email
geometry
analytic geometry
LaTeX
AMC
AIME
rectangle

Sorry. I can't think of a different name. Here's how it works. There is a series of ten problems, and you email me your solution. I score the solution on the following rubric

2 points-correct answer( there could be more than one part, so you can get 1)
0-3 points-elegance of the solution, did you take the really long way, or did you think of a simple, short solution
0-3 points-all steps are shown and clear, everything explained clearly
0-2 points-factual information-is everything written true, or are you making stuff up to get a phony solution? Also, if there's a hole, points are taken from here. (maybe somewhere else as well)
2 points if the solution is submitted in 24 hours or less
1 point  if the solution is submitted in 48 hours or less, more than 24 hours
0 points if it takes more than 48 hours, less than 72

You have 72 hours to do this problem. pm me your solution.

m and n are positive integers such that m>n. point A is defined as point (m,n,). A is reflected over y=x to form B, B is reflected over the y-axis to form C, C is reflected over the x-axis to form D, and D is reflected over the y axis to form E. The area of pentagon ABCDE is 451. Find the coordinates of point A, B, C, D, and E.

Good luck. You have until 11:22 AM on Monday to pm me.

There are going to be five problems, more if neccessary, and we're gonna do some kind of bracket or playoff at the end.

Is anyone going to bother to do this? I'll give everyone a hint.
reflecting (m,n) over y=x just makes it (n,m)
Finding the area of the pentagon isn't as hard as it seems, because it breaks up nicely.
Also, what are the factors of 451? (won't use till end)

I looked at it, but I can't solve it...the shoelace formula says it can't be with integers...I probably am wrong though

edit: I am wrong

you don't need the shoelace formula. Just draw a diagram and see that the pentagon is really easy to find the area of

:( <= he's mad at you because you're lazy

Come on mathgeniusx. It's practice for MathCounts. It's quite easy. It took longer for me to type the solution in LaTeX and email it than to solve it.  :lol:

That's going to be terrible to type up in latex :wacko: . Can I just send you a picture? :)

[quote="kyyuanmathcount"]Wow, this was 2002 AIME, I could just dig that out if I wanted.[/quote]
It better have been the first question. Geometry, just like this year's AIME, I at least. This year's first question was easier, I think. Just Pythagorean twice. :D

3 solutions are in so far, and they're all really good. If you just write the first step, I'll give you some amount of points

The next questions will probably be easier, and you get credit in your index for making feeble attempts at questions.

index-  (points)/(feeble attempts) + 2(feeble attempts)

I don't count it as a feeble attempt if its like, "2+2 is 4 so the answer's 31" by the way.

17 hours left for round 1. Make sure you check back tomorrow, because round 2 will be posted by 6pm.

oops! posted this in the wrong place. I'll post it here to be sure
ere is the solution for problem 1

A(m,n)
B(n,m)
C(-n,m)
D(-n,-m)
E(n,-m)

Break the pentagon up into triangle ABE and rectangle BCDE. The height of the triangle is (m-n), the base length is 2m. Therefore the area is 1/2(m-n)2m = m^2-mn.
The rectangle is 2m by 2n, so the area is 4mn
Therefore, the pentagon's area is m^2+3mn, which factors into
m(m+3n).
Since m(m+3n) = 451, m and n are positive integers, m and m+3n must be 11 and 41, respectively, to make m and n both positive integers. Therefore,

m=11
m+3n=41
n=10

Plugging this in to the above,

A(11,10)
B(10,11)
C(-10,11)
D(-10,-11)
E(10,-11)

Those who submitted solutions:(out of 12)

lotrgreengrapes-12 points
ehehheehee-12 points
mathgeniuse^ln(x)-10.5 points
fireemblem13-9.5 points
Captain Sugar-5.5 points

If I pmed you with a different score, I messed up and these are your real scores
the indexes are just 2 more than these scores

Now, for problem 2 (much easier)

What is the only positive integer n for which 4n-6 is prime? Show that it is indeed the only one.
Hints:
What characteristic will 4n-6 always have?
What characteristics do most primes have?
Is their an exception?

No, this is good, can you please keep it to one thread?

It kinda clogs up the forum when you have one for every round, thanks!

I think you did see me.  We just didn't recognize each other as I saw four mathletes from Michigan...

I got here by winning by margins of 1 so you will crush me

(As a moderator, you can always change his posting of results  :P)

All right. It comes down to this. ehehheehee versus kyyuanmathcount. Michigan vs. Ohio. The east vs. a little further east. I think they border. Allen Yuan vs. Dennis Tseng.

1. TAYLOR HICKS recently became the new American Idol. (w00t) In the finale of American Idol 5, he sung his new single, which took 3 minutes and 12 seconds. 2/3 of the time, he was actually singing (or talking). Each word in the song has an average of 2.1 syllables, and he also said the words "Soul patrol" three times, "Thanks to my family and the fans," once, and "Everybody up, I'm living the American Dream" once. (Note, these quotes do NOT count as words in the actual song.) Taylor Hicks can sing or talk 90 syllables per minute. How many words are in the SONG? Round to the nearest whole number of words. (not very realistic)
2. In the geometric sequence 2, t, a, y, l, o, r, h, 8, find the value of a, expressed in simplest radical form.
3. Kelly, Ruben, Fantasia, Carrie, and Taylor are playing a coin-flipping game. Kelly flips first. To win, a player must flip the same thing that the person directly before him or her flipped. What are the odds that Taylor wins?
4. Quadrilateral IDOL is inscribed in circle P. ID=5, DO=6, OL=9, IL=10. Find the area of the quadrilateral, expressed as a decimal to the nearest hundredth.
5. Katharine McPhee is depressed that she didn't win. She decides to subdue her depression by walking around a cartesian plane. She starts at the origin, and she plans to get to (5,6) via (4,3). If she can only travel along integer gridlines, how many ways can she travel? She can only move up and to the right.
6. Find the coefficient of x^5*y^2 in (x+y)^7.
7. The season is redone. However, I am in charge. Taylor is my favorite, so he has to win. Chris Daughtry is one of my favorites, so he has to make it to the final four again. Also, since I hate Mandisa and Paris Bennett, they have to be the first 2 to go, in either order.(there are 12 at the beginning) How many different orders of finishes are there? There are no ties.

ehehheehee is Dennis Tseng? :O I'm the one who placed EXACTLY below you because of tiebreakers. What was your breakdown? Mine was 16/14

same as you, but since I messed up numbers 1,3 which are 3rd,1st easiest questions on the test (to show my stupidity).  I was also too stupid to realize on #24 that the lowest point must be on the intersection or else it is -/infty...I think I should not think about the test anymore and move on  :roll:

Good Job eheeheehee.....

Carl, you are going to MathPath.  That is good.  I am going too.  I automatically get to go without taking the placement test since this is my second year at this camp.

They haven't told me anything yet. They gave me a 10 day extention because of mathcounts, so I sent in my test a few days ago.

I'm expecting to hear from them on tuesday.

Yeah, you should hear from them around that time.

[quote="jclarkemathL314159"]whoa that was fast. And the winner is......
EHEHHEEHEE
scoreboard:
ehehheehee def. kyyuanmathcount, 5-4
#3 asked for odds, not probability.

1. 71 (neither got it right)
2. 2root2 (both got it right)
3. 2:29 (both wrote the probability, 2/31)
4. 51.96 (both got it right)
5. 130 (both got it right)
6. 21 (both got it right)
7. 241920 (kyyuanmathcount probably forgot to multiply by 2)

It was a flat out tie until the last question. Allen wrote 120960, and Dennis wrote 241920.

Congrats to everyone, and keep an eye out for superduper II, which will be coming out sometime between june 15 and august 15. (depending on whether i have time at mathpath (assuming i made it)) ;)[/quote]

I was screaming to myself.  WRITE THE ODDS!!!!, then I forget  :rotfl: .

Yeah, #7 is characteristic of myself.  If I can make two stupid mistakes on the AIME (I only answered 7), I can probably forget to multiply by two on a MATHCOUNTS question.

[quote="jclarkemathL314159"]

Congrats to everyone, and keep an eye out for superduper II, which will be coming out sometime between june 15 and august 15. (depending on whether i have time at mathpath (assuming i made it)) ;)[/quote]

Reading helps.

Congrats, eheheheheeeheheheheheheehehehehehehe.P.S. I was 10 yrs. old when I got kicked out from the comp. but I turned 11 on the 25th.

i didn't kick you out.  :rotfl:
You just lost.  ;)

Yeah. When you lose, you get kicked out. ;)


Can someone show me the general proof of the AM-GM inequality?

It is obvious by Rearrangement or Jensen's.

In fact it can be greatly generalized into a form equivalent to Jensen's, which I call the [b]Function-Mean Inequality:[/b]

Let the function-mean $M(f)$ of some reals $x_1, x_2, ... x_n$ and of a monotonically increasing or monotonically decreasing continuous function $f(x)$ be defined as

$M(f) = f^{-1} \left( \frac{ \sum_{k=1}^{n} f(x_k) }{n} \right)$

(For example, $M(x)$ is the Arithmetic Mean, $M(\ln x)$ is the Geometric Mean, $M(x^p)$ is the Power Mean.)  Then

$M(f) \ge M(g)$

iff

$f( g^{-1}(x))$

is convex.

You are not a new member and so you should know how to use the SEARCH function. For example, I found this:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48575]http://www.mathlinks.ro/Forum/viewtopic.php?t=48575[/url]

This should be locked, this is discussed plenty of times and can be found from google with lots of results, you just have to put some effort into it and perhaps save other people's time by not repeating the same discussion???

It will be more interesting if we change the topic to "How many ways of proving AM-GM inequality do you know?"

I know about 10 different ways.  :lol:


Is it possible to modify the classroom software so that it parses the file containing the running transcript and outputs it to you as soon as you join a class? It's really useful when someone accidentally closes the window or loses their internet connection in the middle of a class. I know that you can still check the transcript in the link under 'My Classes,' but it would be easier if we had this function.

We are going to implement this feature in the new classroom, which should be available in a couple weeks.


inequalities
rearrangement inequality
combinatorics unsolved
combinatorics

It's a TST; I expect that they give harder problems than a trivial application of Holder.  Even the rearrangement inequality gives the same result.

[quote="al.M.V."]I think the answer is 162: 3 groups of (6*5*1=)30 and 3 groups of (4*3*2=)24. Is this right zhaoli?[/quote]
Yes,you are right. :)

From the conditions of this problem, we know that
(1)  $a_i$, $i=1, 2, 3, 4, 5, 6$, are all positive integers, and pairwise distinct. So does $b_i$, $i=1, 2, 3, 4, 5, 6$; and $c_i$, $i=1, 2, 3, 4, 5, 6$.
(2) They are finite permutations of $1, 2, 3, 4, 5, 6$, there must exist minimum value.

[quote="hardsoul"]so the only possibility is $a_1b_1c_1=5\cdot 3\cdot 1=15$[/quote]

Why is that, hardsoul? I fail to see why  :(  could you explain better?

actually, i think there are more posibilities (like 3,3,3). only a few are under 12. anyway, this problem is much too "computational". i was hoping you had an argument which avoids all that

[quote="perfect_radio"]actually, i think there are more posibilities (like 3,3,3). ..[/quote] It was the last one :D .Although I don't know whether there's another way to avoid comput.,I think that to check if these values sat. the ineq. is not so difficult...

[quote="toumaf"][quote="hardsoul"]
Now suppose that $\sum_{i=1}^{6}a_{i}b_{i}c_{i}=161$ [/quote]

This is a flaw in your proof. There could be an arrangement of $a_i b_i c_i$ with sum less than $162$, and none of sum $161$.

---

Firstly we remark that the product of all terms of the sum is equal to $6!^3$.
Then we see that the numbers that can be written as a product of three positive integers at most $7$ are
\[ 1,2,\cdots,20,24,25,27,30,32,36,\cdots,216.  \]
If one of $a_i b_i c_i$ is less than 24, then it is at most $20$, thus
\[ \sum \left(a_i b_i c_i \right) \geq 20 + 5 \sqrt[5]{\frac{6!^3}{20}} > 20 + 5*28,4 = 162.  \]

If one is at least $36$ then
\[ \sum \left(a_i b_i c_i \right) \geq 36 + 5 \sqrt[5]{\frac{6!^3}{36}} > 36 + 5*26 > 162.  \]

Thus we just have to study the case where all the $a_i b_i c_i$ are in $E=\{24,25,27,30,32\}$.
We want to write $6!^3$ as a product of six numbers of $E$.
These numbers will have to contain three times the factor $5$, so either we write
$6!=30*30*30*x*y*z$ or $6!^3=25*30*w*x*y*z$.

In the first case, to minimize the sum $30+30+30+x+y+z$ we take $x=y=z=24$, which yields $30+30+30+24+24+24=162$.
In the second case $wxyz=2^11*3^5$, and $w$,$x$,$y$ and $z$ are in $\{24,27,32\}$.
We can easily investigate this case, which gives $24*24*27*32=2^11*3^5$, thus finally
$24*24*25*27*30*32=6!^3$ and $24+24+25+27+30+32=162$.

The minimal value of the sum is thus $162$, with a lot of equality cases.[/quote]

Thanks,I see. :)


function
trigonometry
factorial
calculus
algebra
functional equation
calculus computations

Is $ y \equal{} x^2$ the only smooth convex function that passes through all the points $ (k, k^2)$, where $ k$ is any integer?

No. For one simple example, $ f(x)\equal{}x^2\plus{}\frac1{\pi^2}\sin(\pi x)$ also works, as $ f''(x)\equal{}2\minus{}\sin(\pi x)>0$.

Alright, thanks.

