I am in Mathcounts from Indiana.

Who else is?

Edit: 200th Post!

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

[quote="cognos599"]kheg_k what did you get on the indiana mathleague this earth.[/quote]

ummm...is it just me or does that make absolutely no sense (no offense)? i think it's supposed to be "this [i]year[/i]"... :lol:

sorry i don't know what i was thinking when i typed that.
 :blush:

im in 5th grade.........im a hoosier and i wonder if i will get in mathcounts


What is the maximum number of space diagonals can a polyhedron with 10 vertices can have.  A space diagonal is a diagonal with endpoints as vertices of the polyhedron but lies completely inside the polyhedron.

Is there a general case with n sides?

[quote="the future"]Is there a general case with n sides?[/quote]
Maybe not in general, although the following is a sort of attack that might be able to establish some good bounds.

I assume you meant "vertices" for "sides," and that you're only really interested in convex polyhedra (or, at least, polyhedra without holes).  In this case, we may be able to make some arguments via a calculation as follows: suppose we have a polyhedron with $ f_3$ triangular faces, $ f_4$ quadrilateral faces, $ f_5$ pentagonal faces and so on.  Then the total number of faces is $ f_3 \plus{} f_4 \plus{} \ldots$ and the total number of edges is $ \frac {1}{2}(3f_3 \plus{} 4f_4 \plus{} 5f_5 \plus{} \ldots)$, so by Euler's polyhedral formula, $ n \equal{} 2 \plus{} \frac {1}{2}(3f_3 \plus{} 4f_4 \plus{} 5f_5 \plus{} \ldots) \minus{} (f_3 \plus{} f_4 \plus{} \ldots)$.  This equation will have only finitely many solutions in nonnegative integers, and from each solution it's easy to quickly count up the number of interior diagonals (by instead counting the number of edges and face diagonals).  Whether each such solution necessarily leads to a convex polyhedron is not immediately obvious to me, nor is it immediately clear which solution will be optimal.

Later edit:
Okay, direct computation for $ n \leq 10$ suggests that the optimal case is always with $ 2n \minus{} 4$ triangles, arranged as a double-pyramid.


First off I am new here so I am sorry if this is the wrong place to post this (site or category).  If not can someone point me in the right direction?

I was curious if there are any online IQ tests that are fairly accurate?  I know the only way to know your IQ for sure is to take a real test, I'm just looking for something that would give a good idea of where I stand.

Also I seen the High IQ Society website (http://www.highiqsociety.org), if I scored well enough to join would there be any advantage or is that just a scam to get peoples money?  The true meaning of irony, scamming the very people who should be smart enough not to fall it?

Thanks,
Byam

Hi there!  You might try a [url=http://www.artofproblemsolving.com/Forum/search.php]search of this forum[/url] to see what comments people have had about this topic before.  There have definitely been a number of threads concerning IQ, and I think I've seen the High IQ Society mentioned here as well.  (This is the appropriate forum for that sort of question, by the way.)

Well the MENSA test is by far the best but its not online.


arithmetic sequence
algebra proposed
algebra

Let $ {a_n}$ arithmetic progression  such that $ a_1,a_2\in Z$.Prove that $ a_2,a_3...a_{19}\in Z$ either $ a_2,a_3...a_{19}\notin Z$. Q.E.D  :lol:

__________________________________
$ \textbf{Azerbaijan Land of Fire }$

Let $ d$ be the difference. Then $ a_{20}\minus{}a_1\equal{}19d\in\mathbb Z$. If $ d\in\mathbb Z$, we have all $ a_i\in\mathbb Z$.

Otherwise, let $ 2\le i\le 19$.

Suppose $ (i\minus{}1)d\in\mathbb Z$. As $ 19$ is prime, we have $ gcd(i\minus{}1,19)\equal{}1$, and thus there are $ x,y\in\mathbb Z$ such that $ 19dx\plus{}(i\minus{}1)dy\equal{}d$, i.e. $ d\in\mathbb Z$, contradiction.

Thus $ a_i\minus{}a_1\equal{}(i\minus{}1)d\not\in\mathbb Z$,  Thus $ a_i\not\in\mathbb Z$. :)


an organic compund A is a comnponent of essential oils.A on treatment with excess of acidified dichromate solution gives B.B on treatment with Zn dust gives you a compund C.C undergoespolymerization with D giving you E.E is used as a thermal insulation layer.D on treatment with PBr5(excess) gives you a compouynd F.F in the presence of a strong base gives you G.G in redhot iron tune  gives you H.H is also produced by the decarboxylation(excess soda lime) of C.

identify A to H.Also give the common name of A and in which oil is bit a component???

See if this part is right:

[b][i]
D- ethylene glycol
F- ethylene dibromide
G- ethyne
H- benzene
C- Thallic acid
E- Glyptal
[/i][/b]

D,F,G and H are right.....
think abt others..........

Its obvious that C must a dicarboxilic acid, and as it can be decarboxylated to benzene, and as C is not phthalic acid, then there is a good chance of  being terephthalic acid.

yes that's right...........
C is terepthalic acid.....
think of others...

Are you sure there is a single answer for A? I don't think so.

yes i 'm sure that there is a single answer for A.


function
real analysis
integration
calculus
algebra
polynomial
topology

1. Define f on $ \mathbb{R}^{2}: f(x,y) \equal{} x \plus{} y$. Show that $ f^{\minus{}1}(E)$ is a Lebesgue-measurable subset of $ \mathbb{R}^{2}$ for every Lebesgue-measurable subset E of $ \mathbb{R}$.

2. Left f be a Lebesgue-integrable function on $ \mathbb{R}$. Show that $ F(x) \equal{} \sum_{n\equal{} \minus{}\infty}^{\infty} |f(x \plus{}n)|$ is integrable on $ [0,1)$

#2: ... and $ \int_0^1 F(x)\,dx\equal{}\int_{\mathbb{R}}|f(x)|\,dx$. You want to switch the order of sum and integral; it's justified because everything is nonnegative. The Fubini/Tonelli family of theorems applies to any combination of measures, including this instance of Lebesgue in one direction and counting measure in the other.
Even showing that $ F$ is finite almost everywhere is best done by showing that it's integrable.

[quote="jmerry"]The Fubini/Tonelli family of theorems applies to any combination of measures, including this instance of Lebesgue in one direction and counting measure in the other.[/quote]
While that's a reasonable way to put it, I'd feel a little safer on the "qual" if I wrote it up by quoting the Monotone Convergence Theorem.

Fair enough. The monotone convergence theorem is a member of that family, restated in terms of the partial sums.
Since this is a double-sided infinite sum, you probably want to use the sums from $ \minus{}n$ to $ n$, with a side argument that it doesn't matter how we go to $ \infty$ and $ \minus{}\infty$ because of positivity.

The followup question would be to define $ G(x)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}f(x\plus{}n).$

We'd want to show that the sum converges absolutely [i]a.e.,[/i] that $ G\in L^1[0,1],$ and that $ \int_0^1G\equal{}\int_{\mathbb{R}}f.$

This would both quote the previous result about $ F$ and use the other side of the Fubini/Tonelli family (as jmerry would put it) or to use that language I mentioned, the DCT rather than the MCT (with $ F$ as the dominating function.)

Per Kent's Reformulation we can argue as follows:

i) To show that G converges a.e, we note that the integrability of $ F(x) \equal{} \sum_{n\equal{}\minus{}\infty}^{\infty} |f(x \plus{}n)|$  implies that F(x) is finite-valued a.e [for if $ |F| \equal{} \infty$ on [0,1], then $ \int |F| dx > n\cdot m([0,1]) \equal{}n$ for all n, giving a contradiction]. It then follows that G is absolutely convergent a.e.

ii) Also $ G \in L[0,1]$ since $ |G|\leq F$, which is integrable on [0,1].

I think the problem was testing one's familiarity with the DCT and its relatives.

[quote="akech"]1. Define f on $ \mathbb{R}^{2}: f(x,y) \equal{} x \plus{} y$. Show that $ f^{ \minus{} 1}(E)$ is a Lebesgue-measurable subset of $ \mathbb{R}^{2}$ for every Lebesgue-measurable subset E of $ \mathbb{R}$.
[/quote]

So the problem turns out to prove that if $ E\subset\mathbb{R}$ then $ \sigma(E)\equal{}\{(x,y) : x\plus{}y\in E\}$ is measurable.
Consider the following cases :
1. $ E$ is an open set of $ \mathbb{R}$ then $ \sigma(E)$ is open , so it is measurable
2. $ E\equal{}\cap_{i\geq 1} O_{i}$ where $ O_{i}$ is open then $ \sigma(E)\equal{}\cap_{i\geq 1}\sigma(O_{i})$
3. $ E$ is a set of meaure zero
4. $ E$ is measurable set then $ E\equal{}G_{\delta}\cup N$ where $ N$ is a set of measure zero, and consider $ \sigma(E)$

[quote="akech"]I think the problem was testing one's familiarity with the DCT and its relatives.[/quote]
Of course! I would expect any comp or qual on real analysis to have at least one and possibly several questions exploring this territory.

Extending the problem:

Now, suppose $ \int_{\mathbb{R}} f(x)g(x) dx \equal{} 0$ for every bounded measurable function g that satisfies $ g(x\plus{}1) \equal{} g(x)$ for almost all x. Show that for almost all x $ G(x)$ converges to 0, where $ G(x)$ is as defined in Kent's thread.

Can someone give a pointer to the last problem.

Define $ \widehat{f}(\xi)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{\minus{}2\pi i\xi x}\,dx.$ for $ f\in L^1(\mathbb{R})$ and $ \xi\in\mathbb{R}$

and define $ \widehat{G}(k)\equal{}\int_0^1G(x)e^{\minus{}2\pi i kx}\,dx$ for $ G\in L^1(\mathbb{T}_1)$ and $ k\in \mathbb{Z}.$

By $ G\in L^1(\mathbb{T}_1)$ I mean that $ G$ is periodic of period $ 1$ and integrable over any period.

Note: if $ G(x)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}f(x\minus{}n),$ then $ \widehat{G}(k)\equal{}\widehat{f}(k)$ for all $ k\in\mathbb{Z}.$ The proof of this is similar to our previous claims, and requires a DCT-justified interchange of sum and integral.

Since $ e^{\minus{}2\pi i kx}$ is periodic of period $ 1$ for any $ k\in\mathbb{Z},$ the hypotheses of the problem imply that $ \widehat{f}(k)\equal{}0$ for all $ k\in\mathbb{Z}.$ But then $ \widehat{G}(k)\equal{}0.$ By the uniqueness of Fourier series, that implies that $ G\equal{}0$ as an element of $ L^1(\mathbb{T}_1).$

==========

Here's another way to do this proof. It's really the same proof, it just looks a little different. I'll just give hints. We find a trigonometric polynomial $ P$ that approximates $ \text{sgn}(G(x))$ - I'll let you figure out what sense of approximation we want. We have that $ \int_0^1fP\equal{}0,$ so $ \int_0^1GP\equal{}0.$

You can supply the details from here.


function
integration
induction
calculus
calculus computations

Prove that if $f$ is Riemann integrable on $[a,b]$ and $g(x)=f(x)$ for all $x$ except for a finite number, then $g$ is Riemann Integrable.

Rough outline:

Find a partition for which the upper and lower sums for $f$ are within $\epsilon$ (or $\frac{\epsilon}2$ or $\frac{\epsilon}4$ or whatever it takes) of each other.

Refine that partition by adding finitely many points that isolate that "bad" points of $g$ within very skinny subintervals. Note that $g$ is bounded (since $f$ is bounded and $g$ differs from $f$ at finitely many points - and the maximum of a finite number of numbers is bounded.) Use the bound on $g$ to provide a worst-case estimate for the difference between the upper and lower sums for $g$ on those finitely many skinny intervals.

Argue that the difference between the upper and lower sums for $g$ is controlled by the bound for $g$ and by the skinny width of those intervals on those finitely many intervals containing the "bad" points, and the the difference between the upper and lower sums for $g$ is controlled on the rest of the interval by the difference between the upper and lower sums for $f.$

This outline can be and should be formalized, of course - but these are the main ideas.

It follows simply from the Riemann integrability condition for a bounded function.

Observe that since $f$ is Riemann integrable then $f$ is bounded and the set of points at which $f$ is discontinuous has Lebesque measure 0. Now your function $g$ is also bounded and since $f=g$ for all $x$ except for a finite number of them it also follows that the set of points where $g$ is discontinuous has Lebesque measure 0 and hence $g$ is Riemann integrable. It also follows that $\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx$.

It is also possible to prove this result by a, let's say, more elementary result namely by considering Riemann sums.

Define $h: [a,b] \rightarrow \mathbb{R}$ by $h(x) = g(x)-f(x)$.
Then $h(x) = 0$ except for a finite number of points ${c_{1}, c_{2}, ..., c_{n}}$.

Now there are $n$ points where $h(x) \neq 0$, each of which can belong to two subintervals in a given tagged partition $\dot{P}$. But only these terms will make a nonzero contribution to $S(h;\dot{P})$. So let $\delta_{\epsilon}= \frac{\epsilon}{2n}$.

If $\parallel \dot{P}\parallel < \delta_{\epsilon}$, let $\dot{P_{0}}$ be the subset $\dot{P}$ with tags different from ${c_{1}, c_{2}, ..., c_{n}}$, and let $\dot{P_{1}}$ be the subset of $\dot{P}$ with tags at the points ${c_{1}, c_{2}, ..., c_{n}}$.
Since $S(h;\dot{P_{0}})= 0$, we have $S(h;\dot{P}) = S(h;\dot{P_{0}})+S(h;\dot{P_{1}}) = S(h;\dot{P_{1}})$. Since there are at most $2n$ terms in $S(h;\dot{P_{1}})$ and each of which is less than $\delta_{\epsilon}$, we have $0 \leq S(h;\dot{P}) = S(h;\dot{P_{1}}) < 2n\delta_{\epsilon}= \epsilon$. Thus $h \in R[a,b]$ and $\int_{a}^{b}h = 0$.

