A judge ruled that Intelligent Design cannot be taught in public classrooms.

[quote]
 ``To preserve the separation of church and state'' mandated by the First Amendment, the Dover Area School District is barred from maintaining the ID policy in any school, Jones wrote. ``The students, parents and teachers of the Dover Area School District deserved better than to be dragged into this legal maelstrom, with its resulting utter waste of monetary and personal resources.''
[/quote]

[url=http://www.forbes.com/home/feeds/ap/2005/12/20/ap2404791.html]Excerpts[/url]
[url=http://www.bloomberg.com/apps/news?pid=10000103&sid=aHhAnvL4XhjU&refer=us]News story[/url]
[url=http://abcnews.go.com/US/wireStory?id=1424896]Another news story[/url]

What are your reactions?

[quote="bleumoose"]Obviously it shows personal bias.

But it's true. Students... [i]students[/i] don't deserve to be dragged into this crap- they don't deserve their scientific education to be watered down by religious politics.

[size=59]and will someone PLEASE get in a knock-down, drag out fight with me about this?! I'M BORED!!!![/size][/quote]

I'M BORED TOO BUT UNFORTUNATELY I AGREE DARN

[quote="bleumoose"]yeah i got my wisdom teeth out so... i'm bored. erg.[/quote]
Sure? What can we oppose on?

Death penelty?Global warming? Recycling? Politics?

Sorry, but I agree with you on this issue.

[quote="bleumoose"]Obviously it shows personal bias.

But it's true. Students... [i]students[/i] don't deserve to be dragged into this crap- they don't deserve their scientific education to be watered down by religious politics.

[size=59]and will someone PLEASE get in a knock-down, drag out fight with me about this?! I'M BORED!!!![/size][/quote]
I like the small type. :P But unfortunately I agree too - religion has absolutely no place in school.

[quote="bleumoose"]Obviously it shows personal bias.

But it's true. Students... [i]students[/i] don't deserve to be dragged into this crap- they don't deserve their scientific education to be watered down by religious politics.
[/quote]

I don't agree with the way you said it, but I do agree.  I'm not surprised about this ruling at all.

Did anyone hear Stephen Myers talk on KTTH 770 (radio) this morning? It was really interesting.

[quote="MithsApprentice"]Were those excerpts directly taken from the judge's opinion? It shows some very clear personal bias. I'm not sure this is one of those cases where it would be possible to find a judge that's completely neutral, but they should've at least tried for someone more moderate on this issue.[/quote]
mmyeah, from what I've read and heard, the judge appeared a little too biased . . .

I think the wording was a bit harsh, but I enjoyed every single bit of it. lol

Zenny K, are you refering to the discovery institute's accusations about
1) judicial activism
2) censorship

because to 1)
a)separation of church and state is really well established.
b)by the definition of judicial activism, Judge Jones would have to specify implementation, which he doesn't; he merely says teaching intelligent design  is unconstitutional. (btw, this is a contextual definition)
c) just becasue something isn't supported by the majority doesn't mean  it's judicial activism.

to 2)
a)because these are public schools, there are limits to what you can teach.

Oh. We read this article in Social Studies because we were learning about how cases are ruled and why some are important because they violate amendments. I wasn't shocked by the judge's ruling at all.

[quote="catcurio"][quote="MithsApprentice"][quote="matamata"]Ahh yes...  the biases of a christian republican judge... how sad.[/quote]

Er so am I missing something or was that judge liberal?[/quote]

Hmm... I would think that a Christian judge would [i]support [/i]teaching ID in schools...[/quote]

not if the judge remained faithful to 200 years of judicial precedence...

[quote="catcurio"][quote="MithsApprentice"][quote="matamata"]Ahh yes...  the biases of a christian republican judge... how sad.[/quote]

Er so am I missing something or was that judge liberal?[/quote]

Hmm... I would think that a Christian judge would [i]support [/i]teaching ID in schools...[/quote]

I believe that he was appointed by GW.

[quote="PenguinIntegral"][/quote][quote="catcurio"][quote="MithsApprentice"][quote="matamata"]Ahh yes...  the biases of a christian republican judge... how sad.[/quote]

Er so am I missing something or was that judge liberal?[/quote]

Hmm... I would think that a Christian judge would [i]support [/i]teaching ID in schools...[/quote][quote="PenguinIntegral"]

I believe that he was appointed by GW.[/quote]
Yes, he was appointed by the president. And I seriously doubt he is truly a Christian. In any case, he certainly wasn't a scientist.

Not a christian¿ Not a scientist¿ Why¿

Note: ID is [u]not[/u] Christianity.
1. reason: You can believe in the bible and all that but dislike ID (and there are good reasons for that).
2. reason: even ID says it is not related to faith ;)
Any christian that really understood Christianity's idea(l)s and believes in the law too should react like him.
And why he isn't a scientist (except that scientific judges should be rare) I don't see.

But I don't want to start another discussion about ID now...

PS: I also miss the option for :clap:

[quote="ZennyK"]
Yes, he was appointed by the president. And I seriously doubt he is truly a Christian. In any case, he certainly wasn't a scientist.[/quote]

Christians can follow constitutional principles too. Separation of church and state actually began because christians were afraid of government interference in the church.

It's interesting how those advocating the status quo, sometimes engage in harsh fighting against change, then use, as one of the reasons to maintain the status quo that "people shouldn't be dragged trough that crap" or the like.  This shows up in many areas, just as it did here.

Someone mentioned the 200 years of history on this issue.  However, this is another area where the "history" cited by the advocates is not the "history" that one finds when one reads what was written at the time.  The purpose of separation of church and state was to stop the then current practice of states adopting state religions.  This seems so foreign, and thankfully it is  --  to us.  I noted that when a famous tennis player switched his religions in one of the European countries, it was noted by the press that his income would go up 20% or so as the tennis player's new religion had the government withhold and pay the religion only some 10% of the player's income, rather than the 30% or so that the former religion took.  This is similar to our income tax withholding and just as easy to avoid.  Now I probably misstated the amounts, but using governmental power to support organized religions has had a long and current history.  The teaching of ID would take what, part of one week in one class.  Not worth arguing over, absent the religious impact.

I would note in passing that the vast majority of the scientists and others whose works are being cited and supported would have, and currently do, personally believe in ID (and God).  Newton wanted to spend his time with the scriptures, but failed in his attempt and settled for a career that included setting the basics for many things and being the Chancellor of the Exchequer in England (equivalent to a supercharged secretary of the treasury for the US).  All of the founding fathers would be amazed at how far from their view of the appropriate place for God we have moved public education and government.  (Anyone remember that a Bible is still used for witnesses in courts in the US.  The option to swear or affirm is only offered in deference to our Quaker friends who believe that swearing is never allowed.)

ID doesn't mean that one take any religious book as the Gospel (he, he, he).  The classic argument is that one can analogize that finding a watch suggests a watchmaker.  Finding boot prints on my front porch in the snow that fell last night suggests that someone (in boots) was on my porch after the snow fell last night.

We are able to teach the details of the ID beliefs of the Norse, the Booleans, Egyptians, Hebrews, Greeks, Romans (mostly taken from the Greek, but ok).  History is another way that ID can be taught.  The teaching of ID doesn't mean that it is taught as the only truth or the details of any particular religion.  However, a person graduating from school, not knowing about ID will have a seriously distorted view of their history and heritage.  What was it like to be a member of a society where everyone "knew" not only that ID was correct, but the local flavor of ID was correct?  We are in a unusually tolerant society.  One that is worth defending, sometimes even if it is factually wrong.

I have my own view, but it doesn't matter to anyone, or this posting.  One needs to avoid the "religion" of "political correctness" or the current government.  It is a "religion" since there is no compromise or development of facts that change one's views -- they are based on faith (faith that their view is correct).

Sorry that this is too long, but I think it is important for you'll to have a greater appreciation of the question, regardless of whether or not you are ever taught ID.
[/code]


[quote="BobofBobs"]How about doing a Math Jam on the most recent ARML competition?[/quote]

Unfortunately an ARML problem discussion Math Jam is not something we currently have time to prepare.  At least for now the only recent competition discussion Math Jams will be those related toe AMC, AIME, USAMO, and possible the IMO.


algebra
polynomial
Galois Theory
superior algebra
superior algebra theorems

Is there an analogue of Galois groups for multivariable polynomials?  In particular, is there a way to test the solvability in terms of radicals for polynomials of several variables?

What do you mean by that?  Polynomials of several variables don't have finitely many roots; they define varieties instead.  Given a Galois extension of $ \mathbb{Q}$, one can ask whether a polynomial of several variables has points in that extension and consider the action of the corresponding Galois group, but I am not sure this is what you're looking for.

Again, you should probably clarify what it is you're looking for.

[quote="JRav"]I posed this question to a professor and mentioned something called the [url=http://en.wikipedia.org/wiki/Cremona_group]Cremona group[/url] but I'm not sure of the relevance.[/quote]

The Galois group is the "automorphism group" of the set of roots of an equation.  It is also the automorphism group of a field extension defined by that same polynomial.

The Cremona group is basically the same thing: it is the automorphism group of the variety defined by a multivariate polynomial, and it is the automorphism group of a certain field extension.

The roots of a univariate polynomial are just numbers, discrete points.

The roots of a bivariate polynomial form curves, and the roots of a multivariate polynomial form hypersurfaces.  The Cremona group is the automorphism group of the simplest hypersurface.

Since hypersurfaces are so much harder than numbers, people have not made as much progress on their automorphism groups.  The Cremona group is very interesting, even if the curve used to define it is a little dull.

I think a curve like yy=xxx+x+1 might have a more interesting automorphism group, but you'd have to ask a geometer about such things.


calculus
integration
inequalities
function
probability
real analysis
real analysis solved

Let $f(x) = \frac{x^n}{n!}e^{-x}$, where n is a natural number, greater than 1.
Prove that

$\int_0^{2n+2} f(x)\,dx \geq \frac{n}{n+1}$

In general, when n is a real number, we can replace $n!$ by the Gamma function and get a similar result.

The desired result can also be expressed as

$\int_{2n+2}^{\infty}f(x)dx\le\frac1{n+1}.$

Here's one approach. The function $f(x)$ is the density of a random variable. Specfically, it is a random variable from the family of gamma distributions. In particular, it arises as the distribution of the sum of $n+1$ identical independent exponential distributions.

Let $X$ be the random variable. As can be readily computed (or looked up in a probability textbook if necessary), the mean of $X$ is $n+1$ and the variance of $X$ is $n+1,$ which means that the standard deviation is $\sqrt{n+1}.$

Let $Z=\frac{X-(n+1)}{\sqrt{n+1}}.$ Then $Z$ has mean zero and variance one.

We want to estimate

$P\left(X>2n+2\right)=P\left(\frac{X-(n+1)}{\sqrt{n+1}}> \frac{(2n+2)-(n+1)}{\sqrt{n+1}}\right)=P\left(Z>\sqrt{n+1}\right).$

By Chebyshev's inequality (Myth can edit the Russian transliteration if he wants), this probability is $\le\frac1{(\sqrt{n+1})^2}=\frac1{n+1}.$

[quote="vuhung"]In general, when n is a real number, we can replace n! by the Gamma function and get a similar result.[/quote]
The proof I gave works just as well in this case. The only thing we give up is the description as the sum of identical independent exponential distributions, but that was just for mnemonic value, to make it easier to remember what the mean and the variance were. But the gamma family of distributions has the properties we need.

Of course Chebyshev's inequality is a gross overestimate in most cases. In fact,

$\int_{2n+2}^{\infty}f(x)dx$ is always much smaller than $\frac1{n+1},$ and this gets more pronounced as $n\to\infty.$

The Central Limit Theorem suggests that as $n\to\infty$, this integral should go to zero at an exponential rate.

Full score for my teacher Kent  :lol:
My solution is the same to you. It is very nice to make use of strong theorems to get more elementary results, yet beautiful. I posted it here to seek for alternative proofs.

[quote]Of course Chebyshev's inequality is a gross overestimate in most cases...
The Central Limit Theorem suggests that as $n\to\infty$, this integral should go to zero at an exponential rate.[/quote]
How to apply Central Limit Theorem here? It requires, in general N random variables to use C.L.T but in this case, we just have only one.  Can you point it out?

Btw, Chebyshev in Russian: Чебышева, just in case Myth may need :)
TIA

If you view the gamma distribution with $\alpha=n+1$ as the sum of $n+1$ independent exponential distributions, we have the setup for the Central Limit Theorem. That is, let $X_{\alpha}$ have density $\frac1{\Gamma(\alpha)}x^{\alpha-1}e^{-x}$. Now let $Z_{\alpha}=\frac{X_{\alpha}-\alpha}{\sqrt{\alpha}}.$ As $\alpha\to\infty, \, Z_{\alpha}$ converges in distribution to a standard normal distribution.

However, I may have overstated things a little. An exponential is a very skewed distribution, so all of the intermediate stages are somewhat skewed. If we were asking for the probability that $Z_{\alpha}>k$ for some fixed $k$, that wouldn't be a problem, but we are shooting for a moving target - so we can't just quote the CLT. But I still assume that the rate of decay is some kind of exponential.


function
quadratics
algebra
quadratic formula
difference of squares
special factorizations
NSML

junior contest??? 60 perfects???

a few suggestions...

ex. find the hole for the function:

If $n\in\mathbb{N}^{+}$, then find the hole of $f(x)=\frac{x^{2n}+6x^{n}-7}{x-1}$ in terms of $n$.

