A red die has the number 1 on one face, the number 2 on two faces and the number 3 on three faces. Two green dice each has the number 6 on one face and the number 5 on five faces. Calculate the probability of obtaining: (a) a sum of 12, (b) a sum divisible by 3.

Roll 2 dice. 1 dollar is played by you. If the sum is 7 or 11, you win 3 dollars for every dollar played. Is this game worth the while to play?

In your 1st que. how many times are the dices rolled to obtain a sum of 12 and e.t.c?
For your 2nd one(I am guessing that you roll each of the dices once):
The prob. to obtain a 1 in the 1st and a 6 in the 2nd roll are 1/6.So prob. of obtaing a 7 in this way is 1/36.Similarly the otherway of obtaing a 7 has a prob. of 5/18.So the prob. of obtaining a 7 is 11/36.
So it's not worth playing.
If you can roll any dices:
The prob. for obtaining 7 is 11/72.
The prob. of obtainig a 11 is 5/144.(prob. of rolling the 2nd dice twice is 1/4 and obtaining a 6 and a five are respectively 1/6 and 5/6).
So the prob. of winning is 27/144(sum of the 2 prob.s).which is less than losing.
He shouldn't play!
Did I do any wrong steps anywhere??

I got something else for number 2.

The probabilty of rolling a $ 7$ with two dice is $ \frac {1}{6}$.  The probabilty of rolling an $ 11$ with two dice is $ \frac {1}{18}$.  Therefore, the combined probabilty is $ \frac{2}{9}$.  You can pay $ \$9$ for $ 9$ games, but you are expected to only win $ \$6$.  So this game is not worth while playing.

1. (I'm assuming you roll them once each.)

a) First, the green die must either be two 5s or 1 five, 1 six.

To get two 5s, the prob. is (5/6)(5/6)=25/36. Then the probability of getting a 2 on the red die is 1/3. Multiplying, we get

25/108.

To roll one 5 and one 6 on the green dice, the prob. is 2(5/6)(1/6)=5/18. Then the probability of getting a 1 on the red die is 1/6. Multiplying, we get

5/108.

Adding, we get a total of 30/108=5/18.

b) In addition to the above possibilities of getting 12, we can also get 15.

The probability of getting 2 sixes on the green dice, followed by a 3 on the red one is (1/6)(1/6)(1/2)=1/72.

Adding that to the total in Problem 1a), we get 5/18+1/72=20/72+1/72=21/72

=7/24.


function
calculus
integration
calculus computations

Find the general antiderivative of the function below.

f(u) = u^4 + 3(sqrt{u})/u^2.

Antiderivative = integral, right?

\[\int{u^4 + 3\frac{\sqrt{u}}{u^2} \textrm{d}u = \frac{u^5}{5} - \frac{6}{\sqrt{u}} + C}\]


Ok, ridiculous subject.. but hopefully it made you all read it.

I was just wondering the other day.. say you have this point A that starts at (0,0) and moves at a constant speed, say 1 unit per second, along the x axis to positive infinity. Then we have point B that starts at (m,n) and moves at a constant speed k always towards A. What is the graph of B?

[color=cyan]Well, that depends somewhat on k.  For example, if k > 1, B will eventually catch up with A.  Otherwise, you should be able to design a fairly simple differential equation for the position of B, since you always know its velocity.  I'll try the specifics later, but I'm finishing packing for school tomorrow so I should get some sleep to give me ample time and energy to freak out before I go.[/color]

[color=cyan]Actually, scratch the "easy" part of previous post.  Check me on this, but I think the differential equations you want are

y' / x' = y/(x-t)

(x')^2 + (y')^2 = k^2

x(0) = m
y(0) = n

The first one says that the direction is towards (t, 0).  The second one says that the speed is k.  The third and fourth say it starts at (m, n).  But that's some ugly equations to solve.  Have fun.[/color]

I believe this is actually a somewhat famous problem, I think it's called the Lion and the Prey problem or something like that. It comes from imagining the path a lion would take while chasing down a prey that moves in a particular path. There was a very good article written about it in Mathematics Monthly a few years ago - I think when I was in high school, which would have been in the early 90s. If I can find some more info about it I will pass it along.

A good place to ask question like this is on the sci.math newsgroup. Like all newsgroups it can be cluttered with junk, but there are some professional mathematicians that post there who can give good answers to questions like this.

If anyone gets an answer, please post it here! Thanks.


OH MY GOD I CALLED THIS BEAUTIFULLY

FIRST-ROUND LOSS TO VIRGINIA COMMONWEALTH ON PULL-UP JUMPER BY ERIC MAYNOR

DUKE SEASON:

SEASON-ENDING LOSSES TO MARYLAND, CAROLINA

FIRST-ROUND ACC TOURNAMENT LOSS TO 10TH SEEDED NCSU

FIRST-ROUND NCAA TOURNAMENT LOSS TO 11TH SEEDED VCU

COACH K WALKING AROUND AFTER THE GAME LIKE THE SPOILED BRAT HE IS

BET HE'S NOT GOING TO SAY HOW MUCH HE LOVES HIS PLAYERS, HUH

If anyone is in the market for a new team to root for starting Friday, might I suggest Long Beach State?

hmm, I would suggest the tried and true Carolina Tar Heels

I have this weird feeling they'll be in it for longer than DUKE, at least


[quote="Sunkern_sunflora"]What is min value??? You probably wanted to write mean value! :P[/quote]
Sorry for my grammar mistakes,i think it is not important sorry

Sorry Allnames! Now I guess you meant MINIMUM VALUE! Don't you! :blush:

[quote="Allnames"]Any proof,I think it is quite hard and nice
Why no one prove my problem,is it very easy[/quote]
I think this ineq must use lemma :
$ 2(a^3\plus{}b^3\plus{}c^3)\plus{}3abc \ge\ 3(a^2b\plus{}b^2c\plus{}c^2a)$

[quote="onlylove_math"][quote="Allnames"]Any proof,I think it is quite hard and nice
Why no one prove my problem,is it very easy[/quote]
I think this ineq must use lemma :
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc \ge\ 3(a^2b \plus{} b^2c \plus{} c^2a)$[/quote]
Can you post the proof deital? please :wink:

[quote="Allnames"]Find the minimum value of $ P$ there $ a,b,c\geq 0,a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
$ p \equal{} \frac {1}{a^2b \plus{} 2} \plus{} \frac {1}{b^2c \plus{} 2} \plus{} \frac {1}{c^2a \plus{} 2} \plus{} \frac {1}{6} \sqrt [3] {a^2b^2c^2}$[/quote]
[quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Any proof,I think it is quite hard and nice
Why no one prove my problem,is it very easy[/quote]
I think this ineq must use lemma :
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc \ge\ 3(a^2b \plus{} b^2c \plus{} c^2a)$[/quote]
Sorry only love math,this lemma is a very small part in my solution,i think you proved it,post it here.iam wellcome it
With that lemma,any one can prove it?
I think now my ineq is still quite hard![/quote]

 Thank Allnames i think it don't too hard
Here is my proof

We must Prove

 $ \frac {1}{a^2b \plus{} 2} \plus{} \frac {1}{b^2c \plus{} 2} \plus{} \frac {1}{c^2a \plus{} 2} \plus{} \frac {1}{6} \sqrt [3] {a^2b^2c^2} \ge\ \frac {7}{6}$

We have :

$ \frac {1}{2} \minus{} \frac{1}{a^2b \plus{} 2} \equal{} \frac {a^2b}{2(a^2b \plus{} 2)} \le\ \frac {\sqrt [3]{a^4b^2}}{6}$

$ \rightarrow \sum \frac {1}{a^2b \plus{} 2} \ge\ \frac {3}{2} \minus{} \sum \frac {\sqrt [3]{a^4b^2}}{6}$

And with Lemma $ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc \ge\ 3(a^2b \plus{} b^2c \plus{} c^2a)$

$ \rightarrow \frac {1}{9}(a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2})$

$ \rightarrow \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2}) \minus{} \frac {2}{9}$

We have done!

[quote]
 Thank Allnames i think it don't too hard
Here is my proof

We must Prove

 $ \frac {1}{a^2b \plus{} 2} \plus{} \frac {1}{b^2c \plus{} 2} \plus{} \frac {1}{c^2a \plus{} 2} \plus{} \frac {1}{6} \sqrt [3] {a^2b^2c^2} \ge\ \frac {7}{6}$

We have :

$ \frac {1}{2} \minus{} \frac {1}{a^2b \plus{} 2} \equal{} \frac {a^2b}{2(a^2b \plus{} 2)} \le\ \frac {\sqrt [3]{a^4b^2}}{6}$

$ \rightarrow \sum \frac {1}{a^2b \plus{} 2} \ge\ \frac {3}{2} \minus{} \sum \frac {\sqrt [3]{a^4b^2}}{6}$

And with Lemma $ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc \ge\ 3(a^2b \plus{} b^2c \plus{} c^2a)$

$ \rightarrow \frac {1}{9}(a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2})$

$ \rightarrow \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2}) \minus{} \frac {2}{9}$

We have done![/quote]
And Lemma $ a^2b\plus{}b^2c\plus{}c^2a \le\ 2 \plus{} abc$ with $ a^2\plus{}b^2\plus{}c^2\equal{}3$ .
We can prove it a way completely similar

[quote="onlylove_math"][/quote][quote="Allnames"]Find the minimum value of $ P$ there $ a,b,c\geq 0,a^2 + b^2 + c^2 = 3$
$ p = \frac {1}{a^2b + 2} + \frac {1}{b^2c + 2} + \frac {1}{c^2a + 2} + \frac {1}{6} \sqrt [3] {a^2b^2c^2}$[/quote]
[quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Any proof,I think it is quite hard and nice
Why no one prove my problem,is it very easy[/quote]
I think this ineq must use lemma :
$ 2(a^3 + b^3 + c^3) + 3abc \ge\ 3(a^2b + b^2c + c^2a)$[/quote]
Sorry only love math,this lemma is a very small part in my solution,i think you proved it,post it here.iam wellcome it
With that lemma,any one can prove it?
I think now my ineq is still quite hard![/quote][quote="onlylove_math"]

 Thank Allnames i think it don't too hard
Here is my proof

We must Prove

 $ \frac {1}{a^2b + 2} + \frac {1}{b^2c + 2} + \frac {1}{c^2a + 2} + \frac {1}{6} \sqrt [3] {a^2b^2c^2} \ge\ \frac {7}{6}$

We have :

$ \frac {1}{2} - \frac {1}{a^2b + 2} = \frac {a^2b}{2(a^2b + 2)} \le\ \frac {\sqrt [3]{a^4b^2}}{6}$

$ \rightarrow \sum \frac {1}{a^2b + 2} \ge\ \frac {3}{2} - \sum \frac {\sqrt [3]{a^4b^2}}{6}$

And with Lemma $ 2(a^3 + b^3 + c^3) + 3abc \ge\ 3(a^2b + b^2c + c^2a)$

$ \rightarrow \frac {1}{9}(a^2 + b^2 + c^2) + \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2})$

$ \rightarrow \frac {1}{6}\sqrt [3] {a^2b^2c^2} \ge\ \frac {1}{6}(\sum \sqrt [3]{a^4b^2}) - \frac {2}{9}$

We have done![/quote]
Very nice ,only lovemath,here is my proof ,and it is the way that i found it,thanks


How many times does the number $1$ occur in $S$ where

$S= 1+11+111+..............+111.........111$ where the last term has $2002$ digits.

[quote="Sepp"][hide="Quoted Text"]$S = 2002+10+100+...+10^{2001}$[/hide][/quote]

That's wrong. It should be [hide="a banana"] \[S= 2002+10+110+1110+\ldots \] [/hide]
But we can treat it as [hide="an apple"] \[\sum_{k=1}^{2002}\frac{ 10^{k}-1}{9}\] if it helps. [/hide]

[hide="Hint"] Consider $10S$ and $9S$[/hide]

If you have up to 9 1's, then your number is $123456790$.

After this, the digits just start repeating.

So the number of 1's is $\lfloor \frac{2002}{9}\rfloor=222$

Anyone with a solution?

[hide="Answer"] 224[/hide]


You have a triangle, $ABC$.  Draw in the internal angle trisectors.  Let the two trisectors closest to $AB$ intersect at $D$, the two trisectors closest to $BC$ intersect at $E$, and the two closest to $AC$ at $F$.  Prove that $DEF$ is equilateral.

This, of course, is a beautiful theorem[hide="Click to reveal the name"] called "Morley's Theorem"... relativly recent (1899)  , not that well known , (at least among older generation - I did not hear about this Theorem till about 5-10  years ago).

There are quite a few proofs of  the theorem. Using Trignometry  (specially noticing
 $sin(3x) = 4sin(x)sin(x+(\pi/6))(sin(x+2(\pi/6))$ the proof is easy. Without trig, there are few proofs I have seen a simple search in google will give a few interesting proofs..

Also  .[/hide][url=http://www.cs.princeton.edu/~ah/alg_anim/version2/Morley.html]Fun  to play at Princeton site[/url]

If you consider the external trisectors also, togather with the internal trisectors, you can get 27 triangles. 18 of them are equilateral triangles and the rest 9 are not in common conditions.

[quote="Rushil"]Is Morley's Theorm standard to be used in exams???[/quote]
It once appeared in a contest of a city of China in around 1990. I wonder how many students proved it in that exam......

It's proof is indeed difficult to think of yourself!!!

[quote]
It once appeared in a contest of a city of China in around 1990. I wonder how many students proved it in that exam[/quote]
  I  supspect many  would  have proved it....(it is not that difficult to prove)

IMO , (or at least for me) it (the statement of the theorem)  was  a surprise , and result was hard to guess (Guess that's why it was relatively unknown till a  hundred years ago)  but the proof (if nothing else ,  using brute-force), once the theorem is stated, is not difficult.
(My first reaction when  my brother  told me about this was  (actually second reaction - first was  saying something akin to "I don't believe it" )   to calculate  $DE$ and  see for myself that it  was   symetric wrt  variables A B C)

But I agree, it is a nice theorem  something  one may not easily guess...


On Monday, Millie puts a quart of seeds, $ 25\%$ of which are millet, into a bird feeder.  On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left.  Each day the birds eat only $ 25\%$ of the millet in the feeder, but they eat all of the other seeds.  On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

$ \textbf{(A)}\ \text{Tuesday}\qquad \textbf{(B)}\ \text{Wednesday}\qquad \textbf{(C)}\ \text{Thursday} \qquad \textbf{(D)}\ \text{Friday}\qquad \textbf{(E)}\ \text{Saturday}$

Can't [i]all[/i] of the answers choices be correct?

no. the actions described in the problem uniquely determine the result. and the question is clear too: "on which day, *just* after Millie has placed the seeds..."

