Find all $ f:\mathbb{R^{*}}\to\mathbb{R^{*}}$ so that :

$ f(xf(y))f(y)\equal{}f(x\plus{}y),\forall x\in\mathbb{R^{*}}$

( $ \mathbb{R^{*}}$ is the set of all positive real numbers) :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165324

No need for two separate topics.


algebra
polynomial
symmetry
algebra proposed

$ a^{3}c\minus{}a^{3}b\plus{}b^{3}a\minus{}b^{3}c\plus{}c^{3}b\minus{}c^{3}a$

I know the answer already, but want to learn the steps.

Thank you.

notice that $ a\minus{}b,b\minus{}c,c\minus{}a$ must be factors.

[quote="Gib Z"]I cant see why =[ ??[/quote]
Notice that the expression equals $ 0$ if $ a\equal{}b$, $ b\equal{}c$, or $ c\equal{}a$.

A pretty general approach is to treat the expression as a polynomial in one variable (say, $ a$).  This gives

$ a^3 (c \minus{} b) \plus{} a(b^3 \minus{} c^3) \plus{} (c^3 b \minus{} b^3 c) \equal{} (c \minus{} b)(a^3 \minus{} a(b^2 \plus{} bc \plus{} c^2) \plus{} bc(b \plus{} c))$

By symmetry, this implies that $ (a \minus{} b)(a \minus{} c)$ are both also factors.  (We can also confirm that $ a \equal{} b, a \equal{} c$ are both roots of the resulting cubic equation.)  The last factor can only be $ (a \plus{} b \plus{} c)$ by Vieta's.


Cati cai pot sta pe o tabla de sah fara sa se atace? (8 X 8)

$\ Raspuns: \frac{64}{2}=32.$ Problema se gaseste si in excelenta carte a lui Nazar Agahanov si Oleg Podlipski.
  [i] Problema[/i]
   Care e numarul maxim de
a) turnuri
b) nebuni
  care se pot aseza pe o tabla de sah $\ 8x8$ astfel incat sa nu se atace reciproc.

o intrebare...dc 64/2 ? :oops:

Adika ma rog e adevarat ca 32 este raspunsul, pt ca tot timpul un cal ataca un patrat de culoarea opusa...deci daka asezam 32 de cai pe aceasi culoare ei ataca celalalte 32 de culori...:)

rezolvarile voastre sunt putin cam incomplete
intr`adevar, pentru $32$ gaseste oricine exemplu.
dar de ce nu se poate pentru $33$ ?

hai ma ca se face tot cu...se observa ca un cal ataca tot timpu culoarea opusa patratului din care pleaka..si daka punem 33 de cai pe tabla inseamna ca exista cel putzin 2 pe culori diferite...si deci calul aflat pe o culoare dif de ceilaltzi 32 o sa atace un cal dintre cei 32:| ma rog...bla bla :|

PS : nu-mi plac probleme cu cai  :huh:

n`ai de unde sa stii ca intr`o configuratie cu 33 de cai, 32 se afla pe campuri de aceeasi culoare.
ma rog, eu am zis, nu am dat cu parul :)

OK, voi da eu o solutie completa.
  Descompunem tabla de sah in $\ 8$ dreptunghiuri $\ 2x4$.In fiecare dreptunghi marcam patratelele in perechi asa cum se vede in figura fde mai jos.In fiecare pereche nu se poate plasa mai mult de un singur cal.Deci, pe tabla se pot pune cel mult 32 cai care sa nu se atace.Asezand 32 cai pe patratelele negre ale tablei obtinem o situatie care satisface cerinta, fiindca ei ataca numai patratele albe.

mda...maky...:) cand toceam azi pastelu pt testare :) mi-am amintit de problema asta...ca ai zis ca daka zic ca 33 nu merge e incomplet ca nu stiu daka e o configuratzie cu 32 pe aceasi culoare...

Dar nu trebuie sa studiem toate configuratziile...e doar un contraexemplu care sa arate ca 33 nu merge si asa si cu 34 si alea alea...:| deci 33 e max:| era completa:|

nu am inteles ce vrei sa zici
nu mai conteaza acum. raspunsul lui sorin de mai sus este corect

k :( si mi-am pierdut 30 de secunde din viata sa scriu :)

pai editeaza postul, facandu`l mai explicit

Bine, acum ca problema este demonstrata, cum ramane sa aflam numarul maxim de turnuri si nebuni care pot fi asezati pe o tabla de sah $\ 8X8$ astfel incat acestia sa nu se atace reciproc?Cred ca problema este simpla, dar asa, ca sa completam tacamul. :)

Gandindu-ma asa in graba cred ca raspunsul e $8$ la intrabarile cu turul si nebunii,dar nu m-am prea gandit la asta.Cand o sa ma mai gandesc mai bine si daca am facut vreo greseala,ma voi corecta.Motivarea raspunsului $8$ este foarte simpla,avand in vedere traiectoriile bine definite ale turului si nebunului.

Poti sa explici mai bine?Raspunsul este corect in cazul cu turnurile.

A si inca ceva: am gasit ca numarul maxim de [b]regine[/b] care se pot inscrie pe o tabla de sah fara sa se atace este $8$,adik mai exact am gasit o configuratzie in care sunt 8 regine pe o tabla de sah 8X8 care nu se ataca,si in orice loc liber,daca am pune o regina,s-ar ataca cu una.Asta mi-a motivat gasirea raspunsului 8.Pana acum este cel mai mare numar pe care l-am, gasit.Uite si configuratzia:
[size=150]# # # # # # # R[/size]
[size=150]# # # # # R # #[/size]
[size=150]# # # R # # # #[/size]
[size=150]# R # # # # # #[/size]
[size=150]# # # # # # R #[/size]
[size=150]# # # # R # # #[/size]
[size=150]# # R # # # # #[/size]
[size=150]R # # # # # # #[/size]

Configuratzia este cand turnurile sunt pe diagonala
                        si cand nebunii sunt pe o linie
[Scz pt limbajul lacunar si fara demonstratzii clare da` cica am de invatzat..dupa joi e mai bn]

$\ Comentarii$
1)   Spor la invatat.Sa faci bine la examen. :)
2)   Numarul maxim de turnuri este $\ 8$.Uite si demonstratia mea.Pe fiecare orizontala poate sta cel mult un turn, deci pe tabla pot fi asezati cel mult $\ 8$ turnuri fara sa se atace reciproc.Asezand turnurile pe diagonala obtinem o situatie care satisface cerinta.E cam la fel cu ce-ai zis tu, dar...
3)   Din cate inteleg eu, zici ca numarul maxim de nebuni este $\ 8$.(asa inteleg eu)Eu am gasit o configuratie cu 14 nebuni care nu se ataca intre ei.Uite si configuratia:

 $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ #
      #      #       #      #       #      #      #      #
  # # # # # # # #
  # # # # # # # #
  # # # # # # # #
  # # # # # # # #
  # # # # # # # #
 $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ $\ N$ #

intr adevar numarul maxim de turnuri care poate fi asezat pe o tabla de sah este8.avand 8 orizontale nu poate sta mai mult de un turn pe fiecare orizontala.o asezare ar fi sa punem cate un turn in fiecare patratel al unei diagonale mari.
dar numarul de nebuni care poate fi asezat pe o tabla nue 8 cum am fi tentati sa spunem, si 14.ducem toate cele 15  diagonalele posibile(care nu se intersecteaza).punem un nebun in fiecare diagonala cu exceptia uneia mici de patrat unitate.o astfel de asezare este a1,b1,c1,d1,e1,f1 ,g1, a8,b8,c8,d8,e8,f8,g8.


[hide="Solution"]Let $A(x)$ be a non-constant polynomial with integer coefficients. It's well-known that if $a\equiv b\bmod{m}$, then $A(a) \equiv A(b)\bmod{m}$, for all integers $a,b,m$.

Recall that $A(n)$ can't be prime for all integers $n$, because then $A(1)$ would be a prime, say $p$, and as $kp+1\equiv 1\bmod{p}$ for all integers $k$, we would have that $A(kp+1)\equiv A(1) \equiv 0\bmod{p}$. But as $A(kp+1)$ is prime, that means $A(kp+1) = p$. Therefore, $\lim_{k\to\infty}A(kp+1) = p$, which is impossible, as $\lim_{x\to\infty}A(x) = \infty$ for all polynomials $A(x)$.

So let $n_{0}$ be an integer such that $A(n_{0})$ is composite. Let $A(n_{0}) = a$. Now, for all integers $k$ we have $ak+n_{0}\equiv n_{0}\bmod{a}$, so $A(ak+n_{0}) \equiv A(n_{0})\equiv 0 \bmod{a}$. As $a | A(ak+n_{0})$ and $a = A(n_{0})$ is composite, it follows that $A(ak+n_{0})$ is composite. (Unless $A(ak+n_{0}) = 0$, but this occurs a finite number of times.) Therefore, $A(kA(n_{0})+n_{0})$ is composite for all integers $k$, which completes the proof.[/hide]

[quote="trancestorm"]let    f(x)=ax^2 + bx +c   be a nonconstant polynomial and assume a, b, and c are integers.  Prove that there are infinitely many integers n such that f(n) is composite.[/quote]

isn't this homework for EPGY number theory?


linear algebra
matrix
algebra
polynomial
function

An n-by-n  matrix A

A^2+A+I=0

show that (A Invertible) and A^-1=A^2?

[/code]

The minimal polynomial must divide $ \lambda^{2}+\lambda+1=0$ showing that zero can not be an eigenvalue of $ A$, and the matrix is thus invertible.

$ A^{2}+A+I=0 \rightarrow A^{3}+(A^{2}+A)=0 \rightarrow A^{3}-I=0 \rightarrow A^{2}=A^{-1}$

The first line of edoo's post is redundant. The last line - the computation - functions as the entire proof. It shows that $ A$ is invertible by explicitly producing the inverse.

$ A^{2}+A = A(A+I) =-I$

Since A(A+I) is invertible, A must be invertible (and so is A+I). Now multiply the right of both sides by the inverse of A + I.


geometry
geometric transformation
rotation
counting
derangement

Consider $10$ people sitting around a circular table. How many different ways can they exchange seats so that each person has a different neighbor to the right?

I assume that orientations equivalent by rotation are considered distinct.

Interestingly, I got the same number for this as $D_{10}$, the number of derangements for 10 objects.


proof that if X and Y are locally compact spaces, then XxY is also locally compact space.

let $ (x,y)\in X\times Y$. since $ X$ and $ Y$ are locally compact, there exist neighborhoods $ U\subseteq X$ and $ V\subseteq Y$ that are compact. $ U\times V$ is clearly a neighborhood of $ (x,y)$. we'll show that it's also compact. because every open set in the product topology is a union of basic open sets ($ A\times B$, where $ A$ is open in $ X$ and $ B$ open in $ Y$), it suffices to prove the finite subcover criterion for a cover consisting of basic open sets. so let $ U\times V\subseteq \bigcup_{i\in I}U_i\times V_i$, where $ U_i$ is open in $ X$, $ V_i$ open in $ Y$. then $ U\subseteq \bigcup_{i\in I} U_i$ and $ V\subseteq \bigcup_{i\in I} V_i$. since $ U$ and $ V$ are compact, there exist finite $ F, G\subseteq I$ such that $ U\subseteq \bigcup_{i\in F}$ and $ V\subseteq \bigcup_{i\in G}$. then $ U\times V\subseteq \bigcup_{i\in F\cup G} U_i\times V_i$, so that we have a finite subcover.


A convex quadrilateral $ ABCD$ is given for which $ \angle ABC\plus{} \angle BCD < 180^{\circ}$. The common point of the lines $ AB$ and $ CD$ is $ E$. Prove that $ \angle ABC\equal{} \angle ADC$ if and only if $ AC^2\equal{}CD \cdot CE\minus{}AB \cdot AE$.

[quote="moldovan"]A convex quadrilateral $ ABCD$ is given for which $ \angle ABC \plus{} \angle BCD < 180^{\circ}$. The common point of the lines $ AB$ and $ CD$ is $ E$. Prove that $ \angle ABC \equal{} \angle ADC$ if and only if $ AC^2 \equal{} CD \cdot CE \minus{} AB \cdot AE$.[/quote]
There are two cases, so I think it must be $ AC^2 \equal{}|AB \cdot AE \minus{}CD \cdot CE |$

The circumcircle of $\triangle EBC$ cuts $CE$ at $ C,F$ .

