calculus
integration
logarithms
trigonometry
function
calculus computations

The answer is $\ln |x+\sqrt{x^2+1}|-\frac{\sqrt{x^2+1}}{x}+C.$

The standard calculus line starts with the trigonometric substitution $x=\tan\theta.$ Then $dx=\sec^2\theta\,d\theta$ and $\sqrt{x^2+1}=\sec\theta.$ Remember the redundancy of trigonometric functions; change which ones you're using if convenient.


tg a = 0,5 (alpha)
tg β = 1,5 (beta)

If d = 40 cm, find the value of h + x ?

[img]http://img151.imageshack.us/img151/1224/trigonometria.jpg[/img]

By using the acute angle formulas for the tangent function, we have

$ \tan{\alpha}\equal{}\frac{h}{40\plus{}x}\equal{}\frac{1}{2}$
$ \tan{\beta}\equal{}\frac{h}{x}\equal{}\frac{3}{2}$

$ h\equal{}\frac{3}{2}x$

$ \frac{\frac{3}{2}x}{40\plus{}x}\equal{}\frac{1}{2}$
$ \frac{3x}{40\plus{}x}\equal{}1$
$ 3x\equal{}40\plus{}x$
$ 2x\equal{}40$
$ x\equal{}20$
$ h\equal{}\frac{3}{2}(20)\equal{}30$

So $ h\plus{}x\equal{}\boxed{50}$.


algebra
polynomial
function
topology
complex analysis
complex analysis unsolved

Suppose f(z) is analytic in the unit disc {z: |z|<1} and satisfies:

$ \forall a \in C$ , $ \exists n \in N$ such that $ f^{(n)}{(a)}\equal{}0$. Prove that f is a polynomial

Hint: just take some closed interval (for example $ [ \minus{} \frac {1}{2},\frac {1}{2}]$) and prove the same statement for them :wink: (to do this use Baire category theorem , then use uniqueness theorem to get the desired result)

It's easier with analytic functions- we have the strong version of the identity theorem, saying that a function is identically zero if its zeros have an accumulation point.

Hi jmerry,

How does Identity theorem help to prove this, I can not sea it

Let $ S_n\equal{}\{z: f^{(n)}(z)\equal{}0\}$. The union of the $ S_n$ is the whole disk, so one of them (say, $ S_m$) is big. Then $ f^{(m)}$ is identically zero.

[quote="achatter"]Hi jmerry,

Thanks for your help. Could you tell me what do you mean by the word "Big"[/quote]

The Baire theorem sais: if a complete metric space is covered by a sequence $ A_n$ of closed sets, then at least one of  $ A_n$ has non-empty interrior. So "big" here means "with non-empty interrior". There is by the way a little difficulty when one applies the Baire theorem to the problem under consideration. Namely, the open unit disk is not complete in the ordinary metric. I see two solutions of this obstacle:

1. Consider instead of the hole disk some smaller closed disk.

2. Remark that the open unit disk is homeomorphic to the whole complex plane, i.a. it admits a complete metrization.

Actually jmerry does not need Baire category theorem at all (he just need that  some of $ S_n$ has limit point in the disk).  I've proposed an application of that theorem only for the closed interval, hence there are no problems with completeness.

Hi jmerry and leshik

 How do I prove that some of $ S_{n}$ has limit point in the disk? Thanks again

There are uncountably many points in the disk, so there are uncountably many in some $ S_n$. This still works restricted to a compact subset such as a segment, and any infinite subset of a compact set has a limit point.

That's one notion of size. A step up, I could say that some $ S_n$ has positive outer measure, and thus its set of limit points has positive outer measure. A step beyond that, I could invoke Baire, and say that the closure of some $ S_n$ has nonempty interior. That last version lets us use a weaker version of the identity theorem for (real or complex) analytic functions.

I appreciate your time to solve this problem


Solve $(x-2)(x-3)(x-4)^2<0$ for $x$ as an element of real numbers.

[hide]
Note that $(x-4)^2>0$ for all $x$ except for $x=4$.
Therefore, if $(x-2)(x-3)(x-4)^2<0$, then we must have $(x-2)(x-3)<0$.
Solving the inequality gives $\boxed{(2,3)}$
[/hide]

Here's a similar problem.

Solve for real $x$: $(x-1)(x-2)^2(x-3)^2\le 0$

the thing is that I don't understand why $2$ and $3$ would work because wouldn't it just make the entire thing $0$?

No, by $(2,3)$, I meant $2<x<3$. (Interval notation...)

[quote="catcurio"][quote="frt"]Here's a similar problem.

Solve for real $x$: $(x-1)(x-2)^2(x-3)^2\le 0$[/quote]

[hide]The roots of $(x-1)(x-2)^2(x-3)^2=0$ are 1, 2, and 3. So the possible intervals for $(x-1)(x-2)^2(x-3)^2\le 0$ are $(-\infty,1],[1,2],[2,3],\text{and }[3,\infty)$. Testing these in the inequality gives $[2,3]$ as the solution. So $\boxed{2\le x\le3}$.[/hide][/quote]
Not quite...

[quote="catcurio"]Oops, that didn't even make sense... :oops:

[hide]Would it just be $x\le1$, $x=2$, and $x=3$?[/hide][/quote]
Yup, that's correct :)


vector
circumcircle
geometry unsolved
geometry

[quote="u2tommyf"]We have triangle ABC and points D on (BC), E on (AC), F on (AB) such that BD/DC=CE/EA=AF/FB. If the center of the circumscribed circle of DEF and the one of ABC are the same point, then ABC is equal-sided.[/quote]
I don't think, either
IF BD/DC=CE/EA=AF/FB then the Centroid of DEF and the one of DEF are the same point.
Denote G is the Centroid of DEF and ABC, O is the center of the circumscribed circle of DEF and ABC. we have:
$ \vec {OG} \equal{} \frac {\vec{OA} \plus{} \vec{OB} \plus{} \vec{OC}}{3} \equal{} \frac {\vec{OD} \plus{} \vec{OE} \plus{} \vec{OF}}{3}$
And This is clearly true:
$ \vec{DB} \plus{} \vec{FA} \plus{} \vec {CE} \equal{} 0$
Denote $ BD/DC \equal{} CE/EA \equal{} AF/FB \equal{} k$
=> $ k(\vec {CB} \plus{} \vec{BA} \plus{} \vec{AC}) \equal{} 0$
It is true


combinatorics proposed
combinatorics
combinatorial geometry

Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.

why the fact "there are at most $2{n\choose 2}$ isosceles triangles"
does not imply "#scalene triangle + #acute triangle(except isosceles and equilateral triangle)  is at least  ${n\choose 3}-2{n\choose 2}=\frac{n(n-1)(n-8)}{6}$ "  ??

[quote=bodan][b]Lemma.[/b] Among $n$ points in a plane positioned generally (no three collinear) we have at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles.
[b]Proof.[/b] Suppose that $n$ points are fixed and the number of isosceles triangles is $\alpha$. Set $\beta$ to be the number of bases of these triangles (we count three bases for each equilateral and one for each nonequilateral isosceles triangle). Clearly $\beta\geq \alpha$. Then each segment connecting a pair of points can be a base of at most two triangles, as if this is not the case three points will lie on this pairs' perpendicular bisector. Thus there are at most $2{n\choose 2}$ isosceles triangles, yielding that there are at least ${n\choose 3}-2{n\choose 2}=\frac{n(n-1)(n-8)}{6}$ scalene triangles. $\square$

In the problem there are at least $13$ points of the same color and we apply the lemma.[/quote]

Thanks. I had almost the exact same solution, except one question. Why do you care about “ we count three bases for each equilateral and one for each nonequilateral isosceles triangle”? It seems that already 2 bases of triangles at most x n choose 2 pairs of points to form an isosceles triangle is enough.

We solve the following problem first:

Given are $13$ points in the plane, no three of them belonging to a same line. Prove that there are at least $130$ scalene triangles with vertices in the plane.

Notice that any edge between two points in the plane can be a base of at most two isosceles triangles (because otherwise we would have $3$ points on the perpendicular bisector). Hence there are at most $\binom{13}{2} \cdot 2 = 256$ isosceles triangles, so at lesat $\binom{13}{3} - 256 = 130$ scalene triangles.

This solves the original problem because there must exist a color with at least $\left\lceil \frac{50}{4} \right\rceil = 13$ points of that color.

We prove that given 13 points in standard position, there are 130 scalene triangles. This clearly solves the original question.

Note that every base (i.e. two points) gives at most two isosceles triangles, else we have three points along the perpendicular bisector.

Therefore there are at most $\binom{13}{2}\cdot 2=156$ isosceles triangles. At the same time there are $\binom{13}{3}=286$ triangles in total, making 130 non-isosceles AKA scalene triangles. $\blacksquare$

By pigeonhole, there is a color that has at least $13$ points of the same color. This means there are $\binom{13}{3} = 286$ triangles that can be formed.
Let $I$ be the number of isosceles triangles in the $13$ points.
Then each pair of two points can contribute at most $2$ to $I$(due to the collinear condition), so $\binom{13}{2} \cdot 2 = 156$ is the maximum number of isosceles triangles. Then the minimum number of scalene triangles is $130$.

By Pigeonhole, some color has at least $13$ points of that color. So, assume there are exactly $13$ points, we'll show that we can find at least $130$ scalene triangles whose vertices all belong to those $13$. First, the total number of triangles is $\binom{13}{3} = 286$. Also, note that for each pair of points $(A, B)$, there are at most two isosceles triangles $ABC$ which have $\overline{AB}$ as a base, since in order for that to happen, $C$ must lie on the perpendicular bisector of $\overline{AB}$, but if three or more $C$ exist, this is a contradiction to the fact that no three points are collinear. So, there are at most $2 \cdot \binom{13}{2} = 156$ isosceles triangles, thus at least $130$ scalene triangles among those $13$ points (basically at least $130$ of that color), so done.


My public site has been launched!!
The url is [url=http://www.freewebs.com/tzpublicsite/]www.freewebs.com/tzpublicsite/[/url]
It is only starting and doesn't have much things on it.
Please post comments in the guestbook.
I will make the private site soon.

i added a bunch of stuff, which is basically
1991-1995 sprint target
1996 target
1998 sprint target
2001-2002 sprint target team solutions
2003 sprint target

Wow, good job!!!

I'm flattered that I'm one of the poll options for "who made this poll".

Sorry I couldn't contribute -- the sites I know, goodlycatch and mathcounts.saab, were already mentioned.

i guess the poll creator *cough* ran out of people to name

C'mon biffandoc dont be like that, I'm an option too. :D
You might wanna add the toolbox, it's got some useful stuff:[url=http://www.mathcounts.org/Program/toolbox.pdf]www.mathcounts.org/Program/toolbox.pdf[/url]

umm biffanddoc where are the answers to the nationals on it?
Also i can put countdowns in ppt on it from 1998 and on if anyone wants...

Here's a website that's not posted yet on your website.
http://www.yashar.biz/mathcounts/
Credit goldendomer, not my real name.

Thanx goldendomer. Could I pm you tests?

Yeah. I'm a 7th grader so those test would help me a lot!

I can convert anything to PDF.  Unlike most people, I [b][i][u]PAID[/u][/i][/b] for my PDF conversion software, so there is no annoying "Converted by GO2PDF" banner on anything I convert.  If you need me to convert something, PM it to me, or email it at mysmartmouth@gmail.com

If you want to convert to pdf, you can use [url=http://sourceforge.net/projects/pdfcreator/]PDFCreator[/url], which is free and has no ads.

As an interesting side note, some version of Quickbooks came with my computer.  It included pdf converter which works pretty well.

[quote="biffanddoc"][quote="easyas3.14159..."]I have every competition (school, chapter, state, and national) from 1984-1997, so if I could type it up and someone was willing to make it into PDF form...[/quote]
DUDE I WOULD DO ANYTHING FOR THOSE TESTS.
YOU WOULD BE MY HERO FOR LIFE[/quote]

easy as pi is a dudeTTE

I got them from a competition database from this one coach...OMG HE IS MY HERO.It might take awhile to type up 327 pages,  but I'll consider getting a scanner.  ;) JUST KIDDING XD.
I'll start with states and nats since school and chapter are over.

where is the competition database (is it accessible by other peoples)

I believe he got it from this website:[url=http://www.educaide.com/products/databases/mcnt.html]www.educaide.com/products/databases/mcnt.html[/url]

[quote="Zmastr"] HOW DID YOU FIND THOSE?  ALL I KNOW IS...I WANT THEM![/quote]

me too.

Goldendomer's getting '84 states up. As I'm horrible with computers and can't figure out how to use the pdf creator :blush: could I just give you guys some tests to convert?Thanks!

Here's the website you desperate people. 1984 sprint.
http://www.4shared.com/file/971596/40f6f386/1984_State_Sprint.html
Post this on tz's website somebody.

Sorry, the previous download doesnt have pics. 1984 state sprint.
http://www.4shared.com/file/971974/39614640/1984_State_Sprint.html

Thanks for changing them goldendomer!
Are there any tests (name year and level) that people want in particular? I'm trying to give people problems and having them convert it to PDF for the website.

Yeah, thanks much!  Are you working on the 85's now?  I would really appreciate it.

Yup. I think biff's gonna get 84 nats up.

Sweetness.  I want to go to nats so bad and need all of the help I can get.  Thanks again!

easy as pi, do you want me to help you with some other state tests?