...but that is still pretty trivial

or we could steal a problem from an aime...

find a,b such that $\frac{ax^{17}+bx^{16}+1}{(x-1)^{2}}$ is a polynomial...

which both get pwnd by calculus...

hence

[b]lets do quadratic diophantines next year for junior nsml...[/b]

and half of the problems can be like omg complete the square, or quadratic formula, and complete the square, and difference of squares

and one of those that is like (x-2y)^2+(x-4)^2=0

and you can have a pell

and an extended pell

and maybe an infinite descent one...

and a tricky gaussian integers problem like solve in Z x Z... x^2+y^2=3^{20}*5^{20}...

and it will be like woa, people learn something

[sorry if this is kind of obnoxious...but i feel like part of the goal is to make challenging tests...and rational functions is one of those topics were you either know it or you don't]

[quote="Altheman"][quote="nateharman1234"]Ehh, if the regular tests are too easy then try to do orals, its much more fun anyway. Oh, and if you think the junior test was easy, did you look at the senior one?  That took me all of 5 minutes to do perfectly, too bad I was just doing it to kill time after orals so it didn't count.[/quote]

meh...another kid at my school is doing orals...also i want to keep my high scores on individual...and yes the senior contest was easy...double perfects that day...[/quote]

wow double perfects, very nice.  If you can get double perfects and they dont demand you to be oralist your school must be a hell of a lot better than mine, the highest score out of anybody from my school (excluding myself) was 2 correct haha.

eh...i am probably the best overall at my school...but other kids wanted to do orals...so i was like meh...also it is a college app thing...winning the division is better than doing okay at orals...but i made some careless errors...grrr...so i am like 7th, 4th or something now in junior, senior


Does anyone here program games?  I use the Allegro game library and the Mappy map editor, and the mappyal library.(mappyal.h and mappyal.c) :)

yes but really amateur ones, mostly rpgs and applets for finding values of pi.  I use javascript and stuff from clickteam.com  View my first rpg in javascript. look at http://www.angelfire.com/ab7/whodoshoe/games/Forest_RPG_0.05.htm
or download the whole thing at
http://www.angelfire.com/ab7/whodoshoe/games/zip.zip
with no ads

With web design, if you want to build a game pure browser, you're going to have to try Sajax so your Javascript code can work in conjunction with you server-side php code and people can't cheat :)

Ya, I use Visual basic to program small games. VB may be simple and slow, but its GUI makes programming a breeze.

Bah,  I hate Visual Basic :P

Why would you really need much of a GUI for games anyway?  The whole window is usually covered with a rectangular view for the game.

I'm not talking about like giant, 3D, expansive, probably steroid-induced games.  :lol: I just do little stuff. I can't stand text-based programming. Gotta have my OOP!

[quote]I can't stand text-based programming. Gotta have my OOP![/quote]

Are you trying to compare the two?  They refer to completely different things...

Thats up for debate. Both programming, aren't they? Apples and oranges?  :yup:  Just saying I would prefer OOP over text.

But OOP and text are COMPLETELY different.  They aren't even describing the same thing

OOP = Object Oriented Programming (e.g. C++ or Java)
text = Bash, CSH, MS-DOS, ... Text-based programming is simply programming that runs on the aforementioned programs. (eg vim, emacs).

A bit more on topic, the hardest part (at least it seems to me) in making a game is not even the actual programming. It's the graphics, shudders.

The graphics in any sort of game I would make would be stick figures...or worse -- I might not even have the skill for those lol. I really don't understand how people get graphics so good in 3D games!

i use java to program 9-ball, with yahoo style guide line =P


vector
abstract algebra
group theory
linear algebra

Show that if $ V$ is a vector space over $ \mathbb{R}$, the axiom that $ 1x = x$ cannot be deduced from the other axioms.

If $ V$ is an abelian group under addition, $ F$ is our proposed field of scalars, and we define $ ax=0$ whenever $ a\in F,x\in V,$ then this satisfies all of our other axioms.

Question: if you weaken the statement "$ 1x=x$" to "if $ a\ne 0,x\ne 0,$ then $ ax\ne 0$", will that suffice?

[Note the importance in answering this question of the axiom $ a(bx)=(ab)x.$]

Kent, here's my answer to your follow up question.  First let me make sure i understood what you were asking--is it to show that if we replace the axiom $ 1x=x$ with if $ a \neq 0$ and $ x \neq 0$, then $ ax \neq 0$, then we can deduce $ 1x = 1$?  If this is the question, then the answer seems to be yes.

Let $ x \in V$.  Suppose $ 1x = y$.  Then $ 1x = (1 \cdot 1)x = 1(1x) = 1y$, hence $ 1(x-y) = 0$.  From the replaced axiom, we get $ x-y = 0$, or $ y=x$.  So $ 1x = x$.


[quote="Interval"]The mass of an orchid seed is approximately 0.0000035 gram.
Written in scientific notation, that mass is equivalent to 3.5 × 10n.
What is the value of n?

(1) –8      (3) –6
(2) –7      (4) –5[/quote]
Do you mean $10^{n}$?
Just count how many times you have to move the decimal over to the left to see the number. And because you're moving to the right the number is negative. So counting you get -6 or (3).

[quote="Interval"]The mass of an orchid seed is approximately 0.0000035 gram.
Written in scientific notation, that mass is equivalent to 3.5 × 10n.
What is the value of n?

(1) –8      (3) –6
(2) –7      (4) –5[/quote]
I think you mean $3.5 * 10^{n}$
[hide="10^n"]every time $n$ increases, the decimal place moves right (which is a negative power). You want it to move over 6 places to the right so then in scientific notation it is $3.5 * 10^{-}6$ so $\boxed{x=-6}$[/hide]

[quote="Interval"]No, n is not an exponent.

It is 10n not 10^n.

It could be a textbook error but this is what I have to share.[/quote]
Then n would be 0.000001 and that's not one of the choices...

[quote="Interval"]If n is an exponent, the answer is????[/quote]
The answer would be what Quevvy and I put.


inequalities trigonometry induction algorithm symmetry geometry Harvard

Hi,
I'm pretty new to this whole inequality stuff. Terms like power mean and Cauchy-Schwarz are a somewhat intimidating :?: so I think I'll take it step by step. Reading helps, but interacting is better, so perhaps we can kick off a little inequality lesson/mini-marathon. My goal is to eventually learn all of Kiran Kedlaya's notes (and Po-Shen Loh's, if I can find them), some topics of which may be considered too advanced for this forum, but for now this should be the right place. *Nods* :yup:  Besides, who am I to talk about "too difficult" when I can't even understand half the posts that appear as it is! :)

My only request is that if you post a problem, inequality, or solution, please label it appropriately. Since this isn't really a marathon, you don't need to have to post anything at all! But creativity is always a fun thing, right? Feel free to add. ;)

[size=167]Trivial Inequality![/size]

If $x$ is real, then $x^2\ge0$.

I don't know...is there really a proof for this? I mean, it is called "trivial"... :idea:

[size=150]Problems![/size]

1. Find the minimum of $x^2-4x+8$ over all real $x$.

2. (Soucre: AoPS) Prove that $\cos{2\theta}+\sin^2{\theta}$ is nonnegative for all angles $\theta$.

3. Given that $w,x,y,z$ are reals satisfying $w^2+x^2+y^2+z^2+21=w-3x+5y-7z$, find $16wxyz$.

4. (Source: AoPS) Show that $(x^2+1)(y^2+1)\ge(xy+1)^2$ for all $x$ and $y$.

5. Prove the two-variable Arithmetic-Geometric Mean. That is, for positive (real) $x,y$, show that $\frac{x+y}{2}\ge\sqrt{xy}$.

6. Prove that $x^{2n}$ is nonnegative for all integers $n$ over real $x$ whenever it is defined.

Nice work so far.
I should add that $x^{2n}$ is nonnegative whenever it is defined under real $x$. I guess dividing by zero isn't allowed.

I think that we can prove No.6 by mathmetical induction.

Number 4 simplifies to
$x^2y^2+y^2+x^2+1 \ge x^2y^2+2xy+1$
So if we can prove that $x^2+y^2\ge2xy$, then we're done.  So let's do that.  Starting with the expression, we'll simplify it.
$x^2+y^2\ge2xy$
$x^2-2xy+y^2\ge0$
$(x-y)^2\ge0$
The square of any real number$\ge0$, so this statement is clearly true.  Therefore, the original inequality holds.  We can also follow off this to form a proof of number 5.
$x^2+y^2\ge2xy$
$x^2+2xy+y^2\ge4xy$
$(x+y)^2\ge(2\sqrt{xy})^2$
$x+y\ge2\sqrt{xy}$
$\frac{x+y}{2}\ge\sqrt{xy}$

6. $x^{2n}=(x^n)^2$
If n is an integer and x is real, as towersfreak2006 said, then $x^n$ is real.  $Real^2\ge0$

2. Basically the same approach as shobber...
$\cos{2\theta}+\sin^2{\theta}=\sin^2{\theta}+\cos^2{\theta}-\sin^2{\theta}=\cos^2{\theta} \ge 0$
because $Real^2 \ge 0$

And to complete the set, nr. 5:

[hide]We move everything to the left side and factor into squares as such:

$(w-1/2)^2 + (x+3/2)^2 + (y-5/2)^2 + (z+7/2)^2 = 0$.

But since each square must be zero or more, they must all be zero. So:

$w=1/2$, $x=-3/2$, $y=5/2$, $z=-7/2$.

So $16wxyz = (1)(3)(5)(7) = 105$.[/hide]

It appears everything above is done.

[b][size=150]RMS-AM-GM-HM[/size][/b]
$ (\sqrt{\frac{a_1^2+a_2^2+...+a_n^2}{n})}\ge \frac{a_1+a_2+...+a_n}{n}\ge (a_1a_2...a_n)^{\frac 1n}\ge \frac{n}{\frac 1{a_1}+\frac 1{a_2}+...+\frac 1{a_n}}$ (for positive $a_1$, $a_2$, ..., $a_n$) where equality holds when $a_1=a_2=...=a_n$

[b][size=150]Problems[/size][/b]
1. Given the cubic $x^3-ax^2+bx-c$ (a, b, c positive), prove that the equation has two complex roots if $9c>ab$.
2. Prove all of the above in two variables (this uses the above "trivial" inequality).
3. Prove that for positive x, y, z, $x^2(y+z)+y^2(x+z)+z^2(x+y)\ge 6xyz$. [hide="hint"]This is related to #1[/hide]
4. Prove that for positive x, y, z, $2(a+b)(c+d)\le (a+b+c+d)^2$.

Wow, thanks jb05!

[quote="jb05"]
4. Prove that for positive x, y, z, $2(a+b)(c+d)\le (a+b+c+d)^2$.[/quote]

[hide]
I'll assume you meant nonnegative $a,b,c,d$.
Then $2(a+b)(c+d)<4(a+b)(c+d)$.
But by AM-GM, $\frac{(a+b)+(c+d)}{2}\ge\sqrt{(a+b)(c+d)}$.
Then $(a+b+c+d)\ge2\sqrt{(a+b)(c+d)}$.
And $(a+b+c+d)^2\ge4(a+b)(c+d)\ge2(a+b)(c+d)$.
[/hide]

For 3

[hide]I'm not sure what the hint implies. But

Expand the parantheses for 6 summands. Apply AM-GM to complete the inequality.

[/hide]

I think this thread is a great idea.  I am horrible at inequalities and want to learn.  I'm going to try to keep up with this and perhaps post some problems.  Right now it seems that it is at the Intermediate level but I imagine that since we are talking about inequalities, I might have to promote it to Pre Olympiad at some point in time.

[hide="1"]If r, s, and t are the roots of the equation, we have a=r+s+t, b=rs+st+rt, and c=rst. Substituting this in 9c>ab, we have 9rst>(r+s+t)(rs+st+rt), or
$6rst>r^2s+r^2t+rs^2+s^2t+st^2+rt^2$
Dividing by 6 and taking the 6th root of the 6th power of the LHS, we have
$\sqrt[6]{r^6s^6t^6}>\frac{r^2s+r^2t+rs^2+s^2t+st^2+rt^2}{6}$
Which clearly violates AM-GM, so at least two of $r^2s$, $r^2t$, etc., must be negative. (If an odd number of them is negative, the LHS would be complex and wouldn't be either greater than or lesser than the RHS.) Let's say, WLOG, it's $r^2s$ and $r^2t$. r will be complex unless s is negative and t is negative. But that forces b to be negative, which it isn't. And if one root is complex, its conjugate must be a root too, so the equation has two complex roots.[/hide]
I think I just proved that backwards, if I proved it at all. Oh well...
EDIT: Wait, it's wrong anyway. Arg. Can someone else finish this off?

Yeah, that's right.

RMS $\geq$ AM

is due to Power Mean.

[quote="tarquin"]Umm I was reading AoPS volume 2 and it said $RMS \ge AM \ge GM \ge HM$ ... err or maybe I'm reading wrong or something ...[/quote]

You are correct.

