculus
derivative
calculus computations

Let $f(x)=\sqrt[3]{x}=x^{\frac 13}$.  Using a simple application of the power rule:

\[f'(x)=\frac 13 x^{-\frac 23}=\frac{1}{3\sqrt[3]{x^2}}.\]

Btw mods, this probably belongs in the Calculus Computations and Tutorials forum.

[quote="shobber"]The derivative of $f(x)=a^n$ is always: $f'(x)=n\cdot a^{n-1}\text{ where } n\in R $[/quote]

If  $f(x)=a^n$ then $f'(x)=0$ :)


25 points are placed inside a regular hexagon of side length $2$ units.

(a) Prove that there are two points that are no farther than $1$ unit apart.
(b) Show that with $24$ points instead, the points may be placed so that no two are less than $1$ unit apart.

[hide="a"]

Create six equilateral triangles by creating lines from the center of the hexagon to the vertices. Each triangle has side length of 2.

In each triangle, create four smaller triangles my connecting the midpoints of the sides of the triangle. (A triforce type shape)

Each of these 24 triangles has side length 1, so if two points are both inside of one of these triangles, they can be no more than 1 unit apart.

There are 25 points, and by the pigeon hole principle, there must be atleast one triangle with atleast two points.

[/hide]

[hide="b?"]Pretty much the same as for a, but since there aren't more points than areas, the pigeonhole principle doesn't apply?

I don't know if that counts as a proof...[/hide]

well its still possible for one point to to be within one unit of a neighboring triangle. I tried to find a way to position the points in the triangles such that this didn't occur, but i couldn't find a way.

[quote="DanK"]Pretty much the same as for a, but since there aren't more points than areas, the pigeonhole principle doesn't apply?

I don't know if that counts as a proof...[/quote]

Although one particular proof doesn't work, it doesn't mean that *no* proof will work. But, you can use the fact that Pidgeon Hole doesn't work (and hence you can fit one point into each of the small triangles) to solve this.

Sorry for bringing up an old topic but this has been bugging me for quite some time...

I've tried to place 24 points inside the hexagon but failed to satisfy the conditions. What I did is I tried to cover the hexagon with 24 nonoverlapping circles of radius 1/2.

What I am asking about is not about a logical proof that it 'may' be done but I'm just wondering if we can actually construct a way to place 24 points in it. A picture of it would be nice :D


Hi all,

Due to some people's request we have created this forum for discussion of strategies, algorithms, moves and other math-related stuff concerning well known games like [b]Chess[/b], [b]Go[/b], [b]Sudoku[/b], [b]Rubik's Cube[/b] :P, and many others. Feel free to start discussin' them.

woohoo Valentin Vornicu that's so cool  :coolspeak:  :wow:  :10:  :clap:  :clap2:  :omighty:
but a quick question, you know sometimes the russian competitions have math questions about mini-games and strategies, can we discuss them in here as well?

Hmm, yes as long as you are discussing the [b]game[/b] itself. If you are strictly discussing that problem, I'd prefere that you are discussing it in the [url=http://www.mathlinks.ro/Forum/index.php?f=5]Combinatorics Section[/url] of the site ;)

Can I ask why this is under the High School forum? Seems a weird place to put it :?

don't know where this belongs, but long as it's here may as pell use it.

Another math related game is pente/go-moku. I'm really good at it, any1 else play?

[quote="Dsquared"]Don't abuse the forum to get more posts.[/quote]

What do you mean, that doesnt belong here.

PS How about Battleship? Can we talk about that here?

Yes, Battleship is perfectly acceptable.

[quote="Dsquared"]Don't abuse the forum to get more posts.[/quote]

No offense but two users have over 2000 posts here so yeah.
Look at how many posts have been made here fries then you can make your comment :wink:

[quote="fries4guys"]People are not "abusing" the forum, they are simply participating in the forum games they like.[/quote]
Agreed but it is sort of for more posts.
I do the same sometimes (I admit it)
Some users do it a lot some do it a little.
But the point is the word marathon has turned into (post  inflation) marathon

There is no marathon that I know that in the rules says that you try to stop the advancement of words posted.

Well, that's kind of what some people try to do. It makes you think of the most unique word.

[quote="Scamper"]Of course, post count doesn't really matter. You can spam all day, have over 3000 posts, but your reputation in AoPS will go way down. Not to mention that since your posts were spam, they'll probably be deleted. People used to think that post count were equal to the amount of experience you had, how old you are, how smart you are, how popular you are, etc. When I first started AoPS, I used to think that. I saw the first page on the Members List as a sort of "Hall of Fame." Then, I saw that posts didn't really matter anymore.

Eventually, those marathons will get moved to the Fun Factory, die down, and no one will see of them again. That happened before when there were some marathons in the Games Forum and people lost a large amount of posts. I think that will probably happen again in the future. And people will lose a major amount of posts who posted excessively in those marathons. Those marathons are sort of a strategy game though, since you're trying to make a word that cannot be made into another, for example.[/quote]
Hopefully that is so.  But the mods are very inactive in this forum.  Honestly if I was a mod I would:
Move it to FF
Lock it
Delete it (not the best but possible)
People seem to be abusing it and because of how often it is it is beginning to get repetive.

As the new moderator of this forum, I would like to remind you of the purpose of this forum.
[quote="Valentin Vornicu"]Hi all,

Due to some people's request we have created this forum for discussion of strategies, algorithms, moves and other math-related stuff concerning well known games like Chess, Go, Sudoku, Rubik's Cube , and many others. Feel free to start discussin' them.[/quote]

In other words, we shouldn't be seeing word marathons of [b]any[/b] kind (They will be moved to FF). On the other hand, if you want to create a discussion on the finer strategy of word marathons, then feel free.

Regarding the 24 Game Marathon, any discussion pertaining to that would inevitably boil down to the game itself, so I'm letting it stay.

Moved to Fun Factory:
Word Marathon: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=139&t=379045
Palindrome Marathon: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=139&t=373173
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Moved to Linguistics Forum:
Uncommon Words Marathon: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=448&t=362157

so
do card games count?
like pokemon, yugioh, mtg etc.

Not sure, but does othello/reversi count as a "strategy game"? Because some people just play without much thought to the moves.


Find the minximum and maximum of $ y \equal{} \frac{{x^2  \plus{} x \plus{} 1}}{{x^2  \plus{} 1}}$ with $ x \in R$.

How many solution?

$ y \equal{} 1 \plus{} \frac {x}{x^2 \plus{} 1}$
$ y' \equal{} \frac {1*(x^2 \plus{} 1) \minus{} x*2x}{x^4 \plus{} 2x^2 \plus{} 1} \equal{} \frac {1 \minus{} x^2}{x^4 \plus{} 2x^2 \plus{} 1} \equal{} 0$

$ 1 \minus{} x^2 \equal{} 0 \implies x \equal{} \pm1$

idk what you mean by 'how many solution'

but maximum = $ \frac {3}{2}$ when $ x \equal{} 1$
minimum = $ \frac {1}{2}$ when $ x \equal{} \minus{} 1$

I think he might have been referring to local minima and maxima, because then the can be more than one max. and/or min.