Yeah, I would really appreciate it if you had anything before 1996 state-wise. (or even nats for that matter)

Yeah, our coach actually bought the past state comps from the mathcounts site.  I haven't even seen a comp before 2002 before I came to AoPS.  Thanks a ton.  We were shocked when we saw that someone actually had states from 1996 onwards.

the states from pre 1999 arent supposed to be available :huh:

But its cool anyway.
Ill probably have the 1984 state & nats up on the site tonight :)

[quote="goldendomer"]I wish I had a coach like hers who gives tests from 1986. :([/quote]
LOL its not my coach. This one other coach from another school invited me over to his house, and gave me lunch and the comps. Wish me actual coach cared half as much... :(

This should be stickied or put in the resources section

Three new sprints have been added. A few notes.
1) State Sprint (no number) is actually a chapter sprint. It says it at the top of the page.
2) State Target (no number) is a state target. So State Sprint (2) and State Target (2) could be different years.
3) Can anyone identify what years these three are?


algorithm
inequalities
algebra
polynomial
function
graph theory
triangle inequality

Hello everybody!
Let $ G(V,E)$ be a complete graph; we want to solve the travelling salesman problem taking into account also the vertex's weights (each node is labelled with a discrete value among the $ [1, 2, 3]$ set), that indicates the priority assigned to each one of them. The higher is the value, the higher is the priority we want to visit the node with.
It's known that the TSP is a NP-hard problem, but there are some acceptably efficient algorithms based on a cost triangle inequality that approximates the solutions in polynomial time.

We want to design a fairly efficient algorithm that finds the shortest cycle taking such priority in account.

Some solution we are considering includes splitting the main graph into three different graphs according to the vertexes' priority and then apply some solution for the TSP problem (lame); some sort of fitness function that makes more expensive to visit a low-priority node.

Ideas anyone?

Thanks very much
Ian Aaron Davidson
Padua University


superior algebra
superior algebra unsolved

Construct a subfield K of complex numbers such that K:Q is infinite

$ \overline{\mathbb Q}$ (Proof: Construct an infinite chain of subfields), $ \mathbb Q(\alpha)$, where $ \alpha$ is transcendental, $ \mathbb R, \mathbb C$ (Proof: they are both uncountable, or alternately each contain $ \mathbb Q(\pi)$), to name a few.

sory ı forgot to add the condition that all elements of K are algebraic over Q.  :D So including transcendentals are not allowed. I tried to look to include the following set;

$ S \equal{} \{e^{2\mathbb(\pi)ik/m} : (k,m) \equal{} 1, k<m \}$ but I could not prove that they are independent from each other.

Linearly independent over $ \mathbb{Q}$, of course. The problem is that it's false; for $ m\equal{}4$, $ i$ and $ \minus{}i$ aren't independent.

Fortunately, we don't need it. The full set of $ m$ $ m$th roots of unity span a field of degree $ \phi(m)$ over $ \mathbb{Q}$. If we take an increasing union of those (say, all $ m$ that are powers of $ 2$), that union has infinite degree.


1) Show that $ x=lim\ x_{n}$ if and only if every subsequene of $ \langle x_{n}\rangle$ has in turn a subsequence that converges to $ x$.

2) Show that the series $ \langle x_{n}\rangle$ has sum if $ \sum_{n=1}^{\infty}{|x_{n}|}<\infty$.

These problems are well-know theorems in Real Analysis, see in W. Rudin.  :wink:

yea i know i got it from his book. I just want to know the proof of these theorems. <sorry but i'm not quite well in real analysis that's why i'm pushing myself to better on it and searching for other people's ideas>

On the first: I know that as Urysohn's principle, and it works in any topological space. A sequence converges to $ x$ when for any neighborhood $ U$ of $ x$, all but finitely many elements of the sequence are in $ U$; this form of the definition is the way to prove it.


If $ x \plus{}  \dfrac{1}{y}\equal{} 5,\,y \plus{} \dfrac{1}{z} \equal{} 1$ and $ z \plus{} \dfrac{1}{x} \equal{} 2,$ find $ x \plus{} 3y \plus{} 2z$.

$ y \plus{} \frac{1}{z} \equal{}1 < \equal{} > y \plus{} \frac{1}{2\minus{}\frac{1}{x}} \equal{}1 <\equal{}> 2y \plus{} \frac{1\minus{}y}{x}\equal{}1$
$ < \equal{} >$
$ 2y \plus{} \frac{1\minus{}y}{5\minus{}\frac{1}{y}}\equal{}1 < \equal{} > 9y\minus{}6\equal{}\minus{}\frac{1}{y} < \equal{} > (3y \minus{}1)^2\equal{}0 \equal{}> y\equal{}\frac{1}{3}$
in the exact same way we get $ x\equal{}2$ and $ z\equal{}\frac{3}{2}$
so $ x\plus{}3y\plus{}2z\equal{}2\plus{}1\plus{}3\equal{}6$ :)

It is not too hard to solve the system...
(This is pretty much the same as the previous solution)
From the third equation, z=2-(1/x)
If x+(1/y)=5 then y=1/(5-x)
From the second equation, we have z=1/(1-y)
Plug in y and simplify to get z=(5-x)/(4-x).
Set the expressions for z equal to each other.
You will get $ (x \minus{} 2)^2 \equal{} 0$
So x=2, y=1/3, z=3/2, and the solution is 6.


inequalities
search
inequalities proposed

let $ a_i,b_i \ge 0$ .Prove that:
$ \sum_{i \equal{} 1}^n (a_i \plus{} b_i)^2 \ge 4\sqrt {(\sum_{i \equal{} 1}^n a_i^2)(\sum_{i \equal{} 1}^n b_i^2)}$
 :D

c'est trés facile
enlevey les deux membres de l'inequalites
au carres
avant ça ,Tu dois developpez (ai+bi)²=ai²+bi²+2aibi

From $ \sum{(a_i\plus{}b_i)^2}\ge\sum{a_i^2}$ and $ \sum{(a_i\plus{}b_i)^2}\ge\sum{b_i^2}$ the conclusion follows.

This seems nice to search for an improvement.
 :)

i'm sorry, i have made some mistakes
i have correct it :D

Unfortunately  :maybe:  the new inequality you propose is not always truth.

Try when $ a_1^2\plus{}...\plus{}a_n^2\equal{}b_1^2\plus{}...\plus{}b_n^2$. And use AMGM.
 :)


Find the sum $ cos(\theta)+cos(2\theta)+cos(3\theta)+...+cos(n\theta)$

[hide="See"]
*Don't click if you don't want to see the solution.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=929458327&t=50721
[/hide]


abstract algebra
vector
group theory
superior algebra
superior algebra unsolved

In an introduction to ring theory we discussed the definition of a ring from different aspects. My teacher asked us some questions to which I added some other ones (Nice days we had!)
 [b]1.[/b]Can we make a ring out of an arbitrary group? [hide="answer"]As expected not out of every group, but of abelian ones. It suffices to define a trivial product over a given abelian group, $G$.  $\forall a,\;b \in G \;a*b=0$, and according to the definition we are done.[/hide] Now answer the following questions;
[b]2.[/b] Does there exist a nontrivial group, the only multiplication rule we can define over which is the trivial one?
[b]3.[/b] Suppose we are given a multiplicative group, eg $S_{3}$, can we define an addition rule over this group to make it a ring?[hide="Hint"]One of my recent posts maybe helpful.[/hide]
I think it suffices for now. Every one's invited for a [i]discussion[/i] on these questions.

Without looking at answers and with my tiny knowledge about rings.

(1). The group has to be abelian.
(2). What you mean by trivial multiplication?
(3). Don't know.

@3.

No, that is for any nontrivial multiplicative group wrong, because the only ring, whose elements (all of its elements!) form a multiplikative group is the trivial ring.

@2.

This is maybe very difficult. In fact, such a group can't be finitely generated since every (nontrivial) finitely generated group can have a nontrivial ringstructure (because it is a direct product of cyclic groups).

@2: when I look for an abelian group which is neither finitely generated nor a ring nor a vector space, there is only one which comes instantly to my mind. [hide]Q/Z[/hide] owk

Thank you guys. @Gockel You have misunderstood the third question, I'm afraid. Please read it once more and compare it with the first question.
@owk Let me paraphrase the second question; Does there exist a group $(G,+)$ such that $(G,\;+,\circ)$ is a ring with $\forall a,b \in G \;\;a\circ b=0$ but $(G,\;+,*)$ is not a ring for any multiplication rule like $*$ other than $\circ$?

[quote="The Captain"]@owk Let me paraphrase the second question; ...[/quote]
What made you think I could have misinterpreted the question? owk

[quote="owk"]@2:...[b]nor a ring [/b]... Q/Z[/quote]We are looking for a group. Maybe such a group does not exist! In either case you should supply your answer with a proof. (If you are giving a counterexample, don't forget writing the binary operations explicitly.)

I didn't want to give it away, it's really easy. [hide="proof"]The key point is that multiplication must be $\mathbb Z$-bilinear, but $\mathbb Q/\mathbb Z\otimes\mathbb Q/\mathbb Z=0$. (Of course, you can write this down in an elementary way.) [/hide] owk


Hi folks! I want to take the informatics contest for students in my school in about 5 months after.So could you recommend some good books for trainning to get a best result.I have a little time to prepare for the contest,so the book should contain both problems and solutions, and the methods of solving problems as well!
Thanks  :)

I've never used books. Try the USACO gateway instead, it contains many problems starting with really easy ones and also texts that explain the solutions and some general concepts that may come handy:

http://ace.delos.com/usacogate


Use DeMoivre's Theorm to find 4(cos2pi/3 + i sin 2pi/3) in polar form.

Who is Demoivre, and what is his theorm?

DeMoivre's Theorem: $(\cos{(\theta)}+i\sin{(\theta}))^n = \cos{(n\theta)}+i\sin{(n\theta)}$

hm, whats there to find?  I believe that is in polar form.  Are you sure its not that raised to some power?  DeMoivre says that $r(\cos{\theta}+i\sin{\theta})^n = r^n(\cos{n\theta+i\sin{n\theta})}$

or as I like to see it $(r,\theta)^n = (r^n,n\theta)$


:huh:  :huh:  :huh:  :huh:  missing something????
No, uuuurrrrgghh....

[hide]

Is it $-\frac{2abc}{pqr}$ ?

[hide="if so then click to see the solution"]
Let $x=a/p$ , $y=b/q$ , $z=c/r$

Given $\sum x=0$ , $\sum\frac{1}{x}=\frac{\sum xy}{xyz}=1$ :arrow: $\sum xy=xyz$

We wish to find $\sum x^2=(\sum x)^2 -2\sum xy=-2xyz$

[/hide][/hide]

[quote="eminem"]hey riddler , haahha :lol:  this question was Given by YOU to me when we were in 9th  :lol:  seriuosly! dont know whether u remeber , ne way i was stupid that time and i didnt get it that time and YOU had told me the solution!!!!

so u were the last person i expected to get stuck in this !! :)

ne way i am quite sure u'll get it , just do it a bit ....[/quote]

Please try to write your messages in "civilised" english, no chat language please.

[quote="eminem"]hey riddler , haahha :lol:  this question was Given by YOU to me when we were in 9th  :lol:  seriuosly! dont know whether u remeber , ne way i was stupid that time and i didnt get it that time and YOU had told me the solution!!!!

so u were the last person i expected to get stuck in this !! :)

ne way i am quite sure u'll get it , just do it a bit ....[/quote]
do you have to find the integer value eh???? then it is impossible but if above person is right I got the same thing as him.
 :P


search
superior algebra
superior algebra unsolved

[color=green]Find a primitive root for all positive integers k modulo each of the following moduli.
(a) 23 to the power k
(b) 37 to the power k
(c) 29 to the power k[/color]
Thanks guys!!!  :blush:


algebra polynomial function domain algorithm

Why is it that any polynomial over $ \mathbb Q$ of degree 4, which has any $ t_i$ as a zero must be divisible by $ t^4\minus{}4t^3\plus{}6t^2\minus{}4t\minus{}1$? Also, it follows from this that there is no polynomial in one variable over $ \mathbb Q$ which is satisfied by some but not all of the zeros $ t_i$  :o

This sounds absurd! -Is this true?

What do you mean by a polynomial over $ \mathbb{Q}$? A polynomial with domain $ \mathbb{Q}$ and range $ \mathbb{Q}$? And what is $ t_i$?

A polynomial over $ \mathbb{Q}$ probably means one with rational coefficients, that is it is in $ \mathbb{Q}[x]$.

I don't know what $ t_i$ is though. However, this is probably dealing with the irreducible polynomial of it. In general, if $ x$ is a root of a polynomial $ f$ in $ \mathbb{Q}[x]$, then the irreducible polynomial of $ x$ over $ \mathbb{Q}$ divides $ f$. Basically, if you have one root, you have the others too.

Hamster has right. Moreover $ t_i$ is a root of $ t^4\minus{}4t^3\plus{}6t^2\minus{}4t\minus{}1$. Does what you say mean that for $ t^4\minus{}4t^3\plus{}6t^2\minus{}4t\minus{}1\equal{}(t\minus{}1)^4\minus{}2$, if we can find a root of this polynom like $ \alpha\equal{}2^{1/4}\plus{}1$, then any polynomial like $ (t\minus{}\alpha)(t\minus{}\beta)$ where $ \beta$ is any arbitrary number (perhaps rational) [i]will[/i] divide $ t^4\minus{}4t^3\plus{}6t^2\minus{}4t\minus{}1$ evenly?

No I mean that if $ \alpha$ is a root of the polynomial $ P\equal{}t^4\minus{}4t^3\plus{}6t^2\minus{}4t\minus{}1$ (assuming that's irreducible), then if $ \alpha$ is a root of [i]any[/i] polynomial $ f$ in $ \mathbb{Q}[x]$, then $ P | f$.

hm.. so if two polynomials share a root then the polynomial whose degree is fewer will divide the other. (why?)