Step 1: Instead consider 1 more point. By induction claim it works for $n$.

Step 2: As akech suggests define $h(x) = g(x)-f(x)$

Step 3: Note $h(x)=0$ except for some point in $[a,b]$.

Step 4: Show $\int_{a}^{b}h(x) dx = 0$ by definition of integral.

Step 5: So $f(x)+h(x)=g(x)$ and $f$ and $h$ are both integrable on $[a,b]$.

Step 6: $\int_{a}^{b}f(x)+h(x) dx = \int_{a}^{b}g(x) dx$

Step 7: $\int_{a}^{b}f(x)dx+0 = \int_{a}^{b}g(x)dx$ using the additive property.


As you know, we have $ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}}>2$ $ \forall a,b,c\in\mathbb{R}_{\plus{}}^{*}$ but we also have
$ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}}\ge\frac{3\sqrt{2}}{2}$ $ \forall a,b,c\in [1,4]$. Can you prove that? :D

[quote="The soul of rock"]As you know, we have $ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}}> 2$ $ \forall a,b,c\in\mathbb{R}_{\plus{}}^{*}$ but we also have
$ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}}\ge\frac{3\sqrt{2}}{2}$ $ \forall a,b,c\in [1,4]$. Can you prove that? :D[/quote]
I have found the following inequality before some minutes ago.  But I think that it is correct with all positive real numbers.   :D
$ \sum_{cyc}\sqrt{\frac{a}{b\plus{}c}}\ge\frac{3\sqrt{2}}{2}$

No, try with $ a\to 0$, $ b\equal{}c$. Its $ inf\equal{}2$

Actually, we can prove that if $ a,b,c\in [1,7\plus{}4\sqrt 3]$, then

$ \sqrt{\frac{2a}{b\plus{}c}}\plus{}\sqrt{\frac{2b}{c\plus{}a}}\plus{}\sqrt{\frac{2c}{a\plus{}b}}\ge 3$.

[quote="zaizai-hoang"]WLOG we can assume that $ a\plus{}b\plus{}c \equal{} 3$. Using the following result:
\[ \sqrt{\frac{2a}{3\minus{}a}}\ge\frac{3a\plus{}1}{4}\]

\[ \iff\frac{3(a\minus{}1)^{2}(3a\minus{}1)}{16(b\plus{}c)}\ge 0\]
It is true with all $ a,b,c\ge\frac{1}{3}$
I think it is not correct but I can find out the mistake  :D I hope anyone will find out my mistake[/quote]

Don't have $ a\plus{}b\plus{}c\equal{}3$ because $ a;b;c\geq 1$ then $ a\plus{}b\plus{}c\geq 3$

I am so sorry but I didn't use the condition of Vasc  :D  If it is correct, I will try to solve it with all positive real numbers  :D Using a small technique + SOS method or Vornicu Schur's inequalily is the best way possible :wink: I wait for a  reasonable explanation  :P

Due to homogeneity, the condition $ a,b,c\in [1,4]$ is equivalent to $ a,b,c\ge\frac{a\plus{}b\plus{}c}{9}$. If $ a\plus{}b\plus{}c\equal{}3$, then $ a,b,c\ge\frac {1}{3}$ and hence the solution of zaizai-hoang is valid.


i have a problem and i think it's quite hard for me, so if you can, please give me the proof for it.
a,b,c is the sides of a triangle.
max{A,B,C}<pi/2.
prove that
(a+b+c)/(2+ \sqrt 2)>R+r>(a+b+c)/4.

Arosisi,here is my solution:
Your problem is equal to:
 2>\sum sinA/ \sum cosA \geq 1+ \sqrt 2/2.
First,it is easy to see that:sinA(cosB+cosC) \leq cosB+cosC.And we st the same as it.Then plus it together,we have the result we need.
The second inequality is equal to:
  \sum (1-cos(C-pi/4))(sin(A-pi/4)+sin(B-pi/4) \geq 0.
But  notice that:
sin(A-pi/4)+sin(B-pi/4)=2sin(pi/4-C/2)cos(A/2-B/2) \geq 0.
So the inequality is proved. :)


Middle Schoolers - Try your hand at the AGMath Challenge contest. The problems can be found in the file below.
Here are the instructions for the contest:

[b]Quick Instructions:[/b]
Please read everything below before printing and working on the Challenge set.

- Print the problem set below.
- Solve as many problems as you can without a calculator in 20 minutes.
- E-mail your ten answers to jbatterson@agmath.com with the subject:  AGMath Challenge.
- Include your name as you would like for it to appear in the results, your school, and your state.
- Results will be posted on the website http://www.agmath.com shortly after the due date.
- Each week I will be posting challenge sets of ten problems each on this page for an informal competition. These are some of the problem sets I am using with the teams I coach in North Carolina and are similar to the more difficult Warm-Ups found in the MATHCOUNTS school handbooks.

Even if you can't answer all of the questions, it's worth your time to submit your answers.

Thanks!
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310667&start=20]Here[/url] is the AoPS thread for the contest.


I have four 3 cent stamps and three 5 cent stamps.  Using one or more of these stamps, how many different amounts of postage can I make?

Is there a formula for this?  I can't figure out how to apply combinations or permutations.

If there is a formula, I will leave that for students to discover.  Knowing a formula for a problem like this isn't particularly useful anywhere but on a math contest.  However, exploring such a problem in pursuit of a formula can be extremely valuable experience.

[color=indigo]Here is how I did it:
[hide]these are the possible amounts for the 3 cent stamps: 3, 6, 9, 12
and for 5 cent stamps: 5, 10, 15
And of course there could be 0 for either of them.

To check for repeats i first compared 5 and 10. with 5 you can get a ones digit of 5,8,1,4, or 7. with 10 you can get 0,3,6,9, or 2. So there are no repeats there (and the same for 10 and 15, 0 and 5, or 0 and 15).  For 5 and 15 to repeat there would have to be two possibilities for the 3 cent stamps that are 10 apart, like 4 and 14 (this would also have to be true for 0 and 10 to repeat). There are none, so there are no repeats.

  So there are 5 possibilities of the number of 3 cent stamps, and 4 possibilities for the 5 cent stamps. 4*5 = 20 (then you minus one because 0 and 0 isn't allowed). So there are 19 combinations. [/color][/hide]

that's sort of like finding the number of factors of a number with the prime factorization: 3^4 * 5^3, except you can't include the one into the factor list.
Am I right?

[hide="da answer"]find out how many factors 3^4*5^3 has. That's 20 factors.  Subtract one because you can't have

0 of each stamp. so the final answer is $\boxed{19}$[/hide]


