trigonometry
geometry
Pythagorean Theorem
geometry unsolved

[quote="Rushil"]In a convex quadrilateral $PQRS$, $PQ =RS$, $(\sqrt{3} +1 )QR = SP$ and $\angle RSP - \angle SQP = 30^{\circ}$. Prove that $\angle PQR - \angle QRS = 90^{\circ}.$[/quote]
The problem is wrong, not just here, but also under the above link. It should be $\angle RSP - \angle SPQ = 30^{\circ}$. Other than this, the solution is just trigonometry.

Denote $p = PQ, q = QR, r = RS,\ q = QR, s = SP$ and $\angle P = \angle SPQ, \angle Q = \angle PQR, \angle R = \angle QRS, \angle S = RSP$. We are given $p = r,\ (1 + \sqrt 3)q = s$ and $\angle S - \angle P = 30^\circ$ and we have to show that $\angle Q - \angle R = 90^\circ$. Let the perpendicular bisectors of the segments SP, QR meet the lines PQ, RS at points A, B. The triangles $\triangle SPA, \triangle QRB$ are both isosceles with the base angles $\angle P, \angle R$, respectively. Hence, $\angle ASR = \angle S - \angle P = 30^\circ$ and $\angle BQP = \angle Q - \angle R$. To see if this angle is right, we test Pythagorean theorem for the triangle $\triangle AQB$. The legs QA, QB are equal to

$QA = PQ - PA = p - \frac{s}{2\cos \widehat P},\ \ QB = \frac{q}{2\cos \widehat R}$

The supposed hypotenuse AB is obtained by applying the cosine theorem to the triangle $\triangle ASB$ with the angle $\angle ASB = \angle ASR = 30^\circ$ and the sides

$SA = \frac{s}{2 \cos \widehat P},\ \ SB = RS - BS = p - \frac{q}{2 \cos \widehat R}$

$AB^2 = SA^2 + SB^2 - 2\ SA \cdot SB \cos 30^\circ =$

$= \frac{s^2}{4 \cos^2 \widehat P} + p^2 - \frac{pq}{\cos \widehat R} + \frac{q^2}{4 \cos^2 \widehat R} - \frac{2s}{2\cos \widehat P} \left(p - \frac{q}{2 \cos \widehat R}\right) \cdot \frac{\sqrt 3}{2}$

He have to test

$(?)\ \ QA^2 + QB^2 = AB^2$

$(?)\ \ p^2 - \frac{ps}{\cos \widehat P} + \frac{s^2}{4 \cos^2 \widehat P} + \frac{q^2}{4 \cos^2 \widehat R} =$

$= \frac{s^2}{4 \cos^2 \widehat P} + p^2 - \frac{pq}{\cos \widehat R} + \frac{q^2}{4 \cos^2 \widehat R} - \frac{s}{\cos \widehat P} \left(p - \frac{q}{2 \cos \widehat R}\right) \cdot \frac{\sqrt 3}{2}$

$(?)\ \ \frac{ps}{\cos \widehat P} = \frac{pq}{\cos \widehat R} + \frac{ps}{\cos \widehat P} \cdot \frac{\sqrt 3}{2} - \frac{qs}{2 \cos \widehat P \cos \widehat R} \cdot \frac{\sqrt 3}{2}$

Substituting $s = q(1 + \sqrt 3)$ and reducing by q:

$(?)\ \ \frac{p(1 + \sqrt 3)}{\cos \widehat P} = \frac{p}{\cos \widehat R} + \frac{p(1 + \sqrt 3)}{\cos \widehat P} \cdot \frac{\sqrt 3}{2} - \frac{q(1 + \sqrt 3)}{2 \cos \widehat P \cos \widehat R} \cdot \frac{\sqrt 3}{2}$

$(?)\ \ q(3 + \sqrt 3) = 4p \cos \widehat P + 2(1 - \sqrt 3)p \cos \widehat R$

$(?)\ \ \frac q p = \frac{4 \cos \widehat P + 2(1 - \sqrt 3) \cos \widehat R}{3 + \sqrt 3} = \frac{2(3 - \sqrt 3) \cos \widehat P + 2(3 - 2\sqrt 3)\cos \widehat R}{3}$

Using the cosine theorem for the triangles $\triangle SPQ, \triangle QRS$ with the common side QS, one of the diagonals of the quadrilateral PQRS, we get

$QS^2 = PQ^2 + SP^2 - 2\ PQ \cdot SP \cos \widehat P = QR^2 + RS^2 - 2\ QR \cdot RS \cos \widehat R$

$p^2 + s^2 - 2ps \cos \widehat P = q^2 + p^2 - 2pq \cos \widehat R$

$s(s - 2p \cos \widehat P) = q(q - 2p \cos \widehat R)$

Substituting $s = q(1 + \sqrt 3)$ and reducing by q:

$2q(2 + \sqrt 3) - 2p(1 + \sqrt 3) \cos \widehat P = q - 2p \cos \widehat R$

$q(3 + 2\sqrt 3) = 2p(1 + \sqrt 3) \cos \widehat P - 2p \cos \widehat R$

$\frac q p = \frac{2(1 + \sqrt 3) \cos \widehat P - 2 \cos \widehat R}{3 + 2\sqrt 3} = \frac{2(3 - \sqrt 3) \cos \widehat P + 2(3 - 2 \sqrt 3) \cos \widehat R}{3}$

which is exactly the formula we need to successfully complete the Pythagorean theorem test for the triangle $\triangle AQB$.

[b]Another trigonometry masterclass[/b]
Let [XYZ] denote the area of triangle XYZ

$                   [PSR]+[PQR]=[QPS]+[QRS]$
$\Rightarrow sinQ+(\sqrt{3}+1)sinS=sinR+(\sqrt{3}+1)sinP$
$\Rightarrow sinQ-sinR=(\sqrt{3}+1)(sinP-sinS)$
$\Rightarrow 2cos((Q+R)/2)sin((Q-R)/2)=2(\sqrt{3}+1)cos((P+S)/2)sin(-15)^{\circ}$
Note that $cos((Q+R)/2)=-cos((P+S)/2)$
Hence cancelling and simplifications lead us to
$sin((Q-R)/2)=1/\sqrt{2}$
Since PQRS is [b]convex[/b] the only possibility is
$                   ((Q-R)/2)=45^{\circ}$
$\Rightarrow (Q-R)=90^{\circ}$
Isn't Yetti's solution too big! But I think it's fantastic!!


Probably not as natural an approach, but nice as well:

[b]Problem:[/b] Prove that exactly one of the following two statements is true:
A. There is a power of 10 that has exactly 2008 digits in base 2.
 B. There is a power of 10 that has exactly 2008 digits in base 5.

[hide][b]Solution:[/b] We show that for all $ k\in N$, exactly one of the following has a solution for $ n \in N$:
\begin{eqnarray*}
2^k \leq 10^n < 2^{k+1}\\
5^k \leq 10^n < 5^{k+1}
\end{eqnarray*}
Note that the number of integers $ n\in N$ such that $ 10^n < b^k$ for a fixed $ b \in \left\{2,5\right\}$ is equal to the maximum integer $ n$ such that the equation is true, as it will also be true for $ n$ replaced by any number in $ 1, 2, \ldots n$. Thus the inequality has $ \left\lfloor \log{b^k} \right\rfloor$ solutions. Then the number of solutions to $ 10^n < 2^k$ plus the number of solutions to $ 10^n < 5^k$ is $ A = \left\lfloor \log{2^k} \right\rfloor + \left\lfloor \log{5^k} \right\rfloor$. We know that (since $ 2^k, 5^k$ are not perfect powers of 10)
\begin{eqnarray*} \left\lfloor \log{2^k} \right\rfloor + \left\lfloor \log{5^k} \right\rfloor & < & \log{2^k} + \log{5^k} \\
& = & \log{10^k} \\
& = & k \\
\left\lfloor \log{2^k} \right\rfloor + \left\lfloor \log{5^k} \right\rfloor & > & (\log{2^k} - 1) + (\log{5^k} - 1) \\
& = & k - 2 \\
\end{eqnarray*}
Therefore $ A = k - 1$. Similarly, the number of solutions to $ 10^n < 2^{k + 1}$ plus the number of solutions to $ 10^n < 5^{k + 1}$ is $ B = k$. Subtracting, we get that $ B - A = 1$ is the number of solutions to $ 2^k \leq 10^n < 2^{k + 1}$ plus the number of solutions to $ 5^k \leq 10^n < 5^{k + 1}$. This proves the claim. Now setting $ k = 2007$ gives the desired result.[/hide]


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Let $ABCD$ be a convex quadrilateral and $ \omega_1,\omega_2$ denote the incircles of $\triangle ABC$ and $\triangle  ADC,$ the latter one is tangent to $\overline{AC}$ at $ F.$ Assume that there exists a circle $\omega$ (outside ABCD) tangent to lines $ AB,BC,CD$ and $CA.$ Show that

[i]a)[/i] Common external tangents of $ \omega_1,\omega_2$ intersect on $\omega.$

[i]b)[/i] Common external  tangents of $\omega,\omega_1$ intersect at a point $Y$ lying on $BF.$

[i]c) [/i] If $\omega$ and $ \omega_1$ are tangent to line $ BA$ at $ M,N$ and the circle wih diameter $\overline{ MN}$ cuts $\omega$ at $ X,$ such that $ F$ lies on $ XB,$ then show that $ Y$ lies on circle $ \omega_1.$

Questions a) & b)

If $ \omega_1$ is tangent to $AC$ through $E,$ then it's easy to figure out that $ AF \equal{} CE.$ Let $ X$ be the intersection of the center line of $ \omega_1 ,\omega_2$ with line $AC$ $\Longrightarrow$ $X$ is  the insimilicenter of $ \omega_1, \omega_2,$ $ B$ is the exsimilicenter of $ \omega, \omega_1$ and $ D$ is the insimilicenter of $\omega,\omega_2.$ By Monge's theorem, it follows that $ D,X,B$ are collinear. Let  $O,O_1,O_2$ be the centers of $\omega,\omega_1$ and $ \omega_2.$ Since $ AF \equal{} CE,$ then  B-excircle of $ \triangle ABC$ is tangent to $\overline{AC}$ through $ F$ and D-excircle of $ \triangle ADC$ is tangent to $\overline{AC}$ through $ E.$ Rays $ BF$ and $ DE$ pass through the antipodes $ E',F'$ of $ E,F.$ B-excircle is homothetic to $\omega$ through $B.$ Thus, ray $ BF$ cuts $\omega$ at the homologous $ H$ of $ E'$ under such homothety $\Longrightarrow$ $OH \parallel EE'.$ Likewise, D-excircle, $\omega$ and $ \omega_2$ are homothetic through $ D$ $\Longrightarrow$ $ DE$ cuts $\omega$ at the homologous $ H'$ of $ F'$ $\Longrightarrow$ $ OH' \parallel FF'.$ Now, if $ FF' \parallel EE'$ $\Longrightarrow$ $ OH \parallel OH',$ i.e. $ H \equiv H'$ $\Longrightarrow$ Exsimilicenter $ H$ of $ \omega_1,\omega_2$ lies on $\omega.$ Intersection $ Y$ of the common external tangents of $ \omega , \omega_2$ is their exsimilicenter. Hence, by Monge's theorem, we conclude that exsimilicenters of $ \omega_1,\omega_2,\omega$ are collinear $\Longrightarrow$ $Y,B,H$ are collinear.

Question c)

Clearly, $ X \equiv H$ $\Longrightarrow$ $ \angle MHN \equal{} 90^{\circ}.$ Hence, line $ NH$ cuts $\omega$ at the antipode $ M'$ of $ M,$ which is the homologous of $ N$ under the negative homothety $ \omega,\omega_1.$ Since this line passes through $D,$ then it cuts $\omega$ at $ H,$ different from $M',$ and cuts $ \omega_1$ at $ E',$ because of $ EE' \parallel OH.$ In other words, line $ HM'$ goes through $ D,E,N$ $\Longrightarrow$ $ \triangle CNE$ is isosceles with apex $C.$ Circle with diameter $ \overline{MN}$ also passes through $Q',$ which is the inverse of $H$ through the inversion with pole $B,$ thus  $ H,Q,N,M$ are conclyclic. If $ P' \equiv HN \cap QN$ $\Longrightarrow$ $ \angle HQP' \equal{} \angle HNM.$ Since $ \triangle CNE$ is C-isosceles, it follows that

$ \angle CEN \equal{} \angle HMN \equal{}\angle HQP'\Longrightarrow \angle P'EF \equal{} \angle FQ'P' \Longrightarrow FQEP'$ is cyclic.

Which implies that $ P \equiv P'.$ Now, from parallel radii $ O_2P \parallel OM \parallel O_1N$ and $ O_2F \parallel OH,$ it follows that $ Q \equiv PM \cap FH$ is the exsimilicenter of $ \omega_2,\omega$ and our proof is completed.


Let $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}abc$. Prove that: $ \frac{\sqrt{b^2\plus{}2a^2}}{ab}\plus{}\frac{\sqrt{c^2\plus{}2b^2}}{bc}\plus{}\frac{\sqrt{a^2\plus{}2c^2}}{ca} \ge \sqrt{3}$

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} abc$. Prove that: $ \frac {\sqrt {b^2 \plus{} 2a^2}}{ab} \plus{} \frac {\sqrt {c^2 \plus{} 2b^2}}{bc} \plus{} \frac {\sqrt {a^2 \plus{} 2c^2}}{ca} \ge \sqrt {3}$[/quote]

easy but Nice !!