[quote="myyellowducky82"]On which day, [b]just[/b] after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?[/quote]
keyword: just

I think you have to find the first day where it is more than half. There are solutions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1420034#1420034]here[/url].

There was actually a huge [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261252]argument[/url] relating to this question after the test. Some people stated that because the question didn't explicitly state "first" then all of the answers could be correct.

maybe. but you can't use the problem statement to make excuses for getting it wrong. despite the phrasing, it's still VERY clear which answer the question is asking for.

This debate's already happened in the problem's original thread. Look there if you want to see the arguments.


superior algebra
superior algebra unsolved

Let $ K \triangleleft G$ with $ |G/K|\equal{}n$. Show that

a) $ x^n \in K$, $ \forall x \in G$

b) if for some $ x \in G$, $ x^k$ is in $ K$, with $ (k,n)\equal{}1$, show that $ x$ is in $ K$.

Part b) is straightforward: since $ (k,n)\equal{}1$, there are integers $ a$ and $ b$ such that $ ak\plus{}bn\equal{}1$. Because $ x^k$ is in $ K$, so is $ x^{ak}$ in $ K$. From the a) part, $ x^{n}$ is in $ K$ and so is $ x^{bn}$.

So, now $ x^{ak}x^{bn}\equal{}x^{ak\plus{}bn}\equal{}x^1\equal{}x$ is in $ K$.

I'm having trouble with the a) part. Could somebody, please, point out some hints? :)

It's normal. The quotient $ G/K$ is a group. The order of an element divides the order of the group.


Prove that there are infinitely many solutions:
$ a^2b^2\minus{}4b(b\plus{}1)\equal{}c^2$

Edit: It did seem a bit too easy..., I'll see if I can find another one.

Judging by the title, I think he intended for $ a,b,c\in \mathbb Z$.

Edit: Hm the solution is the same. Letting $ b\equal{}\minus{}1$ we arrive at $ a^2\equal{}c^2$ which obviously has infinitely may solutions in integers.

How about adding the condition $ b\neq \minus{}1$?

Yes, solution must be positive integers  :blush:

[hide]
Let $ a = 3$ and $ b = F_{2n - 1}^2$ where $ n\in \mathbb N$. (Note: $ F_n$ is the Fibonacci sequence). Using the well-known fact that $ 5F_{2n - 1}^2 - 4$ is a perfect square for $ n\in \mathbb N$, we have:
\begin{align*} a^2b^2 - 4b(b + 1) & = 9F_{2n - 1}^4 - 4F_{2n - 1}^4 - 4F_{2n - 1}^2 \\
& = 5F_{2n - 1}^4 - 4F_{2n - 1}^2 \\
& = F_{2n - 1}^2(5F_{2n - 1}^2 - 4),\end{align*}
which is a perfect square.

Therefore, since there are infinitely many numbers of the form $ F_{2n - 1}^2$, there are infinitely many integer solutions.[/hide]


How many different seven-digit telephone numbers are available if the only restriction is that the first digit cannot be 0?

And the reason why:

There are 10 digits: 0,1,2,3,4,5,6,7,8, and 9.

However, for our first number we only have 9 choices, since we can't have 0.

For all other 6 digits, we can have any number - so 10 choices.

$ 9 \times 10^6 \equal{} 9000000$


calculus
integration
calculus computations

The problem is dy/dx = sin^2 (x - y). Is this solution correct?

[hide] Let v = x - y such that dv = dx - dy. We rewrite this as the following: dy = dx - dv. Now, we have the original equation in terms of dv

---> (dx - dv)/dx = sin^2(v), or 1 - (dv/dx) = sin^2(v)

---> 1 - sin^2(v) = cos^2(v) = dv/dx

---> dx = dv/(cos^2(v)) = sec^2(v)dv

After integrating both sides, we get:

x + C = tan(v) (note that I include BOTH of the constants of integrations in "C")

therefore v = arctan(x + C) = x - y

Therefore, y = x - arctan(x + C)
[/hide]

yeah...it looks perfect. However if it were not for the title, i would never have known to use a substitution!


number theory
prime numbers
advanced fields
advanced fields theorems

if n is prime prove that 2^n  -  1   is prime

The codes are visible please edit the post!!

I think you wanted the other way. If $ 2^n \minus{} 1$ is prime, then $ n$ must be prime. This class of prime numbers is called Mersenne primes. This result is well known, since if you can factor $ n \equal{} ab$, then

$ 2^{ab} \minus{} 1 \equal{} (2^a \minus{} 1)\left(1 \plus{} 2^a \plus{} 2^{2a} \plus{} \cdots \plus{} 2^{(b \minus{} 1)a}\right)$. So as long as the factoring wasn't trivial i.e. $ a \equal{} 1$ or $ b \equal{} 1$, then we can factor $ 2^n \minus{} 1$ and it isn't a prime.

For the counterexample, there is an even smaller one.

$ 2^{11}\minus{}1\equal{}2047\equal{}(23)(89)$


Fie a,b,c numere pozitive diferite de 0 astfel incat $ abc\equal{}1$ .Demonstrati ca \[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\geq 4(a\plus{}b\plus{}c\minus{}1)\]

Vezi aici una mai tare : http://www.mathlinks.ro/Forum/viewtopic.php?t=128529
Cred ca inegalitatea de mai sus se poate demonstra in aceeasi maniera.

Intradevar...cred ca inegalitatea se poate rezolva si asa...:) si ar iesi chiar frumos...dar sa postez si o alta metoda...
\[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}2abc\plus{}\sum_{cyc}(a^2b\plus{}ab^2)\equal{}\sum_{cyc}(a^2b\plus{}ab^2\plus{}2abc)\minus{}4abc\equal{}\sum_{cyc}(a^2b\plus{}bc^2\plus{}2abc)\minus{}4\equal{}\sum_{cyc}b(a\plus{}c)^2\minus{}4.\]Daca $ a\plus{}b\plus{}c \geq ab\plus{}bc\plus{}ca$ din inegalitatea CBS rezulta ca \[ (a\plus{}b\plus{}c)\sum b(a\plus{}c)^2 \geq \sum ac \sum b(a\plus{}c)^2 \geq(\sum \sqrt{abc}(a\plus{}c))^2\equal{}(\sum(a\plus{}c))^2\equal{}4(\sum a)^2\] deci $ \sum b(a\plus{}c)^2 \geq 4\sum a$  deci $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\geq 4\sum a \minus{} 4$.If $ a\plus{}b\plus{}c<\equal{}ab\plus{}bc\plus{}ca$ then $ (a\plus{}b)(b\plus{}c)(a\plus{}c)\equal{}\sum(a^2b\plus{}ab^2)\plus{}2abc\equal{}\sum(a^2b\plus{}ab^2\equal{}abc)\minus{}abc\equal{}\sum a \cdot (ab\plus{}bc\plus{}ca)\minus{}abc.$ Deci $ (\sum a)(\sum ab)\minus{}1 \geq 4\sum a \minus{} 4$ este echivalenta cu $ (\sum a)(\sum ab)\minus{}4\sum a \plus{}3 \geq 0$ .Cum $ \sum ab \geq \sum a$ ajunge sa demonstram ca $ (\sum a)^2 \minus{} 4\sum a \plus{}3 \geq0$ echivalent cu $ (\sum a \minus{}1)(\sum a \minus{}3)\geq 0$.But $ a\plus{}b\plus{}c \geq 3\sqrt[3]{abc}\equal{}3$ deci inegalitatea este demonstrata.Este oarecum mai greoaie dar e bine de stiut :P


Hello all!

Heres the problem:

Choose a number between 10 to 99. Except 11,22,33,44,55,66,77,88 and 99.

Pick your number, than swap the numbers in it. Than look after the diffrence between your two numbers, then do it again with your new number. Finally youll get one number, 9.

For example:

84---> 48

84-48=36

36--->63

63-36=27

27--->72

72-27=45

45--->54

54-45=9

How do you explain the connection between this numbers because all numbers youll try, youll get the answear, nine!

Please help us, it would be nice if you could show us the answear with algebra.

Regards Fermats heir:P

no=x
new no=y
difference=multiple of nine
now max posible value of the difference=81
81-18=63
63-36=54
54-45=9

value of difference not covered=72,27
72-27=54
54-45=9
so no matter which two digit multiple of nine we pick
after carrying out the given step finite no. of times
we get nine.

but why does one digit minus the inverted value of the first digit become a digit which is multiple with nine?

Example

81=inverted 18

81-18=63

And 63 can be / with 9

63 / 9 = 7.

Regards

[quote="rishabh"]10x+y -(10y+x)=9x-9y[/quote]
yeah...so 9(x-y)...so watevr answer u get has to ALWAYS be the sum of the digits into 9...I guess this is why...

Because
1. the number always decreases.
2. It's always divided by 9.


[hide="hint"]You need find $A$ such that $x^{2}+y^{2}=x+y$ and $xy=A$ has common real root.[/hide]

[quote="chien than"]For $x;y$ are real numbers satising $x^{2}+y^{2}=x+y$.Find the greast value the expression $A=xy$[/quote]
$xy(x+y)^{2}\leq(x^{2}+y^{2})^{2}.$ :wink:


calculus
integration
algebra
function
domain
calculus computations

I have no clue as to how to solve this:

X sin (X+Y) DX DY on the domain R [0,pi/6] X [0,pi/3].

I really have no clue! Help!

Which part are you struggling on?  The integral is $ \int_{0}^{\pi/3}\int_{0}^{\pi/6}x\sin(x\plus{}y)\,dx\,dy$.

Let me ask you a question: how would you evaluate $ \int_{0}^{\pi/6}x\sin(x\plus{}1)\,dx$?  How about $ \int_{0}^{\pi/6}x\sin(x\plus{}10)\,dx$?  Do you see where I'm going?


it has something to do with sets and the (forget the name) overlapping circles
what is it?

I believe the word you're looking for is Venn Diagram, or something of the sort.
The principle of inclusion-exclusion states that:
The cardinality of set A union B is given by:
(# of members in set A) + (# of members in set B) - (# of members in A intersection B).
So basically, the number of members in the union of A and B is given by the number of members in set A + the number of members in set B - the number of members in both sets.
We can extend this to three sets, four sets, etc.

can u explain how it would work in 3 sets and 4 sets?


$F_{n}=F_{n-1}+F_{n-2}$ is the $nth$ term in the  $Fibonacci$ serie.

Lucas numbers are define as follow :
$L_{n}=F_{n+1}+F_{n-1}$

Prove : $F_{2n}=F_{n}*L_{n}$

I try to use induction but not results yet...any hint???...

[hide="Possibly wrong hint."]
Try to prove the following statement by induction over $n$:
$F_{2n}= F_{n-1}F_{n}+F_{n}F_{n+1}, \, F_{2n+1}= F_{n+1}F_{n+2}-F_{n}F_{n-1}$.

Or something like that.
[/hide]

Well actually proving

$F_{2n}=F_{n-1}*F_{n}+F_{n}*F_{n+1}=F_{n}*(F_{n-1}+F_{n+1})=F_{n}*L_{n}$

will be enough 'cause is our problem here...

Don't mind that!

Please try my hint. See if it works out.

I think I need a little help here

proving $F_{2n}=F_{n}F_{n+1}+F_{n}F_{n-1}$ can be done easily if we can prove $F_{2n-1}=F_{n}F_{n+1}-F_{n-1}F_{n-2}$

I tried on induction over $n$ and first I need to prove the second equation to prove the first one as $F_{2n}=F_{2n-1}+F_{2(n-1)}$.

I need some help to prove this second equation

[quote="Altheman"]you could use the explict formulas for those sequences[/quote]
Yes, he can do that (and get a pretty short solution in the process).

But since he wanted to prove it by induction, I wanted to show him that it's possible.

[hide="The computations are straightforward. You just have to have enough patience!!!"]
[color=blue]Let $\left\{ F_{n}\right\}_{n \geq 1}$ be the sequence defined by $F_{1}= F_{2}= 1$ and $F_{n+2}= F_{n+1}+F_{n}$. Prove that $F_{2n}= F_{n-1}F_{n}+F_{n}F_{n+1}, \, F_{2n+1}= F_{n+1}F_{n+2}-F_{n}F_{n-1}$ holds for all $n \geq 2$.[/color]

[i]Proof.[/i] We'll prove the result by induction over $n$.

The base case $n=2$ is trivial:
$F_{4}= 3 = 1 \cdot 1+1 \cdot 2 = F_{1}\cdot F_{2}+F_{2}\cdot F_{3}$;
$F_{5}= 5 = 2 \cdot 3-1 \cdot 2 = F_{3}\cdot F_{4}-F_{1}\cdot F_{2}$.

Now let's suppose that we have proved that $F_{2n}= F_{n-1}F_{n}+F_{n}F_{n+1}, \, F_{2n+1}= F_{n+1}F_{n+2}-F_{n}F_{n-1}$ for some $n \geq 2$.

$1^\circ$. We have that
\begin{eqnarray*}F_{2n+2}&=& F_{2n+1}+F_{2n}\\ \ &=& F_{n-1}F_{n}+F_{n}F_{n+1}+F_{n+1}F_{n+2}-F_{n}F_{n-1}\\ \ &=& F_{n}F_{n+1}+F_{n+1}F_{n+2}.\end{eqnarray*}

$2^\circ$. We want to prove that $F_{2n+3}= F_{n+2}F_{n+3}-F_{n+1}F_{n}$: \[\, \Longleftrightarrow \, \, F_{2n+1}+F_{2n+2}= F_{n+2}F_{n+3}-F_{n+1}F_{n}\]  \[\, \Longleftrightarrow \, \, F_{2n+1}+\left( F_{2n}+F_{2n+1}\right) = F_{n+2}\left( F_{n+1}+F_{n+2}\right)-F_{n+1}F_{n}\]  \[\, \Longleftrightarrow \, \, F_{2n}+2F_{2n+1}= F_{n+1}F_{n+2}+F_{n+2}^{2}-F_{n+1}F_{n}\]  \[\left. \, \Longleftrightarrow \, \, F_{n-1}F_{n}+F_{n}F_{n+1}+2 F_{n+1}F_{n+2}-2 F_{n}F_{n-1}= F_{n+1}F_{n+2}+F_{n+2}^{2}-F_{n+1}F_{n}\right.\]  \[\left. \, \Longleftrightarrow \, \,-F_{n-1}F_{n}+F_{n}F_{n+1}+2 F_{n+1}F_{n+2}= F_{n+1}F_{n+2}+F_{n+2}^{2}-F_{n+1}F_{n}\right.\]  \[\left. \, \Longleftrightarrow \, \,-F_{n-1}F_{n}+2 F_{n}F_{n+1}+2 F_{n+1}F_{n+2}= F_{n+1}F_{n+2}+F_{n+2}^{2}\right.\]  \[\left. \, \Longleftrightarrow \, \,-F_{n-1}F_{n}+2 F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+2}^{2}\right.\]  \[\left. \, \Longleftrightarrow \, \,-F_{n-1}F_{n}+F_{n}F_{n+1}+F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+2}^{2}\right.\]  \[\left. \, \Longleftrightarrow \, \, F_{n}\left(-F_{n-1}+F_{n+1}\right)+F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+2}^{2}\right. \[ \left. \, \Longleftrightarrow \, \, F_{n}^{2}+F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+2}^{2}\right.\]  \[\left. \, \Longleftrightarrow \, \, F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+2}^{2}-F_{n}^{2}\right.\]  \[\left. \, \Longleftrightarrow \, \, F_{n}F_{n+1}+F_{n+1}F_{n+2}= \left( F_{n+2}-F_{n}\right) \left( F_{n+2}+F_{n}\right) \right.\]  \[\left. \, \Longleftrightarrow \, \, F_{n}F_{n+1}+F_{n+1}F_{n+2}= F_{n+1}\left( F_{n+2}+F_{n}\right) , \right.\] which is obviously true.

