AMC AIME algebra polynomial calculus integration trigonometry

1) (AIME 1991) Find $ x^2 \plus{} y^2$ if $ x$ and $ y$ are positive integers such that $ xy \plus{} x \plus{} y \equal{} 71$ and $ x^2y \plus{} xy^2 \equal{} 880$.

2) (AIME 1986) The polynomial $ 1 \minus{} x \plus{} x^2 \minus{} x^3 \plus{} ... \plus{} x^{16} \minus{} x^{17}$ may be written in the form $ a_0 \plus{} a_1y \plus{} a_2y^2 \plus{} ... \plus{} a_{16}y^{16} \plus{} a_{17}y^{17}$, where $ y \equal{} x \plus{} 1$ and the $ a_i$’s are constants. Find the value of $ a_2$.

3)(Putnam 1939) Let $ x^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0$ have integral coefficients and roots $ r$, $ s$, $ t$. Find
a polynomial equation with integer coefficients written in terms of a, b, c, d whose roots are $ r^3$, $ s^3$, $ t^3$.

4) (MOSP) Determine all triples of real numbers $ (x, y, z)$ satisfying the system of equations

$ x \minus{} xz^2 \equal{} 2z$
$ y \minus{} x^2y \equal{} 2x$
$ z \minus{} y^2z \equal{} 2y$.

[hide=" Problem 1, AIME 1991"]
Set $ xy\equal{}a$ and $ x\plus{}y\equal{}b$.  Then, substituting and algebra yields that $ a\plus{}b\equal{}71$ and $ ab\equal{}880\equal{}2^4*5*11$.

It just so happens that the factors 16 and 55 work.  Also, as $ a$ was derived using multiplication and $ x$ and $ y$ are integers, $ a$ will be the larger factor.  Finally, $ x^2\plus{}y^2 \equal{} (x\plus{}y)^2 \minus{}2xy \equal{} b^2\minus{}2a$

Thus, $ x^2\plus{} y^2 \equal{} 16^2\minus{}2(55) \equal{} 256\minus{}110 \equal{} 146$
[/hide]

[hide="Problem 2, AIME 1986"]
Check me on this...
$ \displaystyle 1\minus{}x\plus{}x^{2}\minus{}x^{3}\plus{}...\plus{}x^{16}\minus{}x^{17} \equal{} \frac{1\minus{}x^{18}}{1\plus{}x}$
$ \displaystyle y\equal{}x\plus{}1 \rightarrow \frac{1\minus{}(y\minus{}1)^{18}}{y}$

To find the coefficient of $ y^2$, we must find the coefficient of $ y^3$ in the expression in the numerator.  By the binomial theorum:

$ \displaystyle \minus{}\left(\binom {18}{3} \times (\minus{}1)^3\right)\equal{}816$
[/hide]

hello, solution to problem 4) from the third equation we get $ z\equal{}\frac{2y}{1\minus{}y^2}$ setting this in the first equation we have $ x\minus{}x\left(\frac{2y}{1\minus{}y^2}\right)^2\equal{}\frac{4y}{1\minus{}y^2}$ and from the second equation we have $ y\equal{}\frac{2x}{1\minus{}x^2}$, inserting this in the equation above and simplifying and factorring we get
$ x(x^2\plus{}1)(x^6\minus{}21x^4\plus{}35x^2\minus{}7)\equal{}0$, only $ x\equal{}0$ fulfills our equation, so we have the solution of our system as $ x\equal{}y\equal{}z\equal{}0$.
Sonnhard.

[hide="4"]First, note none of $ x, y, z$ equals $ \pm 1$. Rewrite the equations as
$ x \equal{} \frac {2z}{1 \minus{} z^2}, y \equal{} \frac {2x}{1 \minus{} x^2}, z \equal{} \frac {2y}{1 \minus{} y^2}$
which resemble the tangent double angle formula. Indeed, let $ z \equal{} \tan \theta$, then $ x \equal{} \tan 2\theta$, $ y \equal{} \tan 4\theta$, and $ z \equal{} \tan 8\theta$. Therefore $ \tan \theta \equal{} \tan 8\theta$ which means $ \theta \equiv 8\theta \pmod{\pi} \iff 7\theta \equiv 0\pmod{\pi}$. Thus $ \theta \equal{} \frac {k\pi}{7}$ for integer $ k$. The solution to the original system is $ (\tan \frac {k\pi}{7}, \tan \frac {2k\pi}{7}, \tan \frac {4k\pi}{7})$ for integer $ k$ (particularly $ 0, 1, 2, ..., 6$).
[/hide]

hello, yes, you have right, i have forgotten to solve the equation $ x^6\minus{}21x^4\plus{}35x^2\minus{}7\equal{}0$.
Sonnhard.


Hi, I'm new here just signed up. I got a problem for homework that's been troubling me so I did a quick google for math help forums and came up with this place. Any help you give me is MUCH appreciated, I also hope I posted this in the right area.  :)

Anyway here is the question:

[i]Peter entered a competition in which he got 5 points for every correct answer, but lost 3 points for every wrong answer. At the end of the competition he had no points at all and had answered 16 questions.[/i]

That was basically the first statement. Then we had to work out this problem:

[i]Work out how many questions Peter got right from the 16 questions he answered.[/i]

I did this correctly, I think.

X questions wrong @ -3 points ea.
16 - X questions right @ 5 points ea.

-3x + 5 (16 - x) = 0

-3x + 80 - 5x = 0

-8x = -80

x = 10.

I'm fairly sure that I got that question right, but further down the track came this question;

[i]What numbers of questions could you answer to end up with no points?[/i]

I thought that I could use the same formula in a way as the first question, so I used this:

Let there be N questions where 0 > n [u]>[/u] 20
Let the number of correct answers = X @ 5 points ea.
Therefore, let the number wrong = n - x.  @ -3 points ea.

5x - 3 (n - x) = 0

Now I'm a bit confused. If I keep working it out I'll get something like this:

5x - 3n - 3x = 0

2x - 3n = 0

Now I guess I should take one of these to the other side of the equation.

2x = 3n

Now I'm totally stumped. The teacher said you'd have to use some guesswork to get the answer, but I have no idea how you could do that either.

Aha, I think I've got it. I think I just have to make n and x numbers and find out how many different combinations make 0.

Is this right?

this is how i do these questions

calculate the maximum , in this case, 80
calculate the loss per question, in this case 5--3=8
calculate difference between max and actual, 80
divide by 8, get ten wrong

Since right is 5 and wrong is 3, there need to be 5/3 as many wrongs.

Wrong: 10

Right: 6

That of course, assuming that he answered all of them.

leonisdabomb has a nice solution that'll work for small numbers.

For the second question, I wrote two equations with n being the incorrect number and m being the correct number:-3n+5m=0;n+m=16. As there are two equations with two variables, there can only be one solution. I'm not quite sure what you did with the second problem.

[size=75]p.s.[/size]Welcome to AoPS! :welcome:


ratio
limit
function
logarithms
number theory
prime numbers
number theory open

Hi. Inspired by lecture on Riemann Hypothesis I decided to play a little with prime numbers and Pascal Complilator. I've come up with with a interesting sequence and I've noticed that a certain ratio is repeating very, very often and there must be something in it. Please let me know what you think about that:

Let $ f(n)$ be n-th prime number. $ f(1)\equal{}2, f(2)\equal{}3, f(3)\equal{}5$ and so on. We define a sequence $ a_n$ as follows:

\[ \begin {cases}a_1\equal{}f(k) \\
a_{n\plus{}1}\equal{}f(a_n) \\
 \end {cases}\]

My hypothesis is as follows: For big $ n$

\[ \quad \frac{a_{n\plus{}1}}{a_n} \sim  10\]

\[ \lim _{n\rightarrow \infty} {\frac{a_{n\plus{}1}}{a_n}}\equal{}10\] or maybe:

\[ |\frac{a_{n\plus{}1}}{a_n}|< g(n)\], where g is uknown function approximately equal to 10

From this we can conclude that 1000-th prime number is approximately equal to 10000.

The [url=http://en.wikipedia.org/wiki/Prime_number_theorem]prime number theorem[/url] implies that $ f(n) \sim n \ln n$.  Therefore the ratio $ \frac {f(n)}{n}$ should actually diverge.  The reason it appears constant is that it grows slowly for numbers in the range that an integer in most programming languages can handle.


$4012$ should be replaced by $4013$.  :wink:
[hide="Hint"]Telescoping summation is a prime candidate for such problems.[/hide]


1. Prove that if m is not a prime and phi(m) | m-1, then m has at least three distinct prime factors.

2. Prove that if m is not a prime and phi(m) | m-1, then m has at least four distinct prime factors.

3. Prove that there are infinitely many n for which phi(n) is a perfect square.

If m has one prime factor, it is of the form p^a. Then :phi:(m)=p^(a-1)(p-1). Since p^(a-1) is relatively prime to (p^a-1), the only way that p^(a-1) will divide p^a-1 is if p^(a-1) is 0. This would make a=1 and m a prime, which is contrary to the hyptothesis.

If m has two prime factors, it can be written in the form p^a*q^b. A discussion similar to the preceding paragraph convinces us that a=b=1. This means m=pq, where p and q are distinct primes. So, by the hypothesis, we have :phi:(m)=(p-1)(q-1)=pq-p-q+1|pq-1. This simplifies to pq-p-q+1|p+q-2. Note that (pq-p-q+1)-(p+q-2)=pq-2p-2q+3=(p-2)(q-2)-1 will always be greater than 0 unless either p or q =2, or p=q=3. Now, obviously, something cannot divide something that is smaller than it! (since everything is positive) The latter case is impossible since p and q are distinct, and (WLOG) if p=2, our expression simplifies to q-1|2q-1, which is obviously false for all primes q except for q=2, but p and q must be distinct, so that takes care of that case.

If m has three prime factors, it can be written as p^a*q^b*r^c. We can find that a=b=c=1, by a similar argument to the first paragraph. So now, m=pqr. The statement that :phi:(m)|m-1 means that (p-1)(q-1)(r-1)=pqr-p-q-r+pq+pr+qr-1|pqr-1. This simplifies to pqr-p-q-r+pq+pr+qr-1|p+q+r-pq-qr-pr. Hmmm...about here I get stuck. If I do something similar to what I just did, I might be able to use AM-GM to bound pq+pr+qr on the bottom, but I don't think that would be very nice as then I would have a pqr^(2/3) term, and that's not marvellously helpful.

erg! you edited your post. anyways, i took care of problem 1. you can see i run into problems with problem 2 though...

for problem 3, consider m of the form 2^(2n+1). :phi:(m)=2^2n, which is a perfect square. Obviously there are infinitely many m of the said form.


Is there anyone here who has specialized knowledge about lungs especially their limitations and requirements in the human body? If this is the wrong forum, can someone tell me where to post it? Thanks in advance.

Do you want to know what they do? If so, they supply the blood vessels with blood through the capillaries on the alveoli on the bronchioles(which I assume you would know of.) To go into fuller explanation I'd have to get into the circulatory system. I suppose I'll wait to see if that's what you wanted or not. Sorry if I got any of the terminology wrong.  :blush:


A whole number, $ N$, is chosen so that $ \frac{N}{3}$ is strictly between 7.5 and
8. What is the value of $ N$ ?

$ 7.5<\frac{N}{3}<8 \Rightarrow 22.5<N<24 \Rightarrow N\equal{}23$

note: since this is a countdown problem, we assume there is exactly one number that fits the inequalities.
then, we just simply use its integrality(?) with one inequality.
so just $ 8\cdot 3\minus{}1$ :)

We want a whole number so we might as well just mutiple 7.5 by three which is 22.5 and just use 23.


What is the least amount of numbers you need to make a completable sudoku? Please prove that your answer is correct.

well i think you need at least 2 in each 3x3 box...so 18 i guess??

i can't prove it i don't make sudokus but it's just an estimate

No advertising. And no reviving of old topics.

I'd lock this if I could....

Mods, please lock this.


calculus
integration
inequalities
number theory unsolved
number theory

Find all integral solution of the equation $ x^3\minus{}x\plus{}1\equal{}y^3\plus{}2y^2$.

[quote="MeKnowsNothing"]Find all integral solution of the equation $ x^3 \minus{} x \plus{} 1 \equal{} y^3 \plus{} 2y^2$.[/quote]

After checking, $ y \equal{} 0$ is not a solution and $ x \equal{} 0$ gives solution $ (x,y) \equal{} (0, \minus{} 1)$. and $ x^3 \minus{} x \plus{} 1$ is increasing over all integer values  of $ x$ except $ 0$

For $ y \neq 0$,

$ y^3 \minus{} y \plus{} 1 < y^3 \plus{} 2y^2 \le (y^3 \plus{} 2y^2) \plus{} (y \plus{} 1)^2 \equal{} (y \plus{} 1)^3 \minus{} (y \plus{} 1) \plus{} 1$

So $ y < x \le (y \plus{} 1)$ and equality for the right inequality holds only when $ y \equal{} \minus{} 1$, which gives $ (1, \minus{} 1),( \minus{} 1, \minus{} 1)$ as solutions.

So $ (x,y) \equal{} ( \minus{} 1, \minus{} 1),(0, \minus{} 1),(1, \minus{} 1)$ are the only solutions.


[quote="lovejrz"]I am a Chinese student,could you help me if I can see this BBS in Chinese?[/quote]
Hello.  Can you please explain what BBS means?

[quote="Wikipedia"]
...the term [url=http://en.wikipedia.org/wiki/Bulletin_board_system]BBS[/url] may be used to refer to any online forum or message board.
[/quote]
I don't think you can see this forum in Chinese, though there is a [url=http://www.mathlinks.ro/Forum/index.php?f=252]Chinese subforum[/url].


geometry
3D geometry
pyramid
tetrahedron
convex

[quote="Albanian Eagle"]Find all convex polyhedra which have no diagonals (that is, for which every segment connecting two vertices lies on the boundary of the polyhedron).[/quote]
I can only think polyhedra with the property as the problem states, as follows:

1) - The triangular prism.
2) - The pyramid ( a tetrahedron also, may be considered as a triangular pyramid ).
3) - The truncated triangular pyramid.

Are there any other kind of such that polyhedra?

Kostas Vittas.


Could the CRT(Chinese Remainder Theorem), be useful here?