ARML trigonometry Euler function algebra domain calculus

You have just entered room "armlonlinemathmeet."
Myths Apprentice has entered the room.
altheman1 has entered the room.
jwec05: Okay, here is the thing
jwec05: there is other chatroom
Myths Apprentice: hello
jwec05: but it's really distracted
jwec05: so I'm only going to select few people who will actually work
jwec05: is that all right with you guys?
Myths Apprentice: trying getting joml here
altheman1: no stellar supremacy....
pakman2012 has entered the room.
Myths Apprentice: i just invited pakman2012
towersfreak2006 has entered the room.
BaseballJoe2002 has entered the room.
Myths Apprentice: mind if I provide a problem?
Adunakhor I has entered the room.
towersfreak2006: shoot
jwec05: I have probs here
altheman1: ok, just to keep this reasonable, follow the aops math jam rules, don't post in sentence fragements, but post the whole idea in one post please, and try to make it coherent also
Myths Apprentice: i'm 4everwise by the way
Myths Apprentice: not kithsapprentice
Myths Apprentice: miths*
jwec05: I don't know other people much but if you know any "REASONABLE" people
jwec05: feel free to invite
jwec05: I basically trust people in here but people like Arnav, we don't want here
jwec05: it'll create chaos
jwec05: basic rules:
jwec05: This is ARML chatroom and I'll provide problems in level of Individual Round or Team Round
jwec05: If you know direct way to problems, feel free to tell
jwec05: Just not ANSWER only. That'll create too much chaos
jwec05: It's already 9:07 so let's start
jwec05: 1. (NO CALCULATOR) Determine all positive primes p such that $p^{1994} + p^{1995}$ is a perfect square
altheman1: factor it into p^{1994}* [p+1], we must find then the primes such that p+1=q^2
Myths Apprentice: prime factorization firtst
Myths Apprentice: yeagh
towersfreak2006: then p=q^2 - 1
Myths Apprentice: sorry about the sentence fragments
BaseballJoe2002: p=(q-1)(q+1)
towersfreak2006: which always factors into (q-1)(q+1), unless q-1=1, so then p=3
jwec05: is it only value?
altheman1: we must have that one of the factors is 1, otherwise it would be composite, trying both of those, we get p=3
jwec05: I have answers but I didn't work out these problems either so yeah
jwec05: 3 is indeed only answer
towersfreak2006: yes, for any other value of q, p would not be a prime
jwec05: Problem 2
jwec05: Calculator allowed
Myths Apprentice: there are calculators allowed?
BaseballJoe2002: on teh team round
towersfreak2006: calculators are allowed on team and power rounds?
Myths Apprentice: oh, this is a team round
BaseballJoe2002: yeah
jwec05: 2. Determine all integer values of $X$ with $X$ from 0 to 90 degrees inclusive for $(\cos X + i \sin X)^{75}$ is a real number
jwec05: Team and Power Round, yup
jwec05: This is Team Round
BaseballJoe2002: you need an i in there, right?
towersfreak2006: should there be an i somewhere?
altheman1: you mean cos x+i sin x
jwec05: my bad, it should be + i*sin X
jwec05: yeah. oops
Myths Apprentice: ok, so just demoivre's
altheman1: obviously, we put in into euler's exponential form
Myths Apprentice: we want i to be gone
towersfreak2006: this is equivalent to cis 75x
pakman2012: sin75x= 0
towersfreak2006: or just cos 75x + i sin 75x = 0
towersfreak2006: so sin 75x = -
towersfreak2006: *0
altheman1: e^{ 75x*i}=real
jwec05: I think it's just easy to use what Joe and towersfreak said
jwec05: sin (75x) = 0
towersfreak2006: i am nameless
jwec05: no need to go euler I think
jwec05: I don't know your real name :-P
altheman1: alex
Myths Apprentice: so 75x is a multiple of 180 or 360
jwec05: if it's multiple of 180, it will be for 360 as well right
Myths Apprentice: 75=3*5^2 might help
Myths Apprentice: yes
BaseballJoe2002: So x=k*180/75.
Adunakhor I: it's including zero though, right?
towersfreak2006: 75x = 180n
jwec05: yeah
Myths Apprentice: yes
altheman1: divide it to get x= 60/25 *k
towersfreak2006: 15x = 36n
towersfreak2006: 5x = 12n
Myths Apprentice: we can simplify a bit
towersfreak2006: x is a multiple of 12
Myths Apprentice: yeah, x is a multiple of 12
jwec05: how many integers are multiple of 12 from 0 to 90 inclusive
towersfreak2006: 0 <= x <=90 implies 0, 12, 24, 36, 48, 60, 72, 84
jwec05: Yup
jwec05: That's the correct answer
Myths Apprentice: which year is this? just wondering
jwec05: 1995
jwec05: Next one
jwec05: NO CALCULATOR
pakman2012: Is ARML like AMC... where you can sometimes use calculator, though you shouldn't need it?
Myths Apprentice: yes
Myths Apprentice: i think so
BaseballJoe2002: You usually don't need a calculator.
towersfreak2006: i don't remember a problem that must be solved with a calculator, actually
towersfreak2006: you, joe?
jwec05: 3. Let $f(x) = \sqrt{x+2} + c$. Determine all real values of c such that the graphs of $f(x)$ and its inverse $f^{-1} (x)$ intersect in two distinct points.
BaseballJoe2002: no
pakman2012: oh
altheman1: obviously, it y=x and y=\sqrt{x+2}+c are tangent then there is one solution
pakman2012: sqrt(x+2)+c=x^2-2
altheman1: that is our bound, if we make C larger, then they will intersect 2 times
jwec05: how did you get x^2-2 on RHS Saiem?
Myths Apprentice: inverse function, let c=0?
pakman2012: try to find the inverse of f(x)
jwec05: x = sqrt{y+2} + c isn't it?
Myths Apprentice: y=(x-c)^2-2
jwec05: I thought inverse is just flippin' x and y?
pakman2012: oh wait
pakman2012: i forgot the c
pakman2012: woops
Myths Apprentice: yeah, simplifying gets that
altheman1: yes, but we have to restricy the domain
altheman1: *restrict
altheman1: but there is a better way to do this
altheman1: a function intersects it's inverse if y=x
altheman1: or use y=x to solve for the intersections
altheman1: so...  x-c=\sqrt{x+2}  , there must be 2 solutions, immediately, we must restrict our domain to x>= -2
altheman1: everybody with me?....now what do we do?
jwec05: yeah but we're looking for c
altheman1: (discriminat...)
altheman1: discriminant
jwec05: Oh heh
BaseballJoe2002: yeah
BaseballJoe2002: discriminant > 0
altheman1: squaring gives  x^2-2cx+c^2=x+2   or    x^2 + (-1-2c)x + (c^2-2)=0
BaseballJoe2002: so c>-9/4?
altheman1: i don't know, i don't have any paper :P
BaseballJoe2002: The discriminant is (2c+1)^2 - 4(c^2-2) >0.
jwec05: that's correct
altheman1: wait
BaseballJoe2002: Expand it out and simplify to get c>-9/4.
jwec05: yeah but that's not the only one though
altheman1: how do we know that the solutions will work on the old equation
BaseballJoe2002: hmm
jwec05: to find its MAX, we put MIN of x
jwec05: put x = -2
jwec05: c = -2-0 = -2
jwec05: That's the MAX of c I think..
jwec05: I'm not sure on how to check on interval -2.25 < c <= -2 though
altheman1: do all of those solutions work in the original equation?
altheman1: no
BaseballJoe2002: For the max we want to find where the vertex of the graph lays on y=x.
altheman1: oh wait, maybe
jwec05: so where does the vertex of the graph lays?
BaseballJoe2002: So you want c=-2 as teh max.
altheman1: not necessarily the vertex, this is getting into calculus though, the line is tangent
jwec05: so just put min X -> max C?
BaseballJoe2002: When you change c in the original equation all you do is vertically shift the graph.
BaseballJoe2002: It's always at x=-2.
altheman1: this is the correct answer
altheman1: -9/4 < c<=2
jwec05: Yeah. That's correct interval.
BaseballJoe2002: You want to find where the leftmost point of the graph will lie on y=x.
jwec05: It was #7 on Individual Round so I don't think you would have time to actually plug value to check them
BaseballJoe2002: And the leftmost point (i.e. vertex)  is (-2,c)
jwec05: in case if anyone wanted to attempt it.
altheman1: (graphing it on paper might be helpful if you are not following)
altheman1: only 3 people are posting...
jwec05: #4
jwec05: Calculator allowed
Myths Apprentice: "only 3 people are posting...": sorry, was away
jwec05: 4. Assuming the expression converges, determine the largest integer $n$ with $n \leq 4,000,000$ for which $\sqrt{n+\sqrt{n+\sqrt{n+ \cdots}}}$ is rational
altheman1: it is fine, you don't have to post if you do not want to, i just like to include everybody
BaseballJoe2002: hmm...i already know how to do this one
altheman1: let that expression be equal to x
BaseballJoe2002: so i'll let others do it
pakman2012: wait so do it
pakman2012: so do i *
jwec05: so would this be like sqrt{n+x} = 400000 and solve for n?
altheman1: this is a standard technique that you should recognize immediately
jwec05: wait
Myths Apprentice: do we just do x=\sqrt{n+x}?
jwec05: RHS isn't right my bad
jwec05: yeah
altheman1: x^2-x-n=0
Myths Apprentice: ok
Myths Apprentice: so we want largest n that factors into two numbers
Myths Apprentice: their difference is 1
altheman1: all we need is to have the discrimiant be a perfect square
Myths Apprentice: or that
altheman1: 1+4n^2=q^2
jwec05: Or we can just multiply out two consecutive numbers such that product is less than 4000000
altheman1: for some integer q
BaseballJoe2002: n^2?
Myths Apprentice: the discriminant is 1+4n
towersfreak2006: 4,000,000 is (2,000)^2
Myths Apprentice: so we want that to be a perfect square
altheman1: we had the exact same thing as before, what do we do now?
towersfreak2006: so what's the greatest number less than that which is the product of two consecutive integers?
altheman1: it starts with an F
Myths Apprentice: "so what's the greatest number less than that which is the product of two consecutive integers?" wasn't that my original approach?
altheman1: factor
altheman1: 1=(q-2n)(q+2n)
towersfreak2006: yes, so finish the problem ;-)
altheman1: but wait, is q and integer?
altheman1: no...
altheman1: or is it?
Myths Apprentice: q would be integer
Myths Apprentice: because 1+4n=q^2, right?
Myths Apprentice: and n is integer
altheman1: yes, eventhough the problem says rational, we see from that equation that q is an integer
altheman1: just for ease, let n=2m
Myths Apprentice: wait, is the equation 1+4n or 1+4n^2?
altheman1: 4n^2
Myths Apprentice: ok
altheman1: 1=(q-m)(q+m)
jwec05: wait, you got that from discriminant right?
jwec05: of x^2-x-n = 0
altheman1: correct
Myths Apprentice: so it should be 1+4n, right?
altheman1: oh wait
altheman1: yes, my mistake
jwec05: x = (1 + sqrt{1+4n})/2
Myths Apprentice: let's analyze with modular arithmetic?
altheman1: 1+4n=q^2
Myths Apprentice: what's that called again, quadratic residues?
altheman1: we can still factor
altheman1: 4n=(q-1)(q+1)
Myths Apprentice: ok, so q is odd
altheman1: we can eliminate the 4
altheman1: let m=4n
altheman1: m< 10^6
altheman1: m=(q-1)(q+1)
altheman1: now what?
jwec05: hmm.. Just note that n I have isn't big
Myths Apprentice: uh
jwec05: for answer I mean. It's not really a large number..
Myths Apprentice: can you repaste the problem again?
altheman1: try q=1000, 999 , 1001
altheman1: jwec05: 4. Assuming the expression converges, determine the largest integer n with n <= 4,000,000 for which sqrt{n+sqrt{n+sqrt{n+... is rational
jwec05: Remember this is TEAM so you can use calculators..
towersfreak2006: okay
altheman1: q=1001 is too high
towersfreak2006: altheman1: x^2-x-n=0
Adunakhor I: back, sorry i was gone for so long
altheman1: but q=1000 works
towersfreak2006: n is an integer
altheman1: it gives m=999,999
altheman1: are we done?
jwec05: m = 4n so no
BaseballJoe2002: sorry guys...i have to go
Myths Apprentice: bye
BaseballJoe2002: i'll be on tomorrow night though
towersfreak2006: factor it as (x-t-1)(x+t)
towersfreak2006: then (t-1)(t) is an integer
BaseballJoe2002 has left the room.
towersfreak2006: and we're looking for the greatest n<=4,000,000=(2,000)^2
Jen7math has entered the room.
Myths Apprentice: hello
towersfreak2006: so we want to maximize the product of (n)(n+1) as long as n(n+1) <= (2,000)^2
towersfreak2006: bad variables, sorry
altheman1: n(n+2)
towersfreak2006: no, m(m+1)
Adunakhor I: so that's just 2000 x 1999
altheman1: 4n=(q-1)(q+1)
towersfreak2006: yes
towersfreak2006: so 3,998,000
Adunakhor I: silverfalcon said the an swer was rather small though
altheman1: does that work? tower
jwec05: I'm checking the answers with the source that I got
jwec05: he's looking on the answer book now
towersfreak2006: yes, if n=3,998,000
towersfreak2006: and set the radical expression to 2,000
towersfreak2006: then n+radical=4,000,000
altheman1: yes, that works
towersfreak2006: square root of that is 2,000 again
jwec05: That's correct answer
Myths Apprentice: cool
Myths Apprentice: next problem
altheman1: but it is n*(n+2)
towersfreak2006: no
altheman1: or k(k+2)
towersfreak2006: it is 1999*2000
jwec05: Yes
altheman1: oh
altheman1: huh
jwec05: Myths Apprentice had direct approach
jwec05: note that the question was x^2-x-n = 0
jwec05: all we need is n that can be factored as two consecutive positive integers
jwec05: 1999*2000 < 2000^2
jwec05: so that's our answer
altheman1: k
jwec05: is that okay with you altheman?
jwec05: Ok
jwec05: Unless we do quick, it might be last problem..
jwec05: 5
jwec05: NO CALCULATOR
jwec05: 5. Compute the number of distinct planes passing through at least three vertices of a given cube.
towersfreak2006: a plane is fixed by three points
Myths Apprentice: ok
towersfreak2006: so there are at most 8C3 planes
towersfreak2006: but we overcount some of them
jwec05: yup
towersfreak2006: when four points are coplanar
Myths Apprentice: well, the number of triangles you can form, where the vertices are the vertices of the cube
jwec05: basically yeah
jwec05: now question is: count overcounts
Myths Apprentice: +the number of quadrilaterals
altheman1: if the plane is the face then we over count
towersfreak2006: the 6 faces of the cubes
altheman1: are there any other over counts?
towersfreak2006: yes
towersfreak2006: if you take the diagonal of a face
towersfreak2006: and make a cut into the cube
altheman1: 2 pairs of opposite verticies
towersfreak2006: yeah
Myths Apprentice: just to make sure, is the answer 62?
towersfreak2006: that's what i meant :-!
jwec05: No
Myths Apprentice: k
Adunakhor I: wait, if two planes have different orientation, are they distinct?
altheman1: each plane that has 4 points in it, we count that \binom{4}{3} times
altheman1: yes
altheman1: we have 6 faces
altheman1: how many diagonal planes do we have?
towersfreak2006: 6
towersfreak2006: at least
towersfreak2006: let ABCD be stacked on EFGH
altheman1: let me rephrase
Jen7math: is the answer by any chance 20?
towersfreak2006: so that E is the projection of A
towersfreak2006: no
jwec05: yes Jen
towersfreak2006: that's too small
towersfreak2006: wait what?
jwec05: It's 20
altheman1: how many planes do we have that contain 2 main diagonals
towersfreak2006: what's 8C3?
jwec05: 56
towersfreak2006: what???
towersfreak2006: um
towersfreak2006: i got 280 for some reason
towersfreak2006: -_-
jwec05: Jen, care to explain what you did
altheman1: we over count by 4 on all the planes that have 4 verticies
towersfreak2006: oh dagnabit 56*5=280
jwec05: so we can summarize and maybe add one more final problem
Jen7math: so 8C3 is 56 as said earlier
Jen7math: then we have 6 faces and 6 "other" duplicates
Jen7math: which were counted...
Jen7math: 4 times?
altheman1: yes
towersfreak2006: yeah
pakman2012 has left the room.
Jen7math: something like that
jwec05: what was "other" duplicates?
Jen7math: the diagonal cuts
altheman1: altheman1: how many planes do we have that contain 2 main diagonals
towersfreak2006: there are four of one kind of diagonal cut
towersfreak2006: and two of the other
jwec05: so 6?
towersfreak2006: yeah
towersfreak2006: blah 6 + 6 = 12, so 36 overcounts
altheman1: we have 4 main diagonals, so \binom{4}{2}=6
towersfreak2006: 56 (not 280) - 36 = 20
altheman1: good
altheman1: move on
jwec05: Okay
jwec05: Last problem
jwec05: #6
Myths Apprentice: have time for 1 more silverfalcon?
towersfreak2006: sorry about that
Myths Apprentice: ok, cool
jwec05: 6. Determine all positive integers $k \leq 2000$ for which $x^4+k$ can be factored into two distinct trinomial factors with integer coefficients
jwec05: NO CALCULATOR
altheman1: so it factors into quadratics?
Adunakhor I: yeah
jwec05: it says trinomial..
Adunakhor I: so the squared term, linear, and constant
jwec05: but what about like (x^3+something...)(x+something)?
Myths Apprentice: that's not trinomial
altheman1: there is a technique that i like, basically adding 0
Adunakhor I: x + something is only binomial
altheman1: take x^4+16
jwec05: Oh heh
altheman1: complete the square
Myths Apprentice: oh, then difference of squares
altheman1: x^4+8x+16-8x
altheman1: but i messed up the example
Myths Apprentice: you mean 8x^2, right?
altheman1: the thing that we add must be a perfect square
altheman1: yes
Myths Apprentice: ok
Myths Apprentice: let me think about how to word this
altheman1: what is the thing that we add in terms of k?
Myths Apprentice: so for x^4+c
jwec05: k/2*x^2?
towersfreak2006: i think x^4 + 4k^2 is some identity
Myths Apprentice: i think its \sqrt{k} times 2
towersfreak2006: sophie germain?
Myths Apprentice: it was in an aime problem 14, but that's not important
altheman1: yes, but we don't need the 4 here do we?
altheman1: i saw that topic too
jwec05: I remember hearing it too
jwec05: this is Individual..
jwec05: so there must be quick way
towersfreak2006: well if that's "iff"
altheman1: we add 2*\sqrt{k}x^2
towersfreak2006: then all you need is 4x^2 <= 2000
towersfreak2006: x^2 <= 500
Myths Apprentice: ok, so for x^4+c^4, we complete the square as (x^4+2c^2x^2+c^4)-2c^2x^2
altheman1: 2\sqrt{k}=q^2
Myths Apprentice: which is equal to (x^2+c^2)-2c^2x^2
Myths Apprentice: however, 2c^2x^2 must be a square
Myths Apprentice: so 2c^2 is a square
Myths Apprentice: so we want all those that are squares
altheman1: so k=q^4/2
altheman1: q is divisible by 2
altheman1: q=2m
altheman1: k=4m^2
altheman1: as tower said...
jwec05: remember about integer coefficients..
jwec05: that restricts some answers
altheman1: tower is correct
towersfreak2006: i take it back, iirc it's 4k^4, not 4k^2?
towersfreak2006: (since 16 didn't work earlier)
altheman1: woops...i am not that good with careless errors :P
altheman1: x^4=500
altheman1: x=5 does not work
towersfreak2006: 4^4=256?
towersfreak2006: okay
Myths Apprentice: wait isn't it 2x^2<= 2000
Myths Apprentice: not 4x^2
towersfreak2006: no, 4x^4<=2000
Myths Apprentice: ok
Jen7math: how did we get 4x^4 again?
altheman1: we add 2\sqrt{k}x^2
towersfreak2006: are the answers 4, 64, 324, 1024?
altheman1: that must be a perfect square
jwec05: Yup
Myths Apprentice: cool
towersfreak2006: cool
altheman1: 4k=q^4
jwec05: towersfreak, can you summarize then
jwec05: there were too many ideas
altheman1: are we done? i am not sure that we are
Myths Apprentice: that was the last problem, right?
Myths Apprentice: i think we're done
altheman1: with the problem
Myths Apprentice: because those allo provide integer coefficients
Myths Apprentice: all*
altheman1: in terms of time we are done...lol
Myths Apprentice: oh
Myths Apprentice:
Myths Apprentice: yeah, last problem
jwec05: yeah as soon as he posts summary then we're officially done
altheman1: but we have not shown that is the only way to factor it, but it is if we look into in more
altheman1: it is the only way
towersfreak2006: x^4 + 4k^4 factors into (x^2 - 2kx + 2k^2) (x^2 + 2kx + 2k^2)
altheman1: expand (x^2+bx+c)(x^2+dx+e) and compare coefficients gives b=-d
towersfreak2006: yeah, and afterwards, you fine that c=e, comparing coefficients of the x term
altheman1: this was a great problem solving session, after we got out of the other room, we were productive
Myths Apprentice: I'd say that this arml meet was much better than the first one
Myths Apprentice: yeah
altheman1: i am going to add you guys to my buddy list
Myths Apprentice: same here
jwec05: so wait, just one quick question
jwec05: after your (x^2 - 2kx + 2k^2) (x^2 + 2kx + 2k^2), what do we do again?
jwec05: expand?
towersfreak2006: no
towersfreak2006: that's proving that all n in the form 4k^4 work
jwec05: Oh
altheman1: we get that factorization by completing the square
towersfreak2006: then we "show" that all the other stuff don't work
altheman1: then using difference of squares

Thanks to everyone who participated.

