modular arithmetic
quadratics
function
number theory
totient function

Using index arithmetic, solve the following congruences:

1.) $ 7^x \equiv 6 (mod 17)$
2.) $ x^8 \equiv 5 (mod 14)$

[hide="Solution to Problem 2"]\[ x^8\equiv 5 \pmod{14}\]
Also, we have $ \phi {\left( 14\right)}\equal{}6$:
\[ x^2 \equiv 5 \equiv \pmod{14}\]There are no solutions to this congruence...Looking at the congruence in mod $ 14$, we have that:
\[ 5 \equiv x^2 \pmod{14}\]However, the quadratic residues of $ 14$ are $ 0, 1, 4, 9, 2, 11, 8, 7$. Therefore, there are no solutions.[/hide]

[hide="Solution to Problem 1"]From Euler's Totient Function, we know that $ 7^{16} \equiv 1 \pmod{17}$.
Therefore, all we need to do is check all the residues of $ 7^x$ in mod $ 17$.

EDIT: Since only $ 7^{13} \equiv 6 \pmod{17}$, that is the sole solution.[/hide]

[quote="SCHNOGGINVOGGINSCHWITZERW"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\equiv 13(\text{mod}16)$[/quote]Sorry, I calculated wrong. I will edit my answer.

[quote="Zellex"][quote="SCHNOGGINVOGGINSCHWITZERW"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\equiv 13(\text{mod}16)$[/quote]

How do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]

say $ 7^m \equiv 7^n \mod 17$
then $ 7^{m\minus{}n} \equiv 0 \mod 17$
clearly not true

[quote="CircleSquared"][quote="Zellex"][quote="SCHNOGGINVOGGINSCHWITZERW"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\equiv 13(\text{mod}16)$[/quote]

How do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]

say $ 7^m \equiv 7^n \mod 17$
then $ 7^{m \minus{} n} \equiv 0 \mod 17$
clearly not true[/quote]

Isn't that $ 7^{m\minus{}n}\equiv1\mod17$?

[quote="worthawholebean"][/quote][quote="CircleSquared"][quote="Zellex"][quote="SCHNOGGINVOGGINSCHWITZERW"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\equiv 13(\text{mod}16)$[/quote]

How do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]

say $ 7^m \equiv 7^n \mod 17$
then $ 7^{m - n} \equiv 0 \mod 17$
clearly not true[/quote][quote="worthawholebean"]

Isn't that $ 7^{m - n}\equiv1\mod17$?[/quote]

meh, i knew something seemed wrong with my idea
but the explanation is something simple like that, i'm certain

It's not that simple.

$ 7$ is an example of a [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]primitive root[/url] $ \bmod 17$, and not all residues $ \bmod 17$ have this property.  The powers of $ 4$, for example, only have the residues $ 1, 4, 16, 13 \bmod 17$.


counting distinguishability function LaTeX

Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand.  The order of rings on each finger is significant, but it is not required that each finger have a ring.  Find the leftmost three nonzero digits of $n.$

[hide]First choose the rings. There are $\binom{8}{5}=\binom{8}{3}=\frac{8*7*6}{3*2*1}=56$ ways to do this.

Then choose the ordering of the rings and on what fingers. If 1-5 denote the rings and f denotes a change of fingers, then this is the number of permutations of 12345fff, which is given by $\frac{8!}{3!}=8*7*6*5*4=6720$.

The number of arrangements is then $6720*56=376320\Rightarrow \boxed{376}$.[/hide]

[hide]$\binom{8}{5}$ ways to choose the rings, $5!$ ways to order the rings, and $\binom{8}{3}$ solutions to $a+b+c+d=5$, so the answer is the the first three digits of the product of all these.[/hide]

[hide]
Then there are $\binom{8}{5}=56$ ways to choose the rings, without considering order.
Now, in some arrangment of the rings on the fingers, let the rings be numbered 1 through 5, where 1 is the bottom ring on the first finger, and 5 the top ring on the last finger.  There are $5!$ ways to order the five rings, and the number of ways to distribute five indistinguishable rings between four fingers is the number of nonnegative integer solutions to $a+b+c+d=5$.  There are \[ \binom{4}{1}+4*3+4*3+\binom{4}{1}\binom{3}{2}+\binom{4}{2}\binom{2}{1}+\binom{4}{1}=4+12+12+12+12+4=56 \] such solutions.  So $n=56*56*120=376320$, so our answer is $\boxed{376}$.
[/hide]

The number of solutions to the equation \[ x_1+x_2+x_3+\cdots+x_k = n \] can easily be found by $\binom{n+k-1}{k-1}$, which you can see by looking at the number of ways to place $k-1$ separators in the list $\underbrace {\bullet \bullet \cdots \bullet}_{n \text { bullets}}$.

(Consider the case $a+b+c=10$. You must place 2 separators in ********** for each solution, and there are $\binom{12}{2}$ ways to do this.)

Oh yeah...forgot about that method...
Hmm..but come to think about it, you can place two separators in one space...
Yeah...

[hide]every sequence of 5 rings is $8*7*6*5*4$

we can distribute these 5 items by placing barriers in between them:  _1_1_1_1_1_

there are $\binom{6}{4}+\binom{6}{3}+\binom{6}{2}+\binom{6}{1}$ since we can have rings on 4,3,2,1 fingers

so we get $(15+20+15+6)8*7*6*5*4=376320 \implies \boxed{376}$[/hide]

[quote="chess64"]How could more than one go in the same place? That doesn't make sense.[/quote]

If one of the numbers is $0$...

[quote="chess64"]It would just be ||, with nothing in the middle. ;)[/quote]

Yeah...

Like we have 5 objects and 3 dividers

* * * * *

We could do something like

* * ||| * * *

which gives $(2, 0, 0, 3)$

But yet we know $6^3$ is not the right answer...is there something being repeated?

EDIT: Okay, yeah i found it.  Each possibility is repeated a number of times, because the positions of two dividers can be switched, and it's still the same combination.  Thus, anything with the three dividers in different places is repeated $6$ times; any with exactly two in the same place is repeated $3$ times, and any where they're all in the same place is not repeated.  Thus, our final answer is $\frac{1}{6}(6*5*4)+\frac{1}{3}(3*6*1*5)+6$.  The first term is the number of ways to have all of the dividers in different places, the second the number of ways to have  exactly two in the same place, and the third the number of ways to have all three in the same place.  The answer comes out to $56$.

[quote="chess64"]Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\binom{8}{3}=56$ will still give us the correct answer. ;)[/quote]

* | * | * | ** is (1, 1, 1, 2) while * * ||| * * * is (2, 0, 0, 3).  The first number is the number of stars to the left of the leftmost divider, the second the number of starts between dividers 1 and 2, and so on.

And yeah, now I see why $\binom{8}{3}$ works...8 total items, considering dividers and stars, and 3 ways to choose which ones are dividers.  I was thinking $\binom{6}{3}$ at first, and that's where the having two dividers in one place came up.

[quote="K81o7"]Then again...
The number of solutions of nonnegative integers for $a+b+c+d=5$ is the coefficient of $x^5$ in $(1+x+x^2+x^3+...)^4=(1+2x+3x^2+4x^3+...)^2$ which means the $x^5$ coefficient is $2(1*6+2*5+3*4)=56$.
But the number of ways to put 3 dividers between 5 objects is $6^3$, if more than one divider can go in the same place, and each possibility corresponds to exactly one triple...[/quote]

Or instead of the generating function, you can just use $n+r-1Cr-1$ which will give you the same answer

How do you do the choose bracket thing for combinations in the latex?

EDIT: $\binom{n+r-1}{r-1}$

[quote="chess64"][quote="K81o7"][quote="chess64"]Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\binom{8}{3}=56$ will still give us the correct answer. ;)[/quote]

* | * | * | ** is (1, 1, 1, 2) while * * ||| * * * is (2, 0, 0, 3).  The first number is the number of stars to the left of the leftmost divider, the second the number of starts between dividers 1 and 2, and so on.[/quote]

No, no, I know what it means, (after all, i'm the one who explained it to you :D) but I was asking why you would need different cases for them. :)[/quote]

Oh.

Yeah, I've used this method before...it just never occurred to me to use it here...

[hide]
We have $\binom{8}{5}$ ways to choose the rings. $\binom{8}{3}$ ways to split # of rings among fingers. $5!$ ways for the colors to be there.
$\binom{8}{5}\binom{8}{3}5!=376320$.
Answer:$\boxed{376}$.[/hide]

Sorry to revive.

[quote="AoPSWiki-Balls-and-Urns"]It is used to solve problems of the form: how many ways can one distribute k indistinguishable objects into n bins?.[/quote]

The problem says that the rings are distinguishable, so why can we use balls and urns here?

[quote="4everwise"][hide]There are $\binom{8}{5}$ ways to choose the rings, $\binom{8}{3}$ ways to distribute the rings among the fingers, and $5!$ distinct color arrangements.

Multiplying gives the answer: $\binom{8}{5}\binom{8}{3}5!=376320$, and the three leftmost digits are $\boxed{376}$.[/hide][/quote]
What do you mean by color arrangements?  What are we arranging differently, and how do we know we can do this when the rings are all on different fingers?

Sorry for the revive; but it seems that FlyingWombat's question has not been answered. You can use balls and urns here because, once you choose the number of ways to distribute the rings, there are $5!$ ways to arrange the rings because there are $5$ fixed positions possible for each ring to occupy. Hopefully this clarifies.

First we find the number of ways to pick the rings; this is just $\binom{8}{5}=56$. Now we find the amount of ways to put the rings on $4$ fingers for a total of $5$ rings. By Stars & Bars, this is $\binom{5+4-1}{4-1}=56$. The amount of ways to arrange the five rings is $5!=120$. Multiplying gives $376320 \implies \boxed{376}$.

[quote=vsar0406]Hey... so why wouldn't [hide=this method]There are $\binom{8}{5}$ ways to choose the five rings we are going to use, and since each ring can go on one of $4$ fingers, we multiply this by $4^5$ to get $57344$, or $573$ as the final answer.[/hide] work in this case? Is it because of the fact that the fingers themselves are not distinguishable, since I think I assumed that they were.[/quote]

They are distinguishable.
If the order of the rings on one finger didn’t matter, then your answer would be correct.

Oh, I see... Even the distinguishable-distinguishable method doesn't account for the order of the items in each container, so we have to do a modified stars and bars approach to find the answer. Cool problem :)

You first chose which $5$ of the $8$ rings you want, which we can do in $\binom{8}{5}$ ways. Then we order them, which we can do in $5!$ ways. Lastly, we distribute the rings to the fingers, and this is essentially stars and bars so there are $\binom{8}{3}$ ways. Hence, the answer is the first three digits of $\binom{8}{5} \cdot 5! \cdot \binom{8}{3} = 376320,$ which is $\boxed{376}.$


For how many values of $ 3^{n}\plus{}81$ the square of an integer

[quote="firecricket91"]For how many values of $ 3^{n} \plus{} 81$ the square of an integer[/quote]

I think its one.  ($ 3^4\plus{}81\equal{}18^2$)

None. You can prime factorize the equation to $ 3^n\plus{}3^4$ This cannot be factorized any further.

[quote="hunter34"]None. You can prime factorize the equation to $ 3^n \plus{} 3^4$ This cannot be factorized any further.[/quote]

3^5+81=324 which is 18^2

... well

$ 3^n\plus{}3^4\equal{}3^4(3^{n\minus{}4}\plus{}1)$

For this to be a perfect square $ 3^{n\minus{}4}\plus{}1$ must be a perfect square.

$ 3^{n\minus{}4}\equal{}k^2\minus{}1$
$ 3^{n\minus{}4}\equal{}(k\minus{}1)(k\plus{}1)$
This implies that both $ (k\minus{}1)$ and $ (k\plus{}1)$ have only factors of three.  The only possible value is $ k\equal{}2$

Thus, the future's answer is the only one possible.