No, what it says is that the irreducible polynomial of the root will divide the polynomial. The irreducible polynomial is defined as the monic polynomial of least degree which has it has a root.

I can't recall the proof right now, give me a bit of time.

[quote="Hamster1800"](To fill in any gaps in the proof, here it is in full. It's also contained in t0r's link, but not everyone clicks on links)

Let $ P \in \mathbb{Q}[x]$ be the irreducible polynomial of $ \alpha$. For any $ f \in \mathbb{Q}[x]$ with $ \alpha$ as a root, we can apply the division algorithm to obtain $ f(x) \equal{} P(x)Q(x) \plus{} R(x)$, and $ \deg R < \deg P$. If we plug in $ \alpha$, we get $ f(\alpha) \equal{} P(\alpha)Q(\alpha) \plus{} R(\alpha)$, or $ 0 \equal{} R(\alpha)$, since $ \alpha$ is a root of both $ f$ and $ P$. Thus, $ \alpha$ is a root of $ R$, but we said that $ P$ was the polynomial of minimal degree with $ \alpha$ as a root. Thus, $ R \equiv 0$ (the zero polynomial), which means that $ P | f$.[/quote]

note by the division algorithm, R has rational coefficients, we need that piece of info.


Given triangle $ABC$ and a circle $(O)$. Through $B, C$ draw a circle that touches $(O)$ at $M$. Through $C, A$ draw a circle that touches $(O)$ at $N$. Through $A, B$ draw a circle that touches $(O)$ at $P$.

Prove that $\frac{MB}{MC}\cdot \frac{NC}{NA}\cdot \frac{PA}{PB}= 1$

Many thanks to prowler for such a quick and beautiful proof. My proof is much more complicating. :)

Let's show that $\frac{MB}{MC}$ equals $\frac{\sqrt{\mathcal P_{(O)}(B)}}{\sqrt{\mathcal P_{(O)}(C)}}$, where, in general, $P_\omega(X)$ denotes the power of the point $X$ with respect to the circle $\omega$ (next time, I'll omit the subscript $(O)$ from $\mathcal P_{(O)}$). Clearly, the desired result folows.

Let $B',C'$ be the second intersections between $(O)$ and $BM,CM$ respectively. Then $\frac{MB}{MC}=\frac{MB'}{MC'}=\frac{BB'}{CC'}$, so $\frac{\mathcal P(B)}{\mathcal P(C)}=\frac{BM\cdot BB'}{CM\cdot CC'}=\left(\frac{BM}{CM}\right)^{2}$, and that's that.


Let $ a, b, c$ positive reals. Then

$ \sum(\sqrt{a}\plus{}\sqrt[3]{b}\plus{}\sqrt[6]{c})\leq\frac{1}{3}\sqrt[3]{\frac{\prod[(a\plus{}2b)(a\plus{}2c)]}{abc}}\plus{}6$

.

[quote="Abdek"]Using Muirhead inequality we have:

$ \text{LHS} \equal{} \sum_{cyc}\left(a^{\frac {1}{2}}.1.1 \plus{} a^{\frac {1}{3}}.1.1 \plus{} a^{\frac {1}{6}}.1.1\right) \le \sum_{cyc}(a^1.1.1 \plus{} a^0.1.1 \plus{} a^0.1.1) \equal{} \sum_{cyc}(a \plus{} 2)$

And Thus it sufficent to show that :

$ \prod\left((a \plus{} 2b)(a \plus{} 2c)\right)\ge 27(a \plus{} b \plus{} c)^3.abc$

Which is easy to prove  :)[/quote]
I thought Muirhead works for only symmetric sums. So I will prove $ \sqrt{a}\plus{}\sqrt[3]{a}\plus{}\sqrt[6]{a}\leq a\plus{}2$ in another way  :)
Just using AM-GM we have
\[ \sqrt{a}\plus{}\sqrt[3]{a}\plus{}\sqrt[6]{a}\leq \frac{a\plus{}1}{2}\plus{}\frac{a\plus{}2}{3}\plus{}\frac{a\plus{}5}{6} \\ \equal{}\frac a2\plus{}\frac a3\plus{}\frac a6 \plus{}\frac 12\plus{}\frac 23\plus{}\frac 56 \equal{}a\plus{}2\]

[quote="Mathias_DK"][quote="Potla"]
I thought Muirhead works for only symmetric sums.
[/quote]
Which it is...[/quote]
Yeah; I understand.... sorry....... :(
However, I like this AM-GM work. :)


im new to mathpath,, i was going to go last year but my parents thought wait 1 year because i have piano and im only going to 7th grade. i read the website and all but i want to ask you guys. im pretty sure im going this year though.

It's okay, but the counselors are really strict about unnecessary stuff. I guess I kinda sorta learned a lot.

I kinda have the same question... what's the level of difficulty at the camp?

I know that it defers by courses, but anyways...


AMC
AMC 10
AIME
USA(J)MO
USAMO
AMC 12
search

What score do you need on the AMC10 to make the AIME?  How about on the AIME to make the USAMO?  How does the AMC12 fit in all of this?

Does it go by percentile or score?  How is it scored?  Do yuo have to write proofs for any of them?

AND MOST IMPORTANTLY:  How old do you think Darryl Wu and Bobby Shen will be when they win top 25 in USAMO?   :D

TYIA!!!

I suggest you start by reading the Frequently Asked Questions linked from the AMC starting webpage:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml
Read FAQ AIME #5, then read FAQ USAMO #5.

I also suggest you read:
http://www.unl.edu/amc/e-exams/e5-amc10/amc10.shtml

and

http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml

and

http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml

and

http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml

It would probably also help to read our information brochure:
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Brochure.pdf

For complete details of the rules and procedures, consult the 2008 Teachers' Manual, the 2009 will be nearly identical:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-AIME,2008.pdf
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-USAMO,2008.pdf

A search of this forum will also find existing discussion of your specific questions.

Steve Dunbar
MAA Director, American Mathematics Competitions

AMC 10 and AMC 12 are competitions that students below 10th grade and 12th grade,

respectively, take.

If you take the AMC 10, then you must be in the Top 1% or have more then 120 to

qualify for the AIME (see below)

If you take the AMC 12, then you must be in the Top 5% or have more then 100 to

qualify for the AIME.

Both the AMC 10 and the AMC 12 are 75 minute examinations, with 25 multiple-choice

questions. The answers will be labeled A, B, C, D, and E.

The answer sheet is a bubbling type, so that an opscan reader can check the answers.

If you Qualify for AIME, then you will be taking an exam with 15 questions,

and you will have 3 hours to solve them. These questions are NOT easy (Well, maybe

#! on the 2008 AIME I is an exception).

The questions are no longer multiple choice. Your answers will be integers from 000 to

999. Again, you will bubble these.

About 250,000 students take the AMC 10 and the AMC 12. Of these,

about 7500 (?) take the AIME.

Once you finish the AIME, your paper will be scored, and a FLOOR SCORE will be set.

The FLOOR SCORE is the lowest score that qualifies for USAMO for students taking

the AMC 12. Students taking the AMC 10 must have this AIME score to still have a

chance. Of the students that have this score, they recieve an INDEX.

The INDEX is (AMC score) + 10*(AIME score). A certain Index will be set, and the

people who meet this take the USAMO.

The USAMO is a very difficult competition.

Over 2 days, you will have 9 hours to work on 6 very very difficult problems.

The USAMO requires proofs. Your proofs will be scored out of 7 possible points each.

Then, MOSP scores will be set.

MOSP is the Math Olympiad Summer Program. There are 3 categories:

Black MOSP: This is the top 12 on the USAMO (The score is typically 36+ makes this)
Blue MOSP: These students did very well, but didn't make black (Typically 12-36 or so)
Red MOSP: These are younger students who certainly should try again! (Typically 7-12)

The Black "MOSPers" (Pronounced Moppers), take one final test. THis is called the

Team Selection Test, or the TST. This determines the 6 people that will be representing

the USA at the IMO (International Math Olympiad).

The IMO is in a different country each year, and has the same format as the USAMO.

The IMO problems are very hard, much harder then the USAMO, and are also proofs.

This chain of competitions brings 250,000 students down to the best 6 in the nation.

[quote="BOGTRO"]The IMO problems are very hard, much harder then the USAMO, and are also proofs.[/quote]

Hm...well they actually seem around the same difficulty...

Also, remember the floor score for USAMO qualification also applies to 9th and 10th graders taking the AMC12.  They take something on the order of 330 people by index on the AMC12/AIME (if I recall), and then the remaining 170 or so come on floor score.  They break ties by floor score on USAMO qualification by index.

[quote="BOGTRO"]
MOSP is the Math Olympiad Summer Program. There are 3 categories:

Black MOSP: This is the top 12 on the USAMO (The score is typically 36+ makes this)
Blue MOSP: These students did very well, but didn't make black (Typically 12-36 or so)
Red MOSP: These are younger students who certainly should try again! (Typically 7-12)
[/quote]

These scores you gave are [i]completely[/i] wrong.

I don't think the black MOP cutoff has ever been $ \ge 36$. I think it was 35 [i]once[/i], in 2002 with a very easy USAMO. Typically it's somewhere around 28, so you generally need to solve about 4 problems (plus maybe a bit of partial credit). I believe the cutoffs for the las four years have been: 29,27,23 (that was a very hard year) and 28. Also, sometimes they take a few extra people who weren't in the top 12, if some of the top 12 can't make it, or are not eligible.

I also don't think the blue MOP cutoff has ever been as low as 12. It's typically around 19 or 20. To get in you need to solve about 3 problems. Over the past four years, I think the cutoffs have been: 21,18 (I think a lot of people in the top 30 chose not to go that year), 17 (again, a very hard year), and 20. Also, you neglected to mention that seniors cannot make blue MOP. They are only eligible for black MOP.

The red mop cutoff seems to vary, and tends to be a little higher than what you said. Also you left out the important fact that [i]only[/i] 9th graders are eligible. Over the past 5 years it's been 0,7,9,9,10 (these increases have largely been due to more 9th graders qualifying for the USAMO). So to qualify you usually need one problem completely and a fair amount of partial credit (however, partial credit is pretty hard to come by, so I'd say if you want to be sure to make it, you need to solve 2 problems).

Also, [b]there is no "S" in MOP![/b]

how can the cutoff for red mop be 0? were there $ \le 25$ 9th graders who qualified for USAMO that year?

Yes. That was back in 2004, before the expansions.

Ahh!!!!  So much information!!! So many links!!!!  I'm going to read them all now.

For real:  Thanks to everyone.   :)   I couldn't find a single article about this nd I had looked for a few hours.

However.  I am in eighth grade right now.  Am I right in saying that I can't win any more than a bronze medal on the AMC10, and that I can only win up to a gold if I ace the AMC12s every year from now untill graduation?!

 :(   Penalty for a late start.

Is it even worth it going for the bronze medal?  Could a ninth grader get in the top for the AMC12s?  Do you need to have taken the AMC10s to take the AMC 12s?  How many tests a year are there?

AHHH!!!!!!!  I don't take precalc until end of next year!!!  :what?:  :noo:

[quote="1=2"]Basically yeah. You can only win the bronze. The same for the AMC 12(Which I recommend you take. It's easier to qualify for the AIME, and it helps prepare for the AIME as well, where the AMC 10 is easier than your typical Nats MC.

7th graders can get high scores for the AMC 12. The AMC 12 is basically where it's easier to get the highest in the state. You can start taking the AMC 12's whenever(I took it in 6th grade). You don't need the AMC 10. There is an AMC A and B. You choose one of the AMC A tests(10 or 12), then you choose a B test(10 or 12). If you qualify for at least one of them, you get to take the AIME once that year. After the AIME, there is only one test date for each difficulty.[/quote]

Depends. If you want to qualify for USAMO, take AMC 10. Otherwise, AMC 12.

You can take both AMCs. There are two test dates every year, the A date and the B date. On each date, you may take either the AMC10 or the AMC12 but not both. In other words, your options are:
AMC10A and AMC10B
AMC10A and AMC12B
AMC12A and AMC10B
AMC12A and AMC12B
Juniors and Seniors, of course, may not take either AMC10.

I'm not sure why you would be concerned with winning a medal...

I wouldn't worry about what classes you are taking in school; let's put it this way, to this day (yeah I just finished 9th grade) I still have no idea what a matrix is (yeah you learn that in Algebra I but I didn't take math at school since 6th grade) and I have no problem understanding 99% of USAMO problem solutions, and the 1% that I might not understand can be done in other ways.

[quote="kyyuanmathcount"]I'm not sure why you would be concerned with winning a medal...[/quote]
Yes, qualifying for the USAMO ismore meaningful and interesting than winning a school medal.  In general, it means more to be competitive in a big arena than to be the absolute best in a tiny arena.

[quote]I wouldn't worry about what classes you are taking in school; let's put it this way, to this day (yeah I just finished 9th grade) I still have no idea what a matrix is (yeah you learn that in Algebra I but I didn't take math at school since 6th grade) and I have no problem understanding 99% of USAMO problem solutions, and the 1% that I might not understand can be done in other ways.[/quote]
This can get you through high-school contests, but it's not a healthy world view, in my opinion.  There's so much interesting math beyond what you need for contests---in fact, I think the most incredible thing about the contests is that at their best they provide interesting, challenging, elegant problems with practically no prerequisites (speaking relatively).