You have just entered room "armlpractice."
Fillet Who has entered the room.
Shobhit2005: ahh..towersfreak
Adunakhor I has entered the room.
Shobhit2005: from aops
towersfreak2006: hi
Adunakhor I: hi
Shobhit2005: hello, this is Iversonfan
Fillet Who: eh? it's sunday now?
Shobhit2005: 2005
towersfreak2006: oh ok
Bresident Boosh has entered the room.
towersfreak2006: :-)
Shobhit2005: ok, can everyone introduce themselves?
Fillet Who: dang, okay, i can do american history hw and this at the same
time
Shobhit2005: i need to do ap euro
Shobhit2005: heh
Shobhit2005: but screw it
LessThanEric182: oh
LessThanEric182: im snowstorm
LessThanEric182: sorry
Bresident Boosh: i am asif
Shobhit2005: hi snowstorm
Shobhit2005: hi asif, nice to meet you
Shobhit2005: lmao
LessThanEric182: term paper :-\
Shobhit2005: screw school...arml is more important
Shobhit2005: who's maxharris, repmax, on aops
BaseballJoe2002: yes
Fillet Who: i'm not going this year, but w/e
Shobhit2005: rep123max
Fillet Who: okay, let's start getting things done
elemmnop has entered the room.
Shobhit2005: ok, first problem
Shobhit2005: err...
Shobhit2005: how do we do this?
Fillet Who: Who has the problems?
Shobhit2005: i got problems
BaseballJoe2002: i have problems organized to go over
Shobhit2005: ok
BaseballJoe2002: but you can discuss problems now
Shobhit2005: ok
JeReMy PHam 0 4 has entered the room.
Shobhit2005: here's one: z^6+z^3+1=0
Shobhit2005: find the degree measure
towersfreak2006: roots of unity...
MaxHarris08 has left the room.
Shobhit2005: in the interval 90-180
Shobhit2005: which satisfies this
towersfreak2006: is it cis 160?
Shobhit2005: yep
Shobhit2005: or just 160
Shobhit2005: is acceptable
Fillet Who: holy crap, samson, you're fast
Shobhit2005: just multiply by z^3
Shobhit2005: and subtract
towersfreak2006: i've done this before ;-)
Shobhit2005: who's fillet who?
Fillet Who: philip hu
elemmnop: aops guy
Shobhit2005: on aops that is
Fillet Who: filletwho on aops
towersfreak2006: i think that is his id...
pakman2012 has entered the room.
towersfreak2006: hey saiem
BaseballJoe2002: hmm..should i invite sean?
Shobhit2005: no
BaseballJoe2002: lol
pakman2012: hey
Shobhit2005: unless you want things to get out of control
Shobhit2005: ok, next problem
Shobhit2005: anyone wanna post one?
Adunakhor I: I don't have any arml problems, :-(
Shobhit2005: it's hard to think of non trivial ones
Fillet Who: uh...relay problem...for the first person: let r=1+2x and r=2,
what is r...
tetrahedr0n has entered the room.
Shobhit2005: lmao
Shobhit2005: i'm glad that's non trivial
Perryzheng2001 has entered the room.
Perryzheng2001: lol
Shobhit2005: who are you?
Shobhit2005: on aops
Perryzheng2001: ginasboys
Perryzheng2001: lol
Shobhit2005: ok
Shobhit2005: and i know tetrahedr0n
BaseballJoe2002: if you know any other ppl who would want to be in this
chat...you can invite them given that they won't cause chaos
towersfreak2006: A convex hexagon is inscribed in a circle. If its
successive sides are 2, 2, 7, 7, 11, and 11, compute the daimeter of the
circumscribed circle.
towersfreak2006: *diameter
Shobhit2005: ah nice
elemmnop: you can switch sides around, so switch em up until you get
2,7,11,2,7,11
BaseballJoe2002: after you finish this problem i'll get started with the
problems i have
Shobhit2005: can we put up hints?
mathlete1106 has entered the room.
Shobhit2005: i don't have paper on me...heh
elemmnop: the diamater is the fourth side of the cyclic quadrilateral with
sides 2,7,11 (due to symmetry)
Shobhit2005: yep, look at cyclic quadrilaterals
Xuan768 has entered the room.
Xuan768: someone remind me tomorrow to send my app
Shobhit2005: changing up the triangles around won't affect cyclicity
Shobhit2005: god, can't find paper!
x150151041 has entered the room.
Bresident Boosh: this is harry
x150151041: hi
BaseballJoe2002: oh hey
Shobhit2005: that's asif
Shobhit2005: wtf
Bresident Boosh: no shobhit
x150151041: lol
Bresident Boosh: the person who entered
Bresident Boosh: lol.
Shobhit2005: oh
Shobhit2005: nvm
BaseballJoe2002: ha lol
Shobhit2005: oh oh, dinner calls, be back in 30 mins
Bresident Boosh: towersfreak2006: A convex hexagon is inscribed in a circle.
If its successive sides are 2, 2, 7, 7, 11, and 11, compute the daimeter of
the circumscribed circle.
Shobhit2005 has left the room.
Bresident Boosh: thats the problem btw
Fillet Who: Hmmm...this one is too hard, let's try another one: From
reconstruction to WWI, how were blacks kept in a subordinate position
politically, economically, and socially?
mathlete1106: ....
Bresident Boosh: literacy tests, poll taxes...
Xuan768: lol
Bresident Boosh: lalala
towersfreak2006: grandfather clause
Bresident Boosh: no ap calculus.
Xuan768: those were eliminated by ww1 i berlieve
BaseballJoe2002: ok...are we going to work on that one?
BaseballJoe2002: or do you want me to post my first one?
Fillet Who: go ahead
elemmnop: There's alot of answers for that
towersfreak2006: i want to finish this one
Xuan768: grandfather clause was more of a problem dujring reconstruction
towersfreak2006: elemmnop (sevastian?): why can we move around the sides?
BaseballJoe2002: i dont think we can
BaseballJoe2002: it says successive, right?
towersfreak2006: yeah
elemmnop: Because
elemmnop: say you have a cyclic hexagon ABCDEF...if you switch AB and BC,
then the figure CDEF remains unchanged
elemmnop: angle ABC remains constant, and AB and BC still have the same
lengths, they are just reversed
elemmnop: so the area, 1/2 AB * BC * sin theta is the same
elemmnop: so the total area is equivalent
elemmnop: You're basically flipping a triangle
elemmnop: and clearly you can extend it so that you can rearrange all sides
until you get whatever permutation you want.
JeReMy PHam 0 4 has left the room.
towersfreak2006: okay
towersfreak2006: i don't know what about what you just said...but the same
cyclic quadrilateral part at the beginning convinces me...
Adunakhor I: so the diameter is the fourth side of the quadrilateral with
other three side lengths 2, 7, 11?
elemmnop: Yes.
elemmnop: If you make the sides 2,11,7,2,11,7 successively (or
2,7,11,2,7,11)
Miths Apprentice has entered the room.
elemmnop: you have two pairs of identical cyclic quadrilaterals, which share
a side
elemmnop: this side is the diamater
Adunakhor I: oh ok
pakman2012: now... whats the answer?
BaseballJoe2002: sqrt(174)?
BaseballJoe2002: oh wait
towersfreak2006: it's 14
Adunakhor I: how do we find the diagonal lengths then, of the quadrilateral?
BaseballJoe2002: lol, that was stupid
pakman2012: how do you all figure?
towersfreak2006: the only thing i know about cyclic quadrilaterals is
ptolemy...
pakman2012: me too
pakman2012: how'd you get 14?
Bresident Boosh: well lest all go to mathworld and learn.
towersfreak2006: i checked the answer book
Bresident Boosh: http://mathworld.wolfram.com/CyclicQuadrilateral.html
towersfreak2006: ;-)
pakman2012: oh
pakman2012: haha
pakman2012: nice
towersfreak2006: hi MithsApprentice
BaseballJoe2002: well you have some right angles since that fourth side is
the diameter
towersfreak2006: okay
towersfreak2006: consider ABCD
towersfreak2006: with AB = 2, BC = 7, CD = 11
towersfreak2006: then we want...AD?
elemmnop: yes
Fillet Who: okay...anyone have another problem?
towersfreak2006: m<ABD = m<ACD = 90?
BaseballJoe2002: you can use pythagoras and ptolemy
Adunakhor I: so AD=sqrt{x^2-2^2}
elemmnop: Ptolemy's isn't neccesary.
Miths Apprentice has left the room.
Adunakhor I: so cos ADC = {x^2+2^2-4x}{4x}
towersfreak2006: you can make it a quartic
Adunakhor I: -cosADC= {6^2+2^2}{4x7}
towersfreak2006: using ptolemy and pythagorean
BaseballJoe2002: yeah...
Adunakhor I: and solve that?
towersfreak2006: and then the constant cancels out
towersfreak2006: so there's a cubi
towersfreak2006: x^3 - 49x^2 -125x - 308 = 0
towersfreak2006: if i did everything right...
towersfreak2006: oh no
Adunakhor I: i mistyped earlier
towersfreak2006: x^3 - 174x - 308 =0
towersfreak2006: sorry
elemmnop: Here's what I got: Let one of the cyclic quadrilaterals be ABCD,
with sides AB=2, BC = 11, CD = 7, and we want to find AD. Let BCD = X, so
that BAD = 180-X (cyclic quad).
elemmnop:  Let BD = a, AD = b. By law of cosines, we know that
a^2=b^2+4+2bcosX, and a^2=121+49-77cosX.
elemmnop: In addition, triangle BAD is a right triangle, so we know
a^2+4=x^2.
BaseballJoe2002: yeah, i got that also samson
elemmnop: My bad, a^2+4=b^2.
towersfreak2006: okay cool
towersfreak2006: i hate arithmetic
elemmnop: Substituting b^2=a^2+4 back into our first LoC expression, we
solve for cos X = -4 / b
elemmnop: Substituting a^2=b^2-4 and cos X = -4/b into our second LoC
expression, we get b^2-4 = 170 + (308/b)
elemmnop: Expanding, we get b^3 - 174b - 308 = 0.
BaseballJoe2002: is this from a team test?
towersfreak2006: no
BaseballJoe2002: lol
towersfreak2006: individual 1989 #8
towersfreak2006: >_<
pakman2012: wow this is looking ugly
BaseballJoe2002: ugh
Xuan768: great
towersfreak2006: RRT?
elemmnop: Since we drew our figure SO WELL, we can finger the length of AD
to be between 10 and 20. Using Rational Root Theorem, we get b = 14.
towersfreak2006: it has to be greater than 11
BaseballJoe2002: yeah
elemmnop: And...
elemmnop: That's our answer.
BaseballJoe2002: so you just test 14 i guess
Xuan768: was i the only person who had no clue of what a cyclic quadrateral
was 20 minutes ago?
BaseballJoe2002: ready for the next problem?
pakman2012: probably not.
BaseballJoe2002: If 99log_b(a) = 19log_c(a), then b/c = c^k.  Compute k.
Adunakhor I: yeah
elemmnop: The other quadratic we get from our third degree doesn't produce
solutions cause they are negative, which is bull.
towersfreak2006: it's okay, i'm a senior and i barely know anything about
cyclic quadrilaterals
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1999i5.png
BaseballJoe2002: does that link work?
elemmnop: Yea
towersfreak2006: yeah
BaseballJoe2002: cool
Xuan768: heh... well maybe not okay cuz i'm a senior too
towersfreak2006: [(log a)/(lob b)] / [(log a)/(log c)] = 19/99
towersfreak2006: (log c) / (log b) = 19/99
BaseballJoe2002: Yes, so we use the change of base theorem.
BaseballJoe2002: log_b a = (log b)/(log a)
Bresident Boosh: is it 19/99 + 1...didnt work it out clearly though
elemmnop: -1.
towersfreak2006: i got 99/19 - 1
Bresident Boosh: oh
elemmnop: yep
BaseballJoe2002: yeah
BaseballJoe2002: or 80/19
BaseballJoe2002: everyone understand that?
pakman2012: yep
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1999i5soln.png
towersfreak2006: yay
Bresident Boosh: yeah
towersfreak2006: oh wow nice
pakman2012: logs are nice
pakman2012: haha
BaseballJoe2002: Ok, so we'll move on to the next problem.
elemmnop: there's a typo in there
elemmnop: but
elemmnop: eh
BaseballJoe2002: heh...
BaseballJoe2002: whoops...i didn't proofread
BaseballJoe2002: Let f(x)=4x-x^2.  Consider the sequence x_1,x_2,x_3,� where
x_i=f(x_{i-1}) for i>1.  Compute the number of values of x_1 such that
x_1,x_2, and x_3 are all distinct, but x_i=x_3 for all i>3.
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1997i4.png
elemmnop: x_3 = 0, 3.
LessThanEric182 has left the room.
elemmnop: 0 doesn't work, because then all x_i = 0, but we need teh first
three distinct.
towersfreak2006: no, it could be 4 also, right?
towersfreak2006: 4 ---> 0?
BaseballJoe2002: yeah
towersfreak2006: i would guess each value can come from two possible values
Xuan768: well its 0 or 3 and then you just plug and chug
towersfreak2006: is it individual round?
BaseballJoe2002: yes
BaseballJoe2002: 1994 I - 4
BaseballJoe2002: *1997
towersfreak2006: oh, it's in the url X_X
BaseballJoe2002: lol
BaseballJoe2002: ok, so let's look at x_3 = 0
BaseballJoe2002: What must x_2 then be?
elemmnop: x_2 = 4
BaseballJoe2002: x_2 is 0 or 4
BaseballJoe2002: but we discard 0
BaseballJoe2002: So if x_2 = 4 what can x_1 be?
towersfreak2006: then 4 = 4x - x^2, (x-2)^2 = 0, x= 2?
elemmnop: yields x_1 = 2.
BaseballJoe2002: right
BaseballJoe2002: now if x_3 = 3?
towersfreak2006: then x = 1? (since it can't be 3 again)
elemmnop: x_2 = 3, 1, throw out 3 (uniqueness). If x_2 = 1, then x_1 = 2+-
rt3
BaseballJoe2002: yes
BaseballJoe2002: So what's our answer?
elemmnop: 3.
BaseballJoe2002: yep
BaseballJoe2002: Solution summary:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1997i4soln.png
BaseballJoe2002: (might have typos :P )
BaseballJoe2002: Next Problem:
BaseballJoe2002:  Compute all ordered pairs (x,y) such that: xy  +  9  =
y^2 and xy + 7  =  x^2.
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1998t6.png
towersfreak2006: add the two equations
towersfreak2006: x^2 + y^2 - 2xy = 16
towersfreak2006: actually, don't know if that gets anywhere
BaseballJoe2002: well...what special about x^2+y^2-2xy
Adunakhor I: perfect square
Adunakhor I: (x-y)=+- 4
towersfreak2006: substituting back into an equation should work?
elemmnop: yea.
BaseballJoe2002: that's one way
Fillet Who: subtract the two to get (y-x)(x+y)=2, then substitute x-y
BaseballJoe2002: that's the other i know of ;-)
towersfreak2006: oh nice
BaseballJoe2002: So what answers do we get?
x150151041 has left the room.
Myths Apprentice has entered the room.
BaseballJoe2002: anyone?
elemmnop: (7/4, -9/4) and (9/4, -7/4)
x150151041 has entered the room.
elemmnop: woops
elemmnop: that last one is (-7/4, 9/4).
BaseballJoe2002: lol, yeah
BaseballJoe2002: i had to check my answers...
BaseballJoe2002: Solution summary:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/arml1998t6soln.png
BaseballJoe2002: (there's a typo in the last line of the first paragraph)
BaseballJoe2002: it should say y^2-x^2 and not y^2x^2
BaseballJoe2002: The next problem's a bit harder.
BaseballJoe2002: For  0 < x < 1, let f(x) =
(1+x)(1+x^4)(1+x^16)(1+x^64)(1+x^256)�Compute f^-1(8/(5f(3/8))).
towersfreak2006: ew lol
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/armlfunction.png
BaseballJoe2002: heh
towersfreak2006: bah!
elemmnop: Surely that isn't the nicest form for f(x).
BaseballJoe2002: i didn't type up a solution for this one ^_^
BaseballJoe2002: but anyway...
BaseballJoe2002: any ideas?
towersfreak2006: i've done this problem before, so...
towersfreak2006: :-X
BaseballJoe2002: ok
mathlete1106 has left the room.
elemmnop: There has to be a way to simplify that function.
BaseballJoe2002: i dunno...in my solution i didn't simplify it
elemmnop: hmm
BaseballJoe2002: What if we were to multiply f(x) by something?
BaseballJoe2002: What might we try?
elemmnop: f(1-x)?
BaseballJoe2002: that's not what i did...does that help?
elemmnop: eh
elemmnop: nope
elemmnop: hmm
BaseballJoe2002: ok
BaseballJoe2002: someone describe f(x)
BaseballJoe2002: i mean...the right hand side...what are the terms being
multiplied?
towersfreak2006: ugly
Adunakhor I: lol
towersfreak2006: with powers of 4
BaseballJoe2002: yeah
elemmnop: prod_{i=0}^{infty} (1+x^{4^i})
pakman2012: product of (1+x^(4^z)) from z= 0 to infinity
elemmnop: OH!
elemmnop: no
elemmnop: maybe?
elemmnop: sophie germain's?
BaseballJoe2002: hmm
BaseballJoe2002: I multipled f(x) by something.
towersfreak2006: substitute
towersfreak2006: oh no, jk
Adunakhor I: f(x-1)?
Adunakhor I: nvm
BaseballJoe2002: How might we simplify (1+x)(1+x^2)(1+x^4) if that's what we
had?
elemmnop: x^7 -1 / x-1
towersfreak2006: (multiply by x-1)
x150151041 has left the room.
BaseballJoe2002: right
BaseballJoe2002: we can make it telescope basically
BaseballJoe2002: Can we employ a similar strategy on the problem at hand?
BaseballJoe2002: Let's multiply f(x) by some function g(x) so that it will
telescope nicely.
BaseballJoe2002: What should g(x) be?
elemmnop: (1-x)(1+x^2)(1+x^8)(1+x^32)...
BaseballJoe2002: right!
elemmnop: telescopes to 1
elemmnop: but
elemmnop: g(x) = (1-x)f(x^2)
elemmnop: so we get
BaseballJoe2002: So f(x)g(x)=1.
elemmnop: f(x) = 1 / (1-x)(f(x^2))
elemmnop: or f(x^2) = 1 / (1-x)(f(x)
elemmnop: take the inverse f of all of that
elemmnop: we get x^2 = f^-1 ( 1 / (1-x)(f(x)))
elemmnop: but
elemmnop: the expression on the right with x= 3/8 is exactly what we want to
find.
elemmnop: so 9/64 is our answer?
BaseballJoe2002: yes!
elemmnop: cool.
elemmnop: good problem
BaseballJoe2002: That was a #8 from 2001.
BaseballJoe2002: Any questions before we move on?
towersfreak2006: nope
BaseballJoe2002: Ok, I have one more problem.
BaseballJoe2002: It's another #8....but I think it's a little bit easier.
BaseballJoe2002:  ABCDEFH is a regular heptagon of side 1001.  Compute AC^2
- DE*FB.
BaseballJoe2002: (i didn't think i needed latex for this one ;-))
elemmnop: oh god
elemmnop: a heptagon
BaseballJoe2002: So how should we start?
tetrahedr0n: nice.problem. this one's a classic
BaseballJoe2002: Picture, right?
BaseballJoe2002:
http://img.photobucket.com/albums/v213/joml88/ARML%20Practice/armlheptagon.png
BaseballJoe2002: On paper, you would have to just draw it as accurate as
possible.
BaseballJoe2002: any observations?
elemmnop: We can let the side of the figure and an internal angle be x and y
respectively
elemmnop: easier to work with, and we can plug it all back in later.
elemmnop: plus I'm sure anything with y will cancel out (since 360/7 isn't
nice)
BaseballJoe2002: We could try messy trig and all that.
BaseballJoe2002: But there's a much simpler way.
elemmnop: Yea
elemmnop: Ptolemy's
elemmnop: could work, since we have alot of quads
BaseballJoe2002: Yep, which quad(s) should we look at?
elemmnop: ABCG
elemmnop: BF is equal to CG
elemmnop: since we have a regular heptagon
Xuan768 has left the room.
elemmnop: so in fact all the lengths in our triangle are sides of the
triangle ACG
elemmnop: also, AC = BG, the two diagonals of the quad
elemmnop: since DE = AG
elemmnop: our expression is now ac^2 - AG * CG
elemmnop: or
elemmnop: AC^2 - AB * CG since AB + AG
elemmnop: but Ptolemy's says AB * CG + AG * BC = AC * BG
elemmnop: the RHS is AC^2
elemmnop: so we have AC ^2 - AB * CG = AG * BC = (1001)^2 = 1002001
BaseballJoe2002: correct
BaseballJoe2002: that was the last problem i had
towersfreak2006: wow, nice...
elemmnop: cool
BaseballJoe2002: you all can stay and talk/do more problems
Fillet Who: let's randomly answer some american history problems
towersfreak2006: we're humans; we can't be random
BaseballJoe2002: next week i should be doing this again
towersfreak2006: so let's not
BaseballJoe2002: silverfalcon will be doing one on saturday's as well
BaseballJoe2002: and my computer is about to go crazy...so i'll see you all
later!