No calculus is needed.
We have
$ \frac{1}{x \plus{} 1/x} \plus{} 1$
By AM-GM, $ x \plus{} 1/x \le 2$. So the max is 3/2.
If x is negative, $ \minus{} x \minus{} 1/x \ge \minus{} 2$. So the min is -1.

I think [quote="L_Euler"]How many solution?[/quote]
is mean "how many methods are there for this problem?".
[hide="My method"]We have equation: $ (y \minus{} 1)x^2 \minus{} x \plus{} (y \minus{} 1) \equal{} 0$
Because $ x$ exists, $ \Delta \equal{} \minus{} 4y^2 \plus{} 8y \minus{} 3 \geq 0$
$ \Rightarrow (y \minus{} \frac {3}{2})(y \minus{} \frac {1}{2}) \leq 0$
$ \Rightarrow \frac {1}{2} \leq y \leq \frac {3}{2}$[/hide]


Princeton college Harvard geometry Yale Stanford Duke

I will study mathematics and I am in 11th grade. Whice are the top universities ?
I like combinatorics, graph theory, and also pure mathematics.
Which universities do you recommend I should apply,
I am interested in Princeton, Rutgers, Harvard or Berkeley.

Rutgers!!! The School of Scarlet Knights!!!  :D  :rotfl:
Too bad the football team is... well... too embarrassing to even go to Rutgers Stadium and watch a game... :blush:
:rotfl: :rotfl: :rotfl:

Well, seriously, I know that Rutgers has a comparable program in Operations Research with Stanford. I hope this helps. If you are looking for studies in Topology/Analysis, Rutgers has a strong program. But I recommend you something. It is not the most well-balanced school. So you choose...

Yes I know, th thing is that I likea lot topological graph theory, thats an area dealing with the structure of graphs from a topoligcal point of view, and in Princeton there is the best Professor in this area. But princeton is very expensive and I need a scholarship..

The big-name private universities -- Harvard, Yale, Princeton, Stanford, Duke, etc. -- all have very strong financial aid programs.  So, on paper you may pay 40,000 dollars a year, but a significant portion of students don't pay anywhere near that amount, without needing to have won outside awards or anything.  (The downside of this, of course, is that if you [i]do[/i] have outside awards, they tend to get collapsed into your other financial aid and so they don't help you out very much -- if you pay full tuition, a 5000 dollar award is worth 5000 dollars, but if you have a good financial aid package, you might need a 15,000 dollar award in order to actually pay 5000 dollars less.)

Also, this question or questions like it have been answered many times on this forum -- try a search!  One other suggestion, as well -- if there is a particular professor you want to work with, you might be better off [i]not[/i] going to that professor's school for undergraduate because many graduate schools take very few students from their own undergraduate class and you could be (unintentionally) preventing yourself from working with that professor in grad school (if that's what you think you want to do after college).

[i](Edited by admin, since the forum interprets dollar signs as LaTeX commands)[/i]

I would advise against picking an [i]undergraduate[/i] school based on a specific research speciality and/or an individual professor.  Because, no offense, as a high-school student, you probably don't have a very good picture of the whole world of mathematics, so why limit yourself now?  Why not go to a generally good school and explore different branches of mathematics?  If and when you decide to go to grad school, then you'll want (and need) to specialize more, but for now, don't pigeonhole yourself.

Furthermore, it is believed by many that a big problem in research mathematics today is that there are too many specialists -- people who only know a very small subject-area of mathematics, albeit very well -- and not enough people who see the big picture of mathematics, and the interplay between different branches of mathematics.  (I recommend Gian-Carlo Rota's [i]Indiscrete Thoughts[/i] for, among other things, a discussion of this.)

As far as going different places for undergrad and grad school: I strongly encourage this, not so much for the tactical reasons that JBL mentions (although I think that's true too), but just from the more general reason that it's a good idea to expose yourself to different ideas and different mathematical cultures.  If you have your heart set on doing a Ph.D. with a specific professor, then try to start a dialog with him or her while you're still an undergrad, and then he (assuming that he wants you as a student) can probably help you get into his university as a grad student.

hey manuel! i think(pretty sure) we know each other, from Athens this summer

Just like JBL said, I also think that you shouldn't worry about Princeton's high cost, but you now this - that they have completely need blind and full need policy, right?
i remember you were talking about wanting to go to Princeton, once, in the bus, at imo; Carlos was there, too, as i remember; btw, how is he - doing fine in his freshman year? do you guys still keep in touch? i'm very lazy with mails, i haven't talked with andrei (negut, maybe you remember him - also at P) in a long time, although i suppose i have tons of questions i'd like to ask him.. oh well...

back to universities, i've heard from a number of sources that Princeton offers an excellent undergrad experience - so it's a brilliant choice even if you might surprisingly find out that math isn't the future for you, and also that they have a very strong math department. I know these are generalities, but i just don't have the info to say: they have the best undergrad and the best math department... i have heard quite some opinions saying just that: that Princeton is truly the first choice, for undergrad, and esp for math, but i did hear this from persons who wanted to go P, were currently at P, or had learned there... so... not so sure... Harvard ppl saying the same about harvard, so my conclusion would be that these universities are so good, it doesn't really make a whole lotta difference where you'll go, as an undergrad. (i might be veeery wrong here, anybody who knows about major differences speak out!)

About berkeley i heard that it has an excellent graduate program, nothing more. Haven't heard about undergrad.

Do you know http://www.collegeconfidential.com ? it's a pretty good resource, and parents often post there, so you might get competent answers from people who know a lot about the subject

bye bye for now!

Hey kappa! How are you doing.. Thanks for all that..
Yes carlos is at Princeton... I have talked with him a couple of times and he is doing good.
Right now Princeton is my first choice!

Hola Manuel,

[quote="manuel"]I like combinatorics, graph theory, and also pure mathematics.
Which universities do you recommend I should apply,
I am interested in Princeton, Rutgers, Harvard or Berkeley.[/quote]

Que tal el clima en Puerto Rico, me imagino que mucho mejor que aqui en USA, al menos en esta epoca del anio. :D EXTRANIO EL CARIBE!

Well, now some piece of advise. I don't know how well is your background, but I have seen more than once very strong candidates to be rejected, therefore, if you don't want to spend a year doing anything but college, I would strongly recomend you to balance your options.

I say this because someone I know applied to Yale, Harvard, MIT and Princeton, and nowhere else the same year, and he got four NO!, the guy, who was a very good candidate, but got depressed after that experience,  ended in some college in Las Cruces, New Mexico. I don't know about the quality of the school, but for sure there is a HUGE gap between Ivy Leage and any college in Las Cruces.

If you are not a strong candidate, Universities that big are totally out of your range, that is true, but even if you are a very strong candidate, it turns to be a lotery to get in sometimes.