EDIT: Gah! tetrahedr0n beat me to it.

Could anyone post some more, easier schur problems? And maybe introduce an inequality? Let's get Cheby, Holder, Weighted PM, Karamata, and Muirhead out of the way. I would greatly benefit from intuitive, elementary explanations of the above, with AIME - USAMO level problems. And I cannot tell everyone how much I absolutely love the Inequality-Proof-15 Elementary examples format. Please keep posting more proofs, especially for any inequalities in the future, as understanding the tools one is using is important. And keep the problems easy, please. Or at least not all national math olympiad level. We can't all pick up inequalities immediately from the definitions :(

For those looking for harder problems and more rigourous definitions, I reccomend again the [url=http://web.mit.edu/~tmildorf/www/Inequalities.pdf]Mildorf Packet.[/url] Alas, I need some easier stuff.

[quote="tim1234133"]*realises that this suggestion might be shouted down*

Wouldn't it be easier if we started a new thread along the same lines, and went through from the beginning rather than carrying on from an old thread that hardly anyone is going to know/remember what was discussed in it?[/quote]
Good Idea. The problem is, I don't have enough knowledge to write up things for the advanced inequalities. If you wanted to start a new thread, say "Inequality Lessons 2: Intermediate Inequalities", and wrote up a little description saying the thread should be used for posting instructional material for Schur, Jensen, Cheby, Holder, Weighted PM, Karamata, and Muirhead, and start with a little Schur review and lesson, then I'm sure everyone would just migrate to that.

It's just a matter of who wants to take the time to dig up and latex some easy-but-not-too-easy problems, some hard-but-not-too-hard problems and and proofs of the said inequality. And I suck at inequalities, but I learned most of what I know from here. And if you, or someone else would do the above, then I would be enternally grateful.

Could someone elaborate upon Karamata? I've heard it's effectively God. I'm Hindu, so I'm adding Karamata's Inequality to the list of my polytheistic religion's gods.  :P

If you mean "posting a link to an explanation" by "elaborating", meet your new god [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]here[/url].  :D

Starting off the thread with karamata might be a bit hard... :maybe:

Unless you want to write up another karamata/Jensen chapter, in which case just provide jensen info, link to the karamata thread, then give some practice problems. Let's get the thread moving!  :P

Come on, guys. Someone must have enough knowledge to compile a few lessons :maybe: . I'll do whatever I can to help, but I don't know enough.

[quote="Palytoxin"][quote="xxreddevilzxx"][quote="nr1337"]I would say that the trivial inequality is something that just "is," like how a+b=b+a for all $a,b \in \mathbb{C}$.[/quote]

$a+b=b+a$ is an equation, not an inequality.[/quote]

he meant trivial inequality is a property of real numbers, like commutative property is a property of complex numbers[/quote]

But you do still prove the commutativity of complex. So I think we need a proof for the inequality in question.
There is a property called "The Order Properties of real." There is a nonempty subset P of real, called the set of positive real numbers, that satisfies the following properties:
1. If a, b belong to P, then a + b belongs to P.
2. If a, b belong to P, then ab belongs to P.
3. If a is a real (belongs to R), then exactly one of the following holds:
a belongs to P, or a = 0, or -a belongs to P.

Coming back to $x^{2}\geq 0$.
Proof: If $x = 0$, then the equality holds.
Suppose x <  0, then $x^{2}= (-x).(-x) \in P$ (by 2).
If $x > 0,$ then $x^{2}= (x).(x) \in P (by 2)$.
Thus $x^{2}\geq 0.$ Q.E.D

To say the truth, I'm a little annoyed that you renewed such an old thread :mad:

Also there is another similar thread going on right now.

However, I will keep this up. From a recent "paper" of mine, I reduced Darij's inequality, for positive reals: \[\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\ge\frac{4(x+y+z)}{\sqrt{(x+y)(y+z)(z+x)}}\]
into a final step of:

[i]... Whereupon it remains to prove that [/i]\[\sum_{cyc}\frac{\sum_{cyc}xy}{x}\ge 3\sum_{cyc}x\]

It was the easiest part and quite a relief after the two pages before it :wink:

What do you mean you'll keep this up? What does $cyc$ mean at the bottom of a summation?

cyc means cyclic summation.

I meant that I was going to post here to continue the inequalty lessons idea.

Yeah, but how do you evaluate $cyc$? I've also seen $sym$.. care to explain someone.  :)

Oooh. Sorry about this. It's just a notational thing (I was sure they explained it somewhere in this thread...).

[b]Definition.[/b]

\[\sum_{cyc}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b) \\ \sum_{sym}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)+f(b,a,c)+f(a,c,b)+f(c,b,a)\]
So the first is just adding up what you would get by cyclically switching the variables, the second is the sum of all possible combinations of variables. It's really useful in complicated symmetric algebra (which is basically most inequalities)...

[quote="K81o7"]Number 4 simplifies to
$x^{2}y^{2}+y^{2}+x^{2}+1 \ge x^{2}y^{2}+2xy+1$
So if we can prove that $x^{2}+y^{2}\ge2xy$, then we're done.  So let's do that.  Starting with the expression, we'll simplify it.
$x^{2}+y^{2}\ge2xy$
$x^{2}-2xy+y^{2}\ge0$
$(x-y)^{2}\ge0$
The square of any real number$\ge0$, so this statement is clearly true.  Therefore, the original inequality holds.  We can also follow off this to form a proof of number 5.
$x^{2}+y^{2}\ge2xy$
$x^{2}+2xy+y^{2}\ge4xy$
$(x+y)^{2}\ge(2\sqrt{xy})^{2}$
$x+y\ge2\sqrt{xy}$
$\frac{x+y}{2}\ge\sqrt{xy}$

6. $x^{2n}=(x^{n})^{2}$
If n is an integer and x is real, as towersfreak2006 said, then $x^{n}$ is real.  $Real^{2}\ge0$

2. Basically the same approach as shobber...
$\cos{2\theta}+\sin^{2}{\theta}=\sin^{2}{\theta}+\cos^{2}{\theta}-\sin^{2}{\theta}=\cos^{2}{\theta}\ge 0$
because $Real^{2}\ge 0$[/quote]
Actually, number 4 can be solved by using Cauchy Inequality.
(x^2+1^2)(y^2+1^2)>=(xy+1)^2

The trivial inequality is foundamental:
1. If x=>0 then x.x is a product of 2 nonnegative numbers, so it's nonnegative. We have equality when x=0;
2. If x<0, then x.x is a product of 2 negative numbers, so it's positive.


geometry
3D geometry
induction
modular arithmetic
ratio
AMC
AIME

A collection of 8 cubes consists of one cube with edge-length $k$ for each integer $k,\thinspace 1 \le k \le 8.$  A tower is to be built using all 8 cubes according to the rules:

$\bullet$ Any cube may be the bottom cube in the tower.
$\bullet$ The cube immediately on top of a cube with edge-length $k$ must have edge-length at most $k+2.$

Let $T$ be the number of different towers than can be constructed.  What is the remainder when $T$ is divided by 1000?

There were many good (and difficult) problems on this AIME.  This is one of my favorite combinatorics problems.  As with many combinatorics problems, there is a simple, elegant solution.  [I saw the elegant solution, after brute-forcing the solution to some smaller towers.]

so.. what is the elegant solution anyway?

i ended up brute forcing all the way but somehow did an arithmetic mistake somewhere.

the best way to do this problem is to use Recrusion

i tried PIE, and 2232 ;)

[hide]Let's construct the tower by starting with the cubes with side lengths 1, 2, and 3 in any order. There are $6=2\cdot3^1$ ways to do this. Now, by induction, we claim that there are $2\cdot3^{k-2}$ ways to stack $k$ cubes legally. We already have the base case, and for the inductive step, we just need to show that there are 3 ways to stack $k$ cubes for each stack of $k-1$ cubes. Consider cube $k$. It can be on top of cubes $k-1$ or $k-2$ or on the bottom. Since each of these corresponds to a unique stack, and it covers all legal stacks, induction is complete. That way, with 8 cubes, there are $2\cdot3^6=1458\equiv\fbox{458}\pmod{1000}$ ways to do it.[/hide]

Wow, that problem is really easy if you think of recursion. Why couldn't I have seen that during the contest? :wallbash_red:

[hide="SamE has it."]
I didn't see how to solve the 8-tower, directly.  So I decided to "try a simpler case".  Every 1-, 2-, and 3-tower is permitted.  (So there are 1, 2, and 6, permutations, respectively).  I brute-forced the 4-tower (18 valid towers) and the 5-tower (54 valid towers).  I had begun brute-forcing the 6-tower, but decided to move on.

Later in the evening, when I came back to it, I noticed the ratio in the number of valid towers, as tower size increased by 1, seemed to be a constant 3.  Asking myself why would this be true, I almost immediately realized there were always 3 valid places to insert the n-cube into an existing (n-1) tower.
[/hide]

O, imprecations!

I should've thought of that. O well, glad i didn't spend to much time with it.

Yay! that's what I did. Probably the only elegant solution i had...

I was prodding the entire time for a recursive solution, which I knew existed by the type of problem being asked, but it turned out I was not keen enough to see such a nice solution. I guess this type of thing just isn't my forte... needs practice. :(

This was one of the few problems I didn't screw up by making some small mistake :D  I fiddled around with the 8 case but pretty quickly realized that I should probably test small cases first...I pretty quickly realized that for each arrangement of $n$ blocks, you can place the $n+1$ block in 3 ways.  Honestly, though, I probably wouldn't have noticed that so quickly if I hadn't seen a similar problem before...I think in AoPS Intermediate Counting! ;)

Practice, practice, practice, that's how people do well on these contests. I would know: I don't practice, and make upwards of 3 mistakes every AIME. Do as I say, not as I do. ;)

Here's a motivation for the recursive solution.

There are a few common ways to find solutions to combinatorics problems.  One of the most common kinds of solutions is what I call a "constructive" solution.  For a constructive solution, we must answer the question: "How can we build [i]one[/i] tower?"

Okay, that doesn't alone motivate everything, but it gives us a playing field we don't mind playing on.

The simplest cases (one, two, three blocks) permute for all possible towers, but clearly the fourth case does not.  But we can notice that there are 3 places to insert the fourth cube in any tower we've built in the case for three cubes, and so on.

Constructive solutions to tough combinatorics problems often result in recursion.  Just being aware of this fact can dramatically cut down on the time needed to solve some counting problems -- particularly late on the AIME.

[quote="MCrawford"][quote="dhuang614"]I didn't get this one  :( And I also read about the recursions a few days before the AIME![/quote]

Recursion problems require more practice than most because constructing the recursion itself is an exercise in finding a process that can be expressed mathematically.[/quote]

Yes, MCrawford has it completely right. Recursion is really completely different from all other mathematics. Best way to get better is to practice, practice, practice (or to do USACO =D).

Some people have asked how to know the solution is recursive, or the motivation for it. Well, there are very few weapons in combinatorics, namely sets, combinations, permutations, and PIE. There are also a few slightly more advanced concepts such as "stars and bars." None of these really even remotely do anything to this problem, except possible PIE, but just taking a look at it, PIE will almost certainly require several layers of included and excluded sets, resulting in 20-30 or more calculations, which is infeasible. That leaves us with recursion.

This turns out to be a pretty tough recursion to see, as usually in combinatorics we would tend to think of recursing on the positions in the tower, or something like that. For a while it looks like this is the way to go, but, as I found out the hard way after about 20 minutes of work, it leads to a dead end.

What I tried next was looking at some smaller cases to get a feel for what was going on. Well, after getting 1 2 6 18, I asked myself how I would figure out the k = 5 case. Well, there were 3 places to put the 5 in each case. This inspired the 2*3^k-2 formula given by others.

Sometimes it really is necesarry to try small cases to get a feel for larger cases, especially for the toughest problems like this one. In this case trying small cases ended up [i]providing[/i] the recursion for the larger cases.

We can proceed recursively. Let $f(k)$ be the number of towers for k cubes.

Notice that $f(k+1) = 3f(k)$ for all $k \geq 2$.

If there is 1 block, then there is only 1 tower and if there are only 2 cubes then only 2 towers are possible. So $f(1) = 1$ and $f(2) = 2$.

So $f(k) = 2 * 3^{k-2}$ for all $k \geq 2$ and now we get that $f(8) = 1458 $ mod $1000 = \boxed{458}$.

Literally the same solution as I wrote up to 2015 AIME II/10, but posting for storage.

The idea is to use recursion; given a quasi-increasing sequence of length $n$, the term $n+1$ can be inserted in three places: the beginning of the sequence, or immediately following $n$ or $n-1$. Because $a_2=2$, the answer for general $n$ is $2 \cdot 3^{n-2}$. Thus$$a_8 = 2 \cdot 3^6 = 1\boxed{458}.$$

[quote=amirhtlusa]the best way to do this problem is to use Recrusion[/quote]

I agree; that's how I did this problem.