$\Rightarrow \overline{AF}.\overline{AC}=\overline{AE}.\overline{AB}=AB.AE$ (*)
We can easily see this:
$ (BA,BC)=-(DA,DC) (mod \pi) \Leftrightarrow (FE,FC)=-(DA,DC) (mod \pi) $
$\Leftrightarrow (FE,FC)=(DE,DC) (mod \pi ) \Leftrightarrow E,F,A,D$ are concylic
$\Leftrightarrow \overline{CA}.\overline{CF}=\overline{CD}.\overline{CE}=CD.CE$
$%Error. "Leftirhgtarrow" is a bad command.
\overline{CA}.\overline{CF}-\overline{AF}.\overline{AC}=CD.CE-AB.AE$

So , QED


If $ A \star B$ means $ \frac{A\plus{}B}{2}$, then $ (3 \star 5) \star 8$ is

\[ \textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 30
\]

[hide="Click for solution"]
$ (3 \star 5) \star 8\equal{}\left (\frac{3\plus{}5}{2} \right ) \star 8\equal{}4 \star 8\equal{} \frac{4\plus{}8}{2}\equal{}6$, or $ \boxed{\textbf{(A)}}$.
[/hide]

[hide="Another solution"]
$ a \star b$ determines the average of $ a$ and $ b$, so the average of $ 3$ and $ 5$ is $ 4$, and the average of $ 4$ and $ 8$ is $ 6$, which is $ \boxed{\textbf{(A)}}$.
[/hide]


i don't think they post those on the site, just a few samples sometimes

i took that test today. it was hard. actually it was quite simple. its just that i thought 4*5=7*4 for some odd reason. distress.


real analysis
geometry
geometric transformation
invariant
real analysis unsolved

Let $ \lambda_b(E)\equal{}\sup\{\lambda(X): X \subset E, X \in \mathcal{M}\}$. Is $ \lambda_b$ $ \sigma$-additive,where $ \mathcal{M}$ is set of measurable sets, and $ \lambda$ is Lebesgue measure?

By definition, $ \lambda_b$ is translation invariant and monotone with respect to inclusion. Consider a set $ V$ such that a disjoint countable union of its translated copies covers $ [0,1]$ and is contained in $ [1,3]$. ([url=http://en.wikipedia.org/wiki/Vitali_set]Example[/url].) Now you have a problem with $ \lambda_b(V)$.


Two circles $S_{1}$ and $S_{2}$ touch each other externally at $K$; they also touch a circle $S$ internally at $A_{1}$ and $A_{2}$ respectively. Let $P$ be one point of intersection of $S$ with the common tangent to $S_{1}$ and $S_{2}$ at $K$. The line $PA_{1}$ meets $S_{1}$ again at $B_{1}$, and the line $PA_{2}$ meets $S_{2}$ again at $B_{2}$. Prove that $B_{1}B_{2}$ is a common tangent to $S_{1}$ and $S_{2}$.

Just invert around $P$, with power $PK^{2}$. The inversion fixes the circles $\mathcal S_{1},\mathcal S_{2}$, and turns the circle $\mathcal S$ into the line $B_{1}B_{2}$. Since $\mathcal S$ was tangent to $\mathcal S_{i},\ i=1,2$, the line $B_{1}B_{2}$ will be tangent to $\mathcal S_{i}$ as well.


complex analysis
function
algebra
polynomial
irrational number
advanced fields
advanced fields unsolved

Does anyone know how to solve the following?

Prove that the following is false for any constant $ c > 2:$
[i]Given any irrational number $ \xi,$ there exist infinitely many rational numbers $ h/k$ such that\[ \left|   \xi \minus{} \frac{h}{k}  < \frac{1}{k^c}   \right|.
\]
[/i]

There is theorem that the set of irrational numbers with irrationality measure strictly larger then $ 2$ has Lebersgue measure zero.
I agree with you that for [b]any[/b] irrational number this is not true . Indeed  for example take $ \sqrt {2}$ then by Luiville theorem follows that $ c \equal{} 2$. More generraly this is not true for eny algebraic number also for $ e$ and for all $ e^{t}$ where $ t \in \mathbb{Q}$
but for $ \pi$ it is not known.  :|  last upper bound was $ c < 8.0160450...$It's Hata's  result, he is champion . .

Hello, I dont understand why this is true:

[quote]
Indeed for example take $ \sqrt{2}$ then by Luiville theorem follows that $ c\equal{}2.$ [/quote]

Liouville's theorem says: [hide]
any holomorphic function $ f: \mathbb{C} \to \mathbb{C}$ which satisfies a polynomial bound condition of the form $ \displaystyle \vert f(z)\vert < c \cdot \vert z\vert^n$ for some $ c \in \mathbb{R}$, $ n \in \mathbb{Z}$, and all $ z \in \mathbb{C}$ with $ \vert z\vert$ sufficiently large is necessarily equal to a polynomial function.[/hide]

:lol:
[hide="here"]
http://en.wikipedia.org/wiki/Liouville_number
[/hide]


prove that doesn't exist a number whose square is 5

[hide="If you mean integer then....."]$ 2^2 < 5 < 3^2$, and since no square exists in between two consecutive squares, we are done.[/hide]
[hide="If rational....."]Suppose that $ \frac{a^2}{b^2}\equal{}5$, where $ (a,b)\equal{}1$. Hence $ a^2\equal{}5b^2\implies 5\mid a$. Let $ a\equal{}5a_1\implies 25a_1^2\equal{}5b^2\implies 5a_1^2\equal{}b^2\implies 5\mid b$, contradiction since $ (a,b)\equal{}1$.[/hide]

What about $ \pm \sqrt 5$ ;)
And if you want [b]rational[/b] numbers, then it's too easy for Pre-Olympiad...


inequalities
function
calculus
calculus computations

What is the difference between strictly monotone increasing (or decreasing) and monotone increasing? Is the difference over due to whether or not there's a strict inequality or not?

Yes, that's exactly the difference. For example, $ f(x)\equal{}0$ is monotone increasing (and monotone decreasing, by the way), but it's not strictly monotone increasing, while $ f(x)\equal{}x$ is strictly monotone increasing (and, of course, "just" monotone increasing).

In plain words, if for all $ x<y$, $ f(x)<f(y)$ holds, then the function is strictly monotone increasing. If $ f(x)\leq f(y)$ holds, then the function is monotone increasing. It is easy to notice that if $ f$ is strictly monotone increasing, then it's monotone increasing, but, as our examples show, the converse isn't true.


Prove that for any prime $p$ there is a Fibonacci number divisible by $p$.

I saw this solution due in one of Titu's packets:
[hide]
Any pair of consecutive terms uniquely determines the rest of the terms. Consider the terms mod $p$. Let $F_{0}=0$. If we choose $p^{2}+1$ successive terms, some two pairs are equal. Therefore, all pairs will repeat. In particular, the pair $(0,1)$ repeats, so a number is divisible by $p$.
[/hide]
However, this solution appears to be wrong, because if the repeated pair is not $(0,1)$, then it need not occur again.  [/hide]

[hide="Well,"]
when you find the repeated pair, you will be able to work backwards or forwards to find the pair $(1,1)$, and the term just before that must be $0$. Voila. :) [/hide]

this holds too: for any positive integer $n$ there exists infinitely many fibonacci numbers divisible by $n$. :wink:

The thing is that Fibonacci numbers have a "memory" of one number. If two consecutive numbers repeat, then the whole sequence will repeat from that point. The second thing is all possible pairs will repeat. There is no "tail" of unrepeated numbers. Let's say there was such a tail, and $m,n$ was in the tail. Then $n-m, m$ must also be in the tail. Continuing this argument we see that the tail must either repeat or include every pair.


group theory
abstract algebra
invariant
superior algebra
superior algebra unsolved

$ H$ is subgroup of $ G$, and $ H$ contains no non-trivial normal subgroup.
How to prove: $ \bigcap_{s \in G}sHs^{-1}={e}$. (the converse is also true but is easy to prove).

[quote="-oo-"]You mean that $ H$ contains no nontrivial normal subgroup of $ G$, right?[/quote]
That also tells us that $ H$ [b]is not[/b] a normal subgroup of $ G$. Right?

That intersection of conjugates of the subgroup is itself a subgroup (subgroups work well under intersection), and it is clearly normal; taking any conjugate merely permutes the intersection.
It's a subgroup of $ H$ which is normal in $ G$, and that makes it the trivial group by hypothesis. For the converse, we note that this intersection is actually the largest normal subgroup of $ G$ contained in $ H$; it contains all other conjugation-invariant subsets of $ H$.


parameterization
trigonometry
inequalities
analytic geometry
graphing lines
slope
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14292[/img]

 :)

Which pair of sides are equal length?
* $ AC = AB$:
Denote CH be the altitude of triangle ABC (H on BC). Then C,O,H are collinear.
let $ OH = x (0 < x < R)$, R is radii of circle (O).
$ AB + CH = R + x + 2\sqrt {R^2 - x^2} = R + x + 2\sqrt {(R - x)(R + x)} = R + x + \frac {2}{a}.\sqrt {a^2.(R - x)(R + x)}$ (where $ a$ is an arbitrary parameter, $ a > 1$)
$ \leq R + x + \frac {a^2(R - x) + (R + x)}{a} = R + x + \frac {(a^2 + 1)R - x(a^2 - 1)}{a}$
Choose $ \frac {a^2 - 1}{a} = 1 < = > a = \frac {1 + \sqrt {5}}{2}$ (because a>1)
The sum $ AB + CH$ attains its greatest value when $ a^2(R - x) = R + x < = > R(a^2 - 1) = x(a^2 + 1)$
$ \Rightarrow$ sin$ \angle OAB = \frac {x}{R} = \frac {a^2 - 1}{a^2 + 1}$ $ = \frac {a}{a + 2}$ $ = \frac {1 + \sqrt {5}}{5 + \sqrt {5}}$ $ = \frac {1}{\sqrt {5})}$
Construct an angle $ \alpha$, whose tangents is $ \frac {1}{2}$
Then construct $ \angle OAB = \alpha$

[quote="Jrrbc"][color=darkred] [size=117]Let $ ABC$ be a $ A$ - isosceles triangle which is inscribed in a given circle $ C(O,R)$ .

Denote $ D\in (BC)\cap AO$ . Ascertain the maximum of the sum $ BC + AD$ .[/color][/size] [/quote]
[hide="Proof."]
[color=darkblue]Can suppose w.l.o.g. $ R = 1$ and $ A\le 90^{\circ}$ because if $ A > 90^{\circ}$, then exists the isosceles triangle $ AEF$ , where  $ O\in EF$ ,

$ EF\parallel BC$ and $ EF + AO > BC + AD$ . Thus, in the $ D$ - right triangle $ BOD$ we have $ \|\begin{array}{c} BD = \sin A \\
\ AD = 1 + \cos A\end{array}$ . Therefore,

the sum $ BC + AD$ is maximum iff $ f(A) = 2\sin A + \cos A$ is maximum, where $ A\in (0,\frac {\pi}{2}]$ . Show easily that $ \tan A = 2$ , i.e.

$ \boxed {\ A=\arctan 2\ }$ . You can construct easily the required triangle with the maximum sum $ BC + AD\ !$

[b]Remark.[/b] If denote $ \|\begin{array}{c} BD = x \\
\ OD = y\end{array}$ , then this problem becomes : $ \boxed {\ \begin{array}{c} x^2 + y^2 = 1 \\
 \\
2x + y\ \rightarrow\ \mathrm {max.}\end{array}\ }$ . Indeed, from the [b]Cebytchev's inequality[/b] obtain :

$ (2x + y)^2 = (\frac x2 + \frac x2 + \frac x2 + \frac x2 + y)^2\le$ $ 5(\frac {x^2}{4} + \frac {x^2}{4} + \frac {x^2}{4} + \frac {x^2}{4} + y^2) = 5(x^2 + y^2) = 5$ with equality iff $ \frac x2 = \frac y1 = \frac {\sqrt 5}{5}$ .

Directly, you can use the [b]C.B.S.' inequality[/b] : $ (2x+y)^2\le (2^2+1^2)(x^2+y^2)=5$ , i.e. $ 2x+y\le\sqrt 5$ with equality iff $ \frac x2=\frac y1=\frac {\sqrt 5}{5}$ .

In conclusion, $ \boxed {\ \begin{array}{c}BC = 2x = \frac {4\sqrt 5}{5} \\
 \\
AD = 1 + y = 1 + \frac {\sqrt 5}{5} \\
 \\
BC + AD = 1 + \sqrt 5\end{array}\ }$ . Also you can give an analytical geometrical interpretation : find the tangent $ t$

in $ T(x,y)$ , $ x > 0$ , to the circle $ C(O,1)$ with the slope $ m_t = - 2$ for which the distance from the origin to it is maximum.[/color][/hide]
[quote="Virgil Nicula"] [size=117][color=darkred] [b][u]A similar problem 1.[/u][/b] Let $ ABC$ be a $ A$ - isosceles triangle which is inscribed in a given circle $ C(O,R)$ .

Denote $ E\in AC$ , $ BE\perp AC$ . Ascertain the maximum of the sum $ BC + BE$ .[/color][/size] [/quote]
[hide="Answer."]
$ \boxed {\ A = \arccos\frac {\sqrt {13} - 2}{9}\ }$ .[/hide]
[quote="Virgil Nicula"] [size=117][color=darkred] [b][u]A similar problem 2.[/u][/b] Let $ ABC$ be a $ A$ - isosceles triangle which is inscribed in a given circle $ C(O,R)$ .

Denote $ E\in AC$ , $ BE\perp AC$ . Ascertain the maximum of the sum $ AC + BE$ .[/color][/size] [/quote]
[hide="Answer."]
$ \boxed {\ A =\arctan\frac {21-5\sqrt {17}}{4}\ }$ .[/hide]


MATHCOUNTS
algebra
polynomial
function
articles
number theory
prime numbers

I don't know where else to post this.