Unless I'm mistaken, it's Dirichlet's theorem. A very well-known result.  :D

Yes, you are right!
But proof is very difficult.
You can find the proof on the forum for the particular case $a=p$ and $b=1$, where $p$ is a prime number.

Difficult indeed it is! You need the order of groups and the like, but fun!

haha, this is helpful, I know to give up now, I had been working on this problem for quite some time..... ;)

[quote="jpark"]That must have been frustrating, but who knows? Maybe if we didn't tell you that the proof was difficult, you may have found an extremely elegant proof of it!  :D[/quote]

i don't really think that's possible...
usually this type of problem is tackled by analytic number theory skills and i don't think elementary number theory skills are enough to solve this.

I heard something about a special case $b=1$, which was elementary solved by Selberg (don't know the right writing). But I am not sure.

[quote="probability1.01"]Ahhh it says it's more elementary than another proof... and then goes on to define just about every letter in the English and Greek alphabets, as well as use a variety of words not typically found in your Merriam-Webster's Collegiate Dictionary.  Scary... *shudders*[/quote]

Off topic:

It could be worse.  Someone could write a proof, run out of letters and start having to use smiley faces, squares, scribbles, ... and have to define new words for things...

[quote="Magnara"]I was recently wondering...are there infinitely many primes in the sequences n^2+1, as n goes from 1 to infinity?[/quote]

I think this is still an open problem

I guess it's a good thing I didn't work on it too hard then.

[quote="Scrambled"][quote="seamusoboyle"]Prove there are infinatly many primes in the series,

$ax+b$, with $a,b$ fixed and $\gcd(a,b)=1$[/quote]
pardon my extreme ignorance, but cant we solve this through the same method euclid did to prove the infinity of primes?[/quote]

maybe it works. Can you show me?

Unfortunately, I am 100% sure it doesn't work.
The Dirichlet's theorem proof is quite difficult and uses non-school notions.

...Except, may be in some simple cases (say when a=4, b=1) where the proof is not hard..

The simpilist one is a=4, b=3. Then do a=4, b=1. You need orders and stuff like that. You can then generalise.


Let $ ABC$ be a triangle and $ A'B'C'$ be Cevian triangle of point $ O$. $ M,N,P$ are midpoint of $ AB,BC,OA$. $ d$ is a line through $ B$. $ MB',MP,MN$ intersect $ d$ at $ E,F,G$, resp. $ B'C'\cap BC \equal{} \{I\}$. $ IF\cap GA' \equal{} \{J\}$. Prove that $ E,J,C$ are collinear.

Denote $ L$ the midpoint of $ AB'$ we get $ M,P,L$ are collinear.
But $ AB'//MN$ then $ M(AB'PN)\equal{}\minus{}1 \Rightarrow M(BEFG)\equal{}\minus{}1 \Rightarrow (FGBE)\equal{}\minus{}1$
$ \Rightarrow J(FGBE)\equal{}\minus{}1 \Rightarrow J(IA'BE)\equal{}\minus{}1$
On the other side, $ B'C'\cap BC\equal{}\{I\}$ thus $ (IA'BC)\equal{}\minus{}1$
So $ J,C,E$ are collinear.
We are done.


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

Let $A\in\M_{n}(\mathbb C)$ be a matrix with the property that for any $v\in \mathbb C^{n}$ the vectors
$v,Av,A^{2}v,\ldots,A^{m}v$ are linearly dependent, where $m\ge 1$. Prove then that the matrices $I,A,A^{2},\ldots,A^{m}$ are linearly dependent (i.e. that there is a complex polynomial of degree at most $m$ so that $p(A)=0$ )

I thought it would already be somewhere on this forum but I couldn't find it.
Here is a sketch of a proof:
Denote $\mu_{A}$ the minimal polynomial of $A$.
If $x \in \mathbb{C}^{n}$, denote $I_{x}=\{ P \in \mathbb{C}[X] ~|~ P(A)(x)=0\}$.
$I_{x}$ is an ideal of $\mathbb{C}[X]$,so there exist a unique monic polynomial $P_{x}$ s.t. $I_{x}=P_{x}\mathbb{C}[X]$.
We can find $x$ s.t. $\mu_{A}=P_{x}$ (for instance $x=x_{1}+\cdots+x_{p}$, where $x_{i}\in \mathrm{Ker}(A-t_{i}I_{n})^{\alpha_{i}}$, if $\mu_{A}=\prod_{i=1}^{p}(X-t_{i})^{\alpha_{i}}$).
Then the conclusion follows.


number theory
prime numbers
number theory unsolved

Prove that there are no prime numbers in the infinite sequence $10001,100010001,1000100010001,...$.

(I've got most of the solution already, the only problem I have is how to prove that $10^{4n}-1$ is divisible by $10^{4}-1$.)

[quote="Forgotenland"]prove that $10^{4n}-1$ is divisible by $10^{4}-1$.)[/quote]
I think there's a mistake because $a^{n}-1$ is divisible by $a-1$(easy !) and we put
$a=10^{4}$...

Ah, of course. Now I remember.

Thanks for the assistance.


[quote="am2525"]every body please help me to solve it. find all naturals x,y such that (2^x)-5 = (3^y)[/quote]

Please use modular arithmetic with 2 and 3.

2 solutions, viz x=3, y=1 and x=5 and y=3 are easy to guess, apart from that we shall prove that no other solution exists...
If $ x\equal{}a\plus{}5$ and $ y\equal{}3\plus{}b$ where (a,b)>0 are integers, then congruency modulo 5,9 and 3 gives $ a\equal{}4n$ and $ b\equal{}4m$
Accordingly we have

$ 2^{4n\plus{}5}\minus{}3^{4m\plus{}3}\equal{}5 \\ \implies 32.(16)^n \equiv 3^{4n\plus{}3} (mod5)$

Which evidently has no solutions by virtue of Thue's Lemma....


inequalities
email
inequalities proposed

If you have any question about this ebook, please contact us. Email:
nguyendunghus@gmail.com and vanddth@yahoo.com.vn

Do You know[b] vo thanh van[/b] maybe are similar scripts available on this forum? I'm interested in such e-books apart from 2007 year(because it is easy to find.

on the other hand such collection of inequalities is really great

Awesome! (I also like the fact that two of my solutions showed up :D).

Van and I have edited it from the first version, added chapter 3 (The inequality from IMO 2008) and some national olympiad problems. We give it as a small present for Christmas and New Year 2009 . Hope you enjoy it!

Wow, even my solution is posted here :wink:
Nice, I'm gonna start posting more on that forum from January
Oh, by the way since everybody's gonna party tomorrow - Happy New Year :wink:

P.S. Problem 27 a typo, I believe Maclaren is related to Formula 1 (should be Maclaurin) :wink:


inequalities
calculus
derivative
algebra
polynomial
function
real analysis

Compute
lim{n->oo}\pi{1\leq i<j\leq n} (1+ij/n^3)

I was looking at some of the unsolved problems on various subforums, and I tried to solve this one. What confused me was that I got 00, and that doesn't seem interesting enough:

I took ln of the whole thing, and all we need is the limit of  Sn=\sum{1<=i<j<=n}ln(1+ij/n3). We use the inequality ln(1+x)>=x/(1+x) (you can check it with derivatives and stuff) to get Sn>=\sum{1<=i<j<=n}ij/(n3+ij)>=\sum{1<=i<j<=n}ij/(n3+n2).

It's easy to see that \sum{1<=i<j<=n}ij is written as a 4'th degree polynomial in n, and when we divide it with n3+n2 we get something of degree 1 which tends to 00 as n->00.

Is there a mistake or is the problem just not nice enough?

I don't think there is any mistake and I don't think the problem is interesting. If you want something hard and nice, try the topic " A lovely function"/

Another solution to this uninteresting problem:
   WE can find n_0 such that for all n>=n_0 and all i and j we have ln(1+ij/n^3)>1/2ij/n^3. It's from the fact that ln(1+x)/x tends to 1 when x is close to 0. Then sum ln(1+ij/n^3)>1/2* [ ((n^2+n)/2)^2-n(n+1)(2n+1)/6]/2n^3 and this tends to infinity.

There is a finite limit if we replace n3 with n4 and we can find it by using the double ineq x/(1+x)<=ln(1+x)<=x. I'm a bit too lazy to actually calculate it :).


probability
MATHCOUNTS
email
geometry
conditional probability
AMC 10
AoPS Books
\/closed

We are pleased to announce that our new [i]Introduction to Counting & Probability[/i] textbook is now available!

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3]Click here to see more information, excerpts from the book, and to order your copy.[/url]

[quote="DPatrick"]We are pleased to announce that our new [i]Introduction to Counting & Probability[/i] textbook is now available! [/quote]

Congratulations! Must be exciting...  :D

[quote="biffanddoc"]I am thinking about buying the book, however, I think that the book may be too easy for me (reading the excerpts). Is there a way to know whether it will be too easy or not like "If you can solve the problems from the AOPS volume 1 chapters on robability on counting/probability then this book will only be review", or "It starts from the beginning but gets a lot harder, and I recommend that you buy it"

Thanks[/quote]

Try the post-test for the Intro Counting class (you'll find a link to it on the Enroll page).  If you can do 80+% of the problems easily, the book is too easy for you.

Another metric: if you have no problem with MATHCOUNTS and AMC counting problems, then the book is too easy for you.  If you sometimes have trouble with these, then it isn't too easy for you.

[quote="Raluca"]Will you send the book to Romania?[/quote]

We will send the book to Romania, but unfortunately, we can't accept online orders from Romania.  You would have to use our [url=http://www.artofproblemsolving.com/Books/OrderForm.pdf]printable order form[/url] and email books@artofproblemsolving.com for cost of shipping.  Sadly, our service that clears credit card orders (paypal) does not accept cards from Romania yet.

[quote="DPatrick"][quote="Silverfalcon"]So, does this book cover all counting problems in AMC-12 level (by this, I mean #1 through #25)?[/quote]Not necessarily.  It does not, for example, cover conditional probability or the principle of inclusion/exclusion.  Those topics might appear (in their easiest form) as a hard AMC question.  But the book certainly covers most or all of the material that would be asked on all but the very hardest problems on the AMC 10 or 12.

See the [url=http://www.artofproblemsolving.com/Books/IntroCount/toc.pdf]table of contents[/url] for a look at what topics are covered.[/quote]

Well, I can't really answer much of the all questions anyway so I might just buy the book. I have Mathematics of Choice so I can always look on that book for PIE right?

PIE is also in AoPS Vol2, and in Zeitz.  And it will be in Intermediate Counting & Probability.

[quote="agolsme"]Does the book cover anything that's not in the Intro to Counting and Probability course?[/quote]Not really.  There are a few more examples, and a lot more exercises and problems, but there's not really any substantially new material.  If you've taking our Intro Counting and Probability course, and you feel that you understand the material well, then there's little need for you to pick up this book, unless you want to do additional review and/or a source of additional practice problems.

It seems that I am the first one to buy/read/enjoy the book, and to point out a mistake in the book.

Page 5:  "$20^4=8000$" should be "$20^3=8000$".

When can we read Mr. rrusczyk's book (Introduction to Geometry)?

The Intro Geometry book is in the final draft phase, meaning it should be out early in 2006.

Also, Mr. Crawford, when will YOUR book come out? Will it be around the time of Intro to Geometry? I eagerly await both books!

There should be a discount for those of us planning to buy all [i]three[/i] books!! :D

My book will be available in the Spring if there are no unforeseen hurdles.

We plan discounted rates for people who buy the entire Intro series when it's finished (including an algebra book).

[quote="muralee"]Would you mind listing at least a broad outline of the contents of the soon to be released
"Introduction to Geometry"?

Thanks[/quote]

[url=http://www.artofproblemsolving.com/Books/IntroGeom/geomtoc.pdf]Here's a link to the tentative table of contents.[/url]

[quote="biffanddoc"]Mr. Crawford, you are writing Intro to Number Theory, right?
Who is writing Intro to Algebra?[/quote]

Probably MCrawford and I will collaborate on Algebra.  Naoki Sato is currently working on Intermediate Algebra, which we hope will be out this time next year.

[quote="biffanddoc"]Mr. Crawford, you are writing Intro to Number Theory, right?
Who is writing Intro to Algebra?[/quote]

Yes, I'm writing Intro Number Theory.

Intro Algebra will be sorted out when this round of writing is over, but it's a high priority, so we'll be working on it soon.

I just received the counting book(by the way, absolutely AWESOME book!!!!Blew my expectations away!!) and I will buy the other books as they come out. Will buying the entire intro set be cheaper compared to buying the book with solutions as they come out (I need counting practice now, geometry practice now, etc). Thanks!!

It will be a little less expensive to buy them all at once, but if you're right at the level of each book, it might still make sense to buy them one at a time.

[quote="kyyuanmathcount"]About the book "Introduction to Geometry":

Mr. Rusczyk:  I've finished studying AoPS volume 1 and plan to move on to studying the volume 2.  Apparently the new book "Introduction to Geometry" overlaps considerably with AoPS two books.  So what should I do?  Should I ignore the new book, just go on completing the two AoPS books, or switch to the "Introd to Geometry"?[/quote]

This depends on how you feel about your geometry skills.  If you can answer all the problems in AoPS Vol 1, then you shouldn't need the Intro Geometry book.  If you struggle with at least half of them, then you probably should work through the Intro Geometry.  Much of it will be review, but there will be many challenging problems to help cement your understanding of geometry.

The new book is great.  I have started to use it in my classes on a regular basis.  I can see the improvement in my kids over just a short period. I think that probability is one of the hardest topics for students and teachers.  This book helps both.

A great job!!!  The book is great.   :lol:  :lol:  :lol:  :lol:  :lol:


If n is a positive integer and w =(sqrt5 - 1)/2, prove that [n/w] and [n/w^2] constitute a partition of positive integers.
Sorry if this question is stupdi, but i don't know how to apply beatty theorem...

Try searching "Beatty sequence" on the forum. It's been discussed.. A LOT :).

sorry for my stupidity!!!!!
I mistakenly calculated 1/w + 1/w^2 instead of w+w^2 so i didn't know i can apply beatty theorem directly.
What about this?
w=(1-sqrt5)/2
then w+w^2 is not 1 anymore, is [n/w] and [n/w] a partition of integers anymore?