It's probably not worth the effort in high school to fight your way into classes where you'll be learning math beyond the high school curriculum, but I suggest looking around a little on your own, and maybe digging deeper into something that interests you.

Re 1=2 : If you think AMC 10s are easier than mathcounts, I think your memory deceives you. :wink:

Err... so I should try to take the AMC 10s to get to the AIME, and take the AMC 12s to practice for Junior and Senior years?

I didn't know the medals were school medals.  The more people talk, the more I understand, so thanks.


[quote="bilot4"]Να βρεθούν οι λύσεις της εξίσωσης
.
$ 6x \plus{} 8\sqrt {1 \minus{} x^2} \equal{} 5(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x})$ ,στο διάστημα $ (0.6, 1)$[/quote]
μια προχειρη ιδέα $ x\equal{}cos2u$? και μετά τους τύπους διπλασίου τόξου $ 6cos2u\plus{}8sin2u\equal{}10cos(2u\plus{}a)\equal{}...$ ενὠ $ cosu\plus{}sinu\equal{}\sqrt{2}sin(u\plus{}b)...$

Αν θεσουμε $ x\equal{}cos2t$ και διαιρεσουμε με 10, τοτε στο 1ο μελος εχουμε μια πολυ ωραια μορφη Cauchy-Swartz. Ισος αυτο να βοηθαει καπος, η μπορει να γινεται κατι με την ταυτοτητα του Lagrange...

[quote="bilot4"]Να βρεθούν οι λύσεις της εξίσωσης
.
$ 6x \plus{} 8\sqrt {1 \minus{} x^2} \equal{} 5(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x})$ ,στο διάστημα $ (0.6, 1)$[/quote]

Πολύ προσεγμένη άσκηση! :)

Παρατηρείστε ότι η LHS έχει μέγιστο στο $ 0.6$ (δηλ. είναι φθίνουσα μετά το $ 0.6$). Έτσι, με μονοτονία και κυρτότητα εύκολα προκύπτει ότι θα έχουμε μοναδική λύση στο $ (0.6,1)$.

Λίγες πράξεις, μία παρατήρηση και προκύπτει ότι $ x\equal{}\frac{24}{25}$.

Cheerio,

Durandal 1707

Πολύ ωραίες οι προσεγγίσεις όλων !
Εγώ είχα υπόψην μου την προσέγγιση του κ. Μπόρη :wink: (αν κατάλαβα καλά εννοείτε όπου $ \minus{}a\equal{}\cos^{\minus{}1}0.6$ κι όπου $ \minus{}b\equal{}\frac{\pi}{4}$)

Nίκο μπορείς να εξηγήσεις λίγο τί ακριβώς εννοείς νε κάνουμε με Cauchy-Swartz (εννοώ πώς ακριβώς σχετίζεται)?

[quote="bilot4"]Πολύ ωραίες οι προσεγγίσεις όλων !
Εγώ είχα υπόψην μου την προσέγγιση του κ. Μπόρη :wink: (αν κατάλαβα καλά εννοείτε όπου $ \minus{} a \equal{} \cos^{ \minus{} 1}0.6$ κι όπου $ \minus{} b \equal{} \frac {\pi}{4}$)

Nίκο μπορείς να εξηγήσεις λίγο τί ακριβώς εννοείς νε κάνουμε με Cauchy-Swartz (εννοώ πώς ακριβώς σχετίζεται)?[/quote]
Ακριβώς έτσι αλλά μετά θέλει λίγο κουραστική διερεύνηση στους τύπους με τα 2κπ

[quote="bilot4"]Nίκο μπορείς να εξηγήσεις λίγο τί ακριβώς εννοείς νε κάνουμε με Cauchy-Swartz (εννοώ πώς ακριβώς σχετίζεται)?[/quote]

απλα αν διαιρεσουμε με 10 και υψωσουμε στο τετραγωνο το δεξι μελος γινεται: $ (\frac{6}{10}\cos{2x} \plus{} \frac{8}{10}\sin{2x})^2$, οπου τα τετραγωνα των $ \sin{2x}, \cos{2x}$ αλλα και τα τετραγωνα των συντελεστων τους εχουν αθροισμα 1, αυτο θυμιζει Cauchy-Swartz. Δοκιμασα με Cauchy-Swartz χτες μηπως καταφερω να κανω κατι με την περιπτωση ισοτητας αλλα δεν καταφερα τιποτα, ακομα ομως δεν εχω δοκιμασει την ταυτοτητα του Lagrange απο την οποια προκειπτει η ανισοτητα CS και με την οποια θα μπορουσε να βγει κατι...

[quote="bilot4"]Να βρεθούν οι λύσεις της εξίσωσης
.
$ 6x \plus{} 8\sqrt {1 \minus{} x^2} \equal{} 5(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x})$ ,στο διάστημα $ (0.6, 1)$[/quote]

Λυση:

θετουμε $ z\equal{}\sqrt{1\plus{}x}, y\equal{}\sqrt{1\minus{}x} ,0\leq y \leq \sqrt{0.4}$ ,$ \sqrt{1.6} \leq z \leq \sqrt{2}$
  επομενως η εξισωση μας αναγεται στην επιλυση του συστηματος:
$ 3z^2 \minus{} 3y^2 \plus{}8yz \equal{} 5y\plus{}5z$(1)
$ y^2\plus{}z^2\equal{}2$(2)
η (2) μας επιτρεπει να θεσουμε
$ y\equal{} \sqrt{2}sina ,z\equal{}\sqrt{2}cosa$
και η (1) γραφεται ισοδυναμα
$ 6cos2a\plus{}8sin2a\equal{} 5\sqrt{2} (cosa\plus{}sina)$
$ \leftrightarrow sin(2a\plus{}u)\equal{}sin(a\plus{}\frac{\pi }{4})$ με $ cosu\equal{}\frac{8}{10},sinu\equal{}\frac{6}{10}$
θα διακρινουμε τις περιπτωσεις.(κ ακεραιος)
1st) $ 2a\plus{} u \equal{} 2k \pi \plus{} a\plus{}\frac{\pi }{4} \leftrightarrow \sqrt{2} sina \equal{} \frac{1}{5} \leftrightarrow x\equal{} \frac{24}{25}$
2nd) $ 2a\plus{}u \equal{} 2k \pi \plus{} \pi \minus{} (a\plus{}\frac{\pi }{4}) \leftrightarrow sin3a\equal{} \frac{7 \sqrt{2} }{10} \leftrightarrow 2y^3 \minus{} 3y \plus{} \frac{7}{5}\equal{}0$ σχεση αδυνατη για $ 0\leq y \leq \sqrt{0.4}$
μονη μας λυση λοιπον η $ x\equal{}\frac{24}{25}$


logarithms
Support
inequalities
floor function

[b]1. How do you find the number of digits of a larger products or for a large power..[/b]
for ex: how many digits does 2^2007 has
                             or
how many digits does 12345678 x 98323232323 has???

[b]2. How to find the sum of digits for the same..[/b][/b]

[hide="#1"]If you want to find the number of digits for a big number, take logarithm of base 10, floor it, and add 1.

Let's take $ 12345678\cdot 98323232323$ for an example.

Using [url=http://www90.wolframalpha.com/input/?i=12345678+x+98323232323]WolframAlpha[/url], we firstly note that it has 19 digits.

Now, [url=http://www31.wolframalpha.com/input/?i=floor+%28log_10+%2812345678+x+98323232323%29+%2B1%29]taking logarithm of base 10, flooring it, and adding 1[/url], we indeed have 19 digits.[/hide]

For #2, I do not think there is a shorter way of finding the sum of digits than Brute Force.

[quote="$ LaTeX$"]I think that logico meant: Is there a way to find the number of digits without using anything but your head, and paper and pencil....... Just about anybody could find the numbers of digits with a resource like Wolfram Alpha..[/quote]

yes, that was my point...

using log is easy, but when you get a bigger prime number then it will be difficult....
for ex.  23^2007

There should be a simpler way to do the same without log..

[b]Also any ideas on finding sum of digits???????[/b]

[quote]how many digits does 12345678 x 98323232323 has??? [/quote]

There is a way to find it without using "brute force". And is aplying logharitm laws.

the number of digits is given by:   (log  (12345678  * 98323232323)  ) adding one unit digit and excluding the decimals.

Then    log  (12345678  * 98323232323)  = log 12345678 + log 98323232323

                                                             =7.0915 + 10.992

                                                             =18.0835
Thus, the multiplication has 19 digits.

No, Chalao, nothing you have typed supports that you worked
out the logarithms of those two numbers with pen/pencil and
paper type of method displayed on here.  It appears that you
entered them in a calculator, which was other than point of the
the OP.

For $ 2^{2007}$, if you're like me, you have your basic values of logs memorized, for example, $ \log_{10} 2 \equal{} 0.301029996\approx 0.30103$ and that should be a close enough approximation for powers up to 1 million or so. Then all you have to do is multiply that by 2007, floor it, and add 1. I get 605 digits this way.

[hide="Why this works"]
Note that for any positive integer $ n$, we have $ \log_{10}(10^n) \equal{} n$, and also note that $ 10^n$ has $ n \plus{} 1$ digits. Therefore, any number $ x$ with $ d$ digits satisfies $ 10^{d \minus{} 1}\leq x < 10^d$. Therefore, taking the log of this inequality, we arrive at:
\[ d \minus{} 1\leq \log_{10}x < d,\]
and so we can say $ d \minus{} 1 \equal{} \lfloor \log_{10}x\rfloor$, or $ d \equal{} \lfloor \log_{10}x\rfloor \plus{} 1$.

Now, just plug in $ x \equal{} 2^{2007}$. Then, applying the properties of logarithms, we have $ \lfloor \log_{10}(2^{2007})\rfloor \plus{} 1 \equal{} \lfloor 2007\log_{10} 2\rfloor \plus{} 1 \equal{} 604 \plus{} 1 \equal{} 605$.[/hide]

For $ 12345678 \times 98323232323$, the first number has 8 digits and hence is on the order of $ 10^7$ while the second number has 11 digits and hence is on the order of $ 10^{10}$. Therefore the product will be on the order of either $ 10^{17}$ or $ 10^{18}$, depending on whether this product "overflows." Here, we can easily see that $ 12\times 98$ already overflows, so the original product will certainly overflow, so the answer is on the order of $ 10^{18}$, or 19 digits.

Also, for sum of digits, I do not believe there is any nice answer. The best you can do is calculate it mod 9, which is just the same as the number mod 9, which is easily calculated by number theory.


[quote="simulacrum"]what do you guys think of that book in terms of AMC/AIME preparation?[/quote]
It's problems are of the AIME level difficulty.

[quote="vikas"]What book?[/quote]
Challenging Problems in Algebra

[quote="gighiuhui"][quote="simulacrum"]what do you guys think of that book in terms of AMC/AIME preparation?[/quote]

For AMC/AIME preparation, why not just do AIME problems?

Seriously, there is no need to go all books, books, books.  Just do powrbums.[/quote]

if you mean sample tests, those run out after you've been practicing for a while...

To be honest, I have it and I don't particularly like it. Challenging Problems in Geometry is much better, in my opinion. There are a few nice problems, but most are too easy, tedious, and not altogether very interesting.

i have it as well, and i haven't really looked at it... maybe i should.

Those problems are not aime level, they're a lot easier.

what about for AMC preparation? if not AIME.

it's definetely a good book for amc, but previous tests are probably more reliable.


ARML
trigonometry
inequalities
calculus
geometry
triangle inequality

I wrote up a solutions document [attached]. I realize there are other solutions; if you want, you can write up alternate solutions. I will post results and further thoughts when I have time.

--Alex

Number four: I had written down exactly \[\frac12 =h+(1-h)^{3}\frac12\iff 1=2h+(1-h)^{3}\iff 2h-h+h^{2}-h^{3}=-h(h^{2}-h-1)=0\] where I got stuck since this has no real roots in $[0,1]$... are you sure you factored right? (I'm probably stupid)

lmao i forgot that 0 could be the 2nd digit for #2
eww 3d geo for #3
#7 i wasnt thinking
eww geo for #8

so yeah i did bad.
but it doesnt matter cause im on the CT team

[quote="me@home"]Number four: I had written down exactly
\[\frac12 =h+(1-h)^{3}\frac12\iff 1=2h+(1-h)^{3}\iff 2h-h+h^{2}-h^{3}=-h(h^{2}-h-1)=0 \]
where I got stuck since this has no real roots in $[0,1]$... are you sure you factored right? (I'm probably stupid)[/quote]

There is an algebra mistake...

$1=2h+(1-h)^{3}\implies 0=2h-3h+3h^{2}-h^{3}$

[quote="pianoforte"][hide=" from my comment for #8"]"8. Since O' and O are tangent at A, O' lies on AO=5. From a point on AO, to draw an altitude to BC gives a triangle with an angle twice that of C, and used $x*csc(2C)+x=5$?"

$\sin C=3/5\implies \sin(2C)=\frac{2\cdot3\cdot4}{5\cdot5}=\frac{24}{25}\implies csc(2C)=\frac{25}{24}$.
$\frac{25}{24}x+x=5\implies x=\frac{120}{49}$.[/hide][/quote]
that was my solution

1.ok
2.wow how did i miss this xD
3.i was like "oh my its a 3d geo. screw that" when i had 7 minutes or so left
4.ok
5.ok
6.not ok
7.was too stupid/lazy to do an inequality problem so i guessed 1/2 and spent my time on 8
8.cool prob

4?
i hope i do better on the real thing

A problem like #7 wouldn't be on ARML, or any other competition. It's really easy with calculus, but it's good to be able to solve it without using calculus.

