Could someone help me with the following question:

2. The concentration, C, in nano-grams per milliliter, of a drug in the blood as a function of the time, t, in
hours since the drug was administered is given by
C = 15(t)( (e)^−0.2t).

The area under the concentration curve is a measure of the overall effect of the drug on the body, called
the bioavailability. Find the bioavailability of the drug between t = 0 and t = 3.


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Suppose there are two measurable real functions $ f$ and $ g$ such that $ \int f$ is finite but $ \int g\equal{}\infty$(or $ \minus{}\infty$), can we have $ \int (f\plus{}g)\equal{}\int f\plus{}\int g$?

If $ f\equiv 0$ then the result is obviously true right? What's your claim? True or false? Prove it.

Oh I'm sorry, I must have misunderstood your intentions by your use of the word "can".

I will go and write up a proof.

Does anyone have any ideas? I am still not getting any solution.

By definition, $ \int g \equal{} \infty$ means $ \int g^\plus{} \equal{} \infty$,  $ \int g^\minus{} < \infty$. We want to show $ \int(f\plus{}g) \equal{} \int f \plus{} \int g \equal{} \infty$, i.e.,  $ \int(f\plus{}g)^\plus{} \equal{} \infty$, $ \int(f\plus{}g)^\minus{} < \infty$. This will follow from $ (f\plus{}g)^\plus{} \geq g^\plus{} \minus{} |f|$ and $ (f\plus{}g)^\minus{} \leq f^\minus{} \plus{} g^\minus{}$, which you can prove by considering cases.

[quote="limsup"]That's really good  :)
But what is the motivation for thinking of the following identity?
[quote="WWW"]$ (f \plus{} g)^ \plus{} \geq g^ \plus{} \minus{} |f|$ and $ (f \plus{} g)^ \minus{} \leq f^ \minus{} \plus{} g^ \minus{}$[/quote][/quote]

The motivation is to find something - anything - that will make the problem simpler.
For the first inequality I had to play around a little; the second one should be clear.


I know this is a very well known problem, but I don't know if it is open or not... I'm not only interested in solutions, but also in coments and information about this:
  Decide if there are infinitely many primes $p$ such that $n$ is a primitive root $\mod{p}$.

What I know is:

1. If $n$ is a perfect square it's wrong (not hard to show)
2. It's open for that $n$ not being a perfect square (or more accuratly: it has been proved using a generalisation of the Riemann conjecture)
3. for that $n$ it's called 'Artins conjecture' (Artin also made a conjecture about the density of the numbers fulfilling this property)

PS: you can look here: http://mathworld.wolfram.com/ArtinsConstant.html

yes thanks, i read that yesterday... and i saw that it was proved assuming the Extended Riemann conjecture (something about the least nonquadratic residue $\mod{p}$).. but is it still open for some particular cases of $n$?? for example, are there  only finitely many primes for wich 10 is a primitive root?


The equations $2x+7 = 3$ and $bx-10 = -2$ have the same solution. What is the value of $b$?

A. -8
B. -4
C. -2
D. 4
E. 8

that is because I am not that good at it but I got better recently

[quote="math92"]whoa, you guys have almost the same exact solution, but one doesn't have $\LaTeX$[/quote]

That makes me suspicious.  :ninja:  It isn't right to write b(-2) when you can write it as -2b, but in latex, the /cdot code allows you to write $b\cdot(-2)$.

Thus the difference in the last few lines.  :o

[hide="Click here to see steps"]2x+7=3
Therefore, subtract 7 from both sides. 2x=-4. Then divide 2 on both sides. x=-2.

So...

bx-10=-2
b(-2)-10=-2
b(-2)-10+10=-2+10
(-2)b=8
b(-2)/(-2)=8/(-2)

Therefore...[/hide]
[hide="Look here to see answer"]B[/hide]

[hide]
Solve the first equation, and it gives you that x=-2.
Plugging in x as -2, you get
$-2b-10+10=8$
$b=-4$
So, teh answer is $B$[/hide]


I've been having many problems with the registration online.  First my password was wrong but that got fixed, however yesterday when I was filling out the information about my parents there was an error trying to add phone numbers and today I can't even get to http://www.mathcamp.org.

Try again. [url]http://www.mathcamp.org[/url] It may have been down earlier.

The site is up again however I'm still having trouble with the registration.

Argh, is there any possibility of extending the deadline due to the site dying several times?

Works for me. It seems to me that if you have everything in files, you wont need the website to be up for long. Just keep checking it, and add what you can while it is up.

If the site is flakey, you dont want to be typing in your essay or background directly at the site box - just upload a file.

Laura


OK simple question for you ppl.
whats a limit?
idk.

[quote="CatalystOfNostalgia"]Intuitively speaking, the limit of $ f(x)$ as $ x$ goes to some constant is the value at which $ f(x)$ 'approaches,' as it moves toward $ x$. The limit only exists if this value is the same coming from both sides. If $ f(x)$ is continuous (again, more handwaving, that is, there are no 'breaks' in its graph), then the limit as $ x$ approaches $ c$ of $ f(x)$ is simply $ f(c)$. However, this is pretty much part of the (rigorous) definition of a continuous function.

However, this definition is useless when rigor is needed (I would assume in analysis), but I guess would suffice in cookbook calculus.

An explanation of the epsilon-delta definition, which is rigorous, can be found here: [url]http://www.scottsarra.org/applets/calculus/EpsilonDelta.html[/url].[/quote]
Yes. everything everyone else said was true. But this belongs to calculus, so it shouldn't have been here.

Did you read the pinned thread that says "Please do not post calculus problems in this forum!"?

not necessarily. there's a chapter on limits in aops vol 2, and we discuss problems from that book all the time, don't we  :wink:

Calculus-related issues, regardless of their existence in AoPS Volume 2, do not belong in this forum for two reasons:

1. This is a competition-math concentrated part where more problems regarding competition should be discussed. School problems can be discussed to certain extent that they do NOT include calculus.

2. There IS a calculus board in this forum so why bother to ignore it? That is unnecessary.

Now, as to the original poster, please note that in future, check out the AoPS Wikipedia or even regular Wikipedia or other math book to look for definition of limit because that is fairly common concept and if one actually tried to look over it, it's very easy to be found.

Oh, I'm moving the topic AND locking it.


I've made categories for all of the different problem types and levels.  There are 12 categories based on difficulty level (introductory, intermediate or olympiad) and problem type (algebra, number theory, combinatorics or geometry).  To categorize a problem add the appropriate tag from the following at the end of that page:

[[Category:Introductory Algebra Problems]]
[[Category:Introductory Combinatorics Problems]]
[[Category:Introductory Geometry Problems]]
[[Category:Introductory Number Theory Problems]]
[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Combinatorics Problems]]
[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Number Theory Problems]]
[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Combinatorics Problems]]
[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Number Theory Problems]]

Tell me what other classifications you would like.  Maybe stuff like AIME Geometry, AIME NT, etc?

The advantage of doing this is that you can go to the Category:Number theory problems page and find a large list of problems to browse through, for example.

Fantastic idea!  These categories should be linked to from the main topic pages.

Also, I have to wonder if it wouldn't be best to make three levels of categorization in regards to those problems.  Is it enough to have a list of number theory problems if you don't know which are at your own level of challenge/curriculum?

Thinking about it some more I decided we don't really need the templates.  We can just add the pages to the category.  I also went and added categories for Algebra, NT, Combinatorics, and Geometry for Introductory, Intermediate, and Olympiad levels (thanks for the suggestion Mr. Crawford).


Find all quadruples $(x,y,z,p)$ of positive integers such that $x\geq 3, y,z\geq 2, p$ is prime, and $x^y = p^z+1$

http://mathworld.wolfram.com/CatalansDiophantineProblem.html
this is the problem.
it was a hard problem but it was proved by Preda Mihailescu in 2002
see http://en.wikipedia.org/wiki/Catalan_conjecture

And he probably wants an elementary proof ;)

Ok, let do this one, a easy case of Catalan prob.
  $x^y-1=p^z$ then $x-1=p^{a},x^{y-1}+...+x+1=p^{b}$. Then we have $p|y$ and easily we dinf out that $y=p^{c}$ for some $c$. Now we use $p^a||x-1$ then $p^{a+b}||x^{p^a}-1$. to have :$b=c , x^{p^b}-1=p^{b+a}$ or $x^{m}-1=(x-1)m$ which can be done by calculus.

I think i have seen s.th similar : $x^y-1=z^t$ where $x\equiv\ 1 (mod z)$ in this forum. The author might be Blang

[quote="nguyentrang"][quote="t_c_lhp"] $x^{m}-1=(x-1)m$ which can be done by calculus.[/quote]

How?[/quote]

đạo hàm


probability
integration
geometry
3D geometry
sphere
calculus
trigonometry

A piece is separated at random from each of three identical rods. Find the probability such that an acute triangle can be formed from the three pieces?

given two fixed legs $ a \ge b$, the acute triangles are found between two right triangles

$ c^2\equal{}a^2\plus{}b^2, a^2\equal{}c^2\plus{}b^2$
$ c\equal{}\sqrt{a^2\plus{}b^2}, \sqrt{a^2\minus{}b^2}$

rod: $ [0,1]$

$ P((a,b,c) \in A: a,b)\equal{}P(c \in (\sqrt{a^2\minus{}b^2},\sqrt{a^2\plus{}b^2}))\equal{}\sqrt{a^2\plus{}b^2}\minus{}\sqrt{a^2\minus{}b^2}$

now we integrate for a fixed smaller piece b:

$ P((a,b,c)\in A: b)\equal{}\frac{1}{1\minus{}b}\int_b^1 \sqrt{a^2\plus{}b^2}\minus{}\sqrt{a^2\minus{}b^2} da$

this gets messy quick though, so im not certain this is the smart way to do this

Hi, here is  my view :

Let $ [0, 1]$ be one of the rods. Thus our sample space is $ S = [0, 1] \times [0, 1] \times [0, 1]$. Points $ (x,y,z) \in A$ satisfying $ x^2 + y^2 < z^2 \leq 1$ forms our feasible region which is an eighth sphere.The desired probability :

 ${ \frac {\text{Measure of feasible region}}{\text{Measure of sample space}} = \frac {\text{Volume of eighth sphere}}{\text{Volume of} \ S}} = \frac {\pi}{6}$

I haven't received the correct answer yet. :(

ugly integration got me

$ \frac{1}{3\sqrt 2}\plus{} \frac{\ln|1\plus{}\sqrt 2|}{3}\minus{}\frac{\pi}{6}$

[hide="Solution"]
We have $ 0 \le x,y,z \le 1$ and by cosine rule the three following inequalities,

$ x^2 \plus{} y^2 \ge z^2$

$ x^2 \plus{} z^2 \ge y^2$

$ y^2 \plus{} z^2 \ge x^2$

Which are cones. A simple complementary counting arguments reveals the volume of the region satisfying the above restraints is $ 1 \minus{} \frac{\pi}{4} \approx 0.2146$
[/hide]

That's correct answer.  :) Could you give us your solution in detail?

I think I found the probability being obtuse triangle  :lol:  Therefore I think that the answer is  $ 1 \minus{} \frac{\pi}{6}$. Kunny what is wrong in my solution?

Regrettably your answer is incorrect.

See JavaMan's solution in #6.

[hide="Solution more detail"]
Assume the sticks are of length one, and call the 3 values they take (x,y,z). Then we can interpret the probability as a volume in $ [0,1]^3$.

We have $ 0 \le x,y,z \le 1$,

The cosine of the angles must all be positive.

$ z^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy \cos A \implies 2xy \cos A \equal{} x^2 \plus{} y^2 \minus{} z^2 \ge 0$

and doing this cyclic-ly gives the following inequalities which must be satisfied for the triangle to be acute.

$ x^2 \plus{} y^2 \ge z^2$

$ x^2 \plus{} z^2 \ge y^2$

$ y^2 \plus{} z^2 \ge x^2$

The equation $ z^2 \equal{} x^2 \plus{} y^2$ gives a 1/4th of a cone of height 1 and radius 1. And we're looking for the volume region in $ [0,1]^3$ that is outside each of these three 1/4th cones.

Since the cones are disjoint, you can subtract the volume of the three 1/4th cones from the total volume, which is 1.

$ 1 \minus{} 3 \frac{1}{4} \left( \frac{1}{3} \pi r^2 h \right) \equal{} 1 \minus{} \frac {\pi}{4} \approx 0.2146$
[/hide]


Mr. Sigur will be holding an AIME-a-thon where ARMLers will help others and each other on old AIME problems. Personally I feel it's important if you want to be on the ARML team because the ARML vets will get to knwo you and how well you work together with others. But on the other hand, it's not just for ARMLers, it also helps you score better on the AIME. Before the AIME-a-thon I was scoring 3s on practice tests; after I scored 5s on practice tests; on the real test last year I scored a 7, so it definitely helps.