Now, I don't much about the quality of undergrad education on the US, I came here as a grad student, but you said that you are interested in combinatorics, graph theory (aren't they contained in pure mathematics? or what do you call pure mathematics?) then, why don't you look for universities in those specific topics, and take not only universities in the first five positions of most rankings, but take variety. The advise of someone who spend more than thirty years in admission departments of good universities, is that students who are good candidates  should apply one or two unviersities in the 1-10 positions in the ranking, one or two in the positions 10-20, one or two in the positions 20-50, and one who would accept him for sure. If you don't have exactly what you want, you can always transfer, and who knows, even a No in your favorite school could be a good thing.

For example, at least in grad level, Georgia Tech has a great group in graph theory and combinatorics, but it is, I believe, in the lower 20's or upper 30's possitions in the rankings. I don't know about financial aid for undergrad, but for grad students, they have a pretty decent salaries!

Best,

One thing I do advise if you are looking at graph theory heavily at this point is taking a look at St Olaf's Budapest Semesters in Math program wherever you go - 5 great math classes at the moment here, all serious stuff and a great group of faculty to deal with.  Hungary's strengths are graph theory/combinatorics and I've been really impressed with what we've done so far here - even geometry is grounded in group theory over here.  The courses are great - all under 15 in general (one class above that now, (C&P) but we will weed that down a bit in a couple of weeks as things get harder)

http://www.stolaf.edu/depts/math/budapest/

Do I have to speak hungarian to study in Hungary? :?

To the original poster who wants to study topological graph theory,

This is a very small area of research (in number of researchers) and most of the top US math departments will have nobody who knows about it.  Paul Seymour at Princeton is the only one I can think of.  Lovasz at Yale has other interests for many years.
If you mean combinatorics more broadly, many universities have people who work on it and some have research groups in it, or the functional equivalent in a computer science department.

More to the point, do you really know that this particular area is what you will want to do in the future, rather than some subject you haven't explored yet? If you actually know in high school that in 5-10 years you will want to be a mathematician (don't count on that!), what matters (mostly) is what graduate school you go to, and what Ph.D advisor you have.  Unless you have the option of going to one of the top 5 or so schools for a first degree it is probably better to find a cheap and pleasant place to spend your time for the bachelor's degree and concentrate on longer-range plans: learning what you need for graduate school, and developing enough not-specifically-mathematical skills/experiences/resources that would be useful whether or not you end up as a mathematician.  That, or invest your money in a house and utilize a public library to the maximum....

[quote="tokenadult"]No, the courses in the Hungary program are all taught in English, so if you can read the typical message on AoPS, you should do just fine in Budapest.[/quote]

I think that Hungarian is one of the required courses during the program.

I can't speak about the other colleges you mentioned, but I would recommend against Berkeley. In my opinion (I think a lot of people here agree too) the department sees undergrad math as a cost center and devotes as little effort into maintaining it as possible, so they can put their energy into graduate research, which is what keeps up the ranking. Berkeley being public the tuition is a lot cheaper (for CA residents anyway) but you kind of end up getting what you pay for. There's great professors here but they just don't care about undergrads enough or can't afford to, given the worsening budget situation in recent years. So if you have the choice go elsewhere for your undergrad education, preferably a smallish private school where you'll get all the attention you deserve, instead of being crammed into 250-people classes where you're just a face in the crowd. :?  :?  :(

Manuel has been admitted now, hasn't he? That's what happens when you revive an old thread; sometimes the original poster's question has already been answered in the real world.


Show that cos(x) + cos (sqrt(3)x/2) is not periodic.

[hide]Assume to the contrary.  THe period of $\cos{x}$ is 360.  THe period of $\cos{\sqrt{3}x/2}$ is $240\sqrt{3}$.  Now, at x=0, the combined function would equal 2.  If this function was periodic, then at some point, the function would equal 2 again because both functions equalled 1.  THat would mean that both functions just completed n and m cycles where n and m are positive integers.  However, nat.*nat. can't equal irrational*nat.[/hide]

we must show that there is no $p$ such that

$f(x)=f(x+p)$

i will assume that the $p$ exists, then show it does not exist by contradiction

clearly, $\cos$ is periodic with period $2\pi$, so we have when we add p, it is an integer multiple of $2\pi$, or

$p=2\pi*k_1$ and $\frac{\sqrt{3}}{2}*p=2\pi*k_2$ where $\{k_1,k_2\}\subset\mathbb{Z}$

divide these 2 equations to get

$\frac{k_1}{k_2}=\frac{2\sqrt{3}}{3}$

since $k_1$ and $k_2$ are integers, the left side is rational, but the right side is irrational, contradiction, that means that there is no period


During the local math tournament, there was a tie breaker question.
2,3,5,9,32... find the next number in the series
the right answer was 97.
it has been 10 days now, I still don't know why it is 97. I believe the question might be wrong because no one got the question...
but I have to post it here see if anyone can find the relationship between those numbers.

The [url=http://www.research.att.com/~njas/sequences/?q=2%2C3%2C5%2C9%2C32&language=english&go=Search]OEIS[/url] doesn't know that sequence.

I know... I think they made up some equation and took the first few numbers of it.

[hide="Polynomial"]If the equation is to be determined by those five values, then they could correspond to the values of $ \frac{2}{3}x^4\minus{}\frac{13}{2}x^3\plus{}\frac{137}{6}x^2\minus{}32x\plus{}17$ when $ x\equal{}1,2,3,4,5$.  But if $ x\equal{}6$, then $ \frac{2}{3}x^4\minus{}\frac{13}{2}x^3\plus{}\frac{137}{6}x^2\minus{}32x\plus{}17\equal{}107$.[/hide]


S.T.cosec2x + cosec4x + cosec8x + ........+ cosec2^2009x =cotx - cot2^2009x

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=325470[/url]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group, $ p| |G| \equal{} n$ a prime and $ k$ the biggest number s.t. $ p^k|n$. prove that there exists an elemnt of order a multiple of $ p^k$.

This is certainly not true. Something is wrong with the statement.

yes, sorry, I forgot to write that $ G$ is abelian :oops:

Still not true. 4 divides the order of the Klein V-group but there is no element of order 4

:roll: you are right...I found this as the fundamental argument in the [i]solution[/i] of the following problem:

[i]Prove that in an abelian group $ G$ there exists an element of order the least common divisor of the orders of all the elements except $ 1_G$.[/i]


Let $ S\equal{}\{x\in\mathbb Q|x^2<2\}$. How do you prove $ (\sup S)^2\equal{}2$?

Intuitively, it's clear.  Show that $ \sqrt{2}$ is the supremum/least upper bound.  That is, show that if $ x$ and $ y$ are upper bounds and $ x\equal{}\sqrt{2}$, then $ x\leq y$.

The whole point of this exercise is to prove that $ \sqrt 2$ exists. ;)

Wouldn't that be going back to a construction of the real numbers.?

