Can you say -1 is the recipricol of -1? If not what is the recipricol of -1? Is it undefined?

The reciprocal of x is defined as 1/x, so the only x with undefined reciprocals are -:inf:, 0, and :inf:.

Except that :inf: and -:inf: aren't numbers, so you couldn't take a reciprocal of them anyhow ...

Now, for a related question.  What is the reciprocal of -0?

[hide]pfft.  now, really?[/hide]

[quote="mathfanatic"]Now, for a related question.  What is the reciprocal of -0?[/quote]

Who says there is such a thing as -0?

(I saw the spoiler, by the way.)

There is a -0.  It's equal to 0.  Remember that -x is defined by the equation -x + x = 0.  Because 0 + 0 = 0, that means -0 = 0.

are there any proofs that actually use -0? i cant think of any reason to use it... therefore, no reason to have a conversation about it at all.

It's not hard to show that 0=-0 (in any ring, not just the real numbers or integers or complex numbers or whatever), so there isn't much point in using -0.

There can't be a reciprical of 0 because 1/0 is undefined. (Or you can think of it this way: Since anything times 0 is 0 you can't multiply something to 0 and make it 1)

...

Ok,  there are two things:

1. You revived a thread that has been dead for over a year...that is like one of the greatest sins that an AoPSer (especially one with as many posts as you) can make.

2.  Your post did not add anything new.  You just repeated what Confuted said.

PLEASE don't do these things.  I don't want to discourage usage of this forum...but, honestly, this is not a game to see who can get the most posts.  Don't just post in every thread.  Try to make quality posts.  If you can produce a quality post for just about every thread then you are more than welcome to post them.

Thanks

Thanks joml88.

Like he said, do not revive the old thread.

I'm going to lock this topic for any of not-necessary discussion.

This has been discussed in MC forum by rcv and this applys to same here.


function
quadratics
algebra
domain
quadratic formula

Find the inverse function.

y = 4x - x^2

I replaced x and y.

x = 4y - y^2

I then factored right side.

x = y(4 - y)

Where do I go from there?

Thanks!

$ y \equal{} 4x \minus{} x^2$
By definition $ (f\circ f^{ \minus{} 1})(x) \equal{} x$.

Let $ f^{ \minus{} 1}(x) \equal{} y$.
Then we wish to find $ y$ such that $ 4y \minus{} y^2 \equal{} x$
This is where you are so far.

Assume $ x$ is constant, and then use the quadratic formula to solve for $ y$ in terms of $ x$. I think...

$ f(x)\equal{}4x\minus{}x^2$ is not bijective if the domain of $ f$ is $ \forall x\in\mathbb{R}$. If you want to find the inverse mapping instead, try using the quadratic formula on $ y^2\minus{}4y\plus{}x\equal{}0$.

y=4x-x^2
switch x and y
x=4y-y^2
-x=y^2-4y
complete the square on the right side
4-x=y^2-4y+4
4-x=(y-2)^2

y-2=(4-x)^(1/2)
y=(4-x)^(1/2)+2

but mine has a restriction that x<=4 :(
probably wrong...

and the graph doesnt pass the hoz line test, is there a inverse f ???

You're right about the horizontal line test.  As given, there is no everywhere-defined inverse.  You need to restrict the domain of the original function.


let $ f_n$ be a sequence of Reiman integrable functions on a closed interval $ [a,b]$.  If $ f_n$ converges uniformly to a function $ f$ on $ [a,b]$, show that $ f$ is Reiman integrable.

This is essentially an application of the triangle inequality. Since $ f_n \to f$ uniformly, a step function approximating $ f_n$ well also approximates $ f$ well.

Since $ f_n$ is Riemann integrable, you may find a step function which differs from $ f_n$ by less than $ \epsilon/2$ on all but a set of measure $ <\epsilon$. Now, you can also take $ n$ sufficiently large that $ f_n$ differs from $ f$ by less than $ \epsilon/2$ over the entire interval $ [a, b]$. Now what can you say about the step function (which came from $ f_n$) and the limit function $ f$?


algorithm
modular arithmetic
number theory
prime factorization
relatively prime
number theory proposed

Suppose a number N is a Carmichael number, so

(1) Is there an efficient way of doing this without having to factorize N and proving that p-1 divides n-1 for every prime divisor of n?

(2) If the most efficient method does involve factorizing N, then do we need to know the factorization of n-1 as well?

(3) What is the largest known Carmichael number known to date?

See http://mathworld.wolfram.com/CarmichaelNumber.html .  Erdos worked out a method to calculate large Carmichael numbers.

Thanks. do u have any ideas how to do questions (1) and (2) ?

1)  Do you mean "is there an efficient way of proving that an arbitrary number is a Carmichael number"?  The alternative to prime factorization is to test whether $ a^{n \minus{} 1} \equiv 1 \bmod n$ for every $ a$ relatively prime to $ n$, which is pretty computationally intensive.  I'm not sure if there are other tests.

2)  No.  Calculations can be done with $ n \minus{} 1$ without knowing its prime factorization.

[quote="BanishedTraitor"]When mathematicians say 'large' or 'sufficiently large' what does it actually mean? Is it just 'large' in the intuitive sense?[/quote]

'Sufficiently large' has a very specific meaning.  If a statement is true for sufficiently large $ n$, it means that there exists some integer $ M$ such that the statement is true for all $ n > M$.

'Large,' on the other hand, can mean 'arbitrarily large.'  I believe that is what the MathWorld article means, although I'm not sure.  On the other hand, it can mean a vaguer intuitive notion.

I'm sorry,guys, but i was told that this is not correct. We need to use Pollard rho factorization method in the context of Carmichael numbers. Is anyone familiar with this method?

How to prove that a Carmichael number is squarefree ?
Do use the Korselt's criterion?

If $ p^2|n$, $ p|\phi(n)$, so $ \phi(n)$ can't divide $ n\minus{}1$.

I have another proof for LeoYard:

Suppose $ p^2 | n$, let $ n' \equal{} \frac{n}{p^2}$
Let $ h$ be a primitive root of one mod $ p^2$. By CRT there exist
$ b$ s.t. $ b\equiv h \pmod{p^2}$ and $ b\equiv 1 \pmod {n'}$. If, by absurd,
$ b^{n\minus{}1}\equiv 1 \pmod{n}$ then $ p(p\minus{}1)|(n\minus{}1)$. So this is not the case.
Moreover $ b\equiv 1 \pmod {n'}$ imply $ (b, n) \equal{} 1$. The thesis follows.


If $ a,b,c$ positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$, then

$ \frac{4ab\plus{}3}{1\plus{}c}\plus{}\frac{4bc\plus{}3}{1\plus{}a}\plus{}\frac{4ca\plus{}3}{1\plus{}b}\leq8$

This solution does appear to give the desired result, but a part of it is little weird... ($ \sum_{\text{cyc}} \frac {a^2 \minus{} 2a \plus{} 4}{a \plus{} 1}$ can be greater than 8)

Can anyone point out where is wrong?

[hide="Solution"]$ a \plus{} b \plus{} c \equal{} 1\implies a \plus{} b \equal{} 1 \minus{} c$.

By AM-GM, $ \frac {a \plus{} b}{2}\ge \sqrt {ab}\implies ab\le \frac {(a \plus{} b)^2}{4} \equal{} \frac {c^2 \minus{} 2c \plus{} 1}{4}$ and cyclic.

Applying this result to the inequality, we have $ \frac {4ab \plus{} 3}{1 \plus{} c} \plus{} \frac {4bc \plus{} 3}{1 \plus{} a} \plus{} \frac {4ca \plus{} 3}{1 \plus{} b}\le \frac {a^2 \minus{} 2a \plus{} 4}{a \plus{} 1} \plus{} \frac {b^2 \minus{} 2b \plus{} 4}{b \plus{} 1} \plus{} \frac {c^2 \minus{} 2c \plus{} 4}{c \plus{} 1}$.

It can be easily shown that $ f(x) \equal{} \frac {x^2 \minus{} 2x \plus{} 4}{x \plus{} 1}$ is concave in $ (0,1)$.

Hence, we have $ LHS\le f(a) \plus{} f(b) \plus{} f(c)\le 3f\left(\frac {a \plus{} b \plus{} c}{3}\right) \equal{} 3f\left(\frac {1}{3}\right) \equal{} \frac {31}{4} < 8$, as desired. $ \blacksquare$

Furthermore, the equality never occurs.[/hide]


Ross Mathematics Program
geometry
3D geometry
prism

Assuming the shelf is (roughly) rectangular (probably an inaccurate assumption, but oh well)
The volume of a rectangluar prism is simply area of the base times height. We want to find the height, given the volume and the area of the base:
$80,000 = H * 200,000$
Clearly, H is less than one.
Infact, dividing through by 200,000 gives:
$h = 0.4$ miles thick, on average.

[quote="Anonymous52111"]Good evening guys,

I have found this question from the Internet. If you have successfully answered this question, then would you please give the explanation? Thank you!

The volume of the Ross Ice Shelf above sea level is 80,000 cubic miles. If its area is 200,000 square miles, what is the average thickness of the ice above sea level?[/quote]
Ice shelves aren't usually rectangular, but assuming this one is...

[hide]$A=wl=200000\text{ miles}^{2}$

$V=hwl=80000\text{ miles}^{3}$

$h=\frac{80000\text{ miles}^{3}}{200000\text{ miles}^{2}}=\boxed{\frac{2}{5}\text{ miles}}$[/hide]

It doesn't have to be rectangular. Any 3-D prism's volume is given by $V=bh$.

Perhaps, I am supposed to add multiple choices that can act as hints. Oh well, I'm going to place here. Although, what I did was probably a mistake. I found the answer, but I wasn't sure how to solve this question. For the question, the answer is 2112 feet. So my question is still question. Would you please give an explanation? Thank you!

The volume of the Ross Ice Shelf above sea level is 80,000 cubic miles. If its area is 200,000 square miles, what is the average thickness of the ice above sea level?

A. 880 ft
B. 1056 ft
[u]C. 2112 ft[/u]
D. 2428 ft
E. 2640 ft

We gave you the answer as a decimal part of a mile. Because a mile is 5,280 feet, you want 0.4 of 5280, or $\boxed{2112}$ feet.

Thank you for your help! I'm truly appreciated. :)


I was reading the AoPS vol 1 chapter on probabilities and at the part where the probability of rolling 12 with 3 dice is solved, I am wondering if the solution is incorrect?

The way the book has it is by multiplying the probabilities of rolling a certain number with 2 die by probability of rolling the complement to 12 on the third die, however I think that it has forgotten some cases. This is what the book says:

$P(\text{Rolling 12}) = P(6-6)+P(7-5)+P(8-4)+P(9-3)+P(10-2)+P(11-1)$

But what about $P(5-7)$, $P(4-8)$, $P(3-9)$, etc? Since the sample space of the probability used is $6^{3}= 216$, the permutations of the outcomes. So the above solution only counts certain permutations, but leaves some reversed permutations out.

[quote="breez"]I was reading the AoPS vol 1 chapter on probabilities and at the part where the probability of rolling 12 with 3 dice is solved, I am wondering if the solution is incorrect?

The way the book has it is:

$P(6-6)+P(7-5)+P(8-4)+P(9-3)+P(10-2)+P(11-1)$

But what about $P(5-7)$, $P(4-8)$, $P(3-9)$, etc? Since the sample space of the probability used is 6^3 = 216, the permutations of the outcomes. So the above solution only counts certain permutations, but leaves some reversed permutations out.[/quote]
Well, let's try looking at some combinations in, say, $P(4-8)$. You might have $4, 3, 5$. But this is counted in $P(7-5)$. If you had $4, 6, 2$, this would be counted in $P(10-2)$. In fact, all possible combinations have already been counted in $P(6-6)+P(7-5)+P(8-4)+P(9-3)+P(10-2)+P(11-1)$.

Another way to look at $P(6-6)+P(7-5)+P(8-4)+P(9-3)+P(10-2)+P(11-1)$ is that it goes through all possible cases of what the number the third die will show, which covers every possible situation.


which integers have the following property? If the final digit is deleted, the integer is divisible by the new number.

[hide]Equation to solve is something like :

10A + b = k*A with 0<=b<=9.

So b = A(k-10)

Solutions are :

* k=10, b=0 : every 10-multiple are solutions

* k!=10,b!=0 : then 0< A <= 9

A=1 : b=1,2,..., 9

A=2 : b=2,4,6,8

....

So the trivial solutions are all numbers divisible by 10.

The non trivial solutions (i.e not divisible by 10) are :

10 11 12 13 14 15 16 17 18 19

20 22 24 26 28

30 33 36 39

40 44 48

50 55

60 66

70 77

80 88

90 99

Edited : sorry for poor and confusing spelling[/hide]

wait, explain why they have to be 2 digit numbers, and why there aren't infinite solutions.

He meant to say "non-trivial solutions," that is, those not including all of the infinitely many solutions blahblahblah...blah0, all of which work.


algebra
polynomial
number theory unsolved
number theory

Determine all pairs $(m,n)$ of positive integers such that $a^{n}-1$ is divisible by $m$ for all $a=1,2,...,n$.

First of all, $(m,n)=(m,1)$ works for all positive integers $m$, so let's assume from now on that we're in the non-trivial case $n>1$. Clearly, $m$ must be odd.

Let $p$ be a prime divisor of $m$. We have $p|a^n-1,\ a=\overline{1,n}$, which means that modulo $p$, the polynomials $x^n-1$ and $(x-1)\cdot(x-2)\cdot\ldots\cdot(x-n)\ (*)$ coincide. In particular, $p$ divides the coefficient of $x^{n-1}$ in $(*)$, whih means that $p|n(n+1)$. We cannot have $p|n$, because we get a contradiction from $p|n^n-1$, so $p|n+1$. If $p<n+1,\ x-p$ will appear somewhere in $(*)$, so we'll have $p|p^n-1$, contradiction. We thus have $p=n+1$. In other words, if $n>1$, we must have $m=p^k,\ n=p-1\ (**)$ for some odd prime $p$.