Using the lemma $ a^2\plus{}b^2\plus{}c^2\ge \frac{1}{3}(a\plus{}b\plus{}c)^2$  we have

$ b^2c^2\plus{}a^2c^2\plus{}a^2c^2 \ge \frac{1}{3}(bc\plus{}2ac)^2$

Hence :

$ \sqrt{b^2c^2\plus{}2a^2c^2}\ge \frac{1}{\sqrt{3}}(bc\plus{}2ac)$

Then it follows that :

$ \sum_{cyc}\sqrt{b^2c^2\plus{}2a^2c^2} \ge \sum_{cyc}\frac{1}{\sqrt{3}}(bc\plus{}2ac)\equal{}\sqrt{3}(ab\plus{}bc\plus{}ca)\equal{}\sqrt{3}abc$

Or,equivalently

$ \sum_{cyc}\frac{\sqrt{b^2c^2\plus{}2a^2c^2}}{abc} \ge \sqrt{3}$
which can be written as :

$ \sum_{cyc}\frac{\sqrt{b^2\plus{}2a^2}}{ab} \ge \sqrt{3}$

and we are done !  :P

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} abc$. Prove that: $ \frac {\sqrt {b^2 \plus{} 2a^2}}{ab} \plus{} \frac {\sqrt {c^2 \plus{} 2b^2}}{bc} \plus{} \frac {\sqrt {a^2 \plus{} 2c^2}}{ca} \ge \sqrt {3}$[/quote]

with $ ab \plus{} bc \plus{} ca \equal{} abc \equal{} > \sum \frac {1}{a} \equal{} 1$
$ LHS \equal{} \sum \frac {\sqrt {b^2 \plus{} 2a^2}}{ab} \equal{} \sum \frac {\sqrt {b^2 \plus{} 2a^2}}{\sqrt {a^2b^2}} \equal{} \sum \sqrt {\frac {1}{a^2} \plus{} \frac {2}{b^2}}$
by Mincopki $ LHS \ge \sqrt {(\sqrt {3})^2.(\sum \frac {1}{a})^2} \equal{} \sqrt {3}$

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} abc$. Prove that: $ \frac {\sqrt {b^2 \plus{} 2a^2}}{ab} \plus{} \frac {\sqrt {c^2 \plus{} 2b^2}}{bc} \plus{} \frac {\sqrt {a^2 \plus{} 2c^2}}{ca} \ge \sqrt {3}$[/quote]
$ \iff \sum_{cyc} c \sqrt{\frac{b^2\plus{}2a^2}{3}} \ge ab\plus{}bc\plus{}ca$
But $ \sqrt{\frac{b^2\plus{}2a^2}{3}} \ge \frac{b\plus{}2a}{3}$ by QM-AM, which finishes the problem..


Hello to all! Here goes a problem that I've been proposed with and althought at first it sounded easy, one can complicate thing a bit.

The problem is as follows:

In an international math congress, there are 200 mathematicians. Among those, there are 120 that speak english and 100 that speak french. What's the probability that taken 2 mathematicians, they both speak english and french?

Remember there may be those that speak neither english or french.

good luck!

There is not enough information to answer.

The issue with the problem is that it gives no information about whether speaking English and speaking French are independent events.  The number of mathematicians at the conference who speak both English and French can vary from $ 20$ to $ 100$ in the absence of no other information; however, given that speaking English and speaking French are independent events,

$ P ( \text{English} ) \equal{} \frac{120}{200} \equal{} \frac{3}{5}$
$ P ( \text{French} ) \equal{} \frac{100}{200} \equal{} \frac{1}{2}$
$ P ( \text{English} \cup \text{French} ) \equal{} \frac{3}{5} \cdot \frac{1}{2} \equal{} \frac{3}{10}$

Hence $ 200 \cdot \frac{3}{10} \equal{} 60$ of the mathematicians speak both English and French.  The problem is easy from here.


Let ABC be an isosceles triangle (AB = AC) with angle BAC = 20.  Point D is on side AC such that DBC = 60.  Point E is on side AB such that ECB = 50.  Find  the measure of EDB.

[img]http://www.qbyte.org/puzzles/p022.gif[/img]

Nice problem!

Take point $F$ on $\overline{AB}$ such that $m\angle{BCF} = 60^o$, and let $G$ denote the intersection of $\overline{BD}$ and $\overline{CF}$. Note that triangles $BCG$ and $GDF$ are equilateral, but $BEG$ is isosceles (since $BG = BC = BE$) with $\angle{BGE} \cong \angle{GEB}$. Hence, $m\angle{BGE} = 80^o$ and $m\angle{EGF} = 40^o$. But $m\angle{GFE} = 40^o$ since $m\angle{FBC} + m\angle{BCF} = 140^o$. Since $\angle{EGF} \cong \angle{GFE}$, $DFEG$ is a kite. As $GDF$ is equilateral, $\overline{DE}$ bisects angle $GDF$, hence $m\angle{BDE}$ is $30^o$.


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consider 4 natural consecutive numbers

proove that multiplied($a(a+1)(a+2)(a+3)$) they cannot form a cube($k^3$)

let  $s=a(a+1)(a+2)(a+3)$ then  $a^3|s$ consequently $(a+1)(a+2)(a+3)$ must have factor $a$ ,that it is not true

To Ali: I don't see your point.

To al M.V.: There is a well-known result that 8 and 9 are the only consecutive pair of a cube and a square (a special case of Catalan's conjecture). However I do not know how to prove it; is there a proof using elementary methods? If not, can you finish the proof of the original problem?

I think it's discussed before in this forum

oh (I think) this is a correct solution
 we consider tow case
1. $a$ has prime factor $p$ that is greater than 5. in this case $(a+1)(a+2)(a+3)$ and $p$ are coprime and then
    $a(a+1)(a+2)(a+3)$ hase factor $p$ that is not in cubic form
2.for the $2,3,6$ we check that it cannot be in cubic form . the last case is when  $a=2^n3^m$ thereby $(a+1)$ has
   prime  factor greater than $5$ and I consider this in case 1

There is a wider thorem by Erdos which states that the product of any consecutive positive integers can never be a perfect square of higher power.

Bomb

To Ali: I believe you're still mistaken. For example, in your first argument the fact that $p$ divides $a$ does not mean that the exponent of $p$ in $a$ is not a multiple of 3.

Also to bomb: will you please show the proof of the theorem you mentioned? And the theorem I mentioned above: is there a proof?

i am really waiting for someone to help me solve this problem.
i am in summer holliday,however i have some big goals i want to achieve so i need
to solve as much problems as i can.
The problem stated is from a problem book written by I. Cucurezeanu("Probleme de Teoria Numerelor").
It's problem I.7 i think,because it is from the beggining its not so hard.
So i don't think that it should be that hard to solve.
Also they are talking there about something like (a+1) and (a+2) beeing coprime and taking cases for a =2k
and a=2k+1 also they are talking about a(a+1)(a+3) making some inequalities(if this helps).
also i understood about half of the demonstration,the other half made no sense.
i am really trying hard but i need some help from someone to guide me on what to solve,where to solve from.
this problem book seems quite nice because until now i have understood almost all problems i read and i read some 20 prob from it.
i like this part of maths very much and i am very passionate about it.i believe i can do something about training my mind in this domain altough i'm somewhat old ,i'm 18 now and my main subject of interest was informatics,algorithms since a year ago.

this may help you: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=consecutive+powers&t=22074

spx2, could you post more problems from that book? They seem interesting. Thanks. :)

Just thought I may clarify things up, I think Erdos' conjecture is that the following equation: a^m - b^n = 1 only has solution (a, b, m, n) = (3, 2, 2, 3) in positive integers.

al.M.V.: the conjecture you mention is Catalan's conjecture and was proven recently.

Here is the solution:

In four consecutiv integers one is coprime with all others pairwise (if $a$ is odd then it's $a+2$ else is $a+1$), so that one has to be cube and product of others three has to be cube too. Now if $a$ is odd then $a+2$ is cube and $a(a+1)(a+3)$ is cube to but
$(a+1)^3<a(a+1)(a+3)<(a+2)^3$ (for $a>1$) and in other case when $a(a+2)(a+3)$ is cube then $(a+1)^3<a(a+2)(a+3)<(a+3)^3$ so $a(a+2)(a+3)=(a+2)^3$, $a(a+3)=(a+2)^2$ and that is not possible
I hope i am right

the solution is similar to yours but i really cant understand it

What exactly you can't understand.

You have integers $a,a+1,a+2,a+3$ so if $a$ is even then $(a+1,a)=1$ and $(a+1,a+2)=1$ and if $(a+1,a+3)=d$ then $d|(a+3)-(a+1)=2$ so $d$ is $1$ or $2$ but $a$ is even so $a+1$ and $a+3$ are odd implies that $d$ isn't $2$ so they are comprime too.
Iff $a$ is odd then $(a+2,a+3)=1,(a+2,a+1)=1$ and if $(a+2,a)=d$ then again $d|2$ but it can't be $2$ becouse $a$ and $a+2$ are odd... I hope you understand this part...

Now if $a$ is even then $(a+1)$ is coprime with $a,a+2,a+3$ so it's coprime with $a(a+2)(a+3)$ but $a(a+1)(a+2)(a+3)$ is cube so $a+1$ is cube and $a(a+2)(a+3)$ is cube too (if $xy=p^k$ and $(x,y)=1$ then $x=r^k$ and $y=q^k$. Prove it!) but
$(a+1)^3<a(a+2)(a+3)<(a+2)^3$ so it can't be cube (becouse it is between two consecutive cubes, if it is greater then $(a+1)^3$ and smaller then $(a+2)^3$ so if it is cube, let's say $x^3$ then $a+1<x<a+2$ and that can't be)

If $a$ is odd, everything it's same except, now the cube suposse to be product $a(a+1)(a+3)$ and this is between cubes of $a+1$ and $a+3$ so it has to be cube of $a+2$ so from that $a(a+1)(a+3)=(a+2)^3 \rightarrow contradiction$

I honestly hope that you undestand it know :) :) but if you didn't please be more specific about a part of solution that you don's understand

To ZetaX, can you please show proof of Catalan's conjecture.

Bomb

hello M@re the problem is that in your last post in the eight(8) line the second $a+2$ you wrote should be replaced to an $a+1$ as i see it.If i am mistaking please tell me why

about the cataln's conjecture proof.Google tells me that one of our conationals namely Preda Mihailescu wich i respect now,i never heard of him, (i don't know much yet about number theorists) has prooven this Conjecture and the proof is available if you search for it on google,personally i've tried to understand it but because of lack of theory and beeing a being a beginner in number theory i can't understand it(to me it seems complicated now) .

Try http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf for a prove of Catalan's conjecture.

yes its good now
i have understood all the demonstration,now i am trying to think why you chose this method
out of all the posibilites
also i am searching for other methods to proove it,i could use just some hints but some far away from solution hints like what to use to solve it in another way,i would like it if you could tell me some.Would mordell curve do ?and how complicated is that mordell curve stuff "?
i am going to post yet another problem its trivial just for training ive solved it
 $\text{let k be natural number.proove that }k^2 (mod 16)\text{ is a square}$

now i would want to ask why would calculus be so important for number theory.
i see that calculus would be used to make some inequalities here only in the sense that it is easier
to develop in power series or some sort of series some expressions to make it easier to make inequalities,
other helps from calculus's side i cant see.
could you give me a more detailed view ?

You can prove a lot theorems by concerning
a) some assymptotics (e.g. the prime number theorem)
b) some power series together with identities between them (e.g. my signature implies the strongest version of the four squares theorem)

First I don't know what mordel curve is.. :lol:
Second: You wonder how I get this idea, well, about 2 months ago i was solving this problem : Prove that the product of six consecutive positive integers can't be fifth power.
And the solution is same, but when I was solving that problem I had no idea how to solve it (of course only thng I had in mind was to get rid of backets and then transform that in something like $x^5+y$ but I really didn't think about coprime numbers:))
So there you go... :D

And about that $k^2\equiv_{16}$ well, I think that is very trivial...

If \[k\equiv_{16} \pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm7, 8\] then
\[k^2\equiv_{16} 1, 4, 9, 0, 9, 4, 1, 0\] respectively, so they are all squares...


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find the definite integral of 1/x from x=1 to x=2 using riemann sum method

I like a geometric partition $ x_n\equal{}a^n$ here, with $ a\equal{}2^{\frac1n}$. The left-endpoint sum is then $ \sum_{k\equal{}1}^n \frac1{x_{k\minus{}1}}(x_k\minus{}x_{k\minus{}1})\equal{}1\cdot (a\minus{}1)\plus{}\frac1a\cdot (a^2\minus{}a)\plus{}\cdots\plus{}\frac1{a^{n\minus{}1}}\cdot (a^n\minus{}a^{n\minus{}1})\equal{}n\cdot(a\minus{}1)$
Taking the limit, we get $ \lim_{n\to\infty}n\cdot\left(2^{\frac1n}\minus{}1\right)$. Inverting the variable gives $ \lim_{t\to 0}\frac{2^t\minus{}1}{t}$ (assuming reasonable smoothness).

All right, how exactly did we define the logarithm and exponential functions?

[hide="I type too slowly. jmerry got there first."]There's a huge amount of freedom of choice within the general idea of Riemann sums - just saying that you want use them does not specify the details.

For instance, what sort of partition should you use? You're probably used to seeing partitions with equal-width subintervals - that is to say, partitions that form an arithmetic progression. But you don't have to do that, and in this particular instance, I've got an idea that is better than using an arithmetic progression: use a [i]geometric[/i] progression.