Let $a_{n}$ denote the number of valid stackings of $n$ blocks.
If we have any of the $a_{n-1}$ stackings of $n-1$ blocks then we can place block $n$ on the bottom, the $n-1$th block, or the $n-2$th block.
So for $n\geq3$ $a_{n}=3a_{n-1}$, and $a_{2}=2$
So for $n\geq 2$, $a_{n}=2\cdot3^{n-2}$ with $a_{1}=1$
$a_{8}=2 \cdot 3^{6}=1 \boxed{458}$

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLscelesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

Suppose we have a valid tower of 8 cubes.

Consider the position of cube $8$; it must either be at the bottom, on top of cube $7$, or on top of cube $6$. In either scenario, removing cube $8$ still keeps the tower valid, since any cube can go on top of $7$ or $6$.

Similarly, we can remove cube $7$, then cube $6$, and so on. In each of these steps, we can place the removed cube $k$ in one of three positions (at the bottom, on top of cube $k-1$, or on top of cube $k-2$.)

When there are only two cubes left, there are only two options for the tower. So the number of valid towers is $2\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3=1458$, making the answer $\boxed{458}$.

Let $f(n)$ be the number of towers with $n$ cubes. Notice that if we have a tower with $n$ cubes in it already, the cube with side length $n+1$ can go at the bottom, above the cube with side length $n-1$, or above the cube with side length $n$ if $n \ge 2$. Hence, for $n \ge 2$, we must have

\[f(n+1) = 3f(n).\]

Since $f(2)=2$, our closed form is $f(n)=2 \cdot 3^{n-2}$ for $n \ge 2$. Plugging in $n=8$ yields $1\boxed{458}$.


Initially there are 83 kg of a radioactive material.  This material deays with a half-life of 16 months.  What percent of the original 83 kg will be left after 22 months?
I think I can solve this but what is the rate??

In the 22 month span, after 16 months, there will be 41.5 kg of radioactive material left. The remaining 6 months will be $ \frac{6}{22}$ of another half life.

The problem is a textbook example of exponential decay. Exponential functions may be taught in Algebra 2 courses. They are useful for calculating decay, reaction rates, and continuously compounded interest.
If you have not learned about them, it gets complicated. It's hard to understand and harder to explain.
For each given period, your amount (or percent) remaining changes by the same factor. After 16 months, 0.5 (50%) remains. After another 16 months, 50% of that 50% (25%) is left. After 48 months (3 half-lives), 12.5% of the original radioactivity remains.
If the factor for 2 months is x, and over 16 months (8 times 2 months) the radioactivity changed by a factor of 0.5,
x^8 = 0.5
x = 0.917

This is a Chemistry problem   :D
Radioactive decays are 1st order rns.
The rate is given by 0.693/h where h is the half life .
This can be easily derived  :)

much simply k = (0.693/h) = (2.303/t)*log(a/a-x)
where h = half life , k = rate constant of the rn , t is the time after which you are viewing the system (i.e. present time - initial time) and a is the initial conc and a-x is the conc undecayed at time t .


we cover $R^3$ with some halfspaces prove that we can cover$ R^3$ with  4 of halfspaces.

What are you meaning by covering? Pairwise disjoints?

What are you meaning by halfspaces ? Open or closed? (in the latter, only two suffice
 :D ).

Pierre.

no i mean if some halfspace cover R^3 (of course not necessarily disjoint)then 4 of them is sufficient to cover R^3

I think there's a chance this can be generalized for $n$ dimensions: if $\mathbb R^n$ is covered with some "half-$\mathbb R^n$", then we can choose $n+1$ of them which cover $\mathbb R^n$.

However, I'm not sure whether the problem is Ok if we allow an infinite number of half-planes to cover the space. I think it should be mentioned that we have finitely many half-planes (not just "some" :)). Is that correct, sam-n? We can look at an analogy in one dimension, where we use rays instead of half-planes. We can use infinitely many rays oriented in the same direction to cover the line, but we can't extract a finite set which covers the line.

The following works for any dimension:
On contrary, suppose no n+1 halfspaces cover Rn+1
Consider complementary halfspaces. Then since no n+1 original halfspaces
cover the entire space, any n+1 complementary ones have a common point.
By Helly's theorem, there is a point containing in all complementary halfspaces.
That point is not covered by any of the original halfspaces which gives us a contradiction.

This is just wonderful!

I also have a proof by induction after the dimension, but I don't know if it really works from $3$ to $4$ :). However, it was enough to solve the problem proposed by sam-n.

It's obvious for one dimension, so let's go from $1$ to $2$, and then something similar is done for $2\to 3$.

We want to show that if a finite number of half-planes cover the plane, then we can choose $3$ of them which cover the plane. First of all, take out any redundant half-planes, i.e. look at all the half-planes which are completely covered by unions of other half-planes, choose one of them and eliminate it, then repeat the process. Now consider one of the remaining half-planes, $\pi$, the edge of which is the line $\ell$.

Now apply the result obtained in $1$ dimension; we find that there are points $A,B\in \ell$ s.t. the rays $(AB,(BA$ are contained in two other half-planes, $\alpha,\beta$, respectively. If the intersection point of the two edges of $\alpha,\beta$ lies outside $\pi$, then $\pi$ is redundant, which is a contradiction because we have no more redundant half-planes, so the intersection point lies either inside $\pi$ or at infinity. In both cases $\alpha,\beta,\pi$ cover the plane.

yes the point is helly theorem ,my soln was the same.thanx zortech.

Yes nice argument Zortech. I proved the result in an elementary way for the plane (no induction) but I was thinking about how generalize the argument to the space...
I will try to think about it a little more.

Pierre.


geometry
trapezoid
perimeter
geometry unsolved

It's held just yesterday, November 23rd, so this question is actually brand-new.

Triangle $ABC$ has its base on line segment $PN$ and vertex $A$ on line $PM$. Circles with centers $O$ and $Q$, having radii $r_1$ and $r_2$, repectively, are tangent to the triangle $ABC$ externally and to each of $PM$ and $PN$.

Let $T$ be the point f contract of $BC$ with the circle inscribed in triangle $ABC$. Prove that $(TC)(r_1)+(TB)(r_2)$ is equal to the area of triangle $ABC$.

here is a picture
for part A, there is a very UGLY algebraic solution.

for part B, i guess you have to express BT, CT in terms of other things in the triangle.

Denote a = BC, b = CA, c = AB the sides of the triangle $\triangle ABC$. First, the segment lengths DE = FD = s of the common external tangents of the circles (O), (Q) are equal to the semiperimeter of the triangle $\triangle ABC$ (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39285]Two circles, tangents, triangles[/url]). The distances of the tangency point T of the incircle (I) of the triangle $\triangle ABC$ with the side BC from the vertices B, C are BT = s - b = s - CA, CT = s - c = s - AB. We have to show that

$(?)\ \ |\triangle ABC| = r_1 CT + r_2 BT = r_1 (s - c) + r_2 (s - b) = s (r_1 + r_2) - (r_1 c + r_2 b)$

The area of the congruent trapezoids $OQED \cong OQGF$ is equal to (height times the average of bases)

$|OQED| = |OQGF| = s\ \frac{r_1 + r_2}{2}$

Hence, the area of the hexagon ODEQGF is twice this value, or  $|ODEQGF| = s (r_1 + r_2)$.

The following pairs of right angle triangles are congruent:

$\triangle ODA \cong \triangle OKA,\ \ \triangle OFB \cong \triangle OKB$

$\triangle QEA \cong \triangle QLA,\ \ \triangle QGC \cong \triangle QLC$

It follows that the areas of the pentagons ODABF and QEACG are equal to twice the areas of the triangles $\triangle OAB, \triangle QAC$, respectively, or

$|ODABF| = 2|\triangle OAB| = r_1 c,\ \ |QEACG| = 2|\triangle QAC| = r_2 b$

because $OK = r_1,\ QL = r_2$ are the altitudes of the triangles $\triangle OAB,\ \triangle QAC$ to their bases c = AB, b = CA, respectively. Finally, the area of the triangle $\triangle ABC$ is equal to

$|\triangle ABC| = |ODEQGF| - (|ODABF| + |QEACG|) = s (r_1 + r_2) - (r_1 c + r_2 b)$

which is what we were supposed to show.

The sum of the segment lengths DE + FG of the common external tangents adds up to the perimeter p of the triangle $\triangle ABC$, because AK = AD, AL = AE, BK = BF, CL = CG. Since the 2 segment lengths DE = FG are equal and their sum is p = 2s, each is equal to the triangle semiperimeter s. Then AK + AL = AD + AE = DE = s.

[quote="henry23"]part A of the problem stated
prove the line through K and L divides the perimeter of the problem in half.[/quote]
AK + KL = DA + AE = DE = FG = FB + BC + CG = KB + BC + CL

That's the easiest way, actually this part(part a) cost me one glimpse only on the exam.  :lol:

[u]Remark.[/u] If $AB=AC$ then $r_1+r_2=h_a$ (constant) for any point $P\in BC,\ B\in (PC)$.


Let $ a,b,c>0$ such that $ ab+bc+ca+a+b+c=a^2+b^2+c^2+1$ prove that

\[ \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge 4+\frac{2}{\sqrt{3}}\]

For all $ n\ge 2,\n\in\mathbb{N}$ prove or disprove

\[ \frac{1}{\sqrt[n]{a}}+\frac{1}{\sqrt[n]{b}}+\frac{1}{\sqrt[n]{c}}\ge 2\sqrt[n]{4}+\sqrt[n]{\frac{4}{3}}\]

$ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ac$ , $ a\plus{}b\plus{}c\leq1$ and $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \leq \sqrt{3}$   $ \Longrightarrow$

$ \frac{1}{\sqrt{a}}\plus{}\frac{1}{\sqrt{b}}\plus{}\frac{1}{\sqrt{c}}\geq 3\sqrt{3} \geq 4\plus{}\frac{2}{\sqrt{3}}$

Are you sure about your solution, crazyfehmy?
I see that $ a \plus{} b \plus{} c \ge 1$. Can you redetail your post?

[quote="encyclopedia"]Let $ a,b,c > 0$ such that $ ab + bc + ca + a + b + c = a^2 + b^2 + c^2 + 1$ prove that
\[ \frac {1}{\sqrt {a}} + \frac {1}{\sqrt {b}} + \frac {1}{\sqrt {c}}\ge 4 + \frac {2}{\sqrt {3}}
\]
For all $ n\ge 2,\n\in\mathbb{N}$ prove or disprove
\[ \frac {1}{\sqrt [n]{a}} + \frac {1}{\sqrt [n]{b}} + \frac {1}{\sqrt [n]{c}}\ge 2\sqrt [n]{4} + \sqrt [n]{\frac {4}{3}}
\]
[/quote]
Dear encyclopedia,
I checked your inequality and I think we can compare $ \frac{1}{\sqrt{a}} +\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$ with $ 4+\frac{2}{\sqrt{3}}$. For example, you can try $ (a,b,c)=( 5/4,5/4,7/4+\sqrt{3})$ and $ (a,b,c)=(5/4,5/4,7/3-\sqrt{3})$. Sorry if I got any mistakes. :(

[quote="encyclopedia"]Let $ a,b,c > 0$ such that $ ab + bc + ca + a + b + c = a^2 + b^2 + c^2 + 1$ prove that
\[ \frac {1}{\sqrt {a}} + \frac {1}{\sqrt {b}} + \frac {1}{\sqrt {c}}\ge 4 + \frac {2}{\sqrt {3}}
\]
For all $ n\ge 2,\n\in\mathbb{N}$ prove or disprove
\[ \frac {1}{\sqrt [n]{a}} + \frac {1}{\sqrt [n]{b}} + \frac {1}{\sqrt [n]{c}}\ge 2\sqrt [n]{4} + \sqrt [n]{\frac {4}{3}}
\]
[/quote]
We can have  $ ab + bc + ca + a + b + c = a^2 + b^2 + c^2 + 1$ for $ a\rightarrow \infty$, $ b\rightarrow \infty$ and $ c\rightarrow \infty$, when all of  the inequalities above clearly are false.


A cryptographic devises the following method for encoding positive integers.  First, the integer is expressed in base $ 5$.  Second, a $ 1$-to-$ 1$ correspondence is established between the digits that appear in the expressions in base $ 5$ and the elements of the set {$ V,W,X,Y,Z$}.  Using this correspondence, the cryptographer finds that three consecutive integers in increasing order are coded as $ VYZ$, $ VYX$, $ VVW$, respectively.  What is the base-$ 10$ expression for the integer coded as $ XYZ$?

(A) 48
(B) 71
(C) 82
(D) 108
(E) 113

Thanks in advance!

[quote="aleph0"]W is 0, Z is 3, and X is 4. This makes Y 1 and V 2. XYZ=413 in base five=108 in base ten, so the answer is D.[/quote]

I don't quite follow what you did.  Could you be a bit more detailed, please?  Thanks!

Since the fives digit changes when the units digit goes from X to W, W must be 0. X has to be right before 0, so it must be 4. Z is before X, so Z is 3. This leaves 1 and 2 for V and Y. From the fives digit change, V is after Y, so Y is 1 and V is 2. XYZ=413, and converting into base ten gives 108.


let f :IN*------IN satisfaies :
[color=red]1-f(1)=1, f(p)=1 for any prime number p
2- f(xy)=f(x)y+xf(y) x and y are integers [/color]

[size=150]resolve this equation  f(n)=n[/size]

Put $y=1 \longrightarrow f(x)=f(x)+x \leftrightarrow  \ ABSURD$


$a, a+1, a+k$ is a Pythagorean triple.  Compute all possible values of $k$.

must k be positive?

because if not, there are infinite solutions

Use the pythagorean theorem...