$1^\circ \, \, \& \, \, 2^\circ$ complete the induction step, so we're done$. \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \blacksquare$
[/hide]

Is it clear now? If it's not, then I'm  :(


On a copy machine, when a person selects the option x%, the copy is x% of the original where x must be 50 or larger.   To reduce an original to smaller than 50% requires the user to first reduce the original and then reduce the copy.  To reduce an original to exactly 48% of its size, which of these three sequences of reductions would work?

I. 50%, then 96%	 II.75%, then 64%	    III.  70%, then 88%

A. None of these
B.I only
C. II only
D.  II only
E.I  and II, only

[hide="Solution"]It suffices to check each sequence to see if they result in a valid reduction. $ I$ is valid. $ II$ is valid. $ III$ is not. Answer: $ \boxed{E}$

[/hide]


Find number of divisor $ 5n^{11}\minus{}2n^{5}\minus{}3n$ for all $ n \in N$.

[quote="stephencheng"][quote="mayhem"]Find number of divisor $ 5n^{11} \minus{} 2n^{5} \minus{} 3n$ for all $ n \in N$.[/quote]

$ 5n^{11} \minus{} 2n^{5} \minus{} 3n$ is divisible by $ n$ but we don't know how many divisors $ n$ has, so how can we find out the answer?[/quote]
If you write $ n \equal{} p_1^{a_1}...p_r^{a_r}$ it has $ (a_1\plus{}1)...(a_r\plus{}1)$ divisors. ;)


number theory
prime numbers
number theory unsolved

Let $p$ be a prime number. Prove that there exists infinitely many prime numbers $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.

Define the sequence:
 $x_1=\frac {p^p-1} {p-1},x_{k+1}=\frac {(px_1...x_k)^p-1} {px_1...x_k-1},\forall k\geq 1$
 We have $gcd(x_m,x_n)=1$ and $x_n \equiv 1(mod p)$ ,$x_n \neq 1(mod p^2)$
 Take $q_k$ is a prime divisor of $x_k$,$k=1,2,...$ s.t $q_k\neq 1(mod p^2)$
  We have $(q_k)^{\infty}_{k=1}$ is the sequence follows the given properties.


The figure below consists of four semicircles and the 16-cm diameter of the largest semicircle. What is the total number of square cm in the area of the two shaded regions? Use 3.14 as an approximation for $ \pi$, and express your answer as a decimal to the nearest tenth.

[asy]import graph;
size(200);
picture p;
real h = .25,w = 1,textsize = 10pt,width = 1;
filldraw(Arc((8,0),8,0,180)--cycle,gray(.6),black+linewidth(width));
filldraw(Arc((6,0),6,0,180)--cycle,white,black+linewidth(width));
filldraw(Arc((4,0),4,0,180)--cycle,gray(.6),black+linewidth(width));
filldraw(Arc((2,0),2,0,180)--cycle,white,black+linewidth(width));
draw(p,(0,h)--(0,-h)^^(0,0)--(4,0)^^(4,h)--(4,-h),linewidth(width));
fill(p,(w,h)--(4-w,h)--(4-w,-h)--(w,-h)--cycle,white);
label(p,"4 cm",(2,0),fontsize(textsize));
add(shift(.5*down)*p);
add(shift(4*right)*shift(.5*down)*p);
add(shift(8*right)*shift(.5*down)*p);
add(shift(12*right)*shift(.5*down)*p);[/asy]

Firstly, the area of the whole semi-circle is $ 32\pi$.  Then, we subtract the second-largest semicircle with area $ 18\pi$.  Then, we must add the third-largest semicircle with area $ 8\pi$.  Finally, we must subtract the smallest semicircle with area $ 2\pi$.  So we have $ 32\pi\minus{}18\pi\plus{}8\pi\minus{}2\pi$ or $ 20\pi$, which is $ \boxed{\approx 62.8}$.


Let $ \{f_k\}$ be a sequence of continuous functions on $ [a,b]$.  Show that if $ \{f_k\}$ converges uniformly on $ (a,b)$, then it converges uniformly on $ [a,b]$.

maybe a moderator should move this to real analysis forum.
it is a nice problem i think.

Let $ \{f_k\}$ converge on $ (a,b)$ to $ f$
Let $ \epsilon>0$ be given

We have $ \exists K\in\mathbb{N}$ such that $ n>K$ $ \implies$ $ |f(x)\minus{}f_n(x)|<\frac{\epsilon}{2}\;$ $ \;\forall x\in (a,b)$

Also, $ \exists$ $ \delta>0$ such that $ x\in V_{\delta}(a)\implies\; |f_n(x)\minus{}f_n(a)|<\frac{\epsilon}{2}$

Then, $ \forall\epsilon>0$ $ K\in\mathbb{N}$ and $ \delta>0$ such that $ n>K$ and $ x\in V_{\delta}(a)$ $ \implies$ $ |f(x)\minus{}f_n(a)|<\frac{\epsilon}{2}$

Let $ x_0\in V_{\delta}(a)$. Then, $ \forall$ $ n,m>K$, $ |f_n(x_0)\minus{}f_m(x_0)|<\epsilon$, and hence, $ f_n(a)$ is convergent.

Similarly we can show that $ f_n(b)$ is also convergent


So I made AIME, and I believe I need some preparation to reach my goal of 5, so how should one prepare for a contest like the AIME?

there have been many similar threads posted. you will probably find your answer in one of those threads.

See, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258775]here[/url].

Read my post there if you want advice.


geometry
incenter
circumcircle
perimeter
trigonometry
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14157[/img]

 :(

Suppose that we have constructed the triangle $ ABC$. We extend $ AB$ to the point $ B'$ such that $ BB' = BC$.  We extend $ BA$ to the point $ A'$ such that $ AA' = AC$. Then of course $ A'B' = p$
Now, $ \triangle ACA'$ is isosceles and its base angle is $ \angle AA'C = \frac {1}{2} \angle CAB \ \ (1)$. Similarly $ \angle BB'C = \frac {1}{2}\angle CBA \ \ (2)$.
We bring:
Line parallel to $ AC$ passing through $ A'$
Line parallel to $ BC$ passing through $ B'$
These lines intersect at $ C'$. Then $ \angle B'A'C' = \angle BAC$ and $ \angle A'B'C' = \angle ABC$. So the triangles $ ABC,A'B'C'$ are similar
From $ (1)$ and $ (2)$ it is obvious that $ C$ is the incenter of $ \triangle A'B'C'$
Then we draw the circumcircle $ w$ of  $ \triangle A'B'C'$. Let $ N\in C'C \cap w$. Then it's known that $ NA' = NB' = NC$

[size=150]Construction[/size]
Let $ A'B'$ be a segment with length $ p$. We draw tho arcs (which belongs to the same circle):
The 'upper', which has the property "If $ F$ is on this arc then $ \angle A'FB' = C$
The 'lower', which has the property "If $ G$ is on this arc then $ \angle A'GB' = \pi - C$
Also, on the "upper" half-plane of $ A'B'$ we draw the line $ s$ whose distance from $ A'B'$ is equal to $ h_c$
Let $ N$ be the midpoint of the lower arc. The circle $ (N,NA')$ will (possibly) intersect the line $ s$ at two points. Take one of them, this is point $ C$

Then it's easy to find the points $ A,B \in (A'B')$. The point $ A$ belongs to the perpendicular bisector of $ CA'$. And $ B$ is on the perpendicular bisector of $ CB'$

Now the perimeter of the triangle $ ABC$ is $ p$, the altitude from $ C$ is the desired $ h_c$ and $ \angle ACB = A'C'B'$, which consructed to be equal to the desired angle $ C$

[size=150][hide="Discussion"][/size]The only discussion is weither the circle $ (N)$ intersects the line $ s$
The "higher" point of this circle is the point $ K$, where $ K$ is on the perpendicular bisector of $ A'B'$
So, we have solution iff $ A'N - NM\geq h_c$ where $ M$ is the midpoint of $ A'B'$

But $ \angle MA'N = \frac {C}{2}$

Then $ A'N = \frac {A'M}{\cos \left(\frac {C}{2}\right)} = \frac {p}{2\cos \left(\frac {C}{2}\right)}$

and $ NM = \frac {p}{2}\tan\left(\frac {C}{2}\right)$

$ A'N - NM\geq h_c \iff$

$ \frac {p}{2\cos \left(\frac {C}{2}\right)} - \frac {p}{2}\tan\left(\frac {C}{2}\right)\geq h_c \iff$

$ \frac {1 - \sin\left(\frac {C}{2}\right)}{\cos\left(\frac {C}{2}\right)}\geq \frac {2h_c}{p} \iff$

$ \cdots$

${ \tan\left(\frac {\pi - C}{4}\right)}\geq \frac {2h_c}{p}$[/hide]


calculus
integration
trigonometry
vector
function
derivative
calculus computations

Find the minimum value of $\int_0^{\pi} (a\sin x+b\sin 2x+c\sin 3x-x)^2\ dx$

is there a quick way of solving this ?  :?

[quote="Matnomi"][quote="Jimmy"]$f(a,b,c) = \int_0^{\pi} \left(a\sin x + b\sin 2x + c\sin 3x -x\right)^2 \ dx = \frac{\pi(3a^2-12a+3b^2+6b+3c^2-4c)+2\pi^3}{6}$
[/quote]
Ok, but with the integral ?[/quote]

Ya, it gets a little ugly. :)

Of course, this is really at heart a Fourier analysis - orthogonal projection question.

The key issue: the set $\{\sin x,\sin 2x,\sin 3x\}$ is orthogonal on $[0,\pi].$ What this means is that

 $\int_0^{\pi}\sin nx\sin mx\,dx=0$ if $m\ne n.$

We also have that $\int_0^{\pi}\sin^2mx\,dx=\frac{\pi}2.$

Let's put this in an abstract vector space setting. (Real inner product space, to be specific.) Suppose we have an orthogonal set of vectors $\{v_1,v_2\dots,v_n\}$ Of courses, "orthogonal" means that the inner product $\langle v_j,v_k\rangle=0$ for $j\ne k.$

Suppose we have another vector $w.$ (In the particular problem above, $w$ is the function $f(x)=x.$) We want to find $a_1,a_2,\dots,a_n$ so as to minimize $f(a_1,\dots,a_n)=||a_1v_1+\cdots a_nv_n-w||^2,$ and to determine that minimum value.

We calculate that norm by multiplying out the dot product. Using the orthogonality:

$f(a_1,\dots,a_n)=\langle a_1v_1+\cdots a_nv_n-w,a_1v_1+\cdots a_nv_n-w\rangle$

$=a_1^2||v_1||^2+\cdots+a_n^2||v_n||^2-2a_1\langle v_1,w\rangle-\cdots 2a_n\langle v_n,w\rangle+||w||^2$

(complete the squares)

$=||v_1||^2\left(a_1^2-2a_1\frac{\langle v_1,w\rangle}{||v_1||^2}\right)+\cdots+ ||v_n||^2\left(a_n^2-2a_n\frac{\langle v_n,w\rangle}{||v_1||^2}\right)+||w||^2$

$=||v_1||^2\left(a_1^2-2a_1\frac{\langle v_1,w\rangle}{||v_1||^2}+ \frac{\langle v_1,w\rangle^2}{||v_1||^4}\right)+\cdots+ ||v_n||^2\left(a_n^2-2a_n\frac{\langle v_n,w\rangle}{||v_1||^2}+ \frac{\langle v_n,w\rangle^2}{||v_n||^4}\right)+||w||^2- \frac{\langle v_1,w\rangle^2}{||v_1||^2}-\cdots-  \frac{\langle v_n,w\rangle^2}{||v_n||^2}$

From this we see that the minimum happens when for each $k,$ $a_k=\frac{\langle v_k,w\rangle}{||v_k||^2}.$ Furthermore, the minimum value is

$||w||^2- \frac{\langle v_1,w\rangle^2}{||v_1||^2}-\cdots-  \frac{\langle v_n,w\rangle^2}{||v_n||^2}.$

For the problem at hand, $w$ is the function $x$ and $v_k$ is the function $\sin kx.$ We have that for each $k,$ $||v_k||^2=\frac{\pi}2.$

Thus the minimum value of the function is:

$\int_0^{\pi}x^2\,dx-\frac2{\pi}\left(\int_0^{\pi}x\sin x\,dx\right)^2- \frac2{\pi}\left(\int_0^{\pi}x\sin 2x\,dx\right)^2- \frac2{\pi}\left(\int_0^{\pi}x\sin 3x\,dx\right)^2.$

All that remains is the calculation of those integrals.

[quote="Jimmy"]$ f(a,b,c) \equal{} \int_0^{\pi} \left(a\sin x \plus{} b\sin 2x \plus{} c\sin 3x \minus{} x\right)^2 \ dx \equal{} \frac {\pi(3a^2 \minus{} 12a \plus{} 3b^2 \plus{} 6b \plus{} 3c^2 \minus{} 4c) \plus{} 2\pi^3}{6}$

Thus we want to minimize $ 3a^2 \minus{} 12a \plus{} 3b^2 \plus{} 6b \plus{} 3c^2 \minus{} 4c$.