The reason I ask is that the Gamma function, defined to be the "extension" of the factorial, may or may not be (to my knowledge, at least) the only smooth convex function that passes through the points (n, (n - 1)!) (over the positive reals, at least.) If there are other such functions, why do we use the Gamma function in particular as an extension of the factorial?

I asked this for $ y \equal{} x^2$ as a simpler case.

By the way, if the solution is over my head (I'm not very knowledgable in calculus; I am comfortable with monovar, and that's about it), I'm sorry if I can't understand it. It's just a question that's been bugging me.

[quote="Zhero"]The reason I ask is that the Gamma function, defined to be the "extension" of the factorial, may or may not be (to my knowledge, at least) the only smooth convex function that passes through the points (n, (n - 1)!) (over the positive reals, at least.) [/quote]
The Gamma function is the unique such function that has the additional property of being [url=http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem]log-convex[/url].

Sorry for the double-post, but is $ x^2$ also the only smooth log-convex function that passes through the points $ (k, k^2)$ for all positive integers k (over the positive reals)? The counterexample provided by jmerry is not log-convex.

No, it's not, because $ x^2$ is not log-convex; it's actually log-concave except at zero.

I don't think merely interpolating those points and being log-convex characterizes the Gamma function. You could still add little wiggles between integers, in the general spirit of jmerry's example (although you'll have to work harder on the details of the example).

If you follow t0rajir0u's link, you'll get a theorem which has one additional extremely strong hypothesis - that the function satisfy the gamma recursion $ f(x\plus{}1)\equal{}xf(x).$

It's that algebraic condition (the functional equation) that's really at stake.


geometry
inradius
incenter
geometry unsolved

Prove that:
$IG^2 = r^2 + {1\over 36}\left( 5 \sum_{cyclic}a^2 - 6\sum_{cyclic}ab\right)$

Hereby, $r$ is the inradius, $I$ is the incentre, $G$ is the centroid, $a$, $b$, $c$ are the side lengths of the triangle.

Peng Shi

I don't think it's hard, but I will definitely not solve it here :D.

It looks like all you need to do is write $\overrightarrow{IG}=\frac{\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}}3$, and then apply the dot product and use the fact that $IA^2=r^2+(p-a)^2$ and the like (where $p$ is the semipertimeter).


Does the equation:

                        1/a + 1/b + 1/c + 1/(abc) = m/(a + b + c)

 have infinitely many solutions in positive integers a, b, c
 for any positive integer  m?

Yes. m=12
   Just construct a sequence of solutions.

I think he asked whether if it's true for ALL positive integers, not for SOME positive integer.  :D

of course not :? m=1 gives that the LHS is strictly greater than the RHS...

This has already been posted (a long time ago). The wording of the problem makes it sound as if al.M.V. is right, but it's just bad wording, I think.

Here you can find the solution booklet as distributed at IMO:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15588]IMO 2002 ShortList[/url]

I've looked the question up on Kalva, their solution:

[quote][b]Problem N4[/b]

Does the equation 1/a + 1/b + 1/c + 1/(abc) = m/(a + b + c) have infinitely many solutions in positive integers a, b, c for any positive integer m?

[b]Answer[/b]

Yes, eg m = 12.

[b]Solution[/b]

Consider m = 12. Note that 1/a + 1/b + 1/c + 1/abc - 12/(a+b+c) = p(a,b,c)/(abc(a+b+c)), where p(a,b,c) = a2(b+c) + b2(c+a) + c2(a+b) + a + b + c - 9abc. We can regard this as a quadratic in a. The product of the roots is (b2c + c2b + b + c)/(b+c) = (bc + 1), so if one root is a, then the other is (bc + 1)/a. Moreover if a <= b <= c, then (bc+1)/a > c(b/a) >= c. Thus provided (bc+1)/a is integral, the solution (a,b,c) leads to another solution (b,c,(bc+1)/a).

We can start with the obvious solution (1, 1, 1). So define the sequence a1, a2, a3, ... defined by a1 = a2 = a3 = 1, an+3 = (an+2an+1 + 1)/an. Provided we can show that (an+2an+1 + 1)/an is always integral, we are home, because any three consecutive terms of the sequence give a solution.

It is convenient to enlarge the induction hypothesis to: an divides an+2an+1 + 1 (1); an+1 divides an + an+2 (2); and an+2 divides anan+1 + 1 (3). It is obviously true for n = 1. Suppose it is true for n.

(1) implies an and an+1 are coprime. We have an+2an+3 + 1 = an+2(an+1an+2 + 1)/an + 1 = k/an, where k = (an+2 + an) + an+1an+22. By (2) an+1 divides k, and by (1) an does, so they both do, which establishes (1) for n+1.

Similarly, an+1 + an+3 = an+1 + (an+1an+2 + 1)/an = (an+1an + 1) + an+1an+2 and an+2, an are coprime by (1), so an+2 divides an+1 + an+3, which establishes (2) for n+1. Finally, (3) for n+1 is obvious.

So we have established that a1, a2, a3, ... is an integer sequence. A trivial induction shows it is strictly increasing from a4 onwards, so it generates infinitely many solutions.[/quote]

So they clearly agree with treegoner...


function
integration
number theory
number theory unsolved

I was wondering what's the difference between the following two "mapping" symbols:
$f: A \to B$ vs $f: A \mapsto B$

Also, in how many and in what ways can you read the following notation:
$f \circ g$

$\mapsto$ means morphism but $\to$ means just function
$f \circ g =f(g(x))$ here there is no confusion
if i am wrong somebody please correct me

No, I meant how do say the notation $f \circ g$, not its meaning.  :)

Why $(\frac 12 )!=\sqrt{\pi }?$
It is easy to calculate
\[{ (\frac 12 )!=\frac 12 \Gamma(\frac 12 )=\int_0^{\infty } \frac{1}{\sqrt x }e^{-x}dx=\int_0^{\infty } exp(-t^2}dt =\frac{\sqrt{\pi }}{2}.  \]  :P

f composed with g.
or f applied to g  i think

Maybe you should continue the discussion here : http://www.mathlinks.ro/Forum/viewtopic.php?t=88637 , since I don't think the Number Theory forum is appropriate for this kind of questions. :)


invariant
inequalities
inequalities open

solution1 made by 123eeee.
solution2 made by pppppqqqqq(very nice)

[quote="lzk05"]Let $ a,b,c$ are positive real numbers, prove that

$ \sum{\sqrt {a^3 \plus{} abc}}\geq\sum{\sqrt {bc(b \plus{} c)}}.$[/quote]$ \sum{\sqrt {2a(a^2 \plus{} bc)}}\equiv\sum{x\sqrt {2x^4 \plus{} 2y^2z^2}}$

$ \geq\sum{x(x^2 \plus{} yz)} \equal{} x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz$

$ \geq\sum{yz\sqrt {2y^2 \plus{} 2z^2}}\equiv\sum{\sqrt {2bc(b \plus{} c)}},$

the last inequality see : http://www.mathlinks.ro/viewtopic.php?t=6317

By the way,

$ p\sum{\left(a^3 \plus{} abc\right)^p}\geq p\sum{\left(b^2c \plus{} bc^2\right)^p}$

holds for $ \minus{} 3\leq p\leq 1.$

See also : http://www.mathlinks.ro/viewtopic.php?t=194103


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

If $ A$ is noetherian, $ B$ a finitely generated $ A$-algebra which is flat over $ A$, then $ Spec(B) \to Spec(A)$ is open.

just a special case, but perhaps we can generalize the idea?

assume $ B\equal{}A[x_1,...,x_n]$. if $ b \equal{} \sum_{\mu} a_{\mu} x^\mu \in B$, then clearly $ D_B(b)$ is mapped to the union of the $ D_A(a_\mu)$. conversely if $ \mathfrak{p} \in D_A(a_{\mu})$, the exact sequence $ 0 \to \mathfrak{p}[x_1,...,x_n] \to B \to A/\mathfrak{p}[x_1,...,x_n] \to 0$ shows that $ \mathfrak{p}[x_1,...,x_n]$ is a prime ideal of $ B$. clearly this lies in $ D_B(b)$ and is mapped to $ \mathfrak{p}$. thus $ Spec(B) \to Spec(A)$ is open.


Minneapolis-St. Paul International Airport is $ 8$ miles southwest of downtown St. Paul and $ 10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$ \textbf{(A)}\ 13\qquad
\textbf{(B)}\ 14\qquad
\textbf{(C)}\ 15\qquad
\textbf{(D)}\ 16\qquad
\textbf{(E)}\ 17$

Airport and 2 towns form a right-angled triangle, so the length is $ \sqrt{8^2\plus{}10^2}$ which is closest to 13
A


trigonometry
Vieta
algebra unsolved
algebra

It smells cosines here ...  ;)

First of all, from the radical we are sure that $x_n>0, \ \forall n$

$x_{n+1} = \sqrt{\frac{2x_n}{x_n+1}} \Leftrightarrow$

$x^2_{n+1} = \frac{2x_n}{x_n+1} \Leftrightarrow$

$(\frac{1}{x_{n+1}})^2 = \frac{x_n+1}{2x_n} = \frac{1}{2} + \frac{1}{2x_n} \Leftrightarrow$

$2(\frac{1}{x_{n+1}})^2 = 1 + \frac{1}{x_n}$

and we know the formula

$2cos^2 \theta = 1 + cos2\theta$

set ${\theta}_1 = \frac{\pi}{4}$ and ${\theta}_{n+1} = \frac{1}{2}{\theta}_n, \ \forall n\geq 1$

$\Rightarrow {\theta}_n = \frac{1}{2} \cdot \frac{\pi}{2^n}$

now, notice that $\frac{1}{x_1} = cos{\theta}_1 = \frac{\sqrt 2}{2}$

and $2cos^2 {\theta}_{n+1} = 1 + cos{\theta}_n, \ \forall n$

so we have $\frac{1}{x_n} = cos{\theta}_n , \forall n \Rightarrow$

$\prod_{n=1}^\infty x_n = \frac{1}{\prod_{n=1}^\infty cos{\theta}_n} \ \ \ (1)$

So we need first to find the product $cos(\frac{\pi}{4})cos(\frac{\pi}{8})cos(\frac{\pi}{16})cos(\frac{\pi}{32}) \cdots$

The missing term ${\theta}_0$ of the above sequence $\{{\theta}_n\}$ is of course ${\theta}_0 = \frac{\pi}{2}$. We have:

$sin {\theta}_0 = 2 sin {\theta}_1 cos {\theta}_1$
$sin {\theta}_1 = 2 sin {\theta}_2 cos {\theta}_2$
$\cdots$
$sin {\theta}_{n-1} = 2 sin {\theta}_n cos {\theta}_n$

We multiply all these equations and we get:

$sin {\theta}_0 = (2^n sin {\theta}_n) \cdot cos {\theta}_1 cos {\theta}_2 \cdots cos {\theta}_n \Rightarrow$

$\frac{1}{cos {\theta}_1 cos {\theta}_2 \dots cos {\theta}_n} = 2^n sin {\theta}_n$

$2^n \cdot sin {\theta}_n = 2^n \cdot sin (\frac{\pi}{2^{n+1}}) = \frac{1}{2} \cdot 2^{n+1} \cdot sin (\frac{\pi}{2^{n+1}}) \to \frac{\pi}{2}$

because it is [url=http://www.themathpage.com/aTrig/pi.htm#rule]known[/url] that the sequence $k \cdot sin \frac{\pi}{k}$ tends to $\pi$

So, from $(1)$ we get $\prod_{n=1}^\infty x_n = \frac{\pi}{2}$

[b][color=red]PS[/color][/b]
This formula is strongly related to [url=http://documents.wolfram.com/v5/Demos/Notebooks/CalculatingPi.html]Vieta's Formula[/url] , which is

$\frac{2}{\pi} = (\frac{1}{2} \sqrt{2}) \cdot (\frac{1}{2} \sqrt{2+\sqrt{2}}) \cdot (\frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{2}}}) \cdots$

Indeed, we can see that the terms of this formula are the terms of $\{\frac{1}{x_n}\}$ of our sequence
.....


calculus
integration
calculus computations

Looks like you should be able to do partial fractions on all of these. They'll come out as (somewhat complicated) logarithms.

hello, to your first integral. You have to make the following ansatz:
$ \frac{x}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$
with $ A,B,C \in \mathbb{R}$. The result is $ 2\ln|x+2|-\frac{3}{2}\ln|x+3|-\frac{1}{2}\ln|x+1|+C$.
By the second one we have $ \frac{x}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$. The result is ${ -\frac{1}{2(x+1)}+\frac{1}{4}\ln|x-1|-\frac71}{4}\ln|x+1|+C$.
For the last one we have $ \frac{x^2}{(x-1)^2(x-2)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}$
The result is $ -\frac{4}{-2+x}-\frac{1}{x-1}+4\ln|x-1|-4\ln|x-2|+C$
Sonnhard.


What is the greatest two-digit number that is the product of exactly two prime numbers?

$ 5*19\equal{}95$

Also, this is the 1337th topic...WOOT! :rotfl:  :ninja:  :rotfl:

GameBot is a bot that posts ftw and alumnus problens


Find the smallest positive integer which can be expressed as the sum of distinct positive integers a, b and c such that $ (a\plus{}b), (b\plus{}c)$  and $ (c\plus{}a)$  are perfect squares.

If we look at perfect squares, we can see that they increase in increments of certain pattern:

$ 2^2\equal{}4, 3^2\equal{}9, 4^2\equal{}16, 5^2\equal{}25, 6^2\equal{}36, 7^2\equal{}49,...$

It increases by $ 5,7,9,11,13,...$
Suppose that $ a<b<c$ without loss of generality.
As perfect squares increases by odd increments, if we want to find the smallest set of positive $ a,b,c$, we need $ a\plus{}b\equal{}$odd perfect square and we also need the difference of $ a$ and $ b$ to be odd.