No, since Beatty's theorem states that if $a,b>0$ then the sets $\{ [na] \}$ and $\{ [nb] \}$ form a partition of $\mathbb {N}^*$ if and only if $a,b$ are irrational and $\frac  1 a + \frac 1 b = 1.$
In your case it means $w^2+w = 1.$

Pierre.

1/a+1/b=1 is a must?
if not [n/a],[n/b] can never be a partition of positive integers?
How toprove this?
Thanks!

Have a look here :
http://www.animath.fr/cours/arithm.pdf
This is theorem 1-1-3, p.6-7.
A related result is problem 67 in the same paper.

Pierre.

assume that $a,b$ are irrational positive real numbers and $[na],[nb]$ is a partition of positive integer

let $K$ be an arbitrary positive integer

let $L,M$ be the largest positive integers such that $[La],[Mb] \leq K$

then $La < K+1 < (L+1)a , Mb < K+1 < (M+1)b$

thus $\frac{L+M}{K+1} < \frac{1}{a} + \frac{1}{b} < \frac{L+M+2}{K+1}$

since $[na],[nb]$ is a partition, therefore $L+M=K$

hence $\frac{K}{K+1} < \frac{1}{a} + \frac{1}{b} < \frac{K+2}{K+1} (*)$

$(*)$ is true for any positive integer $K$, therefore if we take $K \rightarrow \infty$, then $\frac{1}{a} + \frac{1}{b} = 1$

This is a similar problem:
Is there positive real numbers   $a,b,...,k$  s.t  {$ [\sqrt na]^2$ , $[\sqrt nb]^2$ ,..., $[\sqrt nk]^2$ } ,n=1,2,...   is a partion of positive integers set?
 Have this one been solved?

[quote="pete_ross"]
since $[na],[nb]$ is a partition, therefore $L+M=K$
[/quote]

may i know why?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beatty&t=25764

Pierre.

Can any one please to give me some hint to my question above(#8) or is that a trivial problem :? ?

Well,... since your sets only contain squares, I do not see how they could cover $\mathbb {N^*}$  ;)

Pierre.

Ofcourse my last question was trivial :blush:
 But I really wonder about the same problem :
  Are there positive real numbers $a,b,...,k$ s.t:
{$[ ( \sqrt {na}+1)^2],[ (\sqrt {nb}+1)^2],...,[ {(\sqrt {nk}+1)^2] }$}
 is a partion of positive integers.


I've been seeing problems in the following pattern a couple times recently:

How many ways can you rearrange [a word] such that no letter remains in its original place?

For some reason I've had difficulty establishing a straightforward method for this sort of thing. Any ideas?

Aren't those called derangements, when no object was in its original place???  Because I called them that in my proof.

I'll try a proof:[hide]Let's start off with having integers 1,2,3,4,5,6,7 where the total amount of permutations is 7!.   First compute the amount of ways the integers can be arranged so that only one object remains in its original place.  That will be (7,1)*6! because there are seven objects that can be fixed and for each one there are 6! permutations.  Then doing the previous steps to find how many arrangements there are for two integers being fixed, there would be (7,2)*5!, because of (7,2) ways to choose the two integers and 5! permutations for each pair chosen.  Following for three, four, five, six, and seven integers being fixed, we get (7,3)*4!, (7,4)*3!, (7,5)*2!, (7,6)*1!, and (7,7)*0! arrangements respectively.   Now PIE can be applied to find the number of ways the integers can be arranged such that none are in their natural position.  7!-(7,1)*6!+(7,2)*5!-(7,3)*4!+(7,4)*3!-(7,5)*2!+(7,6)*1!-(7,7)*0!=7!(1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)=D(7) where D(7) is the number of derangements of seven objects.  This can be extended to n objects also where the number of derangements is n!(1-1/1!+1/2!-1/3!+1/4!-...1/n!).[/hide]

Ian wrote:spoiler: [hide]For me, the easiest way to solve problems like this is by taking the number of permutations of [the word] and then divide by the factor that overcounts by.
For example if we use the word mississippi:

The number of permutations of the letters is 11!

The number of ways that you can make a given string with diifferent permutations of mississippi is 4! * 4! * 2! because the s's p's and i's are interchangeable.

Thus, the number of different strings that can be created is 11!/(4!4!2!).

I think someone else can explain this better, but I might as well post what I've written.[/hide]

I think you misread the question..


Find more practice problems at [url]http://www.imlem.org/meetdoc/index.htm[/url]

Yes, alanchou, they are from Massachusetts

Meet 2 of this school year should be posted sometime soon, it was in late November.

The IMLEM is divided into 5 categories, which are Mystery, Geometry, Number Theory, Arithmetic (which I personally dislike), and Algebra.  The later four are divided into subcategories which are listed below.

Categories can be found at [url]http://www.imlem.org/topics-2007-08.htm[/url]

The arithmetic ones look pretty easy. No wonder you don't like them.

Spider:  Meet 4 of these problems include mod problems.  You might want to look at them.


Find the minimum value of $y=9 \sin{x}+ 16 \csc{x}$ if $x \in \left( 0, \frac{\pi}{2} \right]$.

[quote="kunny"]Just use A.M.-G.M.[/quote]
Note that domain of the $\sin{x}$ is $[-1,+1]$......

Isn't that we have for $0<x\leq \frac{\pi}{2},\ 0<\sin x\leq 1?$

I mean, you cannot use AM-GM in this problem to find the minimum...... $\sin{x} \leq 1$.

Sorry for that I was mistaken, shobber. :blush: From the graph of $y=9t+\frac{6}{t}\ \left(0<t\leq 1\right),$ the desired answer is $25\ (t=1).$

kunny


Prove that i^i,
(ie) i to the power i is real and positive

Just a technical remark: the value Conjurer obtained is just the one of possible values, depending on the $ \text{Log}$ branch you choose. See also [url=http://www.mathlinks.ro/viewtopic.php?t=259248]here[/url].

Hey for complex no.s isnt $ arg(z) \in[\minus{}\pi,\pi]$

Well, it's arguable. Indeed, the most common branch of logarithm is the one with arguments in first and fourth quadrant, so your answer is something we'd call a principal value, but it doesn't make it unique, just one among infinitely many branches.

See http://en.wikipedia.org/wiki/Imaginary_unit#Operations_with_i .

Let, $ z^\alpha \equal{} e^{\alpha \ln z}$.
Thus, $ i^i\equal{}e^{i \ln i}\equal{}\boxed{e^{\minus{}\left(\frac{\pi}{2}\plus{}2k\pi\right)}}. \ \ \left(\because \ \ln i \equal{} i\left(\frac{\pi}{2}\plus{}2k\pi\right)\right)$


In the diagram, the four circles are externally tangent, and each circle has a radius of 1 inch. What is the number of square inches in the shaded region? Express your answer as a decimal to the nearest hundredth.
[asy]draw(Circle((0,0),1)); draw(Circle((0,2),1)); draw(Circle((2,2),1)); draw(Circle((2,0),1));
path p = Arc((0,0),1,0,90)--Arc((0,2),1,270,360)--Arc((2,2),1,180,270)--Arc((2,0),1,90,180)--cycle; fill(p,grey);[/asy]

If we draw a square around the shaded region with side length 2, we see that the area of the shaded region is equal to the area of the square minus the four quarter circles of radius 1.

The area of the square is simply $ 2^2 \equal{} 4$

The area of four quarter circles is equal to the area of one circle, or
$ \pi r^2 \equal{} \pi$

So the area of the shaded region is the difference between the two,

$ \boxed{4 \minus{} \pi}$

It asks for rounded to the nearest hundredth, so the answer is 0.86

If you didn't know, $\pi$ is $3.14159265358979323846264338...$. That should be enough digits. :-D


If $ z_1,z_2,z_3$ are three nonzero complex numbers so that $ z_1^2\plus{}z_2^2\plus{}z_3^2\equal{}0$ then prove that the number $ z\equal{}\frac{z_1^6\plus{}z_2^6\plus{}z_3^6}{z_1^2 z_2^2 z_3^2}$ is real.

Can you explain two lastest phrases ?I can not understand them :(

$ z$ is a complex number $ \Leftrightarrow$ $ z\equal{}a\plus{}b \cdot i$ where $ a,b$ are reals and $ i^2\equal{}\minus{}1.$

If $ z\equal{}a\plus{}bi$ then $ \bar{z}\equal{}a\minus{}bi$. Proving that $ z$ is real is equivalent with proving that $ z\equal{}\bar{z}$.

Now $ z\bar{z}\equal{}|z^2|$ and $ |z^2|$ is real. I will post the solution tomorrow if nobody knows how to solve it.  :)

oooh :P
I know the things that you explained.I thought that the numbers 1 & 2 in front of the z are sth that I do not know them.

please wait . I want to solve it . [size=150]please[/size] untill 2 days  :)

I think this is it..

[hide="Proof"]Let $ z_1^2, z_2^2$ and $ z_3^2$ be the roots of $ P(x) \equal{} x^3 \minus{} ax^2 \plus{} bx \minus{} c$, then since $ z_1^2 \plus{} z_2^2 \plus{} z_3^2 \equal{} a \equal{} 0$,  $ P(x) \equal{} x^3 \plus{} bx \minus{} c$.

We wish to show that

$ z \equal{} \frac {S_3}{c}$ is real, where $ S_3$ is the 3rd Newton's sum.

We know that

$ S_3 \equal{} S_2s_1 \minus{} S_1s_2 \plus{} 3s_3$

$ S_3 \equal{} S_2a \minus{} as_2 \plus{} 3c$

$ S_3 \equal{} 3c$

Therefore, $ z \equal{} \frac {3c}{c} \equal{} 3$ which is real.[/hide]

[quote="moldovan"]If $ z_1,z_2,z_3$ are three nonzero complex numbers so that $ z_1^2 \plus{} z_2^2 \plus{} z_3^2 \equal{} 0$ then prove that the number $ z \equal{} \frac {z_1^6 \plus{} z_2^6 \plus{} z_3^6}{z_1^2 z_2^2 z_3^2}$ is real.[/quote]
hi all
my solution  :play_ball:
[url=http://www.servimg.com/image_preview.php?i=422&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk17.gif[/img][/url]

Oh

I could solve it . your proof are long but If you use Euler`s  identity you can easily find out that it equals to 3 and 3 is real.

$ a^3\plus{}b^3\plus{}c^3\equal{}3abc\plus{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$. Just substitute $ a\equal{}{z_1}^2$, $ b\equal{}{z_2}^2$. $ c\equal{}{z_3}^2$ and we are done.

[quote="keyree10"]$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc \plus{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$. Just substitute $ a \equal{} {z_1}^2$, $ b \equal{} {z_2}^2$. $ c \equal{} {z_3}^2$ and we are done.[/quote]

That's right, So I did. This is very easy problem. :lol:

keyree10

that phrase you said is euler`s identity.


We know that when we rationalize the denominator of $\frac 1{\sqrt a+\sqrt b}$, we can do it like that
\[ \frac 1{\sqrt a +\sqrt b}=\frac{\sqrt a-\sqrt b}{(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)}=\frac{\sqrt a-\sqrt b}{a-b} \]
But how to rationalize the denominator of $\frac 1{\sqrt[3] a+\sqrt[3] b+\sqrt[3] c}$?

[quote="Kurt Gödel"][hide]Substitute $\sqrt[3]{a} = u$, $\sqrt[3]{b} = v$ and $\sqrt[3]{c} = w$. With the formulae $a^3+b^3+c^3-3abc = (a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right)$ and $a^3-b^3 = (a-b)(a^2+ab+b^2)$, we get that

\[ \begin{aligned} \frac{1}{u+v+w} & = \frac{(u-v)^2+(v-w)^2+(w-u)^2}{u^3+v^3+w^3-3uvw} \\ & = \frac{\left((u-v)^2 + (v-w)^2 + (w-u)^2\right)\left((u^3+v^3+w^3)^2 + 3uvw(u^3+v^3+w^3) + (3uvw)^2\right)}{(u^3+v^3+w^3)^3-(3uvw)^3} \\ & = \frac{\left(\left(\sqrt[3]{a}-\sqrt[3]{b}\right)^2 + \left(\sqrt[3]{b}-\sqrt[3]{c}\right)^2 + \left(\sqrt[3]{c}-\sqrt[3]{a}\right)^2\right)\cdot\left(a+b+c+3\sqrt[3]{abc}(a+b+c)+9\sqrt[3]{a^2b^2c^2}\right)}{(a+b+c)^3 - 27abc},\end{aligned}  \]

and we're done.[/hide][/quote]

I think you forgot to square a+b+c in the second factor in the numerator, but wow, nice work. :)

Hehe, thanks. You're right, I'll edit it.


Pro.1: Solve the system: $\{\begin{array}{l}{x\sqrt{(1-y^{2})(1-z^{2})}=x-yz}\\{y\sqrt{(1-z^{2})(1-x^{2})}=y-zx}\\{z\sqrt{(1-x^{2})(1-y^{2})}=z-xy}\end{array}$

Pro.2: Find $(x,y)$ that satisfy:
$\sqrt{35-2\sqrt{45-2x}}=x-5$
(this pro is very easy that we can change it into an equation system :D)

Pro.3: Find $a$ that the follơing equation system:
$\{\begin{array}{l}{x^{3}-ay^{3}=\frac{1}{2}.(a+1)^{2}}\\{x^{3}+ax^{2}y+xy^{2}=1}\end{array}$
have root $(x,y)$ that $x+y=0$

Pro.4: Find m that the system:
$\{\begin{array}{l}y^{2}=x^{3}-4x^{2}+mx\\x^{2}=y^{3}-4y^{2}+my\end{array}$
have only one root

Pro.5: Let $(a,b,c)$ satisfy:
$\{\begin{array}{l}{a^{2}+b^{2}+c^{2}=2}\\{ab+bc+ca=1}\end{array}$
Prove that $\frac{-4}{3}\le a,b,c \le \frac{4}{3}$

Enjoy :D

hello, for 2), the range of definition is $ 5 \le x \le \frac{45}{2}$ ,squaring our equation and sorting the terms we get
$ 2\sqrt{45\minus{}2x}\equal{}10x\plus{}10\minus{}x^2$
squaring again and factoring we get
$ (x\minus{}10)(x\plus{}2)(x^2\minus{}12x\plus{}4)\equal{}0$
and only
$ x\equal{}10$ fulfills our given equation.
Sonnhard.