The AIME-a-thon will be held at the Paideia School (1509 Ponce de Leon Ave, Atlanta, GA) on Sunday March 11 Noon-5PM. All are invited. See you there!

I'm going to see if I can make it. Sounds like fun.

I went last year, and it was helpful, so I'll try to go again this year.


vector
function
trigonometry
integration
complex numbers
inequalities unsolved
inequalities

this was posted in our lovely NYC forum, by user Ravi B.  i thought i'd take it upon myself to post it here too so it could reach a larger audience.

Let $ z_1, \ldots, z_n $ be complex numbers. Prove that there is a set $ S \subseteq \{ 1, \ldots, n \} $ such that

 $ \big|\sum_{j \in S} z_j\big| \geq \frac{1}{\pi} \sum_{j=1}^n |z_j| \, $.

Instead of complex numbers, I'll use vectors v_1, ... , v_n.

We'll find the average length of the projection of a vector v onto a given ray r. If there is an angle theta between v and r, then define the projection function $p_r(v)=|v|\cos\theta$ if $-90^\circ\le\theta\le90^\circ$ and $p_r(v)=0$ otherwise. Then the average length of the projection of $v$ onto the ray r as r varies around a full circle is $\frac{1}{2\pi}\int_{0}^{2\pi}p_r(v)\,d\theta=\frac{|v|}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\theta\,d\theta=\frac{|v|}{\pi}$. Thus, the average length of the sum of the projections of all the vectors onto a ray is $\frac{1}{\pi}\sum_{j=1}^n|v_j|$. Since there must be a ray $r$ so that the length of the projection onto $r$ is at least as great as the average value of the projection, choose $S=\{i\mid p_r(v_i)>0\}$ and we're done.

Yes, nice, zabelman.  It's one of my favorite problems.  I included it in a list of problems using the averaging method [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23507]here[/url].

One further question that I don't know the answer to.  In general, we can't replace the constant $\pi$ by a smaller constant, but for any fixed value of $n$, we can improve the bound a little.  What is the best possible bound in terms of $n$?  For example, perhaps the worst case is when the $n$ vectors are uniformly spaced around the unit circle.


3. The sum of 5 consecutive even integers is 4 less than the sum of the first 8
consecutive odd counting numbers. What is the smallest of the even integers?

$ (A) 6 \qquad (B) 8 \qquad (C) 10 \qquad (D) 12 \qquad (E) 14$

This one is easy if you see the trick
[hide]The first 8 odd counting numbers are 1,3,5,7,9,11,13,15. They sum to 64. If the sum of the five consecutive even integers is 4 less, then the five consecutive even integers sum to 60. Their average must be 12, so the first one would be 8. This checks (8+10+12+14+16=60). [b]The smallest is 8, [size=150]B[/size][/b][/hide]


2. Jim and John plan a dinner party.  The food consists of eight dishes of equal value, five purchased by Jim, and the other three purchased by John.  Jan decides to join them and agrees to pay an equal third of the cost.  The 8 dollars she pays is then split between Jim and John so that everyone contributes the same amount.  How much of Jan's 8 dollars go to Jim?  Express this answer as $a$ (you'll find out later)

23. Find the sum of the following fractions:
$1/2+1/6+1/12+1/20+...+1/9900$
Express your answer as a decimal and refer to it as $b$

26. Alice leaves Denver, driving at a constant speed.  After a while, she passes a mile marker displaying a 2-digit number.  1 hour later, she passes another mile marker with the same two digits in reverse order.  In another hour, she passes a third marker with the same two digits separated by a zero.  What is the rate of Alice's car in miles per hour?  Refer to this answer as $c$

So, based on the information above, evaluate this expression:

$[a+(b*100+1)]/c$
Express your answer as either an improper fraction or as a decimal to the nearest hundreth

[hide]2) Jim pays $\$15,$ John pays $\$9$. So $\$7$ go to Jim.

23) The pattern tells me that the answer is $0.99$.

26), $16, 61, 106$    $45$ mph

Ans. $a+\frac{100b+1}{c}= a+\frac{99+1}{45}= 7+\frac{20}{9}= \frac{83}{9}$

Is the answer $\frac{83}{9}$?
[/hide]
I have mc state tomorrow...

I'm nervous...

actually your answer was right, but i forgot to add the brackets.  don't worry, if you could figure out that problem you'll have no problem at states.  just get a good night sleep and RELAX.  just take it step by step.  take it from me.  im lazy, and i got into nats

#23:
we get
        [hide]$\frac{1}{1\times 2}+\frac{1}{1\times 3}+. . . .+\frac{1}{99\times 100}$

then we notice the pattern 1/2 + 1/6 = 2/3
1/2 + 1/6 + 1/12 = 3/4

and it goes on like that. then you use the telescoping method and you are left with 99/100.[/hide]

yeah, thats 23s answer, but can u answer the whole question above?

you can use partial fractions and you are left with 1-(1/100)

[hide]23. We can rewrite this problem as

$\frac{1}{1*2}+\frac{1}{2*3}+.......$

So what do we do?  We start adding and look for a pattern.  It turns out that we can generalize that for any n,

$\frac{1}{1*2}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$

Since 9900=99X100, our answer is $\frac{99}{100}$  How do we prove the statement above??[/hide]

[hide="Problem 23"]Notice something cool, and rewrite the sum:

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}$
$(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})...+(\frac{1}{99}-\frac{1}{100})$

Now almost everything will cancel out, except for the first and last term, leaving:

$1-\frac{1}{100}= \boxed{\frac{99}{100}}$[/hide]

6. [hide]Let the cost of one dish =X.  So Jim is paying 5X, and John is paying 3X.  When Jan comes in, she is paying 8X/3=8. X=3 dollars.  So $\$$15 for jim, and $\$$9 For john.  So 1 goes to Jim a=1[/hide]
23. [hide]So by telescope, 1/2+1/6+...+1/9900= 1/(1*2)+1/(2*3)+...
Thus the answer is 99/100. B= 99/100[/hide]
26. [hide]If we set Alice's speed to be c, and we let the first mile marker she passes be in the form 10X+Y.  We can see that after C miles, she will pass 10Y+X.  If we add another C, let us assume that we will be on 100X+Y.  So 10Y+X+C=100X+Y. And 10X+Y+C=10Y+X.  From the second, we can see that C=9(Y-X).  From the first C=9(11X-Y).  From the first one, it is obvious to see that X must be 1.  so C=99-9Y and C=9Y-9.  Thus 18Y=108 => Y=6.  So the first marker is at 16, the second is at 61, and the third is at 106.  C=45.[/hide]

FINAL: [a+(100b+1)]/c= 1+100 /45= 101/45


For how many positive integers $ n$, does

$ 1 \plus{} 2 \plus{} 3 \plus{}... n$

evenly divide $ 6n$?

[hide="not sure"] The LHS can be written as $ \frac{n(n\plus{}1)}{2}$, which will be a multiple of 6n if $ \frac{n\plus{}1}{2}$ is a multiple of 6. This is true if $ n\equiv\minus{}1 \mod 12$.

QED?[/hide]

Being a multiple of a number is definitely not the same as dividing that number...

Wait...are there more solutions? Cause all n in satisfying the condition I gave work, since the 1+2+...+n could be written as n(6k) for some positive integer k.

Here's another proof though:

[hide]
$ \frac {n(n \plus{} 1)}{2} \equal{} 6nk$
$ n^2 \plus{} n \equal{} 12nk$
$ n^2 \plus{} n(1 \minus{} 12k) \equal{} 0$
$ 1 \minus{} 12k \equal{} \minus{} n$, since n is a positive integer
$ n \equal{} 12k \minus{} 1$, which is obviously the same as $ n\equiv\minus{}1 \mod 12$

QED[/hide]

Tried to say this before, but...
If $ x|y$, or $ x$ evenly divides $ y$, that means $ y$ is a multiple of $ x$. Or $ kx\equal{}y$.

[hide="half a solution"]$ \frac{kn(n\plus{}1)}{2} \equal{} 6n\\
k(n\plus{}1)\equal{}12$[/hide]

But $ n$ is a positive integer, so
[hide="n="]$ 1,2,3,5,11 \Rightarrow 5$ solutions[/hide]


[img]http://img.photobucket.com/albums/v259/drspyder/d00d.gif[/img]

Mr. D00d goes down the slides of D00m. He can only go down on the slides of D00m. How many paths are there from the top to the bottom?

Umm, there's only $3$ ways, not much of a problem...

[quote="seungsoo"]The answer is obviously 3, and the drawing needs work as well.[/quote]

cmon.... give the guy some credit. I couldn't draw half as good as that (yeah go ahead laugh at me :blush: )

Here is a variation, drawn by, you guessed it, ME!

Same problem: find the number of paths between the red dots, travelling downwards and only turning at the black dots.

Counting Backwards...
[hide="for orig"]3[/hide]
[hide="for new"]17[/hide]
(hope I counted that right...)

P.S.: I think they could all use *some* help (the pictures)

[hide="treethings"]
3
[/hide]
[hide="nosoupforyous"]17[/hide]

[hide="omg this is getting ridiculous"]I get $ 136$ [/hide]

NO. This is ridiculous:

Suppose there is a regular dodecahedron symmetrical along the y-axis, no sides of which are parallel to the x-axis. If all the sides and diagonals are drawn, how many downward paths are there turning only at the vertices?

Your right how many sides in a dodecaherdron?

Actually, I think there is a way to solve my problem. I think the answer is $1+2+6+18+54+162=\boxed{243}$. I think this can be generalized to a formula for any $n$-gon (where $n$ is even): $3^{\frac{n}{2}-1}$

[quote="Treething"]Mr. d00d needs your help =o .

[img]http://img.photobucket.com/albums/v259/drspyder/d00dz0rz.gif[/img][/quote]
[hide]136[/hide]

Continuing my previouis posts, there are actually three cases of $n$-gons to consider:

1) a regular $n$-gon symmetrical along the y-axis, no sides of which are parallel to the x-axis

2) a regular $n$-gon symmetrical along the y-axis, two sides of which are parallel to the x-axis

3) a regular $n$-gon not symmetrical along the y-axis

Examples of squared satisfying criteria 1, 2, and 3 are shown below.

The number of paths drawn from vertex to vertex, traveling downward from top to bottom, of an $n$-gon can be found as follows:

1) $3^{\frac{n}{2}-1}$

2) $4\cdot 3^{\frac{n}{2}-2}$

3) $2^{\frac{n}{2}}$


algebra
polynomial
number theory
least common multiple
logarithms
floor function
modular arithmetic

Let $u$ be a sequence of integers such that $u_m-u_n$ is divisible by $m-n$ and there exist a polinomial $P$, such that $P(n)>|u_n|$. Prove there exist a polinomial $Q$, such that $Q(n)=u_n$ for all $n$.

It has been shown as an exercice in the A.M.M. that the condition $|U_n| < P(n)$ can be replaced by $\limsup |U_n|^{\frac 1 n} < e - 1$ but not by $\limsup |U_n|^{\frac 1 n } \leq e$.

Pierre.

It's some obvious You see that :
$ a_m \equiv  a_0 (\mbox {mod m})$
$ a_m \equiv  a_1 (\mbox {mod m-1})$
$...$
$ a_m \equiv  a_n (\mbox {mod m-n})$
So if we have $a_1,..a_{m-1}$ then $a_m$ is calculated $\mbox {mod lcm }(m,m-1,...,1) \geq 2^{n-1}$
From these easily with lagrange theorem we can find a polynomial with rational coefficients that $a_i=P(i)$ for $1 \leq i \leq n$ And easily you can prove that $p(k)=k$ $\forall k>n$
(Notice there exist $s \in \mathbb{N}$ that  $sP(x) \in \mathbb{Z} [x]$

Sorry. I can't understand you solution.
It is really not that hard problem.

Ok. I will write my own. Though, I guess, Omid meant almost the same.

Suppose $\deg P=n$. Using $a_1$, ..., $a_{n+1}$ we can construct polynomial $R(x)\in\mathbb{Q}[x]$ of degree $n$ s.t. $a_i=R(i)$ for $i=1,...,n$. Let $M$ be such natural number that $MR(x)\in \mathbb{Z}[x]$. We can consider a new sequence $\{Ma_k\}$ (we will assume $a_k\to Ma_k$).