Given the polynomial identity

\[x^6+1 = (x^2+1)(x^2+ax+1)(x^2+bx+1)\],

what is the value of $ab$?

As usual, if you're beyond this level, leave this to other beginning problem solvers.

[hide="solution"]
I would start by just expanding it and finding values of $a$ and $b$ so that the coefficients of $x^5, x^4, x^3, x^2,\text{ and }x$ are $0$.

When we expand it we get:

$x^6 + (a+b)x^5 + (ab+3)x^4 + (2a+2b)x^3 + (ab+3)x^2 + (a+b)x + 1$

so we have the equations:

$a + b = 0$
and
$ab = -3$

Since we want to find $ab$, we are done. The answer is $\boxed{-3}$. However, should we want to find $a$ and $b$, we would solve the two equations to get that one of them is $\sqrt 3$ and the other is $-\sqrt 3$

[/hide]


I would like to know  how  can i  write in  columns, like in  excalibur magazine. But  not  in all pages  of same document,
just  few of them  to  be at columns. Thank you very much!

you should find what you are looking for here:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe]www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe[/url]


calculus
search
function
calculus computations

any interesting problems on continuity and differentiability please!

There are nearly 20000 threads in the calculus section. Look at some of them. Use the search function to narrow it down a little.

Prove that every continuous open mapping from $ R$ to $ R$ is monotonic.


group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a finite cyclic group of order $ n$. Let $ a$ be a generator. Let $ r$ be a nonzero integer and relatively prime to $ n$.

(i) Show that $ a^r$ is also a generator of $ G$.
(ii) Show that every generator of $ G$ can be written in this form.
(iii) Let $ p$ be a prime number, and $ G$ a cyclic group of order $ p$. How many generators does $ G$ have?

i) Use Bezout's theorem, i.e. that for coprime $ a,b$ there are integers $ x,y$ such that $ ax\plus{}by\equal{}1$.

ii) Every element of $ g$ is of type $ a^r$ for some integer $ r$. You have to show that if $ a^r$ is a generator, than $ n$ and $ r$ are coprime. For this, how many different powers of $ a^r$ are there if $ n,r$ have a common divisor $ d>1$.

iii) You have shown that generators come from integers $ r$ coprime to $ n$. Which $ r$ give the same generators¿

alternative for (ii): write $ (a^r)^k \equal{} a$ and use $ n \equal{} ord(a)$.


Find the ratio of the area of an equilateral triangle inscribed in a circle to the area of a square circumscribed about the same circle.

I don't think the area of a equilateral triangle inscribed in a circle belongs in the Basics. I'll move it to the Mathcounts forum.

[hide="Answer"]Drawing three radii from the center of the circle to the vertices of the triangle, and the perpendicular bisectors to each side of the triangle, we have six 30-60-90 triangles, each with hypotenuse $r$. Therefore, the area of the inscribed equilateral triangle is $\frac{3r^2\sqrt{3}}{4}$. The area of the circumscribed square is $4r^2$, so the ratio of the two is $\frac{\frac{3r^2\sqrt{3}}{4}}{4r^2}=\boxed{\frac{3\sqrt{3}}{16}}$.[/hide]


[i]Let a triangle $ABC$ , the median $BD$ , the bisector of $\angle BDA$ $DE$ , and the bisector of  $\angle BDC$ $DF$. Proove that $EF //AC$[/i]

[img]http://img50.imageshack.us/img50/9613/fasfaszq0.png[/img]

[hide]Hint:  angle EDF is a right angle... [/hide]


yes! consider for example a solution consisting of 1 dm^3  with 2 moles of HCl. This gives $pH=-lg[H]=-lg 2 = -0.301...$

Well, 6M HCl is a common concentration; it has pH=-0.778

[quote="Mariag"]it is possible, just not generally available... also, u could have pH >14 (guess that would be too little $H^+$[/quote]
in what conditions can you have pH>14?

[quote] in what conditions can you have pH>14?[/quote]
when $[H^+] < 10^{-14}$

Well, I worked with 6M KOH at camp one year (not USNCO camp; not yet at least).  That has pH=14.778

Once I used 19M NaOH in class.  That breaks 15!

[quote="Mariag"]it is possible, just not generally available... also, u could have pH >14 (guess that would be too little $H^+$[/quote]

8+ M HCl is quite easy to get at hardware stores.  It's cheap, too.

[quote="inteluser"]At USNCO camp, they prepared a 125M solution of KOH.  That breaks 16, though not 16!  (this is a math forum, remember...  :P  )[/quote]

Oops, sorry.

I want to go to USNCO camp!  But I don't think I did so well on the test :(.

i guess i don't have enough experience with crazy concentrated chemicals (i don't recall ever working with any thing with molarity more than 3)  :(
i also hope i was able to do better on the USNCO test, camp sounds like so much fun!

Things are a little bit more complicated.

Measured pH is not -log of H+ concentration, but of H+ activity. The more concentrated solution the higher its ionic strength and the lower activity coefficient - this is described by so called Debye-Huckel theory. And it is not the problem of extremally concentrated solutions - in fact for concentrations above 0.1M even DH theory is not able to properly predict activities. 0.1M HCl should have pH = 1.00, but the result of measurements will be 1.12 - the activity coefficient is 0.75.

This all boils down to the fact that having high concentration of acid (or base) doesn't mean you will have a negative pH or pH higher than 14 - due to low activity coefficients. IIRC they are possible and can be measured, but for much more concentrated solutions that one may expect.

Best,
Borek

What is the difference between pH and activity? Thanks. :blush:

u have to meet KAWKABI ,if u do u will be persuaded :rotfl:

The bottom line is that the activity measures amazingly enough the activity or apparent concentration.  This is what is measured by a pH meter, or pH paper.  The concentration is important if one is reacting a solution to a given endpoint, such as a pH of 7.  Activity is important is one is concerned about strength of the acid/base (will it eat my container?).

Most every "law" or rule that applies math to the real world doesn't work at extreme conditions very well.  These are those conditions when the math would say one has a negative pH.  Even if one got there, it isn't really important.  The only reason to use the log function was to avoid using exponents in talking about acid/base situations.

It's just as critical when the temperature goes negative.  (The then retrograde, now Celsius scale was set using the then known highest and lowest temperatures on earth [knowledge was a bit narrower than it now is].)  Negative simply means that the original folks missed the mark.  Kelvin and Rankin (not spelled correctly) start at absolute zero and have been slid fractions of degrees down, as the initial folks didn't know as much as they thought.)

Even the calendar is wrong, Christ was born a bit off.  It really doesn't matter for anything, but they missed.  The meter was based on a French attempt to measure the distance from the equator to the north pole (or the other way around), which was way off.  As long as we agree and can talk, it really doesn't matter what the units are, or what they are based upon.

[quote="inteluser"]At USNCO camp, they prepared a 125M solution of KOH.  That breaks 16, though not 16!  (this is a math forum, remember...  :P  )[/quote]

They didn't. Solubility of KOH is lower. Even pure melted KOH (if treated as its own solution) will have much lower concentration.

[quote="annornor"]It is possible to have a negative pH?????[/quote]

pH= -log[H+] then ph=f(x) where x=[H+] and f(x)=-logx
Df=R  :)


I remember my fifth grade days when I was in the Cottage, and Jamal would hang around the windows trying to see homeless people. On our field trip to Philadelphia, they saw a homeless person begging for money when he had a huge wallet full of $ 100$ dollar bills, and a leather jacket. (Our group saw plenty of homeless people needing money as well!) Is it right to laugh at them? Is it right to laugh at them privately, in your head or somewhere far away from them? Is it right to walk right past them without giving them any money? It doesn't seem right.

I was homeless for about 5 hours when the extermination people came to our house and we couldn't stay there. We went to Wegmans because it was around lunchtime and we were hungry, when there was plenty of food in our kitchen that we could eat. It seemed kind of sad and I decided that I shouldn't at least laugh with my friends at a funny looking vagrant.

What do you think?

My aunt is the director of Home Aid in Atlanta and that program helps homeless people a lot. I do not think you should laugh at them because 90% of people are born into homelessness and maybe don't know the skills to help themselves. The average age of all homeless people in the U.S. is 9 years old, which I find very sad. I find homelessness to be one of the biggest issues in the U.S. today.

Wait, 90% of all US born kids are homeless? Perhaps this is a world statistic, and not just US.

No, 90% of all current homeless people were born homeless.

9 years old? Wow.

I'd rather really be struggling on the streets, because my friend said she overheard a classmate say, "Mom, can we actually have dinner today?". I'd rather have food in my stomach than a huge apartment where you have some comfort for sleep, that's all.

The problem with homelessness is because of rising home prices, which make people more welcoming to apartments, and the apartment prices rising too. They should make a huge mansion (the kind football players have :no:) and make so many huge apartment- sized rooms for some homeless people.

As part of the organization my aunt directs, they help establish homeless shelters in different communities. I think they are headed in the right direction.

That is definitely a World Statistic, not a U.S. one. In America, homeless kids go to orphanages or foster care and hence are no longer homeless.

If it was a world statistic, Africa takes a large chunk of the percentage  :read:

They aren't homeless so much as tribal at that point. You need a certain amount of development in a country before you can call its people without houses homeless

I'm thinking India.

[quote="dynamo729"]... 90% of people are born into homelessness ... The average age of all homeless people in the U.S. is 9 years old[/quote]  Neither of these claims sounds very plausible to me -- can you cite any source for either of them?

[quote="demonshadow"]The problem with homelessness is because of rising home prices, which make people more welcoming to apartments, and the apartment prices rising too.[/quote]  I don't think this is really true: the lack of affordable housing almost everywhere is a serious problem, but it's not a new problem tied to rising home prices.  In some places rising rental costs could be partially to blame, but purchasing a home has always required either a steady income flow or a large amount of savings or both, and homelessness is contraindicative of both.  I would guess that financial instability leading to temporary inability to pay rent is probably a frequent cause of short-term homelessness.  (Long-term homelessness is a different beast: it's often linked to mental illness, for example.)

[quote="Anaxerzia"]In America, homeless kids go to orphanages or foster care[/quote] I doubt that this is true in any sort of generality.

(Edit: just fixed an errant quote tag.)

My source is my aunt (not biological, but her husband and my dad have known each other from birth and we do a lot of stuff together and just call them aunt and uncle. My uncle was my dads best man and is my brothers godfather), who is the director of the Home Aid, a program in Atlanta (it may be other places, but she is the director in Atlanta) thats helps build shelters and stuff.  She has been in this program for years; may I ask why you find them not very plausible?

I find the first not very plausible because significant portions of the homeless population fall into the following two categories:

1) people who are temporarily homeless, for example because of job loss -- this includes a lot of families with one or multiple children, for example.  Hardly any of these people are born into homelessness, and as individuals they don't stay homeless for very long periods of time (a few months or a year).

2) people who suffer from serious mental illness (including many veterans, for example) make up a lot of the "visible" homeless, and I have no reason to think any significant fraction of them are born into homelessness because in this case there's a different identifiable cause.

90% is a very large number -- it just seems unlikely to me that the two groups I've named are such a small portion of the homeless population.

For the second one, I don't so much doubt it as not understand what it means.  But, I find the following information on the [url=http://www.homeaid.org/index.cfm?tdc=dsp&page=homeless_facts]Facts & Statistics[/url] page of the [url=http://www.homeaid.org/]HomeAid[/url] website:
[quote="HomeAid website"]Officials estimate that, on average, single men comprise 51 percent of the homeless population, families with children 30 percent, single women 17 percent, and unaccompanied youth 2 percent.[/quote]

By the way, HomeAid sounds like a great organization for helping the temporarily homeless.  So, kudos to your aunt for working for such a good cause.

Hm..... either my aunt got it wrong or I misunderstood (the latter most likely)!  Thanks for the clarification!