Suppose we're in the situation $(**)$ with $k\ge 2$. We have $p^2|a^{p-1}-1,\ a=\overline{1,p-1}$. On the other hand, it's been shown before on the forum (try searching) that there is a primitive root of $p^2$ among $1,2,\ldots,p-1$ for all odd primes $p$, so we have a contradiction. This means that $(**)$ we must have $k=1$, and the pairs $(m,n),\ n>1$ we're looking for are $(m,n)=(p,p-1)$ for odd primes $p$ (it's clear that every such pair works).


calculus
derivative
real analysis
real analysis unsolved

how would you get the derivative of:

\[ \frac{\partial}{\partial X} E\big(X-E(X)\big)^{2}\] Where $X$ is a random variable, and $E$ is refer to the expectation.

thanx in advanced  :)

I think you need to give the distribution of X as well because there's no explicit formula for the variance for a general distribution, that I know of.


For which values of $n>1$ does there exists distinct positive integers $x,y$ so that:

$\frac{1}{x}+\frac{1}{y}= \frac{1}{n}$

[hide]
rewriting the equation gives (x-n)(y-n)=n^2

letting x,y>n we have x-n,y-n are and 2 factors of n^2. In fact, one can just let x=n^2+n and y=n+1 for every n>1
[/hide]


What is the units digit in the product of all natural numbers
from 1 to 99, inclusive?

we want the units digit of the product of all natural numbers from 1-99 inclusive

we notice that the numbers 10, 20, 30, etc. are within this set, and when any of these is multiplied with a number, the units digit will be 0

therefore, our answer is 0

The product of the natural numbers form 1 to 99 includes 10 so the last digit will be $\boxed{0}$.


conics
ellipse
geometry
geometric transformation
reflection
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13938[/img]

 :(

Draw a circle $ \mathcal P$ with diameter $ V_1V_2$ (the ellipse pedal circle) and a circle $ \mathcal C_1$ with diameter $ PF_1.$ (You could also use a circle $ \mathcal C_2$ with diameter $ PF_2.$) The circles $ \mathcal P, \mathcal C_1$ intersect at $ P_1, P_2.$ $ PP_1, PP_2$ are the ellipse tangents from $ P.$ To find the tangency point $ T_1$ of $ PP_1$ with the ellipse, reflect $ F_1$ in $ PP_1$ into $ G_1.$ $ G_1F_2$ cuts $ PP_1$ at $ T_1.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13947[/img]

 :thumbup:


The volume of a rectangular prism is 52.  What is the least possible surface area it could have?

Masoud Zargar

$A=a\cdot b\cdot c = 52$
$S=2(ab+bc+ca) \ge 6\sqrt[3]{a^2b^2c^2}= 6A^{\frac23} = 12\sqrt[3]{2\cdot 13^2}$
it is equal iff $a=b=c$, so its a cube ...
(simple am-gm)


I was just curious, I try to get as much done in school and on the bus, probably 30 min a day.

I don't do my homework. I think I stopped in 6th grade  :D

No homework currently; it's near the end of the year and there's really nothing to do.

I'm estimating about next year - approximately 2-3 hours of homework? And then an extra 2 for pure studying. AP World History will be probably the bane of my homework though.

Normally I don't do my homework (or at least the trivial ones) until the lunch period of the day it's due. But then again taking three APs next year as a sophmore might not be good for procrastinating.

normally like .5-1 on a good day, and 1-1.5 on a not good day. 2+ is considered unusually bad.

Strangely, the homework load has skyrocketed towards the end of the year, and now i'm like 3+ each day... ugh

I do many of my homeworks in class...:)...

0-30 minutes...Spelling doesn't take that long, and my math teacher rarely assigns homework.

But next year... :|


How many square units are in the area of the triangle in the first quadrant bounded by the axes and the line with equation $ x\plus{}y \equal{} 10$?

This region is a right triangle. Since $ x \plus{} y \equal{} 10$ intersects the $ y$-axis at the point $ (0,10)$ and intersects the $ x$-axis at the point $ (10,0)$, the lengths of the legs are both 10. That means the area of this triangle is $ 10 \cdot 10 \equal{} 100 \implies 100 \div 2 \equal{} \boxed{50}$.

Find the x and y intercepts which form a right triangle. They are obviously (10,0) and (0,10), so the area is $ \frac {10^2}{2} \equal{} \boxed{50}$.

EDIT: Darn beaten by a couple of seconds..

ill answer this in a series of steps.

[hide="THIS IS MY EXPLANATION "]1.first of all we will want the largest of either [b][i]x[/i][/b] or [b][i]y[/i][/b]

2.and the equation that we want is $ x+y=10 $

3.so why not just have $ x=10 y=0 $ and $ x=0 y=10 $

4.then we multiply $ 10*10 $ and then you get 100

5.then we divide 100/2=50

6.[hide="ANSWER"]50[/hide][/hide]

[hide][b][i][u][size=200][color=#0080FF]PLEASE RATE MY POST THANK YOU SO MUCH AND PLEASE POST ME IF I WAS WRONG !!!!!!!!![/color][/size][/u][/i][/b][/hide]

:10: I HAVE 85% CONFIDENCE AND THE OTHER 15% IS FEAR.


geometry
geometric transformation
analytic geometry
graphing lines
slope
calculus
trigonometry

In an effort to control spam, I ask that you post only one time on this thread.  If you want to discuss a specific book, please start a new thread or use private messenging, instant messengers, word of mouth, etc.

That said, I will ask this question:

[b]What is your favorite book? [/b](You may name more than one book...I know it's hard to decide sometimes.  But if you want to add one, use the edit post feature.)

I personaly enjoyed the entire Ender Series (Ender's Shadow was the best of them), and Asimov's Foundation books.  Among more classical ones, Les Miserables was good (although long-winded, for sure!).  And I'm sure to think of more...

I think [i]Concrete Mathematics[/i] by Graham, Knuth, and Patashnik is my favorite. Some others come close though.

Willard Price's Adventures series, Jane Austen's Pride and Prejudice.

First Eagle by Tony Hillerman and

Silk and Cyanide

Martin Eden by Jack London and The Brothers Karamazov by Dostoevsky

Gentlehands, by M. E. Kerr. Quite good, really, despite its title, and suitable for those 8+.

Hmm . . .

I'm a big fan of Swimming Lessons, by Rohinton Mistry.  Flatland is a fun book.  Brave New World is an excellent distopia -- I preferred it to 1984 on several grounds.  And anything by Agatha Christie is fun, if not exactly highly intellectual.  The Foundation Series by Isaac Asimov had interesting ideas, although I didn't love the books that much.

Lord of the Rings?  I mean, it was good, but, how can you love a book where after 20 pages you already know what the end is, but you have to read so much to get there?  I did love the character of Tom Bombadil, though.

Atlas Shrugged is my favorite.

1984 and Brave New World were both good. Crime and Punishment is good; Plato's Republic isn't bad. Everything by Ayn Rand is excellent.

[quote="shobber"]can comeone introduce the Little Prince?[/quote]

It's pretty difficult for myself to do it without certain level of the bias. I just suggest you to read it yourself. In Chinese it is available [url=http://www.fairydream.net/html/littleprince/]here[/url] or [url=http://www2.hkedcity.net/citizen_files/aa/dw/sc5276/public_html/geocities/little_prince/]here[/url]. I don't know, how good the translation is, however...

[quote="shobber"]can comeone introduce the Little Prince?[/quote]It is one of my wife's favorites, but I still find it TOO BORING!

[quote="nonIMO"][quote="djimenez"]It is one of my wife's favorites, but I still find it TOO BORING![/quote]

On one of the local forums about this book I found, that it is mostly liked by women. I think this once again confirms my belief, that they are more romantic creatures than men are.

Was that [b]still[/b] accidental in your post? :D Actually, I've read the book two times so far, the second time being very different from the first one.[/quote]I may give it a try a second time soon! but not now (end of the semester) :D  :D

[quote="shobber"][quote="nonIMO"][quote="djimenez"]It is one of my wife's favorites, but I still find it TOO BORING![/quote]

On one of the local forums about this book I found, that it is mostly liked by women. I think this once again confirms my belief, that they are more romantic creatures than men are.

Was that [b]still[/b] accidental in your post? :D Actually, I've read the book two times so far, the second time being very different from the first one.[/quote]
yes, i think that reading a book for more than two times will teach me lots of things.
but sometimes, if it is a more than 1000 pages book, it will take you lot of patients, like "the Dream of Red Mansion."![/quote]Well, I have never heard about [i]the Dream of Red Mansion[/i], nevertheless, the [i]Little Prince[/i] is, in the version I have, only 75 pages, the page is just a little bigger than half letter size, has big letter size and has some ilustrations :D :P

I really enjoyed reading the Illiad and the Odyssey

[quote="shobber"][quote="nonIMO"][quote="djimenez"]It is one of my wife's favorites, but I still find it TOO BORING![/quote]

On one of the local forums about this book I found, that it is mostly liked by women. I think this once again confirms my belief, that they are more romantic creatures than men are.

Was that [b]still[/b] accidental in your post? :D Actually, I've read the book two times so far, the second time being very different from the first one.[/quote]
yes, i think that reading a book for more than two times will teach me lots of things.
but sometimes, if it is a more than 1000 pages book, it will take you lot of patients, like "the Dream of Red Mansion."![/quote]

Perhaps your taste is opposite of mine one. I hate to read historical books like "Dream of the Red Mansion" since the time, when I failed to complete 1/3 of "Ivanhoe" by Walter Scott. BTW the first my book, which I had read after an alphabet book at age five, had 600 pages, but IMHO number of pages is not the point.

What about Da Vinci's Code. I found it fascinating and read it in just 2 days.

i didnt like Davinci code partly becuase it was against our catholic beliefs.... but i heard it was well written book....

I didn't care much about the religious views exposed, but the book was actually pretty bad. The plot is predictable, and the characters are just flat and stereotypical.

I don't understand why its a bestseller, if the Holy Grail were really buried at the Louvre, why not just go there and dig it up? In other words, i think the ending is bad.

recently finished 1984. best book ever, no doubt.

I thought the Odyssey and the Illiad were excellent books.
Farenheit 451 was good too.

for the old books, it's gotta be the iliad and the odessey.
for the new books, it is definitely the Hitchhiker's guide to the Galaxy.....


algebra
polynomial
modular arithmetic
function

Let $ f(x)$ be a polynomial,all coefficients of which are integers.When $ x$ is an integer,prove that some values of $ f(x)$ aren't primes.

[quote="nateharman1234"][hide]Let the last coefficient (ie the one not multiplied by x) be A, consider f(A).[/hide]

extension: the same problem only instead of only integer coefficients allow rational coefficients[/quote]There is a possible case that A is prime and f(A)=A.
[hide]In this case, there must be an integer $ n$, such that $ f(nA)\ne f(A)$. So $ f(nA)$ is not a prime.[/hide]

Just to nit pick, this is only true when $ f(x)$ is a non-constant polynomial. Else, we could have $ f(x)\equal{}p$ for prime $ p$ as counterexamples.

Essentially, we have $ f(x\plus{}p)\equiv f(x) \pmod p$, but the equation $ f(x)\equal{}p$ only has finitely many solutions if $ f$ is nonconstant.

The last coefficient of $ f$ can be 1 too, so that argument would be rather difficult to complete.

Let $ f(k) \equal{} p$, where $ p$ is a prime.
Then, $ f(k\plus{}p) \equal{} f(k) \plus{} np \equal{} (n \plus{} 1)p$ is also a prime, therefore $ n \equal{} 0$, and $ f(2k) \equal{} p$

It follows that $ f(k\plus{}mp) \equal{} p$ for $ m \equal{} 1,2,3,...$

Then consider the function $ f(k) \minus{} p$, which, as we just proved, has infinite zeros. Contradiction.

EDIT: Oops, I just kinda repeated what Bertram said. Sorry.


For a positive integer $n$, define:
$A(n)$, which I will abbreviate to $A_{n}$, as the number of digits in the binary representation of $n$, and
$B(n)$, which I will abbreviate to $B_{n}$, as the number $1$'s in the binary representation of $n$.

Now compute:

(1/2)^(A1+B1) + (1/2)^(A2+B2) + (1/2)^(A3+B3) + (1/2)^(A4+B4) + ..........

The same expression in $LATEX$ is shown below.. I hope its correct.

$\frac{1}{2^{(A_{1}+B_{1})}}+\frac{1}{2^{(A_{2}+B_{2})}}+\frac{1}{2^{(A_{3}+B_{3})}}+\cdots$

[hide="This might help:"]
\[A_{x}= n+1 \ \ \forall x \in [ 2^{n}-1, 2^{n+1}-1 ] \]
\[B_{x}= B_{y}+1 \ \ \forall x \in [ 2^{k}+1 , 2^{k+1}] \text{ and }x = y+2^{k}\]
But I don't see a way to sum it right now.   :)

[i]  (EDIT: Actually, my $B_{x}$  is shifted one to the right; I began the sequence with the first term as 0.) [/i][/hide]

[quote="t0rajir0u"][quote="JavaMan"]It's not hard to see [/quote]

I'm sorry, but I don't see it.   :|  Could you explain?[/quote]

[hide]
I grouped 1 digit numbers, 2 digit numbers, 3 digit numbers etc, like so,

$\frac{1}{2^{1}}(\frac{1}{2^{1}})+\frac{1}{2^{2}}(\frac{1}{2^{1}}+\frac{1}{2^{2}})+\frac{1}{2^{3}}(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{2}}+\frac{1}{2^{3}})+...$

Notice what's in the brackets of the n+1th term is 3/2 times what's in the brackets in the nth. This is logical since the number of 1s for each number of digits repeat, but are incremented by 1.

[/hide]


点O是等边三角形ABC内的一点，角AOB=110度，角BOC=135度，试问：
（1）以OA?OB?OC为边能?构?一个三角形？若能，请求出该三角形?角的度数；若?能，请说明?由。
（2）如果角AOB的大?????，那么当角BOC等于多少度时，以OA?OB?OC为边的三角形是一个直角三角形？

My English is not well.So I translated the problem on net.Maybe there are some mistakes in it.

There's a point O in equilateral triangle ABC.Angle AOB =110 degrees. Angle BOC =135 degrees.
Questions:
(1)Can it form a triangle to regard OA , OB , OC as? If can,ask not appearing this triangle reading of every mao; If cannot, please give reasons.
(2)If horn size of AOB keep intact, act as horn when how many degrees BOC mean , is the triangle taking OA , OB , OC as side a right triangle?

[quote="paganinio"]you posted spoiler :(
my photo ONLY shows ONE possible solution[/quote]

Approximations are not accurate!