Choose an integer $ n.$ Let $ r\equal{}\sqrt[n]{2}.$ Let the partition be
$ \{1,r,r^2,r^3,\dots,r^{n\minus{}2},r^{n\minus{}1},r^n\}$ where we note that $ r^n\equal{}2.$ Find the left-endpoint sums on that partition (which are the upper sums for this decreasing function) and find the right-endpoint (lower) sums.

Now take the limit as $ n\to\infty.$

Go ahead - try it![/hide]


a_0=0
a_1=1
a_(n+2)=a_(n+1)+(1/2)*a_n

How can one prove that for any k>=8, a_k is not an integer ??

I don't suppose calculating a_n helps, right? Here it is anyway (it looks like Fibonnacci's sequence, only with 3 instead of 5):

a_n=(1/ \sqrt3)*[((1+  \sqrt3)/2)^n-((1-  \sqrt3)/2)^n]. Nasty right?

using recurrence property it is easy to show a(2n+1) is never an integer :

Since b(n)=a(2n+1)*2^n follows :

b(0)=1  b(1)=3   b(n)=4b(n-1)-b(n-2)  (leaving to the reader!)

Therefore b(n) is always odd and a(2n+1) is never an integer.

But for n even! It is quite complicate  :

See this example :

let  c(n)=a(4n+2)*2^(2n-1) then :

c(0)=1  c(1)=15   c(n)=14c(n-1)-c(n-2)   (still leaving to the reader!)

Then c(n) is always odd and a(4n+2) is never an integer.

This shows how proceed.

Let

d(n)=a(2^m*n+2p)*2^(2^(m-1)*n+p-1)  if m>=2 p odd >=1

Then it can be shown that d(n) always satisfies a recursion of type :

d(0)=1 d(1) odd and  d(n)=E*d(n-1)-d(n-2)  E even

Therefore :

d(n) is always odd and for all n even>=10  and consequently a(n) is never an integer.

a numerical example :

d(0)=1 d(1)=2911  d(n)=a(2^4*n+14)*(2^3*n+6)

then :   d(n)=37634*d(n-1)-d(n-2)

BC


Let triangle ABC. $ S_{ABC}\equal{}1$. M is a point chosen arbitrarily. Prove that
$ aMA\plus{}bMB\plus{}cMC\geq 4$

$ aMA \geq 2S_{ABM}\plus{}2S_{ACM}$
$ \Rightarrow aMA \plus{} bMB \plus{} cMC\geq 4(S_{ABM}\plus{}S_{ACM}\plus{}S_{CBM})\equal{}4$


prove that there are infinitely many integers $ {a,b,c,d}$ such that:

$ {a}^{4}\plus{}{b}^{4}\plus{}{c}^{4}\equal{} 2{d}^{2}$

$ a\equal{}2n,b\equal{}n,c\equal{}n,d\equal{}3n^{2}$ works.

Do you want something stronger, like infinitely many solutions with no common factor?

Just prove that $ 2(ab\plus{}bc\plus{}ca)^{2}\equal{} a^{4}\plus{}b^{4}\plus{}c^{4}$, for all $ a, b, c\in\mathbb{C}$ such that $ a\plus{}b\plus{}c\equal{}0$. :wink:

[quote="s.tringali"]Just prove that $ 2(ab\plus{}bc\plus{}ca)^{2}\equal{} a^{4}\plus{}b^{4}\plus{}c^{4}$, for all $ a, b, c\in\mathbb{C}$ such that $ a\plus{}b\plus{}c \equal{} 0$. :wink:[/quote]
very nice. It give:
$ a^{4}\plus{}b^{4}\plus{}(a\plus{}b)^{4}\equal{}2(a^{2}\plus{}ab\plus{}b^{2})^{2}$.

thanks everyone :)  .
dear jmerry,may i ask something stronger like:
"find [b]all[/b] solutions" ?


Is after Volume 2 a good time to take WOOT or not?

WOOT, as stated in the WOOT Math Jam a couple of days ago (by Mr. Rusczyk), is generally a little bit harder than Volume 2.  If you worked through Volume 2 thoroughly, you probably will be ready.  If you are still not sure, you can take AIME tests from the past and work through other books such as Art and Craft of Problem Solving to help you.

That's right -- if you've finished Volume 2, then you should be ready for WOOT.  (It is also possible to be ready for WOOT without completing Volume 2, of course.)


geometry
3D geometry
AMC
AIME
USA(J)MO
USAMO

Find all triples of positive real numbers $(x,y,z)$ which satisfy the following system:

\begin{eqnarray*} \sqrt[3]{x}-\sqrt[3]{y}-\sqrt[3]{z}=16\\ \sqrt[4]{x}-\sqrt[4]{y}-\sqrt[4]{z}=8\\ \sqrt[6]{x}-\sqrt[6]{y}-\sqrt[6]{z}=4 \end{eqnarray*}

I thought this problem was cool :)

Hold on, I think I've got it.

What book did you find this problem in anyways?

I think I've got it :)
[hide][b]There are no solutions.[/b]

[u]Proof:[/u]
Let:
$\sqrt[12]{x} = a$
$\sqrt[12]{y} = b$
$\sqrt[12]{z} = c$

We have:
$a^4 - b^4 - c^4 = 16$  (1)
$a^3 - b^3 - c^3 = 8$    (2)
$a^2 - b^2 - c^2 = 4$    (3)

Rearrange 1 and 3 as:
$a^4 = b^4 + c^4 + 16$ (1*)
$a^2 = b^2 + c^2 + 4$   (2*)

Substituting (2*) into (1*), we have
$(b^2 + c^2 + 4)^2 = a^4 = b^4 + c^4 + 16$
$b^4 + c^4 +16 + 2b^{2}c^{2} + 8b^2 + 8c^2 = b^4 + c^4 + 16$
$2b^{2}c^{2} + 8b^2 + 8c^2 = 0$

The terms on the LHS are all positive when $b$ and $c$ are not $0$, so this must be true: $b = c = 0$

So we now have the following system:

$a^4 = 16$
$a^3 = 12$
$a^2 = 8$

Clearly, there are no solutions to this. So there exist no real solutions to:

\begin{eqnarray*} \sqrt[3]{x}-\sqrt[3]{y}-\sqrt[3]{z}=16\\ \sqrt[4]{x}-\sqrt[4]{y}-\sqrt[4]{z}=8\\ \sqrt[6]{x}-\sqrt[6]{y}-\sqrt[6]{z}=4 \end{eqnarray*}

Done! Of course, I have a high chance to get the problem wrong, so criticize and critique.
[/hide]
Great problem, chess64!
What number is that, by the way?

I think there is something wrong with the substitution you are using.

Here's my solution (copied from my Independent Study hw)

[hide]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=332[/img]
[/hide]

Arne (or chess64), I don't see what's wrong. Perhaps there something is lost in the substitution?

You should use $\sqrt[12]{x}$, $\sqrt[12]{y}$, $\sqrt[12]{z}$ instead.

I meant that. :oops:  Honestly. I was too busy checking over other parts of the problem. Sorry, Arne.

I edited that. Wow, I feel dumb.

No problem, 1234567890 :)

And chess64, yes, that looks fine to me.

Yes! :D I solved a MOP problem! How hard was that? Like easy AIME, hard AIME, easy USAMO?  :)

I say it was the level of difficulty of a  medium AIME problem. Can't be hard if I solved it. :)

I'd say this is the difficulty of an AIME problem 4/5 or something like that.

Cool. :) I wonder why that was a MOP problem though? It seemed easy enough. And who moved it from Intermediate to Pre-Olympiad?


geometry
rectangle
combinatorics unsolved
combinatorics

Prove or disprove a $3 \times 20$ rectangle can be covered without gaps and without overlaps with pentominoes shows in the picture.(There is one pentomino from each kind)

Yes, it is possible.
Consider the following configuration, the combination of two 'C' shaped pentominoes, and one '+' shaped pentomino
XX  Y  XX
X  YYY  X
XX  Y  XX
This forms a $3 \times 5$ rectangle. Repeating this $4$ times, we get the required configuration.

But what if you want to use each of the 12 pentominoes exactly once?

[quote="I"]There is one pentomino from each kind[/quote]
U have to use each one once.
[size=150]dblues[/size] what are the other Shapes for  :D  :rotfl:
try it once more

Oops... I must say I totally misunderstood the problem...  :blush:   :blush:   :blush:

I think I have solve it ,
I will poste my solution later


Consider the points $1,\frac12,\frac13,...$ on the real axis. Find the smallest value $k \in \mathbb{N}_0$ for which all points above can be covered with 5 [b]closed[/b] intervals of length $\frac1k$.

[quote="Peter VDD"]Current result: 1/8

Who claims this is false and who can make a higher (here lower) bid? :D[/quote]

It is obvious that k=1/5 works, so this should be the immediate upper bound.  Since [1,1/2] is larger than [1/2,1/3] we refine this by moving to intervals which cover all of [0,1/2] and only the point (1).

This can be further refined by moving to intervals of length k=1/9 such that [0, 1/3] are completely covered and (1/2), (1) are covered.

In fact, this is the best we can do, my answer earlier was the product of about 30 seconds of hasty scribbles.

$k = \frac19$ says blahblahblah

anyone topping that? :D

Come on people!! The problem is not yet solved!! :)

[quote="Peter VDD"]doesn't it? well then post some proof it doesn't :)

maths isn't about guessing until you're correct, you need proof for everything!

I edited the message a bit so you all see it's about CLOSED intervals. Blah^3 supposed that but i don't know about liyi.[/quote]

If it involves open intervals, I would imagine we cannot find an answer exactly.

Presumably, we would have to pick the smallest k E R larger than a certain number, but there is no such number.

I know, but then I wonder why you both haven't found the answer yet? And especially why you claim 1/9 is possible and 1/10 is not?

Hmm... actually 1/10 does work: We can cover the points $\displaystyle 1;\;\frac12;\;\frac13;\;\frac14;\;...$ by 5 closed intervals of length $\displaystyle \frac{1}{10}$. In fact, take the interval $A = \displaystyle \left[0;\;0.1\right]$, which contains all numbers $\displaystyle \frac1s$ for $s \geq 10$; then take the interval $B = \displaystyle \left[0.1;\;0.2\right]$, which contains the numbers $\displaystyle \frac1s$ with $9 \geq s \geq 5$; then take the interval $C = \displaystyle \left[0.24;\;0.34\right]$, containing the numbers $\displaystyle \frac14$ and $\displaystyle \frac13$; then take the interval $D = \displaystyle \left[0.5;\;0.6\right]$, where the number $\displaystyle \frac12$ lies, and finally, take the interval $E = \displaystyle \left[0.9;\;1\right]$ containing the number 1. So you see, you can find your five intervals, and moreover, they are even disjoint (except for common boundary points).

I don't know whether 1/10 is the minimum...

  Darij

Yup, 1/10 does work as I said :D

[quote="darij grinberg"]I don't know whether 1/10 is the minimum...[/quote]

Well, then you'll have either to prove or to disprove it for 1/11 :D

[quote="Peter VDD"]Well, then you'll have either to prove or to disprove it for 1/11 :D[/quote]

Not exactly. The number $k$ can be an arbitrary real number, not necessary 1 / something natural.

So I must show that for every real $k$ with $\displaystyle k < \frac{1}{10}$, the numbers $\displaystyle 1;\;\frac12;\;\frac13;\;\frac14;\;...$ cannot be covered by $5$ closed intervals of length $k$. But this is really easy: Assume the contrary, i.e., assume that we can cover numbers $\displaystyle 1;\;\frac12;\;\frac13;\;\frac14;\;...$ by $5$ closed intervals of length $k$. Let $\displaystyle \left[a;\;b\right]$, $\displaystyle \left[c;\;d\right]$, $\displaystyle \left[e;\;f\right]$, $\displaystyle \left[g;\;h\right]$, $\displaystyle \left[i;\;j\right]$ be these $5$ intervals. If the numbers $a$, $c$, $e$, $g$, $i$ were all $> 0$, then we could find a natural number $N$ with $\displaystyle N > \frac1a$, $\displaystyle N > \frac1c$, $\displaystyle N > \frac1e$, $\displaystyle N > \frac1g$, $\displaystyle N > \frac1i$, so that $\displaystyle \frac1N < a$, $\displaystyle \frac1N < c$, $\displaystyle \frac1N < e$, $\displaystyle \frac1N < g$, $\displaystyle \frac1N < i$, and hence, the point $\displaystyle \frac1N$ would not lie in any of our five intervals, contradicting the assumption that the intervals $\displaystyle \left[a;\;b\right]$, $\displaystyle \left[c;\;d\right]$, $\displaystyle \left[e;\;f\right]$, $\displaystyle \left[g;\;h\right]$, $\displaystyle \left[i;\;j\right]$ cover all the points $\displaystyle 1;\;\frac12;\;\frac13;\;\frac14;\;...$. Hence, the numbers $a$, $c$, $e$, $g$, $i$ cannot all be > 0, so that at least one of them is $\leq 0$, and hence, the point $0$ lies in (at least) one of our $5$ intervals. [[b]EDIT:[/b] This argument was not quite correct.]