I read about Goldbach's Conjecture and found it interesting. My book said
"[i]Goldbach's Conjecture[/i] states that every even integer greater than 2 can be expressed as the sum of two prime numbers. Here are some examples.
$ 4\equal{}2\plus{}2$
$ 6\equal{}3\plus{}3$
$ 8\equal{}3\plus{}5$
Goldbach's conjecture was made in 1742. Although it souds simple, no one has ever been able to prove that it is true. "

Can anybody say why it might work? (I know that was a hard question, because no one has proven it) How might someone prove it? Also, is there any pattern in prime numbers? For example, from 5 to 7 to 11 to 17 it goes +2, +4, and +6.

These are ridiculous and spammy answers except for number 3 (but they are the best that I can think of; I think no one in AoPS can answer this).

1) Magic
2) With Math
3) No; mathematicians for many years have been trying to find some sort of pattern.

[quote="PowerOfPi"]I don't know where else to post this.

I read about Goldbach's Conjecture and found it interesting. My book said
"[i]Goldbach's Conjecture[/i] states that every even integer greater than 2 can be expressed as the sum of two prime numbers. Here are some examples.
$ 4 \equal{} 2 \plus{} 2$
$ 6 \equal{} 3 \plus{} 3$
$ 8 \equal{} 3 \plus{} 5$
Goldbach's conjecture was made in 1742. Although it souds simple, no one has ever been able to prove that it is true. "

Can anybody say why it might work? (I know that was a hard question, because no one has proven it) How might someone prove it? Also, is there any pattern in prime numbers? For example, from 5 to 7 to 11 to 17 it goes +2, +4, and +6.[/quote]

This definitely is WAY above mathcounts level.

I'll try to explain some of your questions, though I probably don't have enough background to explain this very well...

First, I'll answer the "is there any pattern in prime numbers." If you mean, is there a polynomial (Or rational function) such that all of it's values are primes, then the answer is no. Primes numbers are distributed weirdly...

The distribution of the prime numbers is related to the roots of the Riemann-Zeta function. Information about the roots translates to information about the distribution of prime numbers.

For example, proving that no roots lie on the line Real(z) = 1 can be used to prove the Prime Number Theorem.

The Riemann Hypothesis conjectures that all nontrivial roots lie on Real(z) = 1/2, also known as the critical line. If we could prove the Riemann Hypothesis, then we would get a lot of information about the distribution of the primes.

"Can anybody say why it might work?"

1. The huge amount of numerical evidence. I believe that every even number from 4 to 10^18 has been proven to be a sum of 2 primes.

2. The probabilistic argument: Using facts about the distribution of prime numbers, you can show that, assuming the primes are distributed randomly, every even number will have a very high chance of being able to be expressed as a sum of 2 primes.

"How might someone prove it?"

This question is pretty hard to answer.

http://en.wikipedia.org/wiki/Goldbach's_conjecture

I don't have the background to know very much about this at all, but...

Eh, primes are a lot more complicated then they appear.

[quote="BOGTRO"]A pattern of primes is a millenium problem (or at least was, idk if it still is). Riemann deals with primes.

Some thigns simply cannot be proven. For example, P vs. NP is widely accepted as unprovable.

It might work because huge numbers have been tested and found to be a sum of two primes. This problem is very hard because there is no [known] pattern to primes.

How might someone prove it?

Start with Riemann, find a pattern for primes, and write a 100+ page proof.[/quote]

1. P vs. NP is not widely accepted as unprovable, the majority of people think that it's false.

On the other hand, the continuum hypothesis has been proven to be unprovable.

2. Well, we don't have a specific formula about the primes, but we know stuff about the distribution of the primes; for example, we know that the nth prime divided by n approaches ln n when n gets large.

(Unfortunately, that's an "approximately n ln n", but there are bounds on the error; in particular, the Riemann Hypothesis is equivalent to a bound on the error term.)

Also, there is the the weak Goldbach conjecture: Every odd number greater than 7 can be written as the sum of 3 odd prime numbers. It has almost been proved, it has been proved that every odd number that is sufficiently large (greater than 10^(some 4-digit number)) is the sum of 3 prime numbers. Note that the strong Goldbach conjecture proves the weak one (add 3 to the even number), but the converse is false. The weak Goldbach conjecture also implies that every even number greater than 6 (I think) can be written as the sum of 4 primes (add 3 to the odd number). Also, Levy's Conjecture (I'm pretty sure its not been proven yet) states that every odd greater than 5 number can be represented as p+2q, p,q are primes (BTW this is a bit stronger than weak Goldbach conjecture). Anyways,

[quote="PowerOfPi"]How might someone prove it?[/quote]
This is not as easy as it looks. Possibilities are proving it using Riemann zeta function, [url=http://en.wikipedia.org/wiki/Chen_prime]Chen Primes[/url], and various other number theoretic functions.

[quote="PowerOfPi"]Also, is there any pattern in prime numbers?[/quote]
Like pythag011 said, there is no polynomial or rational function, even, that generate only primes. (Try proving the "polynomial" part, which shouldn't be hard.) However considering polynomials like n^2+n+41, they are prime for until n=40. This should (I think) work for all numbers d such that Q(sqrt(-(4d-1)) is a UFD. Sadly, there are only nine such d, 41 being the largest. However, the Green-Tao theorem (or something like that) says that for any n, there exist an arithmetic progression such that the first n terms are prime.
BTW, mods, please move this to the appropriate forum.

I don't understand how that proves the Goldback Conjecture. Maybe pythag011 can explain (or some other genius like AIME15)?

The majority of computer scientists believe that P is not equal to NP.  Very few people, I think, seriously believe that it's independent of ZFC, but a few people believe that there's no good reason P couldn't be equal to NP; it's possible that this could be true without implying that NP problems can be solved efficiently if the constants are bad enough.

The weak Goldbach conjecture is a consequence of the generalized Riemann hypothesis.

There's a good writeup of the "heuristic" proof of the Goldbach conjecture [url=http://godplaysdice.blogspot.com/2007/07/riemann-hypothesis-is-probabilistic.html]here[/url].  The idea is to treat the primes as if they were basically random up to "local" obstructions (that is, divisibility), which is essentially the idea that the Riemann hypothesis would allow us to use rigorously.

[quote="Remi"]I believe the answers you are looking for can be found in this link, the proof is still relatively new.

http://www.archive.org/details/BeyondRiemannAndTheRandomDistributionOfPrimes[/quote]

To save you all the trouble of reading this paper, here's a brief summary as far as I can tell.

1. [Philosophical stuff]
2. Odd numbers have no physical meaning.
3. Therefore, all odd numbers are prime.
4. Therefore, the Goldbach Conjecture is true.

If you are reading the paper, parts 3 and 4 happen on page 12.

I don't see any actual math in this.

The writer even claims that all odd numbers are primes.

This DEFINITELY is not a real proof of the Goldbach conjecture.

Well... the writer's definition of prime need not be our definition of prime. Nevertheless it differs from the "standard" definition of prime, so the Goldbach conjecture in its standard sense is not proven.

[quote="Remi"]I guess I shouldn't presume this wasn't entirely over your heads. physics and philosophy explaining math ?????? what happens to the numbers lmfao[/quote]

Do you think that physics and philosophy really constitutes a proof?

I think we can assume Remi = author of the article according to first name.

Your argument is weak. Math is a basis for physics, etc. so physics proving is math = probably not a good idea. Philosophy is even further off the mark.

well at least someone here is observant but frankly I don't think its weak at all.

tell me do you not think if we want our number system to make any sense to us that it must not be governed by the same laws as us. In a Universe of actions and reactions #1 all be it a creator does not exist and every other odd for that matter (actions and there unaccounted for reactions.) not only does the golbach conjecture rain true but what two odds added not become even? hint hint.
the question therefore comes down to: do you suffer from the god delusion.

[b]Do you think that physics and philosophy really constitutes a proof?[/b]
hahahah are you for real: did MATH discover the atom or was it physics and philosophy?

I would have to say you guys have it in reverse math helps to further our understanding not discover it.

[quote="Remi"]did MATH discover the atom or was it physics and philosophy?[/quote]
And the atom relates to the definition of primality how?

Look, it's not an issue of whether physics or mathematics is more important in informing the other; there is a [b]definition[/b] of primality that is implicit in the statement of the Goldbach conjecture and this paper doesn't use it, so it has nothing to say about [b]what we call[/b] the Goldbach conjecture.  What if I said "math discovered the atom, not physics" and then added "by the way, 'the atom' is code for the Leech lattice"?

You're also obviously a troll and should be banned as soon as possible.

I was simply using the atom as an example and as I can recall it was you leaning toward math being more valuable even as you state now the leach lattice which was only discovered in the 1940's.

In 1827, British Botanist Thomas Brown noticed that when a pollen grain floating in a drop of water is examined using a microscope, it can be seen bouncing around in a random motion.

thats when It was discovered and later led to Einsteins work in proving the relationship that Brownian motion has to atomic theory (1909) using math to further our understanding of coarse.

haha a troll, did you feel it necessary in resorting to name calling, quit being so emo.

Again, no one disputes that physics is extremely valuable in informing mathematics.  [b]What does that have to do with the definition of primality?[/b]

"Troll" is the most concise way to describe exactly what you're doing.  Cut it out.

Again, no one disputes that physics is extremely valuable in informing mathematics. What does that have to do with the definition of primality?

if your referring to me using the atom as an example it had nothing to do with primality, it was simply put there to reveal that james4l was a tool in thinking math was more relevant in finding a proof.  I would have much preferred that he would have come to that conclusion on his own because unlike you i'm less direct in the name calling, but hey! you asked why, so he has you to thank. lmfao

if your were referring to the paper i'm simply stating if the numerical system is ever going to make any sense to us, it has to follow the exact same laws that govern us: the rest is physics from there. get it, got it ,good and I hope your finished calling me a troll.

I am posting this to try to stop the argument and let it be. I don't want it to become spam, so I will ask to keep the discussion on topic and not about physics (I don't mean the atom example, but after that it was off topic).


function
real analysis
real analysis unsolved

[color=darkblue][size=150]Whether the series \[ \sum_{n\equal{}1}^{\infty}a_{n}\] convergence or not ,
can we get the conclusion\[ \sum_{n\equal{}1}^{\infty}o(a_{n}) \equal{} o(\sum_{n\equal{}1}^{\infty}a_{n})\] ?
can someone give a prove of it or a counter-example?[/size][/color]

 :arrow:

o for what variable?  Since n is bound in the second one, there must be some other variable intended for the o ??

Perhaps this:

Suppose that $ a, b: {\bf N} \to {\bf R}$ are functions and $ b \equal{} o(a)$.  Let $ f(n) \equal{}\sum_{i \equal{} 1}^n a(i)$ and $ g(n) \equal{} \sum_{i \equal{} 1}^n b(i)$.  Is it true that $ g \equal{} o(f)$?

If it's as JBL stated, it's very obviously false.

An example:

$ 1\plus{}\frac13\plus{}\frac19\plus{}\cdots\plus{}\frac1{3^{n\minus{}1}}\equal{}\frac32\left(1\minus{}\frac1{3^n}\right)$
$ \frac12\plus{}\frac14\plus{}\frac18\plus{}\cdots\plus{}\frac1{2^n}\equal{}1\minus{}\frac1{2^n}$

The second series dominates the first. The partial sums of the first are greater than those of the second.

Well, it becomes true if we assume in addition that $ a>0$ and $ f(n)\to\plus{}\infty$ as $ n\to\infty$. For positive convergent series, it becomes true if we consider the tails $ \sum_{n}^{\infty}$ instead of the partial sums $ \sum_1^n$.

Another funny interpretation (also "obviously false" as stated) is as follows. Assume that $ a_n(x)$ and $ b_n(x)$ are non-negative functions such that $ b_n(x)\equal{}o(a_n(x))$ as $ x\to x_0$ for all $ n$ and that both series $ A(x)\equal{}\sum_n a_n(x)$ and $ B\equal{}\sum_n  b_n(x)$ converge for all $ x$. Then $ B(x)\equal{}o(A(x))$ as $ x\to x_0$.

The task is, of course, not only to produce a counterexample, but also to figure out how to turn this false statement into a true one ;).

Needless to say, neither of these attempts to interpret the original post makes it any less meaningless :(.


analytic geometry
conics
parabola
quadratics

[quote="Interval"]A parabolic arch has a span of 129 feet and a maximum height of 25 feet.  Choose a suitable rectangular coordinate axes and the find the equation of the parabola.  Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center.[/quote]
If you set the "left" side to (0,0), then the x-intersect is (129,0)
When the y is maximum, then x is half of the x-intersect. so the vertex is at (64.5, 25). I'm not sure where to go from there...

[hide] thats a good start but use this equation..

$4a(y-k)=(x-h)^{2}$ where $(h,k)$ is the vertex

So using what Quevvy suggested we see

$4a(y-25)=(x-64.5)^{2}$

so sub in $y=0, x=0$ and solve for $a$

$4a(-25)=(-64.5)^{2}$

$a=\frac{(-64.5)^{2}}{-100}=-\frac{16641}{400}$ (i just converted it into fractions)
so you have..

$-\frac{16641(y-25)}{100}=\left(x-\frac{129}{2}\right)^{2}$

ya so now you can sub in for the rest..