$(a+k)^{2}=(a+1)^{2}+a^{2}$

$a^{2}+(2-2k)a+1-k^{2}=0$

Since $a$ is an integer, the quadratic discriminant is an integer...i.e:

$(2-2k)^{2}-4(1)(1-k^{2})=q^{2}$

$2k^{2}-2k=\left(\frac{q}{2}\right)^{2}$

I will stop here. This is now an extended pell equation. If there is one solution, then there are infinite solutions [note we have: $(k,q)=(2,2)$ as a solution].

[hide="Equivalently"] [b]Case:[/b]  $a = 2mn, a+1 = m^{2}-n^{2}$.  Then we want integer solutions to

$m^{2}-2mn-n^{2}= 1 \implies$
$(m-n)^{2}-2n^{2}= 1$

Which is a Pell's equation with infinite solutions; then $k = (m^{2}+n^{2})-(2mn) = (m-n)^{2}$.  The smallest solution to the Pell's equation is $(m-n, n) = (3, 2)$; generally, the possible values of $k$ are the squares of the integer coefficient of $(3+2 \sqrt{2})^{i}$.  For $i = 1$ we have $k_{1}= 9$ and the triple $20, 21, 29$.

[b]Case:[/b]  $a+1 = 2mn, a = m^{2}-n^{2}$.  Then we want integer solutions to

$n^{2}+2mn-m^{2}= 1 \implies$
$(n+m)^{2}-2m^{2}= 1$

Which is nearly the same Pell's equation; then $k = (m^{2}+n^{2})-(m^{2}-n^{2}) = 2n^{2}$.  These values of $k$ can be computed similarly. [/hide]


excuse me .I want to ask a question.
if for a1>a2>a3......>an,k1>k2>......>kl,and every ai(i=1~n)is prime
a1*a2*a3......*an=k1*k2*....*kl
prove that
2^(a1-1)*3^(a2-1)*5^(a3-1).......*pn^(an-1)<2^(k1-1)*3^(k2-1).......*pl^(kl-1)
thanks a lot. I,ve been thinking for a long time :?

for all n=a1*a2*.....ar(every ai is prime,a1>a2>a3.......>ar) and if 2^ar>=pr
(pr is the rth of prime from 2)
then g(n)=2^(a1-1)*3^(a2-1)*...........pr^(ar-1)

g(n) means the smallest number in the type of 2^(k1-1)*3^(k2-1)*.....pl^(kl-1)
k1*k2*....kl=n


AMC
AIME
geometry
perimeter
trigonometry
3D geometry
tetrahedron

1. ABCD is a convex quadrilateral with AB = CD = 180, perimeter 640, AD ≠ BC, and ∠A = ∠C. Find cos A.

2. Triangle ABC has AC = BC and ∠ACB = 106 degrees. M is a point inside the triangle such that ∠MAC = 7 degrees and ∠MCA = 23 degrees. Find ∠CMB.

Any ideas?

on the second one;
Take a point N inside the triangle ABC such that angleNCB=23 and angleNBC=7 .Observe that CMN is an equaliteral triangle and angleMNB =150.So triangle MNB is similar to triangle CNB. the rest is easy.

[quote="jeffrey"]On the first one you can draw a diagonal and use law of cosines...[/quote]

That was my first approach, where i set the diagonal equal to x and used cosA = cosC in the law of cosines..but I ended up getting an equation with both x and cosC terms even after simplification... maybe my arithmetic was incorrect? I dunno... can I see what your process exactly was?

Heres a solution:

[hide]Draw diagonal BD, apply law of cosines twice to get:

$BD^2=AB^2+AD^2-2ABAD \cos A$
$BD^2=BC^2+CD^2-2BCCD \cos A$

Now, subtract the two equations and move terms to one side. You get

$\cos A(2ABAD-2BCCD)=CD^2+BC^2-AB^2-AD^2$
$\cos A(2)(180)(AB-CD)=BC^2-AD^2=(BC-AD)(AB+CD)$
$\cos A(360)=640-360$
$\cos A=\frac{7}{9}$[/hide]

tetrahedrons solution was the solution i was hinting at. for the second one, try extending AM to a convenient point (perhaps introducing some symmetry) and see what results you get.

Cosiner:
I tried your approach, and it seemed to make the problem simple, but I still can't figure out the answer.

For the second problem there is also a nice trig solution


Mary,Larry, and Harry are going to Disney world together. Mary and Larry drove halfway together that distance was 580miles they figure the gas to where they picked up Harry was 68.50 dollars. find the cents per passenger mile. :huh:

I don't understand your question--could you be more clear?

[hide="best try at a solution from given information"]Assuming that you are asking for the cents per mile if you drove 580 miles and spent $ \$$68.50 on the gas you used, the dollars per mile is clearly $ \dfrac{68.50}{580}$. To convert this to cents per mile, we multiply the top number by 100, as there are 100 cents in a dollar. Our answer is then
\begin{align*} \frac {68.50 \cdot 100}{580} & = \frac {6850}{580} \\
& = \frac{685}{58}. \end{align*}
But we multiply by the number of passengers, giving $ \dfrac{685 \cdot 2}{58}=\boxed{\dfrac{685}{29}}$ as our final answer.
[/hide]

Sorry if this is not what you were asking for, but math154 was right--the problem is unclear. :(

EDIT: Fixed solution; didn't see "passenger" miles.

I happen to like your malapropic title. Are the characters deluded that they're going to Disneyland?


Putnam
algebra
functional equation
number theory proposed
number theory

Can we find two disjoint sets $A,B\subset \mathbb N$ (where $0\in\mathbb N$) s.t. $A\cup B=\mathbb N$ and every positive integer can be written as the sum of two distinct numbers from $A$  in the same number of ways in which it can be written as the sum of tweo distinct numbers from $B$?

It's really nice, but I have the feeling something similar has been discussed before here. Sorry if the exact same problem has already been posted.

Yes, there exists a unique partition with such properties --- characterize the set $A$ with the formal power series $f_A(x) := \sum_{a \in A} x^a$, and verify that the condition of the problem is equivalent to a simple functional equation for $f_A(x)$ which has a unique solution. Btw, this very famous problem was A-6 on Putnam 2003!!! Definitely, last year's Putnam was the weakest one I've ever seen; I previously knew 5 (five) out of the 12 problems!

Yup, this is precisely what I did :). I had no idea it was from a Putnam contest though. I think we can even pinpoint the sets. If I remember correctly, after working that stuff out, we get $A$ to be the set of those numbers which have an even number of $1$'s in their binary representations, and $B$ the rest of the positive integers.

I'm not sure about this..


What is the median number of moons per planet?
\begin{tabular}{c|c}
Planet & $ \#$ of Moons\\
\hline
Mercury&0\\
Venus &0\\
Earth &1\\
Mars &2\\
Jupiter&16\\
Saturn&23\\
Uranus&15\\
Neptune&2\\
Pluto&1\\
\end{tabular}

By definition of a median, just order them by increasing number of moons, then take the middle one.

$ 0,0,1,1,2,2,15,16,23$

Odd number of terms- take middle one only

$ 2$ moons

This info is way off! Really inaccurate an dold!


number theory
greatest common divisor
algebra
polynomial
Vieta

Let $\omega$ be a primitive 49th roots of unity. Compute:

$\displaystyle \prod^{49}_{{k=1, \gcd(k,49) = 1}} (1 - \omega^7)$

You probably mean $1 - \omega ^k$ and the limits on the product seem weird? Why not just write $1 \leq k \leq 49, \text{gcd}(k,49) = 1$?

i think it means $\exp{\frac{2\pi ik}{49}}$, meaning that he does not want the roots that are for $k=7,14,21,35,42,49$

say we have $P(x)=x^{49}-1$, this has the roots or unity, factor out the roots that are not in the product this being $x^7-1$, giving a polynomial with the desired roots of unity being:

$x^{42}+x^{35}+...+x^7+1$

let $y=1-x^7$, or $\sqrt[7]{1-y}=x$

$(1-y)^6+(1-y)^5+...+(1-y)+1$

observe that this polynomial's roots are the terms of the product (the product's roots repeat 6 times, so we will have to take the product to the 6th power at the end),

and by vieta, the product of the roots is the constant term (note the lead coefficient is 1, and since there are an even number of roots, we don't need to worry about the -1)

clearly $f(0)=k=7$, and the total product is $7^6$


analytic geometry
geometry
function
graphing lines
slope

If one diagonal of a square is the portion of the variable line λx + (λ -1)y = λ^2 - λ, λ >0, λ not equal to 1, which is intercepted between the axes, then prove that one of the other two vertices is independent of λ and find its coordiantes. Further if the area of the square is 17/4, find the coordinates of the other vertices of the square.

Hint #1:
The x- and y-intercepts of that line are two opposite vertices of the square, and the segment between them is one of the diagonals. The Algebra II taught in the US should let you get to the answer.[hide="More hints"]The midpoint of that diagonal is the center of the square.
The other diagonal is perpendicular.
It has the same length and the same midpoint.
You can calculate area of a square as a function of the length of a diagonal.[/hide]Maybe there is a more elegant way to solve it, but this is the first solution I could come up with.[hide="My solution"]x-intercept (A):	λx = λ^2 – λ  , so  x = λ – 1
y-intercept (C):	(λ - 1)y = λ^2 - λ  , so  y = λ
midpoint of diagonals (center of the square): ((λ - 1)/2, λ/2)
length of half-diagonal: d = (1/2)sqrt(2 λ^2 – 2λ + 1)
Area of square = 2d^2 = (2 λ^2 – 2λ + 1)/2
To go from the center of the square to vertices A  and C, you move
(λ -1)/2 horizontally and
-λ/2 vertically   (slope = -λ/(λ - 1) if you cared to calculate it).
To go from the center of the square to vertices B  and D, you must move the same distance in a perpendicular direction:
λ/2 horizontally and
(λ -1)/2 vertically   (slope = (λ -1)/ λ if you cared to calculate it).
The other two vertices are:
((λ -1)/2 + λ/2, λ/2 + (λ - 1)/2) = ((2λ - 1)/2, (2λ - 1)/2)   and
((λ -1)/2 - λ/2, λ/2 - (λ - 1)/2) = (-1/2, 1/2)
The point (-1/2, 1/2) is always one of the vertices, independently of the value for λ.
Area of square = (2 λ^2 – 2λ + 1)/2 = 17/4 can be solved to give λ = 5/2 as the positive solution.
The vertices are (-1/2, 1/2) ,  (0, 5/2) , (3/2, 0) and (2, 2).[/hide]


Find the least real number $E$ such that   $\; \; \; \; \; \; \displaystyle \frac{n^n}{n!}\le E^{n}\; \; \; ,\; \; \forall n\in \{1,2,...\}.$

Think Stirling and find $E=e$  :D
 :cool:


I’m throwing 8 dice up in the air. What’s the probability that half land on an even number, while the other half land on 1?

[hide]$ \binom{8}{4}$ ways to choose the 4 that land on 1, then $ 3^4$ ways that the other 4 can land on even numbers. the probability is then $ \frac{3^4\binom{8}{4}}{6^8}$.[/hide]

If they are 6 sided die:
[hide="Solution"]
You need to choose 4 of the 8 die to land on one, then figure out the possible ways to get the remaining to land on an even number.

$ \frac{\binom{8}{4} \cdot 3^4}{6^8} \equal{}  \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 2^4 \cdot 6^4} \equal{} \frac{8 \cdot 7 \cdot 6 \cdot 5}{12^5} \equal{} \frac{35}{5184}$
[/hide]


Prove that if $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}abc$ then $ \frac{1}{a\minus{}1}\plus{}\frac{1}{b\minus{}1}\plus{}\frac{1}{c\minus{}1}\geq \frac{3}{2}$

[quote="arqady"][quote="balan razvan"]Prove that if $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} abc$ then $ \frac {1}{a \minus{} 1} \plus{} \frac {1}{b \minus{} 1} \plus{} \frac {1}{c \minus{} 1}\geq \frac {3}{2}$[/quote]
It's Nesbitt! :D[/quote]
completely  :D , put $ a\equal{} \frac{1}{x}$ ...