Let $ a\plus{}b\equal{}25\equal{}5^2$ and set $ a\equal{}6$ and $ b\equal{}19$. $ b\minus{}a\equal{}13$ and note that $ 7^2\minus{}6^2\equal{}13$.
Letting $ c\equal{}30$, we get our solution.

$ (a,b,c)\equal{}(6,19,30)$ Therefore, the answer will be $ a\plus{}b\plus{}c\equal{}55$


geometry
trapezoid
circumcircle
geometry unsolved

Given trapezoid $ ABCD$ and $ AD \parallel{} BC$. Let $ I$ be the centre of inscribed circle in triangle $ ABC$. Point $ F$  belongs to bisector of $ BDC$ and $ IF$ is perpendicular to $ BC$.  Circumcircle of triangle $ CFD$ intersects line $ BC$ at $ C$ and $ E$. Prove that triangle $ DEF$ is equilateral.

DEF equilateral requires at least <BCD = 60!!

Best regards,
sunken rock


[quote="mathswizard1992"][b] :maybe: you are given a set of 10 positive integers . summing nine of them in ten possible ways we only get nine different sums:86,87,88,89,90,91,93,94,95.find those numbers.[/b][/quote]

[hide="Hint"]
Think of the sums as the total of [i]all[/i] the numbers, minus the one missing
[/hide]

Is there an easier way to do this? And am I right?

First, we have to find out which of the $ 10$ sums was duplicated.

Let our numbers be $ a, b, c, d, e, f, g, i, j, k$. Now we sum up the sums that are given:
\[ 86 \plus{} 87 \plus{} 88 \plus{} 89 \plus{} 90 \plus{} 91 \plus{} 93 \plus{} 94 \plus{} 95 \equal{} 813\equiv {3} \pmod{9}
\]
We need to find the duplicated sum such that it is $ \equiv {6} \pmod{9}$. Of those sums, only $ 87$ works. Thus, are sums are:
\[ 86, 87, 87, 88, 89, 90, 91, 93, 94, 95
\]
Now, we have
\[ 9\left(a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \plus{} g \plus{} h \plus{} i \plus{} j \plus{} k\right) \equal{} 900 \\ \Rightarrow a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \plus{} g \plus{} h \plus{} i \plus{} j \plus{} k \equal{} 100
\]
Now, we subtract each sum from this, so in no particular order, we have:
\[ \boxed{a \equal{} 14, b \equal{} 13, c \equal{} 13, d \equal{} 12, e \equal{} 11, f \equal{} 10, g \equal{} 9, h \equal{} 7, i \equal{} 6, j \equal{} 5}
\]


function
induction
algebra proposed
algebra

Find all functions $f$ from $N$ to $N$ such that:
1) $f(f(n)+n)=f(n)$
2)There exists $n_{0}$ such that $f(n_{0})=1$

Putting $n_0$ in our relation we get $f(n_0+1)=f(n_0)=1$ so by induction we have $f(n)=1$ for $n\geq n_0$.
Now we have $(n_0-1)+f(n_0-1)\geq n_0$ so we have $1=f(n_0-1+f(n_0-1))=f(n_0-1)$ so again by induction we have $f(n)=1$ for $n<n_0$ too.
So for all $n$ we have $f(n)=1$

Why $Omid hatami$ is not in the team IMO 2006 ?

Because he was in IMO 2005 and now he is in university.


Can someone give me a detailed way to solve this problem? Thanks.

Given that a sequence satisfies x0 = 0 and abs(xk) = abs (x(k-1)+3) for all integers k>=1, find minimum possible value of abs (x1 + x2 + ... + x2006).

By the way, x0 is x-sub-zero. xk is x-sub-k. x(k-1) is x-sub-k-1.
abs (...) means absolute value. >= is greater than or equal to. and x2006 is x-sub-2006. thanks.

Past AMCs, AIMEs, USAMOs, etc. almost always have a solution somewhere.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=449185#p449185


Find all hyperbolas with the asymptote $2xy-3y-4x=6$.

[quote="Quevvy"][quote="cincodemayo5590"]Find all hyperbolas with the asymptote $2xy-3-4x=6$.[/quote]
That equation [i]is[/i] a hyperbola, so no hyperbolas can have a hyperbola as an asymptote. :P[/quote]

Oh oops, there's a typo. It should be $3y$. It's fixed now.  :P

[quote="Ubemaya"]how can you have 2 points as an asymptote?[/quote]

They're lines. (One's vertical, the other is horizontal.)

[quote="cincodemayo5590"][quote="Quevvy"]I think that that's still a hyperbola...[/quote]

No, it's degenerate.

[hide="Heres why, dont look unless you dont know what to do"]
Its the same as $(2x-3)(y-2)=0$.[/hide][/quote]
$(2x-3)(y-2)=2xy-3y-4x+6$
$2xy-3y-4x=-6$
The equation that was originally given was $2xy-3y-4x=6$ and obviously, $6\ne-6$ so if you have 6 then it is a hyperbola and -6 makes it 2 lines.
[hide]Then the only hyperbola with those asymptotes is $2xy-3y-4x=+6$[/hide]


function
complex analysis
complex analysis unsolved

Let $ g,f: \mathbb C\longrightarrow\mathbb C$ be two continuous functions such that for each $ z\neq 0$, $ g(z)\equal{}f(\frac1z)$. Prove that there is a $ z\in\mathbb C$ such that $ f(\frac1z)\equal{}f(\minus{}\bar z)$

Hmm, what if f(z)=z? Am I missing something here?

Then $ g$ is not a continuous function on whole $ \mathbb C$

Yes, you can easily combine both maps as a map $F:S^2\rightarrow \mathbb C$ and from Borsuk-Ulam Theorem for such map there is $x\in S^2$ s.t. $F(x)=F(-x)$. Projecting $x$ from $S^2$ to $\mathbb C$ we reach to $ f(\frac{1}z)=f(-\bar z) $


Hello everybody... I need your help!!

I am looking for [b]"Last Christmas"[/b] piano music sheet that is available for pay-per-download online. Most of them are in piano-vocal-guitar arrangements and they are relatively easy to play. I am looking for "intermediate" and better arrangements of the piece for the piano. Would you be able to help me out? Thanks!


algebra
polynomial
logarithms
calculus
calculus computations

Equation
$ x^{n}\equal{} x^{2}\plus{}x\plus{}1$
a)Find all n such that the equation has unique solution $ x>0$
b)For each n $ x_{n}$ is root of equation
Find $ lim n(x_{n}\minus{}1)$

[b]a)[/b] $ n\ge 3$ odd  (and $ n \equal{} 2$).
[b]b)[/b]Easy to see that $ x_{n}> 1$, and $ x_{n}\rightarrow 1$, so $ x_{n}^{n}\rightarrow 3$, so $ n\ln{x_{n}}\rightarrow\ln3$ so $ n(x_{n}\minus{}1) \equal{} n\ln{x_{n}}\frac{x_{n}\minus{}1}{\ln{x_{n}}}\rightarrow\ln3$

OK you have edited it $ x > 0$, so for [b]a)[/b] $ n\ge 3$. Please try to post your problems in the correct form(even if here it wasn't crucial). I mean you make mistakes in something like 50% of your posts.


category theory
advanced fields
advanced fields theorems

does already exist a common term for a functor which is injective on parallel morphisms (i.e. faithful), but also injective on objects as well?

This is unlikely because you can make any functor $ F\colon C\to D$ injective on objects by replacing $ D$ by an equivalent category (objects: disjoint union of $ C$ and $ D$, morphisms: replace objects in $ C$ by their images, then take morphisms in $ D$). owk

P.S. Apparently, some people (not Mac Lane) call those functors [url=http://eom.springer.de/I/i050190.htm]imbeddings[/url]. owk


Euler
inequalities unsolved
inequalities

Let $0< a_0 <a_1 <...<a_k$ be integer numbers. Let $\displaystyle f(x)=\sum_{i=0}^k \frac {x^i}{(x+1)^{a_i}}$.
Prove that $f'(x) <0$ ;$\forall x \in R^+$.

It's quite well-know.I guess it's in AMM and prove it some months ago!

can anyone post the proof?
I cannot access AMM unless i go to university and access it there...
BTW, what's Abel formula?
You are referring to the Abel identity or a kind of generalization of euler summation formula?

[quote="leepakhin"]$\frac{d}{dx}\frac{x^i}{(x+1)^{\alpha_{i}}}=\frac{x^{i-1}}{(x+1)^{\alpha_i+1}}((i-\alpha_i)x+i)$
But when $\alpha_i=i$, I think $f'(x)>0$.[/quote]
but $\alpha_i=i$ can't be ture....


ΠΡΟΤΑΣΗ  4. Τριγώνου ΑΒΓ η πλευρά ΒΓ παραμένει σταθερή και η κορυφή Α κινείται επί ευθείας παράλληλης προς τη ΒΓ. Εάν Δ, Ε, είναι σημεία επί της πλευράς ΒΓ τέτοια ώστε να είναι <ΒΑΔ = <ΑΓΒ  και <ΓΑΕ = <ΑΒΓ, προσδιορίστε τη θέση της κορυφής Α, ώστε να είναι (ΒΓ) = 3(ΔΕ).

Υπάρχει γεωμετρική λύση για το πρόβλημα  :roll: ??

Έχω μία γεωμετρική λύση υπόψη μου και ελπίζω ότι δεν σφάλλω.

     Το πρόβλημα βέβαια δεν έχει πάντοτε λύση. Με δεδομένο το σταθερό μέγεθος της πλευράς ΒΓ, αν η απόσταση της παράλληλης ευθείας (από την ευθεία ΒΓ) επί της οποίας κινείται η κορυφή Α, είναι μεγαλύτερη από κάποια τιμή ( θέμα διερεύνησης ), τότε δεν υπάρχει λύση.

     Κώστας Βήττας.


Galois Theory
Ring Theory
superior algebra
superior algebra unsolved

I got difficulties reading a paper "Finite fields" due to James Ax. There was said that "There are only finitely many subfields $K$ of $N$; moreover, it follows from Chebotarev's theorem that a subfield $K$ of $N$ arises infinitely often if and only if $N$ is a cyclic extension of $K$." Here $N$ is a finite normal extension of $\mathbb{Q}$. Any hints to prove that?

I'm still stuck on this one. Doesanyone knows some good book or article where I can learn some advanced topics about field extensions? However, I found the following theorem which might be helpful:

Let $L/K$ be a finite extension of a given number field and $m\in\mathbb{N}$. Then there exists a cyclic cyclotomic extension $M/K$ of $L/K$, deg $M/K=m$ such that $M\cap L=K$.

What do you mean by $K$? How does it arise?

I guess you consider prime ideals $\mathfrak{P}$ of $N$ and their splitting fields $K$. Then, for unramified primes, $\text{Gal}(N/K)$ is isomorphic to the galois group of the residue field extension $\text{Gal}((N/\mathfrak{P})/(K/(\text{Norm}_{N/K}\mathfrak{P}))$, which is of course cyclic. Since there are only finitely many ramified primes, noncyclic extensions $N/K$ can occur only finitely often.

On the other hand, there is the chebatorev density theorem, which computes the density of those primes $p$ of $\mathbb{Q}$ for which there is a prime ideal $\mathfrak{P}$ of $N$ above $p$ such that the splitting field of $\mathfrak{P}$ is a given extension $K$ with $\text{Gal}(N/K)$ cyclic; the result is always nonzero, which implies that there are infinitely many of them.

I don't think there is some simpler argument avoiding the use of chebatorev's theorem.


Aveti aici atasat fiserele cu rezultatele finale de la ONM 2006 (nationala) + Barajele de Seniori pentru OBM (si OIM).

just wondering: rezultatele de la baraje reprezinta echipa finala pt OIM? ca de abicei se dadeau mai multe baraje cu lotul largit.... :?

Scrie si acolo (numele fisierului): primii 6 sunt echipa de OBM (de Balcaniada) :) Pana la 25 inclusiv (Madalina Persu a intrat cu clauza de cls. 9-a, de aia Galatan Alin nu este in lot desi are acelasi punctaj) sunt in lotul largit si vor continua sa dea baraje la Bucuresti: 1 de Geometrie de 3 probleme, 1 de 4 probleme, dupa care raman (cel mult) 12, si inca 2 baraje in stil OIM in doua zile consecutive, fiecare cate 3 probleme.

Ca si fapt divers, eu sunt Galatan Alin :) Intr-un fel, mai bine ca nu m-am calificat cu punctajul asta, ca cu siguranta nu ajungeam la inter. Si eu am un bac de dat.

La sectiunea de Resources o sa gasesti problemele date in prima zi de la barajul de seniori. Probabil ca si celelalte isi vor face aparitia in curand. :)

[quote="gnosis"]Valentin inca nu am vazut niciunde postate problemele date la baraje. Daca este posibil posteaza-le te rog.[/quote]
http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26&year=2006

O sa le postez si pe celelalte 4 foarte curand. Nu le-am postat de la Iasi deoarece efectiv nu prea am avut access la internet (din motive tehnice) :(

[quote="andyciup"]Cand m-am uitat pe rezultate am fost surprins:Zamorzaev si Boreico (din Moldova) nu au participat la baraje?Sau doar nu le-au fost afisate rezultatele?[/quote]Au participat, dar rezultatele lor au fost separate de restul (ca erau afara din concurs).

Iura a facut tot (55.5 / 56), Zamorzaev a luat 43.5 (7-7-7-6-3.5-0-7-6), Frimu Andrei 21 (6-1-0-0-7-0-7-0), Grecu Mircea 7 (p1 din ziua 2), Timotin Rodica 7 (tot p1 din ziua 2 + 1p la p3 din ziua 2).