Almost, but not quite: the usual way is to assume the completeness axiom in the form of the existence of suprema of bounded sets and proceed from there. The construction of reals allows you to make the completeness axiom a theorem and derive the existence of reals from the existence of rationals and, ultimately, from the existence of integers, but most of the time this construction is not included into analysis courses and the completeness axiom is taken for granted.

Can use continuity and monotonicity of the power function  :lol:

[quote="ifai"]Let $ S = \{x\in\mathbb Q|x^2 < 2\}$. How do you prove $ (\sup S)^2 = 2$?[/quote]
clearly.we must proof supS=$ \sqrt{2}$,due to $ \all x \in S,x<\sqrt{2}$and$ \all \epsilon >0,\exists x\in S,\sqrt{2}>x>sqrt{2}-\epsilon}$,since the full of |R

[quote="sunday"][quote="ifai"]Let $ S = \{x\in\mathbb Q|x^2 < 2\}$. How do you prove $ (\sup S)^2 = 2$?[/quote]
clearly.we must proof supS=$ \sqrt {2}$,due to $ \all x \in S,x < \sqrt {2}$and$ \all \epsilon > 0,\exists x\in S,\sqrt {2} > x > sqrt{2} - \epsilon}$,since the full of |R[/quote]
 i have  a interested book  Kazor's.
And  the problem is solved detaily
http://rapidshare.com/files/149019909/kaczor_i.djvu.html


algebra
polynomial
limit
number theory proposed
number theory

[b]Problem[/b]: Let be given an polynomial $f(x)$ which has integer coefficients, ie... ${f(x)\in\mathbb Z}[x]$  and distince from constant. Prove or disprove that for any given positive integer number $\ell$ then there exist an positive integer $n$ in which $f(n)$ has exactly $\ell$ prime divisors.

Obviously not true in general: $n(n+1)(n+2)$ is always divisible by both 2 and 3 and hence cannot have just one prime divisor.  The question as to whether there are values divisible by $\ell$ distinct primes for all sufficiently large values of $\ell$ seems more interesting.

The affirmative answer for this last question follows easily from the well-known result which states that for any integer polynomial $P$ the set of primes dividing at last one of the $P(k)$ (as $k$ varies over the set of positive integers) is infinite, and from the chinese remainder theorem.

Pierre.

thanks rgep, really it becomes more interesting, let me rewrite it as:
[b]Problem[/b]: Let be given a polynomial $f\in{\mathbb Z}[x]$, not constant. Prove or disprove that there exists an integer $\ell$ so that for any given $s\geq \ell$ then there is an $n$ a positive integer so that $f(n)$ has exactly $s$ prime divisors!

can you explain more clearly Piere?  :)

I think pbornsztein proves that for any $\ell$ there is an $n$ such that $f(n)$ has $\ge \ell$ prime divisors.

ah, this nice result as in my memory, there is a theorem show that in fact the more general is also true: If $f(n,2^n,3^n,...)$ is a polynomial of $n,2^n, ....$ with $\lim_{n\to+\infty}f(...)=+\infty$ then for any $\ell>0$ integer then there is $n$ in which $f(n,2^n,...)$ has more than $\ell$ prime divisors !


Find all sets of four real numbers x1,x2,x3, and x4 such that the sum of any one and the product of the other three is equal to 2.

[hide="Solution"]$ a \plus{} bcd \equal{} 2$
$ b \plus{} acd \equal{} 2$
$ c \plus{} abd \equal{} 2$
$ d \plus{} abc \equal{} 2$

Now let's use the example of the first two: $ a \plus{} b(cd) \equal{} 2$ and $ b \plus{} a(cd) \equal{} 2$. Subtract to get $ (a \minus{} b)(1 \minus{} cd) \equal{} 0$. Hence, $ a \equal{} b$ and/or $ cd \equal{} 1$.

Next look at the pair with lone terms $ b$ and $ c$ to get $ b \equal{} c$ and/or $ ad \equal{} 1$, and similarly with $ c,d$. This brings us to five very similar cases:

[b]Case 1:[/b] [i]From each pairing (a,b) (b,c) (c,d), we chose equality, thus $ a \equal{} b \equal{} c \equal{} d$.[/i]
Substituting, this yeilds one real solution: $ a \equal{} b \equal{} c \equal{} d \equal{} 1$

[b]Case 2:[/b] [i]For two overlapping pairings, we chose equality. Wlog, $ a \equal{} b \equal{} c$ and $ ab \equal{} 1$.[/i]
Substituting this into $ a \plus{} bcd \equal{} 2$ gives $ a \plus{} d \equal{} 2$. By $ b \equal{} 1/b \implies b \equal{} \pm 1$, we must have $ (1,1,1,1)$ or $ ( \minus{} 1, \minus{} 1, \minus{} 1,3)$.

[b]Case 3:[/b] [i]For two non-overlapping pairings, we chose equality. Wlog, $ a \equal{} b$, $ c \equal{} d$, and $ ad \equal{} 1$.[/i]
We know that if $ b \equal{} c$, our problem is solved, so let's strive to make them different. Consider the cases of $ (a,c)$, $ (b,d)$, and $ (a,d)$. For each, we will choose the inverse case. That is, $ bd \equal{} 1$, $ ac \equal{} 1$, and $ bc \equal{} 1$. Using our known equalities, the total system now becomes $ ac \equal{} 1$, $ a^2 \equal{} 1$, $ c^2 \equal{} 1$. Hence $ (a,c) \equal{} \pm(1,1)$. However, substituting negative solution into the original equations reveals it to be extraneous, so we have not achieved anything new here.

[b]Case 4:[/b] [i]We choose equality for one pairing, and inversion for the other two[/i]
Note that this must lead to a second equality. If $ a \equal{} b$, then $ ad \equal{} 1$ and $ ab \equal{} 1$, giving $ (a,b,c) \equal{} \pm (1,1,1)$, and the value of $ c$ follows. From there we conclude that the $ \minus{} 1$'s are extraneous. This works similarly if we use $ c \equal{} d$. If we instead choose $ b \equal{} c$, then, $ ab \equal{} 1$ and $ cd \equal{} 1$. Equating, $ ab \equal{} ac \equal{} cd \implies a \equal{} d$. Thus, we have two equalities and this is the same as either case 2 or 3.

[b]Case 5:[/b] [i]We chose three inverse relationships[/i]
Thus, $ a \equal{} 1/b \equal{} c \equal{} 1/d$, and we have the two equalities $ a \equal{} c$ and $ b \equal{} d$, which is covered in case 3.

[b]Conclusion:[/b] The only ordered triples of reals that solve this system are $ (1,1,1,1)$ and $ (\minus{}1,\minus{}1,\minus{}1,3)$, up to permutations[/hide]

Why is $ (\minus{}1, \minus{}1, \minus{}1, 3)$ extraneous? $ \minus{}1\cdot{}\minus{}1\cdot{}\minus{}1\plus{}3\equal{}2$ and $ \minus{}1\cdot{}\minus{}1\cdot{}3\minus{}1\equal{}2$.  It did not say anything about them being positive, so they would work.