[/hide]

SM4RT: You can't plug in (0, 0) because it's not on the parabola. You have to plug in (64.5, 0) or (-64.5, 0)

i dont get what you mean by center...the origin of the coordinates or the center of the parabola (wtf???)

[hide]i try to not memorize formulas.

we know the vertex is at (0,25)
plug in (64.5, 0) and (-64.5,0) and (0,25) into the quadratic generalization thing $(y=ax^{2}+bx+c)$.  we find that c=25 and then when you plug in $(\pm 64.5, 0)$ into the equation, we get two equatio with 2 variables and then solve for the coefficients.  This gets you the graph of the porabola.  As for the second parts.  I think you should find the intersects between the equation and the circle equation $x^{2}+y^{2}=r^{2}$.[/hide]

It isn't really a formula... it's another equation of a parabola.

For a parabola opening upwards or downwards, the equation is $y-k = \frac{1}{4a}(x-h)^{2}$. $(h, k)$ is the vertex of the parabola, and $a$ is the distance from the vertex to the focus and the distance from the vertex to the directrix.

[hide] cant you just say that the parabola its zero at 0 and 129, so the equation is:

y = kx(x-129) for some k then solving for k with (129/2, 25) gives

25= k (-(129/2)^2) so

k = - 100/(129^2)

then for the other points just plug 129/2 +10, +20, and +40  in.[/hide]

but $(0,0)$ was on the parabola i set the points $(0,0),(129,0)$ and $\left(\frac{129}{2},25\right)$ hmm..cuz the points u are suggesting are differnt..then where i placed them

[quote="Interval"]A parabolic arch has a span of 129 feet and a maximum height of 25 feet.  Choose a suitable rectangular coordinate axes and the find the equation of the parabola.  Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center.[/quote]
[hide]
it would be suitable to either set the center as the origin or 25 units above the origin. Since the last statement asked to calculate height, it'll be easier to set the center as 25 units above the origin.
$y-25=nx^{2}$
$0-25=n(\frac{129}{2})^{2}$
$-25=\frac{129^{2}n}{4}$
$-100=129^{2}n$
$n=\frac{-100}{16641}$
$y=\frac{-100x^{2}+416025}{16641}$
the coordinates of 10 ft, 20 ft, and 40 ft are (10,24.4), (20,22.6), and (40,15.4) respectively.
[/hide]

What is the center of a parabola? Is it the vertex?

it is the vertex, since that is the point given by $(h,k)$ in $(y-h)=n(x-k)^{2}$
also, it is the only intersection between the parabola itself and the axis of symmetry.


search
linear algebra
matrix
LaTeX
linear algebra unsolved

Hi,

My teacher handed us out an assignment, and one of the numbers included in it has me confused. The teacher asks us to simplify:

[img]http://www.crazi.net/linear.jpg[/img]

I don't know how to go about doing the inverse of the sum of two matrices, nor does it say how to in our notes. I tried looking online but don't know exactly what to search for. Can anyone help me out please? At the least, what the rule is with the inverse of the sum of two matrices?

Many thanks!

Well as you should know, the inverse of a matrix (if it exists) doesn't change its size. Neither does adding two matrices.  Therefore, the inverse of the sum is know different from the inverse of any other matrix.  That is, add the matrices, and then take the inverse.  Nothing special.  For this problem, I recommend that you treat these matrices as 2x2's since that has an easy formula for the inverse.  Then you can just do the arithmetic.

OK I think I see the logic, however we are not provided with numerical values for the matrices. We only have matrix A, B, C and D. Could you elaborate please? Thanks so much for the help.  :)

Crazihouse: you can attach your images here or use LaTeX; you don't need to link to other websites for this.

There is no "rule" for this problem, it's more of a trick question.

[hide="Hint"]$ A\left(B \minus{} DC^{ \minus{} 1}A\right)^{ \minus{} 1} \equal{} \left(\left(B \minus{} DC^{ \minus{} 1}A\right)\cdot A^{ \minus{} 1}\right)^{ \minus{} 1}$. Simplify. Modify $ C\left(D \minus{} BA^{ \minus{} 1}C\right)^{ \minus{} 1}$ similarly. Compare.[/hide]

  darij

In general, if we are given an arbitrary matrix, we will its entries with arbitrary $ a_{11}$, $ a_{12}$, ... so in the case of a 2x2, we write
$ A\equal{}\begin{pmatrix}
a_{11}&a_{12}\\
a_{21}&a_{22}
\end{pmatrix}$
Then we do all our operations with those entries.  That's all there is to it.

JRav: Nobody said the matrices are $ 2\times 2$, and I sincerely doubt you would like to do the proof by explicit computation for generic $ n\times n$ matrices.

  darij

I know no one said they are 2x2 but if it holds for a generic matrix, then it should hold for a 2x2 and those are easy to work with.

Thanks for the help! Solved it finally..  :ninja:


This is Leibniz's formular. The proof is an easy induction, you can do it by yourself. I suggest that you use this formular: \[ C^{k}_{n}+C^{k-1}_{n}=C^{k}_{n+1} \]


What was ur state's cutoff?

mine was around a 30

i'll compose a list

State Cutoffs
______________________________________________________
Connecticut ~ 30

I just made this, but with no poll, b/c the range was too big.

WI-32

The message is too small. Please make the message longer before submitting.

KY cut off was actually 37.  I barely made it to nationals.

Really? I find that rather hard to believe...Maybe you didn't actually barely make nats...are you sure?

Definitely. I tied with the third place person with a 37.
Second place got a 39 and the  winner got a 42(and a 30 on the Sprint!!!)

[quote="kstan013"]I'd best look out at nats! Anyway, our cutoff was 32 because of Mr. Slacker footballrocks41237!!! jkjkjkjkjkjkjkjkjkjkjk :) He did good for only having a 21 at chapter[/quote]

Hahaha lol.  Well 4th and 5th got the cutoff too i think..

Arkansas had a cutoff of 26;
It was interesting as the scores varied widely:

1st place: 43 (27/16)  (me)
2nd place: 34 (18/16)
3rd place: 31 (19,12)
4th place: 26 (No idea)

kwazypanda might be right (though i doubt first place got 30 on sprint. he is a 7th grader for those who don't know)

You shouldn't be suprised all these Kentucky scores are a bit higher than what many would expect from a stereotypical redneck/rural state

Kentucky got 13th in nationals in 2005 (highest person was ranked 15th)
Kentucky got 16th I believe in 2006 (highest person was ranked 19th)

This years team from Kentucky has a shot at getting first I believe
Since 3/4 of them came out of nowhere (i assume they just moved to kentucky this year)
Only person I thought that would make nationals for sure was Michael Druggan (who got third)

ya I saw, and I got surspershersh about that. His name's like Evan Miller, right? I'm surprised that KY got their nat competitors posted so quickly for just having state on sat. NY hasn't posted their's yet and they had it 3 weeks ago!

Jorian,

The point of Mr Fischer asking IN kids not to talk about scores every year is probably because he doesn't want people talking about IN scores.  SO STOP talking about them.  You don't KNOW anything.  Find something else to do and DROP IT.

right i don't know anything/shouldn't know anything/shouldn't post anything if i don't know anything/should shut up completely

sorry guys

-jorian

[quote="Stargirl240"]Which state? Indiana?  :o[/quote]
lol, no alaska

[quote="indianamath"]Guam was like 10 probably. lol :rotfl:[/quote]

dude come on have some respect
don't say stuff like that

wow, it's still quoted

...

and you still said it was lower than 24... :rotfl:

jorian

i wonder which state had the lowest cutoff score.  it was probably alaska.  some dude said that their cutoff was really like a 24

[quote="zserf"]4th got 24[/quote]
is it true that Alaska doesn't have pins this year?

speaking of pins, what do the GA pins look like this year?
haha our cutoff was like a 25.

I can tell you that Georgia's cutoff is much higher than 25.

academicsrule, do you read your private messages?


Prove that in any triangle  $\text{ ABC}$  the following relation:
\[\boxed{\text{m}(\hat{\text{A}})=2\sin^{-1}\lambda\Leftrightarrow\text{OI}^2=(2\lambda-1)^2R^2\sin^2\left(\frac 14\left|\text{B-C}\right|\right);\\
\quad\\
\lambda\in(0,1)}\]
is satisfied.

[quote="erdos"]Ehhh.... Sorry but what do you mean by ${m}(\hat{\text{A}})$ ??  :?[/quote]
\[\text{m}(\hat{\text{A}})=\text{measure of the angle}\,\hat{\text{A}}\].

Nene, this doesn't seem right.  Could you clarify or correct?


A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex W (I don't have picture for this)

A. $\displaystyle \frac{1}{12}$
B. $\displaystyle \frac{1}{9}$
C. $\displaystyle \frac{1}{8}$
D. $\displaystyle \frac{1}{6}$
E. $\displaystyle \frac{1}{4}$

hmm.. it would be nice if someone could put the picture up. :D

Heheh... if I remember correctly, the answer is:

[hide="Answer"]$\frac{1}{12}\Rightarrow \boxed{A}$[/hide]

[quote="chess64"]I posted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=unit+cube&t=57322]here[/url] already with the diagram. :)[/quote]

I searched for that thread because I remember you posting it. But you never sent me this link so..

Anyway, fixed.


Find the number of subsets $X$ of $\{ 1,2,...,10 \}$ such that $X$ contains at least two elements and such that no two elements of $X$ differ by $1$.

If there are $n$ elements of the set, and $x$ elements to be chosen (none of which are consecutive numbers), the number of ways they can be arranged is if we take the remaining $n-x$ numbers, and arrange them like this:

$| | | | \cdots | | |$.

The blank spaces is where we can put the $x$ elements, and since there are $n-x+1$ blank spaces, the number of ways is $\binom{n-x+1}{x}$.

Using this formula, we deduce that the answer to this problem is:

$\binom{9}{2}+\binom{8}{3}+\binom{7}{4}+\binom{6}{5}=133$

BTW, if I might be completely wrong since i suck at counting and whatnot...


What is the largest power of $ 30$ which divides $ 1000!$ evenly ?

[hide="Solution"]
$ 30 \equal{} 5\cdot 3\cdot 2$

So we find the number of $ 5$'s that appears in $ 1000!$.

$ \left\lfloor \frac {1000}{5} \right\rfloor \equal{} 200 \\
\left\lfloor \frac {200}{5} \right\rfloor \equal{} 40 \\
\left\lfloor \frac {40}{5} \right\rfloor \equal{} 8 \\
\left\lfloor \frac {8}{5} \right\rfloor \equal{} 1$

We don't need to find the number of powers of $ 2$ or $ 3$ because they are definitely going to have more.

Thus the largest power of $ 30$ that divides $ 1000!$ is $ 30^{249}$ so the answer is $ \fbox{249}$.
[/hide]

[quote="sunehra"]So we find the number of $ 5$'s that appears in $ 1000!$.[/quote]

Would you explain why we do this, and why your method works?

$ 30\equal{}5\cdot3\cdot2$.
Now, numbers containing $ 5$ as a factor are less frequent than those containing $ 3$ or $ 2$ as a factor, obviously, so finding the maximum power of $ 5$ which divides $ 1000!$ is analogous to finding the maximum power of $ 30$.

I'm not sure how his method for finding the number of $ 5$s works, though.

To find the number of $ 5$'s, he first finds all multiples of $ 5$ by doing $ \left \lfloor \frac{1000}{5} \right \rfloor$.

But multiples of $ 25$ still have more, so we do $ \left \lfloor \frac{1000}{25} \right \rfloor$ to weed those out.

Etc.

[quote="grn_trtle"]$ 30 \equal{} 5\cdot3\cdot2$.
Now, numbers containing $ 5$ as a factor are less frequent than those containing $ 3$ or $ 2$ as a factor, obviously, so finding the maximum power of $ 5$ which divides $ 1000!$ is analogous to finding the maximum power of $ 30$.

I'm not sure how his method for finding the number of $ 5$s works, though.[/quote]

$ \frac{200}{5}\equal{}\frac{1000}{5^2}$

$ \frac{40}{5}\equal{}\frac{1000}{5^3}$

$ \frac85\equal{}\frac{1000}{5^4}$

In general, number of 5's in 1000! is $ \sum_{j\equal{}1}^\infty\left\lfloor\frac{1000}{5^j}\right\rfloor$. Note that it 'looks like' an infinite sum but the terms eventually collapse to zero.

[b]Mini-challenge[/b]: Investigate the largest positive integer $ N$ for which $ n^N$ divides $ m!$, $ n\le m$.

If $ a$ is the largest prime factor of $ n$, $ N\equal{}\sum_{k\equal{}1}^{\infty}\left\lfloor\frac{m}{a^k}\right\rfloor$.

So the largest power of 24 that divides $ 24! \equal{} 2^{22}\cdot 3^{10}\cdots$ is what?  :wink:  (In fact, the challenge is not so "mini.")

I believe the difficulty goes away if you replace "prime factor" with "prime power factor."