Now consider the $6$ numbers $1$, $\displaystyle \frac{1}{2}$, $\displaystyle \frac{1}{3}$, $\displaystyle \frac{1}{5}$, $\displaystyle \frac{1}{10}$ and $0$. We know that each of these $6$ numbers lies in (at least) one of our $5$ intervals. Hence, the pigeonhole principle shows that (at least) two of these $6$ numbers must lie in the same interval. But this is impossible, since the pairwise differences of these numbers are all $\displaystyle \geq \frac{1}{10}$, and hence also $> k$, since $\displaystyle k < \frac{1}{10}$. By this contradiction, we get the desired result.

This means that $\displaystyle k = \frac{1}{10}$ is the required smallest number.

Yeah, finally this trivial problem is solved  :D

  Darij

Well, the original problem was with k in IN I think, but it's perfectly solvable for k in IR, as you just showed. Great! :)

Please correct the problem statement. It was not $k \in \mathbb{N}$ but $\frac{1}{k}$ with $k \in \mathbb{N} \backslash \{0\}$.

This thread is a bit of a mess, and the proof in my posts above has a minor gap (an interval starting at a negative number does not per se mean that this interval contains $0$), so let me supplant it by a reference: The problem in its corrected form (namely: find the smallest positive real $\varepsilon$ such that the entire set $\left\{\dfrac{1}{1}, \dfrac{1}{2}, \dfrac{1}{3}, \ldots\right\}$ can be covered by $5$ closed intervals of length $\varepsilon$) is Exercise 6.3.10 in [url=http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf]my notes [i]Math 235: Mathematical Problem Solving[/i][/url] (work in progress, so the numbering can change; if it does, search for "can be covered with $5$ closed intervals" in the notes). The solution I give there is hopefully clear and complete.

I'm wondering what the smallest possible $\varepsilon$ is for other values of $5$, though.

Actually, the precise outcome of Darij's argument above is that we need exactly
\[N(\varepsilon)=\max\{n: \exists 1=k_1<k_2<\dots<k_n: \left\vert \frac{1}{k_i}-\frac{1}{k_{i+1}}\right\vert>\varepsilon\}\]
intervals of length $\varepsilon$ to cover everything. From here we easily find that
$N(\varepsilon)=1$ for $\varepsilon \ge 1$.
$N(\varepsilon)=2$ for $\frac{1}{2} \le \varepsilon<1$.
$N(\varepsilon)=3$ for $\frac{1}{4} \le \varepsilon<\frac{1}{2}$.
$N(\varepsilon)=4$ for $\frac{1}{6} \le \varepsilon<\frac{1}{4}$.
$N(\varepsilon)=5$ for $\frac{1}{10} \le \varepsilon<\frac{1}{6}$.
$N(\varepsilon)=6$ for $\frac{1}{12} \le \varepsilon<\frac{1}{10}$.
$N(\varepsilon)=7$ for $\frac{1}{18} \le \varepsilon<\frac{1}{12}$.
$N(\varepsilon)=8$ for $\frac{1}{22} \le \varepsilon<\frac{1}{18}$.
$N(\varepsilon)=9$ for $\frac{1}{30} \le \varepsilon<\frac{1}{22}$.
$N(\varepsilon)=10$ for $\frac{1}{34} \le \varepsilon<\frac{1}{30}$.
There is exactly one sequence in OEIS which matches the numbers $1,2,4,6,10,12,18,22,30,34$ and it's [url=http://oeis.org/A002491]this[/url] one.
I haven't checked the references there to see whether this makes sense, but if so, this would apparently imply the interesting fact that $\varepsilon \approx \frac{\pi}{N^2}$ for large $N$.

@Tintarn: Why is $N\left(\varepsilon\right)$ achievable? I assume you pick a maximum sequence $1 = k_1 < k_2 < \cdots < k_n$ satisfying $\left\vert \frac{1}{k_i}-\frac{1}{k_{i+1}}\right\vert>\varepsilon$, and define the intervals $\left[\frac{1}{k_i}-\varepsilon, \frac{1}{k_i}\right]$ for all $i$, but is it clear that these intervals cover all $\frac{1}{m}$s?

[b]EDIT:[/b] Ah, the lexicographically smallest maximum sequence $\left(1=k_1<k_2<k_3<\cdots<k_n\right)$ should do the trick.

Also, is it clear that the $\varepsilon$-cutoffs are reciprocals of integers?

I have written some [url=https://sagecell.sagemath.org/?z=eJxlkMEOwiAQRO9-xSReQK0tV6N_4c0Yg7LVBrpt1lZ_XxBTE90T-3YGZnFUg8f21PBA8rDhrrzezBBrnviZBF2NS-ju5DCJEgvE1-EGU3owkSOXbRY7HPzxfX7emkDYy0h5lirEuT0U5ogCZqJNvA_bHaqvMNVZyPqJtNFqok2FpS_CwmuU5bvRk8Subd8TO9VmNu8lhlY2d0LDKJySJzKrOwHHpSCWr6TMCqaqPts3v9_CGttftDT6mze_xKs_n34BYuBhEQ==&lang=sage&interacts=eJyLjgUAARUAuQ==]Sage code[/url] myself, which seems to follow the same approach as you, and the sequence is NOT in the OEIS. But I'm not sure that it is the real maximum.

Yes, my argument was the same as in your EDIT. So do you get the same result for the first $10$ values? I did not do any further computations but I was fairly confident that these are correct, and they did match the OEIS entry...
Also, it is clear that the cutoffs are always rational numbers, but it's not clear to me why they always have to be reciprocals...

[url=https://sagecell.sagemath.org/?z=eJxlkMEOgjAQRO98xSReWhWxV6N-hTdjTLWLkuJCFlA_3xYMGp3T5u3MplNHObi7HQtuSe62bJTXqwRBk8hPJKhynMuqIYfRFFlJfGmvMJkHEzlyQ8xig7059PPjWpSEnXQ07KJqoXuw2H36NkUV-cC3G_iPNeokZP1ImJ5tCBvMoFSfmMJrZBmUR9rf0Hp024Wta2KnYmzAQm0nHN-urE6SvBJwqAWxfCFl5jDL5bt_8fsxrLH-RTOjv6uFjeL5X06_AE-GZHQ=&lang=sage&interacts=eJyLjgUAARUAuQ==]Here is a Sage program[/url] that does exactly what you were doing -- i.e., constructs the lexicographically smallest sequence $\left(1 = k_1 < k_2 < k_3 < \cdots\right)$ satisfying $\left\vert \frac{1}{k_i}-\frac{1}{k_{i+1}}\right\vert>\varepsilon$ for all $i$. The input $k$ is $\dfrac{1}{\varepsilon}$ (assumed to be an integer). The $k$s at which this sequence becomes longer are
\[
1,2,4,6,10,12,18,22,30,34,42,48,56,60,72,82,90,
\]
which is not in the OEIS. But as I said, it might not be the right question to ask, if the cutoffs are not at reciprocals of integers.

OK, thanks, that's interesting.
Meanwhile, I found that
\[\frac{1}{N^2} \lesssim \varepsilon \lesssim \frac{4}{N^2}\]
which is remarkably consistenst with the asymptotics predicted by the (apparently wrong) OEIS entry. Of course the proof of both sides is not very hard and the results can probably be slightly improved.

In fact, I am now pretty sure that both my upper and my lower bound can be improved to meet at
\[\varepsilon \sim \left(\frac{c}{N}\right)^2\]
where
\[c=\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\frac{1}{5\sqrt{4}} \approx 1.86003\]
so that $c^2 \approx 3.45>\pi$. Indeed, this is not very hard:
For the numbers $\frac{1}{k}$ with $k \lesssim \frac{1}{\sqrt{\varepsilon}}$ we have to take singletons.
For those with $\frac{1}{\sqrt{\varepsilon}} \lesssim k \lesssim \sqrt{\frac{2}{\varepsilon}}$ we have to take two-element sets.
For those with $\sqrt{\frac{2}{\varepsilon}} \lesssim k\lesssim \sqrt{\frac{3}{\varepsilon}}$ we have to take three-elements sets.
Etc. This yields a lower bound
\[N \gtrsim \frac{1}{\sqrt{\varepsilon}} \cdot \left(1+\frac{\sqrt{2}-1}{2}+\frac{\sqrt{3}-\sqrt{2}}{3}+\dots\right)\]
which after rearranging amounts to exactly the number claimed above.
On the other hand, up to negligible errors at the ends of the intervals, we can actually make such a construction as described above.
This settles the question of asymptotic growth (and also shows again that it can't be the OEIS entry).

Questions that remain are those of a more arithmetic nature: Is it true that the cutoffs are reciprocals of integers and can we say anything about the sequence of rational cutoff numbers that we get, beyond its growth for $N \to \infty$?

Actually, the cutoffs are [b]not[/b] always reciprocals of integers. The change from $23$ to $24$ intervals happens somewhere between $\dfrac{1}{161}$ and $\dfrac{1}{162}$, as I've just [url=https://sagecell.sagemath.org/?z=eJxlUMtuwyAQvPsrVuoFlMYOPVRq1XxEpN6iqiJh3aDiBWFI_fldsOu-9sIyzAy7Y7CH3nkfxSQfG-C6me_ge9AQdbKetAPKwwkjTC2EiGc7ItghOByQUqW0VTvBHg4Htqq3iClHYg2LkZ34Ewldx4BB8oOlGWoaw0Mw59VSwnjVbhQYxnWc5Wue5-z8iAZWWsEc0lu6AAuAEA2aWaZ5kqN6qf3HxTqE55hxfivFW1yZoo_bhVTK9iBUxwuUVwnb4irhaQ-7b2GpU0T9viKEU2IrBZslSdX9d5ErXbc6BCQjiu5XULyK0Esc-UcEIfLGbMnI7Z-cKsjmTRbqXsn52KhO7b76h7W_k5-Xjova&lang=sage&interacts=eJyLjgUAARUAuQ==]checked with Sage[/url].


calculus
integration
function
trigonometry
analytic geometry
symmetry
algebra

sorry for double posting, but isnt no one trying this one? :maybe:  :(

Sometimes, double integrals or single ones are boring to solve because of big integrands. There're some cases when clever substitutions turn that ugly integrand into a nice one, may be that's why no one wants to solve your problem.

But, you're gonna have to wait, here there're a lot of smart people, so I'm pretty sure you're gonna get a nice solution, or even straightforward.

Here is a somewhat nasty solution:

Let $ f$ be a function of the class $ C^1 [0, a]$. For such $ f$, define $ I(f) = \int_{0}^{2a} \int_{0}^{\sqrt {2ax - x^2}} \frac {x (x^2 + y^2) f'(y)}{\sqrt {4a^2 x^2 - (x^2 + y^2)^2}} \, dydx$. Then

\begin{eqnarray*} I(f) & = & \int_{0}^{\frac {\pi}{2}} \int_{0}^{2a \cos\theta} \frac {r^3 \cos\theta f'(r \sin\theta)}{\sqrt {4a^2 \cos^2\theta - r^2}} \, drd\theta \qquad (\text{polar coordinate}) \\
& = & 8a^3 \int_{0}^{\frac {\pi}{2}} \int_{0}^{1} \frac {s^3 \cos^4\theta f'(2a s \cos\theta \sin\theta)}{\sqrt {1 - s^2}} \, dsd\theta \\
& = & 2a^3 \int_{0}^{1} \int_{0}^{\frac {\pi}{2}} \frac {s^3 (1 + \cos2\theta)^2 f'(a s \sin2\theta)}{\sqrt {1 - s^2}} \, d\theta ds \\
& = & 2a^3 \int_{0}^{1} \int_{0}^{\frac {\pi}{2}} \frac {s^3 (1 + \cos^2 \phi) f'(a s \sin\phi)}{\sqrt {1 - s^2}} \, d\phi ds \qquad (\phi = 2 \theta \text{ and by symmetry}) \\
& = & 2a^3 \int_{0}^{\frac {\pi}{2}} \int_{0}^{\frac {\pi}{2}} \sin^3 \varphi (2 - \sin^2 \phi) f'(a \sin \varphi \sin\phi) \, d\phi d\varphi \qquad (s = \sin\varphi) \end{eqnarray*}

Now let's consider the case when $ f'(x) = x^n$ for $ n = 0, 1, 2, \cdots$. Then by beta function identity, we have

\begin{eqnarray*} I(f) & = & 2a^{n + 3} \left[ \int_{0}^{\frac {\pi}{2}} \sin^{n + 3} \varphi \, d\varphi \right] \left[ \int_{0}^{\frac {\pi}{2}} (2 \sin^n \phi - \sin^{n + 2} \phi) \, d\phi \right] \\
& = & \frac {a^{n + 3}}{2} \beta\left( \tfrac{n + 4}{2}, \tfrac{1}{2} \right) \left[ 2 \beta \left( \tfrac{n + 1}{2}, \tfrac{1}{2} \right) - \beta \left( \tfrac{n + 3}{2}, \tfrac{1}{2} \right) \right] \\
& = & \frac {a^{n + 3}}{2} \frac {\left(\frac {n + 2}{2}\right)!\left( - \frac {1}{2}\right)!}{\left(\frac {n + 3}{2}\right)!} \left[ 2 \frac {\left(\frac {n - 1}{2}\right)!\left( - \frac {1}{2}\right)!}{\left(\frac {n}{2}\right)!} - \frac {\left(\frac {n + 1}{2}\right)!\left( - \frac {1}{2}\right)!}{\left(\frac {n + 2}{2}\right)!} \right] \\
& = & \pi a^2 \cdot \frac {a^{n + 1}}{n + 1}. \end{eqnarray*}

Now, let $ p_n$ be a sequence of polynomials on $ [0, a]$ converging uniformly to $ f'$, and let $ \textstyle P_n(x) = f(0) + \int_{0}^{x} p_n (t) dt$.