Few announcements:

1. I renamed the thread to include ALL transcripts for ARML Online Math Meet.
2. Problems for the ones I did are Latexed.

Please note that unless you're me or joml88, do not post in this thread. I want to keep this thread only with transcripts.


inequalities
function
inequalities proposed

Let  $-\infty<a<b<\infty , x,y,z \in [a,b]$ , and
\[ D(x,y,z)=\frac{ (x-y)^2 +(y-z)^2 +(z-x)^2}{12} \; \; ,\; \; f \in C^2[a,b]. \] Prove that for each $f\in C^2[a,b]$ there exists
 a point $\xi ,\ a\le \xi \le b$ such that
\[ \begin{array}{c} \displaystyle 3f\left(\frac{x+y+z}{3}\right) +f(x)+f(y)+f(z)=\\ \\ =\displaystyle 2f\left(\frac{x+y}{2}\right)+ 2f\left(\frac{y+z}{2}\right) + 2f\left(\frac{z+x}{2}\right) + \displaystyle D(x,y,z)f^{\prime\prime}(\xi) \; . \end{array}  \]

Since $f''$ has the Darboux property, it suffices to show that
$\displaystyle m\leq\frac{6E(f)}{(a-b)^2+(b-c)^2+(c-a)^2}\leq M$,
where
$\displaystyle E(f)=f(a)+f(b)+f(c)+3f(\frac{a+b+c}{3})-2f(\frac{a+b}{2})-2f(\frac{b+c}{2})-
2f(\frac{c+a}{2})$
and $m=inf f''(x)$ and $M=sup f''(x)$.
The functions $\displaystyle g_1=f(x)-\frac{mx^2}{2}$ and $\displaystyle g_2=\frac{Mx^2}{2}-f(x)$  are convex. According to Popoviciu's Inequality, we have $E(g_1)\geq0$ and $E(g_2)\geq0$, whence the desired result come out.


function
inequalities unsolved
inequalities

3 if n=3 and 4 for the other values I think.

Yes,It's not difficult...
$n=3,f(n)=3$ evident.
$n>3$,for  $a_3,a_4,...,a_n \rightarrow 0$ then easy to find $f(n)=4$  and this value satisfy.

[quote="hoaln"]for each positive interger [tex]n(n\ge2)[/tex] find f(n) max(f(n) is function of n) such that for all [tex]a_1;a_2;...;a_n[/tex]positive numbers ;we have:
[tex](a_1 +a_2+...+a_n)^2\ge f(n)(a_1a_2 +a_2a_3+....+a_{n-1}a_n+a_na_1)[/tex][/quote]
I have corrected that problem;i'm sorry!

But the answer is not change.
$f(n)=3$ if $n=3$ and $f(n)=4$ for all $n\ge 4$


Hexagons with side 1 are used to form a large hexagon. The diagram illustrates the case n = 3 with three unit hexagons on each side of the large hexagon. Find the area enclosed by the unit hexagons in the case n = 202.

ok...

[hide]n=3 generates 6(n-1)=6*2=12 hexagons

so n=202 generates 6*201 hexagons on the border

the area enclosed by this is just the sum of the areas of the pattern when n=201,200,199,...,2 and the case when there is just 1 hexagon in the middle

6(1+2+...+200)+1=(6*200*201/2)+1=120601 hexagons which give an area of 120601*3* :rt3: /2=361803 :rt3: /2[/hide]


Define a "Monkey number" to be a number that can be written as the sum of two numbers plus the product of those same two numbers.  Find the probability of picking a monkey number out of a bag of integers from 1 to 100.

Hide all answers and/or solutions.  Also, you can share some monkey number identities you are aware of.

[hide="Answer"]
74%
[/hide]

[hide="Solution"]
A monkey number is anything that can be written as ab + a + b.
ab + a + b
a(b + 1) + b
c = b + 1
ac + c - 1
c(a + 1) - 1
d = a + 1
cd - 1
A monkey number is any number not one less than a prime.  The number of primes between 2 and 101 is 26.  100 - 26 is 74%.
[/hide]

And the product of two monkey numbers is itself a monkey number, was in the superbrain at some stage

The Monkey Identity:
The product of two odd monkey numbers is one less than another odd monkey number and is twice a monkey number.
[hide="Proof"]
Monkey number: ab - 1
Odd monkey number: ab is even and can be written as 2ab.
(2ab - 1)(2cd - 1)
4abcd - 2ab - 2cd - 1
2(2abcd - ab - cd) - 1
2(2ab(cd-.5) - (cd-.5) - .5) - 1
2((cd-.5)(2ab-1) - .5) - 1
2(cd - .5)(2ab - 1) - 2
2(cd - .5)(2ab - 1) - 1      -1
One less than monkey number!
[/hide]


inequalities
quadratics
function
algebra
polynomial
inequalities unsolved

For $ a,b,c\in\mathbb{R}$ such that $ a,b>0$, $ c^2<4ab$, prove that there exists $ t\in (0,\min\{a,b\})\subset \mathbb{R}$ such that $ au^2\plus{}bv^2\plus{}c uv\ge t(u^2\plus{}v^2)\qquad u,v\in\mathbb{R}.$

Your inequality is wrong.

You want to prove there exists $ t\in (0,\min\{a,b\})\subset\mathbb{R}$ that make the quadratic inequality
$ f(u)\equal{}(a\minus{}t)u^2\plus{}cvu\plus{}(b\minus{}t)v^2\geq 0$ holds for arbitrary real number $ u,v$.  $ f$ is a function I set.

 Because $ a\minus{}t\geq 0$, discriminant of $ f(u)$ must be below $ 0$ : $ D\equal{}(cv)^2\minus{}4(a\minus{}t)(b\minus{}t)\leq 0$.
If $ c\equal{}0$, the inequality holds. If $ c\neq 0$, $ D\leq 0$ can be rewritten as $ 0\leq v^2\leq \frac{4(a\minus{}t)(b\minus{}t)}{c^2}$. But there exists $ M\in\mathbb{R}$ such that $ M^2>\frac{4(a\minus{}t)(b\minus{}t)}{c^2}$. If we set $ v$ larger than $ M$, we can find $ u$ that the inequality does not hold.