Ah, thanks, modified :)

I don't know what made me think it was extraneous, I probably added wrong...


geometry
inequalities
trigonometry
inequalities unsolved

[quote="MAT"]Let $ x,y>0$
with : $ x+y+\sqrt{2x^{2}+2xy+3y^{2}}=4$
Proooove:

$ x^{2}y\leq 4$
Thank u :)[/quote]

Check this link: http://www.mathlinks.ro/Forum/viewtopic.php?t=128314
It's not identical to this but it's very similar. But I don't have a proper solution to it.

I think this could be easy to show by analityc geometry.
1st of all we consider only $ x,y>0$
Then we notice that equality $ \sqrt{2x^{2}+2xy+3y^{2}}=4-x-y$ could be only for $ 4-x-y\geq 0$, so we draw area $ 4-x-y\geq 0$ :P
Next we have:
$ 2x^{2}+2xy+3y^{2}=16+x^{2}+y^{2}-8x-8y+2xy$
$ (x+4)^{2}+2(y+2)^{2}=40$, and we know that this is including in $ (x+4)^{2}+(y+2)^{2}\leq 40$, so we draw it... :)
And if we show that $ (x+4)^{2}+(y+2)^{2}=40$ is including in area $ x^{2}y\leq 4$ that'll be end of solution.

[quote="arqady"][quote="MAT"]Let $ x,y>0$
with : $ x+y+\sqrt{2x^{2}+2xy+3y^{2}}=4$
Proooove:

$ x^{2}y\leq 4$
Thank u :)[/quote]
Actually, $ \max_{x+y+\sqrt{2x^{2}+2xy+3y^{2}}=4,x>0,y>0}x^{2}y=32\left(\sqrt{\frac{5}{3}}-1\right)^{3}=0.7885...$ :wink:[/quote]

I didnt understand
please detail

do u mean that:   $ x^{2}y\leq 0788...$?

[quote="FOURRIER"]

do u mean that:   $ x^{2}y\leq 0788...$?[/quote]
Yes!
[quote="FOURRIER"]

I didnt understand
please detail
[/quote]
The hint: use the Holder's inequality. :wink:

[quote="FOURRIER"][quote="arqady"]
The hint: use the Holder's inequality. :wink:[/quote]

No, It's impossible!
Can you please show me how you gonna use Holder's inequality?[/quote]
Only for you, FOURRIER! :lol:
If $ x+y\leq4$ then
$ x+y+\sqrt{2x^{2}+2xy+3y^{2}}=4\Leftrightarrow (x+4)^{2}+2(y+2)^{2}=40.$
Hence, $ 20=\left(2+\frac{x}{2}\right)^{2}+\left(2+\frac{x}{2}\right)^{2}+(2+y)^{2}\geq3\left(\left(2+\frac{x}{2}\right)^{2}(2+y)\right)^{\frac{2}{3}}\geq$
$ \geq3\left(2+\sqrt[3]{\frac{x^{2}y}{4}}\right)^{2}.$
Thus, $ x^{2}y\leq32\left(\sqrt{\frac{5}{3}}-1\right)^{3}.$ The equality holds, when $ y=\frac{x}{2}.$

We have $ (\frac{x+4}{\sqrt{40}})^{2}+(\frac{y+2}{\sqrt{20}})^{2}=1$
so let $ x=2\sqrt{10}\cos \alpha-4, y=2\sqrt{5}\sin\alpha-2$,
I think it will help.


a function f(x) = ax^2 + bx + c    with coefficients E IR

  tq  a,b,c E IR

       f(x) >= 0
       a < b

find the minimun value of : a+b+c /(b-a)

how can i solve it??
thanks! XD

[hide="hint"]
After working it out a little with cuadratic function's stuff you can use AMGM for two numbers on it.

[/hide]

[hide="answer"]The minimun is $ 3$[/hide]

:)

I misunderstand part of the hint.  I used AMGM for three and fall down... SOMEONE SOLVE IT, please.  :oops:

From $ f(x)\equal{}a\left(x\plus{}\dfrac{b}{2a}\right)^2\plus{}\dfrac{4ac\minus{}b^2}{4a}$ it follows that $ a>0$ and $ \dfrac{4ac\minus{}b^2}{4a}\ge 0$. Then $ c\ge\dfrac{b^2}{4a}$.

So $ \dfrac{a\plus{}b\plus{}c}{b\minus{}a}\ge \dfrac{a\plus{}b\plus{}\dfrac{b^2}{4a}}{b\minus{}a}\equal{}\dfrac{(2a\plus{}b)^2}{4a(b\minus{}a)}$

but, by AMGM, $ (2a\plus{}b)^2\ge 4(3a)(b\minus{}a)$ with equality if $ b\equal{}4a$

Then
$ \dfrac{(2a\plus{}b)^2}{4a(b\minus{}a)}\ge\dfrac{3(2a\plus{}b)^2}{4}\dfrac{4}{(2a\plus{}b)^2}\equal{}3$

In conclusion $ \dfrac{a\plus{}b\plus{}c}{b\minus{}a}\ge 3$ with equality if and only if $ b\equal{}4a\equal{}c$. ($ a>0$ of course).

:)


LaTeX
quadratics
articles
geometry
algebra
quadratic formula
special factorizations

I recently installed the stuff necessary to use LaTeX, and everything works fine.  I also made my first document, which is basically a derivation of the quadratic formula.  I want one of you brilliant Hoosiers to find a way to shorten part of my source code.  You'll see why.  Here's the code:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage[margin=1in]{geometry}
\title{The Quadratic Formula}
\author{Wonho Rhee}
\date{August 17, 2008}
\begin{document}
\maketitle
\noindent A method for immediately finding the solutions to a quadratic equation of the form $ax^2+bx+c=0$ is quite a useful tool.  This tool is commonly referred to as the \emph{quadratic formula}.  It says that the solutions to the equation are:
\[
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
$\\$
\indent Fortunately, deriving the quadratic formula is simple when the method of completing the square is used.  The proof is below.
$\\$

First, the expression $ax^2+bx+c$ in the equation is manipulated:
\begin{align*}
0 & = ax^2+bx+c & \Rightarrow\\
& = ax^2+bx+c+\frac{b^2}{4a}-\frac{b^2}{4a} & \Rightarrow\\
& = ax^2+bx+\frac{b^2}{4a}+c-\frac{b^2}{4a} & \Rightarrow\\
& = a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+c-\frac{b^2}{4a} & \Rightarrow\\
& = a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}
\end{align*}

Now, the actual equation is manipulated:

\[
a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} = 0 \qquad\qquad\qquad\qquad\quad\Rightarrow
\]
\[
a(x+\frac{b}{2a})^2 = \frac{b^2}{4a}-c \ \ \qquad\qquad\qquad\qquad\qquad\Rightarrow
\]
\[
(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2} \ \qquad\qquad\qquad\qquad\qquad\Rightarrow
\]
\[
x+\frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a} \ \ \qquad\qquad\qquad\qquad\quad\Rightarrow
\]
\[
\boxed{x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}}
\]
\end{document}[/code]