Then clearly $ I(p_n) = \pi a^2 (P(a) - P(0))$ and $ I(p_n) \to I(f)$.

Since $ P_n$ converges uniformly to $ f$, we have $ I(f) = \pi a^2 (f(a) - f(0))$ as desired.

... But this solution is not so satisfying for me.  :(  Is there someone who has a nice solution or knows some essential reasons why this integral gives rise to a constant multiple of $ \textstyle \int_{0}^{a} f'(x) dx$?

I guess the 'essential reason' is that $ \int_{a - \sqrt {a^2 - y^2}}^{a + \sqrt {a^2 - y^2}}\frac {x(x^2 + y^2)}{\sqrt {4a^2x^2 - (x^2 + y^2)^2}}dx$ is a constant independent of $ y$ for $ y\in \left(0,a\right)$, and the constant happens to be $ \pi a^2$.  Though my solution is still pretty ugly.

If we switch the order of integration the original integral becomes,
\begin{eqnarray*} I & = & \int_0^af'(y)\left[\int_{a - \sqrt {a^2 - y^2}}^{a + \sqrt {a^2 - y^2}}\frac {x(x^2 + y^2)}{\sqrt {4a^2x^2 - (x^2 + y^2)^2}}dx\right]dy \\
& = & \int_0^af'(y)\left[\int_{a - \sqrt {a^2 - y^2}}^{a + \sqrt {a^2 - y^2}}\frac {x(x^2 + y^2 - 2a^2)}{\sqrt {4a^2x^2 - (x^2 + y^2)^2}} + \frac {2a^2x}{\sqrt {4a^2x^2 - (x^2 + y^2)^2}}dx\right]dy \end{eqnarray*}
After laborious algebraic manipulation omitted here, one can find that the first integrand is of the form $ - \frac {u'}{2\sqrt {u}}$ where $ u(x) = \frac {1}{4}\left(4a^2x^2 - (x^2 + y^2)^2\right)$ and the second is of the form $ \frac {a^2v'}{1 + v^2}$ where $ v(x) = \frac {u'}{2x\sqrt {u}}$  (primes are understood to be differentiation with respect to $ x$ with $ y$ taken constant throughout), so we have
\[ \int \frac {x(x^2 + y^2)}{\sqrt {4a^2x^2 - (x^2 + y^2)^2}}dx = a^2\arctan \left( \frac {u'}{2x\sqrt {u}}\right) - \sqrt {u}
\]
It's straightforward to verify that $ u\left(a\pm \sqrt {a^2 - y^2}\right) = 0$, $ u'\left(a + \sqrt {a^2 - y^2}\right) > 0$, and $ u'\left(a - \sqrt {a^2 - y^2}\right) < 0$ for $ y\in (0,a)$, and the conclusion follows.

Wow, I didn't believe that it can be done in rectangular coordinate until I saw your solution, JoeBlow!

For calculation, I think it is inevitable to pass through some nasty work to reach the answer. Well, but don't blame on your solution. This is is even neater than some messy exercises in calculus.

[quote="sos440"] $ = \& 2a^3 \int_{0}^{1} \int_{0}^{\frac {\pi}{2}} \frac {s^3 (1 + \cos^2 \phi) f'(a s \sin\phi)}{\sqrt {1 - s^2}} \, d\phi ds \qquad (\phi = 2 \theta \text{ and by symmetry})$[/quote]
Canyou please say ow this step comes from the previous one? or am i missing something? :maybe:

[quote="Euclidean Geometer"]Canyou please say ow this step comes from the previous one? or am i missing something? :maybe:[/quote]

Using the substitution $ \phi \equal{} 2\theta$, we have

$ 2a^3 \int_{0}^{1} \int_{0}^{\frac {\pi}{2}} \frac {s^3 (1 \plus{} \cos2\theta)^2 f'(a s \sin2\theta)}{\sqrt {1 \minus{} s^2}} \, d\theta ds \equal{} a^3 \int_{0}^{1} \int_{0}^{\pi} \frac {s^3 (1 \plus{} 2\cos\phi \plus{} \cos^2 \phi) f'(a s \sin\phi)}{\sqrt {1 \minus{} s^2}} \, d\theta ds.$

Note that $ f(2a\sin(\pi \minus{} \phi)) \equal{} f(2a\sin \phi)$ but $ \cos(\pi \minus{} \phi) \equal{} \minus{}\cos\phi$. So

$ \int_{0}^{\pi} \cos\phi f'(a s \sin\phi) \, d\theta \equal{} 0$

and

$ \int_{0}^{\pi} (1 \plus{} \cos^2 \phi) f'(a s \sin\phi) \, d\theta \equal{} 2 \int_{0}^{\frac{\pi}{2}} (1 \plus{} \cos^2 \phi) f'(a s \sin\phi) \, d\theta$.

Thus we have the given identity.


geometry
trapezoid
Gauss
conics
geometry unsolved

Let ABCD be a quadrilateral . $ M , N $ are the midpoints of $ AC , BD $ .
" There exists $ I \in MN $ such that : $ d(I,AB) = d(I,CD) $ and $ d(I,AD) = d(I,BC) $ ".
 In which cases , this clause is true ?
 a) ABCD is a trapezium
 b) ABCD is a inscribed quadrilateral
 c) ABCD is a convex quadrilateral  ??

The clause is trivially true in case [b]a)[/b] (or, almost trivially); it also seems true in case [b]b)[/b] (I'll still have to check), and it is generally not true in case [b]c)[/b]. Beautiful problem - remains to prove case [b]b)[/b].

  Darij

Dear Darij
suppose that A,B,C,D lie on a circle (O). Let U = AB ^ CD and V = AD^BC.
As the similtude mapping UAC to UDB maps M to N, it follows that UM/UN = AC/BD and that the pairs of lines (UM,UN) and (AB,CD) have the same bisectors. Same way, we have VM/VN = AC/BD.
As (Gauss theorem) the midpoint of UV lies on the line MN, if follows that, for any point P on the circle with diameter UV, we have PM/PN = AC/BD.
Now, if I is a common point of the circle of diameter [UV] with the line MN, as UM/UN = IM/IN = VM/VN, the line UI bisects (UM,UN), thus UI bisects (AB,CD); same way the line VI bisects (AD,BC).
Thus, we get two solutions for your problem : the common points of the line MN with the circle with diameter [UV].
Kind regards

Dear Darij and nttu, let us try a little conjecture :
There exists I on the line MN such as d(I,AB) = d(I,CD) and d(I,AD) = d(I,BC) if and only if we are in one of the four following cases :
1) AB and CD are parallel
2) AD and BC are parallel
3) A,B,C,D lie on a circle
4) The lines AB, BC, CD, DA touch a circle

I think that it could be interesting to prove - or disprove - this conjecture.
Kind regards

I'm sorry; I've forgotten something.
We know that the property is true in cases 1-2-3.
It is clearly true in case 4 : the line MN is the Newton line of the complete quadrilateral with sidelines AB, BC, CD, DA; thus any conic touching these four lines in centered on MN. Hence, the center of the circle lies on MN.
It remains to see whether there exist other cases.
Kind regards

I have an idea for the reciprocal :
Suppose that I lies on MN such as d(I,AB) = d(I,CD) and d(I,AD) = d(I,BC)
There exist a conic with center I touching AB, BC, CD, DA at A', B', C', D' (because MN is the Newton line of the complete quadrilateral)
Suppose that the conic is not a circle;
then either AB and CD are parallel or A'C' is parallel to an axis of the conic
Same way, either AD and BC are parallel or B'D' is parallel to an axis of the conic
I have checked, by computation, that if A'C' is parallel to an axis andif B'D' is parallel or perpendicular to A'C', then A,B,C,D are concyclic but I didn't find a synthetic proof.
In any case, the conjecture is true.
Kind regards


Mr. James wanted to plant a garden that would be in the shape of a
rectangle. He was given 80 feet of fencing to enclose his garden. He
wants the length to be 10 feet more than twice the width. What are
the dimensions, in feet, for a rectangular garden that will use exactly
80 feet of fencing?

i_like_pie is what I wanted.

I never simply want an answer to any question.

I wanted steps, which i_like_pie provided.


[quote="Ashegh"]tell me why did u choose this name for ure user name.

i choosed Ashegh, because it means lover. and im really lover :!:[/quote]

i choosed bubka, because it means bubka. and im really bubka :!:

cotls18

Lol.

I like the colts....Peyton Manning...
I was bored.
Couldnt think of a better name?

i actually planned my username out years before i actually made it

In 10th grade in high school my friend and I invented a guy named "Xevarion Thoobikishnapov" and ridiculous stuff about him. My username comes from that. (Also "thoobik" which is my user name on some other boards -- but on this board I believe "thoobik" is him!)

my mom made this user.
she made it my korean name for some reason.

ky is my dads first and middle initials.

yuan is my last name.

Mathcount is a misspelling of MATHCOUNTS, which I messed up on.

A. Kalra #1

A is my first initial, Kalra is my last name, #1... no explanation needed ;)

I choose 142857 because it is a great number. See my signature.

lol. :)

[color=white]val;kjdlsdaslkdsjkasdlk[/color]

my explanation's slightly different each time I tell it, it's a satanic reference this time around

ok, i am a poser, this is not my sn, but my friends. mymom told me about this math website when i used to play halo 2 online, any one who's played knwo that stuff is addicting. So i'm in the middle of a team slayer match and it's intense, so my mom asks me for a screen name, i told her I'd do it l8er, she starts namin off these stupid names like mathwizz, or ctmathman or something like that, and i just saw my friend's sn on xbox live, maniacman336, i so i told her to make it maniacman384. and yep, thats how i got my sn, pwned all of you with the longest post. beat that....

my name is Alex....And my school name's initials are I.B.

we can also have disscusion about, why we choosed these pictures for

our user.

i choosed this picture because this is my photo,and i dont like to choose

any picture instead of it :D

Mine is the football (soccer) ref Collina. I like the pic b/c he looks so intense. :D

mine's a schizophrenic club made of 4 people but 2 souls(and surprise i'm in it...!)

Well mine is the way everyone called(and still calls) me at the school I went to in Russia(I live in Canada now) :)


Doing some stuff with Lp spaces, I have something like:

$ \lVert g \rVert_p^{\frac{p}{q}}$

Any suggestions on how to space that better?  It would be nice if the subscript was a bit lower so that it dropped below the line.  But if there is no simple way to do this, I really don't want to get into manually spacing it.

LaTeX puts the subscript as high up as it can without hitting the superscript. This means that you can lower it using phantom tall letters which LaTeX has to make room for without printing anything. So instead of $ \lVert g \rVert_p^{\frac{p}{q}}$ you get $ \lVert g \rVert_{\vphantom{P}p}^{\frac{p}{q}}$ or even $ \lVert g \rVert_{\vphantom{P^P}p}^{\frac{p}{q}}$. You can try out all sorts of things such as $ \lVert g \rVert_{\vphantom{P^{x^2}}p}^{\frac{p}{q}}$

But if you want a bit more fine control use \raisebox to lower the p by whatever amount that you want: $ \lVert g \rVert_{\raisebox{\minus{}0.5ex}{\(\scriptstyle p\)}}^{\frac{p}{q}}$
([i]I had to use \(...\) instead of dollar signs here to stop it being misinterpreted on the forum. In a document you can use dollar signs[/i]).


Complete and balance the following chemical reactions:

a) HCl(aq) + AgNO3(aq) -> AgCl(s) + HNO3(aq)

Is that right, or is there a better answer?

b) K2CO3(aq) + Ba(NO3)2(aq) ->

I'm not sure how to work this one.

c) Pb(NO3)2(aq) + 2KCl(aq) -> PbCL2(s) + 2KNO3(aq)

Once again, is this right, or is there a better answer?

d) Ba(NO3)2(aq) + Na2CrO4(aq) -> BaCrO4(s) + 2NaNO3(aq)

Is this right, or is there a better answer?

Please help me!

I think that a, c, and d are correct.

[hide="b"]$K_2CO_3+Ba(NO_3)_2\rightarrow BaCO_3+2KNO_3$[/hide]

[quote="g1sk8terz"]yea and also...

[hide]Example 2,
the $BaCO_{3}$ is a solid.[/hide][/quote]

Solubility rules!  :)


linear algebra
matrix
analytic geometry
vector
linear algebra unsolved

Let $A: V\rightarrow V$ be a linear map and suppose that A commutes with every other linear map, then show that $A=cI$ for some c in F ( the underlying field)

Choose a basis of V and suppose that for every matrix B AB=BA (this is equivalent to your problem). Now let $c^{1}_{ij}$ mean the ij entry of the matrix AB and $c^{2}_{ij}$ the same for BA. Since AB=BA, we have $c^{1}_{ij}= c^{2}_{ij}$ for every i, j. Using the formula for matrix multiplication we have ${{c^{1}_{ij}= \sum_{k}a_{ik}b_{k}j}= \sum_{k}b_{ik}a_{k}j}= c^{2}_{ij}$. Notice that if i=j, the term $a_{ii}b_{ii}$ appears in both sides of equality and is the only term to do so. By reducing $a_{ii}b_{ii}$ in every equality we get $n^{2}$ equations for $b_{ij}$, and every $a_{ij}$ except for $a_{ii}$ appears as a coefficient in the equations. Since equality AB=BA holds fo r every B, every set of $b_{ij}$ must satisfy our equations, and this is only possible if all $a_{ij}= 0$ (this can be easily deduced from Kronecker-Capelli's theorem). It remains to prove that all $a_{ii}$ are equal, which should be easy now.