[quote="J.Y.Choi"]Your inequality is wrong.

You want to prove there exists $ t\in (0,\min\{a,b\})\subset\mathbb{R}$ that make the quadratic inequality
$ f(u) \equal{} (a \minus{} t)u^2 \plus{} cvu \plus{} (b \minus{} t)v^2\geq 0$ holds for arbitrary real number $ u,v$.  $ f$ is a function I set.

 Because $ a \minus{} t\geq 0$, discriminant of $ f(u)$ must be below $ 0$ : $ D \equal{} (cv)^2 \minus{} 4(a \minus{} t)(b \minus{} t)\leq 0$.
If $ c \equal{} 0$, the inequality holds. If $ c\neq 0$, $ D\leq 0$ can be rewritten as $ 0\leq v^2\leq \frac {4(a \minus{} t)(b \minus{} t)}{c^2}$. But there exists $ M\in\mathbb{R}$ such that $ M^2 > \frac {4(a \minus{} t)(b \minus{} t)}{c^2}$. If we set $ v$ larger than $ M$, we can find $ u$ that the inequality does not hold.[/quote]

Thank you. But
[quote="J.Y.Choi"]If we set $ v$ larger than $ M$, we can find $ u$ that the inequality does not hold.[/quote]

seems unclear. By your argument, $ M$ must be negative. In addition $ 0 < t < \min\{a,b\}$, how can you set $ v > M$ while $ M$ still depends on $ t$? Please give more details! Thanks.

Ps: I am not sure if the proposed inequality is wrong.

[quote="spanferkel"]The error is here:[quote="J.Y.Choi"]
$ D \equal{} (cv)^2 \minus{} 4(a \minus{} t)(b \minus{} t)\leq 0$.
[/quote]
the correct discriminant is $ D \equal{} c^2 \minus{} 4(a \minus{} t)(b \minus{} t)$, up to a factor $ v^2$. You can also consider $ f$ as a polynomial in $ \frac uv$. And because of $ c^2 \minus{} 4ab < 0$, this shows that if we choose $ t > 0$ small enough, we still have $ D < 0$. So the problem is correct and proven. :)[/quote]

Very nice suggestion. So we get the discriminant $ D \equal{} (c^2 \minus{} 4ab) \minus{} 4t [t \minus{} (a \plus{} b)]$. Since $ 0 < t < \min \{a,b\}$ and $ c^2 < 4ab$,
\[ \begin{cases}c^2 \minus{} 4ab < 0, \\
\minus{} 4t [t \minus{} (a \plus{} b)] > 0.\end{cases}\]
I still doubt the existence of $ t$ such that $ D\le 0$?

[quote="man.maiko"]
I still doubt the existence of $ t$ such that $ D\le 0$?[/quote]
:wallbash: but $ D$ as a function of $ t$ is continuous around $ t\equal{}0$ :rotfl:

I had a mistake;; man.maiko's inequality is right. Here is the correct solution.

$ D(of\ f(u))\equal{}v^2(c^2\minus{}4(a\minus{}t)(b\minus{}t))\leq 0$ holds iff $ c^2\minus{}4(a\minus{}t)(b\minus{}t)\equal{}\minus{}4t^2\plus{}4(a\plus{}b)t\plus{}c^2\minus{}4ab\leq 0$.

Let's denote the function $ \minus{}4t^2\plus{}4(a\plus{}b)t\plus{}c^2\minus{}4ab\equal{}g(t)$.
We have $ g(a)\equal{}g(b)\equal{}c^2\geq 0$, $ g(0)\equal{}c^2\minus{}4ab<0$. So, by intermediate value theorem, we can find $ p(0<p\leq\min\left\{a,b\right\})$ such that $ f(p)\equal{}0$. Applying intermediate value theorem again, we can find $ q(0<q<p)$ such that $ f(q)<0$. We have proved $ \exists q : D(of\ f(u))\leq 0$(there are infinite points of $ 'q'$s, frankly), so the question ended.


Find a polynomial $ P(x)$ such that $ P(x)$ is divisible by $ x^{2}+1$ and $ P(x)+1$ is divisible by $ x^{3}+x^{2}+1$

[hide]Euclid's algorithm. Since $ x^{2}+1$ and $ x^{3}+x^{2}+1$ are coprime there exist $ u(x)$ and $ v(x)$ s.t.

$ (x^{3}+x^{2}+1)u(x)-(x^{2}+1)v(x)=1$

Then the all polinomials are: $ P(x)=(x^{2}+1)(v(x)+(x^{3}+x^{2}+1)Q(x))$ where $ Q(x)$ is arbitrary.[/hide]


welcome to my tournament! it is elimination and if u want to join join fast!!!!!!

                            [b]8 PLAYERS!!!!

                                        [/b][i]JOIN NOW

YOU COULD WIN!!!
                                               now cd!!! :first: [/i]

i shall join. and maybe ask others to join

as soon as enough pplz join i will post the time till then u  have to w8
or join nowju
but in advance it would prob be JUNE 20
 :D  :P  :play_ball:

whats this tournament about?

if this tournament is about math(not chess), count me in.

I'll join too! 5 bucks says you won't have an awesome-looking bracket though...

I will not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not join XD

If you are too lazy to count how many not's there are then read this:
[hide][hide][hide]I'll join.. hurray![/hide][/hide][/hide]

PEOPLE WHO JOINED:
1.BRUT3FORC3
2.bob einstein
3.hurdler
4.Fantasylover
5.dynamo
6.mewto55555
7.ac-king

I NEED 1 more!!!

i guess your not joining your own tourney, mathblitz, right?
You could if you wanted to

mathblitz, could you wait to start your tourney until mine is near an end?

I join :O
The message is too small. Please make the message longer before submitting.

i will join

darn i am number 9

but it says 8 minimum so i can still join rite

math, i have to leav on the 26th (or 27th, i forget) , so i might not be able to finish your tournament, which is okay, by me. I'm just letting u know

Not bad, but trust me, filling in names is beastly hard.

these are the players:
1. dynamo
2. bob einstein
3.BRUT3forc3
4.ac king
5.budi
6.mewto
7.fantsylover
8.panjia

TOURNAMENT STARTS june 20th be ready contestants!!!! :first:

 hurdler dropped out

LOL he's seeding by highest ever rating.  :rotfl:  Imitation is the greatest form of flattery... As I said, please wait until mine is over or in its death throes (i.e Final 6)

Dang...

I guess I'm too slow to join, but if you enlarge it to 16 people count me in.

sry super join next time

for those of u who are in u need to pm me your highest ratings so we can get started

I NEED THOSE RATINGS OR IT WILL BE RANDOM those of u who i dont have the ratings(ac-king) (bobeinstein) i need them before we can start  or u 2 wil be playing each other.

For the tournament one person will make the cd while another (as a guest) joins three games pm me the results and i will post them i will also pm who your next will be. good luck and

[b]NOW GO!!!!!!!!!:starwars: [/b]

First to 3 games or best of 3? And i changed my mind, u can start when u want to.

best of 5 games

[list]here are your opponents!!!

Bob einstein vs. Dynamo

Panjia vs. ac-king

Budi vs. Brut3forc3

Mewto55555 vs. FantsyLover
[/list]
Superllamel if anyone wishes to dropp out super takes your place

results must be pmed to me by the 20th of june

 :winner_first:    :winner_second: :winner_third:

[size=200][color=green][b]Mewto55555 dies...at least he kept it competitive...[/b][/color][/size]

...wait then mewto is a first loser (sry, no offense)??

aww... im sry mewto... :(

because of inactivity i had to shut this tournament down

meh this tournery is closed??

ugh... :(

Therefore, I came in last, and everyone tied for 1st but me.... :(

Poor mewto,

math, i think you couldve waited a bit longer...... Some people do not get on everyday......, not to mention play ftw everyday


Let p is a prime and $ p\equal{}4m\plus{}1$. Proof that:
$ \sum_{k\equal{}0}^{p\minus{}1} \sin{\frac{2k^2\pi}{p}}\equal{}0$

It is known Gaus's sum $ \sum_k exp(\frac{2\pi ik^2}{p})$.


calculus
integration
trigonometry
function
calculus computations

$ \int_1^{\infty}\,\frac {dx}{x \plus{} e^{2x}}$

$ \int_{1}^\infty\,\frac {\cos^2(x)}{1\plus{}x^2}dx$
Thanks for the edit

The 2nd integral doesn't look too difficult  :roll:

First is a power rule.

Don't listen to §outh§tar - the first one is not a power rule.

Note that you weren't asked to find the exact values of these - and it's a very good thing you weren't asked that. The only issue is whether these converge (absolutely or otherwise).

First question: for how many different reasons is each of these integrals improper? Each is improper for at least one reason, because of the infinite limit. Are there any other reasons? I'm asking there whether there are any points at which the function being integrated tends to infinity. Let's see - both fractions, but in both cases the denominator is always positive plus positive, hence positive and never zero. (In both cases, by "always" I mean for all $ x\in[1,\infty).$ As far as each problem is concerned, no other values of $ x$ exist.)

So these are each improper for only one reason - that the upper limit is $ \infty.$ Hence the only thing to concern ourselves with is the behavior as $ x\to\infty.$ Also: both are the integrals of nonnegative functions. That means that convergence and absolute convergence are the same question and they're both subject to the "monotone function alternative": either they converge or they tend to $ \infty,$ and its' simply an issue of size.

For the first one, how big is it for large $ x?$ In that denominator, the $ e^2x$ term overwhelms the $ x$ term; the exponential is the only thing that matters. Compare this to $ \int_b^{\infty}e^{\minus{}2x}\,dx,$ which does converge.

For the second one, how big is $ \cos^2x?$ The line I like to use is "It's $ 1,$ only wigglier." In this case, ignore the wiggles: it's no bigger than $ 1.$ And that means that the integrand looks like (or is smaller than) $ \frac1{x^2}.$ So compare this to $ \int_b^{\infty}\frac1{x^2}\,dx.$

Those weren't the integrals he had posted originally Kent..

Yes, I attest SouthStar's plead. And thanks for the response Kent


inequalities geometry trigonometry calculus function MATHCOUNTS probability

Should I say that its currently Geometry...since I just finished reading a LOT about geo so I'd know a lot about it?

[quote="xpmath"]So I assume complex numbers and inequalities are part of algebra? Hmm, I would vote number theory, but since it's in discrete math with combinatorics, I don't really want to vote discrete... Meh, I just voted other because of number theory because I refuse to vote anything that has combinatorics in it :P[/quote]

same here

[quote="isabella2296"]Well, do you LIKE geometry, or simply know lots about it?[/quote]

well for me I like subjects I'm good at...xD

but I'm talking about more like pre-olympiad levlgeo

Hmm....  Well I think my answer to the poll question is that one should come to love all branches of math. :P

@xpmath, well complex numbers could also loosely be in trig as well, meaning it could be geo...  And why dislike combinatorics?  Co and P is fun! :)

Ugh... what disrespect towards discrete math... all bundled up like that... it's like polling:

What's your favorite subject?
english
french
spanish
history
geography
physics, chem, biology, math, comp sci, and other mathish stuff
other

Would you be happy with such a poll?

Oh well... no one seems to like discrete math that much in the MC forum anyways...

i voted calculus because it just seems really cool you can go up and down from infinity back..
but really i don't really like a specific branch (but i kinda don't like Algebra 2 very much.. to much calculators and factoring..)

I pick geometry because it's so interesting, all with the shapes and figures in planes and space.

I personally like Algebra, because it's really fun manipulating numbers.

algebra and geometry, but i voted algebra.

[quote="Poincare"][quote="math154"]Oh; it's just counting probability and number theory and other stuff but for MC just the first 3 things.[/quote]

most people think that discrete math is just counting probability and NT but there a LOT of other stuff too like there is boolean algebra, binary calculus (don't know what the official name maybe something like integer calculus) and so on.....[/quote]

Note how I said 'other stuff' because this is the MC forum and why bother. :)  :P

@isabella:

Boolean algebra is, in fact, related very closely to Boolean values.  The former is simply the use of truth values and the operations upon them.  Dunno how to explain that better, but, that's essentially what it is.

geometry is mostly proving what u did in algebra, which is solving, in geometry u need to memorize :read: , but for algebra there's always a way, :coolspeak:  always a solution and a way to get there
thus i  like algebra :love: .

Hm...I think geometry problems are my favorite problems on contests, but I like counting and probability the best...

i like algebra+number theory+calculus+combinatorics+group and ring theory
i hate probability it doesnt make sense, its useless writing the whole page off for jus a mean
i dislike geometry.

[quote="isabella2296"]The most well-known are counting, probability, and number theory, though...I've never heard of "boolean algebra", but I assume it's similar to the "boolean values" in Java.[/quote]

boolean algebra is like the simplification of boolean equations. i am assuming you know about boolean gates (if not read this : [url]http://en.wikipedia.org/wiki/Logic_gate[/url])  so if you have a big circuit full of gates then you can use boolean algebra to simplify it and in most cases make it faster.

[quote="isabella2296"][quote="binomial_4eva"]
i hate probability it doesnt make sense, its useless writing the whole page off for jus a mean
i dislike geometry.[/quote]

 :yup:

 :clap2:[/quote]
You don't like probability either?

Sometimes it can be fun but challenging too.

P.S. plz don't be posting posts with just smilies instead of actual content, kay?

[quote="gauss1181"]

P.S. plz don't be posting posts with just smilies instead of actual content, kay?[/quote]

Sorry about that.  :oops:

I just felt that in this case, smilies would be sufficient to convey my message.

[quote="ernie"]whoa... what about single digit subtraction?  thats pretty cool if u ask me!  :D[/quote]

Please vote "Arithemetic".  :rotfl:

[quote="isabella2296"]What's your favorite kind of math? I like Algebra. I know this question has been asked before, but this will be more organized as a poll.