1. Same as Altheman

2. Is much simpler if you realize that a number is congruent to its digit sum mod 9, so the number is not just divisible by 11, but by 99 (realizing this makes it very easy to see that 4 digit numbers will not work, and much easier to find a 5 digit number that does).

3. Same

4. Same

5. Same

6. Did essentially the same thing, except in a less lucid and more roundabout way =\ meh.

7. Notice that this is equal to $(21-y)^{2}+(y-x)^{2}+(x-20)^{2}$, then $RMS \geq AM$ will do it.

I don't think calculus is significantly faster - you can fix $y$, differentiate and minmize to $x$, and vice versa, and get $\{ x = \frac{20+y}{2},y=\frac{21+x}{2}\}$, which produces the correct solution.

8. Did what pianoforte did.

WAIT WAIT WAIT for #7 that's what i did but don't those things have to be positive?

so i was like huh i don't want to do all this casework so i just guessed 1/2

gah so i technically could have gotten a 5 (got 1245) had i gone with my first instinct...

for 3 i saw the distance formula thing but didn't see how to proceed from there
there was basically no chance i could have gotten 6 in the given time
but for #8 i didn't even bother reading because i was like EH geometry AND #8 so yeah no chance
didn't turn out to be too bad though
oh well.

[quote="13375P34K43V312"]WAIT WAIT WAIT for #7 that's what i did but don't those things have to be positive?[/quote]
Behold :
\begin{align*}(21-y)^{2}+(y-x)^{2}+(x-20)^{2}&= |21-y|^{2}+|y-x|^{2}+|x-20|^{2}\\ &\ge 3(|21-y|+|y-x|+|x-20|)^{2}\\ &\ge 3|21-y+y-x+x-20|^{2}= 3 \end{align*}
(the triangle inequality)

Equality holds iff. $|21-y|=|x-20|=|y-x|$ and $21-y$, $x-20$, and $y-x$ all have the same sign, i.e., if $21-y = x-20 = y-x$.  It is easy but important to check for the existance of equality cases here, since we want an actual minimum, or else the inequality we proved was too weak.

[quote="13375P34K43V312"]WAIT WAIT WAIT for #7 that's what i did but don't those things have to be positive?

so i was like huh i don't want to do all this casework so i just guessed 1/2
[/quote]

You're absolutely right.

However, consider it as the sum of the squares of the distances between points on a number line (you can even picture it by drawing four points, two anchored and two variable, and drawing squares with one side given by the respective pairs of points)

You would really want $21 \geq y \geq x \geq 20$, or else you'd be making counterproductive motion and your squares would start overlapping and getting really big.

The case in which all are positive is the only case that one would have to look at.

But it's 1/3, rather than 1/2.  See, when x and y are both halfway between 20 and 21, the height is 1/2, while when they are each 1/3 of the way in (on opposite sides) the height is only 1/3.  This is a more coherent version of the reason I originally used to obtain the correct answer.

all problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3733598_2007_mock_arml_i_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


absolute value
combinatorics proposed
combinatorics

we have a array  $n$*$n$ with real number with absolute value at most $1$.and a sum of the all numbers is $0$.prove there exist one column or
one row  such that sum of the absolute value of its entires  are at most $n/2$. ;)

Please continue discussion of this problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=41119 .

  Darij


For what integral values of x and p is (x^2-x+3)(2p+1) an odd number?

$ x^{2}$ and $ x$ are of the same parity, and $ 2p+1$ is always odd, so $ (x^{2}-x+3)(2p+1)$ is odd for all values of x and p.

@mdk: [i]Challenging Problems in Algebra[/i] by Posamentier and Salkind has solutions in the back of the book. I am pretty sure that you have this book because you have posted like 10 consecutive problems from pg. 28-29.


calculus
derivative
function
integration
number theory
prime factorization
calculus computations

If the derivative of a function f is given by f'(x) = sin(x^x), then how many critical points does the function f(x) have on the interval [0.2,2.6]?

If $ 0<x<1,$ then $ 0<x^x<1.$ In particular, the minimum of $ x^x$ on $ [.2,1]$ is some number between $ 0$ and $ 1.$ (And $ 1<\pi.$)

After that, $ x^x$ is an increasing function.

$ (2.6)^{2.6}$ is about $ 12,$ and lies between $ 3\pi$ and $ 4\pi.$ (Calculators do have a few uses.)

The information you need to answer the question is contained in these computations, if you know what you should be looking for.

[quote="ProblemSets"]I'm getting 3 critical points... Is this correct?[/quote]

Well, just to check (not to actually do the problem), you could graph the function on your calc on the given interval and count how many max & min points there are (hopefully there's no straddle points).

Problem sets: that is correct. The critical points are the (unique) solutions to the equations $ x^x\equal{}\pi,x^x\equal{}2\pi,$ and $ x^x\equal{}3\pi.$

Prime factorization: that was a useless post. Displaying the graph of $ f(x)$ is not even readily available to us, since $ f(x)\equal{}\int\sin(x^x)\,dx$ and we can't do that antiderivative in closed form. We could graph $ f'(x)$ and look for its zeros - not that I'd ever completely trust a graphing device on a question like that.


How many numbers less than 1000 are divisible by 3,4,5, and 6?

how do u figure that out?? i could do it the dumb way and take a long... time...

Hint:

You have to find a number that's divisible by 3,4,5, and 6. If the number is divisible by 6, is it divisible by 3? If the number is 6*2, is it divisible by 4? If the number is 6*2*5, is it divisible by 5 and the rest?

If so, what can you say about the largest number that is is divisible by 3,4,5, and 6? Would it be in simliar form as the 6*2*5?


geometry trigonometry circumcircle vector function calculus derivative

Show that the length $d$ of a chord of a circle of radius $r$ is given by the formula
\[d = 2r\sin{\frac{x}{2}}\]
where $x$ is the central angle formed by the radii to the ends of the chord (see the figure). Use this result to derive the fact that $\sin{x}< x$, where $x > 0$ is measure in radians.

..could someone explain how to derive $\sin{x}< x$?

let the chord cut the circle in b&c o is the center of the circle.
take any point on the circumference of circle. call it a
then angle BAC=x/2(any angle substended by a chord at cirum is half that it subs. at the center)
apply sine rule to tri. bac
bc/sinBAC=ab/sinbca=ab/sin acb=2r
where r is the radius of circumcircle of tri. abc
we get d=2r sin (x/2)
hence proved
we know that arc length =angle in radians*radius
let arc bc be s units
then s=rx
but obviously s>d
thus rx>d
or x>d/r
but d/r=2 sin (x/2)
hence we get x/2>sin x/2
hence the problem
a/sin a=2r is proved by vectors
you may assume a functio F(x)=x-sinx
f(0)=0
F`(x)=1-cos x>0 as -1<=cosx= <1
hence the function always increases
hence is always +ve
hence x>=sinx always
hence the problem
by the way how did you draw such a decent diagram?

ok, thanks anant. I'll print that out and try to understand it at school.

lol I used paint to draw that.. not much use in drawing elaborate diagrams, but for simple things like this, it's good enough.

you mean less than or equal to x. What if x=0?

jeez just use calc

all you really need to test is when 0 < x < 1

and since the derivative of sin(x) is cos(x) which is always less than or equal to the derivative of x (1), you win

mhmm.. i get the first method anant described..

I think I sorta get the calculus method treething described..


Καλησπέρα σας!Ήρθε στα χέρια μου ένα πρόβλημα το οποίο πιστεύω αξίζει καποιος να ασχοληθεί μαζί του.
"Σ' έναν κύκλο θεωρούμε 3 διαδοχικά ίσα τόξα (με οποιοδήποτε μήκος) κ εννόνουμε τις γειτονικές άκρες τους.
Στις 3 χορδές που σχηματίζονται εντοπίζουμε τα μέσα τους.Να δειχτεί οτι τα μέσα αυτα είναι κορυφές ισοπλεύρου τριγώνου."
Υπάρχει κ μια ισοδύναμη διατύπωση με ισόπλευρα τρίγωνα που έχουν κοινή κορυφή το κέντρο του κύκλου.
Έχω δει μια λύση με μιγαδικούς αλλά θα ήθελα να με βοηθήσετε αν έχετε στο νου σας καθαρά γεωμετρική λύση.
Ευχαριστώ.

Ισως δεν έχω καταλάβει καλά αλλά αν πάρουμε (για παράδειγμα) σημεία $ A,B,C,D$ διαδοχικά τέτοια ώστε τα τόξα που σχηματίζονται ($ AB,BC,CD$) να ισούνται με 60 μοίρες που σημαίνει ότι $ AD$ διάμετρος τόττε γαι να είναι ισόπλευρο το τρίγωνο που θέλουμε θα έπρεπε να ισχύει $ AD\equal{}BD\equal{}AC$ (ώστε να είναι ίσα και τα μισά τους που θα είναι οι πλευρές του τριγώνου που εξετάζουμε) το οποίο φανερά δεν ισχύει :huh:  Στην πραγματικότητα ισχύει πάντα ότι είναι ισοσκελές (αφού $ AC\equal{}BD$) αλλά όχι ισόπλευρο...

Παρακαλώ διόρθωσε αν εννοείς κάτι άλλο που δεν έλαβα υπόψη μου...

[quote="bilot4"]Ισως δεν έχω καταλάβει καλά αλλά αν πάρουμε (για παράδειγμα) σημεία $ A,B,C,D$ διαδοχικά τέτοια ώστε τα τόξα που σχηματίζονται ($ AB,BC,CD$) να ισούνται με 60 μοίρες που σημαίνει ότι $ AD$ διάμετρος τόττε γαι να είναι ισόπλευρο το τρίγωνο που θέλουμε θα έπρεπε να ισχύει $ AD \equal{} BD \equal{} AC$ (ώστε να είναι ίσα και τα μισά τους που θα είναι οι πλευρές του τριγώνου που εξετάζουμε) το οποίο φανερά δεν ισχύει :huh:  Στην πραγματικότητα ισχύει πάντα ότι είναι ισοσκελές (αφού $ AC \equal{} BD$) αλλά όχι ισόπλευρο...

Παρακαλώ διόρθωσε αν εννοείς κάτι άλλο που δεν έλαβα υπόψη μου...[/quote]

Ίσως δεν το εξήγησα σωστα.Τα τόξα που παίρνουμε δεν είναι διαδοχικά αλλά τυχαία μέσα στον κύκλο.Αρκεί να είναι ίσα.π.χ. ΑΒ, CD, EF.Λάθος δικό μου που χρησιμοποίησα τη λέξη διαδοχικά.Έπειτα εννόνουμε το Β με το C, το D με το E και το F με το Α.Τα τόξα επίσης μπορούν να περιέχουν τμήματα των άλλων τόξων αλλά όχι να ταυτίζονται ούτε να είναι διαδοχικά.Ελπίζω να τη βελτίωσα κάπως την κατάσταση.... :huh:

[quote="silouan"]Η καλύτερη λύση είναι αυτή με τους μιγαδικούς . Για την συνθετική λύση δες εδώ
http://www.mathlinks.ro/viewtopic.php?t=45874[/quote]

Silouan από που βγήκε ο τύπος <ΚΤΜ = (ArcAB + ArcZEDC)/2 ?
Γράφεις ZEDL αλλά λογικά θα εννοείς ZEDC.

Silouan μήπως κοίταξες αυτό που σε είχα ρωτήσει για την <ΚΤΜ?

Νομίζω ότι το πρόβλημα πάλι [b]δεν είναι σωστά διατυπωμένο[/b]:
Θεωρόντας ως ΑΒ, ΓΔ, ΕΖ τα τόξα και Κ, Λ, Μ τα μέσα τους αντίστοιχα και Ο το κέντρο του κύκλου, αρκεί να αποδειχτεί (αναγκαία και απαραίτητη συνθήκη) ότι τα τρίγωνα ΟΚΛ,ΟΛΜ, ΟΜΚ είναι ίσα μεταξύ τους.
Επειδή οι χορδές ΑΒ, ΓΔ, ΕΖ είναι ίσες (επειδή τα τόξα στα οποία βαίνουν είναι ίσα), θα είναι και τα αποστήματα τους ΟΚ, ΟΛ, ΟΜ ίσα.
Άρα αρκεί να δείξουμε (πάλι [u]αναγκαία και απαραίτητη συνθήκη[/u]) ότι οι [u]γωνίες[/u]
ΚΟΛ, ΛΟΜ, ΜΟΚ είναι ίσες, δηλαδή ότι (επειδή αυτές γράφονται ως αθροισμα δύο γωνιών ίσων με το μισό των τόξων η κάθε μία) [b]οι γωνίες μεταξύ των χορδών ΑΒ, ΓΔ, ΕΖ είναι ίσες [/b](δεν το αναφέρει η εκφώνηση).

Μπορεί βέβαια να κάνω λάθος (αν κάνω, ας μου το πει κάποιος). Πάντως, αν έχω δίκιο, η άσκηση δεν είναι ιδιαίτερη.

[quote="silouan"]Κοίτα την εκφώνηση της άσκησης στο link που έδωσα παραπάνω ...
Για την ερώτηση tou biliakos τώρα ... Αυτός ο τύπος είναι στο σχολικό βιβλίο της γεωμετρίας στη σελ 125 .[/quote]

Ευχαριστώ Silouan.Αν έχεις καμιά άλλη σημείωση η κανένα γνωστό link που να κάνει ανφορά κ να έχει λύσεις γεωμετρικές αυτής της άσκησης θα σου ήμουν ευγνώμων αν μου τα έστελνες.Ευχαριστώ κ πάλι εκ των προτέρων.....
(Πάλευα ποσο καιρό να βρω γεωμετρική λύση αυτής της άσκησης....)