We have $m-i\mid a_m-a_i$ for all $i=1,...,n+1$. It follows that $a_m=MR(m)+k\cdot [m-1,m-2,...,m-(n+1)]$. But \[ [m-1,m-2,...,m-(n+1)]\geq \frac{(m-1)...(m-(n+1)}{\prod_{i<j\leq n+1} (m-i,m-j)}\geq \] \[\geq\frac{(m-1)...(m-(n+1)}{\prod_{i<j\leq n+1}(j-i)}>C\cdot m^{n+1}.\]

Therefore $a_m$ must coincide with $MR(m)$ for all sufficietly large $m$.

Pierre, when you have some time and will, could you please post the proof of the problem from AMM? Thanks.

[quote="pbornsztein"]It has been shown as an exercice in the A.M.M. that the condition $|U_n| < P(n)$ can be replaced by $\limsup |U_n|^{\frac 1 n} < e - 1$ but not by $\limsup |U_n|^{\frac 1 n } \leq e$.[/quote]
[quote="harazi"]Pierre, when you have some time and will, could you please post the proof of the problem from AMM?[/quote]
Here is the proof.

Firstly we need to establish few lemmas:

[i]Lemma 1[/i]:

$\ln L_n\sim n$ ($n\to\infty$), where $L_n=LCM(1,\dots,n)$.

[i]Proof[/i]:

$p$ denotes a prime number.

Since \[ L_n=\prod_{p\leqslant n}p^{\lfloor\log_pn\rfloor}<\prod_{p\leqslant n}p^{\log_pn}=\prod_{p\leqslant n}n=n^{\pi(n)} \] it follows that \[ \ln L_n<\pi(n)\ln n\sim n \]
Choose $\frac 1 2<r<1$. From \[ L_n>\prod_{n^r<p\leqslant n}p>\prod_{n^r<p\leqslant n}n^r=n^{r(\pi(n)-\pi(n^r)) }\]
one has that \[ \ln L_n>r(\pi(n)-\pi(n^r))\ln n\sim r\left(\frac n{\ln n}-\frac{n^r}{r\ln n}\right)\ln n=rn-n^r\sim rn \]
$r$ is arbitrarily close to $1$ and therefore \[ \ln L_n\sim n\quad\square \]

[i]Lemma 2[/i]:

If $u_0,u_1,\dots,u_{n-1}$ are fixed, then all of the possible values for $u_n$ such that $n-m\mid u_n-u_m$ for each $m$ have the same remainder when divided by $L_n$.

[i]Proof[/i]:

Follows trivially from Chinese remainder theorem.$\quad\square$

[i]Lemma 3[/i]:

Let $u_0,u_1,\dots,u_{n-1}$ be a sequence which satisfies the condtitions from the problem. Let $r(x)$ be a minimum degree polynomial such that $r_i=r(i)=u_i$ for $0\leqslant i\leqslant n-1$. Then it holds \[ r_n=\sum_{i=1}^n(-1)^{i+1}\binom n iu_{n-i} \] and also \[ n-m\mid r_n-u_m \] for each $0\leqslant m\leqslant n-1$.

[i]Proof[/i]:

$n=1$ is trivial.

For $0\leqslant i\leqslant n-2$ let $u'_i=u_{i+1}-u_i$. Notice that \[ i-j\mid (u_{i+1}-u_{j+1})-(u_i-u_j)=(u_{i+1}-u_i)-(u_{j+1}-u_j)=u'_i-u'_j \] for each $0\leqslant i,j\leqslant n-2$. If $r'(x)$ is a minimum degree polynomials such that $r'(i)=u'_i$ for $0\leqslant i\leqslant n-2$ then $r'(n-1)=r_n-u_{n-1}$.

Let $m>0$. From the inductive assertion it follows that $m\mid(r_n-u_{n-1})-(u_{n-m}-u_{n-m-1})$, and since $m\mid u_{n-1}-u_{n-m-1}$ it follows that $m\mid r_n-u_{n-m}$.

Now, let $m=0$. WLOG, we may assume $u_0=0$ (subracting it from the rest of the sequence if necessary). Therefore
\[ r_n=\sum_{i=1}^{n-1}(-1)^{i+1}\binom n iu_{n-i} \]
Since $(n-1)\binom n i=n\binom{n-1}i$ it follows that $n\mid(n-i)\binom n i$, and from this and $n-i=(n-i)-0\mid u_{n-i}-u_0=u_{n-i}$ one has that $n\mid r_n$.$\quad\square$

[b]Proof that $\limsup |u_n|^{\frac 1 n} < e - 1$ is sufficient[/b]:

Let \[ \limsup |u_n|^{\frac 1 n} < e - 1 \] There is an $\varepsilon>0$ and an integer $n_0$ such that for each $n>n_0$ it holds \[ |u_n|^{\frac 1 n} < e - 1-3\varepsilon \] [i]Lemma 3[/i] now gives that there is an $n_1>n_0$ such that for each $n>n_1$ it holds \[ |r_n|<\sum_{i=n_0}^n\binom n i(e-1-2\varepsilon)^{n-i}<(e-2\varepsilon)^n<\frac{(e-\varepsilon)^n}2 \] [i]Lemma 1[/i] now gives that there is an $n_2>n_1$ such that for each $n>n_2$ it holds $\ln L_n>n\ln(e-\varepsilon)$, what is \[ L_n>(e-\varepsilon)^n \]
Therefore, for $n>n_2$ and $u_0,u_1,\dots,u_{n-1}$ given, $r_n$ is (by [i]Lemma 2[/i]) the only possible value for $u_n$ such that the desired property still holds and that be $|u_n|<\frac{(e-\varepsilon)^n}2$. Since $|u_n| <( e - 1-3\varepsilon)^n$, it must be $u_n=r_n$. But this holds for each $n>n_2$, and therefore $r_n$s are the values of the minimum degree polynomial $r(x)$ such that $r(i)=u_i$ for $0\leqslant i\leqslant n_2$, therefore we may take $Q(x)\equiv r(x)$.$\quad\blacksquare$

[b]Proof that $\limsup |u_n|^{\frac 1 n } \leqslant e$ is not sufficient[/b]:

Let $u_0=0$ and, starting from $n=1$, for $r_n=0$ choose $u_n=L_n$, and for $r_n\neq 0$ choose $u_n$ such that:

1) $u_n\equiv r_n\pmod{L_n}$;
2) $\mathrm{sgn}\ u_n\neq\mathrm{sgn}\ r_n$;
3) $|u_n|<|L_n|$.

Now, $u_n\neq r_n$ for all $n$ and therefore is not given by a polynomial, but since $|u_n|\leqslant |L_n|$ it follows that $\limsup |u_n|^{\frac 1 n } \leqslant e$.$\quad\blacksquare$


complex analysis
function
complex analysis unsolved

A holomorphic map $ f: D \to D$ is said to be proper if the inverse image of any compact subset of $ D$ is compact. Prove that any proper holomorphic map from $ D$ onto itself can be expressed uniquely as a Blaschke product.
Here $ D$ stands for the open unit disk.

Since $ f$ is proper, we have $ |f(z)|\to 1$ as $ |z|\to 1$. Also, $ f$ has finitely many zeroes, so we can factor them out and get a holomorphic function $ g\colon D\to D$ which has no zeroes in $ D$ and satisfies $ |g(z)|\to 1$ as $ |z|\to 1$. Apply the maximum principle to $ 1/g$.


ratio
geometric sequence
number theory unsolved
number theory

Suppose that $ \{b_n\}$ is a geometric sequence of real numbers with ratio $ q$. Determine $ q$, if we know that: $ b_3\minus{}b_4\equal{}\dfrac{16}{9}$ and $ b_6\minus{}b_8\equal{}\dfrac{1}{3}$.

hello, using the formula $ b_n\equal{}b_1q^{n\minus{}1}$ you will get the following system
$ b_1q^2\minus{}b_1q^3\equal{}\frac{16}{3}$
$ b_1q^5\minus{}b_1q^7\equal{}\frac{1}{3}$
solving this you will get $ b_1$ and $ q$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, using the formula $ b_n \equal{} b_1q^{n \minus{} 1}$ you will get the following system
$ b_1q^2 \minus{} b_1q^3 \equal{} \frac {16}{3}$
$ b_1q^5 \minus{} b_1q^7 \equal{} \frac {1}{3}$
solving this you will get $ b_1$ and $ q$.
Sonnhard.[/quote]

Could you post a complete solution?

[quote="Dr Sonnhard Graubner"]hello, what do you mean with a complete solution?
Sonnhard.[/quote]

It means, in more details and with the answers, too.

hello, using the formula $ b_n\equal{}b_1q^{n\minus{}1}$ we get
$ b_3\equal{}b_1q^2$ for $ n\equal{}3$
$ b_4\equal{}b_1q^3$ for $ n\equal{}4$
$ b_6\equal{}b_1q^5$ for $ n\equal{}5$
$ b_8\equal{}b_1q^7$ for $ n\equal{}8$
Sonnhard.

[quote="pco"]And then the equations become :
$ b_1q^2 \minus{} b_1q^3 \equal{} \frac {16}{9}$
$ b_1q^5 \minus{} b_1q^7 \equal{} \frac {1}{3}$

Dividing the second by the first, we get $ q^3(1 \plus{} q) \equal{} \frac 3{16}$ which have a trivial root $ q \equal{} \frac 12$ and a non trivial one $ q \equal{} \minus{} 1,129960525...$[/quote]

How did you find that $ q \equal{} \minus{} 1,129960525..$ is a root of $ 8q^3\plus{}12q^2\plus{}6q\plus{}3\equal{}0$ ?
By hand or by computer?

[quote="Scorpius"][quote="pco"]And then the equations become :
$ b_1q^2 \minus{} b_1q^3 \equal{} \frac {16}{9}$
$ b_1q^5 \minus{} b_1q^7 \equal{} \frac {1}{3}$

Dividing the second by the first, we get $ q^3(1 \plus{} q) \equal{} \frac 3{16}$ which have a trivial root $ q \equal{} \frac 12$ and a non trivial one $ q \equal{} \minus{} 1,129960525...$[/quote]

How did you find that $ q \equal{} \minus{} 1,129960525..$ is a root of $ 8q^3 \plus{} 12q^2 \plus{} 6q \plus{} 3 \equal{} 0$ ?
By hand or by computer?[/quote]

$ 8q^3 \plus{} 12q^2 \plus{} 6q \plus{} 3 \equal{}(2q\plus{}1)^3\plus{}2$ and so the root $ q\equal{}\frac{\minus{}1\minus{}\sqrt[3]2}2$

Thank you very much for the solution and for the patience, pco. :)


There exist or not $P(x)\in \mathbb Z[x]$ satisfying the conditions:
1)$n=\deg P \geq 2005$
2)$P(1),P(2),P(3),\cdots ,P(2n)$ are prime numbers.

If $a,b \in \mathbb{Z}$ are such that $ak+b$ is prime for $0 \leq k \leq (2n)^n$ (we know, by the work of Tao and Green, that such $a,b$ exist), then the polynomial $P(x)=ax^n+b$ has the required properties.  :) However, I don't think that any of us (the forum members) can solve the problem (by [i]solve[/i] I mean presenting a self-contained solution, not just saying that it's a trivial consequence of the Green-Tao theorem). Or am I wrong? Pigfly, since you presented it in the 'Proposed & Own' section, do you know a solution?

[quote="vess"] Or am I wrong? Pigfly, since you presented it in the 'Proposed & Own' section, do you know a solution?[/quote]
I think you're right.When I presented this problem in the 'Proposed &Own' section, I thought I solve it,but actually I had a big mistake  :( .Sorry for my stupid  :D  and of course I  haven't solve it (It's too hard for me ;) )!


How many distinct ways can you arrange the letters in the word "factorial" so that it starts with a?

[quote="math92"][hide]there are 7 other letters than a's in factorial. I think you would do 8!. I don't have a calculator handy so the answer is 8!....i think... :? [/hide][/quote]
No, I think you make mistake in letter $a$. You cannot tell which $a$ is used.
Example:

if $a$'s are different, $\text{fa}_1\text{ctoria}_2\text{l}$, but $\text{a}_1\text{fctoria}_2\text{l}\neq\text{a}_2\text{fctoria}_1\text{l}$
if $a$'s are the same, $\text{afctorial}=\text{afctorial}$

I do not think we consider the same letters distinct, perhaps 236!factorial might clarify this for us.

I didn't realize factorial had two a's  :roll: , change the question to read, how many [b]distinct[/b] arrangements....


Prove that $ n^{5}\minus{}n$ is divisible by $ 10$ if $ n$ is an integer.