Here's a "coordinate-free" way: first show that any non-zero vector $v\in V$ is an eigenvector of $A$.  For otherwise, $w=Av$ would be linearly independent to $v$, so there's a linear map $B$ with $Bv=0$, $Bw=v$, and then $v = Bw = BAv = ABv = 0$, a contradiction.  Now if $v,w$ are any linearly independent vectors, then $Av=\lambda v$ and $Aw=\lambda' w$, and taking a $B$ which switches $v$ and $w$ shows that $\lambda=\lambda'$.

Here's another proof for the case $\mathbf F = \mathbf C$.

Let $J$ be a Jordan canonical form of $A$, then $A = T^{-1}JT$ for some $T \in \mathrm{GL}_{n}(\mathbf C)$. It now follows that $A=J$. Indeed, $JT = TA = AT = T^{-1}JT^{2}$, hence $J = (JT)T^{-1}= T^{-1}JT = A$. Since Jordan blocks may be arranged in arbitrary order, $A$ has a unique eigenvalue $\lambda$.

Note that $J$ cannot contain entries of the form $\begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix}$, because
\[\begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}= \begin{pmatrix}1 & \lambda \\ \lambda & 0 \end{pmatrix}\neq \begin{pmatrix}0 & \lambda \\ \lambda & 1 \end{pmatrix}= \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix}. \]
Thus $A$ is a diagonal matrix with unique eigenvalue, hence $A = \lambda I$.


Rectangle $ ABCD$ lies in a circle $ D$ with $ AB \equal{} 6$ cm and $ CE\equal{}4$ cm. What is the number of centimeters in the length of diagonal $ \overline{AC}$?

[asy]pair A,B,C,D,F;
D=(0,0);
A=(0,12);
B=(16,12);
C=(16,0);
F=(20,0);
draw(Circle(D,20));
draw(D--(0,20));
draw(D--F);
draw(A--B--C--cycle);
label("$A$",A,W);
label("$D$",D,W);
label("$C$",C,S);
label("$B$",B,NE);
label("$F$",F,E);[/asy]

hello, i can't find the point $ E$?
Sonnhard.

[quote="GameBot"]Rectangle $ ABCD$ lies in a circle $ D$ with $ AB \equal{} 6$ cm and $ CE \equal{} 4$ cm. What is the number of centimeters in the length of diagonal $ \overline{AC}$?[/quote]
If, $ CB \equal{} 4$ cm...

$ AC\equal{}\sqrt{AB^2\plus{}CB^2}\equal{}\sqrt{6^2\plus{}4^2}\equal{}\sqrt{52}\equal{}\boxed{2\sqrt{13}}$.

In the AoPS FTW question, it says that it is $CD$, which is basically impossible.

[quote="5849206328x"]I think they meant $ CF$, since for some reason they skipped point $ E$...

Anyways, we have $ CF\plus{}CD\equal{}\text{radius}$, and $ CD\equal{}AB$, so $ \text{radius}\equal{}4\plus{}6\equal{}10$.  Since $ AC\equal{}BD$ and $ BD$ is a radius, $ AC\equal{}\boxed{10}$[/quote]

yes, i did it the same way


number theory
least common multiple
modular arithmetic

There are fewer than 6 dozen rocks in a collection.
    Grouping the rocks by 2s leaves 1 extra.
    Grouping them by 3s leaves 2 extra.
    Grouping by 4s leaves 3 extra.
    Grouping by 5s leaves 4 extra.

How many rocks are in the collection?

[hide="Solution"]
60 means they're all grouped perfectly.
59 means one's left over.

[b]59[/b]
[/hide]

[hide]
just find LMC of 2,3,4,5 and subtract 1
so 59...[/hide]

[hide]x mod 5=-1
x mod 4=-1
etc...
LCM=60
Answer=59[/hide]

[hide]
$N< 72$,

$N \equiv 1 \pmod {2}$
$N \equiv 2 \pmod {3}$
$N \equiv 3 \pmod {4}$
$N \equiv 4 \pmod {5}$

$N$ is odd, so $N=59$.
[/hide]

[hide]I didn't think of the mod arithmetic, that's a good solution. But for the rest of us: :D

Obviously, it's odd as the remainder is 1 when divided by two, and since it's four more than a multiple of five (meaning it ends in 4 or 9) it can only end in 9. Since it has to be three more than a multiple of four, it has to be one less than a multiple of twenty, so 19, 39, 59,... and the first out of these that is two more than a multiple of three is 59. If you are good you can get that in under 10 seconds.[/hide]

[hide]
find the LCM and subtract one...59[/hide]


ratio
geometry
geometric transformation
search
calculus
MIT
college

While I am good at math, my scores usually hover around the 720 range. For some reason, I keep making stupid mistakes. I guess it is because I am just inherantly stupid. How is it that I can do contest math but I can't do SAT math? I mean I can do SAT math, but I always make careless mistakes. Does everyone here have perfect 800s on the math section, or is it just me who does not? Anyway, just thinking about the SAT...

i believe that just cause you get an 800 on the math section youre still not as good as say, oh cant think of it, but it still doesnt mean youre that good at math. yes, youre good. but sat is not a good place to measure mathematical talent.

but getting an 800 is always nice :lol:

just practice them quantitative comparisons. everything else should be a breezzzeee

Just don't rush through the problems, make sure you get each one correct. You have about a minute for each problem, and you know that the problems are easy, so just calm down and be very careful in your calculations.

When I take tests I don't skip problems. (That just makes you think too much about nonmath subjects: Did I skip problem 56 or 64?)

Don't ration time as one minute a problem, since some problems are harder.  (In the SAT they rate problems easy-to-hard, so hards take more time.) Do all the problems in order.

Hope this helps!

Note: My vote in the poll was practice tests. (Wednesday I take another practice--I can't wait  :D :^3:  ) Or maybe I can wait--it's just fake.

dunno why... but I do better on real tests than practice ones...

hmm...realli...%20 of my grade of practice tests just fly right out the window when it comes to real tests...

the exception is the math section...

I should really take another practice test... its been a while.

I haven't done one in ages...and I'm taking the PSAT in a few weeks...

mathnovice-

okay, where the duck is the "Less than 700" category???  okay, just because this is a math forum doesn't mean everyone is gonna have the scores you expect...
why do u assume that just because ppl are on this forum they automatically have above a 700??

anyways, i've personally never taken the SAT I for real...but 800 on both math 1c and 2c

[quote="MithsApprentice"]Mathnovice, it really depends on the person...

Some people like to just take it slow and use their time. I'm the type that does all the problems REALLI REALLI fast. Then I've got bout 2/3 of the time left at the end to go back and catch mistakes. Sometimes I find as many as 2 on the SAT I and up to 6 on the SAT II Math IIC. But this method brings me an 800 almost every time.[/quote]

I agree... I work really fast through the problems and get done about 15 min. early.  Then, I go back and do all the problems over again from scratch.  A lot of the times this helps because you're not looking at your old work which may sidetrack you when you check it.

Only the most competitive colleges even ASK for SAT II tests. I took them for the MIT app.

[quote="JBL"]I would estimate that a high 1C score is more impressive than a medium 2C score, especially for students at schools that don't offer calculus classes.  Significantly smaller percentages of students do very well on the 1C than the 2C, so if you're in a situation where you haven't had the courses that would merit taking the 2C, a 1C score can be perfectly impressive.[/quote]

I meant like if the IC and IIC were the same, the IIC would sound more impressive...but I may be mistaken...it's just that way here...

for me, perfect on IC is FAR more impressive, cuz one wrong means you have like a 760, on 2c, if u miss 1(or 5), nothing happens.....leaves no rooms for stupid mistakes...

i took 1c beginning of soph. year, cuz i was a wuss and didnt want to take the 2c...800 (800 later on the 2c as well)

...but the IIC is harder...so althought it wouldnt matter for most of us, to the more common kid, the IIC score is more impressive...

Be careful about the distinction between "is more difficult" and "covers more advanced material."  The two aren't necessarily synonymous.

Hmm...personally I thought that it was both "more difficult" and "covered more advanced material"...but maybe the "more difficult" is just my opinion...

If it's more difficult, how come an 800 is only 91st percentile (or whatever it is -- it changes from test to test, but the point is many many people do very well on it.)

Those who take the Math IIC are a self-selected group, generally those who are strong at math.  That, together with the more generous curve near the top, explains why there are more 800's on the Math IIC (about 9% as you say, compared to 1% on the Math IC).

I would still say that the Math IIC is the more difficult test.  For most people, if they took both, I would expect their raw score (0 to 50) to be lower on the Math IIC.  For most (average) people, I would also expect their scaled score (200 to 800) to be lower on the Math IIC.

But, and this is a big but, for high-end kids (like those on this site), it is easier to get a high score on the IIC than the IC.  I could probably take the IIC test 100 times in a row and get an 800 each time; what's the odds of getting 5 questions wrong?  But I don't think I could take the IC test say 3 times in a row and be confident of getting an 800 each time; just one wrong answer blows the 800.

I think one reason it is sometimes hard to get an 800 on the SAT I is that there are verbal sections.  You have to change the mode of your brain quickly, and you also might be worn out by the verbal sections.

-interesting_move

im hardly worried about "switching modes"...I'm more worried about my verbal score...

im taking SAT math IIC in december, in a week i think, do you guys have any recommendations for books with realistic questions?  I already have the Real SAT II book with one test, but i dont really feel comfortable with only having taken one test.

I recommend the two books by SparkNotes: "SAT II Math IIC" (with 3 practice tests and review material) and "5 More Practice Tests for the SAT II Math IIC".  You can get them from Barnes and Noble, or online at [url]http://www.sparknotes.com[/url].  Both books have some typos (as do the other books).  But regarding level of difficulty, I have found them closer to the real thing than other commercial books.  They are the books I use when tutoring students for the Math IIC (in addition to my own material).

thanks ravi, i went out and got it today.

Just curious, what else do you tutor people in besides IIC?  SAT I, general math, other?

I tutor students mostly in the Math IC and IIC tests, SAT I Math and Verbal, and academic math and computer science.

Good luck on the test!


videos
linear algebra
superior algebra
superior algebra theorems

I dun know how to find basis, rank, dimenision of a matrix.. can any1 show me example and explaination.. and also what is mean when u say spanning sets and subspace.. how u find them or show that they are subspace of a matrix..
Thanks..

You're asking for explantions of central topics of an "Introduction to Linear Algebra" course. Every textbook that could be used for such a course has explanations of these topics. If you are enrolled in such a course, go ask the instructor. It's too much to ask of the contributors to an on-line message board to ask them to act in place of the instructor of such a course.

go to http://ocw.mit.edu/OcwWeb/Mathematics/18-06Linear-AlgebraFall2002/VideoLectures/index.htm

and watch some video lectures on basic linear algebra courtesy of the nice folks at MIT.


this is a classical question of sequences and series...

find the sum of  1^1+2^2+3^3....n^n

What do you mean? Partical summ? If not, then it's infinity.

sry put some extra dots. now it is edited
i want the sum in terms of n

[url]http://www.research.att.com/~njas/sequences/A001923[/url]

Doesn't give anything too nice.


Let ABC isn't obtuse triangle  such that $ cos^3A\plus{}\sqrt{2}(cosB\plus{}cosC)\geq 2$
Find $ \widehat{A},\widehat{B},\widehat{C}$

We have $ 2\geq cos^3A + \sqrt {2}(cosB + cosC)$
proof
Let $ x = cosA$
$ cosB + cosC = 2cos\frac {B + C}{2}cos\frac {B - C}{2}\leq 2cos\frac {B + C}{2} = 2sin\frac {A}{2} = 2\sqrt {\frac {1 - x}{2}}$
It is suffice to show $ x^{3} + 2\sqrt {1 - x}\leq 2$ or $ x^{3}\leq 2(1 - \sqrt {1 - x)}$
Multiply both side $ 1 + \sqrt {1 - x}$.
Then become  $ x^{3}(1 + \sqrt {1 - x})\leq 2x$. which is true since $ 1 + \sqrt {1 - x}\leq 2,0\leq x^3\leq x$. equal when $ x = 0$,$ B = C$. ie,$ A = 90,B = C = 45$.


vector
inequalities proposed
inequalities

Given a set of $111$ vectos of length $1$ whose sum is the zero vector.
Prove one can select $55$ of them such that their sum has lenght at most $1$

I proved this:
If we have $k$ of this unit vectors whose sum has length less than $1$, then we can choose $1$or $2$ of the remaining unit vectors such that the sum of the $k+1$ or $k+2$ vectors is less than. With this proved the problem becomes obvious because it is the same to prove the statement with 55 vectors and with 56 vectors (55+56=111)
I use this lemmas:
1- If we have $n$ vectors whose sum is zero, then we can construct with then a directed convex polygon.
2- In a triangle the side with its opposite angle less than $60^o$ cannot be the longest side.


Does anyone know when the AMC 8 is? All I know is that it is sometime in November.

You can refer to here if you want to know when a AMC-based contest is: http://www.unl.edu/amc/whatswhen.shtml


Hello. I am not so good with regular euclidean geometry, especially with seeing similarity, congruency, et cetera. I am pretty sure that I know the necessary and basic formulas and principles, except I feel as though I am lacking in practice in identifying stuff. I was wondering if you guys have any tips or know of any places where I can find practice problems. I am looking particularly for questions that involve drawing in your own lines and stuff and are at an AIME difficulty about. Any help is greatly appreciated. Thanks. :)

A great source of problems like that is Challenging Problems in Geometry. The first half covers regular euclidean geometry and the second half goes into more advanced stuff like concurrency/collinearity with Menelaus/Ceva. (It's just a problem/solution book btw, it has no explanation of concepts.)