[color=darkblue]We have $ ab\plus{}bc\plus{}ca\equal{}abc \Leftrightarrow \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}1$. Let $ \frac{1}{a}\equal{}x, \frac{1}{b}\equal{}y, \frac{1}{c}\equal{}z$ and $ x\plus{}y\plus{}z\equal{}1$. Then inequality is equivalent to:
$ \sum \frac{1}{\frac{1}{x}\minus{}1}\equal{}\sum \frac{x}{1\minus{}x}\equal{}\sum \frac{x}{y\plus{}z} \ge \frac{3}{2}$, which is well-known.
[hide="Proof"]
Let $ x\plus{}y\equal{}a, y\plus{}z\equal{}b, x\plus{}z\equal{}c$. Then $ x\equal{}\frac{a\plus{}c\minus{}b}{2}, y\equal{}\frac{a\plus{}b\minus{}c}{2}, z\equal{} \frac{b\plus{}c\minus{}a}{2}$
And we should prove:
$ \sum \frac{a\plus{}c\minus{}b}{2b}\ge \frac{3}{2} \Leftrightarrow \sum \frac{a}{b}\plus{}\frac{c}{b}\ge 6$. But by $ AM\minus{}GM$:
$ \frac{a}{b}\plus{}\frac{c}{b}\plus{}\frac{c}{a}\plus{}\frac{b}{a}\plus{}\frac{a}{c}\plus{}\frac{b}{c}\ge 6\sqrt{frac{a}{b}.\frac{c}{b}.\frac{c}{a}.\frac{b}{a}.\frac{a}{c}.\frac{b}{c}}\equal{}6$, q.e.d.
[/hide][/color]


number theory
prime numbers
number theory unsolved

(a)find all $ p$ prime numbers such that $ \frac {7^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square
(b)find all $ p$ prime numbers such that $ \frac {11^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square


When I checked my report page almost all the subjects in which I had answered questions had disapeared.  I have now completed about 135 problems on Alcumus and I was wondering if the report page resets itself at anytime or if there is a reason it has done this. :?:
Thanks

Thanks. We have confirmed the problem and are investigating it.

It looks like you got some problems wrong which pushed you back into not ready mode for that topic and all related topics. Topics from which you are note considered ready for will not appear on your Report page.


calculus
integration
trigonometry
calculus computations

an integral i just made up :D

show $ \int_0^\infty\int_0^\infty\;\sin\,(x^2\plus{}y^2)\;\;dx\;dy\;\equal{}\;\boxed{\frac{\pi}{4}}$

this problem is very simple

to simplify transform the integral above in the integral below

$ 4 P\equal{}\left(\int _{\minus{}\infty }^{\infty }\int _{\minus{}\infty }^{\infty }\sin
   \left(x^2\plus{}y^2\right)dxdy\right)$

$ \left(\int_{\minus{}\infty }^{\infty } \cos \left(y^2\right) \, dy\right)\left( \int_{\minus{}\infty }^{\infty } \sin  \left(x^2\right) \, dx\right)\plus{}\left(\int_{\minus{}\infty }^{\infty } \cos \left(x^2\right) \, dx\right)   \left(\int_{\minus{}\infty }^{\infty } \sin \left(y^2\right) \, dy\right)$

$ \pi$

Came from the basic fact $ \int_{\minus{}\infty }^{\infty } \cos \left(t^2\right) \, dt\equal{}\sqrt{\frac{\pi }{2}}$

Does anyone know a proof of that integral?

my last integral is:

$ A\equal{}\int_{\minus{}\infty }^{\infty } \cos \left(t^2\right) \, dt\equal{}\int_0^{\infty } \frac{\cos (t)}{\sqrt{t}} \,
   dt\equal{}\Re\left(\frac{1\plus{}i}{\sqrt{2}} \int_0^{\infty } \frac{t^{\minus{}\frac{1}{2}}}{e^t} \, dt\right)$

$ \int_0^{\infty } \frac {\frac {1}{\sqrt {t}}}{e^t} \, dt \equal{} \sqrt {\pi }$


Show that every integer $k>1$ has a multiple less than $k^4$ whose decimal expansion has at most four distinct digits.

For $ k\le10000$ the problem is trivial, since $ k$ itself does the job. For $ k > 10000$, we have an $ n < k$ such that $ 10^{n}< k^{4}< 16^{n}$ (and thus $ 2^{n}> k$). For this $ n$, consider
\[ S\equal{}\{k\in\mathbb{N}|10^{n}\le k<10^{n\plus{}1}\text{ and }k\text{ only consists of digits }0\text{ and }1\text{.}\},\]
then $ |S| \equal{} 2^{n}> k$ and therefor contains two elements $ s_{1}> s_{2}$ with same remainder modulo $ k$. So, $ k|s_{1}\minus{}s_{2}$, and $ s_{1}\minus{}s_{2}< 10^{n}< k^{4}$. On the other hand, differences of two numbers in $ S$ have only digits $ 0,1,8,9$, so we're done.


[hide]the sum of the interior angles in any pentagon is $ 180(5\minus{}2)\equal{}540$. we are given that the first term $ a\equal{}68$. so the arithmetic sequence of the five angles of the pentagon are as follows $ 68, 68\plus{}d, 68\plus{}2d, 68\plus{}3d, 68\plus{}4d$. we must find d, the common difference. using the formula for the n terms of a given sequence $ \frac{n}{2}\cdot(2a\plus{}d(n\minus{}1)$. substituting 5 for n and equating the expression to 540, we get that the common difference is 20. therefore the other four angles are $ 88, 108, 128, 148$.
[/hide]

[quote="mihail911"][hide]the sum of the interior angles in any pentagon is $ 180(5\minus{}2) \equal{} 540$. we are given that the first term $ a \equal{} 68$. so the arithmetic sequence of the five angles of the pentagon are as follows $ 68, 68\plus{}d, 68\plus{}2d, 68\plus{}3d, 68\plus{}4d$. we must find d, the common difference. using the formula for the n terms of a given sequence $ \frac{n}{2}\cdot(2a\plus{}d(n\minus{}1)$. substituting 5 for n and equating the expression to 540, we get that the common difference is 20. therefore the other four angles are $ 88, 108, 128, 148$.
[/hide][/quote]

where does the other parenthesis go on $ \frac{n}{2}\cdot(2a\plus{}d(n\minus{}1)$? (shouldn't there be four) i follow up to there.

It's just after the last parenthesis:
$ (2a\plus{}n(d\minus{}1))$

So, should it look like this when I set it up to solve it?

$ \frac{5}{2}\cdot(2(68)\plus{}d(5\minus{}1)) \equal{} 540$

[hide]$ \theta_{TOTAL}\equal{}\pi(n\minus{}2)$ ; $ n \equal{} 5$
The problem is easier in degrees $ \Longrightarrow\theta_{TOTAL}\equal{}540^{\circ}$
Let the sum of the arithmetic sequence be $ \theta_{TOTAL}$ with $ n$ terms, the first term being $ 68^{\circ}$.

$ T_{1}\plus{}T_{2}\plus{}T_{3}\plus{}T_{4}\plus{}T_{5}\equal{}540\equal{}(68)\plus{}(68\plus{}x)\plus{}(68\plus{}2x)\plus{}(68\plus{}3x)\plus{}(68\plus{}4x)$
Because I'm bored, $ \frac{xn(n\plus{}1)}{2}\equal{} 10x$ ; $ 68(5) \equal{} 340\Longrightarrow 340\plus{}10x\equal{}540$
$ 540\minus{}340\equal{}200$ ; $ \frac{200}{10}\equal{} x$

$ x\equal{} 20$
$ \Longrightarrow T_{1}\equal{} 68^{\circ}$
$ \Longrightarrow T_{2}\equal{} 88^{\circ}$
$ \Longrightarrow T_{3}\equal{} 108^{\circ}$
$ \Longrightarrow T_{4}\equal{} 128^{\circ}$
$ \Longrightarrow T_{5}\equal{} 148^{\circ}$
[/hide]

[quote="linzk3"]The anglos of a pentagon form an arithmetic sequence.  If the smallest angle is 68 degrees, find the measures of the other four angles. i know you use ann arithmetic sequence equation but im not really sure how to even find that.[/quote]

[hide="a little easier"]since the angles are in an arithmetic sequence, the middle one is the same as the angle in a regular pentagon.

the angle of a regular pentagon is $ \frac{540}{5}\equal{}108$

the second smallest term is the average of $ 108$ and $ 68$ so $ 88$.  now we see the common diff. is 20 so the answer is $ 88, 108, 128, 148$.[/hide]

[quote="jli"][hide="a little easier"]since the angles are in an arithmetic sequence, the middle one is the same as the angle in a regular pentagon.

the angle of a regular pentagon is $ \frac{540}{5}\equal{} 108$

the second smallest term is the average of $ 108$ and $ 68$ so $ 88$.  now we see the common diff. is 20 so the answer is $ 88, 108, 128, 148$.[/hide][/quote]

[hide="My guess?"]How do you know that the middle one is the same as the angle in a regular pentagon?

If $ x$ is the middle angle, and $ a$ is the common difference between angles, then...

$ (x\minus{}2a)\plus{}(x\minus{}a)\plus{}(x)\plus{}(x\plus{}a)\plus{}(x\plus{}2a) \equal{} 540$, so...

$ 5x \equal{} 540$
$ x \equal{} 108$, right?[/hide]

The middle one is the avarage of them all, a.k.a. $ \frac{540}{5}\equal{} 108$.

hmmmm. I think this statement is true for all [i]n-gons[/i] for which $ n$ is odd.
[hide]
$ \theta_{INTERIOR}$ must have divisor $ n$. let $ n-1=b$ and $ x$ be the sequential increment. let $ \theta_{INTERIOR}=n\alpha\Longleftrightarrow\alpha =\frac{\theta_{INTERIOR}}{n}$ . A sequence can be formed such that...

$ \theta_{INTERIOR}=(\alpha-bx)+(\alpha-x(b-1))+....+\alpha+....+(\alpha+x(b-1))+(\alpha+bx)$
$ %Error. "Largerigharrow" is a bad command.
\theta_{INTERIOR}=n\alpha$[/hide]
It's just a rearrangement.


:D

 Hello, my name's tom and I'm new to the forums. For my first post I will quickily illustrate change of base.

  124 (root 10)    to     235(root 7)
  First I write the multiples of 7 less-then-or-equal-to the number (root 10)

   7^3      7^2      7^1     7^0
   343        49         7          1

   Now we start from the left and put a number underneath the power of 7 that is <124. We then subract the number and continue to the right.
      49         7          1
       2          3          5

-tom

[quote="hwenterprise"]Isn't this forum for problems, not tutorials?[/quote]
He's new, though, and i don't know whether this forum is for posting problems, tutorials, or both.

[quote="hwenterprise"]Isn't this forum for problems, not tutorials?[/quote]

While problem solving is the most general theme of AoPS, tutorials are absolutely welcome!  They can help other students understand the tools with which we solve problems.

[quote="jeez123"]Try to change 124 (base 10) to base 2 (aka binary)

thanks mathalete for the correction![/quote]

[hide]1111011[/hide]

[color=green][size=84]comment from mod robinhe: please hide answers so that others can solve the problem without first looking[/size][/color]

Can you try to change  1111011 base 2 to BASE *?? :D

Edit: Oopps I'm not sure what base * is either....I had meant Base 8 ;)

well this is how I do it
[hide]
I change it to base 10. Then use the same approach as the other bases.
You just divide the cubes, squares, etc. then find remainder, and go again... (that was unclear)

[/hide]


geometry
3D geometry
combinatorics unsolved
combinatorics

in how many ways a ruby cube can be solved?

What do you mean by how many different ways? Unless you define it precisely I can give you an example where the answer is: infiniite


Can someone show me why this code is depreciated?

[code]
import javax.swing.*;

class FirstFrame extends JFrame
{
	JComboBox bix = new JComboBox();

	FirstFrame()
	{
	super("Hello");
	setSize(500, 500);
	setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
	setVisible(true);
	JPanel pane = new JPanel();
	pane.add(bix);
	bix.addItem("12");
	setContentPane(pane);
	show();
	}

}

public class Main
{
	public static void main(String[] args)
	{
		FirstFrame na = new FirstFrame();
	}
}
[/code]

If I compile this, I get a [code]Note: Main.java uses or overrides a deprecated API.
Note: Recompile with -Xlint:deprecation for details.[/code].

If I do [code]javac Main.java -Xlint [/code]

I get[code]Main.java:18: warning: [deprecation] show() in java.awt.Window has been deprecated
        show();
        ^
Main.java:4: warning: [serial] serializable class FirstFrame has no definition of serialVersionUID
class FirstFrame extends JFrame
^
2 warnings
[/code].

Hmm, I dunno I haven't worked a whole lot with frames and such, but maybe try setVisible instead of show?

ah, lol cant believe i completely missed that, hmm i have no idea then.

You may want to try a couple of things.

(1) Try using your FirstFrame object "na" with your show() method like this:

     na.show();

The show() method is not static, so it requires an object.

(2) import java.awt.*

Many times Java's online documentation will help you understand your errors.

Log onto java.sun.com, then in the J2SE documentation look for the show() method in the java.awt.*  package to see what it does and how it should be used. Happy programming!


In triangle $ABC$, $BN$ and $CM$ are medians and $AD$ is angle bisector. $P = BN\cap CM$, $Q = AD\cap BN$, $R = AD\cap CM$. Find $\frac{S_{PQR}}{S_{ABC}}$ in terms of $a,b,c$.  :D

[color=darkblue]Here is a well-known general problem :[/color]

 [color=darkred]Given are the triangle $ABC$ and the points $M\in [BC]$, $N\in [CA]$, $P\in [AB]$ for which know $\frac{MB}{MC}=x$, $\frac{NC}{NA}=y$, $\frac{PA}{PB}=z$.