Since the parity of $ n^5$ and $ n$ are always the same, $ n^5\minus{}n\equiv0\pmod2$, and we only have to test $ n\equiv0,\pm1,\pm2\pmod5$, each of which gives $ n^5\minus{}n\equiv0\pmod5$.

[hide="Similar"]
1) math154 told us that it's divisible by 2
2)
a) $ 5\mid n$ - it's obvious
b) $ 5\nmid n$ - use FLT
[/hide]

[quote="Z-coli"]What's FLT?
And you too are using casework.
Is there a way that doesn't involve casework?[/quote]

Fermat's little theorem. i think.

That's correct ($ n^4\equiv1\pmod5\implies n^5\minus{}n\equiv n\minus{}n\equiv0\pmod5$). Another way is to realize that the given expression factors into $ n(n \minus{} 1)(n \plus{} 1)(n^2 \plus{} 1)$. This is obviously divisible by $ 5$ for $ n \minus{} 1,n,n \plus{} 1\equiv0\pmod5$, and then we only have $ 2$ more cases to test ($ n\equiv\pm2\pmod5\implies n^2 \plus{} 1\equiv0\pmod5$).

We can also factor this into $ (n\minus{}1)n(n\plus{}1)(n^2\plus{}1)$.

This immediately takes care of $ 0,1,4$ mod 5.

2 and 3 are trivial to check.

And these are clearly divisble by 2, because of the consecutive integers.

Do you realize you essentially repeated my post? :P

[quote="math154"]Another way is to realize that the given expression factors into $ n(n \minus{} 1)(n \plus{} 1)(n^2 \plus{} 1)$. This is obviously divisible by $ 5$ for $ n \minus{} 1,n,n \plus{} 1\equiv0\pmod5$, and then we only have $ 2$ more cases to test ($ n\equiv\pm2\pmod5\implies n^2 \plus{} 1\equiv0\pmod5$).[/quote]

Huh? :huh: Come on, is it that hard to read the whole post?

[quote="math154"]Since the parity of $ n^5$ and $ n$ are always the same, $ n^5 \minus{} n\equiv0\pmod2$, and we only have to test $ n\equiv0,\pm1,\pm2\pmod5$, each of which gives $ n^5 \minus{} n\equiv0\pmod5$.[/quote

Am I missing something  :maybe:

Look at the whole thread. :wink: Above your solution is this post by some person...


Prove that if $ f$ is a polynomial with integer coefficients, and there exists an integer $ k$ such
that none of the integers $ f(1), f(2),..., f(k)$ is divisible by $ k$, then $ f(x)$ has no integer root.

[hide]Suppose that $ f$ has a zero at $ n$, so that $ f(n) \equal{} 0$. Then we can write $ f(x) \equal{} (x \minus{} n)g(x)$ for some polynomial $ g(x)$. In particular, $ x \minus{} n|f(x)$ for all integers $ x$. Also suppose that there exists a positive integer $ k$ such that $ k$ does not divide any of $ f(1),f(2),\dots,f(k)$. However, since $ x \minus{} n|f(x)$ for all integers $ x$, and since $ k$ must divide one of $ (1 \minus{} n),(2 \minus{} n),\dots,(k \minus{} n)$, it follows that $ k$ must divide at least one of $ f(1),f(2),\dots,f(k)$, contradiction.[/hide]


Three girls' birth years are 1986, 1989, and 1992.

Their last names are Kiss, Hunter, and Goldbach.

Their first names are Olga, Maria, and Renata.

It happened one year that they realized that everybody is the same

age as the number of letters in their own full names (first + last name).

What year did this happen and what are their names?

In 2000, this happened to Olga Kiss, Maria Hunter, and Renata Goldbach.

[hide]
girl born in 1986=x+6 years old
girl born in 1989=x+3 years old
girl born in 1992=x years old

You add all the letters in the first and last names to get 33.

So, 3x+9=33
x=8.
Thus, this happend in the year 2000.

girl born in 1986=14 years old
girl born in 1989=11 years old
girl born in 1992=8 years old

The girl born in 1986 must be Renata Goldbach because those are the greatest amount of letters in the first and last names which equal to 14.

The girl born in 1992 must be Olga Kiss because those are both the smallest amount of letters in the first and last name that equal to 8.

Hence, the girl born in 1989 is Maria Hunter.[/hide]


function
integration
calculus
calculus computations

Let $ f$ defined on $ [0,1]$ be a twice differential function such that , $ |f''(x)| <\equal{} 1$ for all $ x$ in $ [0,1]$ . If $ f(0) \equal{} f(1)$ , then show that, $ |f'(x)|<1$ for all $ x$ in $ [0,1]$ .

By Rolle's theorem, there exists $ z\in(0,1)$ such that $ f'(z) \equal{} 0$.  Therefore, for all $ x\in[0,z)$, we have $ |f'(x)|\leq\int_x^z|f''(y)|dy\leq z \minus{} x\leq z < 1$, and similarly for all $ x\in(z,1]$ we have $ |f'(x)|\leq\int_z^x|f''(y)|dy\leq x \minus{} z\leq 1 \minus{} z < 1$.  The conclusion follows.

All right- make it $ |f'(x)|\le \frac12$. Can you prove the stronger version?

Shouldnt it be $ |f'(x)| \equal{} \int_z^x|f''(y)|dy$ and when we change the limit, ie the first integral , we get $ |f'(x)|\leq\int_x^z|f''(y)|dy \equal{} \minus{} |f'(x)|$ . I am sorry for  asking this stupid doubt but plz explain.

OK I think this is better

$ \minus{}f'(x) \equal{} \int_x^zf''(y)dy$
$ \equal{}> |f'(x)| \equal{} |\int_x^zf''(y)dy|  \leq\int_x^z|f''(y)|dy$

Thank you


algorithm
MATHCOUNTS
ceiling function
function
factorial
continued fraction
binomial coefficients

[quote="t0rajir0u"]Generally speaking, a greedy algorithm is any attempt to maximize a certain quantity at the end of $ n$ moves by picking, at each step, the move that maximizes that quantity after that step.  It is by no means guaranteed to be the best strategy for maximizing anything, but it is always a possibility to consider.[/quote]

 :huh: ??????

Here's an example that I think will make the use of the word "greedy" clearer:

Suppose you are traveling through a series of rooms.  In each room there are three coins of different value.  You don't know what the coins are before entering a room.  At each room, you can either take a coin from the room, in which case you will automatically be shoved into the next room, or you can opt to go into the next room, where you can take two coins (and if you skip twice you can take all three coins).  Your goal is to maximize the total worth of your coins.

A greedy algorithm is not a bad way to do it:  all you do is take the best coin in each room and never skip rooms.  Given a random distribution of coins, this is a good way to maximize profit.

A nice case of the greedy algorithm actually being optimal:

Suppose we have a rational number $ r=\frac{m}{n}$ (in lowest terms), and we want to write it as a sum of unit fractions $ \frac1k$ using the fewest terms possible.
The greedy algorithm here is to simply take the first $ k$ as $ \lceil \frac{n}{m}\rceil$; this is the largest $ \frac1k$ less than or equal to $ r$. We then repeat this process with $ r-\frac1k$ in place of $ r$.

Even when it's not the best possible method, a greedy algorithm is typically very reliable, and it's easy to prove that such algorithms work.

Interesting.  So when given a fraction $ \frac{m}{n}$ with $ m>n$, and they ask you to write it as a continued fraction with $ 1$ s in the numerators, you would use the greedy algorithm, right?  I never knew it had a name.

$ (x+0.5)^{20}$.  Expand using biomial theorm.  Find the maximum coefficient of a given $ x^{n}$ where $ 0<n<20$.  Its not 20C10.

is it
[hide] $ \binom{20}{6}* \left(\frac{1}{2}\right)^{6}$?, because for each increase in the bottom number, in the nCr function, the exponent of .5 will also increase, so the last number where an increase in the bottom number will be greater that 2 is 6. [/hide]

yes thats correct.

Its 20C6*.5^6 or  20C7*.5^7 but either way the greatest coefficiant will be 605.625, which goes with x^13 or x^14

[quote="erdogankerem123"]What is a greedy algorithm, and how can it be applied to MATHCOUNTS?[/quote]

how are u suppoesed to steal a avatar :?:

[size=75][color=indigo]LadyKn1ght: Welcome to Art of Problem Solving/Mathlinks, MathSports. But please stay on topic--if you want to talk to erdogankerem about his signature, please PM him instead of putting it in a thread about a math problem.[/color][/size]

Speaking of binomial coefficients, 2 questions:

1) How do we evaluate $ \left(\frac{a}{b}\right)!$ ? How about $ \pi!$ or $ \minus{}3!$?

2) There are 2 notations for "n choose r", one being nCr and the other being $ {n\choose r}$. However, for "n permute r" there is only one notation. Why is that? I find writing $ {n\choose r}\plus{}nPr$ a little bizarre.

[quote="Prime factorization"]1) How do we evaluate $ \left(\frac{a}{b}\right)!$ ? How about $ \pi!$ or $ \minus{}3!$?[/quote]

First, note that the usual definition of factorial really only applies for non-negative integers.

However, we can do what is called an "analytic continuation" of the factorial function.  This leads to a function known as the [url=http://mathworld.wolfram.com/GammaFunction.html]Gamma Function[/url], which is, by the way, still not defined for negative integers.  :)

[quote="Prime factorization"]2) There are 2 notations for "n choose r", one being nCr and the other being $ {n\choose r}$. However, for "n permute r" there is only one notation. Why is that? I find writing $ {n\choose r}\plus{}nPr$ a little bizarre.[/quote]

Permutations have few applications outside of straight counting, whereas combinations are so diverse in application that the field of combinatorics is aptly named after them.  I would personally write $ nPr$ as $ r!{n\choose r}$ because the combination is in fact much more natural.

This is interesting, everyone, just make sure to keep it closer to MathCounts level :) .


How many axes of symmetry does a regular hexagon have?

We can draw axes from vertex to opposite vertex, giving us 3, plus through the midpoints of each side, making another 3, which gives us a total of $ 6$.

 [geogebra]b956ef9ee50b3139ebe8a6234086cb5543e6ead2[/geogebra]


For $x_1 \geq x_2 \geq x_3 \geq ... \geq x_n \geq 0 $ real numbers such that $x_1^2,x_2^2,....,x_n^2$  are the sides of a poligon prove

$ n(\frac{x_1}{x_2} + \frac{x_2}{x_3} +..+\frac{x_n}{x_1} ) \geq (x_1+x_2+...+x_n)(\frac{1}{x_1} + \frac{1}{x_2} +..+\frac{1}{x_n} ) $

Sorry, but I don't know if it is true so you could find at once a counterexample to kill it but if it is true it is fine.

Note that requiring that $x_i$ are sides of a poligon it is not sufficient. As example for $n=4$ take$ (x_1,x_2,x_3,x_4) =(4,3,2,1) $

[quote="manlio"]Sorry, but I don't know if it is true so you could find at once a counterexample to kill it but if it is true it is fine.[/quote]

Sorry for killing... but if I understand you correctly, then the inequality doesn't even hold for n = 4:

With $x_1=33$, $x_2=31$, $x_3=16$, $x_4=14$, the numbers $x_1^2$, $x_2^2$, $x_3^2$, $x_4^2$ are the sides of a quadrilateral, but still

  [tex]4\left( \frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\frac{x_{3}}{x_{4}}+\frac{x_{4}}{x_{1}}\right) -\left( x_{1}+x_{2}+x_{3}+x_{4}\right) \left( \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}\right) =-\frac{11089}{57288}<0[/tex].

On the other hand, the inequality with reversed sign is not generally true, too, since $x_1=21$, $x_2=17$, $x_3=16$, $x_4=14$, again with $x_1^2$, $x_2^2$, $x_3^2$, $x_4^2$ being the sides of a quadrilateral, gives

  [tex]4\left( \frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\frac{x_{3}}{x_{4}}+\frac{x_{4}}{x_{1}}\right) -\left( x_{1}+x_{2}+x_{3}+x_{4}\right) \left( \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}\right) =\frac{10}{119}>0[/tex].

Sorry again, Manlio, I hope at least I've saved your time...

  Darij

A related result since Manlio seems to look for conditions which imply existence of a polygon :

It is known (see reference below) that :

If $3 \leq r \leq n$ and if $a_1, \cdots , a_n$ are positive numbers such that
\[( \sum_{i=1}^n a_i )^2 \geq (n - \frac {(r-2)^2} r ) \sum_{i=1}^n a_i^2\]
then any $r$ of the $a_i$'s are, in any order, the consecutive sides of a convex $r$-gon (possibely degenerated).