[quote="beta"][quote="gsh91best"]点O是等边三角形ABC内的一点，角AOB=110度，角BOC=135度，试问：
（1）以OA?OB?OC为边能?构?一个三角形？若能，请求出该三角形?角的度数；若?能，请说明?由。
[/quote]
translation: Let O be a point inside of equilateral triangle ABC such that angle AOB=110 degrees, angle BOC=135 degrees:

1:Prove or disprove that there exist a triangle with sides OA, OB, OC. If such triangle does exist, find measure of the angles of the triangle

1. 以B为中心，把 A, C, O 转 60 度， A 到 C, C 到 C', O 到 O'.  所以，AO=O'C， BO=BO'. 因为 $\triangle BOO'$ 是等边三角形, BO=OO'. 在 $\triangle COO'$, O'C=OA, OO'=BO, 所以，$\triangle COO'$ 是以OA?OB?OC为边能?构?一个三角形。 $\angle BO'O=\angle BOO'=60^{\circ}$. 所以 $\angle OO'C=110-60=50^{\circ}$，$\angle O'OC=135-60=75^{\circ}$　, $\angle OCO'=55^{\circ}$

[quote="gsh91best"]
（2）如果角AOB的大?????，那么当角BOC等于多少度时，以OA?OB?OC为边的三角形是一个直角三角形？[/quote]
Translation: If the measure of angle AOB does not change and the triangle with sides OA, OB, OC is a right triangle, find the measure of BOC.

如果角AOB的大?????， OO'C 还是 50度。如果 OO'C=90, BOC=90+60=150.
如果 OO'C=40, BOC=40+60=100, 所以 BOC=100, 150.[/quote]

 :) Thank you very much! :)

這個thread真的很老。

[quote=paganinio]边 in English is called 'side'
you seem to know :)

your translation...makes no sense
since its local forum here, you needn't translate, we know Chinese or we wouldn't be here[/quote]

我就是中国人
嗨，其实我是美国人但是，我的家人（除了我的哥哥）都是中国人。我们只不过是在美国出生的。


Pedro chose exactly enough people for his test writing team to provide him 20 good questions. Each team member was asked to write 5 good questions, but he knows that 1/2 of the team will not write any, and 3/5 of the remaining questions won't be good ones. How many people are on his team?

[hide="not sure"]
[quote="xxxxx"]Each team member was asked to write 5 good questions, but he knows that 1/2 of the team will not write any.[/quote] That means you need twice the people you would usually need.
[quote="xxxxx"]And 3/5 of the remaining questions won't be good ones.[/quote] That means that out of every 5 questions written, only 2 will be good. To get 20 good questions, you would need 10 people.
Then, since you need twice the people, the answer would be $\boxed{20}$ people on Pedro's team.[/hide]

[hide]
so 3/5 of the questions will not be good.
therefore,
2x=20
so x=10,
but 1/2 will not write questions,
so 20.
[/hide]


Solve for x, for x is = 0 to 2 pi.

sin 7x = 2sin 5x cos 4x

[hide]One has $ \sin 7x \equal{} 2\sin 5x \cos 4x \equal{} \sin 9x \plus{} \sin x$, so $ \minus{}\sin x \equal{} \sin 9x \minus{} \sin 7x \equal{} 2\sin x \cos 8x$, so $ \sin x (1 \plus{} 2\cos 8x) \equal{} 0$, so $ \sin x \equal{} 0$ or $ \cos 8x \equal{} \minus{}1/2$. From here it's simple.[/hide]

hello, rewrite your equation in the following form
$ (1\plus{}2\cos(2x)\plus{}2\cos(4x))\sin(x)(\minus{}1\plus{}2\sin(2x))(1\plus{}2\sin(2x))\equal{}0$
Sonnhard.


In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $ 23$, $ 14$, $ 11$, and $ 20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $ 18$, what is the least number of points she could have scored in the tenth game?

$ \textbf{(A)}\ 26\qquad
\textbf{(B)}\ 27\qquad
\textbf{(C)}\ 28\qquad
\textbf{(D)}\ 29\qquad
\textbf{(E)}\ 30$

[hide="Click for solution"]
The average for games six through nine was $ \frac{23\plus{}14\plus{}11\plus{}20}{4}\equal{}17$, which exceeded her average for the first five games.  Therefore, she scored at most $ 5 \times 17\minus{}1\equal{}84$ points in the first five games.  Because her average after ten games was more than $ 18$, she scored at least $ 181$ points in the ten games, implying that she score at least $ 181\minus{}84\minus{}68\equal{}29$ points, or $ \boxed{\textbf{(D)}}$, in the tenth game.
[/hide]


algebra
polynomial
partial fractions
linear equation

[quote="linzk3"]Determine A and B such that (12)/((n+4)(n+8)) = ((A)/(n+4)) + ((B)/(n+8)), so far i've multiplied both sides by ((n+4)(n+8)) but I dont know what to do from there[/quote]

Ah... partial fractions

$ B(n\plus{}4)\plus{}A(n\plus{}8)\equal{}12$ How would you normally solve that?

[quote="SimonM"][quote="linzk3"]Determine A and B such that (12)/((n+4)(n+8)) = ((A)/(n+4)) + ((B)/(n+8)), so far i've multiplied both sides by ((n+4)(n+8)) but I dont know what to do from there[/quote]

Ah... partial fractions

$ B(n \plus{} 4) \plus{} A(n \plus{} 8) \equal{} 12$ How would you normally solve that?[/quote]

set up a systems of equations but I can't figure out how..

$ (A\plus{}B)n\plus{}(8A\plus{}4B)\equal{}0\cdot n \plus{} 12$

is it clearer now?

The key here is that $ n$ is not fixed.  $ n$ is arbitrary and varies, and the resulting $ A, B$ must produce an equal expression for [i]any[/i] value of $ n$.  That means both sides must be regarded as polynomials in $ n$, so their coefficients are equal.

In other words, bertram is trying to get you to conclude from

$ (A \plus{} B)n \plus{} (8A \plus{} 4B) \equal{} 0(n) \plus{} 12$

that we have two equations

$ A \plus{} B \equal{} 0$
$ 8A \plus{} 4B \equal{} 12$

$ \frac{12}{(n\plus{}4)(n\plus{}8)}\equal{}\frac{A}{n\plus{}4}\plus{}\frac{B}{n\plus{}8}$

$ \frac{12}{(n\plus{}4)(n\plus{}8)}\equal{}\frac{A(n\plus{}8)}{(n\plus{}4)(n\plus{}8)}\plus{}\frac{B(n\plus{}4)}{(n\plus{}8)(n\plus{}4)}$ Since they all have the same denominator, the denominator cancels out.

$ 12\equal{}A(n\plus{}8)\plus{}B(n\plus{}4)$

$ 12\equal{}An\plus{}8A\plus{}Bn\plus{}4B$

$ 12\equal{}(8A\plus{}4B)\plus{}n(A\plus{}B)$

$ A\plus{}B\equal{}0$
$ 8A\plus{}4B\equal{}12$

$ \fbox{A\equal{}3, B\equal{}\minus{}3}$

[quote="goldendomer"]$ \frac {12}{(n \plus{} 4)(n \plus{} 8)} \equal{} \frac {A}{n \plus{} 4} \plus{} \frac {B}{n \plus{} 8}$

$ \frac {12}{(n \plus{} 4)(n \plus{} 8)} \equal{} \frac {A(n \plus{} 8)}{(n \plus{} 4)(n \plus{} 8)} \plus{} \frac {B(n \plus{} 4)}{(n \plus{} 8)(n \plus{} 4)}$ Since they all have the same denominator, the denominator cancels out.

$ 12 \equal{} A(n \plus{} 8) \plus{} B(n \plus{} 4)$

$ 12 \equal{} An \plus{} 8A \plus{} Bn \plus{} 4B$

$ 12 \equal{} (8A \plus{} 4B) \plus{} n(A \plus{} B)$

$ A \plus{} B \equal{} 0$
$ 8A \plus{} 4B \equal{} 12$

$ \fbox{A \equal{} 3, B \equal{} \minus{} 3}$[/quote]
wait, how do you know A+B=0 and the other equation?

Like I said; he equated the coefficients of both sides as polynomials in $ n$.

Another way to do it is,

$ B(n\plus{}4)\plus{}A(n\plus{}8)\equal{}12$

Let $ n\equal{}\minus{}4$. Then,

$ 4A\equal{}12$
$ A\equal{}3$

Now, let $ n\equal{}\minus{}8$. Then,

$ \minus{}4B\equal{}12$
$ B\equal{}\minus{}3$

So, $ \boxed{A\equal{}3, B\equal{}\minus{}3}$

[quote="mnmath"]Another way to do it is,

$ B(n \plus{} 4) \plus{} A(n \plus{} 8) \equal{} 12$

Let $ n \equal{} \minus{} 4$. Then,

$ 4A \equal{} 12$
$ A \equal{} 3$

Now, let $ n \equal{} \minus{} 8$. Then,

$ \minus{} 4B \equal{} 12$
$ B \equal{} \minus{} 3$

So, $ \boxed{A \equal{} 3, B \equal{} \minus{} 3}$[/quote]

I guess that does work, but since we are dividing by $ n\plus{}4$ and by $ n\plus{}8$, you would just be dividing by zero (which is bad).

In the linear equation this isn't important anymore though, so we can just move arbitrarily close to -4 or -8 and take the limit. ;)

Alternately, we are equating the coefficients of the Laurent series of both sides at $ x \equal{} \minus{}4, \minus{}8$.  Either way, the expressions are equal whenever $ n \neq \minus{}4, \minus{}8$.

Hah. That's funny we just did this in class today.


function
limit
integration
real analysis
real analysis unsolved

What is an example of a function $f\in L^{1}(\mathbb{R})$ s.t. for all $q\in \mathbb{Q}$ there exists $x_{q}\in \mathbb{R}$ such that
\[\lim_{r\to 0+}\frac{1}{2r}\int_{[x_{q}-r,x_{q}+r]}f=q?\]


If you find a bank willing to give you 75% interest coumpounded quarterly for 40 years. How much money to the nearest cent must you put in at the beginning so that you will have one billion dollars at the end of the time period?

How did you get that and it asks to round to the nearest cent so is that still right?

Do $\frac{1000000000}{1.1875^{160}}$ to get approximately $0.11445139026150264247422317815815$ cents. Since it asks to round to the nearest cent, your answer is $\boxed {0}$ cents.


What is the probability that a positive integer $ n \leq 1000$, randomly selected, is divisible by 6? Express your answer as a terminating decimal.

There are $ \lfloor\frac{1000}{6}\rfloor\equal{}166$ multiples of $ 6$ less than or equal to $ 1000$.  There are a total of $ 1000$ numbers less than or equal to $ 1000$.  So our probability is $ \frac{166}{1000}\equal{}\boxed{.166}$

[hide"answer "]we will 1st want to find how many multiples of 6 there are 1000/6=166.666 so there are 166 multiples of 6

there are 1000 #'s in the # so we get the equation

166/1000=.166

so our answer is .166 or $\boxed{.166}$[/hide]

[b][i][u]please rate my post thank you !![/u][/i][/b]

please be fair though !! [=

 :yoda:  :yoda:  :yoda:  :yoda:


The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^2$, $8\text{ in}^2$, and $6\text{ in}^2$ respectively is:

$\textbf{(A)}\ 576\text{ in}^3 \qquad
\textbf{(B)}\ 24\text{ in}^3 \qquad
\textbf{(C)}\ 9\text{ in}^3 \qquad
\textbf{(D)}\ 104\text{ in}^3 \qquad
\textbf{(E)}\ \text{None of these}$

$ lw\equal{}12$
$ lh\equal{}8$
$ wh\equal{}6$
If we multiply all these we get:
$ (lwh)^2\equal{}576$
$ lwh\equal{}volume\equal{}24$
Answer is B

Another way is to find the prime factorization of each number:

6 = 3*2  8 = 2*2*2  12 = 3*2*2

Note that for 8 no factor of 2*2*2 is shared by the other sides, so it must have one side of 2*2 = 4 and the other 2. This means that the other side must be 3 since 12/(2*2) = 3. 4*3*2 = 24. This approach wouldn't work if the sides couldn't be integers.

I think you should know these problems are very, very old. Both in the fact that they are apparently from 1950 (?) but also that this topic is from 2008.


(a)Prove that in the set $ S\equal{}\{2008,2009,. . .,4200\}$ there are $ 5^3$ elements such that any three of them are not in arithmetic progression.

(b)[i]Bonus[/i]: Try to find a smaller integer $ n \in (2008,4200)$ such that in the set $ S'\equal{}\{2008,2009,...,n\}$ there are $ 5^3$ elements such that any three of them are not in arithmetic progression.

ps, from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1260011#1260011]this contest[/url]

(b)
$ 2008 \plus{} \frac {3^7 \minus{} 1}{2} \minus{} 4$
I think...

[hide="To Paolo"]$ \frac {3^7 \minus{} 1}{2}$ is the largest integer having $ < 8$ digits and having no digit $ 2$ in base $ 3$.
$ \minus{} 4$ because $ 2^7 \minus{} 5^3 \equal{} 3$, hence three numbers less.[/hide]

[quote="Akashnil"]$ \frac {3^7 \minus{} 1}{2}$ is the largest integer having $ < 8$ digits and having no digit $ 2$ in base $ 3$.
$ \minus{} 4$ because $ 2^7 \minus{} 5^3 \equal{} 3$, hence three numbers less[/quote]

yes,i thought the same..now i don't know if it can be better :roll:


Show that the map

f : IN² ---> IN
    (p,q) ---> (p+q)(p+q+1)/2  +  q

is bijective .

Write as $ 1 \plus{} 2 \plus{} ... \plus{} (p \plus{} q) \plus{} q$.  Since $ p \plus{} q > q$, the inverse map takes the distance between $ f(p, q)$ and the largest triangular number less than it to be $ q$ and the obvious calculation to be $ p$.

thanks T0rajir0u for your solution ; i just need to draw it up in detail  :roll:


linear algebra
matrix
function
algebra
system of equations
IMO
IMO 1965

Consider the sytem of equations
 \[  a_{11}x_1+a_{12}x_2+a_{13}x_3 = 0 \]\[a_{21}x_1+a_{22}x_2+a_{23}x_3 =0\]\[a_{31}x_1+a_{32}x_2+a_{33}x_3 = 0   \] with unknowns $x_1, x_2, x_3$. The coefficients satisfy the conditions:

a) $a_{11}, a_{22}, a_{33}$ are positive numbers;

b) the remaining coefficients are negative numbers;

c) in each equation, the sum ofthe coefficients is positive.