Well, closing in, but not there yet.  (The largest power of 63 that divides $ 63! \equal{} 7^{10} \cdot 9^{15} \cdots$ is determined by the 7s, not the 9s, for example.)  The "tough" example to understand here is 12.  I feel like I've raised this comment many times on the fora without anyone ever following through.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193286 and the links therein for other such places.  (I found that by searching for posts by me that included the string "12".)

Fair enough.

[hide="Discussion"] Let $ v_p(m)$ denote the power of $ p$ that divides $ m$ and let $ s_p(m)$ denote the sum of the digits of $ m$ in base $ p$.

[b]Lemma:[/b]  $ v_p(m!) \equal{} \frac {m \minus{} s_p(m)}{p \minus{} 1}$.  (The proof is left as an exercise.  One way is to consider how $ v_p(m!)$ would actually be computed in base $ p$, and the other is from the other formula.)

[b]Corollary:[/b]  $ 0 < \frac {m}{p \minus{} 1} \minus{} v_p(m!) \le \log_p(m)$.

The largest power of $ n$ that divides $ m!$ is clearly

$ \text{min}_{p | n} \frac {v_p(m!)}{v_p(n)}$

and we'd like to be able to approximate this by

$ \text{min}_{p | n} \frac {m}{(p \minus{} 1) v_p(n)}$.

This approximation is fine whenever $ \frac {m}{p \minus{} 1}$ is large compared to $ \log_p(m)$, which is pretty much all the time if $ p \ll m$.  In the squarefree case that $ v_p(n) \equal{} 1 \forall p$, this is equivalent to maximizing $ p \minus{} 1$, which is equivalent to maximizing $ p$.  It should "usually" be good enough to maximize $ (p \minus{} 1) v_p(n)$:  in the $ 63$ example, $ (7 \minus{} 1) v_7(63) > (3 \minus{} 1) v_3(63)$, so this is fine.

In certain cases, there are collisions.  For example, in the $ 12$ case $ (3 \minus{} 1) v_3(12) \equal{} (2 \minus{} 1) v_2(12)$, so this is not enough.  The difference is governed by the value of $ s_3(m) \minus{} s_2(m)$.  For example, when $ m \equal{} 12$

$ s_3(12) \minus{} s_2(12) \equal{} s_3(110_3) \minus{} s_2(1100_4) \equal{} 0$

which indicates that the answer should be the same regardless of whether $ 3$ or $ 2$ is chosen.  And indeed, $ 12! \equal{} 2^{10} 3^5 ...$.  Generally, in a collision we want to maximize the value of $ s_p(m)$. [/hide]


trigonometry
vector
parameterization
calculus
calculus computations

how would you find the parametric equations of the curve on which the centre of a wheel radius one is moving on top of the curve y=sinx?

You can use:

1. If $ A$ is on a circle with center $ O$ and $ \ell$ is the line tangent to the circle at $ A$, then $ \overline{OA} \perp \ell$

2. The circle will be tangent to the sine wave at the point of contact.

yes someone told me this..but im completly lost. what do i do exactly??

I have a solution but it uses vectors. How much do you know about vectors? For instance, do you know how to visualize the sum of two vectors, and do you know how to normalize a vector (i.e. rescale it to length one)?

[hide]Answer (solution to come later):

$ (x(t),y(t)) \equal{} (t, \sin t) \plus{} \frac{(\minus{}\cos t,1)}{\sqrt{\cos^2 t \plus{} 1}} \equal{} \left(t \minus{} \frac {\cos t}{\sqrt {\cos^2 t \plus{} 1}}, \sin t \plus{} \frac {1}{\sqrt {\cos^2 t \plus{} 1}}\right)$[/hide]

yes i do know that much about vectors; but i dont know how to arrive at that answer

Note: Gauss202's formula is not correct. $ \sqrt{\cos^2t\plus{}1}$ is the differential element of arclength, not arclength itself. In fact, to create a formula for this curve, you must create a formula for the arclength along the sine curve. Since we cannot do that in closed form, we also cannot write the equations for the cycloid-like curve in closed form.

I think my equation does work, it's just not parameterized with respect to arclength. But he didn't say he needed an equation parameterized by arclength - just an equation that "traces the right path".

Now, if he had asked for the parameterization of the curve traced by a point [i]on[/i] the circle (instead of the center of the circle) as it rolls along the curve $ y \equal{} \sin x$ - then that would have been a much different question, and I probably would have to parameterize it according to the arclength of $ y \equal{} \sin x$.

Oh - I'm sorry, I thought it was the case of the wheel rolling on top of the curve. Now that I re-read the problem, I see that it is ill-defined. How fast is the circle rotating? That's not given.

Since it's talking about the center of the wheel, I don't think it matters how fast the wheel is rotating (or even if it's rotating at all). The center will trace the same parametric curve regardless.


how can you construct a triangle when the lengths of 2 of the medians and an altitude is given?

Can you be specific about which altitude.

I can construct a triangle, given: $ h_a, h_b , m_a$
[hide="Solution"]
Let(according to the figure) $ h_a, h_b, m_a \equal{} AK, BL, AM$ respectively. Drawing $ MN\parallel BL$ such that N is on AC; we see that $ MN \equal{} \frac {1}{2}BL \equal{} \frac {1}{2}h_b$. So we construct the triangle in the following steps:
1. Construct $ \triangle AKM$ from $ AK \minus{} h_a, AM \equal{} m_a, \angle AKM \equal{} \frac {\pi}{2}$
2. With M as centre & AM as diameter we draw a semicircle outwardly.
3. With M as centre we draw an arc of radius $ h_b/2$ that meets the semicircle at point N.
4. Joining AN so that it meets KM at C, we choose a point on MK satisfying MB=MC.
5. Joining AC yields the desired triangle ABC
 [geogebra]50a6ceca8bf1b972e4332cc85731e6fcff4b4420[/geogebra]
[/hide]
Hence proved. :)

You must have Java 6 (Sun microsystems) installed in your computer ;) , because geagabra is based on Java.
For downloading java, go to http://java.sun.com/javase/downloads/index.jsp

Good solution Potla
but is it possible when given only $ m_a, m_b$ and $ h_c$ ?


How would you  read something like $ A'$?

it has different meanings

it may also mean complement of set $ A$

[quote="fishythefish"]Complements are written the same way as conjugates, repeating decimals, and line segments: with a bar on top.

The complement of set A would actually be written as $ \bar{A}$[/quote]
Someone suggests to write $ \overline{A}$ (with \overline) instead of $ \bar{A}$ (with \bar) for complements and for any other "operators" too. \bar is used for "variables" name. It's a matter of taste, anyway.
Bye, Gil

Yeah, I wasn't looking at all the options. I was primarily concerned with just finding the code for a bar. But I'll probably use \overline next time.


modular arithmetic
Diophantine Equations

Solve the equation $28^x =19^y +87^z$, where $x, y, z$ are integers.

I assume that the equation was meant for non-negative integers (for negatives it's nothing interesting).

It's clear that $ x \geq 1$ and $ z \geq 2$ (mod $ 9$, keep in mind that $ 87 \equal{} 3 \cdot 29$). Suppose now that $ y \geq 1$ and $ z \geq 3$.

From the binomial expansion it easily follows that $ 27|(9k\plus{}1)^l \minus{} 1$ iff $ 3|l$. Therefore we must have $ 3|y$ since $ 19 \equal{} 18 \plus{} 1$, $ 28\equal{}27\plus{}1$ and $ 27|87^z$.

Now we look mod $ 29$. Because $ 29|87$ we have the congruence $ (\minus{}1)^x \equiv 19^y \pmod{29}$. It's easy to check that $ 19^y \equiv 1 \pmod{29}$ is equivalent to $ 28|y$ (we have to try $ y\equal{}2, 4, 7, 14$) and it follows that in order to have $ 19^y \equiv \pm 1 \pmod{29}$ we must have $ y \equiv 0 \pmod{28}$ or $ y \equiv 14 \pmod{28}$, anyway we have $ 14|y$.

So we have obtained that $ 42|y$. In particular $ 19^y \equiv 1 \pmod{7}$. $ 87 \equiv 3 \pmod{7}$ and we check that $ 3^z \equiv \minus{}1 \pmod{7}$ is equivalent $ z \equiv 3 \pmod{6}$.

Now mod $ 13$. We have $ 2^x \equiv 6^y \plus{} 9^z \equiv 6^y \plus{} 3^{2z} \pmod{13}$. Since $ 3|z$ and $ 3^3 \equiv 1 \pmod{13}$ it follows that $ 3^{2z} \equiv 1 \pmod{13}$. There are two possibilites: $ y \equiv 6 \pmod{12}$ or $ y \equiv 0 \pmod{12}$. In first case we get that the right side is divisible by $ 13$ which is impossible, and in the second we have $ 2^x \equiv 2 \pmod{13}$ and this is equivalent to $ x \equiv 1 \pmod{12}$.

We finish the solution with mod $ 19$. Since $ 3|z$ and $ 11^3 \equiv 1 \pmod{19}$ we have that $ 87^z \equiv 11^z \equiv 1 \pmod{19}$. It follows that $ 28^x \equiv 1 \pmod{19}$. But since $ x \equiv 1 \pmod{6}$ it follows that $ 28^x \in \{28, 28^7, 28^{13}\} \pmod{19}$ but this numbers give residues: $ 9, 4, 6$ which is a contradiction.

We are left with the cases $ y\equal{}1$ and $ z\equal{}2$. Let's start with the first one.
From the above reasoning the relation $ 3|z$ still holds (mod $ 7$). So let $ z\equal{}3k$, then
$ 28^x \equal{} 87^{3k}\plus{}1 \equal{} (87^{k}\plus{}1)(87^{2k} \minus{} 87^{k} \plus{} 1)$.
It's easy to see that $ 87^{k} \plus{} 1$ and $ 87^{2k}\minus{}87^{k}\plus{}1$ are coprime, so
$ 87^{k} \plus{} 1 \equal{} 4^x$ and $ 87^{2k} \minus{} 87^{k}\plus{}1 \equal{} 7^x$
because the second number is bigger. But:
$ 87^{2k} \minus{} 87^{k} \plus{} 1 \equal{} (4^x \minus{} 1)^2 \minus{} (4^x\minus{}1) \plus{} 1 \equal{} 4^{2x} \minus{} 3 \cdot 4^x \plus{} 3 \equal{} 7^x$,
which is a contradiction mod $ 8$, since obviously $ x \geq 2$.

Now consider the case $ z\equal{}2$. We still have that $ 14|y$ (mod $ 29$), so in particular $ 2|y$. But that's a contradiction mod $ 4$ since the right side is congruent to $ 2 \pmod{4}$.

So the equation doesn't have a solution.

[quote="maxal"]Actually, the equation $ 28^x \equal{} 19^y \plus{} 87^z$ is not solvable modulo $ 889$.[/quote]
Mmm. That is quite interesting. How did you work it out? Only trial? I don't think, there must be something tricky.

[quote=maxal]Actually, the equation $ 28^x \equal{} 19^y \plus{} 87^z$ is not solvable modulo $ 889$.[/quote]

Could you tell me how in the world did you manage to work that out, please ?


analytic geometry
geometry
algebra
polynomial
3D geometry
sphere
tetrahedron

Behold the final problems of the 60th Polish Mathematical Olympiad.
-----------------------------------------
DAY $ |e^{\pi i}|$

[b]Problem 1.[/b]
Each vertex of a convex hexagon is the center of a circle whose radius is equal to the shorter side of the hexagon that contains the vertex. Show that if the intersection of all six circles (including their boundaries) is not empty, then the hexagon is regular.

[b]Problem 2.[/b]
Let $ S$ be a set of all points of a plane whose coordinates are integers. Find the smallest positive integer $ k$ for which there exists a 60-elemen subset of set $ S$ with the folloiwing condition satisfied: for any two elements $ A,B$ of the subset there exists a point $ C$ contained in $ S$ such that the area of triangle $ ABC$ is equal to $ k$.

[b]Problem 3.[/b]
Let $ P,Q,R$ be polynomials of degree at least $ 1$ with integer coefficients such that for any real number $ x$ holds: $ P(Q(x)) \equal{} Q(R(x)) \equal{} R(P(x))$. Show that polynomials $ P,Q,R$ are equal.

-----------------------------------------

DAY $ \sum_{k \equal{} 0}^{\infty}\frac {1}{2^k}$

[b]Problem 4.[/b]
Let $ x_1, x_2, ..., x_n$ be non-negative numbers whose sum is $ 1$. Show that there exist numbers $ a_1, a_2, ..., a_n$ chosen from amongst $ 0,1,2,3,4$ such that $ a_1, a_2, ..., a_n$ are different from $ 2,2,...,2$ and $ 2 \le a_1 x_1 \plus{} a_2 x_2 \plus{} ... \plus{} a_n x_n \le 2 \plus{} \frac {2}{3^n \minus{} 1}.$

[b]Problem 5.[/b]
A sphere is inscribed in tetrahedron $ ABCD$ and tangent to faces $ BCD,ACD,ABD,ABC$ at points $ P,Q,R,S$ respectively. Segment $ PT$ is the sphere's diameter, and lines $ TA,TQ,TR,TS$ meet the plane $ BCD$ at points $ A',Q',R',S'$ respectively. Show that $ A'$ is the center of a circumcircle on the triangle $ Q'R'S'$.