[quote="D3m0n Shad0w"]I'd rather really be struggling on the streets, because my friend said she overheard a classmate say, "Mom, can we actually have dinner today?". I'd rather have food in my stomach than a huge apartment where you have some comfort for sleep, that's all.[/quote]

Wait, what? Are you saying that you'd rather be poor than rich?

er I think he's saying that he'd rather have a full stomach than a comfortable shelter. idk..

If they were asking for some, I'd give them money. It'll at least get them a meal or two, though it would take a whole lot of effort to completely change someone's life. Of course, there are lots of programs that are trying to accomplish just that (i.e. [url=http://www.compassion.com/default.htm]Compassion[/url])

[quote="sailingdog"][quote="JBL"][quote="sailingdog"]or else the gap between rich and poor would be continually widening.[/quote]  I come across a statement like this and I think, "What rock are you living under?"  Income inequality in this country [i]is[/i] growing, and has been for years.[/quote]
XD
I know that. It would accelerate, though.
Besides, I am still 13 - not really old enough to experience this first hand...[/quote]
sailingdog, you're right. that's why we gotta learn how to deal with these kind of things NOW [practicing, kinda], as what...say 11-year-olds (like me, except im 12 in a few days) before moving on to the teen years. that way we have the knowledge about these situations when we face them in real life.

Did BOTH of you just completely miss that a growing income inequality is [i]equivalent[/i] to a widening gap between the rich and the poor?

And more than 50% tax on the rich? Do you realize how high that is already?

Socialism to that degree doesn't work at all. When those who have highly skilled professions earn only a little bit more than those who anybody can do their job, then there is far less incentive to get highly skilled professions. Look at a country like France and its unemployment compared with the U.S.. Even now, many can argue that in America's ghettos people are gaming welfare and having children as teenagers specifically so that they can live solely of off tax dollars. My mom (an OBGYN working in this area) sees this everyday, and this is a TINY community we are talking about.

Granted, I think most welfare can do good, but just needs more regulation so that things like this do not happen.

Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.

I think the best way to get the homeless to work is public infrastructure. America's infrastructure is crumbling, and cheap labor is needed. The homeless should be hungry for such jobs.

[quote="Anaxerzia"]
And more than 50% tax on the rich? Do you realize how high that is already?

Socialism to that degree doesn't work at all. When those who have highly skilled professions earn only a little bit more than those who anybody can do their job, then there is far less incentive to get highly skilled professions.
[/quote]

Most people will prefer to get highly skilled professions even without money as an incentive because "low-level" jobs are usually pretty boring.

Even if I make less money as a mathematician, I will still prefer to be a mathematician than a worker at MacDonalds.

[quote="Anaxerzia"]

Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.
[/quote]

 This is because America is very "socialist" comparing to many other countries in the world.

Not all competitive fields are academic. Also, the poor will have less incentive to peform in school if they shall have a decent life subsidies by the government no matter what.

[quote="Anaxerzia"]And more than 50% tax on the rich? Do you realize how high that is already?

Socialism to that degree doesn't work at all.[/quote]
For most of the post-WW2 era, the US top marginal tax rate has been higher than 50% (and, in fact, higher than 70%).  This happened to coincide with some of the most impressive economic growth anywhere ever, growth that did not happen at the expense of increasing income inequality.

(An x% marginal rate means that for people making more than a certain amount, x% of the [i]additional[/i] amount was taxed.  So, for example, our current top marginal rate is 35% and applies only to those making about 350,000 USD annually.  In order to actually owe one third of your income in income taxes, you'd have to make more than 1.2 million dollars.  These numbers are for a single individual -- other marital statuses would give somewhat different numbers.)

[quote="Anaxerzia"]Look at a country like France and its unemployment compared with the U.S.[/quote]  What about it?  The French have made a societal decision to work shorter hours and be a more generous society than we have.  The result is that they're equally productive on a per-hour-worked basis, have a generally powerful economy, and have vastly lower inequality (as measured by the GINI index) than the US does.  (I also remember vaguely that Europe may compute its unemployment rate differently than we do, but I wasn't able to find a citation for that.)

[quote="Anaxerzia"]Even now, many can argue that in America's ghettos people are gaming welfare and having children as teenagers specifically so that they can live solely of off tax dollars.[/quote]  Yes, one could argue that, but then one would be either an idiot or someone intentionally trying to stir up racial resentment.

[quote="Anaxerzia"]Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]  Um, no:
[quote="[url=http://en.wikipedia.org/wiki/Economic_mobility#Economic_mobility_worldwide]Wikipedia on economic mobility[/url]"]Using the ratio of an individual’s current income to that of their parent’s, the United States has much less relative mobility than other industrialized nations. The income of our parents is a great deal more predictive of our own incomes in the United States than other countries. France, Germany, Sweden, Canada, Finland, Norway, and Denmark all have more relative mobility than the US, while only the United Kingdom is shown to have less mobility. According to this study done by Miles Corak, The United States ratio of relative mobility is 1, whereas the other countries mentioned with more mobility have a range of 1.25 (France) to over 3 (Denmark).[/quote]

[quote="Anaxerzia"]
Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]Woah, hold on. I'm confused: are you saying that it's a bad thing?

[quote="JBL"]

[quote="Anaxerzia"]Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]  Um, no:
[quote="[url=http://en.wikipedia.org/wiki/Economic_mobility#Economic_mobility_worldwide]Wikipedia on economic mobility[/url]"]Using the ratio of an individual’s current income to that of their parent’s, the United States has much less relative mobility than other industrialized nations. The income of our parents is a great deal more predictive of our own incomes in the United States than other countries. France, Germany, Sweden, Canada, Finland, Norway, and Denmark all have more relative mobility than the US, while only the United Kingdom is shown to have less mobility. According to this study done by Miles Corak, The United States ratio of relative mobility is 1, whereas the other countries mentioned with more mobility have a range of 1.25 (France) to over 3 (Denmark).[/quote][/quote]

So what? The way it is matters zilch. Anaxerzia is talking about the opportunity kids have, you are discounting his argument with facts that show what kids end up doing. Just because a kid doesn't take his life as far as a kid in a similar situation in another country, doesn't mean he had less of an opportunity to be successful growing up.

i see. thanks!!

**this is a little off topic, but do posts in the round table count towards your overall post score?

Shouldn't be too hard to figure out...

Also, I found an article some of you might like:

[url=http://www.nytimes.com/2008/05/30/opinion/30jacoby.html?_r=1&ex=1212811200&en=cc94734646d73502&ei=5070&emc=eta1&oref=slogin]Shortened by moderator.[/url]

[quote="sailingdog"]So what? The way it is matters zilch. Anaxerzia is talking about the opportunity kids have, you are discounting his argument with facts that show what kids end up doing. Just because a kid doesn't take his life as far as a kid in a similar situation in another country, doesn't mean he had less of an opportunity to be successful growing up.[/quote]

So, in other words, America is a great place because we have so much opportunity, but Americans are so hugely incompetent that we don't take advantage of this?  This is what I would describe as "highly contrived."  A more natural explanation would be that people everywhere take roughly equal advantage of their opportunities, and there are simply fewer opportunities available in this country than elsewhere.  This is also consistent with the substantially larger levels of economic inequality here than in the rest of the developed world.

Ohh... Who is John Galt?

Are you suggesting that we give away on need instead of ability? That we should sacrifice our own livelihoods to someone who has given up his long ago? That we should pour our minds into alms meant for the people that have lost the capacity to use theirs? "From each according to his ability, to each according to his need?" To spend our cash on someone whom we have no vested interest in only breeds resentment. It is much better to finance something that can bring others up, such as a university, than to aimlessly funnel funds into someone's need. If I give to a charity, I want to convince myself that I am giving because my money may help carry someone forward, to my own stature, not that they will become dependent on my handouts, like a parasite. Call it a Horatio Alger wet dream, but if everyone worked for their own self-interest, rather than in the interest of others, then everyone would fall where their ambition placed them.

<3 rhetorical questions aimed at no one.

[quote="stuck"]if everyone worked for their own self-interest, rather than in the interest of others, then everyone would fall where their ambition placed them.[/quote]  Of course.  After all, only the unworthy get hit by automobiles.  Only the lazy get cancer.  And it's well-documented that the only victims of discrimination and bad luck are those who deserve it.

Meanwhile, those of us who live in the real world seek outcomes that will reduce human suffering, rather than self-serving explanations for why we shouldn't be bothered by it.

Incidentally, increasing income inequality in this country has little to do with availability of higher education and much more to do with structural changes (e.g. to the tax code) that significantly favor the wealthy.  See, for example, http://www.epi.org/content.cfm/ib232

there are people who are just lazy and don't work
i guess those are the ppl who you can despise

[quote="beta"]I agree with calc rulz. Does anyone really NEED >$ \$$500K per year?[/quote]

Yes.


trigonometry
complex numbers
geometric series

[quote="mdk"]Compute the smallest positive angle $ x$ such that $ sin x+sin 3x+...+sin 99x = csc x$,
expressing your answer in radians.[/quote]

[hide="Hint"]Complex numbers, sum of geometric series[/hide]

[hide="hint"]
U can show even without compex numbers that $ \sum_{k=1}^{n}\cos k\phi=\frac{\sin (n+\frac{1}{2})\phi}{2\sin\frac{\phi}{2}}-\frac{1}{2}$[/hide]

[hide="Solution"]
Multiplying the equality with $ 2\sin x$ we have
$ 2\sin^{2}x+2\sin x\sin 3x+\ldots+2\sin x\sin 99x=2$
$ 1-\cos 2x+\cos 2x-\cos 4x+\ldots+\cos 98x-\cos 100x=2$
$ \cos 100x=-1\Rightarrow x=\frac{(2k+1)\pi}{100},k\in\mathbf{Z}$
For $ k=0\Rightarrow x=\frac{\pi}{100}$
[/hide]


[quote="GameBot"]When three fair 6-sided dice are tossed, what is the probability of tossing at least one 6? Express your answer as a common fraction.[/quote]

The probability that we get ZERO 6's is $ \frac{5}{6}^3 \equal{} \frac{125}{216}$. Thus our answer is $ 1 \minus{} \frac{125}{216} \equal{} \frac{91}{216}$.

Or if you were me and decided to add up all the probabilities:
\[3C1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}+3C2\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)+3C3\left(\frac{1}{6}\right)^{3}\]
\[\boxed{\frac{91}{216}}\]


inequalities
absolute value
inequalities proposed

The problem stated:

Let $ a,b,c$ be real numbers with $ a^2\plus{}b^2\plus{}c^2\equal{}2.$ Prove that
$ |a^3\plus{}b^3\plus{}c^3\minus{}abc| \le 2\sqrt{2}.$

I wait for the nicest solution.

Re-homogenizing we see it follows from
\[ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1\implies a^3 \plus{} b^3 \plus{} c^3 \minus{} abc\le 1\]
Now observe that
\[ (a^2 \plus{} b^2 \plus{} c^2)^3 \minus{} (a^3 \plus{} b^3 \plus{} c^3 \minus{} abc)^2 \equal{}\]

\[ \equal{} \sum_{cyc}a^4(b \plus{} c)^2 \plus{} 5a^2b^2c^2 \plus{} \sum_{cyc}a^2b^2(a \minus{} b)^2 \plus{} \sum_{sym}a^4b^2\geq 0\]
:P

Dear friends, thanks for reply the topic. Now, I will post my solutions, I hope you like them

First way: We apply the Cauchy Schwarz Inequality and the AM-GM Inequality, respectively, so that
$ \begin{aligned} (a^3 \plus{} b^3 \plus{} c^3 \minus{} abc)^2 & \equal{} [a(a^2 \minus{} bc) \plus{} b^3 \plus{} c^3]^2 \le (a^2 \plus{} b^2 \plus{} c^2)[(a^2 \minus{} bc)^2 \plus{} b^4 \plus{} c^4] \\
& \equal{} 2(a^4 \plus{} b^4 \plus{} c^4 \plus{} b^2c^2 \minus{} 2a^2bc)\le 2[a^4 \plus{} b^4 \plus{} c^4 \plus{} b^2c^2 \plus{} a^2(b^2 \plus{} c^2)] \le 8,\end{aligned}$
where the last inequality holds because
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} b^2c^2 \plus{} a^2(b^2 \plus{} c^2) \le a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \equal{} (a^2 \plus{} b^2 \plus{} c^2)^2 \equal{} 4.$
Now, from this inequality, we can easily deduce that
$ |a^3 \plus{} b^3 \plus{} c^3 \minus{} abc| \le 2\sqrt {2},$
as desired. This ends our proof.