If you selected other, could you please describe your favorite math?[/quote]
i like algebra.


[quote="pvthuan"]Prove that if $a,b,c\geq 0$, then
\[ \frac a{\sqrt{4a^2+bc}}+\frac b{\sqrt{4b^2+ca}}+\frac c{\sqrt{4c^2+ab}}\geq1.  \][/quote]
It looks very well-known.
Let me guess a solutiuon, but I'm not sure.
Rewrite the ineq:
[quote]Let $x,y,z>0, xyz=1$, prove that:
\[ \frac{1}{\sqrt{4+x}}+\frac{1}{\sqrt{4+y}}+\frac{1}{\sqrt{4+z}} > 1  \][/quote]
Then squaring, we'll get st.
Of course, we need to consider case $a,b,c=0$.
Just I think.

Is this inequality true:
[quote]For $x,y,z>0$ and $xyz=1$, then we have:
\[ \frac{1}{8+x+y}+\frac{1}{8+y+z}+\frac{1}{8+z+x} > \frac{1}{6}  \] [/quote]
If it's right, we are done.

No, It's from a general problem, we can replace 4 by any positive reals.

[quote="pvthuan"]That is kind of brute-force. Any other way?[/quote]
What does [b]kind of brute-force[/b] mean?
PS: brute-force or brutal force?


Given that $ x^2 \minus{} 2y^2 \equal{} 1$ has an infinite amount of positive integral solutions, prove that the equation:

$ a^2 \minus{} 2b^2 \equal{} \minus{} 1$ also has an infinite amount of positive integral solutions.

i am sorry but i am kinda confused here, do u mean that for every $ x,y$ that satisfies the first equation they also satisfy the second equation? (this is meaningless)
or is $ x,y$ some other variables $ a,b$ not equal to $ x,y$ respectively? in this case too why should the first equation be given at all you could have just told to prove infinite solutions for the second. either way i am completely messed up and hope someone will clarify me

Sorry, I meant that given the fact that the equation $ x^2\minus{}2y^2\equal{}1$ has infinite integral solutions, prove that the equation $ a^2\minus{}2b^2\equal{}\minus{}1$ also has infinite integral solutions.

The idea is to rearrange the second equation into something that is of the same form of the first equation, which we already know has infinite integral solutions, thus proving the second does as well.


geometry
circumcircle
incenter
search
parallelogram
geometric transformation
vector

Let a triangle $\triangle ABC $ with $ AB, AC > BC $. On sides $  AB , AC $ we take points $   D , E $  so that :
$  DB = BC = CE $. If  $ K $ is the circumcenter of $\triangle ADE $ and  $ I $ is the incenter of $\triangle ABC $ , prove that $  KI = R $ , where $ R $ is the radius of the circumcircle of  $\triangle ABC$ .

babis

  [i]( it is imposible to me to search in the forum for these problems)[/i]

veryvery nice problem .
many years ago i solved a simillar question to u problem by inversion.
i dont know wether i can solve it by inversion or not.i will try  :( .

Let $O$ be the circumcenter of $ABC$, $T$ the point (different from $A$) where the external angle bisector of $\angle BAC$ cuts the circumcircle of $ABC$, and $P$ the point (again, different from $A$) where the internal angle bisector of $\angle BAC$ cuts the circumcircle of $ABC$.

Since $TB=TC,DB=EC$ and $\angle TBD=\angle TCE$, the triangles $TBD,TCE$  are congruent. In particular, $\angle DTE=\angle BTC=\angle BAC=\angle DAE$, so $T$ is the second intersection between the circles $(ADE),(ABC)$. This means that $OK\perp AT\Rightarrow OK\|IP\ (*)$.

The triangles $TKO,TEC$ are similar ($K$ is turned into $O$ be the spiral similarity centered at $T$ which turns $E$ into $C$ and $D$ into $B$, because $K,O$ are the circumcenters of $TDE,TBC$), so $\frac{OK}R=\frac{OK}{OT}=\frac{CE}{CT}=\frac{BC}{CT}$, which is equal to $\frac{PC}{CO}=\frac{PC}R$, because the triangles $TBC,OPC$ are similar. We have thus found $OK=PC$, which, in turn, is equal to $IP$.

This last result, together with $(*)$ and the fact that $P,O,T$ are collinear and $PO=OT$ shows that $PIO,OKT$ are the image of one another through a translation. In particular, $OIKT$ is a parallelogram, so $KI=OT=R$, Q.E.D.

Fantastic !!!

Many thanks  to [b]all solvers [/b]of Challenges 1 and 2  !!!

Babis

  [i]( By the way  , this was the level of problems in geometry to enter in the Greek Universities( athens technical university) between 1950 and 1978)[/i]

Let P' = CD.BE and let P be the circumcenter of BCP'. Let A'B'C' the translation of ABC of trhe vector PP'.  Is easy to verify that B' and C' are the simmetrical points of P wrt BP' and CP' respectively.  Let D' and E' be the translated points of D and E of the vector P'P. As I is the incenter of ABC is easy to see that is orthocenter  of BCP' and the circumcenter of PD'E'.

As B'BD'D and C'CE'E are two rombus K is the circumcenter of A'B'C'.
If K' is the translated point of K of the vector I'I  it results be the circumcenter of ABC.

If I' is the circumcenter of P'DE by angle chasing one can prove that I' lies on c(ADE) then AI//KK'. Since AK = K'I then AK' = KI.

I have a very nice and strong solution using vector !But now I'm very busy.I will post it soon(when I have enough time)

here is my solution(?):

   Fix side BC  and let vertex A' rotate along the circumcircle of ABC

  (while A'C  and A'B  > BC, naturally).  Circumcenter K' of A'ED  is on circle centered at O

  (circumcenter of ABC) ,  incenter I' is on circle IBC of same radius  centered at P

  (midarc BC).   P is antipode of T  the fixed point of circles A'ED.

  It should be  (?)  clear that K'I' = OP  radius of circumcircle of ABC and remains

  parallel to itself.

   Thank you.

   M.T.

[b]Remark.[/b] If $BD=CE=x\le min\ \{b,c\}$, then $IK^2=R^2+\frac{Rr}{a(p-a)}\cdot (x-a)(b+c+x-a).$


Given $n$ points $X_{1},X_{2},...,X_{n}$ in the interval $[0,1]$. Prove that there is a point $Y$ in $[0,1]$ such that \[\frac{1}{n}\sum_{i=1}^{n}|Y-X_{i}|=\frac{1}{2}.\]

[quote="olorin"]$f: [0,1]\to\mathbb{R}\ ,\ Y\mapsto \frac{1}{n}\sum_{i=1}^{n}|Y-X_{i}|$ is continuous with $f(1)=1-\frac{1}{n}\sum_{i=1}^{n}X_{i}=1-f(0)$.
So the result follows from the intermediate value theorem, as $f(0)\leq{1\over 2}\leq 1-f(0)$ or $f(0)\geq{1\over 2}\geq 1-f(0)$.[/quote]
Yes, $(f(1)-\frac{1}{2})(f(0)-\frac{1}{2})=-(f(0)-\frac{1}{2})^{2}\leq 0$. Nice solution, but my solution isn't so, it long and not nice. Therefore I will not post it! :D


integration
trigonometry
logarithms
limit
real analysis
real analysis unsolved

Calculate

$ \text{a)}\;\sum_{n\equal{}1}^{\infty}\frac{1}{n}\int_{2\pi n}^{\infty}\frac{\sin z}{z}dz$

$ \text{b)}\;\sum_{n\equal{}1}^{\infty}\frac{1}{n}\int_{\pi n}^{\infty}\frac{\sin z}{z}dz$

$ \text{a)}$ is equal to $ \pi ( 1 - \log \sqrt {2\pi} )$. Here is a rough skech of calculation:

\begin{eqnarray*} \sum_{n = 1}^{\infty} \frac {1}{n} \int_{2\pi n}^{\infty} \frac {\sin z}{z} \, dz & = & \sum_{n = 1}^{\infty} \frac {1}{n} \int_{0}^{\infty} \frac {\sin nt}{t + 2\pi} \, dt \\
& = & \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac {1}{n} \int_{0}^{2\pi} \frac {\sin nt}{t + 2m\pi} \, dt \\
& = & \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac {1}{n} \int_{0}^{2\pi} \frac {\sin nt}{t + 2m\pi} \, dt \\
& = & \sum_{m = 1}^{\infty} \int_{0}^{2\pi} \sum_{n = 1}^{\infty} \frac {1}{n} \frac {\sin nt}{t + 2m\pi} \, dt \\
& = & \sum_{m = 1}^{\infty} \int_{0}^{2\pi} \frac {\pi - t}{2(t + 2m\pi)} \, dt \\
& = & \pi \sum_{m = 1}^{\infty} \left[ \left( 1 + \frac {1}{2} \right) \log \left( 1 + \frac {1}{n} \right) - 1 \right] \\
& = & \pi \log \lim_{N \to \infty} e^{ - N} \prod_{n = 1}^{N} \left( \frac {n + 1}{n} \right)^{n + \frac {1}{2}} \\
& = & \pi \log \lim_{N \to \infty} \frac {\sqrt {N + 1} (N + 1)^{N} e^{ - N}}{N!} \\
& = & \pi \log \frac {e}{\sqrt {2\pi}} \\
& = & \pi ( 1 - \log \sqrt {2\pi} ). \end{eqnarray*}

Similar arguments show that $ \text{b})$ is equal to $ \frac {\pi}{2}(1 - \log \pi)$.


[quote="filip"]Given prime number $p$ and $n\in \mathbb{N}$ such that: $p|n^3-1$ and $n|p-1$. Prove that $4p-3$ is perfect square.[/quote]
Easy, if $p|(n^3-1)$ then $p|(n^2+n+1)$, because if $p|(n-1)$ and $n|(p-1)$  we obtain a trivial contraction!, so
\[
\left.\begin{array}{r}
n^2+n+1=Dp, \\
p-1=En
\end{array}
\right\} \ \Rightarrow \ D=1\ mod \ n \ \Rightarrow \ D=1+Fn.
\]
Then $n^2+n+1=Dp=(1+Fp)(1+En)$ and $F\ge 0$, $E\ge 1$, hence $F=0$, so $D=1$ and $4p-3=4n^2+4n+1=(1+2n)^2$ which is a perfect square.

Best regards.
RX the Crazy Math


find all functional f(x) such that:f:R->R and
             f(x^k*f(y)+x^k)=x^k*y+f(x^k);      for all x,y in real and k is natural  >= 1
 :D  :D  :D   good luck to have the solution

I have got that $f(x)=x$ and $f(x)=-x$ in case that k is odd.And I have got  $f(x)=x$ and $f(x)=-x$ in case that k is even and $x>0$.
Put $x=1\Rightarrow{f(1+f(y))=y+f(1)}$, thus f is sjurective. Take y such that $f(y)=0\Rightarrow{y=0}$, so$ f(y)=0$ iff $y=0$. Take y such that $f(y)=-1\Rightarrow{f(x^{k})=-x^{k}y}$, after this put $x=1\Rightarrow y=-f(1)$, so $f(x^{k})=f(1)x^{k} $.If k is odd, then $x^{k}$ varies over all $\mathbb{R}\Rightarrow f(x)=f(1)x$ for all $x$ , putting this into original equation we get $f(1)=\pm{1}$. So we got $f(x)=\pm{x}$ for k is odd. In similar way $f(x)=\pm{x}$ when k is even but $x>0$.
It remains to show that $f(x)=\pm{x}$ for $x<0$.
I haven't finished my solution because I am too tired now. I'll try to finish it tomorrow. :)

P.S. I think it suffices to find solutions for k=1,2, but I am not sure.  ;)

[quote="nhat"]i think it has the solution for all k is possitive integer :D  :D
and your answer is so nice ,i have two solution
do yuo have the another solution :P  :P ?[/quote]
Yeah , of course it can be proved for all k, but I ment that if we find $f$ for $k=1,2$ then we can find it for every k. For example if k is even then we can denote $x^{k}$ by $x^{2}$, and if k is odd then you can  denote $x^{k}$ by $x$.

Here I am finishing my solution for even k and negative x. In my previous post I have  proved that $f(x)=\pm{x}$ for positive x and even k.
[b]Case 1.[/b] Let $f(x)=-x$ , so $f(x^{k}f(y)+x^{k})=x^{k}(y-1)$. Put $y=2$, hence $f(-x^{k})=x^{k}$, so $f(x)=-x$ for negative x-es too.
[b]Case 2. [/b] Let $f(x)=x$ , so $f(x^{k}f(y)+x^{k})=x^{k}(y+1)$. Since f is sjurective, we can find $y_{0}$ such that $f(y_{0})=-1$, putting it in our equation we obtain $x^{k}(y_{0}+1)=0\Rightarrow f(y_{0})=-1$ iff $y_{0}=-1$.Now put $x=1, y=-2\Rightarrow f(f(-2)+1)=-1\Rightarrow f(-2)+1=-1$(since $f(x)=-1$ iff $x=-1$) or $f(-2)=-2$.Now put $y=-2$, so $f(-x^{k})=-x^{k}$,hence $f(x)=x$ for negative x-es too. QED


Describe the set of points (x, y, z) defined by the given equations.

(1) x = 3 and z = 1

(2) x = 1 and y = 2

They're planes.  $ x\equal{}3$, $ z\equal{}1$, $ x\equal{}1$, and $ y\equal{}2$ are all the equations of planes.


function
real analysis
number theory
greatest common divisor
logarithms

Given a positive integer $ k > 1$, show that there are infinitely many integers $ n$ for which $ \tau (n) = k$, but at most finitely many $ n$ with $ \sigma (n) = k$.

Note:  $ \tau(n)$ denotes the number of divisors of $ n$, whereas $ \sigma(n)$ denotes the sum of the divisors of $ n$.