[b]Problem 6.[/b]
Let $ n$ be a natural number equal or greater than $ 3$. A sequence of non-negative numbers $ (c_0,c_1,c_2,...,c_n)$ satisfies the condition: $ c_p c_s \plus{} c_r c_t \equal{} c_{p \plus{} r} c_{r \plus{} s}$ for all non-negative $ p,r,s,t$ such that $ p \plus{} r \plus{} s \plus{} t \equal{} n$. Determine all possible values of $ c_2$ when $ c_1 \equal{} 1$.

Each problem shell be posted in different topic first.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1672082#1672082
Problem 2: http://www.mathlinks.ro/viewtopic.php?t=309750
Problem 3: http://www.mathlinks.ro/viewtopic.php?t=309751
Problem 4: http://www.mathlinks.ro/viewtopic.php?t=309751
Problem 5: http://www.mathlinks.ro/viewtopic.php?p=1672093#1672093
Problem 6: http://www.mathlinks.ro/viewtopic.php?t=309759


function
calculus
derivative
trigonometry
real analysis
real analysis solved

$f$ has intermediate value property on $[a,b]$. Does $f$ necessarily to have a primitive?

Definitely not. You can construct a function which takes all the real values in every interval (call this property $P$). However, a derivative is the pointwise limit of a sequence of continuous functions on $[a,b]$, and such a function has a dense set of continuity points (you've posted this as an unsolved problem; the proof is rather nasty, so I don't think I'll be posting it here :)). A function with property $P$ has no continuity points, so it cannot be a derivative.

[quote="grobber"]However, a derivative is the pointwise limit of a sequence of continuous functions on $[a,b]$, and such a function has a dense set of continuity points (you've posted this as an unsolved problem; the proof is rather nasty, so I don't think I'll be posting it here ). [/quote]
Did I?  :?
Can you help me find it?
I only know that the pointwise limit of a sequence of continuous functions has a $G_\delta$ set of continuity.

Actually, you did not post the exact problem I mentioned, but [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=f%5C%27&t=14527]this one[/url]. This also suffices, and the one I mentioned is also mentioned by Harazi in that topic.

Actually, I remember an old thread from sci.math where someone asked if the intermediate value property implies measurability. I think the answer is negative, and it would solve this problem, since any derivative is measurable.

Let $(a_i)_{i\in (0,1)}\cup\{1\}$ be a basis of $\mathbb R$ over $\mathbb Q$. Also, let $S$ be a subset of $[0,1]$ containing exactly one element in each equivalence class corresponding to the relation $x\equiv y\iff x-y\in\mathbb Q$. Put $f(x)=1,\ \forall x\in S$. After this, put $f(x)=i$ whenever $x\in[0,1]\setminus S$ and $x$ is of the form $r_1+r_2a_i$, where $r_1,r_2\in\mathbb Q$, and $a_i$ is some irrational in our basis $(a_i)$, and give $f$ any value in $(0,1)$ in all other points.

I may be wrong, but this looks like it has the Darboux property (it takes all the values $(0,1)$ in all the intervals, and it takes the value $1$ in some points, but this does not alter the intermediate value property). However, $f^{-1}(1)$ is a non-measurable set $S$ (considering $S$ is the first step in the usual proof that there exist non-measurable real sets), meaning that $f$ is not measurable.

Think it's correct? :? I bet I'm missing something here :).

I think that's right, but it can be even simpler: The sets $a+\mathbb{Q}$ are countable, so when we write $\mathbb{R}$ as a disjoint union of them, we use $c$ such sets. Find a bijection $g$ from such a collection to $\mathbb{R}$, and let our function $f(x)$ be $g(\pi(x))$, where $\pi$ is projection onto $\mathbb{R}/\mathbb{Q}.$ Since $f^{-1}(a)$ is not measurable for any $a$, $f$ is not measurable, but since $\mathbb{Q}$ is dense, $f$ takes on all real values in any interval.

Yeah, it looks like I made it more complicated than it had to be :).

However, to answer liyi's question, we need to go through a lot less trouble: it suffices to find two functions with the intermediate value property whose sum does not have the intermediate value property. We may take, for instance, $f$ to be $\sin \frac 1x$ on $(0,1]$ and $0$ at $0$, while $g$ is $-\sin\frac 1x$ on $(0,1]$ and $1$ at $0$. Their sum is $0$ on $(0,1]$ and $1$ at $0$.

Well its not necessary consider the function f on [0,1] by the following way:

f(x)=x if x is rational & f(x)=1-x if x is irrational.You can check that f has IVT property but f is not differentiable. :lol:

[quote="jmerry"]No, the game, that doesn't work. $f(\frac12)=\frac12$ and $f(1)=1,$ but the only $x$ for which $f(x)=\sqrt{\frac12}$ is $1-\sqrt{\frac12},$ which is not in the interval $\left[\frac12,1\right].$[/quote]
I am sorry that I couldnt see your point.Its true that the above mentioned no donot belong to [.5,1].But what I am saying is that the range of f is [0,1].
Could you please clarify..

The property in question: for any interval $I$, $f(I)$ is an interval. Your example clearly does not have that property.

Hmmm I understand what you are trying to say.


Set up an equation, setting the sum of all of the sides equal to the perimeter (since that's the definition of a perimeter of a polygon):
$ x\plus{}(x\plus{}1)\plus{}17\plus{}8\plus{}13\plus{}13\plus{}15\equal{}95$
$ 2x\plus{}67\equal{}95$
$ 2x\equal{}28$
$ x\equal{}\fbox{14}$


[quote="Pistachio"](x^2-24  )^2     ( x^2-24)
 (----------)      -3(---------)  -10=0
 (      x     )         (    x     )

the dashed lines are supposed to mean they are fractions. The parenthesis is the whole thing is in parenthesis the first is squared minus the next one.

sorry if it looks confusing the way i wrote it.
Also for posting another problem yesterday.[/quote]
$\frac{\left(x^{2}-24\right)^{2}}{x}-\frac{3\left(x^{2}-24\right)}{x}-10=0$

Is that how it is supposed to look like?

Can you clarify? I_like_pie's interpretation evaluates to
$x^{4}-51x^{2}-10x =-648$
which is a quartic polynomial.
Normally, we could solve this as a polynomial of quadratic type, but  because we have a $-10x$ instead of a $-10$, substitution with a dummy variable won't work.

Learn how to use latex. You can type [code](dollarsign) latex goes here (dollarsign) [/code]
using actual dollar signs without parentheses around them.
type this
RIGHT INTO THE POST
and the text comes out looking nice and formatted.

Oh, and use \frac {numerator}{denominator}
That's a backslash. That allows you to see nice looking fractions.
try \frac {3}{5}
That comes out looking like $\frac{3}{5}$

Since there are parentheses around the denominators too, I'd bet the problem is

$\left(\frac{x^{2}-24}{x}\right)^{2}-3\left(\frac{x^{2}-24}{x}\right)-10=0$

which is quite solvable.

[hide="Hint and answer"]Substitute $y=\frac{x^{2}-24}{x}$. Solutions: $x=-6,-3,4,8$[/hide]

EDIT you can delete this post it was on accident

$\left(\frac{x^{2}-24}{x}\right)^{2}-3\left(\frac{x^{2}-24}{x}\right)-10=0$

this is what it looks like sorry guys im new here

i substituted and got factored to get y=5 and y=-2  how did you get four answers?

[quote="Pistachio"]i substituted and got factored to get y=5 and y=-2  how did you get four answers?[/quote]

Now solve ${x^{2}-24\over x}=5$ and ${x^{2}-24\over x}=-2$

This is something called an equation of quadratic type. You'll see this a lot. Effectively, you can substitute a dummy variable.

Farenhajt let y represent the quantity $\left(\frac{x^{2}-24}{x}\right)$
This lets you solve the equation as

$y^{2}-3y-10 = 0$
This factors:
$(y-5)(y+2) = 0$
So our values for y can be 5 or -2.

Now, remember we are trying to solve for x.
Remember our definition of y:
$y=\left(\frac{x^{2}-24}{x}\right)$
use the values 5 and -2 in the place of y, and solve that equation.

You get two values from each quadratic, and we have two y values to test. Therefore you should end up with four solutions.

EDIT: Beat by a second.  :o

And that square in the definition of $y$ has to go as well :)


factorial
number theory proposed
number theory

If $ n$ is a positive integer define $ n!!$ to be $ n!(n\minus{}1)!...1!$. Show that for any $ n\plus{}1$ distinct integers $ \{a_1,a_2,...a_{n\plus{}1}\}$ then
$ (n\minus{}1)!!|\prod(a_i\minus{}a_j)$ where $ i$ is never equal to $ j$

[quote="Ahwingsecretagent"]If $ n$ is a positive integer define $ n!!$ to be $ n!(n \minus{} 1)!...1!$. Show that for any $ n \plus{} 1$ distinct integers $ \{a_1,a_2,...a_{n \plus{} 1}\}$ then
$ (n \minus{} 1)!!|\prod(a_i \minus{} a_j)$ where $ i$ is never equal to $ j$[/quote]

$ (n \minus{} 1)!!\equal{}(n\minus{}1)!(n \minus{} 2)!...1!\equal{}(n\minus{}1)(n\minus{}2)^2(n\minus{}3)^3\cdots2^{n\minus{}2}1^{n\minus{}1}$

Considering mod $ n\minus{}k$, then there are only $ n\minus{}k$ possible residues then there are at least $ k$ pairs with the same residue then $ (n\minus{}k)^k|\prod(a_i \minus{} a_j)$, Actually it is true with $ n$ distinct integer, isn't it? or $ n!!|\prod(a_i \minus{} a_j)$?

[quote="mszew"][quote="Ahwingsecretagent"]If $ n$ is a positive integer define $ n!!$ to be $ n!(n \minus{} 1)!...1!$. Show that for any $ n \plus{} 1$ distinct integers $ \{a_1,a_2,...a_{n \plus{} 1}\}$ then
$ (n \minus{} 1)!!|\prod(a_i \minus{} a_j)$ where $ i$ is never equal to $ j$[/quote]

$ (n \minus{} 1)!! \equal{} (n \minus{} 1)!(n \minus{} 2)!...1! \equal{} (n \minus{} 1)(n \minus{} 2)^2(n \minus{} 3)^3\cdots2^{n \minus{} 2}1^{n \minus{} 1}$

Considering mod $ n \minus{} k$, then there are only $ n \minus{} k$ possible residues then there are at least $ k$ pairs with the same residue then $ (n \minus{} k)^k|\prod(a_i \minus{} a_j)$, Actually it is true with $ n$ distinct integer, isn't it? or $ n!!|\prod(a_i \minus{} a_j)$?[/quote]

Yes it is meant to be $ n$ instead of $ n\plus{}1$.
However I don't think $ (n\minus{}k)^k|\prod(a_i\minus{}a_j)$ for all $ k<n$ implies $ (n\minus{}1)(n\minus{}2)^2...|\prod(a_i\minus{}a_j)$ since they aren't coprime.


calculus
derivative
algebra unsolved
algebra

Which is a critical point of 10x3 - ax 2 +1?

what is a critical point? please explain to me

[quote="Bictor717"]Critical points are all the points at which relative extrema can occur. Set the first derivative to 0 and the solutions for x are the critical points. Values of x that give an undefined value for the derivative are also critical points.

This is elementary calculus, but based on your posts, it doesn't look like you're up to that level. Don't try to learn on your own by jumping into problems. There is no reason to attempt to do stuff that is way out of your level. Master what you're learning now and take it slowly. No one understands everything immediately.[/quote]

not in azery in turkish

turevi al a nin maximum giymetlerini yaz i vse


I've worked on this problem for about half an hour but the answer still elude me.

If $ 60^{a} = 3$ and $ 60^{b}= 5$, then find $ 12^{\left[\frac{(1-a-b)}{2(1-b)}\right]}$

 I rewrote the given information in terms of logarithms and did a great deal of manipulation. However, I am stuck on simplifying $ \left(\frac{\log_{60}4}{\log_{60}144}\right)$.

Please post only HINTS

Thanks in Advance

Remember that $ \frac{\log_a b}{\log_a c}\equal{}\log_c b$ and that $ a^{\log_ab}\equal{}b$.

[quote="mathemagician1729"]I've worked on this problem for about half an hour but the answer still elude me.

If $ 60^{a} = 3$ and $ 60^{b} = 5$, then find $ 12^{\left[\frac {(1 - a - b)}{2(1 - b)}\right]}$

 I rewrote the given information in terms of logarithms and did a great deal of manipulation. However, I am stuck on simplifying $ \left(\frac {\log_{60}4}{\log_{60}144}\right)$.

Please post only HINTS

Thanks in Advance[/quote]

Also note that 4=2^2 and 144=12^2

Ah...