As avea o intrebare: De ce se pun finala si barajele la distanta asa mica unele de altele?
Personal, sunt o fire emotiva si nu rezist niciodata la mai mult de 1 concurs la o perioada. Probabil de aceea la finala intotdeauna ma descurc excelent (premiul I in a VII-a si premiul II in a VIII-a pana in prezent) si la baraje am scoruri penibile (5p in a VII-a si 11,5p in a VIII-a)...

Stormraven, trebuie sa inveti sa rezisti. OIM = 2 zile de concurs, si deci de aia barajele (la seniori) sunt cate doua unul dupa altul. Au fost doua acuma pentru ca Balcaniada este in cateva zile si nu mai era timp pentru o alta selectie la Bucuresti. Juniorii s-au conformat programului ...

Stormraven, cred ca ar trebui sa iei in considerare si dificultatea problemelor. Nu prea poti compara ce se da la finala cu ce se da la baraje... parerea mea.

Umila (si probabil neexperimentata) mea părere e ca problemele de finale sunt radical diferite de cele de baraje. Nu vorbesc de dificultate, vorbesc de tip. Primele le poţi rezolva la o adica daca stai si te gandesti, chiar daca nu ai mai făcut tipul acela de problemă...cât despre celelalte, dacă nu ai o sclipire sau dacă nu iţi vine ideea sau dacă nu ai mai facut tone de probleme de acelasi gen te poti uita lejer ca mâţa-n calendar timp de 3-4 ore la o problemă fără să o poţi rezolva, unele avand rezolvari absolut nedetectabile...Binenţeles, voi ştiţi mai bine decât mine şi mă aştept să mă contraziceţi...
Din câte am înţeles, asta ar fi tradiţia de la prima OIM organizată de Romania până în prezent...Totuşi, cred că n-ar strica nimănui o distanţă cât de mică între finală si baraje...oricine are nevoie de puţina odihnă, nu? Ideea e ca oricum vocea mea nu contează şi va trebui cumva să mă adaptez sau să pier în mediocritate...  :roll:

Intr-adevar ar fi buna o distanta, insa considerentele de ordin tehnic nu ne permit sa punem aceasta distanta.

Oricum nu numai voi elevii faceti un efort: si noi cei din comisia de corectare trebuie sa corectam repede si bine, sa gandim subiectele pentru baraje, sa publicam solutii, sa corectam barajele. Eu cel putin m-am intors frant de la Iasi :)

Doamne fereşte! Nici nu contest asta!
M-am gândit că n-ar fi rea o decalare a Olimpiadei. Eventual locala să fie în noiembrie-decembrie, judeţeana în ianuarie-februarie, finala în martie si barajele în aprilie, la vreo săptămână-două distanţă unul de altul. În felul acesta s-ar evita şi "ruperile de ritm" datorate lipsei de concursuri în lunile de toamnă

Olimpiada nu se face cand vrea oricine. Sunt niste norme stabilite de M.Ed.C. Toate olimpiadele au loc in vacanta de Pasti. Tocmai pentru ca e vacanta de Pasti, de obicei Balcaniada de Seniori e imediat dupa, deci barajele trebuiesc date atunci.

In plus cautam sa selectionam acei elevi care fac fata la stres si multe competitii consecutive din start.

Sa zicem ca ar exista cineva cu adevarat talentat (ipotetic vorbind, nu vorbesc de persoana mea neaparat) care a mostenit o fire panicarda si emotiva. Acel cineva, in pofida valorii, nu s-ar califica niciodata la vreo internationala...Situatia prezenta l-ar nedreptati pe acel cineva...
Asta este, e o randuiala care trebuie pastrata, oricat mi s-ar parea de nedreapta. Omul se adapteaza...

Olimpiada Internationala nu se rezuma numai la talent. Este vorba si de un stres extraordinar in timpul concursului (si acolo sunt doua zile consecutive, catre 4 ore jumatate).

Daca nu poti sa "supravietuiesti" la o nationala de a noastra, ma indoiesc sincer ca ai face mare lucru la un OIM, oricat de talentat ai fi.

Sigur ca nu e vorba doar de talent...e si munca de-a lungul unui an de la o nationala la alta. Oricum, eu am intrebat, n-am dat cu parul  :D . Am facut doar cateva sugestii pe care le-as considera ideale, dar sunt din pacate ireale...asta e, omul se adapteaza...

boreico iurie a facut perfect baraju ?  :o omul acela e un cyborg ... .a iesit si primul la a XI-a , a propus si probleme pt nationala, ( vedeti problema 2 de la clasa a X-a) ... zamorzaev mai e om ... sincer nu stiu ce s-a intamplat cu el de a luat asa de putin la baraj .. putin pt un fenomen , vreau sa zic . pt noi romanii , a luat o gramadaaaaaa . probabil ca nu s-a putut concentra, am cam facut galagie pe holuri eu cu un baiat cu o seara inainte de baraj / myller :P el statea la camera alaturata ...

referitor la olimpiada, as putea spune ca festivitatea de premiere a fost f f f neadecvata... in sensul ca premiatii erau pur si simplu enumerati ... banii au fost f putini , avand in vedere anvergura olimpiadei . medalia de argint seamana izbitor cu cea de aur ... nu s-au organizat excursii in afara orasului , ca sunt totusi multe obiective de vizitat in moldova ... scuzati-ma ca am astfel de pretentii, dar asa se face la chimie, la fiz ... ma gandeam ca o fi asa si pe la mate, dar avand in vedere ca sunt jdemii de probe, se pare ca nu e cazul de asa ceva .

salut! nu sunt chiar omu' potrivit la locul potrivit, fiind clasa a 7-a si neavand nicio participare la nationala de mate (decat calificare, la care am renuntat in favoarea fizicii). vreau sa-ti spun, pureblack, ca indiferent de cate excursii nu s-au organizat, conditiile voastre au fost mult mai bune decat cele de la craiova (onf). deci, chiar nu aveti de ce va plange. cat despre bani, aceeasi is la toti... credeti ca la alte olimpiade s-a dat mai mult s-o din partea ministerului? asta e, inghitim in sec, si invatam mai departe...
cat pentru Stormreaven, cred ca ar trebui sa te obisnuiesti cu efortul. la urma urmei, aceleasi conditii de concurs sunt pentru toti participantii... si la orice nationala exista cel putin doua (fara baraje si probe de departajare) etape.
succes in continuare tuturor! :)

[quote="delia"]nu gasesc fisierul atasat cu rezultatele la onm2006[/quote]Este in primul post. Trebuie sa fii logat(a) ca sa poti downloada fisierele.

Ori in 9 ori in 16 zile, nu stiu exact :D

ai knd e primul baraj la lotul d juniori?

Tot pe atunci :) Cand o sa stiu exact datele o sa le postez si pe pagina principala nu va faceti griji :)

pt cei din lot din clasa a VIIIa s va organiva o sesiune speciala la testarea nationala?

Foarte probabil va fi ca in ceilalti ani. Oricum o sa vi se comunice la Bucuresti.

si cum a fost in ceilalti ani?...eu nu stiu prea multe despre asta....

Anul trecut de exemplu, dupa ce s-a ales echipa pentru JBMO ne-au chemat pe cei din echipa de clasa a 8-a sa dam capacitatea la Cluj. Partea proasta e ca ne-au anuntat cu doar o saptamana, sau chiar mai putin, inaine de examen. Acum doi ani in schimb sesiunea speciala a avut loc la Bucuresti si au putut-o sustine si cei care nu s-au calificat, dar au facut parte din lotul restrans (de 12). Data examenului depinde de data balcaniadei, in general.
 In ceea ce priveste subiectele, daca nu le ai, le gasesti sigur pe net.

ms....;)...am aflat si eu azi de la profu meu de mate....


Let:

$x_1=0,12345678910111213\ldots$

$x_2=0,149162536496481100\ldots$

$x_3=0,182764125216343512\ldots$

$x_4=0,1168125662512962401\ldots$

... and so on.

The decimal expansion of $x_n$ is obtained by writing out the [b]n[/b]th powers of the integers one after the one.

Are any of the $x_i$ rational??

help me, please...  :D

I think this problem should be in the number theory forum,and I think none of the$x_i$is rational.

[quote="ZetaX"]Rational $\iff$ repeating decimal fraction.
But the last one is not the case since there are arbitrary long sequences $1000...000$.[/quote]thanks...  :)


number theory
greatest common divisor
ratio

The gcd of two numbers determined by the process of successive division is
396. Having three ratios that are as small as possible, determine the largest
the two numbers.

Portuguese question:

O mdc de dois números determinado pelo processo das divisões sucessivas é
396. Havendo três quocientes que são os menores possíveis, determine o maior
dos dois números.

what do you mean when you say 3 ratios as small as possible.

Three quotients, in the division.

[url]http://en.wikipedia.org/wiki/Quotient[/url]


A can of dog treats cost $ \$1$ at the local pet store. A blind man goes to buy dog treats for his dog, bringing the guide dog inside the store. Once the blind man brings one can of dog treats to the cashier, the cashier says $ \$1$. The dog thinks, "$ \$7$?! That's a lot!"

Why does the dog think it's $ \$7$?

A random guess, the dog has bad hearing and thought the cashier said 7?

[quote="mathcrazed"]CAUSE 1 IN HUMAN IS 7 IN DOG!!! HAHAHA... lol[/quote]

Haha yes, my problem was an easy one.

I first thought that it had to do with the fact that a 1 and a 7 can look very similar, depending on the font.  I forgot about the fact that the cashier SAID the price.

[quote="mathcrazed"]CAUSE 1 IN HUMAN IS 7 IN DOG!!! HAHAHA... lol[/quote]

If you want to get technical, you could argue that a dog's age in the so-called "dog years" actually depends upon health, fitness, etc, and for each dog, it's completely different(larger dogs age faster than toy dogs), and that the 7 year thing is only a rough estimate, but then again, who wants to be technical?  :P

That is not true. I just went to the vet a few weeks ago with my puppy and he explained they are the human age equivalent of 24 human years after 1 year of their life, and age the equivalent of 4 years every year for the rest of their lives.

Well, it really differs with every dog. I know that to be true with Labradors and various other breeds, but take, for example, my puppy, a Papillion-Pomeranian mix(toy dog). He ages slower than larger dogs. The thing about the dog years is that, for example, the age a dog dies in dog years would be close to a typical human's age when he/she dies.

The Irish Wolfhound lives about 4 years(which is extremely short). If we did the 7 year thing, that would be only 28 years old. If we did what your vet told you, that would be 36 years old. But the average human lives a lot longer, so the dog would age in "dog years" a lot faster.

But anyway, it doesn't really matter. I was just making a comment. It was still a good riddle!  :)

[quote="metafor"]Why would a dog be thinking about prices? O.o[/quote]

Because dogs can care about their owners. Don't forget, it was a blind man.


G. H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started the conversation with: "I came here in a taxi-cab number 1729. That number seems dull to me, which I hope isn't a bad omen." "Nonsense," said Ramanujan. "The number isn't dull at all. It's quite interesting. It's the smallest number that can be expressed as the sum of two cubes in two different ways." Ramanujan had immediately seen that $ 1729\equal{}12^3\plus{}1^3\equal{}10^3\plus{}9^3$. What is the smallest positive integer representable as the sum of the cubes of [i]three[/i] positive integers in two different ways?

Brute force indicates that the answer is

[hide]
$ 251\equal{}125\plus{}125\plus{}1\equal{}216\plus{}27\plus{}8$
[/hide]
but an actual solution would be nice.

can anyone think of a nice solution? (I can't)

Is there something to do with inequalities though?

can someone please post a solution (not involving brute force)?????  :|

The answer is less than or equal to $ 1730$, so brute force isn't so bad since there aren't that many sums of three cubes to check.  You might try reducing the casework by investigating what happens $ \bmod 9$, but it probably won't help much.

How was the answer found using "brute-force"?

In this particular example, just list out the first 6 cubes and notice the sum.  This is probably easier than the rigorous method :)


Consider the equation $ Ax^3\plus{}(2\minus{}A)x^2\minus{}x\minus{}1\equal{}0$, where $ A$ is a real number for which the equation has three real roots, not necessarily distinct. For certain values of $ A$, there is a repeated root $ r$ and a distict root $ s$. List all such values of the triple $ (A,r,s)$

mdk, when will you start to learn Latex? :wink:

You have edited. Don't make excuse as little boy!

[quote="mdk"]what are you talking about?[/quote]

[quote="mdk"]Consider the equation Ax^3 + (2 - A)x^2 - x - 1 = 0, where A is a real number for which the equation has three real roots, not necessarily distinct. For certain values of A, there is a repeated root r and a distict root s. List all such values of the triple (A,r,s)[/quote]


The Hollenbecks own a square lot with area 3600 square
meters. They plan to fence just three sides of this lot. How
many meters of fence must they purchase?

since its a square lot, one side measures $ \sqrt{3600}\equal{}60$

they plan to fence just 3 sides.

60*3=180

answer : 180


linear algebra
matrix
algebra proposed
algebra

Find all $a,b,c,d \in {\mathbb C}$   such that $\; \displaystyle {a\; \; \; b \choose c\; \; \; d}^{17}=\displaystyle {1\; \; \; 7 \choose 0\; \; \; 1}\; .$

You can suppose the matrix is triangular superior so we have to find

all $ a,b,d \in {\mathbb C} $   such that
$ \;\displaystyle {a\; \; \; b \choose 0\; \; \; d}^{17}=\displaystyle
{1\; \; \; 7 \choose 0\; \; \; 1}\; . $

We find $a^{17}=d^{17}=1$ solve it in C and get b with
$b\sum_{k= 0 }^{16}a^{16-k}d^k= 7$

That's not really the answer...
Since  $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)^{17}$ is not diagonalizable, $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$ is not diagonalizable. This means we can write $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)=zI+N$, where $z$ is a complex number and $N^2=0$. (This, of course, depends on the $2 \times 2$ nature of the matrix) We have $(zI+N)^{17}=\sum_{j=0}^{17} {17 \choose j }z^{17-j}I*N^j=z^{17}I+17z^{16}N$.
We therefore have $z^{17}(I+\frac{17}{z}N)=\left(\begin{array}{cc} 1&7 \\ 0&1 \end{array} \right)=I+N^{\prime}$, with $N^{\prime}=\left(\begin{array}{cc} 0&7 \\ 0&0 \end{array} \right)$. To match eigenvalues, we must have $z^{17}=1$ and $\frac{17}{z}N = N^{\prime}$.
Therefore for each $z$ with $z^{17}=1$, we have a solution $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)=\left(\begin{array}{cc}z & \frac{7z}{17} \\ 0 & z \end{array}\right)$ and these are all solutions.