Do geometry problems from the [url=http://www.artofproblemsolving.com/Forum/resources.php]Resources section.[/url]


I live in Japan, and took the AMC 10B at a nearby international school. I got a 132, which I know to be above the qualifying score for the AIME. However, my test proctor emailed me the day before the AIME II that they hadn't received any materials for the AIME. Upon further inquiry, I found out that they hadn't even gotten the AMC results mailed to them (there was an email, though, with the results).

Due to spring break, school had resumed the Monday of the same week of the AIME - I don't think they had much time to check with the MAA or anything..

But could someone suggest what might've happened, or if anyone else has experienced this?

EDIT: and erm fyi, there were students from the school taking the AMC with me, too, though I was the only AIME qualifier.

Too bad I couldn't take it this year.. but oh well  :|

[quote]But could someone suggest what might've happened, or if anyone else has experienced this? [/quote]

What probably happened was the shipment of results, certificates and AIME materials got hung up in customs, or elsewhere.  An email with a UPS tracking number is generated and sent automatically to  the contest manager when the package is sent.  Once the package leaves the building, we have no more ability to find it than the teacher who has the email with the UPS tracking number.

Yes, this happened to others, in the Philipines, Turkey, Switzerland, and Korea that I personally know of and probably elsewhere too.  If the contest manager contacts us in time, we can sometimes figure out a way to expedite the materials to the school.

But I will be honest with you, our schedules are designed to work well with shipping in North America.  We do our best, UPS does its best, usually it is good enough but once the package leaves the US, we have very little control anymore.

Steve Dunbar
MAA Director, American Mathematics Competitions

Ah I see..

In such a case, when the proctor contacts the AMC about it, do you usually resend the certificates? It'd be nice to have those at least, since the AIME doesn't have window dates :lol:

Thank you for the clarification, and thanks in advance!


calculus
derivative
real analysis
real analysis unsolved

Determine $ \sum_{n\equal{}0}^{\infty}n e^{\minus{}nx}$ for each $ x>0$

what is the derivative  of $ f(x) \equal{} \minus{} \sum_{n} e^{ \minus{} nx}$ ?
what is Geometric progression ?
EDIT :

derivetive, gometric, progersion

Please use a dictionary or read first what you have written before you press the "submit" button.

$ nr^n\equal{}\frac{r}{(1\minus{}r)^2}[\{(1\minus{}r)n\plus{}r\}r^{n\minus{}1}\minus{}\{(1\minus{}r)(n\plus{}1)\plus{}r\}r^n]$ follows Telescoping Method.


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra unsolved

Show that, in a unitary ring $R$ of characteristic $2$, the identity $x^6=x$ implies that the ring is Boolean. In other words, show that if $x^6=x$ for all $x \in R$, then $x^2=x$ for all $x \in R$ (the polynomial identity $x^6=x$ implies the polynomial identity $x^2=x$).

First observe that the ring has characteristic $2$, then use $(1+x)^6=1+x$, from which we can deduce $x^4+x^2=0$, meaning that $x^4=x^2$. We thus get $x=x^6=x^4=x^2$, Q.E.D.

[quote="vess"]Good! On the other hand, note that the identity $x^{12}=x$ [b]does not[/b] imply that the ring is Boolean (why?). So, take the following more general problem:

Characterize all $n \in \mathbb{N}$ for which the identity $x^{2n}=x$ in a finite unitary ring (it must obviously have characteristic $2$) implies the identity $x^2=x$.

Enjoy!  :lol:

--Vesselin[/quote]

dear Vess, you can write down the universal one-generator ring of this type and decompose it as direct product of p-torsion subrings.  It is many years since I tried
to work out what this looks like, but I think this procedure should be quite relevant to
the above problem.

I think the following holds:

Given $n\ge 2$, there is a non-boolean finite ring $R$ of characteristic $2$ such that $x^{n}=x,\ \forall x$ iff $n-1$ has a divisor of the form $2^{a}-1,\ a>1$ (notice that my $n$ plays the role of vess’ $2n$).

The sufficiency of that condition:

Take  $R=\mathbb F_{2^{a}}$, the field with $2^{a}$ elements. Since $2^{a}-1|n-1$ and for all $0\ne x\in R$ we have $x^{2^{a}-1}-1=0$, we also have $x^{n-1}-1=0\Rightarrow x^{n}=x$. This also holds for $x=0$, of course, and we’re done (notice that $R$ is not boolean because $a>1$).

The necessity:

There is an element $t$ in $R$ such that $t^{2}\ne t$, and we may assume WLOG that $R$ is generated by $t$ (as an $\mathbb F_{2}$-algebra), and is hence commutative. Notice also that $R$ has no non-zero nilpotent elements. The finite commutative rings with trivial nilradical are precisely the finite direct products of finite fields (this follows from the [url=http://planetmath.org/encyclopedia/WedderburnArtinTheorem.html]Wedderburn-Artin theorem[/url], characterizing the left Artinian rings with trivial Jacobson radical as being the finite direct products of matrix rings over division rings). Since our ring has characteristic $2$, it must be a direct product of the form $\bigotimes_{i=1}^{k}\mathbb F_{2^{a_{i}}}$ with, say, $a=a_{1}>1$, because it's non-boolean. This means, in particular, that all elements $x$ of $\mathbb F_{2^{a}}$ satisfy $x^{n}-x=0$, and this implies $2^{a}-1|n-1$.

P.S.

And it follows from the above or, alternatively, from simple computations, that if $x^{12}=x,\ \forall R$, then the ring $R$ [i]is[/i] boolean :).


AX=0 has a non trivial solution  if and ony if  A is singular.

this can be proved in 2 ways:

 method 1:  non trivial solution $ \Rightarrow$ A us sungular

method 2:   A is singular $ \Rightarrow$ non trivial solution.

can anyone provide me a [b]solution using method 2[/b] ??

That's not two "methods of proof" - that's the two sides of an if and only if.

This is all about definitions. What is the definition of "singular" that you are proceeding from?

the definition with whihc i am proceedng is  :

[b]A square matrix that is not invertible is called singular[/b]


If 1995^10=9986860883748524N5070273447265625 and N is a digit, what digit must N be?

This was in the marathon and I didn't get how to do it...

[quote="13375P34K43V312"][hide]$1995$ is a multiple of 3 so $1995^{10}$ is a multiple of 9, so the sum of all the digits is divisible by 9. The sum of digits is (wow this looks like fun!)

$9+9+8+6+8+6+0+8+8+3+7+4+8+5+2+4+N+5+0+7+0+2+7+3+4+4+7+2+6+5+6+2+5=160+N$, the only value of $N$ such that $160+N$ is clearly $\boxed{2}$.

don't tell me I added wrong...[/hide][/quote]
That agrees with my solution.
You don't actually have to add them all, you can just ignore the 9s and add up the rest, modding out when necessary.

same here, I didn't notice 1995 is divisible by 3 either!!!

[quote="diophantient"]
You don't actually have to add them all, you can just ignore the 9s and add up the rest, modding out when necessary.[/quote]

oh yeah I forgot about the continuous sod function.

i think just keeping track of the sum in mods

jorian

So is it the continious [b]m[/b]od function?

lol, probably (don't trust me on this though)

jorian

lol I noticed that 1995 was divisible by 3 but not the ^10 divisible by nine thingy XD

[quote="magixter"]Oh, that reminds me, does anybody know the divisibility rule for 27? Just wondering...[/quote]
It's not pretty like 9.
Just divide by 9 and check for 3.
The reason 9 is pretty is because 10-1=9
So all factors of 9 work nicely.

[quote="magixter"]Wait, I don't get it. Why is 10-1 so special?[/quote]
Eh, make that $10^{k}-1$ is always divisible by 10-1.

couldn't it also be because 10 is 1 mod 9? im not sure

$10^{k}\equiv 1 \pmod{9}$ so $a10^{k}\equiv a \pmod{9}$. So every digit is as appears mod 9 regardless of the place.

[hide="generalization"]
$x^{k}-y^{k}$ is divisible by $x-y$
Let x=10 and y=1 to get what we have.[/hide]
[hide="proof"]
Let $x^{k}-y^{k}=0$
The roots of this are factors of the LHS
$x^{k}=y^{k}$
$x=y$ is a root
$x-y=0$
So x-y is a factor[/hide]


trigonometry
inequalities
inequalities unsolved

Prove that in any triangle where are angles are $<90$ we have

$\sin 2A + \sin 2B + \sin 2C \leq \frac{3\sqrt{3}}{2}.$

I think it is famous.
for $a_1,a_2,\cdots,a_n.a_i<180$
we have $\sin a_1+\sin a_1+\cdots+\sin a_n  \leq n \sin \frac{a_1+a_2+\cdots +a_n}{n}$
if I did't rememeber wrongly. :)

zhaobin, in vasc's question it's sin2A instead of sinA.
anyway i think that
\[
\begin{array}{l}
 \sin 2A + \sin 2B + \sin 2C \\
  = 4\sin A\sin B\sin C \\
  \le 4\left( {\frac{{\sin A + \sin B + \sin C}}{3}} \right)^3  \\
  \le 4\left( {\frac{{\sqrt 3 }}{2}} \right)^3  \\
  = \frac{{3\sqrt 3 }}{2} \\
 \end{array}
\]

siuhochung, can we use $f(x)=\sin x$ is  $convex$ in $[0,\pi]$
then prove $\sin a_1+\sin a_1+\cdots+\sin a_n \leq n \sin \frac{a_1+a_2+\cdots +a_n}{n}$?

Also  you  can  prove  the  initial  ineq  by  aplying  the well  known  ineq  [tex]sinA+sinB+sinC\leq\frac{3\sqrt{3}}{2}} [/tex]  to  the  triangle  with  angles  [tex]180-2A , 180-2B , 180-2C [/tex]

lets give another proof:

$sin(x)+sin(y)\leq 2sin{{x+y} \over 2}$ for $x,y<\Pi$
so
$sin(2A)+sin(2B)+sin(2C)\leq 3sin({2\Pi \over 3})={3\sqrt{3}  \over 2}$

I can prove this ineq :
[tex]sin(nA)+sin(nB)+sin(nC) \leq \frac{3\sqrt{3}}{2} [/tex] with [tex]n \geq 1 [/tex]


Find the ratio of the energies of the K characteristic x rays of carbon to those of Magnesium.

I did it using Bohr's expression of energy...Is there any other method to do it?

You cannot use Bohr's theory here- the atoms are not hydrogen-like. I do not know if this can be done without using formal QM. You can, however use Moseley's experimental result $ \sqrt\nu\equal{}a(Z\minus{}b)$ with $ b\approx 1$ for $ K_{\alpha}$ frequency.

OK...I see.

Actually Bohr's formula fits to the answer of the MCQ quite nicely...
You are correct.Infact,since the atoms are not Hydrogen like,I posted this problem here...


geometry
rectangle
trigonometry
function
analytic geometry

When changing from a+bi form to r(cos theta + i sin theta), how do you get the theta measurement?

I know to get the reflex angle, you do the tan-1 of (b/a) and that gives you the reflex angle, and then from that I have to get theta, but I'm unsure as to how...

I've never heard the term "reflex angle" before, but theta is it.  Once you take that inverse tangent, you're done.  If you understand the polar and rectangular forms geometrically and draw a picture, it should be clear why this is the case.

You could make a rectangle with sides a and b.  Then you could draw a circle in the cartesian plane such that the rectangle can be inscribed in it in one quadrant.  Then you can draw the diagonal, and you have the angle.  In other words, going from a+bi to $r(\cos{\theta}+i\sin{\theta})$, you can draw a rectange with sides a and b, and then draw the diagonal.  Tan-1 b/a will give you theta, i'm pretty sure...and r=length of the diagonal.

welll....

$a+bi$ is the rectangular form

$r(\cos\theta+i\sin\theta)$ is the polar form, which is formed by this:

$r = \sqrt{a^2+b^2}$, $\theta = \tan^-1 \frac{b}{a}$

thanks---- i dunno about the whole reflex angle thing either. My teacher said that tan-1 of b/a was the reflex angle, and you have to add 180 to get theta...........

but he doesnt make much sense most of the time, and he just turned 22 too....  :?

You mean reference angle. The reference angle is the smaller angle that the terminal side makes with the x-axis. It's useful because the reference angle and the original angle itself have the same values for their trigonometric functions, differing by a sign, usually. I don't see exactly how that would apply here, except to help you with figure out $\tan^{-1} \frac ba$ without a calculator.
A reflex angle is an angle whose degree measure is between 180 and 360, exclusive.


after being thoroughly insulted on the more public threads
(link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218427[/url]

I'm going to keep my focus here.  In fact, I intend on never posting problems in the general threads again...

And now to my question.  Does anybody here attend mathematics competitions in the Panhandle area?

Basically anywhere from Pensacola to Destin..
those are all within 50 miles of my residence.


inequalities
vector
analytic geometry
geometry
inequalities proposed

For the first one:

Let u be the vector having coordinates w and y and let v be the vector having coordinates x and z. Then w^2+y^2=|u|^2=A^2 <=1 (a notation) and x^2+z^2=|v|^2=B^2. We also have wx+yz=uv (dot product of the vectors)=|u|*|v|*cosT (T is the angle between the vectors). The ineq we want to prove is:

(A^2-1)(B^2-1)<=(ABcosT-1)^2, iff A^2+B^2-2ABcosT >= A^2B^2(1-(cosT))^2=A^2B^2(sinT)^2. I assumed B<=1 also, since if it's >=1 we get an obvious ineq, the left part being <=0 and the right one >=0.