[hide="so doing what you guys suggested"] taking the exponents will reduce this to $ \left(\frac{\log_{60}2}{\log_{60}12}\right)$ Now we can use the change of base property backwards, yielding $ \log_{12}2$. I forget to mention that I was working solely on the exponent so we must now add the base of 12 or just $ 12^{\log_{12}2}$. Letting $ x \equal{} \log_{12}2$, our original expression become $ 12^{x}$. Since we know from the log, that 12 raised to the power of $ x$ equals two, we can reuse that fact here. The answer is thus $ 2$.
[/hide]


topology
geometry
real analysis
vector
advanced fields
advanced fields unsolved

Hi. I'm asking for your help as I don't make progress on a problem in smooth manifold theory. In the chapter on Sard's theorem in Bredon's book "Toppology and Geometry" he poses the following problem:

If $C$ is a circle smoothly embedded in $\mathbb{R}^{4}$, show that there exists a three-dimensional hyperplane $H$ such that orthogonal projection of $C$ to $H$ is an embedding.

As tools for a proof one can use the theorems of Sard and Brown:

Let $M^{m}$ and $N^{n}$ be smooth manifolds of dimension $m$ and $n$ respectively.

Sard's Theorem: If $\phi : M^{m}\to \mathbb{R}^{n}$ is a smooth map then the set of critical values of $\phi$ has Lebesgue measure zero.

Brown's Theorem: The set of regular values of a smooth map $\phi : M^{m}\to N^{n}$ is everywhere dense in $N^{n}$.

For a more detailed formulation of Sard's theorem one can look at http://planetmath.org/encyclopedia/SardsTheorem.html .

Please help me!

Let $f: S^{1}\rightarrow R^{4}$ be the embedded circle. Projection on the hyperplane orthogonal to a vector $v$ would not work only if there are two points on the circle that differ by $v$. So define $F: S^{1}\times S^{1}\rightarrow R^{4}$ as: \[F(x,y)=f(x)-f(y)\] This map is smooth but not onto (use the theorems you mentioned) and so one needs to take $v$ outside the image of $F$.


I was reading a while ago but then I suddenly had this really weird question. Consider the following:

$ KNO_{3}_{(aq)}+NaCl_{(aq)}\rightarrow KCl_{(aq)}+NaNO_{3}_{(aq)}$ clearly the ionic equation shows that no reaction has actually taken place.

So my question is that if you evaporate a water solution containing potassium nitrate and sodium chloride what would you get? What's on the left hand side of the reaction, right hand side of the reaction or a mixture of both or what?

[hide="Opinion"]When you dissolve potassium nitrate and sodium chloride in water you will get an aqueous solution composed by water molecules, hydrated potassium cations, hydrated nitrate anions, hydrated sodium cations, and hydrated chloride anions. There is no reaction because all possible ionic compounds are water soluble. When you evaporate the water of this solution, the most stable ionic compounds will crystalize from the solution, and those will be the ones with the highest lattice enthalpies, that is, composed by the smallest and/or highest charged ions. [/hide]

You may get maximum 4 kinds of salt (? not sure what it called in English)
$ KCL$, $ NaCL$, $ KNO_{3}$, and $ NaNO_{3}$.
It also can be 3 of them (it's up to the quantities given)


Let be given a quadrilateral $ABCD$ inscribed in a cricle with center $O$, radius $R$. The lines $AB,CD$ intesect at $P$, the lines $AD,BC$ intesect at $Q$. Prove that $\overrightarrow{OP}\cdot \overrightarrow{OQ}=R^{2}$.

Let circumcircle of triangle $PBC$ intersects $PQ$ at $L$.Then
$\angle CLQ=\angle PBC=\angle ADC \Longrightarrow Q,D,C,L$ cyclic
$\Longrightarrow QO^{2}-R^{2}=QC.QB=QL.QP$ and $PO^{2}-R^{2}=PC.PD=PL.QP$
Adding these two we get $QO^{2}+PO^{2}-2R^{2}=QP^{2}$
$\Longrightarrow \overrightarrow{OP}. \overrightarrow{OQ}=QO.PO.cos \angle QOP =\frac{QO^{2}+PO^{2}-QP^{2}}{2}=R^{2}$


In a class of $ n\geq 4$ some students are friends. In this class any $ n \minus{} 1$ students can be seated in a round table such that every student is sitting next to a friend of him in both sides, but $ n$ students can not be seated in that way. Prove that the minimum value of $ n$ is $ 10$.

Why nobody is interested? I think this is a nice problem. It doesn't involve a special idea I know, but it is enjoying to deal with cases.

Nice problem!
Solution:

Case 1:$ n\le 6$
If n satisfies the conditon,
Because every $ n\minus{}1$ students forms a circle, so each student has at least $ 2$ friends. In addition if a student has only $ 2$ friends, then consider the $ n\minus{}1$ students that only includes $ 1$ friends of him/her, the $ n\minus{}1$ students cannot form a circle, a contradiction. Therefore each student has at least $ 3$ friends.
Choose a student arbitrarily and call him/her $ A$, the other $ n\minus{}1$ students forms a circle. Since $ n\minus{}1\le 5$ and $ A$ has at least $ 3$ friends,  $ A$ must have $ 2$ friends who are adjacent on the circle, and then we can add $ A$ to the circle i.e. we get a circle of the all $ n$ students, a contradiction. Case 1 is done.

Case 2:$ n\equal{}7$
If $ n\equal{}7$ satisfies the conditon,
The same as in Case 1 we have that each student has at least $ 3$ friends. But $ 7$ is an odd number, so there exist a student who has at least $ 4$ friends. Call him $ B$. Consider the circle formed by other $ 6$ students,.Since $ B$ has at least $ 4$ friends, he must have $ 2$ friends who are adjacent on the circle, and then we can add $ B$ to the circle i.e. we get a circle of the all $ 7$ students, a contradiction. Case 2 is done.

Case 3:$ n\equal{}8$
If $ n\equal{}8$ satisfies the conditon,
The same as in Case 1 we have that each student has at least $ 3$ friends. Choose a student arbitrarily and call him/her $ C$, then $ C$ has at least $ 3$ friends. Due to the condition the other $ 7$ students form a circle.Assume these students are $ 1,2,3,4,5,6,7$, respectively. And they are the same order on the circle. If $ C$ has at least $ 4$ friends, then there will be $ 2$ of them who are adjacent on the circle, and then we can add $ C$ to the circle i.e. we get a circle of the all $ 8$ students, a contradiction. So $ C$ has exactly $ 3$ friends. Since $ C$ is chosen arbitrarily, each student has exactly $ 3$ friends.

w.l.o.g we can assume $ C$'s friends are $ 2,4,6$, then $ C,2,4,6$ have no more friends, and $ 1,3,5,7$ each has one more friend. Because $ 1$ and $ 7$ are already friends, so there are only two possible cases:

(1) $ 1$ and $ 5$, $ 3$ and $ 7$ are friends respectively. But then we have a circle of all $ 8$ students: $ 6\minus{}C\minus{}2\minus{}3\minus{}4\minus{}5\minus{}1\minus{}7\minus{}6$, a contradiction.
(2) $ 1$ and $ 3$, $ 5$ and $ 7$ are friends respectively. But then we have a circle of all $ 8$ students: $ 6\minus{}C\minus{}4\minus{}3\minus{}2\minus{}1\minus{}7\minus{}5\minus{}6$, a contradiction, too.
Case 3 is done.

Case 4:$ n\equal{}9$

If $ n\equal{}9$ satisfies the conditon,
we have that each student has at least $ 3$ friends and $ 9$ is an odd number, so there exists a student who has at least $ 4$ friends.Call him/her $ D$.
Due to the condition the other $ 8$ students form a circle.Assume these students are $ 1,2,3,4,5,6,7,8$, respectively. And they are the same order on the circle.
If $ D$ has at least $ 5$ friends, similar to previous case we get that two of them are adjacent on the circle which leads to a contradiction.
So we assume $ D$ has exactly $ 4$ friends, and w.l.o.g we assume that they are $ 1,3,5,7$, respectively. Next color students $ 1,3,5,7,D$ black and color students $ 2,4,6,8$ white. Then all known pair of friends are one black student and one white student. Take $ 5$ black students and $ 3$ white students. Due to the condition they form another circle. Since there are only $ 3$ white students on the latter circle, we have that there exists two adjacent students on this circle who are both black. w.l.o.g there are only two possible relative position of them on the former circle: (1) $ 1$ and $ 3$ ; (2) $ 1$ and $ 5$.
(1) $ 1$ and $ 3$ are friends.But then we have a circle of all $ 9$ students:$ 2\minus{}D\minus{}8\minus{}7\minus{}6\minus{}5\minus{}4\minus{}3\minus{}1\minus{}2$, a contradiction.
(2) $ 1$ and $ 5$ are friends.But then we have a circle of all $ 9$ students:$ 8\minus{}D\minus{}4\minus{}3\minus{}2\minus{}1\minus{}5\minus{}6\minus{}7\minus{}8$, a contradiction, too.
Case 4 is done.

Now we have proved that $ n$ is at least $ 10$.Finally we construct an example to show that $ n\equal{}10$ satisfies the condition.
The example is:
Ten students are $ 1,2,3,4,5,6,7,8,9,10$, respectively. $ 1\minus{}2\minus{}3\minus{}4\minus{}5\minus{}6\minus{}7\minus{}8\minus{}9\minus{}1$ is a circle. All other pairs of friends are $ 10\minus{}3$, $ 10\minus{}6$, $ 10\minus{}9$, $ 1\minus{}5$, $ 2\minus{}7$, $ 4\minus{}8$.It is not hard to check that this example satisfies the condition.

END

Cute and nice problem.
I think I found an alternative that gives more intricate details about the degrees and skips the casework mostly.
Firstly let $n \le 9$.
Let the number of edges in the graph be $e$.
Then consider the vertex $v$ with the greatest degree and the cycle of the remaining $n-1$ vertices.
We see that the our vertex cannot be connected with two adjacent vertices in the $n-1$ cycle.
Thus we see that by the pigeonhole principle, $deg(v) \le [\frac{n+1}{2}]$

Therefore, $e \le [\frac{n}{2}[\frac{n+1}{2}]]$
By considering two chosen cycles that a vertex is a part of ,we see that $\delta(G) \ge 3$.

We now claim that $G$ contains a 4-cycle.
For this consider the vertex with smallest degree $u$.
Call three of its adjacent vertices $A,B,C$.
Now, as $e \le [\frac{n}{2}[\frac{n+1}{2}]] < [\frac{n^2}{4}]$,
$G$ is triangle free.
Thus no two of $A,B,C$ are adjacent.
Now, we prove that at least two $A,B,C$ have a common neighbour apart from $u$ which will give the claim.
For this assume the contrary.
Then we would get that since their neighborhoods apart form $u$ are disjoint
$n \ge 2+2+2+4=10$ ,a contradiction.
Thus $G$ contains a $4-cycle$.
Thus, by a well known result
$e > \frac{n}{4} (1+ \sqrt{4n-3})$
However this just gives that $n$ upto $8$ dont' work.
So,we for $n=9$ we just proceed as above or can adapt an alternative since we got more structure.
This bound just implies $n=9$, then there are atleast $5$ vertices of degree $4$ which is just extra info it gives(may or may not use for the casework ).
FInish by giving a construction,which isn't too difficult.


AMC AIME MIT college USA(J)MO USAMO MATHCOUNTS

So on the MIT application, it asks for AIME scores, something that makes me happy.

However, I don't know where to go about finding said AIME scores. I've heard of some booklet that has the AIME scores of people who have qualified for USAMO. Anybody able to help me out here?

If you had qualified for USAMO, then you probably would have gotten a letter saying so. I believe the proctor/teacher/whoever you took the test with should have recieved your score. However, I don't know my score, either.

Every year, the AMC publishes an AMC 10/12 Summary book for purchase.  The last 6 years' worth have been posted online on the AMC site (see the row near the bottom called "AMC 10/12 Summary Book"):
[url]http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml.[/url]
Near the end of each book is a list of AIME scores of students who did reasonably well on the AIME, usually a score of 6 or more.

The book Ravi mentioned is fairly widely distributed; they sent me a copy, presumably because I hosted the AMC on a college campus.

If Lazarus still doesn't know, he can send me a PM with his name and his school (or location he took the AIME, if different) and I can see if he's in the book.

Try http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC1012/2009_HS_summary.pdf and then just find your name in the AIME section (assuming you got a 6+  :P )

Why did the 1994 AHSME have so many perfects? Was it really easy?

Does anyone have a copy of the 1994 AHSME?  It's not located under the contest tab.

I asked for it under the MathCounts trading forum, and will pass it on to you if I get it.


Does there exist an onto function $ f: \mathbb{R} \to \mathbb{R}$ such that $ g(x): \equal{} f(x)^2 \plus{} f(x)^4$ is monotonous?
If yes, find an example.

As $ f$ is onto, there are $ a, b$ such that $ f(a)\equal{}1, f(b)\equal{}\minus{}1$. But then $ g(a)\equal{}g(b)$, so it is surely not strictly monotonic...