Now, I know that my code is nowhere near professional, but for me, most of it's fine.  The part I want you guys to change is:
[code]\[
a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} = 0 \qquad\qquad\qquad\qquad\quad\Rightarrow
\]
\[
a(x+\frac{b}{2a})^2 = \frac{b^2}{4a}-c \ \ \qquad\qquad\qquad\qquad\qquad\Rightarrow
\]
\[
(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2} \ \qquad\qquad\qquad\qquad\qquad\Rightarrow
\]
\[
x+\frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a} \ \ \qquad\qquad\qquad\qquad\quad\Rightarrow
\]
\[
\boxed{x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}}
\][/code]
Those qquads, quads, and \ look pretty ugly and probably waste space.  Before changing anything, however, I want you guys to copy-paste the code, compile it, and view it.  You'll notice that the document itself doesn't look that bad on a PDF or whatever else.  The only reason I added the qquads, quads, and / is because I wanted the first algebra part to align with the second algebra part.
By "first algebra part," I mean the code that is embedded by:
[code]\begin{align*} and \end{align*}[/code]
By "second algebra part," I mean the code that I want you guys to change, which is embedded by several:
[code]\[ and \][/code]

This forum's been dead anyways... you guys need to stop slacking off! Help me!

I don't know what you mean by "aligning" the first and second algebra parts..

But couldn't you get the same thing using this:

\begin{align*}
&a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} = 0 &\Rightarrow\\
&a(x+\frac{b}{2a})^2 = \frac{b^2}{4a}-c &\Rightarrow\\
&(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2} &\Rightarrow\\
&x+\frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a} &\Rightarrow\\
&\boxed{x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}}
\end{align*}

This is almost irrelevant to your question, since I'm pointing out a flaw (again.. sorry) in your conventions.

I'm not sure if you're using the arrows correctly. Usually, from what I've seen, the thick arrows are used when one shows a short manipulation of an equation, not an expression, [i]inline[/i]. You'd exclude it in an aligned environment.

Also, I like to align equations by the equal sign (it may even be convention), like this:
\begin{align*} a(x + \frac {b}{2a})^2 + c - \frac {b^2}{4a} & = 0 \\
a(x + \frac {b}{2a})^2 & = \frac {b^2}{4a} - c \\
(x + \frac {b}{2a})^2 & = \frac {b^2 - 4ac}{4a^2} \\
x + \frac {b}{2a} & = \pm\frac {\sqrt {b^2 - 4ac}}{2a} \\
x & = \boxed{\frac { - b\pm\sqrt {b^2 - 4ac}}{2a}} \end{align*}
and again, since it's in its own environment, no arrows are necessary.


If a sequence (with a variable in the definition) converges to a value, does it mean all values put it for that variable converges to the value?

Sorry for being confusing.

Like, if I have a recursive formula that is something like $ a_{n}\equal{}2a_{n\minus{}1}$. This isn't really the best formula to explain this with, but if $ a_{1}\equal{}z$, does that mean that if $ a_{n}\equal{}2a_{n\minus{}1}$ converges, then it converges for all values of z? Or will it only converge for specific values of c?

Im not sure if I am understanding what you are asking, but anyway consider the sequence a_n= [a_(n-1)]^2, this converges when |a_1| <= 1, and diverges otherwise.

Well, let's say I have $ a_{n}\equal{}{a_{n\minus{}1}}^{2}\plus{}1$, where $ a_{1}\equal{}p$.

Then, if $ a_{n}\equal{}{a_{n\minus{}1}}^{2}\plus{}1$, how can I determine for which values of p that will happen? Hope this clears things up.

[quote="scorpius119"]One thing is that if a sequence converges to some $ L$ at all, then $ a_{n\minus{}1}$ converges to $ L$ as well. In your particular case, that would mean
\[ L\equal{}L^{2}\plus{}1\Rightarrow L^{2}\minus{}L\plus{}1\equal{}0\]
The above has no real roots, so this particular sequence can't converge for any real $ p$.

In general, there's no magic rule that always says when a sequence converges and when it doesn't, but if you can figure out a closed form for $ a_{n}$ and determine its limit to be $ L$, or figure out a limit $ L$ and show that $ |a_{n}\minus{}L|$ tends to 0, then the sequence will converge to $ L$. A sequence will also converge to something if it is increasing and bounded above (or decreasing and bounded below).[/quote]

So if $ L\equal{}L^{2}\plus{}1\Rightarrow L^{2}\minus{}L\plus{}1\equal{}0$ did have roots, would they be where the sequence converges? And can I prove that only certain values of $ a_{n}$ will converge to a value?

Also, in your first sentence, you said, that if a sequence converges to some $ L$ at all, then $ a_{n\minus{}1}$ converges to $ L$ as well. Is that a theorem? If so, what is the name? Is there anywhere online where I can find a bunch of theorems about convergences and sequences? How did you get $ L\equal{}L^{2}\plus{}1$? How did you know to just set them equal?

The fact that if $ a_{n}$ converges, then $ a_{n\minus{}1}$ converges to the same value follows from the definition of convergence (formally, $ a_{n}$ converges to $ L$ if for all positive $ \epsilon$, there exists $ N$ such that for all $ n > N$, we have $ |a_{n}\minus{}L| <\epsilon$).

For the sequence $ a_{0}\equal{} x, a_{n\plus{}1}\equal{}\sqrt{2}^{a_{n}}$ for some nonnegative $ x$ (involved in a recent post), the equation $ L \equal{}\sqrt{2}^{L}$ has $ L \equal{} 2,4$ as its solutions. But the sequence may:

- converge to 2 ($ x \equal{} 2$ is a trivial example, although  $ x \equal{} 0$, $ x \equal{} 1$, $ x \equal{}\sqrt{2}$, and more generally, any $ x$ such that $ 0\leq x < 4$ work.

To prove this, one can show that for $ x<2$, we have $ a_{n}<2$ and $ a_{n}$ increasing, while for $ 2<x<4$, we have $ a_{n}>2$ and $ a_{n}$ decreasing. So in each case, $ a_{n}$ converges, and cannot converge to 4.)

- converge to 4 (trivially true for $ x \equal{} 4$, but not true anywhere else)

- not converge (true for $ x > 4$: for example, if $ x \equal{} 5$, then $ a_{n}\geq 5$ for all $ n$, so the sequence could not possibly converge to either 2 or 4).

scorpius, so you set $ L\equal{}{sqrt{2}}^{L}$ just because you said $ a_{n\minus{}1}\equal{}a_{n}$ because they converge to the same value? So you can just set them equal like that?

And I still don't understand how to determine when a value of $ a_{0}$ would cause the sequence to converge to the specified value.

Sorry.

This thread has a much more natural home in the calculus forum.  I hope a moderator will move it there.