The sum Moubinool referred to is zero whenever $a \neq d$ and both are 17th roots of unity..


Solve the set of equations:
$\left\{\begin{array}{l}{z^2+2xyz=1}\\{3x^2y^2+3y^2x=1+x^3y^4}\\{z+zy^4+4y^3=4y+6y^2z}\end{array}\right.$
Note:don't forget the subject [b]"use trigonomatric"[/b]


calculus
derivative
function
Asymptote
limit
calculus computations

[quote="kunny"]Let $a\neq 0$ be real numbers. How many tangent lines are there which touch to two curves $y=e^{x}$ and $y=ax^{2}$?[/quote]
I cant understand the problem  :blush:
Can you explain it?
Because I considered $a=1$ and there are none the way I understand it.

Is the question asking how many lines are tangent to both curves?

[hide]
Let this tangent line be tangent to $e^{x}$ at $(b, e^{b})$ and tangent to $ax^{2}$ at $(c, ac^{2})$.  Then it intersects the x-axis at $(b-1, 0)$ which is also $(c/2, 0)$; so we have $2b-2 = c$ and therefore $4ab-4a = e^{b}$.  With $a$ fixed, the left hand side of the equation is a linear graph while the left hand side is an exponential curve.  So with fixed $a$, the number of tangent lines is equal to the number of solutions of $4ax-4a = e^{x}$, which can be 0, 1, or maybe 2??

Anyways to figure that out without a calculator??[/hide]

It's $2ax=e^{x}\;\forall a \in \mathbb{R}, a \not= 0$. whether or not we can obtain a solution set for that...

Assuming pkerichang's equation $4a(x-1)=e^{x}$ works...
[hide]consider the function $f(x)=\frac{e^{x}}{x-1}$. Its derivative is
\[\frac{e^{x}}{x-1}-\frac{e^{x}}{(x-1)^{2}}=\frac{e^{x}(x-2)}{(x-1)^{2}}\]
The derivative is positive for $x>2$ and negative for $x<2$, with local minimum of $e^{2}$ at $x=2$. Also, there is asymptote for $f$ at $x=1$, so we evaluate the limits around the asymptote and at positive and negative infinity:
\[\lim_{x\rightarrow-\infty}f(x)=0,\ \lim_{x\rightarrow 1^{-}}f(x)=-\infty,\]
\[\lim_{x\rightarrow 1^{+}}f(x)=\infty,\ \lim_{x\rightarrow \infty}f(x)=\infty\]
It follows that $f$ is a bijection from $(-\infty, 1)$ to $(-\infty, 0)$, from $(1,2]$ to $[e^{2},\infty)$, and from $(2,\infty)$ to $(e^{2},\infty)$. Therefore, the number of solutions to $f(x)=4a$ is
\[\left\{ \begin{array}{rl}0,&0\leq a<\frac{e^{2}}{4}\\ 1,&a=\frac{e^{2}}{4}\text{ or }a<0\\ 2,&a>\frac{e^{2}}{4}\\ \end{array}\right.\][/hide]


let $K>L>M>N$ be nature number and $KM+LN=(K+L-M+N)(-K+L+M+N)$ prove that $KL+MN$ is composite.
I don't know the solution but I know is has a nice solution with geometry.

I'm moving this thread into Solved Problems (there is already more than enough spam in the Unsolved Problems). Just for the sake of completeness: You can find the solutions of all past IMO problems on http://www.kalva.demon.co.uk/imo.html ; particularly, this one has been solved on http://www.kalva.demon.co.uk/imo/isoln/isoln016.html . You could also look at http://www.mathlinks.ro/Forum/viewtopic.php?t=17474 . I have never heard about a solution using geometry, and, taking into account the date of the first post, I would restrain myself from searching for one.  ;) By the way, it would be quite nice if people would use English to communicate in the Olympiad Section.

  Darij

yes i cant forget that solution but i dont remember its details. i think you kan find it somewhere in the wolfram oficial page for that imo in wich the problem appeared. google imo 2001 problems and solutions or some like that and i think youll find it. :D

LOL you are right!! I just couldn't believe that a number theory problem posted on April 1st with the remark that there exists a geometric solution is not an april fool joke!

http://imo.wolfram.com/problemset/IMO2001_solution6.html

Actually, as one could expect, the application of geometry consists in deducing one algebraic identity from another one. Still it's too cool...

  Darij


for some reason my PM's never get sent, they just stay in my outbox for days.
What is wrong?

Probably the recipient never checks his/her inbox or ignores your message. Either way that means the recipient never read your message (or rather clicked on your message)

oh, so if it is in outbox, it means they never read it, not never recieved it?

Exactly, the message should move to the Sentbox as soon as the recipient actually reads your message.


Find all triplets of primes in the form $ (p, 2p\plus{}1, 4p\plus{}1)$.

[hide]
If $ p\equiv 1\pmod 3$,then $ 2p \plus{} 1\equiv 0\pmod 3$,which is not a prime.
If $ p\equiv 2\pmod 3$,then $ 4p \plus{} 1\equiv 0\pmod 3$,which is not a prime.
so it's necesary $ p\equiv 0\pmod 3$ ->$ p \equal{} 3$,we have the triple $ (3,7,13)$ work.[/hide]


Adam, Babs, Cindy, Debbie, Eve, Fathom, Gerald, and Harold are standing in a random order in a straight line. Find the probability that Adam, Eve, and Fathom are in 3 consecutive positions on that straight line. Express your answer as a common fraction reduced to lowest terms.

[hide]Is it $\binom{8}{3}$?[/hide]

There are $18$ people in a room, and each of th 18 writes down an integer a random from one of the $100$ integers from $1$ to $100$ inclusive. Find the probability that at least two people wrote down the same number. Express your answer as a decimal rounded to $4$ significant digits.
[hide]
First find out the combinations that none of them wrote down the same number and then subtract that from all the possible combinations. For the first part, $100\cdot99\cdot98\cdot...\cdot 82$ and the latter part is $100^{18}$ are those look right so far?[/hide]

my 400th post yay!

[hide="1"]
Total positions is $8!$
Total positions that satisfy the condition is $6(3!)(5!)$
So the answer is $\frac{6(3!)(5!)}{8!}=\frac{6(6)}{8\cdot 7\cdot 6}=\frac{6}{56}=\frac{3}{28}$

I do not know how you got $\binom{8}{3}$ since the probability would have to be between 0 and 1.

EDIT-

You could also just do this:
Since there are $\binom{8}{3}$ possible positions for the three people (consider them indistinguishable) and 6 possibilities that satisfy the conditions, just do $\frac{6}{\binom{8}{3}}=\frac{3}{28}$
[/hide]

[hide="2"]
You need $100\cdot 99\cdot 98\cdot...\cdot 83$ instead of ending it on 82. There should only be 18 terms. You had 19.
[/hide]


function
limit
algebra
rational function

Prove that the exponential function ex   is not a rational function (i.e. it can't be as P(x)/Q(x) , where P,Q are polynomials).
 ...

take x = 1.
P(x)/Q(x) = P(1)/Q(1) must be rational. (if $Q(1) \neq 0$)
but e is irrational.

If Q(1) = 0, we can always find an integer $n\neq 0$ such that $Q(n)\neq 0$,
but e^n is irrational.

Sorry Liyi, the meaning of "rational function" is different !
A rational (or fractional) function f(x) can be written f(x) = P(x)/Q(x) for all x, where P,Q are polynomials.
A rational function can produce a lot of irrational values :
for ex f(x) = (pi *x + 2)/(3*x2 + e)

See you ...

[tex]\lim_{x \to +\infty} e^x \neq \lim_{x \to -\infty} e^x[/tex], while for rational [tex]f[/tex] we always have [tex]\lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(x)[/tex]

[quote="stevem"][quote="luqash"]for rational [tex]f[/tex] we always have [tex]\lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(x)[/tex][/quote]
Are you sure that you mean that? [tex]f(x)=x, f(x)=\frac{1}{x}[/tex] are two simple counterexamples[/quote]

oh, i forgot... the limits are taken in the compactification of [tex]\mathbb R[/tex], thus [tex]-\infty = +\infty = \infty[/tex] for the f(x), but not for the x.

[tex]\lim_{x \to +\infty} x = \lim_{x \to -\infty} x = \infty, \lim_{x \to +\infty} \frac 1x = \lim_{x \to -\infty} \frac 1x = 0[/tex]

The point is that rational functions are continuous at [tex]\infty[/tex], [tex]e^x[/tex] is not.

I thought you must have meant something else  :)
Mind you, I made an awful error with suggesting [tex]f(x)=\frac{1}{x}[/tex] as a counterexample  :blush:

Not too bad Luqash, but how  do you prove that the exponential function ex  is not a rational function, even on a bounded interval [a,b]  ??


your online thing? or your sig? or perhaps mr green?

err..um i spy something that starts with "p"

:wink:  Uhm Bubka, about your first pic :

EDIT : looks like my pic isn't working either.

[quote="Koda"]Let us play I spy. Everyone guesses the last person's spied object and spies something himself. You are also expected to return at some convenient moment to confirm whether what you last spied was guessed correctly.

I start:
I spy something ... green?

[img]http://www.piano-rock.net/koda/11_rutttuke.jpg[/img][/quote]

let me take a wild guess..... tree?

tree

and tree for the next one, and the one after, and the one after, and on for infinity. Now you don't need to do tree anymore bubka.

[quote="Koda"]Ok mustafa got it right!

@deimos: a hepatitic tree?

I spy something tall.
[img]http://www.animated-news.com/archives/bb2c.jpg[/img][/quote]

@bubka: what's yellow in this message? :huh:

[quote="deimos"]koda: or the green thick near "ascending"
funcia: maybe pm or profile, or posts under your avatar?  :|

i spy something yellow. (well this one isn't hard, is it?  :P )[/quote]

yeah it's PM

[quote="mustafa"]Where do you get yellow from?[/quote]

perhaps a  :P , or  :|  or :huh: .


Five pirates and a monkey are shipwrecked on an island. The pirates have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the others, one pirate wakes up during the night and divides the coconuts into five equal parts with one left over, which he gives to the monkey. The pirate then hides his portion of the pile. During the night, each of the other pirates does exactly the same thing by dividing the pile he finds into five equal parts leaving one coconut for the monkey and hiding his portion. In the morning, the pirates gather and split the remaining pile of coconuts into five equal parts and again one is left over for the monkey. What is the smallest number of coconuts the pirates could have collected for their original pile?

Well, (After some tedious calculations) I obtained the following answer :[b] 3121 [/b] (I'm almost sure I'm right, but I feel I might have committed an inadvertent error somewhere, if it's so, do correct me)

See [url=http://www.mathlinks.ro/viewtopic.php?t=185766]here[/url].

Regards. ;)


Very hard
1) ???
[img]http://iqvietnam.com/666/arrrrrrr.JPG[/img]

2)?????
[img]http://iqvietnam.com/666/gggggggggggggg.jpg[/img]

 :roll:  :roll:  :roll:

Did that come from a high school Terra Nova IQ test or something?


The sequence $a_{1},a_{2},...$ is deﬁned in the following way:
$a_{1}= 1, a_{n}= a_{n-1}+(a_{n-1}\;\text{mod}\; n)^{2}, n \geq 2$, where $a\;\text{ mod}\; b$ denotes the remainder of division of $a$ by $b$. What is the maximal possible number of consecutive equal members of this sequence?

Great! Thank you very much  :wink: But what is answer? Who can continuous?

My calculations: $a_{4}=a_{5}=10, \ a_{10}=a_{11}=a_{12}=132$.
For $n\ge 21$ we get $a_{n}=49*(n+1)+7$. Therefore thereare not another solutions $a_{n}=a_{m},n\not =m.$


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group with $ 10$ elements and the elements $ a\neq{b}\neq{e}$ with the property $ a^2\equal{}b^2\equal{}e$. Prove that $ G$ isn't a abelian group.

Suppose that G is abelian.From this we have that $ ab\equal{}ba$, from where it follows that $ H\equal{}\{e,a,b,ab\}$ is a subgroup of $ G$.Therefore $ ord(H) | ord(G)$ $ \rightarrow$ $ 4|10$ contradiction.

An abelian group of order $ pq$ where $ p$ and $ q$ are primes is either cyclic or isomorphic to the direct product of two cyclic groups of order $ p$ and $ q$ respectively. Since $ a^2\equal{}b^2\equal{}e$

Suppose $ G$ is abelian, we have two cases: If $ G$ is cyclic, then $ a\equal{}g^{k_1}, b\equal{}g^{k_2}, e\equal{}g^{k_3}$ then by the equality we have that $ k_1\equal{}k_2\equal{}k_3 mod 10$ which implies some two are equal. But $ a,b,e$ are different.

If $ G$ is a product group then $ a^2\equal{}b^2$ necessarilly implies that $ a^2\equal{}b^2\equal{}1$ then $ e\equal{}i$ but then either $ a$ or $ b$ must be $ 1$ contradiction. Then $ G$ is nonabelian.


trigonometry
geometry
trig identities
Law of Sines

Also known as the law of tangents.