[quote="ZetaX"]As $ f$ is onto, there are $ a, b$ such that $ f(a) \equal{} 1, f(b) \equal{} \minus{} 1$. But then $ g(a) \equal{} g(b)$, so it is surely not strictly monotonic...[/quote]

oh sorry the word strictly shouldn't be here..thank you ZetaX, i have edited it..


trigonometry
algebra proposed
algebra
equation

[hide=*][quote=Yuriy Solovyov]Let
\[ x = \alpha + \beta + \gamma\;\;\textbf{and}\;\;y = \alpha\beta + \beta\gamma + \gamma\alpha
\]
Alsow let $ \alpha\beta\gamma = 1.$ Put this into our equation we get
\[ \alpha^5 + \beta^5 + \gamma^5 = 6
\]

\[ (\alpha\beta)^5 + (\beta\gamma)^5 + (\gamma\alpha)^5 = 9
\]
Recall alsow that
\[ (\alpha\beta\gamma)^5 = 1
\]
Thus, $ \alpha^5,\; \beta^5$ and $ \gamma^5$ are roots of the cubic equation
\[ t^3 - 6t^2 + 9t - 1 = 0
\]
Or setting $ t = w + 2$
\[ w^3 - 3w + 1 = 0
\]
\[ w_k=2\sin\left(\frac{\pi}{18}+\frac{2\pi k}{3}\right),\;k=0,1,2\]
Return to $ t$:
\[ t_k=2+w_k=2+2\sin\left(\frac{\pi}{18}+\frac{2\pi k}{3}\right),\;k=0,1,2\]
So, our solution will be
\[ x=\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}+\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)\right)}+\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)\right)}\]
\[ y=\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)\right)}+\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)\right)}\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)\right)}+\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)\right)}\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}\][/quote]
[/hide]

[b]Let[/b] $ \frac{x^5-6}{x^2-y}=\frac{y^5-9}{y^2-x}=5(xy-1).$ [b]Then[/b]
\[ x=\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}+\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)}+\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)}\]
\[ y=\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)}+\sqrt[5]{2+2\sin\left(\frac{13\pi}{18}\right)}\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)}+\sqrt[5]{2+2\sin\left(\frac{25\pi}{18}\right)}\sqrt[5]{2+2\sin\left(\frac{\pi}{18}\right)}\]

[url]https://artofproblemsolving.com/community/c4h1322724p22425689[/url]


algebra
polynomial
function
topology
linear algebra
complex numbers
superior algebra

I don't know if this belongs here, but I was wondering: Can we find all field morphisms from C to C?

Sure, I assumed as much, but do you have any information on it? Is it a well known problem? I imagine there are people who asked themselves this. Maybe there's no known solution, or maybe it's a famous problem or stuff like that.

I know only that our teacher asked us to give an example different from the trivial ones and we couldn't. And yes, I think it is a famous problem, but I don't know if it is solved.

Did your teacher have a soln, or was he as clueless as mine? :D

The only examples I know are, of course, f(x)=x and f(x)=x' (the conjugate of x).

Maybe we can find something on the Internet.

I suppose he has an answer. I will ask him after the one week-holiday. The answer is positive, there are others, but I don't know how they look.

Hi,

we encountered this problem in linear algebra courses :

we were studying sesquilinear forms, that are linear in one argument en semilinear in the other.  We restricted ourselves to involutive automorphisms.  The professor told us that complex conjugation is the only one.  I asked him, are there any other noninvolutive automorphisms and he said no.

Next course however, he pointed out that this is a common misconception.  There seem to be infinitely many involutive automorphisms of C, but constructing one that differs from complex conjugation is really complicated, so he said giving one example was even too difficult.

Allow me to point out that you can't look for automorphisms that send R to the set R, because the only automorphism of R is the identity.
thus f(a+b*i)=a+b*f(i)
f(i) has to be zero of x*x+1=0, so f(i) is i or -i, and thus we find identity and complex conjugation

the problem of morphisms in general is even more difficult i guess, but you see, there soom te be infinitely many (complicated) automorphisms of C

Some starting thoughts:

If $\phi$ is a field isomorphism from the reals to the reals, then it must fix the rational numbers. Furthermore, if $P$ is any polynomial, $P(\phi(x))=\phi(P(x))$, so any algebraic number must be mapped to an algebraic number. More specifically: an algebraic number must be mapped to a root of the same polynomial.

But that give us a lot of room still to play with, such as permuting roots of polynomials, or even leaving the algebraic numbers fixed and fooling around with the transcendentals.

Warning: any "construction" is very likely to require some version of the Axiom of Choice (Zorn's Lemma, transfinite induction).

Ok, you gave some thoughts but are the morphisms you get different from the trivial ones?
I hope so.

Here's why $\Re$ has no nontrivial field automorphisms:
$x \ge 0 \iff \exists y$ such that $y^2=x$. A monotone function that fixes the rationals must fix $\Re$, and this relation forces each automorphism to be monotone.

On the other hand, the automorphism group of the complex numbers is large; every automorphism of an algebraic field of finite degree can be extended. In other words, the automorphism group of every subfield of the complex numbers is a quotient of the full group.

If you want the full answer, find a graduate algebra text. The well-studied group of automorphisms of the algebraic numbers (which has a standard topology) could be found by looking for "infinte Galois extensions", and the group of automorphisms of C fixing the algebraic numbers by looking for "transcendence bases" or "transcendental extensions.

The full group of automorphisms of C has the same cardinality as the set of functions from C to itself. Any permutation of a transcendence basis induces an automorphism, and a transcendence basis for the complex numbers is an uncountable set.

I could be completely off target here, but it seems easiest to think about the problem in terms of preserving the structure of the complex numbers under the language of addition and multiplication.  In other words, the automorphism f must satisfy f(ab)=f(a)f(b) and f(a+b)=f(a)+f(b) for complex a and b.  After plugging in 0, 1, -1, and i in for a and b, we see that the automorphism must fix 0, 1, -1, and send i to +-i.  Now, it should be easy to see that the the only morphisms are the trivial ones because we can continually add or subtract 1 or i from our original value and see the change in this value is the same under the automorphism.  I hope this helps.  Again, I could be completely off target.


limit
trigonometry
geometry
perimeter
circumcircle
function
geometry proposed

here is a problem that i have thought of on my own...
prove that ${\lim_{n \rightarrow \infty} n* \sin(\frac{180}{n}}) \approx \pi $ (using degrees)

Well...just make a half-curcle then part its perimetre :D

Just to be a bit picky, it should ${\lim_{n \rightarrow \infty} n* \sin(\frac{180}{n}})=\pi$, rather than approximately equal to.
Let O be the center of a regular 2n-sided polygon with unit circumradius. The area of this polygon is $n\sin\frac{\pi}{n}$. When n tends to infinity, the regular polygon tends to be the circumcircle which has area $\pi$. Hence ${\lim_{n\rightarrow\infty}n\sin(\frac{\pi}{n}})=\pi$.

But then, how are you supposed to find those sines? From a calculator that has a built-in sine function? I have two comments about that. First, do you really understand what the calculator is doing? Second - any such calculator (or computer program) also has a built-in approximation to $\pi,$ so what's the point?

Let's make this more arithmetical (and a little more historical), and let's use the half-angle formula:

$\sin\left(\frac{\theta}2\right)=\sqrt{\frac{1-\cos\theta}2}=\sqrt{\frac{1-\sqrt{1-\sin^2(\theta)}}2}$

So: let $m$ be a value for which you know $\sin\left(\frac{\pi}m\right).$ A good place to start would be $m=6.$

Define $s_k=\sin\left(\frac{\pi}{2^km}\right)$ for $k=0,1,\dots.$ By the half-angle formula,

$s_{k+1}=\sqrt{\frac{1-\sqrt{1-s_k^2}}2}.$

We have to compute two square roots with each recursive step, but the sequence does converge fairly quickly. We get four decimal digits of $\pi$ after 6 steps ($n=384$.)

By the argument above, $\pi=\lim_{k\to\infty}2^kms_k$

If you've ever met the technique of Richardson extrapolation (taught in numerical analysis courses), you can considerably speed up the convergence of this sequence. With four steps ($n=96$) and an even-exponent Richardson extrapolation table, I get an estimate of $\pi$ accurate to within $2\times10^{-14}.$

yes, the half angle was what i was thinking of...


Hi,

Is there a way to render summations with several conditions in subscript, i.e.

$\sum_{i<j,\;s\equiv r\;(\mathrm{mod}\;j)}a_ia_j$

looks ugly. I want the modulus condition on a new line just below $i<j$. It's common in texts, I often see it, but it won't let me just hammer a \\ in there. Any ideas?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25664]here[/url] for two ways to do it.


A worker $ A$ make his duty for $ a$ days more than the worker $ B$ and $ B$ make the same duty in $ b$ days more than the worker $ C$.$ A$ and $ B$ working together make this duty in the same time with worker $ C$. Find the time of making this duty by each worker if they work alone. (Hope I am clear)  :)

What is this?    How do you get that solution?

No one want to solve this.
Maybe you don't understand   :maybe:

I think the question might be:

Worker A takes A days more than Worker B to complete a certain job. Worker B takes B days more than the Worker C to complete the same job. If Workers A and B work together they take the same amount of time as worker C. Find the time it takes for each worker, working alone.

If so, aren't there an infinite number of possibilites, like A - 12 days, B - 6 days, C- 4 days?

Yeas but how do you get that numbers ? BY proveing?

I don't think is this the problem's scope. I think we must find the dependence'c formulas of each worker by each other.  :maybe:

Let $ c$ be the number of days in wich worker $ C$ does his duty. So $ B$ does it in $ b \plus{} c$ and $ A$ in $ a \plus{} b \plus{} c$ days. In one day each worker, $ C$, $ B$ and $ A$, does respectively $ \frac {1}{c}$, $ \frac {1}{b \plus{} c}$ and $ \frac {1}{a \plus{} b \plus{} c}$. So if $ A$ and $ B$ together do the duty in the same time $ C$ does, we must have that:
\[ \frac {1}{a \plus{} b \plus{} c} \plus{} \frac {1}{b \plus{} c} \equal{} \frac {1}{c}
\]
Wich can be reduced to
\[ c^2 \equal{} ab \plus{} b^2
\]
As each step is reversible (you will notice if you do the simplification), any three positive number wich satisfy the last equation will work in the problem. This has infinitely many solution, you can set two of the variables and solve for the other (this is always possible even for $ b$)

[hide="If you are looking only for integer solutions"]
Set $ a\plus{}b\plus{}c\equal{}x$ $ b\plus{}c\equal{}y$ we must solve the equation $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{c}$,
Wich is equivalent to $ xy\minus{}c(x\plus{}y)\equal{}0\Leftrightarrow xy\minus{}c(x\plus{}y)\plus{}c^2\equal{}c^2\Leftrightarrow (x\minus{}c)(y\minus{}c)\equal{}c^2$
Let $ d$ be a positive(why?) integer wich divides $ c^2$, all solutions to the equation are of the form $ x\equal{}d\plus{}c$
$ y\equal{}\frac{c^2}{d}\plus{}c$

These are the number of days in wich worker $ A$ and worker $ b$ do their duty alone. (here the variable is $ c$ and $ d$ is a parameter depending on $ c$)
 :)
[/hide]

 :)


In how many ways can 4 on/off switches in a row be set so that no two adjacent switches are on?

[hide]8.  You can just count them[/hide]

I know I've seen this before.

[hide]8

1 is when they're all off, 4 when 3 are off and one is on, and 3 when two are on and two are off. 1+4+3=8. (Is it always half?)[/hide]

EDIT: It is not half when there are only 2 switches, but it is when there are three. I don't think this is a general rule.

Yeah, its not always 1/2. In fact, in general, the number of arrangments is [hide]the fibonacci series, starting at 2 when theres 1 switch..[/hide]

Now, of course, the question (which is much too hard for Mathcounts, but is a good mathematical question anyhow) is to prove the statement that TripleM just made.

By the way, it's not half for three switches.  There are 5 out of 8 possibilities:

000
001
010
100
101

where a 1 is on and a 0 is off.

TripleM's question- does it have anything to do with each Fibonacci being the sum of the 2 previous and there being 2 states to a switch?

just a thought...

Not quite.. well, sort of, in a way.. but let f(n) be the number of arrangements, and then see if you can work out why f(n) = f(n-1) + f(n-2).

Let's say f_1(n) is number of arrangements where switch n is on and f_0(n) is number of arrangements where switch n is off ...

[hide]f(n) = f_0(n) + f_1(n) .... because switch n is either off or on

f_0(n) = f(n-1) ... because any arrangement of n-1 switches can be extended to n switches by adding a switch which is off

f_1(n) = f_0(n-1) ... because only arrangements of n-1 switches where switch n-1 is off can be extended to n switches by adding a switch that is on

but f_0(n-1)=f(n-2), so f(n)=f(n-1)+f(n-2) for any n>2.[/hide]

A similar problem is - how many different ways are there to climb a set of n stairs if you can take either 1 or 2 stairs at each step.

Syntax Error and TripleM -- Congratulations! A few days ago I was trying to come up with a decent contest-like problem that involved Zeckendorf representation and failed miserably. But you managed to make a very nice problem.

Actually, Gandalf, that isn't [i]quite[/i] a proof.  When you say, "any arrangement of n-1 switches can be extended to n switches by adding a switch which is off," isn't the only relevant fact.  You also have to say "any arrangement of n switches ending in a 0 can be converted into a combination of (n - 1) switches by taking off a zero," and then a similar statement for the other case.

Probably the easiest way to state the answer is

[hide]

If the first switch is off, theres f(n-1) ways of arranging the others, if its on, the second one must be off, and theres f(n-2) ways of arranging the others.

[/hide]