Second way: Using the property of absolute value and the Cauchy Schwarz Inequality, we have
$ \begin{aligned} |a^3 \plus{} b^3 \plus{} c^3 \minus{} abc| & \le |a^3| \plus{} |b^3| \plus{} |c^3| \plus{} |abc| \\
&\le \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(a^4 \plus{} b^4 \plus{} c^4)} \plus{} |abc| \\
& \equal{} \sqrt {2(a^4 \plus{} b^4 \plus{} c^4)} \plus{} |abc|. \end{aligned}$
Thus, it suffices to prove that
$ \sqrt {2(a^4 \plus{} b^4 \plus{} c^4)} \plus{} |abc| \le 2\sqrt {2}.$
Because $ 2\sqrt {2} \equal{} (a^2 \plus{} b^2 \plus{} c^2)\sqrt {2},$ this comes down to
$ \sqrt {2} \left( a^2 \plus{} b^2 \plus{} c^2 \minus{} \sqrt {a^4 \plus{} b^4 \plus{} c^4}\right) \ge |abc|,$
which is equivalent to
$ \frac {2\sqrt {2} (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {2(a^4 \plus{} b^4 \plus{} c^4)}} \ge |abc|.$
Note that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {a^4 \plus{} b^4 \plus{} c^4} \le 2(a^2 \plus{} b^2 \plus{} c^2) \equal{} 4,$ one can see that the last inequality was deduced from
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge \sqrt {2} |abc|.$
Now, using the well-known inequality $ (x \plus{} y \plus{} z)^2 \ge 3(xy \plus{} yz \plus{} zx)$ for the triple $ (x,y,z) \equal{} (a^2b^2,b^2c^2,c^2a^2),$ we get
$ (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge 3a^2b^2c^2(a^2 \plus{} b^2 \plus{} c^2) \equal{} 6a^2b^2c^2 \ge 2a^2b^2c^2,$
from which we deduce that
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge \sqrt {2}|abc|,$
as desired. This ends our proof.

The second proof is natural on the aspect of idea. Anyway, they are both wonderful and manageable.  :)

Can I think that your solutions are supernatural  :D , I have also an approach by cauchy shwarz :

denote : $ [x]= abs(x)$

$ [( a^3+b^3+c^3-abc)] = [ \sum{a(a^2-\frac{bc}{3}}] \leq  \sqrt{ (a^2+b^2+c^2)( \sum{(a^2-\frac{bc}{3})^2})}$

and $ \sum{(a^2-\frac{bc}{3})^2 = \sum{a^4} - \frac{2abc\sum{a}}{3} + \frac{\sum{(ab)^2}}{9}  \leq   \sum{a^4)+\sum{(ab)^2} \leq  (\sum{a^2})^2}}$ :)


logarithms
algebra
system of equations
algebra unsolved

sovle this system of equations in $ R$:

$ \begin {cases} \sqrt {8x^3 \plus{} 1} \minus{} \sqrt {y^3 \plus{} 3y^2 \plus{} 3y \plus{} 2} \equal{} y \minus{} 2x \plus{} 1 \\
y^3 \plus{} 4x \plus{} 1 \plus{} \ln{(y^3 \plus{} 2x)} \equal{} 0 \end {cases}$

show that :$ y\plus{}1\equal{}2x$ and $ y\equal{}\minus{}1$ is unique root of equation$ y^3\plus{}2y\plus{}3\plus{}ln( y^2\plus{}y\plus{}1)\equal{}0$

[quote="Vincent Gilbert"]show that :$ y \plus{} 1 \equal{} 2x$ and $ y \equal{} \minus{} 1$ is unique root of equation$ y^3 \plus{} 2y \plus{} 3 \plus{} ln( y^2 \plus{} y \plus{} 1) \equal{} 0$[/quote]

 you have mistake $ y^{3}\plus{}2y\plus{}3\plus{}ln(y^{3}\plus{}y\plus{}1)\equal{}0$

$ y\equal{}\minus{}1$---> $ ln(y^{3}\plus{}y\plus{}1)\equal{}ln(\minus{}1)$--> not solution

^^ .Maybe he had a typo because this problem is exactly alike a problem in a contest at my province


calculus
integration
real analysis
real analysis unsolved

If $f_{n}$ is a convergent sequence in $D'(R)$ (distributions), $F_{n}' = f_{n}$ and if the integral of $\phi_{o}$ is nonzero such that $(F_{n},\phi_{o})$ is bounded. Then show that $F_{n}$ has a convergent subsequence.

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

It's pretty easy to show that $(F'_{n},\phi) \rightarrow (f,\phi)$ but i'm not sure about the rest.

As a first step, you need to show this: if $F'=f$ and $G'=f$ in $\mathcal{D}',$ then there exists a constant $C$ such that $F=G+C.$

The definition of $C$ as a distribution is, of course, $(C,\varphi)=C\int_{\mathbb{R}}\varphi$


floor function
number theory unsolved
number theory

Show that $ \sum_{k\equal{}1}^n \lfloor \dfrac{n}{k} \rfloor \minus{}\lfloor \sqrt{n}\rfloor$ is even for all positive integers $ n$.

[hide]
The quantity $ \sum_{k\equal{}1}^n\left\lfloor\frac{n}{k}\right\rfloor$ counts the number of pairs $ (a,b)$ where $ a\le b\le n$ and $ b$ is divisible by $ a$, since for a given $ a$, it has $ \left\lfloor\frac{n}{k}\right\rfloor$ multiple lss than or equal to $ n$. Now this is equivalent then to $ \sum_{k\equal{}1}^n d(k)$, where $ k$ is the total number of divisors that $ k$ has. This number is even except when $ k$ is a perfect square, so this quantity has the same parity as the number of perfect squares less than or equal to $ n$, or $ \lfloor\sqrt{n}\rfloor$.
[/hide]

Thanks a lot - I just completely algebra bashed this thing with casework n odd, n even. Your way is far more instructive.


superior algebra
superior algebra unsolved

Let A be a ring.

If every finitely generated right ideal is genreated by an idempotent then Jac(A) =0.

Here Jac(A) means jacobson radical.

Proof: Let a be in Jac(A) and pick an idempotent element e such that Ae = Aa, thus a = ae and e=xa for x in A. Hence a =axa so a(1-xa)=0. Since a is in Jac(A) also xa is in Jac(A)
and 1-xa is a unit , hence a=0.

My question is that why is a = ae  please help me with this.

Because Ae = Aa, so I will say that ae = a'a, please help.

$ a\in Aa\equal{}Ae\Rightarrow a\equal{}be\Rightarrow ae\equal{}be^2\equal{}be\equal{}a$.

Thanks, but I was thinking that since Aa =Ae  so I am thinking that a'a =be, what does  elements in Aa look like.

I think that Aa = {a_0a,a_1a,a_2a,......}
so every element in Aa has a form xa for x in A. Byt now we are intresenting only in a so you pick 1a  =a in Aa where 1 is anutral element in A.

Is it  correct what I am saying thanks.


I know a few strong states, Indiana, California, Texas, for instance. The toughest chapter(s) for 2008 is a California chapter, where one needed to score 42/46 to either make chapter or make state (don't remember which one).

And also that Oklahoma is classified as a "weak state" since, for 2008, no one got above 30/46 at the State level. My brother said that Idaho was a "weak state" but surely you know better than even my brother.

Cool, what's the question?

In general, yeah. It's kind of unfair that they take 4 from every state to nats, regardless of intelligence or population. I happen to be in a weaker state. How/can you figure out which states are "good" and "weak" states? The people going to nats last year (including me) had scores of 33,33,32, and 32

This may be of interest, the team champions from 1984-2008 have come from:
VA, FL, Ca, NY, NY, NC, OH, AL, CA, KS, PA, IN, PA, MA, WI, MA, CA, VA, CA, CA, IL, TX, VA, TX, TX.
And Individual winners from:
TX, OK, FL, TN, IL, PA, AK, MA, OH, KA, IL, KY, PA, WI, MA, WI, IL, NJ, IL, WA, IL, LA, NJ, TX, WA.

(This is sourced from Nationals Yearbooks)

Things like this vary from year to year, in general, it's very hard for small states (such as mine) to field 4 incredibly strong members at Mathcounts Nats, but every once and a while we place fairly high.  We've also had several very strong individuals, 3 black group members, 2 IMOers, one 2nd place Mathcounts.  So, it's not impossible.

I could tell a "strong state" from a "weak state" by their state/national cutoff.

But surely it is easier to make nationals if you live in say Guam (I think they have a team) then Texas or California. I have heard that California has the best students but they somehow fail as a team. This may or may not be true.

[quote="cyao15"]If you live in a "weak" state and make the nationals then will the competition be a little harder because you are from a "weak" state?[/quote]

Not necessarily.  You yourself could be beast.  But in general, yes.

You are right. Oklahoma is a horrible state, I'm from Oklahoma. But this year there will be some people (including me) that will probably get a higher score than a 30 at state.
Last year my friend got first with a 29ish...lol :rotfl:

This is a bit off-topic but does anyone know what national cutoff scores (I mean to get in the top 3) for Virginia have been like in recent years?
And BTW, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=561897297&t=84737]This[/url] has some interesting information regarding relative difficulty.

No score discussing; even if data on one chapter is available on a public website, don't provide the link to that website. As far as the Mathcounts forum is concerned, reality is that there are no public discussions of chapter/school scores until the end of the announcement. Reality may be different in different places ;).

Aren't they talking about previous years though?

Only the one who overtly gave the link to the chapter's data for the current year was in direct violation. Everyone else really were discussing goals for this year or past years' scores.

Hmmm, sorry... I thought our chapter coordinator would know better, though.

I would have to say that states with larger populations GENERALLY do better.
(more population pool)
Also those states with higher immigration rates are probably affected in some way (either better or for worse)
We could corroborate data of illegal immigration, high school dropouts, cost of college tuition, median income, etc. and still cannot predict who will win.

Oh, and MOP/IMO/USAMO =/= Mathcounts...

[quote="james4l"]Oh, and MOP/IMO/USAMO =/= Mathcounts...[/quote]

I'm aware of that, I was just making a point that just because your state is small, or in the case of my state, absolutely minuscule, it doesn't mean that there are no smart people or that the team at nats won't do well.

The only reason as to why a larger state would have a better chance to succeed in MATHCOUNTS/USAMO is only because a larger population pool brings about more intense competition. Still, it cannot explain everything, since VT already had one MATHCOUNTS silver medalist (excuse the expression).

Right, I think large states doubly benefit themselves.  The have a largely population to draw from, so statistically they are already at an advantage, and, in addition, more competition pushes people to work harder.

Is West Virginia, Virginia, and Kentucky considered weak, ok, or strong states?

Virginia is a strong state, but I don't know for the other two.

Kentucky ranks 26th in state pop, WV is 38th (including Puerto Rico, which goes to Nationals).
http://en.wikipedia.org/wiki/US_States_by_population

If I could locate my nats sheets for the past 3 yrs, I could tell you if any of those states made the top 22ish (MC only releases up to that), but I can't find them.  Perhaps someone else who has them?