[hide="Proof"]
By the law of sines,
$ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}\equal{}k$

Then $ k\sin A\equal{}a$ and $ k\sin B\equal{}b$

Therefore $ \frac{a\plus{}b}{a\minus{}b}\equal{}\frac{k(\sin A\plus{}\sin B)}{k(\sin A\minus{}\sin B)}\equal{}\frac{\sin A\plus{}\sin B}{\sin A\minus{}\sin B}$

A useful identity to know:
$ \sin A\plus{}\sin B\equal{}2\sin\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}$

Using that,
$ \frac{a\plus{}b}{a\minus{}b}\equal{}\frac{\sin\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}}{\sin\frac{A\minus{}B}{2}\cos\frac{A\plus{}B}{2}}\equal{}\frac{\tan\frac{A\plus{}B}{2}}{\tan\frac{A\minus{}B}{2}}$ $ \Box$
[/hide]


[hide="My guess..."]
would be $3 \cdot 2^{1003}$.[/hide]

i dont think this particular case is hard.

@ cincodemayo5590: if you think your answer is right, prove it.

Oh, oops, I was thinking about something else entirely (highly divisible numbers).  Sorry.  Carry on.

[quote="amirhtlusa"][quote="t0rajir0u"]I know there's a MathWorld article about the general problem, which is not as easy as it seems.  Does anyone know it?[/quote]
i actually think its an easy problem :)
cincodemayo5590, your answer is wrong.

[hide]in general the maximum would be $2^{a}3^{b}$ where a=0,1, or 2.
you cant have any number bigger than 3. this is easy to show; which i can explain further if needed.
in our case the answer is $3^{669}$.[/hide][/quote]

Oh right. Duh, lol. $2^{3}< 3^{2}$, not the other way around. Haha. That's what happens when I post at night...  :P

It can be shown that the numbers used should be as close to e as possible, and since we are restricted to integers, us 3.

[quote="m4467"]Find the largest number which is the product of positive integers with sum 2007.[/quote]

It's the MathPath 2005 "Threes Are Good" theorem.

[hide="Proof"]Write out the numbers that add up to 2007. There's no use having numbers greater than 5 anywhere, because you can break any of them up into smaller numbers that have a greater product. Any 4 can be rewritten as two 2's. Now the only issue remains: whether 3s or 2s are better. Two 3's and three 2's add up to the same number, but their products are 9 and 8 respectively. Thus, 3's are better.[/hide]


All of these problems can be solved using proof by induction.  Some of them are ideas you may have seen before and proved by different methods.

Prove that $ \displaystyle 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} \dots \plus{} N^2 \equal{} \frac{N(N\plus{}1)(2N\plus{}1)}{6}.$

Prove that the sum of the internal angles of a convex $ N$-sided polygon is $ (N \minus{} 2) \cdot 180^\circ$.  (For extra credit, drop the word "convex"!)

Prove that $ \displaystyle F_0 \plus{} F_1 \plus{} F_2 \plus{} \dots \plus{} F_N \equal{} F_{N\plus{}2} \minus{} 1$, where $ F_N$ is the Nth Fibonacci number.

Prove that a set of size $ N$ has exactly $ 2^N$ subsets.

Prove that $ 2^N > N^2$ for sufficiently large $ N$. (The base case is definitely not N=1.)

Prove that you can make any amount of postage greater than 27 cents with 5 cent and 8 cent stamps.  (Hint: try using five base cases.)

Prove that $ \displaystyle \left ( 1 \plus{} \frac{1}{1^3} \right ) \left ( 1 \plus{} \frac{1}{2^3} \right ) \left ( 1 \plus{} \frac{1}{3^3} \right ) \plus{} \dots \plus{} \left (1 \plus{} \frac{1}{N^3} \right ) < 3 \minus{} \frac{1}{N}.$

One last problem for those of you who have seen a bit of graph theory.  Suppose a connected graph in the plane has V vertices and E edges, and it divides the plane into F faces. Prove that V - E + F = 2.  (Hint: use induction on V.)

(If you're unfamiliar with those terms but want to try the problem anyway, you can see some connected graphs in the plane here: http://mathworld.wolfram.com/PlanarGraph.html .  The red dots are vertices, the black lines are edges, and the white regions are faces.)

Please feel free to correct me since I'm a novice!

[hide="#1"]
Lets turn this into a function such that $ F(N) \equal{} 1^{2} \plus{} 2^{2} \plus{} 3^{2} \plus{} \dots \plus{} N^{2}$ and we want to prove that this equals $ \frac {N(N \plus{} 1)(2N \plus{} 1)}{6}$. Testing $ F(1)$, we see that the base case works. Assuming that the rule holds for $ N$, we test
$ F(N \plus{} 1) \equal{} \frac {N(N \plus{} 1)(2N \plus{} 1)}{6} \plus{} (N \plus{} 1)^2$. This simplifies to $ \frac {(N \plus{} 1)(N \plus{} 2)(2N \plus{} 3)}{6}$. Now testing directly with the function for $ N \plus{} 1$: $ \frac {(N \plus{} 1)(N \plus{} 2)(2n \plus{} 3)}{6}$. Since the two quantities are equal, the claim is true.
[/hide]

Solution to the third problem (on Fibonacci numbers). Do correct me if I messed up somewhere.

[hide]We define the function $ P(N) \equal{} F_{0} \plus{} F_{1} \plus{} F_{2} \plus{} \dots \plus{} F_{N}$ and want to prove this equal to $ F_{N\plus{}2} \minus{} 1$.

We can prove a base case, where $ N \equal{} 1$: $ F_{0} \plus{} F_{1} \equal{} 1 \equal{} F_{3} \minus{} 1$ (since $ F_{3} \equal{} 2$). This allows us to assume $ P(N) \equal{} F_{N \plus{} 2} \minus{} 1$ holds for N.

By the Principle of Mathematical Induction, we can see that if $ P(N \plus{} 1) \equal{} F_{N \plus{} 3} \minus{} 1$ is true, then our original statement is true.

We can prove that $ P(N \plus{} 1) \equal{} F_{N \plus{} 3} \minus{} 1$ is true by adding $ F_{N \plus{} 1}$ to both sides of $ P(N) \equal{} F_{N \plus{} 2} \minus{} 1$. This gives us

$ P(N) \plus{} (N \plus{} 1) \equal{} P(N \plus{} 1) \equal{} F_{N \plus{} 2} \minus{} 1 \plus{} F_{N \plus{} 1}$.

Since $ F_{N \plus{} 1} \plus{} F_{N \plus{} 2} \equal{} F_{N \plus{} 3}$ by the definition of Fibonacci numbers, we can substitute $ F_{N \plus{} 3}$ for $ F_{N \plus{} 1} \plus{} F_{N \plus{} 2}$ in the equation $ P(N \plus{} 1) \equal{} F_{N \plus{} 2} \minus{} 1 \plus{} F_{N \plus{} 1}$, producing $ P(N \plus{} 1) \equal{} F_{N \plus{} 3} \minus{} 1$.

Thus, we know that $ P(N \plus{} 1) \equal{} F_{N \plus{} 3} \minus{} 1$ is true, and therefore that $ F_{0} \plus{} F_{1} \plus{} F_{2} \plus{} \dots \plus{} F_{N} \equal{} F_{N\plus{}2} \minus{} 1$.[/hide]

[hide="#2"][A] The sum of the angles in a triangle is 180, by inscribing this in a circle.

[B] We're trying to prove that $ 180(n\minus{}2)\equal{}\text{The sum of the angles in an n\minus{}gon}$. When we make this an (n+1)-gon, this is like taking off one side, and adding two more sides. Or if we ignore diagonals, this would be like attaching another triangle on one side of the triangle. Therefore, this adds 180 to the LHS, and the RHS turns "n-gon" into "(n+1)-gon"

$ 180((n\plus{}1)\minus{}2)\equal{}\text{The sum of the angles in an (n\plus{}1)\minus{}gon}$

Since this works for a triangle, then this formula holds. QED?[/hide]

Someone please correct me if I am wrong. Thanks!

Solution to the fifth problem. I'm not sure about some of the wording here, so (again) correct me if I messed up somewhere.

[hide]First, define $ P(N) \equal{} 2^N$. Experimenting to find the smallest value for which $ P(N) > N^2$  is true, we see that $ 5$ is the lowest value of $ N$ for which this is true. Thus, we will be testing this for $ N$ at least $ 5$.

Our inductive step requires us to prove that $ 2^{N \plus{} 1} > (N \plus{} 1)^2$. We can write that as

$ 2(2^N) \equal{} 2^N \plus{} 2^N > N^2 \plus{} 2N \plus{} 1$.

From the first part, we know that $ 2^N > N^2$, and since $ N^2 > 2N \plus{} 1$ for $ N \equal{} 5$, we can say that $ 2^N > 2N \plus{} 1$. This allows us to add $ 2^N$ to $ 2^N$ and $ 2N \plus{} 1$ to $ N^2$ in the first inequality ($ 2^N > N^2$), producing

$ 2^N \plus{} 2^N > N^2 \plus{} 2N \plus{} 1$,

which is equivalent to

$ 2^{N \plus{} 1} > (N \plus{} 1)^2$,

the statement we needed to prove for this step.

Since both the base case and the inductive step are valid, $ 2^N > N^2$ is true for $ N$ at least $ 5$.[/hide]

[hide="4th"]When an element is introduced into the set. New subsets are formed and each new one is formed by introducing the new element into an existing subset, so the number of subsets doubles.[/hide]

[hide="6th"]For the case of 28: $ 5, 5, 5, 5, 8$
                      29: $ 5, 8, 8, 8$
                      30: $ 5, 5, 5, 5, 5, 5$
                      31: $ 5, 5, 5, 8, 8,$
                      32: $ 8, 8, 8, 8, 8$

By adding a number of $ 5$'s to the above five cases, any amount of postage can be made.[/hide]

[quote="ANieh"]
One last problem for those of you who have seen a bit of graph theory.  Suppose a connected graph in the plane has V vertices and E edges, and it divides the plane into F faces. Prove that V - E + F = 2.  (Hint: use induction on V.)[/quote]This is Euler's Formula.

[hide="Proof"]
[b]Base Case[/b]: This is just the one-vertex planar embedding, which, by definition satisfies $ v = 1 $, $ e = 0 $, and $ f = 1 $. Thus, Euler's Theorem holds for the base case. $ \Box $

[b]Inductive Step[/b]: Suppose $\mathcal{C}$ is a connected graph with a planar embedding, and suppose $a$ and $b$ are distinct, nonadjacent vertices of $\mathcal{G}$ that appear on some discrete faces of the planar embedding. Then, the graph obtained by adding the edge $\overline{ab}$ to the edges of $\mathcal{G}$ has a planar embedding with one more face and one more edge than $\mathcal{G}$.

Thus, $v - e + f$ will remain the same for both graphs, and since by structural induction this quantity is $2$ for G’s embedding, it’s also 2 for the embedding of G with the added edge. So P holds for the constructed embedding.

Now, suppose G and H are connected graphs with planar embeddings and disjoint sets of vertices. Then connecting these two graphs with a bridge merges the two bridged faces into a single face, and leaves all other faces unchanged. So the bridge operation yields a planar embedding of a con nected graph with $v_G + v_H$ vertices, $e_G + e_H + 1$ edges, and $f_G + f_H - 1$ faces. We get: \[ \begin {align*} \left( v_G + v_H \right) - \left( e_G + e_H + 1 \right) + \left( f_G + f_H - 1 \right) &= \left( v_G - e_G + f_G \right) + \left( v_H - e_H + f_H \right) - 2 \\&= 2 + 2 - 2 = 2, \]as desired.

$ \blacksquare $
[/hide]

[quote="ahaanomegas"]
[quote="quinamatics"]Dude I was reviving.[/quote]
Exactly. That's what you shouldn't do. Posts deleted. [size=200][b]There should be no reason to revive.[/b][/size]
[/quote]Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3098814#p3098814

[quote="ahaanomegas"]
There is been no post in this thread for 1 and a 1/2 years, now. It was dead. [b][size=200]Please do not revive dead threads. Thanks.[/size][/b]
[/quote]Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=535&t=359953&p=2591771#p2591771

[quote="ahaanomegas"]
Of course. If you want to take permission, there's PM. In general:
[quote="e_is_cool"]
[b][size=200]Please don't revive old threads.[/size][/b] Thank you.
[/quote][/quote]Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2939581#p2939581

 :whistling:

You can revive if you have a contribution to make. Same rule as Contests Page.

[img]http://www.mememaker.net/static/images/memes/2462153.jpg[/img]

@below: ahaan meant "contests page [b]posts[/b]", but whatever.

Wait, it's most definitely okay to revive posts in the contests page; that's the reason why the contests page exists, after all.

Reviving other posts, however, is a bit more hairy in terms of acceptability.


inequalities
calculus
derivative
Cauchy Inequality
inequalities proposed

[quote="lasha"]Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 25$. Find the minimal possible value of the expression $ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}$.[/quote]
We have : $ M^2\equal{}( \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y})^{2} \geq 3(x^2\plus{}y^2\plus{}z^2)\equal{}75$
$ Min P\equal{}5\sqrt{3} \Leftrightarrow x\equal{}y\equal{}z\equal{}\frac{5}{\sqrt{3}}$

But how do you know that $ \frac{5}{\sqrt3}$    ?
AM GM will not always give you the minimum rite?
BY AM GM, $ (a\plus{}b\plus{}c)^2 \ 3(ab\plus{}bc\plus{}ca)$ for postive reals $ a,b,c$.
I know that this proof can be given using Cauchy but how do you know when one can get the minimum value of the expression?