$ f_a \equal{} 6a \minus{} 12 \equal{} 0 \Rightarrow a \equal{} 2$

$ f_b \equal{} 6b \plus{} 6 \equal{} 0 \Rightarrow b \equal{} \minus{} 1$

$ f_c \equal{} 6c \minus{} 4 \equal{} 0 \Rightarrow c \equal{} \frac {2}{3}$

$ f\left(2, \minus{} 1,\frac {2}{3}\right) \equal{} \frac {\pi(6\pi^2 \minus{} 49)}{18}$[/quote]

Can anyone solve without using partial differentiation?

That's right. The problem is solved by Japanese Highschool Students, so I wanted to see other basic souition. Needless to say, if university students solve the problem, they use partial diffentiation or Fourier Analysis.


[quote="Quevvy"]What are all the real values for $x$ for which $|x^{2}-4|-3x\le0$?
I've tried factoring $x^{2}-4$ into $(x-2)(x+2)$ but I can't get anywhere from there...[url][/url][/quote]

[hide]
first make a restriction that $3x\ge|x^{2}-4|$
then it's $|x^{2}-4|\le3x$
$\pm(x^{2}-4)\le3x$
$x^{2}-3x-4\le0$ $AND$ $x^{2}+3x-4\ge0$
$(x-4)(x+1)\le0$ $AND$ $(x+4)(x-1)\ge0$
critical test points: $-4,-1,1,4$
after testing $-5,-2,0,2,5$, the solution set is found out to be $x=[1,4]$
answers are $x=[1,4]$
i think that's how you do it
[/hide]

You can classify two cases.

case 1: $x^{2}-4\geq 0.$

case 2 :$x^{2}-{4}<0.$

[quote="ProtestanT"][quote="Quevvy"]What are all the real values for $x$ for which $|x^{2}-4|-3x\le0$?
I've tried factoring $x^{2}-4$ into $(x-2)(x+2)$ but I can't get anywhere from there...[url][/url][/quote]

[hide]
first make a restriction that $3x\le|x^{2}-4|$
then it's $|x^{2}-4|\le3x$
$\pm(x^{2}-4)\le3x$
$x^{2}-3x-4\le0$ $AND$ $x^{2}+3x-4\ge0$
$(x-4)(x+1)\le0$ $AND$ $(x+4)(x-1)\ge0$
i think that's how you do it
[/hide][/quote]
How does $3x\le|x^{2}-4|$ become $|x^{2}-4|\le3x$?

the first one is just a restriction. (if the restriction isn't satisfied, then the result does not satisfy the inequality) the second one comes from the original problem

Sketch the graphs of $y=|x^{2}-4|$ and $y=3x.$

[quote="ProtestanT"][quote="bpms"][hide="hmm"]
$|x^{2}-4| \leq 3x$
If x is negative this clearly isn't true. So we can square both sides to get
$(x^{2}-4)-9x^{2}\leq 0$
$(x^{2}+3x-4)(x^{2}-3x-4) \leq 0$
$(x+4)(x-1)(x-4)(x+1) \leq 0$
This can be continued on to get the answer.[/hide][/quote]
what if $x<2$?[/quote]
The reason you can square is because neither side can be negative.

if $0<x<2$, then $x^{2}-4$ would be opposite of $|x^{2}-4|$

Oh you squared both sides...
make it $(x^{2}-4)^{2}-9x^{2}\leq 0$ instead of $(x^{2}-4)-9x^{2}\leq 0$
i think that's what confused me

[quote="mathgeniuse^ln(x)"][quote="Quevvy"]What are all the real values for $x$ for which $|x^{2}-4|-3x\le0$?
I've tried factoring $x^{2}-4$ into $(x-2)(x+2)$ but I can't get anywhere from there...[url][/url][/quote]

[hide]Graphing, we get that the intersection points at only $x=1$ and $x=4$

At this period, we see that the answer is $1 \le x \le 4$?

Check this[/hide][/quote]
[hide]
because $3x$ cannot be less than $|x^{2}-4|$, then only $1$ and $4$ work, the values in between won't work.
edit-nvm you are right, i miscalculated the restriction.
[/hide]

Yeah it is an inequality not an equality.

But nice problem.

my  mistake was i reversed the sign for the restriction so that only the equality worked, any inequality would not work for the restriction.


- That's a broken link, and we'd rather see the problems as text than as an embedded image.
- Duplicate titles are bad. It's very hard to distinguish between your threads.
- See the advice fedja just gave in the other thread.


[hide]
We can divide this into cases.

$\bullet$ Case 1: All $4$ items go in the same bag. There is one possible way to do this.

$\bullet$ Case 2: Three items go in one bag, and the last item goes in another bag. There are $\binom{4}{1}=4$ ways to choose which item goes in a bag by itself.

$\bullet$ Case 3: Two items go in one bag, and the other two go in another bag. There are $\binom{4}{2}=6$ ways to choose which items go in the first bag, but since the bags are identical we must divide by $2$ to correct for overcounting. Therefore, there are $3$ arrangements in this case.

$\bullet$ Case 4: Two items go in one bag, and the other two items each go in a different one of the remaining bags. There are $\binom{4}{2}=6$ ways to choose which two items are put in a bag together, and since the bags are identical it doesn't matter which bags the last two items are put in.

There are a total of $1+4+3+6=\boxed{14}$ different ways to put the items into bags.[/hide]

[quote=stewartmath]I was just assigned this problem directly after "How many ways are there to put 4 distinguishable balls into 3 indistinguishable boxes?", which is just another version of the same question. Should I submit this as an error?[/quote]

no. the question order is randomized.

[quote=DanielHendrycks][hide="Solution Question"]"Case 3: Two items go in one bag, and the other two go in another bag. There are $\binom{4}{2}=6$ ways to choose which items go in the first bag, but since the bags are identical we must divide by 2 to correct for overcounting. Therefore, there are 3 arrangements in this case."

Could someone explain why dividing by 2 fixes the problem? (I don't really see where there was overcounting, either.)[/hide][/quote]

Long time, no replies! :)

Let's say I have a cabbage, tomato, apple, and peach. And my three bags: $\square$ $\square$ $\square$

If the cabbage and tomato go in one bag, then we're left with apple and peach. Usually we'd have 2 choices for where these fruits would go, but since bags are identical, we don't, so we have to divide by 2.

Nope,
The items are distinct, but the bags aren't.
This means that you have to casework bash.

Yes, this had me quite confused! You can't use Stars and Bars.

Oh wait you cant use Stars and Bars?? Why not?

[quote=GameBot]At the grocery store, I bought 4 different items. How many ways are there for the cashier to put the items I bought in 3 identical bags?[/quote]

haha you did this post before I was born. Can there be $0$ items in a bag?

[quote=jbatterson]Frustrating...

I know that there are other problems like this, and this is really a "distinguishable items, indistinguishable boxes" problem and I should have answered it as such, but the answer really should be 6.  No store bagger would put zero items in a bag.  The only way to place four items in 3 grocery bags is 1-1-2, and there are 4x3/2 ways to do this.  I want my point back.[/quote]

whoa, I think this was before he was admin

you should read intro to counting and probability. The last chapter in chapter 5 is distinguishability.

[quote name="pieMax2713" url="/community/p23812285"]
[quote=undiraju]i got 15 using stars and bars. Can someone explain what I did wrong?[/quote]

stars and bars is for putting identical items in distinguishable things, not the other way around
[/quote]
Reverse Stars and Bars
[quote name="weihang" url="/community/p26899648"]
[quote=undiraju]i got 15 using stars and bars. Can someone explain what I did wrong?[/quote]

You have to subtract one because stars and bars counts the case where there is no item in any of the bags.
[/quote]

Can someone explain this?  I don't understand how there could be an empty set.


The same amount of steel used to create eight solid steel balls, each with a radius of 1 inch, is used to create one larger steel ball. What is the radius of the larger ball?

[asy]filldraw(circle((0,0),1),gray);
filldraw(circle((.9,-.8),1),gray);
filldraw(circle((1.8,.9),1),gray);
filldraw(circle((2,0),1),gray);
filldraw(circle((2,-.4),1),gray);
filldraw(circle((3,-.4),1),gray);
filldraw(circle((4.8,-.4),1),gray);
filldraw(circle((3.2,.5),1),gray);

draw((6,.7)--(8,.7),Arrow);

filldraw(circle((11,.2),2),gray);[/asy]

Each of the smaller steel balls has a volume of $ \frac {4}{3}\pi$, so the larger ball has a volume of $ \frac {32}{3}\pi$, or a radius of $ \boxed{2}$.

Note: We found this because the volume of a sphere is $ \frac {4}{3}r^3\pi$.

EDIT: What ernie did is what I did, but he wrote it out.

Or you could set up an equation.

$ 8\pi \cdot 1^3 \equal{} \pi r^3$
$ 8 \equal{} r^3$
$ 2 \equal{} r$

Answer: $ \boxed{2}$

Since it is 3 dimensional, we simply do $ \sqrt[3]{8}\equal{}2$

If we had 81 hyperspheres of radius 1 (is that what they are called?) we could form a hypersphere of radius 3


algebra
function
domain
superior algebra
superior algebra unsolved

I have an interesting problem here, and a proof that I feel a bit dubious about. I was just wondering if you guys could give your opinion on the validity of my solution.

[b]Problem:[/b] Let $ D$ be a Dedekind domain, and let $ I\equal{}(r,s)$ be the ideal of $ D$ generated by elements $ r$ and $ s$. Assume that the principal ideals $ (r)$ and $ (s)$ factor as follows
\[ (r)\equal{}P_1P_2P_3 \\ (s)\equal{}P_1P_2P_4\]
for distinct prime ideals $ P_i$. Prove that $ I\equal{}P_1P_2$.

[hide]
Since this is a Dedekind domain, all our ideals are invertible. In particular, this means $ P_1P_2 = (r)P_3^{ - 1} = (s)P_4^{ - 1}$. We want to show that $ P_1P_2 = I$ which happens if and only if $ IP_1^{ - 1}P_2^{ - 1} = D$. Now, we know that $ I = (r,s) = (r + s)$, so we can substitute to obtain
\[ IP_1^{ - 1}P_2^{ - 1} =\& (r + s)P_1^{ - 1}P_2^{ - 1} \\ =\&(rP_1^{ - 1}P_2^{ - 1}) + (sP_1^{ - 1}P_2^{ - 1}) = (r)(r)^{ - 1}P_3 + (s)(s)^{ - 1}P_4 = P_3 + P_4
\]
However, since in a Dedekind domain, all prime ideals are maximal, this means $ P_3 + P_4 = D$ since they are distinct. Thus $ IP_1^{ - 1}P_2^{ - 1} = D$ which is what we were trying to prove, so we are done.
[/hide]

Thoughts? I'm concerned about the algebraic manipulations of ideals, mostly.

[quote="AdrianP"]
[b]Problem:[/b] Let $ D$ be a Dedekind domain, and let $ I \equal{} (r,s)$ be the ideal of $ D$ generated by elements $ r$ and $ s$. Assume that the principal ideals $ (r)$ and $ (s)$ factor as follows
\[ (r) \equal{} P_1P_2P_3 \\
(s) \equal{} P_1P_2P_4
\]
for distinct prime ideals $ P_i$. Prove that $ I \equal{} P_1P_2$.[/quote]

$ \rm HINT \qquad AB \plus{} AC \equal{} A(B \plus{} C) \equal{} A \quad when \quad B\plus{}C \equal{} 1$


geometry
inequalities
function
inequalities solved

Prove that in any triangle with medians m_a,m_b, m_c and bisectors l_a,l_b,l_c we have m_a*m_b+m_b*m_c+m_c*m_a-l_a*l_b-l_b*l_c-l_c*l_a>=1/4(a^2+b^2+c^2-ab-bc-ca).

indeed it is most un-beautifull :) and I must admit that right now I can't think of anything to do with this ...

I think that it's somehow beautiful .:? OK, I"ll do it .
Just one question: how did you think it out?

Your problem is such a confusing one. Here is my solution:
   It 's easy for you to prove that m_a,m_b,m_c are sides of a triangle and its' area is equal to 3/4*[ABC].
   By using Finsler-Khadivinger inequality, we have
      \sum (m_a*m_b)>=1/2( \summ_a^2 +(3* \sqrt 3)S)=(3/8 \sum a^2+ (3* \sqrt 3)*S) (S is the area of ABC).
    All we have to do is to prove
      1/2p^2 +(3* \sqrt 3)S>=  \sum l_a*l_b
  We have l_a= 2bccos(A/2)/(b+c)
  Therefore l_a*l_b = 4[bac^2cos(A/2)cos(B/2)]/(a+c)(b+c)
    Futhermore, there always exists a triangle XYZ such that
       sinX=  \sqrt[(a+b+c)a/(b+a)(c+a)],......
    and by applying this following inequality
        4* \sum xy*(sinZ)^2<=(x+y+z)^2
    so we have
    \sum l_a*l_b <=  abc[\sum cos(A/2)]^2/(a+b+c) (x=cos(A/2),......)
            And we have this inequality
    1+(3* \sqrt 3)prod[tg(A/2)]>=( \sum 1/(cos(A/2)*cos(B/2) )^2 *pro[sin(A/2)].
      since we have S=p^2* prod[tg(A/2)]
                            abc = 4SR=p^3*prod[tg(A/2)]/prod(cos(A/2))
    By denoting A'= \pi/2 - A/2,........... => A'B'C' is an acute triangle.
    we have 3* \sqrt 3 +  \sum tgA' >= 2*(\sum cotg(A'/2)).(*)  :)
    This inequality is right, I' m sure. But I don't know how to prove it.  :(
    In fact, my friend told methis following inequality created by Mr Titu Andreescu :
          If f is a function such that 1/2[f(x)+f(y)] >=f(1/2*(x+y)) for every x,y in (a,b).
          Then f(x)+f(y)+f(z) +3*f(1/3*(x+y+z)) >=2*( \sum f(1/2*(x+y))
for every x,y,z, in (a,b).
I have no idea to prove this one :D .  Harazi, what about your solution ? I think it's much more interesting than mine, isn't it? :D
      I hope that someone will prove that (*) inequality for me to finish my proof. Thanks. :)

In fact that inequality with f is not created by Titu Andreescu, but by Tiberiu Popoviciu, long time ago. It can be proved using Karamata's inequality. I won' post the solution, since it is very long. I like very much your proof of this inequality. As far as I remember, I used the inequality m_a>=(b+c)/2*cosA/2 to prove that, but I don't remember very well. I will check my notes and I will post my solution.

Unfortunately, I found in my notes exactly NOTHING. And I tried today to prove the inequality, but I failed.


Discussion moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56328]here.[/url]


Are there still prizes for selected people from perfect scorers?