Reference :
B.V. Dekster, 'Assembling $r$-gons out of $n$ given segments', Math. Magazine, 192, p.44-48.

Pierre.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group  and $ F \subset G$ such that $ F$ is not a subgroup, $ e \in F$ and for all $ a \in F,a^{ \minus{} 1} \in F$. Is it true that for all $ a \in F \setminus \{e\}$ there exists $ b \in F$ such that $ a \cdot b \notin F$?

No: let $ G \equal{} {\bf Z}/10{\bf Z}$ under addition and let $ F \equal{} \{0, 1, 4, 5, 6, 9\}$.  We have that $ 5 \plus{} F \equal{} F$ but $ F$ is clearly not a subgroup.

The idea: choose an abelian group.  Choose a nonzero element $ a$ and include in $ F$ the cyclic group generated by $ a$.  Now choose some other element $ b$ not in this subgroup, and include in $ F$ $ b \plus{} na$ and $ \minus{}b \plus{} na$ for all $ n$.  This subset satisfies the needed conditions, and unless you've been incredibly unlucky in your choices it is not a subgroup.


probability
geometry
3D geometry
sphere
probability and stats

A container (box) A contains 6 red and 4 green spheres, whereas a second container B contains 7 red and 3 green spheres. A sphere is chosen by luck from container A and it is put at container B. Then, a sphere is chosen by luck from container B and it is put at A. After the end of the experiment find the probability of the events:
(a) To be chosen a red sphere from container A and a red sphere from container B
(b)No to change the composition of the containers, i.e. container A to contain 6 red and 4 green spheres and container B to contain 7 red and 3 green spheres.

-----------------------------------------------------
Additional Details

When I say "by luck" I mean random. No further explanation. I have the same question about the procedure. It found this in an entry examination for universities.

My basic question is:
What do we mean by saying "After the experiment to be chosen"?

In most of the students solutions it was assumed that the probability we are searching for is the probability of picking red and red DURING the experiment.

In the statements it says AFTER THE END of the experiment. I believe it asks the probability of picking two red balls (one from each box) AFTER the experiment.

What do you think?

doesnt really seem like a math question as much as a diction question.

[quote="leonariso"]In the statements it says AFTER THE END of the experiment. I believe it asks the probability of picking two red balls (one from each box) AFTER the experiment.

What do you think?[/quote]

Your interpretation doesn't make any sense, so it's probably not what was intended :P.  The problem is poorly worded, but there's only one probability it could plausibly be asking for.


number theory
advanced fields
advanced fields theorems

Does any one have any recommendations? I have heard about Newkirch's Algebraic Number Theory as especially good. How is this compared to Lang's Algebraic Number Theory?

What books are good to study Iwasawa Theory, or elliptic curves? I know there are three books by Silverstein. Show I get them all?

[quote="lenny"]

What books are good to study Iwasawa Theory, or elliptic curves? [/quote]

1) The Arithmetic of Elliptic Curves (Graduate Texts in Mathematics) by J.H. Silverman

2) Algebraic Number Theory (Graduate Texts in Mathematics) by Serge Lang

3) Cyclotomic Fields I-II (Graduate Texts in Mathematics) (Hardcover)
by Serge Lang

4)
 Introduction to Cyclotomic Fields (Graduate Texts in Mathematics) by L. C. Washington and Lawrence C. Washington

... ... :(

Just a question. How "hard" is algebraic number theory, once you have the prerequisites in place?

Very much in general.  But it also depends on your "feeling" to a certain subject.  e.g. you have special intuition on algebra, you may find it easier than other subjects like analysis or geometry.


I think $x_i,y_i$ are positive real numbers ;)

I solve the left part in your last question in title (a_i) so the right part  holds ,where you put $a_i^2=\sqrt {x_i^3y_i}$ and $b_i^2=\sqrt {x_iy_i^3}$ in the left part . :D

[quote="Mathx"]$x_i,y_i$ $(i=1,2,...,n)$ are real numbers prove that:
$(\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2)\geq (\sum_{i=1}^n \sqrt{x_i^3y_i})(\sum_{i=1}^n {x_iy_i^3})\geq (\sum_{i=1}^n x_iy_i)^2$[/quote]

are u sure that it isnt $(\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2)\geq (\sum_{i=1}^n \sqrt{x_i^3y_i})(\sum_{i=1}^n \sqrt{x_iy_i^3})\geq (\sum_{i=1}^n x_iy_i)^2$

then the right part is by cauchy as well.
i think the left part can be proven by hölder, poving something like $\sqrt[4]{(x_1^2+...+x_n^2)^3(y_1^2+...+y_n^2)}\ge \sqrt{x_1^3y_1} + ... + \sqrt{x_n^3y_n}$

but i could quite finish it  :oops:


[quote="vedran6"]$ \frac {a \plus{} b \plus{} c} {3} \minus{} \sqrt [3] {abc} \leq max [(\sqrt {a} \minus{} \sqrt {b})^2,(\sqrt {b} \minus{} \sqrt {c})^2,(\sqrt {c} \minus{} \sqrt {a})^2]$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61957
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182745


factorial
Ring Theory
superior algebra
superior algebra unsolved

let us consider a prime $p/1$ in a ring of fractions $S^{-1}A$
why would one need in a proof to say that p has no inverse in $A$ ?

the context of this question is in prooving that if A is factorial (UFD i think)
then $S^{-1}A$ is also factorial(S is considered witout 0 ).

Primes have no inverses. Thus a prime $p \in A$ has no inverse in $A$ (possibly in $S^{-1}A$).

But your question is very vague without the exact context.

ZetaX why do primes have no inverses in A ?
ZetaX why do primes have no inverses in $S^{-1}A$ ?

Primes do have no inverses by definition (see http://en.wikipedia.org/wiki/Prime_element#Divisibility.2C_prime_and_irreducible_elements ).


Does anyone have the state target round problems from two years ago with the cubes??? Those were very good problems...(mebbe it was just one?)...Required lotsa spatial thinking...

Believe it or not, but I do.  That is, if you're talking about State 2001 Target #6.
And in pdf form, actually.
I believe there is no copyright infringement because they were posted on the mathcounts website in 2002 blah blah blah...

I will post the problem here:
[color=red]Four cubes of volume  1 cm:^3:, 8 cm:^3:, 27 cm :^3:, and 125 cm:^3: are glued together at their faces.  What is the number of square centimeters in the smallest possible surface area of the resulting solid figure?[/color]

There was only one about cubes.  If you want the other targets from that year, I can e-mail you the pdf file.

Hope this helps.

the best i could do is when all 4 blocks are squeezed together so that each block touches 3 others.

in this case the surface area exposed is

(25*5+25-9-4-1) + (4*9+9-4-1) + (3*4 + 4-1) + (3) = 194 cm^2.

can anyone do betteR?


geometry
combinatorics unsolved
combinatorics

For a set S of n points in the plane denote by T(S) the set of all triangles whose vertices are in S. Prove that the maximal area can occur at most n times in T(S).

This result is due to Brass, Rote ans Swanepoel in 2000.

see the paper here :
http://www.inf.fu-berlin.de/inst/pubs/tr-b-00-06.abstract.html

Pierre.


function
real analysis
real analysis unsolved

I couldn't solve this equation:

\[ y'\plus{}y\equal{}\tan(x/2)\]

I can't find the particular solution.
 Thanks for your answers.

The integrating factor is $ e^x$. The solutions involve the antiderivative of $ e^x\tan(x/2)$, which is unlikely to be an elementary function.

Jonysatie, are you sure that's the ode you want to solve?

Once, I saw that you may express the non-elementary function like a definite integral.

Anyway, I don't think you can do something else.

I'm sure that's the equation. So the solution can be expressed in terms of an integral.
I had thought that, thanks.


Να βρειτε ολους τους ακεραιους $ (a,b,c,x,y,z)$ ωστε:
$ a\plus{}b\plus{}c \equal{} xyz$
και
$ x\plus{}y\plus{}z\equal{}abc$
και $ a\geq b \geq c \geq 1$, $ x\geq y \geq z \geq 1$

[quote="basted"]Να βρειτε ολους τους ακεραιους $ (a,b,c,x,y,z)$ ωστε:
$ a \plus{} b \plus{} c \equal{} xyz$
και
$ x \plus{} y \plus{} z \equal{} abc$
και $ a\geq b \geq c \geq 1$, $ x \geq y \geq z \geq 1$[/quote]

Σχετικα ευκολη, αλλα θελει πολυ γραψημο...

Για 2 ακεραιους $ m,n > 1$ ισχυει $ mn \geq m \plus{} n \Rightarrow (m \minus{} 1)(n \minus{} 1) \geq 1$ με ισοτητα για $ m \equal{} n \equal{} 2$

Εστω χωρις βλαβη της γενικοτητας $ x \geq y \geq z > 1$ και $ a \geq b \geq c > 1$, τοτε
$ abc \geq ab \plus{} c \geq a \plus{} b \plus{} c \equal{} xyz \geq xy \plus{} z \geq x \plus{} y \plus{} z \equal{} abc$, αρα ισχυουν ολες οι ισοτητες και ειναι $ ab \equal{} a \equal{} b \equal{} c \equal{} 2$, αδυνατο.
Αρα ειναι $ \min\{c,z\} \equal{} 1$, εστω χωρις βλαβη της γενικοτητας $ \min\{c,z\} \equal{} c \equal{} 1$, αλλα $ z > 1$, τοτε ειναι:
$ a \plus{} b \plus{} 1 \equal{} xyz > x \plus{} y \plus{} z \equal{} ab \Rightarrow (a \minus{} 1)(b \minus{} 1) < 2 \Rightarrow (a \minus{} 1)(b \minus{} 1) \equal{} 1 \Rightarrow (a,b,c) \equal{} (2,2,1)$,
και τοτε ειναι: $ xyz \equal{} 5, x \plus{} y \plus{} z \equal{} 4$, το οποιο ειναι αδυνατο. Αρα ειναι $ c \equal{} z \equal{} 1$ και $ a \plus{} b \plus{} 1 \equal{} xy, x \plus{} y \plus{} 1 \equal{} ab$, εστω $ \max\{b,y\} \equal{} y$, αν παλι $ y > 1$ εχουμε:
$ a \plus{} b \plus{} 1 \geq ab \minus{} 1 \Rightarrow (a \minus{} 1)(b \minus{} 1) \leq 3$ απο οπου εχουμε $ (a,b) \in \{(2,2), (3,2), (4,2), (a,1)\}$, τωρα ευκολα επαληθευουμε οτι οι μονες δυνατες περιπτωσεις ειναι οι: $ (a,b) \equal{} (3,2)$ απο οπου  βρησκουμε $ xy \equal{} 6, x \plus{} y \equal{} 5 \Rightarrow (x,y) \equal{} (3,2)$, και $ (a,b) \equal{} (a,1)$ απο οπου εχουμε: $ a \plus{} 2 \equal{} xy, x \plus{} y \plus{} 1 \equal{} a$ αρα $ a \equal{} xy \minus{} 2, (x \minus{} 1)(y \minus{} 1) \equal{} 4 \Rightarrow (a,x,y) \in \{(7,3,3), (8,5,2)\}$.
Αν $ y \equal{} b \equal{} 1$, τοτε καταληγουμε στο: $ a \plus{} 2 \equal{} x, x \plus{} 2 \equal{} a$, το οποιο ειναι αδυνατο.
Αρα οι δοσμενες ισοτητες ισχυουν ταυτοχρονα μονο για $ (x,y,z) \equal{} (a,b,c) \equal{} (3,2,1)$ καi για $ (a,b,c,x,y,z) \in \{(8,1,1,5,2,1), (7,1,1,3,3,1)\}$ και αναποδα.