Στο περιοδικό Απολλώνιος τεύχη 2,3 παρουσιάζονται κοντά στις 5 λύσεις . Υπάρχει μία με μετασχηματισμούς , μία με διανύσματα
και οι υπόλοιπες είναι συνθετικές , αλλά το κλίμα τους βασίζεται πάνω στην απόδειξη που έκανα στο link . Επιπλέον το θέμα όπως ανέφερε και ο darij είναι ειδική περίπτωση ενός γενικέτερου ισχυρού θεωρήματος...


What is the probability that the square root of a randomly
selected two-digit whole number is less than eight? Express
your answer as a common fraction.

Basically, we want to find the number of two-digit integers less than 64 and put it over the number of two-digit integers in total. $ 63 \minus{} 10 \plus{} 1 \equal{} 54$ numbers we want and $ 99 \minus{} 10 \plus{} 1 \equal{} 90$ numbers in total. Our answer is $ \boxed{\frac{3}{5}}$.


Let $n$ be fixed natural number. Consider sequence ${x_n}$ such that:
$\displaystyle 0 < x_1 \leq x_2 \leq \cdot \cdot \cdot \leq x_n$. Find the maximum value of expression: $\displaystyle S=\frac { \sum_{k=1}^n \frac {x_k}{k(k+1)(k+2)}}{ \sum_{k=1}^n \frac {x_k}{(k+1)(k+2)}}$


in the trşangle ABC wıth radıus of the cırcumscrıbed cırcle R and center O and wıth radıus of the ıncrıbed cırcle r and center i where O ıs dıfferent from i.We denote G the ıntersectıon poınt of medıansç Prove that IG ıs perpendıcular to BC ıf and only ıf b=c or b+c=3a

please ı need ıt for today  or better for now

please somebody help me. ı need ıt for today....

[color=darkblue]We use the following identity.[/color]

[color=darkred]Let $ A,B,C,D$ be four points in the plane. Then $ AB^2\minus{}AD^2\equal{}CB^2\minus{}CD^2$ if and only if $ AC\perp BD$.
The "if" part can be easily proved by Pythagorean theorema applied to 4 triangles. "Only if" needs cosine theorem.[/color]
 [color=darkblue]
Therefore is true equivalence $ IG\perp BC\Longleftrightarrow IB^2\minus{}IC^2\equal{}GB^2\minus{}GC^2$ (*).

 From median theorem and property $ BG\equal{}\frac{2}{3}m_b$ we conclude that

$ GB^2\equal{}\dfrac{4}{9}\left(\dfrac{2(a^2\plus{}c^2)\minus{}b^2}{4}\right)\equal{}\dfrac{2a^2\plus{}2c^2\minus{}b^2}{9}$

$ GC^2\equal{}\dfrac{4}{9}\left(\dfrac{2(a^2\plus{}b^2)\minus{}c^2}{4}\right)\equal{}\dfrac{2a^2\plus{}2b^2\minus{}c^2}{9}$

But $ IB^2\equal{}(p\minus{}b)^2\plus{}r^2$ \3i $ IC^2\equal{}(p\minus{}c)^2\plus{}r^2$. Substituting in (*) we obtain

$ IG\perp BC\Longleftrightarrow (p\minus{}b)^2\minus{}(p\minus{}c)^2\equal{}\dfrac{1}{9}\left( b^2\plus{}c^2\right) \Longleftrightarrow b^2\minus{}c^2\plus{}3ac\minus{}3ab\equal{}0 \Longleftrightarrow (b\minus{}c)(b\plus{}c\minus{}3a)\equal{}0$

Therefore $ b\equal{}c$ or $ b\plus{}c\equal{}3a$[/color]


find a,b,c is posive real satisfy  \sum ab=abc
 prove     \sum \sqrt (ab-c) \geq   \sqrt abc + \sum  \sqrt a :(  :)

Is your problem?
Let a,b,c is posive real satisfy  ab=abc
prove   (ab-c)   abc +   a

Is it your problem?

Let a,b,c is posive real satisfy  ab=abc
prove   (ab-c)   abc +   a

thank you
my problem is
       let a,b,c is a positive real satisfy  \sum ab=abc
       Prove:  \sum  \sqrt (ab-c) \geq  \sqrt abc+ \sqrt a+ \sqrt b+ \sqrt c
                                         :D  :?

Can you tell me the source of the problem ?
I think you should tell more clearly about it .
If you don't know the source , write Unknown.


function
calculus
derivative
calculus computations

I cannot seem to find an example of a continuous function such that Newton's Method would diverge if $ f'(x_{1})$ was close to $ 0$. What are all the possible ways for this to occur?

I also do not understand how to determine initial guesses that cause Newton's Method to oscillate between 2 values. As always, any help will be greatly appreciated.

I searched my files and came up with this part of a test question I wrote for a Numerical Analysis course I taught in 1993:

==============================================

(b) I wrote a program to solve $ x^3 \plus{} 6x^2 \minus{} 34 \equal{} 0$ by Newton's method. The basic iterative loop was set to terminate after 40 steps no matter what. I started with a guess of $ x_0 \equal{} \minus{} 3.$  The program terminated without finding a root. When I had it print out the values at each stage, the numbers looked like this:
$ \minus{} 3.00000, \minus{} 3.77778, \minus{} 4.68519, \minus{} 4.15162, \minus{} 3.01766, \minus{} 3.78702, \minus{} 4.72207, \minus{} 4.18392,$
$ \minus{} 3.22693 , \minus{} 3.91157, \minus{} 5.88355, \minus{} 4.98212, \minus{} 4.38708, \minus{} 3.80665, \minus{} 4.81072, \minus{} 4.25720,$
$ \minus{} 3.52233, \minus{} 4.16819, \minus{} 3.13429, \minus{} 3.85269, \minus{} 5.10197, \minus{} 4.47208, \minus{} 3.92856, \minus{} 6.34001,$
$ \minus{} 5.26901, \minus{} 4.58574, \minus{} 4.05714, \minus{} 1.15318, \minus{} 3.95097, \minus{} 7.41696, \minus{} 5.94454, \minus{} 5.02061,$
$ \minus{} 4.41478, \minus{} 3.84981, \minus{} 5.07892, \minus{} 4.45600, \minus{} 3.90768, \minus{} 5.80214, \minus{} 4.93060, \minus{} 4.34928$

Why isn’t it converging? Explain what is going on. (A graph may help.)

=================================================

Notes:

1. Part (a) of the question involved solving the same equation with a first guess of $ x_0\equal{}2,$ in which case it works (converges quickly).

2. If I let the iteration run just a little bit longer, it does fall eventually out of the trap and converge to the same root found in part (a).

From what I can tell:

-The graph's derivative raplidly decreases when approaching its local maximum at $ ( - 4, - 2)$.
-Taking the derivative at a point $ ( - 4 + h,f( - 4 + h))$ (where $ h$ is small and $ > 0$) produces a tangent line whose root is at $ ( - 4 - k,f( - 4 - k))$ (where $ k$ is a similar quantity as $ h$).
-Thus, the graph oscillates in the manner described above over some finite period of iterations until a guess is yielded where $ h$ or $ k$ is small enough to yield a tangent line horizontal enough to get further guesses out of this sensitive interval of values.
-Once the above occurs, Newton's Method will then converge quickly.

This is an intriguing example, but I'm still left wondering if there are functions that never converge no matter how many iterations occur. This of course excludes initial guesses where $ f'(x_{1}) = 0$,  functions that don't converge regardless of the initial guess (eg/ $ x^{1/3)}$), functions whose initial guess is poor enough to cause it to converge on an unintended root, and cycles (eg/ $ f(x) = x^{3} - 2x + 2$ with $ x_{1} = 0$ ).

I guess what I'm asking is, can Newton's Method truly fail to converge for bad initial guesses, or is the iterative process simply going to be inefficient?


Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=2$

prove that

$\abs{a^3+b^3+c^3 - abc} \leq 2 \sqrt{2}$

WLOG:$a \geq b \geq c$
if $abc \geq 0$,then easy to see one of $min \{a^3-abc,b^3-abc,c^3-abc \} \leq 0$,
and $a^3+b^3 \leq 2\sqrt{2}$...the rest is easy.
if $abc \leq 0$,we have two case:
(a) $a \leq 0$ easy.
(b) $a \geq b \geq 0 \geq c$
then let $x=a,y=b,z=-c$,$x^2+y^2+z^2=2$
try to prove $(x^3+y^3+xyz)^2 \leq (x^2+y^2+z^2)^3$(not hard,i think).
sorry for my terrible solution,hope it is right... :)


Prove that the fields $ \mathbb{R}(x)$ and $ \mathbb{R}(x,y)$ aren't isomorphic.

[quote="ZetaX"]They have distinct transcendence degrees.[/quote]

How exactly do you conclude: do you mean that if there were an isomorphisms then it would send $ \mathbb{R}$ to $ \mathbb{R}$ and the image $ \mathbb{R}(x,y)$ must have the same transcendence degree over $ \mathbb{R}$ as $ \mathbb{R}(x)$ over $ \mathbb{R}$ ?

Is there any other more direct way ? still it invokes the proof of the fact that any transcendence bases have the same cardinality.

A better way:

For any field $ F$, define $ F^\infty : \equal{} \bigcap_{n \equal{} 1}^\infty F^n$, where $ F^n \equal{} \{ x^n | x \in F \}$. Now for any field, we have $ \left(F(x_1, ..., x_n) \right)^\infty \equal{} F^\infty$.
Now $ \left(\mathbb R (x_1, ..., x_n) \right)^\infty \equal{} \mathbb R^\infty \equal{} \mathbb R_0^ \plus{}$. Especially, we have that $ K \equal{} \mathbb R (x_1, ..., x_n)$ is generated by $ K^\infty$ and exactly $ n$ indeterminates (that are algebraic independent over the subfield $ \mathbb R \equal{} K^\infty \cup ( \minus{} K^\infty)$. Thus we can describe $ n$ as the transcendence degree over the subfield $ K^\infty \cup ( \minus{} K^\infty)$. The result follows (in it's canonical generalisation).

I now edited it to hopefully describe it more clearly.


Solve in $ M_2(\mathbb{R})$ the equation:

$ X^3\minus{}4X^2\plus{}5X\equal{}\left(
\begin{array}{cc}
10 & 20 \\
5 & 10 \end{array}
\right)$

[hide]The answer is $ X \equal{} \left [\begin{array}{cc} 2 & 4 \\ 1 & 2  \end{array} \right]$. Start by noting that:

1) The eigenvalues of $ \left [\begin{array}{cc} 10 & 20 \\ 5 & 10  \end{array} \right]$ are 20 and 0.

2) Therefore, the eigenvalues of X are 0 and 4.

3) det(X) = 0.

[/hide]

Another approach might be the following :

Since $ \det X\equal{}0$, from Cayley-Hamilton we have that $ X^2\equal{}aX$, $ X^3\equal{}a^2X$, where $ a\equal{}tr(X)$.From the initial equiation we have:

$ tr(X^3\minus{}4X^2\plus{}5X)\equal{}20$ $ \Rightarrow$ $ a^3\minus{}4a^2\plus{}5a\minus{}20 \equal{}0$ $ \Rightarrow$ $ a\equal{}4$.

But $ X^3\minus{}4X^2\plus{}5X\equal{}\begin{pmatrix} 10 & 20 \\ 5 & 10\end{pmatrix}$ $ \Leftrightarrow$ $ X(a^2\minus{}4a\plus{}1)\equal{}\begin{pmatrix} 10 & 20 \\ 5 & 10\end{pmatrix}$.So $ X\equal{}\begin{pmatrix} 2 & 4 \\ 1 & 2\end{pmatrix}$

Here is my solution:

Denote by $ A\equal{}\left(
\begin{array}{cc}
10 & 20 \\
5 & 10 \end{array}
\right)$ Then, we deduce that $ AX\equal{}XA$

Suppose that $ a,b,c,d \in \mathbb{R}$ such that $ X\equal{}\left(
\begin{array}{cc}
a & b \\
c & d \end{array}
\right)$ From $ AX\equal{}XA$ we obtain some equations and after that, the relations: $ a\equal{}d$ and $ b\equal{}4c$

When we use the relation from the hypothesis we obtain after some calculation that $ 2c\equal{}a$, using the fact that $ a,b,c,d \in \mathbb{R}$ and $ a\equal{}d$ and $ b\equal{}4c$, and finally $ c\equal{}1$


A radio transmission tower is 100 feet high.  How long must a guy wire to be if it is to be connected to the top of the tower from a point 30 feet away from the base of the tower?

[hide]
Pythagorean Theorem
$\sqrt{100^{2}+30^{2}}= 10 \sqrt{109}$
[/hide]

Are you saying that the length of the guy wire from a point 30 feet away from the base of the tower is 10 times the square root of 109?

I did not know that square roots were used in terms of measuring.

It's just a value

[hide]
$10\sqrt{109}feet \simeq 104.4 feet$
[/hide]

The length of 104.4 feet makes more sense.

Thank you


Circles with radii $r_1,r_2,r_3$ are externally tangent to each other. Two other circles, with radii $R,r$, are tangent to all previous three circles. Prove that $Rr\ge\frac{r_1r_2r_3}{r_1+r_2+r_3}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=35095

  darij


geometry
circumcircle
symmetry
geometry proposed

Consider the acute-angled triangle $ ABC$, altitude $ AD$ and point $ E$ - intersection of $ BC$ with diameter from $ A$ of circumcircle. Let $ M,N$ be symmetric points of $ D$ with respect to the lines $ AC$ and $ AB$ respectively. Prove that $ \angle{EMC} \equal{} \angle{BNE}$.