Prove that the given system has only the solution $x_1=x_2=x_3=0$.

It seems that this problem needs a solution.

[b]Solution.[/b] Assume without loss of generality that $ |x_{3}|\geq|x_{2}|\geq |x_{1}|$. Assume to the contrary that $ |x_{3}|>0$.  Then it follows, from the given conditions, that
\begin{eqnarray*}0&=&\left| a_{31}\frac{x_{1}}{x_{3}}+a_{32}\frac{x_{2}}{x_{3}}+a_{33}\right|\cdot|x_{3}|\\ &\geq& |x_{3}|(a_{33}-|a_{32}|-|a_{31}|)\\ &=& |x_{3}|(a_{33}+a_{32}+a_{31})>0\end{eqnarray*}
Contradiction! It follows that $ x_{1}=x_{2}=x_{3}=0,$ as required. $ \Box$

This problem can also be solved using determinants  :wink: .

P.S. Sorry for reviving the old thread, I wanted to look at the original problem because this problem is printed wrong in my book (otherwise it make no sense).

[hide="Solution"]We want to prove that the matrix $ A \equal{} [a_{ij}]$ has nonzero determinant. We'll prove that $ \det A > 0$.
We have that $ \det A$ is a linear function in each entry of $ A$. The coefficient of $ a_{11}$ is $ a_{22}a_{33} \minus{} a_{23}a_{32}$, but since $ a_{22} > \minus{} a_{21} \minus{} a_{23} > \minus{} a_{23} > 0$ and $ a_{33} > \minus{} a_{32} > 0$, $ a_{22}a_{33} \minus{} a_{23}a_{32} > 0$.

From this we conclude that $ \det A$ is decreased as $ a_{11}$ is decreased. Analogously, it decreases when $ a_{22},a_{33}$ decrease, so it's greater than the determinant when the sum of each row is $ 0$. But in that case the determinant is zero because the columns are linearly dependent (their sum is zero), so we are done.[/hide]

The system of equations has the unique solution $ x_1 \equal{} x_2 \equal{} x_3 \equal{} 0$ iff $ b_1 \equal{} (a_{11},a_{12},a_{13}), b_2 \equal{} (a_{21},a_{22},a_{23}), b_3 \equal{} (a_{31}, a_{32}, a_{33})$ is linear independent.
Assume they are not. So there exists $ (x_1,x_2,x_3) \neq (0,0,0)$ s.t. $ x_1b_1 \plus{} x_2b_2 \plus{} x_3b_3 \equal{} 0$. c) implies that not all of $ x_1,x_2,x_3$ has the same sign. wlog asume that $ x_1$ has another sign than $ x_2,x_3$ and $ x_1 \neq 0$. Hence the first entry of $ x_1b_1 \plus{} x_2b_2 \plus{} x_3b_3$ has the same sign as $ x_1$ since $ a_{11} > 0$. Contradiction. So $ b_1,b_2,b_3$ are linear independent and the system of equations has the unique solution $ x_1 \equal{} x_2 \equal{} x_3 \equal{} 0$.

Would this solution work? If not, could anyone explain a solution using this approach?

[hide="Solution?"]
Instead of letting $a$ be the coefficients, let $x$ be the coefficients. Now we are trying to find three solutions to the equation $x_1a+x_2b+x_3c=0$, for coefficients $(x_1, x_2, x_3)$, such that

a/b)in each of the equations, one of $a$, $b$ or $c$ is positive while the other two are negative and each variable gets a chance to be positive.

c) When the coefficients are all 1, the sum is positive for all three solutions.

Condition a and b means that one solution must exist in each of the octants where $(x_1,x_2,x_3)$ are $(+, -, -), (-,+,-)$, and $(-,-,+)$. Now all we have to show is that each valid equation can only pass through two octants of those three octants maximum.

If any of the coefficients are negative, then it won't pass through all three octants.

Now, we only consider planes that pass through all three octants and prove that they aren't valid solutions.

Since $x_1a+x_2b+x_3c=0$, by condition 3, any solution must be "behind" the (1,1,1) plane in one of the octants in order to have a sum of less than 0. WLOG, in the $(-,-,+)$ octant, $-(a+b)>c$. However, going back to the original form, $a,b$ correspond to $a_{31}, a_{32}$ and $c$ corresponds to $a_{33}$, and this would mean the sum of the coefficients would be negative, which is a contradiction.

Since no positive or negative values work for $(x_1, x_2, x_3)$, the only thing they can be is all 0, and that works obviously. QED?

[/hide]

Thanks.

Let $$A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}$$
In order to show that $\dim(\ker(A))=0$ we only need to show that $rank(A)=3$.(By rank nullity Theorem.)

Suppose FTSoC the columns $C_i$ ($i=1,2,3$)  of $A$ are linearly dependent.Then there are reals $b_1,b_2,b_3$ not all  0 such that $$b_1C_1+b_2C_2+b_3C_3=\textbf{0}$$

Let $i\in\{1,2,3\}$ be the integer such that $b_i$ has maximum absolute value among $b_1,b_2,b_3$.
But then
$$|a_{ii}|=|\sum_{j\ne i} \frac{b_j}{b_i} a_{ij}|\le \sum_{j\ne i} |\frac{b_j}{b_i} a_{ij}|\le \sum_{j\ne i} | a_{ij}|<|a_{ii}|$$
But this is impossible.Hence $rank(A)=3$.$\blacksquare$


Let $ (a_n)$ be a sequence with $ a_1 \equal{} 1$ satisfying the recursion $ a_{n\plus{}1} \equal{} a_n\minus{}2$ if $ a_n\minus{}2\notin\{a_1; a_2;...; a_n\}$ and $ a_n\minus{}2 > 0$ and $ a_{n\plus{}1} \equal{} a_n \plus{} 3$ otherwise. Prove that for every positive integer $ k$ there is a positive integer $ n$ so that $ a_n \equal{} k^2 \equal{} a_{n\minus{}1} \plus{} 3$.

(need to require $ k > 1$)

If I am understanding this problem correctly, it is straight forward to verify by induction that for all positive integers $ m$.
$ a_{5m\minus{}4} \equal{} 5m \minus{} 4$.
$ a_{5m\minus{}3} \equal{} 5m \minus{} 1$.
$ a_{5m\minus{}2} \equal{} 5m \minus{} 3$.
$ a_{5m\minus{}1} \equal{} 5m$.
$ a_{5m} \equal{} 5m \minus{} 2$.
So the sequence is a reordering of the positive integers, guaranteeing that all squares are in the sequence. The squares mod 5 are $ 0,1,4$. So only $ a_{5m\minus{}4}, a_{5m\minus{}3}, a_{5m\minus{}1}$ can be squares, and we can observe that they are all $ 3$ plus the previous term. I have to be missing something here.


Find $ m\in\mathbb{R}$ :
$ log((m \minus{} 1)^{2} \plus{} 1 \plus{} \cos^{2m}x) \minus{} 2mlog(sinx) \plus{} |cosx| \equal{} 0$


modular arithmetic
number theory
prime numbers
number theory proposed

Find all prime numbers p such that equation $ y^2\equiv x^3\minus{}x (\mod p)$ has exactly p integer solutions such that  $ 0\leq x,y\leq p\minus{}1$ .

Number of solutions to $ y^2 \equiv x^3 - x$ for a fixed $ x$ is $ 1 + \left( \frac {x^3 - x}{p} \right)$. We want to find the sum of this over $ x$:
$ \sum_{x = 0}^{p - 1} (1 + \left( \frac {x^3 - x}{p} \right) ) = p + \sum_{x = 0}^{p - 1}\left( \frac {x^3 - x}{p} \right)$
We want to find the primes $ p$ such that $ \sum_{x = 0}^{p - 1}\left( \frac {x^3 - x}{p} \right) = 0$. This sum is between $ - (p - 1)$ and $ p - 1$, so it is $ p$ iff it is divisible by $ p$. Now,
$ \sum_{x = 0}^{p - 1}\left( \frac {x^3 - x}{p} \right) \equiv \sum_{x = 0}^{p - 1} (x^3 - x)^{\frac {p - 1}{2}} \equiv \sum_{i = 0}^{\frac {p - 1}{2}} ( - 1)^{\frac {p - 1}{2} - i} \binom{\frac {p - 1}{2}}{i} \sum_{x} x^{2i + \frac {p - 1}{2}} \pmod p$

A simple fact is that $ \sum_{x = 0}^{p - 1} x^t$ is divisible by $ p$ when $ t$ is not a multiple of $ p - 1$. In our sum, $ t$ varies from $ \frac {p - 1}{2}$ ($ < p - 1$) to $ 1.5 (p - 1)$ ($ < 2(p - 1)$), so the only relevant term is the one correspending to $ t = p - 1$.

If $ p \equiv 3 \pmod 4$, ${ t = 2i + \frac {p - 1}{2}}$ is never $ p - 1$, so the sum is zero.
If $ p \equiv 1 \pmod 4$, the solution to ${ p - 1 = 2i + \frac {p - 1}{2}}$ is $ i = \frac {p - 1}{4}$, and our sum becomes:
$ ( - 1)^{\frac {p - 1}{2} - \frac {p - 1}{4}} \binom{\frac {p - 1}{2}}{\frac {p - 1}{4}} \sum_{x} x^{p - 1} =$
$ ( - 1)^{\frac {p - 1}{4}} \binom{\frac {p - 1}{2}}{\frac {p - 1}{4}} (p - 1) \neq 0 \pmod p$

When $ p = 2$: $ y^2 \equiv x^3 - x \pmod 2 \leftrightarrow y \equiv 0 \pmod 2$, so the number of solutions is $ 2$.
Conclusion: the number of solutions is $ p$ iff $ p \neq 1 \pmod 4$.

Very good solution ,that 's also my method  to attack this problem :) .


:)
Try this problem!

Find all polynomials f(x) such that f(x)+f(x+2)=2f(x+1)
for all real x.

I don't find it difficult, nevertheless interesting!

um, how many times do you want to post that problem?

although i dont think this problem is apropriate as an intermediate topic. heres what i got:

$f(x+1)=\frac{f(x)+f(x+2)}{2}$

since $x+1$ has a domain of $x\in\mathbb R$ and the equation above satisfies the more genaral equation $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$ with $y=x+2$ this equation is convex and concave, which is defined with $f(\frac{x+y}{2})\le\frac{f(x)+f(y)}{2}$ or $f(\frac{x+y}{2})\ge\frac{f(x)+f(y)}{2}$. therefore the graph of $f$ has to be a straight. so $f$ is either linear or constant which works both ways:
$mx+mx+2m=f(x)+f(x+2)=2f(x+1)=2mx+2m$ which is obviously true
and $c+c=f(x)+f(x+2)=2f(x+2)=2c$ is true also.
done. :ninja:

I didn't see this was posted in Getting Started forum so I locked it.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42946&sid=4fffe3334c32d27efd6a36d5d0d13d27]here[/url]

My function was $f(x) = x$. I'm not sure if that makes sense though.  :roll:


limit integration calculus logarithms derivative trigonometry function

$\int_{0}^{x}\arctan t\ dt=x\arctan x-\ln \sqrt{1+x^{2}}$

[quote="RealTime"]Hi.

I have problem understanding how to solve this kind of thing, where it specifically notes that I should use L`hospital rule...

$\lim_{x\to\infty}\frac{\int_{0}^{x}\arctan(t)dt}{\sqrt{x^{2}+1}}$

Also theres an alternate way, to calculate the integral and then taking the limit right?

Thanks alot.[/quote]
Applying L'Hospital's rule, in fact, makes things a lot easier.

First note that if $I(x) = \int_{g(x)}^{h(x)}f(t)\, dt$ then $I'(x) = f(h(x))h'(x)-f(g(x))g'(x)$.
(This is probably called differentiation under the integral sign -- not sure though about the correct term.)

Let $L = \lim_{x\to\infty}\frac{\int_{0}^{x}\arctan(t)dt}{\sqrt{x^{2}+1}}$ (Note that this has $\frac{\infty}{\infty}$ form.)

So, applying L'Hospital's rule and using differentiation under the integral sign, we get
$L = \lim_{x\rightarrow \infty}\frac{\tan^{-1}(x) \cdot 1-\tan^{-1}(0) \cdot 0}{\frac{1}{2\sqrt{x^{2}+1}}\cdot 2x}= \lim_{x\rightarrow \infty}\frac{\sqrt{x^{2}+1}\tan^{-1}x}{x}$

Now, make a suitable change in variables. We will use the (somewhat obvious) substitution $x=\tan\theta .$

$\therefore L = \lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\sec\theta \cdot \theta}{\tan\theta}= \lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\theta}{\sin\theta}= \boxed{\frac{\pi}{2}}.$

[[b]N.B.[/b]: Even if you find the anti-derivative first, you will still have to differentiate terms later while using L'Hospital's rule, and this, I think, entails a lot of work. If you don't use L'Hospital's rule, then there is a lot of computation involved. In either case, you run the "risk" of making errors while doing all that computation.]

Thanks for the help...

One more small matter...

How do we know its $\frac{\infty}{\infty}$ ?

The denominator is obvious, but the integrand itself? I mean without actually calculating the antiderivative, which ofcourse that was the main reason we used lhospital right?

Thanks.

$\lim_{x\to\infty}\int_{0}^{x}\arctan t\,dt = \infty.$ Why?

Note that since the integrand is positive, this integral is an increasing function of $x.$ As an increasing function, it either converges to a finite limit or goes to infinity. Now note that the integrand, $\arctan t,$ doesn't even go to zero. This improper integral will not converge.

To make the estimate specific: Note that for $t>1,\ \arctan t > \frac{\pi}4.$

Then, for $x>1,$

$\int_{0}^{x}\arctan t\,dt>\int_{0}^{1}0\,dt+\int_{1}^{x}\frac{\pi}4\,dt= \frac{\pi}4(x-1).$

Since this is greater than $x-1,$ it goes to infinity as $x\to\infty.$

[quote="RealTime"]Thanks for the help...

One more small matter...