Denote the intersections $X\in BN\cap CP$, $Y\in CP\cap AM$, $Z\in AM\cap BN$. Then the area $[XYZ]$ of the triangle $XYZ$ is :

$\boxed{\ [XYZ]=\frac{(1-xyz)^{2}}{(1+x+xy)(1+y+yz)(1+z+zx)}\cdot [ABC]\ }\ .$[/color]

[color=darkblue][b]Particular case.[/b] $x=\frac{c}{b}\ ,\ y=z=1$ $\implies$ $[XYZ]=\frac{(b-c)^{2}}{3(b+2c)(2b+c)}\cdot[ABC]\ .$[/color]


trigonometry
function
summer program
Mathcamp
analytic geometry
AMC

For how many values of $ x$ in $ [0,\pi]$ is $ \sin^{\minus{}1}(\sin 6x)\equal{}\cos^{\minus{}1}(\cos x)$?
Note: The functions $ \sin^{\minus{}1}\equal{}\arcsin$ and $ \cos^{\minus{}1}\equal{}\arccos$ denote inverse trigonometric functions.

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

[quote="Darmani"]Yeah, this would be trivial if you could. I meant if you have any doubt that 6 is the right answer, then use a calculator.[/quote]

According to the solutions pamphlet, the correct answer is B.

EDIT: I can't even type the correct letter  :oops:

For anyone who's interested, here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223057]problem[/url] I made up a while ago similar to this. (Ironically I got this problem wrong)

Wait, but neither $ \frac {5\pi}7$ nor $ \frac {4\pi}5$ works! All [b]potential[/b] solutions satisfy $ \sin{6x} \equal{} \sin x$.

My attempt at calculator graphing gives me 4...

Same here...I don't see why the official answer is D.

[quote="Darmani"](Hmmm....three Mathcampers posting in a row.)

[hide="Calc rulz missed a couple"]
Namely, $ \frac {5\pi}{7}$ and $ \frac {4\pi}{5}$. So $ \boxed{D}$.

To verify it, do what Vlad said -- graph $ sin(6x) \minus{} sin(x)$ on your calculator over the interval, and you'll clearly see 6 zeroes.[/hide][/quote]

Yes, $ \frac {5\pi}{7}$ and $ \frac {4\pi}{5}$ do satisfy $ \sin{6x} \equal{} \sin{x}$, but the range of the inverse sine function is the interval $ [ \minus{} \frac {\pi}{2},\frac {\pi}{2}]$, so they do not work  :wink:

By the way, just curious, who are you from Mathcamp, Darmani?

i liked this problem because i found an extraneous solution but missed a correct solution so i still got the right answer  :rotfl:

[quote="abacadaea"][hide="solution"]
We know that the period of sin and cos are $ 2\pi$. Since $ \sin^{ \minus{} 1} a, \cos^{ \minus{} 1} a \in [0, 2\pi]$ for all $ a \in [ \minus{} 1,1]$, we have
$ 6x \equal{} x$
$ 6x \minus{} 2\pi \equal{} x$
$ 6x \minus{} 4\pi \equal{} x$
$ 6x \minus{} 6\pi \equal{} x$
We know we cant have any more than $ 6\pi$ since the maximum value for $ x$ is $ \pi$. Here, we see there are only $ 4$ values of x, so $ \boxed{B}$ is the answer.
[/hide][/quote]

The funny thing is, your fourth solution is greater than pi and your third solution doesn't work on my calculator. :D

I think that the correct answer is $ B$. They haven't posted the official ones anywhere, but graphing the equation in the problems leads to 4 zeroes.

Phew. Some of you had me worried :P

hello, i have also calculated four solutions.
Sonnhard.

[quote] They haven't posted the official ones anywhere, [/quote]

Your contest manager has a copy of the official solutions.  Ask your contest manager for the solutions.

Steve Dunbar
MAA Director, American Mathematics Competitions
[/quote]

Can't we simplify the equation to

$ 6x \equiv x (\mod 2\pi)$

$ 5x \equiv 0 (\mod 2\pi)$

$ 5x \equal{} 2k\pi$

$ x \equal{} \frac {2k\pi}{5}$

THen look for solutions in the interval?
I'm also confused what the inverse function does to the interval of solutions.

ok, um, i don't think that that is the correct method because (obviously) the answer is 4, and those equations only yield 3.

However, the range of possible values of $ cos^{ \minus{} 1}(x)$ is $ [0,\pi]$
The range of possible values of $ sin^{ \minus{} 1}(x)$ is $ [ \minus{} \frac {\pi}{2}, \frac {\pi}{2}]$

They only match at $ [0,\frac {\pi}{2}]$

Thus, $ x$ is between $ 0$ and $ \frac {\pi}{2}$, inclusive (i think)

so i think calc rulz's solution is correct.

But I don't understand how s/he got $ x \plus{} 6x \equal{} 7x \equal{} \pi (mod 2\pi)$ from $ sin 6x \equal{} sin x$

Also, why is doing $ cos 6x \equal{} cos x$ instead, wrong? The graph only yields 3 solutions.

This problem is quite confusing....
Help would be appreciated.

Sorry to revive, but like lightning says, how do you get $ x\plus{}6x \equal{} 7x \equal{}\pi (mod 2\pi)$ from sinx=sin6x?

haha... i'm answering my own question...

ok,

[hide="My_Reasoning"]

$ 6x \equal{} x (mod 2\pi)$

because $ sin y$ has a period of $ 2\pi$

However, if you do the unit circle, one would notice that $ sin y$ occurs twice.
For example,
$ sin (\frac{\pi}{6}) \equal{} \frac{1}{2} \equal{} sin(\frac{5\pi}{6})$

or you can just realize that
$ sin y \equal{} sin (\pi\minus{}y)$

Thus,
$ 6x \equal{} \pi\minus{}x (mod 2\pi)$
$ 7x \equal{} \pi (mod 2\pi)$

and you can solve it from there.[/hide]

[quote="calc rulz"][hide="Solution"]For $ 0 \le x < \pi$, we have $ \cos^{ \minus{} 1}(\cos{x}) \equal{} x$, so the equation becomes $ \sin^{ \minus{} 1}(\sin{6x}) \equal{} x$. This is equivalent to saying that $ \sin6x \equal{} \sin x$ and $ 0 \le x \le \frac {\pi}{2}$. This means that either $ 6x \equal{} x \mod{2\pi}$ or $ x \plus{} 6x \equal{} 7x \equal{} \pi \mod{2\pi}$. The possibilities for $ x$ are $ 0$ and $ \frac {2\pi}{5}$ in the first case and $ \frac {\pi}{7}$ and $ \frac {3\pi}{7}$ in the second. Thus, there are $ \boxed{4}$ solutions.[/hide][/quote]

Sorry for revive this topic,But i don't understand the whole solution,please help!!

[quote="abacadaea"][hide="solution"]
We know that the period of sin and cos are $ 2\pi$. Since $ \sin^{\minus{}1} a, \cos^{\minus{}1} a \in [0, 2\pi]$ for all $ a \in [\minus{}1,1]$, we have
$ 6x\equal{}x$
$ 6x\minus{}2\pi\equal{}x$
$ 6x\minus{}4\pi\equal{}x$
$ 6x\minus{}6\pi\equal{}x$
We know we cant have any more than $ 6\pi$ since the maximum value for $ x$ is $ \pi$. Here, we see there are only $ 4$ values of x, so $ \boxed{B}$ is the answer.
[/hide][/quote]

And how he got four solution?

[quote="luie1168e"]
Sorry for revive this topic,But i don't understand the whole solution,please help!!
[/quote]

Where do you not understand?

[quote="luie1168e"]
And how he got four solution?
[/quote]

The $\sin$ and $\cos$ functions repeat every $2\pi$ (because the period of $\sin$ and $\cos$ are $2\pi$).  From here, we see that one of our solutions would definitely be denoted by the equation $6x = x$.  The next solution must be when $6x$ and $x$ are $2\pi$ away from each other since the period is $2\pi$.  So $6x - 2\pi = x$.  And so on until $6x - 6\pi = x$.  The next equation would be $6x - 8\pi = x$, but then $x = \frac{8\pi}{5}$ which is greater than $\pi$.  So there are $4$ values of $x$ for the $4$ equations we have.

[quote="professordad"][quote="luie1168e"]
Sorry for revive this topic,But i don't understand the whole solution,please help!!
[/quote]

Where do you not understand?

[quote="luie1168e"]
And how he got four solution?
[/quote]

The $\sin$ and $\cos$ functions repeat every $2\pi$ (because the period of $\sin$ and $\cos$ are $2\pi$).  From here, we see that one of our solutions would definitely be denoted by the equation $6x = x$.  The next solution must be when $6x$ and $x$ are $2\pi$ away from each other since the period is $2\pi$.  So $6x - 2\pi = x$.  And so on until $6x - 6\pi = x$.  The next equation would be $6x - 8\pi = x$, but then $x = \frac{8\pi}{5}$ which is greater than $\pi$.  So there are $4$ values of $x$.[/quote]

Why are you talking about $x = \frac{8\pi}{5}$?
I mean why it is over 5?

In response to your second question, we just solve $6x - 8\pi = x$.  This becomes $5x - 8\pi = 0$, so $5x = 8\pi$ and $x = \frac{8\pi}{5}$ which is greater than $\pi$ and so on.

[quote="professordad"]I'm just solving the equation $6x - 8\pi = x$.  We know that all our solutions must be less than or equal to $\pi$.  So if the solution to one of our equations is greater than $\pi$, then we know that we must throw that equation out, and all the equations that follow it.  The solution to $6x - 8\pi = x$ is $x = \frac{8\pi}{5}$ which is greater than $\pi$, so this equation (and therefore all the equations that follow it) should be thrown out.[/quote]

One million thank you to you!!

But for first solution, i don't understand how he got the solution from the mod 2pi thing

[quote="professordad"]Hmm, this will be the hardest to explain.

$\mod 2\pi$ points to the fact that something repeats every $2\pi$.  Indeed, this is true with the $\sin$ and $\cos$ functions.  Saying that $6x = x \mod 2\pi$ is just saying that $6x - 2n\pi = x$, where $n$ is an integer.  The second equation [b]calc rulz[/b] wrote basically says that $x + 6x = n\pi$, where $n$ is an odd number.  Indeed, we find this is true if we graph the $\sin$ and $\cos$ functions.[/quote]

How do we prove the second equation?Thanks in advance...

[quote="professordad"]If we graph one period of the $\sin$ function (sorry, I don't know how to graph functions in Asymptote).  Let's say that we graph the period of $\sin$ function from $0$ to $2\pi$.  We see that this single period is symmetric about the line $x = \pi$, so two points on the $\sin$ function and symmetric on the $x = \pi$ line will have there $x$ coordinates sum to $2\pi$.[/quote]

But why we have $x=n \pi$?
Also,why n must be odd number?
also, why it is symmetrical about the line $x=\pi$?

Sorry to revive but I thought this solution is easier to understand than the official one with intervals and stuff.
From the given information, it is obvious that
$sin(6x)=sin(x)$
$sin(6x)-sin(x)=0$
$sin(3.5x+2.5x)-sin(3.5x-2.5x)=0$
$2sin(2.5x)cos(3.5x)=0$
We also have that:
$-pi/2<=sin^(-1)(x)<=pi/2$
This implies that there are 4 solutions.


integration
algebra
polynomial
trigonometry
calculus
calculus computations

29) $ \int \frac{x\plus{}1}{\sqrt{x^{2} \plus{} 2x \plus{} 3}} dx$

Problem: I have an irreducible polynomial within a root in the numerator which isn't anywhere near the statment $ \sqrt{1\minus{}u^{2}}$ for whatever $ u$ I choose.

30) $ \int \frac{x^{2}}{2 \plus{} x^{6}} dx$

Problem: I'm seeking an answer containing $ \tan^{\minus{}1}(u)$ or something similar, but the 2 in the numerator screws that up.

31) $ \int \frac{dx}{\sqrt{4 \plus{} 2x \minus{} x^{2} } }$

Problem: same as 29.

34)a) $ \int \sin^{4}(t)\cos^{5}(t)$

I used the double angle formula for sin here, then the product to sum formula, but the result is $ \int \frac{1}{16} \sin^{3}(2t) \sin^{3}(t)\sin(t)$, which is hardly easier.

29)

Note that:

$ \left(\sqrt {a x^2 \plus{} b x^2 \plus{} c}\right)' \equal{} \frac {2 a x \plus{} b}{2\sqrt {a x^2 \plus{} b x^2 \plus{} c}}$

and the RHS is quite similar to your problem. Using this you can transform it into something simpler. It will probably be something of the form (actually 100%):

$ \int\frac {dx}{\sqrt {a x^2 \plus{} b x^2 \plus{} c}}$

but you could write this as

$ A\int\frac {dx}{\sqrt {B \plus{} (x \plus{} C)^2}}$

then do $ \sqrt{B} \sin(u) \equal{} x \plus{} C$ and what you get should be a basic integral found in any integral table.

30)

Note:

$ \tan^{ \minus{} 1}(x) \equal{} \frac {1}{1 \plus{} x^2}$

And you have:

$ \frac {x^2}{x^6 \plus{} 2} \equal{} \frac {x^2}{(x^3)^2 \plus{} 2}$

so the $ x^2$ can be seen as a part of the differential of $ x^3$. Play with it.