Nice and easy problem !
Suppose that AB<AC.
$ \angle OAN\equal{}\angle OAB\plus{}\angle BAN\equal{}\angle CAD\plus{}\angle BAD\equal{}\angle BAC$. Similarly, $ \angle OAM\equal{}\angle BAC$, wich imlies AO is the bisector of angle MAN. Denote H,K are the orthogonal projections of E on AN,AM respectively.
Because of symmetry, we have some equalities of angles:
$ \angle ANB\equal{}\angle ADB\equal{}90^o\Rightarrow BN\parallel EH\Rightarrow\angle BNE\equal{}\angle NEH$
Similarly, $ \angle EMC\equal{}\angle MEK$
$ \angle MEK\equal{}\angle NEH$
Hence result.

[color=darkblue]Mine a little different solution..

$ \left\| \begin{array}{cc} \angle{ANB} \equal{} \angle{AMC} \equal{} 90^\circ \\
\angle{NAB} \equal{} 90^\circ \minus{} \angle{B} \\
\angle{MAC} \equal{} 90^\circ \minus{} \angle{C} \\
\angle{BAE} \equal{} 90^\circ \minus{} \angle{C} \\
\angle{EAC} \equal{} 90^\circ \minus{} \angle{B} \end{array} \right\| \Rightarrow \angle{NAE} \equal{} \angle{EAM} \equal{} 180^\circ \minus{} \angle{B} \minus{} \angle{C} \equal{} \angle{A}$

Also

$ \left\|\begin{array}{cc} AN \equal{} AD \\
AM \equal{} AD \end{array} \right\| \Rightarrow AN \equal{} AM$

In $ \triangle{NAE}$ and $ \triangle{EAM}$ we have

$ \left\|\begin{array}{cc} AN \equal{} AM \\
AE \ \minus{} \ \ \mathrm{common \ side} \\
\angle{NAE} \equal{} \angle{EAM} \end{array} \right\| \Rightarrow \triangle{NAE} \equiv \triangle{EAM} \Rightarrow \angle{ANE} \equal{} \angle{AME}$.

But
$ \left\|\begin{array}{cc} \angle{BNE} \equal{} 90^\circ \minus{} \angle{ANE} \\
\angle{EMC} \equal{} 90^\circ \minus{} \angle{AME} \end{array} \right\| \Rightarrow \boxed{\angle{BNE} \equal{} \angle{EMC}}$[/color]

[hide="Image"][img]http://ipicture.ru/uploads/080612/Z9ZVzoeoN9.png[/img][/hide]


[quote="mdk"]show that cosx+cos(xsqrt(2)) is not periodic[/quote]

This problem is cool.
Learn [[LaTeX]].
Embedding websites like this is cool.

Anyway, click below for the solution:
[hide]Let $ f(x)=\cos x+\cos (x\sqrt{2})$
It is clear that at $ x=0$, $ f(x) = 2$. I will prove that it does not equal $ 2$ at any other point, hence the function cannot be periodic.

Assume, to the contrary, that it is possible.
For the function to equal $ 2$, we must have:
$ \cos x = 1$
$ \cos (x\sqrt{2}) = 1$,
because of the bounds on cosine.

These will be satisfied when:
$ x/\pi = 0,\pm 2, \pm 4, \cdots$
$ x/\pi = 0,\pm \sqrt{2}, \pm 2\sqrt{2}, \cdots$

Clearly, $ x/\pi$ must be rational and irrational at the same time, which is impossible.[/hide]

[b]Follow-up[/b] (quite simple, from my solution). Prove that
$ \cos x+\cos (xn)$

is periodic if and only if $ n$ is rational.

Here is a tougher question.
If $ f$ is [b]any[/b] periodic function, then is $ f(x)+f(xn)$ also periodic, where $ n$ is irrational? Rational?

[hide]
if n is rational you just get
$ f(x)+f(nx)=f\left(\frac{bx}{b}\right)+f\left(\frac{ax}{b}\right)$
where nb=a for integeres a and b
which is clearly periodic

also, the reverse is not necessarily true
a trivial counterexpample is f(x)=0
f(x)+f(nx) is periodic for all n
[/hide]


calculus
integration
calculus computations

We write the integrand $\sqrt(1-x^2)$/(x-a)  as  $(1-x^2)$/(x-a).$\sqrt(1-x^2)$

Now writing $x^2$ =$(x-a+a)^2$ =$(x-a)^2 +2a.(x-a) +a^2$ the integrand becomes

$(1-a^2)/(x-a).\sqrt(1-x^2)$. -x/$\sqrt(1-x^2)$. -a/$\sqrt(1-x^2)$
We have thus reduced the integrand to the sum of three standard forms and the remaining steps are routine

The first member of the sum is evaluated using the substitution x-a=1/y


Prove that $ \forall n\geq 3$ there exists $ \sigma\in S_{n}$ so that $ \sigma^{n}\equal{}\sigma, \sigma\neq e$.

In $ \mathfrak{S}_n$, there is always an element $ \sigma$ t.q.  $ \sigma^{n\minus{}1}\equal{}1$.
It is (123...n-1).

[quote]It is (123...n-1).[/quote]
What do you mean by this?
Maybe I used the wrong notation. By $ \sigma$ I meant a permutation of the set ${ M=\{1,2,...,n}\}.$

The permutation whose existence this problem is trying to prove must also be different from the identity permutation.

cnbjy used cycle notation which is the most common form for writing down permutations.

I wasn't aware of such notations.
Now I understand, thanks.


[quote="shobber"]AB is the diameter of circle with center $O$. $C$ is on the circle such that $\triangle{OAC}$ is equilateral. Another circle is consrtucted with diameter $OB$. Let the circle with center $O_1$ be tangent to the two circles and the triangle.
Prove: $O_1O \perp AB$.[/quote]

It already suffices if you require the triangle OAC to be isosceles with OC = AC (it needs not be equilateral). See http://www.mathlinks.ro/Forum/viewtopic.php?t=38681 , http://www.mathlinks.ro/Forum/viewtopic.php?t=20549 or http://www.mathlinks.ro/Forum/viewtopic.php?t=18266 .

  darij


I'm an 8th grader going into 9th grade, I was hoping on making USAMO next year. But i couldn't find any AIME problems anywhere. So can someone tell me what they're like? Are they proofs or maybe AMC-type questions? If they are proofs, I don't think I can make it, since I've just started practicing proofs, but otherwise, I think I might manage it.

You can find past AIME problems at [url]http://www.kalva.demon.co.uk/aime.html[/url], but they are paraphrased. I would recommend purchasing them from the MAA at [url]http://www.unl.edu/amc[/url]. They are very cheap! I think they only cost a dollar or two each.

thx for the help. I think I can handle it, its actually looks pretty easy.

I didn't make it to AIME this year since i didn't pay attention to instructions. I didn't kno blank answers got points, so i answered all of them. And I would have got like 120, mbe more, if i didn't go back and read the instructions. I started erasing my last few answers with like 30 seconds left(I kno i got atleast a few of the ones i erased right) I think my final score was 115 or somewhere around dat. I was so pissed off.


$\displaystyle\frac{x^4-y^4}{x^3-y^3+2} * \displaystyle\frac{x^2-y^2}{x^3+y^3} * \displaystyle\frac{x^6-y^6}{x^2+y^2}$

this can't be completely correct but I got...
[hide]
$\frac{(x-y)(x+y)(x^2+y^2) \dot (x-y)(x+y) \dot (x^3-y^3)(x^3+y^3)}{(x^3-y^3+2)\dot(x+y)(x^2-xy+y^2)\dot(x^2+y^2)}$
$\frac{(x+y)^2(x-y)^3(x^2-xy+y^2)(x^2+xy+x^2)}{x^3-y^3+2}$ With the domains "blah blah blah..." Is there anyway to simplify it further? [/hide]


[quote="E^(pi*i)=-1"]Compute:

$3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{\cdots}}}}}}$[/quote]

[hide]Let $3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{\cdots}}}}}}=S$

$S=3+\frac{1}{1+\frac1{S}}$ (since it repeats forever)

Then solving this equation for $S$ gives $\boxed{\frac{3+\sqrt{21}}{2}}$
[/hide]

[quote="surge"][hide]$x=3+\frac{1}{1+\frac{1}{x}}$

$x=\boxed{\pm\frac{\sqrt{21}+3}{2}?}$[/hide][/quote]

The sum would be positive.

[hide]$x=3+\frac{1}{1+\frac{1}{x}}$
$x=3+\frac{x}{x+1}$
$x^2+2=3x+3+x$
$x^2-3x-3=0$
$x=\frac{3+\sqrt{21}}{2}$, took positive root since cont. fraction is positive.[/hide]


geometry
rhombus
parallelogram
trigonometry
trig identities
Law of Cosines

Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that <APB=<BQC=120degree and an isosceles triangle RAC on the other side of AC such that <ARC=120degree. Show that PQR is an equilateral triangle.

Extend $ AP$ through $ P$ and $ CQ$ through $ Q$, and label the intersection of those two lines $ R'$.  $ \triangle$ $ ARC$ is congruent to $ \triangle$ $ ARC$, and thus we have created rhombus $ ARCR'$.

Let $ AP \equal{} x$, $ PR' \equal{} y$, and $ RC \equal{} z$. Note that $ AP \equal{} QR'$ and $ PR' \equal{} CQ$ (one way to see is looking at parallelogram $ BQR'P$). Also, $ AP \plus{} PR' \equal{} RC$.

A little angle chasing, $ \angle RAP$ = $ \angle RCQ$ $ \equal{} 60$, $ \angle PR'Q$ $ \equal{} 120$.

By law of cosines:

$ RP^2 \equal{} (x \plus{} y)^2 \plus{} x^2 \minus{} (x \plus{} y)x \equal{} x^2 \plus{} xy \plus{} y^2$.
$ RQ^2 \equal{} (x \plus{} y)^2 \plus{} y^2 \minus{} (x \plus{} y)y \equal{} x^2 \plus{} xy \plus{} y^2$.
$ PQ^2 \equal{} x^2 \plus{} y^2 \plus{} xy$.

Therefore, $ RP \equal{} RQ \equal{} PQ$ and $ \triangle$$ PQR$ is equilateral.


your inequality can be write as,
$ 8a_1^2\plus{}4a_2^2\plus{}2a_3^2\plus{}a_4^2\plus{}a_5^2\geq \frac{1}{16}(8a_1\plus{}4a_2\plus{}2a_3\plus{}a_4\plus{}a_5)^2$
wich is Cauchy-Schwarz inequaliti.


Show that $ (T_n)^{1/n} \to \frac{4}{\sqrt{(1\plus{}a)^{1\plus{}a}(1\minus{}a)^{1\minus{}a}}}$
as $ n \to \infty$, where $ 0<a<1$ and $ T_n\equal{}\sum_{k: |k\minus{}\frac{n}{2}|>\frac{an}{2}}\binom{n}{k}$.

[quote="The Wicker Man"]Show that $ (T_n)^{1/n} \to \frac {4}{\sqrt {(1 \plus{} a)^{1 \plus{} a}(1 \minus{} a)^{1 \minus{} a}}}$
as $ n \to \infty$, where $ 0 < a < 1$ and $ T_n \equal{} \sum_{k: |k \minus{} \frac {n}{2}| > \frac {an}{2}}\binom{n}{k}$.[/quote]

$ (T_n)^{1/n} \to \frac {2}{\sqrt {(1 \plus{} a)^{1 \plus{} a}(1 \minus{} a)^{1 \minus{} a}}}$

it's equal to integral like this
http://www.mathlinks.ro/viewtopic.php?t=121692 or something like this. :)


function
inequalities solved
inequalities


[quote="arkhammedos"]solve in the set of reel numbers
$\sin(2arcos(cotan(2arctanx)))=0$
 :?:[/quote]

[hide]This means that $2\arccos{(\cot{\2arctan{x}})}=0$, since if were equal to $\pi$ or $-\pi$ or anything else, it would not exist.

So this means that $\arccos{(\cot{2\arctan{x}})}=0$

So this means that $\cot{2\arctan{x}}=\frac{\pi}{2}$

So $2\arctan{x}=\cot^{-1}{\frac{\pi}{2}}$

Then we get that $\tan^{-1}{x}=\frac{\cot^{-1}{\frac{\pi}{2}}}{2}$

So $x=\tan{\frac{\cot^{-1}{\frac{\pi}{2}}}{2}}$

This can be shown to simplify to

$x=\frac{2}{\pi+\sqrt{\pi^{2}+4}}$.

But when we check back in, this doesnt' work.  So the answer is no solution

I think.[/hide]

[quote="JBL"][quote="mathgeniuse^ln(x)"]This means that $2\arccos{(\cot{\2arctan{x}})}=0$, since if were equal to $\pi$ or $-\pi$ or anything else, it would not exist.[/quote]

You lost me right there.[/quote]

Well, I know I messed up, but if the values of x that produce $\sin{x}=0$, then $x=k\pi$, where k is an integer.

Notice that 2arccos(x) has a maximum value of, dang it, I messed up.

I thought that the range of arccos is -1 to 1, but I was wrong.

So forget my solution.

Well, there was nothing wrong in principle with your approach (although I don't know that it's the best, either): you should still be able to continue through with the other possibilities, right?

[quote="JBL"][quote="mathgeniuse^ln(x)"]Yeah, but then there would be infinitely many possibilities I think that can't be expressed in a cohesive form.[/quote]

Perhaps suprisingly, this isn't true.