Isn't Missouri another example of a weak state...? :maybe:

NORTH CAROLINA!!!!!!!!!

achoo. excuse me. :D

i know that the population principle applies here. for example, in some chapters here, there are entire TEAMS who haven't even heard of mathcounts. but, in other more concentrated places (i.e. Ligon, Lincoln, hopefully Hanes), competition is really intense.

recently texas has pwned mathcounts :roll:

Honestly, it's one of the strongest states, along with California. The weak states are usually the smaller ones or the least densely populated ones like Wyoming, New Hampshire, DC, Department of Defense, Guam, Puerto Rico, Virgin Islands, Hawaii, Alaska, etc. But some small states are still strong b/c of large mathematical institutions, such as Massachusetts, New Jersey, etc. And some medium states are...uh...medium difficulty (illinois, nebraska, wisconsin, arizona, etc.)

[quote="funtwo"]texas is a really weak state.  :D[/quote]

kind of funny that u have a longhorn avatar... :P

I'm not sure if State matters more than Chapter. For example where I live, the chapter I went to was the peninsula chapter, where 30 made state, but down south in Santa Clara chapter, I think every year half the people going to state or so got perfect. At nationals you really need to be good to win, but some people who could make state but are in a difficult chapter could get disqualified for state, even if they probably would've made it from any other chapter in the state.

[quote="gauss1181"]But some small states are still strong b/c of large mathematical institutions, such as Massachusetts, New Jersey, etc. [/quote]

Mass and NJ are 15th and 11th in population...
(or maybe he's referring to area?)
I wonder how state qualification happens for larger states.  My state has only 4 chapters, so, each chapter gets at least two, then the next top w/e teams in the whole state qualify.  Then, top indivs whose teams didn't qualify form a team (of 4) from each chapter.

[quote="Quickster94"]My state has only 4 chapters, so, each chapter gets at least two, then the next top w/e teams in the whole state qualify.  Then, top indivs whose teams didn't qualify form a team (of 4) from each chapter.[/quote]

It seemed just as easy to get to state (or nationals) as it is for Oklahoma.

I live in MA, and its not too difficult. However, the chapter i live in (lexington clarke, diamond, mccall) is really tough. For chap all 8 clarke students made states w/ individual scores, and they had 6 ppl in countdown. their lowest score on chap was a 40 i think

[quote="mathlovers1731"]where can i find practice tests, like 1998 and 1999 state tests?[/quote]

Ask in the resources page.


What exactly are you allowed to take?  I looked at one of the threads for an AMC 10 problem, where someone had arrived at the answer by using a protractor.

I know you are not allowed to use a calculator, but what [i]are[/i] you allowed to use?

EDIT: Fixed spelling.

I think basically anything else?

A compass and a ruler would probably be helpful for geometry.

Front cover of the AMC contest booklets, item 5.
See the Teachers' Manual, pages 20, and 21
http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC1012/2010AMC1012A-TM.pdf

Steve Dunbar
MAA Direcotr, American Mathematics Competitions

It says you can use scratch paper, graph paper, rulers, protractors, and erasers.


An empty cilindrical can floats on water with 6 cm of its height above the level of the water. The base of the can has an area of 50 cm2, and from the center of it a block B of lead (density = 11.2 g/mL, weight = 560 g)) is suspended by a small rope. We have something like this:

[code]                        _
          |          |   |
          |          |   |-  h = 6cm
          |          |  _|
----------|          |------------
  Water   |          |
          |          |
          |__________|
                |
                |
               _|_
              |   |
              | B |
              |___|
[/code]

Next, the lead block B is taken out from the base of the can and is placed inside the can:

[code]                        _
          |          |   |
          |          |   |-  h’ = ?
          |          |  _|
----------|    ___   |------------
  Water   |   |   |  |
          |   | B |  |
          |___|___|__|

[/code]

Compute the new height h' of the can that now stays above the level of the water.

Are you sure that's all the information?

It seems that we need to know the height of the cylinder...
I got that h'=5+6/x, where x is the height of the cylinder, but I'm not sure if that's right.

[quote="Kate25"]Are you sure that's all the information?[/quote]

Yes I am sure: we don't have to know the height of the cylinder. It seems that you are not using all the information that is given. Also, your formula "h'=5+6/x" is wrong.

did you do the free body diagram? in both situations?

if I have time, I'll write the solution tomorrow

Yes, in problems like this its common practice to draw the applied forces on each body.

I made it last night. Send me and e-mail and I will send you the solution.
jonysatie@gmail.com

I don't need the solution, because I know perfectly how to solve this problem. I just posted it because I find it interesting and perhaps other people studying introductory Physics could also like it.

ok ... and yes, it's very interesting, but I have seen better ones.

I wish we had one of these problems when we were studying fluids. It's nice, but simple when you think about it.

[hide="Wrong Solution"]
First, I set up the equation to get the height of the cylinder under water.
$50 \cdot 10^{-3}\cdot 9.8+9.8 \cdot 50 \cdot 10^{-2}\cdot h = 5.488$
This is just equating the force of buoyancy with  the weight of the lead, assuming the can's weight is  negligible.
Therefore, the height under the water is $1.02 m$ That's a lot, so it's possible I messed up somewhere.
This is the more difficult problem of the two, but it gives the total can height as 1.08 m.

For the answer, I set up the equation
$5.488 = 50 \cdot 10^{-2}\cdot 9.8 \cdot h'$
$h' = 1.12 m$ This is larger than the previous answer, so that checks. However, it is also larger than the total can height, so the can sinks.

I think that my order of magnitude may be incorrect somewhere. Can someone please verify my answer?
[/hide]

In the solution below I did not take the mass of the can into account,
which is bad since nothing is said about its weight, but it should not be too hard to
add it to the solution, perhaps I'll do it later.
[hide="my try =)"]
[hide="notations"]
Let the total height of can be $x$ and its bottom area $A$.
Also $\rho_{W}$ is water density and $\rho_{B}$ density of lead.
Accordingly $V_{B}$ is volume of $B$ $\ldots$
[/hide]

Archimedes law gives the buoyancy to be (force)
$\underbrace{V_{B}\rho_{water}g}_{lead thing}+\underbrace{(x-h)Ag\rho_{water}}_{can}=m_{B}g$
solving for $x$ yields
\[x=\frac{m_{B}}{\rho_{W}}\frac{1}{A}\left(1-\frac{\rho_{W}}{\rho_{B}}\right)+h \]
Units are right at least  :lol:

Now the other scenario gives an equation in which $h'$ is unknown but know we know $x$
$(x-h')A\rho_{W}g=m_{B}g$
which gives
$h'=x-\frac{m_{B}}{A\rho_{W}}=\frac{m_{B}}{\rho_{W}}\frac{1}{A}\left(1-\frac{\rho_{W}}{\rho_{B}}\right)+h-\frac{m_{B}}{A\rho_{W}}=h-\frac{m_{B}}{\rho_{W}}\frac{1}{A}\left(\frac{\rho_{W}}{\rho_{B}}\right)$
[/hide]

So It becomes $\approx 5 \rm{cm}$

[hide="with mass of can"]
Just add $m_{can}g$ to $RHS$ and it gives $x=\frac{m_{B}+m_{A}-V_{B}\rho_{W}}{A\rho_{W}}$
now simplify and proceed as above, BUT $m_{A}$ is not given.

[/hide]


logarithms
calculus
integration
inequalities unsolved
inequalities

Let be given a real number $\alpha$.Prove that:
$\alpha>1\Longleftrightarrow \exists k>0$ such that $\sum_{i=2}^n\frac{1}{i (\ln i)^\alpha}<k \forall n$
When $\ln x=\log_e x$

We'll use the contrapositive both ways, and also use implicitly the fact that a sum of positive numbers either converges to a finite number or diverges to infinity.

If $a\leq 1$, the series diverges by comparison to the series $\sum \frac{1}{i\ln i}$ (think integral test or condensation test).

Conversely, if the sum on the left hand side diverges, then $a\leq 1$, since
\[ \sum^{\infty}_{i=1}2^i\cdot\frac{1}{2^i(\log 2^i)^{\alpha}}=\frac{1}{(\log 2)^{\alpha}}\sum^{\infty}_{i=1}\frac{1}{i^\alpha}  \]
and we apply the condensation test again.


Determine the least possible value of the natural number $n$ such that $n!$ ends in exactly $1987$ zeros.

As there are always more factors of 2 than 5 in $n!$ ($n \geq 2$), we need to find $n!$ with exactly $1987$ factors of 5

Now, amongst integers $\leq 8000$, there are-

$1600$ multiples of $5$
$320$ multiples of $5^2=25$
$64$ multiples of $5^3=125$
$12$ multiples of $5^4=625$
$2$ multiples of $5^5=3215$.

This totals 1998, which is 11 too many.

So we have $n$ slightly less than 8000- if we reduce it to $7960$, then we have one fewer multiple of 125, two fewer multiples of 25 and 8 fewer multiples of 5, giving us $1987$ zeros on the end of $7960!$.

We can't reduce this any further, because $7959!$ only has $1986$ zeros on the end, so the answer is

$\boxed{7960}$

we must find n such that
$\sum_{i=1}^{k} [\frac {n}{5^{i}}] =1987$ with $5^{k}\leq n$
use that $[x] \leq x$ we have that $1987\leq \frac {n}{4}.(1- \frac{1}{5^k}) \leq \frac {n-1}{4}$
we have that $n\geq 7949$ then easy find n=7960.

power of 2(say a) in n! is always less than or equal to power of 5(say b) in n!

and 2*5=10 so the number n! will end with exactly b many zeros.

so we need to find n least such that $\ s(n)=[\frac{n}{5}]+[\frac{n}{5^{2}}]+.....=1987$ if at all such n exist.

but then see if $\ n=5^{6},s(n)=\frac{5^{6}-1}{4}=3906 $
and $\ n=5^{5},s(n)=\frac{5^{5}-1}{4}=781 $
so if such n exists it is $\ 5^{5} < n < 5^{6} $
so let we take base 5 expansion for n to get $\ n=a_{5}5^{5}+a_{4}5^{4}+...a_{0} $ where $\  0\leq a_{i} \leq 4 $
then see
$\ s(n)=781a_{5}+156a_{4}+31a_{3}+6a_{2}+a_{1} =1987 $ if possible
to make n least we try to keep a_{i} 's with the larger i's as small as possible.

see a_{5} is at least 1. if we set a_{5}=1 then s(n) is at max 1557 so a_{5} have to be atleast 2.

for a_{5}=2 we see $\ 156a_{4}+31a_{3}+6a_{2}+a_{1} =425 $
see 156*3=468>425 so a_{4} have to be less than 3 but if a_{4}=1 then see
s(n) can be atmost 308 so a_{4}=2
so $ \ 31a_{3}+6a_{2}+a_{1} =113 $
then by similar arguement $ \ a_{3}=3$
so $\ 6a_{2}+a_{1} =20 $
see by same arguement $\ a_{2}=3 $
then $\ a_{1}=2 $
as we want the least no so we set $\ a_{0}=0$
so the desired no is  $\ n=2.5^{5}+2.5^{4}+3.5^{3}+3.5^{2}+2.5+0=7960 $
$ \ n=7960$

[hide="A106"]Note that the number of zeroes at the end of $n!$ is an increasing function. The number of zeroes at the end of $n!$ is $\lfloor frac{n}{5} \floor +\lfloor \frac{n}{5^{2}} \rfloor+..... \approx \frac{\frac{n}{5}}{\frac{4}{5}}$. Therefore, $n$ is about $7948$. Note that $7948$ contains $1589$ $5$s, $317$ additional from $5^2$, $63, 12, 2$ for the rest. This totals $1983$ zeroes. Increasing to $7950$ adds an additional $2$ ($5^2||7950)$, and it is clear that we need 2 more, the first from 7955 and the final from 7960. Therefore, the smallest such number is $7960$.[/hide]


number theory
greatest common divisor
algorithm
algebra
polynomial
Euclidean algorithm

The numbers in the sequence 101, 104, 109, 116, $\dots$ are of the form $a_n = 100 + n^2$, where $n = 1$, 2, 3, $\dots$.  For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$.  Find the maximum value of $d_n$ as $n$ ranges through the positive integers.