31) same as 29)

34)

When you have an integral of the form:

$ \int \sin^n(x) \cos^{2m \plus{} 1}(x)\,dx$

you can write this as:

$ \int \sin^n(x) \cos(x) (\cos^2(x))^m\,dx \equal{}$

$ \equal{} \int \sin^n(x) \cos(x) (1 \minus{} \sin^m(x))^m\,dx$

now use the substitution $ t \equal{} \sin(x)$ and this becomes:

$ \int t^n (1 \minus{} t^2)^m\,dt$

and this is a polynomial $ \rightarrow$ should be easy to solve.

[quote="Prime factorization"]29) $ \int \frac {x \plus{} 1}{\sqrt {x^{2} \plus{} 2x \plus{} 3}} dx$

Problem: I have an irreducible polynomial within a root in the numerator which isn't anywhere near the statment $ \sqrt {1 \minus{} u^{2}}$ for whatever $ u$ I choose.[/quote]

In this case, $ u \equal{} x^2 \plus{} 2x \plus{} 3$ works just fine.

For 30, the closest I get is:

$ \int \frac{du}{3(2\plus{}u^{2})} \equal{} \int \frac{du}{3(1\plus{}u^{2}) \plus{} 3}$

And I don't follow when you say $ \sqrt{B}\sin(u) \equal{} x \plus{} C$

Like I said, instead of looking for $ 1 \plus{} u^2$ you should be looking for $ a^2 \plus{} u^2$.  If you have a table of integrals, one of them should be

$ \int \frac{du}{a^2 \plus{} u^2} \equal{} \frac{1}{a} \arctan \frac{u}{a}$.

Now set $ a \equal{} \sqrt{2}$.  ;)

This is essentially what milin is trying to get you to do with $ \sqrt{B} \sin u \equal{} x \plus{} C$.  This substitution puts the integral into a particularly simple form.

[quote="Prime factorization"]
And I don't follow when you say $ \sqrt {B}\sin(u) \equal{} x \plus{} C$[/quote]

This denotes a substitution.

Here's what you do in 30):

Use the substitution: $ x^3\equal{}u$ and this gives you $ \frac{du}{3 x^2}\equal{}dx$ and the integral becomes:

$ \int \frac{x^2}{2\plus{}x^6}\,dx\equal{}\int \frac{x^2}{2\plus{}u^2}\frac{du}{3 x^2}\equal{}\frac{1}{3}\int \frac{1}{2\plus{}u^2}\,du\equal{}\frac{1}{6}\int \frac{1}{1\plus{}\left(\frac{u}{\sqrt{2}}\right)^2}\,du$

you could do this directly, but to avoid potential confusion, do the substitution $ t \sqrt{2}\equal{}u$ and this gives you ($ \sqrt{2}dt\equal{}du$):

$ \equal{}\frac{1}{6}\int \frac{1}{1\plus{}t^2}\sqrt{2}\,dt\equal{}\frac{1}{3 \sqrt{2}}\int \frac{1}{1\plus{}t^2}\,dt\equal{}\frac{1}{3 \sqrt{2}} \tan^{\minus{}1}(t)\plus{}C\equal{}$

$ \equal{}\frac{1}{3 \sqrt{2}} \tan^{\minus{}1}\left(\frac{u}{\sqrt{2}}\right)\plus{}C\equal{}\frac{1}{3 \sqrt{2}} \tan^{\minus{}1}\left(\frac{x^3}{\sqrt{2}}\right)\plus{}C$

Milin:

I understood that you used $ u \equal{} \frac {\sin^{ \minus{} 1}(x \plus{} c)}{\sqrt {B}}$ so that $ du \equal{} \frac {dx}{\sqrt {B(1 \minus{} (x \plus{} c)^{2})}}$. However, this fact doesn't seem to help me on 31, since I can't get a perfect square in the form $ (x \plus{} c)^{2}$

Edit: Perhaps $ u \equal{} 2x \plus{} 4$ will eliminate this problem?

The first step on 31) is to rewrite as $ \int \frac {dx}{\sqrt {5 \minus{} (x \minus{} 1)^2} }$.  (In other words, to complete the square.)  Can you take it from here?


In problem 2, is 3sin2x - 22y = 0 correct?
Or should it be 3sin2x - 2sin2y = 0?


AMC
USA(J)MO
USAMO
modular arithmetic
number theory
inequalities proposed
inequalities

Yes...I have a solution about that, but please put a complete solucion.

 :wink:

I think it's from a USAMO, but I do not know the year.

[b]Solution[/b]  Assume to the contrary that there is a 14-tuple $(a_{1},\hdots ,a_{14})\in\mathbb{Z}^{14}$ such that $a_{1}^{4}+a_{2}^{4}+a_{3}^{4}+...+a_{13}^{4}+a_{14}^{4}=1599$.
Considering quartic residues modulo 16, we have $x^{4}\equiv 0,1\pmod{16}$.  But $1599\equiv 15\pmod{16}$ and $14<15$.  Contradiction!

And the conclusion follows.$\blacksquare$

how do i know when to use mod 13 or mod 16 or mod 4 ...etc?


The mean of three numbers is 10 more than the least of the numbers and less
than the greatest. The median of the three numbers is 5. What is their sum?

there are three numbers $ a\ge b \ge c$

$ Median \equal{} b \equal{} 5$

and Mean $ \dfrac{a\plus{}5\plus{}c}{3} \equal{} c\plus{}10 \equal{} a\minus{}10$

$ \dfrac{20\plus{}c\plus{}5\plus{}c}{3}\equal{}c\plus{}10$
$ c\equal{}\minus{}5$
$ a\equal{}15$

$ a\plus{}b\plus{}c \equal{} 15$

The mean of all three numbers is halfway between the lowest and highest numbers. But the number halfway between the lowest and highest numbers is also the mean of those two numbers.

Let $ S$ be the sum of all three numbers. Then, $ S\minus{}5$ is the sum of the largest and smallestnumbers. We have:

Mean of all 3 numbers = Mean of high and low

$ \frac{S}{3} \equal{} \frac{S\minus{}5}{2}$

$ \implies S \equal{} 15$

[quote="TZF"]The mean of all three numbers is halfway between the lowest and highest numbers. But the number halfway between the lowest and highest numbers is also the mean of those two numbers.

Let $ S$ be the sum of all three numbers. Then, $ S \minus{} 5$ is the sum of the largest and smallestnumbers. We have:

Mean of all 3 numbers = Mean of high and low

$ \frac {S}{3} \equal{} \frac {S \minus{} 5}{2}$

$ \implies S \equal{} 15$[/quote]

The mean of all three numbers is halfway between the lowest and highest numbers?
Let's say 1,10,100, so the mean is 101/2 = 50.5 here?   Shouldn't the mean be (1+10+100)/3??


Find the sum of the digits in the number $100!$

Oh, $100!$, not $100$[exclamation point]. In that case... $100!$ has $158$ digits. $24$ of them are $0$s, so that leaves us with $134$ (digits). The answer is $608$-ish...
933262154439441837319493572301376159
Here are $36$ of them!

There has to be a nicer way to do that....

I know. Let's look for a pattern:

1!: 1
2!: 2
3!: 6
4!: 6
5!: 3
6!: 9
7!: 9
8!: 9
9!: 27

pattern anyone? If $f(n)$ is the sum of the digits of $n$, then what is $f(kn)$. It's $kf(n)-9*$(the sum of all the carries)

Before we get carried away with patterns, lets agree on the answer first  :maybe:

I said it was 608-ish!

I can see no simple solution. $n!$ is very complicated, and $f(n)$ is more complicated still. I think it's basically impossible to calculate any one digit, other then all the 0s, of $100!$ without a computer or calculator.


[hide]Assuming that the domain is at least the set of integers, we have:
$f(2)+2f(-1)=240$
$f(-1)-f(2)=-120$

Subtracting twice the second equation from the first equation gives $3f(2)=480$, $f(2)=160$[/hide]


pigeonhole principle
geometry
rectangle
combinatorics unsolved
combinatorics

Yes, it [i]is[/i] 15. I'll be concise... Any problem, tell me.

The centers of the crosses are clearly in the "main" $ 8 \times 9$ sub-board, which can be partitioned in three $ 8 \times 3$ sub-board. Now, supposes there are $ 16$ crosses. At least $ 6$ of their centers are in a same $ 8 \times 3$ sub-board (pigeonhole principle). To better understanding, we are going to say the squares in this $ 8 \times 3$ board are ordered pairs $ (a, b)$, with $ 1 \le a \le 3$ and $ 1 \le b \le 8$. It is easy to see the rectangle $ R = {(a, b)|b = 4 \text{ or } b = 5\}}$ must contain at most two centers, as the $ 3 \times 3$ squares above and under it, which gives at most $ 2 + 2 + 2 = 6$ centers. To attain the equality, we must have two centers in each of these figures ($ R$ and the squares). Now, supposing there are centers in $ R$ are $ (1, 4)$ and $ (3, 5)$ (the other case is symmetric), it is easy to check that the centers are $ (1, 1), (3, 2), (1, 4), (3, 5), (1,7)$ and $ (3, 8)$.

This $ 8 \times 3$ rectangle may have one or two ajacent $ 8 \times 3$ rectangles. In the first case, its neighbor rectangle can have at most $ 4$ crosses centers (this is just case analisis). Until now, we have $ 10$ crosses. To get $ 16$, we must have $ 6$ crosess centers in the last rectangle. But, in this case, any configuration of the crosses is going to limit the number of centers in the second rectangle to at most $ 3$.

Case the $ 8 \times 3$ rectangle is the rectangle in the "middle" of the $ 8 \times 9$ rectangle, as I said above, we are going to have at most $ 4$ centers in each of its neighbors, which limits us to at most $ 6 + 4 + 4 = 14$ crosses.

Another solution:

For the sake of contradiction suppose we can place on the table $16$ crosses.

It is obvious that we can’t put more than $3$ crosses to touch one of the first column, last column, first row and the last row. Hence we must have at most $110 -10 +3 -10 +3 -11 +4 -11+4 = 82$ occupied squares. But $16$ crosses need $80$ squares.

Now focus on the following squares:
$(1,3)$, $(2,3)$, $(2,2)$, $(3,2)$, $(3,1)$, $(2,0)$, $(0,2)$.

It is not hard to see that at least three of these squares will be unocuppied by squares, while at most 2 were not in the 82.

Consider the same region for each of the $4$ corners of the table. Hence, at most $82-4 =78\leq 80$ squares can be covered.

Hence the assumption made is false!

[quote=MK4J]Another solution:
Now focus on the following squares:
$(1,3)$, $(2,3)$, $(2,2)$, $(3,2)$, $(3,1)$, $(2,0)$, $(0,2)$.

It is not hard to see that at least three of these squares will be unocuppied by squares, while at most 2 were not in the 82.
[/quote]

What happens if you put the crosses with center $(1,3)$ , $(2,1)$ , $(4,2)$, $(5,3)$?
There is just one square unoccupied, and that could not be in the 82

[quote=April]When the unit squares at the four corners are removed from a three by three squares, the resulting shape is called a cross. What is the maximum number of non-overlapping crosses placed within the boundary of a $ 10\times 11$ chessboard? (Each cross covers exactly five unit squares on the board.)[/quote]

Keep the board horizontally so that there are $10$ rows and $11$ columns. Color the columns [i]black[/i] and [i]white[/i] alternatingly ( first column should be [i]black[/i]). Now there are $6$ black columns and $5$ white columns. A [i]centre[/i] refers to the central cell of a cross. Its clear that a centre either is either [i]white[/i] or [i]black [/i]. Let the number of [i]white centres[/i] be $a$ and the number of [i]black centres[/i] be $b$. Note that the first and  last columns can't have any centre at all and the maximum number of cross in a particular column is $3$ since $\lfloor \frac{10}{3} \rfloor = 3$ . Now note that $a \leq 15$ whereas $b \leq 12$. Therefore can have atleast $15$ crosses. However note that in configuration for $15$ crosses, there cannot be any more crosses ($a$ is already max and black column cells are alternately covered with a [i]cross-cell[/i]). Also note that whenever there is a [i]black-cross[/i] Since it covers $3$ [i]black squares[/i] and $2$ white squares, so it takes up the space for exactly $2$ two [i]white-crosses[/i] (the cells in white column take up center space and the cells in the black column take up other cell space). Even if the [i]black-cross[/i] is centered in the second row (i.e its topmost cell touches the border),  the number of white cells left in the adjacent columns is $8$ (the cross clearly can't be centred in the first or second row of these columns anymore), implying that atmost $2$ [i]white-crosses[/i] can be placed in each of these columns. This observation however implies that whenever there is a [i]black-cross[/i], we can always remove it and instead add $2$ more [i]white-crosses[/i] ( with slight re-arrangements ofcourse). But since the maximum number of [i]white-crosses[/i] can be $15$, we conclude that our final answer is [size=130][b][u][i]15[/i][/u][/b][/size] and we are done :D.

For those wondering about the actual construction for 15, simply take the top $3$ consecutive [i]white-crosses[/i] in the odd [i]white[/i] columns and the bottom three consecutive [i]white-crosses[/i] in the even [i]white[/i] columns. It is clear that this works :D.

Sincerely,
Aayam