[quote="123456789"]And I still don't understand how to determine when a value of $ a_{0}$ would cause the sequence to converge to the specified value.[/quote]

You mean in general or in the case that convergence has been shown to be possible?

You make the assumption that sequences have sensible limit behavior.  In fact, there are interesting examples of very simple recursions that can actually generate chaotic (if bounded) behavior; such sequences are neither periodic (or at least have a very long one) nor have a limit.

[quote="t0rajir0u"][quote="123456789"]And I still don't understand how to determine when a value of $ a_{0}$ would cause the sequence to converge to the specified value.[/quote]

You mean in general or in the case that convergence has been shown to be possible?[/quote]

Well, basically, I want to know how to determine the value of $ a_{0}$ when a convergence has been shown to be possible.

For example, let's say I want to see if $ a_{n}\equal{}{a_{n\minus{}1}}^{2}\minus{}1$ can converge.
I don't know if I'm doing this correctly, but:
So I set up $ x\equal{}{({({({a_{n\minus{}1}}^{2}\plus{}1)}^{2}\plus{}1)}^{2})}^{2}\minus{}1$ and so forth, where x is the convergence. You get the idea.

So I get $ x\equal{}{x}^{2}\minus{}1$.
Then, $ x^{2}\minus{}x\minus{}1\equal{}0$.
So, $ x\equal{}\frac{1\pm\sqrt{5}}{2}$.

So if I can find a value of $ a_{0}$ for which the sequence converges to the golden ratio, then is that the only value of $ a_{0}$ for which the sequence converges to the golden ratio?

Is it possible for a sequence to converge to to values? Such as a sequence converging to both 2 and -2?

[quote="t0rajir0u"]Not in a rigorous sense; limits are unique.  Rather than saying that a sequence $ a_{n}$ has two limits $ 2,\minus{}2$, you could be more specific and say that there exist two monotonic sequences $ p(n), q(n)$ such that

$ \lim_{n\to\infty}a_{p(n)}\equal{} 2$
$ \lim_{n\to\infty}a_{q(n)}\equal{}\minus{}2$

Trivial example:  Say that $ a_{n}\equal{} 2\cdot (\minus{}1)^{n}$.  Then $ \lim_{n\to\infty}a_{n}$ does not exist, but

$ \lim_{n\to\infty}a_{2n}\equal{} 2$
$ \lim_{n\to\infty}a_{2n\plus{}1}\equal{}\minus{}2$[/quote]

So you just pull out $ p(n), q(n)$ from $ a_{n}$, right?

It might not be as clear-cut as that.  $ a_{n}$ could be a sequence which settles down into such an oscillation (with a period of $ 2$), but for other sequences $ p(n), q(n)$ might be complicated.


[quote="Pirkuliyev Rovsen"]Let  a,b,c>0  Prove that: $ \frac {2a^3(a^3 \plus{} 1)^2}{(b \plus{} c)^3} \plus{} \frac {2b^3(b^3 \plus{} 1)^2}{(a \plus{} c)^3} \plus{} \frac {2c^3(c^3 \plus{} 1)^2}{(a \plus{} b)^3}\geqslant3abc$  :idea:  :?:[/quote]
It's easy by Am-Gm
$ LHS \ge \sum \frac{8a^6}{(b\plus{}c)^3} \ge \sum \frac{2a^6}{b^3\plus{}c^3} \ge \sum a^3 \ge 3abc$  :lol:


find solutions to:

$a^4 + b^4 + c^4 + d^4= nabcd$

i would love a solution using infinite descent arguments please  :blush:

I have maybe a beginning for that :

Let $e=pgcd(a,b,c,d)$, Let $a'=\frac{a}{e}$, $b'=\frac{b}{e}$, $c'=\frac{c}{e}$, $d'=\frac{d}{e}$

$a^4+b^4+c^4+d^4=nabcd$

<=> $e^4 (a'^4+b'^4+c'^4+d'^4)=e^4 n a'b'c'd'$

<=> $a'^4+b'^4+c'^4+d'^4=n a'b'c'd'$

So, you can suppose that $pgcd(a,b,c,d)=1$ but after that???

[color=red]The folowing text is absolutlz wrong so don|t waste time reading it. I leave it there, because the there are some  reactions to it.[/color]
Let pgcd(a,b,c,d)=1. There can't be mote than 1 even numbers among a,b,c,d. If one of a,b,c,d was even, right side would be even and left must be odd.
So all the numbers a,b,c,d are odd. It's easy to show that for odd a it's a^4=1 (mod 16), thus left side is 4 (mod 16), 4 must divide right side so it must divide n and 8 can't divide n. So n=8k+4.
For k=1 n=4 from am-gm a=b=c=d is obviously the only solution.  I can't find solutions for any greater k.[

see thats pretty much what i did!! but its wrong because somewhere in the proof we forget a special case or something and the solution $[a,b,c,d] = [k,k,k,3k]$ is missed but where is it? what did i miss? this is driving me crazy because i cant find it!!!!!!!! :mad:

I think that you are making mistake when you suppose that $gcd(a,b,c,d)=1$ then only one of them is even that's not correct $gcd(2,4,3,9)=1$ so you se...

yup so any more replies?? or maybe some argument that says why infinite descent is not aplicable here see what i tried to prove is that if $[a,b,c,d]$ is a solution then $[a/2,b/2,c/2,d/2]$ using only parity and hece the result anyone??

What is $ infinite. descent. arguments$ Is that when you suppose that you have found the smallest solution and you find even smaller so obtain contradiction?
This equation obviously has infinite solutions, but when $(a,b,c,d)=1$ it maybe has only one ore two or whatever it's importante that you can count  them...... I know that nothing I wrote is new but still... :)

I think that you can't aply that as long as you don't know anything about $n$ but maybe I am wrong....
In every problem where i saw that principle used and I didn't see much of such problems you are looking for some divisibility and it usually come from number in witch place is $n$ in this problem ($x^2+y^2+z^2=2004xyz$, or $x^3+3y^3+9z^3=9xyz$)...
I hope you understand what i tried to say....

well i divided my solution in 2 cases :

n odd and n even and i thought i had a solution but i was wrong. still no solution??


geometry
perimeter
geometric transformation
reflection
calculus
symmetry
rotation

prove that among all triangles of assigned base $a$ and area $A$, the one with minimum perimeter is the isosceles one.

I am wondering if there is an algebric method to do that. The geometric way is very easy...

i think you mean that the one with $minimum$ perimeter is the isoceles one?