[quote="NickNafplio"][quote="basted"]
 $ a\geq b \geq c \geq 1$, $ x \geq y \geq z \geq 1$[/quote]
$ (x,y,z) \equal{} (a,b,c) \equal{} (3,2,1)$ και για ολες τις μεταθεσεις αυτων.[/quote]

Χωρίς τις μεταθέσεις αυτών, αφού δίνει διάταξη

[quote="basted"][quote="NickNafplio"]
Αρα οι δοσμενες ισοτητες ισχυουν ταυτοχρονα μονο για $ (x,y,z) \equal{} (a,b,c) \equal{} (3,2,1)$ και για ολες τις μεταθεσεις αυτων.[/quote]
Ισχυει και για (5,2,1,8,1,1) (3,3,1,7,1,1) και συμμετρικα...[/quote]

Σωστος, εκει που βρηκα $ 0 \leq (b \minus{} 1)(a \minus{} 1) \leq 3$ πηρα τις περιπτωσεις: $ (a \minus{} 1)(b \minus{} 1) \equal{} 1,2,3$ και λογο βιασινης παρελειψα την περιπτωση $ (a \minus{} 1)(b \minus{} 1) \equal{} 0$ η οποια δινει τις αλλες δυο λυσεις...
Edited

ΥΓ: Οταν εχεις αργησει στο μαθημα και προσπαθεις να τελειωσεις οσο πιο γρηγορα γινεται αυτα παθαινεις...


invariant
number theory
least common multiple

A game for computing gcd(a,b) and lcm (a,b).
We start with x=a, y=b, u=a, v=b and move as follows:

if x<y then, set y = y-x and v = v+u
if x>y, then set x = x-y and u = u+v

The game ends with x=y=gcd(a,b) and (u+v)/2 = lcm(a,b). Show this.

---------------------------------------------------------------------------------------------
My thoughts:
[hide]
If we start with a=3,b=4.

then we will have
x, y, u, v
3 4 3 4
3 1 3 7
2 1 10 7
1 1 17 7

gcd(x,y) is always going to be gcd(a,b).

also. gcd(a,b) * lcm (a,b) = ab

What do I do next?
[/hide]

>> Invariant problem

Nice clue !

$x$ and $y$ finish as $gcd(a,b)$ by Euclid algorithm.

One invariant is $\frac{x \cdot v + y \cdot u}{2}$ which remains constant $= a \cdot b$

since $a \cdot b = gcd(a,b) \cdot lcm(a,b)$ .....

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

Did you see Valentin, I used \cdot as requested  ;)


hi...i really hate math since my elementary years...i started so bad that i brought it up as i grew...i want to know what are the complete list on which math can be applied to programmers?maybe a list from basic to advance topics will be appreciated...thanks! :)

I think you'll find it difficult to be a good programmer if you hate math.  Anyway, you can start at http://mathworld.wolfram.com/topics/ComputerScience.html .

yah you're right...but i have lately decided that i'll gonna make some changes...so i'm now interested in math but only math that is applicable to my field (programming).thanks for the link. :)


[quote="GameBot"]The average of 11 consecutive even integers is 24. What is the greatest of these integers?[/quote]

$ 24\plus{}2 \cdot 5\equal{}34$.


algebra
polynomial
function
LaTeX
topology
analytic geometry
complex numbers

Let
$P_{t}(x)= \sum_{i=0}^{n}a_{i}(t) x^{i}$
be a complex polynomial respect to both $t$ and $x$
with $a_{n}(t)$ non-vanishing.
let $\{ r_{j}(t)\}_{j=1,2,...,n}$ , be the set of zeros of $P_{t}(x)$ viewed as a polynomial of $x$,
satisfying $Re{r_{j}(t)}\leq Re{r_{i}(t)}$ (if they are equal, then $Im{r_{j}(t)}\leq Im{r_{i}(t)}$) whenever$j \leq i$
(1) Prove or disprove that $r_{j}(t)$ is continuous respect to $t$ for all $j$
(2) if (1) is not true,
if $r_{j}(t)$ are all distinct for $t=t_{0}$, then is it true that there exist sufficiently small $\epsilon$ such that $r_{j}(t)$ are all distinct for all $t$ in $(t_{0}-\epsilon, t_{0}+\epsilon)$

Naturally, the coefficients are supposed to be [b]continuous[/b] functions of $t$. Here is a counterexample to (1): $P_{t}(z)=z^{2}-e^{2it}$ has roots $e^{it}$ and $-e^{it}$. Consider what happens around $t=\pi/2$.

(2) is true. If $P_{t_{0}}'(r_{j}(t_{0}))\ne 0$ for all $j$, then let $\delta=\min_{j}|P_{t_{0}}'(r_{j}(t_{0}))|$ and choose $\epsilon$ depending on $\delta$... OK, this is not enough. Use [url=http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem]Rouche's theorem[/url].

[It seems that you know $\LaTeX$. Why don't you use it?]

Thanks for the reply. The coefficients are assumed to be continuous, as in each of them is a polynomial of t.
(I actually don't know how to use LaTeX without any editor. how does it work on this website?)

Just put dollar signs around the code (math mode only). The forum software does the rest.

oh. thanks. I've fixed it.
This looks much better..:)
For the first part, Rouche's theorem works perfectly for sufficiently small intervals including $t$, which gives $n$ distinct roots.
(take a contour having only one root of $P_{t}(x)$ in it, and then apply Rouche!)
But, I still don't know whether this is also true for the repeated roots or not.

[quote="hankookin"]
For the first part, Rouche's theorem works perfectly for sufficiently small intervals including $t$, which gives $n$ distinct roots.
(take a contour having only one root of $P_{t}(x)$ in it, and then apply Rouche!)[/quote]
In that case, what you have shown is that there is $n$ continuous paths defined on the small interval and that the values of these $n$ paths at $t$ are $n$ distinct roots of polynomial $P_{t}$. But that doesn't mean $r_{1}, \ldots r_{n}$ are continuous on that interval (Consider mlok's counterexample.)

[quote="mlok"][quote="hankookin"](take a contour having only one root of $P_{t}(x)$ in it, and then apply Rouche!)[/quote]
This should work exactly the same for a contour around a multiple root.

The reason why $r_{j}(t)$ are not continuous is not the possibility of multiple roots, but a clumsy attempt to order complex numbers (there is no natural order on $\mathbb C$). It would be more natural to work with unordered set of zeros $Z(P)\subset\mathbb C$ and measure the distance between sets with the [url=http://en.wikipedia.org/wiki/Hausdorff_distance]Hausdorff metric[/url]. The set of all monic polynomials of degree $n$ is naturally identified with $\mathbb C^{n}$. I think that the map $P\mapsto Z(P)$ is continuous from $\mathbb C^{n}$ to the [url=http://en.wikipedia.org/wiki/Hyperspace]hyperspace[/url] of $\mathbb C$. It may be locally Holder continuous with exponent $1/n$. :maybe:[/quote]

I tried to figure out what you said, and reached:
Let $\sim$ be a well defined equivalence relation in $\mathbb C^{n}$ such that $a\sim b$ iff there exists a permutation of coordinates $\sigma$ so that $\sigma (a)=b$.
Then there exists a homeomorphism between $\mathbb C^{n}/ (\sim)$ with appropriate hausdorff topology and the set of monic polynomials of degree $n$ equipped with topology given by a family of seminorms $\{\sup\|. \|_{K}\}$, where $K$ is compact.
This implies that saying (1) (after dropping the crappy order given on roots) is actually equivalent to saying that:
for every continuous function $f$ from $\mathbb R$ to $\mathbb C^{n}/ (\sim)$, there exists a continuous function $g$ from $\mathbb R$ to $\mathbb C^{n}$, such that $\pi g= f$, where $\pi$ is a natural projection map from $\mathbb C^{n}$ to $\mathbb C^{n}/ (\sim)$. (i.e. $\pi(z)=[z]$ for every $\forall z \in \mathbb C^{n}$)

However, is this a true statement? I couldn't prove or disprove this.
(btw, thanks for answering my questions all the time!)

:) Indeed, the space $\mathbb C^{n}/(\sim)$ (with the metric and topology inherited from $\mathbb C^{n}$) works better here than the hyperspace of $\mathbb C$ (in the hyperspace the systems of roots $(1,2,2)$ and $(1,1,2)$ are indistinguishable). We want to prove that the map $\Pi\colon (z_{1},\dots,z_{n})\mapsto \prod_{k=1}^{n}(z-z_{k})$ is a homeomorphism from $\mathbb C/(\sim )$ onto the space of monic polynomials of degree $n$, identified with $\mathbb C^{n}$. The map $\Pi$ is clearly continuous, one-to-one, and onto. We need an extra bit of information to conclude that $\Pi^{-1}$ is continuous: $\Pi$ is [url=http://en.wikipedia.org/wiki/Proper_map]proper[/url].

I've recently found a short proof (& more result!) for this problem.
to simplify, assume $P_{t}(x)$ is an irreducible polynomial
then $\frac{d}{dx}P_{t}$ is relatively prime to $P_{t}(x)$.
Thus we can think about a resultant, ie, $A(t,x)\frac{d}{dx}P_{t}(x)+B(t,x)P_{t}(x)=C(t)$
Noting that $C(t)$ must vanish at each repeated roots, hence except finitely many points $t$, $P_{t}(x)=0$ has $n$ simple roots.


let $ a$ ,  $ b$ , $ c$   positives  reals numbers  with $ abc \equal{} 1$  prove that:

  $ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq \frac {3}{\sqrt2}$

We can use Holder Inequalitie.
 Let $ S \equal{} \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}$ and $ P \equal{} 2(a.b \plus{} b.c \plus{} c.a)$
  We have
     $ S^2 \geq \frac {(a \plus{} b \plus{} c)^3}{P} \geq \frac {9}{2}$

thank's thanhak for your prove :clap2:

i created my inequality by jensen . can you write holder inequality in it's generale ...because i didn't  take it yet ..


LaTeX
factorial
number theory
prime factorization

Alright, this forum is almost dead so i will be holding a contest.  We need about 5-10 people so sign up now!  The problems will be classroom math level.  So if you want to be in just say: i'm in.

IN!!!

this sentence was added so the message wouldn't seem too small.

Me too...

This message is not too small.

I will do it if I have time. If I don't then....sorry.

I'm in, when's it gonna start?

[hide]110[/hide]

Find the distance from the point $(5,7)$ to the center of the circle, $4x^2+8x+4y^2-16y-16=0$

The circle can also be witten as
  $( x+1)^2 + (y-2)^2 = 9$
   which has the centre as $(-1, 2)$
  The reqd. distance is $\sqrt{61}$

 What happened to $math92$ anyway?

[color=green][size=84]LaTeX fixed by mod robinhe[/size][/color]

i dunno but im gonna keep making questions and hopefully this contest will get going.

i'll post scores and next round tommorow.

Scores

1st:Nonie=40 pts
2nd:madatmath=38 pts
3rd:dogseatcheese=35 pts
4th:easyas3.14159...=34 pts
5th:fireemblem13=33pts
6th:ragnorok23=16 pts
7th:b-flat=15 pts
8th chineesechess=12 pts

Next round coming.

ROUND 3
calculators allowed
due saturday

1. A truck uses 3.5 tanks of gas to drive 560 km. How many tanks of gas would this truck use to drive 640 km?

2. Five cars (each of the same length) plus eight trucks (each of the same length) take up the same room in a warehouse as 11 trucks. How many cars can fit into a warehouse which can hold 30 trucks?

3.A chocolate bar is in the shape of an equilateral triangular prism. Each edge of the base is 1 inch, and the length of the bar is 8 inches. What is the number of cubic inches in the volume of the bar? Express your answer in simplest radical form.

4.What is the only pair of integers (x,y) for which twice the square of the first equals three times the square of the second?

5. In the prime factorization of 40!, how many times does the factor 5 occur?

HAVE FUN!!! :ninja:

what happened to this contest? Is it over?

sorry, i just have been overloaded with homework this week.  Next round and scores coming tomorrow. :)

[quote="vicseb"]Is it too late to join?
If it isn't, then I'm in[/quote]
Me, too. :D

I don't know if I can go on with the contest.  I have exams in math, science, english, and german to study for, so i think the contest has come to an end. :(
Hope you understand.

I understand, but I liked the contest (even though I was out of town 3 of the weeks I had answers for only 1 week).  Good job on running it as long as you did! :D

[quote="math92"]thanks....so i guess we shouldn't post here anymore... :ninja:[/quote]

hey did you ever score the last round? I was wondering if I got a few of those wrong...

[quote="Iron Fist"][hide="As the contest is over I might as well post the answers..."]
1. 4
2. 50
3. $2 \sqrt{3}$
4. (0, 0)
5. 9
[/hide][/quote]

Yeah that's what I got too...

anyways good job with the contest for as long as you did it. :)

[quote="ragnarok23"]If $x+\frac{1}{x}=5,\text{then what is } x^3+\frac{1}{x^3}?$
I dunno if this is classroom math, but it is easy enough.[/quote]
(x+1/x)³=x³+3x²(1/x)+3x(1/x)²+(1/x)³
             =x³+3x+3/x+1/x³
             =x³+1/x³+3x+3/x
             =5³=125
x³+1/x³=125-3(5)=110


[quote="tucsonlarry"]Is the Art of Problem Solving Book suitable for AMC/AIME practice?  I am desperate to improve my score after Tuesday's debacle.[/quote]

Definitely.  We have done a number of the problems in math team to prepare for the AMC12.  Most of the problems (from AoPS) that I have looked at are easier than a lot of AIME problems, but the discussions in AoPS are great for introducing techniques that are applicable to the AIME.  If you want to prepare for the AIME, I would recommend doing old AIMEs, which you can find at [url=http://www.kalva.demon.co.uk/aime.html]the kalva site[/url].