A developer has 87 acres and he would like to divide it into smaller lots. Some should be 2 acres, some should be 3 acres, and some should be 5 acres. If the developer must have exactly 25 lots, and at least one lot of each type, how many different ways can he divide up the 87 acres?

I'm looking for a smart, organized, quick way.

[hide="My solution"]$2a+3b+5c=87$
$a+b+c=25$
$2a+2b+2c=50$
$b+3c=37$

From there it's easy to see that there are $12$ solutions for $b+3c+37$, where $b$ and $c$ are natural numbers.[/hide]

First notice that to maximize the number of 3's, you have 6 fives and 19 threes.  Now you can only replace 3's with 2's and 5's (Why?).  The only possible way to do this is to remove 3 threes and add 2 twos and 1 five, and repeat this process for as many times as you like.  You can do this anywhere between 0 and 6 times, giving you 7 possible ways total.

We're both wrong. There are $6$ possible solutions:

$2a+3b+5c=87$
$a+b+c=25$
$2a+2b+2c=50$
$b+3c=37$

From there it's easy to see that there are $12$ solutions for $b+3c=37$, where $b$ and $c$ are natural numbers.

However, if $b=19$, $c=6$, so $a=0$. Increasing the value of $b$, $a$ becomes negative. $a$ has to be positive, so the only solutions are $12,1,12$,    $10,4,11$,    $8,7,10$,    $6,10,9$,    $4,12,8$,    and $2,14,7$

Oops, forgot that you have to have at least one of each.  I don't catch this stuff anymore--some mathcounts skills rot fast :P


quadratics
search
number theory
number theory theorems

Can you help me with some nice materials from where I can learn more about quadratic residues. I have somethnig, but it is more theory then "useful stuff" if you know what I mean. In there, things that I need for problemsolving are literally in footnotes?
Cheers

For some texts, find some suitable number theory textbook (e.g. Burton's, but there are a lot).
For problems, search this forum.
For questions, simply ask here  :)


I was trying to figure out today what the smallest odd integer whose proper factors added up to more then the integer itself.  I was able to find numbers that would work,  but I have no idea how I could find the smallest ones.

Just something I was thinking about today.  Sorry if this is in the wrong category.

According to http://en.wikipedia.org/wiki/Abundant_number , the smallest odd abundant number is $945 = 3^{3}\cdot 5 \cdot 7$.


geometry
parallelogram
conics
ellipse
geometry unsolved

If a parallelogram is circumscribing an ellipse, the point of contacts are the vertices of a parallelogram. How can one prove that the diagonals of the first parallelogram are parallel to the sides of the second parallelogram?

Let ABCD the parallelogram and E,F,G,H the touch point of the ellipse with AB,BC,CD,DA;

it's sufficent to use [url=http://www.mathlinks.ro/Forum/weblog_entry.php?t=195672]this[/url] for get $ \left ( \frac{\overline{AH}^2}{\overline{HD}^2} \cdot \frac{\overline{DG}^2}{\overline{GC}^2} \right )^2 \equal{} 1$ that is equal to $ \frac{\overline{AH}}{\overline{HD}} \equal{} \frac{\overline{GC}}{\overline{DG}}$ that for thales is equal to $ GH \parallel CA$.


Inequaliy is true for any $ a,b,c \in\mathbb{R^*}$

 So, denote $ a^2\equal{}x  , b^2\equal{}y,   c^2\equal{}z$ inequality is equivalent with\[ \sum {\frac{x}{5x\plus{}z}}\le \frac{1}{2}\] or\[ 2\sum {x(5y\plus{}x)(5z\plus{}y)}\le (5x\plus{}z)(5y\plus{}x)(5z\plus{}y)\]or\[ 24xyz\le 5\sum {xy^2}\plus{}3\sum {x^2y}\] wich is true from $ GM\le AM$

[quote="tranvanluan"]Let $ 1\le a;b;c\le 2$
Prove that
1)
$ \frac {a^2}{5a^2 \plus{} c^2} \plus{} \frac {b^2}{5b^2 \plus{} a^2} \plus{} \frac {c^2}{5c^2 \plus{} b^2}\le \frac {1}{2}$
2)[/quote]
Yes,it is quite easy,but why $ 1\le a;b;c\le 2$ not works,tranvanluan.i think it is useful with your 2) :wink:

[quote="Allnames"]i think it is useful with your 2) :wink: [/quote]
But where is the 2) Allnames? Did tranvanluan miss it??


geometry
analytic geometry
trigonometry
similar triangles
power of a point
cyclic quadrilateral
trig identities

Every now and then you hit a road block when solving a problem.  That block came yesterday for me in my further mathematics course.  We have a geometry problem that seems simple enough, but I have been spinning my wheels and cannont solve it.  Here it is:

Two circles intersect at P and Q.  [QP] is produced to R, [RS] is a tangent to one circle and [RT] is a tangent to the other.  Prove RS = RT.

**(Using cyclic quadrilaterals)

Now I have spent several hours on this problem, and there must be something I am not seeing.  This is the first problem in a while that I have had this much trouble solving.  I am fairly sure it is true (I have checked it out on Geometer's Sketchpad).  It should be of note that my students struggled with this problem for several hours and could not solve it either.  I am not going to comment on what I have done, as I feel that since I have not gotten anywhere, it might not help.  Also, [b]you cannot use coordinate geometry in the proof[/b].

I hope someone gets this and can help me out.  I will keep working on it and see where I can get.
Good Luck,
Mr. R.   :chief:

sorry, but I don't understand what "produced to R" means :maybe: Does that mean that $QR$ is just an extention of $PQ$?

continued to some point R, i.e. PQR is a line segment

**you got it
p.s.  Do you have school?

Isn't it just an application of power of point?

$PR \cdot RQ=RT^{2}$ and

$PR \cdot RQ=RS^{2}$

$RS=RT$

School starts in two more days :(

This is before we get to Power of a Point section.  However, that is correct.  I just went back and proved power of a point using similar triangles and then came to the same result.

[b]However[/b], the proof is supposed to involve cyclic quadrilaterals (since it is in that section).   I should have mentioned that.  Also, it does not require triangle similarity (which I used to prove Power of a Point) since it is before that section.  Thank you for your help.  I have been trying to do it without Power of a Point since we are just defining it and proving it tomorrow.  If you see how to use cyclic quadrilaterls, let me know.

well, there is one pretty apparent cyclic quadrilateral since tangents produce $90^\circ$ angles with the radii of the circles and the two $90^\circ$ angles are opposite angles. So the other two opposite angles are also supplementary meaning their $\sin$ values are the same and their $\cos$ values are just opposites. So theoretically, you could just bruteforce trig it (sin area and law of cosines) and then use Ptolemy's Theorem, but that will get pretty complicated.

That's the only way I see of doing it and I'm not sure if there's a nicer way. Of course, maybe they wanted you to prove PoP, since it's not that hard to prove anyways and PoP goes along with cyclic quadrilaterals...

Line $PQ$ is called the radical axis, btw.

So, um, what are we allowed to use?  What properties of cyclic quads are we allowed to use?

Here is the deal.  This problem showed up in the second chapter of a book I am using in further mathematics.  The presumed knowledge is very minimal.  The second chapter deals with cyclic quadrilaterals.  The proof should easily be done using basic properties of cyclic quadrilaterals.  Now, I have figured out several ways to prove the theorem, but I do not have one using a cyclic quadrilateral.  If you have any ideas let me know.

What's the name of the book (just wondering)?

Mathematics (for the International student)
Mathematics HL Options

Haese and Harris Publications

yea it's not possible to prove...too nontrivial if u can't use similar triangles or power of point.

 :noo:  :noo:


calculus
derivative
calculus computations

Find the second derivative of $ f(x)\equal{} x^{\frac{2}{3}} (x\minus{}2)^2$ and possible points of inflection.

This is an exercise in the chain rule, power rule, and product rule.  The first derivative is $ f'(x)\equal{}\frac{2(x\minus{}2)^{2}}{3x^{1/3}}\plus{}2x^{2/3}(x\minus{}2)\equal{}A\plus{}B$.  Now $ B'(x)\equal{}\frac{4(x\minus{}2)}{3x^{1/3}}\plus{}2x^{2/3}$ and $ A'(x)\equal{}\frac{d}{dx}\left(\frac{2}{3}x^{\minus{}1/3}(x\minus{}2)^{2}\right)\equal{}\minus{}\frac{2(x\minus{}2)^{2}}{9x^{4/3}}\plus{}\frac{4(x\minus{}2)}{3x^{\minus{}1/3}}$.  Then the second derivative of the whole thing is $ f''(x)\equal{}\frac{2(x\minus{}2)^{2}}{9x^{4/3}}\plus{}\frac{4(x\minus{}2)}{3x^{\minus{}1/3}}\plus{}\frac{4(x\minus{}2)}{3x^{1/3}}\plus{}2x^{2/3}$.  You may want to check my math as there are  a lot of places to make an arithmetic mistake or forget to apply a product rule and whatnot.  You can find the zeros on your own.

Actually, I was having trouble finding the zeroes.

I got to the second derivative and had simplified a lot:

$ f''(x)\equal{} \frac{(12x^2\minus{}15x)(x^{2/3})\plus{}2x^2\minus{}5x\plus{}2}{9x^2}$...

There is a $ x^{2/3}$ in the numerator, how do I get rid of it?

If you have it, you could stick it in Mathematica and see what it gives you.  Otherwise, you try Newton's Method but that may take even longer.

I put it in mathematica and it came out of something ugly.

I may have made an error in my arithmetic... I will double check.

Thanks for the help, JRav.

$ f(x) \equal{} x^{\frac {2}{3}}(x \minus{} 2)^2\Longrightarrow f'(x) \equal{} \frac{4}{3}x^{ \minus{} \frac {1}{3}}(x \minus{} 2)(2x \minus{} 1)$,

Using Libniz rule: $ \boxed{(fgh)' \equal{} f'gh \plus{} fg'h \plus{} fgh'}$,

$ f''(x) \equal{} \frac {8}{9}x^{ \minus{} \frac {4}{3}}(5x^2 \minus{} 5x \minus{} 1).$

Now it's easy to solve, just solve $ 5x^{2}\minus{}5x\minus{}1$, which gives $ x\equal{}\frac{5\pm\sqrt{45}}{10}\equal{}1\pm\frac{3\sqrt{5}}{5}$.

I knew it, arithmetic error!

Thanks Kunny and JRav!


5n^3+7n^5

Can anyone explain to me why this is always divisible by 12? It's been bugging me for a whole day now.


Let $ a,\ b,\ c$ be real numbers such that  $ 1<a\leq \frac{1\plus{}\sqrt{5}}{2},\ 1<b\leq \frac{1\plus{}\sqrt{5}}{2},\ 1<c\leq \frac{1\plus{}\sqrt{5}}{2}$.

Prove the following inequality.

$ \frac {a^3}{(a^2 \minus{} 1)^3} \plus{} \frac {b^3}{(b^2 \minus{} 1)^3} \plus{} \frac {c^3}{(c^2 \minus{} 1)^3}\geq \frac {9abc}{abc(a \plus{} b \plus{} c) \minus{} (ab \plus{} bc \plus{} ca)}$

I think it is wrong kunny. Try $ a\equal{}b\equal{}c\equal{}2$ or $ \equal{}3$.  :maybe:

:)

You are right, thank you!
I have just edited the problem.
How about this time?


trigonometry
analytic geometry
algebra
polynomial
LaTeX
Pythagorean Theorem
geometry

I haven't graded all the solutions yet. But you guys can talk about what problem was hard and maybe post your own solutions, et cetera. I'll try to post the qualified people sometime today. :)

waaiitt....they're due already?  I thought they were due tonight at midnight...???  :?

Can I send you my answers now before they've been discussed?

My bad, I thought today was Monday.  :blush:

This thread is locked till midnight tonight.

THAT MEANS NO DISCUSSION!

Thanks for reminding eryaman.

You may discuss the solutions to the problems.

I'll post the result tomorrow (I'll try depends on what time I'll be back)

The answers I got were:

1. $\sqrt{433}$
2. $x + 4$
3. $m = \frac{\pi}{2}$
4. $11440$ terms
5. $0$ triangles exist

Could someone confirm or provide different answers?

I got all the same answers except couldn't figure out 4 so I couldn't turn in my solutions :(

Here are the qualifiers!!!  :D

eryaman
Jan
dk2
dakyru
Andreas
JesusFreak197

Congratulation!!

Here are the solutions.

1. Let $AY$ intersect $RE$ at $P$. From Pythagorean Theorem on triangle $RYE$, we find that $RE = 13$. Let the perpendicular segment from $Y$ meet $RE$ at $Q$ and $AC$ at $S$. Now, we have right triangle $AYS$. Since $AS = RQ$ and $SY = YQ + SQ = YQ + 13$, we need to find $RQ$ and $YQ$. $YQ$ is easy since $RY \cdot YE = RE \cdot YQ = [RYE]$ (for convenience, I took out dividing by 2 part on all three sides). Since we know all values, $YQ$ is $\frac{60}{13}$. We can find $RQ$ through proportion:

\[\frac{RQ}{RY} = \frac{RY}{RE}\]

Hence, $RQ = \frac{144}{13}$. Applying the Pythagorean Theorem to $\triangle AYS$, we find $AY = \sqrt{433}$.

2. Let's rewrite $f(x)$ as $q(x) (x^2-4x+3) + r(x)$ where $\deg f(x) > \deg r(x)$. Since $f(x)$ is at least 2 degrees, $\deg r(x)$ is 1. Hence, $r(x) = ax+b$ for some $a$ and $b$. From two division, we have:

$f(1) = 0 q(x) + a + b = a + b = 5\\
f(3) = 0 q(x) + 3a + b = 3a + b = 7$

Solving system of equations show that $a = 1$ and $b = 4$ so the remainder is $x+4$.

3. First, simplify the expression into:

$\frac{\cos (2m) + i \sin (2m)}{\cos (m) + i \sin (m)} = i$

Take the numerator and denominator and compare it with $i$.

$\cos (2m) + i \sin (2m) = i$

OR

$\cos (m) + i \sin (m) = i$

The first equation doesn't make whole fraction equal to $i$. But the second one givese $m = \frac{\pi}{2}$ which works out since this makes the fraction to become:

$\frac{-1}{i} = i$

Hence, the answer is $m = \frac{\pi}{2}$. There are many ways to solve this problem and this is just one "quick" way I thought.