There will be prizes doled out at random among the students who scored a 5 on any problem, and from among the commended solutions.


Let $ a,b,c\geq 0$,no two of wich are zero.Prove that:
$ \frac{a^2}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2}{c^2\plus{}ca\plus{}a^2}\geq 1$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$,no two of wich are zero.Prove that:
$ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} a^2}\geq 1$.[/quote]
Let $ x\equal{}\frac{b}{a},y\equal{}\frac{c}{b},z\equal{}\frac{a}{c}.$

the inequality is equivalent now to:

$ \frac{1}{x^2\plus{}x\plus{}1}\plus{}\frac{1}{y^2\plus{}y\plus{}1}\plus{}\frac{1}{z^2\plus{}z\plus{}1} \ge 1$ with $ xyz\equal{}1$,

which is equivalent after expansion to:

$ x^2\plus{}y^2\plus{}z^2 \ge xy\plus{}yz\plus{}zx$

which is true  :)

By cauchy we have

\[ \sum_{cyc}\frac {a^2}{a^2 \plus{} ab \plus{} b^2}\equal{}\sum_{cyc}\frac {a^2(a\plus{}c)^2}{(a^2 \plus{} ab \plus{} b^2)(a\plus{}c)^2}\ge \frac{(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2}{\sum_{cyc}(a^2 \plus{} ab \plus{} b^2)(a\plus{}c)^2}\]

So it's enough to prove that
\[ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2\ge \sum_{cyc}(a^2 \plus{} ab \plus{} b^2)(a\plus{}c)^2 \Leftrightarrow a^3b\plus{}b^3c\plus{}c^3a\ge abc(a\plus{}b\plus{}c)\]

Wich is clear by AM_GM.

[quote="y-a-s-s-i-n-e"]hi rachid
how did you find the firt equivalent inequality  :?:[/quote]
because:
${ \frac{a^2}{a^2+ab+b^2}=}=\frac{1}{a^2/a^2+ab/a^2+b^2/a^2} = = \frac{1}{1+b/a+b^2/a^2}$
then after substitution we find it  :)


The non-negative integers are divided into three groups:

A= {0,3,6,8,9,...}
B={1,4,7,11,14,...}
C={2,5,10,13,...}

Explain.

[hide]a = any number from set a

b = any number from set b

c = any number from set c

a(mod 3) = 0

b(mod 3) = 1

c(mod 3) = 2[/hide]

hmm h_s...not quite i think...

8 (mod 3) = 2 not 0
11 (mod 3) = 2 not 1
14 (mod 3) = 2 not 1
10 (mod 3) = 1 not 2
13 (mod 3)= 1 not 2

So is there a way to distinguish the sets?

[hide]i think it has to do with the way you write the numbers out, i.e A=curves, B=#'s written with straight lines, C=written with curves+straight lines.  But 12 is in A which doesn't make sense, or is it a mistake?  Because 12 is not in the list, but 13 and 14 are. [/hide]

I don't think that's right.
 A= {0,3,6,8,9,...}
 B={1,4,7,11,14,...}
 C={2,5,10,13,...}

B isn't linear - 7 to 11 isn't the same as 4 to 7.

I tried putting the numbers in binary, squaring the terms, summing digits, etc.  I think there is probably something simple.

[hide]i think you misunderstand what i'm saying.  I'm not talking about the finite differences; I'm talking about how you physically write the numbers.  You can write B with just straight lines[/hide]

i agree with cats... that was my first hunch when i first looked at the problem. i'm not seeing any other pattern.

Wow...I have never thought of that until you mentioned it.  That's a very good brain teaser if cats' answer is correct.

Oops, yeah that looks like it.  Funny problem.

cat's answer can't be right
cause if u look at where number 12 have to fall if they are listed in order.
It will have to fall into A which with cats's answer state it is all just curves and 1 would be a straight line.

Next number in A have to be 12
but the rest I can't tell

[quote="pakagawa"]Why is it only for non-negative intergers anyway?

And the first set is finished by ..., which means that there must be continuous numbers in that sequence. What is the next number?[/quote]

Probably becuase negative numbers always have a straight line, becuase of the negative sign (if your going with cats...'s answer).


Let a positive integer $ c$. The sequence $ (u_n)$ difined by $ u_1\equal{}1, u_2\equal{}c, u_{n\plus{}1}\equal{}2u_n\minus{}u_{n\minus{}1}\plus{}2 (n\ge 2)$. Prove that for each $ m\in\mathbb N$, there exits a natural number $ n$ such that $ u_m.u_{m\plus{}1}\equal{}u_n$

[hide="solution"] The general expression for the sequence is $ u_k \equal{} (k \minus{} 2)^2 \plus{} c(k \minus{} 1)$ so $ u_mu_{m \plus{} 1} \equal{} ((m \minus{} 2)^2 \plus{} c(m \minus{} 1))((m \minus{} 1)^2 \plus{} cm) \equal{} u_{u_m \plus{} m}$

Actually if you extend the sequence to all the integers (i.e. $ u_0 \equal{} 4 \minus{} c,u_{ \minus{} 1} \equal{} 9 \minus{} 2c$,etc.) you find that for each $ m\in \mathbb{Z}$ there are exactly two $ n\in \mathbb{Z}$ such that $ u_mu_{m \plus{} 1} \equal{} u_n$.  The second one is $ n\equal{}4 \minus{} u_m \minus{} m \minus{} c$. [/hide]


[hide]
Carbon suboxide alias $ \ce{C3O2}$ :huh:
[/hide]

Good One,
Try this
RCH=N-OH + acetic anhydride------>

When is the yield max..

RCH=N-OH + acetic anhydride------>  acetic acid + R-CH=N-O-COCH3 .

P.S: I knw this is complete rubbish nevertheless  :D

And the yield is max when they are in a 1:1 ratio  :maybe:

EDIT: Maybe it can also be a type of Beckmann rearrangement

[quote="hash_include"]formation of caprolactam??
[/quote]

Caprolactum?  :huh:  :huh:

just a wild wild guess after u said rearrangement :D

It cant be caprolactum, I have specified any specific  R, for u to guess a 7 membered ring

I thought this would be very easy after the first one...(big clue)

Second Part is slightly tough

Whatever is the product nitish has in mind I think $ \ce{R\minus{}CN}$ will be got with high yield upto maybe 40%. That hydrogen is highly acidic so definitely it is a cyanide. :wink:

You are right actually.......
How do u think can u increase the yield?????

I would think it is 1:2 because first getting rid of acidic proton and then the water formed converts it into acetic acid. :maybe:

yes a cyanide is formed, beckmann rearrangement does not generally occur for aldoximes( if that is the product you have in mind), altho, few cases are known whrer they do when PPA is used.
what yield do you want to increas?


Determine all integer number $n$ such that there exist $x;y$ being rational :
$x+y+\frac{1}{x} +\frac{1}{y}=n$.

Multiplying everything by $xy$...
$x^2y+xy^2+y+x=nxy$
$(x+y)(xy+1)=nxy$
$(x+y)(xy+1)\mid xy$
$gcd(x+y,xy)gcd(xy+1,xy)=xy$
Since $gcd(xy+1,xy)=1$,
$gcd(x+y,xy)=xy$
$x+y\mid xy$
$\displaystyle\frac{x+y}{xy}=n_1$
$\displaystyle\frac{1}{y}+\displaystyle\frac{1}{x}=n_1$
We notice that if this is true, than any multiple of $\displaystyle\frac{1}{x}$ and $\displaystyle\frac{1}{y}$ will also be an integer. Thus, we can say that
$\displaystyle\frac{1}{y}+\displaystyle\frac{1}{x}=1$
This implies that
$x+y=xy$
But $(x,y)=(1,1)$ doesn't work for this, even though it fulfills the beginning requirements. Where did I go wrong?

[quote="gopherhole112"] Where did I go wrong?[/quote]
[quote="anhminh"]Determine all integer number  such that there exist $x,y$ [b]being rational [/b]:
$x+y+\frac{1}{x}+\frac{1}{y}=n$[/quote]
$x,y$ are rational, but not integers


geometry
incenter
parallelogram
geometric transformation
homothety
ratio
Euler

Let $ I$ be the incenter of $ \triangle ABC$. Let $ L_{a}$ be the line parallel to $ IA$ that bisects segment $ BC$. Define $ L_{b}$ and $ L_{c}$ analogously. Prove that $ L_{a}$, $ L_{b}$, and $ L_{c}$ are concurrent.

[hide="Comment:"]There is a one line solution.[/hide]

[color=darkblue]Let $ M,\,N,\,P$ be the midpoint of $ BC,\,CA,\,AB,$ respectively.
We have $ APMN$ is a parallelogram, so $ AI$ is parallel to the angle bisector of angle $ PMN$
Thus $ L_{a}$ is the angle bisector of angle $ PMN$

Similarly, we also have $ L_{b},\,L_{c}$ are the angle bisector of angles $ MNP$ and $ NPM,$ respectively.
Hence $ L_{a},\,L_{b}$ and $ L_{c}$ are concurrent at the incenter of $ \triangle MNP$[/color]

[color=darkred][b]Remark.[/b]
[quote][b]Problem.[/b] Let $ I$ and $ G$ be the incenter and the centroid of the given triangle $ ABC.$ Let $ M,\,N,\,P$ be the midpoint of $ BC,\,CA,\,AB,$ respectively and let $ J$ be the incenter of triangle $ MNP.$
Then we have: $ I,\,G,\,J$ are collinear and $ GI=2\cdot GJ$  :wink: [/quote][/color]
[color=indigo][b]Proof.[/b]
We have $ \triangle ABC\sim\triangle MNP\implies \triangle IAB\sim\triangle JMN\implies \begin{cases}IA\parallel JM\\ IA=2\cdot JM\end{cases}$
Denote $ G'=AM\cap IJ\implies AG'=2\cdot G'M\implies G\equiv G'$
Hence $ I,\,G,\,J$ are collinear and $ GI=2\cdot GJ$[/color]

[size=150]Who can proposed some applications of this result?[/size]

$ L_{a}$ is the radical axis of the $ B$- and $ C$- excircles of $ \triangle ABC$.

or

A homothety of ratio $ -\frac{1}{2}$ about $ G$ maps $ ABC$ to $ MNP$. Likewise, $ AI$ is mapped to $ L_{a}$ since they are parallel and $ A$ maps to $ M$. Consequently, $ I$ maps to the desired point of concurrence.

EDIT:

When I came up with this problem, I was purely thinking of the radical axis solution.

Now I see that homothety lends itself well to this problem.

If we replace $ I$ with any point, $ X$, we get a similar concurrence.

If $ X\equiv H$, we have the euler line.

Another one-line:

Incenter of the medial triangle.

Oddly enough, when at MOP, I was trying to create an easy geo problem and came up with [b]exactly the same problem.[/b]

[quote="Altheman"]Ha thats pretty funny. I still think that the radical center solution is cool. But obviously, this type of question is old. This same technique is used for the euler line, for example.[/quote]

The radical center solution was also what I had in mind at first, but then I realized one didn't even have to create the extriangle (if that's what it's called).


Show that the maximum of a finite set of distinct numbers is unique.

(Corrected slightly.)

I'm not sure if this works:

[hide="Proof?"]Let $x$ be the maximum. Assume that it is not unique. Then there is another element $y$ in the set that is the maximum.

Either $x<y$, $x=y$, or $x>y$.

If $x<y$, then $x$ is not the maximum, for $y$ is greater than $x$. Contradiction.

If $x>y$, then $y$ is not the maximum, for there is an element $x$ that is greater than it. Contradiction.

If $x=y$, contradiction, because we are given that the set has distinct elements.

Thus, $x$ is unique.[/hide]

[quote="Quevvy"][hide]Distinct numbers are always unique so therefore, the maximum of a set of distinct numbers is unique.[/hide][/quote]

Er, you assumed that the maximum was only one of the distinct numbers, which is the claim I'm asking you to prove.

rnwang2, you're correct.

This is a weird question if you ask me, but yeah, use contradiction and trichotomy.


real analysis
function
integration
topology
real analysis unsolved

Hi there, I have a problem about the Lebesgue integrable function, wonder if anyone can help?
Thx!!

Let $ f$ be a Lebesgue integrable function on $ \Re$.
Prove that if $ \int_O f dm \geq 0$ for every open set $ O\subset \Re$, then $ f \geq 0$ $ m$-a.e

We take $ f \in L^1(\mathbb{R}^n)$ such that $ \int_U f \ge 0\,dm$ for every open $ U \subset \mathbb{R}^n$. Let $ V \equal{} \{f < 0\}$. Assume that $ m(V) > 0$. Then $ m(V_r) > 0$ for some $ r > 0$ where $ V_r \equal{} \{f < \minus{}r\}$. Take $ \delta > 0$ so that $ \int_A |f| < (1/2)rm(V_r)$ if $ m(A) < \delta$. Then choose open $ U \supset V_r$ so that $ m(U \setminus V_r) < \delta$. Now $ \int_U f \equal{} \int_{V_r} f \plus{} \int_{U \setminus V_r} f \le \minus{}rm(V_r) \plus{} (1/2)rm(V_r) \equal{} \minus{}(1/2)rm(V_r) < 0$, a contradiction.


integration
calculus
derivative
real analysis
calculus computations

Here's the question.. evaluate the following:  $\displaystyle\sum_{i=1}^{\infty}\frac{i^3}{2^i}  $

We can evaluate $\sum_{k=1}^n \frac {k^3}{2^k}$ using derivatives. Let's denote $S_n^m(x) = \sum_{k=1}^n k^m x^k$. We need to compute the sum $S_n^3\left(\frac 12\right)$. We shall compute $S_n^3(x)$ for all $x \in \mathbb{R}$.

We have \[ \int_0^t \frac 1x \cdot S_n^3 (x) \ dx = \sum_{k=1}^n  k^2 x^k  |_0^t = S_n^2(t). \]
We do the same thing two more times, thus
\[ \int_0^w \frac 1t \cdot S_n^2(t)\ dt = \sum_{k=1}^n kt^k |_0^w = S_n^1 (w) \ \textrm{and} \]
\[ \int_0^z \frac 1w \cdot S_n^1(w) \ dw = \sum_{k=1}^n w^k |_0^z = \frac { z^{n+1} - 1}{z-1 } \]
We now go back using derivatives, so
\[ S_n^1 (x) =  x \left (\frac {x^{n+1}-1}{x-1} \right)' = \frac { nx^{n+2} - (n+1)x^{n+1} +x }{(x-1)^2} \]
\[ S_n^2 (x) = x \left (\frac { nx^{n+2} - (n+1)x^{n+1} +x }{(x-1)^2} \right)' = \ldots \]

In the end we find out $S_n^3\left( \frac 12 \right)$ and we make $n\to \infty$, and easily compute our initial infinite sum. :D

Valentin is making it a little harder than he has to by writing the partial sums rather than the whole series. There's a theorem, taken for granted in most (U.S.) calculus courses and proved in most (U.S.) "advanced calculus" or "real analysis" courses: a power series may be differentiated term by term, as long as you stay strictly within the radius of convergence.