Consider the triangle formed by O (the origin) and the 2 vectors u and v. Then the expression on the left (in the last ineq I wrote) is L^2 where L is the side opposite O in this triangle and the expresion on the right is (2S)^2, where S is the area of this triangle. So the ineq we want is L^2>=(2S)^2=(h*L)^2 where h is the altitude corresponding to L in the triangle we considered. So all we must prove is h<=1. But h<=A (and h<=B), and from A<=1 we get exactly what we want.

The second one seems tough, but maybe it just seems so :D

I'm guessing that the socond one has in the right member a^(2/3)+b^(2/3) not a 2 /3....

You are right, sorry for my mistake.

I have solved it applying Power Mean Inequality.


41 times 9 equals 369

41 times 31 equals 1270

1270 plus 369 equals 1640

A faster way is this:

Factor out a 41, leaving $ 41(9\plus{}31) \equal{} 41*40 \equal{} \boxed{1640}$


calculus
integration
analytic geometry
trigonometry
limit
calculus computations

Let $ f(x,y)\equal{}(x^2\minus{}y^2)(x^2\plus{}y^2)^{\minus{}2}$, and let $ S$ be the unit square $ [0,1]\times[0,1]$.

Show that $ \iint_S|f|dA\equal{}\infty$

$ f$ is homogeneous of degree $ \minus{} 2$, that is, $ f(cx,cy) \equal{} c^{ \minus{} 2}f(x,y)$ for any $ c > 0$ and any $ (x,y)\ne (0,0)$.

Let $ cS$ be the rescaled square $ [0,c]\times [0,c]$. Note that $ S$ contains infinitely many disjoint subsets $ S\setminus (2^{ \minus{} 1}S)$, $ 2^{ \minus{} 1}S\setminus (2^{ \minus{} 2}S)$, $ 2^{ \minus{} 2}S\setminus (2^{ \minus{} 3}S)$, etc. Use the homogeneity of $ f$ (and the change of variables) to show that $ |f|$ has the same integral over all such subsets.

A lower-level argument that uses the same ideas as mlok's:

Let's change this to polar coordinates:

${ \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy=
\int_0^{\frac{\pi}2}(\cos^2\theta-\sin^2\theta)\int_0^{g(\theta)}
\frac1r\,dr\,d\theta}$

(We could easily give the exact formulas for $ g(\theta)$ but that's not critical to the argument).

Note that the inner integral, $ \int_0^{g(\theta)}\frac1r\,dr,$ is divergent for all values of $ \theta.$ Hence the integral diverges.

However, as something akin to a principal value: Let $ S=[0,1]\times[0,1].$ Prove that

$ \lim_{\delta\to0}\iint_{S\setminus B(0;\delta)}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy=0.$


I cannot remember if I have already posted this question here...
However,if that is the case,let it be said that I want to discuss more about it.

Let us define 4 vector by 4 co-ordinates (x1,x2,x3,x4) where (x1,x2,x3) are space components (like x,y,z) and x4 is related to time as x4=ict.Express the following equations in tensor notation.

(i)The continuity equation: div J+(del*rho/del t)=0

(ii)The wave equation del2ø-(1/c2) [del2 ø/del t2]=0

(iii)What will be the value of j4 instead of above?

2. Relevant equations

3. The attempt at a solution

For the first two part I am getting these:

Please check...

Let me know if I am correct.

(i) (∂/∂x_i)(J_i)+ (∂/∂t)ρ=0

(ii) [∂^2/(∂x_i*∂x_i)]ψ-(1/c^2)[(∂^2/∂t^2)ψ]=0

I have changed phi to psi...

Do not mind.

How can I get the 3rd part?

Use $LaTeX$. I dont know how many times have we told you to use it.  :wallbash_red:
Then probably you may get more answers to your post.:whistling:


[quote="Davron"]Isnt this great that in one IMO, winning four gold medals by the students of one teacher from! His name is Muamer Gulen
*he was a teacher of a year (2001).
*his status is only a teacher.
*his age probably (24-29), [I will correct it as soon as possible].

[color=red]proof:[/color] http://imo.wolfram.com/contestants/teamlist.cgi?KAZ

Davron[/quote]
He must be a really good teacher! :D

He must be an excellent teacher, and he must have excellent students, too. Both things are important (maybe 40-60).

[quote="ACET"]He not very good teacher,believe me ,I know![/quote]
Do you know him?We know each other,but i've never heard about him from you  :?:


Yay, 4 has started. For the rules, read at the link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=869371#869371

hurt junggi *2 heal mnmath

1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 23
i_like_pie: 25
ccy: 25
miyomiyo: 25
infinity4ever: 25
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

Hurt infinity4ever*2, heal miyomiyo
1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 23
i_like_pie: 25
ccy: 25
miyomiyo: 26
infinity4ever: 23
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

junggi -- infinity4ever +

1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 21
i_like_pie: 25
ccy: 25
miyomiyo: 26
infinity4ever: 24
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

hurt infinity4ever*2, heal junggi

1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 22
i_like_pie: 25
ccy: 25
miyomiyo: 26
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 22
i_like_pie: 25
ccy: 25
miyomiyo: 26
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

i_like_pie--, junggi+

Miyomiyo forgot to change the numbers.

1=2: 25
xpmath: 25
The Zuton Force: 25
junggi: 23
i_like_pie: 21
ccy: 25
miyomiyo: 27
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

i_like_pie--, miyomiyo+

1=2: 25
xpmath: 23
The Zuton Force: 25
junggi: 23
i_like_pie: 21
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

xpmath--, miyomiyo+

1=2: 25
xpmath: 21
The Zuton Force: 25
junggi: 24
i_like_pie: 21
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

xpmath--, junggi+

[b]+1 junggi
-2 i_like_pie[/b]

1=2: 25
xpmath: 21
The Zuton Force: 25
[color=green][b]junggi: 25[/b][/color]
[color=red][b]i_like_pie: 19[/b][/color]
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

+1 TZF
-2 i_like_pie

1=2: 25
xpmath: 21
The Zuton Force: 26
junggi: 25
i_like_pie: 17
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 25
worthawholebean: 25

[b]+1 cincodemayo
-2 i_like_pie[/b]

1=2: 25
xpmath: 21
The Zuton Force: 26
junggi: 25
[color=red][b]i_like_pie: 15[/b][/color]
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
[color=green][b]cincodemayo5590: 26[/b][/color]
worthawholebean: 25

+1 Zuton
-2 i_like_pie

1=2: 25
xpmath: 21
[color=green][b]The Zuton Force: 27[/b][/color]
junggi: 25
[color=red][b]i_like_pie: 13[/b][/color]
ccy: 25
miyomiyo: 28
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 26
worthawholebean: 25

:D

[b]+1 miyomiyo
-2 i_like_pie[/b]

1=2: 25
xpmath: 21
The Zuton Force: 27
junggi: 25
[color=red][b]i_like_pie: 11[/b][/color]
ccy: 25
[color=green][b]miyomiyo: 29[/b][/color]
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 26
worthawholebean: 25

Yay let's all kill me... I'm out.

1=2: 25
xpmath: 21
The Zuton Force: 27
junggi: 25
ccy: 25
miyomiyo: 29
infinity4ever: 22
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 26
worthawholebean: 25

I don't think you should do that, so as of now, I'm putting you back in:
infinity4ever --, junggi +

1=2: 25
xpmath: 21
The Zuton Force: 27
junggi: 26
i_like_pie:11
ccy: 25
miyomiyo: 29
infinity4ever: 20
BanishedTraitor: 25
perfect628: 25
mnmath: 26
archimedes1: 25
cincodemayo5590: 26
worthawholebean: 25

I'm not sure, but I'm going now

archie +
TZF --

archimedes1: 21
TZF: 10

TZF has low health!

archimedes1:22
TZF: 2 [b]Low health[/b]

Congrats, archie!  :)

(looks like my nickname stuck!)

 :winner_second:

Thank you, everyone, for your support!  (I know, it was just because of my avatar) :D  :P

good job, TZF also for being runnerup  :D

yeah, i meant 23  :oops:

Well I got as far as I could while I had an influence  :D

But as archie (archie!!) said, he does have a cool avatar.

Actually, I probably should've been a bit less aggressive and catalyzing in my proceedings...  :maybe:

Uh, such complicated strategy?

I was under the impression that Hurt and Heal is nearly all luck...


calculus
quadratics
algebra
polynomial
Vieta
calculus computations

this is easy but i dont know why i am not getting it.  Can anyone help?

FInd the inverse of f(x)= (5x^3) + 3x +1

Thank you

This is not easy. Any explicit form for the inverse can only be derived from the Tartaglia-Cardano cubic formulas - in other words, it will be a mess.

It is easy (given some basic calculus) to prove that $f$ is invertible on $\mathbb{R}$ and its inverse is differentiable. That's not the same as having an explicit formula.

hi
download this and see how to obtain it:

There should be $0.6$ instead of $0.6x$ in the formula.

[quote="maokid7"]nice.
where did you get the formula for $x^{3}+px+q=0$?[/quote]
$x^{3}+a^{3}+b^{3}-3abx=(x+a+b)(x+a\omega+b\omega^{2})(x+a\omega^{2}+b\omega)=0$.  Then relate this to $x^{3}+px+q=0$ and construct a quadratic using Vieta and solve for $a,b$. Also Navid, you do not have a correct sign in your $x$ value.

In this part i wanna  to obtain the roots.
in many refrences you can find articles about  finding roots of the polynomials such as :Crux mathematicorum.

Try $x^{3}-2x-1=0$.  This is a counter example to your formula.  From what I showed above you will get $x=\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$ as one of the roots.


I think there should be a BB game on aops. If others want to, I'll run it.

me=in
The message is too small. Please make the message longer before submitting.
w00t

Once I have 14 people I'll post exact rules and stuff. It'll be basically like the show, with a few adjustments.

I'll join, even though I have no idea what Big Brother is :roll:

[quote="junggi"][quote="Koda"]Koda is in! Whoppee!!

[img]http://www.lionking.org/~killer/BrotherBear/Koda01.jpg[/img][/quote]
are you really bubka? :huh:
and the brotherbear thing is...um annoying.

@perfect: i never ever watched big brother before, but i know its kinda like survivor in a house(or something), is that ok?[/quote]

Right on target it's Bubka.
His email id gave him out !

bubka, you don't have to do wierd or even ANNOYING things to hide yourself. Just act regular and that will do the trick.

Competitors:
Ignite168
Klebian
ragnarok23
mustafa
Jashper
Koda(bubka)
moogra
13375P34K43V312
laughinghead505
akalra1
h_s_potter2002
kstan013

The first competition of the week will be the HOH(Head of Household). The way this one works is that everyone will choose a number of people. Every person who you send in, you will give a point value, and their score will go down by that number. After everyone's choices are in, i will add up everyone's scores, and the highest scorer wins HOH.

Now, there will be a nomination by the HOH. He/She will choose two people to put up for eviction.

The POV(Power of Veto) competition is next. All but the HOH will participate. This competition functions similarly to the HOH, but the only choices for people are the other competitors, and you may only help people. You may choose as many people as you want. Occasionally, there will be a different POV comp, but it will be specified.

The winner of the POV will now choose whather or not to veto one of the HOH's nominations, and if so, which one. If used, the HOH would then put up someone else.

Finally, there will be an eviction. Everyone will send in their choice to me by pm. Whoever is evicted then has one post of up to 25 words to "say goodbye", and then can't post in the topic until the competition is over.

Now, there is another week! The previous HOH can't compete in the HOH comp. Points carry over from week to week.

There will be a few surprises along the way, be prepared!

Unforunately, POV over. From now on, I'l do remindewr pms. But now, I'm just gonna declare

[hide="POV"]mustafa[/hide]

Please make your decision.

So what am I doing now? Waiting for Klebian or Hspotter to bribe me to veto them?

Sure-if you want. I want your decision by tomorrow.

mustafa has decided to veto

[hide="..."]no one.[/hide]

Voting on klebian and h_s_potter may now begin.

I'm closing votes. I got [b]3[/b]! By a vothe of 2-1,

[hide="..."]klebian, you have been evicted from the BB house[/hide]

You have 1 25 word post to say goodbye.

All others, HOH choices due Saturday. 13375..., you can't participate.

it's only been 5 hours! He probably doesn't spend all day on this forum

[quote="13375P34K43V312"]I doubt he's still checking this thread...[/quote]
Who me? Bye everyone.  Have a nice day.

The HOH [hide="is"]...[/hide]

[hide="..."]kstan013[/hide]

Please nom. 2 people.

1337... and h_s_potter2002 are up. Please, all but kstan send in POV choices.

Who votes this just stops? No one is doing this.

I am doing it, I just wasn't on yesterday.....


LaTeX
combinatorics unsolved
combinatorics

[quote="Mimoide"]I cannot post the solution because Latex is still a mistery for me, but I think it is not too much difficult to solve it with telescopic sums.[/quote]
For the case you didn't know: there is an introduction at http://www.mathlinks.ro/Forum/viewtopic.php?t=5261
 .

How did you discover that telescoping sum for $\frac{1}{\binom{n}{k}}$?  Was it just previous experience?