[hide="Solution - incomplete"]
After experimenting for a bit, we can confidently make the conjecture that for sequences $x_n$, $\{k+1^2,k+2^2,k+3^2,\ldots\}$, the maximum value of the greatest common divisor of consecutive terms is $\text{gcd}(x_{2k},x_{2k+1})=4k+1$. This suggests that when $k=100$, our answer will be $401$. But how can we prove this conjecture? :?
[/hide]

That is, of course, the right answer.

Here's how I did it

[hide]Solution:
Let $d$ be the divisor of two consecutive terms.
$d| n^2+100$ and $d| n^2+2n+101$.
Therefore, taking the difference, $d| 2n+1$
Squaring, $d| 4n^2+4n+1$

Since $d| n^2+100$, $d| 4n^2+400$

Since $d| 4n^2+400$ and $d| 4n^2+4n+1$, we subtract to achieve $d| 399-4n$

Since $d| 2n+1$, multiply by 2 to get $d| 4n+2$

Since $d| 4n+2$ and $d| 399-4n$, we add to achieve $d| 401$

This means d is a factor of $401$. Since $401$ is prime, d must always be either $1$ or $401$.

In order for $401$ to work, we must find an $n$ such that $401| 2n+1$

Our first instinct is to try $401=2n+1$, which yeilds $n=200$. This works, so the answer is $\boxed{d=401}$
[/hide]

that is in essence the same solution as the one above it...

wow, the solution in Zeits is way longer than that, i suppose he was trying to prove a point though...

:? What's the Eucledian Algorithm?   :fool:

did you guys try out that euclidean algorithm java app? it is funny watching it take 20 steps to figure out that the gcd of two numbers is two.

But if you actually did the polynomial division between the $n$ and $n+1$ terms, you would get the same answer.

I don't understand how repeated subtraction could get rid of the squared terms.

[quote="mathmanman"]What about googling?  :roll:
http://www.google.com/search?q=euclidean+algorithm[/quote]

Yeah! Google loves you

Ah, I see, he taking the remainder of one divided by the other is the same as subtracting repeatedly.

[quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote]

and u did too

[quote="Pineappleperson"][/quote][quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote][quote="Pineappleperson"]

and u did too[/quote]
thank you

[quote="junggi"][/quote][quote="Pineappleperson"][/quote][quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote][quote="Pineappleperson"]

and u did too[/quote][quote="junggi"]
thank you[/quote]

huh

??????

[quote="Elemennop"][/quote][quote="junggi"][/quote][quote="Pineappleperson"][/quote][quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote][quote="Pineappleperson"]

and u did too[/quote][quote="junggi"]
thank you[/quote][quote="Elemennop"]

huh

??????[/quote]
and you just spammed

[quote="chess64"][/quote][quote="Elemennop"][/quote][quote="junggi"][/quote][quote="Pineappleperson"][/quote][quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote][quote="Pineappleperson"]

and u did too[/quote][quote="junggi"]
thank you[/quote][quote="Elemennop"]

huh

??????[/quote][quote="chess64"]
and you just spammed[/quote]

I dont get it

[quote="Elemennop"][/quote][quote="chess64"][/quote][quote="Elemennop"][/quote][quote="junggi"][/quote][quote="Pineappleperson"][/quote][quote="junggi"][quote="Pineappleperson"][quote="pkerichang"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.

This is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]

and we learned it in kindergarden in china.[/quote]
and you just spammed[/quote][quote="Pineappleperson"]

and u did too[/quote][quote="junggi"]
thank you[/quote][quote="Elemennop"]

huh

??????[/quote][quote="chess64"]
and you just spammed[/quote][quote="Elemennop"]

I dont get it[/quote]

you're supposed to say 'and u did too'

so i'll just assume you said that

thank you

[quote="Elemennop"][quote="matt276eagles"][quote="Elemennop"]Let $a_{2k} = 100+4k^2$, and $a_{2k+1}=4k^2+4k+1$. This notation is fine because any two consecutive terms can be expressed as the nth even and nth odd term. By the Euclidean Algorithm, we have

\begin{eqnarray*}(4n^2+100, 4n^2+4n+101) &=& (4n^2+100, 4n+1)\\ &=&(100-n, 4n+1)\\ &=&(3n+101, 4n+1)\\ &=&(3n+101, n-100)\\ &=&(401, n-100)\end{eqnarray*}

So our maximum GCD possible is $401$, the answer. The GCD occurs between the $1002$ and $1003$ terms.[/quote]

A few questions:

Why did you choose to use $2k$ and $2k+1$ rather than simply $k$ and $k+1$?  Also, doesn't the notation assume that the first of the two consecutive integers is even?  It seems to me that the notation overlooks half of the possibilities.[/quote]

I chose $2k$ and $2k+1$ because I already tried $k$ and $k+1$ in my head and it got nowhere (ie. you get stuck with fractions/unremovable variable). You can try it yourself.

Also, it doesn't overlook half the possibilities because the algorithm is commutative, ie. it doesn't matter which one is first. We could just as well have done the odd term first and we would have arrived at the same answer.[/quote]

Correct me if I'm wrong, but I believe it does overlook half the possibilities.  Your expression assumes that the odd term is one more than the even term.  So (2,3), (4,5), (6,7) fit your expression but pairs like (3,4) and (5,6) do not.  Luckily this doesn't really affect the answer because:

\begin{eqnarray*}(4n^{2}+100, 4n^{2}-4n+101) &=& (4n^{2}+100, -4n+1)\\ &=&(100+n, -4n+1)\\ &=&(100 + n, 401)\end{eqnarray*}

 and n cannot be negative.

I have a very similar solution and it's probably worst but oh well this thread seems pretty spammed anyways :P

[hide="solution"]

We apply EA To $(100+n^2, 100+n^2+2n+1) = (100+n^2, 2n+1)$ Note that $2$ is coprime to $2n+1$ meaning that $(100+n^2,2n+1) = (200+2n^2,2n+1)$

Applying EA again $(200+2n^2 - 2n^2 - n ,2n+1) = (200-n,2n+1)$ Again, we see that $2$ is coprime to $2n+1$ and again we have $(200-n,2n+1) = (400-2n,2n+1)$

Applying EA again we get that $(400-2n,2n+1) = (400-2n+2n+1, 2n+1) = (401, 2n+1)$ which is obviously maxed at $401$. To test if this can work we set $n=200$.

We see that $100+200^2 = 40100$ which is obviously divisible by $401$. We use EA again to check for $100+201^2$ as it is a multiple of 401 if and only if $100+201^2 - 100-200^2$ is a multiple of $401$. But we see that this is the same as $201^2 - 200^2 = (201+200)(201-200) = 401$. Thus $100+201^2$ is also a multiple of $401$.

Thus the maximum is $\boxed{401}$

[/hide]

INFORMAL SOLUTION
Let $a_n=1+n^2$ instead. Then, $a_n=2,\mathbf{5,10},17,26,37,50,65,82,101,122...$, and we speculate that $\gcd(a_n,a_{n+1})$ is $5$.

Now, instead let $a_n=2+n^2$.Then, $a_n=3,6,11,\mathbf{18,27},38,51,66,83,102,123...$, and we speculate that $\gcd(a_n,a_{n+1})$ is $9$.

Now, instead let $a_n=3+n^2$.Then, $a_n=4,7,12,19,28,\mathbf{39,52},67,84,103,124...$, and we speculate that $\gcd(a_n,a_{n+1})$ is $13$.

ow, instead let $a_n=4+n^2$.Then, $a_n=5,8,13,20,29,40,53,\mathbf{68,85},104,125...$, and we speculate that $\gcd(a_n,a_{n+1})$ is $17$.

We speculate that that if $a_n = 100 + n^2$, then the maximum $\gcd(a_n,a_{n+1})$ is $\boxed{401}$.

FORMAL SOLUTION
We need to find the maximum value of $\gcd(100 + n^2,100+n^2+2n+1)$. By the Euclidean Algorithm, we can re-write it as $\gcd(2n+1, 100+n^2)$ or $\gcd(2n+1, 100+n^2-100(2n+1))$, which is simplified to $\gcd(2n+1, n^2-200n)$. Noting that $\gcd(n,2n+1)=1$, we can re-write our answer as $\gcd(2n+1, n-200)$, $\gcd(2n+1-2(n-200), n-200)$ or $\gcd(401,n-200)$. Hence, the maximum of $\gcd(100 + n^2,100+n^2+2n+1)$ is $\boxed{401}$.


which is [b]not[/b] a stimulatory plant  hormone?

zeatin
IAA
GA3
ethylene
gibberellins

im thinking is IAA, but not really sure about it...
after reading the other hormone plants i got that conclusion does someone knows more about hormone plants?
thank's

I'd guess GA3 by POE
because
zeatin is the plant hormone for Cytokinine, IAA is the plant hormone for Auxin, ethylene is the gas produced by plants for fruit ripening and gibberlins wake seeds from dormancy.

Indolacetic acid (IAA) was the first auxin discovered in plants.  I've never heard of zeatin (not in campbell's), but I'll take agolsme's word that it's a cytokinin.


trigonometry
function
algebra unsolved
algebra

If $n\in N*$ prove that

$0< \tan (\pi\sqrt {9n^2+n})< \dfrac{1}{\sqrt 3}$

Nice indeed, and easy too :D

Here's my solution:

$\dfrac{1}{\sqrt{3}}=tan(\dfrac{\pi}{6})$, and because the period of the tangent function is $\pi$ and the function is increasing over the period, we need to prove that \[0<\{\sqrt{9n^2+n}\}<\frac{1}{6},\]

where $\{x\}=x-[x]$.

The smallest integer less than $\sqrt{9n^2+n}$ is clearly $\sqrt{9n^2}=3n$. Then all we have to prove is

\[\sqrt{9n^2+n}-3n<\frac{1}{6}    \Leftrightarrow\]
\[\Leftrightarrow    9n^2+n<9n^2+n+\frac{1}{6}\]which is true.

[quote="portoseb"]
 and because the period of the tangent function is $\pi$ and the function is increasing over the period, we need to prove that \[ 0<\{\sqrt{9n^2+n}\}<\dfrac{1}{6}, \]
[/quote]

There is a theorem which says this?

Well, we need to prove that

$\tan(\pi\cdot 0)< \tan (\pi\cdot\sqrt {9n^2+n})< \tan (\pi \cdot\dfrac{1}{6})$.

And since $\tan \pi a+b=\tan b$, if $a \in Z$ and $0\leq b<\pi$, we have

$\tan(\pi\cdot 0)< \tan (\pi\cdot\sqrt {9n^2+n})< \tan (\pi \cdot\dfrac{1}{6}) \Leftrightarrow$

$\Leftrightarrow \tan(\pi\cdot 0)< \tan (\pi\cdot\{\sqrt {9n^2+n}\})< \tan (\pi \cdot\dfrac{1}{6}) \Leftrightarrow$

$\Leftrightarrow 0<\{\sqrt{9n^2+n}\}<\dfrac{1}{6}$, because $\tan x$ is increasing over $[0, \dfrac{\pi}{6}]$ and it is also a bijection over $[0, \pi)$. We are allowed to analyze only the interval $[0, \pi)$ because we applied the above formula: $\tan \pi a+b=\tan b$.


Solve in integers $x^2(y-1)+y^2(x-1)=1$.
In the book, An Introduction to Diophanitine Equations, the author makes the substitutions $x=u+1$ and $y=v+1$, thus the new equation is $(u+1)^2v+(v+1)^2u=1$.
Since the author wants to solve this via decomposition, he says that the equation is equivilent to $(u+v+4)(uv+1)=5$.
Can someone explain how you would know to add 4 to both sides, and somehow factor it?
Since me calculator (TI-89) was able to factor the equation after I added 4, I know there must be some logical way.
Thanks.