How do we know its $\frac{\infty}{\infty}$ ?
.
The denominator is obvious, but the integrand itself? I mean without actually calculating the antiderivative, which ofcourse that was the main reason we used lhospital right?

Thanks.[/quote]
If you draw a graph of $f(x)=\tan^{-1}x,$ you will see exactly why. Note that the graph is an increasing function, and that $y=\frac{\pi}{2}$ is an asymptote. (Of course, there is another asymptote at $y=-\frac{\pi}{2}$.) Now, try finding the area between $y=f(x),$ y-axis and $y=\frac{\pi}{2}.$ You will find that it is finite. (In fact, you don't have to calculate it; just an estimation is enough.) From this, we conclude $\int_{0}^{x}\tan^{-1}t\, dt \rightarrow \infty$ as $x\rightarrow \infty.$

Ah, Prof. Merryfield has answered the question! :)


inequalities
trigonometry
inequalities unsolved

For a fixed $\theta \in \lbrack 0, \frac{\pi}{2} \rbrack$, find the smallest $a \in \mathbb{R}^{+}$ which satisfies the following conditions:

[b]I. [/b] $\frac{\sqrt a}{\cos \theta} + \frac{\sqrt a}{\sin \theta} >
1$.

[b]II.[/b] There exists $x \in \lbrack 1 - \frac{\sqrt a}{\sin \theta},
\frac{\sqrt a}{\cos \theta} \rbrack$ such that $\lbrack (1 -
x)\sin \theta - \sqrt{a - x^2 \cos^{2} \theta} \rbrack^{2} +
\lbrack x\cos \theta - \sqrt{a - (1 - x)^2 \sin^{2} \theta}
\rbrack^{2} \leq a$.


search
combinatorics unsolved
combinatorics

Find the maximum number of subsets $S_{1},S_{2},S_{3},...,S_{x}$ from a set $S$ of $n$ elements such that dont exist $1\leq i,j\leq x$, such that $S_{i}$ is included in $S_{j}$

Try a search about Sperner's theorem on antichains.

Pierre.


[quote="rainer"]if a,b,c are pozitive real numbers ,prove that

1/a2+b2  +  1/a2+ c2   +  1/b2+c2   is greater or equal 3[/quote]
If are you mean that $ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}c^2}\geq3$ then it's wrong.

It's still wrong..
try $ a\equal{}b\equal{}c\equal{}1$


calculus
integration
quadratics
algebra
algebra unsolved

Find all integer numbers $a$, for which $x^3+(a-4)x^2+(a+4)x+9=0$ has 2 different integer roots.

I know it`s not hard. Just post the answer please

Actually , you can make it easy by letting the root of the quadratic equation be $m,n\in\mathbb{N}$ then you have

$m+n=5-a$
$mn=9$

So easily you can get $(m,n)=(1,9),(3,3),(-1,-9),(-3,-3)$ and the reverse .

So for each $(m,n)$ we have its correspoding value of $a$ which have been shown by leepakhin  :)

I will not get maximum points in this math battle  :(  I didn`t solve it correct.

BTW here are the other problems from the math battle

1)Solve in integer numbers:

$m(n^2+36)+n(m^2-36)+m^2(m-12)=0$

2)The difference between any two from five natural numbers is different. Find the biggest  from these five numbers if it`s known that it`s the smallest possible.
[hide]I think it`s 13, becouse - 13,12,10,6,1[/hide]

3)Find the smallest natural number n with these conditions: We have two colours. No matter how we colour the numbers from 1 to n , there always have four numbers with the same colour, for which the sum of 3 of them is equal to the four number.

4)Find the smallest number which can be represented at least at four ways of the form $14a+41b$, where a,b are natural numbers.

[hide] I think that it is 1777[/hide]
And.. the other problems - I can`t translate them :o

(excuse my english)

Similar method works here. (and so I didn't spend much time on finding a nicer solution. :) )
$m(n^2+36)+n(m^2-36)+m^2(m-12)=0$
$mn^2+(m^2-36)n+m(m-6)^2=0$
If $m=0$ then $-36n=0\Rightarrow n=0$
If $m\neq 0$ then the equation is a quadratic.
$\Delta_n=(m^2-36)^2-4(m)[m(m-6)^2]=-3(m-6)^3(m+2)=[(m-6)p]^2$ for some $p\in\mathbb{N}$
$-3(m-6)(m+2)=p^2$
We get $(m-6)(m+2)\leq 0\Rightarrow -2\leq m\leq 6$. (and $m\neq 0$ of course!)
After checking all possibilities, $(m,p)=(-2,0)\text{ or }(4,6)\text{ or }(6,0)$
Correspndingly $n=-8\text{ or }1\text{ or }4\text{ or }0$

Well I don`t think the solution is so bad  :) . Mine is the same  :P
But i`m interested in the second problem which i posted.. Can it be generalized (not for five but for n numbers)?

For the second one , what about fibbonanci number :P 1,2,3,5,8 .  :)


limit
integration
probability
calculus
probability and stats

I'm reading from a  book which quickly presents the [url=http://ro.wikipedia.org/wiki/Teoria_probabilit%C4%83%C5%A3ilor#Teorema_Moivre-Laplace]theorem  deMoivre-Laplace[/url].
it's stated like this :
$ \lim_{n\to \infty} P(a\leq  \frac{\mu \minus{} np}{\sqrt{npq}}  \leq b)  \equal{} \frac{1}{\sqrt{2\pi}} \int_{a}^{b} e^{\minus{}\frac{t^2}{2}}dt$

where n is the number of repetitions of n.
p is the probability of the event occuring.
q is 1-p.

the problem is that I don't know any methods for calculating a and b.
(at least from the book I'm using).
how do I find a and b ?
also,can that integral be calculated ?
(I saw in another book that there is a table that shows the values of this definite integral).

1. This formula holds for every $ a$ and $ b$ ($ a<b$).
2. The integral can be efficiently computed with high precision. Are you interested how to numerically compute the integral in practice? (then you can use either tables or special software).

2)
yes,I'm interested in methods for calculating it.
do you know of any ?

The easiest way to compute the integral is to use the "rentangle method": http://en.wikipedia.org/wiki/Rectangle_method
It is quite efficient. There are even more efficient methods specifically for integrals like this, but they are more complex.


parameterization
geometry unsolved
geometry

[quote="Geómetra"]
[u][b]Proposition 6[/b][/u]

[img]http://img513.imageshack.us/img513/5163/archboolem06ix4.gif[/img]

This is a very interestin' problem, jeje.[/quote]
Yes, very interesting... in fact, being content with just the numerical answer isn't enough... it's all about constructing it! (Don't worry, I've already furnished several proofs myself of the "numericals").

While trying to construct the squished circle, I came up with an interesting locus:

(click twice for better quality:)[img]6894[/img]

It arises from the fact that the brown "line" intersect the big circle is the third and final point of tangency, which trivializes the construction. However, this point is hard to come by - hence the locus intersect idea to develop a conjecture. But what is that locus? (Please don't tell me its a hyperbola... :P ) I remade the locus with different parameters to the left and it certainly looks like a line with a hook. Hope someone gets back with some ideas :)

Could you write precisely how do you define your locus?

:oops: I know I left out something...

Pick variable point on the vertical perpendicular, construct the circle tangent to the two other circles. Now, the intersect of the line thru center of this circle and center of big circle with the circle is our point for the locus. It is the path made by this point from moving the variable point along its perpendicular.


[quote="cezar lupu"]Let $a,b,c>0$.Prove or disprove that:

  $\frac{ab}{c}+\frac{ca}{b}+\frac{ab}{c}+2(a+b+c)\geq 3\cdot\sqrt{3(a^2+b^2+c^2)}$     ;)[/quote]

I think you meant $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}+2\left(a+b+c\right)\geq 3\cdot\sqrt{3\left(a^2+b^2+c^2\right)}$. Anyway, a = 1, b = 1, c = 2 is a counterexample (by the way, for the above inequality with $\frac{ab}{c}$ instead of $\frac{bc}{a}$ as well)...

  darij


Prove that there exists an "n" for each prime number "p" so that if p divides
a(1)a(2)...a(n-1)a(n) (which shows a whole number with n digits)  then it also divides
a(n)a(1)....a(n-1) (that is, it divides the new number formed by transferring the last digit to the begining).

regards,

kcefle

For example, n is 3  when p is 37 and 5 when p is 41.

$A=a_1...a_{n-1}$
$B=a_n$

We need to show that it exists $n$ such that if $10A+B\equiv 0(p)$ then $A+10^{n-1}B\equiv 0(p)$
$A+10^{n-1}B\equiv 0(p)$
$\Leftrightarrow A+10^{n-1}(-10A)\equiv 0(p)$ since $B\equiv -10A 0(p)$
$\Leftrightarrow A(1-10^n)\equiv 0 (p)$
With Fermat $n=p-1$ works if $p\neq 2$ and $p\neq 5$
For $p=2$ and $p=5$ there is no $n$ ...


[quote="Slizzel"]Let $\Delta ABC$ be a equilateral triangle. From $A$,$B$ and $C$ we draw $AA`=2$,$BB`=8$ and $CC`=6$, $\bot$ on the triangle's plane. Find the distance between the medians intersection of triangle $\Delta ABC$ and $\Delta A'B'C'$ (i don't know how to say in english...  :) )[/quote]

So $AA`$,$BB`$ and $CC`$ are perpendicular to the plane that $\Delta ABC$ is in? And $\Delta ABC$ and $\Delta A`B`C`$ do not intersect each other?

[quote="osdes"]So $AA`$,$BB`$ and $CC`$ are perpendicular to the plane that $\Delta ABC$ is in? And $\Delta ABC$ and $\Delta A`B`C`$ do not intersect each other?[/quote]

yes


Let $x,y,z \geq 0such x+y+z=1.$.Prove that : $\sum_{cyc}\frac{x^{2}+1}{y^{2}+1}\leq \frac{7}{2}$

[quote="hunter_cvp_ic"]Let $a,b,c \geq 0$.Prove that : $\sum_{cyc}\frac{x^{2}+1}{y^{2}+1}\leq \frac{7}{2}$[/quote]
What are $a,b,c$? Even if $(a,b,c)=(x,y,z)$, the inequality doesn't hold: just take $x\gg y$.

oh i'm  sorry
i try it now
you can see my problem now[/quote]

[quote="hunter_cvp_ic"]Let $a,b,c \geq 0$.Prove that : $\sum_{cyc}\frac{x^{2}+1}{y^{2}+1}\leq \frac{7}{2}$[/quote]
It contains $x+y+z=3$.

[quote="hunter_cvp_ic"]Let $x,y,z \geq 0$such $x+y+z=1.$.Prove that : $\sum_{cyc}\frac{x^{2}+1}{y^{2}+1}\leq \frac{7}{2}$[/quote]
 Sorry i've mistake


Prove that in a triangle, the sum of the squares of the measures of the medians is $\frac34$ the sum of the squares of the measures of the sides.

[quote="Centy"][hide="Hint"]
Apollonius' Theorem
[/hide][/quote]What's that?  :(

I solved this one using good old Pythagoras. :)

[hide]
By Stewart's or Law of Cosines...

$AD^2 = \frac{2AB^2+2AC^2-BC^2}{4}$.[/hide]

let $m_a$ be the median to side length $a$

whatever you want to call it, (i say median formulas)

$4m_a^2+a^2=2b^2+c^2$

and symmetrically, result follows


algebra
polynomial
calculus
integration
symmetry
number theory unsolved
number theory

I vaguely remember a problem (from an unknown source) that looks something like this. I'm not sure if this is right, though, so feel free to offer corrections.

For a polynomial $f(x)$ with integer coefficients and a positive integer $q$, if $\big(f(k)\big)^{1/q}$ is an integer for [color=red]infinitely many integers[/color] $k$, show that $f(x)=\big(g(x)\big)^q$ where $g(x)$ is a polynomial with integer coefficients.

[color=red]EDIT: replace "infinitely many integers k" with "all integers k"[/color]

Oh, I see you've edited it. I have edited my answer as well :).

Now it's true. I'll use a problem which I have posted in the "Proposed and Own" section, and which I will not prove, since I don't want to spoil the problem for people who might be interested in it:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62269[/url]

Let me restate the result here:

$(*)$ Given a nnon-constant irreducible polynomial $P$ with integer coefficients, for any $k\ge 1$ we can find distinct primes $p_1,\ldots,p_k$ and an integer $n$ such that $p_i|P(n),\ p_i^2\not|P(n),\ \forall i\in\overline{1,k}$.

Let $f=\prod_{j=1}^tf_j^{a_j}$ be the decomposition of $f$ into irreducible integral polynomials of positive degree. Now, for each $j\in\overline{2,t}$, the polynomials $f_1,f_j$ are coprime, so we can find polynomials $\alpha_j,\beta_j$ and a constant $u_j$ such that $\alpha_jf_1+\beta_jf_j=u_j$. By applying $(*)$ to $f_1$ instead of $P$ and a large enough $k$, we can conclude that there is a prime $p$ and an integer $n$ such that $p|f_1(n),p^2\not|f_1(n)$, and $p\not|u_j,\ \forall j\in\overline{2,t}$, which, in turn, means that $p\not|f_j(n),\ \forall j\in\overline{2,t}$. This means that the exponent of $p$ in $f(n)$ is precisely $a_1$, so if $f_n$ is to be a $q$'th power we must have $q|a_1$. By symmetry, $q|a_j,\ \forall j\in\overline{1,t}$, and that's what we were after.

[quote="zabelman"]For a polynomial $f(x)$ with integer coefficients and a positive integer $q$, if $\big(f(k)\big)^{1/q}$ is an integer for [color=red]infinitely many integers[/color] $k$, show that $f(x)=\big(g(x)\big)^q$ where $g(x)$ is a polynomial with integer coefficients.  [/quote]

What is true if you assume the equation for infinitely many (but not necessarily all) integers $k$, is that there is an integer polynomial $K(t)$ such that $f(K(t)) = g(K(t))^q$.    For example, we could have $f(x) = x^6, g(x) = x^{10}, K(x) = x^5$.