4. Now, this problem is our favorite problem.  :P

This problem requires the use of trick. Note that each term in expansion is in form $x_1^{y_1} x_2^{y_2} x_3^{y_3} x_4^{y_4} x_5^{y_5} x_6^{y_6} x_7^{y_7} x_8^{y_8}$ where the sum of the exponents is 9. Hence, the number of term is the solution to the equation

\[y_1+y_2+y_3 + \cdots y_8 = 9\]

in nonnegative numbers and that is $\binom{16}{9} = 11440$.

5. $\cos (C) = 0$ implies $\angle C = 90^\circ$. So, $\triangle ABC$ is right. Let's square the first given equation and make substitution $a^2+b^2 = c^2$:

\[(a+b)^2 = (c+1)^2\\
a^2+2ab+b^2 = c^2+2c+1\\
c^2+2ab = c^2+2c+1\\
2(ab-c) = 1\\
ab - c = \frac{1}{2}\]

This is impossible since $a,b,c$ are integers. Thus, the answer is $0$.

 :)

[quote="Jan"]Woow! I got exactly the same answers! 4 was a killer, I worked really long on that one. Does anybody have an easy way to do it? I did this $n+\displaystyle{n\choose 2}.8+\displaystyle{n\choose 3}.28+\displaystyle{n\choose 4}.56+\displaystyle{n\choose 5}.70+\displaystyle{n\choose 6}.56+\displaystyle{n\choose 7}.28+\displaystyle{n\choose 8}.8$ And then plug in $n=8$
I know this sucks, but it's the only way I could think of...
Also, if you do this my way: the numbers you see, like 28 and 56, is there a way to find them in a more mathematical way? I just counted how many different ways there are...  :?

So does anyone have an elegant and a bit less ugly way to do this question?[/quote]

The way I thought of #4 is like this:

In the expansion of that quantity (which was to the ninth power), each term is composed of nine elements.  Each element can be taken from the set of $x_1,x_2,x_3...x_8$ for $u(7)$.  There are a total of 8 elements to choose from.  In order to find the number of terms, you're looking for how many different ways to form multisets of length 9 from that set of 8 elements.  This would be:

${8 + 9 - 1\choose9} = {16\choose 9} = 11440$

Here are my solutions:

1. I used two methods.

Put it on a coordinate plane, find the equations of the lines that intersect at point $AY$, and find the coordinates of AY, then use distance formula.

Second method was using law of cosines.  Kind of cheap, but it worked.  :D

2. same as silverfalcon's solution

3. First cancel out the 12's.  Then multiply both sides by $\cos m+i\sin m$, so we have $\cos {2m}+i\sin {2m}=i\cos m-\sin m$.  So we set the imaginary and real parts equal so we have $\sin {2m}=\cos m$ and $-\sin m=\sin(-m)=\cos{2m}$.  Use $\sin x=\cos{\left(\frac{\pi}{2}-x\right)}$ and solve the equations.

4. same as silverfalcon's solution

5. Since $\cos C=0$ we know angle C is a right angle.  Therefore we have $a^2+b^2=c^2$.  We also have $a+b=c+1$.  Square this equation to get $a^2+2ab+b^2=c^2+2c+1$.  Since $a^2+b^2=c^2$ we can cancel these out of the equation giving us $2ab=2c+1$.  Since $a+b=c+1$ we can rewrite $2c+1$ as $2(a+b)-1$.  So know we have $2ab=2a+2b-1$.  Now use Simon's Favorite Factoring Trick (subtract 2a and 2b, factor out the 2a, add 2, factor out the two, factor out b-1), to get $(b-1)(2a-2)=1$.  Since $b-1$ and $2a-2$ are integers they must both equal $1$.  Therefore we have $b=1$ and $a=1.5$  But $1.5$ is not an integer, so there are no solutions.

On problem number three is says the range is $\frac{\pi}{7}<m<\frac{4\pi}{7}$.  But the correct answer is $\frac{\pi}{2}$ which is not in that range... :?

[quote="dakyru"][quote="Jan"]Woow! I got exactly the same answers! 4 was a killer, I worked really long on that one. Does anybody have an easy way to do it? I did this $n+\displaystyle{n\choose 2}.8+\displaystyle{n\choose 3}.28+\displaystyle{n\choose 4}.56+\displaystyle{n\choose 5}.70+\displaystyle{n\choose 6}.56+\displaystyle{n\choose 7}.28+\displaystyle{n\choose 8}.8$ And then plug in $n=8$
I know this sucks, but it's the only way I could think of...
Also, if you do this my way: the numbers you see, like 28 and 56, is there a way to find them in a more mathematical way? I just counted how many different ways there are...  :?

So does anyone have an elegant and a bit less ugly way to do this question?[/quote]

The way I thought of 4 is like this:

In the expansion of that quantity (which was to the ninth power), each term is composed of nine elements.  Each element can be taken from the set of $x_1,x_2,x_3...x_8$ for $u(7)$.  There are a total of 8 elements to choose from.  In order to find the number of terms, you're looking for how many different ways to form multisets of length 9 from that set of 8 elements.  This would be:

${8 + 9 - 1\choose9} = {16\choose 9} = 11440$[/quote]

That's essentially same as my solution.

[quote="eryaman"]On problem number three is says the range is $\frac{\pi}{7}<m<\frac{4\pi}{7}$.  But the correct answer is $\frac{\pi}{2}$ which is not in that range... :?[/quote]

Yes it is... $\frac{\pi}{2}=\frac{3.5\pi}{7}<\frac{4\pi}{7}$.

#1. I extended two lines to form a right triangle with $AY$ as hypotenuse, and used that.

#2. Noticed that $\frac{x+4}{x-1}=1R5$ and $\frac{x+4}{x-7}=1R7$, so I figured $(x+4)$ must be the remainder of the other one, too. :P

#3. $\cos \theta +i\sin \theta=e^{i\theta}$, so we have $\frac{e^{2mi}}{e^{mi}}=e^{mi}=\cos m +i\sin m =i$. Therefore, $\sin m=1 \Rightarrow m=\frac{\pi}{2}$.

#4. The exponents must form $n$-digit numbers where all digits sum to 9 and $n$ is the number of variables in the given term. So, we count up all such numbers and multiply by $\binom{8}{n}$ for each $n$.

#5. $\cos C=0 \Rightarrow C=\frac{\pi}{2}$, so $\triangle ABC$ is a right triangle. All three are integers, so we need a Pythagorean triple with $a+b=c+1$, and we know that $c$ is the longest side. So, using the formula for the lengths in a Pythagorean triple, we have $m^2-n^2+2mn=m^2+n^2+1 \Rightarrow (m-n)(m+n)=(m-n)^2+1$. So, this means that $2n(m-n)=1 \Rightarrow n(m-n)=\frac{1}{2}$, which is impossible because $m$ and $n$ must be integers.

1. Law of Cosines...

2. $f(1)=5$ and $f(3)=7$ by Polynomial Remainder Theorem or something, so $f(x)=x+4$, and that's the remainder because the divisor is of greater degree.

3. Solved $e^{mi} = i$ using logs :D

4. :ninja:

5. $a+b=c+1$, $a^2+b^2=c^2$

$(a+b)^2-2ab=c^2$
$(c+1)^2-2ab=c^2$
$(c+1)^2-c^2=2ab$
$2c+1=2ab$
$\text { odd } \ne \text { even. }$

[quote="JesusFreak197"][quote="eryaman"]On problem number three is says the range is $\frac{\pi}{7}<m<\frac{4\pi}{7}$.  But the correct answer is $\frac{\pi}{2}$ which is not in that range... :?[/quote]

Yes it is... $\frac{\pi}{2}=\frac{3.5\pi}{7}<\frac{4\pi}{7}$.[/quote]I feel incredibly stupid right now.  :blush: I was wondering since last night why the only solution seemed to be out of the range...now this. Oops.  :blush:

#4

A nice approach for who doesn't like combinatorics is to note this:
it is possible to make a chart that count the cardinality of the powers.

           2     3     4     5     6     7     8
$k^2$  3     6     10
$k^3$  4    10    20
$k^4$  5    15    35
$k^5$  6    21    56
$k^6$  7    28    84
$k^7$  8    36    120
$k^8$  9    45    165
$k^9$ 10   55    220

Every element is the sum of the above element in the same column with the preceding element in the same line.

After completing this chart you will become very good at sums.  :D

sorry but I'm not good with latex and so this chart looks orrible  :lol:


[quote="Jakey"]Can anyone help me with this?

Prove: (2222^5555)  +  (5555^2222) is divisible by 7

Can i prove this through modular arithmetic, like the euler-phi theorem?

or can anyone provide another solution to this? thanks[/quote]
$ 7\not | 2222$ so $ 2222^{7\minus{}1} \equiv 1\pmod 7 \implies 2222^6 \equiv 1 \pmod 7$ raise both sides to the $ 950$ to get $ 2222^{5550} \equiv 1 \pmod 7$.
Now $ 2222^5 \cdot 2222 \equiv 1 \pmod 7$ so $ 2222^5,2222$ are inverses. Note $ 2222\equiv 3$ and the inverse of $ 3$ is $ 5$. That means $ 2222^5 \equiv 5 \pmod 7$ multiply the two congruences to get $ 2222^{5555} \equiv 5 \pmod 7$ now do the same for the other one which I guess is similar.

[quote="Jakey"]Can i prove this through modular arithmetic[/quote]

Any problem that talks about remainders is modular arithmetic.  In this case, the relevant theorem is a weaker form of Euler's Totient Theorem known as [i]Fermat's Little Theorem.[/i]

[hide="Statement"] For a given prime $ p$ and a number $ a$ not divisible by $ p$, we have $ a^{p \minus{} 1} \equiv 1 \bmod p$. [/hide]
[hide="Discussion of the problem"] The value of modular arithmetic is that any problem involving large numbers can be [i]automatically[/i] reduced to a problem with smaller numbers.  In particular, because

$ 2222 \equiv 3 \bmod 7$

we have

$ 2222^{5555} \equiv 3^{5555} \bmod 7$

and because Fermat's Little Theorem tells us $ 3^{6} \equiv 1 \bmod 7$, we can reduce the exponent $ \bmod 6$.  In particular,

$ 5555 \equiv 5 \bmod 6$

we know that

$ 3^{5555} \equiv 3^5 \bmod 7$

which is a much easier calculation to make.  Similarly, because

$ 5555 \equiv 4 \bmod 7$

we conclude that

$ 5555^{2222} \equiv 4^{2222} \bmod 7$

and because

$ 2222 \equiv 2 \bmod 6$

we conclude that

$ 4^{2222} \equiv 4^2 \bmod 7$.

We have therefore [i]automatically[/i] reduced the statement of the original problem to this:  Prove that $ 3^5 \plus{} 4^2$ is divisible by $ 7$.  Since $ 3^5 \plus{} 4^2 \equal{} 243 \plus{} 16 \equal{} 259 \equal{} 7 \cdot 37$, we are done. [/hide]

oh yeah, thanks....i forgot to use fermat's little theorem because i learned that the euler-phi theorem is the generalization for that...thanks again!:D

Actually isnt it much shorter than that
$ 2222 \equiv 3\pmod 7$
$ 3^3 \equiv \minus{}1\pmod 7$
Therefore $ 2222^{5555} \equiv \minus{}9\pmod 7$
$ 5555 \equiv \minus{}3\pmod 7$
$ (\minus{}3)^3 \equiv 1\pmod 7$
Therefore $ 5555^{2222} \equiv 9\pmod 7$
Hence the result .

oh yeah, that's easier but any will do. thanks alot:D

[quote="madness"]Actually isnt it much shorter than that[/quote]

I was explaining the calculation in a way that made general techniques clear.  The general technique is equally straightforward for large moduli.


Let S(r = 1, n) be notation for a sigma with r = 1 on the bottom and n on the top.

Show that S(r = 1, n){4 - 6r} = n - 3n^2

Hence write down a formula for S(r = 1, 2q){4 - 6r}

Show that S(r = q + 1, 2q){4 - 6r} = q - 9q^2

[quote="sludgethrower"]Show that $\sum_{r=1}^{n}(4-6r) = n-3n^{2}$

Hence write down a formula for $\sum_{r=1}^{2q}(4-6r)$

Show that $\sum_{r=q+1}^{2q}(4-6r) = q-9q^{2}$[/quote]

If you're gonna go to the trouble of coming up with notation, it really wouldn't take much effort for you to pick up the $\LaTeX$ syntax.  Just mouse over anyone's $\LaTeX$ and the browser will show you the code.

[quote="sludgethrower"]I want to learn how to use it, but when i go to latex help, it says the webpage cannot be found...[/quote]

The link "LaTeX Help" on Mathlinks is not available.
The correct link should be:
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

But, it is much better to start with the FAQ in the LaTeX forum here.

For the initial problem $\sum_{r=1}^{n}(4-6r)=4n-6\sum_{r=1}^{n}r=4n-6\frac{n(n+1)}{2}=4n-3n^{2}-3n=n-3n^{2}$
Which was quite obvious. The other two thingies being merely consequences of this.


Hi everyone,

Could anyone help me with the following question, I would really appreciate it! :)

Let summation of a_k be a series with nonnegative terms. Prove that if summation of (a_k)^2 converges, then summation of (a_k/k) converges.

Thanks!

Cauchy-Schwartz:

$\left|\sum a_n\cdot\frac1n\right|\le \sum a_n^2\cdot\sum \frac1{n^2}$

Since $\sum\frac1{n^2}$ converges, the sum on the left converges (absolutely), and we can bound its sum.


For which real numbers $a$ does the sequence defined by the initial condition $u_{0}= a$ and the recursion $u_{n+1}= 2u_{n}-n^{2}$ have $u_{n}> 0$ for all $n \ge 0$?

[hide="Sketch"] Quite similar to that other weird recursion you gave once.

$u_{n+1}= 2u_{n}-n^{2}= 2(2u_{n-1}-(n-1)^{2})-n^{2}= ... = 2^{n+1}a-\left( \sum_{k=1}^{n}2^{n-k}k^{2}\right)$

We evaluate the large sum $S_{k}$ (it's not very hard) and then simply calculate the minimum value of $\frac{S_{k}}{2^{k+1}}$, which will be, y'know, something. [/hide]


:blush:
Seven accidents occur in a week .What is the probability that they occur on the same day?