I'm basically asking, how would one know to proceede in that direction from $(u+1)^2v+(v+1)^2u=1$.

[quote="mna851"]Are you asking about the factoring technique of completing the square?[/quote]

This isn't completing the square....

$\displaystyle (u+1)^2v+(v+1)^2u=1$
$\displaystyle uv(u+v) + 4uv + (u+v)=1$
$\displaystyle (u+v)(uv+1) + 4uv=1$

by this point we know that we have to add something, m,  to factor it with 4uv, either by u+v or uv+1. since 4uv is a multiplication, u+v doesnt seem to work. SO, uv+1 is the factor we are going to factorize 4uv+m with.

$4uv+m = n(uv+1)$
$n=4, m = 4$

therefore, we add four to both sides...

$\displaystyle (u+v+4)(uv+1) = 5$

i hope that kinda explained the concept behind adding numbers to factor.... just like you, i do not know what to add until i factor things out and organize in an according manner.


calculus
integration
trigonometry
logarithms
function
calculus computations

i used brute force to solve it... with substitution $x = \tan^{2}t$.

$\int \frac{x^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}}\, dx$
$= \sqrt{x(x+1)}-\ln ( \sqrt{x}+\sqrt{x+1})+C$

Me too, but it's not so nasty

With trig. substitution you got $2\int{\sec u\tan^{2}udu}= 2\int{\sec^{3}udu}-2\int{\sec udu}$

That's not a nasty integral but with this you got it.

[hide="..."]
I prefer $u$  :rotfl: [/hide]

[quote="Extremal"]I use next substitution $x=sh^{2}(y)$ then $\sqrt{1+x}$=$ch(y)$
and $dsh^{2}(y)=2sh(y)(ch(y)dy$ and our integral is equivalent
$2\int(sh^{2}(y))dy$ which is simpl to find[/quote]

What's $sh(y)$ and $ch(y)$?

Those are hyperbolic functions- the standard abbreviations in the USA are $\sinh$ and $\cosh$.

Ah, I just hadn't seen that notation before.  That's amazing that there is a different notation for them... shouldn't stuff like that be standardized?

Making standards work worldwide across many languages is hard. Following this forum will show you European variations on many things that you thought were standard.

sometimes we use $sinhx$ but it is long and therefore we use $shx$ :lol:


[quote="kam"]Let $ x $ and $y$ be a positive naturel numbers such that: $ 2x^2-1=y{}^1{}^5$
Prove that $5|x$[/quote]
Are you sure?
What about $(x,y)=(1,1)$ ?  ;)

Yes there's a mistake somewhere. I tried to solve it but I only get $x\equiv 1\left( %Error. "func" is a bad command.
{mod}5\right) $ or $x\equiv 4\left( %Error. "func" is a bad command.
{mod}5\right) $  :?
Can you check it again please ?  ;)

well this is the from this years russian olympiad  :D

then it was already solved by Harazi!  :D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=5&t=35316]solution[/url]

it was posted before as ARO 2005.


vector
function
linear algebra
matrix
analytic geometry
calculus
calculus computations

I want to calculate $ \nabla\times[\vec{F}(r)]$ and $ \nabla^2[\vec{F}(r)]$ where F if a function that depends of r, and $ r \equal{} \sqrt {x^2 \plus{} y^2 \plus{} z^2}$ (the magnitude, what I want to say is that it depends of x, y and z)

I know that...
$ 1)\nabla \times \vec A \equal{} \left|\begin{matrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\
 \\
{\frac {\partial}{\partial x}} & {\frac {\partial}{\partial y}} & {\frac {\partial}{\partial z}} \\
 \\
A_x & A_y & A_z \end{matrix}\right|$
$ 2)\nabla^2\vec A \equal{} \nabla(\nabla\cdot\vec A) \minus{} \nabla\times(\nabla\times\vec A)$

My attempt to a solution
$ \left( \frac {\partial}{\partial y}\vec F_{z}(r) \minus{} \frac {\partial}{\partial z}\vec F_{y}(r)\right)\hat{i} \plus{} \left( \frac {\partial}{\partial x}\vec F_{z}(r) \minus{} \frac {\partial}{\partial z}\vec F_{x}(r)\right)\hat{j} \plus{} \left( \frac {\partial}{\partial x}\vec F_{y}(r) \minus{} \frac {\partial}{\partial y}\vec F_{x}(r)\right)\hat{k}$

I need help calculating the partials I tried this:
$ \frac {1}{r}\left(y \minus{} z \right) \plus{} \frac {1}{r}\left(x \minus{} z \right) \plus{} \frac {1}{r}\left(x \minus{} y \right) \Rightarrow \frac {2x \minus{} 2z}{r}$ What happened to the i, j, k? I don't know =P! I suppose I'm wrong for vanishing them...

When I'm trying to calculate the vectorial laplacian I get stuck when I have to calculate $ \nabla\times\vec {F}(r)$ so...

edit:
$ \vec{F}(r)$ is an arbitrary vector function in cartesian form that depends of x, y and z. Maybe I was wrong to say that $ r \equal{}  \sqrt{x^2\plus{}y^2\plus{}z^2}$ maybe $ \left( \frac{\partial}{\partial y}\vec F_{z}(r)\minus{}\frac{\partial}{\partial z}\vec F_{y}(r)\right)\hat{i} \plus{} \left( \frac{\partial}{\partial x}\vec F_{z}(r)\minus{}\frac{\partial}{\partial z}\vec F_{x}(r)\right)\hat{j}
\plus{} \left( \frac{\partial}{\partial x}\vec F_{y}(r)\minus{}\frac{\partial}{\partial y}\vec F_{x}(r)\right)\hat{k}$ could be the general asnwer to calculate the curl of any function that depends of r, for example: $ \left( \frac{r^2\sqrt{csc^2(r)}}{ ln (3r)}\right)\hat i \plus{} \left( 3\sqrt[5]{r^2}\cdot e^{3r^2}\right)\hat j \plus{} \left( \frac{6r^3}{cot^3(r)}\right) \hat k$ (or any other)

Wouldn't it be easier, using spherical coordinates  ?

I'm not very familiar with spherical coordinates, it's easier? i found the gradient, the divergence and the scalar laplacian for that type of functions, and I need to find the curl before I can find the vector laplacian...


LaTeX
linear algebra
matrix
linear algebra unsolved

Question 1:
Proove that if λ is an eigenvalue of [A], then 1/λ is an eigenvalue of [A]T

Question 2
Proove that a square matrices [A] and [A]T have the same Eigenvalues.

Question 3:
Show that |det(A)| is the product of the absolute values of the eigenvalues of
[A]

Question 4:
Let A and B be nonsingular nxn matrices. show that AB^-1 and B^-1A have the same eigenvalues.

I am not asking for complete solutions to the problems, it will suffice if you can point me to some theorem or give a clue that will help toward finding the solution.
I am new to the forum and I don't know how to format the mathematical notations, if there a reference file I can consult?

Use the definition to find the answers to your questions.

And $ \text{\LaTeX}$ is the typesetting language for math on this forum. You can click the "Latex Help" button on top of this forum, if you are new to it.

And shouldn't $ A^T$ be $ A^{-1}$ in Question 1?

Hints/stronger statements:
#1: If $ Av\equal{}\lambda v$, $ A^{\minus{}1}v\equal{}\frac1{\lambda}v$
This works over an arbitrary field.

#2: If $ (A\minus{}\lambda I)v\equal{}0$, $ v^T(A^T\minus{}\lambda I)\equal{}0$. Use the various rank/nullity theorems to show that the left and right eigenspaces of $ A^T$ have the same dimension. This implies that $ A$ and $ A^T$ have the same eigenvalues, with the same geometric multiplicity.
This also works over an arbitrary field.

#3: Drop the absolute values, and count with algebraic multiplicity. This is obviously true if $ A$ is upper triangular. Over an algebraically closed field, every matrix is similar to an upper triangular matrix; for $ \mathbb{C}$, this is Schur's lemma.

#4: $ AB^{\minus{}1}$ and $ B^{\minus{}1}A$ are similar. This works over an arbitrary field.

Why does #3 require an algebraically closed field, where the others don't? #3 needs the existence of a full list of eigenvalues, while the others are true even if there are no eigenvalues at all.


Did anyone take the 3 hour 7 question proof test of the MAML level two yesterday?

I couldn't figure out like a third of the questions, but I'm hoping i get partial credit for a lot of the stuff i wrote down... :|

Btw, did anyone get how to do the first two parts of number 6, or the last part of number 7?

I'm guessing that this is the Massachusetts-specific competition, so perhaps this should be moved to the Massachusetts forum. But yeah, I did the first part of number 6 through contradiction, and I did the second part proving that every number had to be of the form x/999, y/990, or z/900


calculus
derivative
function
limit
calculus computations

Show that the function $ f: \mathbf{R}^2\rightarrow \mathbf{R}$ defined by \[ f(x,y) \equal{} \left\{ \begin{array}{cc}\frac{x^3y}{x^4\plus{}y^2}   & (x,y)\neq (0,0)\\
0& (x,y)\equal{}(0,0) \end{array}\right.\]

has $ \nabla_{0,0}f(u)\equal{}0$ for all $ u$ but is not differentiable at $ (0,0)$.

I'm completely stuck on this... for some reason I can't wrap my head around it.

Thanks.

$ f$ is differentiable at a point $ (x,y)$ iff $ \lim_{(h,k)\to(0,0)}\frac1{\sqrt{h^2\plus{}k^2}}(f(x\plus{}h,y\plus{}k)\minus{}f(x,y)\minus{}f_x(x,y)h\minus{}f_y(x,y)k)\equal{}0.$

Visually, that's the statement that the surface that is the graph of $ f$ has a tangent plane at that point.

In this particular case, that would mean that

$ \lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2\plus{}y^2}}\equal{}0.$

Now, look at points on the curve $ y\equal{}x^2.$ For such points,

$ f(x,y)\equal{}\frac{x^3x^2}{x^4\plus{}x^4}\equal{}\frac{x}2.$

For small $ x,$, we have $ \sqrt{x^2\plus{}y^2}\equal{}\sqrt{x^2\plus{}x^4}\approx \sqrt{x^2}\equal{}|x|.$

That means that the limit we were looking at is $ \frac12$ along the first quadrant portion of this curve and $ \minus{}\frac12$ along the second quadrant portion of the curve; either way, that's not zero.

If you approach the origin along any ray, we do have $ \frac{f(x,y)}{\sqrt{x^2\plus{}y^2}}\to 0.$ That's how we got all the directional derivatives to be zero. But the different curve of approach gives a different result. There's no tangent plane.

Suddenly I remember doing this when I was in Multi, but that was long enough ago that I forgot about the "curve" approach (this problem actually is for analysis hw)

Thanks a lot.

Does this geometric view of "differentiable" expand to $ R^n$ at all, or is it too hard to visualize in the higher dimensions?


Not nearly complete... If you have any more details, please post them here. I only have personal experience with the Montgomery County Middle School Math League and the Univ. of MD competitions. This is a list of competitions exclusive to maryland, so Mathcounts and AMC's and MAO and Mandelbrot, etc. are implied.

Name: Montgomery County Middle School Math League
Since: ?
Type: Individual & team event for middle schools. Used to select ARML.
When: On a wednesday, 4 meets throughout the year
Where: Random middle schools
Eligibility: 1 team of 6 middle schoolers, plus 2 secondary teams. Chosen by coach.
Structure: 6 Individual short answers in 24 minutes. 6 team questions in 20 minutes. 2 relay questions. (questions where one person's result feeds into the next person)
Information and contact: ?