[quote="demji"][hide]
let $h$ be the height of the triangle, thus
$A=\frac{1}{2}ah$
since you can break the triangle of area $A$ into 2 right triangles, possibly if its not right then let $c$ equal the length of the base of the first one along $a$.
So:
$A=\frac{1}{2}ch+\frac{1}{2}(a-c)h$
now we can get the length of the other 2 sides, that is the hypotensuses of the 2 triangles.
$s_{1}=\sqrt{c^2+h^2}$ and $s_{2}=\sqrt{(a-c)^2+h^2}$
So now, perimeter= $a+s_{1}+s_{2}$ and we want to minimize this.  By doing so we need to minimize $s_{1}$ and $s_{2}$ which could be solved by calculus, however taking advantage of symmetry we can see that when $c=\frac{a}{2}$ we get a minimum.  Thus in this case, $s_{1}=s_{2}$ which is the definition of an isocles triangle

Hope that helped
[/hide][/quote]

i am not sure that the symmetry argument really works...saying that "we observe" is not really proving anything...but i do believe that this can be done with inequalities...also i did not think through my hint enough, i got the wrong transformation [hide="..."]rotation[/hide]


algebra
polynomial
number theory unsolved
number theory

if all the complex roots are inside the unit circle, then the polynomial is irriducible in Z!

I didn't understand you very well. The polynomial $ P(x)\equal{}4{x^3}\minus{}2{x^2}\plus{}x$ satisfies your problem? I don't think so. Explain your statement.

just multiply two polynomials which have this property. then you generate a reducible polynomial which has this property.

$ P(x) \equal{} 4x^3 \minus{} x \equal{} x(2x \minus{} 1)(2x \plus{} 1)$ has roots $ x \equal{} 0, \pm\dfrac{1}{2}$ all inside the unit circle but is reducible in $ \mathbb{Z}[x]$.

Are you sure that there aren't any other conditions.


a fly is sitting on a vertex of a cube.  each time the fly moves to one of the three adjacent vetecies near it.  (each with same probability).  in how many ways, can the fly travel through each vertex only once in 7 moves?

[hide]Name the vertices on the top face of the cube clockwise a, b, c, and d.  Name the vertices on the bottom face A, B, C, and D.

Without loss of generality, assume the fly starts on A.  The fly can move to three different vertices from A.  All of these paths are essentially the same, so just consider the path A-B.

From B, the fly can move to two different vertices.  Both of these paths are essentially the same (all form an L shape), so just consider the path A-B-C.

There are only three possible paths starting with A-B-C.

A-B-C-D-d-a-b-c
A-B-C-D-d-c-b-a
A-B-C-c-b-a-d-D

Therefore, there are $3\times 2\times 3=18$ possibilities.[/hide]


Let $f(n)$ be the number of ones that occur in the decimal representations of all the numbers from 1 to $n$.  For example, this gives $f(8)=1$, $f(9)=1$, $f(10)=2$, $f(11)=4$, and $f(12)=5$.  Determine the value of $f(10^{100})$.

[hide="Solution"]
Claim: \[ f(10^k)=1\binom{k}{1}9^{k-1}+2\binom{k}{2}9^{k-2}+\cdots+k\binom{k}{k}9^{k-k}+1. \]

Proof:
Consider all of the numbers less than $10^k$.  They all have $k$ or less digits.  Now consider all the numbers that have $k$ or less digits that have exactly $i$ ones.  There are $\binom{k}{i}$ ways to choose the spots that the ones occupy and the rest of the spots can be filled each with any of the 9 digits that are not one.  Therefore, there are $\binom{k}{i}9^{k-i}$ numbers with exactly $i$ ones.  A number with $i$ ones will contribute $i$ ones to the function.  Summing over all defined $i$ and adding one for $10^k$ gives the desired result.

Now we simplify the formula using $i\binom{n}{i}=n\binom{n-1}{i-1}$ and the binomial theorem \[ 1\binom{k}{1}9^{k-1}+2\binom{k}{2}9^{k-2}+\cdots+k\binom{k}{k}9^{k-k}+1 = n\cdot10^{k-1}+1. \]

So then $f(10^{100})=10^{101}+1.$[/hide]

Nice solution.  I should have thought of that :D

[quote="Peter VDD"]Eum.. I think it can be done easier ;)

Let's first count all 1's from 000...000 to 999...999.
Since we've written $100.10^{100}=10^{102}$ digits and each digit an equal number of times, we've written $10^{101}$ times 1.
So adding the last number, we get $10^{101}+1$[/quote]

i coudnt understand the counting 1's
and why you have written $100.10^{100}=10^{102}$
can you explain it more clearly
 :)

Well, 100=10^2.  By the properties of exponents we have $100\cdot10^{100}=10^2\cdot10^{100}=10^{102}.$

To count them, you just note that you will have written each digit an equal number of times if you write out all those numbers (by symmetry).  For example, considering writing out all the numbers from 00 to 99.  You will write a total of 200 digits and each of the base 10 digits (i.e. 0,1,2,3,4,5,6,7,8,9) will appear exactly 20 times.

A very, very good example of how vertical thinking is often better than horizontal thinking...


If you are being affected by Rita, please post here

also post how you are being affected

just wanted to know

Most of you people are already evacuating

GOOD LUCK!!! :)

Appropriate board please. Round Table it is.

Errr I'm sort of affected...not as much as some though (live in Austin)

Our school has a gymful of evacuees and will be open through the weekend, so I don't know what I'll do in PE tomorrow.  As of yesterday night, Austin (where I live) was going to get hit by a weakened Rita...but now it looks like we might just get a few bands of rain.

The weather guy said that there's a cold front heading to Texas, so when Rita moves up through Texas the cold front might push it back down into Austin.

All the gas stations are out of gas...there's no more water or bread left in grocery stores because everyone bought everything in a big frenzy.

Mother was about to take a connecting flight through Houston to get home. Clearly, that's not going to work. Having to find her a route home, last minute.

Most flights out of Houston (and definitely all INTO Houston) are cancelled. There will be a few more taking off tomorrow...but...that's like it. *shrugs*

it was kinda late to evacuate starting thurs morning.... highways were packed.  my friend spent 5 hours traveling 20 miles. yeahhh...

Lol. Gl there PK. It looks like (and I'm talking bout current projections at the time of landfall) the worse parts of the storm will be skipping over Sugar Land. Lol...so, you guys should be fine.

I'm actually in Austin, TX instead of back home. Honestly though, I'd much rather be back there seeing this storm for myself like many of my friends are.

Yeah, they said the winds in our area could reach a max of 60 mph, but would realistically get up to 40 mph.  I just feel sorry for the people who evacuated to the LA/TX border b/c they're gonna get it hard.  You can hear the wind outside and it's raining a bit, but they said we shouldn't expect anything too bad.


I have a TI-84+ SE.  When I press "Graph, it says "Err: Invalid Dim"  I cant graph anything right now!  Anyone know how to fix it?

Go to Y=, do you have any of the plots (Plot 1, Plot 2, Plot 3) highlighted? If so, unhighlight them.

thanks!  we did that in class, i forgot to remove it.

also sometimes you have to go to windows since it is dimensional error


Prove that the number

          $A = 2003 \cdot 2005^3 - 2004 \cdot 2002^3$

 is a cube of an integer.

[u]Babis[/u]

[hide]$\left (N+1\right )\left (N+3\right )^3-\left (N+2\right ){N}^3 = \left (2 \cdot N+3\right )^{3}$[/hide]