I don't understand what's false in the solution. The author mentions the equality cases as well and only uses Cauchy inequality, which you wrote. (Namely $ (a\plus{}b\plus{}c)^{2}>\equal{}3(ab\plus{}bc\plus{}ca)$).

Yaaeehh...
Easy one.
Just see that
\[ \frac{1}{3}\left( {\frac{{xy}}{z} \plus{} \frac{{yz}}{x} \plus{} \frac{{xz}}{y}} \right) \ge \sqrt {\frac{{x^2  \plus{} y^2  \plus{} ^2 }}{3}}  \equal{} \sqrt {\frac{{25}}{3}} \,\,\,\,\,
\]

which gives us minimum value

\[ \frac{{xy}}{z} \plus{} \frac{{yz}}{x} \plus{} \frac{{xz}}{y} \ge 5\sqrt 3
\]
 :P  :P  :P

The equation $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}25$ is symmetric towards x,y,z. Also, $ \frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{zx}{y}$ is symmetric, therefore: $ \frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{zx}{y}\equal{}3\frac{xy}{z}$. Minumum value of the hemisphere is when x=y=z, therefore $ 3x^2\equal{}25$, $ x\equal{}\frac{5}{\sqrt{3}}$, which makes final answer $ \frac{5}{\sqrt{3}}*3\equal{}5\sqrt{3}$.

lagrange multiplier
$F(x,y,z,L)=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}+L(x^2+y^2+z^2-25)$
by derivatives we obtain
$\frac{x^2y^2+x^2z^2-y^2z^2}{x^3yz}=\frac{x^2y^2+y^2z^2-x^2z^2}{xy^3z}=\frac{x^2z^2+y^2z^2-x^2y^2}{xyz^3}=-2L$
hence$(x^2-y^2)(x^2z^2+y^2z^2-x^2y^2)=0$
let $x^2=a,y^2=b,z^2=c$
then
$a=b or (a+b)c=ab$
$b=c or (b+c)a=bc$
$c=a or (c+a)b=ca$
if at least two of the equations above are the latter one,by generality let
$(a+b)c=ab,(b+c)a=bc$
then$bc>ab>bc$contradiction!
so at least two are the former one,hencde
$a=b=c=\frac{25}{3}$,yielding minimum $5\sqrt{3}$.

suppose $x\geq y\geq z$
\[ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y}= \sum_{cyc} x^2.\frac{y}{xz} \geq \frac{1}{3} (x^2+y^2+z^2)(\sum_{cyc} \frac{y}{xz}) \]
\[ \Rightarrow \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq  \frac{25^2}{3xyz} \geq 5\sqrt{3} \]
equality for $x=y=z=\frac{5}{\sqrt{3}}$

[quote=Alexander333][quote=lasha]Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}25$. Find the minimal possible value of the expression $ \frac{xy}{z} \plus{} \frac{yz}{x} \plus{} \frac{zx}{y}$.[/quote]

Prove that $( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2} \geq 3(x^2+y^2+z^2) = 75$ . If we do that, the answer will be $\sqrt75$ which realises for $x=y=z=\frac {5}{\sqrt3}$.

$( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2}=\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} + 2(x^2+y^2+z^2)$

$2(x^2+y^2+z^2)$ will be cancelled and obtained inequality will be:

$\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} \geq x^2+y^2+z^2$

With the simplyficiation of that expression:

$\frac {\sum_{cyc} x^4y^4}{2} \geq \frac {\sum_{cyc} x^4y^2z^2}{2}$

This is right with Muirhead's inequality.[/quote]
OR AM-GM
$x^4y^4+y^4z^4\geq{2x^2y^4z^2}$....

[quote=SRIDARA]
OR AM-GM
$x^4y^4+y^4z^4\geq{2x^2y^4z^2}$....[/quote]

Muirhead's inequality and AM-GM are same yeah? or not?


logarithms
integration
calculus
function
calculus computations

Solve equality:
x.y'.ln(y/x)=x+y+ln(y/x)
Thank you very much !

hello, do you mean $ xy^'\ln\left(\frac{y}{x}\right)=x+y+\ln\left(\frac{y}{x}\right)$ ?
Sonnhard.

which is a first-order non-linear ordinary differential equation. Can it be solved analytically or do we have to consider numerical methods of solution?

This type of ODE is often called "homogeneous", it is of the form $ y' \equal{} F\left(\frac {y}{x}\right)$.  Making the substitution $ u \equal{} \frac {y}{x}$ we find $ y' \equal{} xu' \plus{} u$ so that our ODE becomes $ xu' \plus{} u \equal{} F(u)\Rightarrow \frac {u'}{F(u) \minus{} u} \equal{} \frac {1}{x}$.  Assuming an initial condition $ u(x_0) \equal{} \frac {y(x_0)}{x_0}$ the solution is, in implicit form,
\[ \int_{u(x_0)}^{u(x)} \frac {dt}{F(t) \minus{} t} \equal{} \log \left(\frac {x}{x_0}\right)\]
where $ y(x) \equal{} u(x)x$ and, in our case, $ F(t) \equal{} \frac {1 \plus{} t \plus{} \log t}{\log t}$.  I don't know if that integral is expressible in closed form, much less that the resulting function of $ u$ is explicitly invertible, but this is a perfectly fine representation of the solution nonetheless in the sense that all possible values of $ u$ (and therefore $ y$) can be computed easily for any given $ x$ just by integrating.


calculus
derivative
LaTeX
integration
function
limit
calculus computations

Here's a problem in my Calc book I'd appreciate some help with.

[quote]If f is a differentiable function and g(x)=xf(x), use the definition of a derivative to show that g'(x)=xf'(x)+f(x).[/quote]

Of course, that makes perfect sense when you use the product rule, but I can't figure out how to do it using the definition of a derivative.  Perhaps we should start with the official definition of a derivative?

I would really appreciate any help with this!

Awesome!  I think I actually understand that!

I was going to download and use Latex to demonstrate how I'm solving it, but I don't think I can figure it out.  I just downloaded and installed both TeXnIcCenter and MiKTeX, but I can't figure out how to use them.

If you click quote on my earlier post and fiddle around with the preview button, you'll (start) to understand how you can use $\LaTeX$ in the forums.  It doesn't require any downloads - it's mostly just enclosing text in dollar signs

e.g, \frac {d}{dx}\int^x_a f'(x)dx=f(x) becomes $\frac{d}{dx}\int^x_a f'(x)dx=f(x)$

Ok, well here goes nothing...

We can start by stating the formal definition of the derivative.  The derivative of a function is the limit of the difference quotient as $h$ approaches $0$.

$\lim_{h \to\ 0}\frac{g(x+h)-g(x)}{h}$

Let's start by working with the difference quotient and then plugging back into the limit.  We can replace $g(x)$ with $xf(x)$:

$\frac{(x+h)f(x+h)-xf(x)}{h}$

Distribute the $(x+h)$:

$\frac{xf(x+h)+hf(x+h)-xf(x)}{h}$

Group and factor out the $x$ from the first and last terms in the numerator:

$\frac{x[f(x+h)-f(x)]+hf(x)}{h}$

So now we have the difference quotient to take the limit of:

$\lim_{h \to \ 0}\frac{x[f(x+h)-f(x)]+hf(x)}{h}$

Then we can break it up:

$\lim_{h \to\ 0}\frac{x[f(x+h)-f(x)]}{h}+\lim_{h \to\ 0}\frac{hf(x+h)}{h}$

We can pull the x out of the first limit and simplify the second limit:

${x\lim_{h \to\ 0}\frac{f(x+h)-f(x)}{h}+\lim_{h \to\ 0}f(x+h)}$

We recognize the first limit as the definition of the derivative of a function, in this case $f(x)$.  The second limit simply becomes $f(x)$.

$xf'(x)+f(x)$

So, is that a good way to go about proving this problem?  Oh, and I think I got the hang of this latex stuff...somewhat.

Thanks!


Assume that all the interior angles of a polygon are all equal. Prove that the lengths of the consecutive sides of the polygon can not form a strictly monotone sequence.

(Edited for spoiler)

[hide]Assume such a polygon exists.

Let 0 < a_0  :le: a_1  :le: ...  :le: a_(n-1) be the sides and v_n=v_0, v_1, ..., v_(n-1) be the vertices in the complex plane, with |v_(k+1) - v_k| = a_k

Rotate/reflect so that v_1 - v_0 = a_0, v_2 - v_1 = a_1*e^(2 :pi: i/n), v_(k+1) - v_k = a_k*e^(2k :pi: i/n)

Then a_0 + a_1*z + ... + a_(n-1)*z^(n-1) = v_n - v_0 = 0 for z = e^(2 :pi: i/n). Multiplying by 1-z,

a_0 + z*(a_1 - a_0) + ... + z^(n-1)*(a_(n-1) - a_(n-2)) = a_(n-1)*z^n = a_(n-1).        (1)

Since  |a_0| + |z*(a_1 - a_0)| + ... + |z^(n-1)*(a_(n-1) - a_(n-2))| =

|a_(n-1)*z^n|, equality in (1) forces the ratio z*(a_1 - a_0)/a_0 to be a positive real number. Since it is a real multiple of z and z is not real, we cannot have equality and the polygon does not exist.

Just a variation on AOPS Challenge 18. If the sequence of sides is nondecreasing and the angles are equal, the polygon is regular.[/hide]

[quote="jmerry"]
Just a variation on AOPS Challenge 18.[/quote]

You got it! Put into spoiler please. It's still fresh. :-)

[quote]If the sequence of sides is nondecreasing and the angles are equal, the polygon is regular.[/quote]

This is better stated than mine!


calculus
integration
function
trigonometry
real analysis
parameterization
real analysis unsolved

How to prove that \[ \int_{\minus{}\infty}^{\infty}\left[\sum_{n\equal{}0}^{\infty}\frac{x^{2 n}}{(2n\plus{}1)!}\right]dx\equal{}\pi\] regards,
Bhupala.

$ \sum_{n\equal{}0}^{\plus{}\infty}\frac{x^{2n\plus{}1}}{(2n\plus{}1)!}\equal{}\sinh x$, and so $ \sum_{n\equal{}0}^{\plus{}\infty}\frac{x^{2n}}{(2n\plus{}1)!}\equal{}\frac{\sinh x}{x}$. Your integral then becomes (noting that this function is even):

$ I \equal{}\int_{\minus{}\infty}^{\plus{}\infty}\frac{\sinh x}{x}\,dx \equal{} 2\int_{0}^{\plus{}\infty}\frac{\sinh x}{x}\,dx \equal{}\int_{0}^{\plus{}\infty}\frac{e^{x}\minus{}e^{\minus{}x}}{x}\,dx$

$ I\equal{}\infty.$ The integral diverges, badly. Go back to the original and interchange sum and integral: every single term diverges to $ \plus{}\infty.$ There's no way to get $ \pi$ out of this.

Sorry the question goes like this\[ \int_{\minus{}\infty}^{\infty}\left[\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^{n}x^{2 n}}{(2n\plus{}1)!}\right]dx\equal{}\pi\]

Then that is $ \int_{\minus{}\infty}^{\infty}\frac{\sin x}{x}\,dx \equal{}\pi.$

(The integral converges as an improper integral but not as a Lebesgue integral.)

This has been posted quite a few times around here. You could try searching.

[quote="Kent Merryfield"]Then that is $ \int_{ \minus{} \infty}^{\infty}\frac {\sin x}{x}\,dx \equal{} \pi.$[/quote]
$ \int_{ \minus{} \infty }^{ \plus{} \infty } {\frac{{\sin x}}{x}\,dx}  \equal{} 2\int_0^\infty  {\frac{{\sin x}}{x}\,dx} .$

The following parameter may be useful: $ \frac{1}{x} \equal{} \int_0^\infty  {e^{ \minus{} ux} \,du} .$

Now we construct a double integral and $ \int_0^\infty  {\frac{{\sin x}}{x}\,dx}  \equal{} \frac{\pi }{2}\,\therefore \,\int_{ \minus{} \infty }^\infty  {\frac{{\sin x}}{x}\,dx}  \equal{} \pi \,\blacksquare$


geometry
trigonometry
area of a triangle
Heron\u0027s formula

[hide="Number 2"]By Heron's, $A=\sqrt{s(s-a)(s-b)(s-c)}, s=\frac{a+b+c}{2}$ plugging in 10 10 12 for a b c, $A=\sqrt{16 \times 6 \times 6 \times 4} = 4 \times 6 \times 2 = 48 \ \text{units}^2$[/hide]

[quote="Silverfalcon"]Here are 4 good but simple problems for getting started forum. Enjoy!

1. Let $S = 1997^2-1996^2+1995^2-1994^2+\cdots +3^2-2^2+1^2$. Then the largest prime factor of $S$ equals:

2. Two of the sides of a triangle have length 10 and the third side has the length 12. What is the area of the triangle?

3. To the nearest degree, the first quadrant angle $x$ that solves the equation $2 \cos x = \sin x$ is:

4. Find the sum of the digits of the square of the twelve digit integer 999,999,999,999.

For the fourth one, you can find the pattern but there is more algebraic way. Can you find it?  :)[/quote]

[hide="number 4"]
I think it's $11664$. The sum of the digits in the twelve digit integer is $108$, and $108^2=11664.$ I'm not sure what you meant by "more algebraic way" but..if you do long multiplication or whatever it's called  :) you have $999,999,999,999\cdot9$ which has a digit sum of $972$ and then you do that $12$ times so $972\cdot12=11664$[/hide]

[hide="number 3"]

Divide both sides by $\cos x$ to get
$2=\tan x$
$x\approx63$[/hide]

[hide="number 1"]

$3989$?
I'm not even sure if that's prime  :|

I used $a^2-b^2=(a-b)(a+b)$ and $(1995-1994)(1995+1994)=3989$
I don't have a methodical way of finding the largest prime [/hide] :?