An easier method: [hide]Instead of peeling the functions off one at a time, consider peeling them off two at a time; in particular, you have $\sin(2\cdot \textrm{something})$, which you can expand; the result gives you something with $\sin \arccos$ and $\cos\arccos$.  These simplify.  Then do the same thing again on the inside.[/hide][/quote]
You mean double angle formulas, right?

[hide="whats the problem with removing a function per time?"]$\sin{(2 \arccos{(\cot{(2 \arctan{x})))}}}= 0$
$\arccos{(\cot{(2\arctan{x}))}}= \frac{n \pi}{2}$, $n \in \mathbb{Z}$
$\cot{(2\arctan{x})}\in \{0,1,-1\}$
$\text{arccot }0=\frac{\pi}{2}+n\pi$
$\text{arccot }1= \frac{\pi}{4}+n \pi$, $n \in \mathbb{Z}$
$\text{arccot}(-1)=-\frac{\pi}{4}+n\pi$, $n \in \mathbb{Z}$
so
$\arctan{x}= \pm \frac{\pi}{8}+\frac{n\pi}{2}\veebar \frac{\pi}{4}+\frac{n\pi}{2}$
$x \in \{1+\sqrt{2}, 1-\sqrt{2},-1+\sqrt{2},-1-\sqrt{2},1,-1\}$
[/hide]

Yes, I did mean the double angle formulas.  Aradesh, you're missing two roots.  (I think you probably lost them when you said, "$\textrm{arccot}0 \not\in \bf R$," which isn't true.)


Simplify: $ \frac{\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}}{\frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}}$ .Express your answer as a common fraction.

[quote="GameBot"]Simplify: $ \frac {\frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4}}{\frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5}}$ .Express your answer as a common fraction.[/quote]

$ \frac{\frac{13}{12}}{\frac{47}{60}}\equal{}\frac{65}{47}$.


does amc 12 b usually have questions on same topics as amc 12 a?
becaues on 12 a i noticed a lot of geometry questions, which i'm not s good at.

The distribution should be similar, but it probably won't be exactly the same.


<asy> tags on Wiki currently do not allow // for comments in the code. /* */ work fine though. Not that it is a big issue but some pictures are screwed because of that.

This was annoying the last time I was trying to debug a piece of code (running each line individually). I didn't realize /* */ worked, though, so it's not a big deal, but it's probably a pretty small issue, so it would be awesome if it was fixed.

I'm automatically transform // into /* */ it's a question of internal parsing. Unfortunately there's nothing I can do about that.


geometry
rectangle
ratio
perimeter
function
Pythagorean Theorem

A rectangle is inscribed in a circle with a radius 5 units long.  The ratio of the dimensions of the rectangle is 0.75.  How many units are in the perimeter of the rectangle?

[hide]It does not matter if the ratio is length/width or width/length, because we are working with a inscribed triangle. Since the ratio is 3/4, the dimensions are 3x and 4x.
 The rectangle touches the circle at each of it's vertices, by definition.

Since it is a rectangle, the center of the circle is in the middle of the triangle, by the last statement.

Therefore, the radius is the diagonal of a triangle with dimensions 3x/2 and 4x/2.

The Pythagorean Theorem gives the radius as a function of x to be $\sqrt{\frac{9x^2}{4} + 4x^2}\rightarrow \sqrt{\frac{25x^2}{4}}\rightarrow \frac{5}{2} x$. Therefore, $5=\frac{5x}{2}\rightarrow 10=5x\rightarrow x=2$ Therefore, the dimensions of the triangle are 6 and 8, so the perimeter is 14*2=28. QED[/hide]

Billy

[hide]It's a simple 6-8-10 triangle, making it too easy (once you realize that the diameter is from one vertice to another)[/hide]


algebra
polynomial
number theory theorems
number theory

Can you tell me what is a cyclotomic polynomial
By the way, can you give me the properties and relevant problems
Sincerely yours

see http://mathworld.wolfram.com/CyclotomicPolynomial.html

or

http://planetmath.org/encyclopedia/CyclotomicPolynomial.html

Searching for "cyclotomic" on this side should also give you something.


Read the AMC's teacher's manuals:

[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-AMC1012A-TM/2007AMC1012TM.pdf[/url]


Prove that direction trnh system has just two the solutions $ x > 0 , y > 0$
 $ e^x \equal{} 2007 \minus{} \frac {y}{\sqrt {y^2 \minus{} 1}}$
 $ e^y \equal{} 2007 \minus{} \frac {x}{\sqrt {x^2 \minus{} 1}}$
Sorry , my English is bad

if $ (x,y)$ is a solution, then $ |x|,|y|>1$. Put $ t\equal{}x\plus{}y, f(z)\equal{}e^{t\minus{}z}\plus{}\frac{z}{\sqrt{z^{2}\minus{}1}}\minus{}2007$
x satisfies: $ f(x)\equal{}f(t\minus{}x)\equal{}0$
$ f'(z)\equal{}\minus{}(e^{t\minus{}z}\plus{}(\frac{1}{(z^{2}\minus{}1)})^{3/2})<0 \implies f$ is strictly decreasing, therefore $ x\equal{}t\minus{}x$, or: $ x\equal{}y$. hence, all the solutions are in the form $ (x,x)$. if we define the following function: $ g(z)\equal{}e^{z}\plus{}\frac{z}{\sqrt{z^{2}\minus{}1}}\minus{}2007$, we need to find all x's such that $ g(x)\equal{}0$. g tends to infinity as x tends to infinity, and g(2) is negative, so we must have at least one solution. but I think we have another one, because when x tends to 1 (from "above"), g tends to infinity. for $ x<\minus{}1<0: g(x)<\frac{1}{e}\minus{}2007<0$.


The Italian flag and Hungarian flag both consist of a green stripe, a white stripe, and a red stripe.  The only difference is that the stripes are vertical for Italy and horizontal for Hungary.

The flags of Russia and the Netherlands have three horizontal stripes of the same color, the only difference being the order of the stripes.

The flags of Indonesia and some other country (Poland, I think) are just inversions of one design, a white stripe and a red stripe. That makes it confusing if a ship is to fly its flag upside down as a distress signal.

ireland and ivory coast are the same, except inverted colors. mexico is also similar, but with the design in the middle. italy too. liberia, malaysia, and the U.S. have very similar designs. libya still has the coolest, though (blank green. try hanging THAT upside down, tokenadult!). peru is kind of like canada, except the design in the middle is different...

actually, it seems like red, white, and green are the most popular colors for flags. can anyone confirm this? is there any historical significance?

How about this, Japan and Bangladish ;)

Both flags are very simple, they just have a red circle on the center. The difference is, Japan has a white background while Bangladish has a green background. I know the color is obviously different, but I think this is still interesting :lol:

The national flags, emblems and anthems, and their meanings.
http://www.lidicity.com/world/index.html

I think one of the most common flags is the one with three horizontal stripes in the colors red white and blue (in this order).
The Dutch flag has a dark shade of blue.
The Luxemburg flag a light shade of blue
The flag of Serbia and Montenegro is upside down (as compared to the Dutch flag).
The Paraguayan flag has a weapon in the middle.
Croatia has another weapon in the middle.

And finally France have vertical stripes in these colors (and order).

Supposedly the five colors of the Olympic rings in the flag of the international Olympics Games were chosen because every national flag in the world uses at least one of those colors, or the white that is the background color of the Olympic flag. Does anyone care to check that assertion exhaustively?

what i found very interesting is that many of the middle eastern countries' flags consist of some combination of green, black, white and red tripes or triangles.
eg egypt, iraq, yemen are almost the same :)

lets see. the five colors are... blue, red, green, black, and... yellow? right? so blue, red, green, black, yellow, white.

botswana? i think it may have some black in the middle, though.
all the ones i looked at at AOPS had those colors, but there were a couple of flags not pictured, so i am not sure about those.

Iraq, Syria, Yemen and Egypt.

Jordan, Palestine, UAE, Kuwait and Sudan.

Well, like [b]Palytoxin[/b], I find very similar flags of the PRC and the USSR:
[img]http://www.olympic.org/common/images/organisation/noc/flags/CHN_flag.gif[/img]
[img]http://www.olympic.org/common/images/organisation/noc/flags/URS_flag.gif[/img]
And also Argentina, Honduras, Nicaragua, El Salvador (all in America, BTW!):
[img]http://www.olympic.org/common/images/organisation/noc/flags/ARG_flag.gif[/img][[img]http://www.mathlinks.ro/Forum/images/flags/honduras.gif[/img]
[img]http://www.olympic.org/common/images/organisation/noc/flags/NCA_flag.gif[/img]
[img]http://www.olympic.org/common/images/organisation/noc/flags/ESA_flag.gif[/img]
Burma and Samoa  :D :
[img]http://www.olympic.org/common/images/organisation/noc/flags/MYA_flag.gif[/img]
[img]http://www.olympic.org/common/images/organisation/noc/flags/SAM_flag.gif[/img]
And what is probably the most different from other flags is the one of Nepal :D :
[img]http://www.olympic.org/common/images/organisation/noc/flags/NEP_flag.gif[/img]

[quote="nonIMO"]And also Argentina, Honduras, Nicaragua, El Salvador (all in America, BTW!)[/quote]
Not coincident, Guatemala, Honduras, Nicaragua, El Salvador  and Costa Rica are the fragments of United States (Provinces) of Central America.

[quote="Palytoxin"]Romania & Republic of Moldova(this is one strange comparison i just found out that made me make this new topic!) [/quote]

Moldova was actually a region of Romania previous to WWII. (There actually still is a region called Moldova, but it used to include the entire nation.) So it isn't a huge surprise that the flags are similar.

Niger and India are really similar too.

Ireland and Italy are really weird though.  They're identical, and they're right next to each other when you put all the countries in alphabetical order.

just look at QATAR and BAHRAIN (middle east) flags!

are there any flags with no difference at all that's actually noticeable?

some european flags ( like norway and iceland, ithink..) are the same with the colors switched

The Union Jack (more properly Union Flag), the national flag of the UK, is also contained within the flags of certain other countries: Australiaand New Zealand spring to mind.  Are there any other countries whose flags are contained within those of an independent country?

[quote="h78reg"]i think romania and chad have almost identical flags as well.[/quote]

And it has caused [url=http://news.bbc.co.uk/1/hi/world/europe/3626821.stm]Flap in Romania[/url][quote]Romanian President Ion Iliescu has said his country will not give up its national flag despite the fact that it appears, to most observers at least, identical to that used by the central African state of Chad.

His comments follow unconfirmed reports in the Romanian media that Chad has called on the UN to examine the issue.

The flags of the two countries converged as far back as 1989 when Romania's dictator, Nicolae Ceasescu, was overthrown and the incoming government dropped the communist-era insignia from the flag's centre.

While both flags have blue, yellow and red stripes, some experts say the blue colour in Chad's flag is darker than that used in Romania's version.....[/quote]

Sorry if it was already posted, but [url=http://www.crwflags.com/fotw/flags/xf-like.html]Some other nearly identical flags[/url]

[quote]While both flags have blue, yellow and red stripes, some experts say the blue colour in Chad's flag is darker than that used in Romania's version.....[/quote]

i think that's true, but, nevertheless, the flags are identical in almost every way.

At IMO 2004 I was asked by several people whether I was Romanian, since the Belgian flag is black-yellow-red and the Romanian one is dark blue-yellow-red. It wasn't easy to see whether the flags on our T-shirts were black or blue. :)

[quote="master cherundolo"]Can anyone here find the distinction between Indonesia's flag and Monaco's flag? :D
Well, I think it's the most undifferent flag in the Earth.[/quote]
You do realize you are bumping this all the way up, it's been like two years. :D
And yes, that is a good example.
But Monaco is like three rocks and two boats, will Indonesia has more than 200 million people.  So yeah, what I'm saying is...

The flag of the (unrecognized) republic of Nagorno Karabach
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Flag_of_Nagorno-Karabakh.svg/125px-Flag_of_Nagorno-Karabakh.svg.png[/img]
and Armenia itself:
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/2/2f/Flag_of_Armenia.svg/125px-Flag_of_Armenia.svg.png[/img]
are also similar, no coincidence of course...

[quote="Palytoxin"]i see many countries with flags that look like almost the same except some minor differences. lets find more of them
NOTE: please no politics, if you want to discuss politics, please leave and make a new topic, my topic is ONLY about flags

i found:
China & USSR(now gone of course)
Romania & Republic of Moldova(this is one strange comparison i just found out that made me make this new topic!)

there are more but i want others to participate :D[/quote]

The similarity between the Romanian flag and the the Moldovan flag is of course quite normal, Moldovan is in fact just the same language as Romanian,and Moldova has been part of Romania. That Moldovan flag is also fairly new, created in 1991 as far as I know.


[quote="arthur eddington"]Any researcher ought to know by heart the value of
'CALCULUS' in mathematics.For anything that changes,that grows,developed into something,calculus is a must.Can anyone
show me where can i find a "FREE" e-print book on 'elementary
calculus?

                                                          Respectfully,
                                                        A.Eddington[/quote]
Hi arthur eddington, and welcome to Aops. I know of an e-book that explains in depth the very basics of calculus. Its called 'The Calculus Bible'. Hope it helps. :)

Check mininova.org and you can find a variety of torrents that have collections of college math books which include all types of calculus.  Most of them are in PDF format.

where I can get math olympiad problem & solution in PDF for high school ? thanks

Very good calculus book for beginners is Gilbert Strang's "Calculus":
[url]http://ocw.mit.edu/ans7870/resources/Strang/strangtext.htm[/url]