Hello,

Extends previous result in two ways :

a)  f(x)  has the form  G(y)  for all integer  x  large enough where G  is a polynomial with integer coefficients, then show f(X) = G(R(X))   where  R  is another polynomial.

b) same problemS with more variables.

lolo


For any nonempty set $\,S\,$ of numbers, let $\,\sigma(S)\,$ and $\,\pi(S)\,$ denote the sum and product, respectively, of the elements of $\,S\,$.  Prove that
\[\sum \frac{\sigma(S)}{\pi(S)}= (n^{2}+2n)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) (n+1), \]
where ``$\Sigma$'' denotes a sum involving all nonempty subsets $S$ of $\{1,2,3, \ldots,n\}$.


geometry
integration
calculus
calculus computations

Find the surface area of the part of the cylinder $ x^2\plus{}z^2\equal{}1$ that is contained within the cylinder $ x^2\plus{}y^2\equal{}1$.

Let's denote the region what you say $ S$. We know orthogonal projection of $ S$ onto $ xy$ plane. It's $ R\equal{}\left\{(x,y): x^2\plus{}y^2\leq 1\right\}$.
Applying the formula computing surface area by its orthoganal projection onto $ xy$ plane, the area $ \left|S\right|$ we want to compute is:
$ \left|S\right|\equal{}2\iint_{R}\sqrt{1\plus{}\left(\frac{\partial z}{\partial y}\right)^2\plus{}\left(\frac{\partial z}{\partial x}\right)^2}dydx$
$ \equal{}2\int_{\minus{}1}^{1}\int_{\minus{}\sqrt{1\minus{}x^2}}^{\sqrt{1\minus{}x^2}}\sqrt{1\plus{}0^2\plus{}\left(\frac{\minus{}x}{\sqrt{1\minus{}x^2}}\right)^2}dydx$
$ \equal{}2\int_{\minus{}1}^{1}\frac{1}{\sqrt{1\minus{}x^2}}\int_{\minus{}\sqrt{1\minus{}x^2}}^{\sqrt{1\minus{}x^2}}dydx$
$ \equal{}2\cdot 2\int_{\minus{}1}^{1}dx\equal{}\boxed 8$.


conics
ellipse
geometry
analytic geometry
inequalities
geometry unsolved

Let $AD,BE,CF$ be the interior angle bisectors of $\bigtriangleup ABC$. $P$ is a point inside $\bigtriangleup ABC$. Denote $h_a,h_b,h_c$ the distance of $P$ to the sides of $\bigtriangleup ABC$. If $\sqrt{h_a},\sqrt{h_b},\sqrt{h_c}$ is the sides of a triangle, then prove
(1) the locus of point $P$ is the interior of an ellipse $\Gamma$, and $BC,CA,AB$ tangents $\Gamma$ at point $D,E,F$ respectively;
(2) the area of $\Gamma$ satisfies $\frac{4\sqrt{3}\pi}{9}S_{\bigtriangleup DEF}\le S_{\Gamma}\le \frac{\sqrt{3}\pi}{9}S_{\bigtriangleup ABC}$.

[b]1)[/b] $ x \equal{} h_a\ ,  y \equal{} h_b \ ,  h_c \equal{} z.$ Thus, trilinear coordinates of $P$ WRT $\triangle ABC$ are $(x: y: z).$ Firstly, let us consider the points $ P$ such that triangle with side length $ \sqrt {x},\sqrt {y},\sqrt {z}$ is degenerate. Squaring $ \sqrt {x}\plus{}\sqrt {y} \equal{} \sqrt {z}$ twice yields

$\mathcal{E} \equiv  x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} yz \plus{} zx) \equal{} 0$

Which is the trilinear equation of the inellipse $\mathcal{E}$ with intouch triangle $ \triangle DEF.$ Cyclic permutation of the squares roots of $ x,y,z$ gives the same inconic. Now, every point inside the inellipse $\mathcal{E}$ satisfies the inequality $ x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} yz \plus{} zx) \ge 0,$ which is clearly equivant to  $ \sqrt {x}\plus{}\sqrt {y} \ge \sqrt {z}$ $ \Longrightarrow$ $ (x,y,z)$ are sides of a non-degenerate triangle. Points outside $\mathcal{E}$ satisfy $ \sqrt {x}\plus{}\sqrt {y} \le \sqrt {z}$ and the triangle does not exist.


I hang a mass m at the end of a massless string, which is attached to a fixed hook in the ceiling, so that the entire system is at rest. it is clear that $ mg\equal{}T$, where T is the tension in the string.This string is then set in motion so that it follows a circular(conical) path. Let the angle that the free end of the string (the one attached to the mass m) makes with the vertical be $ \theta$. we now have $ T\cos\theta\equal{}mg$ and $ T\sin\theta$ is the centripetal force acting. Since mg is constant, we have $ T\equal{}T\cos\theta \Rightarrow \theta\equal{}0$ How is this possible? Where have i gone wrong?

[quote="Carcul"]The tension is not the same in both situations.[/quote]

How come  the tension is not the same? can u explain a bit?

why so u think the tension has to be the same?tension is an intermolecular attempt to prevent the breaking of the system.It's self-adjusting

Plus didn't you just show that if the tension is the same in the conical pendulum and when it's stationary something ridiculous occurs?

I think your problem is with the notation of tension. Use $ T$ and $ T'$ instead.

Remember that $ T$ is constantly varying with the angle and is dependent on the projection of forces.


I'm stuck on a limits question

i have attached a image ... plz c it n help me out ....

I'd appreciate everyone's help

THANKS IN ADVANCE

This really should go in the Calculus forum, but I think these are the answers:

[hide="1"]$\lim_{x\to\infty}\sqrt[3]{\frac{3x+5}{6x-8}}=\lim_{x\to\infty}\frac{\sqrt[3]{3x+5}/\sqrt[3]{x}}{\sqrt[3]{6x-8}/\sqrt[3]{x}}=\lim_{x\to\infty}\frac{\sqrt[3]{\frac{3x+5}{x}}}{\sqrt[3]{\frac{6x-8}{x}}}=\lim_{x\to\infty}\frac{\sqrt[3]{3+\frac{5}{x}}}{\sqrt[3]{6-\frac{8}{x}}}$

$\lim_{x\to\infty}\frac{\sqrt[3]{3+\frac{5}{x}}}{\sqrt[3]{6-\frac{8}{x}}}=\frac{\sqrt[3]{3+0}}{\sqrt[3]{6-0}}=\boxed{\sqrt[3]{\frac{1}{2}}\ \text{or}\ \frac{1}{\sqrt[3]{2}}}$[/hide]

[hide="2"]I assume you mean the limit as x approaces 0.

$\lim_{x\to 0^{-}}f(x)=-1, \lim_{x\to 0^{+}}f(x)=1$

$\lim_{x\to 0^{-}}f(x)\neq\lim_{x\to 0^{+}}f(x)\implies\lim_{x\to 0}f(x)\ \boxed{\text{does not exist}}$[/hide]

I think limits such as these can go here.  Things like $\lim_{x\to0}\frac{x}{\sinh}$ belong in the calculus forum.

For the first one you can simply recall from your pre-calc class that  the horizontal asymptote of an equation of the form: $\frac{ax^{n}+...}{bx^{n}...}$ is simply the ratio of the coefficients, in this case : $\frac{a}{b}$.

Taking the $\lim_{x\to\infty}$ is the same as asking for the horizontal asymptote of the equation (or the imaginary line that the y-value never quite reaches but increasingly approaches).


The set of positive integers is partitioned into finitely many subsets. Show that some subset $S$ has the following property: for every positive integer $n$, $S$ contains infinitely many multiples of $n$.

Consider the set $F = \{1!, 2!, 3!, ... \}$. One of the subsets $S$ must contain infinitely many members of $F$, and for every $n$ all large elements of $F$ are multiples of $n$, giving the result.

induction on the number of sets also kills it :D

Assume $A_1, A_2, .., A_k$ are sets of positive integers that partition $\mathbb{N}$ and assume $A_i$ contains only finitely many multiples of $a_i$. Then $\mathbb{N} = \bigcup_{i=1}^k A_i$ contains only finitely many multiples of $\prod_{i=1}^k a_i$; contradiction

Maybe harder than it needs to be, but lets try induction on the number of sets. If a set contains infinitely many multiples of $n$ for every $n$, call it good. $k=1$: obvious. Assume true for $k-1$. Now consider $k$ sets $A_1,..,A_k$. Define $B$ as the union of $A_{k-1}$ and$ A_k$. Then among $A_1,...,A_{k-2}, B$ there is one good set. If this set is not $B$, we are done.

Suppose the set is $B$. If $A_{k-1}$ is good, done. Else there is a number $m$ such that $A_{k-1}$ does not contain infinitely multiples of m. Then for a sufficiently large $t$, $A_k$ contains $mt,m(t+1),m(t+2),...$. In this sequence we can find a multiple of $n$ for any $n$, so $A_k$ is good, done.


parameterization
calculus
integration
number theory
Diophantine equation

Find the positive integrer numbers $a, b, c, d, e$ such that

$a+b+c+d+e=80$
$5(b+d)=a+c$
$c+d=6(a+b)$
$a+2=b$

[quote="t0rajir0u"]Don't you think this is a little easy for Pre-Olympiad?[/quote]

Actually, it was proposed in an olympiad

Well, it must've been an easy olympiad.  I used linear diophantine equations.  It is very clean, but I don't feel like explaining it now.  I believe the phrase "diophantine equations" should lead you on the right way. :)

Masoud Zargar

That's what I got too. What olympiad is this from?

Masoud Zargar

Another way to do this is say $a=t$ (as a parameter) and you can easily solve for the other variables in terms of $t$ because there are 4 linear equations.

Another way would be to reduce the first equation, using the other equations, into a linear diophantine equation with two unknowns.  Then you can create bounds for the values to obtain integral solutions.  This leads us to checking a few cases; you immediately get decimal values when you have acquired not all integer values.  You will then get the required solution. :)

Masoud Zargar

[hide="my solution"]From (1) we have that $80-e=a+b+c+d$. Plugging (3) here holds $a+b+6a+6b=7(a+b)=80-e$.

From (1) we have that $a+b+c+d=80-e$. Plug (2) in (1) to have $5b+5d+b+d=6(b+d)=80-e$.

Then, $6|80-e$ and $7|80-e$, so $42|80-e$.

We know that $80-e=42k$, so $k=1$, because if not $42k \ge 84 > 80-e$. So, $80-e=42 \Longleftrightarrow e=38$.

Now, $7(a+b)=80-e=42$, but plugging (4) we have $7(2a+2)=42 \Longleftrightarrow a=2$. Then, $b=4$.

Finally, solving the following system
$c+d=36$
$5(d+4)=c+2$ holds that $c=33$ and $d=3$.[/hide]


The difference between the arithmetic mean and the geometric mean of two positive integers $ m$ and $ n$ is 1. Prove that $ \frac {m}{2}$ and $ \frac {n}{2}$ are perfect squares.

$ \frac{m\plus{}n}{2} \equal{} 1\plus{}\sqrt{mn}$

Clearly $ mn$ is a perfect square, therefore $ m\equal{}a^2 t$, $ n\equal{}b^2 t$ for some square free $ t$

Therefore

$ \frac{t(a^2\plus{}b^2)}{2} \equal{} 1\plus{}abt \Rightarrow t(a\minus{}b)^2 \equal{} 2 \Rightarrow t \equal{} 2$ as required

[quote="SimonM"][quote="thomaskurian89"][quote="SimonM"]$ t(a \minus{} b)^2 \equal{} 2 \Rightarrow t \equal{} 2$[/quote]

Sorry I don't follow this part.[/quote]

a and b are integers

Therefore their difference squared can be $ 0,1,4,9,..,,$

So if the product of that and another positive integer is 2, then we can conclude...[/quote]

Of course. Thanks.  :)


trigonometry
function
algebra
polynomial
real analysis
real analysis unsolved

1, Find the equivalence at infinity of the following sequence defined: $u_0\in R^+$ and
i,$u_{n+1}=\sin u_n$
ii,$u_{n+1}=u_n+\sqrt [4]{u_n}$
iii,$u_{n+1}=u_n+u_n^2$
2, Define the sequence of real number $\{u_n\}$: $0<u_0=a\in R$ and
\[ u_{n+1}=u_n+u_n^{\beta_1}+u_n^{\beta_2}+...+u_n^{\beta_k}. \]
Please discuss the equivalence of $u_n$ at infinity when $a,k,\beta_i$ (for $i=1,2,...,k$) vary, $k\in N$, $\beta_i$ are distinct.

i and ii are very classical. Study $v_n=u_n^p$ where $p$ is a well chosen real such that $v_{n+1}-v_n$ tends to a real $L \in \mathbb{R}$

$u_{n+1}=u_n+u_n^2$, $u_0=A>0$.

Clearly $u_n$ tends to infinity, exponentially. Let's study $v_n=ln(u_n)$:
$v_{n+1}=2v_n+ln(1+\frac{1}{u_n})=2v_n+f(n)$ where $f(n)$ tends to $0$.

Hence $v_n$ is like $2^n$. So one can study $w_n=\frac{v_n}{2^n}$:
$w_{n+1}=w_n+\frac{f(n)}{2^{n+1}}$. Summing this equality from $1$ to $n$ we get that:
$w_n=w_1+\sum_1^{n-1} \frac{f(k)}{2^{k+1}}=B+\frac{g(n)}{2^n}$ where $B \in \mathbb{R}$ and $g(n)$ is a function that tends to $0$ because $f(n)$ tends to $0$.

This why: $v_n=B2^n+g(n)$ and thus $u_n$ is equivalent to $(e^B)^{2^n}=D^{2^n}$.

the same proof shows that in the last question, $u_n$ is equivalent to some $D^{b^n}$ where $b=max{\beta_i}$.
And I think (almost the same proof) that if $u_{n+1}=P(u_n)$ where $P$ is a polynomial with leading coefficient $ax^p$, $a>0$, then if $u_n$ tends to $+\infty$, then $u_n$ is equivalent to $A.B^{p^n}$ where $A>0$,$B>0$.


Suppose that u1>u2>u3> ... and that there are infinitely many indices n for which Un >= 1/n.
          Prove that there exists a positive integer N for which
             u1+u2+...+un>2006

First , can anyone pls tell me what "indices" mean? ==  Thxx!

$ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k}=\infty$.
Summing the $ u'_{i}$ for which $ u_{i}\geq \frac{1}{i}$ it follows that this if this series diverges the answer is straightforward.
Otherwise if it were not to diverge than it would coverge. Let $ \epsilon$ be the smallest difference between two consecutive members.
Then $ \sum u_{i}\geq \2\sum u'_{i}+N\epsilon$ which diverges so the answer is also straightforward.
I am not quite sure about the $ \epsilon$ argument because I am in a hurry. I will get back on this.

You can use the following theorem:

[color=blue]If $ \left\{ u_{n}\right\}$ is a decreasing sequence of reals such that $ \lim_{n \to \infty}u_{n}= 0$ and $ \sum_{n=1}^\infty u_{n}$ converges, then $ \lim_{n \to \infty}n u_{n}= 0$.[/color]

You can find a proof of this theorem in the Calculus sub-forum (I think it even has a name, but I forgot it).

That theorem is essentially the limit comparison test for $ u_{n}$ and $ v_{n}= \frac{1}{n}$, is it not?

The index in "$ u_{n}$" is the $ n$ -- it is the subscript which indexes the terms of the sequence.  The plural of index is indices.

This thread belongs in the Calculus Computations and Tutorials forum.

Please use LaTeX.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 .

[quote="JBL"]That theorem is essentially the limit comparison test for $ u_{n}$ and $ v_{n}= \frac1{n}$, is it not?[/quote]
Actually, I can only prove that $ \liminf n u_{n}= 0$ directly from the limit comparison test (to conclude this, we can even drop the monotonicity condition). The other part ($ \limsup$) is the one needed here.

The argument I know about is quite clever (and was first mentioned on the forum by either grobber or harazi). The great thing about it is that it provides yet another proof of the divergence of the harmonic series.


OK.  I want to put my mock mathcounts competition in my signature, but since the url is more than 75 consecutive characters, I can't.  How do I hide the url with a word?  Like when they sey look [color=blue][u]here[/u][/color](but it actually takes you someplace)

[code][url=URL HERE]DISPLAY TEXT HERE[/url][/code]

Are you sure? I tried it, see:
[url=URL HERE]DISPLAY TEXT HERE[/url]

The URL has to have "http://" at the beginning.

Let me test:
[url=http://www.artofproblemsolving.com/Forum/posting.php?mode=reply&t=318483]here[/url]

Thanks.


Please help with this geometry problem:

In the diagram to the right (attached) points $ A$, $ B$, $ C$, and $ P$ are situates so that $ PA \equal{} 2$, $ PB \equal{} 3$, $ PC \equal{} 4$, and $ BC \equal{} 5$.  What is the maximum possible area of $ \triangle ABC$?

Let $ S$ be the area of $ ABC$. We are given the lengths of triangle $ PBC$, so we can fix it and let $ A$ vary. Take $ X$ on $ BC$ as $ A$ varies such that $ AX$ is perpendicular to $ BC$. Also take $ Y$ on $ BC$ such that $ PY$ is perpendicular to $ BC$.

First note $ PBC$ is a right triangle, so $ [PBC] \equal{} \frac {3 \cdot 4}{2} \equal{} 6.$

Now note that $ S \equal{} \frac {(BC)(AX)}{2}$ so, since $ BC$ is fixed, we want to maximize $ AX$. We see that $ AX \le PA \plus{} PY$ with equality if and only if $ X \equal{} Y$, whence the maximum value of $ S$ is
\[ \frac {(BC)(PA \plus{} PY)}{2} \equal{} \frac {(BC)(PA)}{2} \plus{} \frac {(BC)(PY)}{2} \equal{} \frac {5 \cdot 2}{2} \plus{} [PBC] \equal{} 5 \plus{} 6 \equal{} 11.
\]

[hide]clearly $ \triangle BCP$ is right angled with area 6.

let $ \angle APB \equal{} \theta$

then $ [ABC] \equal{} 6 \plus{} 3\sin\theta \plus{} 4\sin(270 \minus{} \theta) \equal{} 6 \plus{} 3\sin\theta \plus{} 4\cos\theta$.

$ \frac {3\sin\theta \plus{} 4\cos\theta}{5} \equal{} \sin(\theta \plus{} x)\le 1$, for x with $ \cos x \equal{} 3/5, \sin x \equal{} 4/5$.

thus, $ 3\sin\theta \plus{} 4\cos\theta \le5$, with equality when $ \theta \equal{} 90 \minus{} x$

so maximum area is 5+6=11.[/hide]


geometry
geometric transformation
rotation
trigonometry
geometry proposed

The distances from a point $ P$ inside an equilateral triangle to the vertices of the triangle are $ 3,4$, and $ 5$. Find the area of the triangle.

Assume that $PA=5,PB=3$ and $PC=4.$ Let $Q$ be the homologous of $P$ under the rotation with center $B$ and rotational angle $60^{\circ}$ counterclockwise. Then $\triangle QBC \cong \triangle PBA$ and $\triangle BPQ$ is equilateral. Consequently, $QC=5$ and $PQ=3$ $\Longrightarrow$ $\triangle PQC$ is right at $P$ $\Longrightarrow$ $\angle BPC=150^{\circ}.$ Hence, by cosine law in $\triangle PBC$ we get

$AB^2=4^2+3^2-2 \cdot 3 \cdot 4 \cdot \cos 150^{\circ}=25+12\sqrt{3} \Longrightarrow  |\triangle ABC|=\frac{25 \sqrt{3}}{4}+9$


geometric series
real analysis
real analysis unsolved

Hi all,

This is a partial solution to a question out of a purported "analysis" problem book but I'm having trouble following two things here. The intent is to find the coefficient of $ x^k$ in the expression $ \left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{(n \minus{} 1)}\right)^2$. So the partial solution is as follows:

$ \left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{(n \minus{} 1)}\right)^2 \equal{} \frac {\left(x^n \minus{} 1\right)^2}{(x \minus{} 1)^2} \equal{} \left(x^{2n} \minus{} 2x^n \plus{} 1\right)\left(1 \plus{} 2x \plus{} 3x^2 \plus{} \ldots \plus{} \text{mx}^{m \minus{} 1} \plus{} \ldots \right)$

And my questions are as follows:

Numero uno: While the first equality is correct, I don't see how the author of the solution would be motivated to put the series on the left into the expression on the right of the first equality sign. Is the manipulation very common?

Number two: I see how for the second equality, $ \left(x^n \minus{} 1\right)^2$ becomes $ \left(x^{2n} \minus{} 2x^n \plus{} 1\right)$, but not the rest. Again, the series $ \left(1 \plus{} 2x \plus{} 3x^2 \plus{} \ldots \plus{} \text{mx}^{m \minus{} 1} \plus{} \ldots \right)$ seems very prototypical, but I seem to not follow the reasoning there.

Thanks folks.

I'm not sure what your first question is, $ \sum_{k = 0}^{n - 1}x^k = \frac {x^n - 1}{x - 1}$ is the sum of a geometric series commonly covered in a precalculus class.

The second equality, though certainly correct for $ |x| < 1$, is indeed very odd.  It seems the author expects you to compute the series for $ \frac {1}{(x - 1)^2}$, but that itself involves computing ${ (\sum_{k = 0}^\infty x^k})^2 = \sum_{k = 1}^\infty kx^{k - 1}$.  Though that isn't very difficult (just match terms), it's almost exactly the same as computing the $ k$th term of $ (\sum_{k = 0}^{n - 1}x^k)^2$, which is the problem you're solving to begin with.

Seems like a very roundabout way to do something very simple; the answer is
\[ (\sum_{k = 0}^{n - 1}x^k)^2 = \sum_{k = 0}^{2(n - 1)}(1 + \min  \lbrace k ,2(n - 1) - k \rbrace) x^k
\]
Moreover this holds for all $ x\in \mathbb{R}$ not just $ |x| < 1$, and doesn't require computing any infinite series, so the calculation the author chooses to make seems very unnatural (unless there is some context I'm missing).

Which book is this?


Positive integer $k$ and $n$ points in a plane are given. Find $s\leqslant k$ of these points such that their convex hull has the greatest area.

The best my friends and I could do is $O(n^3k)$, but we heard that there is also $O(nk)$! Could somebody help?

Well u r given positive integer k and n points on plane. And u need to find s points such that s<=k and the convex hull area of those s points is maximum?

Say u r given 5 points. The square vertices and the center. Then say K=3. then answer will be any of the three vertices of square right?

But I can’t understand one thing why s<=k is needed? If we choose K points what will be wrong?

Btw, what’s ur $n^3k$ algo?


Let $ ABC$ be a triangle and $ \omega(I_A)$,$ \omega(I)$,$ \zeta(O)$,be its $ A$-exincircle,incircle and circumcircle respectively.Let $ M$ be the midpoint of arc $ BC$ and $ K$ be the midpoint of arc $ AM$.Denote $ P\in KI\cap\zeta$,$ Q\in KI_A\cap\zeta$.If $ N\in AM\cap BC$,then prove that $ P,Q,N$ are collinear.

Here is the picture:
[img]http://img227.imageshack.us/img227/5858/asaniq5.png[/img]
I want to believe that it will be easier to work with this picture and to not consider the case of definition of the point $ M$.

Although the solution is simple, it took me some time to find it! Beautiful problem, [i]Erken[/i]!

We'll prove the statement as follows, assume $ P\equal{}KI\cap\zeta$ and $ PN\cap\zeta\equal{}Q$. We show that $ I_A, Q, K$ are collinear.
It's easy to see that $ A,I,N,M,I_A$ are collinear. Also easily we have $ IB\bot I_AB$ and $ IC\bot I_AC$. Then $ IBI_AC$ is an inscribed quadrilateral. Then $ NI\cdot NI_A\equal{}NB\cdot NC\equal{}NP\cdot NQ$, from where $ P,I,Q,I_A$ are concyclic. We'll use the following quite trivial, but not so often used in this formulation
[i]Property:[/i] Let $ ABCD$ be a convex inscribed quadrilateral, with diagonals intersecting at $ X$. Then $ \angle{AXB}\equal{}\dfrac12(\widehat{AB}\plus{}\widehat{CD})$, taken as small arcs ([b]here[/b] $ \widehat{}$ [b]denotes the arc[/b]).
Then $ \angle{PIN}\equal{}\dfrac12(\widehat{PM}\plus{}\widehat{AK})\equal{}\dfrac12(\widehat{PM}\plus{}\widehat{MK})$, since $ K$ is midpoint of arc $ \widehat{AM}$ (containing $ K$). Then $ \angle{PIN}\equal{}\dfrac{\widehat{PK}}2$. But the arc $ \widehat{PK}$ containing $ M$ equals $ 180\minus{}\angle{PMK}\equal{}180\minus{}\angle{PQK}$. Since $ PIQI_A$ is cyclic, $ \angle{PIN}\equal{}\angle{PQI_A}$, so $ \angle{PQI_A}\equal{}180\minus{}\angle{PQK}$, from where $ I_A,Q,K$ are collinear, as desired.


I live about 20 miles out of NYC, so that means I don't fall into the NYC region, but I'm not geographically upstate. What region am I considered part of?

Well it'll probably be easier to go with NYC...I am in a position similar to yours, so Stuyvesant is 60-90 minutes away, but Albany is like 4+ hours...


analytic geometry
geometry
incenter
geometry proposed

$P$ is a point inside the triangle $ABC$. The line through $P$ parallel to $AB$ meets $AC$ $A_0$ and $BC$ at $B_0$. Similarly, the line through $P$ parallel to $CA$ meets $AB$ at $A_1$ and $BC$ at $C_1$, and the line through P parallel to BC meets $AB$ at $B_2$ and $AC$ at $C_2$. Find the point $P$ such that $A_0B_0 = A_1B_1 = A_2C_2$.

Where is $ B_1$ and $ A_2$??
or you want to mean $ A_0B_0\equal{}A_1C_1\equal{}B_2C_2$

I also think that it should be $A_0B_0=A_1C_1=B_2C_2.$  Let $(u:v:w)$ be the normalized  barycentric coordinates of $P$ WRT $\triangle ABC.$ Then lengths of the paralell sections from $P$ to $BC,CA,AB$ are given by

$B_2C_2= a(1-u) \ \ , \ A_1C_1=b(1-v) \ \ , \ A_0B_0=c(1-w)$

Solving $a(1-u)=b(1-v)=c(1-w),$  together with $u+v+w=1$ gives

$P \equiv \left( \frac{1}{b}+\frac{1}{c}-\frac{1}{a}:\frac{1}{a}+\frac{1}{c}-\frac{1}{b}:\frac{1}{a}+\frac{1}{b}-\frac{1}{c} \right),$

which is the anticomplement of the isotomic conjugate of the incenter of $\triangle ABC.$

Hint:
You can use of this lemma :
$(A_0B_0/AB)+(A_1C_1/AC)+(B_2C_2/BC)=2$


trigonometry
quadratics
logarithms
algebra
polynomial
product of roots
AMC

14. What is the units digit of $3^{1001}7^{1002}13^{1003}$?

$\textrm{(A) } 1 \quad \textrm{(B) } 3 \quad \textrm{(C) } 5 \quad \textrm{(D) } 7 \quad \textrm{(E) } 9$

20. If $\tan\alpha$ and $\tan\beta$ are the roots of $x^2 - px + q = 0$ and $\cot\alpha$ and $\cot\beta$ are the roots of $x^2 - rx + s$, then $rs$ is necesarily

$\textrm{(A) } pq \quad \textrm{(B) } \frac{1}{pq} \quad \textrm{(C) } \frac{p}{q^2} \quad \textrm{(D) } \frac{q}{p^2} \quad \textrm{(E) } \frac{p}{q}$

29. Given unit square ABCD and point P in the same plane such that $AP^2 + BP^2 = CP^2$, find the maximum value of $PD$.

$\textrm{(A) } 1 + \sqrt{2} \quad \textrm{(B) } 2\sqrt{2} \quad \textrm{(C) } 2+\sqrt{2} \quad \textrm{(D) } 3\sqrt{2} \quad \textrm{(E) } 3+\sqrt{2}$

[hide="1"]3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81

So, the pattern is 3-9-7-1. Then 1001th power = 1.

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1

So, the pattern is 7-9-3-1. Then 1002th power = 9.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81

So, the pattern is 3-9-7-1. Then, 1003th power = 3.

1*9*3 = 7. So, the 7 is the units digit.[/hide]

Maybe it would've been easier if I made them to lower number but I just like patterns. :P

Number 2:
[hide="click text"]
The only thing we need to note about $\tan$ and $\cot$ is that they are reciprocals.