So let's do this:

$\frac1{1-x}=\sum_{k=0}^{\infty}x^k$. Differentiate:

$\frac1{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k-1}.$ Multiply by $x$:

$\frac{x}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k}.$ Differentiate:

$\frac{1+x}{(1-x)^3}=\sum_{k=1}^{\infty}k^2x^{k-1}.$ Multiply by $x$:

$\frac{x+x^2}{(1-x)^3}=\sum_{k=1}^{\infty}k^2x^{k}.$ Differentiate:

$\frac{1+4x+x^2}{(1-x)^4}=\sum_{k=1}^{\infty}k^3x^{k-1}.$ Multiply by $x$:

$\frac{x(1+4x+x^2)}{(1-x)^4}=\sum_{k=1}^{\infty}k^3x^{k}.$

So $\sum_{k=1}^{\infty}\frac{k^3}{2^k}=\frac{\left(\frac12\right) \left(1+\frac42+\frac14\right)} {\left(\frac12\right)^4} = 26.$

This procedure and yetti's procedure are both capable of reaching the same results; choose whichever one you prefer. This procedure - introducing the variable $x$ into the problem in which it was not initially present - falls under the general heading of "generating functions."

More general theorem:
$a_n:=\sum_{i=1}^{\infty}\frac{i^n}{2^{i+1}}$ then $a_n$ is the number of the mappings $f:S \rightarrow S$ ( $S=\{1,2,...,n\}$ ) that fulfil the condition that when $i$ is in the image of $f$ then also all $j<i$

[quote="ZetaX"]More general theorem:
$a_n: =\sum_{i=1}^{\infty}\frac{i^n}{2^{i+1}}$ then $a_n$ is the number of the mappings $f: S \rightarrow S$ ( $S=\{1,2,...,n\}$ ) that fulfil the condition that when $i$ is in the image of $f$ then also all $j<i$[/quote]

Why?


Let $a,b$ and $c$ be positive reals so that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 2$.
Prove that \[ \sqrt{a+b+c}\geq \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1} \]

http://www.mathlinks.ro/Forum/viewtopic.php?t=89660
http://www.mathlinks.ro/Forum/viewtopic.php?t=56020
http://www.mathlinks.ro/Forum/viewtopic.php?t=49897
http://www.mathlinks.ro/Forum/viewtopic.php?t=21387
http://www.mathlinks.ro/Forum/viewtopic.php?t=25650
http://www.mathlinks.ro/Forum/viewtopic.php?t=13646

and maybe more...

  darij


How many ways can we put 6 different objects in 3 different bags?

Divider time. With 2 dividers, and 7 places to put them, we get 7c2 which is 21.
-edit- nevermind
i remember doing something like this in intro to counting....

[quote="Iversonfan2005"]Why wouldn't it be just $6C3$? That would be equal $20$.[/quote]

You're not [i]picking[/i] 3 of the 6 objects, you're putting the 6 objects into 3 distinct (lets say, different colored) bags. That's a totally different thing.

THERE CAN BE LOTS OF POSSIBILITIES AS THE THINGS ARE ALL DIFFERENT.
BUT IM STILL WORKING ON THE SOLUTION.

[quote="gauss202"][quote="Scrambled"]Divider time. With 2 dividers, and 7 places to put them, we get 7c2 which is 21.[/quote]

The objects are distinct. Your method treats the objects like they are all the same.[/quote]
well then, we have 7*6/2, which is if the objects were all the same, and then *7! because all the objects are distinct\

[quote="Scrambled"][quote="gauss202"]The objects are distinct. Your method treats the objects like they are all the same.[/quote]
well then, we have 7*6/2, which is if the objects were all the same, and then *7! because all the objects are distinct\[/quote]

No, still not quite right. You're counting some configurations of "objects in bags" too many times. For example, if there are two objects in the first bag it doesn't matter which order those two objects appear in the bag, but you're counting that configuration twice.

It'd be better to throw away the dividor approach on this problem and start over again.

Do you have to have at least 1 ball in every bag?

[quote="gauss202"]Give reasons for your calculations.[/quote]

Don't really know why Mr. Neil...but I saw a problem like this before and I thought the formula was like n is the number of objects to put in k bags. Therefore $(n+k-1)C(n)$ or was it $(n+k-1)C(k)$? That's why I think it is either $28$ or $56$.

[quote="Iversonfan2005"][quote="gauss202"]Give reasons for your calculations.[/quote]

Don't really know why Mr. Neal...but I saw a problem like this before and I thought the formula was like n is the number of objects to put in k bags. Therefore $(n+k-1)C(n)$ or was it $(n+k-1)C(k)$? That's why I think it is either $28$ or $56$.[/quote]

That's why its important not to memorize formulas without a sense of what they are [i]for[/i] and under what circumstances to apply them. The formula you used is for when the objects to be distributed are [i]indistinguishable[/i] and the bags are distinct. For example, if you were dividing 7 dollars between 3 people - then the people are distinct, but nobody would care which particular dollar, or dollars, they got. All dollars are the same (i.e. indistinguishable).

But in this case, the objects are are assumed to be distinguisable (or unique). Putting a certain two objects in a bag would be counted differently than putting some other pair of objects in the same bag. So the formula that you used does not apply here. (You should try to figure out why that formula works for [i]indistinguishable[/i] objects though).

For this problem, take a second to slow down and visualize it. Imagine you have 6 unique objects, and you want to put them into three distinct bags, say a green one, a red one, and a blue one.... Can you come up with a way to count the number of ways of doing that?

[hide]How many ways can you put the first object in one of the bags? What about the second object?[/hide]

Well, you can put the first object in 3 different ways into the one of the 3 bags. And the same goes for the other 5 objects, so................
[hide]
3(6)?   I give up :surrender:  :help:  :ewpu:  :wacko: [/hide]

[quote="gauss202"]You are [b]very[/b] close.

[hide]But don't [i]multiply[/i] the 3 and the 6....[/hide][/quote]

But it wouldn't make sense if I added them...? So would it be $3^6$

If we must have at least one item in each bag, we will have 3*2*1 for the first three items in each bag, then 3*3*3 for the remainding objects, giving us ...

[quote="matt276eagles"]Yes, the answer is $3^6$.[/quote]

Okay, yes, because you have 3 bags and it can go to each object 3 different times, and since there are 6 objects, that would be 3x3x3x3x3x3.

[quote="matt276eagles"]Yes, the answer is $3^6$.

Edit:  Never mind.  Now that I think about it, since we are putting six items into three bags, each bag should contain at least one item.  Otherwise, we are not putting the items into three bags, but we are putting the items into one or two bags.  The solution must be less than $3^6$.[/quote]

Generally when you have a problem like this, it's okay to allow some of the bags to be empty unless otherwise stated. A configuration of 6-0-0 counts as a "distribution" of 6 objects into 3 bags.

Let us call the bags $A$, $B$, and $C$.  Let the number of items in each bag be $a$, $b$, and $c$.  Assume $a<b<c$.  Since there are six items, $a+b+c=6$.  Therefore, $(a,b,c)=(1,1,4), (1,2,3), (2,2,2)$.  Every possibility must use one of these three triples.  For the first triple, the number of different ways the items can be arranged is $6\times5=30$.  The number of different ways the bags can be arranged is $3$.  Since $30\times3=90$, the number of possibilities using the first triple is $90$.  Similarly for the second and third triples,
\[6\times\frac{5\times4}{2}=60\\
60\times (3\times 2)=360\\
\frac{6\times5}{2}\times\frac{4\times3}{2}=90\\
90\times1=90\]
Adding the possibilities, $90+360+90=540$.  Therefore, there are $540$ ways that three different items can be placed into three different bags.

[quote="gauss202"]Generally when you have a problem like this, it's okay to allow some of the bags to be empty unless otherwise stated. A configuration of 6-0-0 counts as a "distribution" of 6 objects into 3 bags.[/quote]

Edit:  My solution assumes every bag must contain at least one item.  Taking into account what gauss202 said, the solution would be $3^6$

[quote="matt276eagles"]Let us call the bags $A$, $B$, and $C$.  Let the number of items in each bag be $a$, $b$, and $c$.  Assume $a<b<c$.  Since there are six items, $a+b+c=6$.  Therefore, $(a,b,c)=(1,1,4), (1,2,3), (2,2,2)$.  Every possibility must use one of these three triples.  For the first triple, the number of different ways the items can be arranged is $6\times5=30$.  The number of different ways the bags can be arranged is $3$.  Since $30\times3=90$, the number of possibilities using the first triple is $90$.  Similarly for the second and third triples,
\[6\times\frac{5\times4}{2}=60\\
60\times (3\times 2)=360\\
\frac{6\times5}{2}\times\frac{4\times3}{2}=90\\
90\times1=90\]
Adding the possibilities, $90+360+90=540$.  Therefore, there are $540$ ways that three different items can be placed into three different bags.[/quote]

That is correct assuming all of the bags have to be used. Another way to solve this version of the problem is to use the principle of inclusion-exclusion (PIE).

We already know that the number of ways of distributing 6 distinct objects into 3 distinct bags is $3^6$. By similar logic, the number of ways of distributing 6 distinct objects into 2 bags, and leaving one empty, is ${3 \choose 2} * 2^6$. And the number of ways of distributing 6 distinct objects into 1 bag, and leaving two empty, is ${3 \choose 1} * 1^6$, or simply $3$.

So by the PIE the number of ways of distributing the 6 objects and leaving no bags empty is $3^6 - {3 \choose 2} * 2^6 + 3 = 540.$

Anybody see a general structure here?

:lol: I also tried inclusion-exclusion just like you did, but I was unsure of what to do with the number of possibilities using only one bag ($3$), so I stuck with my first solution.  Why do you add the number of possibilities using only one bag rather than subtract?

[quote="Barnacle"]is this trivial? 3^6? this belongs in middle school forum... -_-[/quote]

A lot of people couldn't get it though...

[quote="Barnacle"]is this trivial? 3^6? this belongs in middle school forum... -_-[/quote]
thats because apparently, everyone was thinking way too hard


[size=150]$46^{th}$ International Mathematical Olympiad[/size]
Mérida, Mexico, Thursday 14 July 2005
Language: Vietnamese

[b]Bài 1.[/b] Trên các cạnh của tam giác đều
$ABC$ chọn sáu điểm: $A_1, A_2$ trên $BC$; $B_1, B_2$ trên $CA$;
$C_1, C_2$ trên $AB$. Các điểm này là đỉnh của một lục giác lồi
$A_1A_2B_1B_2C_1C_2$ có sáu cạnh bằng nhau. Chứng minh rằng các
đường thẳng $A_1B_2$, $B_1C_2$ và $C_1A_2$ đồng qui.

[b]Bài 2.[/b] Cho $a_1, a_2, \ldots$ là một dãy số nguyên
với vô hạn số hạng dương và vô hạn số hạng âm. Biết rằng với mỗi số
nguyên dương $n$, các số $a_1, a_2, \ldots, a_n$ cho $n$ số dư khác
nhau trong phép chia cho $n$. Chứng minh rằng mỗi số nguyên xuất
hiện đúng một lần trong dãy số này.

[b]Bài 3.[/b] Cho $x, y$ và $z$ là các số thực dương sao
cho $xyz \geq 1$. Chứng minh rằng
\[
\frac{x^5 - x^2}
{x^5 + y^2 + z^2}
+
\frac{y^5 - y^2}
{y^5 + z^2 + x^2}
+
\frac{z^5 - z^2}
{z^5 + x^2 + y^2}
\geq 0.\]
[b]
Bài 4.[/b] Xét dãy số $a_1, a_2, \ldots$ được xác định
như sau
\[a_n = 2^n + 3^n + 6^n -1\ \ (n=1,2,\ldots).\]
Tìm tất cả các số nguyên dương nguyên tố cùng nhau với mọi số hạng
của dãy số trên.

[b]Bài 5.[/b]  Cho $ABCD$ là một tứ giác lồi có các cạnh
$BC$ và $AD$ bằng nhau và không song song với nhau. Xét $E$ và $F$
lần lượt là các điểm nằm ở phần trong các cạnh $BC$ và $AD$ sao cho
$BE=DF$. Các đường thẳng $AC$ và $BD$ cắt nhau tại điểm $P$, các
đường thẳng $BD$ và $EF$ cắt nhau tại điểm $Q$, các đường thẳng $EF$
và $AC$ cắt nhau tại điểm $R$. Chứng minh rằng khi $E$ và $F$ thay
đổi, đường tròn ngoại tiếp tam giác $PQR$ luôn đi qua một điểm cố
định thứ hai khác $P$.

[b]Bài 6.[/b] Trong một kỳ thi học sinh giỏi, các thí sinh
phải giải 6 bài toán. Biết rằng với hai bài toán bất kỳ luôn có
nhiều hơn $\frac 25$ số thí sinh dự thi giải được cả hai bài này.
Ngoài ra, không có thí sinh nào giải được cả 6 bài toán. Chứng minh
rằng có ít nhất 2 thí sinh sao cho mỗi người trong họ giải được đúng
5 bài toán.

[i]Thời gian làm bài: 4 giờ 30 phút
Mỗi bài được 7 điểm[/i]

:D  :D

Chi? co' nhu*~ng ai gia?i he^'t ta^'t ca? ba`i tre^n mo*'i co' quye^`n chi? tri'ch ma^'y em thi Olympic.

Ba ba.c, ba ddo^`ng .. not so bad at all ...

Co' chi? tri'ch la` chi? tri'ch  ban la~nh dda.o nha` nu*o*'c ...


Let $ x,y,z$ be positive integers such that $ 1/x \minus{} 1/y \equal{} 1/z$. Let $ d: \equal{}gcd(x,y,z)$. Prove that $ dxyz$ and $ d(y\minus{}x)$ are perfect squares.