Assumptions? If they are that accidents occur independently and each occurs with probability $ 1/7$ on each particular day, then it is clearly $ 7^{\minus{}6}$. What is the texbook answer?

yes it is $ 7^{\minus{}6}$
but my logic was let $ x_{i}$ be the number of accidents occuring on  the ith day $ 1\leq i\leq7$
then we have $ \sum x_{i}\equal{}7$ so in this case number of solutions is $ \binom{13}{7}$ out of this
only 7 cases favour us
assuming each to occur with same probability our required prob is $ \frac{7}{\binom{13}{7}}$

what's the fault? and i don't get the actual answer as well

Let's reduce it to 2 accidents in 2 days. You argue that there are 3 possibilities of equal probability: both accidents on the first day, both on the second day, and one on each day, so the answer is $ 1/3$. I argue that there are 4 possibilities of equal probability: accidents 1 and 2 on the first day, accidents 1 and 2 on the second day, accident 1 on the first day and 2 on the second, and accident 1 on the second day and 2 on the first, which gives me $ 1/4$.

yes that's right
i get it now the problem is to findd the number of ways to distribute 7 dissimilar balls in 7 distincts urns where order doesn't matter and an urn can remain empty :)

Thanks a ton :)


What is harmonic conjugate? How to make use of it?

I don't know why this thread was moved so quickly.  Are you referring to the harmonic conjugates used in geometry or complex analysis?


@CYGAO15 TCJY HAS ASKED TO EXPRESS IN HOURS NOT MINUTES .

THE ANSWER I GOT IS:-

 1 HOUR= 60 MINUTES.

X HOURS=(45+75)MINUTES.

X HOURS=120 MINUTES .

SO, 120 MINUTES = 2 HOURS.

uhh.... but u realize u are posting this in a new topic instead of the old one?


Let $ A$ the set of all polynomials $ p(x)$ of second degree such that $ p(1) \equal{} p(2) \equal{} 0$. Let $ B$ the set of all polynomials $ P(x)$ of second degree such that $ p(1) \equal{} p(2) \equal{} p(3) \equal{} 0$. Hence:

a) $ A \equal{} B$
b) $ A\cup B \equal{} B$
c) $ A\subset B$
d) $ B\subset A$
e) none of the previous answers

[hide="Answer"]

d[/hide]

A polynomial of the second degree can have at most two roots. Therefore $ B \equal{} \emptyset$, and $ A$ contains all polynomials of the form $ a(x \minus{} 1)(x \minus{} 2)$, with $ a \in \mathbb{R}^*$. Therefore, $ B \subset A$ (the empty set is a subset of any other set).


Given a,b,c>0 and abc=1.Prove that
$ \frac {(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)}{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)} \ge \sqrt [3]{\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8}}$
now and future,I will try to solve problem,i dont' creat more ineq.Thank you to all people.

Hi.
$ \frac{x^2 \plus{} 1}{x \plus{} 1}\geq \sqrt[3]{\frac {x^3 \plus{} 1}{2}}$
 $ \Leftrightarrow (x \minus{} 1)^4(x^2 \plus{} x \plus{} 1)\geq 0$
 and $ (a^2 \plus{} bc)(1 \plus{} \frac {bc}{c})\geq (a \plus{} b)^2$
 so $ (a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)\geq abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$
 $ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq(a \plus{} b)(b \plus{} c)(c \plus{} a)$

The stronger is trues and very easy: $ (1\plus{}a)^3(1\plus{}b)^3(1\plus{}c)^3 \geq\ 64(a\plus{}b)(b\plus{}c)(c\plus{}a)$

[quote="arqady"][quote="nguoivn"]The stronger is trues and very easy: $ (1 \plus{} a)^3(1 \plus{} b)^3(1 \plus{} c)^3 \geq\ 64(a \plus{} b)(b \plus{} c)(c \plus{} a)$[/quote]
$ \prod_{cyc}(1 \plus{} a)^{16} \equal{} \left(\prod_{cyc}(1 \plus{} a)(1 \plus{} b)\right)^8 \equal{} \left(\prod_{cyc}(1 \plus{} ab \plus{} a \plus{} b)\right)^8\geq$
$ \geq8^8\left(\prod_{cyc}(1 \plus{} ab)(a \plus{} b)\right)^4 \equal{} 8^8\prod_{cyc}(a \plus{} b)^4\left(\prod_{cyc}(1 \plus{} ab)(1 \plus{} ac)\right)^2 \equal{}$
$ \equal{} 8^8\prod_{cyc}(a \plus{} b)^4\left(\prod_{cyc}(1 \plus{} a \plus{} a(b \plus{} c))\right)^2\geq8^{10}\prod_{cyc}(a \plus{} b)^5\prod_{cyc}(1 \plus{} a),$
which gives your inequality.[/quote]
Very very [b]nice[/b] your proof!I love it :lol:

My full proof: $ (a \plus{} b)(1 \plus{} ab) \leq\ \frac {(1 \plus{} a)^2(1 \plus{} b)^2}{4}$
Because of: $ (1 \plus{} ab)(1 \plus{} bc)(1 \plus{} ca) \equal{} (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)$, (with $ abc \equal{} 1$)we have q.e.d  :)
@babyloverain: [hide]Hi, bai` toan manh. hon nay` thuc ra la` 1 bai` toan cu~ cua anh tu`hoi` lop 12 . Ma` sang tao la` tot', viec gi` em fai? tu` bo? chi? vi` thinh? thoang? trung` y tuong? voi ng` khac' nhi?. Co' len em nhe'  :)  [/hide]


Hello Everyone,

I'm stuck on another one. Sorry about all the questions.
Below is a screen shot of a problem in which I arrive at one answer, and my teacher at another.

What am I missing, or did the teacher just make a blunder?  :D

You have $u=1,2$, but you're solving for $a$. Since $u=a^2$...
Surely you know what to do now.

You found the solutions for $a^2$ not the solutions for $a$. You missed the $a^2 = 1 \quad or \quad a^2 = 2$...


Combinatorics for undergraduates (by Andreescu/Feng I think) is pretty good. Despite the "for undergraduates" part of the title it is just a contest book. Doing problems is always the best way to improve, unless your background is sorely lacking.

i find that doing harder problems first (same concepts or maybe a little more advanced) makes the easier problems a piece of cake :)

[quote="mathemonster"]What is the best combinatorics book for PuMaC / AIME or would u just suggest doing a buttload of problems and that will be fine.[/quote]

Our Intermediate Counting & Probability book (particularly if you already find AMC 10/12 problems easy) is what you're looking for.

The art of problem solving intermediate counting / probability book is a perfect book for the AIME / Pumac problems.

If you want a more challenging USAMO level text, I suggest you go with Richard Stanley's Enumerative Combinatorics (Volume 1 and 2).


calculus
integration
trigonometry
function
LaTeX
calculus computations

:spider:  $Ascertain\ \int \frac{x^2}{(x^2-4)sinx+4xcosx}\ dx,\ where\ x\in (0,\pi).$

$I = \int {\frac{{x^2 }}{{(x^2  - 4)Sinx + 4xCosx}}} dx =$

$\int {\frac{{\frac{{x^2 }}{{(x^2  + 4)}}}}{{\frac{{(x^2  - 4)}}{{(x^2  + 4)}}Sinx + \frac{{4x}}{{(x^2  + 4)}}Cosx}}} dx$

Put $x = 2Cot\frac{\alpha }{2} \Rightarrow dx =  - Co\sec ^2 \frac{\alpha }{2}$
$ \Rightarrow \frac{{(x^2  - 4)}}{{(x^2  + 4)}} = Cos\alpha$    and    $\frac{{4x}}{{(x^2  + 4)}} = \,Sin\alpha$

$ \Rightarrow I =  - \int {\frac{{\frac{{4Cot^2 \frac{\alpha }{2}}}{{4Co\sec ^2 \frac{\alpha }{2}}}Co\sec ^2 \frac{\alpha }{2}}}{{Cos\alpha Sinx + Sin\alpha Cosx}}} d\alpha$

$ \Rightarrow I =  - \int {\frac{{Cot^2 \frac{\alpha }{2}}}{{Sin(\alpha  + 2Cot\frac{\alpha }{2})}}} d\alpha$

Now put $y = \alpha  + 2Cot\frac{\alpha }{2}$

$\Rightarrow dy = (1 - Co\sec ^2 \frac{\alpha }{2})d\alpha  =  - Cot^2 \frac{\alpha }{2}d\alpha$

$\Rightarrow I =  - \int {\frac{{Cot^2 \frac{\alpha }{2}}}{{Sin(\alpha  + 2Cot\frac{\alpha }{2})}}} d\alpha  = \int {Co\sec } ydy$  $= log|Cosecy - Coty|$

Hence the solution.

Some trig functions with $\LaTeX$ codes:
$\sin$,$\cos$,$\tan$,$\sec$,$\cot$,$\csc$.

These are worth using.

[quote="vidyamanohar"]$I = \int {\frac{{x^2 }}{{(x^2  - 4)Sinx + 4xCosx}}} dx =$

$\int {\frac{{\frac{{x^2 }}{{(x^2  + 4)}}}}{{\frac{{(x^2  - 4)}}{{(x^2  + 4)}}Sinx + \frac{{4x}}{{(x^2  + 4)}}Cosx}}} dx$

Put $x = 2Cot\frac{\alpha }{2} \Rightarrow dx =  - Co\sec ^2 \frac{\alpha }{2}$
$ \Rightarrow \frac{{(x^2  - 4)}}{{(x^2  + 4)}} = Cos\alpha$    and    $\frac{{4x}}{{(x^2  + 4)}} = \,Sin\alpha$

$ \Rightarrow I =  - \int {\frac{{\frac{{4Cot^2 \frac{\alpha }{2}}}{{4Co\sec ^2 \frac{\alpha }{2}}}Co\sec ^2 \frac{\alpha }{2}}}{{Cos\alpha Sinx + Sin\alpha Cosx}}} d\alpha$

$ \Rightarrow I =  - \int {\frac{{Cot^2 \frac{\alpha }{2}}}{{Sin(\alpha  + 2Cot\frac{\alpha }{2})}}} d\alpha$

[b][i]Dear sir[/i], it is O.K. Thanks for your nicely solutions. With friendship and respectfully,[/b] [b][i]Virgil Nicula[/i][/b]

Now put $y = \alpha  + 2Cot\frac{\alpha }{2}$

$\Rightarrow dy = (1 - Co\sec ^2 \frac{\alpha }{2})d\alpha  =  - Cot^2 \frac{\alpha }{2}d\alpha$

$\Rightarrow I =  - \int {\frac{{Cot^2 \frac{\alpha }{2}}}{{Sin(\alpha  + 2Cot\frac{\alpha }{2})}}} d\alpha  = \int {Co\sec } ydy$  $= log|Cosecy - Coty|$

Hence the solution.[/quote]


algorithm
number theory proposed
number theory

Hi everybody.
A new arithmetic puzzle : tiercé d'enfer.
Hard to solve, challenging and maybe (I'm not sure) of interest for the number theory.

Here is a list of 10 digits
1 2 2 3 4 5 6 7 7 8

1. Take this list of 10 digits, compute its sum, it is equal to 45.

2. Remove the digit 2, the sum of the 9 remaining digits will be equal to 43.

3. Multiply 2 by 43, your result is 86.

4. Remove the 8 and 6.

5. Compute the sum of the 7 remaining digits: 1 2 3 4 5 7 7

1+2+3+4+5+7+7=29

6. Remove the 7, the remaining sum will be equal to 22.

7. Multiply 7 by 22, you obtain 154.

8. Remove the 1, the 5 and the 4.

9. Compute the sum of the 3 remaining digits: 2 3 7
2+3+7=12

10. Remove the digit 3, the sum of the 2 remaining digits will be equal to 9.

11. Multiply 3 by 9, you obtain 27.

12. Remove the 2 and the 7.

At the end of the process I removed the 10 digits by using the same procedure.

I used 2 (stage 2), 7 (stage 6) and 3 (stage 11) to be able to remove all the 10 figures from the list.

So the triplet used in order is 2-7-3

How much numbers of 3 digits satisfy the algorithm described above?

8-3-2 is one of the solutions

and the 10 digits sequence is

1-1-2-2-3-5-6-8-8-9


abstract algebra
group theory
superior algebra
superior algebra unsolved

(Abstract Algebra, Dummit and Foote: pg. 130, Chapter 4 Section 3 Exercise 6)
Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that <g> is subgroup of C_G (g) = {h in G | hg = gh} to show that there is at most one possible class equation.

By Example (Group of order pq, where p, q are primes with p < q) (pg. 143), I found that if G is of order pq, where p, q are primes with p < q and p does not divide q-1 then G is cyclic. In particular, since 15 = 3*5 and 3 does not divide 5-1 = 4, G should be cyclic: i.e., G should be abelian. I came to conclusion that the problem does not make much sense.

However, is there any way to solve the problem only using the hint provided?

the book's exercise makes no sense: you already cited a more general result, that $ G$ must be cyclic. This contradicts, the statements $ G$ is non-abelian resp. $ Z(G)\equal{}1$ or for an alternative contradiction:

Let $ 1\not\equal{}g\in G$ be arbitrary. Since $ |<g>|$ divides $ |G|\equal{}15$ and $ G$ non-abelian, we must have $ g$ of order 3 or 5. Since $ <g>$ has prime index in $ G$, every subgroup containing $ <g>$ must be equal to $ <g>$ or $ G$. Therefore, $ C_G(g)\equal{} <g>$. Otherwise $ C_G(g)\equal{}G$ and we can choose $ h\in G\setminus <g>$, which commutes with $ g$. Because of $ <g>\subset <g,h>$ we have $ G\equal{}<g,h>$ and $ G$ is abelian, a contradiction. We summarise that for $ 1\not\equal{}g\in G$ we have $ C_G(g)\not\equal{}G\Longleftrightarrow g\not\in Z(G)$. So far so good for the proof of $ Z(G)\equal{}1$.