The roots of the first equation are $(p\pm\sqrt{p^2-4q})/2$. For simplicity, also call these $r_1$ and $r_2$. Then the second equation is $(x-1/r_1)(x-1/r_2)=0$, which expands to $4x^2-(1/r_1+1/r_2)x+1/(r_1r_2)=0$. $rs$ equals $1/(r_1r_2)*(1/r_1+1/r_2)$. Substituting the p's and q's back in, we get $1/(4q/4)*(2/(p+\sqrt{p^2-4q})+2/(p-\sqrt{p^2-4q}))$, which equals $2/q^2*(p-\sqrt{p^2-4q}+p+\sqrt{p^2-4q})$.
This simplifies down to $1/(2q^2)*(2p)$, which equals choice C.
[/hide]

AHSME 1983 had some awesome problems!!!!

[hide="hint for 29"]Label A=(0, 0),  B=(1, 0),  C=(1, 1), D=(0, 1), P=(x, y), what's the locus of P?[/hide]

also try #25

$60^a=3$, and $60^b=5$, then $12^{\frac{1-a-b}{2(1-b)}}$ is?[/hide]

For #1, I go the same patterns as you but I got different answers.  THe way I saw it for 3^1001 everything power that is divisible by 4, the units digit will be a 1.  So 3^1000 will end in a 1, so 3^1001 will end in a 3.  And so on.  Tell me if I am wrong because I know these questions are popular on the AMC 12 which I'm taking.

[quote="mathfanatic"]20. If $\tan\alpha$ and $\tan\beta$ are the roots of $x^2 - px + q = 0$ and $\cot\alpha$ and $\cot\beta$ are the roots of $x^2 - rx + s$, find $rs$.[/quote]

We can do a linear transformation. if $x^2 - rx + s = 0$ has solutions $\cot\alpha$ and $\cot\beta$, setting $x = \frac{1}{y}$, we yield $\frac{1}{y^2} - \frac{r}{y} + s = 0$ has solutions $\tan\alpha$ and $\tan\beta$. Multiplying out, we yield $y^2 - \frac{ry}{s} + \frac{1}{s}$ and $x^2 - px + q$ both have the same roots with the same first term, so they must be identical. Thus $p = \frac{r}{s}$ and $q = \frac{1}{s}$, and so $rs = \frac{p}{q^2}$

PS Now I realise it is better to do the linear transformation on the other one so that we can just blindly multiply it out if needs be. But it is not that much slower.

Also for the quadratic one.  From the point where you have the equation that rs= 1/r1r2 * (1/r1  +  1/r2), you can use sum and product of roots formulas.  After rearranging rs=1/r1r2 * ((r2 +r1)/r2r1) = (r2+r1)/((r2r1)^2).  THen using the sum and product equations from equation 1 you get p/q^2.

[hide]For number one I got 9 for somereason. I used the same method as silverfalcon. Think I musta divided wrong. NUmber two can be solved by noting that r is the product and s is the sum of the roots. It'll simplify nicely. THe one that beta gave define a and b in log base 60 and it'll simplify. I can't figure out the last one.[/hide]

Number 14 is 9. Silverfalcon, you said the pattern for 3's was 3-9-7-1 so 3^1001 = 3^1 = 3 (mod 10), right? And for 13 its the same, so 13^1003 = 13^3 = 3^3 = 7 (mod 10). Then (3)(9)(7) = 9 (mod 10).

Oops.

I wrote the wrong one. Yep, it was a mistake.

Better not make anything like this on Feb.16 :blush:

#25

[hide]All we do is calculate the thing in a very straightforward way.Since $a=\log_{60}3$ and $b=\log_{60}5$, we have \[1-a-b=\log_{60}\left(\tfrac{60}{3\cdot5}\right)=\log_{60}4\] and \[2(1-b)=2\log_{60}\left(\tfrac{60}{5}\right)=2\log_{60}12.\] So \[\frac{1-a-b}{2(1-b)}=\frac{\log_{60}4}{2\log_{60}12}=\tfrac12\log_{12}4.\] Now it's easy to see that \[12^{\tfrac{1}{2}\log_{12}4}=4^{\tfrac12}=\mathbf{2}.\][/hide]

The first is definately 9.

[hide]All computations are done mod 10.

3^1001*7^1002*13^1003 :equiv: 3*7^2*3^3=3^2*(3*7)^2 :equiv: 3^2=9 [/hide][/hide]


As I couldn't find a section of this forum that would be really appropriate for this topic, I decided to post it here. Now for my problem:
I repeatedly sent e-mails to GIL Publishing, because I'm strongly interested in their book "Old and new Inequalities", but my enquiries were always ignored, I never got any reply. Furthermore I have been unable to order the book via their website. Either my browser (Mozilla Firefox) can't cope with that homepage or it is really defective in some parts. All in all I haven't been able to order it in any way. Do you have any suggestions, how I could get hold of that book? I'm not from Romania as you can see, but from Germany, so I need a way to get it from abroad.

Hi! Please tell me the e-mail address that you wrote those enquiries. I must tell you that we answer to anyone, romanian or foreign. Waiting for your answer and then I will give you a solution to your problem. :)


Solve the equation.

[tex] \cos(\cos(\cos(\cos x)))=\sin (\sin (\sin (\sin x)))[/tex]

-----It's a problem that thinking step by step...

      Good luck.. It's a really nice problem.

Indeed it's fun  :)

I think there are no such x, but so far I failed to prove without a computer  :( .

I had to split between two cases. Let 1 > a > 0 (we are looking for such a).

I consider two cases :

1/ 0 <= |sin(x)| <= a < 1
then
0 <= |sin^4(x)| <= |sin^3(a)|

Then sqt(1-a) <= |cos(x)| <= 1
And  cos^3(1-a) >= |cos^4(x)| >= cos^3(1) = 0.6542897905

Condition 1 for a is : sin^3(a) <= cos^3(1).

2/ a < |sin(x)| <= 1
then
|sin^3(a)| < |sin^4(x)| <= sin^3(1)

Also 0 <= cos(x) < sqrt(1-a)
then cos^3(sqrt(1-a)) <= cos^4(x) <= cos(cos(1))

Condition 2 for a is : sin^3(1) <= cos^3(sqrt(1-a)).

I solved both conditions using a calcultor. a = 1/2,  works.

This problem was also discussed in the following topic:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7344[/url]


algebra
difference of squares
special factorizations
number theory unsolved
number theory

Find all prime numbers p for which
a- (7^(p-1)-1)/p is perfect square.
b- (11^(p-1)-1)/p is perfect square

[hide="hint"]The problems are extremely similar to [url=http://www.mathlinks.ro/viewtopic.php?p=341880#341880]this one[/url], so strongly suspect the same difference of squares trick will work $ \frac{a^{p\minus{}1}\minus{}1}{p}\equal{}\frac{(a^{\frac{p\minus{}1}{2}}\minus{}1)(a^{\frac{p\minus{}1}{2}}\plus{}1)}{2}$, but I don't have the time now to post a complete solution :P [/hide]


Lets consider for each natural [tex]n[/tex] similar triangle board as in the picture (theres the situation for [tex]n[/tex]=3, for arbitrary [tex]n[/tex] it only suffices to divide the side of the original (greatest) equilateral triangle in the picture into [tex]n[/tex] same parts). By this we obtain [tex]n^2[/tex] cells in the board, now lets imagine that well inscribe all numbers 1, 2, ..., [tex]n^2[/tex] in the cells so that each cell contains exactly one number and no two cells include the same number. Find out the minimal number of colours [tex]x(n)[/tex] (dependent on [tex]n[/tex]) that are necessary for the colouring of the cells of the triangle board that satisfies these two conditions:
1) Each two neighbouring cells (only by edge, not by vertice) are coloured by different colours.
2) Each two cells that have inscribed any two consecutive natural numbers are coloured by different colours (so the cells with numbers [tex]k[/tex] and [tex](k + 1)[/tex] for each [tex]k < n[/tex] are coloured by different colours).

[img]http://www.email.cz/_shared/getpart.phtml/1/8369521sQcS1qNanlkjS4011173Wzs0ibvR6xhuWydql6r/355785150/image/pjpeg/4356/pic.jpg[/img]

I think I have a solution (and maybe with proof too), but Id reather like to verify it, so my results are:
- [tex]x(1) = 1[/tex],
- [tex]x(2) = 3[/tex],
- if [tex]n > 2[/tex] then [tex]x(n)[/tex] = 4.
Is it true? Thanks for all your replies...

[quote="Dapet"]so the cells with numbers [tex]k[/tex] and [tex](k + 1)[/tex] for each [tex]k < n[/tex] are coloured by different colours.[/quote]
$k<n^2$. Right?

[quote]I think I have a solution (and maybe with proof too), but I’d reather like to verify it, so my results are:
- [tex]x(1) = 1[/tex],
- [tex]x(2) = 3[/tex],
- if [tex]n > 2[/tex] then [tex]x(n)[/tex] = 4.
Is it true? Thanks for all your replies...[/quote]
Quick look gives that, indeed, $x(1)=1$ and $x(2)=3$ ( :) ), and $x(n)\leq 4$. But I couldn't find example for $n=3$, where we need 4 colors.

hmm $x(1)=1$, $x(2)=3$ and $x(n)\geq 4$ for $n\geq 3$ i found. just take 4 triangles i a row and call them 2,4,1,3 in that order. then no two may have the same color. and remember some theorem with which we may deduce $x(n)\leq 4$ but i believe it's simpler...but again, i'm blind.

[quote="Peter Scholze"]just take 4 triangles i a row and call them 2,4,1,3 in that order. then no two may have the same color.[/quote]
Oooppsss...  :blush:
The rest is very simple. Color board as chessboard. Now
if $2k$ is written on white, then recolour it in red;
if $2k$ is written on black, then recolour it in green;
if $2k+1$ is written on white, then recolour it in blue;
and if $2k+1$ is written on black, then recolour it in yellow.
Ok?

AAAHHHHH! I'M SOO BLLLLLIIINNNNNDDDDDDDDD!!!!!!!!!!!

We both didn't see obvious things...  :P

I think that this is what we find for [tex]n=3[/tex]:
[img]http://www.email.cz/_shared/getpart.phtml/1/8369521s9201DWP6hGka4011170nfIXKwvtq0gCEhjRWuW/355904807/image/pjpeg/8189/pic2.jpg[/img]
If "5" is coloured by colour 1, "3" is coloured by colour 2, "6" is coloured by colour 3 ("5", "3" and "6" must be coloured (pairwise) by different colours), then "4" must be coloured by colour 1 - that's dispute that 3 colours suffices ("4" and "5" must be coloured by the same colour). And it's clear that it could be inductively used for each [tex]n>2[/tex]. In the other hand it's obvious that 4 colours always suffice as you've written...


Im just curious, as i dont really know, in the china TST quiz, with the following question:

Given that circle ω is tangent internally to circle Γ at S. ω touches the chord AB of Γ at T .
Let O be the center of ω. Point P lies on the line AO. Show that PB ⊥ AB if and only if PS ⊥ TS.

To prove this, do you have to first assume PB ⊥ AB and then prove PS ⊥ TS.

Then assume  PS ⊥ TS and then prove PB ⊥ AB?

[quote="Arrange your tan"][quote="kurt.math"] Show that PB ⊥ AB >>  if and only if << PS ⊥ TS.

To prove this, >> do you have to first assume << PB ⊥ AB and then prove
 PS ⊥ TS?

Then assume  PS ⊥ TS and then prove PB ⊥ AB?[/quote]

No, because "if and only if" is a double implication, there is no
mandatory order with which direction to start.  You may
or you may not start with the direction you asked about.[/quote]

So how would i prove an "if and only if" situation?

first prove a statement and then prove its converse.......
If you are given statement $ P_a$ & $ P_b$ and are required to prove $ P_a\implies P_b$ iff {..} then you prove : $ P_a\implies P_b$ and then also show that $ P_b\implies P_a$ :rotfl:

But isnt that exactly what i said in my first post? :huh:

[quote="dgreenb801"]Yes, it is. He is just saying is doesnt matter which order you prove the statements. You could show ($ A \implies B$ and then $ B \implies A$), or ($ B \implies A$ and then $ A \implies B$).[/quote]

yeh i know, i probably shouldnt have wrote "firstly", it was just to show that i would do this first (over the other).

Thanks heaps everyone  :lol:


Euler
inequalities
number theory unsolved
number theory

Find all positive interger such that:$ (f(n)\minus{}1)|(n\minus{}2)$

despite solution $ n\equal{}2^k$ and $ n\equal{}p$ be prime number,I see a solution other $ n\equal{}6$ :D

[quote="math10"]Find all positive interger such that:$ (f(n) \minus{} 1)|(n \minus{} 2)$

despite solution $ n \equal{} 2^k$ and $ n \equal{} p$ be prime number,I see a solution other $ n \equal{} 6$ :D[/quote]

Hemm, what is $ f(x)$ ?

Yes, it's nice problem.
If $ n\minus{}2\equal{}f(x)\minus{}1$ then n is a prime.
If $ n\minus{}2>f(x)\minus{}1$ then we denote $ n \equal{} \prod\limits_{i \equal{} 1}^k {p_i^{c_i } }$ and use inequality $ \sqrt[k]{2} \ge \frac{{p_i }}{{p_i  \minus{} 1}}$


A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full price tickets?

We have $ f\plus{}h\equal{}140$ and $ pf\plus{}\frac{ph}2\equal{}2001\implies p(2f\plus{}h)\equal{}4002$

$ \implies p(140\plus{}f)\equal{}4002\implies p\equal{}\frac{4002}{140\plus{}f}$, where $ f,h$ are the number of full and half price tickets, and $ p$ is the price of one full price ticket.

Since $ 4002\equal{}2\cdot3\cdot23\cdot29$, and $ p,f\in\mathbb Z^\plus{}$, we see that $ 0\leq f\leq140\implies140\leq140\plus{}f\leq280$, as well.

Thus, $ 140\plus{}f\equal{}2\cdot3\cdot29\equal{}174\implies f\equal{}34\implies p\equal{}23$ is the only possibility, yielding the answer $ 34\cdot23\equal{}\boxed{782}$.

Checking, $ 23\cdot34\plus{}\frac{23}2\cdot(140\minus{}34)\equal{}2001$.