[quote="Hokkage"]I think you're wrong... If you take 1,000,000,000,000 and square it, you get 1,000,000,000,000,000,000,000,000 which is 25 digits, and must have more digits than 999,999,999,999 squared. If the maximum digits is 24, and the maximum for each digit is 9, then the maximum for their sum is 24*9=216. Maybe I misread the question though.[/quote]

I was calculating the sum of the digits, not the number of digits  :)  sorry if i worded incorrectly

Yes but the sum of the digits cannot exceed the number of digits times 9 as 9 is the largest digit.

[hide="2"]
Drop an altitude to the side of length 12, and you have two 6-8-10 right triangles, each with area 24, making the area of the entire thing 48.
[/hide]

-BLt

[quote="Hokkage"]I think you're wrong... If you take 1,000,000,000,000 and square it, you get 1,000,000,000,000,000,000,000,000 which is 25 digits, and must have more digits than 999,999,999,999 squared. If the maximum digits is 24, and the maximum for each digit is 9, then the maximum for their sum is 24*9=216. Maybe I misread the question though.[/quote]

Hm
your explanation makes SENSE
I just made up a "formula" that would solve for an $x^2$'s sum of digits after one time i was driving back from a math competition and looking over my test  :) I don't know if it's right
But if my answer isn't right, how would you find the right answer?  :|

For number 4, couldn't you use $(n-1)^2=n^2-n+1$, $n=10^12$?

So it'll be $10^24-2(10^{12})+1$.
By logic $10^24-10^{12})$, It would be $99...9990...0$ with 12 $9$s and 12 $0$s.
Subtracting another $10^{12}$ and adding a 1 does nothing, so

[hide]
it equals $9\cdot12=108$?
[/hide]

or was that the pattern? :|

[hide="4"]First, we should notice that $999,999,999,999=10^{12}-1$.

Now, $n=999,999,999,999^2=(10^{12}-1)(10^{12}-1)=10^{24}-2\cdot10^{12}+1$

If we take the common factor out of the first two terms, we get $n=10^{12}(10^{12}-2)+1$

The number $10^{12}-2$ is $999,999,999,998$. Then we have to multiply it by $10^{12}$, i.e. to append $12$ zeros, and add $1$. Therefore, its digits are $11$ times nine, $1$ time eight, $11$ times zero and $1$ time one - in total it makes $108$.[/hide]

[hide="1"]$S=3993+3989+...+5+1=4\left(\frac{998(999)}{2}\right)+999=1997(999)=3^3\cdot 37\cdot 1997$
Largest prime factor is $\boxed{1997}$.[/hide]

[hide="2"]We have two 6-8-10 triangles attached at the 8. Therefore, the area is $\frac{1}{2}(12)(8)=\boxed{48}$.[/hide]

[hide="3"]Calculator yields $\boxed{63}$. :D I assume calculators aren't allowed, though? If so, I really have no idea how to approach it... if not, it's rather easy, though. Divide both sides by $\cos x$, take $\arctan 2$. *shrug*[/hide]

[hide="4"]Because of the way the number multiplies out, we have a $1$ in the units place followed by a string of zeroes, then an $8$ and 11 $9$'s, so our answer is $9\cdot 11+8+1=\boxed{108}$.

[b]Added:[/b] Another way to do it: $(10^{12}-1)(10^{12}-1)=10^{24}-2(10^{12})+1$, so we have an $8$ in the $12^{th}$ place, all nines on the LHS of that, and zeroes and a $1$ on the RHS, so our answer is $11\cdot 9+8+1=\boxed{108}$.[/hide]

Oops, I didn't find the largest prime factor. :oops: Made a small mistake anyway, as well. Going to fix that now.

[quote="robinhe"][hide="For #2"]You could just use heron's formula if you didn't realize anything else or you were lazy to do the thinking.  :sleeping: [/hide][/quote]

The easiest way was to drop the height. then use pythegorenon therom to find it. :juggle:

[hide="3"]you could solve it geometrically...i'm not sure how you'd solve it without a calculator, since it asks for degrees. so i just made a right triangle in the first quadrant where the sides were a(the side opposite the angle), b(side adjacent to the angle) and c(the hypoteneuse). so according to the question, $a/c=(2b)/c$ so a=2b. but that's just tan(2). eh just another method. it's probably be easier just dividing both sides by cosx. [/hide]

 :blush:


How many ordered pairs of integers $(m,n)$ satisfy the equation $m^3+6m^2+5m=27n^3+9n^2+9n+1$?

[hide]This can be simplified to \[ m(m^2+5)=(3n^2+1)(9n+1) \] The right side is $1\mod{3}$, and thus, so is the left side.  If $m \equiv 0\mod{3}$, then the left side is divisible by 3.  For $m \equiv 2\mod{3}$ or $m\equiv 1\mod{3}$, we have $m^2 \equiv 1\mod{3}$, and the left side is divisible by 3.  Thus, the left side is always divisible by 3.  So there are no solutions? $\boxed{0}$[/hide]

Yes, sorry, now the problem is harder. :)

Okay
[hide]
The right hand side is $1\mod{3}$.  The left hand side can be reduced to $m(m+1)(m+5)$.  These three terms are different mod3.  One of them must be $0\mod{3}$.  Thus, the left hand side is always divisible by 3.  Hence, there are still no solutions.  $\boxed{0}$[/hide]

I remember this problem. This was last problem in the second round of Norwegian Math Olympiad. I don't remember the year though. But the solutions gave above are exactly same as how they did it.

[quote]Possibly there's a typo in the LHS? ($6n^2$ should read $6m^2$)[/quote]
[quote]es, sorry, now the problem is harder[/quote]
..Actually m or n does not matter, if you take, as K81o7 did, modulo 3, RHS is 1, and LHS is zero.
$m^3+6 ( does\_not\_matter) +5m$ = $m^3-m = (m-1)m(m+1)$  (Three consecutive integers, one has to be divisible by 3)  :D


Find the integer solutions of:$2005^n+2006^n=x^n$

[quote="seamusoboyle"]Im assuming without FLT[/quote]What?

If x,n are positive integers, then by FLT or Fermat's Last Theorem we have it has no soln, for n>2.
So we check n=1 or 2 or 0. Done.

Bomb

People, forgot this FLT things! Either give proofes for FLT (and post them here, not links  :P ) or solve the question above; in both cases: elementary solutions required (well elementary depends, but Wiles proof for FLT is surely not).


do we have the word $koron$ in music.?

i mean in the notes...


function
geometry
topology
algebra unsolved
algebra

A question which i am VERY bad Stuck in it : :maybe: (Dont laught, its not funny)

Problem: let $A$ be a unit square $[0,1]$x$[0,1]$.
Prove that we dont have a Continuous Function like $\mathcal{F}$ that is dense in $A$.

By dense i mean , if $m,n$ are two different point in  $A$ such that segment $mn$ is not vertical then $f$ cuts the [i]segment [/i]$mn$ in at least one point.

Any not rigorously ideas are also Welcome.

Continuous from $[0,+\infty[$ to $[0,1] \times [0,1]$ for instance ? Easy then, you follow vertical lines dividing your square and distant by $\frac{1}{n}$ ... once the whole square has been walked through from the left to the right, you come back following lines $\frac{1}{n+1}$ apart, etc ...

Right i didnt though of that.

if we want  $f: [0,1] \longrightarrow [0,1]$ and continuous

Do you know any construction in this form?

You mean $f: [0,1] \longrightarrow [0,1]^{2}$ instead¿

As picture of a compact set, it would be compact, thus (we have a Hausdorff space) closed, thus(we have a Hausdorff space)  be the whole set if dense.

No , Though thanks . I actually mean this :

Find a continuous function  $f: [0,1] \longrightarrow [0,1]$ that if $m,n$ are two different point in  $A$ such that segment $mn$ is not vertical then $f$ cuts the [i]segment [/i]$mn$ in at least one point where $A= [0,1]$x$[0,1]$

Hmm ok:

The induced map $g: t \mapsto (t,f(t))$ is continuous, and it is also dense in the square (just take two nonvertical points in the neighbourhood and take an intersection point). Thus again it has to be surjective, impossible.

@Rust: your map is not well defined, where do you map $0.1=0.09999...$¿

[quote="ZetaX"]Hmm ok:

@Rust: your map is not well defined, where do you map $0.1=0.09999...$¿[/quote]
Yes, for correctly define $f(\frac{a}{10^{k}}),a\in N$ we can use only one write x.
 I am wrong. This function is not continiosly in these points.


function
superior algebra
superior algebra unsolved

Let $A$ be a nonempty set with at least 2 elements, $\mathbb N = \{1, 2, 3, ...\}$.
Is there a map $f: A^{\mathbb N}\to A$ (where $A^{\mathbb N}$ is the set of sequences in $A$) with the property that for every $n \in \mathbb N$ and every choice of arguments $a_{i}$ for $i \ne n$, the map $g: A \to A, g(x) = f(a_{1}, a_{2}, ..., a_{n-1}, x, a_{n+1}, ...),$ is injective?

Such a map $f$ exists from $A^{(\mathbb N)}$ to $A$, where $A^{(\mathbb N)}$ is the set of finite sequences in $A$ (or equivalently, the set of sequences in $A$ which differ only in finitely many places from a distinguished element $a_{0}$ of $A$).

This question came to my mind while trying to solve a puzzle which only needs a function on $A^{(\mathbb N)}$.

LiMa

Yes, but you need the Axiom of Choice or an equivalent.

If $A$ is a field, let $S$ be a subspace of $A^{\mathbb{N}}$ which is maximal among those subspaces which intersect $A^{(\mathbb{N})}$ trivially. Such an $S$ exists by Zorn's Lemma.
Now, for any $x\in A^{\mathbb{N}}$, $Ax+S$ intersects $A^{(\mathbb{N})}$ nontrivially; let $ax+s=y$, with $a\neq 0\in A,s\in S,y\in A^{(\mathbb{N})}$. If $ax+s'=y'$, then $s=s'$ and $y=y'$; this representation of $ax$ is unique. Multiplying by $a^{-1}$, we have $A^{\mathbb{N}}=S\oplus A^{(\mathbb{N})}$. With this decomposition, we choose $f(a_{1},a_{2},\dots)=\sum a_{i}$ on $A^{(\mathbb{N})}$ and $f(x)=0$ on $S$, then extend to the whole space by linearity.

If $A$ is not a field (the cardinality of $A$ is finite and not a prime power), we have to work harder. Instead, we write $A=\oplus A_{k}$, with each $A_{k}$ a field. Construct $f$ as above on each $A_{k}$, and extend to all of $A^{\mathbb{N}}$ by linearity: $A^{\mathbb{N}}=\oplus A_{k}^{\mathbb{N}}$.

$\mathbb{N}$ can be replaced by any set in this argument. Its exact size is unimportant.

Thank you.
Why is this always so easy - after you know the solution? *g*

LiMa


I was trying to simplify this fraction:

[hide="The Fraction"]$ \frac{1}{\sqrt{11\minus{}2\sqrt{30}}} \plus{} \frac{3}{\sqrt{7\minus{}2\sqrt{10}}} \plus{} \frac{4}{\sqrt{8\minus{}4\sqrt{3}}}$[/hide]

[hide="My Experience"]After solving this I got a really complex answer, and when I checked the answer key to see if I was wrong or not, I got shocked :o .....the answer key gave a very surprising answer....I'm not going to reveal it because it'll take all the fun away :wink: . [/hide]

[hide="Note"]Phew! That took a long time! A large amounts of previews, edits, Latex reading, etc.[/hide]

A suggestion:

[hide]Multiply numerator and denominator  in the first by  $ {\sqrt{11+2\sqrt{30}}}$,the next one  by $ {\sqrt{7+2\sqrt{10}}}$ and the last one by $ {\sqrt{8+4\sqrt{3}}}$.

My answer is $ \sqrt{6} + 2{\sqrt{5}} + 2{\sqrt{(2+\sqrt{3})}}$.[/hide]

We can simplify the 3rd term in muks' answer by letting $ 2 \plus{} \sqrt{3} \equal{} (a \plus{} b\sqrt{3})^2$. Expanding the RHS gives $ a^2 \plus{} 2ab\sqrt{3} \plus{} 3b^2$. We know $ ab \equal{} \frac{1}{2}$ and $ a^2 \plus{} 3b^2 \equal{} 2$, and we can solve from there for $ a$ and $ b$.

rt(11-2rt30)=rt6-rt5

rt(7-2rt10)=rt5-rt2

rt(8-4rt3)=rt(8-2rt12)=rt6-rt2

Multiplying each dude by its conjugate

rt6+rt5+3(rt5+rt2)/3+4(rt6+rt2)/4=2(rt2+rt5+rt6) and we win.

You may want to refer to AoPS volume 1 for this kind of stuff.

BTW, what test did this problem come from?

No, my calculator doesn't even give that :P

All the terms are positive, so how can it give an answer of 0? Which book did this come from?

I too don't see any way by which we can obtain 0 as the answer.

I would guess it made a mistake and took the negative root for one or something, and ended up with like

rt6-rt2+rt2-rt5+rt5-rt6

[quote="generalmotors"]The answer key says the answer is $ 0$ :wacko:
Did anyone get $ 0$?[/quote]

Sum of three positive numbers is zero?

I would say you wrote the problem incorrectly!

Er, kinda hard to mess up typing something on a book and copying it?  :wink:
And maybe the solution as wrong as mewmew said.


Let f(x) = x^3 + 17. Prove that for each natural number n \geq 2 there is a natural number x for which f(x) is divisible by 3^n but not by 3^(n+1).

I am pretty sure this problem was solved before on the site. Again I am not sure, but I think I solved it.
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