As for the [i]nonsense[/i]: Consider the normalizer $ N_G(<g>)$. The factor group $ N_G(<g>)/C_G(<g>)\equal{}N_G(<g>)/<g>$ divides the index $ |G: <g>|$ and is isomorphic to some subgroup of $ Aut(<g>)$. For both cases of the order of $ g$ we see that $ \gcd(|Aut(<g>),|G: <g>|)\equal{}1$, i.e. $ N_G(g)\equal{}<g>$. Because of trivial intersections and Sylow's theorem, we would have $ 5\cdot(3\minus{}1)\equal{}10$ elements of order $ 3$ and $ 3\cdot (5\minus{}1)\equal{}12$ elements of order $ 5$, a contradiction, i.e. a group with conditions as stated in the exercise does not exist


How many equilateral triangles are there in the figure below?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=660[/img]

[hide]
Go through first and count all the "right side up" equilateral triangles. This is simply going upwards from the bottom row, and counting the number of ways to choose two different points in each row, since that will give you a unique triangle. so we have:
\[ {9\choose 2}+{8\choose 2}+{7\choose 2}+{6\choose 2}+{5\choose 2}+{4\choose 2}+{3\choose 2}+{2\choose 2} = 120 \]

Now we count the number of upside-down triangles. We start from the top row:
\[ {2\choose 2}+{3\choose 2}+{4\choose 2}+{5\choose 2}+ \]
but after that, we have to limit the distance between the two points we choose, so we continue as before, but subtracting the possibilities which will give you a triangle that exceeds the boundaries:
\[ {6\choose 2}-3+{7\choose 2}-10+{8\choose 2}-21 \]

adding everything up, we have:
\[ \boxed{170} \]
[/hide]

Thanks a lot.
Is it possible to find the number by setting up a reccurence relation?
I tried it before but the relation is too complicated and therefore I gave up the approach.

Dear Halluci,
I think the answer shoud be 124 instead.
According to your explanation,  the total number of equilateral triangles should be
 $[{9\choose 2}+{8\choose 2}+{7\choose 2}+{6\choose 2}+{5\choose 2}+{4\choose 2}+{3\choose 2}+{2\choose 2} ] \\ +[ {8\choose 2}+{7\choose 2}+{6\choose 2}+{5\choose 2}+{4\choose 2}+{3\choose 2}+{2\choose 2}] \\ -[{3\choose 2}+{2\choose 2}] \\ -[{5\choose 2}+{4\choose 2}+{3\choose 2}+{2\choose 2}] \\ -[{7\choose 2}+{6\choose 2}+{5\choose 2}+{4\choose 2}+{3\choose 2}+{2\choose 2}]$
which gives the answer 124.

In fact, we can prove that ${2\choose 2} +{3 \choose 2} + ...+{n\choose 2} = {{n+1} \choose 3}.$
Therefore, the above expression can be simplified to
${10 \choose 3} + {9 \choose 3} -{4 \choose 3} -{6\choose 3} -{8 \choose 3} = 124$

Yours, Michael.

the best way that you are going to find a formula for this is to break it up into the sides being even or odd, this makes it much easier to deal with the triangles that are not oriented in the same way as the larger one


inequalities
calculus
calculus computations

Find all x:

$ log_{a}(x^2 \minus{} x \minus{} 2) > log_{a}( \minus{} x^2 \plus{} 2x \plus{} 3)$

hint: check $ x \equal{} \frac {9}{4}$


limit
integration
calculus
function
calculus computations

An exercise for my problem-solving group this week; should be fairly straightforward.

a) Show that $ \sum_{k\equal{}0}^n\frac{(\minus{}1)^k}{2k\plus{}1}\binom{n}{k}\equal{}\prod_{k\equal{}1}^n\frac{2k}{2k\plus{}1}$ for all $ n\in\mathbb{N}.$

b) Compute $ \lim_{n\to\infty}\sqrt{n}\,\prod_{k\equal{}1}^n\frac{2k}{2k\plus{}1}.$

We have

$ \sum\limits_{k\equal{}0}^{n}(\minus{}1)^{k}\frac{1}{2k\plus{}1}\binom{n}{k}\equal{}\int\limits_{0}^{1}\left(\sum\limits_{k\equal{}0}^{n}(\minus{}1)^{k}x^{2k}\binom{n}{k}\right)dx$

$ \equal{}\int\limits_{0}^{1}(1\minus{}x^{2})^{n}dx\equal{}\frac{1}{2}\beta(n,1/2)\equal{}\frac{1}{2}\frac{\Gamma(n\plus{}1)\Gamma(1/2)}{\Gamma(n\plus{}3/2)}$, which can be shown to be equal to that product.

Part b) of the problem can be solved by Stirling's formula.
I hope I made no computational mistakes.

Can you do part b) from the integral representation, without using Stirling?

If we define $ I_n\equal{}\int_0^1(1\minus{}x^2)^n\,dx,$ can you do an elementary, direct calculation to show that $ I_n\equal{}\frac{2n}{2n\plus{}1}I_{n\minus{}1}\ ?$

The recurrence formula can be obtained by integration by parts.

I will concentrate on your first question. We have

$ \sqrt{n}\int\limits_{0}^{1}(1-x^{2})^{n}dx=\sqrt{n}\int\limits_{0}^{1}e^{-nx^{2}}dx-\sqrt{n}\int\limits_{0}^{1}\left(e^{-nx^{2}}-(1-x^{2})^{n}\right)dx$.

Using the substitution $ nx^{2}=y$ we get that
$ \sqrt{n}\int\limits_{0}^{1}e^{-nx^{2}}dx=\frac{1}{2}\int\limits_{0}^{n}e^{-y}y^{1/2-1}dy\rightarrow \frac{1}{2}\Gamma(1/2)=\frac{\sqrt{\pi}}{2}.$

It suffices to prove that the second integral converges to 0.

Let $ f(x)=e^{-nx^{2}}-(1-x^{2})^{n}$.

A calculation shows that the critical points of this function are $ 0$ and $ \alpha_{n}$ where

$ \alpha_{n}$ is the unique root of $ 1-x^{2}=e^{-\frac{n}{n-1}x^{2}}$.

It is not hard to prove that $ \frac{1}{\sqrt{n}}\leq \alpha_{n}\leq \frac{\sqrt{2}}{\sqrt{n}}$.

We note that the maximum of $ f$ is attained at $ \alpha_{n}$ and it is equal to

$ e^{-n\alpha_{n}^{2}}-e^{-\frac{n^{2}}{n-1}\alpha_{n}^{2}}$.

It follows, by using that $ e^{x}-e^{y}=(x-y)e^{\theta}$, that

$ 0<\sqrt{n}\int\limits_{0}^{1}\left(e^{-nx^{2}}-(1-x^{2})^{n}\right)dx\leq \sqrt{n}\left(e^{-n\alpha_{n}^{2}}-e^{-\frac{n^{2}}{n-1}\alpha_{n}^{2}}\right)$

$ \leq \sqrt{n}\alpha_{n}^{2}\frac{n}{n-1}e^{\theta_{n}}$ where

$ \theta_{n}$ is between $ -n\alpha_{n}^{2}$ and $ -\frac{n^{2}}{n-1}\alpha_{n}^{2}$

Thus, by a previous estimate for $ \alpha_{n}$ we get that $ e^{\theta_{n}}$ and $ n\alpha_{n}^{2}$ are bounded, and hence,

$ \sqrt{n}\alpha_{n}=\frac{\sqrt{n}}{{n}}\cdot n\alpha_{n}^{2}\rightarrow 0$ and the problem is solved.

I hope I made no mistakes.

In my own version, I used the substitution $ x\equal{}\frac{u}{\sqrt{n}}$ to get that

$ \sqrt{n}\int_0^1(1\minus{}x^2)^n\,dx\equal{}\int_0^{\sqrt{n}} \left(1\minus{}\frac{u^2}n\right)^n\,du$

We note that $ \left(1\minus{}\frac{u^2}n\right)^n$ tends pointwise to $ e^{\minus{}u^2},$ so that if we could justify the interchange of limit and integral, we would have

$ \int_0^{\infty}e^{\minus{}u^2}\,du\equal{}\frac{\sqrt{\pi}}2.$

The only weak link is justifying the exchange of limit and integral.  There's a Dominated Convergence Theorem argument available, but if I restrict myself to relatively elementary methods, I may not have that tool. We could then split $ [0,\infty)$ into a finite interval on which the sequence of functions converges uniformly and an outer portion which can be estimated to be small.

In all, there are more similarities between this and didilica's argument than there are differences.


limit
function
integration
logarithms
algebra proposed
algebra

Let
$ a_{k}$ is smallest integer greater than
$ \sum\limits_{i=1}^{k}\frac{1}{i}$
Find $ lim\frac{a_{k+1}}{a_{k}}$

[quote="TTsphn"]Let
$ a_{k}$ is smallest integer greater than
$ \sum\limits_{i = 1}^{k}\frac{1}{i}$
Find $ lim\frac{a_{k+1}}{a_{k}}$[/quote]

Since $ \frac{1}{k+1}<1$ $ \forall k>1$, either $ a_{k+1}=a_{k}$, either $ a_{k+1}=a_{k}+1$

So $ 1\leq\frac{a_{k+1}}{a_{k}}\leq 1+\frac{1}{a_{k}}$ $ \forall k>1$

And since $ \lim_{k\rightarrow+\infty}a_{k}=+\infty$, $ \lim_{k\rightarrow+\infty}\frac{a_{k+1}}{a_{k}}=1$

I'm sorry this is my problem
For each k let $ a_{k}$ is the smllest positive integer such that
$ \sum\limits_{i = 1}^{a_{k}}\frac{1}{i}>k$
Find $ lim\frac{a_{k+1}}{a_{k}}$

[quote="TTsphn"]I'm sorry this is my problem
For each k let $ a_{k}$ is the smllest positive integer such that
$ \sum\limits_{i = 1}^{a_{k}}\frac{1}{i}> k$
Find $ lim\frac{a_{k+1}}{a_{k}}$[/quote]
Obviosly e, because from $ H_{a_{k}}= k+O(\frac{1}{a_{k}})$ we get $ a_{k}= e^{-\gamma }(1+O(\frac{1}{k}))$.

lim=?
I don't know function you write because is hasn't in my national's text program?

I think, that e is international symbol ($ e\equal{}\lim_{n\to\infty}(1\plus{}\frac{1}{n})^{n}$).


Find all integral solutions  solutons to $x,y$ if $y^4-2x^4=707$

[hide="Hint"]Try to look $\bmod 5$.[/hide]

[hide]For any integer $n$, $n^4 = 1 \text{ (mod 5)}$

So $y^4-2x^4=1-2(1) \text{ (mod 5)}= 4 \text{ (mod 5)}$ and $707= 2 \text{ (mod 5)}$, so there are no solutions?[/hide]

[hide][quote="jrshoch"]For any integer $n$, $n^4 = 1 \text{ (mod 5)}$

[/quote]
$n^4 \equiv 1$ or $n^4 \equiv 0$ $\bmod 5$. But no solutions.[/hide]


linear algebra
matrix
geometry
superior algebra
superior algebra unsolved

1) if $ R$ is a non-commutative ring and $ ab \minus{} 1$ is invertible for $ a, b \in \Bbb R$, then prove that $ ba\minus{}1$ is invertible as well.

2)Suppose that $ R$ is a ring and $ *$ and $ **$ be the following two properties;

$ *$:  for $ a,b \in R$ such that $ ab \equal{}0$ we have that $ ba\equal{}0$
$ **$ : for $ a,b \in R$ such that $ ab \equal{}1$ we have that $ ba\equal{}1$

Prove that if $ R$ satisfies $ *$ then it satisfies $ **$ as well. To prove that the converse is not true give a counterexample.

1) Over complete fields where it makes sense, $ (1 \minus{} ab)^{ \minus{} 1} \equal{} 1 \plus{} ab \plus{} abab \plus{} ababab \plus{} \ldots$, so $ b(1 \minus{} ab)^{ \minus{} 1}a \equal{} ba \plus{} baba \plus{} \ldots \equal{} (1 \minus{} ba)^{ \minus{} 1} \minus{} 1$.  So we motivate/surmise that the inverse of $ 1 \minus{} ba$ is $ b(1 \minus{} ab)^{ \minus{} 1}a \plus{} 1$, and you can check this easily directly.

2) suppose * holds.  Suppose $ ab \equal{} 1$.  Then $ a(1 \minus{} ba) \equal{} a \minus{} aba \equal{} a \minus{} a \equal{} 0$, so $ (1 \minus{} ba)a \equal{} 0$ by *, and this means $ (1 \minus{} ba) \equal{} (1 \minus{} ba)ab \equal{} 0$ also, so $ ba \equal{} 1$.

3) Matrix ring.

The problem 1) and the formal power series solution is mentioned by Halmos in
his classic expository Intelligencer article [1]. When I asked Paul Cohn (an expert
in this area) what his explanation is for why this works, he replied with the paper [2].

--Bill Dubuque

[1]  Halmos, P. R. Does mathematics have elements?
Math. Intelligencer 3 (1981) 147-153.
http://dx.doi.org/10.1007/BF03022973

[2]  Cohn, P. M. A remark on the quasi-inverse of a product.
Ill. J. Math. 2003. 47, 1–2, 87–88.
http://www.math.uiuc.edu/~hildebr/ijm/spring03/final/cohn.pdf


On Saturday night, Becky goes stargazing. There's an that she will see a shooting star in any given hour. The probability of seeing a shooting star is uniform for the entire hour. What is the probability that Becky will see a shooting star in the first 15 minutes? Express your answer as a percent to the nearest tenth.
There's an 80%..

This is from Introduction to Counting and Probability!

Can you show me the solution to this problem from that book?It's a little bit tricky and I'm confused on interpetation as well as what the problem truly wants to talk about. :oops:

Umm, is this a bit too easy for Pre-Olympiad?

[hide="Hint"]The probability that you will [size=200]not[/size] see the star is uniform at any given time.[/hide]

Your problem is incomplete, by the way.

EDIT: Do you mean, there's an 80% probability?

I don't have the Intro to C and P anymore, but I believe it's supposed to be:

There's an $ 80\%$ chance that at any given moment in any hour a shooting star will appear.

Or something like that. I may be incorrect.

[hide="main idea"]Let p be the probability of seeing a star in the first 15 minutes. Then, $ (1\minus{}p)^4\equal{}\frac{1}{5}$, and you can simplify and stuff, which I'm too lazy to do.[/hide]

This should be in Intermediate, if not High School Basics.

To simplify:

$ \sqrt[4]{(1\minus{}p)^4} \equal{} \sqrt[4]{\frac{1}{5}}$

Thus, our answer is about $ 33.1\%$.