--Dan

[quote="tucsonlarry"]Is the Art of Problem Solving Book suitable for AMC/AIME practice?  I am desperate to improve my score after Tuesday's debacle.[/quote]

The Art of Problem Solving books and online courses are very much aimed at MathCounts/AMC/AIME/USAMO development.

I looked at the table of contents, and it looks like the books are mostly about introducing pretty common formulas and things, like i and the Pythagorean Theorem.  Do they talk about advanced things?  Would they help me do problems 18-25 on the AMC 12 and 10-15 on the AIME?

Yes; the book will certainly help you with those problems.  Knowing the formulas is only one tiny part of the battle in problem solving.  You also have to know how and when to apply them.

For example, problems or concepts necessary for the last dozen or so problems on this year's AMC-12A are in the volumes (such as using geometry for probability & the relationships between roots and coefficients and factoring/telescoping, etc).  All the geometry needed for the geometry problems of the 10A is addressed in volume 1.

plus, i think thats only the first book. the second book contains a lot more useful stuff.

[quote="churchilljrhigh"]are there going to be more art of problem solving books like for more advanced stuff? and other miscelaneous stuff not covered?[/quote]

It's hard to say.  The question is: can we find a better medium and create a model for utilizing it?

An answer is not easy.

[quote="MCrawford"]
It's hard to say.  The question is: can we find a better medium and create a model for utilizing it?

An answer is not easy.[/quote]

How long did it take you folks to write each of the first two volumes?

Sandor and I spent about 1.5 years total on writing the books (1.5 years for each of us, a total of 3 person-years).

I've always wondered..why aren't there any AIME problems in the books?

Good question.   Honestly, I don't even remember why we ended up not putting AIME questions in.  We asked the AMC for permission to use problems, and I can't remember if we didn't ask for AIMEs or if they chose not to give permission because there were so few AIME problems.  My guess is the former, since the AMC is generally very supportive of educational ventures, but I can't remember why we didn't include them.  I suspect we thought we had enough tough problems, but I'm not sure (and in retrospect wish we had included them).

Our math team would like to prepare for some competitions (March 17 and 20) as well as for the AIME (March 23).  If we were to order tomorrow (Feb 23), when can we expect the books to arrive (shipped to Los Angeles)?

We have two shipping options, priority mail and media mail.  We typically ship on Monday and Thursday afternoons - if you order priority mail by noonish today, you'll probably get the book Weds or Thurs.  If you order by media mail, I would expect next week some time (we are in San Diego, so media mail to LA would be considerably shorter than the typical 2 weeks).


Why is 2222^5555=3^5 (mod 7) = 5 (mod 7)? Somehow you use the fact that 5555=5 (mod 6)..

thanks!

By Fermat's Little Theorem, or precisely Eurler's Totient Theorem, $ a^{p\minus{}1}\equiv 1\pmod p$ for $ \gcd (a,p)\equal{}1$.  Here, $ p\equal{}7$ and $ a\equal{}2222$.  Hence, $ 2222^{5555}\equiv 2222^{6\cdot 925\plus{}5}\equiv 2222^5\equiv 3^5\equiv 5\pmod 7$.

Theorem: Let $ p$ be a prime number. If $ a \equiv c \pmod{p}$ and $ b \equiv d \pmod{p\minus{}1}$, then $ a^b \equiv c^d \pmod{p}$.

[hide="Proof"]Fermat's little theorem states that $ a^p \equiv a \pmod{p}$, so using it we can conclude $ a^b \equiv a^d \pmod{p}$.
[hide="Proof of FLT"]If $ a \equiv 0 \pmod{p}$ the result is obvious. Otherwise, we have $ \{a, 2a, \ldots (p\minus{}1)a\} \equal{} \{1,2,\ldots,p\minus{}1\}$ (setwise) when considered $ \mod p$. This implies that $ a \cdot 2a \cdots (p\minus{}1)a \equiv 1 \cdot 2 \cdots (p\minus{}1) \pmod{p}$, or equivalently, $ a^{p\minus{}1} \cdot 1 \cdot 2 \cdots (p\minus{}1) \equiv 1 \cdot 2 \cdots (p\minus{}1) \pmod{p}$. Because we are working in a prime modulus, we have division, so we can cancel stuff out. This leaves $ a^{p\minus{}1} \equiv 1 \pmod{p}$, implying that $ a^p \equiv a \pmod{p}$.[/hide]
Since $ a \equiv c \pmod{p}$, $ a \equal{} c \plus{} pk$ for some integer $ k$, so $ a^d \equal{} (c\plus{}pk)^d \equal{} c^d \plus{} d \cdot c^{d\minus{}1}(pk) \plus{} \ldots \plus{} (pk)^d \equiv c^d \pmod{p}$.[/hide]

[b]last reply[/b]. why do we write $ \{a,2a,...\}\equal{}\{1,2...,p\minus{}1\}$?

and why does this imply $ a \cdot 2a \cdot \dotsm (p\minus{}1) \equiv 1\cdot 2\cdot \dotsm  ( p\minus{}1) (\mod{p})$?

Los

The notation means that the sets contain the same elements ($ \bmod p$), just not necessarily in the same order.  Since the sets contain the same elements, the product of the elements in each set is equal (again, $ \bmod p$).


Draw a pentagon with each of its five diagonals. How many triangles are contained in the figure?
Draw a three by three grid (all of them are squares and they each have the same shape, placed side to side so that it forms a big square) with all of its diagonals. How many squares are contained in the figure? How many triangles are contained in the figure?

There are 35 triangles in the pentagon (7 types, 5-way symmetry)
There are 14 squares in the grid. I don't know about the grid triangles.

Correct, the second one: [hide]since there are three squares horizontally and three squares vertically there are total of 9 triangles. Now try to count all the triangles[/hide]


calculus
integration
Putnam
logarithms
calculus computations

The second is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].

[quote="Carcul"]$ I \equal{}\int^{1}_{0}\frac{\ln(1\plus{}x)}{x}\,dx \equal{}\int^{1}_{0}\left(\sum_{n \equal{} 0}^{\plus{}\infty}(\minus{}1)^{n}\frac{x^{n}}{n\plus{}1}\right)\,dx$
[/quote]

How is this so? Can you elaborate?

[quote="mdk"]How is this so? Can you elaborate?[/quote]

It's just an application of the power series $ \ln(1\plus{}x) \equal{}\sum_{n\equal{}1}^{\infty}(\minus{}1)^{n\minus{}1}\frac{x^{n}}{n}$.


[quote="Valentin Vornicu"][quote="langtupanda"]I see something like IMO, BMO, Aops, ISL... or something like that. What do they mean?[/quote]IMO = International Mathematical Olympiad

BMO = Balkan Mathematical Olympiad
AoPS = Art of Problem Solving
ISL = IMO ShortListed problems[/quote]
Can you tell me a little bit about BMO? and what is difference between ISL and IMO?

You may want to look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1053]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61819]there.[/url]

AoPS is just short for this Art of Problem Solving. If you say you are a AoPS user, you're a user of this site


geometry
parallelogram
trapezoid
perpendicular bisector
geometry unsolved
BritishMathematicalOlympiad

1. Triangle ABC has integer-length sides, and AC = 2007. The internal bisector of angle BAC meets BC at D. Given that AB = CD, determine AB and BC.

2. Let ABC be an acute angled triangle with AB>AC and angle BAC = 60. Denote the circumcentre O and the orthocentre by H and let OH meet AB at P and AC at Q. Prove that PO = HQ.

3. Let ABC be a triangle with AC>AB. The point X lies on the side BA extended through A, and the point Y lies on the side CA in such a way that BX = CA and CY = BA. The line XY meets the perpendicular bisector of side BC at P.
Show that angleBPC + angleBAC = 180.

∠BAC=60  =>  AH=AO  => △AQO≌△APH  =>PH=OQ

Denote D is the point that ABDC is a parallelogram
AX=AY CY=AB=CD  => X,Y,P,D are collinear
Denote Q,Z are the points that QBCY PDBZ are parallelogram respectively
∠BZP=∠BDP=∠BXP  => X,Z,B,P are concyclic
QX=QY  AX=AY  AQ=AQ  => △AXQ≌△AYQ  =>  QA⊥XY  => QA⊥BZ => AZ=QZ
Notice we have ZX=BP => ZQ=BP , ZQBP is a trapezoid => Z,Q,B,P are concyclic  => Z,Q,B,P,X are concyclic  =>△ZXQ≌△PBC
Thus we have ∠BPC+∠BAC=∠QZX+∠QBX=180


[hide]the problem is in the attachment
Thanks for helping me! :)[/hide]

[b]An additional problem 2 :) [/b]

For that second problem you already posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=298963]here[/url] . It is MOSP 2000 . You only need to consider two cases $ a\plus{}b\plus{}c\ge 2$ and $ a\plus{}b\plus{}c\le 2$ .

But a stupid question,What's MOSP?
 :blush:

I post these two mainly beacause find some new method such has $ xy\plus{}yz\plus{}zx\equal{}1$
Can somebody tell me some conclusions or general ideas about $ xy\plus{}yz\plus{}zx\equal{}1$

[quote="admire9898"]But a stupid question,What's MOSP?
 :blush: [/quote]

Mathematical Olympiad Saint Petersburg

if you guys want a different solution see here:
[url]http://www.mathlinks.ro/viewtopic.php?t=205029[/url]


Determine whether there exists a positive integer [i]n[/i] such that [i]n![/i] begins with 2005. Justify you answer.

Yes.

This problem (and similar problems) have been discussed before several times on this forum.

OMG ! Peter VDD . I cant believe you actually type it out !  :rotfl:  :rotfl:

:coolspeak:  :coolspeak:  :thumbup:
Very cool peter .nice

This is the best type of solutions, there are surely no mistakes ;).

that certainly a perfect answer :)
cannot find a better one :)
haha~ :rotfl:

Are you sure there aren't any typos Peter? ;) Think you should check.

By the way, I have seen that particular example elsewhere on the forum (although it wasn't written out completely), Peter is not the first one to find it.

There are no typos in copy-paste Arne ;) I think I finally found a use for maple! :P

Right. As you know you didn't solve the question.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=44064 and http://www.mathlinks.ro/Forum/viewtopic.php?t=23417 . Also I think Peter's solution is more in the spirit of the problem... [img]http://www.smiliemania.de/smilie132/00000087.gif[/img]

  darij

I know my solution is stupid Darij, but I think you'll agree with me that this problem has been posted and solved enough on the forum yet, and so have alike problems.

And hey, if someone writes that on an real life-olympiad you'll put up wide eyes right? :P


I know this has been posted before, but the font is all super small on my computer.  I changed the font size and the computer to the largest it can go and selected the big font size from the right corner of the forum, and everything is still too small to read.

[quote="rrusczyk"]How did you change the font size on your computer?  On IE you can go to View->Text Size.  You can also click the little letters on top of the main Forum page to adjust font size.[/quote]

i tried both.  This only happens on our school's computers

Is it Machintosh? And is the settings only changable by school admins?

That might explain a bit, for I believe there's also a font setting within Control Panel...I forgot, I'll get back to you on that one.


Find all such a, b $ \in Z$ that $ a^2\plus{}b$ and $ a\plus{}b^2$ are both prime.

Just so you know, I wrote a program to look for such pairs, and when $ a$ and $ b$ are upper-bounded by $ 1000$, there are approximately $ 23000$ such pairs. I'm going to guess there are an infinite number of such pairs; do you have additional restrictions?


calculus articles AMC USA(J)MO USAMO AIME trigonometry

So I'm a freshman, and I'm in an enriched (same thing as honors) Algebra 2 class right now.  And it's been one of the most boring, and most stupid (well, it can't be really that stupid if there are so many people that don't really get the material) classes I've taken.  And with other math classes, I've been okay with doing easy stuff I already know, because it wasn't really [i]that[/i] easy.  But this Algebra 2 class is put to the extreme.  And I'm kind of scared that the same thing will happen next year in Pre-Calculus.  And I don't really know what I should do.  I know that I shouldn't fall into the "Calculus Trap", and just try to skip Pre-Cal (I don't know Calculus), but what other options do I have?  I don't think I could take a class at my local University, because my dad teaches math there, and he says that I'd be better off just learning at my school.  But I don't know of anything else I can do that doesn't cost that much money (will schools pay for AoPS classes?  I know in Ohio, any PSEO class for any high schooler can be payed for by the school, but I dunno about AoPS).