[hide="1"] This is simple.  For a prime $ p$, we have $ \tau(p^{k-1}) = k$. [/hide]
[hide="2"] This is also simple.  If $ n > k$ then $ \sigma(n) > n > k$. [/hide]
[b]Follow-up:[/b]  Define the proper divisor function $ p(n) = \sigma(n)-n$.  Do there exist infinitely many $ n$ such that $ p(n) = k$ for a given $ k > 1$?

[b]Second follow-up:[/b]  Define the relative divisor function $ r(n) = \frac{\sigma(n)}{n}$.  Is $ r(n)$ bounded?

Answer (Follow-up 1)

$ p(n)$ gives the sum of all the positive divisors of $ n$ up to, but not including $ n$.

If $ n$ is prime then $ \sigma(n) = n+1 \Rightarrow p(n) = 1$

Hence $ n$ must be composite for $ k > 1$.

There must be a finite number of prime factors in $ n$ as the sum of the prime factors has to be less than $ k$. This is true even if some of the prime factors are the same, as new composite factors  are introduced each of which is greater than the prime itself. Given a finite number of prime  factors, we can only create  finite number of integers, hence there does not exist infinitely many $ n$ such that $ p(n) = k$ for a given $ k > 1$

Answer (Follow-up 2)

Let the positive divisors of $ n$ be $ d_{1},d_{2},d_{3},\ldots,d_{k}$

$ r(n) = \frac{\sigma(n)}{n}= \frac{d_{1}+d_{2}+d_{3}+\ldots+d_{k}}{n}$

$ = \frac{d_{1}}{n}+\frac{d_{2}}{n}+\frac{d_{3}}{n}+\cdots+\frac{d_{k}}{n}$

$ = \frac{1}{\frac{n}{d_{1}}}+\frac{1}{\frac{n}{d_{2}}}+\frac{1}{\frac{n}{d_{3}}}+\cdots+\frac{1}{\frac{n}{d_{k}}}$

But each $ \frac{n}{d_{i}}$ is a unique divisor of $ n$

$ \Rightarrow r(n) = \frac{1}{d_{1}}+\frac{1}{d_{2}}+\frac{1}{d_{3}}+\cdots+\frac{1}{d_{n}}$

Hence $ r(n)$ is the sum of the reciprocals of the positive divisors of $ n$.

$ \sigma(n) \geq n+1 > n$

$ \Rightarrow \frac{\sigma(n)}{n}> 1$

Let the prime factorisation of $ n$ be $ n = p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}\ldots p_{r}^{k_{r}}$

$ \Rightarrow \frac{\sigma(n)}{n}= \frac{\left(\frac{p_{1}^{k_{1}+1}-1}{p_{1}-1}\right) \left(\frac{p_{2}^{k_{2}+1}-1}{p_{2}-1}\right) \cdots \left(\frac{p_{r}^{k_{r}+1}-1}{p_{r}-1}\right)}{p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{r}^{k_{r}}}$

$ \Rightarrow \frac{\sigma(n)}{n}< \frac{\left(\frac{p_{1}^{k_{1}+1}}{p_{1}-1}\right) \left(\frac{p_{2}^{k_{2}+1}}{p_{2}-1}\right) \cdots \left(\frac{p_{r}^{k_{r}+1}}{p_{r}-1}\right)}{p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{r}^{k_{r}}}$

$ \Rightarrow \frac{\sigma(n)}{n}< \left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{2}}{p_{2}-1}\right)\cdots \left(\frac{p_{r}}{p_{r}-1}\right)$

$ \Rightarrow \frac{\sigma(n)}{n}< \frac{p_{r}!}{(p_{r}-1)!}= p_{r}$

Hence:

$ 1 \leq r(n) < \left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{2}}{p_{2}-1}\right)\cdots \left(\frac{p_{r}}{p_{r}-1}\right) \leq p_{r}$

[quote="sludgethrower"]$ \Rightarrow \frac{\sigma(n)}{n}< \left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{2}}{p_{2}-1}\right)\cdots \left(\frac{p_{r}}{p_{r}-1}\right)$

$ \Rightarrow \frac{\sigma(n)}{n}< \frac{p_{r}!}{(p_{r}-1)!}$[/quote]

This step is wrong.  But you're close.  ;)

[quote="sludgethrower"]Question 2

For any positive integer $ n$, show that:

$ \sum_{d|n}\frac{n}{d}\sigma(d) = \sum_{d|n}d\tau(d)$[/quote]

This is a small part of the general theory of arithmetic functions, and there are many such problems.  Here are some useful definitions and lemmas.

[b]Definition:[/b]  A function $ f : \mathbb{N}\to \mathbb{N}$ is said to be multiplicative if $ gcd(a, b) = 1 \implies f(ab) = f(a) f(b)$.  Nearly every common arithmetic function is multiplicative; $ \tau(n), \phi(n), \sigma(n), \mu(n)$...

[b]Corollary:[/b]  If $ f(p^{k}) = g(p^{k})$ for two multiplicative arithmetic functions, then $ f(n) = g(n)$ for all $ n$.

[b]Definition:[/b]  Given two arithmetic functions $ f, g$, the function

$ H(n) = \sum_{d | n}f(d) g \left( \frac{n}{d}\right)$

Is called the [i]Dirichlet convolution[/i] of $ f, g$.

[b]Lemma:[/b]  The Dirichlet convolution of two multiplicative functions is multiplicative.

Can you explain to me why this step in question 1 is wrong?

For question 2, I didn't know the final lemma, so I had to show that both functions were multiplicative. I then applied the corollary since every integer can be expressed as the product of powers of primes.

I guess a proof the lemma would go  as follows:

Let   $ H(n) = \sum_{d|n}f(d)g\left(\frac{n}{d}\right)$ where $ f$ and $ g$ are multiplicative.

Let $ gcd(m,\,n) = 1$

$ \Rightarrow H(mn) = \sum_{d|mn}f(d)g\left(\frac{mn}{d}\right) = \sum_{d_{1}|m, d_{2}|n}f(d_{1}d_{2})g\left(\frac{mn}{d_{1}d_{2}}\right)$

Since $ gcd(d_{1},\,d_{2}) = 1$ then $ gcd\left(\frac{m}{d_{1}},\,\frac{n}{d_{2}}\right) = 1$ since $ \frac{m}{d_{1}}$ and $ \frac{n}{d_{2}}$ are also divisors of $ m$ and $ n$ respectrively.

$ \Rightarrow H(mn) = \sum_{d_{1}|m, d_{2}|n}f(d_{1})f(d_{2})g\left(\frac{m}{d_{1}}\right)g\left(\frac{n}{d_{2}}\right)$

$ \Rightarrow H(mn) = \left(\sum_{d_{1}|m}f(d_{1})g\left(\frac{m}{d_{1}}\right)\right) \left(\sum_{d_{2}|n}f(d_{2})g\left(\frac{n}{d_{2}}\right)\right)$

$ \Rightarrow H(mn) = H(m)H(n)$ as required

[quote="sludgethrower"]Can you explain to me why this step in question 1 is wrong? [/quote]

Well, you claimed that $ p_{1}p_{2}... p_{r}= (p_{r})!$, which is clearly false.  And then you claimed that $ (p_{1}-1)(p_{2}-1)...(p_{r}-1) = (p_{r}-1)!$, which is also false.  But that is the correct product at which you should be looking.

[hide="Lemma 1"]$ \prod_{k=1}^{\infty}(1+a_{k}) \text{ converges}\Leftrightarrow \sum_{k=1}^{\infty}a_{k}\text{ converges}$ [/hide]
[hide="Lemma 2"] $ \sum_{k=1}^{N}\frac{1}{p_{k}}\sim \ln \ln N$ [/hide]
One of the problems with your argument in general, though, is that you established an upper bound, when in fact you should be establishing a lower bound.

[quote]Well, you claimed that $ p_{1}p_{2}... p_{r}= (p_{r})!$, which is clearly false.  And then you claimed that $ (p_{1}-1)(p_{2}-1)...(p_{r}-1) = (p_{r}-1)!$, which is also false.[/quote]

I didn't claim this.

For instance if $ x < 2$ then $ x < 3$

This does not imply that $ 2 = 3$.

What I claim is:

$ \frac{p_{r}!}{(p_{r}-1)!}\geq \left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{2}}{p_{2}-1}\right)\cdots \left(\frac{p_{r}}{p_{r}-1}\right)$

This is true because the right hand side divides the left hand side.

$ \frac{p_{r}!}{(p_{r}-1)!}= \prod_{n = 1}^{p_{r}}\frac{n}{n-1}$

This product contains all of the terms of the right hand side , and perhaps a few more if $ p_{r}> 3$ since $ n$ can be composite (all terms in the product are $ \geq 1$).

Question 3

The Mangoldt function $ \Lambda$ is defined by:

$ \Lambda (n) = \log{p}$, if $ n = p^{k}$, where $ p$ is a prime and $ k \geq 1$ or $ \Lambda (n) = 0$ otherwise.

Prove that    $ \, \, \,\Lambda (n) = \sum_{d|n}\mu \left(\frac{n}{d}\right) \log{d}=-\sum_{d|n}\mu (d) \log{d}$

[quote="sludgethrower"]$ \frac{p_{r}!}{(p_{r}-1)!}\geq \left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{2}}{p_{2}-1}\right)\cdots \left(\frac{p_{r}}{p_{r}-1}\right)$ [/quote]

This is equivalent to

$ p_{r}\ge \prod \frac{p_{k}}{p_{k}-1}$

Which is true, but
- It's a very weak bound, and
- It's a very weak [b]upper[/b] bound, and doesn't help to establish that $ r(n)$ is unbounded.

Yeah...I knew it was a weak bound. I don't know the definition of unbouded however, and I have not studied the properties of divergence and convergence yet. At school we just "assume" that a function converges and apply maclaurin's series.

[quote="sludgethrower"]I don't know the definition of unbouded.[/quote]

[b]Definition:[/b]  A sequence $ a_{n}$ is unbounded if for every real $ N$ we can find $ k$ such that $ a_{k}> N$.

This follows from the definition of a bounded sequence.

[b]Definition:[/b]  A sequence $ a_{n}$ is bounded if there exists a real $ N$ such that for all $ k$ we have $ a_{k}< N$.


quadratics
superior algebra
superior algebra unsolved

Let $R=\{a+bk| a,b\in\mathbb{K}\}\subset\mathbb{M}$, where $\mathbb{M}$ is a field, $\mathbb{K}$ is a ring with unit element(s), $k\in\mathbb{M}, k\not\in\mathbb{K}, k^2\in\mathbb{K}$.

Does it follow from $a+bk=c+dk$ that $a=c,b=d$?

If not true, could someone supply an additional condition to make it true?

I think that's really true and you don't even need all those conditions, just $k\in \mathbb{M}\backslash\mathbb{K}$

suppose $b\neq d$

then $k=\frac{a-c}{d-b}$ which is impossible as k is not in $\mathbb{K}$

if $b=d$ then you use cancellation law and you get immediately $a=c$

there is much more to this, if all powers of k are independent over $\mathbb{K}$ we say the element is transcendental (like $\pi$ over $\mathbb{Q}$)
If not the dimension of the spanned space is exactly the degree of the field extension

[quote="fredbel6"]then $k=\frac{a-c}{d-b}$ which is impossible as k is not in $\mathbb{K}$[/quote]I said a (commutative) [b]ring[/b]. :) It is of course trivial for a field... but in a ring, division by $d-b$ is not always defined.

The conditions are necessary, even the presence of unit elements is necessary, else we have the follow easy counterexample: $\mathbb{M}=\mathbb{R}, \mathbb{K}=4\mathbb{Z}, k=2$.

Edit: Since many definitions are used, I'll list the axioms again for a commutative ring with unit elements. Basicly I just want to know more about possible generalisations of the property for $\mathbb Z$.
- (K,+) is a commutat group
- (K,.) is a commutative monoid, i.e. a group without each element having an inverse. For just "a ring" all we need is a semigroup.
- distributivity of . to +

In some places $(S,\cdot)$ such that $\cdot$ is associative is called a semi-group. If morevoer $\cdot$ has a unit in $S$, then $(S,\cdot)$ is called a monoid.

amfulger, indeed, that's how I mean it. :) Do you know a solution or counterexample?

[quote="fredbel6"]then what is $4\mathbb{Z}$ ?  I'm afraid to say anything now but where is the unity element?[/quote]Nowhere... that's exactly what I'm saying right? The unit condition is necessary, because if we do not require it, then $4\mathbb{Z} = \{4z|z\in\mathbb{Z}\}$ gives a counterexample.

A counterexample: Let $K=\{a+2b\sqrt{7}\}$ with $a,b\in\mathbb{Z}$, and let $k=\sqrt{7}$.

Over nice enough rings, this is a version of the rational root theorem. We need the ring to be pretty nice for that.

[quote="jmerry"]A counterexample: Let $K=\{a+2b\sqrt{7}\}$ with $a,b\in\mathbb{Z}$, and let $k=\sqrt{7}$.[/quote]
Yes, you are right. So I was right this isn't going to lead anywhere. :)

I'll just keep stick to $\mathbb{Z}$ then. Thanks for your comments.

@Arne: and just why would they not exist? ... :huh:

[quote="Arne"]"Ring extensions" don't really exist, do they?

They do not even appear in my Algebra bible (Birkhoff & MacLane), as far as I know.[/quote]

In what context, rings are extended all the  time, think of $\mathbb{Z}[i]$??

I know, I know, but there is no real "general theory" about ring extensions (like there is one about field extensions), right?

I haven't even seen a formal definition of a "ring extension" yet, although it's easy to create one :)

There is even some kind of "weird" Galois Theory for Ring extensions... ;)
See Chapter 7, p. 333, in Lang's "Algebra".

[quote="Peter VDD"]http://mathworld.wolfram.com/ExtensionRing.html[/quote]

That's just a definition, a term. No further theory is developed there.

Such a theory does exist and did develop for example the concept of [b]integrally closed ring[/b]. Just look up the Chapter "Ring Extensions" in Lang where a systematic treatement of ring extensions is given.