[quote="xpmath"][hide="main idea"]Let p be the probability of seeing a star in the first 15 minutes. Then, $ (1 \minus{} p)^4 \equal{} \frac {1}{5}$, and you can simplify and stuff, which I'm too lazy to do.[/hide]

This should be in Intermediate, if not High School Basics.[/quote]
I really don't understand how you got $ (1 \minus{} p)^4 \equal{} \frac {1}{5}$

There is an $ 80\%
$ chance of seeing at least one shooting star in one hour, so there is a $ 20%
$ chance of seeing no shooting stars.  the key is "at least".  Now if we have a chance of $ p$ of seeing a shooting star within a 15-minute period, we have a chance of $ 1 - p$ of seeing no stars each 15 minutes and thus a chance of $ (1 - p)^4$ chance of seeing no shooting stars in an hour.  Thus $ (1 - p)^4 = 20\%
$


trigonometry
quadratics
geometry unsolved
geometry

Prove that

$ \sin{18} \equal{} \textonequarter{(\sqrt{5} \minus{} 1)}$.

Observe that $ 18*5$ = $ 90$.

So let $ 18$ = $ x$. we get $ 2x \plus{} 3x$ = $ 90$.

$ 2x$ = $ 90 \minus{} 3x$.

Take sine on both sides.

$ sin 2x$= cos $ 3x$ ( Because $ sin (90\minus{}y) \equal{} cos y$)

Rest is trivial. Just use double and triple angle formulas and use the fact that $ sin 18$ is a positive value.

I'm sure the rest is trivial but i still dont get it.... can u please complete the proof....

Take isosceles $ \triangle ABC$ with $ a \equal{} BC,$ $ b \equal{} AB \equal{} AC \equal{} 1$ and $ \angle CAB \equal{} 36^\circ.$ Internal bisectors of $ \angle CAB, \angle ABC$ cut $ BC, CA$ at $ D, E.$ Calculating the angles, $ \triangle ABE$ is isosceles with $ BE \equal{} AE$ and $ \triangle EBC$ is isosceles with $ BE \equal{} BC,$ so that $ a \equal{} BC \equal{} AE \equal{} \frac {b^2}{a \plus{} b}.$ With $ b \equal{} 1,$ $ a^2 \plus{} a \minus{} 1 \equal{} 0,$ $ a \equal{} \frac {\sqrt 5 \minus{} 1}{2} < 1$ (the other root is negative).  From the right $ \triangle ABD,$ $ \sin 18^\circ \equal{} \sin \widehat {DAB} \equal{} \frac {BD}{AB} \equal{} \frac {a}{2b} \equal{} \frac {\sqrt 5 \minus{} 1}{4}.$

Hello malik.

Yetti has already given a beautiful proof. For the sake of completion of my proof :

$ sin 2x$ = $ cos 3x$

Expanding $ 2sinx*cosx$ = $ 4 cos^3x \minus{} 3cosx$

Cancelling $ cos x$ throughout you get $ 2 sinx \equal{} 4cos^2x\minus{}3$( Note you can cancel $ cos x$ throughout because you know $ cos 18$ is not zero.)

Now substituting $ cos^2x \equal{} 1\minus{}sin^2x$
u get the quadratic $ 4sin^2x \plus{} 2 sinx \minus{}1 \equal{}0$

Solving you get $ sin x$= $ \sin{18} \equal{} \textonequarter{(\sqrt{5} \minus{} 1)}$ or  $ \minus{}\textonequarter{(\sqrt{5} \plus{} 1)}$

But $ sin 18$ is positive because it is in the first quadrant. So $ sin 18 \equal{} \textonequarter{(\sqrt{5} \minus{} 1)}$

 $ Q.E.D.$


superior algebra
superior algebra unsolved

How many permutations of (1,2,3,....,n) (for some natural number n)
fulfill that the number (i+1) doesn't come right after the number i, for every 1<=i<=(n-1) ?

thanks!

For each subset $A$ of $\{1,2,\dots,n-1\}$, let $S(A)$ be the set of permutations $p$ such that for each $x\in A$, $p(x+1)=p(x)+1$. We can easily show that $|S(A)|=(n-|A|)!$. Also define $T(A)$ to be the set of permutations $p$ such that for each $x\in A$ and no others, $p(x+1)=p(x)$. Obviously, $S(A)=\bigcup_{B\supset A}T(B)$, so $|S(A)|=\sum_{B\supset A}|T(B)|$. By the inclusion-exclusion principle, we have $|T(A)|=\sum_{B\supset A}(-1)^{|B|-|A|}|T(B)|$.
We want to calculate $|T(\emptyset)|$. To do this, we apply the above formula and group the sets by size. There are $\binom{n-1}{k}$ sets of size $k$, so
$|T(\emptyset)|=\sum_{k=0}^{n-1}(-1)^k\cdot\binom{n-1}{k}\cdot(n-k)!= \sum_{k=0}^{n-1}(-1)^k\cdot\frac{(n-1)!}{k!}\cdot(n-k)$.
I have no reason to believe that this sum has a reasonable closed form; the first few values are $1,1,3,11,53$.

OK
there's only one thing i didn't understand in your answer.
why for every subset $A$, $|S(A)|=(n-|A|)!$. ?

I was a little short on time, so I skipped over that. The permutation is completely determined by its values on the complement of $A$ in $\{1,2,\dots,n\}$, since we can work backward from those values to fill in. In fact, it is completely determined by the order of those values; if there are $k$ elements of $A$ immediately before the largest, the second largest value is $n-k-1$ and so on. There are $(n-|A|)!$ possible orders, and we're done.

On something like this, you should always expect to have to do an inclusion-exclusion count; it's much easier to encode "at least this many" in the data.


Can anyone tell me where I can find the Romanian selection tests from 2001?

Hi,

http://www.mathlinks.go.ro/romseltests/romseltests2002.html
-> probably no tests for 2001  :(

http://ajorza.tripod.com/math.htm
-> this should work out properly  :)


geometry
3D geometry
number theory unsolved
number theory

Find all integers x,y,z that

19^x+7^y=z^3.

It's a very nice problem. First of all, it's easy to see that x, y, z are natural (they can't be negative). We check the case x=0 and get 1+7y=z3. I won't solve this here. It's enough to say that the only solution is y=1, z=2, so we have the solution (x,y,z)=(0,1,2).

Now let's see what happens if y=0. We get z3-1=19x, which means that z=19a+1 and the equation is 19x=193a+3*192a+3*19a. We divide it by 19a and we get a contradiction. This means that we're done with xy=0 and we can take x and y to be positive.

The cubes in Z7 (i.e. the residues which are cubes of another residue; we only consider those which are non-zero) are 1 and 6, and in order for 19x to be a cube in Z7 x must be divisible by 3, so x=3a. We get (z-19a)(z2+19a*z+192a)=7y (*), so z=19a+7b with b>=0. My goal is to show that b=0. Let's assume b>0. We replace z in (*) and we get a contradiction by dividing the whole thing with 7b, so b=0, so z=M19+1, so 7y=M19+1. The order of 7 in Z19 is 3, so y must be divisible by 3, so y=3b.

Since x=3a and y=3b, it means that we have a non-trivial solution of the equation (19a)3+(7b)3=z3, which contradicts Fermat's Last Theorem.


Hi,
I was wondering if anyone had the AMC 8 questions for (2006,2007,2008). If it is legal to view them now, I'd appreciate any of you to tell me where I can find them, or give them to me if you have them.

Thank you very much :)

Regards,
limac

I can't see why they wouldn't be legal. If not, tell me. If so, email me at bogtro123@gmail.com


Points $ B$, $ D$, and $ J$ are midpoints of the sides of right
triangle $ ACG$. Points $ K$, $ E$, $ I$ are midpoints of the sides of
triangle $ JDG$, etc.

[asy]/* AMC8 1999 #25 Problem */
draw((0,0)--(8,8)--(8,0)--cycle);
draw((4,0)--(4,4)--(6,4)--(6,6)--(7,6)--(7,7)--(8,7));
fill((4,0)--(8,4)--(8,0)--cycle);
fill((6,4)--(8,6)--(8,4)--cycle);
fill((7,6)--(8,7)--(8,6)--cycle);
label("$A$", (0,0), SW);
label("$J$", (4,4), W);
label("$I$", (6,6), W);
label("$H$", (7,7), W);
label("$G$", (8,8), NE);
label("$F$", (8,7), E);
label("$E$", (8,6), E);
label("$D$", (8,4), E);
label("$C$", (8,0), S);
label("$B$", (4, 0), S);
label(scale(0.8)*"$L$", (7,6.15), S);
label(scale(0.8)*"$K$", (6,4.15), S);[/asy]

 If the dividing and shading process is done 100
times(the first three are shown) and $ AC\equal{}CG\equal{}6$, then the total area
of the shaded triangle is nearest \\ \\
$ \text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ \\ \\
Select the most appropriate letter.

First, note that the total area of the triangle is equal to 6*6/2, or [b]18[/b]
then note that for every shaded triangle, there are two others to the right of it that are not shaded.  Therefore the shaded triangles sum to 1/3 of the total area, or [b]6[/b]
The answer is then the [b]A[/b].

[size=75][color=red]Fixed your tags. -Isabella[/color][/size]

This was on the [url=https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25]1999 AMC 8 (#25)[/url].

We note the largest triangle's area is $\dfrac{9}{2}$.
The total sum is:
$\dfrac{9}{2}+\dfrac{9}{8}+\dfrac{9}{32}+\dots \Rightarrow \dfrac{9}{2}(1+\dfrac{1}{4}+\dfrac{1}{16}+\dots)$.
Using the formula for the sum of a geometric series, we have:
$\dfrac{9}{2} \cdot \dfrac{4}{3}=\boxed{6}$.

Note: since this is a sequence with 100 terms, we can approximate it as an infinite geometric series.


[quote="minhtoan"]Find all triples of positive integers $(x, y, z)$ such that $x^{z+1}-y^{z+1}=2^{100}$[/quote]

Let $n$ be the greatest power of two which divides $x$ and $y$. Then, if $x=2^{n}u$ and $y=2^{n}v$, we have $u^{z+1}-v^{z+1}=2^{a}$. here, we can't have $u$ even and $v$ even together. If $u$ or $v$ is even and the other is odd, then $a=0$ and the equation $u^{z+1}-v^{z+1}=1$ can only have a solution if $z=0$ or $v=0$, two impossible conditions. So $u$ and $v$ are odd.

Then if $u^{z+1}-v^{z+1}=2^{a}$, with $u,v$ odd, Let $p$ be any divisor $>1$ of $(z+1)$. $(u^{p}-v^{p})$ divides $(u^{z+1}-v^{z+1})$, and so is a power of two. We have then :
$u-v=2^{b}$
$u^{p}-v^{p}=2^{c}$
So $(v+2^{b})^{p}-v^{p}=2^{c}$, and  $pv^{p-1}2^{b}+\sum_{i=0}^{p-2}\left(\begin{array}{c}p\\i\end{array}\right)v^{i}2^{b(p-i)}=2^{c}$, and $pv^{p-1}+\sum_{i=0}^{p-2}\left(\begin{array}{c}p\\i\end{array}\right)v^{i}2^{b(p-i-1)}=2^{c-b}$ and, since $v$ is odd, $p$ is even.

So any divisor $>1$ of $z+1$ is even and $z+1=2^{m}$

If $z+1>2$, then $z+1\geq 4$ and we have :
$u-v=2^{b}$
$u^{2}-v^{2}=2^{c}$, hence $u+v=2^{c-b}$, hence $u=2^{c-b-1}+2^{b-1}$ and $v=2^{c-b-1}-2^{b-1}$
$u^{4}-v^{4}=2^{d}$, hence $u^{2}+v^{2}=2^{d-c}$, hence $2^{2(c-b)-1}+2^{2b-1}=2^{d-c}$

But the equation $2^{r}+2^{s}=2^{t}$ implies $r=s$ and so $c=2b$ and so $v=0$, which is impossible.

So $z+1=2$ and we have
$x-y=2^{a}$
$x^{2}-y^{2}=2^{100}$, hence $x+y=2^{100-a}$

and the $49$ only solutions are :

$z=1$
$x=2^{49+k}+2^{49-k}$, for $k\in\{1,2,\cdots,49\}$
$y=2^{49+k}-2^{49-k}$


A real-valued function $ f$ satisfies for all reals $ x$ and $ y$ the equality
\[ f (xy) \equal{} f (x)y \plus{} x f (y).
\]
Prove that this function satisfies for all reals $ x$ and $ y \ne 0$ the equality
\[ f\left(\frac{x}{y}\right)\equal{}\frac{f (x)y \minus{} x f (y)}{y^2}
\]


hello

i think that the q can be done by contradiction method

let \[ x\geq y \]

now it is easy to show that the case \[ x\leq 0 \] is not possible

and the case \[ y\geq 0 \] easily gives \[ x+y>1 \]

so the only case left is \[ x\geq 0\geq y \]

so let \[ y=-z \] where z is a positive number

so by our assumption \[ x<z+1 \]

put this in ur eqn 2

\[ -z^3 + z^2 +2z -3>0 \] must be true but it is not as

let \[ f(z) = -z^3 + z^2 +2z - 3 \]

using maxima minima we get max value of func is at \[ z=\frac{1+\sqrt{7}}{3} \]

putting this value the func gives a negative val

i hope i dint make any mistake  :)

sorry i by mistake sent the same message twice :oops:


invariant
quadratics
superior algebra
superior algebra unsolved

Let $ K$ be a subset of $ C$ be a field, so that $ K$ has only two endomorphism $ f$ and $ g$, and if $ f(x)\equal{}g(x)$ then $ x \in Q$. Prove that there exist a free square integer $ d>1$ such that $ k\equal{}Q[\sqrt {d}]$. Thank you for the solution before.

First, any subfield $ K \subseteq \mathbf{C}$ will contain $ \mathbf{Q}$. Second, since $ K$ only has two endomorphisms (that leave $ \mathbf{Q}$ invariant, since the identity endomorphism is one of them), the degree of $ K/\mathbf{Q}$ is 2. Now any quadratic field extension of $ \mathbf{Q}$ must have the form $ \mathbf{Q}(\sqrt{d})$ for a certain rational $ d$. We then can multiply or divide $ d$ by a suitable square of a rational to make it integer and squarefree.