Let $ a \equal{} \frac {x}{d}, b \equal{} \frac {y}{d}, c \equal{} {z}{d}$. We have (a,b,c)=1 and a,b,c also satisfy the condition
We have $ c\left( {b \minus{} a} \right) \equal{} ab$. Let $ v_p(x)$ be the exponent of $ p$ in the integer $ x$.
Assume that $ v_p(b \minus{} a) \equal{} 2k \plus{} 1$ with $ p$ is a prime.
If $ v_p(a)\ge2k \plus{} 1$, then $ p|b$ and $ p|c$, contradiction
Let $ x \equal{} v_p(a), y \equal{} v_p(b)$, $ x,y < 2k \plus{} 1$. Since $ x\ne y$ $ (x \plus{} y \equal{} 2k \plus{} 1)$, so $ v_p(b \minus{} a) \equal{} max\{v_p(a),v_p(b)\} \equal{} max\{x,y\}$, contradiction
Therefore, $ b \minus{} a$ is a perfect square, impies $ abc \equal{} c^2(b \minus{} a)$ is a perfect square, too. So $ d(y \minus{} x)$ and $ dxyz$ are perfect squares.


I've stumbled upon this cute little problem:

Show that the equation $a^2+b^2+c^2=(a-b)(b-c)(c-a)$ has infinitely many integer solutions. Let's see: maybe there's are lots of different approaches.

This was already solved:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=50[/url]

Thanks, and sorry :(. I did exactly what Galois did in that topic.

Don't worry about it, no one can be expected to have seen and remember everything posted.  Better to take the chance, I say. :cool:


[i] The distance from a point (x,y,z) in space to the origin is sq.rt. of(x^2+y^2+z^2). A lattice point is a point in space which has coordinates which are all integers. Therefore, (3,4,12) is a lattice point but (1.2,3.5,-7) is not. How many lattice points are exactly 7 units away from the origin?[/i]

Does anyone have an answer to this problem? :help:

i found 54, probably more though

[hide="here are the ones I found "]
$ \pm$ {0,0,7}  x6
$ \pm$ {2,3,6}   x48
[/hide]

No. That was a problem straight from the book.

Yes, well, some of the problems in the challenge sets are from the books.

[quote="AIME15"]DO NOT POST HOMEWORK ON THE FORUMS!!!

This was a challenge set problem, wasn't it!?[/quote]

I think it's perfectly fine.  It's kinda like tutoring, but online.  As long as the "tutor" explains his/her answer thoughtfully and the tutoree* understands it, then it is fine.

* tutoree- someone who is being tutored.

uhh so guys what is the answer/solution, im almost certain mine is incorrect.


calculus
integration
probability
function
calculus computations

Hi guys - first time on the forum so I'd like to say hi as well.

Here's my problem:

Given the probability density function p(x) = C for -b<x<b

I must first find the constant C, which be this integral is actually 1/(2b)

Then, I must find the cumulative distribution function, F(x), and by the fundamental theorem of calculus, I get:

(1/2)[ln b - ln (-b)]

And here is my problem, as you may have noticed already, I CAN'T TAKE THE LN OF A NEGATIVE VALUE!

Did I do something wrong along the steps?

Thanks in advance!

$p(x) = \frac{1}{2b}$ is just a constant function, so the cumulative distribution function $F(x)$ is given by

$F(x) = \int_{-b}^{x}p(t) \, dt = \frac{1}{2b}\int_{-b}^{x}dt = \frac{x+b}{2b}$

Oh sorry, I forgot to mention that the boundary is [-b,b] so after solving the integral I got 0

i.e. F(x) = integral of p(x) dx on [-b,b] = 0

We started out with p(x) = C


Well, on my last math lecture I spent about 1,5h on that inequality and I've done practically nothing.
I believe that can be done with some smart SOS mathod, but I have no idea how.

Prove that for all $ a,b,c\in \mathbb{R}_{\plus{}}$ the following inequality holds:
$ \sum_{cyc}\frac{1}{(2a\plus{}b)^{2}}\geq \frac{1}{\sum_{cyc}ab}$
Good luck :wink:

[quote="polskimisiek"][quote="Honey_S"][quote="polskimisiek"]Well, on my last math lecture I spent about 1,5h on that inequality and I've done practically nothing.
I believe that can be done with some smart SOS mathod, but I have no idea how.

Prove that for all $ a,b,c\in \mathbb{R}_{ \plus{} }$ the following inequality holds:
$ \sum_{cyc}\frac {1}{(2a \plus{} b)^{2}}\geq \frac {1}{\sum_{cyc}ab}$
Good luck :wink:[/quote]

See in this attachment, problem 156.

And the stronger is wrong already.
\[ \left( {\frac {1}{{2a \plus{} b}} \plus{} \frac {1}{{2b \plus{} c}} \plus{} \frac {1}{{2c \plus{} a}}} \right)^2 \ge \frac {3}{{ab \plus{} bc \plus{} ca}}
\]
Try with $ a \equal{} 3,b \equal{} 1$ and $ c \equal{} 0$.[/quote]

Thank you very much. I've got a couple of questions. First, that looks like Vietnamese training papers, where'd you get those? :wink:
And what is more important, can someone explain me HOW the author thought of that proof and what were his pattenrs of thought? :wink:[/quote]

Yes it's exactly a Vietnamese training paper, polskimisiek. Maybe you haven't known about it but it's popular and it was posted in many forums.  :wink:
And I'm a Vietnamese.
About the solution, it's not my solution so that I have no idea about it.


[quote="Bi"]Find all pairs (x,y) of integer numbers such that:
\[x^{x+y}=y^{y-x}\][/quote]
First, if $(x,y)$ is a solution then easily is $(-x,-y)$, $x\pm y$ are even, in fact if $x=x_1\cdot x_2$ and $y=x_1\cdot y_2$ with $x_1=\gcd(x,y)$ then clearly $x_2=1$ then
\[
x^{x+y}=y^{y-x} \ \iff \ x_1^{2x_1}=y_2^{y-x} \ \iff \ x_1^2=y_2^{y_2-1},
\]
hence the positive solutions are $(0,0)$ and $((1+2d)^d,(1+2d)^{d+1})$ with $d$ a positive integer, and the negative are $(-(1+2d)^d,-(1+2d)^{d+1})$.

RX


Let two points B, C. Let a point A that it does not belong in the line BC. We create the circles C1( K1, R1) , C2( K2, R2) with chords AB, AC respectivelly. We know that the centers of the circles are exteriors to the triangle ABC. We also know that R1*AC=R2*AB. Let T the point which the circles intersect, which is different from A. IF M is accintental point of the line AT prove that: AC*[MBT]=AB*[MCT]  .( We symbolise with [....] the surface of the triangles.

I wait your solutions.

we have two similar triangles. And then? Let's solve it!

Are you sure the problem is correct?
I prove that [MBT]*CT=[MTC]*BT

Tell me what do you think for the solution!

(AC/sin <CTM)=2*R2
(AB/sin <BTM)=2*R1

so     sin <CTM = sin <BTM

[MCT]/[MBT]=(CT*MT*SIN <CTM)/(BT*MT*SIN <MTB)=CT/BT

(AC/sin <CTM)=(CT/ sin <CAT)
(AB/sin <BTM)=(BT/ sin <TAB)

so     AC/AB  =   (CT* sin <TAB)/(BT*sin <CAT)

IF CT/BT=AC/AB   so   <TAB=<CAT

it is impossible

You solve the exercise! It is not imposible <TAB=<CAT. Because the two angles are equal the two circles are also equal. So the areas are equal and the problem solved!
It is strange?

silouan , are you agree with me that the problem is wrong?
 :?:

My opinion is that is a strange problem but it is not bad to have two equal circles <=>
R1=R2 <=> [MBT]=[MCT].

Take $ K$ and $ L$ diametrally opposite of A on the two circles; from the given relation it follows that, $ \triangle ABK \sim  \triangle ACL$, both of them being right-angled at $ B$ and $ C$ respectively. It follows that $ \measuredangle AKB \equal{} \measuredangle ALC$ and, if we name $ P \equal{} AT \cap BC$, from the cyclic quads $ AKBT$ and $ ATCL$ we get $ \measuredangle BTP \equal{} \measuredangle CTP$, $ AT$ is the bisector of the $ \angle BTC$, hence for any $ M \in AT$ we get $ \frac {S_{[BMT]}}{S_{[CMT]}}\equal{}\frac{BT}{CT}$ and, to bring the question in line, we must re-state the given problem, namely, the required equality exists iff $ \triangle ABC$ is isosceles at $ A$.

Best regards,
sunken rock


algebra
function
domain
integration
trigonometry
calculus
calculus computations

Find the domain of $ f(x)\equal{}\int_0^{\sin x}\sqrt{\frac12\minus{}y}\,dy.$

want $ \sqrt{\frac{1}{2}\minus{}y}$ to exist $ \forall y \in [0,\sin x)$

so $ \frac{1}{2}\ge y \to \frac{1}{2} \ge \sin x \to x \in (\frac{4k\plus{}1}{2}\pi\minus{} \frac{ \pi}{6}, \frac{4k\plus{}1}{2}\pi \plus{}\frac{ \pi}{6})^c$

$ x\notin (\frac{6k\plus{}1}{3}\pi , \frac{6k\plus{}2}{3}\pi)$

$ x \in \cup_{k \in Z} \left( \frac{6k\minus{}4}{3}\pi ,\frac{6k\plus{}1}{3}\pi \right)$


calculus
integration
trigonometry
calculus computations

solve :D

$ \int_0^\infty \frac {\tan^{ \minus{} 1} \; x}{ x^2 \plus{} 2x\; \cos \theta \plus{} 1} \;\; dx$

Put $ I \equal{} \int_{0}^{\infty} \frac{\tan^{\minus{}1} x}{x^2 \plus{} 2x \cos\theta \plus{} 1} \, dx$. By the substitution $ x \equal{} \frac{1}{t}$, we have $ I \equal{} \int_{0}^{\infty} \frac{\tan^{\minus{}1} (1/t)}{t^2 \plus{} 2t \cos\theta \plus{} 1} \, dt$.

Then for $ 0 < \theta < \pi$, we have $ I \equal{} \frac{\pi}{4} \int_{0}^{\infty} \frac{dx}{x^2 \plus{} 2x \cos\theta \plus{} 1} \equal{} \frac{\pi \theta}{4 \sin \theta}.$


geometry perimeter rectangle trapezoid puzzles

I have a query, and can't figure out why it's so......

A square with a side length of 8 has an area of 64.

The same square, rearranged into an isosoles triangle, with:-

h=13
b=10

gives an area of 65.

Why is this so????

maybe you miscalculated the lengths of the sides of the iscocleles triangle. I'm not sure though.

isn't the formula for a triangle base$\times$ height $\times \frac{1}{2}$?

so, substitute the numbers, and 13(10)(0.5) = 65.

[quote="indijane"]The same square, rearranged into an isosoles triangle, with:-

h=13
b=10

[/quote]

How did you rearrange it?

I think this post references a brain teaser puzzle that doesn't translate well without pictures and doesn't belong in this forum.  I'm going to move it to the puzzles forum.

Just because two figures' perimeters are equal, that does not imply that their areas are equal.

---wait... I think this is similar to that question about the triangle which is cut into 4 jigsaw pieces, and when rearranged forms the same triangle with a hole...

A picture would definitely be helpful.

That triangle assumes that the hypotenuse is a straight line, but it's actually slightly different, and therefore technically a quadrilateral.

As for the isosceles triangle, how is it rearranged? It's rather hard to tell how you got the numbers for base and height.

Diagram attached.  Hope it works.  All of the square's area is used to make the triangle.

The question I have is:

Why is the square's area = 64, when triangle is = 65?

the triangle on top has legs 3 and 8, but the triangle on the bottom (if you subtract 3x5 rectangle from trapezoid) has legs 2 and 5, which i dont believe are similar.. so the whole thing isn't really a triange, its a pentagon..

[size=59]fixed hexagon to pentagon...[/size]

This is a famous so-called "mathematical brainteaser" that attempts to prove that $8*8=65$.  The problem in that is that if you draw a line from the upper left hand side of the triangle to the median of the bottom side of the square, it will [i]seem[/i] to be a straight line. It is now easy to prove that it is not a straight line. :)

Masoud Zargar

The same square when rearranged cannot form an isosceles triangle. To be exact it forms a [size=150][i][b]pentagon[/b][/i][/size] (see the attached figure).

The side BE = CF = sqrt (5^2 + 2^2) = sqrt (25 + 4) = sqrt 29 = 5.3851648071345040312507104915403.

angle(ABD) = tan inverse (8/3) = 69.443954780416535691706387252656 degrees
angle(BEG) = tan inverse (5/2) = 68.198590513648188229755133913056 degrees which is greater than angle (ABD). A, B and E are not collinear.

The area of the rearranged pentagon = area of the isosceles triangle, ABC + area of the trapezium, BCFE =(1/2)*6*8 + (1/2)*5*(5 + 3) = 24 + 40 = 64.


Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt));

draw(Circle((0,0),1+sqrt(2)));
draw(Circle((sqrt(2),0),1));
draw(Circle((0,sqrt(2)),1));
draw(Circle((-sqrt(2),0),1));
draw(Circle((0,-sqrt(2)),1));[/asy]$ \textbf{(A)}\ 3\minus{}2\sqrt2 \qquad
\textbf{(B)}\ 2\minus{}\sqrt2 \qquad
\textbf{(C)}\ 4(3\minus{}2\sqrt2) \qquad
\textbf{(D)}\ \frac12(3\minus{}\sqrt2)$
$ \textbf{(E)}\ 2\sqrt2\minus{}2$

[hide]
Let the radius of the large circle be $ R$ and the radius of each of the small circles be $ r$.  The centers of the smaller circles form a square with side length $ 2r$.  The diagonal of this square lies along a diameter of the large circle, encompassing all of it except for two small circle radii.  Thus,

$ 2r\sqrt{2}\equal{}2R\minus{}2r\Rightarrow r(1\plus{}\sqrt{2})\equal{}R$.

The desired ratio is $ \frac{4\pi r^2}{\pi R^2}$, or

$ \frac{4r^2}{r^2(1\plus{}\sqrt{2})^2}\equal{}\frac{4}{3\plus{}2\sqrt{2}}\equal{}12\minus{}8\sqrt{2}\equal{}4(3\minus{}2\sqrt{2})\equal{}C$.
[/hide]

I got this one right after the test ended.  I actually drew the lines between the centers (!) but couldn't seem to connect the dots and solve the problem.  I guess that's called "test stress." :P

BS solution: [hide]knowing the AMC, they would put the answer for 1 smaller circle in addition to the real answer for 4 smaller circles, to screw those who misread and thought the former; those of course differ by a factor of 4, which only applies to answer choices A and C. C is 4A, so it must be the answer.