Name: Montgomery County Interscholastic Mathematics League.
Type: Used to select ARML
Elegibililty: High schoolers chosen by coach
(No Details)

Name: Howard County Math League
Since: 1978
Type: Individual & team event for high schools. Used to select ARML.
When: On a wednesday, 8 meets throughout the year
Where: Random high schools
Eligibility: 2 teams of high schoolers. Chosen by coach
Structure: 6 Individual short answers in 30 minutes. 5 team questions in 15 minutes.
Information and contact: http://www.howard.k12.md.us/math/hcml/hcml.html

Name: Univ. of MD math competition: Part I
Since: 1979
Type: Individual event for high schoolers.
When: October 10 this year
Where: Local high schools
Eligibility: Anyone not graduated from high school
Structure: 25 AMC10 style questions, 75 minutes
Information and contact: http://www.math.umd.edu/highschool/mathcomp/

Name: Univ. of MD math competition: Part II
Since: 1979
Type: Individual event for high schoolers.
When: December 3 this year
Where: Local high schools OR Univ. MD campus
Eligibility: Top scorers (usually around 230) in Part I
Structure: 5 proof style problems, 2 hours
Information and contact: http://www.math.umd.edu/highschool/mathcomp/

There is also a Johns Hopkins Mathematics Competition that is run by some very smart people.

Consider looking into that!  If you have any questions about JHMT, email

mathclub@lists.tripleintegral.org


Even though I'm registered, my name doesn't appear at the bottom of the forum.
I think I am classified "hidden" because I'm not in the registered users nor am I a guest.  How do I become "unhidden"?

Go in your profile (click profile in the links at the top of the forum when you are logged in).  The second line of your preferences sets your hidden status - you can change it there.


If $x,y$ are reals prove that
$4(x^2+xy+y^2)^3-27x^2y^2(x+y)^2\geq 0$

Together your solution is not true because $x,y$ can be negatives , so your squareroots may not exists ;)


The time on a digital clock is 5:55. How many minutes will pass before the clock next shows a time with all digits identical?

[hide]11:11 is the next time.  So there are 5 minutes until 6:00, then 5 hours until 11:00, and finally 11 minutes until 11:11.  So the answer is 5+300+11=[b]316[/b][/hide]

[hide]Since the next time the digits will align is 11:11 (there's no such 6:66, etc.) the time passed will be 316 minutes. From 5:55 to 10:55 is 5 hours or 300 minutes. Add 16 minutes to get to 11:11 to get a total of 316 minutes elapsed from 5:55 to 11:11. (Whew, my first post!)[/hide]


I jusr caome up with a difference equation :

$x_{k+1}=(3k+5)x_{k}+5$ with $x_{0}=\frac{1}{2}$

Can anyone tell me how to solve.

In general, how could we solove $x_{k+1}=f(k)x_{k}+g(k)$?

[quote="bchui"]In general, how could we solove $x_{k+1}=f(k)x_{k}+g(k)$?[/quote]
Consider the special case $f(k)\equiv 1$: you want a formula for the partial sums of a given series. Too much to ask for.

[quote]It is not known how to solve a general recurrence equation to produce an explicit form for the terms of the recursive sequence[/quote] from [url=http://mathworld.wolfram.com/RecursiveSequence.html]Wikipedia[/url]

Of course, some ad hoc approach may work for your sequence. For example, let $b_{k}=\prod_{j=0}^{k-1}(3j+5)$: then $\frac{x_{k+1}}{b_{k+1}}=\frac{x_{k}}{b_{k}}+\frac{5}{b_{k+1}}$.  Therefore, $x_{k}/b_{k}$ can be expressed in terms of the partial sums of $\sum b_{k}^{-1}$; I don't know if they have a closed form.

Here is the solution:

The solution of  ${ x_{k+1}=A_{k}x_{k}+b_{k}~~,~~x_{0}=z}$  is given by

${ x_{k}=\left\lgroup \prod^{k-1}_{i=0}A_{i}\right\rgroup z+\sum_{i=0}^{k-1}\prod_{j=i}^{k-1}A_{j}b_{i}}$    :)

The problem remains on how to compute  ${\prod_{i=0}^{m}(i+\lambda) }$ and ${\sum_{i=0}^{m}\prod_{j=i}^{m}(j+\lambda) }$


All digits of a number $ N$ are equal to $ 1$. Prove that if $ 7$ divides $ N$, then so
does $ 13$.

Write $ N\equal{}\sum_{k\equal{}0}^n10^k$ for some $ n\in\mathbb{N}\cup\{0\}$. Now, $ 0\equiv N\bmod{7}$ implies $ 3^{n\plus{}1}\equiv 1\bmod{7}\Longrightarrow 6|n\plus{}1$. Moreover, we have $ N\equiv\frac{1\minus{}(\minus{}3)^{n\plus{}1}}{4}\equiv 0\bmod{13}$, the last step, because $ (\minus{}3)^6\equiv 1\bmod{13}$.

my solution was completley different (i would be suspicious)

Using the divisibility rule for 7 , we will construct a multiple of 7

starting with one 1 and continuing all the way up to six ones we find that the first multiple is 111111
we can generalize for all the multiples of 7 that; the number of ones will be a multiple of 6.

now, we find that 13 also evenly divides 111111. since it evenly divides 111111, it must also divide all the other numbers in which the number of digits is a multiple of 6(all the digits are 1s)

the ONLY way for 7 to divide N is if the number of digits is a multiple of 6, we found that if the number of digits is a multiple of 6, 13 divides N so, if 7divides N, so does 13.

could you PLZ send feedback on this solution

3q!

Well what you did wasn't exactly rigorous since you only computed the values up to 111111. To generalize you would need to construct the sequence 1, 11, 111,... and take it mod 7, then mod 13, or something similar.

Uhh, repunit with $ n$ ones is equal to $ \frac{10^{n}\minus{}1}{9}$. Consider only the top since the divided by 9 won't matter... $ 10^n\minus{}1\equiv 0 \pmod 7 \Rightarrow 6|n$. Then show $ 10^{6k}\minus{}1\equiv 0 \pmod {13}$, which is pretty easy since $ 10^6 \equiv 1 \pmod {13}$...


Is there any way to draw border around the page using latex?

what should i do?

You posted this message twice, look at your first message.


[hide]$t$ being the constant term, we know that $-t$ is the product of the roots. We also know that $abc=11$ and $a+b+c=-3$, so $t=-(a+b)(b+c)(c+a)=(3+a)(3+b)(3+c)$ after substituting in from $a+b+c$, so after that we foil everything out, we have $27+9c+9b+9a+3bc+3ac+3ab+abc\Longrightarrow 27+9(a+b+c)+3(ac+bc+ab)+abc$, and substituting with everything we know, we have $27+9(-3)+3(4)+11=\boxed{23}$. Note that $ab+bc+ac$ is always equal to the third term in a polynomial.[/hide]

[hide="Answer"]$-t=(a+b)(b+c)(c+a)=ab(a+b)+ac(a+c)+bc(b+c)+2abc$
From the first equation, $abc=11$, $ab+ac+bc=4$, and $a+b+c=-3$, so then $a+b=-(c+3)$, $a+c=-(b+3)$, and $b+c=-(a+3)$. Substituting, we have $ab(c+3)+ac(b+3)+bc(a+3)-2abc=3(ab+ac+bc)+abc=3(4)+11=\boxed{23}$.[/hide]

Let $ y\equal{}b\plus{}c\equal{}a\plus{}b\plus{}c\minus{}a\equal{}\minus{}3\minus{}x$, so substitute $ x\equal{}\minus{}3\minus{}y$. Then $ \minus{}p(\minus{}3\minus{}y)\equal{}y^3\plus{}6y\plus{}13y\plus{}23$...so the answer is $ \boxed{023}$.

hmmm.
couldn't you recognize
$ \minus{}(\minus{}3\minus{}a)(\minus{}3\minus{}b)(\minus{}3\minus{}c)$ as evaluating the initial polynomial at $ x \equal{} \minus{}3$ and then negating it. no foil that way.

$ \minus{}t\equal{}2abc\plus{}\sum a^2b \equal{} 2abc \plus{} \sum(a\plus{}b)ab$

$ \equal{}2abc\plus{}\sum(\minus{}3\minus{}c)ab\equal{}2abc\plus{}\sum(\minus{}3ab\minus{}abc)$

$ \equal{}2*11\plus{}(\minus{}3*4\minus{}3*11)\equal{}\minus{}23$

[hide=My sol]

By vietas, $t$ is the negative product of $(a+b)(b+c)(c+a)$. We will neglect the negative for now for simplicity. Expanding $(a+b)(b+c)(c+a)$ we get $2abc+\sum a^2b$. We can cleverly introduce an extra $abc$ term to make it factorable; which gives $(a+b+c)(\sum ab)-abc$. Since we know all of these three factors/terms from vietas on the given cubic, we plug and get $-3(4)-11$. Notice that $t$ is the negative of this by vietas. Thus, the answer is $-(-23)=\boxed{23}$.
[/hide]

@mathwizarddude AIME answers are non negative.

Oops sorry for bump but I felt that the answer should be corrected.

[hide=My Solution]

By Vieta's Formula, we have that:
$$\implies a + b + c = -3$$
$$\implies ab + bc + ca = 4$$
$$\implies abc = 11$$

By Vieta's Formula:
$\implies t = -{(a + b)(b + c)(c + a)} = -{(-3 - c)(-3 - a)(-3 - b)} = (a + 3)(b + 3)(c + 3) = abc + 3(ab + bc + ca) + 9(a + b + c) + 27 = \boxed{23}$
$\blacksquare$

[hide= Solution]
Quick inspection yields that $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c) - abc = -12 - 11 = -23.$ This yields $t=23$.
[/hide]

[quote=joml88]Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a, b,$ and $c,$ and that the roots of $x^3+rx^2+sx+t=0$ are $a+b, b+c,$ and $c+a.$  Find $t.$[/quote]

why was this on my AMSP test :huh:


[quote="Cauchy_Schwartz"]I couldn't find the way to solve this one too fast...

\[A, B, C, D \in \bigcirc(O, R),\] \[AB=a, BC=b, CD=c, AD=d, \]
\[AB+BC+CD+AD=2p,\]
\[AC=2R.\]
Prove that:
\[(p-a)(p-b)(p-c)(p-d)=\frac{(ad+bc)^2}{4}\][/quote]

Actually, the last formula should be

$\left( p-a\right) \left( p-b\right) \left( p-c\right) \left( p-d\right) =\frac{\left( ab+cd\right) ^{2}}{4}$.

It's not very clear to me whether you have solved this or not, so I post my solution:

Since AC = 2R, the segment AC is a diameter of the circle (O, R), and since the points B and D lie on this circle, we have < ABC = 90 and < ADC = 90. Hence, the triangles ABC and ADC are right-angled, and their areas are

$\left[ ABC\right] =\frac{AB\cdot BC}{2}=\frac{ab}{2}$

and

$\left[ ADC\right] =\frac{CD\cdot DA}{2}=\frac{cd}{2}$,

respectively, where [XYZ] denotes the area of a triangle XYZ. Now, the area of the quadrilateral ABCD is

$\left[ ABCD\right] =\left[ ABC\right] +\left[ ADC\right] =\frac{ab}{2}+\frac{cd}{2}=\frac{ab+cd}{2}$;

hence,

$\left[ ABCD\right] ^{2}=\left( \frac{ab+cd}{2}\right) ^{2}=\frac{\left( ab+cd\right) ^{2}}{4}$.

On the other hand, after the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15973]Brahmagupta formula[/url],

$\left[ ABCD\right] ^{2}=\left( p-a\right) \left( p-b\right) \left( p-c\right) \left( p-d\right) $.

Thus,

$\left( p-a\right) \left( p-b\right) \left( p-c\right) \left( p-d\right) =\frac{\left( ab+cd\right) ^{2}}{4}$

as intended to prove.

Nice problem, thanks!

  Darij

I enjoyed the problem too, but when I had to solve it for a countest, I didn't have any clue... :D