Thanks

What's so bad about the calculus trap?  In math you can a) learn methods and b) learn skills.  One might as well progress with one or the other if they can't get both.  Plus, you'll need to know calculus anyways, as it appears on Putnam.

In the article, ruscyzk says he doesn't believe high schoolers [i]shouldn't[/i] study calculus in high school.  I think he mainly means you shouldn't just take the standard curriclular classes and completely ignore problem solving in math if math is what you want to pursue.  Also,  you want to keep learning math in school and not just waste a period of the day in probably one of (if not the) most favorite class in your schedule.  Being in Algebra II as a freshman, you have to do calculus either next year  (dunno if you want to skip or how your school deals with that ) junior year unless you're unnecessarily holding yourself back.  Plus, most people who do well on the usamo have already studied calculus  (and some much more) so they're continuing to difficult math anyway, not that the calculus is going to be directly useful a lot all the time.  Since you're into competition math it seems, especially the amc/aime/usamo-type tests, i don't believe you're in danger of completely ignoring problem solving.  So, in my opinion, just keep taking more advanced math, calculus or not, and keep yourself challenged in math in school.

[quote="themonster"]I was in a similar situation a year ago.  I was bored in my Algebra 2/Trig class, so I decided to go straight to Ap Calculus.  I felt that studying for the AIME and doing extra curricular stuff I had become familiar with most of the stuff taught in precal (and the small amount of other stuff I didn't really want to spend a lot of time on anyways).  I think that this worked fairly well for me, and I don't feel like I felt into the (maybe mythical) "Calculus trap," as I'm still very interested in problem solving and whatev.[/quote]That is the exact same thing that happened to me this year.  :D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15629&postorder=asc]This may be useful...[/url]

You should go ahead and take AP Calc if you think you're ready (there's really no reason you aren't). I've had experience in being in a math class that I really shouldn't have been in. I took the EPGY Precalculus course during my 8th grade year (at the same time I was taking trig) and it was quite challenging, as my math foundation was a little weak back then. I decided that I should retake the class at my high school. Big mistake. I basically wasted a whole year (well not completely, but we won't go into that) so looking back I definitely shouldn't have retook the class. To make up for it, I self-studied AP Calc AB that year and went into AP Calc BC and now I'm good. So that was really some unnecessary self-studying had I taken the initiative in the first place.


On the solution to problem 13 of the 2009 AIME II in the AoPSWiki, it states:
It can be shown that (sin pi/7)(sin 2pi/7)(sin 3pi/7)= (sqrt 7)/8

Sorry, I don't have LaTeX. Can anybody explain this statement? I'm guessing that it has something to do with the product-to-sum formulas but I can't figure it out. Thanks.

To show this,

[quote="thirdgrademustang"] $ (\sin{\frac{\pi}{7}})(\sin{\frac{2\pi}{7})(\sin{\frac{3\pi}{7})= \sqrt{\frac{7}{8}}}}$ [/quote]

just read the bottom part "[b]Computing the product of sines[/b]" http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_II_Problems/Problem_13
(with the roots of unity)


algebra polynomial calculus derivative topology

$x^5+ax^3+b$ has a double root $\neq0$. Find the relation between $a$ and $b$.

We require that $t^5+at^3+b=0$ and $5t^4+3at^2=0$ (since a polynomial having a double root implies that its derivative vanishes at the root)

The second condition implies that $t^2(5t^2+3a)=0$, or $t=\pm\sqrt{3a/5}$.  Plug back into the first condition to get your answer.

Oh, that's what I was missing. Time to tackle all those multiple roots problems I skipped. I can show the first part too!

$p(x)=(x-r)^2*q(x)$ if it has a double root at $x=r$.
$p'(x)=2(x-r)q(x)+(x-r)^2q'(x)$
$p'(r)=2(0)q(r)+(0)^2q'(r)=0$

Okay, so that was a bit easy, but  :D nevertheless. Thanks blahblahblah.

Hey, no problem.  I'm a bit fuzzy on the details here, but in various algebra texts (Hoffmann and Kunze), a sort of formal derivative (which coincides with the usual derivative, but doesn't use metric-space notions like limits) is defined for polynomials in order to be able to characterize multiple roots and whatnot.

And of course you can make it a two way statement - A polynomial $p(t)$ has a double root at $p=a$ if and only if $p(a)=p'(a)=0$ (with an obvious generalization to roots of multiplicity $n$)  Try to prove that direction as well!

Okay, I think I have it.
$p(a)=0$ implies $p(x)=(x-a)q(x)$ for some polynomial $q(x)$.
$p'(x)=(x-a)q'(x)+q(x)$, so $p'(a)=0$ implies $q(a)=0$ and thus, $(x-a)|q(x)$, so $(x-a)^2|p(x)$.
Yay!

i wonder if one could find a solution if we consider $x^5+ax^3+b = (x-x_0)^2(x^3+px^2+qx+r)$ without calculus...


Find all polynomials $p$ such that $p(x+2)=p(x)+6x^2+12x+8$.

Ryan, please do not just post answers.  That does no help to anyone.

Also, jimhu, you didn't examine it for polynomials with degree greater than 3 so your solution is incomplete.

[quote="Farenhajt"]joml88, there can be no polynomials of a higher degree since $a_n(x+2)^n-a_nx^n$ is of a degree $n-1.$[/quote]

Enough said. :)

Yes, I know.  But you should still mention little things like that as they are very important to the solution.


A real number is placed in each square of an infinite chessboard. Numbers in the same row or column at a distance 1982 apart are equal. Each number is the average either of its two horizontal neighbours or of its two vertical neighbours. Prove that either all the numbers in each row are equal or that all numbers in each column are equal.

Pierre.

Let's assume we have shown that in such a table (we work only on an $n\times n$ table, which can be thought of as being a thorus; here $n=1982$) the maximal number must fill an entire line or column (and the same for the minimal number).

Then we have several lines, each one containing only the maximal number or only the minimal number, and we must show that in the non-trivial case when the numbers in the table are not all equal, each line contains a single number.

If between two lines containing the maximal and minimal number we can find the minimal number again somewhere, consider one such number which is closest to the line with the maximal number in it. Then the minimal number must fill the line containing the respective minimal number. On the other hand, if between two lines containing the minimal number, there are numbers different from the minimal one, we can apply the reasoning we used to prove that the maximal number must lie on a line or column to show that there's an entire line consisting of the maximal number between the lines with minimal number.

In this way, we obtain new lines with the same number until we run out of lines (each time, the minimal number may change its value to the minimal number between the two lines we're working with, not the minimal number in the entire table, and the same for the maximal number).

It's clear from here (I hope; it was kind of hard to explain :)) that all we need to do now is prove that the maximal number must fill an entire line or column. This isn’t hard to do, but even harder to explain, so I’ll leave it out :).


$S=\sum\sqrt{\frac{a^4+b^4}{1+ab}}$
Use Cauchy-Swachtz We have:
$S.S(\sum{(1+ab)})(1+1+1) \ge (\sum{\sqrt[4]{a^4+b^4}})^4$
But $\sqrt[4]{8(a^4+b^4)} \ge a+b$
And
$(a+b+c)^2 \ge 3(ab+bc+ca) \ge \frac{3(3+ab+bc+ca)}{2}$
Then $3S^2 \ge \frac{3}{\sqrt{2}}(\sum{\sqrt[4]{a^4+b^4}})^2$
Easy to have result:
$\sum\sqrt{\frac{a^4+b^4}{1+ab}}\ge \sum\sqrt[4]{\frac{a^4+b^4}{2}}$
It's more tronger than old ineq.

In the fourth line, do you want to say $\sqrt[4]{\frac{a^4+b^4}{2}} \ge \frac{a+b}{2}$?

Oh yes Sorry,small type.
I will edit it.
Thanks.


ratio
geometry
parallelogram
geometry unsolved

Let $ A_1A_2A_3A_4$ be a convex quadrilateral.

Midpoint of $ A_1A_2$ is $ B_1$.
Midpoint of $ A_2A_3$ is $ B_2$.
Midpoint of $ A_3A_4$ is $ B_3$.
Midpoint of $ A_4A_1$ is $ B_4$.

$ C_1$ is an intersection point of $ B_1A_4$ and $ A_1B_2$.
$ C_2$ is an intersection point of $ B_2A_1$ and $ A_2B_3$.
$ C_3$ is an intersection point of $ B_3A_2$ and $ A_3B_4$.
$ C_4$ is an intersection point of $ B_4A_3$ and $ A_4B_1$.

Find the range of $ \frac{[C_1C_2C_3C_4]}{[A_1A_2A_3A_4]}$

[quote="vittasko"]A conjecture about the wanted ratio is $ R \equal{} \frac {(C_{1}C_{2}C_{3}C_4)}{(A_{1}A_{2}A_{3}A_{4})} \leq \frac {1}{5},$ but unfortunately, I have not in mind any proof about it.

The equality holds, when $ A_{1}A_{2}A_{3}A_{4}$ is a parallelogram ( easy to prove ).

Two other well known results, about the configuration of the proposed problem :

[b][size=100](1)[/size][/b] - $ (A_{1}C_{1}C_{4}B_{4}) \plus{} (A_{3}C_{3}C_{2}B_{2}) \equal{} (A_{2}C_{2}C_{1}B_{1}) \plus{} (A_{4}C_{4}C_{3}B_{3})$

[b][size=100](2)[/size][/b] - $ (A_{1}B_{1}C_{1}) \plus{} (A_{2}B_{2}C_{2}) \plus{} (A_{3}B_{3}C_{3}) \plus{} (A_{4}B_{4}C_{4}) \equal{} (C_{1}C_{2}C_{3}C_{4})$

Kostas Vittas.[/quote]

Thanks, I have found it at [url]http://www.ms.uky.edu/~carl/coleman/coleman21.html[/url].
But I don't understand why we can set $ A_1\equal{}(0,1)$ $ A_2\equal{}(0,0)$ $ A_3\equal{}(1,0)$ in [url]http://www.ms.uky.edu/~carl/coleman/coleman21.html[/url].


Find all polynomials $f$  such that  $f(x^{2})+f(x)f(x-1)=0$

$f(x)=-(x^{2}+x+1)^{n}.$ Just factorize over $\mathbb{C}$ after observing that if $f(\alpha)=0$ then $\alpha^{m}=\alpha$ for some $m$ and $1+\alpha$ is also a root of unity.


Source: Massachusetts Mathematics League March 2003

A circle is inscribed in an equilateral triangle, and the circle is circumscribed about another equilateral triangle. If the difference of the two equilateral triangles is $144\sqrt 3,$ what is the circumference of the circle in terms of $\pi$?

If you join the three points where the circle touches the triangle you'll get second equilateral triangle. If the first triangle has a side with measure: $2a$, the sides of the second triangle will measure: $a$

The difference of areas would be: $\frac{3}{4}a^2\sqrt{3}$
and $a=8 \sqrt{3}$

and it's easy to see the radium is: $8$

So the circunference of the circle is $2\pi (8)=16\pi$

[hide]Since the difference in areas is $144\sqrt{3}$, the area of the inner triangle is equivalent to the 3 triangles that surround it if one joins the points of the circle that are tangent to the larger triangle.  So, we find that the area of the smaller triangle is ($\frac{144\sqrt{3}}{3} = 48\sqrt{3}$.  The side length can then be found by $\frac{s^2\sqrt{3}}{4} = \sqrt{192} = 8\sqrt{3}$.  To find the radius I used the formula for the incenter.  The outer triangle has a side length of $16\sqrt{3}$ and thus a semiperimeter of $24\sqrt{3}$.  The total area is $192\sqrt{3}$, thus since sr = area, the radius is 8 and the circumference is $16\pi$[/hide]

[hide]make the smaller equi. triangle touch the sides of the bigger equi. triangle. The smaller equilateral triangle is 1/4 the area of the larger. Dividing the difference $144\sqrt{2}$ by 3, we get the smaller triangle's area = $48\sqrt{3}$. We get the side length = $8\sqrt{3}$, and therefore the altitude $=12$. Since in a equilateral triangle the altitude is the same as the median, and the centroid divides a median into a 2:1 ratio, the distance from a vertex to the centroid is 8. Since the circumcenter is the same point as the centroid, the radius is 8. Therefore the Circumference is $8*2\pi=16\pi$[/hide]