(obviously, not a good recommendation for learning math, but it sure gives a quick and decent guess if you can't solve it)[/hide]

 :rotfl:

@pianoforte: PURE WIN! :D

[hide="my solution"]Draw a segment to each of the four smaller circles from the center. Call the length of each of those segments $ x$. $ (x\plus{}r_s)\sqrt{2}\equal{}2r_s$, where $ r_s$ is the radius of the smaller circle. Solving for $ x$ yields $ x\equal{}(\sqrt{2}\minus{}1)r_s$. Therefore the radius of the bigger circle is $ x\plus{}2r_s\equal{}(\sqrt{2}\plus{}1)r_s$. Thus the ratio is

$ \frac{4r_s^2}{(\sqrt{2}\plus{}1)^2r_s^2}\equal{}\frac{4}{3\plus{}2\sqrt{2}}\equal{}4(3\minus{}2\sqrt{2})$.[/hide]

To pianoforte:  That's actually what I did on the test...

@lowtopology:

The same thing happened to me!  :D  I drew the square from the circle centers, but decided it was not of much use and moved on to the next question...

@pianoforte:

Awesome solution!   :D :rotfl: :D  I should try it sometime.

[hide="Click for solution"]
We don't know anything that we need, so we try to find something that we can use.  Because we are looking for a ratio, we can set the length of the smallest circle to be $ 1$, which is easy to work with.  Connecting the centers of the squares gives:

    [asy]defaultpen(linewidth(.8pt)+fontsize(8pt));
unitsize(6mm);
    draw(Circle((0,0),1+sqrt(2)));
    draw(Circle((sqrt(2),0),1));
    draw(Circle((0,sqrt(2)),1));
    draw(Circle((-sqrt(2),0),1));
    draw(Circle((0,-sqrt(2)),1));
    draw((sqrt(2),0)--(0,sqrt(2))--(-sqrt(2),0)--(0,-sqrt(2))--cycle);[/asy]

Now we can find the diagonal of the square using $ 45\minus{}45\minus{}90$ triangle ratios, and the length of the diagonal is thus $ 2\sqrt{2}$.  Extending the diagonal to the circle, and labeling everything we know:

      [asy]defaultpen(linewidth(.8pt)+fontsize(8pt));
unitsize(6mm);
    draw(Circle((0,0),1+sqrt(2)));
    draw(Circle((sqrt(2),0),1));
    draw(Circle((0,sqrt(2)),1));
    draw(Circle((-sqrt(2),0),1));
    draw(Circle((0,-sqrt(2)),1));
    draw((sqrt(2),0)--(0,sqrt(2))--(-sqrt(2),0)--(0,-sqrt(2))--cycle);
draw((0,1+sqrt(2))--(0,-1-sqrt(2)));
label("$\sqrt{2}$",(0,0),E);
label("$1$",(0,-1/2-sqrt(2)),E);
label("$1$",(0,1/2+sqrt(2)),E);[/asy]

Quite clearly now, the diameter of the large circle is $ 2\plus{}2\sqrt{2}$, so its radius is $ 1\plus{}\sqrt{2}$.  Now the ratio is:

\[ \frac{4(1)^2\pi}{(1\plus{}\sqrt{2})^2\pi}\equal{}\frac{4}{3\plus{}2\sqrt{2}}
\]

We can rationalize the denominator by multiplying both numerator and denominator by the conjugate, which is $ 3\minus{}2\sqrt{2}$.  Thus:

\[ \frac{4}{3\plus{}2\sqrt{2}} \times \frac{3\minus{}2\sqrt{2}}{3\minus{}2\sqrt{2}}\equal{}\frac{4(3\minus{}2\sqrt{2})}{9\minus{}8}\equal{}4(3\minus{}2\sqrt{2})
\]

This corresponds to the answer choice $ \boxed{\textbf{(C)}}$.
[/hide]


Given  a  triangle  ABC , prove  that
R-2r >=(ma-ha)/2.

[i]unknown source[/i]

I'll do it for m(A)<=90  for m(A)>=90 it's in the same way!!!

well if m(A)<=90 thenn we have R(1+cos(A))>=ma and so
ma-ha<=R(1+cosA)-R(cosA+cos(B-C))=2R(sin((B-C)/2))^2
now .. we know that R-2r=R(1-4cos((B-C)/2))sin(A/2)+4(sin(A/2))^2)=R((2sin(A/2)-cos((B-C)/2))^2+(sin((B-C)/2))^2)>=R(sin((B-C)/2))^2 so we get from these 2 relations that:
2(R-2r)>=2R(sin((B-C)/2))^2>=ma-ha

cheers! :D :D

This solution is very nice and I would like to have a proof of the formula:

R-2r=R(1-4cos((B-C)/2))sin(A/2)+4(sin(A/2))^2

Thank you very much Lagrangia.

Sorry for the delay...

R-2r=R(1-4sin(A/2)cosD+4sin^2(A/2)); D=(B-C)/2

[b]Solution:[/b]

here is what I found:
Let a,b,c be the sides of the triangle, so there exist x,y,z pozitive reals so that:
a=y+z;b=z+x and c=x+y.

So R=(y+z)(z+x)(x+y)/(4sqrt((x+y+z)xyz))
r=sqrt(xyz/(x+y+z))
sin(A/2)=sqrt(yz/((x+y)(x+z))
And from Mollweide you get that cosD=sqrt(yz/((x+y)(x+z))*(2x+y+z)/(y+z)

using these it's easy to prove the identity!

cheers! :D :D


I have two AoPS certificates from MathCounts, and would like to buy the Volume Two: And Beyond book using both of them (and a little extra cash). However, it only lets me use one at a time. Is it possible to use both in one checkout?

Most likely not. They never want things to be 100% free. Just like how you usually can't stack coupons at stores.

On an unrelated note, I don't think AoPS is handing them out anymore (at least in CT).


The inscribed circle of ABC touches sides BC, CA, AB at A', B', C' respectively. AA' meet the circle again at M; MB, MC intersect again the circle at P, Q respectively. AP, AQ meet the circle again at I, K respectively. Prove that C'I, B'K, AA' are concurent

[quote="nsato"]It's from Iran, 2002:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2619[/url][/quote]

Unfortunately, it's not the same problem.
I have a proof for this one, too, but I am lacking time to post it...

  darij

You're right; they looked similar, but I should have looked closer.   :(

The problem is to show that two projections of the circle (or conic) onto the line through A are compatible:  certain points projected from B' are supposed to correspond to certain points projected from C'.    This turns out to be an exercise about involutions on a conic with respect to various points.    The criterion for compatibility of two points on the conic (their B' and C' projections onto AA' are the same point) is that the line through them is concurrent with BC and the tangent line at M.  There are several pairs of compatible points in the diagram and the rest is not hard from there.   The Iranian problem can be done in the same way  or as a lemma for this problem.

It is not hard
You can prove that MPA'Q and MIA'K are harmonic quadrilateral (use MB'A'C' be harmonic quadrilateral and Ptolemy 'theorem then use Pascal'theorem for quadrilateral B'IKC' you can be done


geometry
quadratics
trigonometry
area of a triangle
Heron\u0027s formula
algebra
quadratic formula

I just found the Hungarian Problem Books I and II from my school's library (not sure why they have it though.... no one's checked it out in 20 some years).  Anyways, i thought i'd post some problems appropriate for this forum, though i haven't had a lot of time to look through the book... i'll  post more later.

1.  The lengths of the sides of a triangle form an arithmeic progression with difference d.  The area of this triangle is t.  Find the sides and angles of this triangle.  Solve this problem for the case d=1 and t=6.

2.  Prove, for the angles a, b, and c of a right triangle, the following relation: sina*sinb*sin(a-b) + sinb*sinc*sin(b-c) + sinc*sina*sin(c-a) + sin (a-b)*sin(b-c)*sin(c-a) = 0

3.  Let x_1 and x_2 be the roots of the equation x^2-(a+d)x + ad-bc=0.

Show that x_1^3 and x_2 ^3 are the roots of y^2 - (a^3 + d^3 + 3abc + 3bcd)y + (ad-bc)^3=0

For number #1,[hide] it's obvious that 3,4,5 are the sides.  I will just figure out a general formula for the sides. The sides are a-d,a,a+d. Using heron's formula, we have  :sqrt: (3a/2(a/2)((a-2d)/2)(a+2d)/2)=t. so (3a)(a)(a-2d)(a+2d)=16t :^2: , or 3a :^4: -12a :^2: d :^2: -16t :^2: =0. Using quadratic formula, a :^2: =2d :^2:+2 :sqrt: ((9d :^4: +12t :^2: )/3). so a= :sqrt:(2d :^2:+2 :sqrt: ((9d :^4: +12t :^2: )/3)). To figure out the angles, use law of cosine to find cos :theta: , then do a inverse.  [/hide]

Quote:2. Prove, for the angles a, b, and c of a right triangle, the following relation: sina*sinb*sin(a-b) + sinb*sinc*sin(b-c) + sinc*sina*sin(c-a) + sin (a-b)*sin(b-c)*sin(c-a) = 0

Solution:[hide]Using sin(90-x)=cos x, sin(x-y) = -sin(y-x), and letting C=90, the sum becomes sina*sinb*sin(a-b) - sinb*cosb + sina*cosa - sin(a-b)*cosb*sina. Let the triangle ABC have sides x,y,z such that the hypotenuse is z and the angles of the triangle are a,b,90. Then the summands become x/z*y/z*(x^2/z^2 - y^2/z^2) - xy/z^2 + xy/z^2 -xy/z^2*(x^2/z^2 - y^2/z^2) = 0, and we are done.[/hide]

Quote:3. Let x_1 and x_2 be the roots of the equation x^2-(a+d)x + ad-bc=0.

Show that x_1^3 and x_2 ^3 are the roots of y^2 - (a^3 + d^3 + 3abc + 3bcd)y + (ad-bc)^3=0

Solution:[hide]

Note that x_1x_2 = ad-bc, so x_1^3*x_2^3=(ad-bc)^3. Next, we see that x_1^3+x_2^3=(x_1+x_2)^3 - 3x_1^2x_2 - 3x_2^2x_1

=> (x_1+x_2)^3 - 3x_1*x_2(x_1+y_1)

=> (a+d)^3 - 3(ad-bc)(a+d)

=> a^3 + d^3 +3a^2d + 3d^2a - (3a^2 - 3abc + 3d^2a - 3bcd)

=> a^3 + d^3 +3abc +3bcd

Since a monic quadratic is completely determined by the sum of its roots and the product of its roots, and x_1^3 and x_2^3 have sum and product to the corresponding coefficients, they are roots of the second polynomial.[/hide]


Let $\mathcal{C}_1(O_1)$ and $\mathcal{C}_2(O_2)$ be two circles which intersect in the points $A$ and $B$. The tangent in $A$ at $\mathcal{C}_2$ intersects the circle $\mathcal{C}_1$ in $C$, and the tangent in $A$ at $\mathcal{C}_1$ intersects $\mathcal{C}_2$ in $D$. A ray starting from $A$ and lying inside the $\angle CAD$ intersects the circles $\mathcal{C}_1$, $\mathcal{C}_2$ in the points $M$ and $N$ respectively, and the circumcircle of $\triangle ACD$ in $P$.

Prove that $AM=NP$.

Extend CM to meet $C_2$ at E. Then it's easy to show (i)MNDE is an isosceles trapeizum, (ii) $\triangle ACM\sim\triangle DAN$.
From this, we obtain $AM\cdot AN=DN\cdot CM=ME\cdot CM=AM\cdot MP$. The conclusion follows.

Note that $ \bigtriangleup ACM \sim  \bigtriangleup DAN$ and $ \bigtriangleup NPD \sim  \bigtriangleup ACD$. And the result follows.

when I draw the ray i find it out of the obtuse angle CAD, Is it obligatory that it should be lying inside CAD ?

It will be great if some one  sketch the graph of this problem, I really want to try it  :maybe:

Very nice problem with very interesting solution.
[img]http://i012.radikal.ru/0804/93/e0ad4f4f6031.jpg[/img]

Dear Mathlinkers,
just a link with
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=187699
Sincerely
Jean-Louis


Find all real solutions of the system
x^5 + y^5 = 1
x^6 + y^6 = 1

This is a  nice problem !

This is the hint for this problem.

From (1) we get $ x^{30}\equal{}(1\minus{}y^5)^6$

From (2) we get $ x^{30}\equal{}(1\minus{}y^6)^5$

$ \Longrightarrow (1\minus{}y^5)^6\equal{}(1\minus{}y^6)^5$

Use maple we get only $ y\equal{}0$ or $ y\equal{}1$.

So this system have roots $ (0; 1)$ or $ (1; 0)$

$ x^5\plus{}y^5\equal{}1 \qquad(1)$
$ x^6\plus{}y^6\equal{}1 \qquad(2)$

If we consider only real solutions then from (2) we know that $ 0\le x,y \le 1$
Also if we subtract (1) from (2) then $ x^5(x\minus{}1) \equal{} \minus{}y^5(y\minus{}1)$

Let $ f(x) \equal{} x^5(x\minus{}1)$, we need to find $ x,y$ such that $ f(x) \equal{} \minus{}f(y)$
however $ f(x)$ has only two real roots, $ x\equal{}0,1$ and $ f(x)<0, x\in (0,1)$ therefore there can be no solutions

Except, of coarse when $ f(x)\equal{}0$ so $ (x,y) \equal{} (1,0), (0,1)$


There are no real solutions (haven't proven it).

Woops! I made a big mistake!
There are two real solutions
x0=1.26211608596646699...
x1=.295380452552260041...

(I think all of these digits are significant.)

[quote="euler_vn"]nobody solve?[/quote]

Wait Euler_vn i will solve today and post its solution tomorrow... ;)  ...

Kummer


Find the area of the region between the graph of $ y\equal{}\frac{1}{x}$, the x-axis, and the vertical lines $ x\equal{}1$, and $ x\equal{}4$.

This is a calculus problem, but the answer is $ A\equal{}\int_{1}^{4}\frac{dx}{x}\equal{}\ln(4)$.


Let  $ D$  be bounded plane figure, with property that his every point has neighbourhood with nonzero area. Is it possible to divide that figure
with two orthogonal lines at  $ 4$  partes with equal areas?

P.S. It seems to be that's simple problem. Let first line divide $ D$ at two area equal partes. (It's infinitly possabilities for such lines!) Then it's sufficient to move second line trouth $ D$ ortogonal to first, and use theorem of intermediate value for continuos function. Is that thinking correct?

[quote="suadnovic"] every point has neighbourhood with nonzero area[/quote]

can you make this more precise?

Just like that! I supose, condition is taken to prevente $ D$ haz "holes". Precisly,
for every $ x\in D$ exists $ \varepsilon>0$ such that area $ O_{\varepsilon}(x)\cap D>0$